See Winning Ways, p.114. The game is in fact one of the many disguises for Nim. You can state it as `play Nim with heaps in the usual way, but you're also allowed to split a heap when you take from it'. In Guy-Smith code it's the game .77777... In your example the ` heaps' are 1, 2, 3, 4 with nim-values (no surprise) 1, 2, 3, 4. The (only) winning moves must operate on the 4-heap -- either take the whole heap, leaving the P-position (1,2,3) or split it into two equal heaps, here possible in only one way, leaving (1,2,3,1,1). Let's do a more complicated example, say heaps (rows) of 9, 11, 14, 15 which we'll write in binary as 9 = 1 0 0 1 11 = 1 0 1 1 14 = 1 1 1 0 15 = 1 1 1 1 nim-sum 0 0 1 1 = 3 so it's an N-position and the next player should change the nim-value (i.e. the actual size in this case, since we are playing Nim in disguise) of a heap by 3. In ordinary Nim (and in the variant) we can win by taking 3 beans from the 11 or 15 heap, or 1 bean from the 14 heap. In the variant this translates into taking beans from the end of the row. Are there any other good moves? Not from the 11 heap, cos that has to go down to 8, but the 15 has to go to 12 which is also 8 + 4 or 9 + 5 (note that it's nim-addition here) so you can take 3 beans from the interior of the row to leave 8 & 4 or 4 & 8 or 1 bean to leave 9 & 5 or 5 & 9. 14 has to be reduced to 13 (14 \pm 13 = 3 in Nim) which is 8 + 5 or 9 + 4 so there are six different places in the row of 14 from which to take 1 bean & win & confuse your opponent about what strategy you're using. On Wed, 3 Jan 2007, Jeffrey Shallit wrote:
My 10-year-old son introduced me to the following variation on nim. Instead of allowing the players to take any number of counters from a pile on a turn, the counters in each pile are arranged linearly, and then you can only remove A CONTIGUOUS BLOCK of counters. Here contiguous means no empty spaces, too. I am also talking about normal play, not misere play here, so that the person who takes the last block and empties the piles, wins.
I don't know a general strategy for this game. Does anyone know if it has been studied?
In the particular case he plays with his friend, the initial configuration looks like:
Pile 1: 1 Pile 2: 2 3 Pile 3: 4 5 6 Pile 4: 7 8 9 10
where I have numbered the counters for clarity. Pile 1 has 1 counter, on the first line. Pile 2 has 2 counters, on the second line (2 and 3), etc. So on a move you could take counters 8 and 9, leaving 7 and 10, and then on the next move the next player could take 7 or 10 but not both, since there are empty spaces left between them.
It turns out that in this particular case of 10 counters there is a forced win for the 1st player, who can take either 7 8 9 10 or 8 9 on his first move.
Any info about this variation?
Best, Jeff Shallit
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