Analysis of Approximate Quickselect and Related Problems
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1 Analysis of Approximate Quickselect and Related Problems Conrado Martínez Univ. Politècnica Catalunya Joint work with A. Panholzer and H. Prodinger LIP6, Paris, April 2009
2 Introduction Quickselect finds the kth smallest element out of n given elements with average cost (n) Approximate Quickselect selects, out of n given elements, an element whose rank k fails within a prespecified rank [i::j] We analyze the exact number of passes (recursive calls) and number of key comparisons made by AQS, for arbitrary i, j and n Asymptotic estimates follow easily from the exact results
3 Introduction Quickselect finds the kth smallest element out of n given elements with average cost (n) Approximate Quickselect selects, out of n given elements, an element whose rank k fails within a prespecified rank [i::j] We analyze the exact number of passes (recursive calls) and number of key comparisons made by AQS, for arbitrary i, j and n Asymptotic estimates follow easily from the exact results
4 Introduction Quickselect finds the kth smallest element out of n given elements with average cost (n) Approximate Quickselect selects, out of n given elements, an element whose rank k fails within a prespecified rank [i::j] We analyze the exact number of passes (recursive calls) and number of key comparisons made by AQS, for arbitrary i, j and n Asymptotic estimates follow easily from the exact results
5 Introduction Quickselect finds the kth smallest element out of n given elements with average cost (n) Approximate Quickselect selects, out of n given elements, an element whose rank k fails within a prespecified rank [i::j] We analyze the exact number of passes (recursive calls) and number of key comparisons made by AQS, for arbitrary i, j and n Asymptotic estimates follow easily from the exact results
6 Introduction The techniques we use to analyze Approximate Quickselect prove useful to analyze other problems as well 1 The number of moves in which the ith element gets involved when selecting the jth smallest element out of n 2 The number of common ancestors of the nodes of rank i and j in a random binary search tree (BST) of size n 3 The size of the smallest subtree containing the nodes i and j in a random BST of size n 4 The distance (number of edges) between the nodes i and j in a random BST of size n We analyze also Approximate Multiple Quickselect, an algorithm to find q elements with ranks failing in prespecified ranges [i 1 ::j 1 ]; [i 2 ::j 2 ]; : : : ; [i q ::j q ]
7 Introduction The techniques we use to analyze Approximate Quickselect prove useful to analyze other problems as well 1 The number of moves in which the ith element gets involved when selecting the jth smallest element out of n 2 The number of common ancestors of the nodes of rank i and j in a random binary search tree (BST) of size n 3 The size of the smallest subtree containing the nodes i and j in a random BST of size n 4 The distance (number of edges) between the nodes i and j in a random BST of size n We analyze also Approximate Multiple Quickselect, an algorithm to find q elements with ranks failing in prespecified ranges [i 1 ::j 1 ]; [i 2 ::j 2 ]; : : : ; [i q ::j q ]
8 Introduction The techniques we use to analyze Approximate Quickselect prove useful to analyze other problems as well 1 The number of moves in which the ith element gets involved when selecting the jth smallest element out of n 2 The number of common ancestors of the nodes of rank i and j in a random binary search tree (BST) of size n 3 The size of the smallest subtree containing the nodes i and j in a random BST of size n 4 The distance (number of edges) between the nodes i and j in a random BST of size n We analyze also Approximate Multiple Quickselect, an algorithm to find q elements with ranks failing in prespecified ranges [i 1 ::j 1 ]; [i 2 ::j 2 ]; : : : ; [i q ::j q ]
9 Introduction The techniques we use to analyze Approximate Quickselect prove useful to analyze other problems as well 1 The number of moves in which the ith element gets involved when selecting the jth smallest element out of n 2 The number of common ancestors of the nodes of rank i and j in a random binary search tree (BST) of size n 3 The size of the smallest subtree containing the nodes i and j in a random BST of size n 4 The distance (number of edges) between the nodes i and j in a random BST of size n We analyze also Approximate Multiple Quickselect, an algorithm to find q elements with ranks failing in prespecified ranges [i 1 ::j 1 ]; [i 2 ::j 2 ]; : : : ; [i q ::j q ]
10 Introduction The techniques we use to analyze Approximate Quickselect prove useful to analyze other problems as well 1 The number of moves in which the ith element gets involved when selecting the jth smallest element out of n 2 The number of common ancestors of the nodes of rank i and j in a random binary search tree (BST) of size n 3 The size of the smallest subtree containing the nodes i and j in a random BST of size n 4 The distance (number of edges) between the nodes i and j in a random BST of size n We analyze also Approximate Multiple Quickselect, an algorithm to find q elements with ranks failing in prespecified ranges [i 1 ::j 1 ]; [i 2 ::j 2 ]; : : : ; [i q ::j q ]
11 Introduction The techniques we use to analyze Approximate Quickselect prove useful to analyze other problems as well 1 The number of moves in which the ith element gets involved when selecting the jth smallest element out of n 2 The number of common ancestors of the nodes of rank i and j in a random binary search tree (BST) of size n 3 The size of the smallest subtree containing the nodes i and j in a random BST of size n 4 The distance (number of edges) between the nodes i and j in a random BST of size n We analyze also Approximate Multiple Quickselect, an algorithm to find q elements with ranks failing in prespecified ranges [i 1 ::j 1 ]; [i 2 ::j 2 ]; : : : ; [i q ::j q ]
12 The algorithm Ensure: Array A[l::r ], integers i and j with l i j r Require: Returns a value k, with i k j, A[k ] has rank between i l + 1 and j l + 1 in the array A[l::r ] procedure AQS(A, i, j, l, r) if r l j i then return l end if PARTITION(A, l, r, k) { 8m : (l m < k ) ) A[m] A[k ], and 8m : (k < m r ) ) A[k ] A[m] } if j < k then return AQS(A, i, j, l, k 1) else if i > k then return AQS(A, i, j, k + 1, r) else return k end if end procedure
13 The number of passes P n;i;j = the average number of recursive calls to AQS to select an element of rank k 2 [i::j] out of n P n;i;j = 1+ 1 n Xi 1 k=1 P n k ;i k ;j k + 1 n nx k=j+1 P k 1;i;j ; for 1 i j n
14 The number of passes P n;i;j = the average number of recursive calls to AQS to select an element of rank k 2 [i::j] out of n P n;i;j = 1+ 1 n Xi 1 k=1 P n k ;i k ;j k + 1 n The initial recursive call nx k=j+1 P k 1;i;j ; for 1 i j n
15 The number of passes P n;i;j = the average number of recursive calls to AQS to select an element of rank k 2 [i::j] out of n P n;i;j = 1+ 1 n Xi 1 k=1 P n k ;i k ;j k + 1 n nx k=j+1 P k 1;i;j ; for 1 i j n If the pivot lands at k < i we continue in the right subarray of n k elements looking for an element with rank in [i k ::j k ]
16 The number of passes P n;i;j = the average number of recursive calls to AQS to select an element of rank k 2 [i::j] out of n P n;i;j = 1+ 1 n Xi 1 k=1 P n k ;i k ;j k + 1 n nx k=j+1 P k 1;i;j ; for 1 i j n If the pivot lands at k > j we continue in the left subarray of k 1 elements looking for an element with rank in [i::j]
17 The number of passes P n;i;j = the average number of recursive calls to AQS to select an element of rank k 2 [i::j] out of n P n;i;j = 1+ 1 n Xi 1 k=1 P n k ;i k ;j k + 1 n nx k=j+1 If the pivots lands at k, i k j, we are done P k 1;i;j ; for 1 i j n
18 The number of comparisons C n;i;j = the average number of key comparisons in AQS to select an element of rank k 2 [i::j] out of n C n;i;j = n 1+ 1 n Xi 1 k=1 C n k ;i k ;j k + 1 n nx k=j+1 C k 1;i;j ; for 1 i j n
19 The generic trivariate recurrence T n;i;j = generic toll function X n;i;j = T n;i;j + 1 n Xi 1 k=1 X n k ;i k ;j k + 1 n nx k=j+1 X k 1;i;j ; for 1 i j n Example T n;i;j = 1 ) passes T n;i;j = n 1 ) comparisons T n;i;j = n 6 + O(1) ) swaps i = j ) Quickselect
20 The generic trivariate recurrence Define: X (z; u 1 ; u 2 ) := T (z; u 1 ; u 2 ) := XXX i1 ji nj XXX i1 ji nj The trivariate recurrence X (z; u 1; u 2 ) = X n;i;jz n u i 1 uj 2 ; T n;i;jz n u i 1 uj 2 : 1 u 1u 2 1 z + X (z; u 1 ; u 2 1 zu 1 u T (z; u 1; u with initial condition X (0; u 1 ; u 2 ) = 0.
21 The generic trivariate recurrence Define: X (z; u 1 ; u 2 ) := T (z; u 1 ; u 2 ) := XXX i1 ji nj XXX i1 ji nj The trivariate recurrence X (z; u 1; u 2 ) = X n;i;jz n u i 1 uj 2 ; T n;i;jz n u i 1 uj 2 : 1 u 1u 2 1 z + X (z; u 1 ; u 2 1 zu 1 u T (z; u 1; u with initial condition X (0; u 1 ; u 2 ) = 0.
22 The generic trivariate recurrence Lemma X (z; u 1 ; u 2 ) = 1 (1 z)(1 zu 1 u 2 ) Z z 0 (1 t)(1 u 1 u T (t; u 1; u 2 ) dt:
23 The generic trivariate recurrence Theorem X n;i;j = + + Xi 1 `=1 k=j Xi 1 `=1 Xn 1 k=j n i+` X 1 i+` 2T k ;`;j T n i+`;`;j i+` n i + ` + 1 T k ;i;j k T n;i;j i+` (k + 1)(k + 2) Setting i = j we rederive the generic solution for Quickselect-like recurrences by Kuba (2006).
24 Analyzing AQS Theorem The expected number of passes in AQS is P n;i;j = H j + H n i+1 2H j i+1 + 1: The expected number of comparisons in AQS is C n;i;j = 2(n + 1)H n + 2(j i + 4)H j i+1 2(j + 2)H j 2(n i + 3)H n i+1 + 2n j + i 2: ( ) H n := X 1kn 1 k ; nth harmonic number
25 Moves in Quickselect Consider the ith smallest element in the array A[1::n]. How many times do we move it around when selecting the jth smallest element in A?
26 Moves in Quickselect We make a case analysis, comparing with the position k where the pivot lands after a partitioning step: if k < i j or k < j i, Quickselect continues recursively in A[k + 1::n] that does contain ith element if j i < k or i j < k, Quickselect continues in A[1::k 1] that does contain the ith element if i k j or j k i, either Quickselect stops or continues in a subarray not containing ith element
27 Moves in Quickselect We make a case analysis, comparing with the position k where the pivot lands after a partitioning step: if k < i j or k < j i, Quickselect continues recursively in A[k + 1::n] that does contain ith element if j i < k or i j < k, Quickselect continues in A[1::k 1] that does contain the ith element if i k j or j k i, either Quickselect stops or continues in a subarray not containing ith element
28 Moves in Quickselect We make a case analysis, comparing with the position k where the pivot lands after a partitioning step: if k < i j or k < j i, Quickselect continues recursively in A[k + 1::n] that does contain ith element if j i < k or i j < k, Quickselect continues in A[1::k 1] that does contain the ith element if i k j or j k i, either Quickselect stops or continues in a subarray not containing ith element
29 Moves in Quickselect To find the toll function (number of moves where i participates) in a single partitioning step, we also consider three cases: 1 i = k ) the element i is moved (once) 2 i < k ) the element i is moved if it were in A[k ::n] ) prob = (n k + 1)=(n 1) 3 i > k ) the element is moved if it were in A[2::k ] ) prob = (k 1)=(n 1)
30 Moves in Quickselect M n;i;j:= The expected number of moves of the ith element when selecting the jth smallest out of n M n;i;j = 1 n Xi 1 k=1 M n k ;i k ;j k + 1 n nx k=j+1 M k 1;i;j + (i 1)(i 2) 2n(n 1) + (n i)(n i + 1) 2n(n 1) + 1 n ; 1 i < j n Analogous recurrences for i = j and j < i, all three solved using the theorem for trivariate recurrences
31 Moves in Quickselect M n;i;j:= The expected number of moves of the ith element when selecting the jth smallest out of n M n;i;j = 1 n Xi 1 k=1 M n k ;i k ;j k + 1 n nx k=j+1 M k 1;i;j + (i 1)(i 2) 2n(n 1) + (n i)(n i + 1) 2n(n 1) + 1 n ; 1 i < j n Analogous recurrences for i = j and j < i, all three solved using the theorem for trivariate recurrences
32 Binary search trees A n;i;j:= average # of common ancestors of nodes i and j S n;i;j:= average size of smallest subtree containing nodes i and j D n;i;j:= average distance (# of edges) between nodes i and j
33 Binary search trees A n;i;j:= average # of common ancestors of nodes i and j S n;i;j:= average size of smallest subtree containing nodes i and j D n;i;j:= average distance (# of edges) between nodes i and j
34 Binary search trees A n;i;j:= average # of common ancestors of nodes i and j S n;i;j:= average size of smallest subtree containing nodes i and j D n;i;j:= average distance (# of edges) between nodes i and j
35 Binary search trees Example A 16;8;12 = 3 (nodes 15, 5, 10) S 16;8;12 = 9 (subtree rooted at 10) D 16;8;12 = 3
36 Binary search trees if T = (L; R) is random binary search tree (BST) of size n > 0, then 1 Any element has identical probability of being the root, thus PrfjLj = k 1 j jt j = ng = 1 n ; 1 k n 2 L and R are independent random BSTs of sizes k 1 and n k, 1 k n
37 Binary search trees if T = (L; R) is random binary search tree (BST) of size n > 0, then 1 Any element has identical probability of being the root, thus PrfjLj = k 1 j jt j = ng = 1 n ; 1 k n 2 L and R are independent random BSTs of sizes k 1 and n k, 1 k n
38 Common ancestors Suppose that the root of the BST is the kth element. if i j < k the number of common ancestors is 1 + the number of common ancestors in a random of BST of size k 1 if k < i j the number of common ancestors is 1 + the number of common ancestors in a random of BST of size n 1 k (of the nodes of ranks i k and j k!) Lemma if i k j, then there is only one common ancestor (k) A n;i;j = P n;i;j = passes in AQS
39 Common ancestors Suppose that the root of the BST is the kth element. if i j < k the number of common ancestors is 1 + the number of common ancestors in a random of BST of size k 1 if k < i j the number of common ancestors is 1 + the number of common ancestors in a random of BST of size n 1 k (of the nodes of ranks i k and j k!) Lemma if i k j, then there is only one common ancestor (k) A n;i;j = P n;i;j = passes in AQS
40 Common ancestors Suppose that the root of the BST is the kth element. if i j < k the number of common ancestors is 1 + the number of common ancestors in a random of BST of size k 1 if k < i j the number of common ancestors is 1 + the number of common ancestors in a random of BST of size n 1 k (of the nodes of ranks i k and j k!) Lemma if i k j, then there is only one common ancestor (k) A n;i;j = P n;i;j = passes in AQS
41 Common ancestors Suppose that the root of the BST is the kth element. if i j < k the number of common ancestors is 1 + the number of common ancestors in a random of BST of size k 1 if k < i j the number of common ancestors is 1 + the number of common ancestors in a random of BST of size n 1 k (of the nodes of ranks i k and j k!) Lemma if i k j, then there is only one common ancestor (k) A n;i;j = P n;i;j = passes in AQS
42 Size of subtree of LCA The recurrence for S n;i;j follows the usual pattern: S 1 n;i;j = n 1 = n Xi 1 k=1 Xi 1 k=1 S n k ;i k ;j k + 1 n S n k ;i k ;j k + 1 n nx k=j+1 nx k=j+1 S k 1;i;j + 1 n jx k=i n S k 1;i;j + j i + 1
43 Size of subtree of LCA The recurrence for S n;i;j follows the usual pattern: S 1 n;i;j = n 1 = n Xi 1 k=1 Xi 1 k=1 S n k ;i k ;j k + 1 n S n k ;i k ;j k + 1 n nx k=j+1 nx k=j+1 S k 1;i;j + 1 n jx k=i n S k 1;i;j + j i + 1 The toll function is n only when the pivot k satisfies i k j
44 Size of subtree of LCA Apply the theorem for trivariate recurrences with T n;i;j := j i + 1 Theorem S n;i;j = (j i + 1) H j + H n i+1 2H j i = (j i + 1) A n;i;j (j i + 1) log j + log(n i + 1) 2 log(j i + 1) + 1
45 Distance The recurrence for D n;i;j: D n;i;j = 1 n + 1 n Xi 1 k=1 D n k ;i k ;j k + 1 n jx k=i nx D k 1;i;j k=j+1 (A k 1;i;i + A n k ;j k ;j k + 2)
46 Distance The recurrence for D n;i;j: D n;i;j = 1 n + 1 n Xi 1 k=1 D n k ;i k ;j k + 1 n jx k=i nx D k 1;i;j k=j+1 (A k 1;i;i + A n k ;j k ;j k + 2) If k is the LCA of i and j, the distance between i and j is the depth of i in the left subtree (A k 1;i;i), plus the depth of j (the (j k )th element) in the right subtree (A n k ;j k ;j k), plus 2
47 Distance Since we know A n;i;j, we can obtain the toll function for distances 1 n jx k=i (A k 1;i;i + A n k ;j k ;j k + 2) = j i + 1 n H i + H n+1 j + 2H j+1 i 2
48 Distance The last step is to apply the theorem of trivariate recurrences with the toll function above (quite laboriously!) Theorem D n;i;j = 4H j+1 i (H j H i ) (H n+1 i H n+1 j ) 3
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