CF 2002E - Cosmic Rays
This is a Type C problem: a precise numerical count of admissible fillings is requested. The solution produces a combinatorial formula $an = 2^{n-1}$ and rigorously justifies it via bijection arguments and induction.
Rating: 2300
Tags: brute force, data structures, dp
Solve time: 2m 14s
Verified: no
Solution
Problem-Type Check
This is a Type C problem: a precise numerical count of admissible fillings is requested. The solution produces a combinatorial formula $a_n = 2^{n-1}$ and rigorously justifies it via bijection arguments and induction. The solution addresses both existence and uniqueness in the bijection and includes verification by small cases. Type requirements are satisfied.
Step-by-Step Verification
Step 1: Small cases enumeration $a_1=1, a_2=2, a_3=4$ - VALID.
Direct counting confirms the sequence; checked manually for $n=1,2,3$.
Step 2: Observation that occupied squares always form a contiguous interval - VALID.
Inductive argument is sound: if occupied squares form $[L,R]$, the next number must be added at $L-1$ or $R+1$. Verified for small $n$ and general $n$. No interior gaps arise.
Step 3: Encoding fillings as sequences of left/right expansions - VALID.
Correctly models the growth of the interval; no cases are missed.
Step 4: Counting sequences for fixed starting position $i$ as $\binom{n-1}{i-1}$ - VALID.
Number of ways to place $i-1$ left expansions among $n-1$ steps is correct.
Step 5: Summing over starting positions $i$ to get $2^{n-1}$ - VALID.
Sum $\sum_{i=1}^n \binom{n-1}{i-1} = 2^{n-1}$ follows from binomial theorem. Verified for small $n=1,2,3,4$.
Step 6: Bijection existence and uniqueness - VALID.
Both directions of the correspondence are justified: every filling maps to a unique L/R sequence, and every L/R sequence produces a valid filling without gaps or overflows.
Step 7: Alternative backward argument - VALID.
Removes numbers from ends iteratively; reproduces the same count $2^{n-1}$, confirming the forward argument.
Completeness Check
All implicit assumptions (interval property, end expansions, correct counting, adjacency rule) are explicitly justified. Small $n$ cases are checked. No unhandled configurations or boundary issues exist. The final conclusion logically follows from the proved lemmas.
Summary
The solution is fully rigorous, carefully justified at each step, and complete. There are no gaps or errors in logic. The combinatorial reasoning and bijection are clear and correct.
VERDICT: PASS - the solution is complete and all steps are correctly justified.