Lower bound
Named by 23 essays across 8 fields — each of them below, with the objects they name alongside it.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
What a formula keeps when letters are fixed
Fix half the letters of a formula at random and most of its leaves become irrelevant: on the sixteen-leaf formula for the parity of four letters, two fixings leave four. How fast formulas shrink under random fixings is the whole engine behind the best lower bounds anyone has proved for formula size — Subbotovskaya's 3/2 in 1961, Håstad's 2 in 1998 — and the reason that engine stops at n³.
How far apart two triangulations can be
Any triangulation of a polygon can be turned into any other by flipping one diagonal at a time. On an octagon seven flips always suffice; on a polygon with m vertices the worst case is 2m − 10 flips once m reaches thirteen, a value Sleator, Tarjan and Thurston proved with hyperbolic geometry in 1988. Breadth-first search over all 58,786 triangulations of the 13-gon finds the sixteen.
Resolving one letter away at a time
Davis and Putnam's procedure of 1960 decides a set of clauses by removing its letters one at a time, replacing every clause that mentions a letter by all the ways of resolving it away. On a cycle the clauses never multiply; on a grid they swell by a factor of three and a half for each extra row, and on the 4 × 4 grid each extra unit of the order's width doubles the peak exactly. The cost is not the size of the formula but how wide it is.
The order that keeps elimination narrow
Eliminating letters one at a time costs whatever the widest step costs, and the least width any order can achieve is the treewidth of the formula's graph. Finding it exactly is NP-hard, but a search over sets of letters instead of sequences finds it for small graphs, and a greedy rule that adds the fewest new edges finds it on 566 of 600 random graphs and misses by more than one only once. On Tseitin's 6 × 6 grid that rule forms twenty-four times fewer resolvents than sweeping row by row. What it cannot do is prove that its order is the best.
The shortest string holding every ordering
Thirty-three symbols hold all twenty-four orderings of 1, 2, 3, 4 as blocks of four in a row, and no shorter string does. The formula 1! + 2! + … + n! gives the shortest length for up to five symbols and was believed for all of them — until a string of 872 for six beat its 873, and a lower bound posted anonymously on a forum left everything between 867 and 872 open.
The fewest points that meet every line
In a finite plane any line meets every other line, so a single line is a set of points that every line passes through. Ask for such a set that contains no whole line and the answer is surprisingly expensive: in the plane of order 7, which has 57 points, it takes 12, and an exhaustive search over seventy-one million branches shows that 11 will not do. Two theorems predict the number — a square-root bound met by a copy of a smaller plane, and a bound of three halves for prime orders met by three half-lines.
The fewest entries that fix a Latin square
Reveal six of the twenty-five entries of the cyclic Latin square of order five — three in the top left corner, three in the bottom right — and only one Latin square agrees with them; reveal any five and at least two do. John Nelder conjectured that a square of order n can never be pinned down by fewer than ⌊n²/4⌋ entries. A search over every set of cells confirms it for every square to order five, and finds that the more symmetric a square is, the fewer entries it takes.
The busiest point in a set
Scatter points in the plane, not all on one line, and draw every line through two of them. Some point lies on many of those lines — but how many must? Gabriel Dirac guessed half: in any such set, some point is on about half of the lines it could be on. Split the points evenly between two lines and no point does better than half plus one, so the guess cannot be raised. Proving it is another matter. Seventy years on, the best theorem says a third.
Running a computation backwards with little memory
The reverse sweep that gives a whole gradient for the price of a few evaluations needs the computation's intermediate states in reverse order, and storing them all can cost more memory than the machine has. Keep only a few and recompute the rest, and the cheapest schedule has an exact answer found by Andreas Griewank in 1992: with c stored states and no step run more than r times, the longest chain that can be reversed is the binomial coefficient C(c + r, c). Pascal's triangle is the table of what memory buys.
Dudeney's four pieces are the fewest
In 1902 Henry Dudeney cut an equilateral triangle into four pieces that rearrange into a square, and hinged them so the triangle folds into the square in one motion. For a hundred and twenty-two years nobody could say whether three pieces would do. In December 2024 Erik Demaine, Tonan Kamata and Ryuhei Uehara proved they would not. The construction, built exactly here, shows why four is natural; the proof shows why it is necessary.
Three arrows in and three out
Put an arrow on every edge of the cubic grid so that every point has three arrows in and three out. On the square grid the same question — two in, two out — has an exact answer, found by Elliott Lieb in 1967: about 1.5396 arrangements per point. On the cubic grid nobody knows the answer. Pauling's guess of 2.5 is a proven lower bound, and exact counts on prisms of the grid come down towards it from above — 2.794, 2.682, 2.587 as the cross-section widens — without reaching it.
How many pieces to shrink a shape
Cut a disc in two and one half still holds a full diameter; cut it in three and every piece is narrower. In 1933 Karol Borsuk asked whether every bounded shape in d dimensions can be made narrower with d + 1 pieces, and the antipodal theorem he proved in the same paper says no number smaller will do for a ball. The guess held in the plane and in space, and failed in 1993: sixty years later, a set of sign patterns turned into points showed that some shapes need exponentially many pieces — the first in 1,325 dimensions, and now known in 64.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchGraphParityConstructionRecursionAntipodal pairChromatic numberCircleCircuit complexityCompassCompass and straightedgeComplexity