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Berkeley Math Circle: Monthly Contest 6

Combinatorics Difficulty 5.3 AIME, harder Prove it United States

Problem:

We are given a family F\mathcal{F} of functions from the set {1,…,n}\{1, \ldots, n\} to itself. A sequence (f1,…,fk)\left(f_{1}, \ldots, f_{k}\right) of functions in F\mathcal{F} is said to be good⁡\operatorname{good} if fk∘fk−1∘⋯∘f1f_{k} \circ f_{k-1} \circ \cdots \circ f_{1} is a constant function. Prove that if there exists a good sequence, there exists one with k≤n3k \leq n^{3}.

Solution

Solution:

Suppose there exists a good sequence.
Then for any two a,b∈{1,…,n}a, b \in \{1, \ldots, n\} there is some sequence of functions g1,…,gmg_{1}, \ldots, g_{m} such that gm∘gm−1∘⋯∘g1g_{m} \circ g_{m-1} \circ \cdots \circ g_{1} maps aa and bb to the same point. Looking at the images (a,b),(g1(a),g1(b)),(g2(g1(a)),g2(g1(b))),…(a, b), \left(g_{1}(a), g_{1}(b)\right), \left(g_{2}\left(g_{1}(a)\right), g_{2}\left(g_{1}(b)\right)\right), \ldots and so on, by the pigeonhole principle we can find a subsequence with m≤n2m \leq n^{2}.
By appending nn such sequences together, we can collapse all nn inputs to the same output.

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