Auxiliary space — where it appears
Named by 20 essays across 7 fields — each of them below, with the objects they name alongside it.
Measuring what an algorithm keeps
Four counters measure what an algorithm does and none of them measures what it holds. An in-place sort and an out-of-place one with identical comparison counts are different algorithms, and until this phase the site had no way to say so. Two primitives close the gap, and the second of them counts something no array counter can ever see.
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.
The cost is the number of subproblems
The edit-distance recurrence, written down literally, makes 29,737 calls on a six-letter word and a seven-letter word. Written down with a table beside it, it makes 56. Nothing about the arithmetic changed, and the class did.
The same table, filled two ways
Top-down and bottom-up compute identical cells and return identical answers. One of them asks the table half a million questions and recurses four hundred frames deep; the other asks none and recurses none — and on a knapsack it fills twenty-two times as many cells as anything can reach.
The frontier between time and space
The question of which sorting algorithm to use has an honest answer, and it is a shape rather than a name. Comparisons on one axis, peak auxiliary space on the other, and five of the ten algorithms here are on the Pareto frontier while five are dominated — beaten on both counts at once, so that no weighting of the two costs makes them the right choice. Heapsort is one of the five that lose.
In place is a claim, and it is usually wrong about quicksort
Heapsort holds one slot at its peak. Quicksort holds twenty-two at n = 4,096 on random input and 4,097 on a sorted one. Merge sort holds 4,110. All three are described with the same two words, one of the three descriptions is false, and the false one is the algorithm the phrase is most often attached to.
The space the model does not see
A slot is not a byte, a frame is not a slot, sixteen thousand allocations are not one allocation of the same size, and none of these numbers includes the input. The space counters are the newest instrument here and the honest account of what they miss is longer than the account of what they measure — including one bound this phase set out to demonstrate and could not.
The table nobody has to keep
A million-cell table, computed cell for cell in the same order, holding two thousand cells at its peak instead of a million. The saving is exactly (n+1)/2, it costs nothing on any operation counter, and what it buys is paid for with the one thing the table was for.
The alignment that fits in one line
Compute the table twice and hold three rows of it. The factor of two is a geometric series and is predicted exactly; measured, it comes down from 2.269 to 2.052 as the strings grow, and the peak is 3(m+1) cells on the nose.
The depth limit that almost never fires
Introsort counts how deep its recursion has gone and calls heapsort if it passes twice the logarithm. On every input measured, the mechanism handles under a tenth of a per cent of the elements. Then an adversary that answers the comparisons rather than choosing the array drives it to exactly n²/2 with the limit removed, and to one heapsort call with it — a factor of forty-two at n = 8,192, and growing.
What a window costs in bits
The approximate structure grows like the square of a logarithm and the exact one grows like the window, so the approximation wins eventually. Eventually is a window of 6,000 at a 2% tolerance — and below it the summary is larger than the thing it is summarising.
The text that does not have to be kept
The index reproduces its text character for character, in 1,024 mapping steps and zero reads of anything. That is the whole justification for weighing it against the text rather than beside it — and the price is a dial that moves the structure by 3.3 times and the cost of locating one match by 72.
Every substring, in fewer states than substrings
A text of 512 characters has 129,416 distinct substrings. A machine that recognises every one of them, and nothing else, needs 831 states — and the bound it is under, 2n − 1, is reached exactly by a string one line long.
The tuples a summary does not report
A Greenwald–Khanna summary at ε = 0.01 answers `tuples` with seventy-seven. Watched through the run it holds a hundred and thirty-six. The gap is the compression period, it is 1.70 to 1.93 times across every tolerance measured, and it is the number a deployment has to allocate.
A floor under a product
Every lower bound written here so far puts a floor under one quantity. The published time–space bound for sorting puts one under a product: T·S is Ω(n²), so buying time with space and space with time cannot both be done. Measured, heapsort's product is 0.0028 n² — four hundred times under a bound nothing can beat. The bound is not wrong. It is stated in a model where the input may not be written, and every sort here writes to the input, which is where its working memory actually is.
Pieces that fit the buffer
A sorted file with a late tenth appended costs the buffer-limited stable merge sort 2.93 moves a key at a buffer of a 64th, fifteen times the least any sort could move, and two thirds of it is the tail sorting itself. Taking the tail a buffer at a time, inserting each key into a sorted piece in the buffer and merging the piece into the head, was predicted to bring that under two moves a key at small buffers. It never comes within twice the merge sort: insertion costs a quarter of a piece a key, and the smaller the pieces the more often each one's merge crosses the head's top fifth. The same pieces sorted in place by merging do pay, at 2.41 moves a key, and for a reason the proposal did not name: a piece that fits the buffer merges into the head without a single rotation. The gain lasts only while the tail belongs near the top of the file.
How many pieces the head can afford
A sorted file of 65,536 keys with a late tenth appended in random order can be sorted by taking the tail a buffer at a time, sorting each piece, and merging each into the head. It beats the buffered merge sort at a 64th of the file and loses at smaller buffers, because every piece's merge crosses the head's top keys again. Merging the pieces with each other first, while a round costs less than the crossings it removes, was predicted to win at every buffer. It wins at none. The first round always costs more than no rounds — 18.97 moves a key against 16.04 at a 1,024th — because a group larger than the buffer crosses the head by rotation, and the cost falls under the merge sort's only when the rounds run all the way, which is the merge sort. The head can afford thirteen pieces and not twenty-six; beyond that the best a tail can do is be one run.
The rotation that could be priced
A sort that finds a sorted file with a late tail can send the tail's pieces straight into the head or merge them to one group first, and a rule that priced both ends as shifts chose wrongly at a buffer of a 128th, because a group longer than the buffer crosses the head by rotation. Priced instead from measurements of the library merge on synthetic runs, the long crossing comes within 3% of what the sort pays at every buffer and the choice at a 128th comes right. But the rule prices the pieces against the head as it stood before any of them joined it, and each piece also crosses the pieces before it: priced up to 37% low, they send the sort to the dearer end at a 256th.
The head each piece will meet
A rule that priced a sorted file's late tail two ways before moving anything chose the dearer end at a buffer of a 256th, because it priced every piece against the head as it stood before any piece joined it. Each piece also crosses the pieces before it, and those are known in advance: one binary search of each earlier, already sorted piece counts them. Priced against the head each piece will meet, the pieces come within 6% of what they cost at every buffer, and the rule chooses the cheaper end at all seven buffers and all ten reaches of the tail. A third of what each piece crosses turns out to be other pieces, as the arithmetic of the band says it must.
Every number of rounds, priced
A library merge sort given a file whose last tenth arrived late sorts the tail a buffer's worth at a time and then either sends every sorted piece into the head or merges the pieces first. A rule priced from measured tables chose the cheaper of the two ends at every buffer. Priced the same way at every number of rounds in between, it was predicted never to choose a partway number and to price every one within 6%. It chooses the cheapest number of rounds at all seven buffers on that input and at 14 of 15 settings across reaches and buffers; the one partway optimum among them — six rounds of seven, by 1% — it misses by choosing five, 3.8% dear. But the estimates are not all within 6%. At the three smallest buffers the first round is priced 15% to 25% wrong, high at one buffer and low at the others, because its pairs of pieces cross the head in a way no table of single runs describes; the rule survives it because one round is never close to the cheapest.
Named alongside it
The objects these essays reach for when they reach for this one.
Trade offCost modelRotationData movementHonest limitMerge sortPredictionQuicksortStable sortBinary searchDynamic programmingEdit distance