Help, 2 in 5 digital error detecting question (M in N)

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Help, "2 in 5" digital error detecting question (M in N)

Homework Statement



Explain how a "2 in 5" code can detect all single bit errors but only some double errors in the 5-bit coding of a single digit.

give an example where, in one 5-bit code of a single decimal digit

a) Two errors can be detected
b) Two errors cannot be detected

Homework Equations



"m in n" error detection codes in Digital Design

The Attempt at a Solution



if 4 bits are set to 1 and in the total 5 bits, than this can cause confusion when processing.

a) 10101
b) 01111
 
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" m out of n " codes

An alternative method of detecting errors is to use an " m out of n " code where n represents the total number of bits in a binary word, of which m must be set to.

1. if more or less than n bits are set to 1 then errors are present. The error detection circuitry has to count the number of bits set to 1 in a word and compare it with m. This is a relatively simple operation.

e.g.

devise a " 2 in 5 " code to represent the decimal digits 0 to 9. Each codeword must have 2 bits set and be 5 bits long. Valid codewords can be identified by counting in pure binary and using only those words that have 2 bits set.

Count | Action
00000 | Ignore
00001 | Ignore
00010 | Ignore
00011 | Valid code equivalent to 0 (decimal)
00100 | Ignore
00101 | Valid code equivalent to 1 (decimal)

Solution

The full code is

"2 in 5"| Decimal
00011 | 0
00101 | 1
00110 | 2
01001 | 3
01010 | 4
01100 | 5
10001 | 6
10010 | 7
10100 | 8
11000 | 9

________________________________________________________________
Information Source: Digital Logic Techniques by T.J. Stonham (Third Edition)
 


Any ideas Mark44?
 


Tek1Atom said:
" m out of n " codes

An alternative method of detecting errors is to use an " m out of n " code where n represents the total number of bits in a binary word, of which m must be set to.

1. if more or less than n bits are set to 1 then errors are present. The error detection circuitry has to count the number of bits set to 1 in a word and compare it with m. This is a relatively simple operation.
In the above, I think you mean "if more or less than m bits are set to 1". You can't have more than n bits in a word of n bits.
Tek1Atom said:
e.g.

devise a " 2 in 5 " code to represent the decimal digits 0 to 9. Each codeword must have 2 bits set and be 5 bits long. Valid codewords can be identified by counting in pure binary and using only those words that have 2 bits set.

Count | Action
00000 | Ignore
00001 | Ignore
00010 | Ignore
00011 | Valid code equivalent to 0 (decimal)
00100 | Ignore
00101 | Valid code equivalent to 1 (decimal)
Why is the first column labelled Count? Aren't these just example words?
Tek1Atom said:
Solution

The full code is

"2 in 5"| Decimal
00011 | 0
00101 | 1
00110 | 2
01001 | 3
01010 | 4
01100 | 5
10001 | 6
10010 | 7
10100 | 8
11000 | 9

________________________________________________________________
Information Source: Digital Logic Techniques by T.J. Stonham (Third Edition)

There are 5C2 (5 choose 2) ways of setting exactly two bits in a 5-bit word. 5C2 = 5!/(3! * 2!).

I don't know if this is much help, but that's all I can think of. I don't understand how this ties into what you're being asked in the first post.