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Decided by exhaustion — page 7

Questions with finitely many cases, settled by going through all of them — and what changes when a claim about every argument becomes a count.
The postman's route: 24 blocks of street, walked in 28. A street network with its odd-degree vertices marked and the streets a shortest closed route must walk twice drawn doubled, dashed in a second colour, pairing up the odd vertices. Discrete

The streets a postman walks twice

A postman must walk every street of a district and come back. If every corner has an even number of streets, no street needs walking twice. If not, some must — and the ones repeated always join the odd corners in pairs. Pricing every way of pairing them finds the shortest round; pairing the nearest corners first does not.

 Six sentences from two quantifiers, and which imply which. A diagram of the six sentences that can be built from two quantifiers and a relation, arranged from strongest to weakest with arrows for implication, each labelled with how many of the 512 relations on three points satisfy it. Logic

Six sentences from two quantifiers

One relation, two variables, 'for every' and 'there is': there are eight ways to arrange them and six different sentences come out. Which of them imply which is a small, complete diagram, found by checking all 512 relations on three points — and the diagram crosses over in the middle, which is where every confusion about the order of quantifiers lives.

 The pairs from five, joined when disjoint, need three colours. The Petersen graph drawn with its ten vertices labelled by pairs from one to five, edges joining disjoint pairs, and a proper colouring with three colours. Topology

The colours a circle forces

Take every pair from five things and join two pairs when they share nothing. Three colours are enough to colour the result so joined pairs differ, and two are not — but no triangle, no dense cluster and no counting argument explains why. The reason is five points on a circle and a direction that cannot be told apart from its opposite, and the same reason, one sphere at a time, settles Kneser's question for every size.

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