setup-walk

At its defaults it draws the cheapest rebuild period for one drifting sequence, against what a rebuild costs. One sequence, one drift rate, one preconditioner — and six answers, because the answer is not a property of the sequence. Costed in iteration-equivalents, the optimal period runs from every 1 member at a setup worth 5 iterations to every 12 at a setup worth 200. A dense Cholesky at this size is worth 6.7 iterations, which is at the left of the axis, so for it the answer is always to rebuild. The rule that rebuilds when a solve takes 1.5 times what the last fresh one did beats the best fixed period at 1 of the 6 ratios and loses at the rest, because it cannot see the setup cost at all.

setup-walk is one function in lib/figures/reuse.js — reuse — what a kept factorisation or preconditioner is worth, and for how long. 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.

One sequence, one drift rate, one preconditioner — and six answers, because the answer is not a property of the sequence. Costed in iteration-equivalents, the optimal period runs from every 1 member at a setup worth 5 iterations to every 12 at a setup worth 200. A dense Cholesky at this size is worth 6.7 iterations, which is at the left of the axis, so for it the answer is always to rebuild. The rule that rebuilds when a solve takes 1.5 times what the last fresh one did beats the best fixed period at 1 of the 6 ratios and loses at the rest, because it cannot see the setup cost at all.

One sequence, one drift rate, one preconditioner — and six answers, because the answer is not a property of the sequence. Costed in iteration-equivalents, the optimal period runs from every 1 member at a setup worth 5 iterations to every 12 at a setup worth 200. A dense Cholesky at this size is worth 6.7 iterations, which is at the left of the axis, so for it the answer is always to rebuild. The rule that rebuilds when a solve takes 1.5 times what the last fresh one did beats the best fixed period at 1 of the 6 ratios and loses at the rest, because it cannot see the setup cost at all.

One sequence, one drift rate, one preconditioner — and six answers, because the answer is not a property of the sequence. Costed in iteration-equivalents, the optimal period runs from every 1 member at a setup worth 5 iterations to every 12 at a setup worth 200. A dense Cholesky at this size is worth 6.7 iterations, which is at the left of the axis, so for it the answer is always to rebuild. The rule that rebuilds when a solve takes 1.5 times what the last fresh one did beats the best fixed period at 0 of the 6 ratios and loses at the rest, because it cannot see the setup cost at all.

One sequence, one drift rate, one preconditioner — and six answers, because the answer is not a property of the sequence. Costed in iteration-equivalents, the optimal period runs from every 1 member at a setup worth 5 iterations to every 8 at a setup worth 200. A dense Cholesky at this size is worth 6.7 iterations, which is at the left of the axis, so for it the answer is always to rebuild. The rule that rebuilds when a solve takes 1.5 times what the last fresh one did beats the best fixed period at 0 of the 6 ratios and loses at the rest, because it cannot see the setup cost at all.

One sequence, one drift rate, one preconditioner — and six answers, because the answer is not a property of the sequence. Costed in iteration-equivalents, the optimal period runs from every 1 member at a setup worth 5 iterations to every 12 at a setup worth 200. A dense Cholesky at this size is worth 6.7 iterations, which is at the left of the axis, so for it the answer is always to rebuild. The rule that rebuilds when a solve takes 1.5 times what the last fresh one did beats the best fixed period at 1 of the 6 ratios and loses at the rest, because it cannot see the setup cost at all.

One sequence, one drift rate, one preconditioner — and six answers, because the answer is not a property of the sequence. Costed in iteration-equivalents, the optimal period runs from every 1 member at a setup worth 5 iterations to every 8 at a setup worth 200. A dense Cholesky at this size is worth 6.7 iterations, which is at the left of the axis, so for it the answer is always to rebuild. The rule that rebuilds when a solve takes 1.5 times what the last fresh one did beats the best fixed period at 0 of the 6 ratios and loses at the rest, because it cannot see the setup cost at all.

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 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.

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 113 of 238 generators — 98 print a residual and 15 are exempt with a published reason; 125 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.