The collection

Every essay — page 5

Page 5 of 5, continuing through the fields in the same order.

Geometry Analysis Algebra Discrete Topology Probability Number Dynamics Logic Computation What's new Ladders Concepts Search

Computation

A fixed set of operations, and the exact question of what it can and cannot build.

0000010100111001011101112 codewords in the 3-cube, minimum distance 3 — 1 error corrected, 2 detectedthe 2 balls of radius 1 hold 4 words each and cover all 8 exactly once: the code is perfect

Distance is a picture

A message is a corner of a cube and an error is a step along an edge. Everything a code can do is decided by how far apart the corners it uses are — and that is a fact about a drawing.

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four data bits, three parity bitsdataparityweight00000000000101130010110300111014010011140101100301100013011101041000101310011104101001141011000311000103110100141110100411111117sixteen codewords of length seven, minimum distance 3 — measured over all 120pairsweights: 1×0 · 7×3 · 7×4 · 1×7, and the lightest non-zero one weighs 3

Sixteen spheres that fill a cube

A hundred and twenty-eight seven-bit words, sixteen of them chosen, and a ball of eight around each. Sixteen times eight is a hundred and twenty-eight exactly — so the balls tile the space with nothing left over, and the code wastes nothing at all.

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received1011010H111010001110101101001010synsyndromethe bit it names101bit 1111bit 2110bit 3011bit 4100bit 5010bit 6001bit 7000no error1011010 gives syndrome 010, which is column 6 — so bit 6 is wrong and 1011000 is thecodewordthe syndrome is three bits and the message is four: the check finds the error without everrecovering what was sent

Finding the error without reading the message

Three parity checks on a seven-bit word produce three bits. If they are all zero nothing is wrong; otherwise they are the number of the position that broke. The message is never consulted, because the answer does not depend on it.

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012345678910012345678910sentf(0) = 4f(1) = 8f(2) = 7f(3) = 1f(4) = 1f(5) = 7f(6) = 8the message 4, 1, 3 read as 3x² + 1x + 4 over GF(11), evaluated at x = 0…6all 35 ways of keeping 3 of the 7 values were interpolated and every one returned the message

A polynomial through the gaps

Write the message as the coefficients of a polynomial and send its values instead. Any k of them determine the polynomial, so it does not matter which ones are lost — and it does not matter how many, as long as k survive.

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+01αα+101αα+101αα+110α+1ααα+101α+1α10×01αα+101αα+1000001αα+10αα+110α+11α× mod 40000012302020321the 4 elements of GF(4), built as polynomials in α modulo α+1 + α² — every product of two non-zero elements is non-zerobeside it the integers mod 4, where 2 × 2 = 0 and 1 element has no reciprocal

The field with four elements

The integers modulo four are not a field: two times two is zero and two has no reciprocal. There is nevertheless a field with four elements, and building it means giving up on counting as the way to make arithmetic finite.

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1αα²α³α+1α²+αα³+α²α³+α+1α²+1α³+αα²+α+1α³+α²+αα³+α²+α+1α³+α²+1α³+1GF(16)*15 elementspowerelementα⁰1α¹αα²α²α³α³α⁴α+1α⁵α²+αα⁶α³+α²α⁷α³+α+1α⁸α²+1α⁹α³+αα¹⁰α²+α+1α¹¹α³+α²+αα¹²α³+α²+α+1α¹³α³+α²+1α¹⁴α³+1every non-zero element of GF(16) is a power of α — the 15 of them, once each, around the wheel8 of the 15 elements generate the whole group, which is φ(15) = 8, and the product rule waschecked on all 225 pairs

Every element is a power of one of them

Pick the right element of a finite field and its powers run through every other non-zero element exactly once before returning to one. Multiplication becomes addition of exponents, and a table of q − 1 entries replaces the whole multiplication table.

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001010100110101011111point on line?0123456L0L1L2L3L4L5L6seven points, seven lines, three points on every line and three lines through every pointthe drawing was checked against the algebra by searching all 5,040 relabellings — one of them carries GF(2)³ onto thispicture

Seven points, seven lines

A geometry with seven points, in which every two points lie on exactly one line and every two lines meet in exactly one point. There are no parallels, the whole thing is built out of the two-element field, and one of its lines has to be drawn as a circle.

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01234567812 triples0 1 20 3 40 5 60 7 81 3 51 4 71 6 82 3 82 4 62 5 73 6 74 5 89 points, 12 triples, each point in 4 of them — and every one of the 36 pairs appears exactly oncefound by backtracking over the pairs, which decides existence rather than assuming it

A schedule where every pair meets once

Sort n people into groups of three so that every two of them share a group exactly once. Two divisions have to come out whole, that rules out most sizes — and at every size the divisions permit, a schedule exists.

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order 6, symbols 0…5012345123450234501345012450123501234transversals of the cyclic squareorder 33order 40order 515order 60order 7133the cyclic square of order 6 has no transversal at all: all 720 placements were tried and every one repeats asymbola square with no transversal cannot have an orthogonal mate, so the search for one need never begin

The thirty-six officers

Six regiments send six officers each, one of every rank. Arrange all thirty-six in a square so that each row and each column holds every rank once and every regiment once. Euler could not, guessed why, and was wrong about the reason.

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