Concept

Lower bound

A proof that no method can do better than some amount, as opposed to a method that achieves it. Lower bounds are usually harder than upper bounds because they must defeat every possible strategy, including ones nobody has thought of.

Named by 11 essays across 7 fields — each of them below, with the objects they name alongside it.

Even and Paz's halving, for 4 people. The recursive halving procedure run on a cake valued differently by 4 people: each round's marks and cuts, and the final pieces, each worth at least a 1/4 share to its owner. 8 marks are made.

How many cuts a fair share costs

Every person can be guaranteed a share of a cake worth at least one n-th by their own measure, and the oldest rule that does it asks about n²/2 questions. Splitting the people into halves and the cake at a median mark asks about n log n — and a theorem says nothing can ask fewer. Fairness has a price, and it can be counted.

applied · Fair division
Breadth-first levels in a planar triangulation. A Delaunay triangulation of 160 random points with each point coloured by whether it is inside, on or outside the level 4 steps from the centre; that level of 23 points separates 31 from 106.

The few points that cut a flat graph

Any graph that can be drawn without crossings, however large, falls into pieces of at most two thirds once a few points are removed — about the square root of its size, never more than 2.83 times it. A grid shows the square root cannot be beaten, a ring of breadth-first neighbours comes close, and a cycle through a shallow tree finishes the job.

discrete · Planarity
Several colours on every vertex of a Kneser graph. A grid of Kneser graphs on pairs from five to eight points against the number of colours per vertex, each cell giving the fewest colours needed and the counting lower bound.

Several colours on every vertex

Give every pair from six points three colours, so that pairs with nothing in common share no colour. Counting says nine colours might do; ten are needed. Stahl conjectured in 1976 exactly how many colours every such problem needs — a formula that meets Lovász's topological answer at one colour a vertex and the obvious answer at k — and a search over stars and triangles confirms it in every case small enough to run.

topology · Borsuk ulam
Removing one end of an ordinary line from a triangle, its midpoints and its centroid. Two panels. Left: 7 points with all 9 connecting lines, one ordinary line solid and one of its ends ringed. Right: the same points with that end removed, 7 connecting lines left.

At least as many lines as points

Sylvester's theorem says some line through two of the points misses all the rest. Remove one end of that line and the line itself disappears, taking at least one line away with one point. Run that backwards and it proves that n points not all in a line determine at least n lines — and the only sets that manage exactly n are a line of n − 1 points with one point off it.

geometry · Ordinary lines
Every reachable arrangement of the 3×3 sliding puzzle, by moves from solved. A bar chart of the 181440 reachable arrangements of the 3×3 sliding puzzle by the fewest moves that solve them, rising to a peak at 24 and falling to 2 at the maximum distance 31.

Thirty-one moves from solved

Parity settles which half of a sliding puzzle's arrangements can be reached and is silent about how far away any of them is. Searching every reachable arrangement of the three-by-three tray answers the second question exactly — two arrangements sit thirty-one moves out — and parity turns up again, this time as a law about distance.

algebra · Permutation parity
The complete graph on 7 points, drawn straight with 9 crossings. A straight-line drawing of the complete graph on 7 vertices with 9 crossings, the minimum for straight drawings; each crossing comes from four vertices in convex position.

The crossings a graph cannot avoid

Five points all joined need one crossing, six need three, seven need nine. Euler's formula gives a lower bound that grows like the square of the number of points and is badly wrong; a sampling trick turns the same formula into a bound that grows like the fourth power and is right to within a constant. And for drawings with straight edges the count turns out to be something else entirely: the number of quadrilaterals the points make.

discrete · Planarity
The midpoint in the fewest moves: four with a ruler, six with the compass alone. Minimal constructions of the midpoint of AB: 4 moves with ruler and compass, 6 circles compass-only; 55 and 45611 configurations searched.

The fewest moves to draw it

Every construction with ruler and compass is a sequence of moves — draw this line, draw that circle — and it is natural to ask for the shortest. Nobody can answer that by cleverness alone, but a machine can answer it by trying everything: the midpoint of a segment takes four moves and no fewer, the square on it five, a third of it five. Take the ruler away and the compass pays for it: six circles for the midpoint, seven for the square.

computation · Compass-only
Smallest circuits for four functions of three letters. Four small circuit diagrams, each with three inputs feeding a few two-input gates, computing parity, a selector, majority and exactly-one-true with two, three, four and four gates.

How many gates a truth table needs

Every truth function can be built from gates, and the natural measure of a function is the fewest gates that build it. For three letters the whole answer can be computed: no function needs more than four. For n letters, a count shows that almost every function needs about 2ⁿ/n gates, and a construction shows that none needs more. And for any function anyone can actually write down, the best proof in fifty years says it needs 3.1n.

logic · Truth functions
Parity and majority on the cube, and the edges between true and false. Two cubes of three-letter assignments, parity with all twelve edges joining a true and a false corner and majority with 6; Khrapchenko's bound at its best 9.00 against an exact size of 10 and 4.00 against an exact size of 5.

The formula that cannot share

A circuit computes the parity of n letters with a chain of n − 1 exclusive-or gates. A formula in and, or and not, which may use each result only once, needs n² letters at least — a bound Khrapchenko proved by counting the cube's edges between true and false — and exhaustive search over every formula of four letters finds parity sitting exactly on it.

logic · Truth functions
The centre of a circle found with six circles and no straightedge. A circle with two marked points 37.3° apart and the six compass circles that construct its centre.

The centre a compass finds in six circles

A circle is drawn and its centre was never marked. A straightedge alone can never find it again. A compass alone can, from two points on the circle, in six circles — and a search through every construction of five circles or fewer shows that six is the least. The reason six works is an inversion that turns the circle into a straight line; the price of discarding the straightedge is exactly one move.

computation · Compass-only
The price of the centre, arc by arc. 10°: 5, 15°: 4, 20°: 4, 25°: 6, 30°: 3, 35°: 6, 40°: 6, 45°: 5, 50°: 5, 55°: 6, 60°: 2, 65°: 6, 70°: 5, 75°: 4, 80°: 6, 85°: 6, 90°: 5, 95°: 6, 100°: 4, 105°: 4, 110°: 5, 115°: 6, 120°: 6, 125°: 6, 130°: 5, 135°: 5, 140°: 4, 145°: 6, 150°: 3, 155°: 6, 160°: 6, 165°: 4, 170°: 5, 175°: 6.

The arcs that make the centre cheap

Finding a circle's centre with the compass alone, from two points on it, costs six circles when the points are placed with no special relation — and as few as two when the arc between them is 60°. Searching every arc in steps of five degrees maps the price exactly, and it is not a smooth function of the arc: it is a list of coincidences, each one a small number of compass steps that happens to land on a chord equal to the radius.

computation · Compass-only

Named alongside it

The objects these essays reach for when they reach for this one.

Exhaustive searchConstructionParityCircleCircuit complexityCompassCompass and straightedgeEuler formulaPlanar graphRecursionAntipodal pairBoolean circuit

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