In an earlier post I described what’s generally known as the Mutilated Chessboard Puzzle. It goes like this: a chessboard has 2 diagonally opposite corners removed. The challenge is to cover the remaining 62 squares with 31 dominoes, each of which can cover 2 adjacent horizontal or vertical squares. Or, to show that such a coverage is impossible.

Several of you wrote to me about this, in many cases providing the correct solution. Thanks and congratulations to all of you.

The correct solution is that it is impossible to cover the remaining squares of the chessboard this way. But what’s interesting about this puzzle – to me at least – is what it illustrates about mathematical proof.

There would essentially be 2 ways to prove the impossibility of a domino coverage. One way would be to enumerate every possible configuration of the 31 dominoes, and to show that none of these configurations covers the 62 remaining squares on the chessboard. But this takes a lot of time – there are many different ways of laying the dominoes on the chessboard.

The alternative approach is to ‘step back’ and try to reason logically why such a configuration is impossible. This approach won’t always work, but it’s often short and elegant when it does. And with the mutilated chessboard puzzle, it works beautifully…

When you place a domino on a chessboard it will cover 1 black and 1 red square (using the colours in the diagram above). So, 31 dominoes will cover 31 black and 31 red squares. But if you remove diagonally opposite corners from a chessboard, they will be of the same colour, so you’re left with either 32 black squares and 30 red, or vice versa. But you’re never left with 31 squares of each colour, which is the only pattern possible with 31 dominoes. So it’s impossible and the result is proved. Simply and beautifully.

As I mentioned in the previous post the scientific writer Cathy O’Neil cites having been shown this puzzle by her father at a young age as the trigger for her lifelong passion for mathematics. And maybe, even if you don’t have a passion for mathematics yourself, you can at least see why the elegance of this proof might trigger someone’s love for mathematics in the way it did for Cathy.

Having said all that, computer technology now makes proof by enumeration possible in situations where the number of configurations to check might be very large. But structured mathematical thinking is still often necessary to determine the parameters of the search. A good example of this is the well-known four colour theorem. This states that if you take any region that’s been divided into sub-regions – like a map divided into countries – then you only need four colours to shade the map in such a way that no adjacent regions have the same colour.

Here’a an example from the Wiki post:

You can see that, despite the complexity of the sub-regions, only 4 colours were needed to achieve a colouring in which no two adjacent regions have the same colour.

But how would you prove that *any* map of this type would require at most 4 colours? Ideally, as with the mutilated chessboard puzzle, you’d like a ‘stand back’ proof, based on pure logic. But so far no one has been able to find one. Equally, enumeration of all possible maps is clearly impossible – any region can be divided into subregions in infinitely many ways.

Yet a proof has been found which is a kind of hybrid of the ‘stand back’ and ‘enumeration’ approaches. First, a deep understanding of mathematical graphs was used to reduce the infinitely many possible regions to a finite number – actually, around 2000 – of maps to consider. That’s to say, it was shown that it’s not necessary to consider all possible regional mappings – if a 4-colour shading of a certain set of 2000ish different maps could be found, this would be enough to prove that such a shading existed for all possible maps. Then a computer algorithm was developed to search for a 4-colour shading for each of the identified 2000 or so maps. Putting all of this together completed the proof that a 4-colour shading existed for any map, not just the ones included in the search.

Now, none of this is strictly Statistics, though Cathy O’Neil’s book that I referred to in the previous post is in the field of data science, which is at least a close neighbour of Statistics. But in any case, Statistics is built on a solid mathematical framework, and things that we’ve seen in previous posts like the Central Limit Theorem – the phenomenon by which the frequency distributions of many naturally occurring phenomena end up looking bell-shaped – are often based on the proof of a formal mathematical expression, which in some cases is as simple and elegant as that of the mutilated chessboard puzzle.

I’ll stop this thread here so as to avoid a puzzle overload, but I did want to mention that there is an extension of the Mutilated Chessboard Puzzle. Rather than removing 2 diagonally opposite corners, suppose I remove any 2 arbitrary squares, possibly adjacent, possibly not. In that case, can the remaining squares be covered by 31 dominoes?

If the 2 squares removed are of the same colour, the solution given above works equally well, so we know the problem can’t be solved in that case. But what if I remove one black and one red square? In that case, can the remaining squares be covered by the 31 dominoes:

- Always;
- Sometimes; or
- Never?

I already sent this problem to some of you who’d sent me a solution to the original problem. And I should give a special mention to Fabian.Thut@smartodds.co.uk who provided a solution which is completely different to the standard textbook solution. Which illustrates another great thing about mathematics: there is often more than solution to the same problem. If you’d like to try this extension to the original problem, or discuss it with me, please drop me a line.