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Concurrent Search Structure Algorithms Dennis Shasha

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What is a Search Structure? Data structure (typically a B tree, hash structure, R-tree, etc.) that supports a dictionary. Operations are insert key-value pair, delete key-value pair, and search for a key-value pair.

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How to make a search structure algorithm concurrent? Naïve approach: use two phase locking (but then the root is at least read-locked so conflicts occur) Semi-naïve: use hierarchical tree locking: lock root; afterwards lock node n (Still tends to hold locks high in tree.)

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How can we do better? Fundamental insight: In a search structure algorithm, all that we really care about is that we implement the dictionary operations correctly. Operations on structure need not even be serializable provided they maintain certain constraints.

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Train your intuition: parable of the library Bob goes to a library (with books) and looks in the catalogue for a great puzzle book P. He is told it is on stack G. He walks towards G but sees a friend and they have a chat. Alice the librarian moves some books from G to H and leaves a note. She then changes the catalogue. Bob goes to G, sees the note, and finds P on stack H.

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Observe: In no serial execution would Bob visit two stacks – if he had gone after Alice, he would have visited only H; if before, only G. But this is still ok. Why? Intuition: the search is always pointed towards a correct final position.

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Using this for search structures KeySpace = the set of all possible keys (e.g. all possible integers) Inset of a node n = subset of KeySpace that is either in n, a node reachable from n or nowhere in the data structure. Outset of n towards n’ = The subset of KeySpace associated with the edge from n to n’. Depends on the data structure. Keyset of n = inset(n) – U outset(n,n’)

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Binary search tree Root node is 20 Left child is 5 And maybe the left child has descendants. What is the inset of the left child? All values less than 20. Right child of 5 has value 13. What is the outset(5,13) and what is the inset(13)

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Example Suppose the root of a binary search tree has the value 20 and a left child L and right child R. Inset(root) = KeySpace Outset(root,L) = {x| x < 20} Outset(root, R) = {x| x > 20} Keyset(root) = {20}

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Let’s Return to the Library Suppose that, to start, the shelf G has as its inset all books with last name starting with “S”. Alice moves those between “S” and “Si” to shelf H and leaves a note to that effect. Then the keyset of G becomes {x | x begins with “S”} – {x|x <= “Si”} The keyset of G represents the set of books that are in G or nowhere in the structure.

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The Key Invariants If x is in node n, then x is in keyset(n) The keysets partition the KeySpace. If the search for an item x is at node n, then x is in keyset(n) or there is a path from n to node m such that x is in keyset(m) and every edge along the path has x in its outset. The invariants assure that the search, if it terminates, will terminate in the correct place.

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Application: link algorithm Recall splits in B trees. Split n into n and n’ and adjust the parent p to reflect the change. Here is how to do a split: lock(n), move the values from n to n’ as desired and leave a forward pointer to that effect, unlock(n), lock(p), and adjust p, then unlock(p). This is exactly the library scenario.

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Application: give-up algorithm Instead of including a forwarding pointer, just asset explicitly the inset of each node. When splitting, reduce the asserted inset(n), denoted assertinset(n) and establish assertinset(n’). If a search for x arrives at n and x is not in assertinset(n), the search starts over. Happens rarely enough that performance is very good.

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Application: multi-rooted structure Imagine a link B-tree structure with a root at the top of the tree and a root to the left of the leftmost leaf node. Same invariants hold and search can proceed from anywhere.

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Conclusion Simple framework for all search structures. Handful of concepts: KeySpace, inset, outset, keyset. Performs well.

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