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0 sats \ 4 replies \ @satcat 9 Dec \ on: [Logic Puzzle] Chinese Dumplings science

Let me approach this as a game theory problem:

For any move to be valid, a businessman must:
    Take 1, 2, or 3 dumplings (not 0 or >3)
    Not take the last dumpling
Let's analyze what would happen with different moves, assuming optimal play: If first person takes 1:
    8 dumplings remain
    Next person could take 3 → 5 remain
    Next person could take 3 → 2 remain
    Last person can only take 1 → 1 remains
    Back to first person, who can't take the last one
If first person takes 2:
    7 dumplings remain
    Next person could take 3 → 4 remain
    Next person could take 3 → 1 remains
    Last person can't take it (would be the last one)
If first person takes 3:
    6 dumplings remain
    Next person could take 3 → 3 remain
    Next person could take 2 → 1 remains
    Last person can't take it

Looking at these scenarios, taking 2 dumplings is optimal because:

Taking 1 leaves too many for others
Taking 3 doesn't improve the outcome
Taking 2 forces a situation where some future player will be stuck unable to take dumplings (which is rude)

Therefore, the current businessman should take 2 dumplings.

0 sats \ 3 replies \ @SimpleStacker OP 9 Dec

Well you are right to approach it game theoretically, but the answer should depend on N actually...

0 sats \ 2 replies \ @satcat 9 Dec

Ok thanks, for the hint.

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.

0 sats \ 1 reply \ @SimpleStacker OP 9 Dec

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.

0 sats \ 0 replies \ @satcat 9 Dec

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