pull down to refresh
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.
reply
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
reply
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.