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Module 4: Hashing and Tries


Module objectives

 

By the end of this module, you will be able to:

 


4.1   Introduction: Signatures and Addresses

 

We will first consider two aspects of keys:

  • Signature: a function computed on keys, usually resulting in an integer or bit pattern.

  • Addressing: where keys are located in memory and what it takes to reach a particular address.

 

Signatures (of keys):

  • Generally, a signature is a function computed on keys or "texts" that results in an integer or bit pattern for each key.

  • The resulting number or bit pattern is called the "signature" of the key.

  • Example: a signature function on strings:
    • Let s be a string.
    • Suppose signature (s) = "position in alphabet of string's first letter" (from 1,...,26).
      We'll call this signature "first letter of string".
    • Thus, signature("cat") = 3, signature("yadayada") = 25,

    • A different function:
      signature(s) = concatenation of (7-bit) ascii codes of letters in s.
    • Thus, signature("cat") = concatenation of 1100011 (c), 1100001 (a) and 1110100 (t)
      = 110001111000011110100.
    • Similarly, signature("yadayada") = 1111001 1100001 1100100 1100001 1111001 1100001 1100100 1100001
      (with spaces shown only for effect).

  • Desirable properties:
    • Uniqueness: each key should have a unique signature.
    • Brevity: signatures should be shorter than the key.
    • Efficiency: the signature computation should be fast.
    • Order-preservation: key-order should result in signature-order.
      • A signature is strictly order-preserving if:
              key1 > key2 implies signature(key1) > signature(key2).
      • A signature is order-preserving if:
              key1 > key2 implies signature(key1) >= signature(key2).
     

    In-Class Exercise 4.1: Compare the two signatures, "first-letter-of-string" and "concatenation-of-ascii-codes", with regard to uniqueness, brevity, efficiency and order-preservation.
     

  • In practice:
    • We'd like to keep the probability of signature-duplication small.
    • Length: usually 1-8 bytes, depending on application.
    • Usually not order-preserving at all.

  • Signatures can be devised for various kinds of data:
    • Strings (as above).
    • Geometric data (points, polygons, etc).
    • Images.

  • Applications:
    • Computing a "digest":
      • Consider a large text of characters.
      • Suppose Signature (text) = sum of ascii codes of characters.
        ⇒ every text will have a 7-bit signature.
      • If you change the text, the signature is likely to change.
        ⇒ applications: security (documents), change-notification (webpages).
      • In practice: larger signatures (e.g., 8 bytes) are used.
    • Similarity matching:
      • Consider how fingerprints are matched.
      • For each fingerprint, a short feature-based signature is computed.
      • Matching fingerprints should have matching signatures.
    • Data structures: hashing, tries (as we will see).
 

Addresses (of keys):

  • Consider a key (e.g., "abc") stored in a data structure.

  • The key will usually reside at some memory address.

  • To get to this memory address (searching for the key):
    • If data structure is linked:
      • Linked list: walk down list (O(n)).
      • In-order tree: navigate through tree (O(log (n))).
    • If data structure is an array:
      • Unsorted: scan array (O(n)).
      • Sorted: binary search (O(log (n)))
 


4.2   Hashing

 

Key ideas:

  • Hashing combines signatures and addressing.

  • The signature is used as the address itself!

  • Note: Hashing is intended for equality search.
 

Two problems:

  • Valid addresses:
    • If the signature is an integer, how can we use that as a memory address?
    • Some languages (e.g., Java) do not allow direct access to memory.
    • Even in C, the signature may not be a valid address.
    • Solution: use array offsets.

  • Uniqueness:
    • What if two keys have the same signature?
      ⇒ a collision occurs at the address.
    • Solution: store colliding keys together.
 

Details:

  • Create a table (array) with entries 0,...,m-1.

  • Each entry represents an address.

  • Thus, the addresses are 0,...,m-1.

  • Each entry i will contain a linked list used to store keys (and values) whose signature (hash value) is i.

  • Devise a signature function (called hashing function) whose output is one of the values 0,...,m-1:
    • Suppose a signature function f produces values larger than m.
    • Let g(string) = f(string) mod m.
      g produces values in the range 0,...,m-1.

  • Each table entry is sometimes called a bucket.

Example:

  • Suppose we want to store the following 51 strings:
    "catch", "throws", "break", "byte", "volatile", "throw", "protected", "this", "do", "switch", "boolean", "char", "package", "true", "extends", "new", "instanceof", "for", "public", "return", "while", "case", "abstract", "false", "void", "synchronized", "implements", "finally", "null", "try", "const", "default", "native", "continue", "super", "import", "class", "final", "transient", "short", "double", "long", "private", "goto", "int", "if", "else", "static", "interface", "float"

  • We will use the signature (ascii code of the) "first letter of string".

  • Initially, the hashtable is empty: each entry is an empty linked list.

  • Next, when inserting "catch":
    • Compute signature of "catch": signature("catch") = 2.
      (distance from 'a').
    • Compute table index: i = signature mod 26.
      i = 2.
    • Insert into linked list at entry 2:

  • After inserting "throws", "break", "byte", "volatile":

  • ... (remaining steps not shown).
 

Pseudocode:

  • Insertion:
    
    Algorithm: initialize (numBuckets)
    Input: desired number of buckets
    1.   Initialize array of linked lists;
      
    
    Algorithm: insert (key, value)
    Input: key-value pair
         // Compute table entry: 
    1.   entry = key.hashCode() mod numBuckets
    2.   if table[entry] is null
           // No list present, so create one 
    3.     table[entry] = new linked list;
    4.     table[entry].add (key, value)
    5.   else
    6.     // Otherwise, add to existing list 
    7.     table[entry].add (key, value)
    8.   endif
      
  • Search:
    
    Algorithm: search (key)
    Input: search-key
         // Compute table entry: 
    1.   entry = key.hashCode() mod numBuckets
    2.   if table[entry] is null
    3.     return null
    4.   else
    5.     return table[entry].search (key)
    6.   endif
    Output: value, if found; null otherwise.
      
  • Note: the add and search methods for linked lists are not described.
 

About signatures:

  • Signature functions are usually called hash or hashing functions.

  • Hash functions should distribute keys evenly over the buckets:
    • Consider distribution in previous example:
      
      a: 1     ("abstract")
      b: 3     ("boolean", "break", "byte")
      c: 6     ("case", "catch", "char", "class", "const", "continue")
      d: 3     ("default", "do", "double")
      e: 2     ("else", "extends")
      f: 5     ("false", "final", "finally", "float", "for")
      g: 1     ("goto")
      h: 0     
      i: 6     ("if", "implements", "import", "instanceof", "int", "interface")
      j: 0     
      k: 0     
      l: 1     ("long")
      m: 0     
      n: 3     ("native", "new", "null")
      o: 0     
      p: 4     ("package", "private", "protected", "public")
      q: 0     
      r: 1     ("return")
      s: 5     ("short", "static", "super", "switch", "synchronized")
      t: 6     ("this", "throw", "throws", "transient", "true", "try")
      u: 0     
      v: 2     ("void", "volatile")
      w: 1     ("while")
      x: 0     
      y: 0     
      z: 0     
           
    • In contrast, Java's hashing function has the following distribution:
      
      table entry 0:  3
      table entry 1:  0
      table entry 2:  3
      table entry 3:  3
      table entry 4:  0
      table entry 5:  1
      table entry 6:  1
      table entry 7:  1
      table entry 8:  4
      table entry 9:  0
      table entry 10: 0
      table entry 11: 2
      table entry 12: 0
      table entry 13: 2
      table entry 14: 1
      table entry 15: 1
      table entry 16: 0
      table entry 17: 0
      table entry 18: 2
      table entry 19: 2
      table entry 20: 2
      table entry 21: 4
      table entry 22: 1
      table entry 23: 0
      table entry 24: 0
      table entry 25: 0
           
      (A more even, but not perfect distribution).

  • A hashing function should include all parts of the key:
    • "first letter" ignores contribution from remainder of string.
    • "add ascii codes of letters" involves all parts.
    • Java's hash function for String's:
      
            int hashValue = 0;
            for (int i=0; i < length; i++) {
              // Note: 31 = 2**5-1. The string characters are assumed to be  
              // stored in charArray[]. 
              hashValue = 31*hashValue + charArray[i];
            }
            return hashValue;
            
 

In-Class Exercise 4.2: Download HashAnalysis.java and the file words.txt (a dictionary). Print out the distribution of words (only the number) over buckets for each of the following signatures:

  • "first letter"
  • "last letter"
  • Java's hash function for strings (call the string's hashCode() method). Use 26 buckets.
Hint: you do not have to implement a hashtable to compute these numbers; you only need to compute the signature, and record statistics.
 

Sizing a table:

  • If you expect insertion of n keys, how large should the table be?

  • With m buckets and uniform distribution, there will be n / m keys per bucket (average).

  • Hashing has no guarantees: you can get unlucky and have a large bucket.
    (But probability of large bucket is very small)

  • If n is small, consider using a size in the range n to 4n.
    For example: with Java's 51 reserved words and Java's hash function: we need 287 buckets for a perfect distribution (probably excessive).

  • If n is large, use 0.5n or smaller.

  • If n is very large, use a hash tree:
    • Use a fixed number of buckets (e.g. 100 buckets) at each level.
    • Thus, at the root level (level 0) there is the hashtable itself.
    • Each bucket is a "child" in the tree.
           ⇒ There are a 100 children at level 1.
    • Each node at level 1 can have children (the next level)
           ⇒ There are 100 x 100 = 10,000 children at level 2
    • A different hash function is used at each level.
 

Analysis (assuming few keys per bucket):

  • Insertion: O(1) (constant).
  • Search: O(1) (constant).
  • Both search and insertion times are optimal for the search-problem.
 

Variations:

  • Linear probing:
    • Avoid using linked lists: store keys as table entries directly.
    • If a collision occurs, simply scan table looking for next available space.
    • Advantage: avoids space for linked structures.
    • Disadvantage: long scans occasionally required.
  • Double-hashing:
    • Like linear-probing, store keys in table.
    • Instead of scanning table for available space, use second hash function to determine where next to try insertion.
  • Perfect hashing:
    • Given fixed data, design hash functions to ensure perfect distribution over buckets.
    • Use a series of hash functions instead of a single function.
  • External hashing:
    • For hashing when applied to large disk files (that remain on disk).
    • See database course/book for details.
 

In-Class Exercise 4.3: Suppose you had to use a hash table for a collection of floating-point numbers: x1, x2,...,xn. Design a hashing function to hash to the signatures 0,...,m (which are integers).
 


4.3   Geometric Hashing

 

For a collection of points, (x1,y1), (x2,y2), ..., (xn,yn), consider the following queries:

  • Given a query point q, is q in the collection?
  • Given a query point q, what is the nearest point to q from the collection?

One approach:

  • Suppose points are represented by (x,y) values (reals).
  • Use a hash function on x-coordinate of each point.
  • Can work very well for equality search (first query).
  • But nearest-point query?
 

2D hashing:

  • We'll use this example to illustrate:
    • Data:
      p1: (-1.5, 6.25)
      p2: (-0.75, 6.1)
      p3: (-0.1, 6.33)
      p4: (-0.1, 7.1)
      p5: (2.1, 6.48)
      p6: (2.25, 6.8)
    • Query point: q = (0.76, 6.32).

  • Find minimum x and y values:
    xmin = min (x1,...,xn) = -1.5
    ymin = min (y1,...,yn) = 6.1

  • Find maximum x and y values:
    xmax = max (x1,...,xn) = 2.25
    ymax = max (y1,...,yn) = 7.1

  • Then, the rectangle with corners (xmin, ymin), (xmax, ymax) is a bounding rectangle.

  • Divide bounding rectangle into a m x m grid with m2 cells.
    Example with m = 8.

  • Give coordinates to each cell, e.g. when m = 5:

  • Each point lies in some cell: use the cell's coordinates as the 2D-hash value.

    Point Coordinates 2D hash-value
    p1 (-1.5, 6.25) (0,0)
    p2 (-0.75, 6.1) (0,0)
    p3 (-0.1, 6.33) (1,1)
    p4 (-0.1, 7.1) (1,4)
    p5 (2.1, 6.48) (4,1)
    p6 (2.25, 6.8) (4,3)

     

    In-Class Exercise 4.4: Describe how to efficiently compute the hash value for a point p=(x,y), given m and the bounding rectangle.
     

  • Insert all the points using 2D hash-values:
    • Create a 2D table of linked lists (2D hash table).
    • Insert each point in appropriate linked list:

  • For nearest-point query:
    • Hash query point q.
      ⇒ cell (3, 1)

    • Work outwards in a spiral to find nearest non-empty cell.

    • Find closest point in non-empty cell: p5
    • Mark radius around q with distance to current closest point p5:

    • Search all cells that overlap circle:


      ⇒ closest point overall: p3 = (-0.1, 6.33).

 

In-Class Exercise 4.5: The 2D hashtable above is a 2D-geometric analog of the regular 1D hashtable (that we saw earlier for strings). Design a tree-like data structure for 2D point data that is analogous to the binary trees we used for strings.
 


4.4   Tries

 

Consider combinations of "structure" and whether or not signatures are used:

Key ideas:

  • Note: "trie" rhymes with "try".

  • Use bit-string signature for each key.
    Example: "reverse ascii code of first letter in string"
    Thus, signature("C") = 1100001.

  • At level i, use i-th bit of signature.

  • If bit = 0, go left. Otherwise, go right.

  • Compare with binary tree:

  • Note:
    • Trie is not necessarily in in-order.
    • Different signatures will produce different tries.
    • We did not compare with key in node.

  • Varieties of tries:
    • Simple trie: "store keys in interior nodes".
    • Full trie: "store keys in leaves"
    • Compressed trie: "store keys in leaves and compress paths",
    • Patricia trie: "combination of simple trie and compressed-trie".
 


4.5   Simple Tries

 

Key ideas:

  • Note: Simple tries are also called "Digital Search Trees".
  • Insertion: navigate using signature bits until insertion can be made.
  • Search: navigate using signature bits, compare with nodes along the way.
 

Insertion:

  • Example:
    • We will use the signature "reverse ascii code of first letter"
      (The data happens to have only one letter).
    • We will insert the following keys:
      Key Signature
      A 1000001
      B 0100001
      C 1100001
      D 0010001
      E 1010001
      F 0110001
    • Insert "A" (1000001)
      ⇒ Empty trie, place in root.

    • Insert "B" (0100001)
      Compare with "A" ⇒ not equal, so proceed
      First bit = 0 ⇒ go left
      No link ⇒ insert

    • Insert "C"
      Compare with "A" ⇒ not equal, so proceed
      First bit = 1 ⇒ go right
      No link ⇒ insert

    • Insert "D"

    • Insert "E"

    • Insert "F"

  • Pseudocode:
    
    Algorithm: insert (key, value)
    Input: key-value pair
    1.   if trie is empty
    2.     root = create new root with key-value pair;
    3.     return
    4.   endif
         // Start numbering the bits from 0. 
    5.   recursiveInsert (root, key, value, 0)
       
    
    Algorithm: recursiveInsert (node, key, value, bitNum)
    Input: trie node, key-value pair, which bit we are using now   
         // Compare with node key to see if it's a duplicate. 
    1.   if node.key = key
    2.     Handle duplicates;
    3.     return
    4.   endif
         // Otherwise, examine the bitNum-th bit 
    5.   if key.getBit (bitNum) = 0
           // Go left if possible, or insert. 
    6.     if node.left is null
    7.       node.left = new trie node with key-value;
    8.     else
             // Note: at next level we'll need to examine the next bit. 
    9.       recursiveInsert (node.left, key, value, bitNum+1)
    10.    endif
    11.  else
           // Same thing on the right 
    12.    if node.right is null
    13.      node.right = new trie node with key-value;
    14.    else
    15.      recursiveInsert (node.right, key, value, bitNum+1)
    16.    endif
    17.  endif
       

Search:

  • Search is straightforward:
    • Compare the input key with the current node.
    • If equal, the key is found; return.
    • Otherwise, examine i-th bit (at level i) and go left or right accordingly.
    • If next link is null, search ends without finding the key.

  • Pseudocode:
    
    Algorithm: search (key)
    Input: search-key
    1.   node = recursiveSearch (root, key, 0)
    2.   if node is null
    3.     return null
    4.   else
    5.     return node.value
    6.   endif
    Output: value, if key is found
       
    
    Algorithm: recursiveSearch (node, key, bitNum)
    Input: trie node, search-key, which bit to examine
         // Compare with key in node. 
    1.   if node.key = key
           // Found. 
    2.     return node
    3.   endif
         // Otherwise, navigate further. 
    4.   if key.getBit (bitNum) = 0
    5.     if node.left is null
             // Not found => search ends. 
    6.       return null
    7.     else
             // Search left. 
    8.       return recursiveSearch (node.left, key, bitNum+1)
    9.     endif
    10.  else
    11.    if node.right is null
             // Not found => search ends. 
    12.      return null
    13.    else
             // Search right. 
    14.      return recursiveSearch (node.right, key, bitNum+1)
    15.    endif
    16.  endif
    Output: trie node if found, else null.
       
 

In-Class Exercise 4.6: Use the "ascii code of letter" signature to insert "A B C D E F" into a simple trie. Show all your steps. Ascii codes are available here. Notice what happens and explain why we used "reverse ascii" earlier.
 

Analysis:

  • Maximum height = maximum length of bitstring.

  • n keys should need no more than log2(n) bits for signature.
    ⇒ maximum tree height = log (n).
     

    In-Class Exercise 4.7: Why is this true? That is, why is it that n keys need at most log2(n) bits to represent the keys?
     

  • Insertion: O(log (n)).

  • Search: O(log (n)).

  • Note: a signature function should be constructed carefully so that no bits are "wasted"
    • The "ascii code" signature wastes the first few bits (since they are common to all letters).
    • Commonly used hash functions do not waste bits because of the "mod" function.
 


4.6   Full tries

 

Key ideas:

  • Disadvantage of simple trie:
    does not maintain sort-order, even if signature is order-preserving.

  • In Full Trie: maintain sort order
    Note: must use order-preserving signature.

  • To do this: avoid using interior nodes
    ⇒ all keys at leaves.

  • Interior nodes are for navigation only:

  • To sort: thread the leaves in a sorted linked list.

Insertion:

  • Key ideas:
    • Navigate using bits as long as internal nodes exist.
    • If internal node for i-th bit does not exist, create internal node and all other nodes on path to leaf.

  • Note: we will need to know in advance the maximum number of bits.

  • Example:
    • We will insert the keys (strings) "A", "B", "C", "D", "E", "F".
    • Signature: the 5 lowest-order bits of the first letter.
      (Note: the data just happens to have only one letter).
    • Insert "A" (00001)
      • Trie is empty, so create root as well as all internal nodes on path to leaf node for "A".

    • Insert "B" (00010)
      • Traverse left for first three 0's.
      • Bit = 1 at 4-th level.
      • No internal node exists:
        ⇒ create path to leaf.

    • Insert "C" (00011)
      • All internal nodes on path exists
        only navigation required (and leaf node).

    • Insert "D" (00100)

    • "E" (00101)

    • "F" (00110)

  • Pseudocode:
    
    Algorithm: initialize (maxBits)
    Input: maximum number of bits
    1.   Store maximum number of bits to use;
        
    
    Algorithm: insert (key, value)
    Input: key-value pair
    1.   if trie is empty
    2.     root = create empty internal node;
           // Start bit-numbering at 0 and create path to leaf: 
    3.     extendBranch (root, key, value, 0)
    4.     return
    5.   endif
    6.   recursiveInsert (root, key, value, 0)
        
    
    Algorithm: extendBranch (node, key, value, bitNum)
    Input: internal trie node, key-value pair, bit number
    1.  Create path of internal nodes from level bitNum to maxBits-1;
    2.  if key.getBit (maxBits) = 0
    3.    create left leaf at end of path;
    4.  else
    5.    create right leaf at end of path;
    6.  endif
        
    
    Algorithm: recursiveInsert (node, key, value, bitNum)
    Input: trie node, key-value pair, bit number
    1.  if key.getBit (bitNum) = 0
          // Check left side. 
    2.    if node.left is null
            // Grow a branch of internal nodes and append leaf. 
    3.      extendBranch (node, key, value, bitNum)
    4.    else
    5.      recursiveInsert (node.left, key, value, bitNum+1)
    6.    endif
    7.  else
          // Check right side. 
    8.    if node.right is null
            // Grow a branch of internal nodes and append leaf. 
    9.      extendBranch (node, key, value, bitNum)
    10.   else
    11.     recursiveInsert (node.right, key, value, bitNum+1)
    12.   endif
    13. endif
        

Search:

  • Similar to simple trie.

  • One difference: compare with keys only at leaves.

Sort-order scan:

  • Must use order-preserving signature.

  • Since all keys are at leaves
    ⇒ in-order traversal will result in sort-order.

  • Optimization: link leaves together.
 

Analysis:

  • Again, maximum height = number of bits in signature.
    log2(n).

  • Insertion: O(log (n)).

  • Search: O(log (n)).

  • Note:
    • Simple trie: a key-comparison occurs at each interior node in search path.
    • Full trie: only bit-evaluations occur at internal nodes
      ⇒ only ONE key comparison (at leaf)
      ⇒ faster search (especially if keys are long).
    • Simple trie: One node per key.
    • Full trie: multiple (wasted) internal nodes.
      O(n) extra storage. (Why?)
 

In-Class Exercise 4.8: Why is it that the full trie wastes O(n) storage? In other words, explain why the number of internal nodes is O(n).
 


4.7   Compressed tries

 

Key ideas:

  • Suppose "D" and "E" are the only keys to the left of the root.

  • Common internal nodes are "wasted":

  • Compressed trie: compress common sections.

  • Idea: during insertion, extend path only as much as needed:

  • Note: sort-order is still maintained in leaves.

Insertion:

  • Example:
    • Insert "A" (00001)
      • Trie empty ⇒ insert as root.

    • Insert "B" (00010)
      • "B" differs from "A" in fourth bit
        ⇒ path of length 3 required.

    • Insert "C" (00011)
      • "C" differs from "B" in last bit
        ⇒ full path required

    • Insert "D" (00100)
      • "D" differs at third bit, no path-creation required.

    • Insert "E" (00101)
      • "E" differs from "D" in last bit
        ⇒ path extension required.

    • Insert "F" (00110)
      • No path creation required.

  • Pseudocode:
    
    Algorithm: insert (key, value)
    Input: key-value pair
    1.   if trie is empty
    2.     root = new root node with key-value;
    3.     return
    4.   endif
         // Start with level 0 (bit number 0): 
    5.   recursiveInsert (root, key, value, 0)
        
    
    Algorithm: recursiveInsert (node, key, value, bitNum)
    Input: trie node, key-value pair, bit number
    1.   if node contains a key
           // Need to create a path to distinguish key and node.key 
    2.     extendSmallestBranch (node, node.key, node.value, key, value, bitNum)
    3.     return
    4.   endif
         // Otherwise, node is an empty interior node for navigation only. 
    5.   if key.getBit (bitNum) = 0
           // Check left. 
    6.     if node.left is null
    7.       node.left = new node containing key-value;
    8.     else
    9.       recursiveInsert (node.left, key, value, bitNum+1)
    10.    endif
    11.  else
           // Check right. 
    12.    if node.right is null
    13.      node.right = new node containing key-value;
    14.    else
    15.      recursiveInsert (node.right, key, value, bitNum+1)
    16.    endif
    17.  endif
        
    
    Algorithm: extendSmallestBranch (node, key1, value1, key2, value2, bitNum)
    Input: node from which to build branch, two key-value pairs, bit number.
         // Examine bits from bitNum to maxBits. 
         // As long as the bits are equal in the two keys, extend branch. 
         // When the bits differ, stop and create children with key1 and key2. 
        
 

In-Class Exercise 4.9: Use the "reverse ascii code of first letter" signature to insert "A B C D E F" into a compressed trie. Show all your steps. Ascii codes are available here.
 


4.8   Handling Duplicates (in any Data Structure)

 

Motivation:

  • In practice, data often contains duplicates.

  • In some applications, you want to store multiple values for a single key.
    ⇒ data structure should be able to store all values for a key.

Solutions (using Binary Search Tree as data structure):

  • Duplicate lists:
    • Use the "attached-list" idea from hashing.
    • Store duplicates in a linked-list off of tree nodes.

    • Advantages: flexible, efficient (only those keys with duplicates need the list).
    • Disadvantages: extra space required in each node.

  • Duplicate cache:
    • If duplicates are few (O(1)), store in separate table.

    • Constant-cost added to each search (to search cache).
    • Advantages: simple, efficient for few duplicates, interchangeable with data structures.
    • Disadvantages: inefficient for many duplicates.

A Java programming trick for handling duplicates: (source file)


import java.util.*;

public class Duplicate {

  // Any Map can be used, e.g., TreeMap. 
  static Map originalDataStructure = new HashMap();


  // Insertion. 

  static void insert (Object key, Object value)
  {
    // 1. Attempt a direct insertion. 
    Object oldValue = originalDataStructure.put (key, value);

    // 2. If the value wasn't already there, nothing to be done. 
    if (oldValue == null) {
      // No duplicates. 
      return;
    }

    // 3. Otherwise, check if duplicates already exist. 
    if (oldValue instanceof HashSet) {
      // 3.1 There are, so add the new one. 
      HashSet duplicates = (HashSet) oldValue;
      duplicates.add (value);
    }
    else {
      // 3.2 This is the first duplicate => create a HashSet to store duplicates. 
      HashSet duplicates = new HashSet ();
      // 3.3 Place old value and duplicate in hashset. 
      duplicates.add (oldValue);
      duplicates.add (value);
      // 3.4 Add the hashset itself as the value. 
      originalDataStructure.put (key, duplicates);
    }
  }


  // Enumerate and print all values. 

  static void printAll ()
  {
    Collection values = originalDataStructure.values();
    for (Iterator i=values.iterator();  i.hasNext(); ) {
      Object obj = i.next();
      // If the value is a HashSet, we have duplicates. 
      if (obj instanceof HashSet) {
        // Extract the different values. 
        HashSet duplicates = (HashSet) obj;
        for (Iterator j=duplicates.iterator();  j.hasNext(); ) {
          System.out.println (j.next());
        }
      }
      else {
        // If it's not a HashSet, the extract obj is a value. 
        System.out.println (obj);
      }
    }
  }

  public static void main (String[] argv)
  {
    // Add data with duplicates. 
    insert ("Ali", "Addled Ali");
    insert ("Bill", "Bewildered Bill");
    insert ("Bill", "Befuddled Bill");     // Duplicate. 
    insert ("Chen", "Cantankerous Chen");
    insert ("Dave", "Discombobulated Dave");
    insert ("Dave", "Distracted Dave");     // Duplicate. 
    insert ("Dave", "Disconcerted Dave");      // Duplicate. 
    insert ("Ella", "Egregious Ella");
    insert ("Franco", "Flummoxed Franco");
    insert ("Gita", "Grumpy Gita"); 
    insert ("Gita", "Grouchy Gita");       // Duplicate. 

    // Print. 
    printAll();
  }

}
 



© 2001, Rahul Simha