Series

Structure — the series

8 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 110100257amortised 2.000256512append numbercost of that append (log scale)growth factor 2, cost = 1 write + a copy of the array when it resizes9 resizes in 512 appends

    What amortised means

    Appending to a dynamic array is O(1) amortised. It is also, on 512 appends, an operation that costs one unit 503 times and 257 units once. The amortised bound is a true statement about the sequence and a false one about any append in it, and the picture that shows why is a sawtooth nobody draws.

    part 1 · structures
  2. sorted insertion — height 62shuffled insertion — height 10truncated at depth 1663 keys, identical set, different arrival order62 deep against 10

    The tree that is a list

    A binary search tree gives logarithmic lookup. Build one from 128 keys in sorted order and it has height 127 — every node has one child, and a lookup is a linear scan. The failure is not gradual and it happens on the input people try first, which makes "O(log n) lookup" a claim about the insertion order rather than about the structure.

    part 1 · structures
  3. 0%25%50%75%12345×1.125×1.25×1.5×2×3×4capacity left unused at the endamortised cost per append20,000 appends, cost = 1 write + a copy on resizeneither end wins

    Choosing a growth factor

    When a dynamic array fills up, how much bigger should the new one be? Doubling costs 2.02 units per append and leaves 39% of the allocation empty. Growing by an eighth costs 9.89 and leaves 10%. Every factor is a trade between time and space, no factor wins on both, and real implementations disagree about the answer for reasons that are measurable.

    part 2 · structures
  4. 10³10⁴10³10⁴10⁵ncomparisonsbottom-up (Floyd)repeated insertionn from 128 to 65,536, random input, seeded1.21× between the two at the right-hand edge

    Building a heap from the bottom

    Bottom-up heap construction is Θ(n) and repeated insertion is Θ(n log n), and the second of those is a worst case quoted as a behaviour. On random input, repeated insertion measures linear too — 2.22 comparisons per element against 1.87 — and the famous logarithmic factor never appears. On ascending input it appears in full, and it is a factor of six.

    part 3 · structures
  5. root, priority 0.9651.00.0prioritykey24 keys, sorted insertion, seed 20260810height 8 against an ideal of 4

    The priority nobody supplied

    Insert 4,096 sorted keys into a binary search tree and it reaches height 4,095, costing 8,386,560 comparisons to build. Give every key a second, random key and keep the tree heap-ordered on that instead, and the same insertion reaches height 26 for 32,750 comparisons. Nothing detected the imbalance, and nothing rebalanced.

    part 4 · structures
  6. three-entry rule (as shipped, 2002–2015)1414101410314103141054+14+10+3+2+4final stack: 14, 10, 5, 414 is not > 10 + 5 — the invariant is brokenfour-entry rule (Java, after the proof)141410141031410333+14+10+3+2+4final stack: 33every triple satisfies the invariantrun lengths 14, 10, 3, 2, 4both outputs are correctly sorted

    The invariant that was wrong for seven years

    Timsort's merge policy is supposed to keep its run stack shallow, and the rule that enforces it inspects the top three entries. In 2015 a group of formal-methods researchers proved that the rule does not imply what it was believed to imply. Thirty-three elements are enough to break it, the array still comes out perfectly sorted, and the defect is in a structure that nothing about the output can show.

    part 5 · wrong
  7. 04812162024283236404448slots from homepale: linear probing · dark: Robin Hoodmean 2.384identical for bothworst 48 → 12var 37 → 7435 keys, 512 slots, seed 20260811displacements, counted exactly

    The probe nobody waits for

    Robin Hood hashing makes an inserting key steal a slot from a key that has probed less far. The mean number of probes afterwards is 4.817, and before it was 4.817 — identical, and it cannot be otherwise, because the total displacement is fixed by the hash. What changes is the worst case, from 114 slots from home to 19, and a table reported by its average lookup cost shows no difference at all.

    part 6 · structures
  8. column height = keys in that bucket · line = threshold of 8a well-spread hashlongest 44 → 4worst lookupthe low bits onlylongest 3131 → 5worst lookupevery key collideslongest 192192 → 8worst lookup192 keys, 256 buckets, seed 20260811threshold 8, 64 buckets shown

    A bucket that becomes a tree

    Java's HashMap converts a chained bucket into a red-black tree once it holds eight entries. The comment in the source computes the probability of that happening under a decent hash at about six in a hundred million, so the mechanism is written never to run. Under a hash that fails, the worst lookup falls from 192 comparisons to 8 — and the whole value of the tree is in a case its author does not control.

    part 7 · structures

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