问题
I really don't know how to find out the result of nested loops. For example in the following pseudo-code, I can't sort out what will be given at the end of execution. I'll be so glad if anyone gives me a simple solution.
r <- 0
for i <- 1 to n do
for j <- 1 to i do
for k <- j to i+j do
r <- r + 1
return r
Question is:
What is the result of the code and give the result r in terms of n?
I write it but every time I get confused.
回答1:
In your pseudo-code, Inner most loop, k <- j to i+j can be written as k <- 0 to i (this is by removing j). Hence your code can be simplified as follows:
r <- 0
for i <- 1 to n do
for j <- 1 to i do
for k <- 0 to i do // notice here `j` removed
r <- r + 1
return r
Based on this pseudo-code, I have written a C program(as below) to generate sequence for N = 1 to 10. (you originally tagged question as java but I am writing c code because what you wants is independent of language constraints)
#include<stdio.h>
int main(){
int i =0, k =0, j =0, n =0;
int N =0;
int r =0;
N =10;
for (n=1; n <= N; n++){
// unindented code here
r =0;
for (i=1; i<=n; i++)
for (j=1; j<=i; j++)
for (k=0; k<=i; k++)
r++;
printf("\n N=%d result = %d",n, r);
}
printf("\n");
}
Output of this program is something like:
$ ./a.out
N=1 result = 2
N=2 result = 8
N=3 result = 20
N=4 result = 40
N=5 result = 70
N=6 result = 112
N=7 result = 168
N=8 result = 240
N=9 result = 330
N=10 result = 440
Then, Tried to explore, How does it work? with some diagrams:
Execution Tree For N=1:
1<=i<=1, (i=1)
|
1<=j<=i, (j=1)
/ \
0<=k<=i, (K=0) (K=1)
| |
r=0 r++ r++ => r = 2
( 1 + 1 )
That is (1*2) = 2
Tree For N=2:
1<=i<=2, (i=1)-----------------------(i=2)
| |---------|------|
1<=j<=i, (j=1) (j=1) (j=2)
/ \ / | \ / | \
0<=k<=i, (K=0) (K=1) (K=0)(K=1)(k=2) (K=0)(K=1)(k=2)
| | | | | | | |
r=0 r++ r++ r++ r++ r++ r++ r++ r++ => 8
-------------- ---------------------------------
( 1 + 1) ( 3 + 3 )
That is (1 + 1) + (3 + 3) = 8
Similarly I drawn a tree for N=3:
1<=i<=3, (i=1)-----------------------(i=2)--------------------------------------------(i=3)
| |---------|------| |----------------------|----------------------|
1<=j<=3, (j=1) (j=1) (j=2) ( j=1 ) ( j=2 ) ( j=3 )
/ \ / | \ / | \ / | | \ / | | \ / | | \
0<=k<=i, (K=0) (K=1) (K=0)(K=1)(k=2) (K=0)(K=1)(k=2) / | | \ / | | \ / | | \
| | | | | | | | (K=0)(K=1)(k=2)(k=3) (K=0)(K=1)(k=2)(k=3) (K=0)(K=1)(k=2)(k=3)
r=0 r++ r++ r++ r++ r++ r++ r++ r++ r++ r++ r++ r++ r++ r++ r++ r++ r++ r++ r++ r++
That is (1 + 1) + (3 + 3) + (4 + 4+ 4)= 20
N = 1, (1 + 1) = 2
N = 2, (1 + 1) + (3 + 3) = 8
N = 3, (1 + 1) + (3 + 3) + (4 + 4 + 4)= 20
N = 4, (1 + 1) + (3 + 3) + (4 + 4 + 4) + (5 + 5 + 5 + 5) = 40
N = 5, (1 + 1) + (3 + 3) + (4 + 4 + 4) + (5 + 5 + 5 + 5) + (6 + 6 + 6 + 6 + 6) = 70
N = 6, (1 + 1) + (3 + 3) + (4 + 4 + 4) + (5 + 5 + 5 + 5) + (6 + 6 + 6 + 6 + 6) + (7 + 7 + 7 + 7 + 7 + 7)= 112
For N=6 we can also be write above sequence as:
(1*2) + (2*3) + (3*4) + (4*5) + (5*6) + (6*7)
Finally, I could understand that sum of N in three loop is:
(1*2) + (2*3) + (3*4) + (4*5) + (5*6) + ... + (N * (N+1))
With help from math.stackexchange.com, I could simplify this equation:
I asked here: How to simplify summation equation in terms of N?
So as I commented to your question, Result in term of N is ( ((N) * (N+1) * (N+2)) / 3 ).
And, I think its correct. I cross checked it as follows:
N = 1, (1 * 2 * 3)/3 = 2
N = 2, (2 * 3 * 4)/3 = 8
N = 3, (3 * 4 * 5)/3 = 20
N = 4, (4 * 5 * 6)/3 = 40
N = 5, (5 * 6 * 7)/3 = 70
回答2:
Try using some code like this to work it out... i.e. code up what it is and what you think it should be and test it.
EDIT: updated based on comment above.
public class CountLoop{
public static void main(String[] args){
for(int i=1;i<=10;i++)
System.out.println("It's "+run(i)+" and I think "+guess(i));;
}
public static int run(int n){
int r = 0;
for(int i=1;i<=n;i++)
for(int j=1; j <= i;j++)
for(int k=j; k <= i+j; k++)
r += 1;
return r;
}
public static int guess(int n){
// taken from the comments
int r = ((n * (n+1) * (n+2)) /3);
return r;
}
}
Running this gets
It's 2 and I think 2
It's 8 and I think 8
It's 20 and I think 20
It's 40 and I think 40
It's 70 and I think 70
It's 112 and I think 112
It's 168 and I think 168
It's 240 and I think 240
It's 330 and I think 330
It's 440 and I think 440
so we're happy.
回答3:
I am getting it something like this :
n = 1: r = 2
n = 2: r = 8
n = 3: r = 20
n = 4: r = 40
n = 5: r = 70
n = 6: r = 112
n = 7: r = 168
n = 8: r = 240
n = 9: r = 330
n = 10: r = 440
lets say for n = 10,
r = 2 + 6 + 12 + 20 + 30 + 42 + 56 + 72 + 90 + 110 = 440
=> r = 2(1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55)
Intuitively, I think
n = sum(n-1) + n * (n + 1).
where
sum(n-1) = value of r for n-1
来源：https://stackoverflow.com/questions/17019807/nested-loops-result