Prove that in a self-complementing code the sum of the
weights must be 9?
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Answer Posted / aamod joshi
Lets assume that the weights are w1,w2,w3 and w4.
Assume the first code is X1X2X3X4 and second number is
Y1Y2Y3Y4 where X1X2X3X4 + Y1Y2Y3Y4 = 9. e.g. if the weights
are 2,4,2,1 then 1011 = (1*2)+(0*4)+(1*2)+(1*2) = 5 and
0100 = (0*2)+(1*4)+(0*2)+(0*2)= 4 i.e. 1011+0100=4+5=9.
Putting this in mathematical equation,
(W1X1 + W2X2 + W3X3 + W4X4)
+ (W1Y1 + W2Y2 + W3Y3 + W4y4)
= 9
Which means
W1(X1+Y1) + W2 (X2+Y2) + W3 (X3+Y3) + W4 (X4+Y4) = 9
Now, as X1X2X3X4 and Y1Y2Y3Y4 are self-complementing, at a
time either of X1 and Y1 can be 1, the other will be zero.
Same for X2,Y2 and X3,Y3 and X4,Y4. Which further means
that X1+Y1 = 1, X2+Y2=1, X3+Y3=1, X4+Y4=1.
So,
W1(1) + W2(1) + W3(1) + W4(1) = 9
i.e. W1+W2+W3+W4 = 9 thus it proves that sum of the weights
has to be 9.
| Is This Answer Correct ? | 86 Yes | 13 No |
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