CS 312: Algorithm Analysis

This work is licensed under a Creative Commons Attribution-Share Alike 3.0 Unported License. CS 312: Algorithm Analysis Lecture #6: Divide and Conquer Slides by: Eric Ringger, with contributions from Mike Jones, Eric Mercer, Sean Warnick

Announcements • Project #1 • Due: today (5pm) • Holiday: Monday • Project #2 • Instructions and software distribution available from the schedule • We do have one improvement to release later today, though • Help session on Tuesday • 5pm in 1066 TMCB • HW #4 due Wednesday

Objectives • Introduce Divide and Conquer • Apply to Multiplication • Analyze by defining a recurrence relation • Introduce the Convex Hull problem

Thought “Nothing is particularly hard if you divide it into small jobs.” -- Henry Ford

Divide and Conquer ... ... ... Solve the smaller pieces.

Recall Efficiency of Multiplication Algorithms • Problem: Multiplication • Algorithms: • American • British • a la Francaise / Russe • Arabic • … • Efficiency so far:

Divide and Conquer 0502 5001 05  50 shift 4 = 2500000 05  01 shift 2 = 500 02  50 shift 2 = 10000 02  01 shift 0 = 2 2510502

Divide and Conquer 0502 5001 Is it correct? 05 50 shift 4 = 2500000 05 01 shift 2 = 500 02 50 shift 2 = 10000 02 01 shift 0 = 2 2510502

Divide and Conquer 0502 5001 Is it correct? 5001 x 502 10002 0 2500500 2510502 05 x 50 shift 4 = 2500000 05 x 01 shift 2 = 500 02 x 50 shift 2 = 10000 02 x 01 shift 0 = 2 2510502 How long does it take on an input of size n?

Is it Faster? No.

Can we do better? • Yes. We shall see …

Divide and Conquer w x 05 x 50 shift 4 = 2500000 05 x 01 shift 2 = 500 02 x 50 shift 2 = 10000 02 x 01 shift 0 = 2 2510502 wy104 + wz102 + xy102 + xz100 0502 5001 y z product = wy104 + (wz + xy)102 + xz

Divide and Conquer w x product = wy104 + (wz + xy)102 + xz 0502 5001 Let r = (w + x)(y + z) = wy + wz + xy + xz Thenwz + xy = r – wy - xz y z = wy104 + ((w+x)(y+z)– wy - xz)102 + xz product = wy104 + (r – wy - xz)102 + xz

Multiply only three times Original product = wy104 + (wz + xy)102 + xz100 p = wy q = xz r = (w + x)(y + z) p104+ (r-p-q)102 + q = wy104 + (wy + wz + xy + xz - wy - xz)102 + xz = wy104 + (wz + xy)102 + xz

Recursive application ... ... ...

Algorithm

How long does it take? • Characterize running time using a recurrence relation. • Let be the amount of time to compute an answer on an input of size . • Then: * • where , the cost of dividing and reassembling sub-results • Answering a question with a question! * Actually: -- solution in same asymptotic order of growth

Master Theorem:Analysis of Divide & Conquer Define: = number of sub-instances that must be solved = original instance size (variable) * = size of sub-instances = polynomial order of , where = cost of dividing and recombining Then: We’ll prove this! * Assume that is a power of .

Analysis of DC Multiplicationwith Karatsuba’s Insight a = 3 there are 3 instances of multiplication b = 2 each is 1/2 the size of the original multiplication d = 1 the divide and the reassemble has linear complexity

Another Logarithm Identity Consequence:

Essentials ofDivide and Conquer • Given problem instance of size • Divide into sub-instances of size • Recursively solve these sub-instances • Appropriately combine their answers • Analyze with the Master Theorem

General: Divide and Conquer functionDC(x) : answer iflength(x) < threshold then returnadhoc(x) decompose x into a sub-instances x1, x2 … xa of size n/b for i  1 to adoyiDC(xi) recombine the yi’s to get a solution y for x return y Where : adhoc(x) = is the basic algorithm for small instances a = the number of divisions at each level n/b = the fraction of the whole for a sub-instance

Choosing a Threshold • Threshold = t means:“stop dividing when problem size = t” • Threshold depends on • Algorithm • and implementation • and platform • If t is too big? • If t is too small?

Recursive vs. Iterative Algorithms • Tail Recursion can be written iteratively easily • When results of recursive call aren’t used in an assignment • In fact, all recursive algorithms can be written as iteration. How? • By maintaining your own stack • Is the iterative implementation better? • Saves on stack allocation and function call • Is the recursive implementation better? • More “elegant” to some tastes:more natural, more readable, more maintainable. • Same algorithmic complexity

Convex Hull (Proj. #2)

Convexity

Assignment • Read: Recurrence Relations Notes, Part 1 • HW #4: 1.27, 2.1, 2.5(a-e) using the Master Theorem • Due Wednesday

Extra Slides

Possible DC Solution: Step 1 • Given set of n points • Divide into two subsets • L containing the leftmost points • R containing the rightmost points • All points with same x coordinate • Assign to the same subset • Even if this makes the division not exactly into halves

Possible DC Solution: Step 2 Compute the convex hulls of L and R recursively

Possible DC Solution: Step 3 • Combine the left hull CH(L) and the right hull CH(R): • Find the two edges known as the upper and lower common tangents (shown in red) • Common tangent: a line segment in the exterior of both polygons intersecting each polygon at a single vertex or a single edge. • The line containing the common tangent does not intersect the interior of either polygon

Possible DC Solution: Tips • Find the upper common tangent: • Scan around the left hull in a clockwise direction and around the right hull in a counter-clockwise direction • Come up with the details of finding the common tangents – hints available in the guidelines document • The tangents divide each hull into two pieces • Delete the right edges belonging to the left hull and the left edges belonging to the right hull

CS 312: Algorithm Analysis