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Decided by exhaustion — page 2

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 subgrouper²r · the subgrouprr³m₁ · the subgroupm₁m₃m₂ · the subgroupm₂m₄2 elements in the subgroup, 4 blocks, 2 × 4 = 8 in the group — which is Lagrange'stheorem, and it is arithmetic about the picture rather than a theorem quoted at iteach block is the subgroup with one element composed onto the front of it, so it is thesame size; two blocks that share an element are the same block, which is why theycannot overlap Algebra

The blocks a subgroup cuts out

Take any part of a group that is closed under composition, and it slices the whole group into blocks of its own size that do not overlap. Everything Lagrange's theorem says is arithmetic about that picture — and whether the blocks can be multiplied is a separate question with a surprising answer.

 -6-4-202460.000.100.200.300.400.50valueprobability1.5σ2σ3σthe meanmean 0.300, standard deviation 1.847; beyond 1.5 of them lies 0.2000 against a bound of 0.4444, beyond 2 ofthem lies 0.1000 against a bound of 0.2500, beyond 3 of them lies 0.0000 against a bound of 0.1111the bound knows only the variance — not the shape, not the number of values, not whether the distribution issymmetric — which is why it is so far from tight here and cannot be improved in general Probability

How far from the average a thing can be

Knowing only an average and a spread — nothing about the shape, nothing about the number of outcomes, nothing about symmetry — the chance of landing three standard deviations out is at most one in nine. And there is a distribution that lands there exactly that often, so the bound cannot be improved.

 01234560.000.100.200.300.400.50how many are looked at and passed overchance of taking the best14.3%35.0%41.4%40.7%35.2%26.2%14.3%every one of the 5,040 orders, for every threshold: looking at 2 of 7 and then taking the first that beats them wins2,088 times, which is 41.4%taking the first one wins 14.3% and so does taking the last, because both amount to choosing without looking —the rule beats them by a factor of 2.90 Probability

When to stop looking

Candidates arrive one at a time in a random order. Each must be accepted or rejected on the spot, with no going back and no way to know what is still to come. The best possible rule is to look at about a third of them and then take the first one that beats everything seen — and it works about a third of the time, however many there are.

 K4no crossingsK51 crossing, and no drawing has noneK3,31 crossing, and no drawing has noneK4: 4 points, 6 edges, and at most 6 allowed; K5: 5 points, 10 edges, and at most 9 allowed; K3,3: 6 points, 9 edges,and at most 8 allowed6,000 layouts were tried for each; the best found is drawn, and the bound is what rules out anything better Discrete

Two graphs that will not lie flat

Five points, every pair joined: no matter how the points are placed or how the lines are drawn, two of the lines cross. The proof is not about drawing at all — it counts edges against faces and finds one edge too many.

 78°54°48°the outer triangle's angles are 78°, 54°, 48° and have nothing equal about them; the innertriangle's three sides are 0.152693, 0.152693, 0.152693the three corners are found by intersecting the trisectors, and the three lengths are thenmeasured off those corners — nothing in the construction asked for them to be equal Geometry

Three trisectors and a triangle nobody expected

Cut every angle of a triangle into three. The trisectors nearest each side meet in three points, and those three points are always the corners of an equilateral triangle — for every triangle there is, with no exceptions and no reason anybody finds obvious.

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