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The thread: The model has parameters — page 2

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depth 0190.1%depth 19642.9%depth 24,92915.0%depth 37,00921.4%depth 48,47325.9%depth 56,24119.0%depth 63,31010.1%depth 71,2503.8%depth 84221.3%depth 91240.4%depth 10270.1%positions32,768 characters · 1172 phrasesmean 3.98 What the libraries do

The number that would choose a cap

A depth histogram is one linear pass — 3.16 operations a character over thirty-two thousand of them — and it says the whole text sits at a mean depth of 3.98 with a worst of ten. Nobody prints it, and every choice of cap in this collection was made without it.

trusts the estimateinsured ×2insured ×8always the index11.5235expected regret, logarithmic · label: worst within three standard deviationsrho 14.42.51.11.0rho 215.17.82.22.0rho 413.57.54.04.0rho 1616.016.016.016.0σ = 1.5, median error e^-1, 65,536 rowsexact over the error distribution When it does not fit

What insurance against an estimate costs

A planner that trusts its row estimate expects to pay 1.057 times the better plan and risks 7.76. One that insures itself by halving its estimate before it decides expects 1.057 and risks 4.10 — the insurance is free. At a read ratio of sixteen the same insurance costs six per cent in expectation and makes the worst case worse. Whether a conservative planner is paying a sensible premium depends on two numbers the planner can measure and usually does not — its device's read ratio and the direction its own errors run.

alignments on a tiedistinct costs · predictionwhole bits633 of 6 · 0.86half bits793 of 6 · 0.91quarter bits414 of 6 · 0.93a grain of 0.2235 of 6 · 0.89eighth bits156 of 6 · 0.93a grain of 0.05205 of 6 · 0.95unrounded156 of 6 · 0.93200 test pairs, 16 directionsbar: alignments within 0.01 of a tie When the algorithm is a table

The lattice that decides the ties

Rounding a fitted substitution matrix to whole bits puts 63 of 200 alignments on a tie where the exact fit puts 15. Rounding to half bits — a finer grain, and the obvious repair — puts 79. What tracks the ties is not how fine the lattice is but how many of the six fitted costs it keeps apart: whole and half bits both leave three, an eighth of a bit leaves all six, and matches the exact fit exactly.

10010³10⁴10⁵10⁶⌊1/2ε⌋ = 50151025501002005001,000tuples, and tuples examinedcompression period, in updatestuples examinedpeak tuplesresident tuplesworst rank errorε = 0.01 · 20,000 arrivalspeak 10× · work 72× · answer 1.21× One pass, and no room

The period that is not a promise

Greenwald–Khanna's ε appears twice — once as the rank tolerance the structure promises, and once as ⌊1/2ε⌋, the number of updates between compressions. Unhook the second from the first and sweep it across a thousand-fold range. The tuples held move by 10%, the worst rank error by 21%, the peak by ten times and the housekeeping by seventy.

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.

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.

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.

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.

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.

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.

234681016110size ratioblock transfers · levelsa range of 100 keyslevelsan absent point lookup1,048,576 keys, 5 bits a keyfilters answer one of these two When it does not fit

The read a filter has no key for

A Bloom filter on every run of a log-structured store turns a lookup for a missing key from a read of every level into a fraction of one — 0.72 transfers across eight levels at five bits a key, and 0.00027 at twenty. A range query over the same store reads nine transfers at five bits and nine at twenty, because a filter answers whether one named key is in a run and a range has no key to name.

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.

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.

0396,049792,0981,188,1481,584,1970326496128queries answeredcounted work, cumulativebreak-even at 2.7 queriesBellman–Ford from each sourceone reweighting, then Dijkstra256 vertices, 2,009 arcsanswers compared entry by entry Two parameters

How long a reweighting stays true

Johnson's one Bellman–Ford run costs under one per cent of an all-pairs computation because it is divided over every source. Asked one query at a time it is divided over nothing, and it still repays itself after 2.7 queries — because the preparation is one Bellman–Ford and every query saves a third of another. What decides the trade is not the query count but whether the graph holds still: at half a per cent of arcs redrawn between queries the stored potential is worth exactly nothing, and its life is geometric at a per-arc failure rate of 6.6%.

how far the top k movedthe order warninghow far the floors are from doublingthe regime warningstationary Zipf0.160.53 (54%)one key floods a stretch0.170.59 (54%)a heavy hitter that stops0.170.65 (52%)the popular keys drift0.9134.00 (0%)k = 32 · 40,000 arrivalsin brackets: the leaf model at sixty-four shards What is taught wrongly

The warning that is silent for the right reason

The statistic shipped to warn that a merge prediction is about to fail reads 0.160 on a stationary stream, 0.172 on a bursty one and 0.909 on a drifting one. It was asked to be looked at again because it does not catch a burst. It does not, and the reason is that on a burst there is nothing to catch.

0.010.11errors allowed, kshare of windows proposed01234q = 3q = 4q = 5a shaded dot is a collapsed thresholdm = 24 · n = 4,000 · 6 planted1 collapsed rows What is taught wrongly

The threshold that reaches zero

At q = 5 and four errors on a twenty-four-character pattern the filter demands zero shared q-grams, proposes all 2,977 windows, and computes 986,266 table cells where filling the whole table would have cost 96,000. The failure is arithmetic and is knowable before a character is read.

a text that repeats itself2.96xchain 13 · z 110English-like1.31xchain 10 · z 604four symbols, uniform1.28xchain 10 · z 816periodic, period 171.11xchain 9 · z 1,285phrases under a cap of four, against no cap4,096 characters each2.96x to 1.11x What is taught wrongly

The cap that binds on one text and not another

A periodic text looks like the one made of chains and pays 1.11 times the phrases for a cap of four. A text that repeats itself pays 2.96. The guess is backwards, and the reason is that depth measures nesting rather than repetition.

0102030255075100125one sampled position in every …of both halves saved, per cent"about a sixth"the marks alone: 14.0%8,192 characters16.7% at one in 32 What is taught wrongly

A sixth of what, exactly

The saving from a counting-only reverse half is 30.5% at one sampled position in four and 14.0% at one in a hundred and twenty-eight. A number quoted without its sampling rate is a number about a setting somebody chose.

46812windowswindow lengthspanning a joinin no documentwith a separatorEnglish-like · 8 documents44 invented at m = 12 What is taught wrongly

The occurrences a join invents

Eight documents run together hold forty-nine eight-character windows that span a join, twenty-three of which occur in no document at all. Every index built over the concatenation reports them, and five essays of this collection paid that cost silently.

binary · sigma 22.00x2 against 4dna · sigma 42.67x6 against 16protein · sigma 204.76x42 against 200latin · sigma 264.81x54 against 26064-position interval2.00x to 4.81x What a bound is

Two at binary, five at twenty-six

The saving is a factor in the alphabet, so a two-symbol alphabet gets two. Approximate matching in this field is mostly done on DNA, which sits near the bottom of the list at 2.7.

1.0%10.0%0.11fraction of characters changedper characterthe real runsthe real phrasesgenerated runsgenerated phrasesten revisions of one file · 24,580 characters2.32% against 3.24% What is taught wrongly

The dial that has no setting

The generator has one parameter. The real version history's run count asks it for 2.3% and its phrase count asks for 3.2% — and the reason is not that the dial is badly calibrated. Real edits average 7.3 characters a block and generated ones average 1.04.

05e+31e+41.5e+40246bits kept in the low partbits⌊log₂(n/m)⌋ = 2low partshigh vector2,049 marks in 16,385 positions1 bits between the two roundings What a bound is

The flat bottom of a shallow curve

The low width is chosen as the floor of log of the universe over the count. Rounding it up instead costs one bit on five thousand, because the total is m·w plus n over two to the w and the minimum is where those two are equal.

025507510000.50011.502errors permittedfactor against the published loop64.5%55.8%74.7%the walk: 11xthe descentthe label is the shareof dead extensions6 patterns · sigma 2198x to 105x What a bound is

Flat in the budget, and not

One saving is eleven times at every error budget, because it is a property of the alphabet. The other moves between ninety-eight and a hundred and five, because it follows the share of extensions that find nothing. Two savings, two shapes, and neither line crosses the other.

110100k, the operators after the alternationstatesits DFA: 512its NFA: 30a literal's DFA: 11alphabet ab2^(k+1), exactly What a bound is

The exponential is in the expression

The subset construction on one family reaches two to the k plus one states, exactly and not approximately. A literal of the same length gives eleven. Both are regular expressions and the difference is that one of them asks the machine to remember something.

02040608000.50011.502errors permittedbranches that find nothing, per cent5.75x6.98x9.22xthe label is thesaving at that budget8 patterns · 8 characters31.3% to 74.9% What the libraries do

A looser budget wastes a larger share

More errors permitted means more work, and the fraction of that work which was never going to help rises with it — from thirty-one per cent at no errors to seventy-five at two. The saving is worth most where the search is most expensive.

020406080255075100125one sampled position in every …of the plain index, per centsize: 84.7%67.5 steps16,384 characters · 8-character patternssize falls, the walk lengthens Structures

Bits and steps on one frame

The size falls from ninety-nine per cent to eighty-six as the sampling thins, and the walk to a sampled position rises from two and a half steps to sixty-four. Neither line is the answer; the answer is a point on the pair.

10010³10⁴10³10⁴10⁵characters of textoperations, construction includedcrosses at 256the NFAthe DFA64 DFA statescrosses at 256 characters What the libraries do

Where the table starts paying

Five thousand and sixty-five operations before the first character, then one per character. Against nothing before the first character and thirty-nine per character. They cross at two hundred and fifty-six characters, and that crossing is what an engine's compile decision actually is.

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