Tuesday, March 29, 2016

[LintCode][System Design] Consistent Hashing - 简书


[LintCode][System Design] Consistent Hashing - 简书
一般的数据库进行horizontal shard的方法是指,把 id 对 数据库服务器总数 n 取模,然后来得到他在哪台机器上。这种方法的缺点是,当数据继续增加,我们需要增加数据库服务器,将 n 变为 n+1 时,几乎所有的数据都要移动,这就造成了不 consistent。为了减少这种 naive 的 hash方法(%n) 带来的缺陷,出现了一种新的hash算法:一致性哈希的算法――Consistent Hashing。这种算法有很多种实现方式,这里我们来实现一种简单的 Consistent Hashing。
  1. 将 id 对 360 取模,假如一开始有3台机器,那么让3台机器分别负责0~119, 120~239, 240~359 的三个部分。那么模出来是多少,查一下在哪个区间,就去哪台机器。
  2. 当机器从 n 台变为 n+1 台了以后,我们从n个区间中,找到最大的一个区间,然后一分为二,把一半给第n+1台机器。
  3. 比如从3台变4台的时候,我们找到了第3个区间0~119是当前最大的一个区间,那么我们把0~119分为0~59和60~119两个部分。0~59仍然给第1台机器,60~119给第4台机器。
  4. 然后接着从4台变5台,我们找到最大的区间是第3个区间120~239,一分为二之后,变为 120~179, 180~239。
假设一开始所有的数据都在一台机器上,请问加到第 n 台机器的时候,区间的分布情况和对应的机器编号分别是多少?

Clarification
If the maximal interval is [x, y], and it belongs to machine id z, when you add a new machine with id n, you should divide [x, y, z] into two intervals:[x, (x + y) / 2, z] and [(x + y) / 2 + 1, y, n]
for n = 1, return
[
  [0,359,1]
]
represent 0~359 belongs to machine 1.
for n = 2, return
[
  [0,179,1],
  [180,359,2]
]
for n = 3, return
[
  [0,89,1]
  [90,179,3],
  [180,359,2]
]
for n = 4, return
[
  [0,89,1],
  [90,179,3],
  [180,269,2],
  [270,359,4]
]
for n = 5, return
[
  [0,44,1],
  [45,89,1],
  [90,179,3],
  [180,269,2],
  [270,359,4]
]
模拟的方法,每次选取最大的range,然后插入新的range。

    vector<vector<int>> consistentHashing(int n) {
        // Write your code here
        vector<vector<int>> ret;
        vector<int> range;
        range.push_back(0);
        range.push_back(359);
        range.push_back(1);
        ret.push_back(range);

        for(int i = 2; i <= n; i++) {
            int maxRange = INT_MIN;
            int index = -1;
            for(int j = 0; j < ret.size(); j++) {
                if (maxRange < ret[j][1] - ret[j][0]) {
                    maxRange = ret[j][1] - ret[j][0];
                    index = j;
                }
            }

            int mid = (ret[index][1] + ret[index][0]) / 2;
            int end = ret[index][1];
            ret[index][1] = mid;
            range[0] = mid + 1;
            range[1] = end;
            range[2] = i;
            ret.push_back(range);
        }

        return ret;
    }
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