fiedler-sweep

At its defaults it draws one real vector, 23 candidate cuts, and the best of them is number 12. The Fiedler vector of the barbell 12, its entries sorted, drawn as the pale rising curve against the right-hand scale; and against the logarithmic left-hand scale, the conductance of the cut that takes the first k vertices in that order. λ₂ = 0.14435. The eigenvector is a real vector and the answer wanted is a subset, so something has to round it: the sweep takes every prefix and keeps the best, which here is k = 12 at a conductance of 0.007519 against a worst prefix of 1 — a factor of 133 between the best cut this vector offers and the worst. Nothing in the eigenvalue problem chose k; the sorting did.

fiedler-sweep is one function in lib/figures/graphlap.js — the matrix a graph makes — row sums that are exactly zero, a count that is a threshold, and a partition with no vector. Everything below came out of it during this build, at arguments taken from the essays rather than invented for this page. A figure here is the figure a reader meets in an essay, and if the generator changes, this page changes with it.

The Fiedler vector of the barbell 12, its entries sorted, drawn as the pale rising curve against the right-hand scale; and against the logarithmic left-hand scale, the conductance of the cut that takes the first k vertices in that order. λ₂ = 0.14435. The eigenvector is a real vector and the answer wanted is a subset, so something has to round it: the sweep takes every prefix and keeps the best, which here is k = 12 at a conductance of 0.007519 against a worst prefix of 1 — a factor of 133 between the best cut this vector offers and the worst. Nothing in the eigenvalue problem chose k; the sorting did.

The Fiedler vector of the barbell 12, its entries sorted, drawn as the pale rising curve against the right-hand scale; and against the logarithmic left-hand scale, the conductance of the cut that takes the first k vertices in that order. λ₂ = 0.14435. The eigenvector is a real vector and the answer wanted is a subset, so something has to round it: the sweep takes every prefix and keeps the best, which here is k = 12 at a conductance of 0.007519 against a worst prefix of 1 — a factor of 133 between the best cut this vector offers and the worst. Nothing in the eigenvalue problem chose k; the sorting did.

The Fiedler vector of the cycle 40, its entries sorted, drawn as the pale rising curve against the right-hand scale; and against the logarithmic left-hand scale, the conductance of the cut that takes the first k vertices in that order. λ₂ = 0.024623. The eigenvector is a real vector and the answer wanted is a subset, so something has to round it: the sweep takes every prefix and keeps the best, which here is k = 20 at a conductance of 0.05 against a worst prefix of 1 — a factor of 20 between the best cut this vector offers and the worst. Nothing in the eigenvalue problem chose k; the sorting did.

The Fiedler vector of the two blocks 60, its entries sorted, drawn as the pale rising curve against the right-hand scale; and against the logarithmic left-hand scale, the conductance of the cut that takes the first k vertices in that order. λ₂ = 2.5538. The eigenvector is a real vector and the answer wanted is a subset, so something has to round it: the sweep takes every prefix and keeps the best, which here is k = 30 at a conductance of 0.08982 against a worst prefix of 1 — a factor of 11.1 between the best cut this vector offers and the worst. Nothing in the eigenvalue problem chose k; the sorting did.

The Fiedler vector of the path 30, its entries sorted, drawn as the pale rising curve against the right-hand scale; and against the logarithmic left-hand scale, the conductance of the cut that takes the first k vertices in that order. λ₂ = 0.010956. The eigenvector is a real vector and the answer wanted is a subset, so something has to round it: the sweep takes every prefix and keeps the best, which here is k = 15 at a conductance of 0.03448 against a worst prefix of 1 — a factor of 29 between the best cut this vector offers and the worst. Nothing in the eigenvalue problem chose k; the sorting did.

The Fiedler vector of the grid 6×6, its entries sorted, drawn as the pale rising curve against the right-hand scale; and against the logarithmic left-hand scale, the conductance of the cut that takes the first k vertices in that order. λ₂ = 0.26795. The eigenvector is a real vector and the answer wanted is a subset, so something has to round it: the sweep takes every prefix and keeps the best, which here is k = 18 at a conductance of 0.1 against a worst prefix of 1 — a factor of 10 between the best cut this vector offers and the worst. Nothing in the eigenvalue problem chose k; the sorting did.

Every figure above asserted its own claims on the way to being drawn, and a claim that failed would have failed the build rather than drawn a wrong picture. Those assertions used to leave no trace at all: a passing one returned true and the only evidence the figure had checked anything was that nothing crashed. The list below is what they actually said, collected by running this generator with an observer installed — not a description of what it is believed to check.

10 distinct claims across 6 sets of arguments, grouped below by shape — because most of them are one sentence with a different number in it, and how many separate times that sentence was put to the test is the informative part.

It draws a decomposition and prints its residual. It calls fiedler, and every figure above carries the badge — which residualcheck verifies by looking for it in the emitted SVG rather than by finding the call that builds one. A badge that is constructed and then left out of the body is the failure that check exists for.

Across the library: the rule bites on 214 of 382 generators — 194 print a residual and 20 are exempt with a published reason; 168 factorise nothing. Read from lib/residual-rule.js, which is the same body the gate enforces from, and the gate's last check fails the build if this page and it disagree about any generator.