Cuckoo Hashing – Worst case O(1) Lookup!

Background :
There are three basic operations that must be supported by a hash table (or a dictionary):

• Lookup(key): return true if key is there in the table, else false
• Insert(key): adds the item ‘key’ to the table if not already present
• Delete(key): removes ‘key’ from the table

Collisions are very likely even if we have a big table to store keys. Using the results from the birthday paradox: with only 23 persons, the probability that two people share the same birth date is 50%! There are 3 general strategies towards resolving hash collisions:

• Closed addressing or Chaining: store colliding elements in an auxiliary data structure like a linked list or a binary search tree.
• Open addressing: allow elements to overflow out of their target bucket and into other spaces.

Although above solutions provide expected lookup cost as O(1), the expected worst-case cost of a lookup in Open Addressing (with linear probing) is Ω(log n) and Θ(log n / log log n) in simple chaining (Source : Standford Lecture Notes). To close the gap of expected time and worst case expected time, two ideas are used:

• Multiple-choice hashing: Give each element multiple choices for positions where it can reside in the hash table
• Relocation hashing: Allow elements in the hash table to move after being placed

Cuckoo Hashing :
Cuckoo hashing applies the idea of multiple-choice and relocation together and guarantees O(1) worst case lookup time!

• Multiple-choice: We give a key two choices h1(key) and h2(key) for residing.
• Relocation: It may happen that h1(key) and h2(key) are preoccupied. This is resolved by imitating the Cuckoo bird: it pushes the other eggs or young out of the nest when it hatches. Analogously, inserting a new key into a cuckoo hashing table may push an older key to a different location. This leaves us with the problem of re-placing the older key.
• If alternate position of older key is vacant, there is no problem.
• Otherwise, older key displaces another key. This continues until the procedure finds a vacant position, or enters a cycle. In case of cycle, new hash functions are chosen and the whole data structure is ‘rehashed’. Multiple rehashes might be necessary before Cuckoo succeeds.

Insertion is expected O(1) (amortized) with high probability, even considering the possibility rehashing, as long as the number of keys is kept below half of the capacity of the hash table, i.e., the load factor is below 50%.

Deletion is O(1) worst-case as it requires inspection of just two locations in the hash table.

Illustration :

Input:

{20, 50, 53, 75, 100, 67, 105, 3, 36, 39}

Hash Functions:

h1(key) = key%11
h2(key) = (key/11)%11 Let’s start with inserting 20 at its possible position in the first table determined by h1(20): Next: 50 Next: 53. h1(53) = 9. But 20 is already there at 9. We place 53 in table 1 & 20 in table 2 at h2(20) Next: 75. h1(75) = 9. But 53 is already there at 9. We place 75 in table 1 & 53 in table 2 at h2(53) Next: 100. h1(100) = 1. Next: 67. h1(67) = 1. But 100 is already there at 1. We place 67 in table 1 & 100 in table 2 Next: 105. h1(105) = 6. But 50 is already there at 6. We place 105 in table 1 & 50 in table 2 at h2(50) = 4. Now 53 has been displaced. h1(53) = 9. 75 displaced: h2(75) = 6. Next: 3. h1(3) = 3. Next: 36. h1(36) = 3. h2(3) = 0. Next: 39. h1(39) = 6. h2(105) = 9. h1(100) = 1. h2(67) = 6. h1(75) = 9. h2(53) = 4. h1(50) = 6. h2(39) = 3.

Here, the new key 39 is displaced later in the recursive calls to place 105 which it displaced. Implementation :
Below is the implementation of Cuckoo hashing

C++

 // C++ program to demonstrate working of Cuckoo // hashing. #include    // upper bound on number of elements in our set #define MAXN 11    // choices for position #define ver 2    // Auxiliary space bounded by a small multiple // of MAXN, minimizing wastage int hashtable[ver][MAXN];    // Array to store possible positions for a key int pos[ver];    /* function to fill hash table with dummy value  * dummy value: INT_MIN  * number of hashtables: ver */ void initTable() {     for (int j=0; j

Java

 // Java program to demonstrate working of  // Cuckoo hashing. import java.util.*;    class GFG  {    // upper bound on number of elements in our set static int MAXN = 11;    // choices for position static int ver = 2;    // Auxiliary space bounded by a small multiple // of MAXN, minimizing wastage static int [][]hashtable = new int[ver][MAXN];    // Array to store possible positions for a key static int []pos = new int[ver];    /* function to fill hash table with dummy value * dummy value: INT_MIN * number of hashtables: ver */ static void initTable() {     for (int j = 0; j < MAXN; j++)         for (int i = 0; i < ver; i++)             hashtable[i][j] = Integer.MIN_VALUE; }    /* return hashed value for a key * function: ID of hash function according to which     key has to hashed * key: item to be hashed */ static int hash(int function, int key) {     switch (function)     {         case 1: return key % MAXN;         case 2: return (key / MAXN) % MAXN;     }     return Integer.MIN_VALUE; }    /* function to place a key in one of its possible positions * tableID: table in which key has to be placed, also equal   to function according to which key must be hashed * cnt: number of times function has already been called   in order to place the first input key * n: maximum number of times function can be recursively   called before stopping and declaring presence of cycle */ static void place(int key, int tableID, int cnt, int n) {     /* if function has been recursively called max number     of times, stop and declare cycle. Rehash. */     if (cnt == n)     {         System.out.printf("%d unpositioned\n", key);         System.out.printf("Cycle present. REHASH.\n");         return;     }        /* calculate and store possible positions for the key.     * check if key already present at any of the positions.     If YES, return. */     for (int i = 0; i < ver; i++)     {         pos[i] = hash(i + 1, key);         if (hashtable[i][pos[i]] == key)         return;     }        /* check if another key is already present at the        position for the new key in the table     * If YES: place the new key in its position     * and place the older key in an alternate position     for it in the next table */     if (hashtable[tableID][pos[tableID]] != Integer.MIN_VALUE)     {         int dis = hashtable[tableID][pos[tableID]];         hashtable[tableID][pos[tableID]] = key;         place(dis, (tableID + 1) % ver, cnt + 1, n);     }     else // else: place the new key in its position     hashtable[tableID][pos[tableID]] = key; }    /* function to print hash table contents */ static void printTable() {     System.out.printf("Final hash tables:\n");        for (int i = 0; i < ver; i++, System.out.printf("\n"))         for (int j = 0; j < MAXN; j++)             if(hashtable[i][j] == Integer.MIN_VALUE)                  System.out.printf("- ");             else                 System.out.printf("%d ", hashtable[i][j]);        System.out.printf("\n"); }    /* function for Cuckoo-hashing keys * keys[]: input array of keys * n: size of input array */ static void cuckoo(int keys[], int n) {     // initialize hash tables to a dummy value      // (INT-MIN) indicating empty position     initTable();        // start with placing every key at its position in     // the first hash table according to first hash     // function     for (int i = 0, cnt = 0; i < n; i++, cnt = 0)         place(keys[i], 0, cnt, n);        // print the final hash tables     printTable(); }    // Driver Code public static void main(String[] args)  {     /* following array doesn't have any cycles and     hence all keys will be inserted without any     rehashing */     int keys_1[] = {20, 50, 53, 75, 100,                      67, 105, 3, 36, 39};        int n = keys_1.length;        cuckoo(keys_1, n);        /* following array has a cycle and hence we will     have to rehash to position every key */     int keys_2[] = {20, 50, 53, 75, 100,                      67, 105, 3, 36, 39, 6};        int m = keys_2.length;        cuckoo(keys_2, m); } }     // This code is contributed by Princi Singh

C#

 // C# program to demonstrate working of  // Cuckoo hashing. using System;        class GFG  {    // upper bound on number of // elements in our set static int MAXN = 11;    // choices for position static int ver = 2;    // Auxiliary space bounded by a small  // multiple of MAXN, minimizing wastage static int [,]hashtable = new int[ver, MAXN];    // Array to store  // possible positions for a key static int []pos = new int[ver];    /* function to fill hash table  with dummy value * dummy value: INT_MIN * number of hashtables: ver */ static void initTable() {     for (int j = 0; j < MAXN; j++)         for (int i = 0; i < ver; i++)             hashtable[i, j] = int.MinValue; }    /* return hashed value for a key * function: ID of hash function  according to which key has to hashed * key: item to be hashed */ static int hash(int function, int key) {     switch (function)     {         case 1: return key % MAXN;         case 2: return (key / MAXN) % MAXN;     }     return int.MinValue; }    /* function to place a key in one of  its possible positions * tableID: table in which key  has to be placed, also equal to function according to which key must be hashed * cnt: number of times function has already  been called in order to place the first input key * n: maximum number of times function  can be recursively called before stopping and declaring presence of cycle */ static void place(int key, int tableID,                    int cnt, int n) {     /* if function has been recursively      called max number of times,     stop and declare cycle. Rehash. */     if (cnt == n)     {         Console.Write("{0} unpositioned\n", key);         Console.Write("Cycle present. REHASH.\n");         return;     }        /* calculate and store possible positions      * for the key. Check if key already present      at any of the positions. If YES, return. */     for (int i = 0; i < ver; i++)     {         pos[i] = hash(i + 1, key);         if (hashtable[i, pos[i]] == key)         return;     }        /* check if another key is already present      at the position for the new key in the table     * If YES: place the new key in its position     * and place the older key in an alternate position     for it in the next table */     if (hashtable[tableID,                pos[tableID]] != int.MinValue)     {         int dis = hashtable[tableID, pos[tableID]];         hashtable[tableID, pos[tableID]] = key;         place(dis, (tableID + 1) % ver, cnt + 1, n);     }     else // else: place the new key in its position     hashtable[tableID, pos[tableID]] = key; }    /* function to print hash table contents */ static void printTable() {     Console.Write("Final hash tables:\n");        for (int i = 0; i < ver;              i++, Console.Write("\n"))         for (int j = 0; j < MAXN; j++)             if(hashtable[i, j] == int.MinValue)                  Console.Write("- ");             else                 Console.Write("{0} ",                         hashtable[i, j]);        Console.Write("\n"); }    /* function for Cuckoo-hashing keys * keys[]: input array of keys * n: size of input array */ static void cuckoo(int []keys, int n) {     // initialize hash tables to a      // dummy value (INT-MIN)     // indicating empty position     initTable();        // start with placing every key      // at its position in the first      // hash table according to first      // hash function     for (int i = 0, cnt = 0;              i < n; i++, cnt = 0)         place(keys[i], 0, cnt, n);        // print the final hash tables     printTable(); }    // Driver Code public static void Main(String[] args)  {     /* following array doesn't have      any cycles and hence all keys      will be inserted without any rehashing */     int []keys_1 = {20, 50, 53, 75, 100,                      67, 105, 3, 36, 39};        int n = keys_1.Length;        cuckoo(keys_1, n);        /* following array has a cycle and      hence we will have to rehash to      position every key */     int []keys_2 = {20, 50, 53, 75, 100,                      67, 105, 3, 36, 39, 6};        int m = keys_2.Length;        cuckoo(keys_2, m); } }    // This code is contributed by PrinciRaj1992

Output:

Final hash tables:
- 100 - 36 - - 50 - - 75 -
3 20 - 39 53 - 67 - - 105 -

105 unpositioned
Cycle present. REHASH.
Final hash tables:
- 67 - 3 - - 39 - - 53 -
6 20 - 36 50 - 75 - - 100 -

Generalizations of cuckoo hashing that use more than 2 alternative hash functions can be expected to utilize a larger part of the capacity of the hash table efficiently while sacrificing some lookup and insertion speed. Example: if we use 3 hash functions, it’s safe to load 91% and still be operating within expected bounds (Source : Wiki)

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