DATA COMPRESSION AND ITS TYPES
Data Compression, also known as source coding, is the process of encoding or converting data in such a way that it consumes less memory space. Data compression reduces the number of resources required to store and transmit data.
It can be done in two ways- lossless compression and lossy compression. Lossy compression reduces the size of data by removing unnecessary information, while there is no data loss in lossless compression.

WHAT IS SHANNON FANO CODING?
Shannon Fano Algorithm is an entropy encoding technique for lossless data compression of multimedia. Named after Claude Shannon and Robert Fano, it assigns a code to each symbol based on their probabilities of occurrence. It is a variable length encoding scheme, that is, the codes assigned to the symbols will be of varying length.

HOW DOES IT WORK?
The steps of the algorithm are as follows:

1.      Create a list of probabilities or frequency counts for the given set of symbols so that the relative frequency of occurrence of each symbol is known.

2.      Sort the list of symbols in decreasing order of probability, the most probable ones to the left and least probable to the right.

3.      Split the list into two parts, with the total probability of both the parts being as close to each other as possible.

4.      Assign the value 0 to the left part and 1 to the right part.

5.      Repeat the steps 3 and 4 for each part, until all the symbols are split into individual subgroups.

The Shannon codes are considered accurate if the code of each symbol is unique.

EXAMPLE:
Given task is to construct Shannon codes for the given set of symbols using the Shannon-Fano lossless compression technique.

Step:

Tree:

Solution:

Let P(x) be the probability of occurrence of symbol x:

1.      Upon arranging the symbols in decreasing order of probability:

P(D) + P(B) = 0.30 + 0.2 = 0.58

and,

P(A) + P(C) + P(E) = 0.22 + 0.15 + 0.05 = 0.42

And since thealmost equally split the table, the most is dividedit the blockquote table isblockquotento

{D, B} and {A, C, E}

and assign them the values 0 and 1 respectively.

Step:

Tree:

2.      Now, in {D, B} group,

P(D) = 0.30 and P(B) = 0.28

which means that P(D)~P(B), so divide {D, B} into {D} and {B} and assign 0 to D and 1 to B.

Step:

Tree:

3.      In {A, C, E} group,

P(A) = 0.22 and P(C) + P(E) = 0.20

So the group is divided into

{A} and {C, E}

and they are assigned values 0 and 1 respectively.

4.      In {C, E} group,

P(C) = 0.15 and P(E) = 0.05

So divide them into {C} and {E} and assign 0 to {C} and 1 to {E}

Step:

Tree:

Note: The splitting is now stopped as each symbol is separated now.

The Shannon codes for the set of symbols are:

As it can be seen, these are all unique and of varying lengths.

Below is the implementation of the above approach:

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// C++ program for Shannon Fano Algorithm  // include header files#include <bits/stdc++.h>using namespace std;  // declare structure nodestruct node {      // for storing symbol    string sym;      // for storing probability or frquency    float pro;    int arr[20];    int top;} p[20];  typedef struct node node;  // function to find shannon codevoid shannon(int l, int h, node p[]){    float pack1 = 0, pack2 = 0, diff1 = 0, diff2 = 0;    int i, d, k, j;    if ((l + 1) == h || l == h || l > h) {        if (l == h || l > h)            return;        p[h].arr[++(p[h].top)] = 0;        p[l].arr[++(p[l].top)] = 1;        return;    }    else {        for (i = l; i <= h – 1; i++)            pack1 = pack1 + p[i].pro;        pack2 = pack2 + p[h].pro;        diff1 = pack1 – pack2;        if (diff1 < 0)            diff1 = diff1 * -1;        j = 2;        while (j != h – l + 1) {            k = h – j;            pack1 = pack2 = 0;            for (i = l; i <= k; i++)                pack1 = pack1 + p[i].pro;            for (i = h; i > k; i–)                pack2 = pack2 + p[i].pro;            diff2 = pack1 – pack2;            if (diff2 < 0)                diff2 = diff2 * -1;            if (diff2 >= diff1)                break;            diff1 = diff2;            j++;        }        k++;        for (i = l; i <= k; i++)            p[i].arr[++(p[i].top)] = 1;        for (i = k + 1; i <= h; i++)            p[i].arr[++(p[i].top)] = 0;          // Invoke shannon function        shannon(l, k, p);        shannon(k + 1, h, p);    }}  // Function to sort the symbols// based on their probability or frequencyvoid sortByProbability(int n, node p[]){    int i, j;    node temp;    for (j = 1; j <= n – 1; j++) {        for (i = 0; i < n – 1; i++) {            if ((p[i].pro) > (p[i + 1].pro)) {                temp.pro = p[i].pro;                temp.sym = p[i].sym;                  p[i].pro = p[i + 1].pro;                p[i].sym = p[i + 1].sym;                  p[i + 1].pro = temp.pro;                p[i + 1].sym = temp.sym;            }        }    }}  // function to display shannon codesvoid display(int n, node p[]){    int i, j;    cout << “\n\n\n\tSymbol\tProbability\tCode”;    for (i = n – 1; i >= 0; i–) {        cout << “\n\t” << p[i].sym << “\t\t” << p[i].pro << “\t”;        for (j = 0; j <= p[i].top; j++)            cout << p[i].arr[j];    }}  // Driver codeint main(){    int n, i, j;    float total = 0;    string ch;    node temp;      // Input number of symbols    cout << “Enter number of symbols\t: “;    n = 5;    cout << n << endl;      // Input symbols    for (i = 0; i < n; i++) {        cout << “Enter symbol ” << i + 1 << ” : “;        ch = (char)(65 + i);        cout << ch << endl;          // Insert the symbol to node        p[i].sym += ch;    }      // Input probability of symbols    float x[] = { 0.22, 0.28, 0.15, 0.30, 0.05 };    for (i = 0; i < n; i++) {        cout << “\nEnter probability of ” << p[i].sym << ” : “;        cout << x[i] << endl;          // Insert the value to node        p[i].pro = x[i];        total = total + p[i].pro;          // checking max probability        if (total > 1) {            cout << “Invalid. Enter new values”;            total = total – p[i].pro;            i–;        }    }      p[i].pro = 1 – total;      // Sorting the symbols based on    // their probability or frequency    sortByProbability(n, p);      for (i = 0; i < n; i++)        p[i].top = -1;      // Find the shannon code    shannon(0, n – 1, p);      // Display the codes    display(n, p);    return 0;}

Output:

Enter number of symbols    : 5

Enter symbol 1 : A

Enter symbol 2 : B

Enter symbol 3 : C

Enter symbol 4 : D

Enter symbol 5 : E

Enter probability of A : 0.22

Enter probability of B : 0.28

Enter probability of C : 0.15

Enter probability of D : 0.3

Enter probability of E : 0.05

    Symbol    Probability    Code

    D        0.3    00

    B        0.28    01

    A        0.22    10

    C        0.15    110

    E        0.05    111