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The thread: The model has parameters

In the external-memory model nothing is optimal until a block size and a memory size are named: the base of every logarithm is M/B, and an algorithm tuned for one block size is ordinary at another. The parameters are part of the claim rather than part of the hardware.
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 When it does not fit

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.

10³10⁴10010³n (elements)block transfersM = 256 · fan-in 7M = 1024 · fan-in 31B = 32, M as labelled4 passes against 3 When it does not fit

Sorting what will not fit

Merge sort's Θ(n log n) is a statement about comparisons and says nothing about a file larger than memory. Counted in transfers the answer is (n/B)·log_{M/B}(n/B), and the base of that logarithm is the number of blocks that fit in memory — so doubling the memory does not halve the work, it moves a staircase. The measured cost jumps by 32,768 transfers at one step and by nothing for the next four.

hash rowh1h2h3h4h1 → cell 30: 10,433h2 → cell 29: 10,624h3 → cell 19: 10,575h4 → cell 4: 10,895key 0 occurred 9,579 times · the minimum of the four is 10,433 · over by 854the additive bound at this width is e/w × N = 5,0974×32 counters · 4,096 bits · Zipf s = 1.1exact would take 162,288 bits One pass, and no room

A count that is never under

The Count-Min sketch holds four rows of counters and answers how often a key occurred. Its error is one-sided with no probability attached — the estimate is never below the truth on any stream — and the probabilistic half of its guarantee is only about how far above.

1010010B (elements per block)block transfers per searchtuned for B = 64Sorted arrayB-tree tuned for B = 64van Emde Boas — told nothingM = 16,384, B as drawnone layout, 7 block sizes, no parameter When it does not fit

The layout that is told nothing

A B-tree is built around a block size somebody looked up. The van Emde Boas layout is given neither the block size nor the memory size, and across seven block sizes spanning a factor of 64 it tracks the best structure that was told them. An algorithm with no parameters making a claim at every level of the hierarchy at once is a strange thing to be able to measure, and this is what it costs.

16326412825610244096window length W, in arrivals (D = W × the mean gap)share of the counts they disagree onevenPoissonburstydrifting1 ms clock · mean gap 10 ticks · 30,000 arrivalseven: 0% at every length One pass, and no room

A window that is a duration

Nobody asks for the error rate over the last four thousand and ninety-six requests. They ask for the last five minutes. The two are the same question exactly when the arrivals are evenly spaced, and on a stream whose rate drifts they disagree about fifty-seven per cent of the counts.

least recently usedthe offline optimumA loop over 7 blocks35 accesses35113.18×Random over 7 blocks35 accesses971.29×A straight sweep35 accesses35351.00×B = 8, M = 48 (M/B = 6)3.2× on the loop What is taught wrongly

The block that is not a block

This field's model has one memory, one block size, one processor, no prefetcher, no queue and no clock, and every number in it is a count of transfers rather than a duration. A real machine has five levels, reads ahead, issues a hundred requests at once, and charges four orders of magnitude more for one kind of transfer than another. What survives the difference is worth stating exactly, and so is what does not.

exact-11.2%-3.3%0.0%3.3%11.2%rmse 3.61%worst 9.71%23 of 60outside the band5,120 bits · 60 seeds · relative error of one runpredicted ±3.25% One pass, and no room

The correction that makes it work

HyperLogLog and LogLog read the same registers and differ only in how they average them. The harmonic mean is worth 30% of the error for nothing, and below two and a half registers' worth of keys the estimator both are built on is 137% high and has to be abandoned.

10⁴10010³n (elements)block transfersmeasured sortthe bound3.20×2.67×2.29×2.94×2.67×2.40×B = 32, M = 512 (M/B = 16)3.20× the floor at worst The floors

The floor under moving data

The information-theoretic floor for comparison sorting is log₂(n!) and it says nothing about a file on a disk. In the external model the floor is (n/B)·log_{M/B}(n/B), it is a bound on every algorithm rather than on merge sorts, and a measured external sort sits 2.40 to 2.97 times above it. Both numbers are computable, and the gap between them is what a real implementation costs.

1100ε — the exponent the fanout is B toelements written per key33445679ε = 1 — the B-tree ·fanout 256 · 3transfers a queryε = 0.5 · fanout 16 · 5a querythe number above eachpoint is what a querycostsB = 256, M = 16,384 (M/B = 64)131,072 random keys When it does not fit

One dial between two structures

A B-tree writes 226 elements of block for every key stored and a log-structured store writes two. They are presented as rival designs. They are one design at two settings of an exponent that nothing in either description mentions, and every setting between them is available.

71 s60 ticks11100 ms600 ticks1410 ms6,000 ticks171 ms60,000 ticks270.001 ms60,000,000 ticksclock resolutionbits per stamp⌈log₂ 2D/r⌉no arrival rateappears in itD = 60 s · key 32 bitscomputed, not measured The other axis

The clock that cannot see the burst

A stream generator asked for a burst ten times faster than its mean rate, on a clock whose resolution was the mean gap, produced a perfectly even stream — index of dispersion 0.00, for something called bursty. Nothing had gone wrong except that the instrument could not represent what it was being asked to measure.

1,00010,00010³10⁴bits · runs12481632characters in the collection · copies aboven·H₃, bitsr, runs in thetransformEnglish-like · base 512 · divergence 0n·H₃ x15.8 · r x1.00 The data that is not a number

The entropy that cannot see a copy

Two copies of a text have exactly the same symbol statistics as one, so every entropy on this site doubles when the second copy arrives and the second copy carries no information at all. The number of runs in the Burrows-Wheeler transform is 224 at two copies and 224 at thirty-two.

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 When it does not fit

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.

stay 0.9, 400 pairsacgtacgt0525505225045240stay 0.5, 400 pairsacgtacgt0212202112022120cost of aligning the row letter with the column letterpseudocount 1 When the algorithm is a table

The matrix a corpus wrote

A substitution matrix is not a property of an alphabet. Fit one to four hundred pairs of sequences that rarely change and the dearest substitution costs five; fit the same model to four hundred pairs that often change and it costs two. Two hundred test pairs aligned under each matrix give different alignments in 115 cases — and a matrix fitted to eight pairs of the first kind moves 79 of them, from sampling alone.

levels on the path (outline) · transfers actually paid (filled)B = 411 levels · 6.99 transfers · 16,384 blocks residentB = 166 levels · 3.23 transfers · 4,096 blocks residentB = 644 levels · 1.99 transfers · 1,024 blocks residentB = 2563 levels · 1.30 transfers · 256 blocks residentB = 10243 levels · 1.01 transfers · 64 blocks residentbinary search over the same 4,194,304 keys: 20 transfersB = 1024, M = 65,536 (M/B = 64)20× between binary search and the widest tree Structures

A tree with nodes the size of a block

A B-tree is a binary search tree that has read the hardware manual. Its node holds as many keys as fit in one transfer, so the height falls from log₂ n to log_B n — and the measured cost falls further still, to 1.01 transfers over four million keys, because the top of the tree is small enough to stay in memory. The comparison count goes up.

above 0.01 of the stream · 13 really areone pass: 18 candidatestwo passes: 13 keys, exactly the heavy ones123456789101112131415161718128 counters · carry 1,376 bits5 of 18 spurious What a bound is

The pass that was never a parameter

One pass is the streaming model's defining restriction, and this collection has enforced it by making a second read throw. A prohibition cannot be swept. Turn it into a dial and the first thing it says is that a second pass turns eighteen candidate heavy hitters, six of them wrong, into twelve that are exactly right — for 3,680 bits carried across the boundary and no extra space at all.

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 When it does not fit

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.

10×10³10⁴the key side, as a multiple of memoryblock transfersblock nested looppartitioned hash joinsort–merge joinB = 64, M = 512 (M/B = 8)the other relation: 16,384 rows When it does not fit

Two ways to join, and the ratio that decides

The same join costs 260 transfers one way and 1,040 the other; at eight times the memory the same two costs are 2,880 and 1,280, the other way round. Neither number is a property of how large the tables are. The quantity that decides is how the smaller of them compares to memory, and a rule of thumb phrased in rows is a rule about somebody's machine.

123456123456cost to open a gapcost to extend a gapintentionexecution571 settingsinte---ntion---execution3 settingsinte-ntion-execution1 setting-intentionexec-ution1 settingintention against execution, affine costseach colour is one optimal alignment When the algorithm is a table

The parameter plane has few answers

Sweep the cost of opening a gap against the cost of extending one over five hundred and seventy-six settings, and the optimal alignment of intention against execution takes four values — one of them at 571 of the settings. Under a linear model the plane divides into three wedges through the origin, because doubling every cost changes nothing and only the ratio is a parameter. Tuning an aligner is choosing a region, and most of the plane is one.

11010010³10⁴12510rows the query actually matchescost ÷ the better plan's costrho plus the descent, 4.01estimate ÷64estimate ÷8estimate ×8estimate ×64n = 65,536, B = 64, M = 4,096 (M/B = 64), scattered read ×4dotted: the estimate switches at 512 rows When it does not fit

The estimate a plan rests on

A planner chooses between an index and a scan on how many rows it thinks will match, and the number it has is wrong by a factor. Guess sixty-four times too many on a narrow query and the scan it picks costs 13.5 times the index. Guess sixty-four times too few on a wide one and the index costs at most 4.01 times the scan — a ceiling that can be named before any query runs.

1,000248163264high-biased, α = 0.56low-biased, α = 0.56none-biased, α = 0.74tuples keptshards mergedlog-normal, σ = 1.2 — a latency distribution · 20,000 valuesα 0.56 / 0.56 / 0.74 · worst residual 6.5% The other axis

The cheap tail and the expensive merge

A summary whose tolerance tightens towards the tail keeps seven times the tuples of a plain one on a single pass, and after merging sixty-four shards it keeps three and a half times as many. The error function that buys a useful tail promise is also the one that pays most for never having the values in one place.

10×20×50×00.40.811.2skew of the join columnratio, logarithmicestimate off byregret, decided partregret, whole planR 4,000, S 40,000, T 2,000 rows, B = 64, M = 4,096 (M/B = 64)every order writes the same output When it does not fit

The join order is a guess

Three tables, two orders, and an estimate of the first intermediate result that assumes the join column is uniform. When the column is skewed the estimate is out by seventy-two times, and the plan chosen on it costs 1.49 times the better one — which sounds tolerable until the shared output is taken away, and the part of the cost the order actually decided turns out to be 43.9 times worse.

010020030040021φ 382φ 7203φ 17434φ 42885φ 1012486φ 242measured 409counts chargedlevel of the foldcharged at this levelrunning total64 shards · k = 32 · hashedcharged 409 · measured 409 The floors

The floor charged at every level

A key surviving a fold of sixty-four shards is charged 2, then 8, then 20, then 43, then 88, then 248 — the floor of whatever summary it was merged against, level by level. They sum to 409, and the damage read off the merged table is 409. The model that charged sixty-three copies of the leaf floor said 222.

1,00010,00010³10⁴characters of textsteps, precomputation plus scancrossing at 8,000published rulesexact rules2 patterns · four symbolscrossing 8,000 · was 32,000 What the libraries do

Where the exact rules pay now

With a construction as cheap as the published one, the exact shift rules pay for themselves past eight thousand characters of text at two patterns, four thousand at four, and never at thirty-two — because by thirty-two patterns the two rules make identical decisions.

010010³10⁴interval extensions48.1%68.5%69.8%errors allowed · share removed belowno pruningpruned on D4,000 characters · m = 1669.8% removed at k = 3 What a bound is

The branch that cannot reach an answer

Seventy-two rank operations over the pattern remove 27,906 of the 39,957 interval extensions a bounded-error index walk performs — 70% of the tree, at a budget of three. The share grows with the budget, which is what a pruning has to do to be worth its cost.

0.020.050.10.20.5121235810filter bits a keywasted reads per absent lookupno filters: all 3 levelssame rate every levelrates sized to levelsn = 65,536, size ratio 4, first run 1,024 keysthe same memory at every point When it does not fit

The filter each run carries

A log-structured store turns every lookup for a missing key into a read of every level, and a Bloom filter on each run buys those reads back with memory. Spread five bits a key evenly across three levels and a missing key still wastes 0.279 reads. Give the small levels more bits and the large one fewer — the same memory — and it wastes 0.201. At four levels the gap is 0.382 against 0.209, because sized filters stop the waste growing with the number of levels.

10×20×50×100×200×00.250.50.7511.251.51.752skew of the join columnregret of the decided part, logarithmicuniform estimate4 counters a side16 counters a side64 counters a side256 counters a sideR 4,000, S 40,000, T 2,000 rows · 64-record blocks, 4,096 in memoryMisra–Gries on each side of the join column When it does not fit

The skew a few counters cannot repair

A join order chosen on the textbook estimate costs 243.9 times the better order at a Zipf exponent of two, and two counters a side are enough to fix it. At an exponent of one half the estimate is out by less than a factor of two, the plan it picks costs 1.37 times the better one, and no number of counters up to 256 changes that. The easy case is the extreme one, and the reason the moderate one is hard is a series that stops converging at exactly one half.

rounded to whole bitsunroundeda resample, 400 near pairs0.860.9340 pairs at stay 0.90.820.928 pairs at stay 0.90.710.88400 pairs at stay 0.70.680.86400 pairs at stay 0.50.620.790.5: no prediction200 test pairs, 16 directionsdashed: a coin flip When the algorithm is a table

The ties a rounded matrix makes

Measure how far each optimal alignment is from a tie — the smallest change to any one cost that makes another alignment win — and it predicts which alignments a refitted substitution matrix will move. A resample of the same corpus moves 30 of the 63 test alignments that sit on a tie and 3 of the other 137. A matrix fitted to a different divergence moves alignments far from a tie as well, and the prediction weakens to a chance of 0.62. And a third of the alignments were on a tie only because the matrix was rounded to whole bits — fitted without rounding, 15 of 200 are, and every prediction improves.

0.111010010³10³10⁴floor, in countsarrivals in the shard, nround-robin — n^1.02hashed — fit refusedresidual 2.7%slope 5.2 → 1.19k = 32 · 40,000 arrivalsthe table holds 1.33 of a hashed shard's keys and 0.01 of the stream's The floors

A floor with two variables in it

Under round-robin a Space-Saving summary's floor is 0.0203·n^1.018 over a hundred-and-twenty-eight-fold range of shard size, worst residual 2.7%. Under hashing the same measurement has no exponent at all — the local slope runs from n^5.17 to n^1.19 — and a least-squares line through it reports n^1.73 at a 441% residual.

100.0010.010.1fraction of the text proposedlog_4 n = 7.101234,56,7seed length, characters · errors allowed abovepattern 24 · text 20,000 · four symbolsrings: the closed form What is taught wrongly

The filter that proposes everything

Seed-and-extend saves two thousand times the work at zero errors and costs more than doing nothing at four. Between them the selectivity falls through the floor, and where it falls is set by two numbers that can be computed before the filter is run — one of which does not contain the length of the text at all.

124816nonephrasescap on the copy chainphrasesworst chainEnglish-like · 8,192 charactersz 156 to 7,351 The data that is not a number

A parse that will not follow a long chain

A greedy self-referential parse bounds the copy depth by nothing at all — thirty-two copies of a text give a position costing twenty-two phrase follows. Restricting every phrase to sources no deeper than D holds it at D, and the whole question is what that costs.

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