使用加密/兰特生成带有rand.Perm的排列

Question

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?

Answer 1

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 struct, rather than allocating a new slice on every call.

type cryptoSource struct {
    buf [8]byte
}

func (s *cryptoSource) Int63() int64 {
    cryptoRand.Read(s.buf[:])
    return int64(binary.BigEndian.Uint64(s.buf[:]) & (1<<63 - 1))
}

Or just make the source a [8]byte itself

type cryptoSource [8]byte

func (s *cryptoSource) Int63() int64 {
    cryptoRand.Read(s[:])
    return int64(binary.BigEndian.Uint64(s[:]) & (1<<63 - 1))
}