Answer / 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.

Answer / 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

Answer / dp

It is related to Digital Logic Design

Answer / vivek

I really dint understand this question or the answer. can u
please tell me which subject is this question related to ??

Answer / poojitha

In general 2421,5211,642-3,84-2-1 and excess-3 codes are satisfies the condition of self complementing code.All the above codes gives sum as 9 as shown below:
2+4+2+1=9
5+2+1+1=9
6+4+2-3=9
8+4-2-1=9
so,in the self complementing code sum of all weights is always 9.

Answer / munjal

if I am talking about 8421 code , then the sum must be 15 ,,
the question is limited to only 2421 code......

Answer / ankit agrawal

the sum of the number and its complements should be 9999

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