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Using crypto/rand for generating permutations with rand.Perm

Tags:

random

go

Go has two packages for random numbers:

  • crypto/rand, which provides a way to get random bytes
  • math/rand, which has a nice algorithm for shuffling ints

I want to use the Perm algorithm from math/rand, but provide it with high-quality random numbers.

Since the two rand packages are part of the same standard library there should be a way to combine them in a way so that crypto/rand provides a good source of random numbers that is used by math/rand.Perm to generate a permutation.

Here (and on the Playground) is the code I wrote to connect these two packages:

package main

import (
    cryptoRand "crypto/rand"
    "encoding/binary"
    "fmt"
    mathRand "math/rand"
)

type cryptoSource struct{}

func (s cryptoSource) Int63() int64 {
    bytes := make([]byte, 8, 8)
    cryptoRand.Read(bytes)
    return int64(binary.BigEndian.Uint64(bytes) >> 1)
}

func (s cryptoSource) Seed(seed int64) {
    panic("seed")
}

func main() {
    rnd := mathRand.New(&cryptoSource{})
    perm := rnd.Perm(52)
    fmt.Println(perm)
}

This code works. Ideally I don't want to define the cryptoSource type myself but just stick together the two rand packages so that they work together. So is there a predefined version of this cryptoSource type somewhere?

like image 623
Roland Illig Avatar asked Oct 17 '22 21:10

Roland Illig


1 Answers

That's basically what you need to do. It's not often that you need a cryptographically secure source of randomness for the common usage of math/rand, so there's no adaptor provided. You can make the implementation slightly more efficient by allocating the buffer space directly in the value, rather than allocating a new slice on every call. However in the unlikely event that reading the OS random source fails, this will need to panic to prevent returning invalid results.

type cryptoSource [8]byte

func (s *cryptoSource) Int63() int64 {
    _, err := cryptoRand.Read(s[:])
    if err != nil {
        panic(err)
    }
    return int64(binary.BigEndian.Uint64(s[:]) & (1<<63 - 1))
}
like image 154
JimB Avatar answered Oct 20 '22 20:10

JimB