MathLabs

Problem 3

A social network has 20192019 users, some pairs of whom are friends (friendship is a symmetric relation). Events of the following kind may happen repeatedly, one at a time: three users AA, BB, and CC such that AA is friends with both BB and CC, but BB and CC are not friends, change their friendship statuses such that BB and CC are now friends, but AA is no longer friends with BB, and no longer friends with CC. All other friendship statuses remain unchanged. Initially, 10101010 users have 10091009 friends each, and 10091009 users have 10101010 friends each. Prove that there exists a sequence of such events after which each user is friends with at most one other user.
Step 6 of 6: Terminate at a matching
∣Et+1∣=∣Et∣−1,no legal toggle⟺ deg⁡(v)≤1for every v|E_{t+1}|=|E_t|-1,\qquad \text{no legal toggle}\Longleftrightarrow\ \deg(v)\le 1\quad\text{for every }v
Detailed analysis

Continue making any legal toggle in the current forest. Each one reduces the edge count by 11, so the finite process must terminate. At termination, if some vertex AA had two neighbors B,CB,C, acyclicity would imply BCBC is absent, giving another legal toggle; hence every vertex has degree at most 11. Thus the final friendship graph is a matching together with isolated vertices, exactly as required.