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A simple example

A simple example. finding the maximum of a set S of n numbers . Time complexity. Time complexity: Calculation of T(n): Assume n = 2 k , T(n) = 2T(n/2)+1 = 2(2T(n/4)+1)+1 = 4T(n/4)+2+1 : =2 k-1 T(2)+2 k-2 + … +4+2+1 =2 k-1 +2 k-2 + … +4+2+1 =2 k -1 = n-1.

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A simple example

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  1. A simple example • finding the maximum of a set S of n numbers

  2. Time complexity • Time complexity: • Calculation of T(n): Assume n = 2k, T(n) = 2T(n/2)+1 = 2(2T(n/4)+1)+1 = 4T(n/4)+2+1 : =2k-1T(2)+2k-2+…+4+2+1 =2k-1+2k-2+…+4+2+1 =2k-1 = n-1

  3. A general divide-and-conquer algorithm Step 1: If the problem size is small, solve this problem directly; otherwise, split the original problem into 2 sub-problems with equal sizes. Step 2: Recursively solve these 2 sub-problems by applying this algorithm. Step 3: Merge the solutions of the 2 sub- problems into a solution of the original problem.

  4. Time complexity of the general algorithm • Time complexity: where S(n) : time for splitting M(n) : time for merging b : a constant c : a constant • e.g. Binary search • e.g. quick sort • e.g. merge sort

  5. Divide and Conquer • Divide-and-conquer method for algorithm design: • Divide: if the input size is too large to deal with in a straightforward manner, divide the problem into two or more disjoint subproblems • Conquer: use divide and conquer recursively to solve the subproblems • Combine: take the solutions to the subproblems and “merge” these solutions into a solution for the original problem

  6. Recurrences • Running times of algorithms with Recursive calls can be described using recurrences • A recurrence is an equation or inequality that describes a function in terms of its value on smaller inputs • Example: Merge Sort

  7. Solving Recurrences • Repeated substitution method • Expanding the recurrence by substitution and noticing patterns • Substitution method • guessing the solutions • verifying the solution by the mathematical induction • Recursion-trees • Master method • templates for different classes of recurrences

  8. Repeated Substitution Method • Let’s find the running time of merge sort (let’s assume that n=2b, for some b).

  9. Repeated Substitution Method • The procedure is straightforward: • Substitute • Expand • Substitute • Expand • … • Observe a pattern and write how your expression looks after the i-th substitution • Find out what the value of i (e.g., lgn) should be to get the base case of the recurrence (say T(1)) • Insert the value of T(1) and the expression of i into your expression

  10. Integer Multiplication • Algorithm: Multiply two n-bit integers I and J. • Divide step: Split I and J into high-order and low-order bits • We can then define I*J by multiplying the parts and adding: • So, T(n) = 4T(n/2) + n, which implies T(n) is O(n2). • But that is no better than the algorithm we learned in grade school.

  11. An Improved Integer Multiplication Algorithm • Algorithm: Multiply two n-bit integers I and J. • Divide step: Split I and J into high-order and low-order bits • Observe that there is a different way to multiply parts: • So, T(n) = 3T(n/2) + n, which implies T(n) is O(nlog23), by the Master Theorem. • Thus, T(n) is O(n1.585).

  12. Matrix multiplication • Let A, B and C be n  n matrices C = AB C(i, j) = A(i, k)B(k, j) • The straightforward method to perform a matrix multiplication requires O(n3) time.

  13. Divide-and-conquer approach • C = AB C11 = A11 B11 + A12 B21 C12 = A11B12 + A12 B22 C21 = A21 B11 + A22 B21 C22 = A21 B12 + A22 B22 • Time complexity: (# of additions : n2) We get T(n) = O(n3)

  14. Strassen’s matrix multiplicaiton • P = (A11 + A22)(B11 + B22) Q = (A21 + A22)B11 R = A11(B12 - B22) S = A22(B21 - B11) T = (A11 + A12)B22 U = (A21 - A11)(B11 + B12) V = (A12 - A22)(B21 + B22). • C11 = P + S - T + V C12 = R + T C21 = Q + S C22 = P + R - Q + U

  15. Time complexity • 7 multiplications and 18 additions or subtractions • Time complexity: T(n) = an2 + 7T(n/2) = an2 + 7(a(n/2)2 + 7T(n/4)) = an2 + (7/4)an2 + 72T(n/4) = … : = an2(1 + 7/4 + (7/4)2+…+(7/4)k-1+7kT(1))  cn2(7/4)log2n+7log2n, c is a constant = cnlog24+log27-log24 +nlog27 = O(nlog27)  O(n2.81)

  16. Median Finding • Given a set of "n" unordered numbers we want to find the "k th" smallest number. (k is an integer between 1 and n).

  17. A Simple Solution • A simple sorting algorithm like heapsort will take Order of O(nlg2n) time. Step Running Time Sort n elements using heapsort O(nlog2n) Return the kth smallest element O(1) Total running time O(nlog2n)

  18. Linear Time selection algorithm • Also called Median Finding Algorithm. • Find k th smallest element in O (n) time in worst case. • Uses Divide and Conquer strategy. • Uses elimination in order to cut down the running time substantially.

  19. Steps to solve the problem • Step 1: If n is small, for example n<6, just sort and return the kth smallest number in constant time i.e; O(1) time. • Step 2: Group the given number in subsets of 5 in O(n) time.

  20. Step3: Sort each of the group in O (n) time. Find median of each group. • Given a set (……..2,5,9,19,24,54,5,87,9,10,44,32,21,13,24,18,26,16,19,25,39,47,56,71,91,61,44,28………) having n elements.

  21. Arrange the numbers in groups of five 2 54 44 4 25 ……………….. ……………….. ……………….. 5 32 18 39 ……………….. 5 ……………….. 47 21 26 9 87 ……………….. 13 16 56 19 9 ……………….. ……………….. 2 19 71 ……………….. 24 10 ………………..

  22. Find median of N/5 groups 2 5 2 4 25 ……………….. ……………….. ……………….. 5 13 16 39 ……………….. 9 ……………….. 9 10 18 47 21 ……………….. 32 19 56 19 54 ……………….. ……………….. 44 26 71 ……………….. 24 87 ……………….. Median of each group

  23. Find the Median of each group 2 5 2 4 25 ……………….. ……………….. 3.n/10 ……………….. 5 13 16 39 ……………….. 9 ……………….. 47 18 21 9 10 ……………….. 32 19 56 19 54 ……………….. ……………….. 44 26 71 ……………….. 24 87 ……………….. Find m ,the median of medians

  24. Find the sets L and R • Compare the n-1 elements with the median m and find two sets L and R such that every element in L is smaller than m and every element in R is greater than m. m L R 3n/10<L<7n/10 3n/10<R<7n/10

  25. Description of the Algorithm step • If n is small, for example n<6, just sort and return the k the smallest number.(time- 7) • If n>5, then partition the numbers into groups of 5.(time n) • Sort the numbers within each group. Select the middle elements (the medians). (time- 7n/5) • Call your "Selection" routine recursively to find the median of n/5 medians and call it m. (time-Tn/5) • Compare all n-1 elements with the median of medians m and determine the sets L and R, where L contains all elements <m, and R contains all elements >m. Clearly, the rank of m is r=|L|+1 (|L| is the size of L) (time- n)

  26. Contd…. • If k=r, then return m • If k<r, then return k th smallest of the set L .(time T7n/10) • If k>r, then return k-r th smallest of the set R.

  27. Recursive formula • T (n)=O(n) + T(n/5) +T(7n/10) We will solve this equation in order to get the complexity. We assume that T (n)< C*n T (n)= a*n + T (n/5) + T (7n/10) <= C*n/5+ C*7*n/10 + a*n <= C*n if 9*C/10 +a <= C Which implies C >= 10*a Hence 10*a*n satisfies the recurrence and so T(n) is O(n)

  28. 2-D ranking finding • Def: Let A = (a1,a2), B = (b1,b2). A dominates B iff a1> b1 and a2 > b2 • Def: Given a set S of n points, the rank of a point x is the number of points dominated by x. D B C A E rank(A)= 0 rank(B) = 1 rank(C) = 1 rank(D) = 3 rank(E) = 0

  29. Straightforward algorithm: compare all pairs of points : O(n2)

  30. Divide-and-conquer 2-D ranking finding Input: A set S of planar points P1,P2,…,Pn Output: The rank of every point in S Step 1: (Split the points along the median line L into A and B.) a. If S contains only one point, return its rank as 0. b. Else, choose a cut line L perpendicular to the x-axis such that n/2 points of S have x-values L (call this set of points A) and the remainder points have x-values L(call this set B). Note that L is a median x-value of this set. Step 2: Find ranks of points in A and ranks of points in B, recursively. Step 3: Sort points in A and B according to their y-values. Scan these points sequentially and determine, for each point in B, the number of points in A whose y-values are less than its y-value. The rank of this point is equal to the rank of this point among points in B, plus the number of points in A whose y-values are less than its y-value.

  31. 2-D maxima finding problem • Def : A point (x1, y1) dominates (x2, y2) if x1 > x2 and y1 > y2. A point is called a maxima if no other point dominates it • Straightforward method : Compare every pair of points. Time complexity: O(n2)

  32. Divide-and-conquer for maxima finding The maximal points of SL and SR

  33. The algorithm: • Input: A set of n planar points. • Output: The maximal points of S. Step 1: If S contains only one point, return it as the maxima. Otherwise, find a line L perpendicular to the X-axis which separates the set of points into two subsets SLand SR , each of which consisting of n/2 points. Step 2: Recursively find the maximal points of SL and SR. Step 3: Find the largest y-value of SR. Project the maximal points of SL onto L. Discard each of the maximal points of SL if its y-value is less than the largest y-value of SR.

  34. Time complexity: T(n) Step 1: O(n) Step 2: 2T(n/2) Step 3: O(n) Assume n = 2k T(n) = O(n log n)

  35. The closest pair problem • Given a set S of n points, find a pair of points which are closest together. • 1-D version : Solved by sorting Time complexity : O(n log n) 2-D version

  36. at most 6 points in area A:

  37. The algorithm: • Input: A set of n planar points. • Output: The distance between two closest points. Step 1: Sort points in S according to their y-values and x-values. Step 2:If S contains only two points, return infinity as their distance. Step 3:Find a median line L perpendicular to the X-axis to divide S into two subsets, with equal sizes, SL and SR. Step 4:Recursively apply Step 2 and Step 3 to solve the closest pair problems of SL and SR. Let dL(dR) denote the distance between the closest pair in SL (SR). Let d = min(dL, dR).

  38. Step 5: For a point P in the half-slab bounded by L-d and L, let its y-value by denoted as yP . For each such P, find all points in the half-slab bounded by L and L+d whose y-value fall within yP +d and yP -d. If the distance dbetween P and a point in the other half-slab is less than d, let d=d . The final value of d is the answer. • Time complexity: O(n log n) Step 1: O(n log n) Steps 2~5: T(n) = O(n log n)

  39. concave polygon: convex polygon: The convex hull problem • The convex hull of a set of planar points is the smallest convex polygon containing all of the points.

  40. The divide-and-conquer strategy to solve the problem:

  41. The merging procedure: • Select an interior point p. • There are 3 sequences of points which have increasing polar angles with respect to p. (1) g, h, i, j, k (2) a, b, c, d (3) f, e • Merge these 3 sequences into 1 sequence: g, h, a, b, f, c, e, d, i, j, k. • Apply Graham scan to examine the points one by one and eliminate the points which cause reflexive angles.

  42. e.g. points b and f need to be deleted. Final result:

  43. Divide-and-conquer for convex hull • Input : A set S of planar points • Output : A convex hull for S Step 1: If S contains no more than five points, use exhaustive searching to find the convex hull and return. Step 2: Find a median line perpendicular to the X-axis which divides S into SL and SR; SL lies to the left of SR. Step 3: Recursively construct convex hulls for SL and SR. Denote these convex hulls by Hull(SL) and Hull(SR) respectively.

  44. Step 4:Apply the merging procedure to merge Hull(SL) and Hull(SR) together to form a convex hull. • Time complexity: T(n) = 2T(n/2) + O(n) = O(n log n)

  45. The Fast Fourier Transform

  46. Outline • Polynomial Multiplication Problem • Primitive Roots of Unity • The Discrete Fourier Transform • The FFT Algorithm

  47. Polynomials • Polynomial: • In general,

  48. Polynomial Evaluation • Horner’s Rule: • Given coefficients (a0,a1,a2,…,an-1), defining polynomial • Given x, we can evaluate p(x) in O(n) time using the equation • Eval(A,x): [Where A=(a0,a1,a2,…,an-1)] • If n=1, then return a0 • Else, • Let A’=(a1,a2,…,an-1) [assume this can be done in constant time] • return a0+x*Eval(A’,x)

  49. Polynomial Multiplication • Given coefficients (a0,a1,a2,…,an-1) and (b0,b1,b2,…,bn-1) defining two polynomials, p() and q(), and number x, compute p(x)q(x). • Horner’s rule doesn’t help, since where • A straightforward evaluation would take O(n2) time. The “magical” FFT will do it in O(n log n) time.

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