Concept

Working set — where it appears

The memory an algorithm touches within a short window, which decides which level of the hierarchy the run actually lives in. It decides which level of the hierarchy a run actually lives in, which is a property of the access pattern rather than of the data's total size.

Named by 9 essays across 4 fields — each of them below, with the objects they name alongside it.

0%25%50%75%100%645124,09665,536cache holds 512array size n (elements)miss ratefully associative · 64 lines × 8 elements · LRU20,000 random accesses per point

The cliff where the data stops fitting

Below the cache's capacity, almost every access hits. A factor of eight above it, almost every access misses. The transition is not gradual and it is not a property of any algorithm — it is a property of how much data there is, and an algorithm's complexity class says nothing about which side of it a program is working on.

machine · Machine
Merge sort49% sequential · 7,540 accesses2560Heapsort15% sequential · 14,044 accesses2560time (accesses, left to right) · index (bottom to top)one run each, n = 256every access plotted

Where an algorithm looks

Plotted as index against time, every array access an algorithm makes becomes a picture that no count contains. Merge sort's is a set of sweeps. Heapsort's is a spray. Quicksort's is a narrowing triangle. These shapes decide how fast the algorithms run and they are entirely absent from the analysis that says all three are Θ(n log n).

machine · Machine
executionintention87777776569888888765one unit = one subproblem given a value100 cells computed, 20 held at once

The table nobody has to keep

A million-cell table, computed cell for cell in the same order, holding two thousand cells at its peak instead of a million. The saving is exactly (n+1)/2, it costs nothing on any operation counter, and what it buys is paid for with the one thing the table was for.

space · Table
01234567801234567891724303539424410182531364043111926323741122027333813212834142229152316one unit = one subproblem given a valueeach number is a storage offset, of 45 slots

A triangle stored in a square

An interval table has a cell for every range of keys and nothing below its diagonal, and it can be stored as a square array, as packed rows, or as packed diagonals — the last matching the order it is filled in. On sixty-four keys, with every read replayed through a small cache, the square misses 39.7% of its reads, packed rows 38.8%, and packed diagonals 78.4%. Storing a table in the order it is written is storing it in the order it is not read.

tables · Table
10⁵10⁶10⁷10³10⁴inversions in the permutationblock transfers to carry it outsort by destination, 1,536w 8w 32w 128w 512w 2048w 819216 swaps64 swaps256 swaps1024 swapsshuffled inside windowsa few pairs swapped farn = 16,384, B = 64, M = 512 (M/B = 8)inversions do not order the cost

The permutation that moves almost nothing

Two ways to scramble sixteen thousand elements. Shuffling them inside windows of five hundred and twelve puts two million pairs out of order and costs 3,095 block transfers to carry out. Swapping a thousand pairs across the whole array puts seven million out of order and costs 1,189. Inversions are the textbook measure of disorder, and on a disk they rank these two backwards.

applied · Transfer
cache misses per split point consideredsquare array, by length1.1063,128,465 missestwo copies, by rows0.212598,455 missessquare array, split scans0.095268,386 missesfully associative · 32 lines × 8 elements · LRU256 keys, 32,896 cells

The split scan cut into blocks

Every way of filling an interval table one cell at a time stops at about one cache miss per split point considered once the table outgrows the cache — 1.01 at 128 keys, whether the cells go by length, by rows, or in a recursive tiling. Cut each cell's scan into blocks instead, and apply a block of split points to a block of cells whose inputs are all in hand, recursively at every scale, and the same 357,760 split points cost 0.094 misses each. The fill is told nothing about the cache, blocks of one and of four do equally well, and it needs no extra memory, where storing the table twice gets to 0.151 by doubling it.

tables · Table
rounds, if every ready cell ran at oncerow by row65,793column by column65,793anti-diagonal by anti-diagonal513reads that miss the cacherow by row6.3%column by column28.2%anti-diagonal by anti-diagonal31.1%fully associative · 32 lines × 8 elements · LRU263,169 reads and writes per order

The order with the best depth

An edit-distance table can be filled row by row, column by column, or one anti-diagonal at a time, and the anti-diagonal order is the one that needs the fewest rounds — 513 against 65,793 on two strings of 256 characters, because every cell on an anti-diagonal is independent of the others. Stored the usual way, row by row, it also misses the cache on 31.1% of its reads, where row order misses 6.3%. The order that is best for parallel work is worst for the memory it runs on.

machine · Machine
rounds, if every ready cell ran at oncerow order, stored by rows65,793anti-diagonal order, stored by rows513anti-diagonal order, stored by diagonals513reads that miss the cacherow order, stored by rows6.3%anti-diagonal order, stored by rows31.1%anti-diagonal order, stored by diagonals8.7%fully associative · 32 lines × 8 elements · LRU263,169 reads and writes per order

The table stored the way it is filled

Store an edit-distance table by anti-diagonals instead of by rows, and the anti-diagonal fill keeps its 513 rounds while its cache misses fall from 31.1% of reads to 8.7%. It does not fall to row order's 6.3%, and the gap is not noise — on caches of four and eight lines the two rates are 9.4% and 6.3%, exactly three to two, because a cell reads from two earlier diagonals and only one earlier row. The same layout turns row order into the order that strides, at 28.3%. How a table is stored and the order it is filled in are one decision, and its price is the number of earlier fronts the recurrence reads.

machine · Machine
124816326410step width, cellscache lines a stepstored by diagonalsby diagonals, each line-alignedstored by rows512 × 512, lines of 8 cellsone step of an anti-diagonal

Eight cells at once

The anti-diagonal fill order exists because its cells do not depend on one another, and every table filled here has been walked one cell at a time anyway. Computed eight at a time, a step touches 5.71 cache lines on the layout that stores the table by diagonals and 10.87 on the one that stores it by rows — and per cell the first keeps falling to 0.42 while the second stops at 1.27. The prediction that a diagonal step would touch three or four lines was wrong, and line-aligning each diagonal only takes it to 4.94.

machine · Machine

Named alongside it

The objects these essays reach for when they reach for this one.

CacheLocalityAccess patternEvaluation orderMemory layoutMiss rateDynamic programmingEdit distanceTrade offDepthParallelismCompulsory miss

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