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

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