Tom White: Consistent Hashing
The basic idea behind the consistent hashing algorithm is to hash both objects and caches using the same hash function. The reason to do this is to map the cache to an interval, which will contain a number of object hashes. If the cache is removed then its interval is taken over by a cache with an adjacent interval. All the other caches remain unchanged.
public class ConsistentHash<T> {
private final HashFunction hashFunction;
private final int numberOfReplicas;
private final SortedMap<Integer, T> circle =
new TreeMap<Integer, T>();
public ConsistentHash(HashFunction hashFunction,
int numberOfReplicas, Collection<T> nodes) {
this.hashFunction = hashFunction;
this.numberOfReplicas = numberOfReplicas;
for (T node : nodes) {
add(node);
}
}
public void add(T node) {
for (int i = 0; i < numberOfReplicas; i++) {
circle.put(hashFunction.hash(node.toString() + i),
node);
}
}
public void remove(T node) {
for (int i = 0; i < numberOfReplicas; i++) {
circle.remove(hashFunction.hash(node.toString() + i));
}
}
public T get(Object key) {
if (circle.isEmpty()) {
return null;
}
int hash = hashFunction.hash(key);
if (!circle.containsKey(hash)) {
SortedMap<Integer, T> tailMap =
circle.tailMap(hash);
hash = tailMap.isEmpty() ?
circle.firstKey() : tailMap.firstKey();
}
return circle.get(hash);
}
}
https://github.com/checkcheckzz/system-design-interview
Read full article from Tom White: Consistent Hashing
The basic idea behind the consistent hashing algorithm is to hash both objects and caches using the same hash function. The reason to do this is to map the cache to an interval, which will contain a number of object hashes. If the cache is removed then its interval is taken over by a cache with an adjacent interval. All the other caches remain unchanged.
Here's a picture of the circle with a number of objects (1, 2, 3, 4) and caches (A, B, C) marked at the points that they hash to:
public class ConsistentHash<T> {
private final HashFunction hashFunction;
private final int numberOfReplicas;
private final SortedMap<Integer, T> circle =
new TreeMap<Integer, T>();
public ConsistentHash(HashFunction hashFunction,
int numberOfReplicas, Collection<T> nodes) {
this.hashFunction = hashFunction;
this.numberOfReplicas = numberOfReplicas;
for (T node : nodes) {
add(node);
}
}
public void add(T node) {
for (int i = 0; i < numberOfReplicas; i++) {
circle.put(hashFunction.hash(node.toString() + i),
node);
}
}
public void remove(T node) {
for (int i = 0; i < numberOfReplicas; i++) {
circle.remove(hashFunction.hash(node.toString() + i));
}
}
public T get(Object key) {
if (circle.isEmpty()) {
return null;
}
int hash = hashFunction.hash(key);
if (!circle.containsKey(hash)) {
SortedMap<Integer, T> tailMap =
circle.tailMap(hash);
hash = tailMap.isEmpty() ?
circle.firstKey() : tailMap.firstKey();
}
return circle.get(hash);
}
}
The circle is represented as a sorted map of integers, which represent the hash values, to caches (of type
When a
To find a node for an object (the
T
here).When a
ConsistentHash
object is created each node is added to the circle map a number of times (controlled by numberOfReplicas
). The location of each replica is chosen by hashing the node's name along with a numerical suffix, and the node is stored at each of these points in the map.To find a node for an object (the
get
method), the hash value of the object is used to look in the map. Most of the time there will not be a node stored at this hash value (since the hash value space is typically much larger than the number of nodes, even with replicas), so the next node is found by looking for the first key in the tail map. If the tail map is empty then we wrap around the circle by getting the first key in the circle.https://github.com/checkcheckzz/system-design-interview
Read full article from Tom White: Consistent Hashing