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The thread: Measured, not assumed — page 6

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right to left8302.74xleft to right8772.89xone per piece4471.48xthe scheme3031.00xinterval extensionsall four found the same 6 positions15 characters · k = 22.74x apart The data that is not a number

The search that starts in the middle

The same pattern, the same six occurrences, the same index — and 830 interval extensions, or 303, according to which end the search begins at. A pattern cut into three pieces has an error-free one, and only a search with two ends can start there.

11010³cap on the depthphrases9.8% more phrasesevery occurrence consideredthe earliest only4,096 charactersworst at cap 4 What a bound is

The cap an automaton cannot see

A state of a suffix automaton stands for a set of occurrences and hands back one of them. So a capped parse driven by it can ask whether the earliest occurrence is shallow enough and cannot ask whether any occurrence is — which costs up to 9.8% of the phrases, and only at the caps that bind.

steps cost one to ninesteps cost one01,0002,0003,000cells expanded, mean over the gridsA* from one endalways shortestalways shortesttwo-ended Dijkstraalways shortestalways shortesttwo A*, separate estimateswrong on 12always shortesttwo A*, stop on either keyalways shortestalways shortesttwo A*, averaged potentialalways shortestalways shortest40 grids a bar, 2,500 cellsestimate: straight-line cells Two parameters

Two estimates that must agree

Run A* from both ends of a query at once, each search guided by its own straight-line estimate, and stop by the rule that is correct for two-ended Dijkstra. On 40 weighted grids it expands 1,002 cells on average and returns a longer path than the shortest on 12 of them. Give both searches one potential, half of one estimate minus half of the other, and the same rule is correct again — on all 40 grids, for 1,041 cells. Two estimates that measure different things cannot share a stopping rule until they are made to measure the same thing.

probe 1: 0–1 absent9 still to ask10 of 10 as good as anyprobe 2: 0–2 absent8 still to ask9 of 9 as good as anyprobe 3: 0–3 absent7 still to ask8 of 8 as good as anyprobe 4: 0–4 present6 still to ask7 of 7 as good as anyprobe 5: 1–2 absent5 still to ask6 of 6 as good as anyprobe 6: 1–3 absent4 still to ask5 of 5 as good as anyprobe 7: 1–4 present3 still to ask4 of 4 as good as anyprobe 8: 2–3 absent2 still to ask3 of 3 as good as anyprobe 9: 2–4 present1 still to ask2 of 2 as good as anyprobe 10: 3–4 absentdecided1 of 1 as good as any5 vertices · 10 pairs · 59,049 states solvedsolid: present · dotted: absent · coloured: this probe The floors

Every pair must be asked

Ask whether a six-vertex graph is connected, one pair of vertices at a time, and the best possible algorithm needs all fifteen questions on its worst graph. The claim that this holds for every monotone property of graphs is a conjecture fifty years old. At four vertices it can be settled completely: all 2,046 properties that do not depend on vertex names need every pair. Name one vertex, and the count drops from ten to four.

0.000.250.500.751.00stationarydepartingburstydriftingprediction ÷ measurement, as a factorshare of the top k that moves between halves8 shards · round · k = 327.1× out where the statistic reads 1.00 What is taught wrongly

The histogram that cannot see the order

A prediction accurate to one per cent on three streams is seven times out on the fourth, and the input that fails is the one every capacity plan is built from. A statistic computed from the same input says in advance which case is in hand — and misses one of the two ways it can go wrong.

folded in one at a time2,616 tuples17 ranks outcombined pairwise, in a tree3,637 tuples17 ranks outfolded in, last shard first2,615 tuples17 ranks outtuples kept, and worst rank error against a promise of 10032 shards · ε = 0.01 · high-biased · round1.39× the space, 0 ranks of answer Structures

The shape that moves the bill

Thirty-two quantile summaries combined pairwise keep 3,637 tuples and the same thirty-two folded in one at a time keep 2,616, for answers that differ by nothing at all. The counter tables measured for the same thing do the opposite — their order moves the answer and leaves the space alone.

plain51,600block-coded47,6687.6%run-coded60,805-17.8%plain, uniform51,600block-coded, uniform50,2422.6%run-coded, uniform67,121-30.1%bits · the lower three are a permutation with no structure3,612 points7.6% against 2.6% The index that replaces the text

A block, a class and an offset

Replacing each block of a bit vector by how many ones it holds and which arrangement it is takes 7.6% off the grid. On a permutation with no structure at all it takes 2.6%, so five of the seven points are the data and two of them are the encoding.

1,00010⁴10⁵10⁶charactersoperationscharacter comparisonstransitions and linksrepetitive · 8 copies346 phrases at 8,192 The data that is not a number

The parse in one pass of the text

The same parse, phrase for phrase, from 1,324,336 character comparisons or from 25,420 transitions and suffix-link steps. One of those numbers grows with the text and the other grows with its square, and the difference is why every measurement about a depth cap here was taken on a few thousand characters.

24816326412825510⁶10⁷average out-degreecounted workBellman–Ford from every sourceJohnson's reweightingFloyd–Warshall256 vertices, every answer comparedwork: relaxations + heap comparisons Two parameters

One Bellman–Ford buys every Dijkstra

A directed graph of 256 vertices with a third of its arcs negative needs shortest paths between every pair. Running Bellman–Ford from every source costs 25.8 million counted operations on the densest graph drawn; running it once, repricing every arc by what it found, and then running Dijkstra from every source costs 13.1 million, and the one Bellman–Ford is under one per cent of that. Floyd–Warshall's 16.8 million is never the cheapest count on the plate. On the sparsest graphs the repeated Bellman–Ford wins, because its early exit makes nine passes rather than 255.

length of the longer list, nshorter, m1234561234561floor 12floor 22floor 23floor 33floor 33floor 33floor 34floor 45floor 45floor 56floor 55floor 56floor 67floor 67floor 77floor 78floor 79floor 89floor 810floor 911floor 10the optimum is one above the floorthe floor is reachedevery cell solved exactly · m ≤ n ≤ 6large: the optimum · small: ⌈log₂ C(m+n, m)⌉ The floors

The floor a merge cannot reach

Merging two sorted lists of five keys each has 252 possible outcomes, so counting says eight comparisons might do. Solving the game says nine are needed, and on equal lengths the shortfall keeps growing, as half the logarithm of the length. Averaged over random inputs, though, the same count is missed by a tenth of a comparison. The counting floor is nearly exact on average and wrong in the worst case.

1.821251.951.431752.731.302253.521.292503.911.183755.861.005128.00longest block ÷ shortestarrivals per block, and bursts per blockpart of a bursta whole numberbursts of 64 · 8 blocks · 20,000 arrivalsdispersion 0.57 and a ratio of 1.00 What is taught wrongly

The boundary that hides the burst

A window whose blocks hold five hundred and twelve arrivals reports a perfectly even stream — every block the same duration to the tick — while the arrivals it is retiring have an index of dispersion of 0.57. Move the block to a hundred and twenty-five and the same stream varies by 1.8 times.

weighted path lengthΣ wᵢdᵢ — what Huffman minimisescuts takenΣ over the mergesdamageworst error leftchaintreesmallest-firstlargest-first543k147k123k738k309254259388323148183403Misra-Gries · 32 shards · hashedleast path smallest · least damage balanced Structures

The fold that minimises the wrong thing

A fold charges per level and a survivor pays the cuts on its path, so the bill looks like a weighted external path length — and Huffman's construction minimises that quantity by proof. Built and measured on thirty-two uneven shards it does minimise it, 181,407 against a balanced tree's 200,000, and leaves more damage than the tree does.

twelve essays — runs0.491generated, matched0.109twelve essays — phrases0.213generated, matched0.046eight source modules — runs0.461generated, matched0.122eight source modules — phrases0.210generated, matched0.050ten revisions of one file — runs0.126generated, matched0.113ten revisions of one file — phrases0.067generated, matched0.047per character24,576 characters a partdivergence 0.02 The data that is not a number

A corpus that was not generated

Every collection in five strands has been copies of a generated text with a fraction of its characters replaced — three numbers, one dial. Here is one that was not — twelve essays, eight source modules, ten revisions of one file — measured beside the model of it.

run together867 bits a characterone separator877 bits a charactera separator each897 bits a charactergenerated, run together215 bits a characterdistinct symbolstwelve essays0.0% premium The index that replaces the text

Documents that are not the same length

A separator per document costs a whole bit per character — on a generated collection whose alphabet is 21 symbols, where adding a few crosses 32. Real prose has 87 symbols and sits 41 short of the next power of two, so the same separators cost 0.04%.

0200400600800cells expanded per query, meanno landmarks897four near the centre423four at random209farthest-first141the corners1268 maps × 150 queriesdots: each map's mean Two parameters

Where the landmarks stand

Four tables of exact distances, each from a chosen cell, turn a straight-line estimate that barely helps on rough terrain into one that cuts a search by a factor of six. Averaged over 1,200 queries on eight maps, the same four tables expand 126 cells a query when their cells are the map's corners and 423 when they are near its centre. The standard choice, each landmark as far as possible from the ones before, expands 141 and loses to the corners on all eight maps. Moving four landmarks to the right places buys more than doubling their number.

0%25%50%75%Merge sort0.0% · 1.03× the floorQuicksort, random pivot0.0% · 1.16× the floorQuicksort, first-element pivot0.0% · 1.18× the floorMerge sort with an insertion cutoff0.0% · 1.31× the floorInsertion sort0.0% · 9.83× the floorShellsort9.8% · 1.31× the floorQuicksort, median-of-three15.4% · 1.18× the floorHeapsort26.2% · 1.45× the floorBubble sort49.2% · 9.83× the floorSelection sort59.9% · 7.77× the floorshare of comparisons whose answer was already implied256 elements, random orderlight: repeats nothing Counting

The questions a sort asks twice

Selection sort makes 32,640 comparisons on 256 elements and 19,561 of them have answers it already holds. Remove every one and it still makes 7.8 times the information floor, because a question can be new and nearly worthless: insertion sort repeats nothing at all and removes 0.72 of a bit per comparison where merge sort removes 0.96. And bubble sort, less its repeats, makes exactly insertion sort's comparisons — at every size.

01234characters of context the sort reads, kbits a symbol after move-to-front012481632the full transform, 1.802ties kept in text orderties sorted by symbol: the stream alonethe text's own floormodel: order 0 after move-to-front · Words from a fixed vocabularyexactly the transform from k = 24 The floors

The order inside a tie

Sort the rotations of a text by their first four characters rather than by everything that follows, and the output clusters slightly better than the full Burrows–Wheeler transform: 1.780 bits a symbol against 1.802, from 63% of the character reads. It also costs nothing to undo. The prediction that a shorter context leaves ties for the inverse to pay for was wrong. The cost to undo comes from how a tie is ordered, not from how long the context is, and at k = 0 the wrong tie rule is exactly the sort.

counter tablesworst error, ratio to bestquantile summariestuples kept, ratio to bestchain2.18× (323)1.10× (2,807)tree1.00× (148)1.26× (3,211)smallest-first1.24× (183)1.28× (3,278)largest-first2.72× (403)1.00× (2,556)32 shards · hashed · k = 32each column against its own best shape Structures

The shape one structure will not fold

Folding thirty-two shards largest-pair-first keeps 2,556 quantile tuples against a balanced tree's 3,211 — a fifth of the space saved. The same fold on the counter tables beside them leaves 403 counts of error against the tree's 148. A deployment holding both cannot fold once and be right twice.

1 — no differencethe whole array reviewed1.03×only this batch reviewed1.22×a block window, which does alias1.48×20,000 arrivals · period 50 · cyclethe instrument reads 1.48 where an alias is known to be What is taught wrongly

The sampler that cannot alias

A block window's boundary fires on an arrival count, and a stream whose burst repeats every sixty-four arrivals is reported as perfectly even by a block of five hundred and twelve. A quantile summary compresses on an update count, which is the same arrangement. Swept against three periodic value processes and their shuffles, it does not alias — and the reason is one line of arithmetic rather than a lucky sweep.

parentheses131,07216.7%select support115,71214.8%superblock minima16,3982.1%superblock table80,97010.3%block minima147,46518.8%block tables291,25537.2%the partsbitstotal 782,872 · the segment tree 2,228,224shared lookup table 82,944, not counted65,536 values · openings in index order, found by select2.85x smaller The other axis

Two bits a value, and what undoes them

The parentheses of a range minimum over 65,536 values are 131,072 bits. Everything that makes them answerable is 651,800 more — five times the payload — and one encoding choice nobody quotes accounts for a sixth of it on its own.

① prose144,617 chars12 docs · median 12,335① code106,283 chars8 docs · median 13,150① revisions75,658 chars10 docs · median 7,774② essays1,004,822 chars82 docs · median 12,335② captions549,210 chars2,214 docs · median 230② history174,282 chars14 docs · median 12,963① the first freeze · ② the second6 parts The data that is not a number

A million characters of the same thing

Every measurement this collection has published about real text was taken on twenty-four thousand characters, because the phrase count was quadratic. It is linear now, so here is the same corpus at forty times the size — and what forty times does to its own numbers.

125102050100200100observed queries the selection sawcells expanded, meanon fresh querieson the sample it was chosen fromthe cornersfarthest-first8 maps × 150 fresh queriesflat lines read no queries Two parameters

What the queries know that the map does not

A greedy rule that chooses landmark cells by rerunning a sample of past queries needs two hundred of them to draw level with a rule that reads only the map — and what it finally chooses, on map after map, is the four corners. Give the queries a destination instead of scattering them, and twenty are enough to beat the corners by 29% on eight maps out of eight. A query log is worth reading exactly to the extent that it is not uniform.

32641282565121010010³10⁴elements sortedword operations a comparison must cost to break evenSelection sort, pairsSelection sort, closureBubble sort, pairsBubble sort, closureHeapsort, pairsHeapsort, closureQuicksort, median-of-three, pairsQuicksort, median-of-three, closuresolid: pairs asked · dashed: everything impliedrandom input · counted exactly Counting

The price of remembering an answer

Give selection sort a table of the pairs it has already compared and it makes 16,805 comparisons on 256 elements instead of 32,640. The table pays for itself once a comparison costs more than 3.1 word operations, and that price stays near three at every size measured. Give it the full closure of everything its answers imply and it makes 13,079. The closure pays only past 425 operations a comparison at that size, and past 1,659 at twice the size. The cheap memory is worth having on long keys. The complete one is worth having almost nowhere.

01002003004005006007008009001000roundloads 1.0×blockedloads 1.0×hashedloads 17.6×worst error over the heaviest keyschaintreesmallest-firstlargest-first32 shards · k = 32 · 40,000 arrivalseven 1.00× · uneven 2.7× What is taught wrongly

A parameter that waits for another

Four merge fold shapes over thirty-two evenly loaded shards leave errors of 665, 667, 665 and 667 — a fifth of a per cent apart. Give the same four shapes shards whose loads span seventeen-fold and they leave 148, 183, 323 and 403. The parameter did nothing until a second parameter moved, and every measurement that fixed the second one saw nothing.

one index35,3354.31 bits/charthe forward half35,3354.31 bits/charthe reverse half35,3354.31 bits/charboth, which is the structure70,6708.63 bits/charbits8,192 characters · sample 322.00x one index The other axis

The structure that was supposed to halve

A bidirectional index holds the transform of the text and the transform of its reversal — 35,335 bits each, 70,670 together, exactly twice one index. The deferral that named it hoped it would stop the index doubling. It does not remove the doubling; it reuses it.

10⁴10⁵10⁶characters in the collectionshuffled ÷ real, phraseswhat the order is wortha million characters of prose · phrases per characterx1.37 over the sweep What is taught wrongly

The half of a fall that is the logarithm

Phrases per character on a real collection of essays fall by a factor of 2.34 as it grows. A shuffle of the same characters falls by 1.70. Nearly three quarters of the movement is arithmetic, and no definition of the measure says so.

0.000.250.500.751.0005001,0001,500by key60% forward70% forward80% forward90% forwardone-endedhow the expansions are sharedcells expandedseparate key ÷ route40 grids, 2,500 cellsdashed: how close the separate key got to firing Two parameters

A stop that is correct and never sooner

A two-ended search can stop when the two frontiers' keys together reach the best route found, and it can also stop when either frontier's own estimate reaches it alone. Both rules are safe, so a search may use whichever fires first. On forty weighted grids the second never fires: at the moment the first one stops the search, the larger of the two own-keys stands at 64% of the route. The extra rule costs 60% more counted work and a second priority queue to find that out.

11010010³10⁴10⁵10⁶k, the number handed backcomparisonstournament, measuredfloor that names the outputsbest earlier floorn = 65,536, one shuffled inputthe floors are worst-case, the count is one input Counting

The comparisons that name the answer

Returning the 4,000 smallest of 65,536 keys in order needs 61,536 comparisons to eliminate the rest and 42,100 to order the ones returned. That was the floor, 103,636, and a tournament made 125,388. What the floor never charged is saying which 4,000 come back. Charge that, and the floor is 125,341. On the same input the tournament is 47 comparisons above it, and for every k up to a hundred it is exactly on it.

6080100how far a full leaf looks for room, in siblingsmean leaf fill, per cent12481664randomascendingdescendingascending, 1% late131,072 keys · leaves of 64 · fan-out 64a reach of 1 is the adjacent-sibling rule When it does not fit

A key passed along the row

A full B+-tree leaf that offers a key to its immediate neighbours before splitting leaves an ascending stream 67.2% full. Let it look one sibling further and the same stream fills its leaves completely, 2,048 leaves where there were 3,048, for about the same key moves. On random keys each doubling of the reach adds a few points of fill and more writes than it saves, and at the whole parent a key travels sixteen leaves on average. On a stream that is mostly in order the same reach costs almost nothing.

0.60.81.01.21.41.61.84φ share 0.908φ share 0.8516φ share 0.8032φ share 0.7564φ share 0.70predicted ÷ measuredshards, mlevel floors, uncorrectedleaf floorslevel floors, correctedk = 32 · hashed · 40,000 arrivalsworst 22% against 73% and 46% What is taught wrongly

The floor a merge does not settle at

Compute a fold's level floors from the shard histograms and the prediction over-shoots by 1.73. A merged summary's floor is not the floor a summary settles at on the same arrivals — it is 0.90 of it at four shards and 0.70 at sixty-four, straight in log₂ m at a 3% residual, because merging preserves the heavy counters and never runs their eviction cascade.

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