## My attempt at Question 19 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 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: 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 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.

Seven Bridges of Königsberg

Eulerian path

Hamiltonian path

Route inspection problem

Graph (discrete mathematics)

Graph theory

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.

Posted in Education

## My attempt at Question 2 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 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: 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 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.

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.

Posted in Education

## My attempt at Question 21 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 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: 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 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 24, 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. Posted in Education

## My attempt at Question 16 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 16 from this test.

The question is about the maximum number of pieces a donut (doughnut) can be sliced with three simultaneous plane cuts. Well, the “maximum” part is implied but not stated clearly. And by “donut”, they mean “torus”.

Hmmm… some question! Ask any policeman, they’ll know.

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 16.

Actually, this is a good question and tests your ability to think in 3D. If you are so inclined, you could do the imaging in your head. In this case, I would suggest to imaging the torus being very big. This will help, because there are some small pieces produced as well.

Or you could use your favorite 3D modelling software to draw a torus and three planes cutting it. You should choose a 3D modelling program that not only is your favorite bit that it also allows you to view everything in 3D and rotate and translate the model in real time. This is because you will want to see your model from many different angles.

I would also recommend that you alternate between making the objects (the torus and the three planes) transparent, semi-transparent and opaque, in order to design and study your model better.

I like POV-Ray, but for this purpose I would strongly recommend something along the lines of Blender. Anyway, if you use POV-Ray, here is how you could create a torus:

```torus
{
1, 0.4

rotate  &lt; 90 , 0, 0&gt;

pigment
{
rgbt &lt;0,1,0,0.0&gt;
}
}
```

and here is how you could create a plane:

```polygon {
4,
&lt;2, -2&gt;, &lt;2, 2&gt;, &lt;-2, 2&gt;, &lt;-2, -2&gt;

rotate  &lt; 0 , 60, -20&gt;
translate &lt;2,0,0&gt;

pigment {
rgbt &lt;1,0,0,0.0&gt;
}
}
```

With the last decimal in the pigment rgbt you can control the opacity of the object.

OK. Now, let;’s get to the answer. A few paragraphs above I suggested that you ask a policeman. That was, of course, a joke. Who wants to address a pig? But, when it comes to a cool guy like Martin Gardner, well I would certainly want his opinion.

Turns out, 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. Crazy, huh? Well things are about to get way more crazy and ridiculous. This same book contains the answer for another of the Haselbauer-Dickheiser test’s questions! And other books in the series contain the answers to some other of the test’s questions.

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 not answer, I will let *the* man Martin Gardner to do it. He is way to the infinity better than me.

Below, I took a screenshot of pages 149 and 150 from Martin Gardner’s book. The answer is that we can create as many as 13 pieces, albeit some of them very small compared to the others.

Another, more refined picture of the above, which I found online: and another: Let me help you better understand what is going on. The reason that we are able to create an odd number of pieces is because of the somewhat vertical plane not being in “alignment”. In the image below, I highlighted two intersections and you should note that the blue line is not parallel to the red line. Anyway, rather than looking at still images, it is better to create a 3D model and view it from all perspectives.

So, the answer is 13 pieces.

Posted in Education

## My attempt at Question 24 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 24 from this test.

The question is about a probability calculation concerning two coins, one of which is heads on both sides.

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 24.

This question is highly inappropriate as well, as are most (if not all) of the questions on this test. The reason this question is inappropriate is that it does not test cleverness but knowledge. If you have been taught probability theory and the concept of sample space, then you can find the solution. Otherwise, you are clueless. Probability theory and the concept of sample space is one of the domains that the following brilliant comic is right on the spot. So, we have two coins. One coin is heads on both sides (um…. who comes up with these scenarios?). The other coin is heads on one side and tails on the other side. Ok, that’s the normal coin. The first coin is the abnormal one, the “rebel”.

If the question just stated that one coin is selected at random and we examine one of its faces at random, then we would have the following 4 cases (the following sample space):

Coin H-H is selected and its first face H is examined.
Coin H-H is selected and its second face H is examined.
Coin H-T is selected and its first face H is examined.
Coin H-T is selected and its second face T is examined.

But the question states something more: That the face that we examine is heads. So the sample space ends up having the following 3 cases:

Coin H-H is selected and its first face H is examined.
Coin H-H is selected and its second face H is examined.
Coin H-T is selected and its first face H is examined.

So, only the 3 cases above could have happened.

Now, from the 3 above cases, only the first 2 correspond to the other face being heads.

Thus, the probability we are looking for is 2/3.

Posted in Education

## My attempt at Question 13 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 13 from this test.

The question informs us that we have eight identical squares in size and we have to find the order in which the y were place to provide the configuration depicted.

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 13.

Actually, I find that this is the least obnoxious of the question. Still, I do not consider the question fit to examine a person’s intelligence. But it is not completely hopeless, either, unlike the rest of the question on this test.

When I encountered this question, I thought that the solver did not need to make any assumptions. So, this is what is right about this question.

So, I observed and observed and observed the configuration. And since I did not come up with anything conclusive at first, I thought about importing the picture in PowerPoint and drawing the vertical and horizontal lines that you see in the following image. And I immediately understood what was going on and the order that the squares were placed. And I wrote the order from bottom to top at the comments section at the bottom of PowerPoint, as you can see in the image above.

So, the question asks us to write the order from top to bottom, but it helps us to understand how the squares were placed by going from bottom to top. Then it is straightforward to reverse the order.

So, I solved the question, but I wanted to make absolutely sure and to prove that was the answer. So, I used Excel. And I drew the squares one by one as they formed the configuration. My work using Excel is depicted in the following image. So, the order from top to bottom is A D G H F B E C.

Posted in Education

## My attempt at Question 22 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 22 from this test.

The question gives a 10 X 10 square grid. Each square has a color and a single digit number. Also, a number appears after each row and each column, except after the third row, where there is a question mark. The question makes obvious that we have to find the number that goes in the place of the question mark.

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 22.

When I first encountered the question, I saw the numbers outside the grid and they looked like sums to me. They looked like a the result of a calculation that involved the corresponding row or column. In hindsight, yes, I was correct. But the question does not state this. You have to guess. Guessing is not a good thing. Guessing is a highly inappropriate thing. I am talking in the context of an IQ test. It is so easy for me to create an IQ test like that. I will name it “Guess What I Am Thinking!”.  And you will have to guess whatever rule I coined. That’s not right.

So, as it turns out, these numbers are sums. But if they are sums, is the question ill-conceived? And does the question allow you to cheat?

What I mean is that if these numbers are sums and each square is summed depending on its color and its number, then the sums of the column sums will be the same as the sum of the row sums.

In other words, we should have:

95 + 93 + 62 + 77 + 106 + 100 + 102 + 100 + 78+ 97 =
= 80 + 92 + x + 89 + 98 + 91 + 78 + 99 + 83 + 88

And from the above equation, we have x = 112.

Indeed, in hindsight, this is the correct answer. In this blog post I will explain why I am certain that this is the correct answer.

But I do not think that the examiner wanted the solution to be found this way. Again I am guessing. There is and will be a lot of guessing involved in this test. And this is one of the things that is bad about it.

So the correct answer to life, the universe and everything, or just the missing number in this question is 112. But if the test was administered in an environment where you could type your answer and receive the result of whether your answer was true or false, you could easily game the system and try 112 or other different numbers and then, when you would have been informed that your answer was correct, you could use this information to trace things backwards.

So, one of the problems pertains to the way the test is administered. Answering 112 does not amount to much. You have to state why you gave the answer that you gave. But I am afraid that the test is not administered in this way, but it is administered in a way that allows you to cheat and game the system.

OK. Let us suppose that we are not certain that the numbers outside the grid are sums and we are not certain that the sum of the numbers in the bottom is equal to the sum of the numbers on the right. How will we proceed to answer this question?

Here is what we have to do. We will assume that each number outside the grid is the sum of the numbers in the corresponding row or column. But each number will have to be differentiated according to the color of the square that it resides on.

Any normal person would assume that each color corresponds to a number, and in hindsight, this is correct. Any normal person would also assume that the number of the color multiplies the number of the digit. But in hindsight, this is not correct.

Anyway, I did not know this and I assumed that the number corresponding to the color in each square multiplied the digit in the square. So I had to find the number that corresponded to each color.

Here is what I did. I observed and observed and observed the grid. And I noticed that the sixth column from the left contains only four colors (pink, green, blue, red) of the five colors in the grid (pink, green, blue, red, yellow). Specifically, in the sixth column we have the following:

The sum of the red squares is 9 + 2 + 6 + 4 = 21.
The sum of the pink squares is 5 + 3 = 8.
The sum of the green squares is 4.
The sum of the blue squares is 4 + 8 + 4 = 16.

And the sum (I loosely refer to it as sum, it is really the result of an unknown calculation) of the whole column is 100.

Since I was assuming that each color multiplies the digit by a constant number, I immediately knew  that the red color corresponded to an even number. Why? Because the sum was an even number (100) and the sums from the other colors each was an even number. So, even if their color corresponded to an odd number, the multiplication would end up in an even number. Since the whole sum had to be even, it was without doubt that the red color was an even number.

So, red had to be 2 or 4, because had it been 6 or higher, 21 * red would be more than 100. Actually, red could only be 2, because if red was 4, then even if the other colors were each 1, the sum would end up more than 100. Also, each color had to be a different number, because it would be illogical not to.

So, it should be that red =2 and pink, green and blue had to have different values.

The problem was that the equation

21 * red + 8 * pink + 4 * green + 16 * blue = 100 =>
=> 21 * 2 + 8 * pink + 4 * green + 16 * blue = 100 =>
=> 42 + 8 * pink + 4 * green + 16 * blue = 100 =>
=> 8 * pink + 4 * green + 16 * blue = 58

does not hold for any integer values we may try for pink, green and blue, given that we have to choose from the integers 1,3,4,… and all three should be different. (I remind you that 2 is missing, since red =2).

So, I struggled with this for a while, until I gave up. So, it seemed that the multiplication concept was not what was going on in the calculation. So, how where the “sums” produced? I thought that if there was not a multiplication of the color number, it would be an addition of the color number.

So, I assumed that each color corresponded to a different integer and that integer was added to the digit that was in the square. And, in hindsight, that was the correct assumption.

So, I had to test my assumption. I had to assign a different integer number to each one of the five colors and calculate the sum of each column and row, where for each square I would add the color number to the digit.

But I had to do a what-if analysis. I wanted to try different values for each color to get the sums to match those given. So, here is what I did: I used Excel.

I created five names, one for each color, and I would set integers as their values, to see if the sums would match those that the question gave.

Each cell was the sum off the digit and the color. The digit was a number and the color was an Excel name. So, the cell A1 would have the formula =3+red. The cell B1 would have the formula =4+green. And so on.

In hindsight, it would have been easier for me to do all the checks I made, if I did not use names but instead I would use cells. This is because I had to open name manager each time I wanted to make a change, whereas a cell would always remain open in front of me, thus making any changes and what-if analysis easier.

Anyway, I struggled a little bit, trying different values, always being careful to use different values for different colors and doing the analysis primarily on columns which contained fewer than the five colors. Whenever I got a sum that was correct in one row or column, another sum would be off, so I struggled a little bit, I tried different permutations of integers and I finally got one that produced all sums correctly. The is depicted in the following screenshot. I am certain that this is the correct answer, since it would be highly improbable for the examiner to have something else in mind. Each row and column sum is validated and checked to be in agreement with what the question provides.

So now we know that the numbers outside the grid are the sums of each row and column, where by sum we mean the calculation that takes each square and adds the color to the digit and then adds all these sums. Blue = 7, Green = 2, Pink = 6,  Red = 4, Yellow = 3. The missing number is 112.

Posted in Education