The most common types of error-correcting codes used in RAM are based on the codes devised by R. W. Hamming. In the Hamming code, k parity bits are added to an n-bit data word, forming a new word of n+k bits. The bit positions are numbered in sequence from 1 to n+k. Those positions numbered with powers of two are reserved for the parity bits. The remaining bits are the data bits. The code can be used with words of any length.

## General Algorithm of Hamming code

The Hamming Code is simply the use of extra parity bits to allow the identification of an error.

1. Write the bit positions starting from 1 in binary form (1, 10, 11, 100, etc).
2. All the bit positions that are a power of 2 are marked as parity bits (1, 2, 4, 8, etc).
3. All the other bit positions are marked as data bits.
4. Each data bit is included in a unique set of parity bits, as determined by its bit position in binary form.
• Parity bit 1 covers all the bit's positions whose binary representation includes a 1 in the least significant position (1, 3, 5, 7, 9, 11, etc).
• Parity bit 2 covers all the bit's positions whose binary representation includes a 1 in the second position from the least significant bit (2, 3, 6, 7, 10, 11, etc).
• Parity bit 4 covers all the bit's positions whose binary representation includes a 1 in the third position from the least significant bit (4–7, 12–15, 20–23, etc).
• Parity bit 8 covers all the bits positions whose binary representation includes a 1 in the fourth position from the least significant bit bits (8–15, 24–31, 40–47, etc).
• In general, each parity bit covers all bits where the bitwise AND of the parity position and the bit position is non-zero.
5. Since we check for even parity set a parity bit to 1 if the total number of ones in the positions it checks is odd.
6. Set a parity bit to 0 if the total number of ones in the positions it checks is even.

## Determining the position of redundant bits

These redundancy bits are placed at the positions which correspond to the power of 2. As in the above example:

1. The number of data bits = 7
2. The number of redundant bits = 4
3. The total number of bits = 11
4. The redundant bits are placed at positions corresponding to power of 2: 1, 2, 4, and 8

Suppose the data to be transmitted is 1011001, the bits will be placed as follows:

## Determining the Parity bits –

1. R1 bit is calculated using parity check at all the bits positions whose binary representation includes a 1 in the least significant

R1: bits 1, 3, 5, 7, 9, 11

To find the redundant bit R1, we check for even parity. Since the total number of 1’s in all the bit positions corresponding to R1 is an even number the value of R1 (parity bit’s value) = 0

2. R2 bit is calculated using parity check at all the bits positions whose binary representation includes a 1 in the            second position from the least significant

R2: bits 2,3,6,7,10,11

To find the redundant bit R2, we check for even parity. Since the total number of 1’s in all the bit positions corresponding to R2 is an odd number the value of R2(parity bit’s value)=1

3. R4 bit is calculated using parity check at all the bits positions whose binary representation includes a 1 in the            third position from the least significant

R4: bits 4, 5, 6, 7

To find the redundant bit R4, we check for even parity. Since the total number of 1’s in all the bit positions corresponding to R4 is an odd number the value of R4(parity bit’s value) = 1

4. R8 bit is calculated using parity check at all the bits positions whose binary representation includes a 1 in the            fourth position from the least significant

R8: bit 8,9,10,11

To find the redundant bit R8, we check for even parity. Since the total number of 1’s in all the bit positions corresponding to R8 is an even number the value of R8(parity bit’s value)=0.

Thus, the data transferred is:

## Error detection and correction

Suppose in the above example the 6th bit is changed from 0 to 1 during data transmission, then it gives new parity values in the binary number:

The bits give the binary number as 0110 whose decimal representation is 6. Thus, the bit 6 contains an error. To correct the error the 6th bit is changed from 1 to 0.