【发布时间】：2012-03-29 12:58:41
【问题描述】：
我的作业涉及大 O 分析，我想我已经掌握了窍门，但我不能 100% 确定。如果我走在正确的轨道上，你们可爱的人会介意看看并告诉我吗？
作业如下。对于问题 1 和 3，我的分析和答案在右边，在 // 标记之后。对于问题2，我的分析和答案在算法类型下面。
提前感谢您的帮助！ :-)
1.For each of the following program fragments, give a Big-Oh analysis of the running time in terms of N:
(a) // Fragment (a)
for ( int i = 0, Sum = 0; i < N; i++ ) // N operations
for( int j = 0; j < N; j++ ) // N operations
Sum++; // Total: N^2 operations => O(N^2)
(b) // Fragment (b)
for( int i = 0, Sum = 0; i < N * N; i++ ) // N^2 operations
for( int j = 0; j < N; j ++ ) // N operations
Sum++; // Total: N^3 operations => O(N^3)
(c) // Fragment (c)
for( int i = 0, Sum = 0; i < N; i++ ) // N operations
for( int j = 0; j < i; j ++ ) // N-1 operations
Sum++; // Total: N(N-1) = N^2 – N operations => O(N^2)
(d) // Fragment (d)
for( int i = 0, Sum = 0; i < N; i++ ) // N operations
for( int j = 0; j < N * N; j++ ) // N^2 operations
for( int k = 0; k < j; k++ ) // N^2 operations
Sum++; // Total: N^5 operations => O(N^5)
2. An algorithm takes 0.5 milliseconds for input size 100. How long will it take for input size 500 if the running time is:
a. Linear
0.5 *5 = 2.5 milliseconds
b. O( N log N)
O (N log N) – treat the first N as a constant, so O (N log N) = O (log N)
Input size 100 = (log 100) + 1 = 2 + 1 = 3 operations
Input size 500 = (log 500) + 1= 2.7 + 1 = 3.7 ≈ 4 operations
Input size 100 runs in 0.5 milliseconds, so input size 500 takes 0.5 * (4/3) ≈ 0.67 milliseconds
c. Quadratic
Input size 100 in quadratic runs 100^2 operations = 10,000 operations
Input size 500 in quadratic runs 500^2 operations = 250,000 operations = 25 times as many
Input size of 100 runs in 0.5 milliseconds, so input size of 500 takes 25 * 0.5 = 12.5 milliseconds
d. Cubic
Input size 100 in quadratic runs 100^3 operations = 1,000,000 operations
Input size 500 in quadratic runs 500^3 operations = 125,000,000 operations = 125 times as many
Input size of 100 runs in 0.5 milliseconds, so input size of 500 takes 125 * 0.5 = 62.5 milliseconds
3. Find the Big-O for the following:
(a) f(x) = 2x^3 + x^2 log x // O(x^3)
(b) f(x) = (x^4 – 34x^3 + x^2 -20) // O(x^4)
(c) f(x) = x^3 – 1/log(x) // O(x^3)
4. Order the following functions by growth rate: (1 is slowest growth rate; 11 is fastest growth rate)
__6_ (a) N
__5_ (b) √N
__7_ (c) N^1.5
__9_ (d) N^2
__4_ (e) N log N
__2_ (f) 2/N
_11_ (g) 2^N
__3_ (h) 37
_10_ (i) N^3
__1_ (j) 1/ N^2
__8_ (k) N^2 /log N
* My logic in putting (j) and (f) as the slowest is that as N grows, 1/N^2 and 2/N decrease, so their growth rates are negative and therefore slower than the rest which have positive growth rates (or a 0 growth rate in the case of 37 (h)). Is that correct?
【问题讨论】：
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虽然与编程密切相关，但这对cs.stackexchange.com 来说是一个更好的问题。如果您以后有这样的问题，请直接在那里。