written 7.8 years ago by | modified 2.7 years ago by |

**Mumbai University > Computer Engineering > Sem 3 > Discrete Structures**

**Marks:** 8 Marks

**Year:** May 201

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Show that the (2, 5) encoding function e: $B^2-> B^5$ defined by e(00) = 00000 e(01) = 01110 e (10) = 10101 e (11) = 11011 is a group code. How many errors will it detect and correct?

written 7.8 years ago by | modified 2.7 years ago by |

**Mumbai University > Computer Engineering > Sem 3 > Discrete Structures**

**Marks:** 8 Marks

**Year:** May 201

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written 7.8 years ago by |

To prove, it is a group code, we have to examine the following steps:

**Identity Element**: Since 00000 belongs to the set, hence check for identity is satisfied.**For Closer**:(00000) $\oplus$ (01110) = 01110, which belongs to the set.

(00000) $\oplus$ (10101) = 10101, which belongs to the set.

(01110) $\oplus$ (10101) = 11011, which belongs to the set.

(01110) $\oplus$ (11011) = 10101, which belongs to the set.

(10101) $\oplus$ (11011) = 01110, which belongs to the set.

By checking this way, we found that if x,y belong to the set, then x $\oplus$ y also belong to the set. It can be easily verified that $\oplus$ is associative. Also, for each element inverse exists. Hence, it is a group code.

A code will detect *k* or fewer errors if its minimum distance is at least *k+1*

Since here the minimum distance is 2, so we have

$\hspace{7cm}2 \ge k+1 \\ \Rightarrow \hspace{7cm} k \le1.$

So, the code will detect one or fewer errors.

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