The thread: The answer is an estimate — page 2
What a planner pays to find out what to pay
Insurance against a row estimate is set from the error's median and spread, and a running system knows neither — it has to fit them from executed queries. Fitted from one query the divisor costs 1.263 times the better plan against 1.215 for a planner that never insures at all, so learning is worse than not learning until about sixteen queries have run. The tail, though, is bought immediately: one observation already holds the worst case to 24.7 against 36.6.
One pass, and no roomA 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.
One pass, and no roomWhat a heavier tail actually buys
Four decays, each tuned until it is equally quiet on a stream with nothing happening, then given one burst. At a burst of three hundred they see it for 54, 63, 68 and 104 seconds — indistinguishable. At nine thousand the exact polynomial reaches 384 seconds and the exponential 156. The shape of a tail does not decide how far back a scheme can see. It decides how fast that distance grows with the size of the event, and the exchange rate is the reciprocal of the exponent.
When the algorithm is a tableA cost built from two properties
A substitution matrix fitted as a partition gives each class of pairs one cost, and the best number of classes can be chosen from the data. When the true costs are a sum of two terms — a transversion penalty and a graded difference between the letters — every pair costs something different, and the chosen partition's number of classes becomes a statement about the corpus: two classes at 25 sequence pairs, four at 200, five at 1,600. A model with the two terms and three parameters predicts a fresh corpus better than any partition at every size, and gets 191 of 200 alignments right from 50 pairs where the chosen partition needs 3,200. Told the wrong property it is worse than the plain two-class matrix, and nothing in its own fit says so.
When it does not fitA summary's silence is a bound, not a count
When both sides of a join are skewed towards the same values the textbook estimate is 399 times too low and the order it picks costs 244 times the better one; two counters a side repair it. Skew them towards different values and the same estimate becomes 254 times too high — and the counters have nothing to say, because no value is in both summaries. Using their silence instead repairs the middle of the sweep and makes the far end six times worse.
One pass, and no roomThe moments past the second
Past the second moment no sign trick works, and the estimator that does — pick one position of the stream, count how often its key comes again, report m(r^k − (r − 1)^k) — is unbiased for every moment on every stream. Its variance can be computed exactly rather than sampled, and on real streams it is nowhere near its guarantee: 103 copies give F3 to ten per cent on a flat stream of 2,048 keys, and 2,630 on a moderately skewed one. The guarantee asks for 48,381, and it is not loose. One heavy key among 2,047 keys seen once needs 11,799.
One pass, and no roomA detector that learns its own quiet
A burst detector built from a fast decayed counter needs a threshold, and the one it was measured with came from sixty independent runs of the background — an oracle no deployed system has. Two counters and a running mean square of their difference can supply it from the stream itself, and on a stream that does not move they hit a 5% false-alarm rate at every setting tried. What they lose is the burst: a scale that learns every reading learns the burst as noise and forgets it in 20 to 40 seconds whatever its size. Holding the scale while alarmed brings the oracle's horizon back for small bursts, turns a slow drift into an alarm that will not stop, and still cannot see past the lag at which the slow counter holds more of the burst than the fast one.
One pass, and no roomThe clock a burst cannot outlast
A burst detector that holds its scale while alarmed keeps a burst visible to a ceiling set by two half-lives, and cannot stop alarming on a slow drift. The repair proposed for it was a schedule: learn again once an excursion has outlasted anything a burst could produce. Timed by how long the alarm has lasted, the schedule changes almost nothing, because on a gentle rise the alarm breaks every few seconds and the clock never reaches the ceiling. Timed by how long the difference between the counters has stayed positive, it keeps every burst to within about a second of the ceiling and falls below the target false-alarm rate on steep drifts — and one quantity, the drift's offset measured in spreads of the difference, says where it still fails.
When the algorithm is a tableA line the corpus finds
Substitution costs built as a transversion penalty plus the distance between the two letters' values on a graded property beat every partition of the pairs, when the model is told the property, and fall below the plainest matrix when told the wrong one. Fitting each letter's position on a line instead was predicted to find the property's order from about four times the told model's data and then behave like it. It finds the order on every corpus of 100 sequence pairs or more but one, and every miss starts with the two letters whose values differ by a fifth of a bit. But knowing the order is not knowing the costs. Its alignments are no better than six free classes' until 800 pairs, sixteen times the data the told model needed for the same score, and on a generator with no graded property at all it finds a line anyway, a quarter as long as a real one at 200 pairs, and pays for it against the plain partition.
When it does not fitA planner that knows how sure it is
A planner that fits its insurance divisor from a handful of executed queries under-insures, because a spread fitted to a small sample is biased low. The textbook repair is to use the upper end of a confidence interval on the spread instead of its centre. At a read ratio of four it makes the planner worse than trusting its estimate for longer — until sixty-four queries rather than sixteen — and it buys the worst case immediately, from 22.4 times the better plan to 6.2 after two queries. The bound is not a correction towards the truth. It is a dial between two different planners.
One pass, and no roomThe keys worth counting before sampling
The sampling estimator for the third frequency moment spends most of its copies on keys that barely contribute to the answer. Finding the heavy keys first with a Count-Sketch, counting them, and sampling only the rest cuts the state for ten per cent from 5,260 words to about 1,500 on a moderately skewed stream, and from 23,598 to under a thousand on the stream built to be worst for the sampler. It costs five times more than it saves on a flat stream, where there is no heavy key to find. What decides it is how the frequencies are shaped, not how long the stream is, and the frequencies must be counted, not read off the sketch, because the sketch's error on a heavy key is cubed.
When the algorithm is a tableA line with a prior toward the classes
Substitution costs fitted as the letters' positions on a free line learn the letters' order quickly, and on a generator with no graded property they invent a line anyway. Charging the line for its length, at a rate growing with the corpus as the partitions' charge does, was predicted to collapse the invented line from 50 pairs and to cost the real one little. It does neither. On the generator with no property the charged line still has a median length of 0.12 bits at 3,200 pairs. On the generator with a property it takes the letters' order away at every size up to 400 pairs where the free line had it on sixteen seeds of sixteen. Charging the line for the number of gaps it keeps, the partitions' own rule, collapses the invented line on every seed at every size and matches the chosen partition to the bit — and then needs 1,600 pairs before it believes a real line is there.
One pass, and no roomThe slope the slow counter already shows
A burst detector built from a fast and a slow decayed counter reads a steady rise in its background as an offset in their difference, β(R − 1)H/ln 2, and alarms on it. Every repair so far waited the offset out. This one subtracts it: the slow counter's own change over one slow half-life is a reading of the slope, and the offset follows from it. At R = 16 the corrected detector alarms on 5.5% to 7.8% of readings on rises that make the held scale alarm on 11% to 63%, and it costs a burst at most 3.7 seconds of detection. The estimate has to be quick: four times slower, it has recovered only 71% of the offset when alarms start to count.
Two parametersA placement after the traffic moved
Four landmarks fitted to twenty queries that all end at one cell expand two thirds of the corners' cells there. Move the destination ten cells and the advantage is gone; move it thirty and the fitted placement expands nearly twice the corners' cells. Choosing again costs 317,000 cells of search — the saving of 8,600 queries — so on streams whose destination drifts, a policy that refits on a counter never beats fitting once and keeping the result, and past a drift of about ten cells nothing fitted beats the corners.
One pass, and no roomThe rise the correction kept subtracting
A burst detector that subtracts a rising background's offset, read off its own slow counter, goes on subtracting it after the rise has stopped. The prediction was a deficit for about one slow half-life that would cost bursts of 300 arrivals. Measured, the deficit is deeper and longer: down to 1.46 spreads below zero, below half a spread for seventeen minutes after a 40% rise. Bursts of 300 are never missed. What the deficit costs is small bursts — 30 arrivals caught 27% of the time where the rise saw 77% — and a fifth of a large burst's visible time.
Two parametersA refit that changes one landmark
Choosing four landmarks again after traffic moves costs hundreds of thousands of cells of search, and on drifting streams it never paid. Keeping three and re-choosing the fourth costs a fifth as much — 0.19 of a full selection, not the twelfth the earlier page guessed. On drifting streams, refitting on the counter goes from a loss of up to 13% to within 2% of fitting once, and it wins where the destination moves six cells. When the destination jumps, it is the cheapest policy measured, 12% under the corners at one jump in 10,000 queries — unless the jumps come as often as one in 2,500.
One pass, and no roomA share of a list of equals
Estimating the third frequency moment by finding the heavy keys first and sampling the rest works best with one counted candidate on steep streams and sixteen at Zipf 0.5, and a system that does not know its stream's skew has to pick. Keeping a long list and counting a candidate only if it holds a stated share of the list's own moment was predicted to come within a few per cent of the best fixed list on every exponent. The share does choose: at 3% it keeps one candidate at Zipf 2, three at Zipf 1, five at Zipf 0.5. But a list the stream shortens is not a short list. Its sixty-four candidates cost 192 words on every stream, four times the whole of the best estimator on the steep ones. On a flat stream every candidate is a near-equal share of a list of equals, and no share can tell that none of them is heavy. No list length and share comes within 10% everywhere; at Zipf 0.5 alone, the chosen list beats every fixed one.
One pass, and no roomA list that gives back its places
An estimator of the third frequency moment keeps a list of candidate heavy keys beside a Count-Sketch and samples the rest, and cutting a long list at the end of the pass to the candidates that hold a share of its moment picked the right keys and paid for the whole list. Dropping a candidate during the pass, as soon as it falls below that share, was predicted to make the list's memory follow the stream: under four candidates on steep streams, within a fifth of the best fixed list, and about ten on Zipf 0.5 with most of the earlier win kept. The list ends the pass the right length — three candidates on Zipf 1 — but its memory is its peak, and the peak is set by the twenty-fifth arrival, before any key has stood out: nineteen candidates on Zipf 1, and on flatter streams the most a share θ allows, one over θ. Charged its peak, it beats the end-of-pass cut on every stream from Zipf 0.75 up, 208 words against 364 at Zipf 1, and never reaches the best fixed list, 159.
One pass, and no roomA check as slow as what it checks
A burst detector that subtracts a rising background's offset goes on subtracting it after the rise ends, and loses small bursts in the trough that opens. Capping the subtraction by what the difference itself shows, averaged over one slow half-life, was predicted to close the trough to under half a spread. It lifts it from 1.46 spreads to 1.24, because that average is as late as the estimate it checks. Averaged over a quarter of the time it lifts it to 0.70 and catches small bursts in the trough more often than any other design but the one that alarms on a fifth of its readings. Neither cap helps after a large burst, which every slope correction reads as a rise.
One pass, and no roomA variance that had seen one hash
A budget of 40,960 bits split between a HyperLogLog and a bottom-k sample answers an intersection best by weighting the two estimates by their variances, except in one corner: a sample of 160 hashes asked about sets sharing one key in a thousand, where the weighting's worst regret was 4.46. The sample was said to see nothing, report a variance of nought and be believed completely. Its variance was never nought — it was taken at one shared hash whenever none was seen — and it was believed too little, not too much. Half a shared hash in the variance gives the sample more weight and cuts the corner's regret to 3.08, not under the 2.5 predicted; half a hash in the estimate too gives 3.74 there, and 1.61 at five eighths of the budget in registers, the lowest worst regret any split and rule has reached. No weighting measured gets the corner under the sample's own 2.30: 160 hashes expecting a sixth of a shared one is a limit, not a formula.
The index that replaces the textA count that names every point
A grid that can report the points in a rectangle can be no smaller than log₂(z!) bits, because reporting them recovers the permutation. A grid that only counts them was left open as possibly smaller. Exactly, it cannot be: counts alone decode every point, and on the phrase index's 3,612 points 42,860 count questions recover the whole permutation, so an exact counter pays the same floor of 37,485 bits. Approximately it can — and the rectangles a phrase index asks are where an approximate count is worst. They hold 2.6 points on average and a third of them hold exactly one; the wavelet grid cut to eight of its twelve levels, 8% under the floor, counts 46% of them right.
Two parametersA margin the count sets itself
A landmark watch switched to the corners when a count of the traffic's stretch ends read fast saved 3.5 cells a query at a moderate pace once given a second, lower threshold for switching off — but only on a count of three, and at a cost at the fast pace. A fixed gap is the same at every count, so the gap was set by the count instead: switch off only when the reading plus Z of its own standard errors is under 1,600. At one standard error on three stretch ends it makes exactly the fixed gap's choices, 117.8 cells a query at one jump in 2,500 and 144.5 at one in 1,000 — losing 1.9 to the override at the fast pace where under one was predicted, because fast spacings are bunched, not similar. Averaged over six streams every switch on three or five stretch ends lies within about a cell and a half of the best, 113.8: the off threshold, however it is chosen, trades the moderate pace against the fast one, and the count's length decides more than its margin.
Two parametersThe count asks and the price answers
A landmark watch on a weighted maze returns to its four corner landmarks when a count of the traffic's stretch ends reads fast, and every rule for when to switch off has traded the moderate pace against the fast one. The count reads the traffic and never the money, while the watch already prices placements on a sample of the queries it has served. Letting the count propose a return to the corners and the watch's last price confirm it was predicted to keep the margin rule's 117.8 cells a query at the moderate pace and recover the override's 142.6 at the fast pace. It costs 118.8 and 145.1. The price agrees with the count three times in four, a quarter of those because the watch had not yet left the corners; the tolerance, swept from nought to thirty cells, changes nothing; and pricing the corners at every look costs about a cell a query on every stream, including the slow ones where the count never asks.