Generate all possible n-character passwords

Question

As part of a learning-Go exercise, I'm writing a simplistic brute-force password cracker.

To generate all possible 2-character passwords that use the characters A-E in Python, I would use itertools.product():

from itertools import product
for permutation in product('ABCDE', repeat=2):
  print permutation

However, I'm struggling to do this in Go.

Other questions seem to be about permutations, which isn't quite what I want. And while the Python docs include a sample implementation of the function, I don't know how to translate yield into Go.

I suppose I should mention two restrictions:

1. I'd like the length of the password to be variable. That is, I may want to do 8-character passwords, or 6-character, or something else. This means we can't just nest n loops.
2. I don't want to have all of them in memory at once.

Answer 1

What you want is basically the n-ary cartesian product of a set with itself. So for all 3-character passwords you want Prod(set,set,set). This can be constructed iteratively. First construct the n-1 product, then for each product and each element of the initial set, add the element. So for instance all 2 character passwords -> 3 character passwords where the only valid characters are 'a' or 'b'.

"ab" = {a,b} -> {(a,a),(a,b),(b,a),(b,b)} -> {(a,a,a),(a,a,b),(a,b,a),(a,b,b),(b,a,a),(b,a,b),(b,b,a),(b,b,b)}

func NAryProduct(input string, n int) []string {
    if n <= 0 {
        return nil
    }

    // Copy input into initial product set -- a set of
    // one character sets
    prod := make([]string, len(input))
    for i, char := range input {
        prod[i] = string(char)
    }

    for i := 1; i < n; i++ {
        // The bigger product should be the size of the input times the size of
        // the n-1 size product
        next := make([]string, 0, len(input)*len(prod))

        // Add each char to each word and add it to the new set
        for _, word := range prod {
            for _, char := range input {
                next = append(next, word + string(char))
            }
        }

        prod = next
    }

    return prod
}

Playground version: http://play.golang.org/p/6LhApeJ1bv

It should be noted that there's a lot of room for improvement on this solution. If you want to construct all passwords of length, say, 6-18, calling this method independently for each one will recalculate previously computed sets. I'll leave writing the better version up to you. Given what I've shown you, it shouldn't be too difficult to modify the code to take an arbitrary (n-m)ary product and compute the n-ary product from it. (Hint: think about how you'd do this recursively)

Answer 2

For example, satisfying your restrictions,

package main

import "fmt"

func nextPassword(n int, c string) func() string {
    r := []rune(c)
    p := make([]rune, n)
    x := make([]int, len(p))
    return func() string {
        p := p[:len(x)]
        for i, xi := range x {
            p[i] = r[xi]
        }
        for i := len(x) - 1; i >= 0; i-- {
            x[i]++
            if x[i] < len(r) {
                break
            }
            x[i] = 0
            if i <= 0 {
                x = x[0:0]
                break
            }
        }
        return string(p)
    }
}

func main() {
    np := nextPassword(2, "ABCDE")
    for {
        pwd := np()
        if len(pwd) == 0 {
            break
        }
        fmt.Println(pwd)
    }
}

Output:

AA
AB
AC
AD
AE
BA
BB
BC
BD
BE
CA
CB
CC
CD
CE
DA
DB
DC
DD
DE
EA
EB
EC
ED
EE