Re: [math-fun] Re: Kepler conjecture & Flyspeck
John McCarthy wrote: << When a long computer calculation is an important part of the proof of a mathematical theorem, we should ask for proof (preferably both computer-checked and hand checked) that the program does the advertised calculation. For example, we want a proof that if the calculation comes out, the four color conjecture is true.
This is of course desirable, but brings up a new question: where does this stop? A typical "verification of correctness" of a sizeable computer program involves the application of another program. Then we need to know that the verification program is valid and was used properly. Etc. --Dan
'<< When a long computer calculation is an important part of the proof of a mathematical theorem, we should ask for proof (preferably both computer-checked and hand checked) that the program does the advertised calculation. For example, we want a proof that if the calculation comes out, the four color conjecture is true.
This is of course desirable, but brings up a new question: where does this stop? A typical "verification of correctness" of a sizeable computer program involves the application of another program. Then we need to know that the verification program is valid and was used properly. Etc.' I think that the main problem with computer proves is proving the exhaustion of all possible cases. With some problems, e.g. Ramsey Theory on small graphs, this is possible and proofs are acceptable and accepted. With problems that do not fall into simple enumeration patterns, e.g. the four colour conjecture and Kepler's conjecture, the basic axiom of truth is that a deterministic model such as the computer is incapable of such a proof, in the same way proving the primality of numbers with 10^100 digits will remain out of deterministic computing's reach for eternity. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com
Let there be n buttons. A player may take 1 or 2 buttons MORE than his/her opponent took. Player 1 can take either 1 or 2 buttons. e.g. 28->27->25->21->15->7->0 (the last player must take either 8 or 9, as only 7 remain, they must take them all). What are the strategies for various n, with either last button wins/loses? Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com
I assume normal play, so that if there are insufficient buttons for you to take an increased number, then you can't move and lose. Then, for n = 0, 1, 2, ... if you are in the range 3n^2 + 2n to 3n^2 + $n, e.g., 0; 5-6; 16-20; 33-39; ... P-positions (previous player winning). Nim-value 0. Remoteness 2n. 3n^2 + 4n + 1 to 3n^2 + 5n, e.g., empty; 8; 21-22; 40-42; ... N-positions (next player winning). Nim-value 2. Remoteness 2n+1. Win by taking 1. 3n^2 + 5n + 1 to 3n^2 + 6n + 1, e.g. 1; 9-10; 23-25; 43-46; ... N- 1. 2n+1. Win by taking 1. 3n^2 + 6n + 2 e.g., 2; 11; 26; 47; ... N- 1. 2n+1. Win by taking 1 or 2. 3n^2 + 6n + 3 to 3n^2 + 7n + 2, e.g., empty; 12; 27-28; 48-50; ... N- 1. 2n+1. Win by taking 2. 3n^2 + 7n + 3 to 3n^2 + 8n + 4, e.g., 3-4; 13-15; 29-32; ... N- 2. 2n+1. Win by taking 2. When you CAN win, I suspect that a winning strategy is to increase by 1 when your opponent has increased by 2, and to increase by 2 when your opponent has increased by 1. As the values are only 0, * and *2, I guess that the mis`ere play is analyzable. The game with more than one heap may be quite tricky to analyze, if you assume that when it is your turn to move you must take 1 or 2 more buttons than before, regardless of which heap you take them from. R. On Tue, 29 Jul 2003, Jon Perry wrote:
Let there be n buttons. A player may take 1 or 2 buttons MORE than his/her opponent took. Player 1 can take either 1 or 2 buttons.
e.g. 28->27->25->21->15->7->0
(the last player must take either 8 or 9, as only 7 remain, they must take them all).
What are the strategies for various n, with either last button wins/loses? Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com
Still doesn't answer the variable range problem. Surely maths is capable of greater things? Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com]On Behalf Of Richard Guy Sent: 05 August 2003 04:47 To: math-fun Cc: David Wolfe; John Conway; Richard Nowakowski; Aviezri Fraenkel; Elwyn Berlekamp; Sally Smith; John Conway Subject: Re: [math-fun] Nim-like game I assume normal play, so that if there are insufficient buttons for you to take an increased number, then you can't move and lose. Then, for n = 0, 1, 2, ... if you are in the range 3n^2 + 2n to 3n^2 + $n, e.g., 0; 5-6; 16-20; 33-39; ... P-positions (previous player winning). Nim-value 0. Remoteness 2n. 3n^2 + 4n + 1 to 3n^2 + 5n, e.g., empty; 8; 21-22; 40-42; ... N-positions (next player winning). Nim-value 2. Remoteness 2n+1. Win by taking 1. 3n^2 + 5n + 1 to 3n^2 + 6n + 1, e.g. 1; 9-10; 23-25; 43-46; ... N- 1. 2n+1. Win by taking 1. 3n^2 + 6n + 2 e.g., 2; 11; 26; 47; ... N- 1. 2n+1. Win by taking 1 or 2. 3n^2 + 6n + 3 to 3n^2 + 7n + 2, e.g., empty; 12; 27-28; 48-50; ... N- 1. 2n+1. Win by taking 2. 3n^2 + 7n + 3 to 3n^2 + 8n + 4, e.g., 3-4; 13-15; 29-32; ... N- 2. 2n+1. Win by taking 2. When you CAN win, I suspect that a winning strategy is to increase by 1 when your opponent has increased by 2, and to increase by 2 when your opponent has increased by 1. As the values are only 0, * and *2, I guess that the mis`ere play is analyzable. The game with more than one heap may be quite tricky to analyze, if you assume that when it is your turn to move you must take 1 or 2 more buttons than before, regardless of which heap you take them from. R. On Tue, 29 Jul 2003, Jon Perry wrote:
Let there be n buttons. A player may take 1 or 2 buttons MORE than his/her opponent took. Player 1 can take either 1 or 2 buttons.
e.g. 28->27->25->21->15->7->0
(the last player must take either 8 or 9, as only 7 remain, they must take them all).
What are the strategies for various n, with either last button wins/loses? Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com
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