Concept

Presortedness — where it appears

How close an input already is to sorted order, measured by a stated count such as inversions, runs or displacement. Adaptive sorts are fast on presorted input, but which measure applies depends on what the algorithm pays for, and different measures rank inputs differently.

Named by 7 essays across 3 fields — each of them below, with the objects they name alongside it.

11010010³10⁴10⁵10⁶10⁷natural runs r in the inputcomparisonsn / minrun = 256TimsortMerge sortInsertionn + n log₂ rn = 8,192, runs built exactlycomparisons, counted exactly

A run is a property of the input

A benchmark that says "nearly sorted" never says how nearly. It is a recipe with a seed, not a measurement, and an adaptive bound stated against it is a bound with an undefined second parameter. Counting the natural runs turns the shape of an input into a number — and then Timsort's bound becomes something that can be fitted rather than quoted.

practice · Practice
10⁵10⁶10⁷10³10⁴inversions in the permutationblock transfers to carry it outsort by destination, 1,536w 8w 32w 128w 512w 2048w 819216 swaps64 swaps256 swaps1024 swapsshuffled inside windowsa few pairs swapped farn = 16,384, B = 64, M = 512 (M/B = 8)inversions do not order the cost

The permutation that moves almost nothing

Two ways to scramble sixteen thousand elements. Shuffling them inside windows of five hundred and twelve puts two million pairs out of order and costs 3,095 block transfers to carry out. Swapping a thousand pairs across the whole array puts seven million out of order and costs 1,189. Inversions are the textbook measure of disorder, and on a disk they rank these two backwards.

applied · Transfer
0.0010.010.11125102050fraction of positions reshuffled, pmean comparisons, in multiples of the mean on random inputFirst-element quicksort, 81.9×Median-of-three quicksort, 41.4×Insertion sort, 2.0×Merge sort, 1.0××: unshuffled2,048 elements · 12 shuffles a point1 = the mean on random input

A worst case ten positions wide

Sorted input costs first-element quicksort 2,096,128 comparisons on 2,048 elements, 82 times its average. Reshuffle about eleven of the 2,048 positions and the cost halves — and it takes about ten at 128 elements, and between ten and thirteen at every size between. Reversed input costs insertion sort twice its average, and reshuffling half the positions still leaves 71% of the work. A worst case is a place in the space of inputs, and the two famous ones are places of very different sizes.

counting · Count
1M2M4M16M64Mmean run length, in memoriesrandom33 runssorted1 runreversed64 runssorted, 1% arriving late2 runssorted, 10% arriving late7 runs262,144 records, 4,096 in memorydashed: two memories

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.

applied · Transfer
0%25%50%75%100%ln 2mean leaf fillrandom, even splits2,906 leavesrandom, rightmost-split rule2,949 leavesascending, even splits3,971 leavesascending, rightmost-split rule2,048 leavesdescending, even splits4,095 leavesdescending, rightmost-split rule4,095 leavesbulk-loaded from sorted keys2,048 leaves131,072 keys, leaves of 64each bar names its rule

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.

applied · Transfer
50%60%70%80%90%100%0%1%2%5%10%25%50%share of keys arriving latemean leaf filleven splitsrightmost-split rulesibling first, two into threeln 2131,072 keys, leaves of 64late keys arrive at a random later point

The sibling a full leaf asks first

The rule databases use to fix ascending inserts fills their leaves completely and collapses to 53.4% when one key in a hundred arrives late. A leaf that offers a key to a sibling before it splits, and splits two full leaves into three when neither will take one, holds 84.2% on the same stream — and is better with a trickle of late keys than without one, because a perfectly ascending stream has no sibling with room.

applied · Transfer
6080100how far a full leaf looks for room, in siblingsmean leaf fill, per cent12481664randomascendingdescendingascending, 1% late131,072 keys · leaves of 64 · fan-out 64a reach of 1 is the adjacent-sibling rule

A key passed along the row

A full B+-tree leaf that offers a key to its immediate neighbours before splitting leaves an ascending stream 67.2% full. Let it look one sibling further and the same stream fills its leaves completely, 2,048 leaves where there were 3,048, for about the same key moves. On random keys each doubling of the reach adds a few points of fill and more writes than it saves, and at the whole parent a key travels sixteen leaves on average. On a stream that is mostly in order the same reach costs almost nothing.

applied · Transfer

Named alongside it

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

Block transferTrade offWorst caseB-treeExternal-memory modelInversionsAccess patternAdaptive sortDesign parameterHeapIndex maintenanceInsertion order

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