For this array, trying something like this:
void rollover(int val,int count) {
if(count==0) {
return;
}
printf("%d ",val);
count--;
rollover(val,count);
}
int main() {
int arr[]={0,1};
for(int i=0;i<=1;i++) {
rollover(arr[i],4);
}
printf("\n");
return 0;
}
Expected output using recursion method:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Can't understand how to write that rec function. I have spent several hours to solve it. Can someone assist to write that function?
I am/was trying to do something like G_G posted below. How can i write such recursion function? Do i have to use one for loop to call that recursion function or two for loop with recursion or should i call the recursion function twice? For example:
void rollover(int val,int count) {
if(count==0) {
return;
}
printf("%d ",val);
count--;
rollover(val,count);
//.. do something if necessary ..
rollover(val,count);
//.. do something if necessary ..
}
There is obviously a problem with this since 4 bits gives 16 possible combinations, and there are only 10 unique decimal digits, 0 to 9.
The function convert() is used to convert a number decimal system to binary system using recursion.
Simplest solution : binary conversion, no recursion
for(int i = 0; i < 16: ++i) {
printf("%u%u%u%u", i/8%2, i/4%2, i/2%2, i%2);
}
See MOHAMED's answer for a recursive version of this loop
Binary recursion used by the following solutions
_ 000
_ 00 _/
/ \_ 001
0 _ 010
\_ 01 _/
\_ 011
_ 100
_ 10 _/
/ \_ 101
1 _ 110
\_ 11 _/
\_ 111
Recursive solution using char*
buffer, no binary conversion
void char_buffer_rec(char number[4], int n) {
if(n > 0) {
number[4-n] = '0';
char_buffer_rec(number, n - 1);
number[4-n] = '1';
char_buffer_rec(number, n - 1);
}
else {
printf("%s\n", number);
}
}
usage :
char number[5] = {0};
char_buffer_rec(number, 4);
Recursive solution using only int
, no buffer, no binary conversion
void int_ten_rec(int number, int tenpower) {
if(tenpower > 0) {
int_ten_rec(number, tenpower/10);
int_ten_rec(number + tenpower, tenpower/10);
}
else {
printf("%04u\n", number);
}
}
usage :
int_ten_rec(0, 1000);
Another version of this solution replacing tenpower
width bitwidth
, replacing the printf width
with a variable padding depending on the length variable. length
could be defined as a new parameter, a program constant, etc.
void int_rec(int number, int bitwidth) {
static int length = bitwidth;
int i, n;
if(bitwidth > 0) {
int_rec(number, bitwidth-1);
/* n := 10^(bitwidth-2) */
for(i=0,n=1;i<bitwidth-1;++i,n*=10);
int_rec(number + n, bitwidth-1);
}
else {
/* i := number of digit in 'number' */
for(i=1,n=number;n>=10;++i,n/=10);
/* print (length-i) zeros */
for(n=i; n<length; ++n) printf("0");
printf("%u\n", number);
}
}
usage :
int_rec(0, 4);
Tree Solution, recursive using char*
buffer, no binary conversion
struct Node {
int val;
struct Node *left, *right;
};
void build_tree(struct Node* tree, int n) {
if(n > 0) {
tree->left = (Node*)malloc(sizeof(Node));
tree->right= (Node*)malloc(sizeof(Node));
tree->left->val = 0;
build_tree(tree->left, n - 1);
tree->right->val = 1;
build_tree(tree->right, n - 1);
}
else {
tree->left = tree->right = NULL;
}
}
void print_tree(struct Node* tree, char* buffer, int index) {
if(tree->left != NULL && tree->right != NULL) {
sprintf(buffer+index, "%u", tree->val);
print_tree(tree->left, buffer, index+1);
sprintf(buffer+index, "%u", tree->val);
print_tree(tree->right, buffer, index+1);
}
else {
printf("%s%u\n", buffer, tree->val);
}
}
usage :
char buffer[5] = {0};
Node* tree = (Node*)malloc(sizeof(Node));
tree->val = 0;
build_tree(tree, 4);
print_tree(tree, buffer, 0);
But this would print an additional 0
at the begining of each line, to avoid this, build two smaller trees :
Node* tree0 = (Node*)malloc(sizeof(Node));
Node* tree1 = (Node*)malloc(sizeof(Node));
tree0->val = 0;
tree1->val = 1;
build_tree(tree0, 3);
build_tree(tree1, 3);
print_tree(tree0, buffer, 0);
print_tree(tree1, buffer, 0);
Recursive solution using int* array
#define MAX_LENGTH 32
int number[MAX_LENGTH];
void int_buffer_rec(int n, int length) {
if(n > 0) {
number[length - n] = 0;
int_buffer_rec(n - 1, length);
number[length - n] = 1;
int_buffer_rec(n - 1, length);
}
else {
for(int i = 0; i < length; ++i) {
printf("%u", number[i]);
}
printf("\n");
}
}
usage :
int_buffer_rec(4, 4);
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