Huffman coding

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Revision as of 11:48, 24 December 2019 by rosettacode>Yuriy Chumak (added Ol)
Task
Huffman coding
You are encouraged to solve this task according to the task description, using any language you may know.

Huffman encoding is a way to assign binary codes to symbols that reduces the overall number of bits used to encode a typical string of those symbols.

For example, if you use letters as symbols and have details of the frequency of occurrence of those letters in typical strings, then you could just encode each letter with a fixed number of bits, such as in ASCII codes. You can do better than this by encoding more frequently occurring letters such as e and a, with smaller bit strings; and less frequently occurring letters such as q and x with longer bit strings.

Any string of letters will be encoded as a string of bits that are no-longer of the same length per letter. To successfully decode such as string, the smaller codes assigned to letters such as 'e' cannot occur as a prefix in the larger codes such as that for 'x'.

If you were to assign a code 01 for 'e' and code 011 for 'x', then if the bits to decode started as 011... then you would not know if you should decode an 'e' or an 'x'.

The Huffman coding scheme takes each symbol and its weight (or frequency of occurrence), and generates proper encodings for each symbol taking account of the weights of each symbol, so that higher weighted symbols have fewer bits in their encoding. (See the WP article for more information).

A Huffman encoding can be computed by first creating a tree of nodes:

  1. Create a leaf node for each symbol and add it to the priority queue.
  2. While there is more than one node in the queue:
    1. Remove the node of highest priority (lowest probability) twice to get two nodes.
    2. Create a new internal node with these two nodes as children and with probability equal to the sum of the two nodes' probabilities.
    3. Add the new node to the queue.
  3. The remaining node is the root node and the tree is complete.


Traverse the constructed binary tree from root to leaves assigning and accumulating a '0' for one branch and a '1' for the other at each node. The accumulated zeros and ones at each leaf constitute a Huffman encoding for those symbols and weights:


Task

Using the characters and their frequency from the string:

  this is an example for huffman encoding

create a program to generate a Huffman encoding for each character as a table.

Ada

Works with: Ada 2005

huffman.ads: <lang Ada>with Ada.Containers.Indefinite_Ordered_Maps; with Ada.Containers.Ordered_Maps; with Ada.Finalization; generic

  type Symbol_Type is private;
  with function "<" (Left, Right : Symbol_Type) return Boolean is <>;
  with procedure Put (Item : Symbol_Type);
  type Symbol_Sequence is array (Positive range <>) of Symbol_Type;
  type Frequency_Type is private;
  with function "+" (Left, Right : Frequency_Type) return Frequency_Type
    is <>;
  with function "<" (Left, Right : Frequency_Type) return Boolean is <>;

package Huffman is

  -- bits = booleans (true/false = 1/0)
  type Bit_Sequence is array (Positive range <>) of Boolean;
  Zero_Sequence : constant Bit_Sequence (1 .. 0) := (others => False);
  -- output the sequence
  procedure Put (Code : Bit_Sequence);
  -- type for freqency map
  package Frequency_Maps is new Ada.Containers.Ordered_Maps
    (Element_Type => Frequency_Type,
     Key_Type     => Symbol_Type);
  type Huffman_Tree is private;
  -- create a huffman tree from frequency map
  procedure Create_Tree
    (Tree        : out Huffman_Tree;
     Frequencies : Frequency_Maps.Map);
  -- encode a single symbol
  function Encode
    (Tree   : Huffman_Tree;
     Symbol : Symbol_Type)
     return   Bit_Sequence;
  -- encode a symbol sequence
  function Encode
    (Tree    : Huffman_Tree;
     Symbols : Symbol_Sequence)
     return    Bit_Sequence;
  -- decode a bit sequence
  function Decode
    (Tree : Huffman_Tree;
     Code : Bit_Sequence)
     return Symbol_Sequence;
  -- dump the encoding table
  procedure Dump_Encoding (Tree : Huffman_Tree);

private

  -- type for encoding map
  package Encoding_Maps is new Ada.Containers.Indefinite_Ordered_Maps
    (Element_Type => Bit_Sequence,
     Key_Type     => Symbol_Type);
  type Huffman_Node;
  type Node_Access is access Huffman_Node;
  -- a node is either internal (left_child/right_child used)
  -- or a leaf (left_child/right_child are null)
  type Huffman_Node is record
     Frequency   : Frequency_Type;
     Left_Child  : Node_Access := null;
     Right_Child : Node_Access := null;
     Symbol      : Symbol_Type;
  end record;
  -- create a leaf node
  function Create_Node
    (Symbol    : Symbol_Type;
     Frequency : Frequency_Type)
     return      Node_Access;
  -- create an internal node
  function Create_Node (Left, Right : Node_Access) return Node_Access;
  -- fill the encoding map
  procedure Fill
    (The_Node : Node_Access;
     Map      : in out Encoding_Maps.Map;
     Prefix   : Bit_Sequence);
  -- huffman tree has a tree and an encoding map
  type Huffman_Tree is new Ada.Finalization.Controlled with record
     Tree : Node_Access       := null;
     Map  : Encoding_Maps.Map := Encoding_Maps.Empty_Map;
  end record;
  -- free memory after finalization
  overriding procedure Finalize (Object : in out Huffman_Tree);

end Huffman;</lang>

huffman.adb: <lang Ada>with Ada.Text_IO; with Ada.Unchecked_Deallocation; with Ada.Containers.Vectors; package body Huffman is

  package Node_Vectors is new Ada.Containers.Vectors
    (Element_Type => Node_Access,
     Index_Type   => Positive);
  function "<" (Left, Right : Node_Access) return Boolean is
  begin
     -- compare frequency
     if Left.Frequency < Right.Frequency then
        return True;
     elsif Right.Frequency < Left.Frequency then
        return False;
     end if;
     -- same frequency, choose leaf node
     if Left.Left_Child = null and then Right.Left_Child /= null then
        return True;
     elsif Left.Left_Child /= null and then Right.Left_Child = null then
        return False;
     end if;
     -- same frequency, same node type (internal/leaf)
     if Left.Left_Child /= null then
        -- for internal nodes, compare left children, then right children
        if Left.Left_Child < Right.Left_Child then
           return True;
        elsif Right.Left_Child < Left.Left_Child then
           return False;
        else
           return Left.Right_Child < Right.Right_Child;
        end if;
     else
        -- for leaf nodes, compare symbol
        return Left.Symbol < Right.Symbol;
     end if;
  end "<";
  package Node_Vector_Sort is new Node_Vectors.Generic_Sorting;
  procedure Create_Tree
    (Tree        : out Huffman_Tree;
     Frequencies : Frequency_Maps.Map) is
     Node_Queue : Node_Vectors.Vector := Node_Vectors.Empty_Vector;
  begin
     -- insert all leafs into the queue
     declare
        use Frequency_Maps;
        Position : Cursor      := Frequencies.First;
        The_Node : Node_Access := null;
     begin
        while Position /= No_Element loop
           The_Node :=
             Create_Node
               (Symbol    => Key (Position),
                Frequency => Element (Position));
           Node_Queue.Append (The_Node);
           Next (Position);
        end loop;
     end;
     -- sort by frequency (see "<")
     Node_Vector_Sort.Sort (Node_Queue);
     -- iterate over all elements
     while not Node_Queue.Is_Empty loop
        declare
           First : constant Node_Access := Node_Queue.First_Element;
        begin
           Node_Queue.Delete_First;
           -- if we only have one node left, it is the root node of the tree
           if Node_Queue.Is_Empty then
              Tree.Tree := First;
           else
              -- create new internal node with two smallest frequencies
              declare
                 Second : constant Node_Access := Node_Queue.First_Element;
              begin
                 Node_Queue.Delete_First;
                 Node_Queue.Append (Create_Node (First, Second));
              end;
              Node_Vector_Sort.Sort (Node_Queue);
           end if;
        end;
     end loop;
     -- fill encoding map
     Fill (The_Node => Tree.Tree, Map => Tree.Map, Prefix => Zero_Sequence);
  end Create_Tree;
  -- create leaf node
  function Create_Node
    (Symbol    : Symbol_Type;
     Frequency : Frequency_Type)
     return      Node_Access
  is
     Result : Node_Access := new Huffman_Node;
  begin
     Result.Frequency := Frequency;
     Result.Symbol    := Symbol;
     return Result;
  end Create_Node;
  -- create internal node
  function Create_Node (Left, Right : Node_Access) return Node_Access is
     Result : Node_Access := new Huffman_Node;
  begin
     Result.Frequency   := Left.Frequency + Right.Frequency;
     Result.Left_Child  := Left;
     Result.Right_Child := Right;
     return Result;
  end Create_Node;
  -- fill encoding map
  procedure Fill
    (The_Node : Node_Access;
     Map      : in out Encoding_Maps.Map;
     Prefix   : Bit_Sequence) is
  begin
     if The_Node.Left_Child /= null then
        -- append false (0) for left child
        Fill (The_Node.Left_Child, Map, Prefix & False);
        -- append true (1) for right child
        Fill (The_Node.Right_Child, Map, Prefix & True);
     else
        -- leaf node reached, prefix = code for symbol
        Map.Insert (The_Node.Symbol, Prefix);
     end if;
  end Fill;
  -- free memory after finalization
  overriding procedure Finalize (Object : in out Huffman_Tree) is
     procedure Free is new Ada.Unchecked_Deallocation
       (Name   => Node_Access,
        Object => Huffman_Node);
     -- recursively free all nodes
     procedure Recursive_Free (The_Node : in out Node_Access) is
     begin
        -- free node if it is a leaf
        if The_Node.Left_Child = null then
           Free (The_Node);
        else
           -- free left and right child if node is internal
           Recursive_Free (The_Node.Left_Child);
           Recursive_Free (The_Node.Right_Child);
           -- free node afterwards
           Free (The_Node);
        end if;
     end Recursive_Free;
  begin
     -- recursively free root node
     Recursive_Free (Object.Tree);
  end Finalize;
  -- encode single symbol
  function Encode
    (Tree   : Huffman_Tree;
     Symbol : Symbol_Type)
     return   Bit_Sequence
  is
  begin
     -- simply lookup in map
     return Tree.Map.Element (Symbol);
  end Encode;
  -- encode symbol sequence
  function Encode
    (Tree    : Huffman_Tree;
     Symbols : Symbol_Sequence)
     return    Bit_Sequence
  is
  begin
     -- only one element
     if Symbols'Length = 1 then
        -- see above
        return Encode (Tree, Symbols (Symbols'First));
     else
        -- encode first element, append result of recursive call
        return Encode (Tree, Symbols (Symbols'First)) &
        Encode (Tree, Symbols (Symbols'First + 1 .. Symbols'Last));
     end if;
  end Encode;
  -- decode a bit sequence
  function Decode
    (Tree : Huffman_Tree;
     Code : Bit_Sequence)
     return Symbol_Sequence
  is
     -- maximum length = code length
     Result   : Symbol_Sequence (1 .. Code'Length);
     -- last used index of result
     Last     : Natural     := 0;
     The_Node : Node_Access := Tree.Tree;
  begin
     -- iterate over the code
     for I in Code'Range loop
        -- if current element is true, descent the right branch
        if Code (I) then
           The_Node := The_Node.Right_Child;
        else
           -- false: descend left branch
           The_Node := The_Node.Left_Child;
        end if;
        if The_Node.Left_Child = null then
           -- reached leaf node: append symbol to result
           Last          := Last + 1;
           Result (Last) := The_Node.Symbol;
           -- reset current node to root
           The_Node := Tree.Tree;
        end if;
     end loop;
     -- return subset of result array
     return Result (1 .. Last);
  end Decode;
  -- output a bit sequence
  procedure Put (Code : Bit_Sequence) is
     package Int_IO is new Ada.Text_IO.Integer_IO (Integer);
  begin
     for I in Code'Range loop
        if Code (I) then
           -- true = 1
           Int_IO.Put (1, 0);
        else
           -- false = 0
           Int_IO.Put (0, 0);
        end if;
     end loop;
     Ada.Text_IO.New_Line;
  end Put;
  -- dump encoding map
  procedure Dump_Encoding (Tree : Huffman_Tree) is
     use type Encoding_Maps.Cursor;
     Position : Encoding_Maps.Cursor := Tree.Map.First;
  begin
     -- iterate map
     while Position /= Encoding_Maps.No_Element loop
        -- key
        Put (Encoding_Maps.Key (Position));
        Ada.Text_IO.Put (" = ");
        -- code
        Put (Encoding_Maps.Element (Position));
        Encoding_Maps.Next (Position);
     end loop;
  end Dump_Encoding;

end Huffman;</lang>

example main.adb: <lang Ada>with Ada.Text_IO; with Huffman; procedure Main is

  package Char_Natural_Huffman_Tree is new Huffman
    (Symbol_Type => Character,
     Put => Ada.Text_IO.Put,
     Symbol_Sequence => String,
     Frequency_Type => Natural);
  Tree         : Char_Natural_Huffman_Tree.Huffman_Tree;
  Frequencies  : Char_Natural_Huffman_Tree.Frequency_Maps.Map;
  Input_String : constant String :=
    "this is an example for huffman encoding";

begin

  -- build frequency map
  for I in Input_String'Range loop
     declare
        use Char_Natural_Huffman_Tree.Frequency_Maps;
        Position : constant Cursor := Frequencies.Find (Input_String (I));
     begin
        if Position = No_Element then
           Frequencies.Insert (Key => Input_String (I), New_Item => 1);
        else
           Frequencies.Replace_Element
             (Position => Position,
              New_Item => Element (Position) + 1);
        end if;
     end;
  end loop;
  -- create huffman tree
  Char_Natural_Huffman_Tree.Create_Tree
    (Tree        => Tree,
     Frequencies => Frequencies);
  -- dump encodings
  Char_Natural_Huffman_Tree.Dump_Encoding (Tree => Tree);
  -- encode example string
  declare
     Code : constant Char_Natural_Huffman_Tree.Bit_Sequence :=
       Char_Natural_Huffman_Tree.Encode
         (Tree    => Tree,
          Symbols => Input_String);
  begin
     Char_Natural_Huffman_Tree.Put (Code);
     Ada.Text_IO.Put_Line
       (Char_Natural_Huffman_Tree.Decode (Tree => Tree, Code => Code));
  end;

end Main;</lang>

Output:
  = 101
a = 1001
c = 01010
d = 01011
e = 1100
f = 1101
g = 01100
h = 11111
i = 1110
l = 01101
m = 0010
n = 000
o = 0011
p = 01110
r = 01111
s = 0100
t = 10000
u = 10001
x = 11110
1000011111111001001011110010010110010001011100111101001001001110011011100101110100110111110111111100011101110100101001000101110000001010001101011111000001100
this is an example for huffman encoding

BBC BASIC

This example is incorrect. Please fix the code and remove this message.
Details: Huffman code can not contain another code as a prefix

<lang bbcbasic> INSTALL @lib$+"SORTSALIB"

     SortUp% = FN_sortSAinit(0,0) : REM Ascending
     SortDn% = FN_sortSAinit(1,0) : REM Descending
     
     Text$ = "this is an example for huffman encoding"
     
     DIM tree{(127) ch&, num%, lkl%, lkr%}
     FOR i% = 1 TO LEN(Text$)
       c% = ASCMID$(Text$,i%)
       tree{(c%)}.ch& = c%
       tree{(c%)}.num% += 1
     NEXT
     
     C% = DIM(tree{()},1) + 1
     CALL SortDn%, tree{()}, tree{(0)}.num%
     FOR i% = 0 TO DIM(tree{()},1)
       IF tree{(i%)}.num% = 0 EXIT FOR
     NEXT
     size% = i%
     
     linked% = 0
     REPEAT
       C% = size%
       CALL SortUp%, tree{()}, tree{(0)}.num%
       i% = 0 : WHILE tree{(i%)}.lkl% OR tree{(i%)}.lkr% i% += 1 : ENDWHILE
       tree{(i%)}.lkl% = size%
       j% = 0 : WHILE tree{(j%)}.lkl% OR tree{(j%)}.lkr% j% += 1 : ENDWHILE
       tree{(j%)}.lkr% = size%
       linked% += 2
       tree{(size%)}.num% = tree{(i%)}.num% + tree{(j%)}.num%
       size% += 1
     UNTIL linked% = (size% - 1)
     
     FOR i% = size% - 1 TO 0 STEP -1
       IF tree{(i%)}.ch& THEN
         h$ = ""
         j% = i%
         REPEAT
           CASE TRUE OF
             WHEN tree{(j%)}.lkl% <> 0:
               h$ = "0" + h$
               j% = tree{(j%)}.lkl%
             WHEN tree{(j%)}.lkr% <> 0:
               h$ = "1" + h$
               j% = tree{(j%)}.lkr%
             OTHERWISE:
               EXIT REPEAT
           ENDCASE
         UNTIL FALSE
         VDU tree{(i%)}.ch& : PRINT "  " h$
       ENDIF
     NEXT
     END</lang>
Output:
   101
n  000
e  1110
f  1101
a  1100
i  1011
s  0110
m  0101
h  0100
o  0011
c  0010
l  0001
r  0000
x  11111
p  11110
d  11101
u  11100
g  11011
t  11010

Bracmat

<lang bracmat>( "this is an example for huffman encoding":?S & 0:?chars & 0:?p & ( @( !S

    :   ?
        ( [!p %?char [?p ?
        & !char+!chars:?chars
        & ~
        )
    )
 | 
 )

& 0:?prioritized & whl

 ' ( !chars:?n*%@?w+?chars
   & (!n.!w)+!prioritized:?prioritized
   )

& whl

 ' ( !prioritized:(?p.?x)+(?q.?y)+?nprioritized
   & (!p+!q.(!p.0,!x)+(!q.1,!y))+!nprioritized:?prioritized
   )

& 0:?L & ( walk

 =   bits tree bit subtree
   .   !arg:(?bits.?tree)
     &   whl
       ' ( !tree:(?p.?bit,?subtree)+?tree
         & (   !subtree:@
             & (!subtree.str$(!bits !bit))+!L:?L
           | walk$(!bits !bit.!subtree)
           )
         )
 )

& !prioritized:(?.?prioritized) & walk$(.!prioritized) & lst$L & :?encoded & 0:?p & ( @( !S

    :   ?
        ( [!p %?char [?p ?
        & !L:?+(!char.?code)+?
        & !encoded !code:?encoded
        & ~
        )
    )
 | out$(str$!encoded)
 )

& ( decode

 =   char bits
   .       !L
         : ?+(?char.?bits&@(!arg:!bits ?arg))+?
       & !char decode$!arg
     | !arg
 )

& out$("decoded:" str$(decode$(str$!encoded)));</lang>

Output:
(L=
  (" ".101)
+ (a.1001)
+ (c.01010)
+ (d.01011)
+ (e.1100)
+ (f.1101)
+ (g.01100)
+ (h.11111)
+ (i.1110)
+ (l.01101)
+ (m.0010)
+ (n.000)
+ (o.0011)
+ (p.01110)
+ (r.01111)
+ (s.0100)
+ (t.10000)
+ (u.10001)
+ (x.11110));
1000011111111001001011110010010110010001011100111101001001001110011011100101110100110111110111111100011101110100101001000101110000001010001101011111000001100
decoded: this is an example for huffman encoding

C

This code lacks a lot of needed checkings, especially for memory allocation.

<lang c>#include <stdio.h>

  1. include <stdlib.h>
  2. include <string.h>
  1. define BYTES 256

struct huffcode {

 int nbits;
 int code;

}; typedef struct huffcode huffcode_t;

struct huffheap {

 int *h;
 int n, s, cs;
 long *f;

}; typedef struct huffheap heap_t;

/* heap handling funcs */ static heap_t *_heap_create(int s, long *f) {

 heap_t *h;
 h = malloc(sizeof(heap_t));
 h->h = malloc(sizeof(int)*s);
 h->s = h->cs = s;
 h->n = 0;
 h->f = f;
 return h;

}

static void _heap_destroy(heap_t *heap) {

 free(heap->h);
 free(heap);

}

  1. define swap_(I,J) do { int t_; t_ = a[(I)]; \
     a[(I)] = a[(J)]; a[(J)] = t_; } while(0)

static void _heap_sort(heap_t *heap) {

 int i=1, j=2; /* gnome sort */
 int *a = heap->h;
 while(i < heap->n) { /* smaller values are kept at the end */
   if ( heap->f[a[i-1]] >= heap->f[a[i]] ) {
     i = j; j++;
   } else {
     swap_(i-1, i);
     i--;
     i = (i==0) ? j++ : i;
   }
 }

}

  1. undef swap_

static void _heap_add(heap_t *heap, int c) {

 if ( (heap->n + 1) > heap->s ) {
   heap->h = realloc(heap->h, heap->s + heap->cs);
   heap->s += heap->cs;
 }
 heap->h[heap->n] = c;
 heap->n++;
 _heap_sort(heap);

}

static int _heap_remove(heap_t *heap) {

 if ( heap->n > 0 ) {
   heap->n--;
   return heap->h[heap->n];
 }
 return -1;

}

/* huffmann code generator */ huffcode_t **create_huffman_codes(long *freqs) {

 huffcode_t **codes;
 heap_t *heap;
 long efreqs[BYTES*2];
 int preds[BYTES*2];
 int i, extf=BYTES;
 int r1, r2;
 memcpy(efreqs, freqs, sizeof(long)*BYTES);
 memset(&efreqs[BYTES], 0, sizeof(long)*BYTES);
 heap = _heap_create(BYTES*2, efreqs);
 if ( heap == NULL ) return NULL;
 for(i=0; i < BYTES; i++) if ( efreqs[i] > 0 ) _heap_add(heap, i);
 while( heap->n > 1 )
 {
   r1 = _heap_remove(heap);
   r2 = _heap_remove(heap);
   efreqs[extf] = efreqs[r1] + efreqs[r2];
   _heap_add(heap, extf);
   preds[r1] = extf;
   preds[r2] = -extf;
   extf++;
 }
 r1 = _heap_remove(heap);
 preds[r1] = r1;
 _heap_destroy(heap);
 codes = malloc(sizeof(huffcode_t *)*BYTES);
 int bc, bn, ix;
 for(i=0; i < BYTES; i++) {
   bc=0; bn=0;
   if ( efreqs[i] == 0 ) { codes[i] = NULL; continue; }
   ix = i;
   while( abs(preds[ix]) != ix ) {
     bc |= ((preds[ix] >= 0) ? 1 : 0 ) << bn;
     ix = abs(preds[ix]);
     bn++;
   }
   codes[i] = malloc(sizeof(huffcode_t));
   codes[i]->nbits = bn;
   codes[i]->code = bc;
 }
 return codes;

}

void free_huffman_codes(huffcode_t **c) {

 int i;
 for(i=0; i < BYTES; i++) free(c[i]);
 free(c);

}

  1. define MAXBITSPERCODE 100

void inttobits(int c, int n, char *s) {

 s[n] = 0;
 while(n > 0) {
   s[n-1] = (c%2) + '0';
   c >>= 1; n--;
 }

}

const char *test = "this is an example for huffman encoding";

int main() {

 huffcode_t **r;
 int i;
 char strbit[MAXBITSPERCODE];
 const char *p;
 long freqs[BYTES];
 memset(freqs, 0, sizeof freqs);
 p = test;
 while(*p != '\0') freqs[*p++]++;
 r = create_huffman_codes(freqs);
 for(i=0; i < BYTES; i++) {
   if ( r[i] != NULL ) {
     inttobits(r[i]->code, r[i]->nbits, strbit);
     printf("%c (%d) %s\n", i, r[i]->code, strbit);
   }
 }
 free_huffman_codes(r);
 return 0;

}</lang>

Alternative

Using a simple heap-based priority queue. Heap is an array, while ndoe tree is done by binary links. <lang c>#include <stdio.h>

  1. include <string.h>

typedef struct node_t { struct node_t *left, *right; int freq; char c; } *node;

struct node_t pool[256] = Template:0; node qqq[255], *q = qqq - 1; int n_nodes = 0, qend = 1; char *code[128] = {0}, buf[1024];

node new_node(int freq, char c, node a, node b) { node n = pool + n_nodes++; if (freq) n->c = c, n->freq = freq; else { n->left = a, n->right = b; n->freq = a->freq + b->freq; } return n; }

/* priority queue */ void qinsert(node n) { int j, i = qend++; while ((j = i / 2)) { if (q[j]->freq <= n->freq) break; q[i] = q[j], i = j; } q[i] = n; }

node qremove() { int i, l; node n = q[i = 1];

if (qend < 2) return 0; qend--; while ((l = i * 2) < qend) { if (l + 1 < qend && q[l + 1]->freq < q[l]->freq) l++; q[i] = q[l], i = l; } q[i] = q[qend]; return n; }

/* walk the tree and put 0s and 1s */ void build_code(node n, char *s, int len) { static char *out = buf; if (n->c) { s[len] = 0; strcpy(out, s); code[n->c] = out; out += len + 1; return; }

s[len] = '0'; build_code(n->left, s, len + 1); s[len] = '1'; build_code(n->right, s, len + 1); }

void init(const char *s) { int i, freq[128] = {0}; char c[16];

while (*s) freq[(int)*s++]++;

for (i = 0; i < 128; i++) if (freq[i]) qinsert(new_node(freq[i], i, 0, 0));

while (qend > 2) qinsert(new_node(0, 0, qremove(), qremove()));

build_code(q[1], c, 0); }

void encode(const char *s, char *out) { while (*s) { strcpy(out, code[*s]); out += strlen(code[*s++]); } }

void decode(const char *s, node t) { node n = t; while (*s) { if (*s++ == '0') n = n->left; else n = n->right;

if (n->c) putchar(n->c), n = t; }

putchar('\n'); if (t != n) printf("garbage input\n"); }

int main(void) { int i; const char *str = "this is an example for huffman encoding";

       char buf[1024];

init(str); for (i = 0; i < 128; i++) if (code[i]) printf("'%c': %s\n", i, code[i]);

encode(str, buf); printf("encoded: %s\n", buf);

printf("decoded: "); decode(buf, q[1]);

return 0; }</lang>

Output:
' ': 000
'a': 1000
'c': 01101
'd': 01100
'e': 0101
'f': 0010
'g': 010000
'h': 1101
'i': 0011
'l': 010001
'm': 1111
'n': 101
'o': 1110
'p': 10011
'r': 10010
's': 1100
't': 01111
'u': 01110
'x': 01001
encoded: 0111111010011110000000111100000100010100001010100110001111100110100010101000001011101001000011010111000100010111110001010000101101011011110011000011101010000
decoded: this is an example for huffman encoding

C#

<lang csharp>using System; using System.Collections.Generic;

namespace Huffman_Encoding {

   public class PriorityQueue<T> where T : IComparable
   {
       protected List<T> LstHeap = new List<T>();
       public virtual int Count
       {
           get { return LstHeap.Count; }
       }
       public virtual void Add(T val)
       {
           LstHeap.Add(val);
           SetAt(LstHeap.Count - 1, val);
           UpHeap(LstHeap.Count - 1);
       }
       public virtual T Peek()
       {
           if (LstHeap.Count == 0)
           {
               throw new IndexOutOfRangeException("Peeking at an empty priority queue");
           }
           return LstHeap[0];
       }
       public virtual T Pop()
       {
           if (LstHeap.Count == 0)
           {
               throw new IndexOutOfRangeException("Popping an empty priority queue");
           }
           T valRet = LstHeap[0];
           SetAt(0, LstHeap[LstHeap.Count - 1]);
           LstHeap.RemoveAt(LstHeap.Count - 1);
           DownHeap(0);
           return valRet;
       }
       protected virtual void SetAt(int i, T val)
       {
           LstHeap[i] = val;
       }
       protected bool RightSonExists(int i)
       {
           return RightChildIndex(i) < LstHeap.Count;
       }
       protected bool LeftSonExists(int i)
       {
           return LeftChildIndex(i) < LstHeap.Count;
       }
       protected int ParentIndex(int i)
       {
           return (i - 1) / 2;
       }
       protected int LeftChildIndex(int i)
       {
           return 2 * i + 1;
       }
       protected int RightChildIndex(int i)
       {
           return 2 * (i + 1);
       }
       protected T ArrayVal(int i)
       {
           return LstHeap[i];
       }
       protected T Parent(int i)
       {
           return LstHeap[ParentIndex(i)];
       }
       protected T Left(int i)
       {
           return LstHeap[LeftChildIndex(i)];
       }
       protected T Right(int i)
       {
           return LstHeap[RightChildIndex(i)];
       }
       protected void Swap(int i, int j)
       {
           T valHold = ArrayVal(i);
           SetAt(i, LstHeap[j]);
           SetAt(j, valHold);
       }
       protected void UpHeap(int i)
       {
           while (i > 0 && ArrayVal(i).CompareTo(Parent(i)) > 0)
           {
               Swap(i, ParentIndex(i));
               i = ParentIndex(i);
           }
       }
       protected void DownHeap(int i)
       {
           while (i >= 0)
           {
               int iContinue = -1;
               if (RightSonExists(i) && Right(i).CompareTo(ArrayVal(i)) > 0)
               {
                   iContinue = Left(i).CompareTo(Right(i)) < 0 ? RightChildIndex(i) : LeftChildIndex(i);
               }
               else if (LeftSonExists(i) && Left(i).CompareTo(ArrayVal(i)) > 0)
               {
                   iContinue = LeftChildIndex(i);
               }
               if (iContinue >= 0 && iContinue < LstHeap.Count)
               {
                   Swap(i, iContinue);
               }
               i = iContinue;
           }
       }
   }
   internal class HuffmanNode<T> : IComparable
   {
       internal HuffmanNode(double probability, T value)
       {
           Probability = probability;
           LeftSon = RightSon = Parent = null;
           Value = value;
           IsLeaf = true;
       }
       internal HuffmanNode(HuffmanNode<T> leftSon, HuffmanNode<T> rightSon)
       {
           LeftSon = leftSon;
           RightSon = rightSon;
           Probability = leftSon.Probability + rightSon.Probability;
           leftSon.IsZero = true;
           rightSon.IsZero = false;
           leftSon.Parent = rightSon.Parent = this;
           IsLeaf = false;
       }
       internal HuffmanNode<T> LeftSon { get; set; }
       internal HuffmanNode<T> RightSon { get; set; }
       internal HuffmanNode<T> Parent { get; set; }
       internal T Value { get; set; }
       internal bool IsLeaf { get; set; }
       internal bool IsZero { get; set; }
       internal int Bit
       {
           get { return IsZero ? 0 : 1; }
       }
       internal bool IsRoot
       {
           get { return Parent == null; }
       }
       internal double Probability { get; set; }
       public int CompareTo(object obj)
       {
           return -Probability.CompareTo(((HuffmanNode<T>) obj).Probability);
       }
   }
   public class Huffman<T> where T : IComparable
   {
       private readonly Dictionary<T, HuffmanNode<T>> _leafDictionary = new Dictionary<T, HuffmanNode<T>>();
       private readonly HuffmanNode<T> _root;
       public Huffman(IEnumerable<T> values)
       {
           var counts = new Dictionary<T, int>();
           var priorityQueue = new PriorityQueue<HuffmanNode<T>>();
           int valueCount = 0;
           foreach (T value in values)
           {
               if (!counts.ContainsKey(value))
               {
                   counts[value] = 0;
               }
               counts[value]++;
               valueCount++;
           }
           foreach (T value in counts.Keys)
           {
               var node = new HuffmanNode<T>((double) counts[value] / valueCount, value);
               priorityQueue.Add(node);
               _leafDictionary[value] = node;
           }
           while (priorityQueue.Count > 1)
           {
               HuffmanNode<T> leftSon = priorityQueue.Pop();
               HuffmanNode<T> rightSon = priorityQueue.Pop();
               var parent = new HuffmanNode<T>(leftSon, rightSon);
               priorityQueue.Add(parent);
           }
           _root = priorityQueue.Pop();
           _root.IsZero = false;
       }
       public List<int> Encode(T value)
       {
           var returnValue = new List<int>();
           Encode(value, returnValue);
           return returnValue;
       }
       public void Encode(T value, List<int> encoding)
       {
           if (!_leafDictionary.ContainsKey(value))
           {
               throw new ArgumentException("Invalid value in Encode");
           }
           HuffmanNode<T> nodeCur = _leafDictionary[value];
           var reverseEncoding = new List<int>();
           while (!nodeCur.IsRoot)
           {
               reverseEncoding.Add(nodeCur.Bit);
               nodeCur = nodeCur.Parent;
           }
           reverseEncoding.Reverse();
           encoding.AddRange(reverseEncoding);
       }
       public List<int> Encode(IEnumerable<T> values)
       {
           var returnValue = new List<int>();
           foreach (T value in values)
           {
               Encode(value, returnValue);
           }
           return returnValue;
       }
       public T Decode(List<int> bitString, ref int position)
       {
           HuffmanNode<T> nodeCur = _root;
           while (!nodeCur.IsLeaf)
           {
               if (position > bitString.Count)
               {
                   throw new ArgumentException("Invalid bitstring in Decode");
               }
               nodeCur = bitString[position++] == 0 ? nodeCur.LeftSon : nodeCur.RightSon;
           }
           return nodeCur.Value;
       }
       public List<T> Decode(List<int> bitString)
       {
           int position = 0;
           var returnValue = new List<T>();
           while (position != bitString.Count)
           {
               returnValue.Add(Decode(bitString, ref position));
           }
           return returnValue;
       }
   }
   internal class Program
   {
       private const string Example = "this is an example for huffman encoding";
       private static void Main()
       {
           var huffman = new Huffman<char>(Example);
           List<int> encoding = huffman.Encode(Example);
           List<char> decoding = huffman.Decode(encoding);
           var outString = new string(decoding.ToArray());
           Console.WriteLine(outString == Example ? "Encoding/decoding worked" : "Encoding/Decoding failed");
           var chars = new HashSet<char>(Example);
           foreach (char c in chars)
           {
               encoding = huffman.Encode(c);
               Console.Write("{0}:  ", c);
               foreach (int bit in encoding)
               {
                   Console.Write("{0}", bit);
               }
               Console.WriteLine();
           }
           Console.ReadKey();
       }
   }

}</lang> File:CSharpHuffman.jpg

C++

This code builds a tree to generate huffman codes, then prints the codes.

<lang cpp>#include <iostream>

  1. include <queue>
  2. include <map>
  3. include <climits> // for CHAR_BIT
  4. include <iterator>
  5. include <algorithm>

const int UniqueSymbols = 1 << CHAR_BIT; const char* SampleString = "this is an example for huffman encoding";

typedef std::vector<bool> HuffCode; typedef std::map<char, HuffCode> HuffCodeMap;

class INode { public:

   const int f;
   virtual ~INode() {}

protected:

   INode(int f) : f(f) {}

};

class InternalNode : public INode { public:

   INode *const left;
   INode *const right;
   InternalNode(INode* c0, INode* c1) : INode(c0->f + c1->f), left(c0), right(c1) {}
   ~InternalNode()
   {
       delete left;
       delete right;
   }

};

class LeafNode : public INode { public:

   const char c;
   LeafNode(int f, char c) : INode(f), c(c) {}

};

struct NodeCmp {

   bool operator()(const INode* lhs, const INode* rhs) const { return lhs->f > rhs->f; }

};

INode* BuildTree(const int (&frequencies)[UniqueSymbols]) {

   std::priority_queue<INode*, std::vector<INode*>, NodeCmp> trees;
   for (int i = 0; i < UniqueSymbols; ++i)
   {
       if(frequencies[i] != 0)
           trees.push(new LeafNode(frequencies[i], (char)i));
   }
   while (trees.size() > 1)
   {
       INode* childR = trees.top();
       trees.pop();
       INode* childL = trees.top();
       trees.pop();
       INode* parent = new InternalNode(childR, childL);
       trees.push(parent);
   }
   return trees.top();

}

void GenerateCodes(const INode* node, const HuffCode& prefix, HuffCodeMap& outCodes) {

   if (const LeafNode* lf = dynamic_cast<const LeafNode*>(node))
   {
       outCodes[lf->c] = prefix;
   }
   else if (const InternalNode* in = dynamic_cast<const InternalNode*>(node))
   {
       HuffCode leftPrefix = prefix;
       leftPrefix.push_back(false);
       GenerateCodes(in->left, leftPrefix, outCodes);
       HuffCode rightPrefix = prefix;
       rightPrefix.push_back(true);
       GenerateCodes(in->right, rightPrefix, outCodes);
   }

}

int main() {

   // Build frequency table
   int frequencies[UniqueSymbols] = {0};
   const char* ptr = SampleString;
   while (*ptr != '\0')
       ++frequencies[*ptr++];
   INode* root = BuildTree(frequencies);
   
   HuffCodeMap codes;
   GenerateCodes(root, HuffCode(), codes);
   delete root;
   for (HuffCodeMap::const_iterator it = codes.begin(); it != codes.end(); ++it)
   {
       std::cout << it->first << " ";
       std::copy(it->second.begin(), it->second.end(),
                 std::ostream_iterator<bool>(std::cout));
       std::cout << std::endl;
   }
   return 0;

}</lang>

Output:
  110
a 1001
c 101010
d 10001
e 1111
f 1011
g 101011
h 0101
i 1110
l 01110
m 0011
n 000
o 0010
p 01000
r 01001
s 0110
t 01111
u 10100
x 10000

Clojure

(Updated to 1.6 & includes pretty-printing). Uses Java PriorityQueue <lang clojure>(require '[clojure.pprint :refer :all])

(defn probs [s]

 (let [freqs (frequencies s) sum (apply + (vals freqs))]
   (into {} (map (fn k v [k (/ v sum)]) freqs))))

(defn init-pq [weighted-items]

 (let [comp (proxy [java.util.Comparator] []
               (compare [a b] (compare (:priority a) (:priority b))))
       pq (java.util.PriorityQueue. (count weighted-items) comp)]
   (doseq [[item prob] weighted-items] (.add pq { :symbol item, :priority prob }))
   pq))

(defn huffman-tree [pq]

 (while (> (.size pq) 1)
   (let [a (.poll pq) b (.poll pq)

new-node {:priority (+ (:priority a) (:priority b)) :left a :right b}]

     (.add pq new-node)))
 (.poll pq))

(defn symbol-map

 ([t] (symbol-map t ""))
 ([{:keys [symbol priority left right] :as t} code]
   (if symbol [{:symbol symbol :weight priority :code code}]
     (concat (symbol-map left (str code \0))
             (symbol-map right (str code \1))))))

(defn huffman-encode [items]

 (-> items probs init-pq huffman-tree symbol-map))

(defn display-huffman-encode [s]

 (->> s huffman-encode (sort-by :weight >) print-table))

(display-huffman-encode "this is an example for huffman encoding")</lang>

Output:
| :symbol | :weight |  :code |
|---------+---------+--------|
|         |    2/13 |    111 |
|       n |    4/39 |    011 |
|       a |    1/13 |   1001 |
|       e |    1/13 |   1011 |
|       i |    1/13 |   1100 |
|       f |    1/13 |   1101 |
|       h |    2/39 |   0001 |
|       s |    2/39 |   0010 |
|       m |    2/39 |   0100 |
|       o |    2/39 |   0101 |
|       d |    1/39 |  00000 |
|       t |    1/39 |  00001 |
|       c |    1/39 |  00110 |
|       x |    1/39 |  00111 |
|       u |    1/39 |  10000 |
|       l |    1/39 |  10001 |
|       r |    1/39 |  10100 |
|       g |    1/39 | 101010 |
|       p |    1/39 | 101011 |

Alternate Version

Uses c.d.priority-map. Creates a more shallow tree but appears to meet the requirements. <lang clojure>(require '[clojure.data.priority-map :refer [priority-map-keyfn-by]]) (require '[clojure.pprint :refer [print-table]])

(defn init-pq [s]

 (let [c (count s)]
   (->> s frequencies

(map (fn k v [k {:sym k :weight (/ v c)}])) (into (priority-map-keyfn-by :weight <)))))

(defn huffman-tree [pq]

 (letfn [(build-step

[pq] (let [a (second (peek pq)) b (second (peek (pop pq))) nn {:sym (str (:sym a) (:sym b)) :weight (+ (:weight a) (:weight b)) :left a :right b}] (assoc (pop (pop pq)) (:sym nn) nn)))]

   (->> (iterate build-step pq)

(drop-while #(> (count %) 1)) first vals first)))

(defn symbol-map [m]

 (letfn [(sym-step

[{:keys [sym weight left right] :as m} code] (cond (and left right) #(vector (trampoline sym-step left (str code \0)) (trampoline sym-step right (str code \1))) left #(sym-step left (str code \0)) right #(sym-step right (str code \1)) :else {:sym sym :weight weight :code code}))]

   (trampoline sym-step m "")))

(defn huffman-encode [s]

 (->> s init-pq huffman-tree symbol-map flatten))

(defn display-huffman-encode [s]

 (->> s huffman-encode (sort-by :weight >) print-table))

(display-huffman-encode "this is an example for huffman encoding")</lang>

Output:
| :sym | :weight | :code |
|------+---------+-------|
|      |    2/13 |   101 |
|    n |    4/39 |   010 |
|    a |    1/13 |  1001 |
|    i |    1/13 |  1101 |
|    e |    1/13 |  1110 |
|    f |    1/13 |  1111 |
|    m |    2/39 |  0000 |
|    o |    2/39 |  0001 |
|    s |    2/39 |  0010 |
|    h |    2/39 | 11001 |
|    g |    1/39 | 00110 |
|    l |    1/39 | 00111 |
|    t |    1/39 | 01100 |
|    u |    1/39 | 01101 |
|    c |    1/39 | 01110 |
|    d |    1/39 | 01111 |
|    p |    1/39 | 10000 |
|    r |    1/39 | 10001 |
|    x |    1/39 | 11000 |

CoffeeScript

<lang coffeescript> huffman_encoding_table = (counts) ->

 # counts is a hash where keys are characters and
 # values are frequencies;
 # return a hash where keys are codes and values
 # are characters
 
 build_huffman_tree = ->
   # returns a Huffman tree.  Each node has
   #   cnt: total frequency of all chars in subtree
   #   c: character to be encoded (leafs only)
   #   children: children nodes (branches only)
   q = min_queue()
   for c, cnt of counts
     q.enqueue cnt,
       cnt: cnt
       c: c
   while q.size() >= 2
     a = q.dequeue()
     b = q.dequeue()
     cnt = a.cnt + b.cnt
     node = 
       cnt: cnt
       children: [a, b]
     q.enqueue cnt, node
   root = q.dequeue()
   
 root = build_huffman_tree()
 
 codes = {}
 encode = (node, code) ->
   if node.c?
     codes[code] = node.c
   else
     encode node.children[0], code + "0"
     encode node.children[1], code + "1"
 
 encode(root, "")
 codes

min_queue = ->

 # This is very non-optimized; you could use a binary heap for better
 # performance.  Items with smaller priority get dequeued first.
 arr = []
 enqueue: (priority, data) ->
   i = 0
   while i < arr.length
     if priority < arr[i].priority
       break
     i += 1  
   arr.splice i, 0,
     priority: priority
     data: data
 dequeue: ->
   arr.shift().data
 size: -> arr.length
 _internal: ->
   arr

freq_count = (s) ->

 cnts = {}
 for c in s
   cnts[c] ?= 0
   cnts[c] += 1
 cnts
 

rpad = (s, n) ->

 while s.length < n
   s += ' '
 s

examples = [

 "this is an example for huffman encoding"
 "abcd"
 "abbccccddddddddeeeeeeeee"

]

for s in examples

 console.log "---- #{s}"
 counts = freq_count(s)
 huffman_table = huffman_encoding_table(counts)
 codes = (code for code of huffman_table).sort()
 for code in codes
   c = huffman_table[code]
   console.log "#{rpad(code, 5)}: #{c} (#{counts[c]})"
 console.log()
</lang>
Output:
> coffee huffman.coffee 
---- this is an example for huffman encoding
000  : n (4)
0010 : s (2)
0011 : m (2)
0100 : o (2)
01010: t (1)
01011: x (1)
01100: p (1)
01101: l (1)
01110: r (1)
01111: u (1)
10000: c (1)
10001: d (1)
1001 : i (3)
101  :   (6)
1100 : a (3)
1101 : e (3)
1110 : f (3)
11110: g (1)
11111: h (2)

---- abcd
00   : a (1)
01   : b (1)
10   : c (1)
11   : d (1)

---- abbccccddddddddeeeeeeeee
0    : e (9)
1000 : a (1)
1001 : b (2)
101  : c (4)
11   : d (8)

Common Lisp

This implementation uses a tree built of huffman-nodes, and a hash table mapping from elements of the input sequence to huffman-nodes. The priority queue is implemented as a sorted list. (For a more efficient implementation of a priority queue, see the Heapsort task.)

<lang lisp>(defstruct huffman-node

 (weight 0 :type number)
 (element nil :type t)
 (encoding nil :type (or null bit-vector))
 (left nil :type (or null huffman-node))
 (right nil :type (or null huffman-node)))

(defun initial-huffman-nodes (sequence &key (test 'eql))

 (let* ((length (length sequence))
        (increment (/ 1 length))
        (nodes (make-hash-table :size length :test test))
        (queue '()))
   (map nil #'(lambda (element)
                (multiple-value-bind (node presentp) (gethash element nodes)
                  (if presentp
                    (incf (huffman-node-weight node) increment)
                    (let ((node (make-huffman-node :weight increment
                                                   :element element)))
                      (setf (gethash element nodes) node
                            queue (list* node queue))))))
        sequence)
   (values nodes (sort queue '< :key 'huffman-node-weight))))

(defun huffman-tree (sequence &key (test 'eql))

 (multiple-value-bind (nodes queue)
     (initial-huffman-nodes sequence :test test)
   (do () ((endp (rest queue)) (values nodes (first queue)))
     (destructuring-bind (n1 n2 &rest queue-rest) queue
       (let ((n3 (make-huffman-node
                  :left n1
                  :right n2
                  :weight (+ (huffman-node-weight n1)
                             (huffman-node-weight n2)))))
         (setf queue (merge 'list (list n3) queue-rest '<
                            :key 'huffman-node-weight)))))))1

(defun huffman-codes (sequence &key (test 'eql))

 (multiple-value-bind (nodes tree)
     (huffman-tree sequence :test test)
   (labels ((hc (node length bits)
              (let ((left (huffman-node-left node))
                    (right (huffman-node-right node)))
                (cond
                 ((and (null left) (null right))
                  (setf (huffman-node-encoding node)
                        (make-array length :element-type 'bit
                                    :initial-contents (reverse bits))))
                 (t (hc left (1+ length) (list* 0 bits))
                    (hc right (1+ length) (list* 1 bits)))))))
     (hc tree 0 '())
     nodes)))

(defun print-huffman-code-table (nodes &optional (out *standard-output*))

 (format out "~&Element~10tWeight~20tCode")
 (loop for node being each hash-value of nodes
       do (format out "~&~s~10t~s~20t~s"
                  (huffman-node-element node)
                  (huffman-node-weight node)
                  (huffman-node-encoding node))))</lang>

Example:

> (print-huffman-code-table 
   (huffman-codes "this is an example for huffman encoding"))
Element   Weight    Code
#\t       1/39      #*10010
#\d       1/39      #*01101
#\m       2/39      #*0100
#\f       1/13      #*1100
#\o       2/39      #*0111
#\x       1/39      #*100111
#\h       2/39      #*1000
#\a       1/13      #*1010
#\s       2/39      #*0101
#\c       1/39      #*00010
#\l       1/39      #*00001
#\u       1/39      #*00011
#\e       1/13      #*1101
#\n       4/39      #*001
#\g       1/39      #*01100
#\p       1/39      #*100110
#\i       1/13      #*1011
#\r       1/39      #*00000
#\Space   2/13      #*111

D

<lang d>import std.stdio, std.algorithm, std.typecons, std.container, std.array;

auto encode(alias eq, R)(Group!(eq, R) sf) /*pure nothrow @safe*/ {

   auto heap = sf.map!(s => tuple(s[1], [tuple(s[0], "")]))
               .array.heapify!q{b < a};
   while (heap.length > 1) {
       auto lo = heap.front; heap.removeFront;
       auto hi = heap.front; heap.removeFront;
       lo[1].each!((ref pair) => pair[1] = '0' ~ pair[1]);
       hi[1].each!((ref pair) => pair[1] = '1' ~ pair[1]);
       heap.insert(tuple(lo[0] + hi[0], lo[1] ~ hi[1]));
   }
   return heap.front[1].schwartzSort!q{ tuple(a[1].length, a[0]) };

}

void main() /*@safe*/ {

   immutable s = "this is an example for huffman encoding"d;
   foreach (const p; s.dup.sort().group.encode)
       writefln("'%s'  %s", p[]);

}</lang>

Output:
' '  101
'n'  010
'a'  1001
'e'  1100
'f'  1101
'h'  0001
'i'  1110
'm'  0010
'o'  0011
's'  0111
'g'  00000
'l'  00001
'p'  01100
'r'  01101
't'  10000
'u'  10001
'x'  11110
'c'  111110
'd'  111111

Eiffel

Adapted C# solution. <lang eiffel> class HUFFMAN_NODE[T -> COMPARABLE] inherit COMPARABLE redefine three_way_comparison end create leaf_node, inner_node feature {NONE} leaf_node (a_probability: REAL_64; a_value: T) do probability := a_probability value := a_value is_leaf := true

left := void right := void parent := void end

inner_node (a_left, a_right: HUFFMAN_NODE[T]) do left := a_left right := a_right

a_left.parent := Current a_right.parent := Current a_left.is_zero := true a_right.is_zero := false

probability := a_left.probability + a_right.probability is_leaf := false end

feature probability: REAL_64 value: detachable T


is_leaf: BOOLEAN is_zero: BOOLEAN assign set_is_zero

set_is_zero (a_value: BOOLEAN) do is_zero := a_value end

left: detachable HUFFMAN_NODE[T] right: detachable HUFFMAN_NODE[T] parent: detachable HUFFMAN_NODE[T] assign set_parent

set_parent (a_parent: detachable HUFFMAN_NODE[T]) do parent := a_parent end

is_root: BOOLEAN do Result := parent = void end

bit_value: INTEGER do if is_zero then Result := 0 else Result := 1 end end feature -- comparable implementation is_less alias "<" (other: like Current): BOOLEAN do Result := three_way_comparison (other) = -1 end

three_way_comparison (other: like Current): INTEGER do Result := -probability.three_way_comparison (other.probability) end end

class HUFFMAN create make feature {NONE} make(a_string: STRING) require non_empty_string: a_string.count > 0 local l_queue: HEAP_PRIORITY_QUEUE[HUFFMAN_NODE[CHARACTER]] l_counts: HASH_TABLE[INTEGER, CHARACTER] l_node: HUFFMAN_NODE[CHARACTER] l_left, l_right: HUFFMAN_NODE[CHARACTER] do create l_queue.make (a_string.count) create l_counts.make (10)

across a_string as char loop if not l_counts.has (char.item) then l_counts.put (0, char.item) end l_counts.replace (l_counts.at (char.item) + 1, char.item) end

create leaf_dictionary.make(l_counts.count)

across l_counts as kv loop create l_node.leaf_node ((kv.item * 1.0) / a_string.count, kv.key) l_queue.put (l_node) leaf_dictionary.put (l_node, kv.key) end

from until l_queue.count <= 1 loop l_left := l_queue.item l_queue.remove l_right := l_queue.item l_queue.remove

create l_node.inner_node (l_left, l_right) l_queue.put (l_node) end

root := l_queue.item root.is_zero := false end feature root: HUFFMAN_NODE[CHARACTER] leaf_dictionary: HASH_TABLE[HUFFMAN_NODE[CHARACTER], CHARACTER]

encode(a_value: CHARACTER): STRING require encodable: leaf_dictionary.has (a_value) local l_node: HUFFMAN_NODE[CHARACTER] do Result := "" if attached leaf_dictionary.item (a_value) as attached_node then l_node := attached_node from

until l_node.is_root loop Result.append_integer (l_node.bit_value) if attached l_node.parent as parent then l_node := parent end end

Result.mirror end end end

class APPLICATION create make

feature {NONE} make -- entry point local l_str: STRING huff: HUFFMAN chars: BINARY_SEARCH_TREE_SET[CHARACTER] do l_str := "this is an example for huffman encoding"

create huff.make (l_str)

create chars.make chars.fill (l_str)

from chars.start until chars.off loop print (chars.item.out + ": " + huff.encode (chars.item) + "%N") chars.forth end end end </lang>

Output:
 : 101
a: 1001
c: 01110
d: 01111
e: 1111
f: 1100
g: 01001
h: 11101
i: 1101
l: 10001
m: 0010
n: 000
o: 0011
p: 10000
r: 11100
s: 0110
t: 01000
u: 01011
x: 01010

Erlang

The main part of the code used here is extracted from Michel Rijnders' GitHubGist. See also his blog, for a complete description of the original module. <lang erlang>-module(huffman).

-export([encode/1, decode/2, main/0]).

encode(Text) ->

   Tree  = tree(freq_table(Text)),
   Dict = dict:from_list(codewords(Tree)),
   Code = << <<(dict:fetch(Char, Dict))/bitstring>> || Char <- Text >>,
   {Code, Tree, Dict}.

decode(Code, Tree) ->

   decode(Code, Tree, Tree, []).

main() ->

   {Code, Tree, Dict} = encode("this is an example for huffman encoding"),
   [begin 
       io:format("~s: ",Key),
       print_bits(Value)
    end || {Key, Value} <- lists:sort(dict:to_list(Dict))],
   io:format("encoded: "),
   print_bits(Code),
   io:format("decoded: "),
   io:format("~s\n",[decode(Code, Tree)]).

decode(<<>>, _, _, Result) ->

   lists:reverse(Result);

decode(<<0:1, Rest/bits>>, Tree, {L = {_, _}, _R}, Result) ->

   decode(<<Rest/bits>>, Tree, L, Result);

decode(<<0:1, Rest/bits>>, Tree, {L, _R}, Result) ->

   decode(<<Rest/bits>>, Tree, Tree, [L | Result]);

decode(<<1:1, Rest/bits>>, Tree, {_L, R = {_, _}}, Result) ->

   decode(<<Rest/bits>>, Tree, R, Result);

decode(<<1:1, Rest/bits>>, Tree, {_L, R}, Result) ->

   decode(<<Rest/bits>>, Tree, Tree, [R | Result]).

codewords({L, R}) ->

   codewords(L, <<0:1>>) ++ codewords(R, <<1:1>>).

codewords({L, R}, <<Bits/bits>>) ->

   codewords(L, <<Bits/bits, 0:1>>) ++ codewords(R, <<Bits/bits, 1:1>>);

codewords(Symbol, <<Bits/bitstring>>) ->

   [{Symbol, Bits}].

tree([{N, _} | []]) ->

   N;

tree(Ns) ->

   [{N1, C1}, {N2, C2} | Rest] = lists:keysort(2, Ns),
   tree([{{N1, N2}, C1 + C2} | Rest]).

freq_table(Text) ->

   freq_table(lists:sort(Text), []).

freq_table([], Acc) ->

   Acc;

freq_table([S | Rest], Acc) ->

   {Block, MoreBlocks} = lists:splitwith(fun (X) -> X == S end, Rest),
   freq_table(MoreBlocks, [{S, 1 + length(Block)} | Acc]).

print_bits(<<>>) ->

 io:format("\n");

print_bits(<<Bit:1, Rest/bitstring>>) ->

 io:format("~w", [Bit]),
 print_bits(Rest).</lang>
Output:
 : 111
a: 1011
c: 10010
d: 100111
e: 1010
f: 1101
g: 100110
h: 1000
i: 1100
l: 00001
m: 0101
n: 001
o: 0100
p: 00000
r: 00011
s: 0111
t: 00010
u: 01101
x: 01100
encoded: 0001010001100011111111000111111101100111110100110010110101000000000110101111101010000011111100001101110111010101101100111110100011001001001001111100001100110
decoded: this is an example for huffman encoding

F#

Translation of: OCaml

<lang fsharp>type 'a HuffmanTree =

   | Leaf of int * 'a
   | Node of int * 'a HuffmanTree * 'a HuffmanTree

let freq = function Leaf (f, _) | Node (f, _, _) -> f let freqCompare a b = compare (freq a) (freq b)

let buildTree charFreqs =

   let leaves = List.map (fun (c,f) -> Leaf (f,c)) charFreqs
   let freqSort = List.sortWith freqCompare
   let rec aux = function
       | [] -> failwith "empty list"
       | [a] -> a
       | a::b::tl ->
           let node = Node(freq a + freq b, a, b)
           aux (freqSort(node::tl))
   aux (freqSort leaves)

let rec printTree = function

 | code, Leaf (f, c) ->
     printfn "%c\t%d\t%s" c f (String.concat "" (List.rev code));
 | code, Node (_, l, r) ->
     printTree ("0"::code, l);
     printTree ("1"::code, r)

let () =

 let str = "this is an example for huffman encoding"
 let charFreqs =
   str |> Seq.groupBy id
       |> Seq.map (fun (c, vals) -> (c, Seq.length vals))
       |> Map.ofSeq
        
 let tree = charFreqs |> Map.toList |> buildTree
 printfn "Symbol\tWeight\tHuffman code";
 printTree ([], tree)</lang>
Output:
Symbol	Weight	Huffman code
p	1	00000
r	1	00001
g	1	00010
l	1	00011
n	4	001
m	2	0100
o	2	0101
c	1	01100
d	1	01101
h	2	0111
s	2	1000
x	1	10010
t	1	100110
u	1	100111
f	3	1010
i	3	1011
a	3	1100
e	3	1101
 	6	111

Factor

<lang factor> USING: kernel sequences combinators accessors assocs math hashtables math.order sorting.slots classes formatting prettyprint ;

IN: huffman

! ------------------------------------- ! CLASSES ----------------------------- ! -------------------------------------

TUPLE: huffman-node

   weight element encoding left right ;

! For nodes

<huffman-tnode> ( left right -- huffman )
   huffman-node new [ left<< ] [ swap >>right ] bi ;

! For leafs

<huffman-node> ( element -- huffman )
   1 swap f f f huffman-node boa ;


! -------------------------------------- ! INITIAL HASHTABLE -------------------- ! --------------------------------------

<PRIVATE

! Increment node if it already exists ! Else make it and add it to the hash-table

huffman-gen ( element nodes -- )
   2dup at
   [ [ [ 1 + ] change-weight ] change-at ] 
   [ [ dup <huffman-node> swap ] dip set-at ] if ;

! Curry node-hash. Then each over the seq ! to get the weighted values

(huffman) ( nodes seq -- nodes )
   dup [ [ huffman-gen ] curry each ] dip ;

! --------------------------------------- ! TREE GENERATION ----------------------- ! ---------------------------------------

(huffman-weight) ( node1 node2 -- weight )
   [ weight>> ] dup bi* + ;

! Combine two nodes into the children of a parent ! node which has a weight equal to their collective ! weight

(huffman-combine) ( node1 node2 -- node3 )
   [ (huffman-weight) ]
   [ <huffman-tnode> ] 2bi
   swap >>weight ;

! Generate a tree by combining nodes ! in the priority queue until we're ! left with the root node

(huffman-tree) ( nodes -- tree )
   dup rest empty?
   [ first ] [
       { { weight>> <=> } } sort-by
       [ rest rest ] [ first ]
       [ second ] tri
       (huffman-combine) prefix
       (huffman-tree)
   ] if  ; recursive

! -------------------------------------- ! ENCODING ----------------------------- ! --------------------------------------

(huffman-leaf?) ( node -- bool )
   [ left>>  huffman-node instance? ]
   [ right>> huffman-node instance? ] bi and not ;
(huffman-leaf) ( leaf bit -- )
   swap encoding<< ;

DEFER: (huffman-encoding)

! Recursively walk the nodes left and right

(huffman-node) ( bit nodes -- )
   [ 0 suffix ] [ 1 suffix ] bi
   [ [ left>> ] [ right>> ] bi ] 2dip
   [ swap ] dip
   [ (huffman-encoding) ] 2bi@ ;
(huffman-encoding) ( bit nodes -- )
   over (huffman-leaf?)
   [ (huffman-leaf) ]
   [ (huffman-node) ] if ;

PRIVATE>

! ------------------------------- ! USER WORDS -------------------- ! -------------------------------

huffman-print ( nodes -- )
   "Element" "Weight" "Code" "\n%10s\t%10s\t%6s\n" printf
   { { weight>> >=< } } sort-by
   [  [ encoding>> ] [ element>> ] [ weight>> ] tri
      "%8c\t%7d\t\t" printf  pprint "\n" printf ] each ;
huffman ( sequence -- nodes )
   H{ } clone (huffman) values 
   [ (huffman-tree) { } (huffman-encoding) ] keep ;

! --------------------------------- ! USAGE --------------------------- ! ---------------------------------

! { 1 2 3 4 } huffman huffman-print ! "this is an example of a huffman tree" huffman huffman-print

! Element Weight Code ! 7 { 0 0 0 } ! a 4 { 1 1 1 } ! e 4 { 1 1 0 } ! f 3 { 0 0 1 0 } ! h 2 { 1 0 1 0 } ! i 2 { 0 1 0 1 } ! m 2 { 0 1 0 0 } ! n 2 { 0 1 1 1 } ! s 2 { 0 1 1 0 } ! t 2 { 0 0 1 1 } ! l 1 { 1 0 1 1 1 } ! o 1 { 1 0 1 1 0 } ! p 1 { 1 0 0 0 1 } ! r 1 { 1 0 0 0 0 } ! u 1 { 1 0 0 1 1 } ! x 1 { 1 0 0 1 0 } </lang>

Fantom

<lang fantom> class Node {

 Float probability := 0.0f

}

class Leaf : Node {

 Int character
 new make (Int character, Float probability)
 {
   this.character = character
   this.probability = probability
 }

}

class Branch : Node {

 Node left
 Node right
 new make (Node left, Node right)
 {
   this.left = left
   this.right = right
   probability = this.left.probability + this.right.probability
 }

}

class Huffman {

 Node[] queue := [,]
 Str:Str table := [:]
 new make (Int[] items)
 {
   uniqueItems := items.dup.unique
   uniqueItems.each |Int item|
   {
     num := items.findAll { it == item }.size
     queue.add (Leaf(item, num.toFloat / items.size)) 
   }
   createTree 
   createTable
 }
 Void createTree ()
 {
   while (queue.size > 1)
   {
     queue.sort |a,b| {a.probability <=> b.probability}
     node1 := queue.removeAt (0)
     node2 := queue.removeAt (0)
     queue.add (Branch (node1, node2))
   }
 }
 Void traverse (Node node, Str encoding)
 {
   if (node is Leaf)
   {
     table[(node as Leaf).character.toChar] = encoding
   }
   else // (node is Branch)
   {
     traverse ((node as Branch).left, encoding + "0")
     traverse ((node as Branch).right, encoding + "1")
   }
 }
 Void createTable ()
 {
   if (queue.size != 1) return // error!
   traverse (queue.first, "")
 }
 override Str toStr ()
 {
   result := "Huffman Encoding Table:\n"
   table.keys.sort.each |Str key|
   {
     result += "$key -> ${table[key]}\n"
   }
   return result
 }

}

class Main {

 public static Void main ()
 {
   example := "this is an example for huffman encoding"
   huffman := Huffman (example.chars)
   echo ("From \"$example\"")
   echo (huffman)
 }

} </lang>

Output:
From "this is an example for huffman encoding"
Huffman Encoding Table:
  -> 101
a -> 1100
c -> 10000
d -> 10001
e -> 1101
f -> 1110
g -> 11110
h -> 11111
i -> 1001
l -> 01101
m -> 0011
n -> 000
o -> 0100
p -> 01100
r -> 01110
s -> 0010
t -> 01010
u -> 01111
x -> 01011

Fortran

<lang fortran>! output: ! d-> 00000, t-> 00001, h-> 0001, s-> 0010, ! c-> 00110, x-> 00111, m-> 0100, o-> 0101, ! n-> 011, u-> 10000, l-> 10001, a-> 1001, ! r-> 10100, g-> 101010, p-> 101011, ! e-> 1011, i-> 1100, f-> 1101, -> 111 ! ! 00001|0001|1100|0010|111|1100|0010|111|1001|011| ! 111|1011|00111|1001|0100|101011|10001|1011|111| ! 1101|0101|10100|111|0001|10000|1101|1101|0100| ! 1001|011|111|1011|011|00110|0101|00000|1100|011|101010| ! module huffman implicit none type node

 character (len=1 ), allocatable :: sym(:)
 character (len=10), allocatable :: code(:) 
 integer                         :: freq

contains

 procedure                       :: show => show_node

end type

type queue

 type(node), allocatable :: buf(:)
 integer                 :: n = 0

contains

 procedure :: extractmin
 procedure :: append
 procedure :: siftdown

end type

contains

subroutine siftdown(this, a)

 class (queue)           :: this
 integer                 :: a, parent, child
 associate (x => this%buf)
 parent = a
 do while(parent*2 <= this%n)
   child = parent*2
   if (child + 1 <= this%n) then 
     if (x(child+1)%freq < x(child)%freq ) then
       child = child +1 
     end if
   end if
   if (x(parent)%freq > x(child)%freq) then
     x([child, parent]) = x([parent, child])
     parent = child
   else
     exit
   end if  
 end do      
 end associate

end subroutine

function extractmin(this) result (res)

 class(queue) :: this
 type(node)   :: res
 res = this%buf(1)
 this%buf(1) = this%buf(this%n)
 this%n = this%n - 1
 call this%siftdown(1)

end function

subroutine append(this, x)

 class(queue), intent(inout) :: this
 type(node)                  :: x
 type(node), allocatable     :: tmp(:)
 integer                     :: i
 this%n = this%n +1  
 if (.not.allocated(this%buf)) allocate(this%buf(1))
 if (size(this%buf)<this%n) then
   allocate(tmp(2*size(this%buf)))
   tmp(1:this%n-1) = this%buf
   call move_alloc(tmp, this%buf)
 end if
 this%buf(this%n) = x
 i = this%n
 do 
   i = i / 2
   if (i==0) exit
   call this%siftdown(i)
 end do

end subroutine

function join(a, b) result(c)

 type(node)             :: a, b, c
 integer                :: i, n, n1
 n1 = size(a%sym)
 n = n1 + size(b%sym)  
 c%freq = a%freq + b%freq
 allocate (c%sym(n), c%code(n))
 do i = 1, n1
   c%sym(i) = a%sym(i)
   c%code(i) = "0" // trim(a%code(i))
 end do
 do i = 1, size(b%sym)
   c%sym(i+n1) = b%sym(i)
   c%code(i+n1) = "1" // trim(b%code(i))
 end do

end function

subroutine show_node(this)

 class(node) :: this
 integer     :: i
 write(*, "(*(g0,'-> ',g0,:,', '))", advance="no") &
  (this%sym(i), trim(this%code(i)), i=1,size(this%sym))
 print *

end subroutine

function create(letter, freq) result (this)

 character :: letter
 integer   :: freq
 type(node) :: this
 allocate(this%sym(1), this%code(1))
 this%sym(1) = letter ; this%code(1) = ""
 this%freq = freq

end function end module

program main

 use huffman
 character (len=*), parameter   :: txt = &
  "this is an example for huffman encoding"
 integer                        :: i, freq(0:255) = 0
 type(queue)                    :: Q
 type(node)                     :: x
 do i = 1, len(txt)
   freq(ichar(txt(i:i))) = freq(ichar(txt(i:i))) + 1 
 end do
 do i = 0, 255
   if (freq(i)>0) then
     call Q%append(create(char(i), freq(i)))
   end if
 end do
 do i = 1, Q%n-1
   call Q%append(join(Q%extractmin(),Q%extractmin()))
 end do
 x = Q%extractmin()
 call x%show()
 do i = 1, len(txt) 
   do k = 1, size(x%sym)
     if (x%sym(k)==txt(i:i)) exit
    end do
    write (*, "(a,'|')", advance="no")  trim(x%code(k))
 end do
 print *

end program </lang>

Go

Translation of: Java

<lang go>package main

import (

   "container/heap"
   "fmt"

)

type HuffmanTree interface {

   Freq() int

}

type HuffmanLeaf struct {

   freq  int
   value rune

}

type HuffmanNode struct {

   freq        int
   left, right HuffmanTree

}

func (self HuffmanLeaf) Freq() int {

   return self.freq

}

func (self HuffmanNode) Freq() int {

   return self.freq

}

type treeHeap []HuffmanTree

func (th treeHeap) Len() int { return len(th) } func (th treeHeap) Less(i, j int) bool {

   return th[i].Freq() < th[j].Freq()

} func (th *treeHeap) Push(ele interface{}) {

   *th = append(*th, ele.(HuffmanTree))

} func (th *treeHeap) Pop() (popped interface{}) {

   popped = (*th)[len(*th)-1]
   *th = (*th)[:len(*th)-1]
   return

} func (th treeHeap) Swap(i, j int) { th[i], th[j] = th[j], th[i] }

func buildTree(symFreqs map[rune]int) HuffmanTree {

   var trees treeHeap
   for c, f := range symFreqs {
       trees = append(trees, HuffmanLeaf{f, c})
   }
   heap.Init(&trees)
   for trees.Len() > 1 {
       // two trees with least frequency
       a := heap.Pop(&trees).(HuffmanTree)
       b := heap.Pop(&trees).(HuffmanTree)
       // put into new node and re-insert into queue
       heap.Push(&trees, HuffmanNode{a.Freq() + b.Freq(), a, b})
   }
   return heap.Pop(&trees).(HuffmanTree)

}

func printCodes(tree HuffmanTree, prefix []byte) {

   switch i := tree.(type) {
   case HuffmanLeaf:
       // print out symbol, frequency, and code for this
       // leaf (which is just the prefix)
       fmt.Printf("%c\t%d\t%s\n", i.value, i.freq, string(prefix))
   case HuffmanNode:
       // traverse left
       prefix = append(prefix, '0')
       printCodes(i.left, prefix)
       prefix = prefix[:len(prefix)-1]
       // traverse right
       prefix = append(prefix, '1')
       printCodes(i.right, prefix)
       prefix = prefix[:len(prefix)-1]
   }

}

func main() {

   test := "this is an example for huffman encoding"
   symFreqs := make(map[rune]int)
   // read each symbol and record the frequencies
   for _, c := range test {
       symFreqs[c]++
   }
   // build tree
   tree := buildTree(symFreqs)
   // print out results
   fmt.Println("SYMBOL\tWEIGHT\tHUFFMAN CODE")
   printCodes(tree, []byte{})

}</lang>

Output:
SYMBOL	WEIGHT	HUFFMAN CODE
n	4	000
m	2	0010
o	2	0011
s	2	0100
u	1	01010
p	1	01011
h	2	0110
d	1	01110
c	1	01111
t	1	10000
l	1	10001
x	1	10010
r	1	100110
g	1	100111
i	3	1010
e	3	1011
 	6	110
f	3	1110
a	3	1111
Translation of: Python

<lang go>package main

import (

   "container/heap"
   "fmt"

)

type coded struct {

   sym  rune
   code string

}

type counted struct {

   total int
   syms []coded

}

type cHeap []counted

// satisfy heap.Interface func (c cHeap) Len() int { return len(c) } func (c cHeap) Less(i, j int) bool { return c[i].total < c[j].total } func (c cHeap) Swap(i, j int) { c[i], c[j] = c[j], c[i] } func (c *cHeap) Push(ele interface{}) {

   *c = append(*c, ele.(counted))

} func (c *cHeap) Pop() (popped interface{}) {

   popped = (*c)[len(*c)-1]
   *c = (*c)[:len(*c)-1]
   return

}

func encode(sym2freq map[rune]int) []coded {

   var ch cHeap
   for sym, freq := range sym2freq {
       ch = append(ch, counted{freq, []codedTemplate:Sym: sym})
   }
   heap.Init(&ch)
   for len(ch) > 1 {
       a := heap.Pop(&ch).(counted)
       b := heap.Pop(&ch).(counted)
       for i, c := range a.syms {
           a.syms[i].code = "0" + c.code
       }
       for i, c := range b.syms {
           b.syms[i].code = "1" + c.code
       }
       heap.Push(&ch, counted{a.total + b.total, append(a.syms, b.syms...)})
   }
   return heap.Pop(&ch).(counted).syms

}

const txt = "this is an example for huffman encoding"

func main() {

   sym2freq := make(map[rune]int)
   for _, c := range txt {
       sym2freq[c]++
   }
   table := encode(sym2freq)
   fmt.Println("Symbol  Weight Huffman Code")
   for _, c := range table {
       fmt.Printf("     %c    %d    %s\n", c.sym, sym2freq[c.sym], c.code)
   }

}</lang>

Groovy

Implemented and tested with Groovy 2.3.

<lang groovy> import groovy.transform.*

@Canonical @Sortable(includes = ['freq', 'letter']) class Node {

   String letter
   int freq
   Node left
   Node right
   boolean isLeaf() { left == null && right == null }    

}

Map correspondance(Node n, Map corresp = [:], String prefix = ) {

   if (n.isLeaf()) {
       corresp[n.letter] = prefix ?: '0'
   } else {
       correspondance(n.left,  corresp, prefix + '0')
       correspondance(n.right, corresp, prefix + '1')
   }
   return corresp

}

Map huffmanCode(String message) {

   def queue = message.toList().countBy { it } // char frequencies
       .collect { String letter, int freq ->   // transformed into tree nodes
           new Node(letter, freq) 
       } as TreeSet // put in a queue that maintains ordering
   
   while(queue.size() > 1) {
       def (nodeLeft, nodeRight)  = [queue.pollFirst(), queue.pollFirst()]
       
       queue << new Node(
           freq:   nodeLeft.freq   + nodeRight.freq,
           letter: nodeLeft.letter + nodeRight.letter,
           left: nodeLeft, right: nodeRight
       )
   }
   
   return correspondance(queue.pollFirst())

}

String encode(CharSequence msg, Map codeTable) {

   msg.collect { codeTable[it] }.join()

}

String decode(String codedMsg, Map codeTable, String decoded = ) {

   def pair = codeTable.find { k, v -> codedMsg.startsWith(v) }
   pair ? pair.key + decode(codedMsg.substring(pair.value.size()), codeTable)
        : decoded   

} </lang> Usage: <lang groovy> def message = "this is an example for huffman encoding"

def codeTable = huffmanCode(message) codeTable.each { k, v -> println "$k: $v" }

def encoded = encode(message, codeTable) println encoded

def decoded = decode(encoded, codeTable) println decoded </lang>

Output:
g: 00000
l: 00001
h: 0001
m: 0010
o: 0011
n: 010
p: 01100
r: 01101
s: 0111
t: 10000
u: 10001
a: 1001
 : 101
e: 1100
f: 1101
i: 1110
x: 11110
c: 111110
d: 111111
1000000011110011110111100111101100101010111001111010010010011000000111001011101001101101101000110001110111010010100101010111000101111100011111111111001000000
this is an example for huffman encoding

Haskell

Credits go to huffman where you'll also find a non-tree solution. Uses sorted list as a priority queue. <lang haskell>import Data.List (group, insertBy, sort, sortBy) import Control.Arrow ((&&&), second) import Data.Ord (comparing)

data HTree a

 = Leaf a
 | Branch (HTree a)
          (HTree a)
 deriving (Show, Eq, Ord)

test :: String -> IO () test =

 mapM_ (\(a, b) -> putStrLn ('\ : a : ("' : " ++ b))) .
 serialize . huffmanTree . freq

serialize :: HTree a -> [(a, String)] serialize (Branch l r) =

 (second ('0' :) <$> serialize l) ++ (second ('1' :) <$> serialize r)

serialize (Leaf x) = [(x, "")]

huffmanTree

 :: (Ord w, Num w)
 => [(w, a)] -> HTree a

huffmanTree =

 snd .
 head . until (null . tail) hstep . sortBy (comparing fst) . fmap (second Leaf)

hstep

 :: (Ord a, Num a)
 => [(a, HTree b)] -> [(a, HTree b)]

hstep ((w1, t1):(w2, t2):wts) =

 insertBy (comparing fst) (w1 + w2, Branch t1 t2) wts

freq

 :: Ord a
 => [a] -> [(Int, a)]

freq = fmap (length &&& head) . group . sort

main :: IO () main = test "this is an example for huffman encoding"</lang>

Output:

<lang haskell>'p' : 00000 'r' : 00001 'g' : 00010 'l' : 00011 'n' : 001 'm' : 0100 'o' : 0101 'c' : 01100 'd' : 01101 'h' : 0111 's' : 1000 'x' : 10010 't' : 100110 'u' : 100111 'f' : 1010 'i' : 1011 'a' : 1100 'e' : 1101 ' ' : 111</lang>

Using Set as a priority queue

(might be worth it for bigger alphabets): <lang haskell>import qualified Data.Set as S

htree :: (Ord t, Num t, Ord a) => S.Set (t, HTree a) -> HTree a htree ts | S.null ts_1 = t1

        | otherwise = htree ts_3
          where
            ((w1,t1), ts_1) = S.deleteFindMin ts
            ((w2,t2), ts_2) = S.deleteFindMin ts_1
            ts_3 = S.insert (w1 + w2, Branch t1 t2) ts_2

huffmanTree :: (Ord w, Num w, Ord a) => [(w, a)] -> HTree a huffmanTree = htree . S.fromList . map (second Leaf)</lang>

A non-tree version

This produces the output required without building the Huffman tree at all, by building all the trace strings directly while reducing the histogram: <lang haskell>import Data.List (sortBy, insertBy, sort, group) import Control.Arrow (second, (&&&)) import Data.Ord (comparing)

freq :: Ord a => [a] -> [(Int, a)] freq = map (length &&& head) . group . sort

huffman :: [(Int, Char)] -> [(Char, String)] huffman = reduce . map (\(p, c) -> (p, [(c ,"")])) . sortBy (comparing fst)

 where add (p1, xs1) (p2, xs2) = (p1 + p2, map (second ('0':)) xs1 ++ map (second ('1':)) xs2)
       reduce [(_, ys)]  = sortBy (comparing fst) ys
       reduce (x1:x2:xs) = reduce $ insertBy (comparing fst) (add x1 x2) xs

test s = mapM_ (\(a, b) -> putStrLn ('\ : a : "\' : " ++ b)) . huffman . freq $ s

main = do

   test "this is an example for huffman encoding"</lang>

Icon and Unicon

<lang Icon>record huffnode(l,r,n,c) # internal and leaf nodes record huffcode(c,n,b,i) # encoding table char, freq, bitstring, bits (int)

procedure main()

s := "this is an example for huffman encoding"

Count := huffcount(s) # frequency count Tree := huffTree(Count) # heap and tree

Code := [] # extract encodings CodeT := table() every x := huffBits(Tree) do

  put( Code, CodeT[c] := huffcode( c := x[-1], Count[c].n, b := x[1:-1], integer("2r"||b) ) )


Code := sortf( Code, 1 ) # show table in char order write("Input String : ",image(s)) write(right("char",5), right("freq",5), " encoding" ) every write(right(image((x := !Code).c),5), right(x.n,5), " ", x.b )

end

procedure huffBits(N) # generates huffman bitcodes with trailing character if \N.c then return N.c # . append leaf char code suspend "0" || huffBits(N.l) # . left suspend "1" || huffBits(N.r) # . right end


procedure huffTree(T) # two queue huffman tree method local Q1,Q2,x,n1,n2

Q1 := [] # queue of characters and weights every x := !T do # ensure all are huffnodes

  if type(x) == "huffnode" then put(Q1,x) else runerr(205,x)

Q1 := sortf(Q1,3) # sort by weight ( 3 means by .n )

if *Q1 > 1 then Q2 := [] while *Q1+*\Q2 > 1 do { # While there is more than one node ...

  n1 := if Q1[1] & ( ( Q1[1].n <= Q2[1].n ) | not Q2[1] ) then get(Q1) else get(Q2)  # lowest weight from Q1 or Q2
  n2 := if Q1[1] & ( ( Q1[1].n <= Q2[1].n ) | not Q2[1] ) then get(Q1) else get(Q2)  # lowest weight from Q1 or Q2
  put( Q2, huffnode( n1, n2, n1.n + n2.n ) )   # new weighted node to end of Q2

}

return (\Q2 | Q1)[1] # return the root node end

procedure huffcount(s) # return characters and frequencies in a table of huffnodes by char local c,T

T := table() every c := !s do {

  /T[c] := huffnode(,,0,c) 
  T[c].n +:= 1	  
  }

return T end</lang>

Output:
Input String : "this is an example for huffman encoding"
 char freq encoding
  " "    6 101
  "a"    3 1100
  "c"    1 10000
  "d"    1 10001
  "e"    3 1101
  "f"    3 1110
  "g"    1 11110
  "h"    2 11111
  "i"    3 1001
  "l"    1 01101
  "m"    2 0011
  "n"    4 000
  "o"    2 0100
  "p"    1 01100
  "r"    1 01110
  "s"    2 0010
  "t"    1 01010
  "u"    1 01111
  "x"    1 01011

The following Unicon specific solution takes advantage of the Heap priority queue implementation found in the UniLib Collections package and implements the algorithm given in the problem description. The program produces Huffman codes based on each line of input. <lang Unicon>import Collections

procedure main(A)

   every line := !&input do {
       every (t := table(0))[!line] +:= 1           # Frequency table
       heap := Heap(sort(t), field, "<")            # Initial priority queue
       while heap.size() > 1 do {                   # Tree construction
           every (p1|p2) := heap.get()
           heap.add([&null, p1[2]+p2[2], p1, p2])
           }
       codes := treeWalk(heap.get(),"")             # Get codes from tree
       write("Huffman encoding:")                   # Display codes
       every pair := !sort(codes) do
           write("\t'",\pair[1],"'-> ",pair[2])
       }

end

procedure field(node) # selector function for Heap

   return node[2]  # field to use for priority ordering

end

procedure treeWalk(node, prefix, codeMap)

   /codeMap := table("")
   if /node[1] then {  # interior node
       treeWalk(node[3], prefix||"0", codeMap)
       treeWalk(node[4], prefix||"1", codeMap)
       }
   else codeMap[node[1]] := prefix
   return codeMap

end</lang>

A sample run:

->huffman
this is an example for huffman encoding
Huffman encoding:
        ' '-> 111
        'a'-> 1001
        'c'-> 00110
        'd'-> 00000
        'e'-> 1011
        'f'-> 1101
        'g'-> 101010
        'h'-> 0001
        'i'-> 1100
        'l'-> 10001
        'm'-> 0100
        'n'-> 011
        'o'-> 0101
        'p'-> 101011
        'r'-> 10100
        's'-> 0010
        't'-> 00001
        'u'-> 10000
        'x'-> 00111
aardvarks are ant eaters
Huffman encoding:
        ' '-> 011
        'a'-> 10
        'd'-> 0010
        'e'-> 010
        'k'-> 0011
        'n'-> 0001
        'r'-> 110
        's'-> 1111
        't'-> 1110
        'v'-> 0000
->

HuffStuff provides huffman encoding routines

J

Solution (drawn from the J wiki):

<lang j>hc=: 4 : 0

if. 1=#x do. y
else. ((i{x),+/j{x) hc (i{y),<j{y [ i=. (i.#x) -. j=. 2{./:x end.

)

hcodes=: 4 : 0

assert. x -:&$ y           NB. weights and words have same shape
assert. (0<:x) *. 1=#$x    NB. weights are non-negative
assert. 1 >: L.y           NB. words are boxed not more than once
w=. ,&.> y                 NB. standardized words
assert. w -: ~.w           NB. words are unique
t=. 0 {:: x hc w           NB. minimal weight binary tree
((< S: 0 t) i. w) { <@(1&=)@; S: 1 {:: t

)</lang>

Example:<lang j>  ;"1":L:0(#/.~ (],.(<' '),.hcodes) ,&.>@~.)'this is an example for huffman encoding'

t    0 1 0 1 0
h    1 1 1 1 1
i    1 0 0 1  
s    0 0 1 0  
     1 0 1    
a    1 1 0 0  
n    0 0 0    
e    1 1 0 1  
x    0 1 0 1 1
m    0 0 1 1  
p    0 1 1 0 0
l    0 1 1 0 1
f    1 1 1 0  
o    0 1 0 0  
r    0 1 1 1 0
u    0 1 1 1 1
c    1 0 0 0 0
d    1 0 0 0 1
g    1 1 1 1 0</lang>

Java

This implementation creates an actual tree structure, and then traverses the tree to recover the code. <lang java>import java.util.*;

abstract class HuffmanTree implements Comparable<HuffmanTree> {

   public final int frequency; // the frequency of this tree
   public HuffmanTree(int freq) { frequency = freq; }
   // compares on the frequency
   public int compareTo(HuffmanTree tree) {
       return frequency - tree.frequency;
   }

}

class HuffmanLeaf extends HuffmanTree {

   public final char value; // the character this leaf represents
  
   public HuffmanLeaf(int freq, char val) {
       super(freq);
       value = val;
   }

}

class HuffmanNode extends HuffmanTree {

   public final HuffmanTree left, right; // subtrees
  
   public HuffmanNode(HuffmanTree l, HuffmanTree r) {
       super(l.frequency + r.frequency);
       left = l;
       right = r;
   }

}

public class HuffmanCode {

   // input is an array of frequencies, indexed by character code
   public static HuffmanTree buildTree(int[] charFreqs) {
       PriorityQueue<HuffmanTree> trees = new PriorityQueue<HuffmanTree>();
       // initially, we have a forest of leaves
       // one for each non-empty character
       for (int i = 0; i < charFreqs.length; i++)
           if (charFreqs[i] > 0)
               trees.offer(new HuffmanLeaf(charFreqs[i], (char)i));
       assert trees.size() > 0;
       // loop until there is only one tree left
       while (trees.size() > 1) {
           // two trees with least frequency
           HuffmanTree a = trees.poll();
           HuffmanTree b = trees.poll();
           // put into new node and re-insert into queue
           trees.offer(new HuffmanNode(a, b));
       }
       return trees.poll();
   }
   public static void printCodes(HuffmanTree tree, StringBuffer prefix) {
       assert tree != null;
       if (tree instanceof HuffmanLeaf) {
           HuffmanLeaf leaf = (HuffmanLeaf)tree;
           // print out character, frequency, and code for this leaf (which is just the prefix)
           System.out.println(leaf.value + "\t" + leaf.frequency + "\t" + prefix);
       } else if (tree instanceof HuffmanNode) {
           HuffmanNode node = (HuffmanNode)tree;
           // traverse left
           prefix.append('0');
           printCodes(node.left, prefix);
           prefix.deleteCharAt(prefix.length()-1);
           // traverse right
           prefix.append('1');
           printCodes(node.right, prefix);
           prefix.deleteCharAt(prefix.length()-1);
       }
   }
   public static void main(String[] args) {
       String test = "this is an example for huffman encoding";
       // we will assume that all our characters will have
       // code less than 256, for simplicity
       int[] charFreqs = new int[256];
       // read each character and record the frequencies
       for (char c : test.toCharArray())
           charFreqs[c]++;
       // build tree
       HuffmanTree tree = buildTree(charFreqs);
       // print out results
       System.out.println("SYMBOL\tWEIGHT\tHUFFMAN CODE");
       printCodes(tree, new StringBuffer());
   }

}</lang>

Output:
SYMBOL	WEIGHT	HUFFMAN CODE
d	1	00000
t	1	00001
h	2	0001
s	2	0010
c	1	00110
x	1	00111
m	2	0100
o	2	0101
n	4	011
u	1	10000
l	1	10001
a	3	1001
r	1	10100
g	1	101010
p	1	101011
e	3	1011
i	3	1100
f	3	1101
 	6	111



JavaScript

Translation of: Ruby
Works with: SpiderMonkey
for the print() function.

First, use the Binary Heap implementation from here: http://eloquentjavascript.net/appendix2.html

The Huffman encoder <lang javascript>function HuffmanEncoding(str) {

   this.str = str;
   var count_chars = {};
   for (var i = 0; i < str.length; i++) 
       if (str[i] in count_chars) 
           count_chars[str[i]] ++;
       else 
           count_chars[str[i]] = 1;
   var pq = new BinaryHeap(function(x){return x[0];});
   for (var ch in count_chars) 
       pq.push([count_chars[ch], ch]);
   while (pq.size() > 1) {
       var pair1 = pq.pop();
       var pair2 = pq.pop();
       pq.push([pair1[0]+pair2[0], [pair1[1], pair2[1]]]);
   }
   var tree = pq.pop();
   this.encoding = {};
   this._generate_encoding(tree[1], "");
   this.encoded_string = ""
   for (var i = 0; i < this.str.length; i++) {
       this.encoded_string += this.encoding[str[i]];
   }

}

HuffmanEncoding.prototype._generate_encoding = function(ary, prefix) {

   if (ary instanceof Array) {
       this._generate_encoding(ary[0], prefix + "0");
       this._generate_encoding(ary[1], prefix + "1");
   }
   else {
       this.encoding[ary] = prefix;
   }

}

HuffmanEncoding.prototype.inspect_encoding = function() {

   for (var ch in this.encoding) {
       print("'" + ch + "': " + this.encoding[ch])
   }

}

HuffmanEncoding.prototype.decode = function(encoded) {

   var rev_enc = {};
   for (var ch in this.encoding) 
       rev_enc[this.encoding[ch]] = ch;
   var decoded = "";
   var pos = 0;
   while (pos < encoded.length) {
       var key = ""
       while (!(key in rev_enc)) {
           key += encoded[pos];
           pos++;
       }
       decoded += rev_enc[key];
   }
   return decoded;

}</lang>

And, using the Huffman encoder <lang javascript>var s = "this is an example for huffman encoding"; print(s);

var huff = new HuffmanEncoding(s); huff.inspect_encoding();

var e = huff.encoded_string; print(e);

var t = huff.decode(e); print(t);

print("is decoded string same as original? " + (s==t));</lang>

Output:
this is an example for huffman encoding
'n': 000
's': 0010
'm': 0011
'o': 0100
't': 01010
'x': 01011
'p': 01100
'l': 01101
'r': 01110
'u': 01111
'c': 10000
'd': 10001
'i': 1001
' ': 101
'a': 1100
'e': 1101
'f': 1110
'g': 11110
'h': 11111
0101011111100100101011001001010111000001011101010111100001101100011011101101111001000111010111111011111110111000111100000101110100010000010010001100100011110
this is an example for huffman encoding
is decoded string same as original? true

Julia

<lang julia> abstract type HuffmanTree end

struct HuffmanLeaf <: HuffmanTree

   ch::Char
   freq::Int

end

struct HuffmanNode <: HuffmanTree

   freq::Int
   left::HuffmanTree
   right::HuffmanTree

end

function makefreqdict(s::String)

   d = Dict{Char, Int}()
   for c in s
       if !haskey(d, c)
           d[c] = 1
       else
           d[c] += 1
       end
   end
   d

end

function huffmantree(ftable::Dict)

   trees::Vector{HuffmanTree} = [HuffmanLeaf(ch, fq) for (ch, fq) in ftable]
   while length(trees) > 1
       sort!(trees, lt = (x, y) -> x.freq < y.freq, rev = true)
       least = pop!(trees)
       nextleast = pop!(trees)
       push!(trees, HuffmanNode(least.freq + nextleast.freq, least, nextleast))
   end
   trees[1]

end

printencoding(lf::HuffmanLeaf, code) = println(lf.ch == ' ' ? "space" : lf.ch, "\t", lf.freq, "\t", code)

function printencoding(nd::HuffmanNode, code)

   code *= '0'
   printencoding(nd.left, code)
   code = code[1:end-1]
   code *= '1'
   printencoding(nd.right, code)
   code = code[1:end-1]

end

const msg = "this is an example for huffman encoding"

println("Char\tFreq\tHuffman code")

printencoding(huffmantree(makefreqdict(msg)), "")

</lang>
Output:

Char    Freq    Huffman code
p       1       00000
c       1       00001
g       1       00010
x       1       00011
n       4       001
s       2       0100
h       2       0101
u       1       01100
l       1       01101
m       2       0111
o       2       1000
d       1       10010
r       1       100110
t       1       100111
e       3       1010
f       3       1011
a       3       1100
i       3       1101
space   6       111

Kotlin

Translation of: Java

This implementation creates an actual tree structure, and then traverses the tree to recover the code. <lang kotlin>import java.util.*

abstract class HuffmanTree(var freq: Int) : Comparable<HuffmanTree> {

   override fun compareTo(other: HuffmanTree) = freq - other.freq

}

class HuffmanLeaf(freq: Int, var value: Char) : HuffmanTree(freq)

class HuffmanNode(var left: HuffmanTree, var right: HuffmanTree) : HuffmanTree(left.freq + right.freq)

fun buildTree(charFreqs: IntArray) : HuffmanTree {

   val trees = PriorityQueue<HuffmanTree>()
   charFreqs.forEachIndexed { index, freq ->
       if(freq > 0) trees.offer(HuffmanLeaf(freq, index.toChar()))
   }
   assert(trees.size > 0)
   while (trees.size > 1) {
       val a = trees.poll()
       val b = trees.poll()
       trees.offer(HuffmanNode(a, b))
   }
   return trees.poll()

}

fun printCodes(tree: HuffmanTree, prefix: StringBuffer) {

   when(tree) {
       is HuffmanLeaf -> println("${tree.value}\t${tree.freq}\t$prefix")
       is HuffmanNode -> {
           //traverse left
           prefix.append('0')
           printCodes(tree.left, prefix)
           prefix.deleteCharAt(prefix.lastIndex)
           //traverse right
           prefix.append('1')
           printCodes(tree.right, prefix)
           prefix.deleteCharAt(prefix.lastIndex)
       }
   }

}

fun main(args: Array<String>) {

   val test = "this is an example for huffman encoding"
   val maxIndex = test.max()!!.toInt() + 1
   val freqs = IntArray(maxIndex) //256 enough for latin ASCII table, but dynamic size is more fun
   test.forEach { freqs[it.toInt()] += 1 }
   val tree = buildTree(freqs)
   println("SYMBOL\tWEIGHT\tHUFFMAN CODE")
   printCodes(tree, StringBuffer())

}</lang>

Output:
SYMBOL	WEIGHT	HUFFMAN CODE
d	1	00000
t	1	00001
h	2	0001
s	2	0010
c	1	00110
x	1	00111
m	2	0100
o	2	0101
n	4	011
u	1	10000
l	1	10001
a	3	1001
r	1	10100
g	1	101010
p	1	101011
e	3	1011
i	3	1100
f	3	1101
6	111

Lua

Translation of: Lua

This implementation proceeds in three steps: determine word frequencies, construct the Huffman tree, and finally fold the tree into the codes while outputting them. <lang lua>local build_freqtable = function (data)

 local freq = { }
 for i = 1, #data do
   local cur = string.sub (data, i, i)
   local count = freq [cur] or 0
   freq [cur] = count + 1
 end
 local nodes = { }
 for w, f in next, freq do
   nodes [#nodes + 1] = { word = w, freq = f }
 end
 table.sort (nodes, function (a, b) return a.freq > b.freq end) --- reverse order!
 return nodes

end

local build_hufftree = function (nodes)

 while true do
   local n = #nodes
   local left = nodes [n]
   nodes [n] = nil
   local right = nodes [n - 1]
   nodes [n - 1] = nil
   local new = { freq = left.freq + right.freq, left = left, right = right }
   if n == 2 then return new end
   --- insert new node at correct priority
   local prio = 1
   while prio < #nodes and nodes [prio].freq > new.freq do
     prio = prio + 1
   end
   table.insert (nodes, prio, new)
 end

end

local print_huffcodes do

 local rec_build_huffcodes
 rec_build_huffcodes = function (node, bits, acc)
   if node.word == nil then
     rec_build_huffcodes (node.left,  bits .. "0", acc)
     rec_build_huffcodes (node.right, bits .. "1", acc)
     return acc
   else --- leaf
     acc [#acc + 1] = { node.freq, node.word, bits }
   end
   return acc
 end
 print_huffcodes = function (root)
   local codes = rec_build_huffcodes (root, "", { })
   table.sort (codes, function (a, b) return a [1] < b [1] end)
   print ("frequency\tword\thuffman code")
   for i = 1, #codes do
     print (string.format ("%9d\t‘%s’\t“%s”", table.unpack (codes [i])))
   end
 end

end


local huffcode = function (data)

 local nodes = build_freqtable (data)
 local huff = build_hufftree (nodes)
 print_huffcodes (huff)
 return 0

end

return huffcode "this is an example for huffman encoding"

</lang>

frequency	word	huffman code
        1	‘g’	“01111”
        1	‘p’	“01011”
        1	‘d’	“01100”
        1	‘c’	“01101”
        1	‘t’	“01010”
        1	‘r’	“10000”
        1	‘u’	“11110”
        1	‘x’	“10001”
        1	‘l’	“01110”
        2	‘o’	“11111”
        2	‘m’	“0011”
        2	‘h’	“0010”
        2	‘s’	“0100”
        3	‘i’	“1101”
        3	‘f’	“1110”
        3	‘a’	“1100”
        3	‘e’	“1001”
        4	‘n’	“000”
        6	‘ ’	“101”

M2000 Interpreter

<lang M2000 Interpreter> Module Huffman {

     comp=lambda (a, b) ->{
           =array(a, 0)<array(b, 0)
     }
     module InsertPQ (a, n, &comp) {
           if len(a)=0 then stack a {data n} : exit
           if comp(n, stackitem(a)) then stack a {push n} : exit
            stack a {
                 push n
                 t=2: b=len(a)
                  m=b
                  While t<=b {
                        t1=m
                       m=(b+t) div 2
                       if m=0 then  m=t1 : exit 
                       If comp(stackitem(m),n) then t=m+1:  continue
                       b=m-1
                       m=b
                 }
                 if m>1 then shiftback m
           }
     }
     
     a$="this is an example for huffman encoding"
     
     inventory queue freq
     For i=1 to len(a$)   {
           b$=mid$(a$,i,1)
           if exist(freq, b$) then Return freq, b$:=freq(b$)+1 : continue
           append freq, b$:=1
     }
     sort ascending freq
     b=stack
     K=each(freq)
     LenA=len(a$)
     While k {
           InsertPQ b, (Round(Eval(k)/lenA, 4), eval$(k, k^)), &comp
     }
     While len(b)>1 {
           Stack b {
                Read m1, m2
                InsertPQ b, (Array(m1)+Array(m2), (m1, m2) ), &comp
           }
     }
     Print  "Size of stack object (has only Root):"; len(b)
     Print "Root probability:";Round(Array(Stackitem(b)), 3)
     inventory encode, decode
     Traverse(stackitem(b), "")
     message$=""
     For i=1 to len(a$)
     message$+=encode$(mid$(a$, i, 1))
     Next i
     Print  message$
     j=1
     check$=""
     For i=1 to len(a$)
           d=each(encode)
           While d {
                 code$=eval$(d)
                 if mid$(message$, j, len(code$))=code$ then {
                       check$+=decode$(code$)
                       Print decode$(code$); : j+=len(code$)
                 }
           }
     Next i
     Print
     Print len(message$);" bits ", if$(a$=check$->"Encoding/decoding worked", "Encoding/Decoding failed")
     
     
     Sub Traverse(a, a$)
           local b=array(a,1)      
           if type$(b)="mArray"  Else {
                 Print  @(10); quote$(array$(a, 1));" "; a$,@(20),array(a)
                 Append decode, a$ :=array$(a, 1)
                 Append encode, array$(a, 1):=a$
                 Exit Sub   
           }
           traverse(array(b), a$+"0")
           traverse(array(b,1), a$+"1")
     End Sub

} Huffman </lang>

Output:
"p" 00000      0,0256
"l" 00001      0,0256
"t" 00010      0,0256
"r" 00011      0,0256
"x" 00100      0,0256
"u" 00101      0,0256
"s" 0011       0,0513
"o" 0100       0,0513
"m" 0101       0,0513
"n" 011        0,1026
"h" 1000       0,0513
"c" 10010      0,0256
"g" 100110     0,0256
"d" 100111     0,0256
"e" 1010       0,0769
"a" 1011       0,0769
"i" 1100       0,0769
"f" 1101       0,0769
" " 111        0,1538
0001010001100001111111000011111101101111110100010010110101000000000110101111101010000011111100000101110111010101101101111110100111001001001001111100011100110
this is an example for huffman encoding
      
157 bits    Encoding/decoding worked

Mathematica / Wolfram Language

<lang mathematica>huffman[s_String] := huffman[Characters[s]]; huffman[l_List] := Module[{merge, structure, rules},

  (*merge front two branches. list is assumed to be sorted*)
  merge[k_] := Replace[k, {{a_, aC_}, {b_, bC_}, rest___} :> {{{a, b}, aC + bC}, rest}];
      
  structure = FixedPoint[
     Composition[merge, SortBy[#, Last] &],
     Tally[l]]1, 1;
      
  rules = (# -> Flatten[Position[structure, #] - 1]) & /@ DeleteDuplicates[l];
  {Flatten[l /. rules], rules}];</lang>

Nim

<lang nim>import tables, seqUtils

const sampleString = "this is an example for huffman encoding"

type

   # Following range can be changed to produce Huffman codes on arbitrary alphabet (e.g. ternary codes)
   CodeSymbol = range[0..1]
   HuffCode = seq[CodeSymbol]
   Node = ref object
       f: int
       parent: Node
       case isLeaf: bool
       of true:
           c: char
       else:
           childs: array[CodeSymbol, Node]

proc `<`(a: Node, b: Node): bool =

   # For min operator
   a.f < b.f

proc `$`(hc: HuffCode): string =

   result = ""
   for symbol in hc:
       result &= $symbol

proc freeChildList(tree: seq[Node], parent: Node = nil): seq[Node] =

   # Constructs a sequence of nodes which can be adopted
   # Optional parent parameter can be set to ensure node will not adopt itself
   result = @[]
   for node in tree:
       if node.parent == nil and node != parent:
           result.add(node)

proc connect(parent: Node, child: Node) =

   # Only call this proc when sure that parent has a free child slot
   child.parent = parent
   parent.f += child.f
   for i in parent.childs.low..parent.childs.high:
       if parent.childs[i] == nil:
           parent.childs[i] = child
           return

proc generateCodes(codes: TableRef[char, HuffCode], currentNode: Node, currentCode: HuffCode = @[]) =

   if currentNode.isLeaf:
       let key = currentNode.c
       codes[key] = currentCode
       return
   for i in currentNode.childs.low..currentNode.childs.high:
       if currentNode.childs[i] != nil:
           let newCode = currentCode & i
           generateCodes(codes, currentNode.childs[i], newCode)

proc buildTree(frequencies: CountTable[char]): seq[Node] =

   result = newSeq[Node](frequencies.len)
   for i in result.low..result.high:
       let key = toSeq(frequencies.keys)[i]
       result[i] = Node(f: frequencies[key], isLeaf: true, c: key)
   while result.freeChildList.len > 1:
       let currentNode = new Node
       result.add(currentNode)
       for c in currentNode.childs:
           currentNode.connect(min(result.freeChildList(currentNode)))
           if result.freeChildList.len <= 1:
               break

var sampleFrequencies = initCountTable[char]() for c in sampleString:

   sampleFrequencies.inc(c)

let

   tree = buildTree(sampleFrequencies)
   root = tree.freeChildList[0]

var huffCodes = newTable[char, HuffCode]() generateCodes(huffCodes, root) echo huffCodes</lang>

Output:
{ : 101, a: 1001, c: 01010, d: 01011, e: 1100, f: 1101, g: 01100, h: 11111, i: 1110, l: 01101, m: 0010, n: 000, o: 0011, p: 01110, r: 01111, s: 0100, t: 10000, u: 10001, x: 11110}

Oberon-2

Works with: oo2c

<lang oberon2> MODULE HuffmanEncoding; IMPORT

 Object,
 PriorityQueue,
 Strings,
 Out;

TYPE

 Leaf = POINTER TO LeafDesc;
 LeafDesc = RECORD
   (Object.ObjectDesc)
   c: CHAR;
 END;
 Inner = POINTER TO InnerDesc;
 InnerDesc = RECORD
   (Object.ObjectDesc)
   left,right: Object.Object;
 END;
 

VAR

 str: ARRAY 128 OF CHAR;
 i: INTEGER; 
 f: ARRAY 96 OF INTEGER;
 q: PriorityQueue.Queue;
 a: PriorityQueue.Node;
 b: PriorityQueue.Node;
 c: PriorityQueue.Node;
 h: ARRAY 64 OF CHAR;
 

PROCEDURE NewLeaf(c: CHAR): Leaf; VAR

 x: Leaf;

BEGIN

 NEW(x);x.c := c; RETURN x

END NewLeaf;

PROCEDURE NewInner(l,r: Object.Object): Inner; VAR

 x: Inner;

BEGIN

 NEW(x); x.left := l; x.right := r; RETURN x

END NewInner;


PROCEDURE Preorder(n: Object.Object; VAR x: ARRAY OF CHAR); BEGIN

 IF n IS Leaf THEN
   Out.Char(n(Leaf).c);Out.String(": ");Out.String(h);Out.Ln
 ELSE
   IF n(Inner).left # NIL THEN 
     Strings.Append("0",x);
     Preorder(n(Inner).left,x);
     Strings.Delete(x,(Strings.Length(x) - 1),1)
   END;
   IF n(Inner).right # NIL THEN 
     Strings.Append("1",x);
     Preorder(n(Inner).right,x);
     Strings.Delete(x,(Strings.Length(x) - 1),1)
   END
 END

END Preorder;

BEGIN

 str := "this is an example for huffman encoding";
 
 (* Collect letter frecuencies *)
 i := 0;
 WHILE str[i] # 0X DO INC(f[ORD(CAP(str[i])) - ORD(' ')]);INC(i) END;

 (* Create Priority Queue *)
 NEW(q);q.Clear();
 
 (* Insert into the queue *)
 i := 0;
 WHILE (i < LEN(f)) DO
   IF f[i] # 0 THEN 
     q.Insert(f[i]/Strings.Length(str),NewLeaf(CHR(i + ORD(' ')))) 
   END;
   INC(i)
 END;
 
 (* create tree *)
 WHILE q.Length() > 1 DO 
   q.Remove(a);q.Remove(b);
   q.Insert(a.w + b.w,NewInner(a.d,b.d));
 END;
 (* tree traversal *)
 h[0] := 0X;q.Remove(c);Preorder(c.d,h);

END HuffmanEncoding. </lang>

Output:
D: 00000
T: 00001
H: 0001
S: 0010
C: 00110
X: 00111
M: 0100
O: 0101
N: 011
U: 10000
L: 10001
A: 1001
R: 10100
G: 101010
P: 101011
E: 1011
I: 1100
F: 1101
 : 111

Objective-C

Translation of: Java

This is not purely Objective-C. It uses Apple's Core Foundation library for its binary heap, which admittedly is very ugly. Thus, this only builds on Mac OS X, not GNUstep. <lang objc>#import <Foundation/Foundation.h>


@interface HuffmanTree : NSObject { int freq; } -(instancetype)initWithFreq:(int)f; @property (nonatomic, readonly) int freq; @end

@implementation HuffmanTree @synthesize freq; // the frequency of this tree -(instancetype)initWithFreq:(int)f { if (self = [super init]) { freq = f; } return self; } @end


const void *HuffmanRetain(CFAllocatorRef allocator, const void *ptr) { return (__bridge_retained const void *)(__bridge id)ptr; } void HuffmanRelease(CFAllocatorRef allocator, const void *ptr) { (void)(__bridge_transfer id)ptr; } CFComparisonResult HuffmanCompare(const void *ptr1, const void *ptr2, void *unused) { int f1 = ((__bridge HuffmanTree *)ptr1).freq; int f2 = ((__bridge HuffmanTree *)ptr2).freq; if (f1 == f2) return kCFCompareEqualTo; else if (f1 > f2) return kCFCompareGreaterThan; else return kCFCompareLessThan; }


@interface HuffmanLeaf : HuffmanTree { char value; // the character this leaf represents } @property (readonly) char value; -(instancetype)initWithFreq:(int)f character:(char)c; @end

@implementation HuffmanLeaf @synthesize value; -(instancetype)initWithFreq:(int)f character:(char)c { if (self = [super initWithFreq:f]) { value = c; } return self; } @end


@interface HuffmanNode : HuffmanTree { HuffmanTree *left, *right; // subtrees } @property (readonly) HuffmanTree *left, *right; -(instancetype)initWithLeft:(HuffmanTree *)l right:(HuffmanTree *)r; @end

@implementation HuffmanNode @synthesize left, right; -(instancetype)initWithLeft:(HuffmanTree *)l right:(HuffmanTree *)r { if (self = [super initWithFreq:l.freq+r.freq]) { left = l; right = r; } return self; } @end


HuffmanTree *buildTree(NSCountedSet *chars) {

CFBinaryHeapCallBacks callBacks = {0, HuffmanRetain, HuffmanRelease, NULL, HuffmanCompare}; CFBinaryHeapRef trees = CFBinaryHeapCreate(NULL, 0, &callBacks, NULL);

// initially, we have a forest of leaves // one for each non-empty character for (NSNumber *ch in chars) { int freq = [chars countForObject:ch]; if (freq > 0) CFBinaryHeapAddValue(trees, (__bridge const void *)[[HuffmanLeaf alloc] initWithFreq:freq character:(char)[ch intValue]]); }

NSCAssert(CFBinaryHeapGetCount(trees) > 0, @"String must have at least one character"); // loop until there is only one tree left while (CFBinaryHeapGetCount(trees) > 1) { // two trees with least frequency HuffmanTree *a = (__bridge HuffmanTree *)CFBinaryHeapGetMinimum(trees); CFBinaryHeapRemoveMinimumValue(trees); HuffmanTree *b = (__bridge HuffmanTree *)CFBinaryHeapGetMinimum(trees); CFBinaryHeapRemoveMinimumValue(trees);

// put into new node and re-insert into queue CFBinaryHeapAddValue(trees, (__bridge const void *)[[HuffmanNode alloc] initWithLeft:a right:b]); } HuffmanTree *result = (__bridge HuffmanTree *)CFBinaryHeapGetMinimum(trees); CFRelease(trees); return result; }

void printCodes(HuffmanTree *tree, NSMutableString *prefix) { NSCAssert(tree != nil, @"tree must not be nil"); if ([tree isKindOfClass:[HuffmanLeaf class]]) { HuffmanLeaf *leaf = (HuffmanLeaf *)tree;

// print out character, frequency, and code for this leaf (which is just the prefix) NSLog(@"%c\t%d\t%@", leaf.value, leaf.freq, prefix);

} else if ([tree isKindOfClass:[HuffmanNode class]]) { HuffmanNode *node = (HuffmanNode *)tree;

// traverse left [prefix appendString:@"0"]; printCodes(node.left, prefix); [prefix deleteCharactersInRange:NSMakeRange([prefix length]-1, 1)];

// traverse right [prefix appendString:@"1"]; printCodes(node.right, prefix); [prefix deleteCharactersInRange:NSMakeRange([prefix length]-1, 1)]; } }

int main(int argc, const char * argv[]) {

   @autoreleasepool {

NSString *test = @"this is an example for huffman encoding";

// read each character and record the frequencies NSCountedSet *chars = [[NSCountedSet alloc] init]; int n = [test length]; for (int i = 0; i < n; i++) [chars addObject:@([test characterAtIndex:i])];

// build tree HuffmanTree *tree = buildTree(chars);

// print out results NSLog(@"SYMBOL\tWEIGHT\tHUFFMAN CODE"); printCodes(tree, [NSMutableString string]);

   }
   return 0;

}</lang>

Output:
SYMBOL	WEIGHT	HUFFMAN CODE
g	1	00000
x	1	00001
m	2	0001
d	1	00100
u	1	00101
t	1	00110
r	1	00111
n	4	010
s	2	0110
o	2	0111
p	1	10000
l	1	10001
a	3	1001
 	6	101
f	3	1100
e	3	1101
c	1	11100
h	2	11101
i	3	1111

OCaml

Translation of: Standard ML

We use a Set (which is automatically sorted) as a priority queue.

Works with: OCaml version 4.02+

<lang ocaml>type 'a huffman_tree =

 | Leaf of 'a
 | Node of 'a huffman_tree * 'a huffman_tree

module HSet = Set.Make

 (struct
    type t = int * char huffman_tree (* pair of frequency and the tree *)
    let compare = compare
      (* We can use the built-in compare function to order this: it will order
         first by the first element (frequency) and then by the second (the tree),
         the latter of which we don't care about but which helps prevent elements
         from being equal, since Set does not allow duplicate elements *)
  end);;

let build_tree charFreqs =

 let leaves = HSet.of_list (List.map (fun (c,f) -> (f, Leaf c)) charFreqs) in
 let rec aux trees =
   let f1, a = HSet.min_elt trees in
   let trees' = HSet.remove (f1,a) trees in
   if HSet.is_empty trees' then
     a
   else
     let f2, b = HSet.min_elt trees' in
     let trees = HSet.remove (f2,b) trees' in
     let trees' = HSet.add (f1 + f2, Node (a, b)) trees in
     aux trees
 in
 aux leaves

let rec print_tree code = function

 | Leaf c ->
     Printf.printf "%c\t%s\n" c (String.concat "" (List.rev code));
 | Node (l, r) ->
     print_tree ("0"::code) l;
     print_tree ("1"::code) r

let () =

 let str = "this is an example for huffman encoding" in
 let charFreqs = Hashtbl.create 42 in
 String.iter (fun c ->
     let old =
       try Hashtbl.find charFreqs c
       with Not_found -> 0 in
     Hashtbl.replace charFreqs c (old+1)
   ) str;
 let charFreqs = Hashtbl.fold (fun c f acc -> (c,f)::acc) charFreqs [] in
 let tree = build_tree charFreqs in
 print_string "Symbol\tHuffman code\n";
 print_tree [] tree</lang>

Ol

<lang scheme> (define phrase "this is an example for huffman encoding")

prepare initial probabilities table

(define table (ff->list

  (fold (lambda (ff x)
           (put ff x (+ (ff x 0) 1)))
     {}
     (string->runes phrase))))
just sorter...

(define (resort l)

  (sort (lambda (x y) (< (cdr x) (cdr y))) l))
...to sort table

(define table (resort table))

build huffman tree

(define tree

  (let loop ((table table))
     (if (null? (cdr table))
        (car table)
        (loop (resort (cons
           (cons
              { 1 (car table) 0 (cadr table)}
              (+ (cdar table) (cdadr table)))
           (cddr table)))))))
huffman codes

(define codes

  (map (lambda (i)
        (call/cc (lambda (return)
           (let loop ((prefix #null) (tree tree))
              (if (ff? (car tree))
                 (begin
                    (loop (cons 0 prefix) ((car tree) 0))
                    (loop (cons 1 prefix) ((car tree) 1)))
                 (if (eq? (car tree) i)
                    (return (reverse prefix))))))))
     (map car table)))

</lang>

Output:

<lang scheme> (print "weights: ---------------------------") (for-each (lambda (ch)

     (print (string (car ch)) ": " (cdr ch)))
  (reverse table))

(print "codes: -----------------------------") (map (lambda (char code)

     (print (string char) ": " code))
  (reverse (map car table))
  (reverse codes))

</lang>

weights: ---------------------------
 : 6
n: 4
i: 3
f: 3
e: 3
a: 3
s: 2
o: 2
m: 2
h: 2
x: 1
u: 1
t: 1
r: 1
p: 1
l: 1
g: 1
d: 1
c: 1
codes: -----------------------------
 : (0 0 0)
n: (1 1 0)
i: (0 1 0 0)
f: (0 1 0 1)
e: (0 0 1 0)
a: (0 0 1 1)
s: (0 1 1 1)
o: (1 0 1 0)
m: (1 0 1 1)
h: (1 0 0 0)
x: (0 1 1 0 1)
u: (0 1 1 0 0 0)
t: (0 1 1 0 0 1)
r: (1 1 1 1 0)
p: (1 1 1 1 1)
l: (1 1 1 0 0)
g: (1 1 1 0 1)
d: (1 0 0 1 0)
c: (1 0 0 1 1)

Perl

<lang perl>use 5.10.0; use strict;

  1. produce encode and decode dictionary from a tree

sub walk { my ($node, $code, $h, $rev_h) = @_;

my $c = $node->[0]; if (ref $c) { walk($c->[$_], $code.$_, $h, $rev_h) for 0,1 } else { $h->{$c} = $code; $rev_h->{$code} = $c }

$h, $rev_h }

  1. make a tree, and return resulting dictionaries

sub mktree { my (%freq, @nodes); $freq{$_}++ for split , shift; @nodes = map([$_, $freq{$_}], keys %freq);

do { # poor man's priority queue @nodes = sort {$a->[1] <=> $b->[1]} @nodes; my ($x, $y) = splice @nodes, 0, 2; push @nodes, [[$x, $y], $x->[1] + $y->[1]] } while (@nodes > 1);

walk($nodes[0], , {}, {}) }

sub encode { my ($str, $dict) = @_; join , map $dict->{$_}//die("bad char $_"), split , $str }

sub decode { my ($str, $dict) = @_; my ($seg, @out) = ("");

# append to current segment until it's in the dictionary for (split , $str) { $seg .= $_; my $x = $dict->{$seg} // next; push @out, $x; $seg = ; } die "bad code" if length($seg); join , @out }

my $txt = 'this is an example for huffman encoding'; my ($h, $rev_h) = mktree($txt); for (keys %$h) { print "'$_': $h->{$_}\n" }

my $enc = encode($txt, $h); print "$enc\n";

print decode($enc, $rev_h), "\n";</lang>

Output:
'u': 10000
'd': 01111
'a': 1101
'l': 10001
'i': 1110
'g': 11110
'h': 0100
'r': 01110
' ': 101
'p': 01100
't': 01101
'n': 000
'm': 0011
'x': 01011
'f': 1100
'c': 01010
'o': 0010
's': 11111
'e': 1001
0110101001110111111011110111111011101000101100101011110100110110010001100110111000010011101010100100001100110000111101000101100100001010001001111111000011110
this is an example for huffman encoding

Perl 6

By building a tree

This version uses nested Arrays to build a tree like shown in this diagram, and then recursively traverses the finished tree to accumulate the prefixes.

Works with: rakudo version 2015-12-17

<lang perl6>sub huffman (%frequencies) {

   my @queue = %frequencies.map({ [.value, .key] }).sort;
   while @queue > 1 {
       given @queue.splice(0, 2) -> ([$freq1, $node1], [$freq2, $node2]) {
           @queue = (|@queue, [$freq1 + $freq2, [$node1, $node2]]).sort;
       }
   }
   hash gather walk @queue[0][1], ;

}

multi walk ($node, $prefix) { take $node => $prefix; } multi walk ([$node1, $node2], $prefix) { walk $node1, $prefix ~ '0';

                                        walk $node2, $prefix ~ '1'; }</lang>

Without building a tree

This version uses an Array of Pairs to implement a simple priority queue. Each value of the queue is a Hash mapping from letters to prefixes, and when the queue is reduced the hashes are merged on-the-fly, so that the last one remaining is the wanted Huffman table.

Works with: rakudo version 2015-12-17

<lang perl6>sub huffman (%frequencies) {

   my @queue = %frequencies.map: { .value => (hash .key => ) };
   while @queue > 1 {
       @queue.=sort;
       my $x = @queue.shift;
       my $y = @queue.shift;
       @queue.push: ($x.key + $y.key) => hash $x.value.deepmap('0' ~ *),
                                              $y.value.deepmap('1' ~ *);
   }
   @queue[0].value;

}

  1. Testing

for huffman 'this is an example for huffman encoding'.comb.Bag {

   say "'{.key}' : {.value}";

}

  1. To demonstrate that the table can do a round trip:

say ; my $original = 'this is an example for huffman encoding';

my %encode-key = huffman $original.comb.Bag; my %decode-key = %encode-key.invert; my @codes = %decode-key.keys;

my $encoded = $original.subst: /./, { %encode-key{$_} }, :g; my $decoded = $encoded .subst: /@codes/, { %decode-key{$_} }, :g;

.say for $original, $encoded, $decoded;</lang>

Output:
'x' : 11000
'p' : 01100
'h' : 0001
'g' : 00000
'a' : 1001
'e' : 1101
'd' : 110011
's' : 0111
'f' : 1110
'c' : 110010
'm' : 0010
' ' : 101
'n' : 010
'o' : 0011
'u' : 10001
't' : 10000
'i' : 1111
'r' : 01101
'l' : 00001

this is an example for huffman encoding
1000000011111011110111110111101100101010111011100010010010011000000111011011110001101101101000110001111011100010100101010111010101100100011110011111101000000
this is an example for huffman encoding

Phix

Translation of: Lua

<lang Phix>function store_nodes(object key, object data, integer nodes)

   setd({data,key},0,nodes)
   return 1

end function constant r_store_nodes = routine_id("store_nodes")

function build_freqtable(string data) integer freq = new_dict(),

       nodes = new_dict()
   for i=1 to length(data) do
       integer di = data[i]
       setd(di,getd(di,freq)+1,freq)
   end for
   traverse_dict(r_store_nodes, nodes, freq)
   destroy_dict(freq)
   return nodes

end function

function build_hufftree(integer nodes) sequence lkey, rkey, node integer lfreq, rfreq

   while true do
       lkey = getd_partial_key({0,0},nodes)
       lfreq = lkey[1]
       deld(lkey,nodes)
       rkey = getd_partial_key({0,0},nodes)
       rfreq = rkey[1]
       deld(rkey,nodes)

       node = {lfreq+rfreq,{lkey,rkey}}
       if dict_size(nodes)=0 then exit end if

       setd(node,0,nodes)
   end while
   destroy_dict(nodes)
   return node

end function

procedure build_huffcodes(object node, string bits, integer d)

   {integer freq, object data} = node
   if sequence(data) then
       build_huffcodes(data[1],bits&'0',d)
       build_huffcodes(data[2],bits&'1',d)
   else
       setd(data,{freq,bits},d)
   end if

end procedure

function print_huffcode(integer key, sequence data, integer /*user_data*/)

   printf(1,"'%c' [%d] %s\n",key&data)
   return 1

end function constant r_print_huffcode = routine_id("print_huffcode")

procedure print_huffcodes(integer d)

   traverse_dict(r_print_huffcode, 0, d)

end procedure

function invert_huffcode(integer key, sequence data, integer rd)

   setd(data[2],key,rd)
   return 1

end function constant r_invert_huffcode = routine_id("invert_huffcode")

procedure main(string data)

   if length(data)<2 then ?9/0 end if
   integer nodes = build_freqtable(data)
   sequence huff = build_hufftree(nodes)
   integer d = new_dict()
   build_huffcodes(huff, "", d)
   print_huffcodes(d)
   string encoded = ""
   for i=1 to length(data) do
       encoded &= getd(data[i],d)[2]
   end for
   ?encoded

   integer rd = new_dict()
   traverse_dict(r_invert_huffcode, rd, d)
   string decoded = ""
   integer done = 0
   while done<length(encoded) do
       string key = ""
       integer node = 0
       while node=0 do
           done += 1
           key &= encoded[done]
           node = getd_index(key, rd) 
       end while
       decoded &= getd_by_index(node,rd)
   end while
   ?decoded

end procedure

main("this is an example for huffman encoding")</lang>

Output:
' ' [6] 101
'a' [3] 1001
'c' [1] 01010
'd' [1] 01011
'e' [3] 1100
'f' [3] 1101
'g' [1] 01100
'h' [2] 11111
'i' [3] 1110
'l' [1] 01101
'm' [2] 0010
'n' [4] 000
'o' [2] 0011
'p' [1] 01110
'r' [1] 01111
's' [2] 0100
't' [1] 10000
'u' [1] 10001
'x' [1] 11110
"1000011111111001001011110010010110010001011100111101001001001110011011100101110100110111110111111100011101110100101001000101110000001010001101011111000001100"
"this is an example for huffman encoding"

PHP

Works with: PHP version 5.3+
Translation of: Python
(not exactly)

<lang php><?php function encode($symb2freq) {

   $heap = new SplPriorityQueue;
   $heap->setExtractFlags(SplPriorityQueue::EXTR_BOTH);
   foreach ($symb2freq as $sym => $wt)
       $heap->insert(array($sym => ), -$wt);
   while ($heap->count() > 1) {
       $lo = $heap->extract();
       $hi = $heap->extract();
       foreach ($lo['data'] as &$x)
           $x = '0'.$x;
       foreach ($hi['data'] as &$x)
           $x = '1'.$x;
       $heap->insert($lo['data'] + $hi['data'],
                     $lo['priority'] + $hi['priority']);
   }
   $result = $heap->extract();
   return $result['data'];

}

$txt = 'this is an example for huffman encoding'; $symb2freq = array_count_values(str_split($txt)); $huff = encode($symb2freq); echo "Symbol\tWeight\tHuffman Code\n"; foreach ($huff as $sym => $code)

   echo "$sym\t$symb2freq[$sym]\t$code\n";

?></lang>

Output:
Symbol	Weight	Huffman Code
n	4	000
m	2	0010
o	2	0011
t	1	01000
g	1	01001
x	1	01010
u	1	01011
s	2	0110
c	1	01110
d	1	01111
p	1	10000
l	1	10001
a	3	1001
 	6	101
f	3	1100
i	3	1101
r	1	11100
h	2	11101
e	3	1111

PicoLisp

Using a cons cells (freq . char) for leaves, and two cells (freq left . right) for nodes. <lang PicoLisp>(de prio (Idx)

  (while (cadr Idx) (setq Idx @))
  (car Idx) )

(let (A NIL P NIL L NIL)

  (for C (chop "this is an example for huffman encoding")
     (accu 'A C 1) )                  # Count characters
  (for X A                            # Build index tree as priority queue
     (idx 'P (cons (cdr X) (car X)) T) )
  (while (or (cadr P) (cddr P))       # Remove entries, insert as nodes
     (let (A (car (idx 'P (prio P) NIL))  B (car (idx 'P (prio P) NIL)))
        (idx 'P (cons (+ (car A) (car B)) A B) T) ) )
  (setq P (car P))
  (recur (P L)                        # Traverse and print
     (if (atom (cdr P))
        (prinl (cdr P)  " " L)
        (recurse (cadr P) (cons 0 L))
        (recurse (cddr P) (cons 1 L)) ) ) )</lang>
Output:
n 000
m 0100
o 1100
s 0010
c 01010
d 11010
g 00110
l 10110
p 01110
r 11110
t 00001
u 10001
a 1001
  101
e 0011
f 1011
i 0111
x 01111
h 11111

PL/I

<lang pli>*process source attributes xref or(!);

hencode: Proc Options(main);
/*--------------------------------------------------------------------
* 28.12.013 Walter Pachl  translated from REXX
*-------------------------------------------------------------------*/
Dcl debug Bit(1) Init('0'b);
Dcl (i,j,k) Bin Fixed(15);
Dcl c Char(1);
Dcl s Char(100) Var Init('this is an example for huffman encoding');
Dcl sc Char(1000) Var Init();
Dcl sr Char(100)  Var Init();
Dcl 1 cocc(100),
     2 c  Char(1),
     2 occ Bin Fixed(31);
Dcl cocc_n Bin Fixed(15) Init(0);
dcl 1 node,
     2 id      Bin Fixed(15),         /* Node id               */
     2 c       Char(1),               /* character             */
     2 occ     Bin Fixed(15),         /* number of occurrences */
     2 left    Bin Fixed(15),         /* left child            */
     2 rite    Bin Fixed(15),         /* right child           */
     2 father  Bin Fixed(15),         /* father                */
     2 digit   Pic'9',                /* digit (0 or 1)        */
     2 term    Pic'9';                /* 1=terminal node       */
node=;
Dcl 1 m(100) Like node;
Dcl m_n Bin Fixed(15) Init(0);
Dcl father(100) Bin Fixed(15);
Dcl 1 t(100),
     2 char Char(1),
     2 code Char(20) Var;
Dcl t_n Bin Fixed(15) Init(0);
Do i=1 To length(s);               /* first collect used characters */
  c=substr(s,i,1);                 /* and number of occurrences     */
  Do j=1 To cocc_n;
    If cocc(j).c=c Then Leave;
    End;
  If j<= cocc_n Then
    cocc(j).occ+=1;
  Else Do;
    cocc(j).c=c;
    cocc(j).occ=1;
    cocc_n+=1;
    End;
  End;
Do j=1 To cocc_n;                     /* create initial node list   */
  node.id+=1;
  node.c=cocc(j).c;
  node.occ=cocc(j).occ;
  node.term=1;
  Call add_node;
  End;
If debug Then
  Call show;
Do While(pairs());  /* while there is more than one fatherless node */
  Call mk_node;                       /* create a father node       */
  If debug Then
    Call show;
  End;
Call show;                            /* show the node table        */
Call mk_trans;                        /* create the translate table */
Put Edit('The translate table:')(Skip,a);
Do i=1 To t_n;                        /* show it                    */
  Put Edit(t(i).char,' -> ',t(i).code)(Skip,a,a,a);
  End;
Call encode;                          /* encode the string s -> sc  */
Put Edit('length(sc)=',length(sc))    /* show it                    */
        (Skip,a,f(3));
Do i=1 By 70 To length(sc);
  Put Edit(substr(sc,i,70))(Skip,a);
  End;
Call decode;                          /* decode the string sc -> sr */
Put Edit('input : ',s)(skip,a,a);
Put Edit('result: ',sr)(skip,a,a);
Return;
add_node: Proc;
/*--------------------------------------------------------------------
* Insert the node according to increasing occurrences
*-------------------------------------------------------------------*/
il:
  Do i=1 To m_n;
    If m(i).occ>=node.occ Then Do;
      Do k=m_n To i By -1;
        m(k+1)=m(k);
        End;
      Leave il;
      End;
    End;
  m(i)=node;
  m_n+=1;
End;
show: Proc;
/*--------------------------------------------------------------------
* Show the contents of the node table
*-------------------------------------------------------------------*/
Put Edit('The list of nodes:')(Skip,a);
Put Edit('id c oc  l  r  f d  t')(Skip,a);
Do i=1 To m_n;
  Put Edit(m(i).id,m(i).c,m(i).occ,
           m(i).left,m(i).rite,m(i).father,m(i).digit,m(i).term)
          (Skip,f(2),x(1),a,4(f(3)),f(2),f(3));
  End;
End;
mk_node: Proc;
/*--------------------------------------------------------------------
* construct and store a new intermediate node or the top node
*-------------------------------------------------------------------*/
Dcl z Bin Fixed(15);
node=;
node.id=m_n+1;                /* the next node id                   */
node.c='*';
ni=m_n+1;
loop:
Do i=1 To m_n;                /* loop over node lines               */
 If m(i).father=0 Then Do;    /* a fatherless node                  */
   z=m(i).id;                 /* its id                             */
   If node.left=0 Then Do;    /* new node has no left child         */
      node.left=z;            /* make this the lect child           */
      node.occ=m(i).occ;      /* occurrences                        */
      m(i).father=ni;         /* store father info                  */
      m(i).digit=0;           /* digit 0 to be used                 */
      father(z)=ni;           /* remember z's father (redundant)    */
      End;
    Else Do;                  /* New node has already left child    */
      node.rite=z;            /* make this the right child          */
      node.occ=node.occ+m(i).occ;  /* add in the occurrences        */
      m(i).father=ni;         /* store father info                  */
      m(i).digit=1;           /* digit 1 to be used                 */
      father(z)=ni;           /* remember z's father (redundant)    */
      Leave loop;
      End;
    End;
  End;
Call add_node;
End;
pairs: Proc Returns(Bit(1));
/*--------------------------------------------------------------------
* Return true if there are at least 2 fatherless nodes
*-------------------------------------------------------------------*/
Dcl i   Bin Fixed(15);
Dcl cnt Bin Fixed(15) Init(0);
Do i=1 To m_n;
  If m(i).father=0 Then Do;
    cnt+=1;
    If cnt>1 Then
      Return('1'b);
    End;
  End;
Return('0'b);
End;
mk_trans: Proc;
/*--------------------------------------------------------------------
* Compute the codes for all terminal nodes (characters)
* and store the relation char -> code in array t(*)
*-------------------------------------------------------------------*/
Dcl (i,fi,fid,fidz,node,z) Bin Fixed(15);
Dcl code Char(20) Var;
Do i=1 To m_n;     /* now we loop over all lines representing nodes */
  If m(i).term Then Do;   /* for each terminal node                 */
    code=m(i).digit;      /* its digit is the last code digit       */
    node=m(i).id;         /* its id                                 */
    Do fi=1 To 1000;      /* actually Forever                       */
      fid=father(node);   /* id of father                           */
      If fid>0 Then Do;   /* father exists                          */
        fidz=zeile(fid);  /* line that contains the father          */
        code=m(fidz).digit!!code;    /* prepend the digit           */
        node=fid;         /* look for next father                   */
        End;
      Else                /* no father (we reached the top          */
        Leave;
      End;
    If length(code)>1 Then /* more than one character in input      */
      code=substr(code,2); /* remove the the top node's 0           */
    call dbg(m(i).c!!' -> '!!code); /* character is encoded this way*/
ti_loop:
    Do ti=1 To t_n;
      If t(ti).char>m(i).c Then Do;
        Do tj=t_n To ti By -1
          t(tj+1)=t(tj);
          End;
        Leave ti_loop;
        End;
      End;
    t(ti).char=m(i).c;
    t(ti).code=code;
    t_n+=1;
    Call dbg(t(ti).char!!' -> '!!t(ti).code);
    End;
  End;
End;
zeile: Proc(nid) Returns(Bin Fixed(15));
/*--------------------------------------------------------------------
* find and return line number containing node-id
*-------------------------------------------------------------------*/
Dcl (nid,i) Bin Fixed(15);
do i=1 To m_n;
  If m(i).id=nid Then
    Return(i);
  End;
Stop;
End;
dbg: Proc(txt);
/*--------------------------------------------------------------------
* Show text if debug is enabled
*-------------------------------------------------------------------*/
Dcl txt Char(*);
If debug Then
  Put Skip List(txt);
End;
encode: Proc;
/*--------------------------------------------------------------------
* encode the string s -> sc
*-------------------------------------------------------------------*/
Dcl (i,j) Bin Fixed(15);
Do i=1 To length(s);
  c=substr(s,i,1);
  Do j=1 To t_n;
    If c=t(j).char Then
      Leave;
    End;
  sc=sc!!t(j).code;
  End;
End;
decode: Proc;
/*--------------------------------------------------------------------
* decode the string sc -> sr
*-------------------------------------------------------------------*/
Dcl (i,j) Bin Fixed(15);
Do While(sc>);
  Do j=1 To t_n;
    If substr(sc,1,length(t(j).code))=t(j).code Then
      Leave;
    End;
  sr=sr!!t(j).char;
  sc=substr(sc,length(t(j).code)+1);
  End;
End;
End;</lang>
Output:
The list of nodes:
id c oc  l  r  f d  t
19 g  1  0  0 20 0  1
18 d  1  0  0 20 1  1
17 c  1  0  0 21 0  1
16 u  1  0  0 21 1  1
15 r  1  0  0 22 0  1
12 l  1  0  0 22 1  1
11 p  1  0  0 23 0  1
 9 x  1  0  0 23 1  1
 1 t  1  0  0 24 0  1
23 *  2 11  9 24 1  0
22 *  2 15 12 25 0  0
21 *  2 17 16 25 1  0
20 *  2 19 18 26 0  0
14 o  2  0  0 26 1  1
10 m  2  0  0 27 0  1
 4 s  2  0  0 27 1  1
 2 h  2  0  0 28 0  1
24 *  3  1 23 28 1  0
13 f  3  0  0 29 0  1
 8 e  3  0  0 29 1  1
 6 a  3  0  0 30 0  1
 3 i  3  0  0 30 1  1
27 *  4 10  4 31 0  0
26 *  4 20 14 31 1  0
25 *  4 22 21 32 0  0
 7 n  4  0  0 32 1  1
28 *  5  2 24 33 0  0
30 *  6  6  3 33 1  0
29 *  6 13  8 34 0  0
 5    6  0  0 34 1  1
32 *  8 25  7 35 0  0
31 *  8 27 26 35 1  0
33 * 11 28 30 36 0  0
34 * 12 29  5 36 1  0
35 * 16 32 31 37 0  0
36 * 23 33 34 37 1  0
37 * 39 35 36  0 0  0
The translate table:
  -> 111
a -> 1010
c -> 00010
d -> 01101
e -> 1101
f -> 1100
g -> 01100
h -> 1000
i -> 1011
l -> 00001
m -> 0100
n -> 001
o -> 0111
p -> 100110
r -> 00000
s -> 0101
t -> 10010
u -> 00011
x -> 100111
length(sc)=157
1001010001011010111110110101111101000111111011001111010010010011000001
1101111110001110000011110000001111001100010010100011111101001000100111
01101101100101100
input : this is an example for huffman encoding
result: this is an example for huffman encoding

PowerShell

Works with: PowerShell version 2

<lang PowerShell> function Get-HuffmanEncodingTable ( $String )

   {
   #  Create leaf nodes
   $ID = 0
   $Nodes = [char[]]$String |
       Group-Object |
       ForEach { $ID++; $_ } |
       Select  @{ Label = 'Symbol'  ; Expression = { $_.Name  } },
               @{ Label = 'Count'   ; Expression = { $_.Count } },
               @{ Label = 'ID'      ; Expression = { $ID      } },
               @{ Label = 'Parent'  ; Expression = { 0        } },
               @{ Label = 'Code'    ; Expression = {        } }

   #  Grow stems under leafs
   ForEach ( $Branch in 2..($Nodes.Count) )
       {
       #  Get the two nodes with the lowest count
       $LowNodes = $Nodes | Where Parent -eq 0 | Sort Count | Select -First 2

       #  Create a new stem node
       $ID++
       $Nodes +=  |
           Select  @{ Label = 'Symbol'  ; Expression = {        } },
                   @{ Label = 'Count'   ; Expression = { $LowNodes[0].Count + $LowNodes[1].Count } },
                   @{ Label = 'ID'      ; Expression = { $ID      } },
                   @{ Label = 'Parent'  ; Expression = { 0        } },
                   @{ Label = 'Code'    ; Expression = {        } }

       #  Put the two nodes in the new stem node
       $LowNodes[0].Parent = $ID
       $LowNodes[1].Parent = $ID

       #  Assign 0 and 1 to the left and right nodes
       $LowNodes[0].Code = '0'
       $LowNodes[1].Code = '1'
       }
  
   #  Assign coding to nodes
   ForEach ( $Node in $Nodes[($Nodes.Count-2)..0] )
       {
       $Node.Code = ( $Nodes | Where ID -eq $Node.Parent ).Code + $Node.Code
       }

   $EncodingTable = $Nodes | Where { $_.Symbol } | Select Symbol, Code | Sort Symbol
   return $EncodingTable
   }

  1. Get table for given string

$String = "this is an example for huffman encoding" $HuffmanEncodingTable = Get-HuffmanEncodingTable $String

  1. Display table

$HuffmanEncodingTable | Format-Table -AutoSize

  1. Encode string

$EncodedString = $String ForEach ( $Node in $HuffmanEncodingTable )

   {
   $EncodedString = $EncodedString.Replace( $Node.Symbol, $Node.Code )
   }

$EncodedString </lang>

Output:
Symbol Code 
------ ---- 
       101  
a      1100 
c      01011
d      01100
e      1101 
f      1110 
g      01110
h      11111
i      1001 
l      11110
m      0011 
n      000  
o      0100 
p      10001
r      01111
s      0010 
t      01010
u      01101
x      10000


0101011111100100101011001001010111000001011101100001100001110001111101101101111001000111110111111011011110111000111100000101110100001011010001100100100001110

Prolog

Works with SWI-Prolog <lang Prolog>huffman :- L = 'this is an example for huffman encoding', atom_chars(L, LA), msort(LA, LS), packList(LS, PL), sort(PL, PLS), build_tree(PLS, A), coding(A, [], C), sort(C, SC), format('Symbol~t Weight~t~30|Code~n'), maplist(print_code, SC).

build_tree(R1], [V2|R2]|T], AF) :- V is V1 + V2, A = [V, [V1|R1], [V2|R2, ( T=[] -> AF=A ; sort([A|T], NT), build_tree(NT, AF) ).

coding([_A,FG,FD], Code, CF) :- ( is_node(FG) -> coding(FG, [0 | Code], C1)  ; leaf_coding(FG, [0 | Code], C1) ), ( is_node(FD) -> coding(FD, [1 | Code], C2)  ; leaf_coding(FD, [1 | Code], C2) ), append(C1, C2, CF).

leaf_coding([FG,FD], Code, CF) :- reverse(Code, CodeR), CF = FG, FD, CodeR .

is_node([_V, _FG, _FD]).

print_code([N, Car, Code]):- format('~w :~t~w~t~30|', [Car, N]), forall(member(V, Code), write(V)), nl.

packList([], []). packList([X], 1,X) :- !. packList([X|Rest], [XRun|Packed]):-

   run(X, Rest, XRun, RRest),
   packList(RRest, Packed).

run(V, [], [1,V], []). run(V, [V|LRest], [N1,V], RRest):-

   run(V, LRest, [N, V], RRest),
   N1 is N + 1.

run(V, [Other|RRest], [1,V], [Other|RRest]):-

   dif(V, Other).</lang>
Output:
 ?- huffman.
Symbol          Weight        Code
c :             1             01010
d :             1             01011
g :             1             01100
l :             1             01101
p :             1             01110
r :             1             01111
t :             1             10000
u :             1             10001
x :             1             11110
h :             2             11111
m :             2             0010
o :             2             0011
s :             2             0100
a :             3             1001
e :             3             1100
f :             3             1101
i :             3             1110
n :             4             000
  :             6             101

PureBasic

Works with: PureBasic version 4.50

<lang PureBasic> OpenConsole()

SampleString.s="this is an example for huffman encoding" datalen=Len(SampleString)

Structure ztree

 linked.c
 ischar.c
 char.c
 number.l
 left.l
 right.l

EndStructure

Dim memc.c(datalen) CopyMemory(@SampleString, @memc(0), datalen * SizeOf(Character))

Dim tree.ztree(255)

For i=0 To datalen-1

 tree(memc(i))\char=memc(i)
 tree(memc(i))\number+1
 tree(memc(i))\ischar=1

Next

SortStructuredArray(tree(),#PB_Sort_Descending,OffsetOf(ztree\number),#PB_Integer)

For i=0 To 255

 If tree(i)\number=0
   ReDim tree(i-1)
   Break
 EndIf

Next

dimsize=ArraySize(tree()) Repeat

 min1.l=0
 min2.l=0
 For i=0 To dimsize
   If tree(i)\linked=0
     If tree(i)\number<min1 Or min1=0
       min1=tree(i)\number
       hmin1=i
     ElseIf tree(i)\number<min2 Or min2=0
       min2=tree(i)\number
       hmin2=i
     EndIf
   EndIf
 Next

 If min1=0 Or min2=0
   Break
 EndIf

 dimsize+1
 ReDim tree(dimsize)
 tree(dimsize)\number=tree(hmin1)\number+tree(hmin2)\number
 tree(hmin1)\left=dimsize
 tree(hmin2)\right=dimsize
 tree(hmin1)\linked=1
 tree(hmin2)\linked=1

ForEver

i=0 While tree(i)\ischar=1

 str.s=""
 k=i
 ZNEXT:
 If tree(k)\left<>0
   str="0"+str
   k=tree(k)\left
   Goto ZNEXT
 ElseIf tree(k)\right<>0
   str="1"+str
   k=tree(k)\right
   Goto ZNEXT
 EndIf
 PrintN(Chr(tree(i)\char)+" "+str)
 i+1

Wend Input()

CloseConsole() </lang>

Output:
  110
n 000
e 1010
f 1001
a 1011
i 1110
h 0010
s 11111
o 0011
m 0100
x 01010
u 01011
l 01100
r 01101
c 01110
g 01111
p 10000
t 10001
d 11110

Python

A slight modification of the method outlined in the task description allows the code to be accumulated as the heap is manipulated.

The output is sorted first on length of the code, then on the symbols.

<lang python>from heapq import heappush, heappop, heapify from collections import defaultdict

def encode(symb2freq):

   """Huffman encode the given dict mapping symbols to weights"""
   heap = [[wt, [sym, ""]] for sym, wt in symb2freq.items()]
   heapify(heap)
   while len(heap) > 1:
       lo = heappop(heap)
       hi = heappop(heap)
       for pair in lo[1:]:
           pair[1] = '0' + pair[1]
       for pair in hi[1:]:
           pair[1] = '1' + pair[1]
       heappush(heap, [lo[0] + hi[0]] + lo[1:] + hi[1:])
   return sorted(heappop(heap)[1:], key=lambda p: (len(p[-1]), p))

txt = "this is an example for huffman encoding" symb2freq = defaultdict(int) for ch in txt:

   symb2freq[ch] += 1
  1. in Python 3.1+:
  2. symb2freq = collections.Counter(txt)

huff = encode(symb2freq) print "Symbol\tWeight\tHuffman Code" for p in huff:

   print "%s\t%s\t%s" % (p[0], symb2freq[p[0]], p[1])</lang>
Output:
Symbol  Weight  Huffman Code
    6   101
n   4   010
a   3   1001
e   3   1100
f   3   1101
h   2   0001
i   3   1110
m   2   0010
o   2   0011
s   2   0111
g   1   00000
l   1   00001
p   1   01100
r   1   01101
t   1   10000
u   1   10001
x   1   11110
c   1   111110
d   1   111111

An extension to the method outlined above is given here.


Racket

<lang racket>

  1. lang racket

(require data/heap

        data/bit-vector)

A node is either an interior, or a leaf.
In either case, they record an item with an associated frequency.

(struct node (freq) #:transparent) (struct interior node (left right) #:transparent) (struct leaf node (val) #:transparent)

node<=?
node node -> boolean
Compares two nodes by frequency.

(define (node<=? x y)

 (<= (node-freq x) (node-freq y)))

make-huffman-tree
(listof leaf) -> interior-node

(define (make-huffman-tree leaves)

 (define a-heap (make-heap node<=?))
 (heap-add-all! a-heap leaves)
 (for ([i (sub1 (length leaves))])
   (define min-1 (heap-min a-heap))
   (heap-remove-min! a-heap)
   (define min-2 (heap-min a-heap))
   (heap-remove-min! a-heap)
   (heap-add! a-heap (interior (+ (node-freq min-1) (node-freq min-2))
                               min-1 min-2)))
 (heap-min a-heap))

string->huffman-tree
string -> node
Given a string, produces its huffman tree. The leaves hold the characters
and their relative frequencies.

(define (string->huffman-tree str)

 (define ht (make-hash))
 (define n (sequence-length str))
 (for ([ch str])
    (hash-update! ht ch add1 (λ () 0)))
 (make-huffman-tree
  (for/list ([(k v) (in-hash ht)])
    (leaf (/ v n) k))))

make-encoder
node -> (string -> bit-vector)
Given a huffman tree, generates the encoder function.

(define (make-encoder a-tree)

 (define dict (huffman-tree->dictionary a-tree))
 (lambda (a-str)
   (list->bit-vector (apply append (for/list ([ch a-str]) (hash-ref dict ch))))))

huffman-tree->dictionary
node -> (hashof val (listof boolean))
A helper for the encoder
maps characters to their code sequences.

(define (huffman-tree->dictionary a-node)

 (define ht (make-hash))
 (let loop ([a-node a-node]
            [path/rev '()])
   (cond
     [(interior? a-node)
      (loop (interior-left a-node) (cons #f path/rev))
      (loop (interior-right a-node) (cons #t path/rev))]
     [(leaf? a-node)
       (hash-set! ht (reverse path/rev) (leaf-val a-node))]))
 (for/hash ([(k v) ht])
   (values v k)))

make-decoder
interior-node -> (bit-vector -> string)
Generates the decoder function from the tree.

(define (make-decoder a-tree)

 (lambda (a-bitvector)
   (define-values (decoded/rev _)
     (for/fold ([decoded/rev '()]
                [a-node a-tree])
               ([bit a-bitvector])
       (define next-node 
         (cond
           [(not bit)
            (interior-left a-node)]
           [else
            (interior-right a-node)]))
       (cond [(leaf? next-node)
              (values (cons (leaf-val next-node) decoded/rev) 
                      a-tree)]
             [else
              (values decoded/rev next-node)])))
   (apply string (reverse decoded/rev))))


Example application

(define msg "this is an example for huffman encoding")

(define tree (string->huffman-tree msg))

We can print out the mapping for inspection

(huffman-tree->dictionary tree)

(define encode (make-encoder tree)) (define encoded (encode msg))

Here's what the encoded message looks like

(bit-vector->string encoded)

(define decode (make-decoder tree))

Here's what the decoded message looks like

(decode encoded)</lang>

Red

<lang Red>Red [file: %huffy.red]

message to encode

msg: "this is an example for huffman encoding"

map to collect leave knots per uniq character of message

m: make map! []

knot: make object! [ left: right: none  ;; pointer to left/right sibling code: none  ;; first holds char for debugging, later binary code count: depth: 1  ;;occurence of character - length of branch ]

-----------------------------------------

set-code: func ["recursive function to generate binary code sequence" wknot wcode [string!]] [

-----------------------------------------

either wknot/left = none [ wknot/code: wcode ] [ set-code wknot/left rejoin [wcode "1"] set-code wknot/right rejoin [wcode "0"] ] ] ;;-- end func

-------------------------------

merge-2knots: func ["function to merge 2 knots into 1 new" t [block!]][

-------------------------------

nknot: copy knot  ;; create new knot nknot/count: t/1/count + t/2/count nknot/right: t/1 nknot/left: t/2 nknot/depth: t/1/depth + 1 tab: remove/part t 2 ;; delete first 2 knots insert t nknot  ;; insert new generated knot ] ;;-- end func

count occurence of characters, save in map
m

foreach chr msg [ either k: select/case m chr [ k/count: k/count + 1 ][ put/case m chr nknot: copy knot nknot/code: chr ] ]

create sortable block (=tab) for use as prio queue

foreach k keys-of m [ append tab: [] :m/:k ]

build tree

while [ 1 < length? tab][ sort/compare tab function [a b] [ a/count < b/count or ( a/count = b/count and ( a/depth > b/depth ) ) ] merge-2knots tab ;; merge 2 knots with lowest count / max depth ]

set-code tab/1 "" ;; generate binary codes, save at leave knot

display codes

foreach k sort keys-of m [ print [k " = " m/:k/code] append codes: "" m/:k/code ]

encode orig message string

foreach chr msg [ k: select/case m chr append msg-new: "" k/code ]

print [ "length of encoded msg " length? msg-new] print [ "length of (binary) codes " length? codes ]

print ["orig. message: " msg newline "encoded message: " "^/" msg-new] prin "decoded: "

decode message (destructive! )

while [ not empty? msg-new ][

 foreach [k v] body-of   m [
   if  t: find/match msg-new v/code   [
     prin k
     msg-new: t
   ]
 ]
]

</lang>

Output:
   =  111
a  =  1101
c  =  00101
d  =  00100
e  =  1011
f  =  1100
g  =  10010
h  =  1000
i  =  1010
l  =  00000
m  =  0001
n  =  011
o  =  0101
p  =  00001
r  =  00111
s  =  0100
t  =  100111
u  =  100110
x  =  00110
length of encoded msg  157
length of (binary) codes  85
orig. message:  this is an example for huffman encoding 
encoded message:  
1001111000101001001111010010011111010111111011001101101000100001000001011111110001010011111110001001101100110000011101011111101101100101010100100101001110010
decoded: this is an example for huffman encoding

REXX

<lang rexx>/* REXX ---------------------------------------------------------------

  • 27.12.2013 Walter Pachl
  • 29.12.2013 -"- changed for test of s=xrange('00'x,'ff'x)
  • 14.03.2018 -"- use format instead of right to diagnose size poblems
  • Stem m contains eventually the following node data
  • m.i.0id Node id
  • m.i.0c character
  • m.i.0o number of occurrences
  • m.i.0l left child
  • m.i.0r right child
  • m.i.0f father
  • m.i.0d digit (0 or 1)
  • m.i.0t 1=a terminal node 0=an intermediate or the top node
  • --------------------------------------------------------------------*/

Parse Arg s If s= Then

 s='this is an example for huffman encoding'

Say 'We encode this string:' Say s debug=0 o.=0 c.=0 codel.=0 code.= father.=0 cl= /* list of characters */ do i=1 To length(s)

 Call memorize substr(s,i,1)
 End

If debug Then Do

 Do i=1 To c.0
   c=c.i
   Say i c o.c
   End
 End

n.=0 Do i=1 To c.0

 c=c.i
 n.i.0c=c
 n.i.0o=o.c
 n.i.0id=i
 Call dbg i n.i.0id n.i.0c n.i.0o
 End

n=c.0 /* number of nodes */ m.=0 Do i=1 To n /* construct initial array */

 Do j=1 To m.0                        /* sorted by occurrences      */
   If m.j.0o>n.i.0o Then
     Leave
   End
 Do k=m.0 To j By -1
   k1=k+1
   m.k1.0id=m.k.0id
   m.k1.0c =m.k.0c
   m.k1.0o =m.k.0o
   m.k1.0t =m.k.0t
   End
 m.j.0id=i
 m.j.0c =n.i.0c
 m.j.0o =n.i.0o
 m.j.0t =1
 m.0=m.0+1
 End

If debug Then

 Call show

Do While pairs()>1 /* while there are at least 2 fatherless nodes */

 Call mknode         /* create and fill a new father node           */
 If debug Then
   Call show
 End

Call show c.=0 Do i=1 To m.0 /* now we loop over all lines representing nodes */

 If m.i.0t Then Do   /* for each terminal node                 */
   code=m.i.0d       /* its digit is the last code digit            */
   node=m.i.0id      /* its id                                      */
   Do fi=1 To 1000   /* actually Forever                            */
     fid=father.node           /* id of father                      */
     If fid<>0 Then Do         /* father exists                     */
       fidz=zeile(fid)         /* line that contains the father     */
       code=m.fidz.0d||code    /* prepend the digit                 */
       node=fid                /* look for next father              */
       End
     Else                      /* no father (we reached the top     */
       Leave
     End
   If length(code)>1 Then      /* more than one character in input  */
     code=substr(code,2)       /* remove the the top node's 0       */
   call dbg m.i.0c '->' code   /* character is encoded this way     */
   char=m.i.0c
   code.char=code
   z=codel.0+1
   codel.z=code
   codel.0=z
   char.code=char
   End
 End

Call show_char2code /* show used characters and corresponding codes */

codes.=0 /* now we build the array of codes/characters */ Do j=1 To codel.0

 z=codes.0+1
 code=codel.j
 codes.z=code
 chars.z=char.code
 codes.0=z
 Call dbg codes.z '----->' chars.z
 End

sc= /* here we ecnode the string */ Do i=1 To length(s) /* loop over input */

 c=substr(s,i,1)      /* a character                                */
 sc=sc||code.c        /* append the corresponding code              */
 End

Say 'Length of encoded string:' length(sc) Do i=1 To length(sc) by 70

 Say substr(sc,i,70)
 End

sr= /* now decode the string */ Do si=1 To 999 While sc<>

 Do i=codes.0 To 1 By -1              /* loop over codes            */
   cl=length(codes.i)                 /* length of code             */
   If left(sc,cl)==codes.i Then Do    /* found on top of string     */
     sr=sr||chars.i                   /* append character to result */
     sc=substr(sc,cl+1)               /* cut off the used code      */
     Leave                            /* this was one character     */
     End
   End
 End

Say 'Input ="'s'"' Say 'result="'sr'"'

Exit

show: /*---------------------------------------------------------------------

  • show all lines representing node data
  • --------------------------------------------------------------------*/

Say ' i pp id c f l r d' Do i=1 To m.0

 Say format(i,3) format(m.i.0o,4) format(m.i.0id,3),
         format(m.i.0f,3) format(m.i.0l,3) format(m.i.0r,3) m.i.0d m.i.0t
 End

Call dbg copies('-',21) Return

pairs: Procedure Expose m. /*---------------------------------------------------------------------

  • return number of fatherless nodes
  • --------------------------------------------------------------------*/
 res=0
 Do i=1 To m.0
   If m.i.0f=0 Then
     res=res+1
   End
 Return res

mknode: /*---------------------------------------------------------------------

  • construct and store a new intermediate or the top node
  • --------------------------------------------------------------------*/

new.=0 ni=m.0+1 /* the next node id */ Do i=1 To m.0 /* loop over node lines */

 If m.i.0f=0 Then Do    /* a fatherless node                        */
   z=m.i.0id            /* its id                                   */
   If new.0l=0 Then Do  /* new node has no left child               */
     new.0l=z           /* make this the lect child                 */
     new.0o=m.i.0o      /* occurrences                              */
     m.i.0f=ni          /* store father info                        */
     m.i.0d='0'         /* digit 0 to be used                       */
     father.z=ni        /* remember z's father (redundant)          */
     End
   Else Do              /* New node has already left child          */
     new.0r=z           /* make this the right child                */
     new.0o=new.0o+m.i.0o  /* add in the occurrences                */
     m.i.0f=ni          /* store father info                        */
     m.i.0d=1           /* digit 1 to be used                       */
     father.z=ni        /* remember z's father (redundant)          */
     Leave
     End
   End
 End

Do i=1 To m.0 /* Insert new node according to occurrences */

 If m.i.0o>=new.0o Then Do
   Do k=m.0 To i By -1
     k1=k+1
     m.k1.0id=m.k.0id
     m.k1.0o =m.k.0o
     m.k1.0c =m.k.0c
     m.k1.0l =m.k.0l
     m.k1.0r =m.k.0r
     m.k1.0f =m.k.0f
     m.k1.0d =m.k.0d
     m.k1.0t =m.k.0t
     End
   Leave
   End
 End

m.i.0id=ni m.i.0c ='*' m.i.0o =new.0o m.i.0l =new.0l m.i.0r =new.0r m.i.0t =0 father.ni=0 m.0=ni Return

zeile: /*---------------------------------------------------------------------

  • find and return line number containing node-id
  • --------------------------------------------------------------------*/
 do fidz=1 To m.0
   If m.fidz.0id=arg(1) Then
     Return fidz
   End
 Call dbg arg(1) 'not found'
 Pull .

dbg: /*---------------------------------------------------------------------

  • Show text if debug is enabled
  • --------------------------------------------------------------------*/
 If debug=1 Then
   Say arg(1)
 Return


memorize: Procedure Expose c. o. /*---------------------------------------------------------------------

  • store characters and corresponding occurrences
  • --------------------------------------------------------------------*/
 Parse Arg c
 If o.c=0 Then Do
   z=c.0+1
   c.z=c
   c.0=z
   End
 o.c=o.c+1
 Return

show_char2code: /*---------------------------------------------------------------------

  • show used characters and corresponding codes
  • --------------------------------------------------------------------*/

cl=xrange('00'x,'ff'x) Say 'char --> code' Do While cl<>

 Parse Var cl c +1 cl
 If code.c<> Then
   Say '   'c '-->' code.c
 End

Return</lang>

Output:
We encode this string:
this is an example for huffman encoding
  i   pp  id   c   f   l r d
  1    1   1  20   0   0 0 1
  2    1   9  20   0   0 1 1
  3    1  11  21   0   0 0 1
  4    1  12  21   0   0 1 1
  5    1  15  22   0   0 0 1
  6    1  16  22   0   0 1 1
  7    1  17  23   0   0 0 1
  8    1  18  23   0   0 1 1
  9    1  19  24   0   0 0 1
 10    2  23  24  17  18 1 0
 11    2  22  25  15  16 0 0
 12    2  21  25  11  12 1 0
 13    2  20  26   1   9 0 0
 14    2   2  26   0   0 1 1
 15    2   4  27   0   0 0 1
 16    2  10  27   0   0 1 1
 17    2  14  28   0   0 0 1
 18    3  24  28  19  23 1 0
 19    3   3  29   0   0 0 1
 20    3   6  29   0   0 1 1
 21    3   8  30   0   0 0 1
 22    3  13  30   0   0 1 1
 23    4  27  31   4  10 0 0
 24    4  26  31  20   2 1 0
 25    4  25  32  22  21 0 0
 26    4   7  32   0   0 1 1
 27    5  28  33  14  24 0 0
 28    6  30  33   8  13 1 0
 29    6  29  34   3   6 0 0
 30    6   5  34   0   0 1 1
 31    8  32  35  25   7 0 0
 32    8  31  35  27  26 1 0
 33   11  33  36  28  30 0 0
 34   12  34  36  29   5 1 0
 35   16  35  37  32  31 0 0
 36   23  36  37  33  34 1 0
 37   39  37   0  35  36 0 0
char --> code
     --> 111
   a --> 1101
   c --> 100110
   d --> 100111
   e --> 1010
   f --> 1011
   g --> 10010
   h --> 0111
   i --> 1100
   l --> 00011
   m --> 0101
   n --> 001
   o --> 1000
   p --> 00010
   r --> 00000
   s --> 0100
   t --> 01100
   u --> 00001
   x --> 01101
Length of encoded string: 157
0110001111100010011111000100111110100111110100110111010101000100001110
1011110111000000001110111000011011101101011101001111101000110011010001
00111110000110010
Input ="this is an example for huffman encoding"
result="this is an example for huffman encoding"

Ruby

Uses a
Library: RubyGems
package PriorityQueue

<lang ruby>require 'priority_queue'

def huffman_encoding(str)

 char_count = Hash.new(0)
 str.each_char {|c| char_count[c] += 1}
 
 pq = CPriorityQueue.new
 # chars with fewest count have highest priority
 char_count.each {|char, count| pq.push(char, count)}
 
 while pq.length > 1
   key1, prio1 = pq.delete_min
   key2, prio2 = pq.delete_min
   pq.push([key1, key2], prio1 + prio2)
 end
 
 Hash[*generate_encoding(pq.min_key)]

end

def generate_encoding(ary, prefix="")

 case ary
 when Array
   generate_encoding(ary[0], "#{prefix}0") + generate_encoding(ary[1], "#{prefix}1")
 else
   [ary, prefix]
 end

end

def encode(str, encoding)

 str.each_char.collect {|char| encoding[char]}.join

end

def decode(encoded, encoding)

 rev_enc = encoding.invert
 decoded = ""
 pos = 0
 while pos < encoded.length
   key = ""
   while rev_enc[key].nil?
     key << encoded[pos]
     pos += 1
   end
   decoded << rev_enc[key]
 end
 decoded

end

str = "this is an example for huffman encoding" encoding = huffman_encoding(str) encoding.to_a.sort.each {|x| p x}

enc = encode(str, encoding) dec = decode(enc, encoding) puts "success!" if str == dec</lang>

[" ", "111"]
["a", "1011"]
["c", "00001"]
["d", "00000"]
["e", "1101"]
["f", "1100"]
["g", "00100"]
["h", "1000"]
["i", "1001"]
["l", "01110"]
["m", "10101"]
["n", "010"]
["o", "0001"]
["p", "00101"]
["r", "00111"]
["s", "0110"]
["t", "00110"]
["u", "01111"]
["x", "10100"]
success!

Rust

Adapted C++ solution.

<lang rust> use std::collections::BTreeMap; use std::collections::binary_heap::BinaryHeap;

  1. [derive(Debug, Eq, PartialEq)]

enum NodeKind {

   Internal(Box<Node>, Box<Node>),
   Leaf(char),

}

  1. [derive(Debug, Eq, PartialEq)]

struct Node {

   frequency: usize,
   kind: NodeKind,

}

impl Ord for Node {

   fn cmp(&self, rhs: &Self) -> std::cmp::Ordering {
       rhs.frequency.cmp(&self.frequency)
   }

}

impl PartialOrd for Node {

   fn partial_cmp(&self, rhs: &Self) -> Option<std::cmp::Ordering> {
       Some(self.cmp(&rhs))
   }

}

type HuffmanCodeMap = BTreeMap<char, Vec<u8>>;

fn main() {

   let text = "this is an example for huffman encoding";
   let mut frequencies = BTreeMap::new();
   for ch in text.chars() {
       *frequencies.entry(ch).or_insert(0) += 1;
   }
   let mut prioritized_frequencies = BinaryHeap::new();
   for counted_char in frequencies {
       prioritized_frequencies.push(Node {
           frequency: counted_char.1,
           kind: NodeKind::Leaf(counted_char.0),
       });
   }
   while prioritized_frequencies.len() > 1 {
       let left_child = prioritized_frequencies.pop().unwrap();
       let right_child = prioritized_frequencies.pop().unwrap();
       prioritized_frequencies.push(Node {
           frequency: right_child.frequency + left_child.frequency,
           kind: NodeKind::Internal(Box::new(left_child), Box::new(right_child)),
       });
   }
   let mut codes = HuffmanCodeMap::new();
   generate_codes(
       prioritized_frequencies.peek().unwrap(),
       vec![0u8; 0],
       &mut codes,
   );
   for item in codes {
       print!("{}: ", item.0);
       for bit in item.1 {
           print!("{}", bit);
       }
       println!();
   }

}

fn generate_codes(node: &Node, prefix: Vec<u8>, out_codes: &mut HuffmanCodeMap) {

   match node.kind {
       NodeKind::Internal(ref left_child, ref right_child) => {
           let mut left_prefix = prefix.clone();
           left_prefix.push(0);
           generate_codes(&left_child, left_prefix, out_codes);
           let mut right_prefix = prefix;
           right_prefix.push(1);
           generate_codes(&right_child, right_prefix, out_codes);
       }
       NodeKind::Leaf(ch) => {
           out_codes.insert(ch, prefix);
       }
   }

} </lang>

Output:

 : 110
a: 1001
c: 101010
d: 10001
e: 1111
f: 1011
g: 101011
h: 0101
i: 1110
l: 01110
m: 0011
n: 000
o: 0010
p: 01000
r: 01001
s: 0110
t: 01111
u: 10100
x: 10000

Scala

Works with: scala version 2.8

<lang scala>object Huffman {

 import scala.collection.mutable.{Map, PriorityQueue}
 
 sealed abstract class Tree
 case class Node(left: Tree, right: Tree) extends Tree
 case class Leaf(c: Char) extends Tree
 
 def treeOrdering(m: Map[Tree, Int]) = new Ordering[Tree] { 
    def compare(x: Tree, y: Tree) = m(y).compare(m(x))
 }
 def stringMap(text: String) = text groupBy (x => Leaf(x) : Tree) mapValues (_.length)
 
 def buildNode(queue: PriorityQueue[Tree], map: Map[Tree,Int]) {
   val right = queue.dequeue
   val left = queue.dequeue
   val node = Node(left, right)
   map(node) = map(left) + map(right)
   queue.enqueue(node)
 }
 def codify(tree: Tree, map: Map[Tree, Int]) = {
   def recurse(tree: Tree, prefix: String): List[(Char, (Int, String))] = tree match {
     case Node(left, right) => recurse(left, prefix+"0") ::: recurse(right, prefix+"1")
     case leaf @ Leaf(c) => c -> ((map(leaf), prefix)) :: Nil
   }
   recurse(tree, "")
 }
 def encode(text: String) = {
   val map = Map.empty[Tree,Int] ++= stringMap(text)
   val queue = new PriorityQueue[Tree]()(treeOrdering(map)) ++= map.keysIterator
   
   while(queue.size > 1) {
     buildNode(queue, map)
   }
   codify(queue.dequeue, map)
 }
 
 
 def main(args: Array[String]) {
   val text = "this is an example for huffman encoding"
   val code = encode(text)
   println("Char\tWeight\t\tEncoding")
   code sortBy (_._2._1) foreach { 
     case (c, (weight, encoding)) => println("%c:\t%3d/%-3d\t\t%s" format (c, weight, text.length, encoding)) 
   }
 }

}</lang>

Output:
Char    Weight          Encoding
t:        1/39          011000
p:        1/39          011001
r:        1/39          01101
c:        1/39          01110
x:        1/39          01111
g:        1/39          10110
l:        1/39          10111
u:        1/39          11000
d:        1/39          11001
o:        2/39          1010
s:        2/39          1101
m:        2/39          1110
h:        2/39          1111
f:        3/39          0000
a:        3/39          0001
e:        3/39          0010
i:        3/39          0011
n:        4/39          100
 :        6/39          010

Scala (Alternate version)

Works with: scala version 2.11.7

<lang scala> // this version uses immutable data only, recursive functions and pattern matching object Huffman {

 sealed trait Tree[+A]
 case class Leaf[A](value: A) extends Tree[A]
 case class Branch[A](left: Tree[A], right: Tree[A]) extends Tree[A]
 // recursively build the binary tree needed to Huffman encode the text
 def merge(xs: List[(Tree[Char], Int)]): List[(Tree[Char], Int)] = {
   if (xs.length == 1) xs else {
     val l = xs.head
     val r = xs.tail.head
     val merged = (Branch(l._1, r._1), l._2 + r._2)
     merge((merged :: xs.drop(2)).sortBy(_._2))
   }
 }
 // recursively search the branches of the tree for the required character
 def contains(tree: Tree[Char], char: Char): Boolean = tree match {
   case Leaf(c) => if (c == char) true else false
   case Branch(l, r) => contains(l, char) || contains(r, char)
 }
 // recursively build the path string required to traverse the tree to the required character
 def encodeChar(tree: Tree[Char], char: Char): String = {
   def go(tree: Tree[Char], char: Char, code: String): String = tree match {
     case Leaf(_) => code
     case Branch(l, r) => if (contains(l, char)) go(l, char, code + '0') else go(r, char, code + '1')
   }
   go(tree, char, "")
 }
 def main(args: Array[String]) {
   val text = "this is an example for huffman encoding"
   // transform the text into a list of tuples.
   // each tuple contains a Leaf node containing a unique character and an Int representing that character's weight
   val frequencies = text.groupBy(chars => chars).mapValues(group => group.length).toList.map(x => (Leaf(x._1), x._2)).sortBy(_._2)
   // build the Huffman Tree for this text
   val huffmanTree = merge(frequencies).head._1
   // output the resulting character codes
   println("Char\tWeight\tCode")
   frequencies.foreach(x => println(x._1.value + "\t" + x._2 + s"/${text.length}" + s"\t${encodeChar(huffmanTree, x._1.value)}"))
 }

}</lang>

Char    Weight  Code
x       1/39    01100
t       1/39    01101
u       1/39    00010
g       1/39    00011
l       1/39    00000
p       1/39    00001
c       1/39    100110
r       1/39    100111
d       1/39    10010
s       2/39    0111
m       2/39    0100
h       2/39    0101
o       2/39    1000
e       3/39    1100
f       3/39    1101
a       3/39    1010
i       3/39    1011
n       4/39    001
        6/39    111

Scheme

<lang scheme>(define (char-freq port table)

 (if
  (eof-object? (peek-char port))
  table
  (char-freq port (add-char (read-char port) table))))

(define (add-char char table)

 (cond
  ((null? table) (list (list char 1)))
  ((eq? (caar table) char) (cons (list char (+ (cadar table) 1)) (cdr table)))
  (#t (cons (car table) (add-char char (cdr table))))))

(define (nodeify table)

 (map (lambda (x) (list x '() '())) table))

(define node-freq cadar)

(define (huffman-tree nodes)

 (let ((queue (sort nodes (lambda (x y) (< (node-freq x) (node-freq y))))))
   (if
    (null? (cdr queue))
    (car queue)
    (huffman-tree
     (cons
      (list
       (list 'notleaf (+ (node-freq (car queue)) (node-freq (cadr queue))))
       (car queue)
       (cadr queue))
      (cddr queue))))))

(define (list-encodings tree chars)

 (for-each (lambda (c) (format #t "~a:~a~%" c (encode c tree))) chars))

(define (encode char tree)

 (cond
  ((null? tree) #f)
  ((eq? (caar tree) char) '())
  (#t
   (let ((left (encode char (cadr tree))) (right (encode char (caddr tree))))
     (cond
      ((not (or left right)) #f)
      (left (cons #\1 left))
      (right (cons #\0 right)))))))

(define (decode digits tree)

 (cond
  ((not (eq? (caar tree) 'notleaf)) (caar tree))
  ((eq? (car digits) #\0) (decode (cdr digits) (cadr tree)))
  (#t (decode (cdr digits) (caddr tree)))))

(define input "this is an example for huffman encoding") (define freq-table (char-freq (open-input-string input) '())) (define tree (huffman-tree (nodeify freq-table))) (list-encodings tree (map car freq-table))</lang>

Output:
t:(1 0 0 1 1)
h:(1 0 0 0)
i:(0 0 1 1)
s:(1 0 1 1)
 :(0 0 0)
a:(0 0 1 0)
n:(1 1 0)
e:(0 1 0 1)
x:(1 0 0 1 0)
m:(1 0 1 0)
p:(1 1 1 0 1)
l:(1 1 1 0 0)
f:(0 1 0 0)
o:(0 1 1 1)
r:(1 1 1 1 1)
u:(1 1 1 1 0)
c:(0 1 1 0 0 1)
d:(0 1 1 0 0 0)
g:(0 1 1 0 1)

SETL

<lang SETL>var forest := {}, encTab := {};

plaintext := 'this is an example for huffman encoding';

ft := {}; (for c in plaintext)

 ft(c) +:= 1;

end;

forest := {[f, c]: [c, f] in ft}; (while 1 < #forest)

 [f1, n1] := getLFN();
 [f2, n2] := getLFN();
 forest with:= [f1+f2, [n1,n2]];

end; addToTable(, arb range forest);

(for e = encTab(c))

 print(c, ft(c), e);

end;

print(+/ [encTab(c): c in plaintext]);

proc addToTable(prefix, node);

 if is_tuple node then
   addToTable(prefix + '0', node(1));
   addToTable(prefix + '1', node(2));
 else
   encTab(node) := prefix;
 end;

end proc;

proc getLFN();

 f := min/ domain forest;
 n := arb forest{f};
 forest less:= [f, n];
 return [f, n];

end proc;</lang>

Sidef

<lang ruby>func walk(n, s, h) {

   if (n.contains(:a)) {
       h{n{:a}} = s
       say "#{n{:a}}: #{s}"
       return nil
   }
   walk(n{:0}, s+'0', h)
   walk(n{:1}, s+'1', h)

}

func make_tree(text) {

   var letters = Hash()
   text.each { |c| letters{c} := 0 ++ }
   var nodes = letters.keys.map { |l|
       Hash(a => l, freq => letters{l})
   }
   var n = Hash()
   while (nodes.sort_by!{|c| c{:freq} }.len > 1) {
       n = Hash(:0 => nodes.shift, :1 => nodes.shift)
       n{:freq} = (n{:0}{:freq} + n{:1}{:freq})
       nodes.append(n)
   }
   walk(n, "", n{:tree} = Hash())
   return n

}

func encode(s, t) {

   t = t{:tree}
   s.chars.map{|c| t{c} }.join

}

func decode (enc, tree) {

   var n = tree
   var out = ""
   enc.each {|bit|
       n = n{bit}
       if (n.contains(:a)) {
           out += n{:a}
           n = tree
       }
   }
   return out

}

var text = "this is an example for huffman encoding" var tree = make_tree(text) var enc = encode(text, tree)

say enc say decode(enc, tree)</lang>

Output:
n: 000
s: 0010
o: 0011
h: 0100
l: 01010
g: 01011
x: 01100
c: 01101
d: 01110
u: 01111
p: 10000
t: 10001
i: 1001
 : 101
f: 1100
a: 1101
e: 1110
r: 11110
m: 11111
1000101001001001010110010010101110100010111100110011011111110000010101110101110000111111010101000111111001100111111101000101111000001101001101110100100001011
this is an example for huffman encoding

Standard ML

Works with: SML/NJ

<lang sml>datatype 'a huffman_tree =

        Leaf of 'a
      | Node of 'a huffman_tree * 'a huffman_tree

structure HuffmanPriority = struct

 type priority = int

(* reverse comparison to achieve min-heap *)

 fun compare (a, b) = Int.compare (b, a)
 type item = int * char huffman_tree
 val priority : item -> int = #1

end

structure HPQueue = LeftPriorityQFn (HuffmanPriority)

fun buildTree charFreqs = let

   fun aux trees = let
       val ((f1,a), trees) = HPQueue.remove trees
   in
       if HPQueue.isEmpty trees then
           a
       else let
               val ((f2,b), trees) = HPQueue.remove trees
               val trees = HPQueue.insert ((f1 + f2, Node (a, b)),
                                           trees)
           in
               aux trees
           end
   end
   val trees = HPQueue.fromList (map (fn (c,f) => (f, Leaf c)) charFreqs)

in

   aux trees

end

fun printCodes (revPrefix, Leaf c) =

   print (String.str c ^ "\t" ^
          implode (rev revPrefix) ^ "\n")
 | printCodes (revPrefix, Node (l, r)) = (
   printCodes (#"0"::revPrefix, l);
   printCodes (#"1"::revPrefix, r)
   );

let

   val test = "this is an example for huffman encoding"
   val charFreqs = HashTable.mkTable
                       (HashString.hashString o String.str, op=)
                       (42, Empty)
   val () =
       app (fn c =>
               let val old = getOpt (HashTable.find charFreqs c, 0)
               in HashTable.insert charFreqs (c, old+1)
               end)
           (explode test)
   val tree = buildTree (HashTable.listItemsi charFreqs)

in

   print "SYMBOL\tHUFFMAN CODE\n";
   printCodes ([], tree)

end</lang>

Swift

Rather than a priority queue of subtrees, we use the strategy of two sorted lists, one for leaves and one for nodes, and "merge" them as we iterate through them, taking advantage of the fact that any new nodes we create are bigger than any previously created nodes, so go at the end of the nodes list.

Works with: Swift version 2+

<lang swift>enum HuffmanTree<T> {

 case Leaf(T)
 indirect case Node(HuffmanTree<T>, HuffmanTree<T>)
 
 func printCodes(prefix: String) {
   switch(self) {
   case let .Leaf(c):
     print("\(c)\t\(prefix)")
   case let .Node(l, r):
     l.printCodes(prefix + "0")
     r.printCodes(prefix + "1")
   }
 }

}

func buildTree<T>(freqs: [(T, Int)]) -> HuffmanTree<T> {

 assert(freqs.count > 0, "must contain at least one character")
 // leaves sorted by increasing frequency
 let leaves : [(Int, HuffmanTree<T>)] = freqs.sort { (p1, p2) in p1.1 < p2.1 }.map { (x, w) in (w, .Leaf(x)) }
 // nodes sorted by increasing frequency
 var nodes = [(Int, HuffmanTree<T>)]()
 // iterate through leaves and nodes in order of increasing frequency
 for var i = 0, j = 0; ; {
   assert(i < leaves.count || j < nodes.count)
   // get subtree of least frequency
   var e1 : (Int, HuffmanTree<T>)
   if j == nodes.count || i < leaves.count && leaves[i].0 < nodes[j].0 {
     e1 = leaves[i]
     i++
   } else {
     e1 = nodes[j]
     j++
   }
   
   // if there's no subtrees left, then that one was the answer
   if i == leaves.count && j == nodes.count {
     return e1.1
   }
   
   // get next subtree of least frequency
   var e2 : (Int, HuffmanTree<T>)
   if j == nodes.count || i < leaves.count && leaves[i].0 < nodes[j].0 {
     e2 = leaves[i]
     i++
   } else {
     e2 = nodes[j]
     j++
   }
   // create node from two subtrees
   nodes.append((e1.0 + e2.0, .Node(e1.1, e2.1)))
 }

}

func getFreqs(seq: S) -> [(S.Generator.Element, Int)] {

 var freqs : [S.Generator.Element : Int] = [:]
 for c in seq {
   freqs[c] = (freqs[c] ?? 0) + 1
 }
 return Array(freqs)

}

let str = "this is an example for huffman encoding" let charFreqs = getFreqs(str.characters) let tree = buildTree(charFreqs) print("Symbol\tHuffman code") tree.printCodes("")</lang>

Output:
Symbol	Huffman code
u	00000
t	00001
d	00010
r	00011
c	00100
l	00101
o	0011
m	0100
s	0101
n	011
h	1000
g	10010
p	100110
x	100111
f	1010
a	1011
i	1100
e	1101
 	111

Tcl

Library: Tcllib (Package: struct::prioqueue)

<lang tcl>package require Tcl 8.5 package require struct::prioqueue

proc huffmanEncode {str args} {

   array set opts [concat -dump false $args]
   
   set charcount [dict create]
   foreach char [split $str ""] {
       dict incr charcount $char
   }
   
   set pq [struct::prioqueue -dictionary] ;# want lower values to have higher priority
   dict for {char count} $charcount {
       $pq put $char $count
   }
   
   while {[$pq size] > 1} {
       lassign [$pq peekpriority 2] p1 p2
       $pq put [$pq get 2] [expr {$p1 + $p2}]
   }
   
   set encoding [walkTree [$pq get]]
   
   if {$opts(-dump)} {
       foreach {char huffCode} [lsort -index 1 -stride 2 -command compare $encoding] {
           puts "$char\t[dict get $charcount $char]\t$huffCode"
       }
   }
   $pq destroy
   
   return $encoding

}

proc walkTree {tree {prefix ""}} {

   if {[llength $tree] < 2} {
       return [list $tree $prefix]
   }
   lassign $tree left right
   return [concat [walkTree $left "${prefix}0"] [walkTree $right "${prefix}1"]]

}

proc compare {a b} {

   if {[string length $a] < [string length $b]} {return -1}
   if {[string length $a] > [string length $b]} {return  1}
   return [string compare $a $b]

}

set str "this is an example for huffman encoding"

set encoding [huffmanEncode $str -dump true]

puts $str puts [string map $encoding $str]</lang>

Output:
n	4	000
 	6	101
s	2	0010
m	2	0011
o	2	0100
i	3	1001
a	3	1100
e	3	1101
f	3	1110
t	1	01010
x	1	01011
p	1	01100
l	1	01101
r	1	01110
u	1	01111
c	1	10000
d	1	10001
g	1	11110
h	2	11111
this is an example for huffman encoding
0101011111100100101011001001010111000001011101010111100001101100011011101101111001000111010111111011111110111000111100000101110100010000010010001100100011110

Ursala

following the algorithm given above <lang Ursala>#import std

  1. import nat
  2. import flo

code_table = # takes a training dataset to a table <char: code...>

-+

  *^ ~&v?\~&iNC @v ~&t?\~&h ~&plrDSLrnPlrmPCAS/'01',
  ~&itB->h fleq-<&d; ^C\~&tt @hthPX ^V\~&lrNCC plus@bd,
  ^V(div@rrPlX,~&rlNVNC)^*D(plus:-0.@rS,~&)+ *K2 ^/~&h float+ length+-
  1. cast %csAL

table = code_table 'this is an example for huffman encoding'</lang> a quick walk through the code starting from the bottom:

  • *K2 ^/~&h float+ length compute character frequencies by partitioning the input list of characters by equality, and transforming each equivalence class to a pair containing its member and its cardinality represented as a floating point number
  • ^V(div@rrPlX,~&rlNVNC)^*D(plus:-0.@rS,~&) construct a list of unary trees, one for each character class, with its normalized frequency in the root, and the character in the leaf
  • ~&itB->h while the list contains more than one tree, do the following, and when done take the head of the list
  • fleq-<&d; sort the trees in increasing order by their roots
  • ^C\~&tt @hthPX ^V\~&lrNCC plus@bd change the first two trees in the sorted list to a single binary tree whose root is the sum of their roots
  • *^ visit the following function on each node of the tree obtained from the loop and propagate the results upward from the leaves
  • ~&v?\~&iNC if the node is a leaf, construct a singleton list containing the pair of its root (a character) and the empty string (of bits)
  • @v ~&t?\~&h if there is only a single subtree, propagate the result already obtained for it
  • ~&plrDSLrnPlrmPCAS/'01' otherwise there are two subtrees, hence two lists previously computed results propagating upward, so insert a zero into all of the bit strings in the results on the left, and a one into all the ones on the right, concatenate the left and right results, and propagate the contatenation upward
Output:
<
   `r: '00000',
   `l: '00001',
   `c: '00010',
   `u: '00011',
   `n: '001',
   `m: '0100',
   `h: '0101',
   `g: '01100',
   `d: '01101',
   `o: '0111',
   `s: '1000',
   `t: '10010',
   `p: '100110',
   `x: '100111',
   `a: '1010',
   `f: '1011',
   `i: '1100',
   `e: '1101',
   ` : '111'>

zkl

This code was adapted from Perl, Python and most of the other examples. <lang zkl>fcn buildHuffman(text){ //-->(encode dictionary, decode dictionary)

  ft:=Dictionary();
  foreach c in (text){ ft[c]=ft.find(c,0)+1 }  // leafs w/count
  // build the tree, which is a list of lists of ...
  tree:=ft.pump(List,fcn([(c,cnt)]){  //-->L( (cnt, ((sym,code))), ...)
     L(cnt, L(L(c,"")))
  }).copy();	// make it writable
  while(tree.len()>1){  // fake up a [lame] priorty queue
     tree=tree.sort(fcn(a,b){ a[0]>b[0] }); //prioritize high to low
     a,b:=tree.pop(-2,2);	        //remove 2 least frequent symbols
     mc:=fcn(n,c){ n[1] = c + n[1]; };	//(sym,code),"0"|"1"
     a[1].apply2(mc,"0"); b[1].apply2(mc,"1"); // mc(a[1],"0")
     tree.append( L(a[0]+b[0],a[1].extend(b[1])) ); //(a,b)-->new node
  }//-->L(L(39, L( L(" ","000"),L("e","0010"),L("a","0011") ...
  tree=tree[0][1].pump(List,fcn(i){ // flatten rather than traverse

if(T.isType(i))return(Void.Recurse,i,self.fcn); i });

  encodeTable:=tree.toDictionary();  // symbol:Huffman code
  decodeTable:=encodeTable.pump(Dictionary(),"reverse"); // code:symbol
  return(encodeTable,decodeTable);

}</lang> <lang zkl>fcn encode(text,table){ text.pump(String,table.get) } fcn decode(bits,table){ // this is a horrible decoder, for testing only

  w:=bits.walker(); sink:=Sink(String);
  try{ s:=""; while(1){
     s+=w.next(); if(c:=table.find(s)) { sink.write(c); s=""; }
  }}catch(TheEnd){}
  sink.close();

}</lang> <lang zkl>text:="this is an example for huffman encoding"; encodeTable,decodeTable := buildHuffman(text); encodeTable.pump(Console.println,fcn(kv){"%s : %s".fmt(kv.xplode())});

e:=encode(text,encodeTable); "Encode %d characters (%d bits) to %d bits (%d bytes):"

  .fmt(text.len(),text.len()*8,e.len(),(e.len()+7)/8).println();

println(e);

0'|Bits decoded to: "%s"|.fmt(decode(e,decodeTable)).println();</lang>

Output:
a : 0011
c : 10101
d : 10100
e : 0010
f : 0110
g : 10111
h : 1000
i : 0101
l : 10110
m : 1001
n : 110
o : 01000
p : 11111
r : 11100
s : 0111
t : 01001
u : 11101
x : 11110
  : 000
Encode 39 characters (312 bits) to 157 bits (20 bytes):
0100110000101011100001010111000001111000000101111000111001111111011000100000110010001110000010001110101100110100100111100000010110101010100010100010111010111
Bits decoded to: "this is an example for huffman encoding"