Chess and Mathematics Conference London Olympia 6-7 Dec

[Matthew Turner]

I refer the honourable gentleman to my previous answer

[benedgell]

Okay, I give up.

[Matt Fletcher]

Actually on re-combination, eg from the corner am I just missing half a 5-node tree - and similarly for the other starting points?

So from h8 it would be 2^7 (total) - 2^6 (missing from RHS) - 2^4 (missing on re-combination) = 48

And from b8 it would be 2^7 - 2^5 (LHS) - 2^3 (re-combination) - 1 (RHS) = 87

And from d8 it would be 2^7 - 2^3 (LHS) - 2^2 (RHS) - 2 (LHS re-comb) - 1 (RHS re-comb) = 113

And from f8 it would be 2^7 - 2^4 (RHS) - 2^2 (RHS r-c) - 2 (LHS) = 106

Again doing this too quickly so could very easily be wrong.

[Matthew Turner]

Matt
I think this looks like the same methodology to the one I used. I am not sure what is going wrong, but you are quite some way off.
Matt

[Michael Farthing]

I think it's 98 for a start on f1

My reasoning is that if the board extended laterally for ever we would have 2^7 possibilties and the actual landing squares would be given by the 7th line of Pascal's triangle, which is:

1 7 21 35 35 21 7 1

the two '35' values correspond to the two squares on files close to the bishop's start point, namely e8 and g8
Putting the files above the numbers above we then have

Code: Select all

  a  c  e   g           
1 7 21  35  35  21  7  1

So the total paths are 7+21 +35 +35 =98.

The reader can repeat the last step for the other starting points to get a grand total!

[MartinCarpenter]

The only way I can see which I'd trust would be starting from 2^7 and working out precisely which chunks of the tree were lost from that each time you reached an edge. That'd give an answer in 10-15 minutes (or more?) but the fact is we (seemingly) know there is an answer which can be had in 2.

That has to be something hugely more elegant than I can see at the moment, so brute force doesn't count

[Michael Farthing]

I think it's 98 for a start on f1

My reasoning is that if the board extended laterally for ever we would have 2^7 possibilties and the actual landing squares would be given by the 7th line of Pascal's triangle, which is:

1 7 21 35 35 21 7 1

the two '35' values correspond to the two squares on files close to the bishop's start point, namely e8 and g8
Putting the files above the numbers above we then have

Code: Select all

  a  c  e   g           
1 7 21  35  35  21  7  1

So the total paths are 7+21 +35 +35 =98.

The reader can repeat the last step for the other starting points to get a grand total!

Well that was a load of rubbish

[Michael Farthing]

What about:

Code: Select all

 
---------------------------------
| 35|   | 89|   |103|   | 69|   |
---------------------------------
|   | 35|   | 54|   | 49|   | 20|
---------------------------------
| 10|   | 25|   | 29|   | 20|   |
---------------------------------
|   | 10|   | 15|   | 14|   |  6|
---------------------------------
|  3|   |  7|   |  8|   |  6|   |
---------------------------------
|   |  3|   |  4|   |  4|   |  2|
---------------------------------
|  1|   |  2|   |  2|   |  2|   |
---------------------------------
|   |  1|   |  1|   |  1|   |  1|
---------------------------------

Starting from the first rank each square shows the number of ways in which it can be reached.
Adding the numbers on the eight rank gives a total number of 35+89+103+69 = 296

Drawing the rather inelegant diagram has taken ages longer than the problem, so I hope its correct this time.

[MartinCarpenter]

That seems to have a very good chance of being right. Still leaves the question of what those numbers actually are I guess. Maybe nothing elegant

[Neill Cooper]

What about:

Code: Select all

 
---------------------------------
| 35|   | 89|   |103|   | 69|   |
---------------------------------
|   | 35|   | 54|   | 49|   | 20|
---------------------------------
| 10|   | 25|   | 29|   | 20|   |
---------------------------------
|   | 10|   | 15|   | 14|   |  6|
---------------------------------
|  3|   |  7|   |  8|   |  6|   |
---------------------------------
|   |  3|   |  4|   |  4|   |  2|
---------------------------------
|  1|   |  2|   |  2|   |  2|   |
---------------------------------
|   |  1|   |  1|   |  1|   |  1|
---------------------------------

Starting from the first rank each square shows the number of ways in which it can be reached.
Adding the numbers on the eight rank gives a total number of 35+89+103+69 = 296

Drawing the rather inelegant diagram has taken ages longer than the problem, so I hope its correct this time.

You can make it a bit more elegant by only going half way and then using symmetry properties.
ps You find the original problem at http://www.bmoc.maths.org/home/bmo1-2009.pdf

[Matthew Turner]

296 is indeed the right answer
I find this interesting because if the board was infinitely wide there would be 2^9 possibilities (7 moves and 4 initial starting squares).
296 = 2^9 - 6^3 I find it incredibly hard to believe that is a coincidence. However, I am assured by very eminent mathematicians that this is indeed the case and that you cannot develop a (simple) formula for an x by y board.

[Arshad Ali]

For some reason I get 385. The BMO problem is a bit different and I think using symmetry, the answer should be 385*2 (assuming my answer is correct).

Postecript: 385 is wrong, I'm double counting some paths. Whatever the answer, multiply by 2 for the BMO problem.

Postscript 2: Multiply by 2, then take away 1, otherwise the long diagonal is being counted twice.

[Stewart Reuben]

Justin >Quite unforgivable after playing in a simul against him (Mestel) at Imperial.
Who gave the simul?

I think people who consider a correlation between mathematical and chess ability overlook an important point. Many people who are good at one have the mind-set to enjoy the other. Thus you get a high correlation which doesn't at all prove that studying chess improves one's results at mathematics - or science.
I remember a teacher at school saying, 'It is often thought that studying Latin improves your academic results. So Why not study chess? That must have a similar effect, but be more fun.'
Stuart Margulies did a study that showed people who did a chess course achieved higher grades in their Stats in New York than those who didn't. It was deeply flawed. It is very likely that a student who went to the trouble of following any course, perhaps in motor car maintenance or knitting, would do better than somebody who undertook no such project.

Chess is
Inexpensive
Language free. That is one reason why it was popularised in the USSR.
Accessible to people who aren't sports orientated.
Fun for some people.
Provides a quanitifiable rating which means the prowess of a player can be determined approximately without using an exam.

The first 4 apply equally well to mathematics.

[Matthew Turner]

Arshad,
I have read the BMO question a couple of times but I still cannot see why you are multiplying by 2.
Matt

[JustinHadi]

Justin >Quite unforgivable after playing in a simul against him (Mestel) at Imperial.
Who gave the simul?

Stewart, I'm sure if Imperial College had consulted you earlier you'd be a far superior simul giver to either of us.