Data Structures and Algorithms (CSCI 340)

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1 University of Wisconsin Parkside Fall Semester 2008 Department of Computer Science Prof. Dr. F. Seutter Data Structures and Algorithms (CSCI 340) Homework Assignments The numbering of the problems refers to the exercises of Baase, van Gelder, Computer Algorithms. Dynamic Programming 10.5 Suppose the dimensions of the matrices A, B, C, and D are 20 2, 2 15, 15 40, and 40 4, respectively, and we want to know how best to compute A B C D. Show the arrays cost, last and Order computed by Algorithms 10.1 and Solution: Dimensions array: D = (20, 2, 15, 40, 4) low high k cost[low, k] cost[k, high] cost[low, high] last[low, high] Ref (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) >2800 (11) (12) >1680 (13) >1680 (14) 1

2 cost = last = Ref. (1) (2) (3) (4) (5) (6) (7) 2

3 cost = last = Ref. 0 1 (8) (9) (10) (11) (12) (13) (14) Order[1] = last[1, 3] = 2 Order[2] = last[1, 4] = 3 Order[3] = last[0, 4] = 1 The optimal order is (2, 3, 1). It corresponds to the optimal factoring A((BC)D) which has the optimal cost of cost[0, 4] =

4 10.6 Let A 1,...,A n be matrices where the dimensions of A i are d i 1 d i,fori = 1,...,n. Here is a proposal for a greedy algorithm to determine the best order in which to perform the matrix multiplications to compute A 1 A 2... A n. Algorithm 10.6 (MatrixMultiplicationOrder) Input: Dimension D =(d 0,d 1,...,d n )ofn matrices Output: An array Order in which entry i contains the ith multiplication MaMuOrd(D) 1 At each step, choose the largest remaining dimension 2 (among d 1,...,d n 1 ), and multiply two adjacent matrices 3 that share that dimension return Order Observe that this strategy produces the optimal order of multiplications for the matrices in the example discussed in class. (a) What is the order of the running time of this algorithm (only to determine the order in which to multiply the matrices, not including the actual multiplications)? (b) Either give a convincing argument that this strategy will always minimize the number of multiplications, or give an example where it does not do so. Solution: (a) Algorithm 10.6 (MatrixMultiplicationOrder) Input: Dimension D =(d 0,d 1,...,d n )ofn matrices Output: An array Order in which entry i contains the ith multiplication MaMuOrd(D) 1 i is a global variable initialized by 0 2 if length(d) 3 3 then k IndexOfMax(d 1,...,d n 1 ) 4 D max(d 0,...,d k 1,d k+1,...,d n ) 5 i i +1 6 Order[i] k 7 MaMuOrd(D) 8 return Order The running time is in Θ(n 2 ). There are Θ(n) recursive calls, and the search of the maximum dimension needs Θ(n) comparisons. (b) Let A, B, andc be (a b), (b c), and (c d) matrices, respectively. The multiplication order (AB)C results in abc + acd multiplications and A(BC) inabd + bcd. Assume b = max(b, c), then (AB)C is executed and this inequality should hold: abc + acd = ac(b + d) <bd(a + c) =(abd + bcd) But this is false for a =9,b=10,c=9andd = 8, for example. (AB)C results in 1458 multiplications, and A(BC) in

5 10.7 Construct an example with only three or four matrices where the worst multiplication order does at least 100 times as many element-wise multiplications as the best order. Solution: Let A, B, andc be (a b), (b c), and (c d) matrices, respectively. The multiplication order (AB)C results in abc + acd multiplications and A(BC) inabd + bcd. Now this inequality should hold: abc + acd = ac(b + d) > 100bd(a + c) = 100(abd + bcd) This is true for a = c = 1000 and b = d = 5, for example. (AB)C results in multiplications, and A(BC) in

6 10.16 The binominal coefficients can be defined by the recurrence equation: C(n, k) = C(n 1,k 1) + C(n 1,k) for n>0andk>0 C(n, 0) = 1 for n 0 C(0,k) = 0 for k>0 C(n, k) is also called n choose k and denoted ( n k). It is the number of ways to choose k distinct objects from a set of n objects. Consider the following four ways to compute C(n, k) forn k. 1. A recursive function as suggested by the recurrence relation given for C(n, k). 2. A dynamic programming algorithm. 3. The formula 4. The formula C(n, k) = Evaluate these methods as follows. n(n 1) (n k +1). 1 2 k C(n, k) = n! k!(n k)!. (a) Write out an outline of each method to make it clear you understand what work is to be done for each. (b) Compare the amount of work done by each method. Indicate what operations you are counting. Compare the amount of space used by each method. (c) Are there any other strong advantages or disadvantages of any of the four methods? (For example, is one of them more likely to cause an arithmetic overflow error? What about truncations caused by integer division?) 6

7 Solution: 1. (a) Algorithm 10.1 (Binominal Coefficients) Input: n, k N 0 and n k Output: c = ( ) n k BinCoef(n, k) 1 if n>0 k>0 2 then return BinCoef(n 1,k 1)+BinCoef(n 1,k) 3 else if k =0 4 return 1 5 else return 0 (b) Basic nonrecursive operation is the addition in line 2, i.e. f(n) = 1. The complexity of the algorithm in total results from the number of recursive calls. The shortest path down to the base case is of length k (from BinCoef(n 1,k 1)), and the longest path down to the base case is of length n (from BinCoef(n 1,k)). Estimation of complexity by Chip and be Conquered -formula: I. T (n, k) T (k): II. T (n, k) T (n): T (k) = 2T (k 1) + f(1) k = 2 i i=0 = 2 k+1 1 Ω(2 k ) T (n) = 2T (n 1) + f(1) n = 2 i i=0 = 2 n+1 1 O(2 n ) Thus the algorithm has an exponential complexity. A constant amount of space is needed per activation frame, therefore the space complexity also is exponential. (c) Arithmetic causes no trouble, only 1 s are added up. But the exponential complexity in time and space is pretty bad. 7

8 2. (a) Algorithm 10.2 (Binominal Coefficients) Input: n, k N 0 and n k Output: c = ( ) n k BinCoef(n, k) 1 for j 0 to k 2 do for i j to n 3 do if j =0 j = i 4 then T [i, j] 1 5 else T [i, j] T [i 1,j 1] + T [i 1,j] 6 return T [n, k] (b) Basic operations of BinCoef are the assignments in lines 4 and 5 and the addition in line 5 which are executed at most (n + 1)(k +1) times, once per table element. Thus, the algorithm has complexity Θ(nk). Space is needed for table T which is of size (n + 1)(k + 1), i.e. the space complexity is Θ(nk). Remark: For not all elements of the table are needed to compute the binomial coefficient in question, a more tricky index calculation could speed up the run time. But the worst-case complexity of Θ(nk) is kept. (c) Arithmetic causes no trouble and complexity in time and space is not bad. 3. (a) Algorithm 10.3 (Binominal Coefficients) Input: n, k N 0 and n k Output: c = ( ) n k BinCoef(n, k) 1 c 1 2 for i 1 to k 3 do c c (n i +1)/i 4 return c (b) Basic operation is the evaluation of the expression in line 3 which is executed k times. Thus the algorithm has complexity Θ(k). Space is only needed for the four variables n, k, i and c, i.e. the space complexity is Θ(1). (c) Truncations because of integer divisions do not occur. The product of i consecutive integers always is divisible by i, 1 i k. Let the binominal coefficient be in the range of type integer. An overflow error of intermediate products will not occur. Only possible exception, before the last division. 8

9 4. (a) Algorithm 10.4 (Binominal Coefficients) Input: n, k N 0 and n k Output: c = ( ) n k BinCoef(n, k) 1 c Fact(n)/(Fact(k) Fact(n k)) 2 return c Fact(n) 1 f =1 2 for i 1 to n 3 do f f i 4 return f (b) Basic operation is the multiplication in line 3 of algorithm Fact for computing the factorial. That operation is executed n+k+n k =2n times. Thus the algorithm has complexity Θ(n). Space is only needed for the variables n, k, i, f and c, i.e. the space complexity is Θ(1). (c) An overflow error of intermediate products will occur for pretty small values of n because factorial is a very fast growing function. 9

Assignment 1 (concept): Solutions

CS10b Data Structures and Algorithms Due: Thursday, January 0th Assignment 1 (concept): Solutions Note, throughout Exercises 1 to 4, n denotes the input size of a problem. 1. (10%) Rank the following functions