Concept

Locality — where it appears

How near in memory an algorithm's successive accesses fall, which decides its cache misses and therefore its duration at a fixed operation count. It decides the cache misses and therefore the duration at a fixed operation count, and it is invisible to every counter that charges one per access.

Named by 21 essays across 5 fields — each of them below, with the objects they name alongside it.

block transfersIn order, 0 to n−11,0241.0× a scan · 64.0 elements per transferEvery B-th element (B = 64)65,53664.0× a scan · 1.0 elements per transferUniformly at random61,40760.0× a scan · 1.1 elements per transfera scan of this array is 1,024 transfersB = 64, M = 4,096 (M/B = 64)64× between the cheapest order and the dearest

One access, eight kilobytes

Every count on this site charges one for an array access. A machine charges for a block. The same 65,536 accesses cost 1,024 transfers in one order and 65,536 in another, with nothing about the algorithm's work changed — a factor of 64, which is exactly the number of elements in a block, and which no counter here could see until now.

applied · Transfer
10⁵10⁶10³10⁴10⁵comparisonscache misses (modelled)Insertion sortSelection sortBubble sortMerge sortHeapsortQuicksort, firstQuicksort, median-3Quicksort, randomShellsortMerge + cutofffully associative · 64 lines × 8 elements · LRUa modelled count, not a time

The count is not the time

An operation count is exact, machine-independent, and not a running time. The gap between them is mostly memory, and it is large enough to reorder the rankings. This site carries a second count — modelled cache misses from the same runs — and asserts that the two disagree, because if they agreed the second one would carry no information.

machine · Machine
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
10³10³10⁴Vmodelled missesadjacency listCSR array96% miss27% miss64 lines × 8 elements, fully associative, LRU3.6× between two layouts of one graph

A list and a block of memory

The same traversal, over the same graph, examining the same edges in the same order, laid out two ways. Twelve thousand two hundred and eighty-eight edge slots either way; 11,812 modelled cache misses against 3,258. This is the site's largest gap between two counts of one run, and it exists because one of the layouts is a pointer chase and the other is a sweep.

graphs · Graph
Sorted array (binary search)2 blocksLevel order4 blocksvan Emde Boas2 blocksmemory address, left to right · alternating outlines are blocksB = 8, M = 64 (M/B = 8)4 blocks against 2, for the same 6 comparisons

Two searches, one comparison count

Three arrangements of the same binary search tree over the same million keys, walking the same path, making the same twenty comparisons. One costs 15 block transfers, one costs 13, and one costs 3. Nothing about the algorithm differs between them — only where the nodes were put — and no counter this site had before this phase could tell them apart.

machine · Machine
sittingkitten111111101111110-1111110-1-111110-1-1-11110-1-1-10110-1-1-10-11one unit = one subproblem given a value-1, 0, 1 — 3 values, 2 bits each

A column computed in machine words

Adjacent cells of a distance table differ by at most one, so a whole column is two bits per cell — and thirty-two of them fit in one register. Fifteen word operations per character replace three cell evaluations per cell, and below a pattern of fifteen characters the trade is a loss.

machine · Machine
1010010³10⁴B — elements to the blockblock transfersthey cross near B = 4one element at a timesorted by destination, 2passesM = 4,096, B as drawnthe lower bound is the smaller of the two, and it is proved rather than measured

Permuting is the harder problem here

Rearranging 65,536 elements into a stated order costs 63,601 transfers one at a time and 4,096 by sorting them into place. In the model every other field on this site uses, the first is the cheap method and beats the second by a factor of eight. The two models disagree about which problem is easy, and they disagree by about the same factor in opposite directions.

applied · Transfer
11010010³10⁴1010010³rows matching the queryblock transfersn/B rows — where the arithmetic says they meetindex, rows scatteredindex, file in key orderread the whole fileB = 64, M = 4,096 (M/B = 64)plus 3 transfers to descend the index

The index that is not worth reading

An index turns a query over 65,536 rows from 1,024 transfers into four. At a thousand matching rows it costs 654 and still wins; at sixteen thousand it costs 1,027 and has lost. Where it turns is decided by the block size — a number the query does not contain, the schema does not mention, and nobody writing either has seen.

applied · Transfer
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
1632649612810³10⁴10⁵table sizesplit points appliedevery splitbounded per blockbounded per cell, by lengthweights satisfying the quadrangle inequalityall three compute the same table

The bound a block can and cannot have

Knuth's condition turns an interval table's cubic fill into a quadratic one by bounding each cell's best split between its two neighbours'. A blocked fill cannot use it a cell at a time, and the two cells that bound a block lie outside the block — one to its left, one below it. The schedule has finished both for ten per cent of the blocks, the bound then removes eleven per cent of the splits, and it removes half a per cent of the cache misses, because the splits it skips are the ones already in the cache.

tables · Table
0%1.1%2.1%3.2%4.2%the whole filterblocks of 64blocks of 5120123456distinct 512-bit lines a lookup readsabsent keys answered yes16,384 bits, 2,048 keys, k = 6one line is what a block buys

Positions confined to one line

A Bloom filter lookup reads 5.55 cache lines because its six positions are scattered across the whole filter. Confining them to a 512-bit block makes it exactly one, and costs 7% more false positives at eight bits a key. At sixteen bits a key the same block costs 91%, and the two-value trick that is free across a whole filter costs another 135% inside one — because a block is a small filter, and small filters are where the penalty lives.

randomness · Randomness
48163264128110entry width, bytescache lines a lookup readsentries inlinea line of tags in front8,192 slots, load 0.9, buckets of 8keys the table holds

A lookup that stops caring how wide an entry is

Buckets of eight entries aligned to a cache line read 1.20 lines a lookup when an entry is eight bytes and 19.25 when it is 128, because the bound was arithmetic about alignment and the arithmetic stops holding. Keeping one byte of each key's hash in a separate array and the entries in a parallel one reads 2.21 lines at every width from four bytes to sixty-four — and for a key the table does not hold, 2.05 against 31.98.

machine · Machine
8121620240.000010.00010.0010.01bits a keyfalse-positive ratethe whole filterone block of 512 bitsthe emptier of two 512-bit blocksone block of 1,024 bits2,048 keys · 24 filters a pointdashed: no blocks

Two blocks and the chances they add

Send each key to the emptier of two 512-bit blocks and the busiest block of a filter at sixteen bits a key holds 38 keys instead of 55. The false-positive rate does not move: 0.100% against 0.095%. At eight bits a key it doubles. A lookup cannot tell which block a key went to, so it has to ask both, and asking twice is two chances to be wrong. The repair that tames a hash table's worst bucket buys a filter nothing that a block twice as wide does not.

randomness · Randomness
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
linear probingchainedcuckoo, two tables0240.10.20.30.40.50.60.70.80.9load factorentries read per lookup0120.10.20.30.40.50.60.70.80.9load factorcache misses per lookup8,192 slots, 64 cache lines of 8a probe is an entry read; a miss is a line fetched

Two probes are two misses

Cuckoo hashing's lookup reads at most two slots, and at a load of 0.45 it reads 1.27 on average where linear probing reads 1.39. Replayed through a cache, it misses 1.18 times a lookup where linear probing misses 0.98. The table that wins the count the analysis uses loses the count the machine charges, because two slots in unrelated places are two cache lines, and a run of adjacent slots is usually one.

machine · Machine
linear probingcuckoo, two tablescuckoo, buckets of 8024680.30.450.60.750.850.95load factorentries read per lookup00.511.50.30.450.60.750.850.95load factorcache misses per lookup8,192 slots, 64 cache lines of 8a probe is an entry read; a miss is a line fetched

The bucket that fits a line

Make each of a cuckoo table's two candidates a bucket of eight slots laid out on one cache line, and no lookup ever touches more than two lines, the table builds past a load of 0.95, and at that load it misses 1.21 times a lookup where linear probing misses 1.79. The prediction that it would lose to linear probing at low loads was wrong — it misses less at every load measured, 0.94 against 0.96 at 0.3 — because a key it holds almost never lives in its second bucket. The guarantee belongs to the alignment, not the bucket; eight slots on lines of four put a lookup on four lines.

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.

CacheMemory layoutAccess patternMiss rateWorking setCost modelBlock transferDesign parameterDynamic programmingEdit distanceEvaluation orderExternal-memory model

All concepts