Structures Data and Algorithms Homework 1 Solution (Problem 1.)
(a) In Figure 1, in each node we have two numbers: the black number indicates n, and the red one indicates the return value for this node. (b) X(n) = 2n. This can be proved by induction on n. (In the induction step, it is convenient to distinguish between two cases: n = 2k and n = 2k + 1.)
Solution (Problem 2.)
(a) log(9 n) = (log n + 9) One justification: To begin, 1 log(9 n) = log 9 + log n. 2 There is a positive integer n0 such that log n0 2 log 9 and log n0 9. (When log = log2 as in the book, we may set n0 = 512.) With this n0 , we have 1 1 1 (log n + 9) log n < log 9 + log n log n < log n + 9 4 2 2 whenever n n0 , and so we may set c = 1/4, c = 1 in the definition on p. 170. (b) log2 n = (7 log n) One justification: We have 1 log2 n (7 log n) 7 for all positive integers n, and so we may set n0 = 1, c = 1/7 in the definition on p. 170. (c) 5(n + log n) = O(n log n + n) One justification: We have 5(n + log n) 5(n + n log n) for all positive integers n, and so we may set n0 = 1, c = 5 in the definition on p. 167.
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X(1),2 X(3),6 X(2),4 X(1),2 X(7),14 X(2),4 X(1),2 X(4),8 X(1),2 X(2),4 X(1),2 X(15),30 X(1),2 X(2),4 X(1),2 X(4),8 X(1),2 X(2),4 X(1),2 X(8),16 X(1),2 X(2),4 X(1),2 X(4), 8 X(1),2 X(2),4 X(1),2 Figure 1: Recursion Trace for X(15) X(1),2 X(1),2
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(d) 4n3/2 = (6n log n) One justification: There is a positive n0 integer such that n1/2 log n whenever n n0 . (When log = log2 as in the book, we may set n0 = 16.) With this n0 , we may set c = 2/3 in the definition on p. 170. (e) 5n2 = (2n2 - 5n) One justification: We may set n0 = 5, c = 1, c = 5 in the definition on p. 170.
Solution (Problem 3.)
Note: By Equation 4 in Proposition 4.1 on p. 155, we have (logb n) = (logd n) for all choices of bases b and d, and so we may write simply (log n) instead of (log2 n), (log3 n), etc. Maximum and minimum number of times each loop is repeated Loop 1 is repeated at least (log n) times and at most (n) times. Loop 3 is repeated (log n) times.
Loop 2 is repeated at least (log n) times and at most (n) times.
Loop 4 is repeated n/k times (see p. 161 for the definition of the ceiling function x) Lower and upper bound of the run-time of f : Each iteration of Loop 4 runs in time (1). Each iteration of Loop 3 runs in time n (1) k . Each iteration of Loop 2 runs in time (1).
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Each iteration of Loop 1 runs in time at least (log n) +
k
n (1) k n (1), k
and at most (n) +
k
where k in the sum runs through all the powers 30 , 31 , 32 , . . . which are smaller than n. Since n n k = O(log n) +
k
k
n k
and
k
n < k
s=0
n = O(n) 3s
(see Proposition A.14 on p. 676 for the last equation), it follows that each iteration of Loop 1 runs in time (n). The run-time of f is at least (n log n) and at most (n2 ).
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