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WSPD Applications

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Agenda Reminders Spanner of bounded degree Spanner with logarithmic spanner diameter Applications to other proximity problems

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Reminder: Well Separated Pairs Let s > 0 be a real number and A, B are two finite sets of points in R d. A and B are well separated with respect to s if there are 2 disjoint d-dimensional balls Ca and Cb such that: Ca and Cb have the same radius. Ca contains the axes parallel bounding box R(A) of A. Cb contains the axes parallel bounding box R(B) of B. The distance between Ca and Cb is greater than or equal to s times the radius of Ca.

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Reminder: Well Separated Pairs

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Reminder: WSPD Let S be a set of n points in R d, and let s > 0 be a real number. A WSPD for S in respect to s is a sequence {A1,B1},{A2,B2},…,{A m, B m } of pairs of nonempty subsets of S, for some integer m such that: For each 1≤i≤m, A i and B i are well separated with respect to s. For any two distinct points p and q, there is exactly one index 1≤i≤m such that p Ai and q Bi or the opposite. The integer m is called the size of the WSPD.

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Reminder: Split Tree R(S) is the bounding box of the set S. The split tree for S is a binary tree containing the points of S in its leaves. If S=1, the node consists of the single point. Else, the node consists of S, and it’s 2 sons consist of S’s 2 equal halves S1 & S2. The split tree is used to find WSPD.

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Reminder: Split Tree S S S2 S1 S4S3 S6S5 S3S4 S5S6

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Main Idea of spanner construction Let S be a set of n points in R d Compute a split tree T for S. Construct a WSPD from T. Choose the pairs’ representatives wisely. Connect the representatives. We get a t-spanner with special properties.

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Agenda Reminders Spanner of bounded degree Spanner with logarithmic spanner diameter Applications to other proximity problems: ◦T◦T he closest pair problem ◦C◦C omputing k-closest pairs ◦T◦T he all-nearest neighbors problem ◦C◦C omputing an approximate MST

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Spanner of bounded degree Let S be a set of n points in R d Compute a split tree T for S. Construct a WSPD with s=4(√t + 1) (√t - 1). As shown last week (section 9.2), G=(S,E), where E:= {{a i,b i } : 1≤i≤m}, is a √t-spanner for S. We will choose a i,b i from A i, B i wisely, using the split tree T. Construction

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Spanner of bounded degree For each leaf u in T, let r(u) be the point of S stored in it. For each internal node u of T, let r(u) be the point of S that is stored in the rightmost leaf of the left sub-tree of u. Since every internal node of T has 2 children, r(u) is distinct for every internal node. Representatives selection

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Spanner of bounded degree Representatives selection S S2 S1 S4S3 S6S5 How does S relate to S4 ?

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Spanner of bounded degree Representatives selection Hence, in O(n) total time we can compute r(u) for each node u of T. Let i be 1≤i≤m, the pair A i, B i is represented implicitly by 2 nodes of T, u i & v i, respectively. A i, B i are the set of points stored in the sub-trees of u i & v i. We take r(u i ) A i to be it’s representative. We do the same for r(v i ) B i.

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Spanner of bounded degree Let p be a point of S. There are at most 2 nodes u in T for which r(u) = p. Follows from the way we selected the representatives. <> For each i,1≤i≤m, we direct the pair {A i,B i} as shown last week (lemma 9.4.4). We denote the directed graph as G`. Claim Proof

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Spanner of bounded degree The out degree of of each vertex in G` is ≤ 2((2s + 4)√d +4) d Let p be an arbitrary vertex in G`. We’ve seen that there are at most 2 nodes u in T such that r(u) = p. For each directed pair {A i,B i}, where Ai is the set of points stored in u’s sub-tree, p has one out going edge. Claim Proof

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Spanner of bounded degree All out going edges of p are obtained by the same way. As shown last week (lemma 9.4.5 – the packing lemma), for each node u in T, there are at most ((2s + 4)√d +4) d nodes v in T such that (Su,Sv) is a pair in WSPD(T,s). <> Now we have a directed graph of bounded out degree. We want to transform it into an undirected t-spanner of bounded degree. Proof cont.

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Spanner of bounded degree For each point q in S, let Vq be all the points r S, such that (r,q) is a directed edge in G`. We replace all edges (r,q), by the edges of a q-sink (t/√t)-spanner. Now, we replace all the directed edges in the graph with un-directed edges. We got an undirected t-spanner with bounded degree of O(1/(t-1) 2d-1 )). Proof => Out Of Scope. The transformation

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Spanner of bounded degree WSPD construction: O(nlogn + s d n) as shown last week (Theorem 9.4.6). Representative selection: O(n). Undirected t-spanner of bounded degree transformation: O(nlogn). Proof => Out Of Scope (section 4.2.2). All in all : O(nlogn). Time complexity

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Agenda Reminders Spanner of bounded degree Spanner with logarithmic spanner diameter Applications to other proximity problems: ◦T◦T he closest pair problem ◦C◦C omputing k-closest pairs ◦T◦T he all-nearest neighbors problem ◦C◦C omputing an approximate MST

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Spanner with log. spanner diameter Let S be a set of n points in R d Compute a split tree T for S. Construct a WSPD with s=4(t + 1)/(t - 1). As shown last week (section 9.2), G=(S,E), where E:= {{a i,b i } : 1≤i≤m}, is a t-spanner for S. We will choose a i,b i from A i, B i wisely, using the split tree T. Construction

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Spanner with log. spanner diameter Let u be any internal node of T. Let u l & u r be u’s left and right sons. Let e l & e r be the edges that connect u to u l & u r respectively. Let n l & n r be number of leaves in the sub- trees rooted by u l & u r respectively. If nl ≥ nr, we label el as ‘heavy’, and e r as ‘light’. If nl < nr, we label er as ‘heavy’, and el as ‘light’. Representatives selection

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Spanner with log. spanner diameter Each internal node is connected by exactly one ‘heavy’ edge to one of it’s children. Each internal node has a unique chain of heavy edges down to the bottom of T. Let l(u) be the leaf whose chain contains u. Let r(u) be the point of S stored in l(u). Representatives selection

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Spanner with log. spanner diameter Representatives selection S S2 S1 S6S5 S4 S3 S10 S9S8S7 S12S11

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Spanner with log. spanner diameter Representatives selection Let i be 1≤i≤m, the pair A i, B i is represented implicitly by 2 nodes of T, u i & v i, respectively. A i, B i are the set of points stored in the sub-trees of u i & v i. We take r(u i ) A i to be it’s representative. We do the same for r(v i ) B i. Now G=(S,E) is a t-spanner. We need to prove it’s diameter is logarithmic.

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Spanner with log. spanner diameter For each i,1≤i≤m, and for each point p A i there is a t-spanner path in G between p and the representative a i of A i, that contains at most log|A i | edges. From construction, there is a t-spanner path between p and a i. If p = a i, we are done. Assume otherwise. Claim Proof

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Spanner with log. spanner diameter Let j be the index such that: (i) p A j and a i B j. Or (ii) ai A j and p B j. WLOG assume (i). Recursively construct a path Q1 between p and it’s representative a j. Recursively construct a path Q2 between a i and it’s representative b j. Return Q = Q1, {a j, b j }, Q2. Proof cont.

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Spanner with log. spanner diameter We will show that |Q| ≤ log|A i | (p A i ). Induction on the size of set A i. Base: Ai = {a i } => p = a i => we are done. Let i be an index such that |A i | > 1, p ≠ a i. Assume, for all k such that |A k | < |A i |, and for all x A k, the algorithm constructs a t-spanner path between x and a k with at most log|A k | edges. Proof cont.

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Spanner with log. spanner diameter Let u i, u j, v j be the nodes of T whose sub- trees store the sets A i, A j, B j respectively. u i is a common ancestor of the leaves storing p and a i. u j lies on the path from root to the leaf of p. v j lies on the path from root to the leaf of a i A j, B j are disjoint => u j, v j are both in the sub-tree of u i, and neither is an ancestor of the other. Proof cont.

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Spanner with log. spanner diameter Proof cont. S Ui Vj Uj p aiai => a i = b j. => |Q2|=0

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Spanner with log. spanner diameter |A j | ≤ |A i |/2. Otherwise all the edges between u i and u j would be ‘heavy’. But then, the representative of Ai would be an element of Aj, but it’s not (a i Bj represents Ai). By induction hypothesis, the path Q1 between p and a j contains at most log|A j | edges, which is ≤ log|A i | - 1. We add {a j, b j } and get that |Q| ≤ log|A i |. <> Proof cont.

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Spanner with log. spanner diameter G is a t-spanner for S, whose diameter is less than equal to 2logn – 1. Let p and q be 2 distinct points in G. We will show that there is a t-spanner path between them with at most 2logn–1 edges. Claim Proof

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Spanner with log. spanner diameter Let j be the index such that: (i) p A j and q B j. Or (ii) q A j and p B j. WLOG assume (i). Recursively construct a path Q1 between p and it’s representative a j. Recursively construct a path Q2 between a i and it’s representative b j. Return Q = Q1, {a j, b j }, Q2. Proof cont.

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Spanner with log. spanner diameter We proved that |Q1| ≤ log|A j |. Similarly, |Q2| ≤ log|B j |. Hence, |Q| ≤ log|A j | + log|B j | + 1. A j, B j are disjoint => |A j | + |B j | ≤ n. log|A j | + log|B j | ≤ log|A j | + log|n - A j | ≤ 2log(n/2) = 2logn – 2. Hence, Q contains at most 2logn – 1 edges.<> Proof cont.

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Spanner with log. spanner diameter WSPD construction: O(nlogn + s d n) as shown last week (Theorem 9.4.6). Representative selection: O(m). All in all : O(nlogn). Time complexity

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