Question

a)  If x,y\in F_{2,}^{n} show that

wt(x+y)=wt(y)-2wt(x\cap y)

where x\cap y is the vector in F_{2}^{n} which has 1s precisely at those positions where x and y have 1s. 

(Hint: Let x=(x_1,x_2,.....,x-n)\, and\, y=(y_1,y_2,....,y_n). SupposeImage ignouassignments-ignouacademy-com--p-that-42424Observe that wt(x)=n_1+n_2 \: and \: wt (y)=n_1+n_3.

08 Mar 2024
Answer :
Word Count : 723
To solve the problem, we'll analyze the weights of the vectors \( x \), \( y \), and \( x + y \) in the vector space \( \mathbb{F}_2^n \). ### Definitions: 1. Weight (wt): The weight of a vector is the number of its non-zero (i.e., 1) entries. 2. Addition in \( \mathbb{F}_2^n \): The sum \( x + y \) is computed component-wise modulo 2. This means: - \( (x + y)_i = 1 \) if \( x_i \neq y_i \). - \( (x + y)_i = 0 \) if \( x_i = y_i \). ### Step-by-Step Solution: 1. Partition the Indices: - Let \( n_1 \) be the number of positions where both \( x \) and ________ __________ _____ ________ ________ __________ ________ ____.
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