Six people, and the trio that cannot be avoided
complete-graph is one function. Everything below came out of it during this
build, at parameters taken from the essays rather than invented for this page — so a figure
here is the same figure a reader meets in an essay, and if the generator changes, this page
changes with it.
With nothing chosen
All 16 trees on 4 labelled points
Four trees on seven points, gracefully labelled
All twenty-three trees on eight points, gracefully labelled
Every tree up to twelve points is graceful
How many graceful labellings trees have
What it checks while it draws
Collected by running the family and recording what it asserted, not written here. The count is how many separate times the claim was put to the test while these drawings were made.
- a tree with a cherry has a symmetry, so the shares are ordered at 4 ×13
- the branch sets of 0 and 1 are joined by an edge ×13
- asymmetric trees on 1 points ×12
- the labellings on 1 points add to Cayley's count ×12
- the number of trees on 2 points ×12
- every tree on 2 points is graceful ×11
- the 2+1-part family meets the formula at edge density 0.560 ×9
- the 2-point witness has no four-cycle ×8
- functions with 1 cyclic points and marked trees with 1 points on the path ×7
- on 3 points, avoiding a complete graph on 3 allows Turán's count ×7
- at the Turán density 1 − 1/2 the curve touches Goodman's bound ×6
- K5: a subdivision of K5 is also a minor ×6
- 2^2 functions, 2^0 trees, 2² pairs of marks ×5
- measured cyclic points agree with the sum at 10 ×5
- no two functions on 2 points give the same marked tree ×5
- Ramanujan's sum is √(πn/2) − 1/3 to within a fraction at 10 ×5
- the plane over 2 gives a graph with no four-cycle ×5
- there is some n at which the expected count of 4-sets is below one ×5
- and the codes are all 3 sequences of length 1 ×4
- every marked tree on 2 points reads back to its function ×4
- on 3 points no two trees share a code ×4
- the search reaches the known fewest crossings for 7 points ×4
- 10 to 120 points ×3
- K4: the counting bound and the best drawing found agree about whether it lies flat ×3
- K5, drawn flat as a solid's skeleton, has neither ×3
- K5: an edge joins two different points of the graph ×3
- K5: the minor test and the subdivision test agree about planarity ×3
- the best weighting scores ½(1 − 1/ω) with ω = 2 ×3
- and 6 is below the known R(4,4) of 18 ×2
- cube: a subdivision of K5 is also a minor ×2
- every vertex has degree 7 ×2
- K5 with a point split: a subdivision of K5 is also a minor ×2
- K5: the best drawing found has one crossing ×2
- octahedron: a subdivision of K5 is also a minor ×2
- Petersen graph: a subdivision of K5 is also a minor ×2
- the grid side is between 6 and 20 ×2
- the most triangle-free edges on 6 points is ⌊n²/4⌋ ×2
- the table runs to between 8 and 12 points ×2
- there are exactly n to the n minus two trees on 4 labelled points ×2
- Wagner graph: a subdivision of K5 is also a minor ×2
- — the midpoint of the two corners they hang from ×1
- 12 edges on 7 points can avoid every triangle ×1
- 175 edges: 50 × 7 / 2 ×1
- 2 to 4 cars ×1
- a caterpillar has a graceful labelling ×1
- a named graph ×1
- a path has a graceful labelling ×1
- a plane over the field with 2, 3 or 5 elements ×1
- a plane small enough to draw its pairs ×1
- a point's degree is one more than its count in the Prüfer sequence ×1
- a spider has a graceful labelling ×1
- a star has a graceful labelling ×1
- a tree is given as a list of pairs of points ×1
- a tree with more than one edge always has a leaf to strip ×1
- a vertex meets five others ×1
- after contracting the spokes, all ten pairs of the five merged points are joined ×1
- after deletion the random graph has no four-cycle ×1
- almost every large random tree has a cherry ×1
- an asymmetric tree has no cherry ×1
- and each exactly once ×1
- and every pair gets one colour either way round ×1
- and far above Euler's ×1
- and in fact at least two, which is Goodman's bound ×1
- and it contains a subdivision of K3,3, so it is not planar ×1
- and it has at most 2d + 1 points for a tree of depth d ×1
- and it is reached at a balanced split ×1
- and it is the edge count of the balanced two-part graph ×1
- and meets the Moore bound ×1
- and no move creates the forbidden clique ×1
- and no tree is listed twice ×1
- and none has fewer than the minimum curve allows ×1
- and on this many points it is the most any such graph has ×1
- and respects the cherry bound ×1
- and so are its point labels ×1
- and that is more than two thirds of the grid ×1
- and the bound for graphs with no triangle rules out K3,3 ×1
- and the layout kept is one a reader can follow ×1
- and the next size up is the first with an expectation of at least one ×1
- and the second has d(d − 1), every one reached exactly once ×1
- and the trees, built one by one, number the same ×1
- and the two colours take half the pairs each ×1
- at least one colouring is tried, and at most twenty thousand ×1
- at least one level is balanced ×1
- between 30 and 200 points ×1
- between three and seven nested triangles ×1
- branch sets are disjoint ×1
- cars is read only by the parklist view ×1
- counting every tree gives the binomial law for the degree ×1
- cube, drawn flat as a solid's skeleton, has neither ×1
- cube: an edge joins two different points of the graph ×1
- cube: the minor test and the subdivision test agree about planarity ×1
- curved drawings never need more crossings than straight ones ×1
- each clique size counted is between 3 and 8 ×1
- each clique size is a whole number between 3 and 8 ×1
- each is triangle-free and not bipartite ×1
- each pair of neighbours of a point is a pair with that common neighbour, and none is counted twice ×1
- each path runs along edges of the graph ×1
- each piece holds at most half of its parent ×1
- each triangle is a fixed fraction of the one outside it ×1
- enough well-spaced points were placed ×1
- EVERY colouring of six people contains a monochromatic triangle ×1
- every count of labellings is whole ×1
- every crossing of a straight drawing is the pair of diagonals of four points in convex position, and every such four gives one ×1
- every edge the extremal graph is missing would complete a triangle ×1
- every free row joins the free column's component, so that component holds at least k points per free row ×1
- every lower bound is below the known value ×1
- every move strictly adds edges ×1
- every pair has at most one common neighbour ×1
- every pair is coloured exactly once ×1
- every point at the average of its neighbours, with the outer face convex, gives a drawing with no crossings ×1
- every point gets an image ×1
- every point has three neighbours ×1
- every ratio lies inside the drawn range ×1
- every tree has at least one graceful labelling ×1
- every tree is drawn for between 3 and 5 points ×1
- every tree on eight points is graceful ×1
- every triangle-free graph is bipartite up to four vertices; at nine fewer than half are ×1
- every vertex is within two steps of every other, checked from another vertex too ×1
- every vertex is within two steps of the first ×1
- exactly one of the n + 1 rotations of a list empties the last spot ×1
- fewer removed points than rows leave free rows and a free column ×1
- five edges in the pentagon ×1
- five to nine points ×1
- for large complete graphs the sampling bound is within a factor of about eight of the conjectured value ×1
- K5 has ten edges ×1
- K5 with a point split contains a subdivision of K3,3 ×1
- K5 with a point split contains no subdivision of K5 ×1
- K5 with a point split, drawn flat as a solid's skeleton, has neither ×1
- K5 with a point split: an edge joins two different points of the graph ×1
- K5 with a point split: the minor test and the subdivision test agree about planarity ×1
- K6 has fifteen edges ×1
- Mantel: at most 16 edges on eight vertices, reached only by the 35 balanced complete bipartite graphs ×1
- minus one is itself a square, so the rule does not depend on which way round the pair is taken ×1
- moving a point to a smaller part adds edges ×1
- no cage is smaller than the Moore bound ×1
- no graph has more triangles than a clique of the same edge density ×1
- no triangle has all three edges the same colour ×1
- no two edges of the triangulation cross ×1
- no two points share two neighbours, so there is no four-cycle ×1
- no weighting beats it ×1
- non-bipartite triangle-free graphs need at least five edges ×1
- octahedron, drawn flat as a solid's skeleton, has neither ×1
- octahedron: an edge joins two different points of the graph ×1
- octahedron: the minor test and the subdivision test agree about planarity ×1
- on eight vertices every non-bipartite triangle-free graph has a shortest odd cycle of five or seven ×1
- on seven vertices every non-bipartite triangle-free graph becomes bipartite on deleting one edge ×1
- on the circle each preference is one of the n + 1 spots ×1
- on the circle every list parks every car and leaves exactly one spot empty ×1
- one edge more forces ⌊n/2⌋ triangles — Rademacher's theorem ×1
- one, one and three asymmetric trees on seven, eight and nine points ×1
- only degrees 2, 3, 7 and 57 give whole-number multiplicities ×1
- Otter's formula agrees with the shapes found from every Prüfer code ×1
- Petersen graph contains a subdivision of K3,3 ×1
- Petersen graph contains no subdivision of K5 ×1
- Petersen graph, drawn flat as a solid's skeleton, has neither ×1
- Petersen graph: an edge joins two different points of the graph ×1
- Petersen graph: the minor test and the subdivision test agree about planarity ×1
- points minus edges, plus one for a flipped edge, is one ×1
- prefs is a list of 3 to 7 preferred spots, each between 1 and the number of cars ×1
- prefs is read only by the parking and parkcircle views ×1
- q + 1 points are orthogonal to themselves ×1
- Razborov's curve never dips below Goodman's bound ×1
- sampled trees have the predicted share of leaves ×1
- six to fifteen points ×1
- so the cherries are at most the pairs ×1
- so the graph has exactly 1 + d + d(d − 1) = d² + 1 vertices, the most a diameter of two allows ×1
- so the parking functions are (n + 1)^n / (n + 1) of the lists ×1
- some fundamental cycle leaves at most two thirds on each side ×1
- some level leaves at most two thirds on each side ×1
- stretching the drawing to fill its panel changes no crossing ×1
- t(n) α⁻ⁿ n^(5/2) approaches 0.53495 ×1
- the alkane counts begin 1, 1, 1, 2, 3, 5, 9, 18, 35, 75 ×1
- the averaging converges ×1
- the bijection is checked up to between 3 and 6 points ×1
- the bipartite count's exponent approaches n²/4 from above ×1
- the bound is of the order of 2 to the k over 2 ×1
- the cherries' swaps are symmetries, so their count divides the total ×1
- the code is two shorter than the number of points ×1
- the colouring is drawn on 5, 13 or 17 points, each a prime one more than a multiple of four ×1
- the colouring treats its two colours alike, so the largest sets match ×1
- the complete graph has between 5 and 9 points ×1
- the converged drawing has no crossings ×1
- the decoded tree has one edge fewer than points ×1
- the dial has a radius between 40 and 400 ×1
- the drawing has no crossings ×1
- the drive and the sorting test agree on whether every car parks ×1
- the end is Turán's graph and Turán's count ×1
- the enumeration and the generating function agree on the bipartite graphs ×1
- the enumeration matches Rényi's count of trees with k leaves ×1
- the even weighting on a largest clique reaches the same score ×1
- the exhaustive search finds a K5 minor as well ×1
- the expectation is swept to between 20 and 200 points ×1
- the extremal graph splits the points in two with every edge crossing ×1
- the first complete graph the bound rules out is K5 ×1
- the first layer is the neighbours ×1
- the forbidden clique has three to five points ×1
- the forbidden complete graphs have between three and five points ×1
- the free-tree counts begin 1, 1, 1, 2, 3, 6, 11, 23, 47, 106, … ×1
- the function permutes its cyclic points ×1
- the function sends every point to a point ×1
- the graph has q(q + 1)²/2 edges ×1
- the graph is connected ×1
- the graph is one of the three-connected ones the view knows ×1
- the graph is small enough for bitmask search ×1
- the graph searched is the Petersen graph, the split K5 or the Wagner graph ×1
- the growth constant is 2.95576528565… ×1
- the K3,3 columns never part ×1
- the K5 columns part on exactly the split K5 and the Petersen graph ×1
- the labelled points number between 3 and 7 ×1
- the labelling's edge differences are all different ×1
- the labellings of all shapes on 7 points add to Cayley's count ×1
- the largest count is the balanced split's ×1
- the last spot stays empty exactly when the list is a parking function on the line ×1
- the majority colour is one of the two ×1
- the mean is Ramanujan's sum Σ n!/((n − k)! n^k) ×1
- the mean number of leaves is n(1 − 1/n)^(n − 2) ×1
- the pairs of rooted halves come in twos ×1
- the Paley colouring is drawn on 5, 13 or 17 points ×1
- the parking functions number (n + 1)^(n − 1) ×1
- the parking functions of length n number the labelled trees on n + 1 points ×1
- the paths share no interior point, and pass through no branch point ×1
- the Petersen graph has girth 5 ×1
- the Petersen graph has K5 as a minor and not as a subdivision ×1
- the Petersen graph's shortest cycle has five edges ×1
- the Petersen or the Hoffman–Singleton graph ×1
- the plane beats the random graph and stays under the bound at every size ×1
- the plane has q² + q + 1 points ×1
- the process ends near n^(3/2)√(ln n)/(2√2) edges, far from bipartite ×1
- the random tree is graceful ×1
- the ratio with Otter's power removed is close to α long before the plain ratio is ×1
- the reading gives n − 1 edges ×1
- the recursion is shallow enough to colour ×1
- the result is complete multipartite ×1
- the rooted-tree recurrence divides exactly ×1
- the rotated copies reach every edge of K₇ ×1
- the scallops rise at most a few hundredths above the parabola ×1
- the search found a layout that is not degenerate ×1
- the search runs on between three and six points ×1
- the search runs to between five and nine points ×1
- the seed is a whole number the colouring can be drawn from ×1
- the shortest cycle has five edges ×1
- the stages are two or three sweep counts ×1
- the sweep runs to between 20 and 200 points ×1
- the table or the check runs to a size the enumeration can reach ×1
- the table runs to between four and six points ×1
- the thinnest balanced level grows like √n: its ratio to √n stays within a narrow band ×1
- the thinnest balanced level is within the square-root scale ×1
- the three ends make three pairs, and any one of them closes a trio ×1
- the tree drawn has one fewer edge than it has points ×1
- the tree encoded has between 4 and 7 points ×1
- the tree, read back, is the same function ×1
- the triangle bound allows it and the girth bound does not ×1
- the two points hung between opposite corners are pulled to the same place ×1
- the view is one the family draws ×1
- the Wagner graph has no K5 minor ×1
- the Wagner graph has no K5 minor but a K3,3 minor ×1
- there are 23 tree shapes on eight points, by search and by Otter's formula ×1
- there are 23 trees on eight points ×1
- they are exactly the lists whose sorted form never exceeds its positions ×1
- this colouring has a monochromatic triangle ×1
- two colours over five edges forces three of one ×1
- two of the three cannot be drawn flat ×1
- Wagner graph, drawn flat as a solid's skeleton, has neither ×1
- Wagner graph: an edge joins two different points of the graph ×1
- Wagner graph: the minor test and the subdivision test agree about planarity ×1
- while it allows the two that can be drawn flat ×1
- while the level itself grows several-fold ×1
- whitening the point set leaves every four in the position it was ×1
- with at most one part fewer than the forbidden clique ×1
Where it is called
Every figure on this list is drawn by the same rule, so a change to the rule changes all of them at once. That is why the list is published.
A determinant that counts trees
Write down a graph's Laplacian, strike out one row and its column, take the determinant. The answer is the number of spanning trees — and the minus signs in the determinant are what cancel every subset of edges that is not one.
DiscreteA labelling every tree seems to have
Number the points of a tree 0 to n − 1 and write on each edge the difference of the numbers at its ends. The labelling is graceful if the edges then carry 1 to n − 1, each exactly once. Every tree anyone has ever checked — every one of the 551 trees on twelve points, and every tree up to about thirty-five — has such a labelling, and no one knows why. A graceful tree also tiles a complete graph by rotation, which is why the question was asked.
DiscreteA random tree is one part in e leaves
Choose a labelled tree on n points uniformly at random. A point is a leaf exactly when its label never appears in the tree's Prüfer code, so the share of leaves is (1 − 1/n)^(n − 2) — half the points for a tree on four, 36.8% for a large one, the reciprocal of e. The whole degree distribution follows the same way: one plus a Poisson count with mean one.
DiscreteAlmost every tree can be turned over
Cayley's n^(n−2) counts trees with labels on their points. Take the labels off and the count has no formula, because a symmetric shape absorbs labellings: the star on seven points can be labelled only seven ways, the one asymmetric shape 5,040. The bookkeeping that reconciles the two counts says something unexpected — almost every tree, labelled or not, can be turned over onto itself, where almost every graph cannot.
DiscreteAs many points as two steps allow
In a graph where every point has d neighbours and every point is within two steps of every other, there can be at most d² + 1 points — one, its d neighbours, and d(d − 1) more reached through them. Graphs that meet the bound exactly are rare to the point of absurdity. The pentagon does it for d = 2, the Petersen graph for d = 3, a fifty-point graph found in 1960 for d = 7, and an eigenvalue argument proves there is nothing else — except possibly one graph with 3,250 points and 57 neighbours each, which nobody has found or ruled out.
DiscreteCars that park, and trees that grow
Three cars arrive at a one-way street with three spaces; each has a favourite space, drives to it, and takes the first free one from there on. Of the 27 lists of favourites, exactly 16 let every car park — the same 16 as the labelled trees on four points. The reason is a circular street with one extra space, on which every list parks and exactly one rotation of it leaves the extra space empty.
DiscreteCounting the colourings
Asking whether a graph can be coloured with four colours gives a yes or a no. Asking how many ways there are gives a polynomial — and the polynomial answers the first question, and several others nobody asked.
DiscreteEighteen people, and the seventeen that escape
Among any eighteen people, four are mutual acquaintances or four are mutual strangers. Seventeen can be arranged so that neither happens, and the arrangement is not a lucky find — it is a rule about squares.
DiscreteEvery function is a tree with two marks
There are n to the n functions from n points to themselves, and n to the n − 2 trees on those points. André Joyal noticed in 1981 that the missing factor of n² is a choice of two points — a head and a tail — and that a function, read the right way, simply is a tree with a head and a tail. The reading turns Cayley's formula into one line and hands over a fact about random trees from the birthday problem.
DiscreteEvery point at the average of its neighbours
A planar graph can be drawn without crossings, but finding such a drawing looks like a search. Tutte found in 1963 that it is not: pin one face to a convex polygon, put every other point at the average of its neighbours, and the drawing that results has straight edges, convex faces and no crossings at all — provided no two points can cut the graph apart. It is the position a network of equal springs settles into, and it is also where a random walk expects to leave.
DiscreteFive spokes squeezed into K5
The Petersen graph has no point with four neighbours, so no stretched copy of K5 can sit inside it. Contract its five spokes and K5 appears anyway. Kuratowski's theorem forbids stretched copies and Wagner's forbids squeezed ones, the two notions disagree on this graph — and they still name exactly the same planar graphs.
DiscreteHow many shapes a tree can take
There are 262,144 labelled trees on eight points and only twenty-three shapes. Labels make counting easy — Cayley's nⁿ⁻² — and shapes make it hard, because a symmetric shape absorbs labellings. The count of shapes comes instead from a recurrence for trees with a root and a theorem of Richard Otter's for taking the root away: in every tree there is exactly one more class of points than of edges. Run to two hundred points, the counts grow like 0.535 × 2.9558ⁿ ÷ n^(5/2), and the same arithmetic counts the alkanes.
DiscreteMoves that only ever add edges
Turán's theorem says the densest graph avoiding a complete graph on r + 1 points is the balanced r-part graph. Zykov's proof finds it by a sequence of moves — turn a point into a copy of a better-connected one it is not joined to — each of which adds edges and none of which can create the forbidden clique. A second proof spreads a unit of weight over the points and gets the same bound from a maximum.
DiscreteSeven regions on a doughnut
A map on a torus can need seven colours, and the proof is a picture — seven regions, each sharing a border with all six others. The plane needed a computer and eighty-six years; the harder surface was settled in 1890 by drawing something.
DiscreteSix people at a party
Among any six people, three are mutual acquaintances or three are mutual strangers. Five is not enough, and the arrangement that saves five is a pentagon. Beyond that the numbers become unknowable.
DiscreteSixteen trees on four points
How many ways are there to connect n labelled points into a single tree? The answer is n to the power n minus two, which is a strange enough formula to demand an explanation — and the explanation is a code that turns every tree into a short list of numbers, and every short list of numbers back into a tree.
DiscreteThe colouring nobody has ever seen
Count the monochromatic sets a random colouring is expected to contain. If the average is below one, some colouring has none — and the argument is finished, having produced nothing anyone can look at.
DiscreteThe 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.
DiscreteThe densest graph without a square
Forbid four points joined in a cycle and a graph can keep only about ½n^(3/2) of its edges — far fewer than the quarter of all pairs a triangle-free graph keeps. Counting pairs of neighbours proves the ceiling in two lines. What reaches it is not a random graph but a finite geometry: the points of a projective plane, joined when they are orthogonal.
DiscreteThe edge that forces a triangle
A graph on six points can carry nine edges with no three of them closing a triangle. It cannot carry ten. The bound is n²/4, the graphs that achieve it are all the same shape, and both facts fall out of examining every graph there is.
DiscreteThe 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.
DiscreteThe fewest triangles an edge density allows
Half of all possible edges can be drawn without a single triangle. One more, and triangles appear — not one but several at once. Push the density further and the question becomes a curve: for every share of edges, the fewest triangles a large graph can hold. The answer is a string of scallops, touching a simple parabola at the densities of the balanced multipartite graphs and bulging above it between them. It was guessed in the 1980s and proved in 2008 by a method that turns counting into positive-definite matrices.
DiscreteTriangle-free, and two-sided eventually
Every graph with no triangle and the most edges possible is bipartite. Erdős, Kleitman and Rothschild proved that almost every triangle-free graph is bipartite too — and yet among the 246,348,115 triangle-free graphs on nine labelled vertices, fewer than half are. The share falls before it rises, and the reason is that small graphs are sparse.
DiscreteTwo 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.
TopologyTwo trees, and every edge in exactly one of them
Euler's formula is usually proved by deleting things until nothing is left. There is a better argument that deletes nothing — a tree through the corners and a tree through the faces, which between them use every edge once and can therefore be counted.