COM347J1/COM552J1 Networks and Data Communications

http://www.eej.ulst.ac.uk/~ian/modules/COM347J1 COM347J1_L4C 1/12

COM347J1/COM552J1Networks and Data Communications

Ian McCrum Room 5D03B

Tel: 90 366364 voice mail on 6th ring

Email: [email protected]

Web site: http://www.eej.ulst.ac.uk

Lecture 4C: Further examples of Hamming codes (Data Correction)


The Hamming Code• k parity bits are added to n data bits giving a new word

(k+n) bits long• The word is numbered from bit 1 to (k+n) for more

convenience when working things out• The parity bits are inserted throughout the word – at

positions given by powers of two. I.e bits 1,2,4,8, 16 etc• Each parity bit is calculated from a subset of the data bits,

these overlap• Comparing the parity bits received with those calculated

at the receiver will give all zeroes if there is no discrepancy (which is why we number from one)

• If the comparison is non-zero, e.g 101 then bit 5 is in error – this is very handy, easy to correct in hardware


Hamming applied to n bit data words• In general the total number of bits that can be accommodated by

the method is 2k-1• So (n+k) = 2k-1• If k is 3 we can accommodate 7 bits (4 data)• If k is 4 we can accommodate up to 15 bits (11 data)• If k is 5 we can accommodate up to 31 bits (26 data)• There are better codes, hamming only fixes single bit errors (c.f

convolutional codes although we won’t cover them here). • If a 100 character block needs protected, it is better to use

hamming on the 100 bits vertically through the block, so all bit 0s are protected, then all bit 1s etc., this extends each 100 bits to 107 bits since 7 hamming bits can protect 127 data bits. This is good for bursts of noise of a few bits…


Hamming applied to 8 bit data[0010 0011] msb…lsb (b7..b0)

• We reserve the bit positions that are powers of two to hold the special parity bits, usually called check bits

b1: b2 : b3 : b4 : b5 : b6 : b7 : b8 : b9 : b10: b11:b12

P1: P2 : 1 : P4 : 1 : 0 : 0 : P8 : 0 : 1 : 0 : 0

• We apply parity to subsets of the original 8 data bits, each parity bit tests 4 or 5 data bits, there is a pattern as to which!

3,5,7,9,11

3,6,7,10,11

5,6,7,12

9,10,11,12


• The new 12 bit word can now be sent…

P1: P2 : 1 : P4 : 1 : 0 : 0 : P8 : 0 : 1 : 0 : 0

3,5,7,9,11

3,6,7,10,11

5,6,7,12

9,10,11,12

P1 must be 0 to make {0, 1,1,0,0,0,} even

P2 must be 0 to make {0, 1,0,0,1,0,} even

P4 must be 1 to make {1, 1,0,0,0 } even

P8 must be 1 to make {1, 0,1,0,0 } even

0: 0 : 1 : 1 : 1 : 0 : 0 : 1: 0 : 1 : 0 : 0 = 001110010100


Picking the subsets to apply parity to• The parity bits are labelled P1, P2, P4 and P8

• Data bits are in positions 3,5,6,7,9,10,11,12

• For each data bit, write its position as powers of 2 so position 10 is 8 + 2

• We use P2 and P8 to “cover” 10

• 3=>P1 and P2, 5=>P1 and P4, 6=>P4 & P2

• 7=>P1 & P2 & P4, 9=>1001 => P8,P4,P2,P1

• Hence you can draw the arrows and work out the parity bits.

• A better way is to arrange a 2D table, see the website for a good link showing this (http://candle.ctit.utwente.nl/wp5/tel-sys/exercises ) or the answers to the tutorial questions later in this presentation.

http://candle.ctit.utwente.nl/wp5/tel-sys/exercises


Using Hamming on 001110010100 bit positions… 123456789012

• 001110010100 arrives into the receiver• The receiver calculates four parity bits, it

uses the original grouping but includes the parity bits themselves.

• So C1 = {b1, b3, b5, b7, b9, b11 } =011000• So C2 = {b2, b3, b6, b7, b10, b11 } =010010• So C4 = {b4, b5, b6, b7, b12 } =11000

• So C8 = {b8, b9, b10, b11, b12 } =10100• These should all be zero 0000 (even parity)• The next slide puts an error in bit 5


Using Hamming on 001100010100 bit positions… 123456789012• So C1 = {b1, b3, b5, b7, b9, b11 } =010000• So C2 = {b2, b3, b6, b7, b10, b11 } =010010• So C4 = {b4, b5, b6, b7, b12 } =10000

• So C8 = {b8, b9, b10, b11, b12 } =10100

• Now C1 and C4 are set, The checkbits arranged as a binary number C8 C4 C2 C1 are 0101

• This is five in binary, so invert bit 5 to correct the code

• 001100010100 = > 001110010100


Tutorial questions on Hamming• [1] A 12 bit Hamming code containing 8 bits of data

and 4 parity bits is received, what was the original?• [1](a) 0000 1110 1010• [1](b) 1011 1000 0110• [1](c) 1011 1111 0100• [2] Given the 8 bit data word 0101 1011 give the 12

bit that can correct a single bit error• [3] Given a 11 bit data word, generate the 15 bit

hamming code word


Answers to tutorial questions from Lecture 4C

000011101010

C1=0_0_1_1_1_1_

C2=_00__11__01_

C4=___0111____0

C8=_______01010So C8421 = 0110

I.e bit 6 is bad, invert it giving

000010101010

NB C1={1,3,5,7,9,11}, C2={2,3,6,7,10,11}, C4={4,5,6,7,12}, C8={8,9,10,11,12}

And these groups should have an even number of ones, if not then set that C bit

101110000110

C1= 1_1_1_0_0_1_

C2= _01__00__11_

C4= ___1100____0

C8= _______00110So C8421 = 0010

I.e bit 2 is bad, invert it giving

111110000110

101111110100

C1=1_1_1_1_0_0_

C2=_01__11__10_

C4=___1111____0

C8=_______10100So C8421 = 0000

I.e No bits are bad so code is

101111110100


Answers to lecture 4C page 2

•[2] Given the 8 bit data word 0101 1011 give the 12 bit that can correct a single bit error

Code is :P1 :P2 :b3 :P4 :b5 :b6 :b7 :P8 :b9 :b10:b11:b12:i.e : ? : ? : 1 : ? : 1 : 0 : 1 : ? : 1 : 0 : 1 : 0 :Take b3 :[1]:[1]: : (3 = 1+2)Take b5 :[1]: : :[1]: (5 = 1+4) Take b6 : :[0]: :[0]: (6 = 2+4)Take b7 :[1]:[1]: :[1]: (7 = 1+2+4) Take b9 :[1]: : : : : : :[1]: (9 = 1+8)Take b10: :[0]: : : : : :[0]: (10= 2+8)Take b11:[1]:[1]: : : : : :[1]: (11= 1+2+8)Take b12: : : :[0]: : : :[0]: (12= 4+8)

Parity Pn[1] [1] [0] [0] (make column even)12 bits = 1 1 1 0 1 0 1 0 1 0 1 0i.e [1110010101010]


Answers to Lecture 4C Tutorials Page 3

Code is :P1 :P2 :b3 :P4 :b5 :b6 :b7 :P8 :b9 :b10:b11:b12:b13:b14:b15:P16:b17:i.e : ? : ? : 1 : ? : 1 : 0 : 1 : ? : 1 : 0 : 1 : 0 : 1 : 1 : 0 : ? : 0 : Take b3 :[1]:[1]: : (3 = 1+2)Take b5 :[1]: : :[1]: (5 = 1+4) Take b6 : :[0]: :[0]: (6 = 2+4)Take b7 :[1]:[1]: :[1]: (7 = 1+2+4) Take b9 :[1]: : : : : : :[1]: (9 = 1+8)Take b10: :[0]: : : : : :[0]: (10= 2+8)Take b11:[1]:[1]: : : : : :[1]: (11= 1+2+8)Take b12: : : :[0]: : : :[0]: (12= 4+8)Take b13:[1]: : :[1]: : : :[1]: (13= 1+4+8)Take b14: :[1]: :[1]: : : :[1]: (14= 2+4+8)Take b15:[0]:[0]: :[0]: : : :[0]: (15= 1+2+4+8)Take b17:[0]: : : : : : : : : : : : : : :[0]: (17= 1+16)Take b18: :[0]: : : : : : : : : : : : : :[0]: (18= 2+16)

Parity Pn[0] [0] [0] [0] [0] (make column even) 12 bits = 0 0 1 0 1 0 1 0 1 0 1 0 1 1 0 0 0

i.e 00101010101011000

•[3] Given a 11 bit data word, generate the 15 bit hamming code word, I pick an arbitrary 001 1010 11011

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COM347J1/COM552J1 Networks and Data Communications