Timeline for Dim - A Game with Pebbles
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 25, 2015 at 18:29 | vote | accept | Mike Earnest | ||
| May 25, 2015 at 16:37 | comment | added | Rand al'Thor | @KateGregory Optimal play is defined recursively as follows. 1) n=1 is (in this game) a losing position by definition. 2) A position is winning if there exists a single move from it that gets to a losing position. 3) A position is losing if every single move from it gets to a winning position. | |
| May 25, 2015 at 13:31 | comment | added | Kate Gregory | I understand the question only asked what n was given optimal play, but it seems to me that a simple definition of that optimal play belongs in the answer, especially since non-optimal play is provided as an example in the question. | |
| May 25, 2015 at 13:28 | comment | added | Mark N | @KateGregory Alice keeps making it an odd number for Bob (i.e subtract 1), then Bob must then make it even again for Alice (since he can't take all the pebbles, and there is no odd number whose divisor when subtracted will have another odd number). Eventually Alice will make it equal to 3 then 1 (odd) to which Bob losses. | |
| May 25, 2015 at 11:50 | comment | added | Kate Gregory | I don't follow this. Take the example in the question, they start with 12 which is even. So Alice is going to win. She removes 2 leaving 10. This is the same as starting with 10 (which is even) and Bob going first. So Bob is going to win? Or is the "optimal play" that Alice must choose a number of stones to remove that leaves Bob with an odd number? | |
| May 25, 2015 at 8:46 | history | answered | Rand al'Thor | CC BY-SA 3.0 |