Regime — where it appears
Named by 25 essays across 7 fields — each of them below, with the objects they name alongside it.
What O-notation does not say
Big-O is a statement about a limit. It does not say how fast, it does not say which is better, it does not say anything at all about any particular n, and it discards precisely the factor that usually decides the answer. Knowing exactly what it claims is the difference between using it and being misled by it.
Sorting what will not fit
Merge sort's Θ(n log n) is a statement about comparisons and says nothing about a file larger than memory. Counted in transfers the answer is (n/B)·log_{M/B}(n/B), and the base of that logarithm is the number of blocks that fit in memory — so doubling the memory does not halve the work, it moves a staircase. The measured cost jumps by 32,768 transfers at one step and by nothing for the next four.
Two parameters, one bound, no order
With one size parameter the candidate classes are ordered — n beats n log n beats n², always, and comparing two bounds is reading them. With two, E log V and V² have no order at all, and which one is smaller is a property of the graph. Sweeping V at fixed degree and at fixed density are different experiments, and the same algorithm fits different classes in the two.
A limit is not a prediction
Measured from n = 64 to n = 4,096, this site's hybrid merge sort fits a linear class better than n log n. Measured out to n = 65,536, the ranking reverses. Nothing changed but the range — and this is not a flaw in the method, it is the method finding the exact place where measurement stops being able to help.
The probe formula nobody checks
The expected number of probes to insert into a hash table under linear probing is ½(1 + 1/(1−α)²). It is quoted constantly, it is correct, and applied to a table of 256 slots at 95% load it overstates the measured cost by nearly half — because it is an asymptotic result and a real table is not asymptotic.
Two parameters are not enough either
Two graphs on 1,024 vertices with 3,072 edges each — identical in both numbers every bound in this field is written in. Enumerating every pair of neighbours of every vertex costs 18,480 examinations on one and 42,076 on the other. The quantity that separates them is a third parameter, it is computable in linear time, and it appears in no statement of the problem.
One dial between two structures
A B-tree writes 226 elements of block for every key stored and a log-structured store writes two. They are presented as rival designs. They are one design at two settings of an exponent that nothing in either description mentions, and every setting between them is available.
The cells that were never worth having
Two three-hundred-character strings over twenty-six letters give a table of 90,601 cells, and 3,421 of them are pairs of positions whose characters agree. Only those can lengthen anything. A method that enumerates exactly those computes a twenty-sixth of the table — and on a two-letter alphabet it computes half of it and is worse than the table it replaced.
Permuting is the harder problem here
Rearranging 65,536 elements into a stated order costs 63,601 transfers one at a time and 4,096 by sorting them into place. In the model every other field on this site uses, the first is the cheap method and beats the second by a factor of eight. The two models disagree about which problem is easy, and they disagree by about the same factor in opposite directions.
The index that is not worth reading
An index turns a query over 65,536 rows from 1,024 transfers into four. At a thousand matching rows it costs 654 and still wins; at sixteen thousand it costs 1,027 and has lost. Where it turns is decided by the block size — a number the query does not contain, the schema does not mention, and nobody writing either has seen.
Two ways to join, and the ratio that decides
The same join costs 260 transfers one way and 1,040 the other; at eight times the memory the same two costs are 2,880 and 1,280, the other way round. Neither number is a property of how large the tables are. The quantity that decides is how the smaller of them compares to memory, and a rule of thumb phrased in rows is a rule about somebody's machine.
The parameter plane has few answers
Sweep the cost of opening a gap against the cost of extending one over five hundred and seventy-six settings, and the optimal alignment of intention against execution takes four values — one of them at 571 of the settings. Under a linear model the plane divides into three wedges through the origin, because doubling every cost changes nothing and only the ratio is a parameter. Tuning an aligner is choosing a region, and most of the plane is one.
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.
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.
Runs twice as long as memory
Feed 262,144 random records through a heap that holds 4,096 and the sorted runs that come out average 1.94 memories — the snowplow's famous factor of two. At a fan-in of 63 that saves a merge pass at 262,144 records, and at two of fourteen sizes in all. Feed the same heap a sorted file with one record in a thousand out of place and it writes two runs instead of sixty-four. And it spends 19 comparisons a record doing so, on every input, where sorting the chunks spends five on sorted data. The factor of two is the least of what the method does.
The keys that arrive late
Insert 131,072 keys into a B+-tree in random order and its leaves end up 70.5% full; in ascending order, 51.6%; in descending order, 50.0%. The rule databases use to fix ascending inserts — split a full leaf at its right-hand end — fills them completely, and it does nothing for descending keys. Let one key in a hundred arrive late in an otherwise ascending stream and the rule's leaves fall from 100% to 53.4% full. How much of an index is empty is decided by the order its keys arrived in, and a trickle of disorder undoes the fix.
Two hash values and the keys they copy
A Bloom filter that makes its k bit positions from two hash values, as h₁ + i·h₂, answers yes to 1.6% of absent keys on a 64-bit filter where k independent hashes answer 0.69%. The penalty is not the one expected. With the step forced odd no key ever repeats a bit, while a quarter of independent keys do. What costs the filter is a query whose start and step reproduce a stored key's whole progression, which happens with probability 4n/m², measured to within a few per cent from 64 bits to 4,096. The penalty fades as the filter grows and returns as the hash count rises — 1.13 times at seven positions on 1,024 bits, 5.35 times at thirteen.
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.
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%.
A 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.
A 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.
A switch that hears the traffic stop
A landmark watch switched off by a count of the traffic's stretch ends cannot see traffic slow down, because a slowing is the events stopping, and a count of events keeps its last reading. Letting the open spacing since the last stretch end into the reading, once it is longer than the recorded ones, was predicted to turn the watch back on within four thousand queries of a slowing and cost nothing on steady traffic. It costs nothing on steady traffic, and on ten stretch ends it hears the slowing late, after three to eleven thousand queries, because the silence is averaged with nine fast spacings. On five or three it hears it in time, and the switch on five comes within two cells of the best known policy on every stream.
A silence that overrules the count
A landmark watch switched to the corners by a count of the traffic's stretch ends heard a slowing only once the silence outweighed the fast spacings it was averaged with, about 5,900 queries on ten stretch ends. Letting an open spacing of 1,600 queries overrule the count outright was predicted to make the count's length stop mattering on the stream that slows. It hears the slowing within a thousand queries at every count from three to ten, and keeps the corners 8% of the slow half where the count kept them 65 to 75%; the counts still differ, by 3.4 cells a query, because the fast half is where a short count keeps the corners that pay. On the stream that speeds up the override is the cheapest switch at every count, and on steady traffic it is within a cell and a half of the silence.
A gap between off and on
A landmark watch switched to the corners whenever a count of the traffic's last stretch ends read under 1,600 queries, and back on when a silence of 1,600 overruled the count. A count that only decides when traffic is fast need not use the silence's threshold, so a second, lower one was given it: keep the corners only once the count reads under 1,200. At one jump of the destination in 2,500 queries, on three stretch ends, that saves 3.5 cells a query — 117.8 against the override's 121.3 — but keeps the corners 13% of the time, not under 5%, and stays 1.4 cells above the watch rather than within one. At one jump in 1,000, where the corners are right, the gap costs 1.7 cells; on five stretch ends it saves nothing anywhere; and at slower paces, on traffic that changes pace and on drift, it changes nothing. The two jobs do want two numbers, but only one count length wants them.
A 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.
Named alongside it
The objects these essays reach for when they reach for this one.
Measured countParameter choiceShortest pathBlock transferExternal-memory modelLandmarkHonest limitQuery distributionTrade offCost modelHeuristic searchBreak-even