dtvgo28624
dtvgo28624
2016-12-07 14:15

映射迭代是否足够随机以随机选择密钥?

Can I rely on the random iteration order of maps to implement a random "pairing" of clients in a web application? I've tried looking around but can't seem to find a breakdown of how random this randomness is.

The algorithm would look something like:

var clients map[Client]struct{}

func PairClient(c Client) (Client, error) {
    for m := range clients {
        if m != c {
            return m, nil
        }
    }
    return nil, fmt.Errorf("lobby: insufficient number of clients")
}

Would this be sufficient when there are >1000 connected clients, or should I maintain a separate slice of clients and randomly select from that?

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2条回答

  • duanlu0075 duanlu0075 5年前

    Even though it said to be random (randomized) (spec, blog, hashmap source, another blog, SO), the distribution is far from being perfect.

    Why? Because we like maps being fast, and better random distribution tends to require more computation and/or bigger delays. A compromise had to be made. And because the intention is not to provide a quality "shuffle" functionality by for range, but only to prevent developers relying on stable iteration order (because it could change even without explicit randomization).

    But "how good" may this distribution be? Easy to get a "taste". Let's create a map of 10 pairs, and start iterating over it lot of times. And let's count the distribution of the very first index (key):

    m := map[int]int{}
    for i := 0; i < 10; i++ {
        m[i] = i
    }
    
    dist := make([]int, 10)
    for i := 0; i < 100000; i++ {
        for idx := range m {
            dist[idx]++
            break
        }
    }
    
    fmt.Println("Distribution:", dist)
    

    Output (try it on the Go Playground):

    Distribution: [25194 24904 6196 6134 6313 6274 6297 6189 6189 6310]
    

    The first 2 keys (0 and 1) were roughly encountered 4 times more than the rest which had roughly the same probability.

    You can tell it's pretty bad for being true (or even good) random, but that's not the point. It's good enough to provide varying iteration order (and importantly: it's fast).

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  • dongre1907 dongre1907 5年前

    From the spec:

    1. The iteration order over maps is not specified and is not guaranteed to be the same from one iteration to the next. (...)

    Which essentially means that the order in which maps are iterated is undefined. Even if it's random right now with the default Go compiler, other compilers or other versions might differ.

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