我是线程的新手，我正在尝试使用c中的线程将旅行商问题（动态编程）的顺序程序更改为并行程序。
#include <stdio.h>
#include <limits.h>
#define size 10 //maximum 10 cities
#define min(a,b) (a > b ? b : a)
#define sizePOW 1024 // 2^10
//Space complexity: O(n * 2^n)
//Time complexity: O(n^2 * 2^n)
int n; npow;
int g[size][sizePOW];
int p[size][sizePOW];
int adj[size][size];
int compute(int start, int set) {
int masked, mask, temp, i;
int result = INT_MAX; //result stores the minimum
if (g[start][set] != -1) //memoization DP top-down,check for repeated subproblem
return g[start][set];
for (i = 0; i < n; i++) { //npow-1 because we always exclude "home" vertex from our set
mask = (npow - 1) - (1 << i); //remove ith vertex from this set
masked = set & mask;
if (masked != set) { //in case same set is generated(because ith vertex was not present in the set hence we get the same set on removal) eg 12&13=12
temp = adj[start][i] + compute(i, masked); //compute the removed set
if (temp < result)
result = temp,
p[start][set] = i; //removing ith vertex gave us minimum
}
}
return g[start][set] = result; //return minimum
}
void getpath(int start, int set) {
if (p[start][set] == -1)
return; //reached null set
int x = p[start][set];
int mask = (npow - 1) - (1 << x);
int masked = set & mask; //remove p from set
printf("%d ", x);
getpath(x, masked);
}
void TSP() {
int i, j;
//g(i,S) is length of shortest path starting at i visiting all vertices in S and ending at 1
for (i = 0; i < n; i++)
for (j = 0; j < npow; j++)
g[i][j] = p[i][j] = -1;
for (i = 0; i < n; i++)
g[i][0] = adj[i][0]; //g(i,nullset)= direct edge between (i,1)
int result = compute(0, npow - 2);//npow-2 to exclude our "home" vertex
printf("Tour cost:%d\n", result);
printf("Tour path:\n0 ");
getpath(0, npow - 2);
printf("0\n");
}
int main(void) {
int i, j;
printf("Enter number of cities\n");
scanf("%d",&n);
npow=(int)pow(2, n);//bit number required to represent all possible sets
printf("Enter the adjacency matrix\n");
for(i = 0; i < n; i++)
for(j = 0; j < n; j++)
scanf("%d", &adj[i][j]);
TSP();
return 0;
}
这是来自ideone code的顺序程序。 这是我的
的并行代码#include <stdio.h>
#include <math.h>
#include <pthread.h>
#include <signal.h>
#include <errno.h>
#include <unistd.h>
#include<limits.h>
#define size 10 //maximum 10 cities
#define min(a,b) a > b ? b:a
#define sizePOW 1024 // 2^10
struct threadargs {
int a, b;
int *c;
};
//Space complexity: O(n * 2^n)
//Time complexity: O(n^2 * 2^n)
int n, npow;
int g[size][sizePOW];
int p[size][sizePOW];
int adj[size][size];
void printMatrix() {
int i, j;
for (i = 0; i < 4; i++) {
for (j = 0; j < 16; j++) {
printf("%d ",g[i][j]);
printf("\n");
}
printf("\n\n");
}
void *compute(void *args) {
int masked, mask, i, start, set;
int result = INT_MAX; //result stores the minimum
struct threadargs *recvargs = (struct threadargs *) args;
start = recvargs->a;
set = recvargs->b;
int *retval = recvargs->c;
if (g[start][set] != -1) { //memoization DP top-down,check for repeated subproblem
*retval += g[start][set];
return;
}
printMatrix();
//sleep(1);
int temp[n];
for (i = 0; i < n; i++)
temp[i] = INT_MAX;
pthread_t threads[n];
struct threadargs arguments[n];
int running_thread_count = 0;
for (i = 0; i < n; i++)
threads[i] == -1;
for (i = 0; i < n; i++) { //npow-1 because we always exclude "home" vertex from our set
mask= (npow - 1) - (1 << i); //remove ith vertex from this set
masked = set & mask;
//printf("hello world");
if (masked != set)//in case same set is generated(because ith vertex was not present in the set hence we get the same set on removal) eg 12&13=12
{
temp[i] = adj[start][i];
arguments[i].a = i;
arguments[i].b = masked;
arguments[i].c = &temp[i];
pthread_create(&threads[i], NULL, compute, (void *)&arguments[i] );
running_thread_count++;
}
}
for (i = 0; i < n; i++) {
if (pthread_kill(threads[i], 0) != ESRCH)
pthread_join(threads[i], NULL);
}
int ith = 0;
result = temp[0];
for (i = 1; i < n; i++) {
if(temp[i] < result) {
result = temp[i];
ith = i;
}
}
p[start][set] = ith;
if (result != INT_MAX)
g[start][set] = result; //return minimum
*retval += g[start][set];
}
void getpath(int start,int set)
{
if (p[start][set] == -1)
return; //reached null set
int x = p[start][set];
int mask= (npow - 1) - (1 << x);
int masked = set & mask;//remove p from set
printf("%d ",x);
getpath(x, masked);
}
void TSP()
{ int i, j;
//g(i,S) is length of shortest path starting at i visiting all vertices in S and ending at 1
for(i=0; i < n; i++)
for( j = 0; j < npow; j++)
g[i][j] = p[i][j] = -1;
for (i = 0; i < n; i++)
g[i][0] = adj[i][0]; //g(i,nullset)= direct edge between (i,1)
int result;
struct threadargs arguments;
arguments.a = 0;
arguments.b = npow-2;
arguments.c = &result;
compute((void *) &arguments);//npow-2 to exclude our "home" vertex
printf("Tour cost:%d\n",result);
printf("Tour path:\n0 ");
getpath(0,npow-2);
printf("0\n");
}
int main(void) {
int i, j;
printf("Enter number of cities\n");
scanf("%d", &n);
npow=(int)pow(2, n);//bit number required to represent all possible sets
printf("Enter the adjacency matrix\n");
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
scanf("%d", &adj[i][j]);
TSP();
return 0;
}
但是我在尝试执行此代码时遇到了分段错误。以下是gdb的输出
Program received signal SIGSEGV, Segmentation fault.
0x00007ffff78bf63e in pthread_join (threadid=4196128,
thread_return=0x0) at pthread_join.c:85
85 pthread_join.c: No such file or directory.
(gdb) backtrace
#0 0x00007ffff78bf63e in pthread_join (threadid=4196128,
thread_return=0x0) at pthread_join.c:85
#1 0x0000000000400b36 in compute (args=0x7fffffffde90) at tsp3.c:69
#2 0x0000000000400db2 in TSP () at tsp3.c:107
#3 0x0000000000400ec8 in main () at tsp3.c:120
(gdb)
我知道这不会给任何明显的性能提升，但我想试试这个。提前谢谢。
**编辑：** 我已经纠正了错误，但现在我遇到了新的错误。我正在程序运行时得到正确的答案，但如果我删除了行'printMatrix（） ，我得到分段错误。 gdb日志如下
(gdb) backtrace
#0 __pthread_kill (threadid=0, signo=0)
at ../nptl/sysdeps/unix/sysv/linux/pthread_kill.c:42
#1 0x0000000000400c85 in compute (args=0x7ffff74eede0) at tsp3.c:79
#2 0x00007ffff78be182 in start_thread (arg=0x7ffff64ed700)
at pthread_create.c:312
#3 0x00007ffff75eaefd in clone ()
at ../sysdeps/unix/sysv/linux/x86_64/clone.S:111
(gdb)
为什么会发生这种情况。请解释。提前谢谢。
答案 0 :(得分：0)
You are creating threads in compare function, and you pass compare as thread function, that's chaos. You maybe exceed available number of threads.