Hashing Vishnu Kotrajaras, PhD

What do we want to do? Insert Delete find (constant time) No sorting No Findmin findmax

Hash table We have key and value. The key is an argument of our hash function. The result of a hash function is an index that we will store our value. Therefore a hash function should: –Be easy to calculate. –Different keys must give you different index. This is difficult to achieve, but it can be done.

Hash function We use it to try to distribute values evenly throughout our table. We may use: –Key number % tableSize But if tableSize is 10, 20, 30, … we cannot use this function. –What if keys are Strings? Let ’ s see some example.

Hash function (1 st example) Sum the ASCII values of all alphabets public static int hash(String key, int tableSize){ int hashVal = 0; for(int i =0; i<key.length(); i++) hashVal += key.charAt(i); return hashVal%tableSize; }

The method in the last page is not good if the table is large: –Whet if each key is short (e.g. 8 alphabets?) –An ASCII normally has a maximum value of 127. Therefore the sum of all 8 alphabets will not exceed 127*8. –If the table is big, data will not be distributed evenly. The 10,000th member Indices will concentrate at the front.

Hash function (2 nd example) Assume we have a big table, and each key is made from at least 3 random alphabets. We look at the first 3 alphabets only. public static int hash(String key, int tableSize){ return (key.charAt(0) +27*key.charAt(1) +729* key.charAT(2))%tableSize; } All alphabets, including space 27*27 This distributes well in a table of size (10007 is the first prime after 10000, we will use this number. You will see why).

Wait, any actual key will never be random like this: –There will be a lot of repetition.

Hash function (3rd example) We calculate a polynomial function of 37, using Horner ’ s Rule. We can calculate k k *37k 2 by using [(k 2 *37)+k 1 ]*37 +k 0 Horner rule is to repeat this -> n times. In fact, it is a calculation of:

public static int hash(String key, int tableSize){ int hashVal = 0; for(int i =0; i<key.length(); i++) hashVal= 37*hashVal+key.charAt(i); hashVal %= tableSize; if(hashVal<0) hashVal += tableSize; return hashVal; } Possible overflow

May not be very well distributed, but it ’ s easy to calculate. But if a key is long, the corresponding calculation will take some time. –We solve it by not using every alphabet. –We may chose alphabets from important parts of the key. In any case a hash function cannot distribute items into a table with 100% uniquely different indices. When 2 or more values fall in the same slot we say it is a collision. How do we fix a collision?

Fixing collision: separate chaining Store repeated elements in a linked list. If you want to search for an element, use hash function, then search in the list given by that hash function. If you want to insert an element, –use hash function to find a list to put that element in. –After that, check the list to see whether it already contains the element. If the list does not have that element then insert the element at the front. –Statistically, a newly inserted element is often accessed again soon after the insertion.

Code for an object that has a hash function. 1. public interface Hashable 2. { 3. /** 4. * Compute a hash function for this object. 5. tableSize the hash table size. 6. (deterministically) a number between 7. * 0 and tableSize-1, distributed equitably. 8. */ 9. int hash( int tableSize ); 10. }

static method from our HashTable class. How we use a Hashable object. Public class Student implements Hashable{ private String name; private double number; private int year; public int hash(int tableSize){ return SeparateChainingHashTable.hash(name, tableSize); } public boolean equals(Object rhs){ return name.equals(((Student)rhs).name); }

1. public class SeparateChainingHashTable 2. { 3. /** 4. * Construct the hash table. 5. */ 6. public SeparateChainingHashTable( ) 7. { 8. this( DEFAULT_TABLE_SIZE ); 9. } 10. /** 11. * Construct the hash table. 12. size approximate table size. 13. */ 14. public SeparateChainingHashTable( int size ) 15. { 16. theLists = new LinkedList[ nextPrime( size ) ]; 17. for( int i = 0; i < theLists.length; i++ ) 18. theLists[ i ] = new LinkedList( ); 19. }

20. /** 21. * Insert into the hash table. If the item is 22. * already present, then do nothing. 23. x the item to insert. 24. */ 25. public void insert( Hashable x ) 26. { 27. LinkedList whichList = theLists[ x.hash( theLists.length ) ]; 28. LinkedListItr itr = whichList.find( x ); 29. if( itr.isPastEnd( ) ) 30. whichList.insert( x, whichList.zeroth( ) ); 31. } 32. /** 33. * Remove from the hash table. 34. x the item to remove. 35. */ 36. public void remove( Hashable x ) 37. { 38. theLists[ x.hash( theLists.length ) ].remove( x ); 39. } We use Student here

40. /** 41. * Find an item in the hash table. 42. x the item to search for. 43. the matching item, or null if not found. 44. */ 45. public Hashable find( Hashable x ) 46. { 47. return (Hashable)theLists[ x.hash( theLists.length ) ].find( x ).retrieve( ); 48. } 49. /** 50. * Make the hash table logically empty. 51. */ 52. public void makeEmpty( ) 53. { 54. for( int i = 0; i < theLists.length; i++ ) 55. theLists[ i ].makeEmpty( ); 56. }

57. /** 58. * A hash routine for String objects. 59. key the String to hash. 60. tableSize the size of the hash table. 61. the hash value. 62. */ 63. public static int hash( String key, int tableSize ) 64. { 65. int hashVal = 0; 66. for( int i = 0; i < key.length( ); i++ ) 67. hashVal = 37 * hashVal + key.charAt( i ); 68. hashVal %= tableSize; 69. if( hashVal < 0 ) 70. hashVal += tableSize; 71. return hashVal; 72. }

73. private static final int DEFAULT_TABLE_SIZE = 101; 74. /** The array of Lists. */ 75. private LinkedList [ ] theLists; 76. /** 77. * Internal method to find a prime number at least as large as n. 78. n the starting number (must be positive). 79. a prime number larger than or equal to n. 80. */ 81. private static int nextPrime( int n ) 82. { 83. if( n % 2 == 0 ) 84. n++; 85. for( ; !isPrime( n ); n += 2 ) 86. ; 87. return n; 88. }

89. /** 90. * Internal method to test if a number is prime. 91. * Not an efficient algorithm. 92. n the number to test. 93. the result of the test. 94. */ 95. private static boolean isPrime( int n ) 96. { 97. if( n == 2 || n == 3 ) 98. return true; 99. if( n == 1 || n % 2 == 0 ) 100. return false; 101. for( int i = 3; i * i <= n; i += 2 ) 102. if( n % i == 0 ) 103. return false; 104. return true; 105. }

106. // Simple main 107. public static void main( String [ ] args ) 108. { 109. SeparateChainingHashTable H = new SeparateChainingHashTable( ); 110. final int NUMS = 4000; 111. final int GAP = 37; 112. System.out.println( "Checking... (no more output means success)" ); 113. for( int i = GAP; i != 0; i = ( i + GAP ) % NUMS ) 114. H.insert( new MyInteger( i ) ); 115. for( int i = 1; i < NUMS; i+= 2 ) 116. H.remove( new MyInteger( i ) ); 117. for( int i = 2; i < NUMS; i+=2 ) 118. if( ((MyInteger)(H.find( new MyInteger( i ) ))).intValue( ) != i ) 119. System.out.println( "Find fails " + i ); 120. for( int i = 1; i < NUMS; i+=2 ) 121. { 122. if( H.find( new MyInteger( i ) ) != null ) 123. System.out.println( "OOPS!!! " + i ); 124. } 125. } 126.}

Definition Load factor It is an average length of linked list. Search time = time to do hashing + time to search list = constant + time to search list Unsuccessful search Search time == average list length == load factor

Successful search –In a list that we will search, there is one node that contains an object that we want to find. There are other nodes too (0 or more). –in a table, if we have N members, distributed into M lists. There are N-1 nodes that do not have what we want. If we distribute these nodes evenly among the lists. Each list will have (N-1)/M nodes. = lambda- (1/M) = lambda, because M is large. On average, half the list will be searched before we find what we want. That is, lambda/2 steps will be executed. Therefore the average time to find the required element is 1 + (lambda/2) steps. The tableSize is not important. What really matters is the load factor.

Fixing collision by using Open addressing No list. If there is a collision, then keep calculating a new index until an empty slot is found. –The new index is at h 0 (x), h 1 (x), … –h i (x)=[hash(x)+f(i)]%tableSize, f(0)=0 Every data must be put into our table. Therefore the table must be large enough to distribute data. –Load factor <=0.5

Open addressing: linear probing F is a linear function of i. Normally we have -> f(i)=i It is “ looking ahead one slot at a time. ” –This may take time. There will be consecutive filled slots, called primary clustering. If a new collision takes place, it will take some time before we can find another empty slot.

Open addressing: quadratic probing There is no primary clustering by this method. We usually have -> f(i)=i 2 h i (x)=[hash(x)+f(i)]%tableSize a if b collides with a, we add 1 2 to find a new empty slot. If c also collides with a, we add 1 2 to find b. We need to go further by adding 2 2 instead.

However, if our table is more than half full or the tableSIze is not prime, this method does not guarantee an empty slot. But if the table is not yet half full and the tableSize is prime, it is proven that we can always find an empty slot for a new value.

Proof Let the tableSize be a prime number greater than 3. Let (h(x)+i 2 ) mod tableSize (h(x)+j 2 ) mod tableSize Prove by contradiction –Assume both positions are the same and i !=j. Be 2 empty slot positions.

i-j =0 is impossible because we assumed they are not equal. i+j=0 is also impossible, Therefore our assumption that the two positions are the same is wrong. Thus the two positions are always different. So there is always a slot for a new value, if the table is not yet half full and the tableSize is prime.

Why prime? If not, the number of available slots will greatly reduce. Example: tableSize == 16. Assume a normal hashing gives index ==0. (quadratic probing) You can see that they fall in the same positions.

We cannot use ordinary deletion. If we remove, then later attempt to find another value, we may encounter an empty space and think that we cannot find the value (in fact the value is in the table, but requires jumping from a collision point) Use lazy deletion -> mark a deleted slot without actually removing its element.

Open addressing implementation class HashEntry { Hashable element; // the element boolean isActive; // false means -> deleted public HashEntry( Hashable e ){ this( e, true ); } public HashEntry( Hashable e, boolean i ){ element = e; isActive = i; }

1. public class QuadraticProbingHashTable{ 2. private static final int DEFAULT_TABLE_SIZE = 11; 3. /** The array of elements. */ 4. private HashEntry [ ] array; // The array of elements 5. private int currentSize; // The number of occupied cells public QuadraticProbingHashTable( ){ 8. this( DEFAULT_TABLE_SIZE ); 9. } 10. /** 11. * Construct the hash table. 12. size the approximate initial size. 13. */ 14. public QuadraticProbingHashTable( int size ){ 15. allocateArray( size ); 16. makeEmpty( ); 17. } nullactive nonactive

18. /** 19. * Internal method to allocate array. 20. arraySize the size of the array. 21. */ 22. private void allocateArray( int arraySize ){ 23. array = new HashEntry[ arraySize ]; 24. } 25. /** 26. * Make the hash table logically empty. 27. */ 28. public void makeEmpty( ){ 29. currentSize = 0; 30. for( int i = 0; i < array.length; i++ ) 31. array[ i ] = null; 32. }

33. /** 34. * Return true if currentPos exists and is active. 35. currentPos the result of a call to findPos. 36. true if currentPos is active. 37. */ 38. private boolean isActive( int currentPos ){ 39. return array[ currentPos ] != null && array[ currentPos ].isActive; 40. }

41. /** 42. * Method that performs quadratic probing resolution. 43. x the item to search for. 44. the position where the search terminates. 45. */ 46. private int findPos( Hashable x ) { 47./* 1*/ int collisionNum = 0; 48./* 2*/ int currentPos = x.hash( array.length ); 49./* 3*/ while( array[ currentPos ] != null && 50. !array[ currentPos ].element.equals( x ) ){ 51./* 4*/ currentPos += 2 * ++collisionNum - 1; // Compute ith probe 52./* 5*/ if( currentPos >= array.length ) // Implement the mod 53./* 6*/ currentPos -= array.length; 54. } 55./* 7*/ return currentPos; 56. } f(i)=i 2 =f(i-1)+2i-1

57. /** 58. * Find an item in the hash table. 59. x the item to search for. 60. the matching item. 61. */ 62. public Hashable find( Hashable x ){ 63. int currentPos = findPos( x ); 64. return isActive( currentPos ) ? array[ currentPos ].element : null; 65. }

66. /** 67. * Insert into the hash table. If the item is 68. * already present, do nothing. 69. x the item to insert. 70. */ 71. public void insert( Hashable x ) 72. { 73. // Insert x as active 74. int currentPos = findPos( x ); 75. if( isActive( currentPos ) ) 76. return; //x is already inside, so do nothing 77. array[ currentPos ] = new HashEntry( x, true ); 78. // Rehash; see Section if( ++currentSize > array.length / 2 ) 80. rehash( ); 81. }

82. /** 83. * Expand the hash table. 84. */ 85. private void rehash( ) 86. { 87. HashEntry [ ] oldArray = array; 88. // Create a new double-sized, empty table 89. allocateArray( nextPrime( 2 * oldArray.length ) ); 90. currentSize = 0; 91. // Copy table over 92. for( int i = 0; i < oldArray.length; i++ ) 93. if( oldArray[ i ] != null && oldArray[ i ].isActive ) 94. insert( oldArray[ i ].element ); 95. return; 96. } recalculate index because this is a new array. O(N) because there are N members to be rehashed. This is not done often because the table has to be half filled first.

rehashing Rehash can be done due to 3 situations. –Do it immediately when the table is half full. –Do it when our insert starts to fail. –Do it when a load factor is up to some value (Does not have to be 0.5) Do not forget that the more the load factor value, the more difficult it is to insert.

hash, nextPrime, isPrime are the same as before. 97. /** 98. * Remove from the hash table. 99. x the item to remove */ 101. public void remove( Hashable x ) 102. { 103. int currentPos = findPos( x ); 104. if( isActive( currentPos ) ) 105. array[ currentPos ].isActive = false; 106. }

Downside of quadratic probing Secondary clustering Fixed by double hashing: –f(i) = i*hash 2 (x) –We find hash 2 (x), 2 *hash 2 (x), … and so on. Must be careful when choosing a function. –If our array has 9 slots and hash 2 (x) = x%9 -> if we insert 99, we will always get 0. –hash 2 (x) must not give 0.

Example of hash 2 Assume hash(x) = x%tableSize hash 2 (x)=R-(x%R), R is prime and R< tableSize Let our tableSize be 16. We insert 9, 25, 26, 41, 42, 58 respectively collides, so we add 13-(25%13)=1 26 collides, so we add 13-(26%13)=13

collides, so we add 13-(41%13)=11 42 collides, so we add 13-(42%13)=10 but 42 still collides, so we add 2*10 from its original index.

collides, so we add 13-(58%13)=7