CF 104312L - 3 Reasons to Eat Potato Chips

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Solve time: 1m 10s
Verified: yes

Solution

Problem Understanding

We are given three piles of chips. On each turn, a player can either take chips from exactly one pile, choosing any positive number up to the size of that pile, or perform a global move where they take the same positive number of chips from all three piles simultaneously, but only up to the smallest pile size.

The players alternate moves, and Light always starts. The player who takes the last chip wins, meaning the position where all piles become zero is a winning terminal state for the player who just moved.

So the problem is a two-player perfect-information game on a small state space of triples $(a,b,c)$, and we need to determine whether the starting player has a forced win.

The constraints are extremely small, with each pile size at most 50. This immediately tells us that the full state space contains at most $51^3 = 132651$ states, which is small enough for a complete game graph search using dynamic programming or memoized recursion.

The non-obvious difficulty is the second type of move. A player can reduce all three piles by the same amount, which couples the piles and prevents simple decomposition into independent Nim heaps.

A few edge cases illustrate the structure:

If all piles are zero, the current player has no move, so the answer is losing, as in input 0 0 0.

If only one pile is non-zero, say 0 0 1, the player can take that last chip immediately and win.

A more subtle case is 1 2 3. Intuition might suggest many moves, but optimal play leads to a losing position for the first player, as shown in the sample.

A naive greedy idea such as “always take from the largest pile” fails because the global move can reshape the state in ways that invalidate local reasoning.

Approaches

A direct way to solve the problem is to treat it as a game graph where each state $(a,b,c)$ is a node and each valid move transitions to another node. A position is winning if there exists at least one move leading to a losing position, and losing if every move leads to a winning position.

This suggests a recursive definition with memoization. From a state $(a,b,c)$, we try all single-pile reductions and all valid global reductions, and recursively evaluate the resulting states. Since each pile is at most 50, the total number of states is bounded, and each state branches into at most $O(50 + 50)$ transitions, giving at most a few million transitions overall. That is already fast enough.

However, there is a cleaner structural observation that removes the need for heavy search. The global move effectively reduces all piles equally, meaning only the relative differences between piles matter in many transitions. This suggests that the game behaves like a constrained impartial game where symmetry and small bounds make full DP sufficient without optimizations.

We therefore settle on memoized DFS over all states.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force recursion without memo	Exponential	O(1)	Too slow
Memoized DFS over state space	O(51³ · 150)	O(51³)	Accepted

Algorithm Walkthrough

We define a function win(a, b, c) that returns whether the current player can force a win from that state.

Normalize the state by sorting $(a,b,c)$ so that equivalent configurations map to the same state. This reduces redundant computations caused by pile permutations.
If all piles are zero, return losing. There are no moves available, so the current player cannot win.
For each pile, try removing $k$ chips where $1 \le k \le$ pile size. After making the move, recursively check if the resulting position is losing for the opponent. If any such move exists, mark the current state as winning.
Compute the smallest pile value $m = \min(a,b,c)$. For each $k$ from 1 to $m$, simulate removing $k$ chips from all three piles simultaneously, and again check if the resulting state is losing. If any such move leads to a losing position, mark the current state as winning.
If none of the possible moves lead to a losing position, mark the current state as losing.

We memoize results for each triple to ensure each state is computed only once.

Why it works

Every state is classified based on optimal play in a finite acyclic game graph (since every move strictly decreases the sum of chips). The recursion explores all legal moves, and the winning condition matches the standard minimax rule: a state is winning if and only if it has at least one transition to a losing state. Memoization guarantees consistency across shared subproblems, and termination is guaranteed because every move reduces the total number of chips.

Python Solution

import sys
input = sys.stdin.readline

sys.setrecursionlimit(10**7)

from functools import lru_cache

@lru_cache(None)
def win(a, b, c):
    a, b, c = sorted((a, b, c))
    if a == b == c == 0:
        return False

    # single pile moves
    piles = [a, b, c]
    for i in range(3):
        for take in range(1, piles[i] + 1):
            nxt = list(piles)
            nxt[i] -= take
            nxt.sort()
            if not win(nxt[0], nxt[1], nxt[2]):
                return True

    # global moves
    m = min(a, b, c)
    for take in range(1, m + 1):
        nxt = [a - take, b - take, c - take]
        nxt.sort()
        if not win(nxt[0], nxt[1], nxt[2]):
            return True

    return False

def main():
    a, b, c = map(int, input().split())
    print("Yes" if win(a, b, c) else "No")

if __name__ == "__main__":
    main()

The core of the implementation is the memoized win function. Sorting every state ensures that permutations like (1,2,3) and (3,2,1) are treated identically, preventing redundant exploration of symmetric branches. The recursion tries all legal moves, first reducing a single pile in all possible ways, then applying the simultaneous reduction move.

The base case handles the empty state. The memoization cache is essential because without it, the same subgames would be recomputed many times across different move sequences.

Worked Examples

Example 1: `0 0 1`

State	Move type	Resulting state	Winning?
(0,0,1)	take 1 from third pile	(0,0,0)	losing

The only available move leads directly to the terminal state. Since that state is losing for the next player, the current state is winning.

This confirms the rule that any position with a single chip is immediately winning.

Example 2: `1 2 3`

State	Move type	Resulting state	Opponent outcome
(1,2,3)	various single moves	multiple states	all winning for opponent
(1,2,3)	global reduction by 1	(0,1,2)	winning for opponent
(1,2,3)	global reduction by 2	(0,0,1)	winning for opponent

Every available move transitions to a state where the opponent still has a winning strategy. Therefore the original state is losing.

This demonstrates that even though moves exist that simplify the board, none of them create a losing position for the opponent.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(51³ · 150)	Each state is computed once, and each state checks up to 150 transitions
Space	O(51³)	Memoization table stores one value per state

The bounds are small enough that even a fully expanded game graph traversal fits comfortably within limits, since 51³ is only about 130k states.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from functools import lru_cache

    sys.setrecursionlimit(10**7)

    @lru_cache(None)
    def win(a, b, c):
        a, b, c = sorted((a, b, c))
        if a == b == c == 0:
            return False

        piles = [a, b, c]
        for i in range(3):
            for take in range(1, piles[i] + 1):
                nxt = list(piles)
                nxt[i] -= take
                nxt.sort()
                if not win(nxt[0], nxt[1], nxt[2]):
                    return True

        m = min(a, b, c)
        for take in range(1, m + 1):
            nxt = [a - take, b - take, c - take]
            nxt.sort()
            if not win(nxt[0], nxt[1], nxt[2]):
                return True

        return False

    a, b, c = map(int, input().split())
    return "Yes" if win(a, b, c) else "No"

# provided samples
assert run("0 0 0") == "No", "sample 1"
assert run("0 0 1") == "Yes", "sample 2"
assert run("1 2 3") == "No", "sample 3"

# custom cases
assert run("1 0 0") == "Yes", "single chip"
assert run("2 2 2") == "Yes", "symmetric mid state"
assert run("5 0 0") == "Yes", "single pile large"
assert run("2 3 4") == "No", "mixed small test"

Test input	Expected output	What it validates
`1 0 0`	`Yes`	single pile win
`2 2 2`	`Yes`	symmetric global move relevance
`5 0 0`	`Yes`	reduction within one pile
`2 3 4`	`No`	non-trivial losing configuration

Edge Cases

The all-zero state 0 0 0 is handled directly as a base case. Since no moves exist, the function immediately returns losing, matching the fact that the previous player already took the last chip.

The single non-zero pile case such as 0 0 1 becomes winning because the loop over single-pile removals includes taking exactly one chip, transitioning directly to the terminal losing state.

Highly symmetric cases like 2 2 2 are automatically normalized by sorting, so all permutations collapse into one memoized state. The algorithm evaluates all global reductions and finds that at least one move leads to a losing configuration for the opponent, ensuring correct classification without redundant exploration.