Prove that in a self-complementing code the sum of the
weights must be 9?
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Answer Posted / nobody
In self complementing code, 9's complement in decimal is the
1's complement in binary.
Now, assume a code with weights are W1,W2,W3 and W4.
We want to prove that,
W1 + W2 + W3 + W4 = 9
Let 'N' be a number in decimal.
Let N's binary equivalent in given code be
X1X2X3X4
Lets represent complement of N as COMP(N).
Let COMP(N)'s binary equivalent in given code be
Y1Y2Y3Y4
Therefore,
N = W1(X1) + W2(X2) + W3(X3) + W4(X4) ***************[1]
Now 9's complement of N is
[9's COMP(N)] = 9 - N
[9's COMP(N)] = 9 - [W1(X1) + W2(X2) + W3(X3) + W4(X4)]
***** (from eq.1) *******[2]
Now, COMP(N)'s binary equivalent in given code is Y1Y2Y3Y4.
COMP(N) = W1(Y1) + W2(Y2) + W3(Y3) + W4(Y4)
********************[3]
Now as it is a self complementing code, 9's complement in
decimal is equal to the 1's complement in binary.
Therefore,
[9's COMP(N)] = COMP(N)
From [2] and [3]
9 - [W1(X1) + W2(X2) + W3(X3) + W4(X4)] = W1(Y1) + W2(Y2) +
W3(Y3) + W4(Y4)
[W1(X1) + W2(X2) + W3(X3) + W4(X4)] + [W1(Y1) + W2(Y2) +
W3(Y3) + W4(Y4)] = 9
W1(X1+Y1) + W2(X2+Y2) + W3(X3+Y3) + W4(X4+Y4) = 9
****************************[4]
Now, as (X1,Y1) (X2,Y2), (X3,Y3), (X4,Y4) are complements of
each other, their sums will always be 1.
i.e
(X1 + Y1) = (X2 + Y2) = (X3 + Y3) = (X4 + Y4) = 1.
Putting this in eq. [4], we get,
W1(1) + W2(1) + W3(1) + W4(1) = 9
i.e
W1 + W2 + W3 + W4 = 9
| Is This Answer Correct ? | 27 Yes | 4 No |
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