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The thread: Measured, not assumed — page 4

Page 4 of 9, continuing the same thread in the same order.
10³10⁴capacity Wsubproblems given a valueBottom-up table · 1.00Top-down, reachable only · 0.14one unit = one subproblem given a valuesubproblems given a value, W from 200 to 3200 What is taught wrongly

A table wider than its input

The knapsack table has (n+1)(W+1) cells and is called polynomial. Adding one character to the input doubles it — across four settings the table grows sixty-four times while the input it is written from grows by half.

0%25%50%75%100%1.5×shareheadroom over the mean occupancyoverflowingstanding idledrifting · 4.0 s window · 100 Hz47% overflow at the mean What the libraries do

Sized for a rate that does not hold still

A four-second window on a stream at a hundred arrivals a second holds four hundred items on average and between 105 and 2,169 when the rate moves. An allocation set at that average overflows at 47 per cent of instants while 35 per cent of it stands empty, which is the same decision failing in both directions at once.

key 05,416short by 4,163key 1462short by 4,154key 31short by 2,099key 71short by 943key 19401short by 3counter held (bar) against true count (tick)12 counters · 768 bits · no randomnessbound N/(k+1) = 4,615 Structures

The items that survive k counters

Misra-Gries keeps k counters, decrements all of them on a miss, and never returns a count above the truth — with no hashing, no randomness and no failure probability. At equal state it is more accurate than the randomised sketch on the question both are usually asked, at every size measured.

1,00010,000110occurrences12481632characters in the collection · copies aboveall occurrencessecondaryprimarypattern "ss is un" · z = 1561 primary · 31 secondary The data that is not a number

The occurrences that cross a boundary

One pattern, thirty-two copies of a text, thirty-two occurrences. The search finds one of them and produces the other thirty-one by arithmetic, and the count it finds is the same one at two copies, at eight and at thirty-two — the searching does not grow when the answer does.

10010³10⁴10⁵errors allowed, kacts01234index walkthe whole table4,000 characters · m = 16 · 4 symbolscrossing at k = 4 What a bound is

The branches an error opens

The tree multiplies by 19.6 for the first error, 12.3 for the second, 10.3 for the third and 8.8 for the fourth. A branching factor of four on a sixteen-character pattern would predict sixty-four, and the gap between sixty-four and eight is the intervals emptying.

0000200121200the parenthesis sequencethe minimum excess of each block of 213 blocks · lookup table 72 bits, sharedblock 2 · 24 parentheses13 blocks What the machine does

The table that fits inside a block

A block of six parentheses has sixty-four possible shapes and twenty-eight questions can be asked about each, so all 1,792 answers fit in a table of 9,408 bits — computed once, shared by every structure of that block length, and never counted in any of their sizes.

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 When it does not fit

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.

every orderthe named inputsmean0102030comparisonsInsertion sort7 to 28 · named inputs reach 28Merge sort12 to 17 · named inputs reach 16Heapsort21 to 29 · named inputs reach 27Quicksort, first-element pivot13 to 28 · named inputs reach 28Quicksort, median of three25 to 29 · named inputs reach 2540,320 orders of 8 distinct elements3 worst cases unnamed Counting

A count over every input

Run five sorts on every one of the 40,320 orderings of eight elements and read off each one's best, mean and worst comparison count. Then mark where the inputs a benchmark generator names — sorted, reversed, nearly sorted, random, few unique — land. For merge sort, heapsort and quicksort with a median-of-three pivot, the worst case is an ordering none of them produces, and for the last of the three every named input lands on its best case.

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 When the algorithm is a table

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.

universe 0…11 · prefixes of 6 · candidate keeps 9 bitsthe two prefixes that collideA01234567891011B01234567891011first differencesame state — the candidate stores "7" after boththen both read the same suffix -1, -2, 12, 13, 14true median of A3true median of B4and one answer for both924 prefixes · floor ⌈log₂ C(12,6)⌉ = 10 bits10 bits collide on none The floors

A floor one pass cannot get under

An exact one-pass selector must reach a different memory state for every prefix it might have read, and the pigeonhole that proves it is small enough to perform — nine hundred and twenty-four prefixes through a nine-bit state, the collision produced, the suffix that separates it, and two true medians it cannot both return. Ten bits collide on none, so the bound is exact — and a second pass walks under it by a factor of ninety.

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.

HyperLogLog1024/1024 identicalCount-Min256/256 identicalbottom-k128/128 identicalMisra-Gries20/36 identicalfraction of the state that merged to the identical value→ 1,974→ 10,291→ 2,322→ 8,766two streams of 30,000 · 2,007 distinct keys in the union3 of 4 merge exactly Structures

The summaries that add

Two sketches built over two streams and merged are, for three of the four structures here, byte for byte the summary the concatenated stream would have produced. For the fourth the guarantee survives and the state does not, and calling both properties mergeability hides the difference that matters.

k = 0k = 1k = 2k = 3k = 4k = 5k = 6q = 223211917151311q = 3221916131074q = 4211713951-3q = 520151050-5-10q = 6191371-5-11-17q = 81791-7-15-23-31a shaded cell is a threshold of zero or less: every window proposedt = m + 1 − q(k+1) · m = 2411 collapsed cells The data that is not a number

The q-grams an error cannot destroy

A pattern of twenty-four characters holds twenty-one four-grams. Two errors can destroy at most eight of them, so any occurrence with two errors still shares thirteen — and a filter that keeps only the windows sharing thirteen proposes 104 of 3,977 and computes 16,744 table cells instead of 96,000.

101001,00010010³10⁴10⁵characters of pattern in the setcomparisons before the scanboth rules, exactlybad character onlythe 1979 shift functionsfour symbols · patterns of 10137.35x at 128 patterns What a bound is

The table that walks every pair

The exact shift rules cost 769,724 character comparisons to build for 128 patterns and the published ones cost 5,604. The scan they are both built for reads 41,580 characters, so one of the two constructions is eighteen times the work it is there to save.

1,00010,00010³10⁴characters of textcomparisons, built and scannedcrossing at n = 32,000both rules, exactlythe 1979 tables2 patterns of 10 · four symbolscrossing n = 32,000 What the libraries do

The rule that pays on a long enough text

With two patterns, the cheap tables cost 106 steps and the scan reads 13,084 characters; the exact tables cost 594 and the scan reads 12,306. Below thirty-two thousand characters the cheap tables win the total, above it the extra skipping pays for them, and with thirty-two patterns there is no crossing at all.

40%60%80%100%8121624324864elements sortedshare of the known worst case the climbs reach, on averageInsertion sort · 20 of 24Merge sort · 24 of 24First-element quicksort · 0 of 24hollow: none reached it24 climbs a size · 100 swaps per elementworst cases known exactly Counting

The worst case found by climbing

A search that swaps two elements at a time and keeps whatever does not lower the count finds the worst case of all five sorts at eight elements, where every answer can be checked. At sixty-four it finds merge sort's worst case every time and reaches 39% of first-element quicksort's — whose worst case is sorted input, the most famous bad input there is. Checking a search where the answer is known certifies it only there.

10³10⁴10⁵words in the vocabularysubproblems given a valueevery table in full · 1.16best so far · 0.99answer known · 0.96one unit = one subproblem given a valuesubproblems given a value, n from 250 to 2424 When the algorithm is a table

The bound the search finds for itself

A spelling checker that computes the full edit-distance table against every word in a 2,424-word vocabulary fills 156,714 cells for each misspelt query. Bound each table by the best distance found so far, and abandon it the moment a whole row exceeds that bound, and the same search fills 40,273 and finds the same words. Meet the candidates nearest in length first and it fills 26,203, starting a table for exactly the words a search that knew the answer in advance would start. The last factor of 1.7 is the price of not knowing, and it is largest when the misspelling is smallest.

1234152abcbcbccbsolid: a transition on a character · dashed: a suffix or failure link8 states ≤ 9, 9 transitions ≤ 11 Structures

Every substring, in fewer states than substrings

A text of 512 characters has 129,416 distinct substrings. A machine that recognises every one of them, and nothing else, needs 831 states — and the bound it is under, 2n − 1, is reached exactly by a string one line long.

the same 16 patterns, one of them shortenedshortest 1658.4%mean shift 3.15shortest 1262.8%mean shift 2.99shortest 869.7%mean shift 2.64shortest 680.5%mean shift 2.35shortest 493.4%mean shift 1.98shortest 399.2%mean shift 1.71shortest 2100.0%mean shift 1.3716 patterns, 15 of them 16 characterstext 20,000 What is taught wrongly

The ceiling the shortest pattern sets

A matcher that skips is described as faster than one that reads every character, and the description leaves out what decides it. No shift can exceed the shortest pattern in the set, so adding one two-character pattern to fifteen of sixteen characters takes a run from reading fifty-eight per cent of the text to reading all of it twice.

shard 1 · 120 → 119shard 2 · 121 → 120shard 3 · 120 → 120shard 4 · 121 → 120shard 5 · 119 → 118shard 6 · 124 → 123shard 7 · 121 → 120shard 8 · 122 → 121floor, in arrivalsnaive: tail ÷ kfixed pointmeasuredstationary Zipf · round · k = 321.006× the measured floor One pass, and no room

The floor a histogram already knows

A summary of thirty-two counters settles at a smallest counter of 119, and the number can be computed from the shard's key frequencies before a single counter is allocated. The obvious way to compute it is wrong by a factor of two, and the reason is that the heavy counters carry no error at all.

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.

ε = 0.0515 → 29 (1.93×)ε = 0.0238 → 73 (1.92×)ε = 0.0177 → 136 (1.77×)ε = 0.005152 → 270 (1.78×)ε = 0.002397 → 674 (1.70×)reportedoccupied at the peak20,000 arrivals · lognormalpeak = resident + period, to 14% The other axis

The tuples a summary does not report

A Greenwald–Khanna summary at ε = 0.01 answers `tuples` with seventy-seven. Watched through the run it holds a hundred and thirty-six. The gap is the compression period, it is 1.70 to 1.93 times across every tolerance measured, and it is the number a deployment has to allocate.

·1, 4b1, 3b1, 2b4, 2a4, 2a2, 2a4, 2b4, 2b4, 2b4, 2patterns abbb, baab, bbab · shortest 4d₁, d₂ under each node · a filled node ends a patternthe reversed-pattern trie10 nodes The data that is not a number

The shift somebody published

The exact rules for shifting a multi-pattern window are a definition that quantifies over every pattern at every offset. The 1979 rules are two tables read off the trie's own failure links, they are computed in one pass, and on this pattern set they agree with the definition at every node.

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.

0.0010.010.11125102050fraction of positions reshuffled, pmean comparisons, in multiples of the mean on random inputFirst-element quicksort, 81.9×Median-of-three quicksort, 41.4×Insertion sort, 2.0×Merge sort, 1.0××: unshuffled2,048 elements · 12 shuffles a point1 = the mean on random input Counting

A worst case ten positions wide

Sorted input costs first-element quicksort 2,096,128 comparisons on 2,048 elements, 82 times its average. Reshuffle about eleven of the 2,048 positions and the cost halves — and it takes about ten at 128 elements, and between ten and thirteen at every size between. Reversed input costs insertion sort twice its average, and reshuffling half the positions still leaves 71% of the work. A worst case is a place in the space of inputs, and the two famous ones are places of very different sizes.

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.

10³10⁴10⁵words in the vocabularysubproblems given a valueevery table in full · 1.16a trie, no bound · 0.99best so far · 0.99a trie, best so far · 0.82one unit = one subproblem given a valuesubproblems given a value, n from 250 to 2424 When the algorithm is a table

The columns the candidates share

Three thousand tables against one query, and most of them begin the same way. Stored as a trie, the 2,424-word vocabulary has 7,710 distinct prefixes holding 17,239 letters, and a search that computes one column per prefix reads 61,449 cells against 156,714 — before it applies any bound at all. Apply the bound at a prefix instead of at a word and it reads 16,958, beating a list search that was told the answer in advance.

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.

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