The collection

Every essay — page 17

Essays 385 to 408 of 586, in the same order.

The matrix that is a graph

Every matrix in the fields before this one arrives as a matrix. A graph does not: it arrives as vertices and edges, two different matrices can be built from it that answer different questions, and what anybody wants back is a partition, a ranking or a count — a discrete answer, computed by continuous arithmetic. So a rounding here does not make the answer slightly wrong, it makes it a different answer: a connected graph reported as two pieces, a leaderboard whose middle places are decided by one uncrawled link, a determinant that is known in advance to be a whole number and stops being one at sixteen vertices while the elimination that produced it stays backward stable throughout. Two things this field has that nothing else on the site does. The Laplacian's null vector is exact — the rows are cancellations of integers, so L·1 is zero rather than nearly zero. And there is a two-sided bound that is proved rather than estimated, from one matrix–vector product.

10⁻³10⁻²10⁻¹110¹12345678910conductance, and the two bounds on itpathcyclegridbarbelltwo blockshypercubepreferentialstarcompletethe bar is the inequalitythe dot is the graph

A bound with a square root in it

Cheeger's inequality brackets a graph's best cut between λ₂/2 and √(2λ₂).

4 figures · Spectral partition, essay 2
14710131619110¹indexeigenvalue of Lan eigenvalue with no vectorλ₂0.35λ₃0.35gap1.7·10⁻¹⁵partitions found3best conductance0.045worst0.045two eigenvalues, one valueand the answer is not a function of the graph

A partition decided in the last digit

On a graph with a symmetry there is no Fiedler vector — there is a plane, and every vector in it is an exact eigenvector.

4 figures · Spectral partition, essay 3
01428425670849810⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²iterationchange between iteratestwo rates, one curveα0.85λ₂(P)0.9predicted rate0.77measured0.77iterations108against the solve10⁻¹⁶the upper dashed line is αᵏthe curve is on the other one

A ranking that is an eigenvector

PageRank is the stationary vector of a walk that follows links with probability α and jumps at random otherwise.

6 figures · Random walk, essay 1
0173451688510211910⁻⁶10⁻⁵10⁻⁴10⁻³10⁻²10⁻¹1steps of the walkdistance from stationarya rate read as a timeλ₂ of the walk0.9steps measured122steps predicted131ratio0.93the dashed line is the eigenvaluethe curve is the walk

The rate is the second eigenvalue

A walk forgets where it started at a rate the graph's second eigenvalue names exactly.

6 figures · Random walk, essay 2
02244668811013215417610⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²iterationchange between iteratestwo rates, one curveα0.85λ₂(P)1predicted rate0.85measured0.84iterations176against the solve1.4·10⁻¹⁶the upper dashed line is αᵏthe curve is on the other one

A chain with no stationary vector

A page with no outgoing links loses forty per cent of the walker's probability in six hundred steps.

4 figures · Random walk, essay 3
1611162126313610⁻²²10⁻¹⁹10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹vertex, clique first then tail|entry| relative to the largestone rounding of the largest entrya bound that is provedPerron root11bracket, low11bracket, high11bracket width1.2·10⁻⁴smallest entry-1.5·10⁻²⁰entries below zero4every entry is positiveand the picture disagrees

An eigenvector that must not change sign

Perron's theorem says the leading eigenvector of a connected nonnegative matrix is strictly positive.

5 figures · Perron frobenius, essay 1
places settled, of 14the arithmetic14 of 14one missing link3 of 14α from 0.80 to 0.9011 of 14entries move by 4.05·10⁻¹¹entries move by 0.00443entries move by 0.00123three kinds of doubtcomparisons14smallest gap6.3·10⁻⁵largest gap0.0024arithmetic band4.1·10⁻¹¹one-link band0.0044α band0.0012the arithmetic settles nearly all of themthe data settles three

A ranking whose order is not determined

Of the fourteen adjacent comparisons in a top-fifteen, fourteen survive perturbing the arithmetic at the rounding level, eleven survive moving the teleportation parameter across its usual range, and three survive removing one link.

5 figures · Perron frobenius, essay 2
1357911131510⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1vertexrelative error against the exact answerthree routes, one numberworst, by elimination4.8·10⁻¹⁶worst, by spectrum10·10⁻¹⁶solve residual1.5·10⁻¹⁵Foster's sum15n − 115difference3.6·10⁻¹⁵the answer is a ratio of integersso the error is known

A distance computed by a solve

Effective resistance is the one quantity in this field with no combinatorial route to it — it is defined by a linear system.

6 figures · Effective resistance, essay 1
19172533414911.251.51.752indexsample ÷ originalkept, and not keptedges kept344of1225eigenvalue ratio, low0.43high1.7degree ratio, low0.56high1.5diameter, after3filled dots are eigenvaluesopen dots are degrees

A graph with a tenth of the edges

Keeping 344 of 1,225 edges, sampled by effective resistance and reweighted, preserves every eigenvalue of the Laplacian to within a factor of 1.7.

6 figures · Effective resistance, essay 2
05101520253035059118177236295vertices eliminatededges of fill so farminDegree: 71natural: 125reverse: 125random: 160maxDegree: 293fill, by orderingminDegree71natural125reverse125random160maxDegree293edges to start60eliminating a vertex makes a cliqueand the order decides how big

Eliminating a vertex is a graph operation

Gaussian elimination on a Laplacian deletes a vertex and joins its neighbours into a clique with conductances wᵢwⱼ over Σw.

6 figures · Graph elimination, essay 1
10⁻¹110¹10²10³1234eigenvalue of the preconditioned Laplacian, and the stretchbreadth-firstκ = 34.62randomκ = 55.17random, secondκ = 37.11the bar is the spectrumthe tick is a combinatorial bound

A preconditioner that is a tree

Every eigenvalue of a tree-preconditioned Laplacian is at least one and at most the total stretch — a combinatorial integer with no arithmetic in it.

6 figures · Graph elimination, essay 2
57911131517192110⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹vertices of the complete graphrelative sizethe algorithm is not the problemsizes counted18worst relative error9.4·10⁻¹⁵first wrong at n =16digits in the answer17spacing of doubles there8largest integer a double holds9·10¹⁵the elimination stays at the rounding levelthe format runs out

A count that comes out of a determinant

The number of spanning trees of a graph is the determinant of its grounded Laplacian, so it is a whole number known in advance.

6 figures · Graph elimination, essay 3
a triangle and a square, joinedK₂,₃ with a pendant edge1234560123456indexeigenvaluethe same, and not the samevertices each6edges each7spanning trees12spectra differ by5.3·10⁻¹⁵highest degree, left4highest degree, right3one has a trianglethe other is bipartite

The spectrum is not the graph

Two graphs on six vertices with the same Laplacian characteristic polynomial — as integer polynomials, not to fourteen digits.

7 figures · Graph invariant, essay 1
Re λIm λwhat survives the arrowsvertices18arcs18worst row sum0worst column sum0largest |Im λ|0.98asymmetry1the null vector is still exactand nothing else about the spectrum is real

A Laplacian that is not symmetric

Point the edges and the matrix stops being symmetric.

9 figures · Directed laplacian, essay 1
Cheeger's band, for the circulationcirculation · arcsvertices on the smaller sidewhich of the two the theorem is aboutλ₂ of 𝓛0.066λ₂/20.033√(2λ₂)0.36circulation Φ0.12arc conductance0.04their ratio3.1the inequality holdsabout a quantity nobody counts

A conductance the arcs do not measure

Symmetrising a directed Laplacian with respect to its walk recovers everything the arrows took — a real spectrum, a sweep cut, a Cheeger inequality.

7 figures · Directed laplacian, essay 2
separated by the groupboth groups cyclic: cannot bethe same non-cyclic group6 vertices, 2 pairs117 vertices, 65 pairs3023128 vertices, 1022 pairs435361226a cyclic group is fixed by its order, and the order is the tree countthe group heard less than half the time

A finer invariant that hears less

The Laplacian spectrum fixes a graph's number of spanning trees and not the group those trees form, so the Smith normal form of the grounded Laplacian is a strictly finer integer invariant — and on the six-vertex pair the spectrum cannot separate, it does: ℤ₁₂ against ℤ₂ ⊕ ℤ₆.

6 figures · Graph invariant, essay 2
top bar: same Laplacian spectrum · bottom bar: same adjacency spectrum0%25%50%75%100%share of cospectral pairs the invariant separatesLaplacian spectrum0%100%adjacency spectrum100%0%signless Laplacian spectrum89%100%spanning-tree count0%98%critical group43%98%degree sequence35%83%triangle count34%0%red: Laplacian-cospectral pairs · blue: adjacency-cospectraleach spectrum hears the other's pairs

Each spectrum hears the other's pairs

Pairs of graphs with the same Laplacian spectrum share a spanning-tree count, so the critical group could only tell them apart by how that count factors — and did on two pairs in five.

5 figures · Graph invariant, essay 3
the critical group, alonequartic pairs separated181of quartic pairs587cubic pairs separated014 vertices, degree 30 of 216 vertices, degree 30 of 2218 vertices, degree 30 of 5020 vertices, degree 30 of 4412 vertices, degree 436 of 12614 vertices, degree 497 of 30016 vertices, degree 448 of 161both groups cyclicseparated by the groupnot separatedbars: distinct pairs, split three waysthree in ten quartic, no cubic

The group hears a switch only at two

On a regular graph the Laplacian is the degree times the identity less the adjacency matrix, so the two spectra are one piece of information and the tree count comes with them; between cospectral regular graphs the critical group is the only exact invariant left.

5 figures · Graph invariant, essay 4
10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²10⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1coupling ε between the clustersworst relative entry errorGTHGaussian eliminationpower iterationthe stationary vector is computed against an exact rational solutiononly the elimination with no subtraction holds

No subtraction, and no waiting

The power iteration keeps a positive vector's small entries correct because it never subtracts, and it pays for that in steps.

5 figures · Perron frobenius, essay 3
10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²110¹10²10³10⁴10⁵coupling ε between the clusterssteps to every entry within 10⁻¹²neverpower iterationblock Gauss–Seidelpower smoothingblock Gauss–Seidel, no subtraction'never': stalled above 10⁻¹²aggregation does not wait for the coupling

The stall was in the block

On two dense clusters joined by a coupling ε, the power iteration needs about 7/ε steps and then stalls at about u/(4ε).

5 figures · Perron frobenius, essay 4
0%25%50%75%100%degree 3, independent0 of 118degree 4, independent181 of 346degree 4, independent, 1 full0 of 10degree 4, a matching inside0 of 231degree 5, independent0 of 6degree 5, independent, 1 full0 of 2degree 5, independent, 2 full0 of 2degree 5, a matching inside31 of 73degree 5, a matching inside, 1 full7 of 11degree 5, a 4-cycle inside0 of 4set null modulo twonot nullshare of switches whose critical groups differcubic and quartic: distinct pairs; quintic: every switching setthe group hears only a null set

The set has to be even outward

Switching on four vertices makes cospectral regular graphs, and the critical group separated 181 of 346 quartic pairs whose switching set had no edges inside and none of 118 cubic ones.

5 figures · Graph invariant, essay 5
03103010030012 vertices, degree 4, 1 insideoutward 3, divisible by 3403413 vertices, degree 4, 1 insideoutward 3, divisible by 33412 vertices, degree 5, 2 insideoutward 3, divisible by 315432214 vertices, degree 5, 1 insideoutward 406214 vertices, degree 5, 2 insideoutward 3, divisible by 302switches giving a new graph, logarithmicdark: heard · light: silent1,500 graphs a caseonly when divisible, not always then

Three is where it can hear

A Godsil–McKay switch on four vertices conjugates by a matrix of halves, and the critical group hears it at the prime two exactly when the switching set's indicator is a Laplacian null vector modulo two.

5 figures · Graph invariant, essay 6

Exact arithmetic, and what it costs instead

An integer matrix eliminated over the rationals commits no rounding anywhere, so the forward error is zero, the residual is the zero vector, and the identity the rest of this site is built on has nothing left to say. The cost does not disappear with the error — it moves into the length of the numbers, where three eliminations returning the same integer differ by four orders of magnitude in the widest one they form. What is left to decide is which route pays: divide by the previous pivot and stay in the integers, reduce every fraction and pay a gcd per operation, or work modulo primes where nothing can grow and put the answer back together afterwards.