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Variations on Balanced Trees Lazy Red-Black Trees Stefan Kahrs

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Overview some general introduction on BSTs some specific observations on red-black trees how we can make them lazy - and why we may want to conclusions

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Binary Search Trees commonly used data structure to implement sets or finite maps (only keys shown): 56 33 227

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A problem with ordinary BSTs on random data searching or inserting or deleting an entry performs in O(log(n)) time where n is the number of entries, but... if the data is biased then this can deteriorate to O(n)...and thus a tree-formation can deteriorate to O(n 2 )

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Therefore... people have come up with various schemes that make trees self-balance the idea is always that insertion/deletion pay a O(log(n)) tax to maintain an invariant the invariant guarantees that search or insert or delete all perform in logarithmic time

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Well-known invariants for trees Braun trees: size of left/right subtree vary by at most 1 – too strong for search trees O(n 0.58 ) AVL trees: depth of left/right subtree vary by at most 1 2-3-4 trees: a node has 1 to 3 keys, and 2 to 4 subtrees (special case of B-tree) Red-Black trees: an indirect realisation of 2-3- 4 trees

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Red-Black Tree BST with an additional colour field which can be RED or BLACK invariant 1: red nodes have only black children, root/nil are black thus, a non-empty black node has between 2 and 4 black children invariant 2: all paths to leaves go through the same number of black nodes

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Example 68 12 83 7 43 75 96 98 94 7076

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Perceived Wisdom Red-Black trees are cheaper to maintain than AVL trees, though they may not be quite as balanced pretty balanced though: average path-length for a Red-Black tree is in the worst case only 5% longer that that of a Braun-tree

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Aside: a problem with balanced trees an ordinary BST has on random data an average path length of 2*ln(n) this is only 38% longer than the average path length of a Braun tree thus: most balanced tree schemes lose against ordinary BST on random data, because they fail to pay their tax from those 38% red-black trees succeed though

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Algorithms on RB trees search: unchanged, ignores colour insert: – insert as in BST (a fresh red node) – rotate subtrees until color violation goes away – colour root black delete (more complex than insert): – delete as in BST – if underflow rotate from siblings until underflow goes away

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Example 68 12 83 7 43 75 96 98 94 7076 69

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Example 68 12 83 7 43 75 96 98 94 7076 69

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Example 68 12 83 7 43 75 96 98 94 7076 69

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Standard Imperative Algorithm find the place of insertion in a loop check your parent whether you’re a naughty child, and correct behaviour if necessary, by going up the tree

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Problem with this Question: how do you go up the tree? Answer: children should know their parent. Which means: trees in imperative implementations are often not proper trees, every link consists of two pointers

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Functional Implementations in a pure FP language such as Haskell you don’t have pointer comparison and so parent pointers won’t work instead we do something like this: insert x tree = repair (simplInsert x tree) simplInsert inserts data in subtree and produces a tree with a potential invariant violation at the top, repair fixes that the ancestors sit on the recursion stack

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Recursion actually, nothing stops us from doing likewise in an imperative language, using recursive insertion (or deletion) cost: recursive calls rather than loops benefit: no parent pointers – saves memory and makes all rotations cheaper is still more expensive though...

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Can we do better? problem is that the recursive insertion algorithm is not tail-recursive and thus not directly loopifiable: we repair after we insert what if we turn this around? newinsert x tree = simplinsert x (repair tree) this is the fundamental idea behind lazy red- black trees

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What does that mean? we allow colour violations to happen in the first place these violations remain in the tree we repair them when we are about to revisit a node this is all nicely loopifiable and requires no parent pointers

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In the imperative code where we used to have... n = n.left;...to continue in the left branch we now have: n = n.left = n.left.repair();

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Invariants? the standard red-black tree invariant is broken with this (affects search) in principle, we can have B-R-R-B-R-R-B-R-R paths, though these are rare but this is as bad as it gets, so we do have an invariant that guarantees O(log(n)) average path lengths are similar to RB trees

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Performance? I implemented this in Java, and the performance data were initially inconclusive (JIT compiler, garbage collection) after forcing gc between tests, standard RB remains faster (40% faster on random inputs), though this may still be tweakable so what is the extra cost, and can we do anything about it?

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Checks! most nodes we visit and check are fine especially high up in the tree, as these are constantly repaired...and the ones low down do not matter that much anyway so we could move from regular maintenance to student-flat maintenance, i.e. repair trees only once in a blue moon

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What? yes, the colour invariant goes to pot with that we do maintain black height though......and trust the healing powers of occasional repair: suppose we have a biased insertion sequence and don’t repair for a while...

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Example 12 83 7 43 96 70 suppose the tree has this shape, and now we insert a 5 in repair-mode

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Result 12 83 7 4396 70 5

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Findings on random data, performance of lazy red- black trees is virtually unaffected, even if we perform safe-insert only 1/100 on biased data works a bit better under student-flat, but still loses to RB (15% slower for this bias) average tree depth: 1.5 longer than RB – on random inputs – also on biased inputs (where BST falls off the cliff)

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Conclusions Ultimately: failure! Lazy RB trees are not faster than normal ones. On random inputs, Lazy RB perform very similarly to plain BST Some small room for improvement – I doubt though the gap to plain RB can be closed Perhaps other algorithms would benefit more from lazy invariant maintenance?

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