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CS 341 (26)
Lecture

# Lecture #5 - Closest pair and Linear Time Selection Fall 2009

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School
University of Waterloo
Department
Computer Science
Course
CS 341
Professor
Forbes Burkowski
Semester
Fall

Description
Closest Pair and Linear Time Selection by Brancusi Closest Pair • Problem definition: –We are given n points p = (x, y), i = 1,…, n. i i i –How far apart are the closest pair of points? • We need the i and j values that minimize the Euclidean distance: 2 (xi− xj) +(yi− yj)2 • Then return this distance. –Brute force: compute distances for all n(n-1)/2 pairs and find the minimum. • This algorithm runs in time Θ(n 2). 1 Closest Pair by Divide and Conquer • Sort by x-coordinate, then : • Divide: – Find vertical line splitting points in half. • Conquer: – Recursively find closest pairs in each half. • Combine: – Check vertices near the border to see if any pair straddling the border is closer together than the minimum seen so far. • Our goal: – Θ(n) overhead so that the total run time is Θ(n log n). Implementation Details • Input: a set of points P sorted with respect to the x coordinate. • Initially, P = all points, and we pay Θ(n log n) to sort them before making the first call to the recursive subroutine. • Given the sorted points, it is easy to find the dividing line. – Let PLbe points to the left of the dividing line. – Let PRbe points to the right of the dividing line. 2 Closest Pair Implementation Details • Recursively: – Find closest pair in P : Let δ beLtheir separation distance – Similarly find closest pair in P , Rith separation distance δ R – Clever observation: If the closest pair straddles the dividing line, then each point of the pair must be within δ = min{δ ,Lδ }Rof the dividing line. • This will usually specify a fairly narrow band for our “straddling”search. Implementation Details – Suppose p and q, d(p,q)= δ δ are candidate closest points, with p on the left and q on the right of the dividing line. • q not to the right oδband. δ • if p = (x, y) then with y coords interval [y δ, y +δ] can be p successfully paired with p. – So, we need only look at points within δ above and below a horizontal line through p. δ – Since the points in this rectangle must be separated by at least δ we have at most 6 points to investigate.(Diagram shows thisDividingLinePossible point q “worst case” situation). 3 Closest Pair Pseudocode • Let P be a global array storing all the points with P Rnd P deLined as described earlier. • Let QL be the subset of points in P Lhat are at most δ (delta in the code) to the left of the dividing line. • Let QR be the subset of points in P Rhat are at most δ to the right of the dividing line. //-------- main --------- // P contains all the points sort P by x-coordinate; return closest_pair(1, n); Closest Pair Pseudocode function closest_pair(l,r) // Find the closest pair in P[l..r] (sorted by x-coordinate) if size(P) < 2 then return infinity; mid := (l + r)/2; midx := P[mid].x; dl := closest_pair(l, mid); dr := closest_pair(mid + 1, r); // Side effect: P[l..mid] and P[mid+1..r] are now sorted // wrt the y-coordinate delta := min(dl, dr); QL := select_candidates(l, mid, delta, midx); QR := select_candidates(mid + 1, r, delta, midx); dm := delta_m(QL, QR, delta); // Use merge to make P[l..r] sorted by y-coordinate Merge(l, mid, r); // Merge as in MergeSort return min(dm, dl, dr); 4 Closest Pair Pseudocode function select_candidates(l,r,delta,midx) // From P[l..r] select all points which are // at most delta from midx line create empty array Q; for i := l to r do if (abs(P[i].x - midx) <= delta) add P[i] to Q; return Q; Closest Pair Pseudocode function delta_m(QL,QR,delta) // Are there two points p in QL, q in QR such that // d(p,q)<=delta? Return distance of closest pair. // Assume QL and QR are sorted by the y coordinate. j := 1; dm := delta; for i := 1 to size(QL) do p := QL[i]; // find the bottom-most candidate from QR while (j <= size(QR) and QR[j].y < p.y-delta) do j := j+1; // check all candidates from QR starting with j k := j; while (k <= size(QR) and QR[k].y <= p.y + delta) do dm := min(dm, distance(p, QR[k])); k := k+1; return dm; 5 Closest Pair Analysis – Let T(n) be the time required to solve the problem for n points: • Divide: Θ(1) • Conquer: 2T (n/2) • Combine: Θ(n) – The precise form of the recurrence is: T(n) =T ( n/2 ) ( n/2 )+Θ(n) which we can approximate by: T(n) =2T n(2 + )(n) – Solution: Θ(n log n). Linear-Time Selection • Problem statement: – Given an array A of n numbers A[1..n] and a number i (1 < i < n), find i smallest number in A. th – Definition: Median of A is the n/2 element in A. • Example: If A = (7, 4, 8, 2, 4); then |A| = 5 and the 3 smallest element (and median) is 4. – Trivial solutions for our problem: th • Sort the array and find the i element. – Execution time: (n log n) • If i is small we can find i using a linear scan similar to finding a minimum or maximum in the array. Idea: keep current top i rather than current max only. 6 Linear-Time Selection (page 2) • Strategy: Partition-based (divide and conquer) selection – Choose one element p from array A (pivot element) – Split input into three s
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