CS 3343 -- Spring 2009 Powering a number Problem: Compute a n , where n ∈ N . Naive algorithm: Θ ( n ). Divide-and-conquer algorithm: (recursive squaring) a n/ 2 ⋅ a n/ 2 if n is even; a n = More Divide & Conquer a ( n– 1) / 2 ⋅ a ( n– 1) / 2 ⋅ a if n is odd. Carola Wenk T ( n ) = T ( n /2) + Θ (1) ⇒ T ( n ) = Θ (log n ) . Slides courtesy of Charles Leiserson with small changes by Carola Wenk 2/5/09 CS 3343 Analysis of Algorithms 1 2/5/09 CS 3343 Analysis of Algorithms 2 Computing Fibonacci Fibonacci numbers numbers Recursive definition: Naive recursive squaring: F n = φ n / 5 rounded to the nearest integer. 0 if n = 0; • Recursive squaring: Θ (log n ) time. 1 if n = 1; F n = F n– 1 + F n –2 if n ≥ 2. • This method is unreliable, since floating-point arithmetic is prone to round-off errors. 0 1 1 2 3 5 8 13 21 34 L Bottom-up (one-dimensional dynamic programming) : Naive recursive algorithm: Ω ( φ n ) • Compute F 0 , F 1 , F 2 , …, F n in order, forming ( + (exponential time), where φ = 1 5 ) / 2 each number by summing the two previous. is the golden ratio . • Running time: Θ ( n ). 2/5/09 CS 3343 Analysis of Algorithms 3 2/5/09 CS 3343 Analysis of Algorithms 4 1
Convex Hull Convex Hull: Divide & Conquer • Preprocessing: sort the points by x- • Given a set of pins on a pinboard 4 coordinate 5 3 • And a rubber band around them • Divide the set of points into two sets A and B : • How does the rubber band look 6 • A contains the left n/2 points, when it snaps tight? 2 • B contains the right n/2 points • Recursively compute the convex • We represent convex hull as the 1 0 hull of A sequence of points on the convex A B • Recursively compute the convex hull polygon, in counter-clockwise order. hull of B • Merge the two convex hulls 2/5/09 CS 3343 Analysis of Algorithms 5 2/5/09 CS 3343 Analysis of Algorithms 6 Merging Finding the lower tangent 3 • Find upper and lower tangent a = rightmost point of A 4=b • With those tangents the convex hull b = leftmost point of B 4 2 of A ∪ B can be computed from the while T=ab not lower tangent to both convex hulls of A and the convex hull 3 convex hulls of A and B do{ 5 of B in O(n) linear time 5 a=2 while T not lower tangent to 6 1 convex hull of A do{ 7 a=a-1 1 } 0 0 while T not lower tangent to A B A B convex hull of B do{ b=b+1 } right turn or can be checked } left turn? in constant time 2/5/09 CS 3343 Analysis of Algorithms 7 2/5/09 CS 3343 Analysis of Algorithms 8 2
Convex Hull: Runtime Convex Hull: Runtime • Preprocessing: sort the points by x- • Runtime Recurrence: O(n log n) just once coordinate T(n) = 2 T(n/2) + cn • Divide the set of points into two O(1) sets A and B : • A contains the left n/2 points, • Solves to T(n) = Θ (n log n) • B contains the right n/2 points • Recursively compute the convex T(n/2) hull of A • Recursively compute the convex T(n/2) hull of B O(n) • Merge the two convex hulls 2/5/09 CS 3343 Analysis of Algorithms 9 2/5/09 CS 3343 Analysis of Algorithms 10 Matrix multiplication Standard algorithm Input: A = [ a ij ], B = [ b ij ]. for i ← 1 to n i , j = 1, 2,… , n . Output: C = [ c ij ] = A ⋅ B . do for j ← 1 to n do c ij ← 0 c c c a a a b b b L L L 11 12 1 n 11 12 1 n 11 12 1 n for k ← 1 to n c c c a a a b b b L L L 21 22 2 n 21 22 2 n 21 22 2 n = ⋅ do c ij ← c ij + a ik ⋅ b kj M M O M M M O M M M O M c c c a a a b b b L L L n 1 n 2 nn n 1 n 2 nn n 1 n 2 nn Running time = Θ ( n 3 ) n ∑ = ⋅ c a b ij ik kj = k 1 2/5/09 CS 3343 Analysis of Algorithms 11 2/5/09 CS 3343 Analysis of Algorithms 12 3
Divide-and-conquer algorithm Analysis of D&C algorithm I DEA : T ( n ) = 8 T ( n /2) + Θ ( n 2 ) n × n matrix = 2 × 2 matrix of ( n /2) ×( n /2) submatrices: r s a b e f work adding = ⋅ # submatrices t u c d g h submatrices submatrix size ⋅ C = A B n log ba = n log 2 8 = n 3 ⇒ C ASE 1 ⇒ T ( n ) = Θ ( n 3 ). r = a·e+b·g s = a·f+ b·h 8 recursive mults of ( n /2) ×( n /2) submatrices 4 adds of ( n /2) ×( n /2) submatrices No better than the ordinary algorithm. t = c·e+d·g u = c·f +d·h 2/5/09 CS 3343 Analysis of Algorithms 13 2/5/09 CS 3343 Analysis of Algorithms 14 Strassen’s idea Strassen’s idea • Multiply 2 × 2 matrices with only 7 recursive mults. • Multiply 2 × 2 matrices with only 7 recursive mults. P 1 = a ⋅ ( f – h ) P 1 = a ⋅ ( f – h ) r = P 5 + P 4 – P 2 + P 6 r = P 5 + P 4 – P 2 + P 6 P 2 = ( a + b ) ⋅ h P 2 = ( a + b ) ⋅ h s = P 1 + P 2 = ( a + d )( e + h ) P 3 = ( c + d ) ⋅ e P 3 = ( c + d ) ⋅ e t = P 3 + P 4 + d ( g – e ) – ( a + b ) h P 4 = d ⋅ ( g – e ) P 4 = d ⋅ ( g – e ) u = P 5 + P 1 – P 3 – P 7 + ( b – d )( g + h ) P 5 = ( a + d ) ⋅ ( e + h ) P 5 = ( a + d ) ⋅ ( e + h ) = ae + ah + de + dh 7 mults, 18 adds/subs. 7 mults, 18 adds/subs. P 6 = ( b – d ) ⋅ ( g + h ) P 6 = ( b – d ) ⋅ ( g + h ) + dg –de – ah – bh Note: No reliance on Note: No reliance on P 7 = ( a – c ) ⋅ ( e + f ) P 7 = ( a – c ) ⋅ ( e + f ) + bg + bh – dg – dh commutativity of mult! commutativity of mult! = ae + bg 2/5/09 CS 3343 Analysis of Algorithms 15 2/5/09 CS 3343 Analysis of Algorithms 16 4
Strassen’s algorithm Analysis of Strassen 1. Divide: Partition A and B into T ( n ) = 7 T ( n /2) + Θ ( n 2 ) ( n /2) × ( n /2) submatrices. Form P -terms n log ba = n log 2 7 ≈ n 2.81 ⇒ C ASE 1 ⇒ T ( n ) = Θ ( n log 7 ). to be multiplied using + and – . 2. Conquer: Perform 7 multiplications of The number 2.81 may not seem much smaller than ( n /2) × ( n /2) submatrices recursively. 3, but because the difference is in the exponent, the 3. Combine: Form C using + and – on impact on running time is significant. In fact, ( n /2) × ( n /2) submatrices. Strassen’s algorithm beats the ordinary algorithm on today’s machines for n ≥ 30 or so. T ( n ) = 7 T ( n /2) + Θ ( n 2 ) Best to date (of theoretical interest only): Θ ( n 2.376 L ). 2/5/09 CS 3343 Analysis of Algorithms 17 2/5/09 CS 3343 Analysis of Algorithms 18 Conclusion • Divide and conquer is just one of several powerful techniques for algorithm design. • Divide-and-conquer algorithms can be analyzed using recurrences and the master method (so practice this math). • Can lead to more efficient algorithms 2/5/09 CS 3343 Analysis of Algorithms 19 5
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