Floor — the series
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The floor under every comparison sort
No algorithm that sorts by comparing pairs of elements can average fewer than log₂(n!) comparisons. Not one that exists, and not one that ever will. The argument takes three sentences, it is about counting leaves in a tree, and it is one of the few results in this subject that is genuinely about every possible algorithm rather than about a particular one.
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How close anything gets to the floor
The interesting question about a sorting algorithm is not its complexity class but its distance from the bound nothing can cross. Merge sort comes within 2.2% of the information-theoretic floor. Heapsort uses 96% more than it needs to. Selection sort uses nineteen times. Those three numbers say more than the classification does.
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The floor moves when the question does
Sorting 4,096 elements needs at least 43,250 comparisons. Finding one element among the same 4,096, already sorted, needs at least 13. The difference is a factor of 3,300 and it comes entirely from how many different answers the algorithm has to be able to give. A lower bound is a property of the question, not of any algorithm.
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The adversary who hides the edge
The floor under comparison sorting comes from counting outputs — n! of them, so log₂(n!) comparisons. Connectivity has two outputs, so the same argument gives a floor of one comparison, which is useless. A different kind of argument gives Ω(E), and having both on the site is the point: lower bounds are not one technique.
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The floor when the values repeat
log₂(n!) counts orderings of distinguishable things. Two hundred and fifty-six values drawn from eight distinct ones have 1,684 bits of permutation entropy and 739 bits of distinguishability, so the real floor is less than half the one every table quotes — and merge sort, which sits exactly on the quoted floor, is 2.3 times above the one that applies.
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A floor on the bits
Answering membership for n keys with a false-positive rate of 1% and no false negatives requires at least 6.64 bits per key, whatever the structure. A Bloom filter uses 9.59. The gap is 44.27% at that rate and at every other rate, and it is the first bound on this site that a real structure comes close to.
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The floor under moving data
The information-theoretic floor for comparison sorting is log₂(n!) and it says nothing about a file on a disk. In the external model the floor is (n/B)·log_{M/B}(n/B), it is a bound on every algorithm rather than on merge sorts, and a measured external sort sits 2.40 to 2.97 times above it. Both numbers are computable, and the gap between them is what a real implementation costs.
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The floor charged at every level
A key surviving a fold of sixty-four shards is charged 2, then 8, then 20, then 43, then 88, then 248 — the floor of whatever summary it was merged against, level by level. They sum to 409, and the damage read off the merged table is 409. The model that charged sixty-three copies of the leaf floor said 222.
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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.
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Every pair must be asked
Ask whether a six-vertex graph is connected, one pair of vertices at a time, and the best possible algorithm needs all fifteen questions on its worst graph. The claim that this holds for every monotone property of graphs is a conjecture fifty years old. At four vertices it can be settled completely: all 2,046 properties that do not depend on vertex names need every pair. Name one vertex, and the count drops from ten to four.
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The floor a merge cannot reach
Merging two sorted lists of five keys each has 252 possible outcomes, so counting says eight comparisons might do. Solving the game says nine are needed, and on equal lengths the shortfall keeps growing, as half the logarithm of the length. Averaged over random inputs, though, the same count is missed by a tenth of a comparison. The counting floor is nearly exact on average and wrong in the worst case.
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The merge that can say equal
When two sorted lists draw their keys from a few values, equal keys can go on either side of each other, so far fewer answers are needed and the counting floor falls. Merging four keys into four over four values needs only 27 answers, not 70, and the floor falls from seven comparisons to five. A merge whose comparisons answer only 'at most' still needs seven on its worst input, exactly as many as distinct keys, because it can never learn that two keys are equal. A merge whose comparisons can answer 'equal' needs five and meets the floor. Solved exactly over every input, the third answer is the whole difference, and on average it is worth more than an optimal strategy.
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The runs a short list cannot have
A merge whose comparator can say 'equal' meets the counting floor on lists of repeated values, but it takes the gain only when one list's head happens to meet an equal head in the other. Comparing keys inside a list would find runs of equal keys on purpose, and was predicted to leave the worst case alone and help a great deal on average when the values are few. Solved exactly at four keys a list, it leaves the worst case alone and helps by under a hundredth of a comparison: a list of four has no run worth finding, and an equal answer already places two keys at once. On long lists the same idea, galloping over each run before comparing across again, does pay — a twenty-sixth of the textbook merge's comparisons with one value at 1,024 keys a list — and at every length it stops paying at the same place: when a value's run averages fewer than about thirty-two keys a list.
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A merge that decides per run whether to gallop
Merging two sorted lists of repeated values, a merge that gallops along every equal run beats the textbook merge while the runs average more than about thirty keys a list and loses below, a rule about the whole input. A merge that learns the runs as it goes was predicted to match the better of the two on uniform values and beat both on skewed ones. The proposal's rule — gallop when the last few runs averaged long — cannot learn a run that has not ended: at one, two or four values it walks every key. Timsort's own rule, applied to equal answers — walk until a bar of consecutive equal answers, then gallop, the bar moving by one as galloping pays or fails — does what was predicted. On uniform values it stays within a hundredth of a comparison a key of the better merge; under Zipf 1, at 4,096 keys a list and 1,024 values, it makes 0.447 comparisons a key against 0.595 walking and 0.594 galloping, a quarter under both.
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A bar that wins rarely meet
A merge of two sorted lists of repeated values saves most of a textbook merge's comparisons by galloping along a run of equal answers once a bar of them have come in a row, with the bar lowered after a gallop that pays and raised after one that does not. Timsort keeps the same kind of bar for the other kind of stretch, one list winning repeatedly, and a merge of repeated values meets both. One count and one bar for either kind was predicted to save a further tenth under Zipf 1 and change nothing on uniform values. It saves at most 1.0% under Zipf 1, costs up to 1.3% on uniform values, and two separate bars do no better. In two lists drawn from the same distribution a stretch of wins long enough to gallop over is rare: under Zipf 1 at most 18% of the shared bar's gallops are over wins, and where wins dominate the gallops, at a thousand uniform values, the runs of either kind are too short for any gallop to pay.