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Counting the same thing twice — page 6

One collection, counted by two different methods, and an identity that falls out because both answers have to agree. The proof is the pair of counts.

The two squares of 97, produced by division. A table of the division chain on 97 and a square root of minus one modulo it, with each row's quotient and remainder, the point at which the remainder falls below the square root marked, and the two squares that add to 97.

Number

The two squares actually produced

Three proofs say a prime one more than a multiple of four is a sum of two squares, and not one of them hands over the squares. Running the Euclidean algorithm half-way does — and where to stop is the whole of the correctness argument.

Waiting for all 6 when they are not equally likely. One bar per kind giving the expected wait for that kind on its own, with the rarest much the tallest, and the expected wait for the whole collection printed above them.

Probability

The one that hardly ever comes up

Make the kinds unequally likely and the tidy decomposition into stages fails, because a stage's rate now depends on which kinds turned up rather than on how many. What replaces it is an alternating sum over every subset — and the rarest kind turns out to be nearly the whole answer.

Six lists of cycle shapes, and how many coverings each has. A table of lists of cycle shapes over a sphere, each with the Euler characteristic the Riemann–Hurwitz count gives, the number of lists of permutations with that product, and the number of those that connect all the sheets.

Topology

A count that can say zero

The branched count ends on a list of cycle shapes that passes every test and describes no covering. There is an exact formula for how many coverings a list has — a sum over the character table of a symmetric group — and it returns nought without giving any reason why.

A circle trapped between two 12-sided polygons. A circle with a regular polygon of 12 sides inscribed in it and another circumscribed about it, beside a table of the bounds on pi obtained by doubling the side count.

Geometry

Pinned between two sequences

The ring dissection makes the answer obvious and proves nothing. Archimedes' method proves it and makes nothing obvious — it never exhibits the area at all, it rules out every other value — and the recursion that drives it computes π by hand with one square root a step.

Geometry

The slice that has to match

The same slicing one dimension up gives the sphere's volume in a line, once one comparison is noticed: at every height a hemisphere's disc has exactly the area of a cylinder's slice with a cone's taken out of it. The principle that licenses that comparison also returns a false answer the moment the slices are not parallel.

2 independent rows and 2 independent columns. An array of 3 rows and 4 columns beside its transpose, with the independent rows of each shaded, showing the same count on both.

Algebra

Counted across and counted down

A rectangular array has a number of independent rows and a number of independent columns. The two are counted in different spaces, from different objects, by computations that share nothing — and they are always the same number, which is why 'rank' is one word.

2 independent cycles and 3 independent cuts, on 5 edges. A small graph beside its incidence matrix, with the matrix's rank and nullity given and shown to be the number of independent cuts and the number of independent cycles.

Algebra

The cycles and the cuts

The count that splits a map's source into what dies and what survives has nothing to do with graphs. Apply it to a matrix built from a graph's edges and points and it says that a graph's independent cycles and its independent cuts add to its number of edges — a theorem about drawings, obtained from an array.

A design on 7 points cannot have fewer than 7 blocks. The incidence matrix of a design on 7 points and 7 blocks beside the product of it with its own transpose, which has a constant off the diagonal and a determinant computed exactly.

Computation

More blocks than points

A schedule in which every pair meets once cannot use fewer groups than it has people. Nothing about the counting conditions says so, and the proof is not combinatorial at all — it is a determinant, computed over a field the schedules have nothing to do with.

A plane of 13 points from a list of 4 numbers. A ring of 13 points with one block of 4 of them drawn as a closed path, beside the table of the 13 blocks its shifts produce.

Computation

A plane in a list of numbers

A projective plane of order three has thirteen points and thirteen lines and fifty-two incidences. All of it is in the four numbers 0, 1, 3, 9 — because their pairwise differences hit every non-zero residue modulo thirteen exactly once, and the plane is that list's thirteen shifts.

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