Google doodle pizzas solutions

Today (2021-12-06) Google published a doodle that has to do with cutting pizzas.

In this blog post, I will provide the questions and their answers. Each of my answers depicts the last cut that is about to be done. I do not provide the result of the last cut, because after the last cut is performed, the resulting pizza pieces are removed almost immediately. But I assure you that each last cut that I depict leads to a perfect score.

Question 1

q1

Solution for question 1

s1

Question 2

q2

Solution for question 2

s2

Question 3

q3

Solution for question 3

s3

Question 4

q4

Solution for question 4

s4

Question 5

q5

Solution for question 5

s5

Question 6

q6

Solution for question 6

s6

Question 7

q7

Solution for question 7

s7

Question 8

q8

Solution for question 8

s8

Question 9

q9

Solution for question 9

s9

Question 10

q10

Solution for question 10

s10

Question 11

q11

Solution for question 11

s11

End result

zzz

Posted in Science | Comments Off on Google doodle pizzas solutions

My attempt at Question 10 from the Haselbauer-Dickheiser Test

The Haselbauer-Dickheiser Test can be found at http://matrix67.com/iqtest/.

In this blog post, I will study Question 10 from this test.

The question is about an analogy.

Below I present the original question for your convenience:

Please do not read the rest of this article, if you want to attempt to solve this question on your own. The rest of this article describes my attempt at solving this question and you should not read it, unless you want to or you do not mind coming across relevant ideas, spoilers, hints, solutions, and strong opinions concerning this test.

You have been warned and I now consider that you continue to read knowing that what you come across for the rest of this article may forever spoil things for you and/or present strong opinions against this test.

Last warning: please do not read this blog post, unless you are certain that you know what you are doing. If you are not sure, then it would be best if you stopped reading at this point.

OK. If you are here, it means that you want to know my opinion. Well, ok then!

This is one of the most idiotic questions ever posed. But is it the most idiotic question ever posed? Or is there a question that tops it in the idiocity department? The fact that I am contemplating this, does not really vouch for this question. As I said, this is my opinion.

So, a bunch of red shapes corresponds to 58. And, in a corresponding way, a bunch of yellow shapes corresponds to something. What is this something? Probably another number, and probably a two digit one, like 58. Is it 58 or “5” and “8” stringed together? Whatever. We are dealing with idiots, remember?

You can try and come up with ways to relate the number 58 (or the numbers 5 and 8) to the red shapes. After that, you can use the same thinking to relate another number to the yellow shapes. This is not the correct way to approach this problem, but before I talk about the correct way, let me provide you with one (out of infinite many ways) to relate the red shapes to number 58.

If you count the number line segments used to draw the red shapes, you can find a sum of 58 line segments, if you try to count in specific ways. One of these ways is the following:

or the following:

The difference is in the way the third shape’s line segments are counted. But in both cases, they amount to 13 line segments.

These attempts seperate the line segments in a way as to add them up to 58. This line segment separation and count leads to the corresponding way for the yellow shapes, perhaps as follows:

You may add the numbers and get a sum. And say that this is the answer. But it may not be, really. And we can argue which is the most consistent way of counting the yellow shapes’ line segments, so that they relate exactly to the red shape’s line segments counting. But it does not matter, since this may not be the correct way to solve Question 9.

And we can find other ways like that. We must be very careful though. Each way of solving that we suggest, must be unambiguous. It must function in an unambiguous manner. So, let us say that we are counting. The way we are counting must leave no ambiguity. And also, and this is very important, even if a way is specific for the red shapes, it is of no value if it is not specific for the yellow shapes as well. So, if you found a non-ambiguous way to produce 58 from the red shapes, but this way is ambiguous (due to the different case) for the yellow shapes, then your method is not valid.

But these ways, and any others like that, do not seem to be the correct way to proceed. In this blog post, I will explain to you what the correct way to proceed is.

When I was looking at these shapes, I was alert to the fact that we must process all information that these shapes provide. We may not pick and choose. And first of all, we have to understand what these shapes are. Where do they come from? What is their story? What do they depict? Why do the yellow shapes not all have the same length? Why do the yellow shapes have smaller height than the red shapes? What is going on here? Am I on candid camera?

There are 6 red shapes. And there are 6 yellow shapes. Let us focus on the 6 red shapes. Why are there 6 of them? And they all have the same dimensions. Like the 6 sides of a cubic box. But these cannot be the 6 sides of a box. So, what can they be? At some point, I understood exactly that. Do you know what they are? They are projections. They are projections of a cubic solid. A cube has 6 sides, so we have 6 projections: front and back, left and right, top and bottom. 6 if you count correctly. So, this is why we are given 6 red shapes. Each one corresponds to the way the solid appears from this side. Of course, since the solid can be oriented any way in space, naming front as front, is an arbitrary convention. We will just pick a convention, when we get to that point.

The Internet, especially Pinterest, is full of projections like these. Search for orthographic projections, orthometric projections, isometric projections, vistas, or any other relevant term you may find. Among others, I found a very nice link:

https://med.se-todo.com/other/47422/index.html

I took the liberty to copy one of the images from there to here, for your convenience:

So, we consider a 3X3X3 cube, which is built with 1X1X1 cubes as elements. We may remove some of its elements. We want to describe the resulting cubic solid. So, we take three projections: top, front, side. Now, this can be a nice puzzle. Starting from the cubic solid, we want to draw the three projections. Or, starting from the three projections, we want to find the cubic solid. Well, this second case is the one Question 10 is about.

Let me explain how the cubic solid is formed. It is like a Rubik’s cube, if a Rubik’s cube was comprised of 27 cubic elements and it was solid, i.e. no turning sides.

Except in our cubic solid, some of the 27 elements may be missing. So, imagine a 3X3X3 cubic solid, constructed from at most 27 smaller cubes, all identical in dimensions. At most 27. And, as I said, some of the smaller cubes may be missing. So, there is a huge number of possible 3X3X3 cubic solids than can be constructed this way. But, I guess that each one of these cubic solids should be “connected”.

“Connected” is a term that I coined. Or rather, I copied from Graph Theory, a branch of Mathematics that deals with graphs. A graph has nodes and edges (which are lines connecting the nodes). If every node in a graph is connected to at least one other node, then the graph is “connected”. No node left behind. So, in our cubic solid, a smaller cubic element is always connected to at least one other smaller cubic element. Or else, the “shape” will no longer be one shape, but two or more independent ones.

As you see from the example icons I posted above, the projections (or, equivalently, views) are usually 3: front, top, one of the sides. But here, in Question 10, we have 6 projections. So, this got me thinking: We either talking about two cubic solids, or we are talking about one cubic solid, but then, the projections are: front and back, left and right, top and bottom. So I said to myself: I will try to come up with only one cubic solid. If I cannot, I will look for two seperate cubic solids. But I was quick to find out a single cubic solid that corresponded to these 6 projections and I had the problem solved.

Here is how I did it.

I was planning to get physical 1X1X1 cubes to build such 3X3X3 cubic solids, or draw them in some way in my computer in 3D, but I did not need to. After thinking for a little while, I had the problem solved using just a 2D piece of paper.

Here is how I ended up relating the projections to the cubic solid:

I drew the three layers of the cubic solid as follows. Each 1X1X1 element would have a question mark, until I decided whether it existed, in which case I would color it, or it would not exist, in which case I would leave it white by erasing the question mark.

And I imagined that the base layer sits at the bottom, on top of the base layer we place the middle layer, and on top of the middle layer we place the top layer.

Now all, that is left is to decide which of the 27 1X1X1 cubes exist and which do not.

I started with the sides that have the most flatness. Those are the ones form the left, which have the largest uninterrupted areas. Because when lines are involved, this means that we change depth.

So, I took the first projection from the left. I decided to evaluate it as the back side, since I could visualize it without seeing it and since I needed the front unobscured so I could keep working. And I colored the 1x1X1 elements that were needed to have this projection at the back, as follows:

Quite an improvement, don’t you think? Please notice the middle layer. Do you see all the white elements on both sides, having no question marks? This means that they do not exist. This is because the first red projection from the left, which I evaluated in the back, has a thin waste! You can see throught the sides!

Alright, let us proceed. I then examined the second red projection from the left and I evaluated it at the top layer. The previous projection was quite symmetrical to pose any objections! Yes, evaluating it on the top layer, it would obscure things, but I was now feeling very confident. In the top layer, I already had a beginning (three elements drawn red), so I placed it accordingly, as follows:

Please notice that I did not yet place the two-element area. These two red squares are in a different layer (either the base layer or the middle layer).

In order to place these two elments, I considered two things. The first was the bottom side. It had a large 7 element area. The second was the left and right side. They had to be see-through in two middle elements.

And thus, finally, I ended up with the following cubic solid:

This is it. This is the solution. It is a 27 element cubic solid with 15 elements existing and 12 elements non-existing. The middle layer connects the base layer and the top layer with only one element, which is not good for the structural integrity of the whole thing. Not that it matters, just saying.

Ok, let us prove that this is the correct cubic solid. We will examine each of the 6 projections given, to see if we get the correct view and explain the areas of each projection. Here we go:

So, we have the correct cubic solid. All projections (and all individual areas of each projection) are accounted for. I will again present the cubic solid for your convenince:

How is it related to 58? Well, I cannot look inside the sick little minds of those who posed the question. But there are a lot of things that we can consider. I will only provide a brief elementary analysis.

Just by looking at the layers vertically, we see three columns of 5 red (existing) elements each and the largest white (non-existing) number of elements is in the middle layer and their count number is 8.

But let us be a little bit more methodical.

First of all, we have a cubic element with 27 elements, 15 red (exsiting) and 12 white (non-existing). We can study percentages of these, if we wish.

Another analysis that can be done is to list and count the elements, either from base to top, or from left to right, or from back to front. Here is this analysis, which is meant to be read vertically:

These numbers may as well be taken into consideration.

So, the cubic solid or its projections are somehow related to 58. The analysis above may help in finding this relation, which could be virtually anything.

Now that we have finished our analysis about the red projections, let us begin to study the yellow projections.

It is immediatelly obvious that we have to deal with a 3X2X2 rectangular cuboid solid. The cuboid has 3X2X2 = 12 elements, and the middle ones are elongated in length, whereas the outer ones are not. Some of these 12 elements will exist, and some will be missing. And I guess, this yellow cuboid will have to be “connected” somehow, since the red one was too. Below I present what I have in mind, with a view from above looking down.

So, let us find the cuboid in question. In the end, we will have something like the following relation:

We will begin by creating the layers. We can choose to have two 3X2 layers, the base layer and the top layer, and we can imagine the base layer to be at the bottom and the top layer to go at the top. In the beginning, all elements have questionmarks, since we do not yet know if they exist or not.

We will begin by choosing the back side, because it has a large area and this will help. By placing it in the back, we have the following configuration. Please note that we deleted the two questionmarks in the top layer, because these elements are non-existing.

Great! We progressed quite a bit with just one projection. If we take into account all the other porojections, we arrive easily at the following cuboid:

Let us prove that this cuboid is the actual cuboid we are looking for, by deriving the 6 projections from it.

Again I present the 3X2X2 cuboid here, for your convenience.

Let us do a little rudimentary analysis. We have a cuboid and its 6 projections. The cuboid is 3X2X2, so it has 12 elements, of which 6 are existing and 6 are non-existing. I also provide the following analysis, which is meant to be read vertically:

I guess, that is all I have to say. I hope I shone a little light as far as Question 10 is concerned. It is about the 6 projections a cuboid has and that these projections can be used to find it and describe it.

Since my answers are quite lengthy, one might be inclined to assume that the questions of this test may have some merit. This is the point I want to clarify. They do not. The way I see it, I have to make lengthy answers, because the questions are not well posed. In the world I want to live in, these questions would be a book’s length. Then the questions would have merit and the answers would be of reasonable length.

Update, March 15. 2021: I have tried to find how the red cuboid is related to the number 58. I have found some ways, but they do not seem to be robust. They seem to be ambiguous.

But then, after some consideration, I found a way that seems robust and unambiguous enough. This is counting the external tiles of the reb cuboid. In other words, how many squares do the external sides have? If we need to count the external tiles or if we need to cover the cuboid with new tiles, how many tiles would we need? If we were to paint it (with red or any other color), how many tiles would we need to paint?

I counted the external tiles and since the count number is 58, this seems that it is a robust answer.

Let us do this counting for the red cuboid.

We are counting the external tiles.

We begin with the top layer. We have 7 top tiles. We have 6 botton tiles, because the seventh one is hidden between the top and middle layer. We have 16 side tiles.

We continue with the middle layer. We have 4 side tiles. We do not have any top or bottom tiles, since the only top one is hidden between the top and middle layer, and the only bottom one is hidden between the middle and base layer.

We finish with the base layer. We have 6 top tiles, because the seventh one is hidden between the base and middle layer. We have 7 bottom tiles. We have 12 side tiles.

Let us add all these numbers: 7 + 6 + 16 + 4 + 6 + 7 + 12 = 58.

This seems to be the correct relation between the red cuboid and the number 58. So, we are not talking about two seperate numbers, five and eight. We are talking about the number fifty eight, which corresponds to the number of external tiles of the red cuboid.

Since this is the correct relation, we can extrapolate this thinking in order to find the number of external tiles of the yellow cuboid. But here, a problem arises. Not all of the yellow tiles have the same dimensions. The middle ones are elongated.

We will discuss about this, but, first, let us do the counting for the yellow cuboid.

We are counting the external tiles.

We begin with the top layer. We have 2 top tiles. We have 8 side tiles. We have no bottom tiles, since both are hidden between the top layer and the base layer.

We finish with the base layer. We have 2 top tiles, since the other two are hidden between the top layer and the base layer. We have 4 bottom tiles. We have 10 side tiles.

Let us add all these numbers: 2 + 8 + 2 + 4 + 10 = 26.

Now, this is a valid answer. But if we want to account for the 4 elongated tiles around the middle of the base layer, we need to find a way to do it. If we assume that their area is twice the area of a small yellow tile, which is a big “if”, then we need to add 4 more tiles to the count. Thus, we end up with 26 + 4 = 30.

So, I have all but proven that Question 10 is ambiguous, which means that it is open to interpretation. Some of the questions in this test are. For example, another question, Question 8 is also ambiguous, and I have discussed about its ambiguity in my corresponding blog post.

As far as Question 10 is concerned, we can view the yellow cuboid as a 3X2X2 cuboid with the middle elements elongated, in which case the number of external tiles is 26. Or we can view the yellow cuboid as a 4X2X2 cuboid with all elements identical, in which case the number of external tiles is 30. I depict this second case below:

This second case seems to be even more robust, but only if we assume that the middle tiles are twice the length of the other tiles. And, as I said earlier, this is a big “if”.

Let us again count the external tiles for this 4X2X2 cuboid.

We begin with the top layer. We have 2 top tiles. We have 8 side tiles. We have no bottom tiles, since both are hidden between the top layer and the base layer.

We finish with the base layer. We have 3 top tiles, since the other two are hidden between the top layer and the base layer. We have 5 bottom tiles. We have 12 side tiles.

Let us add all these numbers: 2 + 8 + 3 + 5 + 12 = 30.

So, now we have all the data that we need to make an informed evaluation of this question.

Posted in Education | Comments Off on My attempt at Question 10 from the Haselbauer-Dickheiser Test

My attempt at Question 9 from the Haselbauer-Dickheiser Test

The Haselbauer-Dickheiser Test can be found at http://matrix67.com/iqtest/.

In this blog post, I will study Question 9 from this test.

The question is about deciphering a given ciphertext (referred to as “cryptogram”).

Below I present the original question for your convenience:

Please do not read the rest of this article, if you want to attempt to solve this question on your own. The rest of this article describes my attempt at solving this question and you should not read it, unless you want to or you do not mind coming across relevant ideas, spoilers, hints, solutions, and strong opinions concerning this test.

You have been warned and I now consider that you continue to read knowing that what you come across for the rest of this article may forever spoil things for you and/or present strong opinions against this test.

Last warning: please do not read this blog post, unless you are certain that you know what you are doing. If you are not sure, then it would be best if you stopped reading at this point.

OK. If you are here, it means that you want to know my opinion. Well, ok then!

Suppose we want to encrypt a message. We call the message “plaintext”. We call the encrypted message “ciphertext”. The encryption process takes the plaintext and produces the ciphertext. The decryption process takes the ciphertext and produces the plaintext. There are an infinite number of ways to do encryption/decryption. In this article, I use the words “decryption” and “deciphering” interchangeably. The same goes for “ciphertext” and “cryptogram”.

In this question, we are given the ciphertext and we are asked to do decryption, thus producing the plaintext. We are also told that the plaintext is a common phrase.

Just by looking at the ciphertext, I immediately noticed that it is composed of only 6 characters, the characters A D F G X V, which I list here in alphabetical order. And I immediately realized that 1 letter of the plaintext corresponds to 2 letters of the ciphertext. This is obvious, because the English language has 26 letters and with 6 available characters, we need 2 characters to represent each of the 26 letters. I immediately thought of the following square, which would help me represent a possible correspondence between the characters A D F G V X and the 26 letters of the alphabet. I also found cool that this 6 character encoding lets us fit the 10 numerical digits inside as well.

So, in the sample square above, the plaintext “g” is encoded as the ciphertext “DA”. And the ciphertext “DA” is decoded as the plaintext “g”. The plaintext “yes” is encoded as the ciphertext “VAAVGA”. And the ciphertext “VAAVGA” is decoded as the plaintext “yes”. And so on.

Of course, the letters and numbers may be shuffled inside the square to produce a different new square and thus, there are 36! such squares that can be produced. 36 factorial. Do not try to count so high. So, just by guessing, it is vitually impossible to find the particular square that was used to encrypt the plaintext. And brute-force calculations are out of the question, too, because of how big this number is. Actually, my analysis is not entirely acurate. The placement of the numbers should not matter, since I supposed we only have letters in the common phrase, and in hindsight I was correct. (Also, not all letters of the alphabet need to exist in the common phrase). Still, even though the placement of the numbers does not matter, the rearrangements of the letters are so many, that it is virtually impossible to guess or brute-force (which means: to examine each possible rearrangement one by one).

Anyway, I was pretty sure that this was the way the encrypted the plaintext. I was pretty sure they wrote the plaintext using only small letters (or capital letters, but not a mix) with no spaces and they encoded it using such a square. In hindsight, I was right.

Since the ciphertext is 90 characters long and two characters correspond to a letter from the plaintext, I immediatley knew that the plaintext was 45 letters long. So, the common phrase was 45 letters long. But what common phrase is 45 letters long? I could not recall any such phrase.

“rosesareredvioletsareblue”? Less than 45 letters.
“achainisonlyasstrongasitsweakestlink”? Still less than 45 letters.
“whenthegoinggetstoughthetoughgetgoing”? Still less than 45 letters.
“ajourneyofathousandmilesbeginswithasinglestep”? Exactly 45 letters! Bingo! No, in hindsight, this is not the phrase that was encrypted.

So, I was working from both ends. From one end, I was trying to analyze the ciphertext in order to decrypt it. From the other end, I was trying to find common phrases that were exactly 45 letters long, in the hope that one of them might be the one that was encrypted. And that could possibly help me.

After a short while, I discovered that this square that I coined is “a thing”. Is is called “the Polybius square”:

https://en.wikipedia.org/wiki/Polybius_square

Not only that, I also discovered that the encryption scheme that was used in this question is also “a thing”. It is called “the ADFGVX cipher”:

https://en.wikipedia.org/wiki/ADFGVX_cipher

Please read the contents of the above link carefully, because this is the theory that the question is based on. Of course, I should have expected it. In the Haselbauer-Dickheiser Test, virtually everything is borrowed and copied from other sources.

The Wikipedia article describes the ADFGVX cipher in considerable detail. It also shows how the encryption and the decryption is done. Last but not least, it discusses the cryptanalysis of the ADFGVX cipher, which is the ways that the cipher may be broken. But to say that to break this cipher is difficult, is quite an understatement. In other words, it is extremely difficult to break the ADFGVX cipher. But, please, get involved in doing so, if this is something that you like. In this blog post, I will not get into the cryptanalysis of the ADFGVX cipher.

“But why?”, you may ask. If I am not going to study the cryptanalysis of the ADFGVX cipher, then how am I going to provide the amswer to the question? Well, here is the thing! I was saved by the bell (or, in this case, the Internet). I came across an article in a web site that was about a person that solved this question. The site was a local site or an educational site (community, university, or something like that). The web page was discussing the achievemenst of this person, but the article was not technical in nature. But here is the thing: On the bottom of the web page, with no other relevant context, there was the phrase:

“inlifethereareonlytwocertaintiesdeathandtaxes”

45 letters! Bingo! I immediately knew that this was the common phrase. And since the question wants us to put the last word of the phrase in the answer line, the answer is

“taxes”

I checked the HTML source of the web page to see if I could find anything more, but I could not. After a few months, the site changed the appearance of its web pages, but each page kept the same content as before. With the exception of the web page in question, which lost the common phrase that had been previously lying on the bottom of the page. And now, I can no longer find the site itself.

But, now I know the common phrase. And the problem becomes easier.

So, the problem I had at hand was the following: Given the cryptogram of Question 9, and given that the common phrase is “inlifethereareonlytwocertaintiesdeathandtaxes”, how do these two correspond with each other? In other words, given this common phrase, what is the encryption process that produces the cryptogram? And given the cryprogram, what is the decryption process that reveals the common phrase? From now on, I will refer to this problem as the “reduced problem” and I will refer to Question 9 as the “original problem”.

OK, I know that tackling this reduced problem (instead of the original problem posed in Question 9) is cheating. But I will address non-cheating later. For now, I would like to address and solve this reduced problem. But before I give you the solution, dear reader, I would like you to pause and ponder. Can you solve the reduced problem on your own? Well, if you have solved the original problem already, then you do not need to answer this question! But if you have not solved the original problem as of yet, it might be beneficial for you to think about solving this reduced probelm. This way, you will be in a good postion to judge my writings, whereas if you just keep reading casually, you might not be in a good position to evaluate the rest of this blog post.

Anyway, here is how I solved the reduced problem. It only took me a few minutes. And I did not need to write any programs or do frequency analysis or any of the advanced techniques cryptanalysts use.

I viewed the ciphertext as a matrix with 6 rows and 15 columns. And I obtained the corresponding transpose matrix (also known as transposition matrix).

https://en.wikipedia.org/wiki/Transpose

The transposition matrix is obtained very easily by turning each row of the original matrix into a column. So, the first row of the original matrix becomes the first column of the transposition matrix. The second row of the original matrix becomes the second column of the transposition matrix. And so on.

Ciphertext:

A F F D X V X A A A G X X D F
X X F G G A F A F A G G X F G
A A X A A X X F X X F F F G G
A F X G G V G G A F F A A F G
V G G A X F X G X X F F A A F
X D F F F A V F D X F D D X F

Transposition matrix:

A X A A V X
F X A F G D
F F X X G F
D G A G A F
X G A G X F
V A X V F A
X F X G X V
A A F G G F
A F X A X D
A A X F X X
G G F F F F
X G F A F D
X X F A A D
D F G F A X
F G G G F F

And we are almost done! All that remains is to rearrange the 6 columns so that they correspond to the common phrase.

There are 6! = 720 permutations (arrangements) of the 6 columns. Only one of these arrangements corresponds to the common phrase. But since we know the common phrase, it is easy to find this unique arrangement. Here is how I did it. I opened Excel and copied the common phrase and the transposition matrix, as shown in the picture below.

The reason I made such a construct is the following: each letter from the common phrase corresponds to two characters from the ciphertext. After we rerrange the columns correctly, each letter from the common phrase will correspond to two characters from the ciphertext that will emerge, as follows: the first letter from the common phrase will correspond to the first two characters in the first row of the emerged ciphertext. The second letter from the common phrase will correspond to the third and fourth characters in the first row of the emerged ciphertext. And so on.

Actually, there are two possible “winning” rearrangenents instead of one. If a reaarangement is found, then by mutually switching columns 1st and 2nd, 3rd and 4th, 5th and 6th at the same time, a second equally valid rearrangement is produced.

Let us find one of these two rearrangements.

Looking at the Excel construct, it is easy to find the rearragnement that corresponds to the common phrase. Only a few tries of reordering (by cutting and pasting) columns are needed. Here is how I proceeded:

The first row and the the second row have the same letter “i” in the same place: first. So, the first column and the second column can only be the columns with first elements AA and XX, and since we are just starting we can choose the order these two columns will have, i.e. which will be first and which will be second.

Then I looked at the 4th and 5th lines. The have the letters “r” and “e” in common and in the same place. We already have the first two columns matching, so the third and fourth column (no order yet between them) will be those with FF and GG in these positions. Well, the order is the X and A for the first characters of these columns, since the order A and X was reserved for the letter “i”.

Line 2 and line 3 have the letter “e” in common. Looking at the 4 th and 5th lines, which contain the letter “e” in the second position, we can find the order of the 5th and 6 th column: A and then V.

Anyway, it is very easy to do and here is the solution.

But, as I said, there is one more equally valid solution, which is produced if we switch the 1st and 2nd, the 3rd and 4th, and the 5th and 6th columns. Here is the other solution:

As you saw, for each of the two equivalent solutions, I also provided the correspondence between letters and characters, both as a list and as a square (Polybius square or, could I possily call it “pivot table”?). Of course, the depiction of the list and the square is not needed, but I provided them for your convenience.

And this concludes the solution of the reduced problem.

I wrote that it took me a few minutes to solve the reduced problem. Well, at most. Perhaps it took me only a minute. I did not measure. But this was after I thought to obtain the transpose matrix and decipher it. Because my original thought was not that.

“Well, what was you original thought, Dimitrios? The ADFGVX cipher encyption/decryption process clearly states that transpose matrices are involved. This is the norm. This is the convention. How come you did not think of that?”, I think I hear you say to me, dear reader.

Well, I did think of transpose matrices. But, at first, I did not imagine that the test would be so easy as to demand the straightforward transpose matrix of the original ciphertext. The way I imagined it, I thought that the cretators of the question would want us to either consider the matrix as it is (with no transposition) and rearrange the 15 columns, or to obtain the transpose matrix but get it in a 15-column format as well. I would think that this last configuration is the actual convention, anyway.

Then, having obtain the transposed matrix but in a form that would still had 6 rows and 15 columns, we would have to rearrange the 15 columns. Since the rearrangement is done with a word key, we could have it easy! The word key had to be 15 letters wide and all letters had to be unique so they can be sorted alphabetically and unambiguously. And there are only two words in the Englsih langage, as far as I know, that have these properties: “uncopyrightable” and “dermatoglyphics”. So, one of these two words should be the key that was used to rearrange the 15 columns and the question would be about applying the decryption process without doing frequency analysis or any advanced cryptanalysis stuff.

But, in hindsight, I was incorrect. The creators of the question did not want us to think this way. They meant the ciphertext to be transposed to a matrix with 15 rows and 6 columns, and they wanted us to be very creative with cryptanalysis in order to find the correct rearrangement of the 6 columns. They meant the permutations to be only as low a number as 6! = 720. They meant the column number to be even, which helps a lot in deciphering this particular cipher. And they meant each line of 6 characters to correspond to 3 letters from the common phrase. So, they meant to throw in some conveniency in the ciphertext.

This would have been a question that I would have liked, if it was posed this way: We are given the transposition matrix (15 rows, 6 columns) and we are asked which rearrangement of the 6 columns corresponds to a common phrase. But since this question was not given in this form, I cannot “upvote” it.

I would like to clarify that the solution of this question, in the form that I propose here, is beyond the scope of this blog post. But it is an interesting question and one that you might want to tackle, if you like code breaking, dear reader. There are a lot of ways you might want to approach this problem. I would recommend the use computers and programming. To begin with, there are, as we have seen, 6! = 720 arrangements of the 6 columns and one of them (actually, two of them, as we have seen) is the “winning” arrangement. To find it, you need to be creative. Read the cryptanalysis section from the Wikipedia article I provided on the ADFGVX cipher. Takes cues and ideas from it. Create programs that produce all 720 arrangements and for each arrangement calculate statistics and frequency analysis. And try to match character combinations with letters and words (using a word list). There is room for many more ideas and I am willing to bet that you will come up with many more ever so brilliant ideas, dear reader.

But even if you do not want to use programming, there is another way, which I do not recommend. There are 6! = 720 arrangements. But, as I explained, each arrangement has an equivalent unique twin which is formed by switching the 1st and 2nd, the 3rd and 4th, and the 5th and 6th columns. So, arrangements go by twos, and you need to solve only one of the two, because due the symmetry of the problem, they are completely equivalent. So, the arrangements you need to consider are 720 / 2 = 360. Take each one, one day at a time. It will take you 360 days. For each of the 360 different arrangements you will have a problem equivalent to solving a code word crossword. But solving it with no letters as clues. Usually, code word crosswords give you a few letters as clues. But, I like to solve them without any such help. This was the reason behind my blog post

https://dkalemis.wordpress.com/2004/11/29/an-excel-vba-macro-for-cipher-crosswords/

in whch I created an Excel macro that would allow me to solve code word crosswords without any clues and without it being tedious. But doing that for 360 such code word crosswords, when only one (actually two, remember?, which are equivalent) is the “winning” one that actually corresponds to having real words, well… it is tedious. You better not go with this approach. Go with a programming apporach. It is more cool and glamorous. Do not be “old school” on that front. And this comes from me, who usually likes being “old school”.

Now that I have finished my analysis, I would like to present you with my failed approaches. I thought that the creators of the question wanted as to rearrange a 15 column matrix with a 15-letter word as key. In hindsight, I was incorrect, and the resulting 6 row and 15 column matrix does not correspond to the solution we are seeking. Still, I present my work, in case you might find interest in it, dear reader. But, again, the result of this decryption process is not what we are after, and this is because the decryption process (and especially how I obtained the transposition matrix) is not what the creators of the question had in mind.

So, here is my failed attempt using the keyword “uncopyrightable”:

CIPHER TEXT

A F F D X V X A A A G X X D F
X X F G G A F A F A G G X F G
A A X A A X X F X X F F F G G
A F X G G V G G A F F A A F G
V G G A X F X G X X F F A A F
X D F F F A V F D X F D D X F

TRANSPOSITION MATRIX

A X A A V X F X A F G D F F X 
X G F D G A G A F X G A G X F
V A X V F A X F X G X V A A F
G G F A F X A X D A A X F X X 
G G F F F F X G F A F D X X F
A A D D F G F A X F G G G F F

a b c e g h i l n o p r t u y

A X A A V X F X A F G D F F X 
X G F D G A G A F X G A G X F
V A X V F A X F X G X V A A F
G G F A F X A X D A A X F X X 
G G F F F F X G F A F D X X F
A A D D F G F A X F G G G F F

u n c o p y r i g h t a b l e 

F A A F G X D F V X F A X X A
X F F X G F A G G A G X G A D
A X X G X F V X F A A V A F V
X D F A A X X A F X F G G X A
X F F A F F D X F F X G G G F
F X D F G F G F F G G A A A D

And here is my failed attempt using the keyword “dermatoglyphics”:

CIPHER TEXT

A F F D X V X A A A G X X D F
X X F G G A F A F A G G X F G
A A X A A X X F X X F F F G G
A F X G G V G G A F F A A F G
V G G A X F X G X X F F A A F
X D F F F A V F D X F D D X F

TRANSPOSITION MATRIX

A X A A V X F X A F G D F F X 
X G F D G A G A F X G A G X F
V A X V F A X F X G X V A A F
G G F A F X A X D A A X F X X 
G G F F F F X G F A F D X X F
A A D D F G F A X F G G G F F

a c d e g h i l m o p r s t y

A X A A V X F X A F G D F F X 
X G F D G A G A F X G A G X F
V A X V F A X F X G X V A A F
G G F A F X A X D A A X F X X 
G G F F F F X G F A F D X X F
A A D D F G F A X F G G G F F

d e r m a t o g l y p h i c s

A A D A A F F V X X G X F X F
F D A F X X X G A F G A G G G
X V V X V A G F F F X A X A A
F A X D G X A F X X A X A G F
F F D F G X A F G F F F X G X
D D G X A F F F A F G G F A G

I also tried other approaches, but I was always aiming for a 6-row 15-column matrix. so, I would consider the original ciphertext with no transposition, or the original ciphertext transposed, for one example, as follows:

CIPHER TEXT

A F F D X V X A A A G X X D F
X X F G G A F A F A G G X F G
A A X A A X X F X X F F F G G
A F X G G V G G A F F A A F G
V G G A X F X G X X F F A A F
X D F F F A V F D X F D D X F

TRANSPOSITION MATRIX

A X A A V X F X A F G D F F X 
X G F D G A G A F X G A G X F
V A X V F A X F X G X V A A F
G G F A F X A X D A A X F X X 
G G F F F F X G F A F D X X F
A A D D F G F A X F G G G F F

And I tried many other approaches, as well. And I would use the keywords “uncopyrightable” and “dermatoglyphics” in many of them. And I would try at least two methods: one where I would sort the letters first and then rearrange the letters to form the word, and another where I would first use the letters in the order the appear in the word and then sort the letters. I would transpose, or not transpose. And when I would transpose, I would use rows to columns, columns to rows, and other constructs, sometimes unconventional amd/or crazy. So, yes, I tried many different approaches, but I would always strive to end up with a 6 row 15 column matrix. And in hindsight, this was not what the question wanted. So, this is why I believe that this question would be better, if it were posed as follows:

A common phrase has been encrypted. The objective is to rearrange the 6 columns of the following cryptogam in order to decipher it and type the last word of this phrase in the answer box.

A X A A V X
F X A F G D
F F X X G F
D G A G A F
X G A G X F
V A X V F A
X F X G X V
A A F G G F
A F X A X D
A A X F X X
G G F F F F
X G F A F D
X X F A A D
D F G F A X
F G G G F F

And finally, I would like to point out that I did not expect that the authors of this question would have the audacity to use a common phrase that includes such a shocking and morbid word as the word “taxes”.

Posted in Education | Comments Off on My attempt at Question 9 from the Haselbauer-Dickheiser Test

My attempt at Question 25 from the Haselbauer-Dickheiser Test

The Haselbauer-Dickheiser Test can be found at http://matrix67.com/iqtest/.

In this blog post, I will study Question 25 from this test.

The question is about population dynamics in Biology.

Below I present the original question for your convenience:

Please do not read the rest of this article, if you want to attempt to solve this question on your own. The rest of this article describes my attempt at solving this question and you should not read it, unless you want to or you do not mind coming across relevant ideas, spoilers, hints, solutions, and strong opinions concerning this test.

You have been warned and I now consider that you continue to read knowing that what you come across for the rest of this article may forever spoil things for you and/or present strong opinions against this test.

Last warning: please do not read this blog post, unless you are certain that you know what you are doing. If you are not sure, then it would be best if you stopped reading at this point.

OK. If you are here, it means that you want to know my opinion. Well, ok then!

The first thing I would like to say is that you do not need to read my analysis. You can simply go to the small course on Behavioral Ecology on sparknotes.com and find the answer to this question there. I will give you the links:

https://www.sparknotes.com/biology/animalbehavior/behavioralecology/
https://www.sparknotes.com/biology/animalbehavior/behavioralecology/summary/
https://www.sparknotes.com/biology/animalbehavior/behavioralecology/section2/
https://www.sparknotes.com/biology/animalbehavior/behavioralecology/section2/page/2/
https://www.sparknotes.com/biology/animalbehavior/behavioralecology/problems_2/

The first link contains links to the whole of the small course. The last link contains answers to questions as well as the answer to the specific question we are discussing here. In the list of links, I have included all links that are relevant to the question and answer. The first four links correpsond to the underlying theory and the last link corresponds to questions and their answers. The reason I gave you the five links instead of only the first link, is that there is a slight problem in the sparknotes.com site, and the link navigation does not work perfectly, because some links are incorrect. You will also find that some images are missing. That does not matter that much though, because you can use the links I provide as guidance. Also, the missing images from the theoretical pages do not present that much hinderance. In addition, the answer provided in the last link is correct.

So, everything is settled, but I would recommend that you read my analysis, that I present below. In my analysis, I will explain many apsects as of to why the answer is this and not some other and I will provide a comprehensive overview of the subject.

My only inhibition is that my analysis will be a bit lengthy. I though of ways of compacting it, but then I would not be really giving you the full picture. And this is a subject that demands that every detail is mentioned. No shortcuts. Actually, my analysis is a story which has many beginnings. And the most difficult thing for me was to decide which beginning to choose. After a lot of consideration, I decided to start from *my* beginning.

And so I begin.

I am definitely not a biologist. I would know if I were! But when I was studying Signal Processing, Signals and Systems, and Systems Analysis, Design and Control, I was known in my class for my fascination about the study of biological systems. Here I was, studying Electronic Systems, and realizing that the same mathematic formalism, the same concepts, the same linear and non-linear modeling could describe population dynamics, intraspecific competition, interspecific competition, and predator-prey dynamics as well.

Population dynamics, intraspecific competition (competition among a species) and interspecific competition (competition between two or more species), are modelled extensively all over the world’s institutions. For these studies, the concepts involved are: the Logistic Function and Equation, Carrying Capacity, r/K selection theory, the Lotka–Volterra model and other related topics.

Of course, I knew enough about these subjects, so I could not fall for some misconceptions that other people seemed to have when they tried to tackle this question. A few people wrote to me, asking about the answer to this question, offering their opinions. Unfortunatlely, they all seem to misunderstand this question. And there were a lot of misundertsandings. If I could pick the most prominent of all, I would pick the “maximization of offspiring” fallacy. To me, it was obvious from the get-go that this question has nothing to do with “maximization of offspring”. The goal is not the “maximization of offsping”. The goal is the “stability of the ecosystem”. While each species may want to maximize their offsping number, this is irrelevant to what we are trying to model. We are trying to model a system that, first and foremost, is stable. And I would also not fall to the “insufficient parameters” fallacy. On the contrary, the question was well posed and no information was missing. Perhaps some parameters were mising because they were simply not needed in order for us to construct our model, or they would cancel out eventually. Same thing. But I wasn’t entirely sure about this second fallacy yet, that is, until I constructed a model and found that everyting I needed was indeed in place.

The approach I was contemplating was that of the predator-prey dynamics. We are talking about Furbles and to help you establish a visual representation, imagine a Furble being something like if Hillary Clinton and Nancy Pelosi had a child together. Although we are talking about Furbles, which is only one species, we have Dominator Furbles and Sharer Furbles. And Dominators are way different than Sharers in their behavior. This difference made the problem resemble the predator-prey population dynamics.

And what better model for predator-prey dynamics than the Lotka–Volterra equations! Actually, hmmm… they say that the Lotka–Volterra model is a bad model but profoundly important. Anyway, a very wise person once said that all models are wrong, but some of them are useful. And I say that the Lotka–Volterra model is up there with the most useful models of all.

Using the Lotka–Volterra models… this *is* the way to approach these problems. This *is* the way to model predetor-prey dynamics. So, for someone like me, coming from a systems background, the Lotka–Volterra formalism was the road that would lead to the answer to Question 25 from the Haselbauer-Dickheiser Test.

If you know a bit about the Lotka–Volterra model, you realize that for two competing species, we have a system of two differential equations. Now you might ask, how was I contemplating differential equations in a test that assumes no such advanced mathematical knowledge. Oh, but I had all things figured out! You see, once you construct the two differential equations, and since you are looking for a stable state, you set the derivatives to be equal to zero. By doing so, you end up with the equations of two lines. These are called the isoclines of the Lotka-Volterra model. If your model is correct for this particular question, you expect these two lines to intersect. The attributes of the intersection point will give us the answer. And the equations of the two lines, even though they are deduced from a purely mathematical standpoint, they can also come from pure thinking. I would have only been able to produce them from the differential equations, but someone more clever than me (vitrually everyone else) could have certainly been able to produce them just by logical thinking.

Ok, so I had everything figured out. But talk is cheap. “Come up with the model, Dimitrios! And show us the results!”, I think I heard you say. Well, here is the thing… I did not have everything figured out. There were some “details” that bothered me.

One such “detail”, although ever so subtle, was the phrase “Individual Furbles cannot switch strategies.” This phrase bothered me to no end. Well, of course a Furble cannot switch strategy. Or, could it? The Lotka-Volterra models do not model behavioral switching. Although this is a thing in Biology, Lotka-Volterra models are a different cup of tea. But here, the most common thing occurs: no switching in behavior. Foxes are foxes, hares are hares. Predators are Predetors, Prey is Prey, Shareres are Shareres. Why discuss about behavior switching? You might argue that the examiner wanted to be as specific as possible. But here is the thing: The whole Haselbauer-Dickheiser test is so lacking in explanations, that adding such an explanation, which was really not needed, looked very strange to me.

And what is the deal with the word “strategy”? Strategies? We do not discuss about strategies as far as the Lotka-Volterra model is concerned. We discuss behaviors. And this was someting else that was puzzling me.

The last thing that bothered me was that, for the Question 25 scenario, I could not really create a model that would make me content. Somehow, the data that were given, were not matching those of other scenarios that I had encounter. In other words, I could not see how I could create robust parameters for the Lotka-Volterra model from the values given in Equation 25. And so I kept struggling.

Little did I know that there is another approach to population dynamics, and it does not have to do with such modeling. This other approach has to do with Game Theory. And this is the key to understanding how we should approach this question. This question uses Game Theory jargon and requires Game Theory formalism for its solution. This is why we have the use of the word “strategy” and this is why we need the added assurance that “Individual Furbles cannot switch strategies.”

So, yes, this question is about populations dynamics, but there are two different formalisms to approach these problems. One is the Lotka-Volterra equations, the other is Game Theory. Obviously, Question 25 is of the second kind. We need to think in terms of Game Theory in order to utilize the givens and find the answer to the question.

John Maynard Smith and George Robert Price published their seminal paper titled “The Logic of Animal Conflict” on November 2, 1973. This was the beginning of the application of Game Theory to the study of population dynamics in Biology. This new branch of science, which combines together Game Theory and population dynamics into something new, is now called “Evolutionary Game Theory”. The particular piece of the theory we need to know to solve Question 25 is called the Hawk-Dove game.

Here are some interesting links:

https://blogs.bl.uk/untoldlives/2020/03/john-maynard-smith-evolutionary-biology-and-the-logic-of-animal-conflict.html

https://www.nature.com/scitable/knowledge/library/game-theory-evolutionary-stable-strategies-and-the-25953132/

https://en.wikipedia.org/wiki/Evolutionary_game_theory

https://en.wikipedia.org/wiki/Chicken_(game)

But the most useful link for the purposes of our discussion is the third one, which discusses the Evolutionary Game Theory. This Wikipedia article provides an example of the Hawk-Dove game that is closer to Question 25 that any other analysis I have come across. So it is best to begin from this example, because in this case we will have to make the least of adaptations and modifications in order to get to our answer.

Below, I provide a screenshot of the particular piece of the theory we are going to use:

And so, we can now begin to solve Question 25.

Here, the Hawks are the Dominator Furbles and the Doves are the Sharer Furbles.

We will construct the so called “payoff matrix” from the numbers given in the question.

If a Dominator encounters a Dominator, one Domintator gets to have 10 children, the other Dominator suffers the cost of 10 children. The net result is zero for each Dominator.

If a Dominator encounters a Sharer, the Dominator gets to have 10 children, the Sharer gets to have 3 children.

If a Sharer encounters a Dominator, the Dominator gets to have 10 children, the Sharer gets to have 3 children.

If a Sharer encounters a Sharer, each Sharer gets to have 5 children.

From the payoff matrix we can produce two equation with two unknowns, the population dencities. Let D be the population density of the Dominators and S be the population Density of the Sharers. The following two equations hold:

D + S = 1
0 * D + 10 * S = 3 * D + 5 * S

The first equation is deduced from the fact that we have population dencities. Each of the two populations is a fraction of the whole population. The whole population is equal to 1, so the population of the Dominators as well as the population of the Sharers is a fraction (more than 0 and less than 1) and the sum of these two populations is 1.

The second equation is deduced from the fact that the payoff for each encounter is proportional to the population density of the one that we will encounter, i.e. the chance, the posssibility that this encounter has in order to be materialized. So, the left part of the equation gives us the total payoff for the encounters of a Dominator (i.e. the first row of the payoff matrix). And the right part of the equation gives us the total payoff for the encounters of a Sharer (i.e. the second row of the payoff matrix). This equation holds because we assume a stable system, i.e. one with equal total payoffs.

This is a system of two equations than can easily be solved. From the firat equation we have D = 1 – S. Then from the second equation we have 0 + 10 * S = 3 * (1 – S) + 5 * S => 10 * S = 3 – 3 * S + 5 * S => 10 * S + 3 * S – 5 * S = 3 => 8 * S = 3 => S = 3/8.

Thus, the population densities are S = 3/8 and D = 5/8. So, when the total population is 1, the individual populations are 5/8 Dominators and 3/8 Sharers. Now that the total populationn is 2000 Furbles, the Dominator Furbles are (5/8) * 2000 = 1250 and the Sharer Furbles are (3/8) * 2000 = 750.

Thus, the answer is that we expect 1250 Dominator Furbles.

Perhaps you have calculated other values for the payoff matrix. Be careful though. Before you ask if your values are correct, try to solve the system of the two equations and see what results you get. Remember, the population densities that you obtain must each be from 0 to 1. So, each population density must be a positive number less than 1. Otherwise, it has no biological meaning.

Let me give you an example. Suppose you calculate the payoff of a Dominator-Dominator encounter. And suppose you forget or do not want to include the cost, but only the gain. So, instead of (10-10)/2=0 you might enter (10+0)/2=5 as the payoff. Let us see what we get in this case: 5 * D + 10 * S = 3 * D + 5 * S => 5 * (1 – S) + 10 * S = 3 * (1 – S) + 5 * S => 5 – 5 * S + 10 * S = 3 – 3 * S + 5 * S => – 5 * S + 10 * S + 3 * S – 5 * S = 3 – 5 => 3 * S = -2 => S = -2/3. So, S is negative and this is an absurd result, biologically speaking.

To recapitulate, the question is correct and well posed. It is a question about population dynamics, and it is meant to be solved as a Hawk-Dove game. Actually, the question is so well posed, that I would not be surprised if the authors of the test asked John Maynard Smith himself to author the question. Anyway, although the question is an excellent question on the Hawk-Dove game, it seems to me that it has no place in an IQ test.

Posted in Education | Comments Off on My attempt at Question 25 from the Haselbauer-Dickheiser Test

My attempt at Question 6 from the Haselbauer-Dickheiser Test

The Haselbauer-Dickheiser Test can be found at http://matrix67.com/iqtest/.

In this blog post, I will study Question 6 from this test.

The question is about physicists’ names as strings. Particularly, each alphabet letter is assigned a number and we have to find which number each letter has assigned to it, given the sums of strings, where each string is the surname of a famous physicist.

Below I present the original question for your convenience:

Please do not read the rest of this article, if you want to attempt to solve this question on your own. The rest of this article describes my attempt at solving this question and you should not read it, unless you want to or you do not mind coming across relevant ideas, spoilers, hints, solutions, and strong opinions concerning this test.

You have been warned and I now consider that you continue to read knowing that what you come across for the rest of this article may forever spoil things for you and/or present strong opinions against this test.

Last warning: please do not read this blog post, unless you are certain that you know what you are doing. If you are not sure, then it would be best if you stopped reading at this point.

OK. If you are here, it means that you want to know my opinion. Well, ok then!

To cut a long story short, my opinion is that the test is highly inappropriate. In this blog post, I will focus on the study of question 6.

I have not solved this question, but I promise you, dear reader, that if you bare with me, I will make this blog post worthwhile for you to read and I will provide interesting information about this question.

This is a question that is really hideous and I am against it. It is a very inappropriate question. The question makes you think that you need  to solve a system of linear equations and this is true to some extent. But if you look carefully, FEYNMAN has the letter Y in his name, and the letter Y does not appear anywhere else in the names given. What is worse, the same happens for the letters J, Q, X. Thus, four letters (J,Q,X,Y) do not appear in the 25 names for which we are given the sum of the letters that comprise them.

So, the question is about finding the value of each letter, but the value cannot be solely derived from the sums. We also have to guess. And this is what makes this question inappropriate.

First of all, we have 25 equations with 22 unknowns. The 25 equations are the following:

r+o+2*e+2*n+t+g=104
l+o+r+e+n+t+z=102
c+u+r+i+e=69
m+i+c+h+e+l+s+o+n=109
l+i+2*p+m+a+n=88
m+a+r+c+o+n+i=90
k+a+m+e+r+l+i+n+g+h=120
p+l+a+n+c+k=96
s+t+a+r+k=60
2*e+2*i+2*n+s+t=94
b+o+h+r=47
m+2*i+2*l+k+a+n=103
s+i+e+g+b+a+h+n=87
p+e+2*r+i+n=98
2*r+i+c+h+a+d+s+o+n=155
h+3*e+i+s+n+b+r+g=118
s+c+h+2*r+o+d+i+n+g+e=168
2*c+h+a+d+w+i+k=114
a+n+d+e+r+s+o+n=127
d+a+v+i+2*s+o+n=103
f+e+r+m+i=57
s+t+e+r+n=70
b+l+o+c+k=73
z+2*e+r+n+i+k=99
c+h+2*e+r+n+k+o+v=109

And the 22 unknowns are the following:

a
b
c
d
e
f
g
h
i
k
l
m
n
o
p
r
s
t
u
v
w
z

Please note that the letters j, q, x, y are missing.

I pasted the 25 equations and the 22 unknowns in a form on the site https://quickmath.com. Specifically, I used the Equations -> Solve -> Advanced page https://quickmath.com/webMathematica3/quickmath/equations/solve/advanced.jsp as shown below:

Immediately the site produced the following three images. The first image contains the 25 equations:

The second image contains the 22 unknowns:

The third image contains the solution to this system of linear equations:

Below I present the above image in text form, in case you need to copy it.

a = 6
b = 11
c = 20
d = 24
e = 7
f = 16
g = 12
h = 5
i = 9
k = 13
l = 21
m = 2
n = 22
o = 8
p = 14
r = 23
s = 15
t = 3
u = 10
v = 4
w = 17
z = 18

Below you can see how I used Excel to check the validity of the results. The results are correct. The value for FEYNMAN is calculated by setting Y=0, since we do not obtain its value from solving the system of linear equations.

Below you can see the results listed in Excel, first sorted by letter and then sorted by number.

Below are the same results, in text form, should you need to copy them.

? M T V H A E O I U B G K P S F W Z ? C L N R D ? ?

06 11 20 24 07 16 12 05 09 ? 13 21 02 22 08 14 ? 23 15 03 10 04 17 ? ? 18

We can see that solving the system of linear equations gives us the values of 22 letters of the alphabet, and we miss the values for the letters j, q, x, y. If we assume that each letter is assigned a different value from 1 to 26, then the missing numbers are 1, 19, 25, 26. The previous assumption, that each latter is assigned a different value from 1 to 26, should be valid.

But what are the values of the missing letters j,q,x,y? Which is 1, which is 19, which is 25 and which is 26?

I do not know. We have to guess. We can either consider the pattern of the letters (sorted by their corresponding number) ? M T V H A E O I U B G K P S F W Z ? C L N R D ? ?, the pattern of the numbers (sorted by their corresponding letter) 6 11 20 24 07 16 12 05 09 ? 13 21 02 22 08 14 ? 23 15 03 10 04 17 ? ? 18, or the relationship between each letter and its corrsesponding number. Any pattern that might be found, might be the pattern that was used for the assignments. Another psossibility might be that a clever hash function was used, a function that provides a rearrangement of the numbers from 1 to 26.

Anything goes. But to assume is not a mature thing to ask from someone. Actually, I believe that most people who solved this question gamed the system. They did not really find the pattern of number assignment and instead tried all four possibilities, one by one, to see which one the system accepted as correct. And if they were ever asked how come they proposed their solution, they would just say the truth, or they would just provide a tongue-in-cheek answer that they would know that it would be deemed incorrect.

So the four possibilities are:
Y=1 => FEYNMAN = 75 + 1 = 76.
Y=19 => FEYNMAN = 75 + 19 = 94.
Y=25 => FEYNMAN = 75 + 25 = 100.
Y=26 => FEYNMAN = 75 + 26 = 101.

I believe that the whole point of this question is for us to find the pattern that was used for the assigment of numbers to the alphabet letters. This is why I skimmed throught the solution of the system of linear equations. Indeed one may spend time and reduce the system of the 25 equations to smaller systems and find clever ways to solve it. Or one may solve it manually. Or one may create a program that solves it. But this is not what the question is about. The question is about guessing the value of the letter Y after we have found the values of the 22 letters. And, to me, this is an inappropriate thing to ask.

Update, April 2, 2021: A neat way to assign the remaining letters to numbers might be what I will now describe.

I assume that we have an one-to-one correspondence between the numbers from 01 to 26 and the letters of the alphabet.

We can assume the default order for either the numbers (01, 02, …, 26) or the letters (A, B, …, Z).

Wikipedia has a nice article on permutations, their notation, and the cycles of a permutation. Here is the link:

https://en.wikipedia.org/wiki/Permutation

Taking the above under consideration, I will now present an analysis of both the permutation of the numbers and the permuation of the letters that occurs from the system of equations that we solved. For both the numbers and the letters, I will present a two-line notation for the corresponding permutation, following with the depiction of the cycles that occur.

For the permutation of the numbers, we have the following analysis:

Two-line notation:

01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 
06 11 20 24 07 16 12 05 09  ? 13 21 02 22 08 14  ? 23 15 03 10 04 17  ?  ? 18

Missing numbers: 01 19 25 26

Cycles of the permutation, from shorter to longer:

09 -> 09 

03 -> 20 -> 03

02 -> 11 -> 13 -> 02

26 -> 18 -> 23 -> 17 -> ?

01 -> 06 -> 16 -> 14 -> 22 -> 04 -> 24 -> ?

19 -> 15 -> 08 -> 05 -> 07 -> 12 -> 21 -> 10 -> ?

All numbers appear in the cycles, including the numbers 01 19 26. 
The only number missing is the number 25.

For the permutation of the letters, we have the following analysis:

Two-line notation:

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
? M T V H A E O I U B G K P S F W Z ? C L N R D ? ?

Missing letters: J Q X Y

Cycles of the permutation, from shorter to longer:

I -> I

C -> T -> C

B -> M -> K -> B

Q -> W -> R -> Z -> ?

X -> D -> V -> N -> P -> F -> A -> ?

J -> U -> L -> G -> E -> H -> O -> S -> ?

All letters appear in the cycles, including the letters J Q X. 
The only letter missing is the letter Y.

From the get-go, we knew that the numbers 01 19 25 26 and the letters J Q X Y are missing.

But any one of the four numbers can correspond to any one of the four letters. The analysis of the permutation does not guarantee anything more. We can create any one-to-one correspondence between these 4 numbers and 4 letters and the full 26-element permutation analysis will be updated accordingly.

What I mean is that for any one-to-one correspondence between 01 19 25 26 and the letters J Q X Y, we will end up with a two-line notation and list of cycles. Now, one correspondence might, for example, change the number of cycles, so that the three incomplete cycles might end up as only one very long cycle.

So, the actual 26-element correspondence determines the cycles, and not the other way around. But I had the idea that there is one one-to-one correspondence between 01 19 25 26 and the letters J Q X Y that ends up in a neat analysis. This is of, course, quite subjective, to say the least. Thus, I do not consider this a robust argument. Again, it is the permutation that should dectate the analysis and the number and compostion of the cycles, and not the other way around. But, judging from the way the cycles are composed thus far, it only takes a small step to arrive at the following final configuations.

For the permutation of the numbers:

01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 
06 11 20 24 07 16 12 05 09 19 13 21 02 22 08 14 25 23 15 03 10 04 17 01 26 18

Missing numbers: 01 19 25 26

09 -> 09 

03 -> 20 -> 03

02 -> 11 -> 13 -> 02

26 -> 18 -> 23 -> 17 -> 25 -> 26

01 -> 06 -> 16 -> 14 -> 22 -> 04 -> 24 -> 01

19 -> 15 -> 08 -> 05 -> 07 -> 12 -> 21 -> 10 -> 19

For the permutation of the letters:

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
X M T V H A E O I U B G K P S F W Z J C L N R D Q Y

Missing letters: J Q X Y

I -> I

C -> T -> C

B -> M -> K -> B

Q -> W -> R -> Z -> Y -> Q

X -> D -> V -> N -> P -> F -> A -> X

J -> U -> L -> G -> E -> H -> O -> S -> J

As we can see from the above, the number 25 goes into the 4th cycle and we just end each of the last three cycles without connecting them with each other. Thus, the proposed correspondence between the numbers 01 19 25 26 and the letters J Q X Y is:

J -> 19
Q -> 25
X -> 01
Y -> 26

Again, we need not have these cycles. We can create any other one-to-one correspondence between the numbers 01 19 25 26 and the letters J Q X Y, and then study the resulting analysis for the full 26-element permutation. It may not be as neat as the one presented above, but, then again, this is not a definitive argument.

Posted in Education | Comments Off on My attempt at Question 6 from the Haselbauer-Dickheiser Test

My attempt at Question 15 from the Haselbauer-Dickheiser Test

The Haselbauer-Dickheiser Test can be found at http://matrix67.com/iqtest/.

In this blog post, I will study Question 15 from this test.

The question is about packing spheres in a rectangular box.

Below I present the original question for your convenience:

Please do not read the rest of this article, if you want to attempt to solve this question on your own. The rest of this article describes my attempt at solving this question and you should not read it, unless you want to or you do not mind coming across relevant ideas, spoilers, hints, solutions, and strong opinions concerning this test.

You have been warned and I now consider that you continue to read knowing that what you come across for the rest of this article may forever spoil things for you and/or present strong opinions against this test.

Last warning: please do not read this blog post, unless you are certain that you know what you are doing. If you are not sure, then it would be best if you stopped reading at this point.

OK. If you are here, it means that you want to know my opinion. Well, ok then!

To cut a long story short, my opinion is that the test is highly inappropriate. In this blog post, I will focus on the study of question 15.

In my previous blog post, I promised that this blog post would be about this question. This is because both the question I examined in the previous blog post and the question I examine in this blog post are asked and answer in the same book and the same chapter. And their answers are given on the same page! Imagine that! The authors of this test appear to be completely deranged!

The book I am talking about is Martin Garner’s “New Mathematical Diversions”. The chapter I am talking about is chapter 7, “Packing Spheres”. The page this question is asked is page 86. The page this question is answered is 90. Below I provide the whole chapter, for your convenience:

The whole chapter is fascinating, but before I discuss the question further, I would like to draw attention to page 86, right after the question is posed, where Martin Gardner answers a very interesting question: how come honeycombs are hexagonal.

Ok, back to the question at hand. The answer is that the maximum number of spheres is 594 and Martin Gardner provides this answer and the explanation on page 90. The explanation Martin Gardner provides is extremely brief, to the point of it being difficult to be understood. In the rest of this blog post, I would like to make this answer a little more understandable.

Here is the answer provided by Martin Gardner:

Turn the box on its side and form the first layer by making a row of five, then a row of four, then of five, and so on.

It is possible to make eleven rows (six rows of five each, five rows of four each), accommodating 50 balls and leaving a space of more than .3 inch to spare.

The second layer also will take eleven rows, alternating four and five balls to a row, but this time the layer begins and ends with four-ball rows so that the number of balls in the layer is only 49.

(The last row of four balls will project .28+ inch beyond the edge of the first layer, but because this is less than .3 inch, there is space for it.)

Twelve layers (with a total height of 9.98+ inches) can be placed in the box, alternating layers of 50 balls with layers of 49, to make a grand total of 594 balls.

The above answer is provided by *the* man Martin Gardner, and I use bold italic font for this. I took the liberty to divide the answer in paragraphs. The first two paragraphs talk about the first layer of spheres, the second two paragraphs talk about the second layer of spheres and the fifth and last paragraph talk about all layers of spheres. Below, I will divide the answer into these three steps: studying the first layer, then the second layer, then all layers.

So, let me begin. All lengths below are in inches. I will use the Pythagorean theorem and the fact that when two spheres touch, the line segment that connects their centers and is the distance between their centers is equal to the sum of their radii. And since we are dealing with spheres of the same radius, when two spheres touch, their distance is twice their radius.

I will first study the first layer. We have the box sitting flat on a horizontal table and the box’s dimensions are length=10, width = 5 and height = 10. So the height is one of the two large edges. In the following image, I put some spheres, not all, only the first five rows. The image depicts the first layer, as we look down from above.

In the following image, which is the same as the image above, I calculate the distance between two rows of spheres. The distance is the red vertical line segment. Each dotted line passes through the centers of the spheres in a row, so the red vertical line segment is indeed the distance between two consecutive rows.

Let me denote with x the vertical line segment which is the distance between two rows of spheres. From the right triangle we have:

x^2 + r^2 = (2r)^2 =>
x^2 = (2r)^2 – r^2 =>
x^2 = (2r + r) (2r – r) =>
x^2 = 3r^2 =>
x = r * sqrt(3) =>
x = 0.5 * 1.7321 =>
x = 0,866

So how many spheres can be packed in the first layer? I made the following table, adding distances, remembering to account for the initial and final 0.5 inches of the first and last row, and being careful not to exceed the 10 inches which is the length of the box. The table is the following:

As far as the previous table is concerned, in the first column I denote lengths  and in the second column I denote number of spheres in a row. Red color denotes a sum.

This table explains what Martin Gardner meant when he wrote:

Turn the box on its side and form the first layer by making a row of five, then a row of four, then of five, and so on.

It is possible to make eleven rows (six rows of five each, five rows of four each), accommodating 50 balls and leaving a space of more than .3 inch to spare.

Martin Gardner was describing the arrangement of the first layer as we look down on the box from above and I just explained these two paragraphs of Martin Gardner’s text.

Moving on, I will now study the second layer as we look down from above. The following image presents the layout of the second layer of spheres as we look down from above.

I will now superimpose the second layer on top of the first layer, thus producing the following image.

In the image above, the first layer is depicted containing blue spheres and the second layer is depicted containing white translucent spheres. So, the white translucent spheres are above the blue spheres and we are looking from above the white translucent spheres.

To calculate distances, I made the following image. In this image, we have three blue spheres from the first layer and a red sphere from the second layer. I changed the color of the second layer sphere from white translucent to red, so it is easier to see.

Here, I want to calculate the horizontal displacement that the center of the second layer sphere has in relation to the center of the first row of spheres from the first layer. This can be found from the green equilateral triangle formed by the centers of the three spheres on the first layer. The barycenter of this equilateral triangle is exactly below the center of the red sphere from the second layer. Thus, the displacement I am looking for is the distance of this barycenter from a side of this equilateral triangle.

If I denote h as one of the heights of the equilateral triangle (all three height being equal in an an equilateral triangle), then the distance of the barycenter (the point where all three heights meet) from one of the sides is (1/3)h.

The displacement I am trying to calculate is shown with the black line segment in the following image:

From the right triangle that is formed from one of the edges of the green equilateral triangle and one of its heights, we can calculate this displacement.

h^2 + r^2 = (2r)^2 =>
h^2 = 4r^2 – r^2 =>
h^2 = 3r^2 =>
h = r * sqrt(3) =>
(1/3)h = r * sqrt(3) / 3 =>
(1/3)h = 0.5 * sqrt(3) / 3 =>
(1/3/)h = 0,2887, almost equal to 0,289.

Adding this horizontal displacement for the second layer, my table for the second layer is as follows:

And this explains the following two paragraphs of what Martin Gardner wrote:

The second layer also will take eleven rows, alternating four and five balls to a row, but this time the layer begins and ends with four-ball rows so that the number of balls in the layer is only 49.

(The last row of four balls will project .28+ inch beyond the edge of the first layer, but because this is less than .3 inch, there is space for it.)

So I have explained the first two layers, and I can now talk about all layers. To do this, I need to calculate the vertical displacement that the second layer has in relation to the first layer, i.e. how much higher are the centers of the spheres of the second layer in relation to the centers of the first layer.

Up until now, we were looking the structure from above and we studied horizontal geometrical shapes and horizontal projections. Now we will need to think vertically. We will need to look the box from the side.

So, I want you to consider the same arrangement of the first and second layer as before:

But this time, imagine that we do not want to look this arrangement from above, but we want to look at it from the side, in order to find how much is the center of the red sphere on the second layer is elevated in relation to the centers of the blue spheres of the first layer.

This elevation, this vertical distance, let’s call x. We can find x from the following right triangle.

This right triangle is formed as follows: The side with length 2r is formed by the center of one of the spheres of the first layer and one of the spheres of the second layer. The side with length x is completely perpendicular. And the side with length (2/3)h is completely horizontal. Imagine that this is a triangle that we can see when we look at the arrangement in the image before that, from the side.

From the right triangle in the last image we have:

x^2 + ((2/3)h)^2 = (2r)^2 =>
x^2 + (4/9)h^2 = 4 r^2

But from the discussion about the second layer, we know that h^2 = 3r^2.

So we have:

x^2 + (4/9) (3r^2) = 4r^2 =>
x^2 + (4/3)r^2 = 4r^2 =>
x^2 = 4r^2 – (4/3)r^2 =>
x^2 = (12/3)r@ – (4/3)r^2 =>
x^2 = (8/3)r^2 =>
x = r * sqrt(8/3) =>
x = 0.5 * sqrt(8/3) =>
x= 0.8165.

And so the table for all the layers that can fit on the box is the following:

And this is what the following last paragraph of Martin Gardner’s text is about:

Twelve layers (with a total height of 9.98+ inches) can be placed in the box, alternating layers of 50 balls with layers of 49, to make a grand total of 594 balls.

Now, if you see at the image provided with the question, the spheres are arranged in a cubic manner, and this in no way will provide the maximum packing. This manner will provide a 5*10*10=500 spheres as the number of spheres that can fit in the box. The authors of the test are setting us up to fail. This is the behavior of a very bad person.

So I am asking: Would you want to be friends and belong to a group of people whose mentality is to judge people on their intelligence instead of their character, and that keep knowledge hidden, and that set others for failure?

Posted in Education | Comments Off on My attempt at Question 15 from the Haselbauer-Dickheiser Test

My attempt at Question 4 from the Haselbauer-Dickheiser Test

The Haselbauer-Dickheiser Test can be found at http://matrix67.com/iqtest/.

In this blog post, I will study Question 4 from this test.

The question is about spheres stacked up to create tetrahedral pyramids.

Below I present the original question for your convenience:

Please do not read the rest of this article, if you want to attempt to solve this question on your own. The rest of this article describes my attempt at solving this question and you should not read it, unless you want to or you do not mind coming across relevant ideas, spoilers, hints, solutions, and strong opinions concerning this test.

You have been warned and I now consider that you continue to read knowing that what you come across for the rest of this article may forever spoil things for you and/or present strong opinions against this test.

Last warning: please do not read this blog post, unless you are certain that you know what you are doing. If you are not sure, then it would be best if you stopped reading at this point.

OK. If you are here, it means that you want to know my opinion. Well, ok then!

To cut a long story short, my opinion is that the test is highly inappropriate. In this blog post, I will focus on the study of question 4.

This question is great, if you plan to submit it to an elementary school student. But for a test for high intelligence, it is ridiculous. Not only that, but this question (question 4) and another question from the test concerning spheres (question 15) are asked and answered in Martin Gardner’s book titled “New Mathematical Diversions”.

These are not the only questions that are blatantly copied word for word from Martin Gardner’s books, but here the situation becomes extreme: Two questions, from the same book and also …  wait for it… from the same chapter: chapter 7, titled “Packing Spheres”. What more can I say? All I can ask is: Are the people who created this test completely deranged?

I do not know where to begin. First of all, the answer is really easy to find. I would pose this question to an elementary school student. (By the way, speaking about “school”, who came up with this invention? Some invention they made. Give me the person who invented school and I will sent her to a group of students to lynch her.)

Before I present the answer I came up for this question, I would like to present the pages from Martin Gardner’s book where this question and the corresponding answer are mentioned. I promise that my next blog post will be about the other question from the test that is asked and answered in the same chapter.

So, question 4 from the test can be found in page 84:

and the corresponding answer can be found in page 90:

And yes, underneath the answer for this question, the answer for the other question is also found. One page, page 90, contains the answers to two questions from the test. And, yes, the creators of the test want to preserve the integrity of the test. And yes, all the above make no sense at all.

I will now present an answer to this question, by thinking as an elementary school student would. With simple thinking and the use of Excel, I found the answer in a few minutes.

I opened Excel. I created three columns with numbers. The first column had the natural numbers 1, 2, 3, … that denote the layer of a tetrahedral pyramid starting from the top. The second column had the number of spheres that comprise the layer. This number is equal to n(n+1)/2, where n is the number of the layer. The third column is the running total of the second column, thus it is the number of spheres of the layer and all the layers above it. Thus , the third column depicts the total number of spheres that a tetrahedral pyramid contains.

Now, the problem we want to solve is to take the numbers from the third column and, from these numbers, find the minimum number that is the sum of two others. In order to do that, I put the numbers in a horizontal line and in a vertical line, thus forming a square. For each pair of numbers, I had Excel calculate their sum. Then, all that was left to do, was to search in these sums to find a number from the third column.

I needed to only search either above or below the diagonal, since the sums below the diagonal are the same as those above the diagonal. Searching is easy because the value of the numbers is increasing from top to bottom and from left to right.

So, I found that the answer is 680.

A tetrahedral pyramid with 8 layers contains 120 spheres. A tetrahedral pyramid with 14 layers contains 560 spheres. A tetrahedral pyramid with 15 layers contains 680 spheres. 120+560=680 and 680 is the smallest number for which this occurs. That is, for which the sum of the spheres of two different tetrahedral pyramids is equal to the number of spheres of another tetrahedral pyramid.

 

Posted in Education | Comments Off on My attempt at Question 4 from the Haselbauer-Dickheiser Test

My attempt at Question 8 from the Haselbauer-Dickheiser Test

The Haselbauer-Dickheiser Test can be found at http://matrix67.com/iqtest/.

In this blog post, I will study Question 8 from this test.

The question is about discovering a pattern among line segments. There are many crossings among these line segments and each crossing is made by only two line segments.

Below I present the original question for your convenience:

Please do not read the rest of this article, if you want to attempt to solve this question on your own. The rest of this article describes my attempt at solving this question and you should not read it, unless you want to or you do not mind coming across relevant ideas, spoilers, hints, solutions, and strong opinions concerning this test.

You have been warned and I now consider that you continue to read knowing that what you come across for the rest of this article may forever spoil things for you and/or present strong opinions against this test.

Last warning: please do not read this blog post, unless you are certain that you know what you are doing. If you are not sure, then it would be best if you stopped reading at this point.

OK. If you are here, it means that you want to know my opinion. Well, ok then!

To cut a long story short, my opinion is that the test is highly inappropriate. In this blog post, I will focus on the study of question 8.

I do not know the answer to this question, but I promise you, dear reader, that if you continue reading the rest of this blog post, I will make every effort to make it useful to you.

First of all, I will make a brief discussion about the question. The question wants us to find the pattern behind the letter sequence that stems from the image.

It is the same as the person who posed the question is asking us to find what she is thinking. “What am I thinking?” This question may come from a child, since a child is immature. It may also come from a mentally retarded person. It may also come from a person who has the incorrect impression that intelligence can be measured and that people that happened to be good at guessing are intelligent. It may also come from sociopaths. Knowing and not revealing may be a good indication of a sociopath.

How can we handle such a question? First of all, we should not be handling it. It is not a proper question. It is not a valid question. It is a question a child would ask because of her immaturity. I always say to my students that there are no stupid questions, that any honest question is valid, has merit and deserves to be answered honestly. I say this to my students because no student ever asked me: “What am I thinking?” If a student ever asked me this question, I would certainly tell her that she is being silly.

This question wants us to find the reason behind the letter sequence and the reason stems from the image. The image depicts line segments. Not lines, but line segments. They each have a name, each is named by a different letter of the alphabet from A to Q. And there are a lot of crossings, each crossing by two of the segments. No three or more segments cross at a point. The closed areas that are formed are painted and I think that this is very important. This is done in order to exclude line segment L from the sequence.

Let me explain this important point.

As you can see, line segment L crosses line segment J, but no closed area is defined there, thus no color is applied. This is the clue that denotes that we should not take this line segment under consideration and that our attention should instead focus on the colored areas.

But I am not expressing it correctly. What I am really trying to say is this: The pattern that is given to us is Q P O N M. So, these are the letters backwards. So we would continue the sequence as follows: Q P O N M L K J. And this would be correct, because we would think that all line segments participate in the image, and the simplest explanation is that we take the letters in reverse order.

Occam’s razor. We will accept the simplest explanation. But here, line segment L is obviously the odd one out. It creates a crossing that does not correspond to a painted area. And this is shown in a very conspicuous way. Thus, the creator of the question definitely wants us to steer away from choosing L as the next line segment.

It is important to point out that the simplest explanation that takes all evidence into account is the best explanation. Thus, if we find two models that explain all evidence, then we should choose the simplest one. But of course, we must not discard any evidence. So, the alphabet letters in reverse order is too simple a model and does not encompass all evidence from the image. So, we should discard it.

Let us see if we can find other models that account for all evidence. If we can, then the simplest ones should be candidates for the solution.

Oh, and I have to stress another important point. The pattern we find has to produce a sequence starting with Q P O N M and it has to do it in a unique manner for any member of the sequence. It has to be unambiguous what each member in the sequence should be, starting from Q and going all the way to the last member of the sequence.

Let me begin by counting the number of crossings that each line segment makes.
A 7
B 9
C 9
D 8
E 7
F 7
G 9
H 2
I 2
J 8
K 4
L 1
M 8
N 10
O 10
P 10
Q 10

Below I present the same sequence, sorted from the most crossings to the least crossings and with a secondary sort of reverse alphabetic order.

Q 10
P 10
O 10
N 10
G 9
C 9
B 9
M 8
J 8
D 8
F 7
E 7
A 7
K 4
I 2
H 2
L 1

If we take the even numbers from the above sequence, we have the following sequence: Q P O N M J D K.

Let us see if we can find any other pattern that produces a sequence beginning with Q P O N M and unambiguously defines each member of the sequence up to the last.

We start from the last alphabet letter that exists: Q. What line segments does Q cross? We can find them easily and we can order the alphabetically and we can take the last one. It is  P.

Now we take P. What line segments does P cross? We order them alphabetically and take the last one not already in the sequence. It is O.

Now we take O. What line segments does O cross? We order them alphabetically and take the last one not already in the sequence. It is N.

Now we take N. What line segments does N cross? We order them alphabetically and take the last one not already in the sequence. It is M.

Now we take M. What line segments does M cross? We order them alphabetically and take the last one not already in the sequence. It is G.

Now we take G. What line segments does G cross? We order them alphabetically and take the last one not already in the sequence. It is K.

Now we take K. What line segments does K cross? We order them alphabetically and take the last one not already in the sequence. It is J.

So the sequence produced is Q P O N M G K J.

Ok, we found two different and simple patterns that produce an unambiguous sequence. The first pattern sorts the line segments by decreasing number of crossings, then in reverse alphabetical order and discards the odd numbers. The second pattern begins from the last letter assigned and for each line segment takes the biggest letter whose line segment it crosses. Which of these two patterns is the simplest? It is subjective. Is there a simpler pattern that requires even less creativity from our part? Hard to say.

Here is another pattern: Let us begin from Q. Again we will find the line segments that Q crosses. We will order them alphabetically and will take the last two and list them in reverse alphabetical order. So from Q we get P O.

Then we take O. We find the line segments that it crosses and are not yet part of the sequence, we order them alphabetically and take the last two.

So from O we get N M. And so on.

So, this is another pattern that produces Q P O N M unambiguously and can continue to produce unambiguous members of the sequence. And other patterns such as the above can be found. And I stress the fact that they can produce the sequence members in an  unambiguous manner. But since they are more elaborate, more creative and more complex, we should discard them.

The same goes for taking into consideration the number of areas attached to each line segment or the corners of the areas attached to each line segment. They are quite complex and do not correspond well to the sequence Q P O N M. But, if you want to try for yourself, below I present an image depicting the number of edges for each one of the areas.

Since there are already many simpler models producing the pattern Q P O N M in an unambiguous and unique manner, I believe that the image above should be of no help to you, but I provide it in case you want to study it.

Update January 1st, 2019: Some people who read this blog post, understood it incorrectly. When I say that I have not solved this question, I am being polite.

I have solved this question.

Or rather, with this blog post, I proved that the question is subjective, and thus, incorrect.

In this blog post, I revealed and provided answers and contradictions from the research I did.

Please do not mistake my politeness for inability.

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My attempt at Question 17 from the Haselbauer-Dickheiser Test

The Haselbauer-Dickheiser Test can be found at http://matrix67.com/iqtest/.

In this blog post, I will study Question 17 from this test.

The question is about cube nets covering an area.

Below I present the original question for your convenience:

Please do not read the rest of this article, if you want to attempt to solve this question on your own. The rest of this article describes my attempt at solving this question and you should not read it, unless you want to or you do not mind coming across relevant ideas, spoilers, hints, solutions, and strong opinions concerning this test.

You have been warned and I now consider that you continue to read knowing that what you come across for the rest of this article may forever spoil things for you and/or present strong opinions against this test.

Last warning: please do not read this blog post, unless you are certain that you know what you are doing. If you are not sure, then it would be best if you stopped reading at this point.

OK. If you are here, it means that you want to know my opinion. Well, ok then!

To cut a long story short, my opinion is that the test is highly inappropriate. In this blog post, I will focus on the study of question 17.

This question shows most everything that is wrong and bad about the whole test. I do not know where to begin. Honestly.

Ok, I guess I will have to begin from somewhere. Any start I think of choosing, I can find a worst one. So I guess I will start from the first thing that comes to my mind: discrimination and unfair advantage. This question about cube nets. Hexominoes and cube nets as their subsets are know to elementary school children in the USA. But not in Greece. Here, a child will be lucky if she only encounters and learns about one of the cube nets. But in the USA, children learn about the 11 kinds of cubes nets. They are even given all kinds of hexominoes and asked to discern which ones are cube nets. A child in Greece, nay, any person in Greece has absolutely no notion of what I am writing about. So, how come do I know these things? Well, I am not like other Greeks. Although I was born and raised in Greece, my culture resembles that of an American. Blame it on TV and radio. Blame it on all the scientific books in English I have read throughout my life. Blame it on me having interests way different that other Greeks. Well, blame it on anything you want, my culture has nothing to do with a usual Greek person’s culture. I have listened to the great radio programs from AFRTS one too many times. I have watched American movies and TV series one too many times. I have read American text books one too many times. So, the point I am trying to make is that any normal but also highly intelligent Greek would not understand what this question is about and/or be at a great disadvantage compared to an American.

Another thing that really bothers me is that this question allows you to cheat. The grid has 96 squares. Each cube net has 6 squares. 16 X 6 = 96. So, the least amount of cube nets that would cover the grid with no overlap is 16. So, it is probable that the answer is 16 cube nets. If not, then the test taker can start adding one more to the answer, until she finds the correct answer. In hindsight, and as I will show in the last part of this blog post, the answer is indeed 16, meaning that the grid can be covered with cube nets with no overlap among them.

This really bothers me because one may go ahead and guess that the answer is 16. Is is easy to guess that. Coming up with a configuration that achieves it is a little bit harder. But the test does not demand an actual solution, an actual configuration. It just demands the number. This is unacceptable to me. And, in any event, I cannot consider this question and the corresponding answer as indicators of high intelligence. All they are is a fun way to spent an afternoon. And this is something that unfortunately I cannot say for the rest of the questions of this test, because most of them I consider boring and tedious. In these series of bog posts, I want to prove that this test is detrimental to people and mankind, but I am faced with boredom, since for many of the questions, although I know how to solve them, I find it tedious to get involved with them. I have always been honest with you, dear reader, and I am not willing to change that. So, in all honesty, if I had a girlfriend, I would not have gone to the trouble of solving the questions. I would have devoted my time to said girlfriend. But since I have spare time, I do this, in case you find it helpful. And still, each time I try to tackle a question, I say to myself that this is stupid and pointless and there are way better and more sensible problems I can solve in my spare time.

As for this problem, I previously said that it is a fun way to spent an afternoon. Well, yes but for toddlers. Mental note: Next time toddles are in the house, (toddlers are frequently here, children of friends and neighbors) I would give this puzzle to them to solve it. And some others puzzles from this test. Because, honestly, I am grown person and I should not be involved in this nonsense. Am I certain that some of these questions may pass as legitimate to a child. But if I tell them that this is test that measures high intelligence, I know what they would tell me. I have been around long enough with toddlers to know their come back answer. If I tell them that this is test that measures high intelligence, they will tell me: “I don’t believe you.” And the last thing you want is to lose your credibility with toddlers. They are harsh and unforgiving judges.

I can say a lot more about how bad this question is and how atrocious the whole test is, but let’s cut this short and instead focus on the solution of this particular question. I was talking about hexominoes. They are constructs made from six squares. From all possible hexominoes, 11 are cube nets. You can look up “hexominoes” and you can look up “cube nets”. To help you, I present the following two links:

Hexomino

Cube

For your convenience, I present a drawing containing the 11 cube nets.

The question wants us to use these to cover the grid given. We are not allowed to place a cube net that is partly outside the grid, and the word “outside” includes the white squares that are inside the grid. The cube nets must not cover any part of those as well. Each cube net must cover blue squares of the grid and only them. And we want to cover the grid with the least amount of cube nets. If we can achieve no overlap among the placement of the cube nets, then this is the configuration with the least amount of cube nets. And yes, we can achieve such a configuration. And since there are 96 squares in the grid and each cube net has 6 squares, a total if 16 cube nets cover the grid and it is the least number of cube nets that can achieve this.

Let me give you such a configuration and tell you how I came up with it. I prsent my solution below:

No, it is not some satanic drawing of symbols, I just opened the grid in Paint and started drawing cube nets with a yellow color line. I noticed that the grid has a 4-way symmetry. So, if I was able to fill one side with 4 cube nets, then the other 3 sides would be exactly the same. So, I had to find a configuration 0f 4 cube nets which would fit with itself in its adjacent areas, in its borders.

So, what I did is that started placing two cube nets on a side and I would do the exact same arrangement in the other 3 sides. Then I had to be extremely careful so that the third cube net would be such is such a way that it allowed the fourth cube net to cover exactly the remaining space. This was easier said than done, bit I finally came up with the above configuration. I hope you notice the 4-way symmetry. To help you, you can imagine any four cube nets in series. When you repeat this series, each time rotating it 90 degrees, you will fill the entire grid. Each of the four sides of the grid is filled in exactly the same way and with exactly the same cube nets in exactly the same arrangement.

One last thing I would like to point out about unfair advantages is that although I solved this problem with just thinking about it, there are people who are involved with hexomino tiling as scientific research. And computer programs also exist that allow you to specify a grid with forbidden areas and to specify which hexominoes to use then and perform the tilling of the grid for you. Test takers with such knowledge and/or programs are at a huge advantage over other test takers.

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My attempt at Question 11 from the Haselbauer-Dickheiser Test

The Haselbauer-Dickheiser Test can be found at http://matrix67.com/iqtest/.

In this blog post, I will study Question 11 from this test.

The question is about a game played in a hexagonal grid.

Below I present the original question for your convenience:

Please do not read the rest of this article, if you want to attempt to solve this question on your own. The rest of this article describes my attempt at solving this question and you should not read it, unless you want to or you do not mind coming across relevant ideas, spoilers, hints, solutions, and strong opinions concerning this test.

You have been warned and I now consider that you continue to read knowing that what you come across for the rest of this article may forever spoil things for you and/or present strong opinions against this test.

Last warning: please do not read this blog post, unless you are certain that you know what you are doing. If you are not sure, then it would be best if you stopped reading at this point.

OK. If you are here, it means that you want to know my opinion. Well, ok then!

To cut a long story short, my opinion is that the test is highly inappropriate. In this blog post, I will focus on the study of question 11.

Even though I have not really understood this question, I promise you, dear reader, that if you are patient enough to read this blog post, I am sure that you will find it useful nonetheless.

Question 11 is about a fictional game: Quess (I guess the name is coined so that it roughly reminds us of the game of chess). Quess is played on a specific hexagonal grid. In Quess, there may be many kinds of pieces, but we are informed about two of them: Warriors and Warp Gates. From now on, I will denote Warriors as red circles (disks) and WarpGates as blue circles (disks). Warriors may move up to seven spaces (hexagons) in a straight line, unless they encounter a WarpGate, in which case they warp to another WarpGate and continue their path in the same direction. WarpGates may move up to one space (hexagon) in any direction.

We are asked to find the minimum number of WarpGates that allow only four Warriors to attack the entire grid.

Already, a few questions come to mind:

When a Warrior encounters a WarpGate, the Warrior has to warp to another WarpGate. But can the Warrior choose which WarpGate it should warp to? Of course, if we use up to two WarpGates, this question does not apply. But what if we have more than two WarpGates? The rest of this blog post will assume that the Warrior is able to choose which WarpGate it will warp to.

We would like to use four Warriors to attack the whole grid. But why? Only one Warrior can attack the whole grid just as fine. A single Warrior can move only in a straight line. Once it stops, it can then move a gain to a different direction, thus visiting the whole grid. Do we want to have four Warriors, so that whatever space (hexagon) in the grid we choose, it can be immediately attacked by just moving one of the Warriors only once? The rest of this blog post will assume that.

If we assume that we want to have four Warriors able to reach any space (hexagon) in a single move, and we want to place the least number of WarpGates to achieve this, why does the question state that each WarpGate moves by only one space? Are we going to be moving WarpGates?

The rest of this blog post will assume that we will place four Warriors on the grid, trying to have them reach any one space (hexagon) in a single move. We will place the least amount of WarpGates in order to achieve that and we will not move any Warp Gate. Now, the question states how each WarpGate moves, but why it does so is a mystery to me. So, perhaps I have not understood the question. So, in the rest of this blog post, I might be solving the wrong problem.

For the following discussion,I found  the following hexagon grid from Pratt & Larson.

I used the above grid to create the following Quess grid. You can use this Quess grid to try to solve the question.

The first thing I did when I created the Quess grid was to study it. For this, I drew straight lines, arriving at the following image. Please note that the following image contains no Warriors or WarpGates, just straight lines that can be drawn. This image helped me understand the way Warriors move on the grid.

From then on, I began to solve the problem. Keep in mind that I might be solving the wrong problem, since I have not understood the question.

Also, please note that in the following images, red circles denote Warriors, blue circles denote WarpGates, red lines denote paths a Warrior takes and blue lines denote paths a Warrior tales after it emerges from a WarpGate. Red paths (lines) always begin from a red circle. Blue paths (lines) always begin from a blue circle.

In addition please note that, in order to not clutter the images, I do not draw many parts of the red paths (especially if the hexagons where the paths lead have already been covered by another path) and I do not draw many blue paths. I do not draw only all lines and I do not draw them to their full extent. This way, it is easier to see the logic.

So, the first attempt I made is the following image, where there are four Warriors and they cover the whole grid except 6 hexagons. Now, if we place a WarpGate in each of these hexagons, then we have the whole grid covered with 6 WarpGates. But these 6 WarpGates are not used for warping, since they are not reachable by Warriors. These 6 WarpGate are only used to fill hexagons. Are we allowed to do that? That’s another question.

Moving on, the following attempt uses 2 WarpGates, but leaves 3 hexagons uncovered. May we fill them with more WarpGates?

The following attempt still uses 2 WarpGates, but leaves 2 hexagons uncovered. Better, but not adequate.

The following attempt uses 4 WarpGates. Here, the whole grid is covered and I present this as a solution.

Here is how the above grid is covered:

If the Warrior from above enters the middle left WarpGate, it will warp to the right WarpGate and continue downwards.

If the Warrior from above enters the middle right WarpGate, it will warp to the left WarpGate and continue downwards.

If the Warrior from below enters the middle left WarpGate, it will warp to the right WarpGate and continue upwards.

If the Warrior from below enters the middle right WarpGate, it will warp to the left WarpGate and continue upwards.

So, with 4 WarpGates, we have covered the whole grid. Can we do it with less?

I am not certain. I doubt that it can be done with 0 WarpGates. What about with 1 WarpGate? Well, once a Warrior enters a WarpGate, it has to emerge from another WarpGate and not the same. So, if there would be a single WarpGate, it would have to be unreachable to the Warriors and it will be used just to fill a hexagon. What about with 2 or 3 WarpGates? I cannot exclude this possibility. All I was able to do is to solve the problem with 4 WarpGates and the solution is in the image above.

Of course, as I have already mentioned, I might be solving the wrong problem.

Now, this is one of the many questions in the test that the test taker can find the answer without understanding or solving the question, thus gaming the system. The test taker can answer 0, then 1, then 2 and so on until the system rewards her.

I guess this is why the creators of the test say that the average person answers less than two questions correctly. They look at the answer submissions. But these submissions are made frantically from people who try to game the system to reveal the answers. So, a person would submit again and again, giving purposely wrong answers to all questions except the one she is trying to guess. This way, using the system as a oracle, she can deduce the answer to many questions that do not have a free form. And most of the questions do not. So, the system can be gamed easily. What the creators of the test witnessed when they reviewed the submission logs were not honest attempts by average people but coordinated attempts by people who were gaming the system and were successful at that, as well. Each attempt had as target to find the answer to only one question, thus all other answers were purposely incorrect. To the creators of the test, this seemed like the people that were submitting were trying but could only get one correct answer at most. The creators of the test got the wrong impression, because they thought that the submissions in their logs were honest ones. Little did they know that a large number of test takers succeeded in gaming and abusing their system.

At least, that’s my theory. When the system was live I knew nothing about it. I never used it. So, I can only speculate as to how it worked and how the submissions of the answers were made.

Anyhow, I do not like the fact that I tried to answer this problem (question 11), only to find that I had questions about it and no one to answer them. And I do not like the fact that the answers are not provided. Knowing but not telling: I find it to be a childish behavior.

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