Best algorithm to find all possible permutation of given binary bits - binary

Question

I am looking for an optimal algorithm to find out remaining all possible permutation
of a give binary number.
For ex:
Binary number is : ........1. algorithm should return the remaining 2^7 remaining binary numbers, like 00000001,00000011, etc.
Thanks,
sathish

Answer 1

The example given is not a permutation!
A permutation is a reordering of the input.
So if the input is 00000001, 00100000 and 00000010 are permutations, but 00000011 is not.

Answer 2

If this is only for small numbers (probably up to 16 bits), then just iterate over all of them and ignore the mismatches:
int fixed = 0x01; // this is the fixed part
int mask = 0x01; // these are the bits of the fixed part which matter
for (int i=0; i<256; i++) {
if (i & mask == fixed) {
print i;
}
}

Answer 3

to find all you aren't going to do better than looping over all numbers e.g. if you want to loop over all 8 bit numbers
for (int i =0; i < (1<<8) ; ++i)
{
//do stuff with i
}
if you need to output in binary then look at the string formatting options you have in what ever language you are using.
e.g.
printf("%b",i); //not standard in C/C++
for calculation the base should be irrelavent in most languages.

Answer 4

I read your question as: "given some binary number with some bits always set, create the remaining possible binary numbers".
For example, given 1xx1: you want: 1001, 1011, 1101, 1111.
An O(N) algorithm is as follows.
Suppose the bits are defined in mask m. You also have a hash h.
To generate the numbers < n-1, in pseudocode:
counter = 0
for x in 0..n-1:
x' = x | ~m
if h[x'] is not set:
h[x'] = counter
counter += 1
The idea in the code is to walk through all numbers from 0 to n-1, and set the pre-defined bits to 1. Then memoize the resulting number (iff not already memoized) by mapping the resulting number to the value of a running counter.
The keys of h will be the permutations. As a bonus the h[p] will contain a unique index number for the permutation p, although you did not need it in your original question, it can be useful.

Answer 5

Why are you making it complicated !
It is as simple as the following:
// permutation of i on a length K
// Example : decimal i=10 is permuted over length k= 7
// [10]0001010-> [5] 0000101-> [66] 1000010 and 33, 80, 40, 20 etc.
main(){
int i=10,j,k=7; j=i;
do printf("%d \n", i= ( (i&1)<< k + i >>1); while (i!=j);
}

Answer 6

There are many permutation generating algorithms you can use, such as this one:
#include <stdio.h>
void print(const int *v, const int size)
{
if (v != 0) {
for (int i = 0; i < size; i++) {
printf("%4d", v[i] );
}
printf("\n");
}
} // print
void visit(int *Value, int N, int k)
{
static level = -1;
level = level+1; Value[k] = level;
if (level == N)
print(Value, N);
else
for (int i = 0; i < N; i++)
if (Value[i] == 0)
visit(Value, N, i);
level = level-1; Value[k] = 0;
}
main()
{
const int N = 4;
int Value[N];
for (int i = 0; i < N; i++) {
Value[i] = 0;
}
visit(Value, N, 0);
}
source: http://www.bearcave.com/random_hacks/permute.html
Make sure you adapt the relevant constants to your needs (binary number, 7 bits, etc...)

Answer 7

If you are really looking for permutations then the following code should do.
To find all possible permutations of a given binary string(pattern) for example.
The permutations of 1000 are 1000, 0100, 0010, 0001:
void permutation(int no_ones, int no_zeroes, string accum){
if(no_ones == 0){
for(int i=0;i<no_zeroes;i++){
accum += "0";
}
cout << accum << endl;
return;
}
else if(no_zeroes == 0){
for(int j=0;j<no_ones;j++){
accum += "1";
}
cout << accum << endl;
return;
}
permutation (no_ones - 1, no_zeroes, accum + "1");
permutation (no_ones , no_zeroes - 1, accum + "0");
}
int main(){
string append = "";
//finding permutation of 11000
permutation(2, 6, append); //the permutations are
//11000
//10100
//10010
//10001
//01100
//01010
cin.get();
}

Answer 8

If you intend to generate all the string combinations for n bits , then the problem can be solved using backtracking.
Here you go :
//Generating all string of n bits assuming A[0..n-1] is array of size n
public class Backtracking {
int[] A;
void Binary(int n){
if(n<1){
for(int i : A)
System.out.print(i);
System.out.println();
}else{
A[n-1] = 0;
Binary(n-1);
A[n-1] = 1;
Binary(n-1);
}
}
public static void main(String[] args) {
// n is number of bits
int n = 8;
Backtracking backtracking = new Backtracking();
backtracking.A = new int[n];
backtracking.Binary(n);
}
}