Arithmetic cost — where it appears
Named by 8 essays across 2 fields — each of them below, with the objects they name alongside it.
The last digit is the cheapest
Every cost curve measured here has the same shape: the first digits are cheap and the last ones are not. One method inverts it. Doubling the work buys twice as many digits as the previous doubling did, so the price of a digit halves every time it is paid.
A ceiling is not a target
Asked to hold the smallest possible intermediate, an evaluation order costs a median of 1.55 times the cheapest order's arithmetic. Asked to hold no more than that same amount, it costs 1.15. The two answers hold exactly the same number of numbers, and on one network they are 5.17 times apart in work.
The plan that was right at rank four
An evaluation order is chosen once and paid for thousands of times, and the dimensions it was chosen at are not the dimensions it runs at. Compiled at rank four and run at rank 256 it costs 8.01 times the order that rank deserves; compiled at 256 and run at 2 it costs 301 times. A one-line rule recomputed on arrival costs 2.92 and 1.11.
The search that got worse as it widened
Keeping two candidate orders instead of one removes a third of the cheapest-product rule's excess, and by eight it has removed all of it — the two greedy rules this field separated become the same rule. On twenty-four of sixty networks a wider search returns a dearer order than a narrower one, and it is the better rule that it more often makes worse.
The circle between two eigenvalues
A contour count's error is not approximately governed by the nearest eigenvalue; it is exactly one closed-form term per eigenvalue, and summing those terms reproduces the quadrature to a millionth at 1,720 radius and point pairs. Three things follow. The rate is the ratio of the two moduli the circle sits between, not a distance, so two circles 0.2 from their nearest eigenvalue converge three times apart. Ten digits cost about 21 points divided by log₁₀ of that ratio, 71 points with twelve eigenvalues inside and 3,476 with six. And the best circle is not halfway: at the geometric mean of two eigenvalues on one ray their two terms are equal and opposite, and ten digits cost 27 points where the midpoint needs 84.
An order found once knows what the rows know
Searching for a fraction-free elimination's least-cost row order costs forty to two hundred and sixty times the elimination, so it can only pay if one order serves a whole family. The prediction was that it cannot: the best order depends on which rows happen to have short entries, so a reused order should do no better than the natural one. On independent entries and on a shared zero pattern that holds — another member's best order is an ordinary order, at the middle of the 40,320. On a family that shares its row scales it fails completely: another member's order costs a median 1.13 of the least where the natural order costs 1.88. But on every family, including that one, the rule that just takes the smallest pivot entry is cheaper than any reused or trained order — 1.24, 1.08 and 1.01. What an order carries from one matrix to the next is what the family shares — and only when that is a particular matrix, not a visible structure, does an order trained on the family beat the rule.
A centre that chooses which pair cancels
A contour count's error is one closed-form term per eigenvalue, and two real eigenvalues on one ray cancel at their geometric mean — about the origin. Moved along the real axis, the centre was predicted to buy a further factor of two by putting real eigenvalues on two rays, the far one cancelling with a sign flip. The flip is parity: an eigenvalue behind the centre changes sign only at odd point counts. The factor is not two. Counting the delay problem's first two and three real eigenvalues, the best centre costs 22 and 21 points against the origin's 41 and 78; counting all four, 32 against 341, because about the origin the circle has to pass between a real eigenvalue and a complex pair and about 2.3 it does not. The centre does not add a cancellation. It chooses which pair cancels and pushes everything else away.
Count the upper half, read the lower
Moving a contour's centre off the real axis was predicted never to pay for a count closed under conjugation: one member of each pair comes closer, the other goes away, and the slower one always gets slower. For one circle the prediction holds at every offset measured, from 52 points on the axis to 567 three units above it. But one circle is not the only way to count a set like that. On the delay problem, the four real eigenvalues on a circle of their own and each band's upper cluster on a small circle of its own cost 55 points against 52 for one band, 68 against 97 for two, and 115 against 275 for six. What pays is the mirror: for a problem with real coefficients the count about the conjugate centre is the conjugate of the count already made, equal to 1.6·10⁻¹⁵, so the lower half of the plane costs nothing.
Named alongside it
The objects these essays reach for when they reach for this one.
Contour integralConvergence rateHeuristicNonlinear eigenvalue problemQuadratureContour eigensolverContraction orderExact ground truthFlop countSymbolic phaseTensor networkElimination order