Here is the code.what i did is implement linear search on some elements of the array and the push searched elements in stack,afterwards I print the popped elements from stack and print them.But in search function it displays two index values.
using namespace std;
int searched[10];
int stack[100], n=100, top=-1;
void push(int val) {
if(top>=n-1)
cout<<"Stack Overflow"<<endl;
else {
top++;
stack[top]=val;
}
}
void pop() {
if(top<=-1)
cout<<"Stack Underflow"<<endl;
else {
cout<<"The popped element is "<< stack[top] <<endl;
top--;
}
}
void display() {
if(top>=0) {
cout<<"Stack elements are:";
for(int i=top; i>=0; i--)
cout<<stack[i]<<" ";
cout<<endl;
} else
cout<<"Stack is empty";
}
int search(int arr[], int n, int x)
{
int i;
for (i = 0; i < n; i++)
if (arr[i] == x)
cout<<"The element is found at the index"<<i<<"\n\n";
return x;
}
int main(void)
{
int arr[15] = { 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 };
int x = 0;
for(int i=0;i<8;i++)
{
x++;
int result = search(arr, n, x);
cout << "searched Element is " << result<<"\t\t";
push(result);
pop();
}
return 0;
}```
There are two issues that lead to this confusing result.
First, if I am not mistaken the search function, which was written like this:
int search(int arr[], int n, int x)
{
int i;
for (i = 0; i < n; i++)
if (arr[i] == x)
cout<<"The element is found at the index"<<i<<"\n\n";
return x;
}
is parsed similarly to the following:
int search(int arr[], int n, int x)
{
int i;
for (i = 0; i < n; i++) {
if (arr[i] == x) {
cout<<"The element is found at the index"<<i<<"\n\n";
}
}
return x;
}
Presumably you meant this:
int search(int arr[], int n, int x)
{
int i;
for (i = 0; i < n; i++) {
if (arr[i] == x) {
cout<<"The element is found at the index"<<i<<"\n\n";
return x;
}
}
}
Second, since the global n starts at the value 100, the search loop runs off the end of the length 15 array, into other memory. This is probably undefined behavior.
I am struggling with this code. It's about counting some patterns from a text file. I tried to use thread(divide and conquer) processing, but it return a wrong value. I used mutex value to synchronize critical section.. The code is below. First main argument is number of threads, second is the name of a text file I want counting patterns from, and following patterns I want to look up on the text.
please save me..
Code is below
char *buffer;
int fsize, count;
char **searchword;
int *wordcount;
int *strlength;
typedef struct _params{
int num1;
int num2;
}params;
pthread_mutex_t mutex;
void *childfunc(void *arg)
{
int size, i, j, k, t, start, end, len, flag = -1;
int result;
params *a = (params *)arg;
start = a->num1;
end = a->num2;
while(1){
if(start == 0 || start == fsize)
break;
if(buffer[start]!= ' ' && buffer[start] != '\n' && buffer[start] != '\t')
start++;
else
break;
}
while(1){
if(end == fsize)
break;
if(buffer[end] != ' ' && buffer[end] != '\n' && buffer[end] != '\t')
end++;
else
break;
}
for(i = 0; i < count; i++){
len = strlength[i];
for(j = start; j<(end - len + 1); j++){
if(buffer[j] == searchword[i][0]){
flag = 0;
for(k = j +1; k<j + len; k++){
if(buffer[k] != searchword[i][k-j])
{
flag = 1;
break;
}
}
if(flag == 0){
pthread_mutex_lock(&mutex);
wordcount[i]++;
pthread_mutex_unlock(&mutex);// mutex unlocking
sleep(1);
flag = -1;
}
}
}
}
}
int main(int argc, char **argv){
FILE *fp;
char *inputFile;
pthread_t *tid;
int *status;
int inputNumber, i, j, diff, searchstart, searchend;
int result = 0;
count = argc -3;
inputNumber = atoi(argv[1]);
inputFile = argv[2];
searchword = (char **)malloc(sizeof(char *)*count);
tid = malloc(sizeof(pthread_t)*inputNumber);
strlength = (int *)malloc(4*count);
status = (int *)malloc(4*inputNumber);
wordcount = (int *)malloc(4*count);
for(i = 0; i < count; i++)
searchword[i] = (char*)malloc(sizeof(char)*(strlen(argv[i+3]) + 1));
for(i = 3; i < argc; i++)
strcpy(searchword[i-3], argv[i]);
fp = fopen(inputFile, "r");
fseek(fp, 0, SEEK_END);
fsize = ftell(fp);
rewind(fp);
buffer = (char *)malloc(1*fsize);
fread(buffer, fsize, 1, fp);
diff = fsize / inputNumber;
if(diff == 0)
diff = 1;
for(i = 0; i < count ; i++){
strlength[i] = strlen(searchword[i]);
wordcount[i] = 0;
}
for(i = 0; i < inputNumber; i++){
searchstart = 0 + i*diff;
searchend = searchstart + diff;
if(searchstart > fsize)
searchstart = fsize;
if(searchend > fsize)
searchend = fsize;
if( i == inputNumber -1)
searchend = fsize;
params a;
a.num1 = searchstart;
a.num2 = searchend;
pthread_mutex_init(&mutex, NULL);
result = pthread_create(&tid[i], NULL, childfunc, (void *)&a);
if(result < 0){
perror("pthread_create()");
}
}
//스레드 받는 부분
for(i = 0; i < inputNumber; i++){
result = pthread_join(tid[i], (void **)status);
if(result < 0)
perror("pthread_join()");
}
pthread_mutex_destroy(&mutex); // mutex 해제
for(i = 0; i < count; i++)
printf("%s : %d \n", searchword[i], wordcount[i]);
for(i = 0; i < count; i++) //동적메모리해제
free(searchword[i]);
free(searchword);
free(buffer);
free(tid);
free(strlength);
free(wordcount);
free(status);
fclose(fp);
return 0;
}
params a; // new a for each loop, previous a no longer exists
a.num1 = searchstart;
a.num2 = searchend;
pthread_mutex_init(&mutex, NULL);
result = pthread_create(&tid[i], NULL, childfunc, (void *)&a);
if(result < 0){
perror("pthread_create()");
}
} // a goes out of scope here
You pass each thread the address of a, but then a goes out of scope immediately after you create the thread. So the thread now has the address of some random leftover junk on the stack.
You need to have some conception of ownership of any object that's accessed by more than one thread like a is here. It can be owned by the thread that called pthread_create, owned by the newly-created thread, or jointly owned. But you have to be consistent. You have neither of these, why?
Is a owned only be the thread that called pthread_create? No, because the newly-created thread has a pointer to it and accesses it through that pointer. So the thread that called pthread_create cannot destroy it.
Is a owned only be the thread created by pthread_create? No, because it's on the stack of the thread that called pthread_create and will cease to exist when the next loop comes around.
Is a jointly owned? Well, no, because the thread that called pthread_create can destroy the object before the newly-created thread accesses it.
So no sane model of multi-thread use is followed by a. It's broken.
One common solution to this problem is to allocate a new structure (using malloc or new) for each thread, fill it in, and pass the thread the address of the structure. Let the thread free (or delete) the structure when it's done with it.
Given an infinite stream of characters and a list L of strings, create a function that calls an external API when a word in L is recognized during the processing of the stream.
Example:
L = ["ok","test","one","try","trying"]
stream = a,b,c,o,k,d,e,f,t,r,y,i,n,g.............
The call to external API will happen when 'k' is encountered, again when the 'y' is encountered, and again at 'g'.
My idea:
Create trie out of the list and navigate the nodes as you read from stream in linear time. But there would be a bug if you just do simple trie search.
Assume you have words "abxyz" and "xyw" and your input is "abxyw".In this case you can't recognize "xyw" with trie.
So search should be modified as below:
let's take above use case "abxyw". We start the search and we find we have all the element till 'x'. Moment you get 'x' you have two options:
Check if the current element is equal to the head of trie and if it is equal to head of trie then call recursive search.
Continue till the end of current word. In this case for your given input it will return false but for the recursive search we started in point 1, it will return true.
Below is my modified search but I think it has bugs and can be improved. Any suggestions?
#define SIZE 26
struct tri{
int complete;
struct tri *child[SIZE];
};
void insert(char *c, struct tri **t)
{
struct tri *current = *t;
while(*c != '\0')
{
int i;
int letter = *c - 'a';
if(current->child[letter] == NULL) {
current->child[letter] = malloc(sizeof(*current));
memset(current->child[letter], 0, sizeof(struct tri));
}
current = current->child[letter];
c++;
}
current->complete = 1;
}
struct tri *t;
int flag = 0;
int found(char *c, struct tri *tt)
{
struct tri *current = tt;
if (current == NULL)
return 0;
while(*c != '\0')
{
int i;
int letter = *c - 'a';
/* if this is the first char then recurse from begining*/
if (t->child[letter] != NULL)
flag = found(c+1, t->child[letter]);
if (flag == 1)
return 1;
if(!flag && current->child[letter] == NULL) {
return 0;
}
current = current->child[letter];
c++;
}
return current->complete;
}
int main()
{
int i;
t = malloc(sizeof(*t));
t->complete = 0;
memset(t, 0, sizeof(struct tri));
insert("weathez", &t);
insert("eather", &t);
insert("weather", &t);
(1 ==found("weather", t))?printf("found\n"):printf("not found\n");
return 0;
}
What you want to do is exactly what Aho-Corasick algorithm does.
You can take a look at my Aho-Corasick implementation. It's contest-oriented, so maybe not focused on readability but I think it's quite clear:
typedef vector<int> VI;
struct Node {
int size;
Node *fail, *output;
VI id;
map<char, Node*> next;
};
typedef pair<Node*, Node*> P;
typedef map<char, Node*> MCP;
Node* root;
inline void init() {
root = new Node;
root->size = 0;
root->output = root->fail = NULL;
}
Node* add(string& s, int u, int c = 0, Node* p = root) {
if (p == NULL) {
p = new Node;
p->size = c;
p->fail = p->output = NULL;
}
if (c == s.size()) p->id.push_back(u);
else {
if (not p->next.count(s[c])) p->next[s[c]] = NULL;
p->next[s[c]] = add(s, u, c + 1, p->next[s[c]]);
}
return p;
}
void fill_fail_output() {
queue<pair<char, P> > Q;
for (MCP::iterator it=root->next.begin();
it!=root->next.end();++it)
Q.push(pair<char, P> (it->first, P(root, it->second)));
while (not Q.empty()) {
Node *pare = Q.front().second.first;
Node *fill = Q.front().second.second;
char c = Q.front().first; Q.pop();
while (pare != root && !pare->fail->next.count(c))
pare=pare->fail;
if (pare == root) fill->fail = root;
else fill->fail = pare->fail->next[c];
if (fill->fail->id.size() != 0)
fill->output = fill->fail;
else fill->output = fill->fail->output;
for (MCP::iterator it=fill->next.begin();
it!=fill->next.end();++it)
Q.push(pair<char,P>(it->first,P(fill,it->second)));
}
}
void match(int c, VI& id) {
for (int i = 0; i < id.size(); ++i) {
cout << "Matching of pattern " << id[i];
cout << " ended at " << c << endl;
}
}
void search(string& s) {
int i = 0, j = 0;
Node *p = root, *q;
while (j < s.size()) {
while (p->next.count(s[j])) {
p = p->next[s[j++]];
if (p->id.size() != 0) match(j - 1, p->id);
q = p->output;
while (q != NULL) {
match(j - 1, q->id);
q = q->output;
}
}
if (p != root) {
p = p->fail;
i = j - p->size;
}
else i = ++j;
}
}
void erase(Node* p = root) {
for (MCP::iterator it = p->next.begin();
it != p->next.end(); ++it)
erase(it->second);
delete p;
}
int main() {
init();
int n;
cin >> n;
for (int i = 0; i < n; ++i) {
string s;
cin >> s;
add(s, i);
}
fill_fail_output();
string text;
cin >> text;
search(text);
erase(root);
}
Here is the entire program, please help me, I've tried everything to find out what exactly is going with the memory. The problem is everything runs perfectly, but there are some extra characters printed with output.
Here is the .h file:
class MyString
{
public:
MyString();
MyString(const char *message);
MyString(const MyString &source);
~MyString();
const void Print() const;
const int Length() const;
MyString& operator()(const int index, const char b);
char& operator()(const int i);
MyString& operator=(const MyString& rhs);
bool operator==(const MyString& other) const;
bool operator!=(const MyString& other) const;
const MyString operator+(const MyString& rhs) const;
MyString& operator+=(const MyString& rhs);
friend ostream& operator<<(ostream& output, const MyString& rhs);
const int Find(const MyString& other);
MyString Substring(int start, int length);
private:
char *String;
int Size;
};
istream& operator>>(istream& input, MyString& rhs);
The .cpp file:
MyString::MyString()
{
char temp[] = "Hello World";
int counter(0);
while(temp[counter] != '\0')
{
counter++;
}
Size = counter;
String = new char [Size];
for(int i=0; i < Size; i++)
String[i] = temp[i];
}
//alternate constructor that allows for setting of the inital value of the string
MyString::MyString(const char *message)
{
int counter(0);
while(message[counter] != '\0')
{
counter++;
}
Size = counter;
String = new char [Size];
for(int i=0; i < Size; i++)
String[i] = message[i];
}
//copy constructor
MyString::MyString(const MyString &source)
{
int counter(0);
while(source.String[counter] != '\0')
{
counter++;
}
Size = counter+1;
String = new char[Size];
for(int i = 0; i <= Size; i++)
String[i] = source.String[i];
}
//Deconstructor
MyString::~MyString()
{
delete [] String;
}
//Length() method that reports the length of the string
const int MyString::Length() const
{
int counter(0);
while(String[counter] != '\0')
{
counter ++;
}
return (counter);
}
/*Parenthesis operator should be overloaded to replace the Set and Get functions of your previous assignment. Note that both instances should issue exit(1) upon violation of the string array bounaries.
*/
MyString& MyString::operator()(const int index, const char b)
{
if(String[index] == '\0')
{
exit(1);
}
else
{
String[index] = b;
}
}
char& MyString::operator()(const int i)
{
if(String[i] == '\0')
{
exit(1);
}
else
{
return String[i];
}
}
/*Assignment operator (=) which will copy the source string into the destination string. Note that size of the destination needs to be adjusted to be the same as the source.
*/
MyString& MyString::operator=(const MyString& rhs)
{
if(this != &rhs)
{
delete [] String;
String = new char[rhs.Size];
Size = rhs.Size;
for(int i = 0; i < rhs.Size+1 ; i++)
{
String[i] = rhs.String[i];
}
}
return *this;
}
/*Logical comparison operator (==) that returns true iff the two strings are identical in size and contents.
*/
bool MyString::operator==(const MyString& other)const
{
if(other.Size == this->Size) {
for(int i = 0; i < this->Size+1; i++)
{
if(&other == this)
return true;
}
}
else
return false;
}
//Negated logical comparison operator (!=) that returns boolean negation of 2
bool MyString::operator!=(const MyString& other) const
{
return !(*this == other);
}
//Addition operator (+) that concatenates two strings
const MyString MyString::operator+(const MyString& rhs) const
{
char* tmp = new char[Size + rhs.Size +1];
for(int i = 0; i < Size; i++)
{
tmp[i] = String[i];
}
for(int i = 0; i < rhs.Size+1; i++) {
tmp[i+Size] = rhs.String[i];
}
MyString result;
delete [] result.String;
result.String = tmp;
result.Size = Size+rhs.Size;
return result;
}
/*Addition/Assigment operator (+=) used in the following fashion: String1 += String2 to operate as String1 = String1 + String2
*/
MyString& MyString::operator+=(const MyString& rhs)
{
char* tmp = new char[Size + rhs.Size + 1];
for(int i = 0; i < Size; i++) {
tmp[i] = String[i];
}
for(int i = 0; i < rhs.Size+1; i++)
{
tmp[i+Size] = rhs.String[i];
}
delete [] String;
String = tmp;
Size += rhs.Size;
return *this;
}
istream& operator>>(istream& input, MyString& rhs)
{
char* t;
int size(256);
t = new char[size];
input.getline(t,size);
rhs = MyString(t);
delete [] t;
return input;
}
ostream& operator<<(ostream& output, const MyString& rhs)
{
if(rhs.String != '\0')
{
output << rhs.String;
}
else
{
output<<"No String to output\n";
}
return output;
}
/*MyString::Find that finds a string in a larger string and returns the starting location of the substring. Note that your string location starts from 0 and ends at length -1. If the string is not found, a value of -1 will be returned
*/
const int MyString::Find(const MyString& other)
{
int nfound = -1;
if(other.Size > Size)
{
return nfound;
}
int i = 0, j = 0;
for(i = 0; i < Size; i++) {
for(j = 0; j < other.Size; j++) {
if( ((i+j) >= Size) || (String[i+j] != other.String[j]) )
{
break;
}
}
if(j == other.Size)
{
return i;
}
}
return nfound;
}
/*MyString::Substring(start, length). This method returns a substring of the original string that contains the same characters as the original string starting at location start and is as long as length.
*/
MyString MyString::Substring(int start, int length)
{
char* leo = new char[length+1];
for(int i = start; i < start + length+1; ++i)
{
leo[i-start] = String[i];
}
MyString sub;
delete [] sub.String; sub.String = leo; sub.Size = Size;
return sub;
}
//Print() method that prints the string
const void MyString::Print() const
{
for(int i=0; i < Size; i++)
{
cout<<String[i];
}
cout<<endl;
}
The main.cpp file:
int main (int argc, char **argv)
{
MyString String1;
const MyString ConstString("Target string"); //Test of alternate constructor
MyString SearchString; //Test of default constructor that should set "Hello World".
MyString TargetString (String1); //Test of copy constructor
cout << "Please enter two strings. ";
cout << "Each string needs to be shorter than 256 characters or terminated by /\n." << endl;
cout << "The first string will be searched to see whether it contains exactly the second string. " << endl;
cin >> SearchString >> TargetString; // Test of cascaded string-extraction operator
if(SearchString.Find(TargetString) == -1) {
cout << TargetString << " is not in " << SearchString << endl;
}
else {
cout << TargetString << " is in " << SearchString << endl;
cout << "Details of the hit: " << endl;
cout << "Starting position of the hit: " << SearchString.Find(TargetString) << endl;
cout << "The matching substring is: " << SearchString.Substring(SearchString.Find(TargetString), TargetString.Length()-1)<<"\n";
}
return 0;
}
Running the program you get this:
Please enter two strings. Each string needs to be shorter than 256 characters or terminated by /
.
The first string will be searched to see whether it contains exactly the second string.
firstly
real
realt World is not in firstly
Please Help!!
try adding a '\0' at the end of your strings in your MyString::MyString(const char *message) constructor
#Sam's answer is correct. I'm going to add on to it to help you learn what's happening.
C and C++ strings are really character arrays that follow a convention that the string is terminated with \0, sometimes called NUL (not null), which is a character where all bits are 0.
Your code gets the first part right in that it creates an array of characters. However, you do not apply the convention that the string must be NUL terminated.
You then pass a string that does not follow the NUL termination convention to cout, which does follow that convention. In other words, it runs through the string, printing each character to stdout, until it happens across the character \0 in memory. It's actually fairly lucky that it terminates. If there were not a \0 in the character array it is outputing, it would just keep on going until reaching a memory address that does not belong to your program and failing with a segmentation fault.
As a programming exercise, I just finished writing a Sudoku solver that uses the backtracking algorithm (see Wikipedia for a simple example written in C).
To take this a step further, I would like to use Snow Leopard's GCD to parallelize this so that it runs on all of my machine's cores. Can someone give me pointers on how I should go about doing this and what code changes I should make? Thanks!
Matt
Please let me know if you end up using it. It is run of the mill ANSI C, so should run on everything. See other post for usage.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
short sudoku[9][9];
unsigned long long cubeSolutions=0;
void* cubeValues[10];
const unsigned char oneLookup[64] = {0x8b, 0x80, 0, 0x80, 0, 0, 0, 0x80, 0, 0,0,0,0,0,0, 0x80, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0x80,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
int ifOne(int val) {
if ( oneLookup[(val-1) >> 3] & (1 << ((val-1) & 0x7)) )
return val;
return 0;
}
void init_sudoku() {
int i,j;
for (i=0; i<9; i++)
for (j=0; j<9; j++)
sudoku[i][j]=0x1ff;
}
void set_sudoku( char* initialValues) {
int i;
if ( strlen (initialValues) != 81 ) {
printf("Error: inputString should have length=81, length is %2.2d\n", strlen(initialValues) );
exit (-12);
}
for (i=0; i < 81; i++)
if ((initialValues[i] > 0x30) && (initialValues[i] <= 0x3a))
sudoku[i/9][i%9] = 1 << (initialValues[i] - 0x31) ;
}
void print_sudoku ( int style ) {
int i, j, k;
for (i=0; i < 9; i++) {
for (j=0; j < 9; j++) {
if ( ifOne(sudoku[i][j]) || !style) {
for (k=0; k < 9; k++)
if (sudoku[i][j] & 1<<k)
printf("%d", k+1);
} else
printf("*");
if ( !((j+1)%3) )
printf("\t");
else
printf(",");
}
printf("\n");
if (!((i+1) % 3) )
printf("\n");
}
}
void print_HTML_sudoku () {
int i, j, k, l, m;
printf("<TABLE>\n");
for (i=0; i<3; i++) {
printf(" <TR>\n");
for (j=0; j<3; j++) {
printf(" <TD><TABLE>\n");
for (l=0; l<3; l++) { printf(" <TR>"); for (m=0; m<3; m++) { printf("<TD>"); for (k=0; k < 9; k++) { if (sudoku[i*3+l][j*3+m] & 1<<k)
printf("%d", k+1);
}
printf("</TD>");
}
printf("</TR>\n");
}
printf(" </TABLE></TD>\n");
}
printf(" </TR>\n");
}
printf("</TABLE>");
}
int doRow () {
int count=0, new_value, row_value, i, j;
for (i=0; i<9; i++) {
row_value=0x1ff;
for (j=0; j<9; j++)
row_value&=~ifOne(sudoku[i][j]);
for (j=0; j<9; j++) {
new_value=sudoku[i][j] & row_value;
if (new_value && (new_value != sudoku[i][j]) ) {
count++;
sudoku[i][j] = new_value;
}
}
}
return count;
}
int doCol () {
int count=0, new_value, col_value, i, j;
for (i=0; i<9; i++) {
col_value=0x1ff;
for (j=0; j<9; j++)
col_value&=~ifOne(sudoku[j][i]);
for (j=0; j<9; j++) {
new_value=sudoku[j][i] & col_value;
if (new_value && (new_value != sudoku[j][i]) ) {
count++;
sudoku[j][i] = new_value;
}
}
}
return count;
}
int doCube () {
int count=0, new_value, cube_value, i, j, l, m;
for (i=0; i<3; i++)
for (j=0; j<3; j++) {
cube_value=0x1ff;
for (l=0; l<3; l++)
for (m=0; m<3; m++)
cube_value&=~ifOne(sudoku[i*3+l][j*3+m]);
for (l=0; l<3; l++)
for (m=0; m<3; m++) {
new_value=sudoku[i*3+l][j*3+m] & cube_value;
if (new_value && (new_value != sudoku[i*3+l][j*3+m]) ) {
count++;
sudoku[i*3+l][j*3+m] = new_value;
}
}
}
return count;
}
#define FALSE -1
#define TRUE 1
#define INCOMPLETE 0
int validCube () {
int i, j, l, m, r, c;
int pigeon;
int solved=TRUE;
//check horizontal
for (i=0; i<9; i++) {
pigeon=0;
for (j=0; j<9; j++)
if (ifOne(sudoku[i][j])) {
if (pigeon & sudoku[i][j]) return FALSE;
pigeon |= sudoku[i][j];
} else {
solved=INCOMPLETE;
}
}
//check vertical
for (i=0; i<9; i++) {
pigeon=0;
for (j=0; j<9; j++)
if (ifOne(sudoku[j][i])) {
if (pigeon & sudoku[j][i]) return FALSE;
pigeon |= sudoku[j][i];
}
else {
solved=INCOMPLETE;
}
}
//check cube
for (i=0; i<3; i++)
for (j=0; j<3; j++) {
pigeon=0;
r=j*3; c=i*3;
for (l=0; l<3; l++)
for (m=0; m<3; m++)
if (ifOne(sudoku[r+l][c+m])) {
if (pigeon & sudoku[r+l][c+m]) return FALSE;
pigeon |= sudoku[r+l][c+m];
}
else {
solved=INCOMPLETE;
}
}
return solved;
}
int solveSudoku(int position ) {
int status, i, k;
short oldCube[9][9];
for (i=position; i < 81; i++) {
while ( doCube() + doRow() + doCol() );
status = validCube() ;
if ((status == TRUE) || (status == FALSE))
return status;
if ((status == INCOMPLETE) && !ifOne(sudoku[i/9][i%9]) ) {
memcpy( &oldCube, &sudoku, sizeof(short) * 81) ;
for (k=0; k < 9; k++) {
if ( sudoku[i/9][i%9] & (1<<k) ) {
sudoku[i/9][i%9] = 1 << k ;
if (solveSudoku(i+1) == TRUE ) {
/* return TRUE; */
/* Or look for entire set of solutions */
if (cubeSolutions < 10) {
cubeValues[cubeSolutions] = malloc ( sizeof(short) * 81 ) ;
memcpy( cubeValues[cubeSolutions], &sudoku, sizeof(short) * 81) ;
}
cubeSolutions++;
if ((cubeSolutions & 0x3ffff) == 0x3ffff ) {
printf ("cubeSolutions = %llx\n", cubeSolutions+1 );
}
//if ( cubeSolutions > 10 )
// return TRUE;
}
memcpy( &sudoku, &oldCube, sizeof(short) * 81) ;
}
if (k==8)
return FALSE;
}
}
}
return FALSE;
}
int main ( int argc, char** argv) {
int i;
if (argc != 2) {
printf("Error: number of arguments on command line is incorrect\n");
exit (-12);
}
init_sudoku();
set_sudoku(argv[1]);
printf("[----------------------- Input Data ------------------------]\n\n");
print_sudoku(1);
solveSudoku(0);
if ((validCube()==1) && !cubeSolutions) {
// If sudoku is effectively already solved, cubeSolutions will not be set
printf ("\n This is a trivial sudoku. \n\n");
print_sudoku(1);
}
if (!cubeSolutions && validCube()!=1)
printf("Not Solvable\n");
if (cubeSolutions > 1) {
if (cubeSolutions >= 10)
printf("10+ Solutions, returning first 10 (%lld) [%llx] \n", cubeSolutions, cubeSolutions);
else
printf("%llx Solutions. \n", cubeSolutions);
}
for (i=0; (i < cubeSolutions) && (i < 10); i++) {
memcpy ( &sudoku, cubeValues[i], sizeof(short) * 81 );
printf("[----------------------- Solution %2.2d ------------------------]\n\n", i+1);
print_sudoku(0);
//print_HTML_sudoku();
}
return 0;
}
For one, since backtracking is a depth-first search it is not directly parallelizable, since any newly computed result cannot be used be directly used by another thread. Instead, you must divide the problem early, i.e. thread #1 starts with the first combination for a node in the backtracking graph, and proceeds to search the rest of that subgraph. Thread #2 starts with the second possible combination at the first and so forth. In short, for n threads find the n possible combinations on the top level of the search space (do not "forward-track"), then assign these n starting points to n threads.
However I think the idea is fundamentally flawed: Many sudoku permutations are solved in a matter of a couple thousands of forward+backtracking steps, and are solved within milliseconds on a single thread. This is in fact so fast that even the small coordination required for a few threads (assume that n threads reduce computation time to 1/n of original time) on a multi-core/multi-CPU is not negligible compared to the total running time, thus it is not by any chance a more efficient solution.
Are you sure you want to do that? Like, what problem are you trying to solve? If you want to use all cores, use threads. If you want a fast sudoku solver, I can give you one I wrote, see output below. If you want to make work for yourself, go ahead and use GCD ;).
Update:
I don't think GCD is bad, it just isn't terribly relevant to the task of solving sudoku. GCD is a technology to tie GUI events to code. Essentially, GCD solves two problems, a Quirk in how the MacOS X updates windows, and, it provides an improved method (as compared to threads) of tying code to GUI events.
It doesn't apply to this problem because Sudoku can be solved significantly faster than a person can think (in my humble opinion). That being said, if your goal was to solve Sudoku faster, you would want to use threads, because you would want to directly use more than one processor.
[bear#bear scripts]$ time ./a.out ..1..4.......6.3.5...9.....8.....7.3.......285...7.6..3...8...6..92......4...1...
[----------------------- Input Data ------------------------]
*,*,1 *,*,4 *,*,*
*,*,* *,6,* 3,*,5
*,*,* 9,*,* *,*,*
8,*,* *,*,* 7,*,3
*,*,* *,*,* *,2,8
5,*,* *,7,* 6,*,*
3,*,* *,8,* *,*,6
*,*,9 2,*,* *,*,*
*,4,* *,*,1 *,*,*
[----------------------- Solution 01 ------------------------]
7,6,1 3,5,4 2,8,9
2,9,8 1,6,7 3,4,5
4,5,3 9,2,8 1,6,7
8,1,2 6,4,9 7,5,3
9,7,6 5,1,3 4,2,8
5,3,4 8,7,2 6,9,1
3,2,7 4,8,5 9,1,6
1,8,9 2,3,6 5,7,4
6,4,5 7,9,1 8,3,2
real 0m0.044s
user 0m0.041s
sys 0m0.001s