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Generating All Possible N-Character Passwords in Go
In Python, one can employ the itertools.product() function to generate all possible passwords of a fixed length. For example, to create 2-character passwords using characters A-E, one could use:
from itertools import product for permutation in product('ABCDE', repeat=2): print(permutation)
However, in Go, the task of generating passwords can be achieved through a different approach:
The desired operation is essentially an n-ary Cartesian product of a set with itself. For example, to create all 3-character passwords, we need Prod(set,set,set). This can be constructed iteratively.
Initially, we create our n-1 product, then for each product and each element of the original set, we append that element. Let's demonstrate with a 2-characters to 3-characters password generation:
"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)}
In Go, we can implement this as:
func NAryProduct(input string, n int) []string { if n <= 0 { return nil } prod := make([]string, len(input)) for i, char := range input { prod[i] = string(char) } for i := 1; i < n; i++ { next := make([]string, 0, len(input)*len(prod)) for _, word := range prod { for _, char := range input { next = append(next, word+string(char)) } } prod = next } return prod }
This solution can be further optimized by leveraging lazy evaluation, reducing memory usage. Here's a playground link for reference: http://play.golang.org/p/6LhApeJ1bv
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