In a parliament, each member has at most three enemies.Prove that the house can be separated into two houses, such that each member has at most one enemy in his house.
Can some one give a proof of this question, the book gives one, but I can't understand it...
#A problem from Arthur Engelman's problem solving strategy
faint adder
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In a parliament, each member has **at most three enemies**.Prove that the house ...
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OK so I will give you a hint imagine 2 boxes and u take a member and choose to put him in one of the box now take another member and put him in a box as you keep doing this notice that no matter what there is always one box you can put such that only one enemy is there
tribal wyvern
In a parliament, each member has **at most three enemies**.Prove that the house ...
It's a relatively trivial proof by contradiction. Suppose, towards a contradiction, that for some member X this is not possible. What this would mean is that no matter which house you put X into, X has two or more enemies already in that house. But this would require X to have four enemies.
sand isle
OK so I will give you a hint imagine 2 boxes and u take a member and choose to p...
@faint adder since Ido forgot to ping you when they responded
sand isle
It's a relatively trivial proof by contradiction. Suppose, towards a contradicti...
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tribal wyvern
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They asked for a proof, and said that the book gives one, but they don't understand it.
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@faint adder
faint adder
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