Timeline for A little number theoretic game
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 20, 2023 at 19:06 | comment | added | Gro-Tsen | @StevenStadnicki Actually it appears Peter Taylor was talking about the “modified” game whereas I. J. Kennedy's computations are for the original game. I hope the comment-in-form-of-answer I just posted clears up this confusion. | |
| Apr 20, 2023 at 17:19 | comment | added | Steven Stadnicki | @Gro-Tsen I think at this point the folks looking at it are primarily working with the 'modified' version in which a player has to win if they can — i.e., the only legal move from a prime is the (winning) one to 1; this is much more amenable to analysis for exactly these sorts of reasons. | |
| Apr 20, 2023 at 13:13 | comment | added | Gro-Tsen | Even though it was proved by OP that the game always terminates in finite time if at least one player tries to win, it still has a loopy part, so I'm a bit worried about nim values: for me, the game $G$ has nim value $n$ iff $G+(*n)$ is a second-player win (where $*n$ is a single nim row of $n$ sticks), but it can conceivably happen that the game $G+(*n)$ is a draw: computation with loopy nim values is more complicated than just taking a mex, see Siegel, Combinatorial Game Theory (2013), definition IV.4.12. ❧ Is there an argument I missed to explain that this subtlety can't happen here? | |
| Apr 20, 2023 at 3:39 | history | edited | I. J. Kennedy | CC BY-SA 4.0 |
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| Apr 19, 2023 at 14:19 | comment | added | Michael Lugo | @GregMartin but not quite - the coin flipping process you describe would lead to an average gap of exactly 3. | |
| Apr 19, 2023 at 8:27 | comment | added | Peter Taylor | @TimothyChow, nimber frequencies in the first $10^7$ values of $n$: 0: 3261996, 1: 3390181, 2: 1978115, 3: 1105841, 4: 242523, 5: 20995, 6: 349. | |
| Apr 19, 2023 at 7:15 | comment | added | Leif Sabellek | @AlessandroDellaCorte Not a strict argument, but an indication: The number of possible replys to the number n is (1 + number of primes that divide n). Since larger number will be divisible by more primes, there will on average be more possible replys for large n. But n is only winning if all possible replies are losing. So the "chance" that n is winning becomes smaller for larger n. EDIT: Joachim König answered the same thing, just a little more formal, in a comment to the original question. | |
| Apr 19, 2023 at 0:51 | comment | added | Timothy Chow | Each number n has an associated nimber. Can you easily adapt your code to compute these nimbers? That would give more information than just whether n is winning or losing. | |
| Apr 18, 2023 at 21:54 | comment | added | Alessandro Della Corte | The argument by the OP shows that any game is decided in finitely many moves, that is any number is either winning or losing. Is there any “easy” reason for which winning numbers should be much fewer than losing ones? (Let alone having density 0…) | |
| Apr 18, 2023 at 21:12 | comment | added | Greg Martin | The $\frac1{2^{g-1}}$ phenomenon is consistent with the winning positions being distributed at random other than not having two of them next to each other. So this suggests to me that there's no simple formula: the numbers are just acting like they're flipping coins. | |
| Apr 18, 2023 at 18:09 | comment | added | Michael Lugo | In the first ten consecutive blocks of one million integers, the number of winning numbers is 327483, 325969, 326438, 326090, 326242, 326203, 325602, 325987, 325875, 326107. If the frequency of winning numbers drops off it does so very slowly, although per Joachim König's comments this could be tied to the very slow growth of the number of prime divisors. | |
| Apr 18, 2023 at 17:33 | history | edited | I. J. Kennedy | CC BY-SA 4.0 |
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| Apr 18, 2023 at 17:06 | history | answered | I. J. Kennedy | CC BY-SA 4.0 |