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 images 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 ? ?

6 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 linera 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.

]]>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?

]]>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.

]]>

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:

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 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.

]]>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:

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:

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.

]]>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:

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 awards 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.

]]>In this blog post, I will study Question 3 from this test.

The question is about the maximum number of regions that a circle can be divided if we draw chords connecting points on its circumference.

Below I present the original question for your convenience:

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 3.

Well, any question from this test leaves me speechless, but this is not a compliment to the test. I find that the test is pointless, tedious, unintelligent and lacks explanation. I cannot get it. I cannot understand it. I do not know what to say. I am speechless.

Here we have this question. Question number 3. What does it want from us? Well, it is very clear what it wants. I wants us to find the maximum number of regions that are created when there are seven points on the circumference of a circle and we connect them with straight lines.

It is a nice problem and a known one, with a known solution. Since the authors of the test want it to maintain its integrity, why pose a well known problem? Points for originality: Nada! Since the authors of the test want to test intelligence, why ask a question that in this blog post I will prove that a toddler can answer? Points for intelligence: Nada! What is the purpose of such a question in a test for high intelligence? Points for sense and sensibility: Nada!

When I try to solve a question from the test, I feel to encounter two S’s. Solitude and Stupidity. Am I the Sole person who thinks that this test is Stupid?

Before we go any further, I would like to repeat that is a well known problem with a well known solution.

I do not where to begin, because this problem can be solved by a toddler, but more on this later.

First of all, this problem is discussed in the Martin Gardner’s book “Mathematical Circus”. Below I present the pages of this book that pertain to this problem. Also, at the end of the book there are references pertaining to this problem.

There are quite a few questions from this exam that are depicted almost word-for-word in Martin Gardner’s books. As I said: Points for originality: Nada! I will let the reader decide who is the unoriginal: Martin Gardner or the authors of the test. At least, Martin Gardner names his sources!

So, here is the material from Martin Gardner’s book.

Below, I present two links that provide the solution (and the rationale behind it) to this well-known and well-studied problem.

But this problem can be solved by a toddler as well. The toddler may not be able to provide a general formula, but she will be able to count the number of regions. To prove to myself that a toddler could do it, I put myself in a toddler’s position. Although I did not crawl on all fours, (do not be surprised, that would be ridiculous for anyone else, but not for me!), I drew the relevant shape and counted the numbers of regions. Below, I present all cases from 2 points on the circumference up to 7 points on the circumference. Obviously, since we want the maximum number of regions, care has to be taken that each crossing point is made by only 2 lines. If 3 or more lines pass through a crossing point, this means that 1 or more regions are collapsed to this point, so we do not have the maximum number of regions.

I drew each shape in PowerPoint. As I drew each line (chord), I would count it, so that I knew how many lines there were, even if the question did not ask for it. Then I opened each image in Paint. I took the area painter (the bucket) and started clicking inside each region, thus counting them. Each region I would click and count was colored by my click, so I knew not to double count it.

And I did not stop there. When I had counted the regions by coloring them with the same color, in the colored shape the lines and the intersection points were still shown. So I took the eraser and I put a little erasing smudge on each intersecting point, thus counting the intersecting points as well, even though the question did not ask for it.

I created the following table which depicts my results. “Points” are the number of points on the circumference. “Chords” are the line segments, where each line segment is drawn between two of the points in the circumference. “Crossings” are the inner crossings, the crossings where two line segments cross inside the circle. “Regions” are the empty areas that are formed inside the circle, “empty” meaning they contain no lines inside them.

The above table shows that the answer to the puzzle’s question is 57.

Let us see how we can generalize our findings. The following analysis is widely known to all mathematicians and also physicists like me know this theory very well. Dare I say, all students know this theory very well? But this measures knowledge or intelligence? I say that it only measures knowledge. Anyway, the test question only asks for the number of regions, so the answer is 57 and that is all. By merely counting the regions we arrived at the answer.

First of all, let us denote the number of points on the circumference with the letter n. This is the column named “Points”.

In order to have the maximum number of regions, each point is connected to each other point with a chord. To find the total number of chords, we can think in one of three ways.

First way: We have n points. Each point is connected with each one of the other n-1 points. So we have n*(n-1) lines (line segments). But in this number, we have accounted for each line twice, from the perspective of both points in each line. So we have n*(n-1)/2 lines.

Second way: From the theory of combinations. The number of lines is

(n choose 2) = n!/[2!*(n-2)!] = n*(n-1)/2.

Third way: We can say that the first point made n-1 connections with the other points and left the building. Then the second point made n-2 connections with the remaining points and left the building. And so on, until the second from last point made 1 connection with the last point at left the building. Thus, the last point is now alone and can make no connection. Thus, the number of connections (lines) made is

(n – 1) + (n – 2) + … + 2 + 1 = n*(n-1)/2.

So, the formula for the column named “Chords” is (n choose 2) = n*(n-1)/2.

Now let us find the formula for the column named “Crossings”.

For 2 lines to create a crossing, the 2 points of one line have to be different than the 2 points of the other line. In other words, to have a crossing, 4 different points on the circumference have to be involved, two of them defining one of the lines and the other two defining the other line.

This concept is easier to grasp if you look at the shape I provide for the simplest case. This is the shape with 4 points on the circumference which corresponds to only 1 crossing. This sole crossing exists because the 4 points on the circumference are 4 different points. Any 4 different points define two lines and 1 crossing.

Thus, the formula for the column named “Crossings” is (n choose 4) = n!/[4!*(n-4)!].

Now let us find the formula for the column named “Regions”, which is what the puzzle describes as regions.

The Wikipedia article I mentioned above rigorously describes the different approaches that explain and provide the corresponding formula. Because someone, like me, may lose themselves in mathematical rigor, before reading the full rigorous explanations and proofs, it is best to have a high level overview of them.

So, as an introduction to these rigorous proofs and explanations, I would like to give a simplistic overview.

Imagine (you don’t have to imagine, you can actually draw what I describe) that we have the circle and nothing else. So, we only have 1 region, which comprises the full circle. Then imagine that we draw parallel lines. As each new line is drawn, another region is created. Now imagine that the lines we already drew start intersecting. As each intersection (crossing) occurs, another region is created.

So, the formula for the column named “Regions” is

1 + Chords + Crossings = 1 + (n choose 2) + (n choose 4).

So, I can present you with the following theoretical table which coincides exactly with the empirical table I gave you above.

]]>

In this blog post, I will study Question 19 from this test.

The question is about giving numerical values to symbols, so that the given equations hold.

Below I present the original question for your convenience:

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 19.

This is not a bad question, but I fail to see what it has to do with a test that measures intelligence. It is easy for someone to do some trial and error attempts in order to find the solution. I suppose the test had to had an easy question to tackle. But there are some sinister things going on concerning this question and I am about to bring them forth.

But first, let me provide the answer.

This is a nice question that allows the solver to draw in a piece of paper and try out things and make different attempts. But I do not know how the test was administered. You will shortly understand why I am saying this. This question seeks an answer and it makes clear that the answer is higher than 17. Any attempt, no matter how naive, at solving the problem will not provide a number much bigger than 17. So, by trying different numbers like 18, 19, … the solver is about to find the solution, thus gaming the system.

But since I do not know how the test was actually administered, I will leave this point at that.

But even then, the test expects a number. Giving a number willy nilly, does not say anything. Giving a picture with a path drawn, writing an answer that proves your thinking, posting a blog post like this one, does say something. So, again how was this test administered? I do not know, but I am afraid that the answer might be highly disturbing.

Back to the answer. Below, I provide two images.

First image:

Second image:

From the two images above, we can see that the answer is 19. The minimum numbers of streets to walk on in order to cross all of them is 19. And this is the answer and if you need a little more explaining (do not worry about the word “little”, I am going to provide a whole lot more of explaining in the rest of this blog post) you can observe the two images.

Indeed, you might observe that I provide two equivalent solutions. Two, because there are two images. But wait! In each image, you see that I start from the left and I stop at the right. Or vice versa. The path may be the same but does it count as one or two solutions depending on the direction of travel? I should say that it does not. Your opinion may differ and it is all right. We will discuss more about this later on, anyway.

In both images, the start and stop points are on the left and right middle buildings. But on the first image we observe a point where the path crosses itself. This is like the center of a swastika. When you reach this point, you can go straight or turn 90 degrees. Does this count as two separate solutions or one? Either way, we all know what we are talking about and what we see here concerning the path taken.

But this goes for the second image as well. See that in the same building that is the center of the swastika, the path does not cross itself? I made it so that you can understand the path more clearly. But in reality, it can cross itself, and so different paths and solutions arise, each with 19 steps.

So, in the first image, when you reach the node in the second row and second column from the left, then you can go straight or turn 90 degrees. In the second image, when you reach this node, you can go up or down.

So, the answer is 19, all is ok and we can go and watch Chesapeake Shores (or is it Cheesypeake Shores? Chesapeake Whores? I can never remember!)

And you can stop reading here and all is well. But If you want to go the distance, well, please stay a little longer, dear reader. There are some things that need more explaining and there are some sinister things going on as well.

How am I sure that the minimum is 19?

I will answer, but first let me say something else, that will help me arrive at the answer for the question I just posed.

One of the sinister things concerning this question is that it does not state if we are allowed to choose where to start and where to end our path. Are we allowed? The question supposes that we are, but someone would not know that. But then, how do I know that the question allows that?

I know because this question is about the mathematical subject called Graph Theory and Graphs. And the problem presented here is a common problem in Graph Theory and all common assumptions apply.

And this gives an unfair advantage to someone who knows Graph Theory compared to someone who doesn’t. A huge chasm. A hugely unfair advantage.

Graph Theory makes me certain that the minimum is 19. Someone who does not know Graph Theory would have a hard time to prove that 19 is the minimum, whereas it is easy for me. As I said: Unfair advantage. And I have a problem with unfair advantages. A very big problem.

Anyway, this puzzle pertains to graphs and graphs have their terminology and their common assumptions that make sense in this area of Mathematics.

Here are some links that may help, especially the one about the Seven Bridges of Konigsberg. That Wikipedia article alone is enough to explain what is going on in this question.

To explain something and/or to tell a story, sometimes the most difficult thing is to find where to begin. Our story begins when Euler visited the town of Konigsberg. There, the town folk had a pet peeve. Or a question. Or both. They had healthy inquiring minds. Or not. Anyway, their town is depicted in the map below and the good people of Konigsberg wanted to know if one could go to all four land areas by crossing each of the seven bridges exactly once.

I said that the town folk had healthy inquiring minds, but I am not so sure. To me, it seems obvious that this cannot be done. I cannot find a path that would go to all land areas and traverse each bridge only once. But the town folk either did not have a map like the above (highly improbable), or they were way more inquiring than me and needed a mathematical proof for the possibility or impossibility of such a feat.

Anyway, Euler heard about the problem and thought that this was an interesting problem for him to solve. And solve it he did. He proved that this is impossible. He proved that it is impossible to visit all four land areas by traversing each bridge only once.

How Euler thought and went about solving this problem is crucial for us in order to understand why the answer I gave in the test puzzle is 19 and it is certain that it is 19.

But before I get to that, I must stress that the town folk cared about a path but did not care about which land would be the beginning and which land would be the end of the path. They just wanted a path that would traverse all bridges, each bridge exactly once and visit all land areas.

This notion is an important notion in Graph Theory and is named in honor of Euler as Eulerian path. So, in retrospect, the town folk wanted to know if a Eulerian path exists. And the question of this puzzle is almost the same. It states that there is no Eulerian path and asks us to find the minimum number of bridge traversals we have to make. In the Konigsberg problem, we have land areas and bridges. In this puzzle, we have buildings and streets, respectively.

but the common understanding is that when we are dealing with Eulerian paths, we are not choosy about which is the starting point and which is the finishing point. Of course, in the street cleaner example that the puzzle is all about, let me tell you, the street cleaner would definitely like to have a saying as to where her work would start and where it would end: I guess as close to her home as possible. But the puzzle assumes what is usually assumed in such graph problems. Thus, it is inappropriate for “general consumption”.

But I guess that it would make a great question for a mathematician interested in graphs. Below I will try to explain what is going on in this puzzle, but it will help if you know a few things about graphs already, like studying the links that I provided above.

So, in this puzzle, what we have in front of us is a graph. The nodes (or vertices) of the graph are the buildings. The edges of the graph are the streets. This graph is an undirected connected graph. “Undirected” means that the edges have no direction (for each edge there are two nodes, but not a beginning and an ending node, just two nodes). “Connected” means that no “islands” exist. There is always a path from a node to another node, but it might be through another node. This is why I did not characterize the graph as “complete”. It is not “complete”. Not all possible edges exist. This graph is just “connected”.

So we have an undirected connected graph and we want to study the Eulerian paths. Actually, we know that there are no Eulerian paths, but we want to find the minimum next best path.

Before we get to that, I would just like to bring your attention to the difference between Hamiltonian and Eulerian paths. A Hamiltonian path is one that visits each node once. A Eulerian path is one that visits each edge once. And in both, it does not matter where we begin and where we end.

When Euler studied the Seven Bridges of Konigsberg, he understood a few very important concepts. He understood that what plays the defining role is the number of bridges (edges) that each land (node) has attached. The number of edges that are connected to a node is called the degree of the node. And it is the most important thing in order to study the existence of the Eulerian paths.

You see, Euler understood that during a walk, the path you go along traversing land areas and bridges, will bring you in and out of land areas. Let us exclude the beginning and ending land area for a moment. All other land areas will be visited by going in one bridge and leaving in another (a different one, a different bridge) since we do not want to traverse the same bridge twice. And so he realized that for a Eulerian path to exist, all intermediate nodes in the path have to have an even degree, that is, an even number of edges connected to them. Now for the beginning and ending node, they have to be both of either odd or even degree. If one of them is of odd degree and the other one is of even degree, then a Eulerian path does not exist. If there are two nodes of odd degree, then anyone of them will be the beginning node and then the other will have to be the ending node of the Eulerian path. If there are zero nodes of odd degree, then a Eulerian path definitely exists.

So, for a Eulerian path to exist, all nodes have to be of even degree, or two nodes have to be of odd degree, In the latter case, each one of them can be the beginning node of the path, but once we chose the beginning node, the other node has to be the ending node of the path.

Now we have all the concepts we need to solve the puzzle in the most mathematical and precise way there is.

The Seven Bridges of Konigsberg did not have a Eulerian path because the corresponding graph contained 4 nodes and each one of those had an odd degree.As you can see in the map, three land areas had three bridges connected to them and one land area had five bridges connected to them. So, all land areas had an odd number of bridges connected to them. I wish I could go to Konigsberg and scream to the ear of each town folk: “Hey baster, there is no Eulerian path, because in this connected graph, there are four nodes and all have an odd degree. Get it? No? You need to have zero or two nodes of odd degree in order to have a Eulerian path! Get it now? You stupid %@^&!”.

The same goes for the corresponding graph for this puzzle. There is no Eulerian path, because there are six nodes of odd degree. The following image depicts these six nodes by having circles around them. I circled them and I also used the green color for the nodes on the far left and far right and the red color for the nodes in the upper an lower part. I did this in order to facilitate the discussion that is going to follow.

In the solutions that I gave, I always start and finish at the green nodes. And I always traverse twice the two pairs of red nodes, the pair in the upper part and the pair in the lower part. This is not coincidental. By traversing the red nodes twice, it is as if I create another edge among them. It is as if I create another street. Please note that in the two images that depict the solutions that I gave, the pairs of the nodes have another red line going backwards and in a slight angle. This means that I traversed this path twice. And it can equivalently mean that I created another street between these two buildings. And by doing so, I changed the degree of the nodes (buildings) from odd to even.

Thus, it only takes one more edge between the two upper middle nodes and one more edge between the lower middle modes to turn this graph in to one with two nodes of odd degree (the nodes circled with green color). And this is why all solutions start and end in these two green nodes. Because, by traversing twice the red nodes edge, it is as if I created another edge, thus making them into even degree nodes, thus leaving the green circled nodes to be the only two odd degree nodes, so they had to be the beginning and ending node of the Eulerian path that was just created by the virtual addition of the two new nodes that my double traversing made.

Now, one may ask if we can have another pair of the six odd degree nodes as the beginning and ending node. The answer is that we cannot. Actually we can create an edge between any two of them and another edge between any other two of them, thus leaving again only two nodes with an odd degree. But the edges that we will draw will not be 1 kilometer long. There will be lengthier. So, although we can create a Eulerian path by leaving any two odd degree nodes, the only case that we can do so and with the minimum distance is if we connect the adjacent pairs of the red nodes.

And this is how I know that the solutions that I gave are optimal, that the beginning and ending building are those on the middle far left and right and that the minimum path needed to traverse all streets is 19 kilometers long.

]]>In this blog post, I will study Question 2 from this test.

The question is about giving numerical values to symbols, so that the given equations hold.

Below I present the original question for your convenience:

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 2.

This question is one of the most inappropriate questions on this test. Let me begin my analysis, where it will become very clear why I have this bad opinion about this question.

I will arrange the equations given as follows and I will begin to solve them in that order:

When I began solving this puzzle, I was skeptical. Part of me said that this would be a question that would be solvable with reasonable assumptions and part of me said that this would be a question that needed unreasonable guesses. In hindsight, I cannot believe how right my latter part was and actually this is also an understatement.

I began with trying to guess the first three equations, i.e. those that had numerical values on their right side. Since 23 and 29 are prime numbers, I assumed that symbols next to each other were simply added (as opposed to multiplied), and, in hindsight, I was correct.

I tried to give values to the symbols in the first equation, so that the second equation would hold.

So, I came up with the following for the first equation:

2 + 3 + 3 + 3 + 3 + 4 = 18 or 4 + 3 + 3 + 3 + 3 + 2 = 18

so that the components of the second equation would be like:

3, 2^4=16 or 4^2=16, 2*2=4 or 2^2=4

and the second equation would be 3 + 16 + 4 = 23.

Things were looking up and I had options. Unfortunately, none of my options seemed to satisfy the third equation. I struggled a lot, until I discovered that the only way to satisfy all first three equations was to have GreenCircle = 2, RedSquare = 3, YellowTriangle = 4 and, and here is the kicker, BaseYellowSquare representing the raise to the second power.

So, when the yellow triangle is alone as a symbol, it is equal to 4. When it is under another symbol, it raises the value of the symbol to the power of 2. And when a yellow square is underneath another yellow square, the yellow square which is above represents the number 4, whereas the yellow square which is underneath raises said 4 to the second power.

For the yellow square to have two different interpretations, depending on whether it was alone (or on the very top of a structure) and whether it was used as base with a symbol on top, is absurd. But since it was the only thing that could make the first three equations working, I went along with it, half-heartedly and with doubt, because of Occam’s razor. “Surely there must be a more sensible interpretation”, I was thinking to myself. In hindsight, things were to become way more absurd, to the final point of complete absurdity. You will see why later on.

So, at this point, things look as follows:

First equation: 2 + 3 + 3 + 3 + 3 + 4 = 18.

Second equation: 3 + 2^2 + 4^2 = 23.

Third equation: 3^2 + 4^2 + 2 + 2 = 29.

From this point, the fourth and fifth equations can be studied independently. From the fourth equation we can get the value of BlueRhombus and from the fifth equation we can understand what the BaseGreenCircle represents. Because, if we are to learn something from our previous experience, symbols here, when used as a base, represent operations rather than numbers.

Let us continue by studying the fourth equation. Let x denote the value of BlueRhombus. We have:

x^2 + (x+3)^2 + 3 = (x+4)^2 + 3^2 + 2 =>

=> x^2 + x^2 + 2*3*x + 3^2 + 3 = x^2 + 2*4*x + 4^2 + 9 + 2 =>

=> x^2 + 6*x + 12 = 8*x + 27 =>

=> x^2 – 2*x – 15 = 0 =>

=> x = -3 or x = 5, and I keep the positive value x = 5.

Thus BlueRhombus = 5. Nothing absurd here. We are at a very sensible point.

Let us continue by studying the fifth equation. The right side is equal to:

3^2 + 4 + 4 + 2 = 19. Now let us study the left side. There is a green circle with 3 red squares on top and a yellow triangle with another yellow triangle on top and on top of the second yellow triangle there is a green circle.

Obviously, the yellow triangle structure is (2^2)^2 = 4^2 = 16. This is because the triangle that is on top along with the circle that is on top of it amount to 2^2 = 4. So the bottom triangle contains the value 4, thus it amounts to 4^2 = 16.

So, if the triangle composite is 16, this leaves the green square structure to be 3. But the 3 red squares that are on top are equal to 3 + 3 + 3 = 9. So, having a green circle as base corresponds to calculating the square root. The square root of 9 is 3 and the green circle with the 3 red squares is equal to 3, because the 3 red squares amount to 9 and 3 is the square root of nine.

Thus, BaseGreenCircle corresponds to the square root calculation.

So, the absurdity continues. When a green circle is used alone (or on the very top of a structure) it is a 2 but it is used as a base it is the square root calculation.

What I found most absurd in the above was that the square root calculation had entered the picture. I was unsure and uneasy with my interpretation, since I believed that the test was about judging intelligence and not mathematical knowledge. And I consider the square root to be a more advanced concept that multiplication or raising to a power. But all evidence pointed to the fact that indeed the BaseGreenSquare was the operation of the square root calculation.

So, I accepted all the above with some uncertainty and skepticism. Little did I know that this was just the tip of the iceberg.

So, we are the following point:

It seems that we are close to the end, and indeed we are, but we are in for a big surprise. Before we proceed, let us recapitulate:

GreenCircle=2.

RedSquare=3.

YellowTriangle=4.

BlueRhombus=5.

BaseYellowTriangle produces the power of 2 of whatever is on top of it.

BaseGreenSquare produces the square root of whatever is on top of it.

I could not find anything more simple to interpret the first five equations, so I was content that I was near the end. I just had to interpret the sixth and seventh equation and then I would have to calculate the eight expression. To interpret the sixth and seventh equation meant to interpret what the BaseBlueRhombus meant, i.e. what operation the blue rhombus performed when it was used as a base for something.

This is because both the sixth and the seventh equation feature the blue rhombus as a base in their left side.

Let us study the sixth equation. The second part is 5. The first part is the blue rhombus as the base and a blue rhombus on top. So we need to find what operation, let us call this operation f, the blue rhombus denotes. From the sixth equation we have that this operation transforms a 5 to a 5.

Let us now study the seventh equation. The right part is equal to:

4^2 + SQRT(4) + 3 = 16 +2 + 3 = 21.

The left part is a blue rhombus with two yellow triangles on top. Tow yellow triangles are equal to 4 + 4 =8. So, from the seventh operation we have that a blue rhombus transforms an 8 to 21.

Let us recapitulate. We are trying to find the mathematical operation f that BaseBlueRhomus corresponds to. And we have that: f(5) = 5 and f(8) = 21. Once we know what f is, it will be straightforward to calculate the eighth and final expression. This is because the left side of the eighth equation is f(f5+2)) = f(f(7)).

This is easy to see. The left side of the eighth equation is a blue rhombus with another blue rhombus on top and the second blue rhombus has a blue rhombus and a green circle on top. The blue rhombus and the green circle on top are equal to 5 + 2 = 7. So the blue rhombus that is on top is equal to f(7) and the whole structure is equal to f(f(7)).

So, we are the following point:

So, what is f? What is this mathematical operation that sends a 5 to a 5 and an 8 to a 21? From the moment I encountered this puzzle, I thought of the Fibonacci numbers. (Here indeed, 5 and 21 belong to the Fibonacci number sequence.) But the moment I thought of the Fibonacci numbers, I dismissed them. Here in Greece, this is an advanced concept that students do not learn as part of their mathematical education. Here in Greece, I guess that concepts such as the Golden Ratio, Fibonacci numbers and the squaring of the circle are considered esoteric, advanced, unscientific and they are not taught as part of the curriculum. So, my instinct was to avoid Fibonacci numbers, because they are way too advanced for a puzzle that measures intelligence by balancing equations.

Let me explain this a little bit further, so there are no misunderstandings. I am the only one that thinks pi, the Fibonacci number sequence, numbers in general are mystical, esoteric entities. Scientists scoff at notions like mine. Although I treat the mathematical concepts as divine, all others have completely demystified them. I agree with their interpretations, although I think there is something more hidden underneath all our understanding. Anyway, although I find concepts such as those to be esoteric, I vouch for them. This is in contrast with main science which finds these concepts to either be non-esoteric. Because the Fibonacci sequence is a concept that some people like me believe to be esoteric, the Greek curriculum avoids it. Or this is what I think that the reason is.

So, I ignored the Fibonacci number sequence and all other esoteric mathematical entities and I tried to find a mathematical operation that turned a 5 into a 5 and an 8 into a 21.

I thought about decimals, fractions, factorials, but I sort of focused with modulo arithmetic and mostly division. What if this strange elusive operation f was modulo division 5? After all, 21 = 2*8+5, thus 21 mod 2 = 5. I tried desperately to make a 5 go to a 5 and an 8 go to 21 with all sorts of modulo operations, but I couldn’t find any that would persuade me that this is what the authors had in mind.

Occam’s razor. Again. Always.

So, I am back to the Fibonacci sequence. What was Sherlock Holmes saying? “Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth.” Then, that is that. I made sure to eliminate all other mathematical operations. But I was still uneasy. The Fibonacci sequence is not an operation. It is a number sequence.

Let me give you the final part of the solution and here you will understand why I am still uneasy.

So, the Fibonacci sequence is produced by starting with 0 and 1 and from then on each number in the sequence is the sum of the previous numbers in the sequence:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, …

If we assign an index of 0 to the first element, then for the function f that supplies a number for the sequence, we have:

f(0)=0, f(1)=1, f(2)=1, f(3)=2, f(4)=3, f(5)=5, f(6)=8, f(7)=13, f(8)=21, f(9)=34, f(10)=55, f(11)=89, f(12)=144, f(13)=233, f(14)=377, f(15)=610, f(16)=987, f(17)=1597, f(18)=2584, f(19)=4181, f(20)=6765, …

So, from the above, we have that:

f(5)=5, thus the sixth equation holds

and

f(8)=21, thus the seventh equation holds.

So let us compute the eight expression:

f(f(5+2))=f(f(7))=f(13)=233.

Thus, the answer is 233.

Now, what makes me uneasy is that the Fibonacci sequence is not an operation. OK, I admit that if we want to stress things, an operation can be seen as a function and here we have a function f that returns a number from the Fibonacci sequence.

But this is a puzzle concerning the balancing of symbol equations. Only a crazy person would assign the symbol of a blue rhombus to be the function of the Fibonacci sequence and the symbol that is on top of it to be the index of the Fibonacci sequence. You see, I have seen many puzzles concerning the balancing of symbol equations and to me this is crazy thinking.

This leads me to another point I would like to make. I believe that this test is highly inappropriate, because it wants you to be knowledgeable and it wants you to guess. The latter is way worse than the first. The mentality of this test is: “Guess what I am thinking and we will see if you are lucky enough”. And also “I know of this mathematical theory ad discovery and paper that if you too have happened to come across, then you will be lucky enough to be able to answer this question”.

The whole morality of this text is off. Also, I found things there that made me question my sanity.

So, here what I will assume. I assume that this test was put forth to see who objects. I imagine the authors being somewhere shaking their heads in contempt. “No one opposed it!” they will cry in dismay. “No one had the decency to object! Where has this world gone to! Only one person had the integrity to question us. And this person is Dimitrios Kalemis.”

Of course, I do not really believe the above, that this test was put forth to fool us and test us, but it would be the only sensible explanation for me.

I believe that the authors wanted to find highly intelligent individuals to help each other towards a good purpose. I do not know. But this is what I believe. The problem is that the authors claimed to have created a test that measures high intelligence, whereas I disagree.

But the main issue is the following and it reminds me of the P vs NP problem:

Suppose I pose a question that you understand. And suppose that you devote time and energy to try to answer this question. And suppose that you fail to answer this question. And suppose that the answer to this question is an answer than when I give it to you, you will understand it. Then, I say that it is my moral obligation to give you the answer.

The authors of the test do not provide the answers to this test.

If the answers to this test can be understood, then my claim is that the authors have the moral obligation to provide them.

Since the answer of a question is not available, and I happen to know it, I have the moral obligation to provide it.

I am scientist and scientists publish the results of their research. And I just published here my research and these are my results. And this is how the question is answered and this is what it means for the validity of the test.

]]>In this blog post, I will study Question 21 from this test.

The question is about the construction of a square from 24 non-overlapping smaller squares, each with a side of unique length.

Below I present the original question for your convenience:

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 21.

Actually, this one of the good questions of the exam. I am saying this because the exam contains some highly inappropriate questions as well. But there is still a problem with this question. As with some other questions in this test, the answer can be found in one of Martin Gardner’s books.

So, why would Haselbauer and Dickheiser copy (not just get inspiration, but blatantly copy) questions from Gardner’s books, questions that are fully answered in those same books, when they claim, and I believe them, that they want the integrity of the test to be high?

I said and I am planning to prove that the Haselbauer-Dickheiser Test is inappropriate (and this for many reasons), but this is ridiculous!

Anyway, back to answering the question. I will provide information about Martin Gardner’s book later on in this blog post. But first, I would like to present my attempt at solving this question, which I did before I read the corresponding chapter of the book.

So here is how I solved this question without reading the corresponding chapter.

I observed the image and I thought that I should tackle the problem beginning from the smallest square. I named each of the 24 squares with a number from 1 to 14, counting in my order of preference that I chose loosely based on the order in which I imagined that my calculations would proceed. Anyway, giving a name to each square is arbitrary, because the actual name does not matter, what matters is that we need to have a way to refer to each square. So, I chose the names that are depicted in the following image:

Please note that image above does not depict the size of the side of each square. It depicts the name of each square. I made this so that I can refer to each square and so that I can calculate the size of the side of each square.

So, imagine that the smallest square is named 1, or rather A1, because I am going to use Excel and this way, I will easily create the spreadsheet so that the size of each square will correspond to the cell that has the name of that square.

Here is the Excel spreadsheet that I created.

And here is the way I worked, what I was seeing when I was working at the solution:

So, I had these two windows side-by-side as I was working with the spreadsheet.

Let me describe the spreadsheet. Next to each cell is another cell that depicts the formula used in the first cell. I only enter numbers in the four white cells. The numbers in the yellow cells are automatically calculated by the corresponding formulas in those cells.

Cell A1 corresponds to the side of square 1, the smallest square. Cell A2 corresponds to the side of square 2, the one on the left of square 1. I do a trial and error analysis, where I guess the values of the sides of squares 1, 2, 14, and 15. And I let the formulas that I entered produce the values of the sides of the other squares.

So, the side of square 3 is on the cell A3, which contains the formula =A1+A2, because the side of square 3 is the sum of the sizes of squares 1 and 2. And so on for the other squares.

So, by guessing the sizes of four squares, I calculate the sizes of the rest of the squares. The values I choose have to be small for the squares 1, 2 and 14, and the value for the square 1 has to be the smallest of the three. Also, by looking at the value that is produced from square 1 and square 2 for square 10, I choose a somewhat smaller value for square 15. This is because the image gives me this hint, if I roughly compare square 15 and square 10.

In order to know if the values I enter by trial and error produce the correct result, I check that the sizes of the all encompassing square are all equal. This is why I have created the column D. Each cell is the resulting calculation of one of the four sizes of the all encompassing square. By entering values in the four white cells in column A, I do not really care about the values produced in the yellow squares of column A. I really care about the four values in column D. They all have to be equal to each other. So this is how I know if I have the correct values in the four white cells in column A.

By trial and error, I found that when A1=1, A2=3, A14=2, and A15=29, all values in column D match, i.e. they are equal to each other and equal to 175. So, this means that I solved the puzzle. So, the all encompassing square has a side equal to 175. And the square in question, square A10, has a side equal to 38. And this is the answer: 38.

The way I solved the puzzle, I was very cautious that the numbers I input might have been coprime, especially A1 and A2. When I first saw this puzzle and understood that I would solve it using trial and error, basing my guesses and beginning from the smallest square and the one next to it, I was extra careful not to assume any relationship between these two numbers. In hindsight, things were not that bad, since A1=1 and A2=3, but I could not have known. So I was extra careful to account for a case like A1=2, A2=5 and so on.

Another thing is that the question should have been stated in a way that acknowledges that there are infinite such rectangles and we are searching for the smallest one. Indeed, any integer multiple of the solution is a solution as well. For example, multiplying by 2, we get A1=2, A2=6, A14=4, A15= 58 which also produces a solution. And multiplying by 3, we get A1=3, A2=9, A14=6, A15= 87 which also produces a solution. And so on to infinity. But the question forgets, or rather assumes that we correctly assume it. Good assumption from our part, bad assumption from the question’s part.

As I said, the answer is depicted in one of Martin Gardner’s books. Martin Gardner’s book titled “The Second Scientific American Book of Mathematical Puzzles and Diversions”, which is the 2nd book out of a 15-book series has the answer to this puzzle. Below, I present a screenshot of the pages 205 and 206 from this book.

]]>