Divide and Conquer

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Divide and Conquer 10∗ If n ≠ 2 k , we can add enough rows and columns to make it satisfy our assumption– Each of A, B, and C is partitioned into submatrices of size n 2such that[ ] [ ] [ ]A11 A 12 B11 B 12 C11 C=12A 21 A 22 B 21 B 22 C 21 C 22– This yields eight multiplications of n 2 × n 2 matrices (T n ), and four additions of the same size of matrices (assume2cn 2 for some constant c, starting with 4 · ( n 2 )2 )– The recurrence is{ b n ≤ 2T n =8T n + cn2 n > 22– A solution of this recurrence yields T n = O(n 3 ) (no improvement)– Observation: Matrix multiplications are more expensive (O(n 3 )) than matrix additions (O(n 2 ))– Strassen’s method relies on devising a clever workaround to minimize the number of multiplications while increasingthe number of additions∗ 7 multiplications and 18 additions/subtractions– The intermediate computations are– The elements of C are given asP 1 = (A 11 + A 22 )(B 11 + B 22 )P 2 = (A 21 + A 22 )B 11P 3 = A 11 (B 12 − B 22 )P 4 = A 22 (B 21 − B 11 )P 5 = (A 11 + A 12 )B 22P 6 = (A 21 − A 11 )(B 11 + B 12 )P 7 = (A 12 − A 22 )(B 21 + B 22 )C 11 = P 1 + P 4 − P 5 + P 7C 12 = P 3 + P 5C 21 = P 2 + P 4C 22 = P 1 + P 3 − P 2 + P 6– The recurrence relation iswhere a and b are constants– Solving the recurrence{ b n ≤ 2T n =7T n + an2 n > 22T n = an 2 [1 + 7 ( 2 ( ) k−1 7 74 4) + + · · · + ] + 7 k T (1)4( lg n 7≤ cn4) 2 + 7 lg n , c is a constant= cn lg 4+lg 7−lg 4 + n lg 7= O(n lg 7 )≈ O(n 2.81 )Convex hull
Divide and Conquer 11• Structure (smallest convex polygon) used in the construction of geometric structures• Polygon: A piecewise-linear, closed curve in a plane– Curve composed of a sequence of straight line segments, or sides of polygon– Curve starts and ends at the same point– A point that is common to two sides is called a vertexDefinition 1 A polygon is defined to be convex if for any two vertices p 1 and p 2 inside the polygon, the directed linesegment from p 1 to p 2 (denoted as 〈p 1 , p 2 〉) is fully contained in the polygon.Definition 2 The convex hull of a set S of points in the plane is defined to be the smallest convex polygon containing allthe points of S.• Vertices of the convex hull of a set S of points form a [not necessarily proper] subset of S• Two variants of the convex hull problem1. Obtain the vertices (extreme points) of the convex hull2. Obtain the vertices of the convex hull in some order, such as clockwise• Obtaining extreme points of a given set S of points in a planefor each p in S{for each possible triplet of points in Sif p is not inside the triangle formed by any tripletmark p as an extreme point;}– Testing for p being inside a given triangle is performed in Θ(1) time– Number of possible triangles is Θ(n 3 )– Since there are n points, the above algorithm runs in Θ(n 4 ) time• Divide and conquer allows us to solve the convex hull problem (either form) in O(n log n) time• Geometric primitives– Let A be an n × n matrix with elements denoted by a ijDefinition 3 The ijth minor of A, denoted by A ij , is defined to be the submatrix of A obtained by deleting the ithrow and jth column.Definition 4 The determinant of A, denoted by |A|, is given by{a11 n = 1|A| =a 11 · |A 11 | − a 12 · |A 12 | + · · · + (−1) n−1 · |A 1n | n > 1– Consider the directed line segment 〈p 1 , p 2 〉 from some point p 1 = (x 1 , y 1 ) to some other point p 2 = (x 2 , y 2 ). Ifq = (x 3 , y 3 ) is another point, we say q is to the left [right] of 〈p 1 , p 2 〉 if the angle p 1 p 2 q is a left [right] turn.∗ An angle θ is said to be a left turn if θ ≤ 180; otherwise, it is considered to be a right turn∗ We can check whether q is to the left or right of 〈p 1 , p 2 〉 by evaluating the determinant of their coordinatesx 1 x 2 x 3y 1 y 2 y 3∣ 1 1 1 ∣
Page 1 and 2: Divide and ConquerGeneral methodDiv
Page 3 and 4: Divide and Conquer 3- Pitfall∗ T
Page 5 and 6: Divide and Conquer 5· Case 2: n lo
Page 7 and 8: Divide and Conquer 7return ( r );el
Page 9: Divide and Conquer 9k−1∑= 2 k
Page 13 and 14: Divide and Conquer 13p3 = p, for wh

Divide and Conquer

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