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0 sats \ 3 replies \ @SimpleStacker OP 9 Dec 2024 \ parent \ on: [Logic Puzzle] Chinese Dumplings science
Well you are right to approach it game theoretically, but the answer should depend on N actually...
Ok thanks, for the hint.
Data for the first several N values:
N | Optimal Move | Position Type
0 | can't move | Losing
1 | can't move | Losing
2 | 1 | Winning
3 | 2 | Winning
4 | 3 | Winning
5 | 1 | Winning
6 | 1 | Winning
7 | 2 | Winning
8 | 3 | Winning
9 | 1 | Winning
10 | 1 | Winning
11 | 2 | Winning
12 | 3 | Winning
Looking at winning positions:
For N = 2,3,4: take N-1 dumplings
For N = 5,6,7: take (N-4) dumplings
For N = 8,9,10: take (N-8) dumplings
For N = 11,12,13: take (N-10) dumplings
Key observations:
The pattern repeats every 4 numbers
For any N, we can find our position in the cycle using modulo 4
But we need to handle N = 0,1 as special cases
Therefore, if N ≥ 2:
Let k = N mod 4
If k = 1, take 1
If k = 2, take 2
If k = 3, take 3
If k = 0, take 3
Or more concisely, for N ≥ 2:
optimal move = k if k ∈ {1,2,3}
optimal move = 3 if k = 0
where k = N mod 4
For N < 2, no valid move exists.
This ensures you always force your opponents into a losing position eventually, while never breaking the etiquette rules.
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You're on the right track, but the solution is not quite accurate yet.
For example, if N=3, your solution says take 3. But that's rude because you take the last dumpling.
Another example, if N=5, your solution says take 1. But if N=5, you can take 3, then the next person will take 1, leaving the next person with the last dumpling. This maximizes your dumplings without being rude.
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Ah yes, how about this:
optimal_move = if N ≤ 2: invalid
if N ≤ 4: N-1
if N mod 4 ∈ {1,3}: 2
if N mod 4 ∈ {0,2}: 3
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