Every essay — page 7
When the problem arrives again
Every other field here solves one system and measures how wrong the answer is. Almost no computation is shaped like that: a solve is one step of an outer loop, and the vector it returns is an input rather than a deliverable. That changes what its accuracy is for. An outer iteration recomputes its residual from the matrix at every step, so whatever the inner solve got wrong is measured again rather than carried — and four quantities that looked like accuracy requirements turn out to be assets with a shelf life, priced here against a root that is known exactly.
A factorisation kept past its date
One Cholesky factor can serve five members of a drifting sequence and save 44 per cent of the work. Kept for twenty it does not lose accuracy — it stops converging altogether. The optimum and the cliff are four members apart, both move with the drift, and a rule written in a ratio the iteration has already computed finds them without being told what the drift is.
Where the drift lands
The standing rule for when a preconditioner has gone stale is to rebuild it once the matrix has changed by more than some fraction of itself. Two drifts of exactly the same relative size cost 19 iterations and 5 on the same matrix, and the quantity that separates them is not in the rule at all — the perturbation is divided by the eigenvalue it lands on.
What a rebuild is worth
One sequence, one drift, one preconditioner — and six different right answers, because the cheapest rebuild period depends on what a rebuild cost to build and on nothing else. The optimum walks from every member to every twelfth as the setup gets dearer, and the free rule that reads the iteration count beats it in the middle of that range and loses at both ends.
Stable once, and three thousand times
A sliding window adds a row and removes one at every step and never looks at the data again. No single step of it amplifies by more than 2.72, no downdate fails, and after three thousand steps the triangular factor in memory is 3.9·10⁻¹⁴ from the matrix it is supposed to be a factor of — six hundred times growth from a per-step bound that says nothing about chains.
Where the flop count stopped predicting the time
Every cost claim in the other eleven fields is a count of arithmetic, and those counts have not decided which of two implementations is faster for about thirty years. A blocked and an unblocked elimination perform the identical multiplications in a different order, return a factorisation identical to the last bit, and move a factor of two different amounts of data. What is counted here is words moved between a fast memory and a slow one, with the size of the fast memory named on every figure — because nothing in the model is optimal until it is.
The same arithmetic at a different price
A blocked and an unblocked elimination perform 72,568 operations each — the same operations, associated differently — choose the same pivots, and return a factorisation identical to the last bit: ‖PA − LU‖/‖A‖ = 4.487946226420872·10⁻¹⁶ in both. One of them moves 41,332 words between fast and slow memory and the other moves 19,476.
A block size is a property of the machine
Three lines of counting say the best block size is √(M/3). Scanned over every integer at five fast memories, the measured optimum is √M − 2 — exactly, at all five. The count has the right scaling and the wrong constant, low by a factor of 1.56, and the wrong form: the answer is affine in √M rather than proportional to it.
A reduction that changes the order
A tall-skinny QR computed as a tree of independent block factorisations touches a 512×12 matrix once instead of twelve times, computes a completely different sequence of roundings from the sweep it replaces, and returns ‖AᵀA − RᵀR‖/‖AᵀA‖ = 1.65·10⁻¹⁵ against the sweep's 9.95·10⁻¹⁵. On the same matrix classical Gram–Schmidt returns 4.6·10⁻¹⁰.
The message and the word
Three factorisations of one matrix on sixteen processors: 48 communication rounds, 4, and 4. The words sent are 1,170, 1,170 and 2,160 — so the method with the fewest rounds sends the most words, and the count that separates the three is the one no operation count can see.
Doing it twice
Cholesky QR squares the condition number — a fitted slope of 1.95 in κ against the Householder sweep's 1.00. Run the identical routine a second time on the Q it returned and the slope is 0.93, the orthogonality is at or below the sweep's at every κ, and the price is one more all-reduce.
Memory bought with messages
Holding four copies of the data instead of one is supposed to cut a matrix multiplication's communication by √4. Measured on a machine of 64 processors it costs 14% more traffic; at 576 it saves 44%, which is 72% of what the law promises. The memory is exactly four times, and that part is not asymptotic.
Where the format starts paying
A hierarchical solve costs 1.48 times a dense factorisation at 64 unknowns and 0.16 times it at 512. The crossover is between 64 and 128, it walks right when the accuracy is tightened, and the exponent between consecutive sizes is 2.13, 1.93, 1.74 — falling towards one and never arriving.