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I need to find the answer of this kind of expression $a_1 + a_2 + a_3 + (a_1\oplus a_2) + (a_2\oplus a_3) + (a_1\oplus a_3) + (a_1\oplus a_2\oplus a_3) $.

After observing I found this pattern in above expression $ X + (X \oplus Y)$. But I don't know how to proceed further.

ujjwal_bansal
  • 123

1 Answers1

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$p \oplus q = (p+q)\cdot(p'+q') = p'q + pq'$

You can simplify XOR operations using the above relation, and further simply them using the laws of boolean algebra.

$a_1 + (a_1\oplus a_2) = a_1 + (a_1'a_2 + a_1a_2')$

$= a_1 + a_1a_2' + a_1'a_2 = a_1 + a_1'a_2$

Similarly, you can simplify other terms.

After simplification, the expression would be,

$= a_1 + a_2 + a_3 + (a_1 \oplus a_2 \oplus a_3)$

$= a_1 + a_2 + a_3 + (a_1'a_2 + a_1a_2)'a_3 + (a_1'a_2 + a_1a_2)a_3'$

$= a_1 + a_2 + a_3 + (a_1'a_2 + a_1a_2)a_3'$

$= a_1 + a_2 + a_3 + a_1'a_2a_3c + a_1a_2a_3'$

The law of boolean algebra used here is ABSORPTION LAW

The simplified expression is $a_1 + a_2 + a_3$

User 123
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  • After substituting the result in given expression I get $ a_1 + a_2 + a_3 + a_1'a_2 + a_2'a_3 + a_3'a_1 + (a_1 \oplus a_2 \oplus a_3) $. I still don't get. – ujjwal_bansal Dec 05 '21 at 18:30
  • Now, you can simplify your expression further by rearranging the terms like this : $a_1 + a_3'a_1 + a_2 + a_1'a_2 + a_3 + a_2'a_3 + (a_1\oplus a_2\oplus a_3)$ – User 123 Dec 05 '21 at 18:32
  • Use ABSORPTION LAW to simplify – User 123 Dec 05 '21 at 18:34
  • If I take $a_1 = 1, a_2 = 2, a_3 = 3 $ and put the values in my expression $a_1 + a_2 + a_3 + (a_1\oplus a_2) + (a_2\oplus a_3) + (a_1\oplus a_3) + (a_1\oplus a_2\oplus a_3) $ and my answer is 12 but if I follows what you said then I am getting 6. So what's the issue here. – ujjwal_bansal Dec 05 '21 at 18:39
  • This is boolean algebra. You can use only 0 and 1. – User 123 Dec 05 '21 at 18:41