Concept

Estimator — where it appears

A rule turning a summary's state into an answer, which can be varied without changing the state and is part of the structure's specification. Two estimators over one state can be optimal on different parts of a distribution, so which end of a bracket to report is a design choice.

Named by 31 essays across 6 fields — each of them below, with the objects they name alongside it.

1,00010,0000.1bits of state heldrelative errorHyperLogLog (-0.49)LogLog (-0.44)bottom-k (-0.38)50,000 distinct keys · 14 runs per point · truth counted exactlybest: 2.00% at 10,240 bits

The answer that is allowed to be wrong

Every algorithm on this site so far was checked for correctness before it was measured. A summary of a stream cannot be — the data goes past once and does not fit — so the error becomes a resource, bought with bits, at an exchange rate that is a measurement.

streaming · Sketch
one summary, k = 32769one summary, k = 25608 summaries of 32, merged536worst error over the top keys, in arrivalsconcentration 1.00 — the mean share of a heavy key held by one shard8 shards · hashed · balancedmerge 536 against matched 0

The state a merge is standing in for

A merge of eight summaries of thirty-two counters is wrong by 536 where one summary of thirty-two is wrong by 769, which reads as merging helping. One summary of two hundred and fifty-six counters — exactly what the eight were holding between them — is wrong by nothing at all.

streaming · Merge
0.1bits the register neededrelative errora = 2a = 1.5a = 1.2a = 1.1a = 1.05a = 1.02√((a−1)/2), predicted200 runs per base · n = 20,000 · exact counter needs 15 bits72.6% at 5 bits

Counting past what the register holds

Morris's counter counts ten million events in five bits by incrementing with probability 2 to the minus c. The estimate is exactly unbiased at every n, its relative error is 71%, and the base is a dial that trades one against the other at a rate of the square root of half of a minus one.

streaming · Sketch
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

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.

streaming · Sketch
01325leading-zero rank keptregister, 0 to 255estimate 7,107truth 7,368error -3.54%0 still empty256 registers × 5 bits = 1,280 bitspredicted ±6.50%

A count read off the leading zeros

Hash every key and watch for the longest run of leading zeros. Seeing k of them is evidence of about two to the k distinct keys — an estimator with a variance so large it is worthless, and the two devices that fix it are the whole of what a cardinality sketch is.

streaming · Cardinality
quantilepromisedone summarymerge of 8q = 0.5100.01418q = 0.920.0183q = 0.992.005overq = 0.9990.214overrank error, in items out of 20,000ε = 0.01 · high-biased · 8 shards, round2 of 4 quantiles over the single summary's promise

The promise that does not survive the tree

A high-biased quantile summary at ε = 1% answers the 99th percentile of twenty thousand values to within two ranks. Merge thirty-two of them and the answer is seventeen ranks out — inside the merged bound, eight times outside the one anybody was quoted.

wrong · Rank
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%

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.

streaming · Cardinality
0.010.10.11Jaccard index of the two setsrelative error of the intersectionas wrong as the answerinclusion–exclusionthe minimum valuesHyperLogLog p = 12, 20,480 bits · bottom-4,096, 131,072 bitstruth from a hash map over 200,000 keys

The error of a difference

Three sketches, each within a per cent or two of its own answer, subtracted into an intersection. At a Jaccard index of 0.82 the answer is 1.3% out. At 0.005 it is 146% out — the same three sketches, the same accuracy, a different question. The error never grew: it stayed a fixed fraction of the union, and the union stopped being the thing being asked about.

streaming · Cardinality
+1−1keys, most frequent firstΣ s(x)·f(x) = 9,034squared: 81,613,156true F2: 36,931,352121.0% outa polynomial of degree 4 · cash-register model · counts exactone register, 32 bits

The estimate that squares the stream

The length of a stream is a counter and the number of distinct keys is a register bank. The sum of the squared frequencies has nothing obvious to count — and one number, one sign per key, and a squaring get within 4% of it in a fortieth of the space.

streaming · Moment
answered 19.90%25%50%75%100%0.36420.52,170value, logarithmicfraction of the stream at or belowrank ±2%value 19.3–21.9answered 2.9% out20,000 values · log-normal, σ = 1.2 — a latency distribution · Greenwald–Khanna, ε = 0.02rank 1.02% · value 2.9%

The error that is on the rank

A summary of 77 tuples answers eight quantiles of a stream of 20,000 values, and every answer is guaranteed to sit within 0.9% of the stream from where it was asked for. The guarantee is deterministic, it holds on every distribution, and it is not about the numbers it returns.

streaming · Rank
10,0000.1total bits of stateruns outside ±15%1 × 483 × 485 × 487 × 489 × 48240 runs per design · tolerance ±15% · truth 12,00067 → 3 failures

The estimate that is a median of means

An estimator with a 70% spread is not usable and an estimator with a stated failure probability is. The construction that turns the first into the second is two lines long, it is where every delta in this field comes from, and its exponential is measured here by counting failures rather than by evaluating a bound.

randomness · Sketch
stay 0.9, 400 pairsacgtacgt0525505225045240stay 0.5, 400 pairsacgtacgt0212202112022120cost of aligning the row letter with the column letterpseudocount 1

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.

tables · Cost
decayedwindowed060120secondscount attributed to key 0half-life 4 s · window 5.77 s · burstythe same key, two questions

The counter with no window in it

A counter that fades by half every H settles, on a steady stream, at exactly the count of a window of 1.44H. That correspondence holds in the mean, on a steady stream, and nowhere else — and it is the reason a decayed counter is not an estimate of a windowed count for any window.

streaming · Decay
1 coefficient13 members2 coefficients169 members3 coefficients2,197 members4 coefficients28,561 members1 key2 keys3 keys4 keys5 keysexact92% gone99% gone100% gone100% goneexactexact92% gone99% gone100% goneexactexactexact92% gone99% goneexactexactexactexact92% goneevery member walked · GF(13) · no tolerance and no seeddegree 1, 2, 3, 4

The independence an estimator spends

Every sketch's analysis begins by assuming a truly random hash, and nobody comes back to that line. Independence has a degree, the degree is enumerable over a small field, and an estimator's mean and its variance spend different amounts of it.

randomness · Moment
1101001,0000.111010010³true count of the keyrelative errorheaviest keyrarest key4×64 counters · 8,192 bits · Zipf s = 1.1 · 3,528 distinct keys at 702 positions4.11% to 34100%

The guarantee that is one query wide

A sketch described as accurate to within a per cent is accurate to within a per cent of the whole stream, not of the number asked about. On a skewed stream the same sketch is 4% wrong about its heaviest key and 34,100% wrong about one of its rarest, and both figures satisfy the bound.

wrong · Sketch
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

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.

applied · Transfer
queries answered yesabsent key, the AND0.190%absent key, built on the intersection0.025%in one set only, the AND1.800%in one set only, built on it0.000%bits set: A 6,351, B 6,294, AND 3,232, direct 1,859no common key is ever denied

The intersection two filters cannot report

Two Bloom filters over sets that share five hundred keys, ANDed bit by bit. The result never denies a shared key, and it looks like a filter of the intersection. It is not one — a key in only one of the sets passes it 1.8% of the time where a real filter of the intersection passes none, and reading the intersection's size off its bits gives 900.

randomness · Randomness
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

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.

structures · Sketch
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

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.

structures · Cardinality
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

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.

applied · Transfer
answered 4,1900%25%50%75%100%11.794,190value, logarithmicfraction of the stream at or belowrank ±2%value 24.9–4,190answered 1553.5% out20,000 values · Pareto, α = 1.2 — a heavy tail · Greenwald–Khanna, ε = 0.02rank 0.10% · value 1553.5%

A promise about the rank is not a promise about the value

A quantile summary asked for the 99th percentile of a log-normal stream returns the largest value it ever saw — 2,169 against a true 318, six times too high — and its rank error is 1.00% against a promised 2%. The guarantee held. It was never about the number.

wrong · Rank
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

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.

streaming · Merge
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

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.

applied · Transfer
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

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.

tables · Cost
Count-Min 4×64exact0 under200 overrms 303Count-Sketch 4×64-2,774-1,38701,3872,774120 under80 overrms 256cash register · insertions only, Zipf s = 1.1 · counts exact8,192 bits each

A sketch that is allowed to be under

Count-Min's estimate is never below the truth, and it pays for that with an error proportional to the whole stream. Give every key a sign and take a median instead, and the same table is 2.7 times more accurate on the keys anybody asks about — and wrong in both directions.

structures · Moment
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

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.

applied · Transfer
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

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.

tables · Cost
f = 6,3628/8 holdingf = 3,0748/8 holdingf = 1,9638/8 holdingf = 1,3738/8 holdingf = 1,0988/8 holdingf = 9353/8 holdingf = 7690/8 holdingf = 6560/8 holdingpredicted damage, in arrivals — every row totals 967Space-Saving's shareMisra-Gries's share8 shards · round · k = 32bill 967 arrivals

The bill a partition only divides

The two predicted damages for any key sum to the same number under every partition — 967 arrivals here, whatever the arrangement. Round-robin hands nearly all of it to Misra-Gries and hashing hands most of it to Space-Saving, and neither of them is paying more than the other in total.

structures · Merge
00.2500.5000.7501010203040age of an arrival, secondsweight it carries nowexponential, half-life 4.0 sbackward polynomial, α = 2, scale 4.0 sforward, β = 2, landmark 20 s backforward, β = 2, landmark 160 s backdotted: half weightweights at the moment of reading

A decay measured from where it started

An exponentially decayed counter is one number because its weights fade by elapsed time alone. Forward decay keeps a polynomial weight in one number too, by measuring each arrival from a fixed landmark. Its memory is then a share of the time since that landmark: at β = 2 an arrival counts half at 29% of that time. Anchored at the start of a stream, it takes 5.3 seconds to register a fourfold rise twenty seconds in and 83 seconds when the rise comes at five minutes.

streaming · Decay
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

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.

wrong · Merge
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

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.

wrong · Merge

Named alongside it

The objects these essays reach for when they reach for this one.

State bitsSketchRelative errorGuaranteeHeavy hitterZipf distributionHonest limitMergeable summaryMisra–GriesOne passTrade offUnbiased estimator

All concepts