I want to generate permutations of string of 5 0s followed by the permutations of 4 0s and a single 1, followed by the permutations of 3 0s with 2 1s etc? My code is as follows:
#include<stdio.h>
int main(){
int i,j,k,l,s[5];
for(i=0;i<5;i++)
s[i]=0;
for(k=0;k<5;k++)
printf("%d ",s[k]);
printf("\n");
printf("---------------------------------------------\n");
for(i=0;i<5;i++){
for(j=0;j<5;j++)
if(i==j)
s[j]=1;
else
s[j]=0;
for(k=0;k<5;k++)
printf("%d ",s[k]);
printf("\n");
}
printf("---------------------------------------------\n");
for(i=0;i<5;i++){
for(k=0;k<5;k++)
s[k]=0;
s[i]=1;
for(j=i+1;j<5;j++){
s[j]=1;
for(k=0;k<5;k++)
printf("%d ",s[k]);
printf("\n");
for(k=j;k<5;k++)
s[k]=0;
}
}
printf("---------------------------------------------\n");
for(i=0;i<5;i++){
for(j=i+1;j<5;j++){
for(k=0;k<5;k++)
s[k]=0;
s[i]=1;
s[j]=1;
for(l=j+1;l<5;l++){
s[l]=1;
for(k=0;k<5;k++)
printf("%d ",s[k]);
printf("\n");
for(k=l;k<5;k++)
s[k]=0;
}
}
}
}
So output is
0 0 0 0 0
---------------------------------------------
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
---------------------------------------------
1 1 0 0 0
1 0 1 0 0
1 0 0 1 0
1 0 0 0 1
0 1 1 0 0
0 1 0 1 0
0 1 0 0 1
0 0 1 1 0
0 0 1 0 1
0 0 0 1 1
---------------------------------------------
1 1 1 0 0
1 1 0 1 0
1 1 0 0 1
1 0 1 1 0
1 0 1 0 1
1 0 0 1 1
0 1 1 1 0
0 1 1 0 1
0 1 0 1 1
0 0 1 1 1
Output is ok. However in my code I use
different for loops for different cases.
Is it possible to use better approach so
that length of the code is reduced?
One approach follows. This solution needs O(n) space and each output string requires O(n) time.
#include <stdio.h>
#include <stdlib.h>
char *buf;
// Print combinations of m 1's in a field of n 0/1's starting at s.
void print_combinations(char *s, int n, int m)
{
// If there is nothing left to append, we are done. Print the buffer.
if (m == 0 && n == 0) {
*s = '\0';
printf("%s\n", buf);
return;
}
// Cut if there are more 1's than positions left or negative numbers.
if (m > n || m < 0 || n < 0) return;
// Append a 0 and recur to print the rest.
*s = '0';
print_combinations(s + 1, n - 1, m);
// Now do the same with 1.
*s = '1';
print_combinations(s + 1, n - 1, m - 1);
}
int main(void)
{
int n = 5;
buf = malloc(n + 1);
for (int m = 0; m <= n; m++) {
print_combinations(buf, n, m);
printf("-----\n");
}
return 0;
}
You could use a recursive function like so - you don't have to print the result when finished, you could add it to a list etc.
The function works by starting with an empty string. At each step you add one more character - in this case you add either a 0 or a 1.
If a 1 is added we account for this by decrementing the ones value on the next call to the function. (In a more general case you could pass a list of all the elements to be permuted - then the process would be to pick from this list, add it to your permutation and remove it from the list. You repeat that until the list is empty and you have permuted all of the elements in the list.)
When the string reaches the desired length we have finished and so we return.
#include <stdio.h>
void recurse(char *str, int length, int maxLength, int ones)
{
if (length == maxLength)
{
// we are finished
printf("%s\n", str);
return;
}
if (ones > 0)
{
// put a 1 into the new string
str[length] = '1';
recurse(str, length + 1, maxLength, ones - 1);
}
if (ones < maxLength - length)
{
// there are still spaces for 0s
// put a 0 into the string
str[length] = '0';
recurse(str, length + 1, maxLength, ones);
}
}
int main()
{
const int maxLength = 5;
char buffer[maxLength + 1];
buffer[maxLength] = 0;
int ones;
for (ones = 0; ones <= maxLength; ones++)
{
printf("Ones: %i\n", ones);
recurse(buffer, 0, maxLength, ones);
printf("\n");
}
return 0;
}
The output looks like this:
Ones: 0
00000
Ones: 1
10000
01000
00100
00010
00001
Ones: 2
11000
10100
10010
10001
01100
01010
01001
00110
00101
00011
Ones: 3
11100
11010
11001
10110
10101
10011
01110
01101
01011
00111
Ones: 4
11110
11101
11011
10111
01111
Ones: 5
11111
Finally, unless you really want to/need to learn/use C, I would recommend using C++ because you get really nice features like std::vector and std::set and so many other things which will make your life so much easier. I would have written this completely different in C++.
Related
I found here an implementation for Hoshen-Kopelman Algorithm, But it checks neighbors only up and left, meaning that a diagonal connection is not considered a connection.
How can I improve this code so that even a diagonal connection will be considered a connection?
In the following example I expect 1 object and not 7 objects:
4 5
1 0 1 0 1
0 1 0 1 0
1 0 1 0 0
0 0 1 0 0
--input--
1 0 1 0 1
0 1 0 1 0
1 0 1 0 0
0 0 1 0 0
--output--
1 0 2 0 3
0 4 0 5 0
6 0 7 0 0
0 0 7 0 0
HK reports 7 clusters found
This is the implementation (full code can be found here):
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
/* Implementation of Union-Find Algorithm */
/* The 'labels' array has the meaning that labels[x] is an alias for the label x; by
following this chain until x == labels[x], you can find the canonical name of an
equivalence class. The labels start at one; labels[0] is a special value indicating
the highest label already used. */
int* labels;
int n_labels = 0; /* length of the labels array */
/* uf_find returns the canonical label for the equivalence class containing x */
int uf_find(int x)
{
int y = x;
while (labels[y] != y)
y = labels[y];
while (labels[x] != x)
{
int z = labels[x];
labels[x] = y;
x = z;
}
return y;
}
/* uf_union joins two equivalence classes and returns the canonical label of the resulting class. */
int uf_union(int x, int y)
{
return labels[uf_find(x)] = uf_find(y);
}
/* uf_make_set creates a new equivalence class and returns its label */
int uf_make_set(void)
{
labels[0] ++;
assert(labels[0] < n_labels);
labels[labels[0]] = labels[0];
return labels[0];
}
/* uf_intitialize sets up the data structures needed by the union-find implementation. */
void uf_initialize(int max_labels)
{
n_labels = max_labels;
labels = calloc(sizeof(int), n_labels);
labels[0] = 0;
}
/* uf_done frees the memory used by the union-find data structures */
void uf_done(void)
{
n_labels = 0;
free(labels);
labels = 0;
}
/* End Union-Find implementation */
#define max(a,b) (a>b?a:b)
#define min(a,b) (a>b?b:a)
/* print_matrix prints out a matrix that is set up in the "pointer to pointers" scheme
(aka, an array of arrays); this is incompatible with C's usual representation of 2D
arrays, but allows for 2D arrays with dimensions determined at run-time */
void print_matrix(int** matrix, int m, int n)
{
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
printf("%3d ", matrix[i][j]);
printf("\n");
}
}
/* Label the clusters in "matrix". Return the total number of clusters found. */
int hoshen_kopelman(int** matrix, int m, int n)
{
uf_initialize(m * n / 2);
/* scan the matrix */
for (int y = 0; y < m; y++)
{
for (int x = 0; x < n; x++)
{
if (matrix[y][x])
{ // if occupied ...
int up = (y == 0 ? 0 : matrix[y - 1][x]); // look up
int left = (x == 0 ? 0 : matrix[y][x - 1]); // look left
switch (!!up + !!left)
{
case 0:
matrix[y][x] = uf_make_set(); // a new cluster
break;
case 1: // part of an existing cluster
matrix[y][x] = max(up, left); // whichever is nonzero is labelled
break;
case 2: // this site binds two clusters
matrix[y][x] = uf_union(up, left);
break;
}
}
}
}
/* apply the relabeling to the matrix */
/* This is a little bit sneaky.. we create a mapping from the canonical labels
determined by union/find into a new set of canonical labels, which are
guaranteed to be sequential. */
int* new_labels = calloc(sizeof(int), n_labels); // allocate array, initialized to zero
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (matrix[i][j])
{
int x = uf_find(matrix[i][j]);
if (new_labels[x] == 0)
{
new_labels[0]++;
new_labels[x] = new_labels[0];
}
matrix[i][j] = new_labels[x];
}
int total_clusters = new_labels[0];
free(new_labels);
uf_done();
return total_clusters;
}
/* This procedure checks to see that any occupied neighbors of an occupied site
have the same label. */
void check_labelling(int** matrix, int m, int n)
{
int N, S, E, W;
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (matrix[i][j])
{
N = (i == 0 ? 0 : matrix[i - 1][j]);
S = (i == m - 1 ? 0 : matrix[i + 1][j]);
E = (j == n - 1 ? 0 : matrix[i][j + 1]);
W = (j == 0 ? 0 : matrix[i][j - 1]);
assert(N == 0 || matrix[i][j] == N);
assert(S == 0 || matrix[i][j] == S);
assert(E == 0 || matrix[i][j] == E);
assert(W == 0 || matrix[i][j] == W);
}
}
/* The sample program reads in a matrix from standard input, runs the HK algorithm on
it, and prints out the results. The form of the input is two integers giving the
dimensions of the matrix, followed by the matrix elements (with data separated by
whitespace).
a sample input file is the following:
8 8
1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 1
1 0 0 0 0 1 0 1
1 0 0 1 0 1 0 1
1 0 0 1 0 1 0 1
1 0 0 1 1 1 0 1
1 1 1 1 0 0 0 1
0 0 0 1 1 1 0 1
this sample input gives the following output:
--input--
1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 1
1 0 0 0 0 1 0 1
1 0 0 1 0 1 0 1
1 0 0 1 0 1 0 1
1 0 0 1 1 1 0 1
1 1 1 1 0 0 0 1
0 0 0 1 1 1 0 1
--output--
1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 1
2 0 0 0 0 2 0 1
2 0 0 2 0 2 0 1
2 0 0 2 0 2 0 1
2 0 0 2 2 2 0 1
2 2 2 2 0 0 0 1
0 0 0 2 2 2 0 1
HK reports 2 clusters found
*/
int main(int argc, char** argv)
{
int m, n;
int** matrix;
/* Read in the matrix from standard input
The whitespace-deliminated matrix input is preceeded
by the number of rows and number of columns */
while (2 == scanf_s("%d %d", &m, &n))
{ // m = rows, n = columns
matrix = (int**)calloc(m, sizeof(int*));
for (int i = 0; i < m; i++)
{
matrix[i] = (int*)calloc(n, sizeof(int));
for (int j = 0; j < n; j++)
scanf_s("%d", &(matrix[i][j]));
}
printf_s(" --input-- \n");
print_matrix(matrix, m, n);
printf(" --output-- \n");
/* Process the matrix */
int clusters = hoshen_kopelman(matrix, m, n);
/* Output the result */
print_matrix(matrix, m, n);
check_labelling(matrix, m, n);
printf("HK reports %d clusters found\n", clusters);
for (int i = 0; i < m; i++)
free(matrix[i]);
free(matrix);
}
return 0;
}
I tried to change the function hoshen_kopelman as described below, but I still get 2 objects instead of 1:
int hoshen_kopelman(int** matrix, int m, int n)
{
uf_initialize(m * n / 2);
/* scan the matrix */
for (int y = 0; y < m; y++)
{
for (int x = 0; x < n; x++)
{
if (matrix[y][x])
{ // if occupied ...
int up = (y == 0 ? 0 : matrix[y - 1][x]); // look up
int left = (x == 0 ? 0 : matrix[y][x - 1]); // look left
// ----------- THE NEW CODE -------------
if (x > 0)
{
if (up == 0 && y > 0) // left+up
up = matrix[y - 1][x - 1];
if (left == 0 && y < m - 1) // left+down
left = matrix[y + 1][x - 1];
}
// ---------- END NEW CODE --------------
switch (!!up + !!left)
{
case 0:
matrix[y][x] = uf_make_set(); // a new cluster
break;
case 1: // part of an existing cluster
matrix[y][x] = max(up, left); // whichever is nonzero is labelled
break;
case 2: // this site binds two clusters
matrix[y][x] = uf_union(up, left);
break;
}
}
}
}
/* apply the relabeling to the matrix */
/* This is a little bit sneaky.. we create a mapping from the canonical labels
determined by union/find into a new set of canonical labels, which are
guaranteed to be sequential. */
int* new_labels = calloc(sizeof(int), n_labels); // allocate array, initialized to zero
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (matrix[i][j])
{
int x = uf_find(matrix[i][j]);
if (new_labels[x] == 0)
{
new_labels[0]++;
new_labels[x] = new_labels[0];
}
matrix[i][j] = new_labels[x];
}
int total_clusters = new_labels[0];
free(new_labels);
uf_done();
return total_clusters;
}
The following output is now obtained (I am expecting 1 and got 2):
4 5
1 0 1 0 1
0 1 0 1 0
1 0 1 0 0
0 0 1 0 0
--input--
1 0 1 0 1
0 1 0 1 0
1 0 1 0 0
0 0 1 0 0
--output--
1 0 1 0 1
0 1 0 1 0
2 0 1 0 0
0 0 1 0 0
HK reports 2 clusters found
What is the correct way to correct the code to check all 8 neighbors?
I led you astray saying to check down-left. The algorithm relies on the current node it is examining being after all the neighbors it checks. So you need to check left, up, up-left, and up-right. You can use this in place of your new code:
if (y > 0)
{
if (left == 0 && x > 0) // left+up
left = matrix[y - 1][x - 1];
if (up == 0 && x < n-1) // right+up
up = matrix[y - 1][x + 1];
}
My input looks like this :
15 5
0 0 1 0 0
0 0 1 0 0
0 0 1 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 1 0 1 0
0 0 1 0 0
0 0 0 0 1
0 0 0 0 0
0 0 0 0 0
0 1 0 0 0
1 0 0 0 1
0 0 1 0 0
0 0 0 1 0
The first row contains the number of rows and columns of my array. And basically I want to find out where these 1s are in the array.
So in the first 3 rows i want to get 3, in the 7th 1, and in the 8th, i want to get 2 3 etc..
My code looks like this so far
#include <stdio.h>
int main() {
int row, column;
FILE* input;
FILE* output;
input = fopen("input.txt", "r");
if (input == 0) {
printf("ERROR couldn't open input.txt");
return 1;
}
if (! fscanf(input, "%d %d", &row, &column)) {
printf("ERROR not recognised value");
fclose(input);
return 2;
}
output = fopen("output.txt", "w");
int meteor[row][column];
for (int i = 0; i < row; i++) {
for (int j = 0; j < column; j++) {
fscanf(input, "%d", &meteor[i][j]);
}
}
int sum;
int loc[row];
for (int i = 0; i < row; i++) {
sum = 0;
for (int j = 0; j < column; j++) {
sum += meteor[i][j];
if (meteor[i][j] == 1) {
loc[i] = (j + 1);
}
}
printf("%d %d\n", loc[i], sum);
}
fclose(input);
fclose(output);
return 0;
}
My output is this :
3 1
3 1
3 1
0 0
-1 0
1 1
4 2
3 1
5 1
0 0
4214921 0
2 1
5 2
3 1
4 1
The first column shows some of the locations, the second shows how many 1s are in the row, but it all fails when there are only 0s in the row or there is more than one 1. And also I would like to store these values.
You need to initialize loc[].. If you look carefully at your code, you only populate it on the condition of if (meteor[i][j] == 1)... but you print it for every index of i. which would print uninitialized memory (i.e. unknown).
To answer the second part of your question, if you want to store the "sum". Simply make loc[] a 2d array just like meteor, but with 2 columns (row and sum). i.e. int loc[row][2].. making sure to initialize both columns of course :)
I have some dificulties in creating the following array. My task is to fill using recursion a 2D array with all the possible combinations of 0 and 1 taken m times in lexical order. Mathematically speaking there are 2 ^ m combinations.My program just fills the first 3 rows of the array with the same order 0 1 0 1 and then just prints for the rest of the rows 0 0 0 0.
Example
m=4
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
This is my code so far and I appreciate if someone could correct it and explain me what I am doing wrong as I can't spot the mistake myself
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
void *safeMalloc(int n) {
void *p = malloc(n);
if (p == NULL) {
printf("Error: malloc(%d) failed. Out of memory?\n", n);
exit(EXIT_FAILURE);
}
return p;
}
void combine(int** arrTF,int m,int n,int row,int col){
if(m==0){
if(row<pow(2,m)){
row++;
combine(arrTF,n,n,row,0);
}else{
return;
}
}else{
arrTF[row][col]=0;
col++;
combine(arrTF,m-1,n,row,col);
arrTF[row][col]=1;
col++;
combine(arrTF,m-1,n,row,col);
}
}
int main(int argc, char *argv[]) {
int m
scanf("%d",&m);
int** arrTF;
arrTF = safeMalloc(pow(2,m)*sizeof(int *));
for (int r=0; r < pow(2,m); r++) {
arrTF[r] = safeMalloc(m*sizeof(int));
}
for(int i=0;i<pow(2,m);i++){
for(int j=0;j<m;j++){
arrTF[i][j]=0;
}
}
combine(arrTF,m,m,0,0);
for(int i=0;i<pow(2,m);i++){
for(int j=0;j<m;j++){
printf("%d ",arrTF[i][j]);
}
printf("\n");
}
return 0;
}
You want all the possible (2^m) combinations of 0's and 1's taken m times in lexical order and you are using a 2D array to store the result.
Things would be very easy if you just want to print all the possible combination of 0's and 1's instead of storing it in 2D array and printing array later.
Storing a combination of 0's and 1's to 2D array is a little bit tricky as every combination is one element of your 2D array.
You want to generate the combination of 0's and 1's in accordance with the recursive algorithm.
So, let's say, at some stage if your algorithm generates the combination 0010 which is stored in an element in 2D array.
And the next combination would be 0011 which the recursive algorithm will generate just by changing the last number from 0 to 1 in the last combination (0010).
So, that means everytime when a combination is generated, you need to copy that combination to its successive location in 2D array.
For e.g. if 0010 is stored at index 2 in 2D array before the algorithm starts computing the next combination, we need to do two things:
Copy the elements of index 2 to index 3
Increase the row number so that last combination will be intact
(Say, this is 2D array)
|0|0|0|0| index 0
|0|0|0|1| index 1
|0|0|1|0| index 2 ---> copy this to its successive location (i.e. at index 3)
|0|0|1|1| index 3 ---> Last combination (index 2) and the last digit is changed from 0 to 1
.....
.....
.....
This we need to do for after every combination generated.
Now, I hope you got where you are making the mistake.
Few practice good to follow:
If you want to allocate memory as well as initialized it with 0, use calloc instead of malloc.
Any math function you are calling again and again for the same input, it's better to call it once and store the result in a variable and use that result where ever required.
Do not include any header file which is not required in your program.
Once done, make sure to free the dynamically allocated memory in your program.
I have made the corrections in your program:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
void *safeMalloc(size_t n, size_t size) {
void *p = calloc(n, size);
if (p == NULL) {
printf("Error: calloc(%zu) failed. Out of memory!\n", n);
exit(EXIT_FAILURE);
}
return p;
}
void deallocate(int ** ptr, int row) {
for(int i = 0; i<row; i++)
free(ptr[i]);
free(ptr);
}
void combine(int **arrTF, int m, int max_col, int max_row) {
static int row;
if(m==0){
int i;
if (row<(max_row - 1))
{
for(i=0; i<max_col; i++)
arrTF[row+1][i] = arrTF[row][i];
}
row++;
return;
} else {
arrTF[row][max_col-m] = 0;
combine(arrTF, m-1, max_col, max_row);
arrTF[row][max_col-m] = 1;
combine(arrTF, m-1, max_col, max_row);
}
}
int main(int argc, char *argv[]) {
int** arrTF;
int m, max_row;
printf ("Enter number: \n");
scanf("%d", &m);
max_row = pow(2, m);
arrTF = safeMalloc(max_row, sizeof(int *));
for (int r=0; r<max_row; r++) {
arrTF[r] = safeMalloc(m, sizeof(int));
}
combine(arrTF, m, m, max_row);
for(int i=0; i<max_row; i++) {
for(int j=0; j<m; j++) {
printf("%d ", arrTF[i][j]);
}
printf("\n");
}
deallocate(arrTF, max_row);
return 0;
}
Output:
$ ./a.out
Enter number:
2
0 0
0 1
1 0
1 1
$ ./a.out
4
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
Hope this helps.
When main runs, I've been getting this output. The first columns being somehow confused with the last one, even though it's specifically initialized.
0 0 0 0 0 0 4 3 13 3 3 0 0 0 0 0 0
0 0 0 0 0 4 0 0 5 0 0 2 0 0 0 0 0
0 0 0 0 4 0 0 0 5 0 0 0 2 0 0 0 0
0 0 0 18 0 0 0 0 5 0 0 0 0 17 0 0 0
0 0 4 0 2 0 7 12 19 12 3 0 8 0 2 0 0
0 4 0 0 0 0 0 0 5 0 0 0 0 0 0 2 0
0 0 0 0 5 0 0 0 5 0 0 0 5 0 0 0 2
2 0 0 0 11 0 0 0 5 0 0 0 11 0 0 0 1
1 7 7 7 20 7 7 7 14 7 7 7 20 7 7 7 1
1 0 0 0 11 0 0 0 5 0 0 0 11 0 0 0 1
1 0 0 0 1 0 0 0 5 0 0 0 1 0 0 0 0
0 6 0 0 0 0 0 0 5 0 0 0 0 0 0 8 0
0 0 6 0 4 0 7 12 19 12 3 0 6 0 8 0 0
0 0 0 15 0 0 0 0 5 0 0 0 0 16 0 0 0
0 0 0 0 6 0 0 0 5 0 0 0 8 0 0 0 0
0 0 0 0 0 6 0 0 5 0 0 8 0 0 0 0 0
0 0 0 0 0 0 7 7 7 7 8 0 0 0 0 0 0
motion.h
struct square{
int directions;
int isRobotHere;
int isMultipleDirections;
int printable;
};
typedef struct square Square;
struct robot{
int robotx;
int roboty;
int robotz;
int destinationx;
int destinationy;
int destinationz;
};
typedef struct robot Robot;
enum direction{North, NorthWest, West, SouthWest, South, SouthEast, East, NorthEast};
Square firstfloor[16][16];
Square secondfloor[16][16];
void printbothfloors();
void initializeArrays();
initializeArrays.c
#include <stdio.h>
#include "motion.h"
void initializeArrays(){
firstfloor[6][0].directions=5;
firstfloor[7][0].directions=5;
firstfloor[8][0].directions=14;
firstfloor[9][0].directions=5;
firstfloor[10][0].directions=6;
firstfloor[11][1].directions=6;
firstfloor[12][2].directions=6;
firstfloor[13][3].directions=15;
firstfloor[14][4].directions=6;
firstfloor[15][5].directions=6;
firstfloor[16][6].directions=7;
firstfloor[16][7].directions=7;
firstfloor[16][8].directions=7;
firstfloor[16][9].directions=7;
firstfloor[16][10].directions=8;
firstfloor[15][11].directions=8;
firstfloor[14][12].directions=8;
firstfloor[13][13].directions=16;
firstfloor[12][14].directions=8;
firstfloor[11][15].directions=8;
firstfloor[10][16].directions=1;
firstfloor[9][16].directions=1;
firstfloor[8][16].directions=1;
firstfloor[7][16].directions=1;
firstfloor[6][16].directions=2;
firstfloor[5][15].directions=2;
firstfloor[4][14].directions=2;
firstfloor[3][13].directions=17;
firstfloor[2][12].directions=2;
firstfloor[1][11].directions=2;
firstfloor[0][10].directions=3;
firstfloor[0][9].directions=3;
firstfloor[0][8].directions=13;
firstfloor[0][7].directions=3;
firstfloor[0][6].directions=4;
firstfloor[1][5].directions=4;
firstfloor[2][4].directions=4;
firstfloor[3][3].directions=18;
firstfloor[4][2].directions=4;
firstfloor[5][1].directions=4;
firstfloor[1][8].directions=5;
firstfloor[2][8].directions=5;
firstfloor[3][8].directions=5;
firstfloor[4][8].directions=19;
firstfloor[5][8].directions=5;
firstfloor[6][8].directions=5;
firstfloor[7][8].directions=5;
firstfloor[8][8].directions=14;
firstfloor[9][8].directions=5;
firstfloor[10][8].directions=5;
firstfloor[11][8].directions=5;
firstfloor[12][8].directions=19;
firstfloor[13][8].directions=5;
firstfloor[14][8].directions=5;
firstfloor[15][8].directions=5;
firstfloor[8][1].directions=7;
firstfloor[8][2].directions=7;
firstfloor[8][3].directions=7;
firstfloor[8][4].directions=20;
firstfloor[8][5].directions=7;
firstfloor[8][6].directions=7;
firstfloor[8][7].directions=7;
firstfloor[8][9].directions=7;
firstfloor[8][10].directions=7;
firstfloor[8][11].directions=7;
firstfloor[8][12].directions=20;
firstfloor[8][13].directions=7;
firstfloor[8][14].directions=7;
firstfloor[8][15].directions=7;
firstfloor[7][4].directions=11;
firstfloor[9][4].directions=11;
firstfloor[4][7].directions=12;
firstfloor[4][9].directions=12;
firstfloor[12][7].directions=12;
firstfloor[12][9].directions=12;
firstfloor[7][12].directions=11;
firstfloor[9][12].directions=11;
firstfloor[4][4].directions=2;
firstfloor[4][6].directions=7;
firstfloor[4][10].directions=3;
firstfloor[4][12].directions=8;
firstfloor[6][4].directions=5;
firstfloor[6][12].directions=5;
firstfloor[10][4].directions=1;
firstfloor[10][12].directions=1;
firstfloor[12][4].directions=4;
firstfloor[12][6].directions=7;
firstfloor[12][10].directions=3;
firstfloor[12][12].directions=6;
firstfloor[11][0].directions=0;
firstfloor[5][16].directions=0;
}
printbothfloors.c
#include <stdio.h>
#include <stdlib.h>
#include "motion.h"
void printbothfloors(){
// printf("printfloor is running");
int upper, lower, i, j;
printf("%4d %4d %4d %4d %4d", firstfloor[6][0].directions, firstfloor[7][0].directions, firstfloor[8][0].directions, firstfloor[9][0].directions, firstfloor[10][0].directions);
printf("%4d %4d %4d %4d %4d", firstfloor[6][16].directions, firstfloor[7][16].directions, firstfloor[8][16].directions, firstfloor[9][16].directions, firstfloor[10][16].directions);
printf("FIRST FLOOR");
printf("\n-");
/* The next for loop prints out the upper edge */
for (upper = 0; upper < 18; upper++){
printf("----");
}
printf("\n");
/*The next for loop prints out the floor, every element is 4 digits wide */
for (i = 0; i <= 16; i++){
printf("|");
for(j = 0; j <= 16; j++){
printf("%4d", firstfloor[i][j].directions);
}
printf(" |\n");
}
/* The next for loop prints out the lower edge */
for (lower = 0; lower < 18; lower++){
printf("----");
}
printf("-\n");
/*
printf("SECOND FLOOR");
// printf("printfloor is running");
printf("\n-----");
// The next for loop prints out the upper edge
for (upper = 0; upper < 17; upper++){
printf("----");
}
printf("\n");
//The next for loop prints out the floor, every element is 4 digits wide
for (i = 0; i <= 16; i++){
printf("|");
for(j = 0; j <= 16; j++){
printf("%4d", secondfloor[i][j].directions);
}
printf(" |\n");
}
// The next for loop prints out the lower edge
for (lower = 0; lower < 17; lower++){
printf("----");
}
printf("-----\n");
*/
}
prog2.c
#include <stdio.h>
#include <stdlib.h>
#include "motion.h"
//#include "printbothfloors.h"
int main(){
int row = 0;
int column = 0;
initializeArrays();
// printbothfloors();
/*
for (row=0; row < 17; row++){
for (column=0; column < 17; column++){
// firstfloor[row][column].directions=0;
// secondfloor[row][column].directions=5;
//firstfloor[i][j].isRobotHere=0;
//secondfloor[i][j].isRobotHere=0;
//firstfloor[i][j].isMultipleDirections=0;
//secondfloor[i][j].isMultipleDirections=0;
}
}
firstfloor[6][0].directions=5;
firstfloor[7][0].directions=5;
firstfloor[8][0].directions=14;
firstfloor[9][0].directions=5;
firstfloor[10][0].directions=6;
firstfloor[11][1].directions=6;
firstfloor[12][2].directions=6;
firstfloor[13][3].directions=15;
firstfloor[14][4].directions=6;
firstfloor[15][5].directions=6;
firstfloor[16][6].directions=7;
firstfloor[16][7].directions=7;
firstfloor[16][8].directions=7;
firstfloor[16][9].directions=7;
firstfloor[16][10].directions=8;
firstfloor[15][11].directions=8;
firstfloor[14][12].directions=8;
firstfloor[13][13].directions=16;
firstfloor[12][14].directions=8;
firstfloor[11][15].directions=8;
firstfloor[10][16].directions=1;
firstfloor[9][16].directions=1;
firstfloor[8][16].directions=1;
firstfloor[7][16].directions=1;
firstfloor[6][16].directions=2;
firstfloor[5][15].directions=2;
firstfloor[4][14].directions=2;
firstfloor[3][13].directions=17;
firstfloor[2][12].directions=2;
firstfloor[1][11].directions=2;
firstfloor[0][10].directions=3;
firstfloor[0][9].directions=3;
firstfloor[0][8].directions=13;
firstfloor[0][7].directions=3;
firstfloor[0][6].directions=4;
firstfloor[1][5].directions=4;
firstfloor[2][4].directions=4;
firstfloor[3][3].directions=18;
firstfloor[4][2].directions=4;
firstfloor[5][1].directions=4;
firstfloor[1][8].directions=5;
firstfloor[2][8].directions=5;
firstfloor[3][8].directions=5;
firstfloor[4][8].directions=19;
firstfloor[5][8].directions=5;
firstfloor[6][8].directions=5;
firstfloor[7][8].directions=5;
firstfloor[8][8].directions=14;
firstfloor[9][8].directions=5;
firstfloor[10][8].directions=5;
firstfloor[11][8].directions=5;
firstfloor[12][8].directions=19;
firstfloor[13][8].directions=5;
firstfloor[14][8].directions=5;
firstfloor[15][8].directions=5;
firstfloor[8][1].directions=7;
firstfloor[8][2].directions=7;
firstfloor[8][3].directions=7;
firstfloor[8][4].directions=20;
firstfloor[8][5].directions=7;
firstfloor[8][6].directions=7;
firstfloor[8][7].directions=7;
firstfloor[8][9].directions=7;
firstfloor[8][10].directions=7;
firstfloor[8][11].directions=7;
firstfloor[8][12].directions=20;
firstfloor[8][13].directions=7;
firstfloor[8][14].directions=7;
firstfloor[8][15].directions=7;
firstfloor[7][4].directions=11;
firstfloor[9][4].directions=11;
firstfloor[4][7].directions=12;
firstfloor[4][9].directions=12;
firstfloor[12][7].directions=12;
firstfloor[12][9].directions=12;
firstfloor[7][12].directions=11;
firstfloor[9][12].directions=11;
firstfloor[4][4].directions=2;
firstfloor[4][6].directions=7;
firstfloor[4][10].directions=3;
firstfloor[4][12].directions=8;
firstfloor[6][4].directions=5;
firstfloor[6][12].directions=5;
firstfloor[10][4].directions=1;
firstfloor[10][12].directions=1;
firstfloor[12][4].directions=4;
firstfloor[12][6].directions=7;
firstfloor[12][10].directions=3;
firstfloor[12][12].directions=6;
firstfloor[11][0].directions=0;
firstfloor[5][16].directions=0;
*/
// for (i = 0; i < 17; i++){
// firstfloor
// printbothfloors();
// pbf_entrypoints();
/*
row = 0;
column = 0;
while (i < 17){
firstfloor[i][8] = 1;
secondfloor[i][8] = 1;
firstfloor[8][i] = 1;
secondfloor[8][i] = 1;
i++;
}
*/
printf("Function got here");
printbothfloors();
return 0;
}
Valid indices for Type arr[N] are between 0 and N-1.
This goes for any Type and for any number of dimensions.
Indexes in C start at 0, so if you declare an array to have 16 elements, valid indexes start from 0 and end at 15.
16 boxes, numbered 0 to 15:
---------------------------------------------------------------------------------
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
---------------------------------------------------------------------------------
Accessing an element outside of this range (e.g. by using the index -1 or 16) invokes undefined behaviour.
Both the first answers are correct, I'll add some examples from your code.
firstfloor[6][16].directions
firstfloor[x][y], has been initialized as firstfloor[16][16] meaning the values x and y can can from 0 to 15. The numbering system is 0 based, a hang-over from Java's C language ancestry. 0-15 covers 16 elements.
/*The next for loop prints out the floor, every element is 4 digits wide */
for (i = 0; i <= 16; i++){
printf("|");
for(j = 0; j <= 16; j++){
printf("%4d", firstfloor[i][j].directions);
}
printf(" |\n");
}
The two for loops, start correctly at i = 0 and should only continue while i < 16 or if you prefer i <= 15.
At the moment I'm personally jumping between Java, C and Python, so someone might correct me in this, but calling firstfloor[6][16] if it does not generate an array out-of-bounds error, will give you the same result as firstfloor[7][0]. This might explain why you are observing the first columns being confused with the last.
Good luck :-)
I have a problem with this piece of code in C.
#include <stdio.h>
#include <stdint.h>
typedef uint64_t bboard;
// Accessing a square of the bitboard
int
get (bboard b, int square)
{
return (b & (1ULL << square));
}
void
print_board (bboard b)
{
int i, j, square;
for (i = 7; i >= 0; i--) // rank => top to bottom
{
for (j = 0; j < 8; j++) // file => left to right
printf ("%d ", get (b, j+8*i) ? 1 : 0);
printf ("\n");
}
}
int
main ()
{
bboard b = 0xffffffffffffffff;
print_board (b);
}
// result that I have
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
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
Ok, why the bitboard is not set with all bit at 1?
For any question please add a comment. Ty :D
get returns an int, but (b & (1ULL << square)) is a uint64_t. When (b & (1ULL << square)) is greater than INT_MAX, the result is undefined; in this case it truncates and returns 0.
If get returns a bboard instead, this works as expected (verified here: http://codepad.org/zEZiJKeR).