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Exercise 1245Combinatorics › Graph theoryOlympiadAdded 4 Nov 2026

Splitting a parliament into two houses

#monovariant#graphs#extremal#process

Statement

In a parliament, each member has at most 33 enemies, and enmity is mutual. Prove that the members can be split into two houses so that every member has at most 11 enemy in his own house.

Hint

Start with an arbitrary split. If somebody has at least 22 enemies in his own house, move him to the other house. Consider the number of pairs of enemies who sit in the same house.

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Solution

Idea. Use a monovariant: a nonnegative integer that strictly decreases at each step.

Process. Start with any split of the members into two houses. Let EE be the number of pairs of enemies belonging to the same house. If every member has at most one enemy in his own house, we are done. Otherwise, take a member xx with at least 22 enemies in his own house.

Effect of the move. Since xx has at most 33 enemies in total, he has at most 11 enemy in the other house. Move xx to the other house.
- The pairs {x,y}\{x,y\} with yy an enemy in his former house are no longer in the same house: at least 22 pairs are removed from the count.
- The pairs {x,y}\{x,y\} with yy an enemy in the other house now count: at most 11 new pair.
- All other pairs are unaffected.

Hence EE decreases by at least 11.

Termination. EE is a nonnegative integer, so the process cannot continue forever. When it stops, nobody has 22 or more enemies in his own house.

Lesson. Monovariant: to prove a configuration with some property exists, start anywhere and improve a nonnegative integer quantity until it cannot be improved.