Chess and Mathematics Conference London Olympia 6-7 Dec

[Neill Cooper]

A common feature of maths (particularly at University level) and games such as chess is the requirement for abstract thinking. This is discussed in the interesting book 'Games and Mathematics' by David Wells who is one of the speakers at the conference.

[Matthew Turner]

Neill,
You may want to have a look at this/contact this Canadian Organisation

http://www.chess-math.org/about.htm

They certainly seem to be quite high profile in Canadian Chess

Matt

[Matthew Turner]

Some quite tricky Maths

http://mathoverflow.net/questions/63423 ... 250#122250

[MartinCarpenter]

Fun thanks. Can even more or less understand what they're going on about without epic amounts of background.
(Mostly just a bit of Cantor's stuff really, which is more very abstract rather than impossible to understand.).

I do like the position where its 'a win for white if the players are required to play according to a deterministic computable strategy, but which is a draw without that restriction.'. That'd make writing the laws really fun

[Matthew Turner]

Martin,
If you enjoy this sort of question then you might like this one

A Bishop begins on a White square on the back rank. It moves one square diagonally forward on each turn. How many different paths are there to get to the other side of the board.

It took me about an hour, one of the sixth form did it in less than two minutes and a couple of weeks ago one of the year 8's (12 years old) did it in under ten minutes.

Matt

[Christopher Kreuzer]

Is it 2 to the power of 8 (i.e. 256)? I'm guessing not if it took you an hour to work it out... Ah, the sides of the board provide some limits? By the time you get to move four, you've expanded across the whole board. So, um, 2 to the power of 4 plus something? (Back to the infinite board, much easier).

[MartinCarpenter]

That's more arithmetic mind Depends how you define forwards/the board though surely?

If forwards means towards the back rank then even starting in the middle of an 8 * 8 board truncates it, and the corners are even worse. If any square not yet visited then somewhat different of course. Come to think of it, its easiest to define forwards in the most basic usage and give the answer as 1 in a corner facing the right way or 0 otherwise......

[Matthew Turner]

The Bishop will make seven moves in total, but it could start on b1, d1, f1 or h1.
I hope that makes sense.
Matt

[Matt Fletcher]

A Bishop begins on a White square on the back rank. It moves one square diagonally forward on each turn. How many different paths are there to get to the other side of the board.

It took me about an hour, one of the sixth form did it in less than two minutes and a couple of weeks ago one of the year 8's (12 years old) did it in under ten minutes.

Matt

It depends which square you start from - I think I've worked it out based on an infinite board (7 moves from end to end, not then working out how many moves are lost due to the boundaries.

The total number of moves on an infinitely wide board with 8 ranks would be 2^7 (128). On an 8x8 board, I lose exactly half of the tree if I start on h1; if I start n squares away from the corner, I lose half of an n-node tree on one side and half of a (8-n)-node tree on the other side.

I think the answer is something like;
b1 95
d1 112
f1 110
h1 64

[numbers could well be wrong but think the principle is about right]

[MartinCarpenter]

Must be close but can't be quite right. The corner of a 4*4 board is 3 not 4. 5*5 is 6.

You're also losing the chunk of the tree that would involve crossing back onto the actual board, which doesn't seem trivial to quantify to me because it can get very convoluted, esp when starting in the middle. Could count when it happens of course but that wouldn't do 2 minutes!

[Matt Fletcher]

Must be close but can't be quite right. The corner of a 4*4 board is 3 not 4. 5*5 is 6.

You're also losing the chunk of the tree that would involve crossing back onto the actual board, which doesn't seem trivial to quantify to me because it can get very convoluted, esp when starting in the middle. Could count when it happens of course but that wouldn't do 2 minutes!

Yes of course, there's some more factors missing there somewhere - slightly trickier than I'd thought.

[Simon Brown]

Is the total for all four starting squares 499?

Edit: No it isn't. Is it 468 (using 2^7 less the number of times they will end up on the a or h files)?

[Arshad Ali]

Suppose it starts on b1 and want to find how many paths to g8. Work backwards from g8. From g8 it can go back to h7 or f7. If h7, there is only one path to b1. If f7, there are two paths back -- one through g6 and on through e6. If, g6, there is only one path to b1. If e6, then two paths back, through f5 and d5 respectively. If f5, just one path back to b1. Move in this manner backwards, and keep a running total. Then do the same for e8, et cetera.

Alternatively, start on b1 and see that the only white squares it can reach on the fifth rank are b5, d5, and f5. There is only one path to f5. See how many paths there are from f5 to c8, e8, and g8 (e.g., there are only three paths to g8). From d5 there is only one path to g8. But since there are four paths from b1 to d5, the total number of paths from b1 to g8 is 3+ 4 =7. Likewise there are only one path from b5 to e8, three paths from d5 to e8 and three paths from f5 to e8. But there are five paths from b1 to b5 and four paths from b1 to d5, so we get (5*1) + (4*3) + (1*3) = 20 paths from b1 to e8. Use similar reasoning for c8 and a8. I'm sure somewhere one can use symmetry arguments to simplify the enumeration.

Alternatively, using combinatorial reasoning, observe that to get from b1 to g8 involves six movements right and one left. There are no constraints for this case. Get 7!/6!1! = 7, agreeing with the answer achieved by brute force methods above. To get from b1 to e8 involves five right moves and two left moves, but now there are constraints: we can't have two initial moves left. So use the expression (7!/5!2!) - 1 = 20, agreeing again with brute force. Using similar reasoning, 32 paths from b1 to c8 and 14 paths from b1 to a8. So a total of 7+20+32+14= 73 paths from b1 to the other side. To calculate paths from d1, the same exercise again, but can use symmetry reasoning to cut down calculation (e.g., number of paths from d1 to g8 will be the same as number from b1 to e8).

[Matthew Turner]

No answers that are particularly close so far. Sorry

[benedgell]

Is it 50?