Algorithms & Data Structures
the bound is a claim, and it is checkable
Concept

Peak and total — where it appears

Two ways to charge for memory — the most held at once and the sum of everything taken — which rank the same structures differently. The two rank the same structures differently, and which one matters depends on whether the constraint is a memory limit or an allocator's throughput.

Named by 4 essays across one field — each of them below, with the objects they name alongside it.

peak slots held at once, logarithmic18645124096Insertion sort11Selection sort11Bubble sort11Heapsort11Shellsort11Quicksort, median of three22log nQuicksort, random pivot28log nQuicksort, first-element29log nMerge sort with a cutoff4,106nMerge sort4,110nn = 4,096, random inputone slot = one array element or one stack frame

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.

space · Space 10010³1010010³nstack framesa stack of 512 framespivot: first-elementpivot: median of threepivot: random pivotMerge sortalready sorted input, n from 64 to 4,096one frame charged as one slot

The stack nobody counts

Merge sort makes 8,192 calls to sort 4,096 elements and holds fourteen of them at once. Depth-first search on a grid holds twelve vertices, or sixty-six, or a hundred and forty-four, depending on which of three equally standard implementations is running. The stack is a resource, it is the one that fails hard rather than slowly, and nothing that watches the data can see it.

space · Space ε = 0.0515 → 29 (1.93×)ε = 0.0238 → 73 (1.92×)ε = 0.0177 → 136 (1.77×)ε = 0.005152 → 270 (1.78×)ε = 0.002397 → 674 (1.70×)reportedoccupied at the peak20,000 arrivals · lognormalpeak = resident + period, to 14%

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.

space · Space 1,00010,00010phrases followed12481632characters in the collection · copies aboveworst depthmean depthEnglish-like · z = 156median 18 · 90th 31

The character that costs a chain

The index over thirty-two copies is 8,892 bits and does not grow. Producing one character of the text it indexes costs 17.89 phrase-follows on average and 38 in the worst case, against 2.38 and 7 at one copy — the size stopped growing and the price of reading it did not.

space · Parse

Named alongside it

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

Recursion depthAllocatorAuxiliary spaceMeasurementQuicksortSpace overheadTrade offTraversalAdjacencyCall stackCompression scheduleCopy depth

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