LCB-04: Roulettes
The Task
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The Code
Claude Opus 4.6 with max-effort extended thinking. Exact outputs. Side by side.
Baseline
FAIL866ch / 188simport sys
def solve():
data = sys.stdin.read().split()
idx = 0
N, M = int(data[idx]), int(data[idx+1])
idx += 2
wheels = []
for i in range(N):
C = int(data[idx])
P = int(data[idx+1])
S = [int(data[idx+2+j]) for j in range(P)]
idx += 2 + P
wheels.append((C, P, S))
dp = [0.0] * (M + 1)
for j in range(M - 1, -1, -1):
best = float('inf')
for C, P, S in wheels:
zero_count = 0
non_zero_sum = 0.0
for s in S:
if s == 0:
zero_count += 1
else:
non_zero_sum += dp[min(j + s, M)]
val = (C + non_zero_sum / P) / (1.0 - zero_count / P)
if val < best:
best = val
dp[j] = best
print(f"{dp[0]:.15f}")
solve()Scroll to read full code
Augmented
PASS830ch / 317simport sys
def main():
data = sys.stdin.read().split()
idx = 0
N = int(data[idx]); idx += 1
M = int(data[idx]); idx += 1
wheels = []
for _ in range(N):
C = int(data[idx]); idx += 1
P = int(data[idx]); idx += 1
S = []
for _ in range(P):
S.append(int(data[idx])); idx += 1
wheels.append((C, P, S))
E = [0.0] * (M + 1)
for s in range(M - 1, -1, -1):
best = float('inf')
for C, P, S in wheels:
z = 0
sf = 0.0
for v in S:
if v == 0:
z += 1
elif s + v < M:
sf += E[s + v]
cost = (C * P + sf) / (P - z)
if cost < best:
best = cost
E[s] = best
print(f"{E[0]:.15f}")
main()Scroll to read full code
Blind Evaluation
Independent evaluator. Randomized A/B labels. No knowledge of which solution used the injection.
Augmented 41, Baseline 38
| Dimension | Baseline | Augmented |
|---|---|---|
| Correctness | 10 | 10 |
| Efficiency | 8 | 8 |
| Code Structure | 7 | 8 |
| Readability | 5 | 7 |
| Robustness | 8 | 8 |
Blind Analysis
Both solutions implement the same dynamic programming approach. Define E[s] as the minimum expected cost to reach M points when currently at s points. Base case: E[M] = 0. For each state s < M, iterate over all wheels and pick the one minimizing expected cost. The self-referential term from zero-valued outcomes (which leave the state unchanged) is resolved algebraically. The two formulations are mathematically identical.
Blind Observation
The two solutions are nearly isomorphic — same algorithm, same loop structure, same complexity. The differences are confined to two narrow choices: how to express the formula and how to name variables. The baseline's variable naming (j for state, s for outcome) suggests the author wrote the DP loop mechanically without mapping variable names back to the problem domain. The augmented solution shows one additional pass of refinement.
Source: LiveCodeBench Hard benchmark. Full code outputs: baseline · augmented · runner script & methodology