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CS473-Algorithms I

CS473-Algorithms I. Lecture X 1 Properties of Ranks. Analysis Of Union By Rank With Path Compression When we use both union-by-rank and path-compression the worst case running time is O( mα ( m , n )) where α ( m , n ) is the very slowly growing inverse of the Ackerman’s function.

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CS473-Algorithms I

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  1. CS473-Algorithms I Lecture X1 Properties of Ranks Lecture X

  2. Analysis Of Union By Rank With Path Compression When we use bothunion-by-rank and path-compression the worst case running time is O(mα(m,n)) whereα(m,n) is the very slowly growing inverse of the Ackerman’s function. In any conceivable application of disjoint-set data structure α(m,n) ≤ 4. Thus, we can view the running time as linear as in practical situations. Lecture X

  3. Analysis Of Union By Rank With Path Compression We shall examine the function α to see just how it slowly grows. Then, rather than presenting the very complex proof ofO(mα(m,n)) we shall offer a simpler proof of a slightly weaker upper boundO(mlg*n)on the running time. We will prove an O(mlg*n) bound on the running time of the disjoint set operations with union-by-rank & path-compression in order to prove this bound. In order to prove this bound: We first prove some simple properties of ranks. Lecture X

  4. Properties of Ranks • LEMMA 1: For all nodes x • rank[x] ≤ rank[p[x]] • Ifx ≠ p[x] thenrank[x] ≤ rank[p[x]] • (b) rank[x] is initially 0 • Increases through time until x ≠ p[x], then, it does not change value • (c) rank[p[x]] is a monotonically increasing function of time Lecture X

  5. Properties of Ranks • Proof of (b) and (c) is trivial • After a MAKE-SET(x) operation x is a root node & rank[x] = 0 • Rank of only root nodes may increase during a LINK operation • A non-root node may never become a root node • Q.E.D • Proof of (a) is by induction on the number of LINK & FIND-SET operations. Lecture X

  6. Inductive Proof Of Lemma 1 • BASIS: True before the first LINK & FIND-SET operations sincerank[x] = rank[p[x]] = 0 for all x’s • INDUCTIVE STEP: Assume that the lemma holds before performing the operation LINK(x,y). • Let rank denote the rank just before the LINK operation. • Let rank' denote the rank just after the LINK operation. Lecture X

  7. Inductive Proof Of Lemma 1 • CASE: rank[x] ≠ rank[y] • Assume without loss of generality that rank[x] < rank[y] • Node y becomes the root of the tree formed by the link • No rank change occurs rank'[x] = rank[x] • rank'[y] = rank[y] • Only the parent of x changes from p[x] = x to p[x] = y but rank'[x] < rank'[y = p[x]] due to case assumption Lecture X

  8. Inductive Proof Of Lemma 1 • CASE: rank[x] = rank[y] • Node y becomes the root of the new tree • Only the rank of node y changes as • rank'[y] = rank[y] + 1 • Only the parent of x changes from p[x] = x to p[y] = y • rank'[y] = rank[y] + 1 = rank[x] + 1 = rank'[x] + 1 • so rank'[y = p[x]] > rank'[x] Lecture X

  9. Proving the Time Bound We will use the aggregate method of amortized analysis to prove the O(mlg*n) time bound. We will assume that we invoke the LINK operations rather than the UNION operation i.e. assume that appropriate FIND-SET operations are performed. The following lemma shows that, even if we count the extra FIND-SET operations the asymptotic running time remains unchanged. Lecture X

  10. Proving the Time Bound LEMMA: Suppose we convert a sequence S' of m'MAKE-SET, UNION and FIND-SET operations into a sequence S of mMAKE-SET, LINK and FIND-SET operations by turning each UNION into two FIND-SET operations followed by a LINK. If sequence S runs in O(mlg*n) time, then, sequence S' runs in O(m'lg*n) time. Lecture X

  11. Proving the Time Bound PROOF: We have m' ≤ m ≤ 3m' since each union operation in S' is converted to 3 operations in sequence S. Since m = O(m') O(mlg*n) for the converted sequence S implies O(m'lg*n) for the original sequence S' of m' operations. Lecture X

  12. Proving the Time Bound • THEOREM: • A sequence of mMAKE-SET, LINK and FIND-SET operations, n of which areMAKE-SEToperations, can be performed on a disjoint-set with union-by-rank and path-compression in worst case time O(mlg*n). • PROOF: • We assess charges corresponding to the actual cost of each set operation then, compute the total number of charges assessed. • Once the entire sequence of operations has been performed, this total gives the actual cost of all the set operations. Lecture X

  13. Proving the Time Bound • The charges assessed to MAKE-SET and LINK operations are simpler one charge per operation. • Since these operations each take O(1) actual time, the charges assessed are equal to the actual costs. • Before discussing charges assessed to the FIND-SET operations we partition node ranks into blocks by putting rank r into block lg*r for r = 0,1,…, . • Recall that is the maximum rank. Lecture X

  14. Proving the Time Bound The blocks are numbered as Bj for j = 0,1,2,…,jmax where jmax = lg*(lg n) = lg*n – 1 For notational convenience, we define for integers j ≥ -1 Lecture X

  15. Proving the Time Bound Then, for j = 0,1,…, (lg*n – 1) the j-th block Bj consists of the set of ranks {B(j-1) + 1, B(j-1) + 2, …, B(j)} Note that two ranks r1 & r2 belong to the same blockiff lg*r1 = lg*r2 Note thatB(j) = 2B(j-1) for j ≥ 0 Lecture X

  16. Proving the Time Bound Examples to the block concept Let n = 64K = 216 = rmax = = = 16 ; r = {0,1,2,3,…,16} jmax = lg*n - 1 = lg - 1 = (3 + 1) - 1 = 3 ; j = {0,1,2,3} B0 = {B(-1)+1,…,B(0)} = { ,…,1} = {0,1} B1 = {B(0)+1,…,B(1)} = {1+1,…,2} = {2} B2 = {B(1)+1,…,B(2)} = {2+1,…,22} = {3,4} B3 = {B(2)+1,…,B(3)} = {22+1,…, } = {5,…,16} = {5,6,7,8,…,16} Lecture X

  17. Proving the Time Bound Examples to the block concept Let n = 264K = rmax = = = 64K = 216; r = {0,1,2,3,…,65536} jmax = lg*n - 1 = lg - 1 = (4 + 1) - 1 = 4 ; j = {0,1,2,3,4} B0 = {0,1} B1 = {2} B2 = {3,4} B3 = {5,6,7,8,…,16} B4 = {17,18,19,20,21,…,65536} Lecture X

  18. Proving the Time Bound We use two types of charges for a find-set operation: Block Charges and Path Charges Consider the FIND-SET(x0) operation Assume that find path where xl is the root of the tree containing node x0 p[xi] = xi+1 for i = 0,1,2,3,…l-1 We assess one block charge to the last node with rank in block j on the find path to the child of the root, i.e. Lecture X

  19. Proving the Time Bound Because ranks strictly increase along any find path we effectively assess block charge to each node where p[xi] = xl (xi is the root or its child) lg*(rank[xi]) < lg*(rank[xi+1]) We assess one path charge to each node on the find path for which we do not assess a block charge. Lecture X

  20. Proving the Time Bound FACT: Once a node other than the root or its child is assessed block charges, it will never again be assessed path charges. PROOF: Each time path compression occurs, the rank of a node for xi which p[xi] ≠ xl remains the same but, a new parent of xi has a rank > rank of its old parent. The difference between the ranks of xi & its parent is a monotonically increasing function of time. Lecture X

  21. Proving the Time Bound Once xi & p[xi] have ranks in different blocks they will always have ranks in different blocks. So, xiwill never again be assessed a path charge. Since we charge once for each node visited in each FIND-SET total number of charges assessed = total number of nodes visited in all FIND-SET operations. Thus, total number of charges represents actual cost of all FIND-SETs we wish to show that this total is O(mlg*n) Lecture X

  22. Proving the Time Bound Bound on the number of block charges At most one block charge assessed for each block number on the FP plus one block charge for the child of the root. Since block numbers change from 0 to lg*n – 1 at most lg*n+1 block charges assessed for each FIND-SET Thus, at most m(lg*n + 1) block charges assessed over all FIND-SET operations. Lecture X

  23. Proving the Time Bound Bound on the number of path charges If a node xi is assessed a path charge, then p[xi] ≠ xl so that xi will be assigned a new parent during path compression. Moreover, xi’s new parent has a higher rank than its old parent. Suppose that, node x’s rank is in block j. Lecture X

  24. Proving the Time Bound How many times can xi be assigned a new parent and thus assessed a path charge before xi is assigned a parent whose rank is in a different block after which xi will never again be assessed a path charge. This number of times is maximized if xi has the lowest rank in its block, namely B(j-1) + 1 and its parents’ rank successively on values B(j-1) + 2, B(j-1) + 3,…,B(j) We conclude that a vertex can be assessed at most B(j) – B(j – 1) – 1 path charges while its rank is in block j. Lecture X

  25. Proving the Time Bound Bound on the number of path charges…continued Hence, we should bound the number of nodes N(j) whose ranks are in block j. By Lemma 4: For j = 0; Lecture X

  26. Proving the Time Bound For j ≥ 1; N(j) ≤ = = ≤ = = = Thus, for all integers j ≥ 0. Lecture X

  27. Proving the Time Bound • Bound on the number of path charges…continued • We can bound the total number of path charges P(n) by summing over all blocks the product of • The maximum number of nodes with ranks in the block, and • The maximum number of path charges per node in that block Lecture X

  28. Proving the Time Bound P(n) ≤ = ≤ = = Thus, the total number of charges incurred by FIND-SET operations is O(m(lg*n + 1) + nlg*n) = O(mlg*n) since m ≥ n Since there are O(n) MAKE-SET & LINK operations with one charge each, the total time is O(mlg*n). Lecture X

  29. Properties of Ranks Define size(x) to be the number of nodes in the tree rooted at x, including node x itself LEMMA2:For all tree roots xsize(x)  2rank[x] PROOF:By induction on the number of LINK operations NOTE:FIND-SET operations change neither the rank not the size of a tree Lecture X

  30. Properties of Ranks • BASIS:True for the first link since size(x)=1 & rank[x]=0 initially • INDUCTION:Assume that lemma holds before performing LINK(x,y) • Let: • rank & size denote the rank & size just beforethe LINK operation • rank' & size' denote the rank & size just afterthe LINK operation Lecture X

  31. Properties of Ranks CASE: rank[x] rank[y] rank'[x] rank[x] & rank'[y] rank[y] Assume without loss of generality that: rank[x]<rank[y] Node y becomes the root of the tree formed by the LINK operation Lecture X

  32. Properties of Ranks size'(y) size(x) +size(y)  2rank[x]+ 2rank[y]DUE TO INDUCTION  2rank[y]  2rank'[y] since rank'[y] rank[y] in this case No ranks or size changes for any nodes other than y. Lecture X

  33. Properties of Ranks CASE: rank[x] rank[y] node y becomes the root of the new tree size'(y) size(x) +size(y)  2rank[x]+2rank[y] 2rank[y]+12rank'[y] Lecture X

  34. Properties of Ranks INDUCTIVE STEP:Assume that the lemma holds before performing the operation FIND-SET(x) Let rank denote the rank just before the FIND-SET operation and rank'denote the rank just after the FIND-SET operation Consider the find-path of node x: FPx {a0=x, a1, a2, ..., aq} Lecture X

  35. Properties of Ranks WHERE:ai for i1, 2,..., q are the ancestors of node xa0 a1p[a0x], a2p[a1], ..., aip[ai-1] ... aq p[aq-1]and aq is the root of the tree containing node x Only the parents of node x andin FPx change as p[x] aq and p[di] aq for i 1, 2, ....q-1 Lecture X

  36. a 9 a 9-1 a Ta 9 i a i a a x = i-1 0 a a a a 9-2 9-1 i-1 i x a = 0 Ta Sa Sa = i i i Properties of Ranks Lecture X

  37. Properties of Ranks • Due to induction; • rank[a0=x] < rank[a1] < … < rank[aq-2] < rank[aq-1] < rank[aq] • However, FIND-SET operation does not make any rank update. Therefore rank'[y] = rank[y] for any node • rank'[x] < rank'[aq=p[x]] • rank'[a1] < rank'[aq=p[a1] • . • . • . • rank'[aq-2] < rank'[aq=p[aq-2]] Lecture X

  38. Properties of Ranks Analysis of FIND-SET(x) operations…continued Let denote the tree rooted at ai just before the FIND-SET, and denote the tree rooted at ai just before the FIND-SET Tree rooted at nodes a2, a3,…,aq-1,aq change after FIND-SET, i.e. = for i = 2, 3, 4,…,q Hence, height of these nodes may change after FIND-SET. In fact, their height may decrease after a FIND-SET. Lecture X

  39. Properties of Ranks Since, FIND-SET operation does not update rank values; (old) rank values are still upper bounds on the heights of trees rooted at these nodes after FIND-SET. In fact, only the rank value of the root nodeaq is important since rank value only of this node may be checked during a future LINK operation. Hence, a tree with greater height may be linked to a tree with smaller height during a LINK operation. However, the important point is the following fact. Lecture X

  40. Properties of Ranks FACT: Ifsize(aq) ≥ 2rank[aq]beforeFIND-SET(x) operation thensize'(aq) ≥ 2rank'[aq] afterFIND-SET(x) operation whereaq is the root of tree containing node x PROOF: Trivial since FIND-SET operation does not change the size and the rank of the tree. Lecture X

  41. x x x x 1 3 5 7 x x x x 2 4 6 8 Properties of Ranks Consider the Following Sequence MAKE-SET(x1) . . . MAKE-SET(x8) rank=1 1 1 1 UNION(x2, x1) rank[x1]=rank[x2]=1 UNION(x4, x3) UNION(x6, x5) UNION(x8, x7) Lecture X

  42. x x 1 5 x x x x 2 6 3 7 x x 4 8 Properties of Ranks Consider the Following Sequence UNION(x3, x1) rank[x1]=rank[x5]=2 UNION(x7, x5) Lecture X

  43. 1 5 4 3 2 8 7 6 1 5 4 3 2 8 7 6 Properties of Ranks Consider the Following Sequence UNION(x8, x4) x5← FIND-SET(x8) x1 ← FIND-SET(x4) LINK(x5,x1) Lecture X

  44. Properties of Ranks Consider the last UNION operation UNION(x8,x4) size(x1) = 4; size(x5) = 4; size'(x1) = 8 23 = 8 rank(x1) = 2; rank(x5) = 2; rank'(x1) = 3 height(x1) = 2; height(x5) = 2; height'(x1) = 2 Hence, rank'[x1] is a loose upper bound on the height of . However, rank'[x1] is a tight lower bound on the log of size of . Lecture X

  45. Properties of Ranks • LEMMA 3: For any integer r ≥ 0 there are at most n/2r nodes of rank r • PROOF: Trivial for r = 0; there are at most n/20 = n nodes of rank 0 • Fix a particular value of r > 0 • Suppose that when we assign a rank r to a node x we attach a label x to all nodes in the tree rooted at x • Lemma 2 assures that at least 2r nodes are labeled each time since size(x) ≥ 2r for a node x with rank r. • Note that, a new rank > 0 is assigned to a node Lecture X

  46. x y y x T T x y • Properties of Ranks • Only we link two trees of equal rank during a UNION • rank[x]=r-1 rank[y]=r-1 rank[y]=r • size[x] nodes are • labeled with x • size[y] ≥2r-1 size[y] ≥2r-1 • size[x] = size[Tx] ≥ 2r Lecture X

  47. Properties of Ranks • Consider the nodes in Tx • Whenever their roots change due to a LINK during a future UNION Lemma 1 ensures that the rank of the new root > r. • Thus, each node is labeled at most once when its root is first assigned the rank r. • Since there are n nodes there are at most n labeled nodes with at least 2r labels assigned for each node of rank r. • If there were more than n/2r nodes of rank r then more than 2r∙n/2r = n nodes would be labeled. • CONTRADICTION Lecture X

  48. Properties of Ranks COROLLARY: Every node has rank at most PROOF: If we let max. rank r > lg n due to Lemma 3, there are at most n / 2r nodes of rank r. However in this case n/2r < n/2lg n = 1 Hence, the number of nodes of rank r > lg n is < 1 Q.E.D. Lecture X

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