Logic Riddles for Chess Players (5)

by Arne Kaehler
7/10/2020 – Part five of our “Logic Riddles” series is here! This time it is all about a chess-related riddle, pertaining colours. We recently posted a philosophical brain-teaser which can be solved in more than one way. One of the many possible answers is shared in this article!

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How many colours?

A young boy was playing chess against his father. The kid just learned the rules and the father was a very good player, so of course the boy lost all his games against him. The father didn’t want to lose on purpose and encourage some false hope for his son.

One day, the father asked his son if he wanted to play another game of chess against him. But the child said he didn’t want to play with him because he would lose anyway. This answer left the father a bit sad. He didn't want his son to feel bad about chess. It is such a nice game, after all.

The father thought for a moment and said: “How about I give you a riddle instead? If you can solve it, I will count it as a win against me”.

This idea didn’t fail and caught the son’s interest, so he went to his father. When the boy arrived, he saw that the chessboard was set up with nothing else but a king.

 

“Listen carefully”, the father started, “if you colour all the squares of the chessboard so that the king enters a different-coloured square in each move, how many different colours do you need?”

The son thought about the riddle for a while. After a couple of hours, he ran to his father to tell him the correct answer. Now it is your turn:


How many different colours are needed?


Note that the very first thought might be that we only need black and white square. But keep in mind: the king can reach a lot of different squares in each move!

I hope that there are people out there who haven’t heard of this puzzle before.

It would be interesting to know how much time you needed to solve the riddle. It varied a lot while testing!


The white-always-wins solution

In this philosophical puzzle, practically everyone who wrote a comment to the solution got it right! Well done and thank you once again for your interest, analysis and thoughts.

Nonetheless, here is the intentional solution:

Fritz knows about both paper rolls, so he will “eat” one of them. The other paper roll will show the word “black”, leading to the conclusion that the paper roll he ate read “white”.

Houdini knows that Fritz knows about the paper rolls, so he will change both rolls to read “white” instead of “black”.

Fritz knows that Houdini knows that Fritz knows about the paper rolls. So Fritz won’t eat one of the paper rolls, but instead take it, read it, play with white and win the tournament.

The end.

The riddle was very philosophical, because this back and forth game could have continued forever, leading to a 50/50 situation for both players. Since an intentional manipulating plan was involved, the outcome leaves space for more speculation. How well do the chess players know each other? What if Fritz never knew about the plan in the first place? If he never knew about the plan, would he have eaten one of the paper rolls?


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Arne Kaehler, a creative thinker who is passionate about board games in general was born in Hamburg and learned how to play chess at a very young age. Through teaching chess to youth teams and creating chess content on YouTube, Arne was able to extend this passion onto others and has even made an online chess course for anyone who wants to learn how to play this game. Currently, Arne blogs for the English news page of ChessBase and focuses on creating promotional and entertaining articles.
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SPARSH SHARMA SPARSH SHARMA 7/21/2020 07:41
@Michael Jones
Yeah! The grammar of the question is weak and incorrect.
SPARSH SHARMA SPARSH SHARMA 7/12/2020 01:10
@Timothy Chow
Oh, yes, it differs in relation with the 4 colour theorem. Even a few rules does not match with this question. But the desired answer is same, that is, 4 colours are required which I verified physically making a chessboard with a pen and marking each square as abbreviation of colour.
Michael Jones Michael Jones 7/11/2020 09:55
The question is poorly worded: "so that the king enters a different-coloured square in each move"... different from what? If it means "different from the colour of any of the squares it entered in preceding moves", then provided the king cannot move back to a square which it has previously visited (which would presumably be the same colour the second time as it was the first), the answer would be 64, or 63 if the square on which it started is not considered one which it entered in a move.

If it means "so that any two squares on the board between which the king can move are of different colours", it's fairly straightforward to prove that the answer is four. Consider a 2*2 square: a king can move from any of the four squares to any other, so all of them must be different colours. Make 16 copies of that 4-coloured 2*2 square and stick them together (keeping them all in the same orientation) to make the standard 8*8 board. A quick glance tells you that no two squares of the same colour are adjacent, so the king can't move between them and the puzzle is solved. There is nothing special about 8*8 in this respect: four colours are sufficient for any size of board.
serdale serdale 7/11/2020 07:55
4 couleurs en 3 minutes
Timothy Chow Timothy Chow 7/11/2020 04:19
This is *not* an example of the "Four Colour Map Theorem." In that theorem, two regions that touch only at a corner are not considered to be adjacent. (For those who know some graph theory, the 3x4 king graph contains a subgraph homeomorphic to K_5 and hence is not planar.)
JoshuaVGreen JoshuaVGreen 7/11/2020 02:59
@Flávio Patricio Doro, that merely shows that four colors suffice. Fortunately, it isn't hard to show that four colors are necessary.
Flávio Patricio Doro Flávio Patricio Doro 7/11/2020 02:42
Agreed. Four colours are needed.
Say each colour is a digit: 1, 2, 3, 4. Let each digit repeat itself every two rows and files. The first rank would read 12121212, the second 34343434, the third 12121212 again and so on.
JoshuaVGreen JoshuaVGreen 7/11/2020 01:59
@SPARSH SHARMA, the "Four Colour Map Theorem" doesn't directly apply since corner connections count here.
SPARSH SHARMA SPARSH SHARMA 7/11/2020 12:06
@Mike Magnan
Your solution is for the at most colours needed to colour the chessboard. But the question asks for fewest colours needed to colour the chessboard.
SPARSH SHARMA SPARSH SHARMA 7/11/2020 12:04
Answer is 4.
This is an example of a well known 'Four Colour Map Theorem'.
It's method and proof: https://www.mathsisfun.com/activity/coloring.html
I once visited this website and remembered it deeply that I began to go for some advanced research for the same.
Mike Magnan Mike Magnan 7/10/2020 10:56
63 would work without figuring anything out.
Zvi Mendlowitz Zvi Mendlowitz 7/10/2020 05:11
The answer is the same as the answer for a famous mathematical problem, solved in 1976.
Also, the answer can be found here:
https://mathworld.wolfram.com/KingGraph.html
JoshuaVGreen JoshuaVGreen 7/10/2020 03:39
The new riddle isn't so clear. Is it fair to rephrase it as

"Color the chessboard so that no two squares connected by a King's move are the same color. How many distinct colors are needed?"

?
Erdmundr Erdmundr 7/10/2020 03:39
Symmetry is your friend here. Once you have the right shape in mind, it is straightforward to see the the number of colors needed locally is both necessary and sufficient.
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