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CSC165H1 (167)
Lecture

# tut07solution.pdf

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Department
Computer Science
Course
CSC165H1
Professor
Nathalie Fournier
Semester
Fall

Description
CSC165H1F Tutorial # 7 | Sample Solutions Fall 2011 1. Write a detailed, structured proof that 8f : N ! R ;8g : N ! R ;g 2 O(f) ) g 2 O(f ) 2 (where f and g are de▯ned in the obvious way: 8n 2 N;f (n) = f(n) ▯ f(n), and similarly for g). (I show only the ▯nished proof here, not its development.) + + Assume f : N ! R and g : N ! R . Assume g 2 O(f). + Then 9c 2 R ;9B 2 N;8n 2 N;n > B ) g(n) 6 c ▯ f(n). # de▯nition of O Let c 0 R + and B 20N be such that 8n 2 N;n > B ) g(n0 6 c ▯ f(n).0 2 2 # Show that g 2 O(f ): Let c = c . Then c 2 R . + # because c 2 R + 1 0 1 0 Let B 1 B . 0hen B 2 N.1 # because B 20N Assume n 2 N and n > B = B . 1 0 Then g(n) 6 c 0 f(n) (because n > B ),0 so g (n) = g(n) ▯ g(n) 6 (c ▯ f(n)) ▯ (c ▯ f(n)) = c ▯ f(n) ▯ f(n) = c ▯ f (n). 0 0 0 1 Hence, 8n 2 N;n > B ) 1 (n) 6 c ▯ f (1). 2 + 2 2 Then 9c 2 R ;9B 2 N;8n 2 N;n > B ) g (n) 6 c ▯ f (n). Thus, g 2 O(f ).2 # by de▯nition of O 2 2 Therefore, g 2 O(f) ) g 2 O(f ). Then, 8f : N ! R ;8g : N ! R ;g 2 O(f) ) g 2 O(f ). 2 2. Prove that T (n) 2 ▯(n ), where BFT is the algorithm below. BFT BFT(E;n): 1. i n ▯ 1 2. while i > 0: 3. P[i] ▯1 4. Q[i] ▯1 5. i i ▯ 1 6. P[0] n 7. Q[0] 0 8. t 0 9. h 0 10. while h 6 t: 11. i 0 12. while i < n: 13. if E[Q[h]][i] 6= 0 and P[i] < 0: 14. P[i] Q[h] 15. t t + 1 16. Q[t] i 17. i i + 1 18. h h + 1 (Although this is not directly relevant to the question, this algorithm carries out a breadth-▯rst traversal of the graph on n vertices whose adjacency matrix is stored in E.) 2 2 2 We show that T BFT (n) 2 ▯(n ) by proving T BFT (n) 2 O(n ) and T BFT (n) 2 (n ). Dept. of Computer Science, University of Toronto, St. George Campus Page 1 of 4 CSC165H1F Tutorial # 7 | Sample Solutions Fall 2011 TBFT (n) 2 O(n ) : + Let c = 16 and B = 1. Then, c 2 R and B 2 N. Assume n 2 N, n > B = 1, and E is an arbitrary input of size n. One of the tricky features of this algorithm is that the main loop depends on the values of h and t, but the algorithm does not explicitly bound either value. To prove an upper bound on T BFT(n), we start by proving a bound on the value of t. Namely, we show that at any point during the execution of the algorithm, t 6 n. From lines 1{9, when the main loop (lines 10{18) begins execution, h = t = 0, P[0] = n, Q[0] = 0, and P[i] = Q[i] = ▯1 for i = 1;2;:::;n ▯ 1. Note that the value of t is changed only on line 15, and this line is executed only when P[i] < 0 (among other conditions). Moreover, each time t is incremented, the value of Q[t] is set to a natural number (on line 16), so that at any point during the execution of the algorithm, Q[0:::t] 2 N and Q[t + 1:::n ▯ 1] = ▯1. Since h 6 t (from line 10), this means that Q[h] > 0 is always true inside the main loop. Hence, on line 14, the assignment P[i] = Q[h] guarantees that P[i] > 0 from that point on. This means that the value of t can increase at most once for each value of i = 0;1;:::;n ▯ 1 (it increases only when P[i] < 0, at which point P[i] is set to a natural number), i.e., t 6 n. From the algorithm, ▯ line 1 takes 1 step; ▯ lines 2{5 take 4 steps for one iteration, and are executed exactly n ▯ 1 times (once for each value of i = n ▯ 1;n ▯ 2;:::;1), plus 1 more step for the last execution of line 2, for a total of 4(n ▯ 1) + 1 = 4n ▯ 3 steps; ▯ lines 6{9 take 4 steps; ▯ lines 12{17 take at most 6 steps for one iteration (if the condition of the if statement is true at every iteration), and are executed exactly n times (once for each value of i = 0;1;:::;n ▯ 1), plus 1 more step for the last execution of line 12, for a total of at most 6n + 1 steps; ▯ lines 10{18 take at most 6n + 1 + 3 = 6n + 4 steps for one iteration, and are executed at 2 most n times (since t 6 n, as shown above), for a total of at most 6n + 4n steps; ▯ so in total, the algorithm takes at most 1 + 4n ▯ 3 + 4 + 6n + 4n = 6n + 8n + 2 steps. Since n > 1, this means that the number of steps executed by the algorithm on input (E;n) is 6 6n + 8n + 2 6 6n + 8n + 2n = 16n . 2 Since (E;n) was arbitrary, 8n 2 N;n > 1 ) T BFT (n) 6 16n . 2 Therefore, TBFT (n) 2 O(n ). TBFT (n) 2 (n ): + Let c = 1 and B = 1. Then, c 2 R and B 2 N. Assume n 2 N and n > B = 1. Consider an input (E;n) such that E[i][j] = 1 for all indices 0 6 i < n;0 6 j < n. The ▯rst time that lines 12{17 are executed, the condition of the if statement will be true for all values of i = 0;1;:::;n ▯ 1 so at the end of the loop, t will have value at least n (since t starts at 0 and gets incremented n times). Since lines 12{17 always get executed exactly n times (once for each value of i = 0;1;:::;n ▯ 1), they take at least n steps. This means that lines 10{18 will get executed for every value of h = 0;1;:::;n ▯ 1 (at least), 2 and take at least n steps at each iteration, for a total of at least n steps. So the number of steps on input (E;n) is > n . 2 2 Hence, 8n 2 N;n > 1 ) T BFT (n) > n . Therefore, TBFT (n) 2 (n 2). Dept. of Computer Science, University of Toronto, St. George Campus Page 2 of 4 CSC165H1F Tutorial # 7 | Sample Solutions Fall 2011 3. Find a tight bound on the worst-case running time of the following algorithm. (This example was started during lecture, but it was not ▯nished.) # Precondition: L is a list that contains n
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