## How can i generate 4 bit binary combination using recursion in C for 0,1?

binary combination generator

generate all strings of length n

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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 .. }

**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);

**Generate all the binary strings of N bits,** Examples: Input: 2 Output: 0 0 0 1 1 0 1 1 Input: 3 Output: 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 using namespace std; int n = 4;. int arr[n];. // Print all binary strings. generateAllBinaryStrings(n, arr, 0);. return 0; Article Tags : Backtracking · C++ · Recursion · binary-string. Practice Tags : Most visited in C++. In order to add larger binary numbers, the carry bit must be incorporated as an input. This is accomplished by combining 2 Half Adder circuits to generate a Full Adder. Full Adders can then be cascaded together to add larger binary numbers. In my project I cascaded 4 Full Adders which enabled me to have 4 bit inputs.

the recursion could be done with `+1`

void f(unsigned int x) { printf("%u%u%u%u\n", (x>>3)&0x1, (x>>2)&0x1, (x>>1)&0x1, x&0x1); if(x==0xF) return; else f(x+1); } int main(void) { f(0); }

**Execution:**

$ ./test 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

**Print all possible combinations of r elements in a given array of size ,** Given an array of size n, generate and print all possible combinations of r For example, if input array is {1, 2, 3, 4} and r is 2, then output should be {1, 2}, {1 We first fix 1 at index 0 in data[], then recur for remaining indexes, then we fix Following diagram shows recursion tree for same input. Print all combination using. Leaf nodes from Preorder of a Binary Search Tree (Using Recursion) Find the node with maximum value in a Binary Search Tree using recursion; Maximum decimal equivalent possible among all connected components of a Binary Valued Graph; Program to Convert BCD number into Decimal number; Sum of digit of a number using recursion

just traverse DFS a binary tree of depth 4, going left is 0, going right is 1.

tr(int dep, int val) { if(dep == 4) { printf("\n"); } else { printf("%d", val); tr(dep+1, 0); // going left tr(dep+1, 1); // going right } return; } int main() { tr(0,0); }

**How does recursion stack work in generating an all n-bit binary string?,** 2-bit binary string and we are going to use the below function to generate the combinations [code]void Binary(int n){ if(n<1) printf("%s ", A); else{ A[n-1] = '0'; Binary(n-1); makestring(curr+"0",n);; makestring(curr+"1",n);; }; int main() {; int n=4;; string in C to convert the decimal number to a binary number using recursion? Generate all binary permutations such that there are more or equal 1's than 0's before every point in all permutations Bitwise AND of N binary strings Generate string with Hamming Distance as half of the hamming distance between strings A and B

I tried to limit my solution using to the same arguments but I would definitively add an extra argument to know the initial value of count.

void rec(int val, int count) { if (count <= 1) { int i; int f = 0; for (i = sizeof(int) * 8; i >= 0; i--) { f |= (val >> i) & 1; if (f) { printf("%d", (val >> i) & 1); } } printf("\n"); } else { rec(val * 2, count - 1); rec(val * 2 + 1, count - 1); } }

Output:

1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111

In order to add the leading 0, I added an argument :

#include <stdio.h> void rec2(int val, int count, int b) { if (count <= 1) { int i; for (i = b - 1; i >= 0; i--) { printf("%d", (val >> i) & 1); } printf("\n"); } else { rec2(val * 2, count - 1, b); rec2(val * 2 + 1, count - 1, b); } } void rec(int val, int count) { rec2(val, count, count); } int main() { rec(0, 4); rec(1, 4); return 0; }

Output:

0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

**Algorithms for Permutations and Combinations,** Aside from the array itself, which consumes (n) storage, we have recursion consuming stack with bit numbers running from 7 down to 0, represents the set: { 7 6 4 2 0 } If we want to generated all n C k combinations of n integers from 0..n-1 taken k at a time, we can just generate all binary numbers with exactly k 1-bits. Well let us take a simple example and consider a 2-bit binary string and we are going to use the below function to generate the combinations [code]void Binary(int n

Let us start by designing the prototype of the recursive function. Hopefully, it'll make sense from there. Take a look at a non-recursive version of this code, and you'll need the same variables. You don't need to pass *any* of them as arguments, but I'd prefer to pass them all, and make the solution as flexible and modular as possible. Consider the return value, too. That should probably indicate some sort of *success*, in order to mimic consistency with the C standard library.

int count_r(char *destination, /* The storage for the function to store * * the 0s and 1s as we count. */ size_t length, /* The number of digits in the number. */ char *digit); /* The set of digits */

Now let us focus on designing the first iteration. Like in primary school, we start by defining our `count_r`

to iterate only single digits at a time. Once we can prove that it knows how to count from `0`

to `9`

, we introduce it to double digits... but for now, one step at a time.

Let us assume `destination`

is initialised to contain `length`

bytes of `digits[0]`

, prior to the first call. This initialisation is done by the caller, and the caller would presumably output that pre-initialised array before calling. The first iteration should modify only one byte: The one indicated by `length`

, and then return to the caller.

int count_r(char *destination, size_t length, char *digit) { /* The position of the right-most digit is before the '\0' in destination, * * so we need to decrement length */ length--; /* Find the digit at the very end of destination, within our "digit" parameter */ char *d = strchr(digit, destination[length]); /* d[1] points to the next digit (or '\0') */ destination[length] = d[1]; return 0; }

The caller then presumably prints the array, and calls `count_r`

again with the same buffer to increment the counter. This works with different bases, and by reversing the `digit`

string we can decrement instead of incrementing. However, as we'll soon see, it fails after it reaches the highest number it can count to: `'F'`

in the example below.

int main(void) { char num[] = "0"; do { puts(num); } while (count_r(num, strlen(num), "0123456789ABCDEF") == 0); }

When the time comes for counting higher, d[1] will be `'\0'`

as it will have iterated beyond the set of digits and into the null terminator for the string. Let us consider adding code to handle our second iteration.

A bit of code is needed to set `destination[length]`

back to the first `digit`

and recursively move left onto the next digit. This occurs when `d[1] == '\0'`

, so we can write an `if (...) { ... }`

branch to handle that.

There is a problem when `length`

is passed in as 0, which we would discover after implementing the change mentioned just now. Here is where the function should return `1`

to indicating that counting has finished, because it has moved as far left as possible and reached the highest number possible.

void count_r(char *destination, size_t length, char *digit) { /* The position of the right-most digit is before the '\0' in destination, * * so we need to decrement length */ if (length-- == 0) { return 1; } /* Find the digit at the very end of destination, within our "digit" parameter */ char *d = strchr(digit, destination[length]); /* d[1] points to the next digit (or '\0') */ if (d[1] == '\0') { /* Set destination[length] to the first digit */ destination[length] = digit[0]; /* Recurse onto the next digit. We've already decremented length */ return count_r(destination, length, digit); } destination[length] = d[1]; return 0; }

After adding a few `assert`

ions (eg. `assert(strlen(digit) > 1);`

) and writing some testcases, we might then decide that this function is ready for production. I hope I was able to help. :)

**How to Calculate the Probability of Combinations,** How do you find all the combinations of a number? n-bit Gray Codes can be generated from list of (n-1)-bit Gray codes using following steps. 1) Let the list of (n-1)-bit Gray codes be L1. Create another list L2 which is reverse of L1. 2) Modify the list L1 by prefixing a ‘0’ in all codes of L1. 3) Modify the list L2 by prefixing a ‘1’ in all codes of L2. 4) Concatenate L1 and L2. The

**Algorithm to generate all combinations of a string,** instr, StringBuffer outstr, int index) { for (int i = index; i < instr. length(); i++) { outstr. A Computer Science portal for geeks. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions.

**Cracking Your PIN Code: Easy as 1-2-3-4,** 4 . 1 INTEGER COMPUTATIONS Enumeration of combinatorial objects , such as permutations and combinations , requires the computation of integer factorials . bi € { 0 , 1 } and n = Lig ' b ; 2 . ( See $ 4 . 1 . 3 . ) Each bi is a binary digit ( bit ) . the computation of a binomial coefficient into a recursive loop using C ) = k ! Changing a bit’s value to 0 is referred to as resetting a bit. How to display binary values. To best make sense of the C language’s binary manipulation operators, it helps to see a binary number in action. The printf() function lacks a binary conversion character, and the C library doesn’t host a binary output function.

**Handbook of Discrete and Combinatorial Mathematics,** 0 1 2 0 1 3 0 1 4 0 2 3 0 2 4 0 3 4 1 2 3 1 2 4 1 3 4 2 3 4 With combinations and permutations generation tasks. Nice algorithm without recursion borrowed from C. Recursion is elegant but iteration is C DS 64X array of 8 bit integers "Generate the combinations of n elements from a list of [0..m)" [m n] Generate all permutations of given length such that every permutation has more or equal 1’s than 0’s in all prefixes of the permutation. Examples: Input: len = 4 Output: 1111 1110 1101 1100 1011 1010 Note that a permutation like 0101 can not be in output because there are more 0's from index 0 to 2 in this permutation.

##### Comments

- What have you achieved so far?
- so with the code above, you call
`rec(0, 4)`

and`rec(1, 4)`

, what exactly do you expect? - @Howard, I have added more code. Please notice.
- @ExP, for rec(0,4) expecting to print first 8 row starting with zero and with rec(1,4) expecting to print last 8 row starting with one.
- Thank You. But can you rewrite that code without making string or without using characters. Can you use integer for recursion instead of characters?
- You could use an integer (or more logical, boolean) array as a buffer, but you would have to print it manually at the end which is less convenient than a char array. In any case you can't do this without some sort of accumulation buffer, if you want to use only
`int`

as recursion parameters (as in MOHAMED's answer), you would have to convert it to binary before printing it and a simple loop would do the same job. - Can you rewrite that code to make a binary tree(depth=4), left side 0 and right side 1 ?
- Well that's basically what this algorithm is doing, except it doesn't actually store the tree in memory. Building a tree won't be more helpful because while traversing the tree in depth you won't be able to print before reaching the leafs so you'll still need a buffer to store the path you've taken in the tree. See my last edit for an example using a binary tree, you'll see my initial solution is a simplified version of the last one.
- Thank you for your reply. But i am looking for a answer like G_G posted. Tree looks complex here for this. Using recursion only can do this.
- //print here the first 4 bits of x = right side 4 bits = less significant bits ? That means, i have to convert from decimal to binary ?
- But that's more a stack-loop than a real recursion.
- @guru I updated my answer concerning the print of 4 first bits of x
- Actually i don't want to convert from decimal to binary or to use shift operator.