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*This is a explanation of this problem from USACO's training website. I have converted it to markdown. Please do not just copy code; you will not learn anything; at least type it out and understand so you can do it yourself in the future!*

Given N, B, and D: Find a set of N codewords (1 <= N <= 64), each of length B bits (1 <= B <= 8), such that each of the codewords is at least Hamming distance of D (1 <= D <= 7) away from each of the other codewords.

The Hamming distance between a pair of codewords is the number of binary bits that differ in their binary notation. Consider the two codewords 0x554 and 0x234 and their differences (0x554 means the hexadecimal number with hex digits 5, 5, and 4)(a hex digit requires four bits):

```
0x554 = 0101 0101 0100
0x234 = 0010 0011 0100
Bit differences: -XXX -XX- ----
```

Since five bits were different, the Hamming distance is 5.

### PROGRAM NAME: hamming

### INPUT FORMAT

N, B, D on a single line

### SAMPLE INPUT (file hamming.in)

```
16 7 3
```

### OUTPUT FORMAT

N codewords, sorted, in decimal, ten per line. In the case of multiple solutions, your program should output the solution which, if interpreted as a base 2^B integer, would have the least value.

### SAMPLE OUTPUT (file hamming.out)

```
0 7 25 30 42 45 51 52 75 76
82 85 97 102 120 127
```

## CODE

## Java

## C++

## Pascal

## ANALYSIS

##
Brian Jacokes

There are only a few tools we need to solve this problem. First of all, we can use basic techniques to find the Nth bit of a number M: counting the least significant bit as bit 0, the next bit as bit 1, etc., we can find the value of that bit through the expression

```
int Nth_bit = (1 << N) & M;
```

In other words, we are shifting the number 1 left by N spaces, and then performing a binary AND on the resulting number with our original number, which will leave either a 1 in the Nth bit or a 0. So the first thing we have to do is find out the distance between each pair of numbers within the set of all numbers with B bits (0..2B-1). We also know that 0 can be one of the numbers in the set, because if the minimum number in the set were N instead of 0, we could XOR N to each number in the set and they would still be the same distance apart. The limits on the problem are low enough that we can do a DFS, and as soon as we hit a solution we can output it and exit.

```
#include <stdio.h>
#include <stdlib.h>
#include <iostream.h>
#define MAX (1 << 8 + 1)
#define NMAX 65
#define BMAX 10
#define DMAX 10
int nums[NMAX], dist[MAX][MAX];
int N, B, D, maxval;
FILE *in, *out;
void findgroups(int cur, int start) {
int a, b, canuse;
char ch;
if (cur == N) {
for (a = 0; a < cur; a++) {
if (a % 10)
fprintf(out, " ");
fprintf(out, "%d", nums[a]);
if (a % 10 == 9 || a == cur-1)
fprintf(out, "\n");
}
exit(0);
}
for (a = start; a < maxval; a++) {
canuse = 1;
for (b = 0; b < cur; b++)
if (dist[nums[b]][a] < D) {
canuse = 0;
break;
}
if (canuse) {
nums[cur] = a;
findgroups(cur+1, a+1);
}
}
}
int main() {
in = fopen("hamming.in", "r");
out = fopen("hamming.out", "w");
int a, b, c;
fscanf(in, "%d%d%d", &N, &B, &D);
maxval = (1 << B);
for (a = 0; a < maxval; a++)
for (b = 0; b < maxval; b++) {
dist[a][b] = 0;
for (c = 0; c < B; c++)
if (((1 << c) & a) != ((1 << c) & b))
dist[a][b]++;
}
nums[0] = 0;
findgroups(1, 1);
return 0;
}
```

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