Generator

split-rule

One function in the branchbound library, called 8 times across 8 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 5 claims it put to the test while drawing them, and where it stands against the rule this site is named for.

At its defaults it draws two subdivision rules on roots 0.01 apart, one of which sits on the first cut. Six bars. Cutting a box at its midpoint puts the root at x = 1 on a boundary of every box the search ever asks about, and the operator cannot verify a root it never sees the interior of: 0 verified after 211 evaluations. Cutting at 0.485 of the width verifies both roots in 87.

split-rule is one function in lib/figures/branchbound.js — from a verdict to a search — where the box is cut, and what that costs. 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.

At its defaults

Drawn even though every essay passes arguments — which on this site is every essay, at 100% of placements since the standard pass. A default nothing exercises is a trap for the next essay to call this with none, and this is the page where a default that has drifted from the figures around it becomes visible.

Two subdivision rules on roots 0.01 apart, one of which sits on the first cutSix bars. Cutting a box at its midpoint puts the root at x = 1 on a boundary of every box the search ever asks about, and the operator cannot verify a root it never sees the interior of: 0 verified after 211 evaluations. Cutting at 0.485 of the width verifies both roots in 87.both rules see the same problem and the same operatormidpoint: verified0midpoint: undecided6midpoint: evaluations211off-centre: verified2off-centre: undecided0off-centre: evaluations87where the box is cutroots verified, midpoint0roots verified, off-centre2evaluations saved124a root on a boundary is never in an interiorand 0.485 is the whole repair

Six bars. Cutting a box at its midpoint puts the root at x = 1 on a boundary of every box the search ever asks about, and the operator cannot verify a root it never sees the interior of: 0 verified after 211 evaluations. Cutting at 0.485 of the width verifies both roots in 87.

delta: 0.01

The arguments are the ones A bound that is proved passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

Two subdivision rules on roots 0.01 apart, one of which sits on the first cutSix bars. Cutting a box at its midpoint puts the root at x = 1 on a boundary of every box the search ever asks about, and the operator cannot verify a root it never sees the interior of: 0 verified after 211 evaluations. Cutting at 0.485 of the width verifies both roots in 87.both rules see the same problem and the same operatormidpoint: verified0midpoint: undecided6midpoint: evaluations211off-centre: verified2off-centre: undecided0off-centre: evaluations87where the box is cutroots verified, midpoint0roots verified, off-centre2evaluations saved124a root on a boundary is never in an interiorand 0.485 is the whole repair

Six bars. Cutting a box at its midpoint puts the root at x = 1 on a boundary of every box the search ever asks about, and the operator cannot verify a root it never sees the interior of: 0 verified after 211 evaluations. Cutting at 0.485 of the width verifies both roots in 87.

delta: 0.1

The arguments are the ones A coin flip that fixes the average passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

Two subdivision rules on roots 0.1 apart, one of which sits on the first cutSix bars. Cutting a box at its midpoint puts the root at x = 1 on a boundary of every box the search ever asks about, and the operator cannot verify a root it never sees the interior of: 0 verified after 163 evaluations. Cutting at 0.485 of the width verifies both roots in 45.both rules see the same problem and the same operatormidpoint: verified0midpoint: undecided6midpoint: evaluations163off-centre: verified2off-centre: undecided0off-centre: evaluations45where the box is cutroots verified, midpoint0roots verified, off-centre2evaluations saved118a root on a boundary is never in an interiorand 0.485 is the whole repair

Six bars. Cutting a box at its midpoint puts the root at x = 1 on a boundary of every box the search ever asks about, and the operator cannot verify a root it never sees the interior of: 0 verified after 163 evaluations. Cutting at 0.485 of the width verifies both roots in 45.

delta: 0.001

The arguments are the ones An answer that is known passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

Two subdivision rules on roots 0.001 apart, one of which sits on the first cutSix bars. Cutting a box at its midpoint puts the root at x = 1 on a boundary of every box the search ever asks about, and the operator cannot verify a root it never sees the interior of: 0 verified after 267 evaluations. Cutting at 0.485 of the width verifies both roots in 125.both rules see the same problem and the same operatormidpoint: verified0midpoint: undecided6midpoint: evaluations267off-centre: verified2off-centre: undecided0off-centre: evaluations125where the box is cutroots verified, midpoint0roots verified, off-centre2evaluations saved142a root on a boundary is never in an interiorand 0.485 is the whole repair

Six bars. Cutting a box at its midpoint puts the root at x = 1 on a boundary of every box the search ever asks about, and the operator cannot verify a root it never sees the interior of: 0 verified after 267 evaluations. Cutting at 0.485 of the width verifies both roots in 125.

What it checked while drawing

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.

5 distinct claims across 4 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.

a gap the search can be asked to resolve

and the off-centre rule verifies both roots

in fewer evaluations

LU is for square matrices

the midpoint rule verifies nothing

Against the rule

The rule does not apply to it. It factorises nothing, so there is no residual it could be withholding. That is worth stating rather than leaving blank: a site that reported the rule as satisfied by every generator would be counting mostly generators the rule never reached.

Across the library: the rule bites on 70 of 151 generators — 55 print a residual and 15 are exempt with a published reason; 81 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.

Where it is called

Changing this generator changes every figure on this list. That is what makes the list worth publishing rather than keeping in a check script.

The arithmetic underneath

A bound that is proved

Every error statement on this site so far is a measurement of one run. Interval arithmetic makes a different kind of claim — the answer lies in this set, for this input, with no probability attached — and its failure mode is that it returns nothing at all. On a Hilbert system it proves a bound 23 times the error it bounds, and one size later it refuses.

The arithmetic underneath

A coin flip that fixes the average

Add 0.1 to 256 a thousand times at eight significand bits and the answer is 256. Not approximately — the total never moves, not once, and no error bound says so. Round up one time in twenty instead of never, and it arrives at 348 against a true 356.

Two errors, and whose fault they are

An answer that is known

Almost every demonstration of numerical error estimates the error by computing the same thing more carefully. The Hilbert matrix does not need that: its inverse is a closed form in integers, so the true answer is available exactly and the error is measured rather than approximated.

The arithmetic underneath

Cancellation takes the answer, not a digit

Subtracting two nearly equal numbers is exact. That is what makes it dangerous — the subtraction introduces no error at all, it exposes error the operands were already carrying, and the exposure can consume every significant figure at once.

The arithmetic underneath

Proving the answer is in the box

Every other method here computes a number and estimates how wrong it is. This one returns a verdict: there is exactly one solution in this box, or there is none, or — the honest third outcome — nothing can be said. Two of the three are proofs about infinitely many points from finitely many operations.

The arithmetic underneath

The numbers below the smallest one

Below the smallest normal number the spacing stops halving and stays put, all the way to zero. That is what gradual underflow is, and the thing it buys is the sentence every algorithm assumes without being told — x minus y is zero only when x equals y.

Elimination, and the swap

The swap that is not optional

Run elimination without a row interchange on a matrix that needs one and nothing announces a failure. There is no division by zero, no warning, and an answer of the right shape. It is simply wrong, and how wrong depends on a number you did not look at.

The arithmetic underneath

Where the box is cut

A branch-and-bound with an interval operator settles a whole square — two roots proved unique, forty-two regions proved empty, nothing left undecided, in 87 evaluations. Move the roots so one lands on the first bisection and it proves nothing at all, at any depth. Cutting at 0.485 instead of 0.5 finds both, in a quarter of the work.

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