Permutations without repetition C
-1
I have this code for permutation WITH repetition. Would anyone help to modify it to be WITHOUT repetition. I can't figure it out.
int array1 = {1, 2, 3}; //array can be {1,1,2,3} for example also
int array2[3];
void permWithRep (int array1, int array2, int last, int index){
int i, len = last+1;
enter code here
for ( i=0; i<len; i++ )
{
array2[index] = array1[i] ;
if (index == last){
for(int i = 0; i < 3; i++) {
printf("%d ", array2[i]);
}
printf("n");
}
else // Recur for higher indexes
permWithRep (array1, array2, last, index+1);
}
}
int main()
{
int len = sizeof(array1)/sizeof(int);
permWithRep (array1, array2, len-1, 0);
return 0;
}
c permutation repetition
asked Nov 12 '18 at 21:19
Krivi21
82
add a comment |
-1
I have this code for permutation WITH repetition. Would anyone help to modify it to be WITHOUT repetition. I can't figure it out.
int array1 = {1, 2, 3}; //array can be {1,1,2,3} for example also
int array2[3];
void permWithRep (int array1, int array2, int last, int index){
int i, len = last+1;
enter code here
for ( i=0; i<len; i++ )
{
array2[index] = array1[i] ;
if (index == last){
for(int i = 0; i < 3; i++) {
printf("%d ", array2[i]);
}
printf("n");
}
else // Recur for higher indexes
permWithRep (array1, array2, last, index+1);
}
}
int main()
{
int len = sizeof(array1)/sizeof(int);
permWithRep (array1, array2, len-1, 0);
return 0;
}
c permutation repetition
asked Nov 12 '18 at 21:19
Krivi21
82
Would you define what you mean by "repetition"? That may sound like a strange question, but I do not see any repeated code. Perhaps you are talking about the loops. Do you want to remove all loops? Is this a homework exercise?
– paddy
Nov 12 '18 at 21:26
For anyone unclear on the terminology of the question - en.wikipedia.org/wiki/Permutation#Permutations_with_repetition
– Wumpus Q. Wumbley
Nov 12 '18 at 21:49
What have you tried? What don't you understand about the problem?
– chb
Nov 12 '18 at 23:01
add a comment |
-1
-1
-1
I have this code for permutation WITH repetition. Would anyone help to modify it to be WITHOUT repetition. I can't figure it out.
int array1 = {1, 2, 3}; //array can be {1,1,2,3} for example also
int array2[3];
void permWithRep (int array1, int array2, int last, int index){
int i, len = last+1;
enter code here
for ( i=0; i<len; i++ )
{
array2[index] = array1[i] ;
if (index == last){
for(int i = 0; i < 3; i++) {
printf("%d ", array2[i]);
}
printf("n");
}
else // Recur for higher indexes
permWithRep (array1, array2, last, index+1);
}
}
int main()
{
int len = sizeof(array1)/sizeof(int);
permWithRep (array1, array2, len-1, 0);
return 0;
}
c permutation repetition
asked Nov 12 '18 at 21:19
Krivi21
82
I have this code for permutation WITH repetition. Would anyone help to modify it to be WITHOUT repetition. I can't figure it out.
int array1 = {1, 2, 3}; //array can be {1,1,2,3} for example also
int array2[3];
void permWithRep (int array1, int array2, int last, int index){
int i, len = last+1;
enter code here
for ( i=0; i<len; i++ )
{
array2[index] = array1[i] ;
if (index == last){
for(int i = 0; i < 3; i++) {
printf("%d ", array2[i]);
}
printf("n");
}
else // Recur for higher indexes
permWithRep (array1, array2, last, index+1);
}
}
int main()
{
int len = sizeof(array1)/sizeof(int);
permWithRep (array1, array2, len-1, 0);
return 0;
}
c permutation repetition
c permutation repetition
asked Nov 12 '18 at 21:19
Krivi21
82
asked Nov 12 '18 at 21:19
Krivi21
82
edited Nov 12 '18 at 22:01
asked Nov 12 '18 at 21:19
Krivi21
82
asked Nov 12 '18 at 21:19
Krivi21
82
asked Nov 12 '18 at 21:19
Krivi21
82
82
Would you define what you mean by "repetition"? That may sound like a strange question, but I do not see any repeated code. Perhaps you are talking about the loops. Do you want to remove all loops? Is this a homework exercise?
– paddy
Nov 12 '18 at 21:26
For anyone unclear on the terminology of the question - en.wikipedia.org/wiki/Permutation#Permutations_with_repetition
– Wumpus Q. Wumbley
Nov 12 '18 at 21:49
What have you tried? What don't you understand about the problem?
– chb
Nov 12 '18 at 23:01
add a comment |
Would you define what you mean by "repetition"? That may sound like a strange question, but I do not see any repeated code. Perhaps you are talking about the loops. Do you want to remove all loops? Is this a homework exercise?
– paddy
Nov 12 '18 at 21:26
For anyone unclear on the terminology of the question - en.wikipedia.org/wiki/Permutation#Permutations_with_repetition
– Wumpus Q. Wumbley
Nov 12 '18 at 21:49
What have you tried? What don't you understand about the problem?
– chb
Nov 12 '18 at 23:01
Would you define what you mean by "repetition"? That may sound like a strange question, but I do not see any repeated code. Perhaps you are talking about the loops. Do you want to remove all loops? Is this a homework exercise?
– paddy
Nov 12 '18 at 21:26
Would you define what you mean by "repetition"? That may sound like a strange question, but I do not see any repeated code. Perhaps you are talking about the loops. Do you want to remove all loops? Is this a homework exercise?
– paddy
Nov 12 '18 at 21:26
For anyone unclear on the terminology of the question - en.wikipedia.org/wiki/Permutation#Permutations_with_repetition
– Wumpus Q. Wumbley
Nov 12 '18 at 21:49
For anyone unclear on the terminology of the question - en.wikipedia.org/wiki/Permutation#Permutations_with_repetition
– Wumpus Q. Wumbley
Nov 12 '18 at 21:49
What have you tried? What don't you understand about the problem?
– chb
Nov 12 '18 at 23:01
What have you tried? What don't you understand about the problem?
– chb
Nov 12 '18 at 23:01
add a comment |
1 Answer
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I tested this on few inputs and it works.
There is certainly a better way to do it, but that's what i did:
#include <stdio.h>
int array1 = {1, 2, 3}; //array can be {1,1,2,3} for example also
int array2[3];
int fixed[3];
void permWithRep(int array1, int array2,int fixed_index, int size , int level){
if(level == size) return print_array(array2,size);
for(int k = 0; k< size; k++){
if(!is_present_in_fixed(fixed_index,level,k)){
fixed_index[level] = k;
array2[level] = array1[k];
permWithRep(array1,array2,fixed_index,size,level+1);
}
}
}
void print_array(int array, int size){
for(int k = 0; k < size; k++)printf("%d ",array[k]);
printf("n");
}
int is_present_in_fixed(int array, int size, int element ){
for(int k =0; k<size;k++)
if(element == array[k]) return 1;
return 0;
}
int main()
{
int len = sizeof(array1)/sizeof(int);
permWithRep (array1, array2, fixed, len, 0);
return 0;
}
What I did:
- Add an array (
fixed_index) that keep track of the already visited indexes for one permutation. - In the loop, check if the index of current element belongs to the index already visited in that permutation (
is_present_in_fixed(fixed_index,level,k)). - If so, it just skip that element (it does not enter the
ifstatement). - Otherwise adds that element to the visited for that permutation and go on with recursive call.
- When all array has been iterated on (
level == size), just print the results contained inarray2.
answered Nov 12 '18 at 23:05
Ollaw
940517
add a comment |
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1 Answer
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1 Answer
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0
I tested this on few inputs and it works.
There is certainly a better way to do it, but that's what i did:
#include <stdio.h>
int array1 = {1, 2, 3}; //array can be {1,1,2,3} for example also
int array2[3];
int fixed[3];
void permWithRep(int array1, int array2,int fixed_index, int size , int level){
if(level == size) return print_array(array2,size);
for(int k = 0; k< size; k++){
if(!is_present_in_fixed(fixed_index,level,k)){
fixed_index[level] = k;
array2[level] = array1[k];
permWithRep(array1,array2,fixed_index,size,level+1);
}
}
}
void print_array(int array, int size){
for(int k = 0; k < size; k++)printf("%d ",array[k]);
printf("n");
}
int is_present_in_fixed(int array, int size, int element ){
for(int k =0; k<size;k++)
if(element == array[k]) return 1;
return 0;
}
int main()
{
int len = sizeof(array1)/sizeof(int);
permWithRep (array1, array2, fixed, len, 0);
return 0;
}
What I did:
- Add an array (
fixed_index) that keep track of the already visited indexes for one permutation. - In the loop, check if the index of current element belongs to the index already visited in that permutation (
is_present_in_fixed(fixed_index,level,k)). - If so, it just skip that element (it does not enter the
ifstatement). - Otherwise adds that element to the visited for that permutation and go on with recursive call.
- When all array has been iterated on (
level == size), just print the results contained inarray2.
answered Nov 12 '18 at 23:05
Ollaw
940517
add a comment |
0
I tested this on few inputs and it works.
There is certainly a better way to do it, but that's what i did:
#include <stdio.h>
int array1 = {1, 2, 3}; //array can be {1,1,2,3} for example also
int array2[3];
int fixed[3];
void permWithRep(int array1, int array2,int fixed_index, int size , int level){
if(level == size) return print_array(array2,size);
for(int k = 0; k< size; k++){
if(!is_present_in_fixed(fixed_index,level,k)){
fixed_index[level] = k;
array2[level] = array1[k];
permWithRep(array1,array2,fixed_index,size,level+1);
}
}
}
void print_array(int array, int size){
for(int k = 0; k < size; k++)printf("%d ",array[k]);
printf("n");
}
int is_present_in_fixed(int array, int size, int element ){
for(int k =0; k<size;k++)
if(element == array[k]) return 1;
return 0;
}
int main()
{
int len = sizeof(array1)/sizeof(int);
permWithRep (array1, array2, fixed, len, 0);
return 0;
}
What I did:
- Add an array (
fixed_index) that keep track of the already visited indexes for one permutation. - In the loop, check if the index of current element belongs to the index already visited in that permutation (
is_present_in_fixed(fixed_index,level,k)). - If so, it just skip that element (it does not enter the
ifstatement). - Otherwise adds that element to the visited for that permutation and go on with recursive call.
- When all array has been iterated on (
level == size), just print the results contained inarray2.
answered Nov 12 '18 at 23:05
Ollaw
940517
add a comment |
0
0
0
I tested this on few inputs and it works.
There is certainly a better way to do it, but that's what i did:
#include <stdio.h>
int array1 = {1, 2, 3}; //array can be {1,1,2,3} for example also
int array2[3];
int fixed[3];
void permWithRep(int array1, int array2,int fixed_index, int size , int level){
if(level == size) return print_array(array2,size);
for(int k = 0; k< size; k++){
if(!is_present_in_fixed(fixed_index,level,k)){
fixed_index[level] = k;
array2[level] = array1[k];
permWithRep(array1,array2,fixed_index,size,level+1);
}
}
}
void print_array(int array, int size){
for(int k = 0; k < size; k++)printf("%d ",array[k]);
printf("n");
}
int is_present_in_fixed(int array, int size, int element ){
for(int k =0; k<size;k++)
if(element == array[k]) return 1;
return 0;
}
int main()
{
int len = sizeof(array1)/sizeof(int);
permWithRep (array1, array2, fixed, len, 0);
return 0;
}
What I did:
- Add an array (
fixed_index) that keep track of the already visited indexes for one permutation. - In the loop, check if the index of current element belongs to the index already visited in that permutation (
is_present_in_fixed(fixed_index,level,k)). - If so, it just skip that element (it does not enter the
ifstatement). - Otherwise adds that element to the visited for that permutation and go on with recursive call.
- When all array has been iterated on (
level == size), just print the results contained inarray2.
answered Nov 12 '18 at 23:05
Ollaw
940517
I tested this on few inputs and it works.
There is certainly a better way to do it, but that's what i did:
#include <stdio.h>
int array1 = {1, 2, 3}; //array can be {1,1,2,3} for example also
int array2[3];
int fixed[3];
void permWithRep(int array1, int array2,int fixed_index, int size , int level){
if(level == size) return print_array(array2,size);
for(int k = 0; k< size; k++){
if(!is_present_in_fixed(fixed_index,level,k)){
fixed_index[level] = k;
array2[level] = array1[k];
permWithRep(array1,array2,fixed_index,size,level+1);
}
}
}
void print_array(int array, int size){
for(int k = 0; k < size; k++)printf("%d ",array[k]);
printf("n");
}
int is_present_in_fixed(int array, int size, int element ){
for(int k =0; k<size;k++)
if(element == array[k]) return 1;
return 0;
}
int main()
{
int len = sizeof(array1)/sizeof(int);
permWithRep (array1, array2, fixed, len, 0);
return 0;
}
What I did:
- Add an array (
fixed_index) that keep track of the already visited indexes for one permutation. - In the loop, check if the index of current element belongs to the index already visited in that permutation (
is_present_in_fixed(fixed_index,level,k)). - If so, it just skip that element (it does not enter the
ifstatement). - Otherwise adds that element to the visited for that permutation and go on with recursive call.
- When all array has been iterated on (
level == size), just print the results contained inarray2.
answered Nov 12 '18 at 23:05
Ollaw
940517
answered Nov 12 '18 at 23:05
Ollaw
940517
answered Nov 12 '18 at 23:05
Ollaw
940517
answered Nov 12 '18 at 23:05
Ollaw
940517
940517
add a comment |
add a comment |
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Would you define what you mean by "repetition"? That may sound like a strange question, but I do not see any repeated code. Perhaps you are talking about the loops. Do you want to remove all loops? Is this a homework exercise?
– paddy
Nov 12 '18 at 21:26
For anyone unclear on the terminology of the question - en.wikipedia.org/wiki/Permutation#Permutations_with_repetition
– Wumpus Q. Wumbley
Nov 12 '18 at 21:49
What have you tried? What don't you understand about the problem?
– chb
Nov 12 '18 at 23:01