Question

For binary data, the L1 distance corresponds to the Hamming distance; that is, the number of bits that are different between two binary vectors. For the following two binary vectors, compute:

X = 10011011

Y = 01011000

1 - Hamming distance

2 - Jaccard Similarity Coefficient (JSC) and Simple Matching Coefficient (SMC)

3 - Cosine Similarity

4 - L2 (Euclidean) and L∞ (Supermum) distances

5 - Correlation between X and Y

Answer 1

Answer- (the question is lengthy, so as per guidelines i am answering first four parts)

given,

X = 10011011

Y = 01011000

1.

Hamming distance- (the number of bits that are different between two binary vectors)  = 4

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2.

Jaccard Similarity Coefficient (JSC) and Simple Matching Coefficient (SMC)

For computing the similaritie we need following quantities,

B₀₁ = no. of attributes where X is 0 and Y is 1

B₁₀ = no. of attributes where X is 1 and Y is 0

B₀₀ = no. of attributes where X is 0 and Y is 0

B₁₁ = no. of attributes where X is 1 and Y is 1

X	1	0	0	1	1	0	1	1
Y	0	1	0	1	1	0	0	0

B₀₁ = 1 B₁₀ = 3 B₁₁= 2 B₀₀ = 2

Jaccard Similarity Coefficient (JSC)= J = no. of 11 matches / no. of not-both-zero attributes values

J= (B₁₁) / (B₀₁ + B₁₀+ B₁₁) = 2/6 = 0.33

Simple Matching Coefficient (SMC)= number of matches / number of attributes

SMC = (B₁₁ + B₀₀) / (B₀₁ + B₁₀+ B₁₁ + B₀₀) = 4/8 = 0.5

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3.

Cosine Similarity-If X and Y are two binary vectors ,then

Cos (X, Y) = (X.Y) / ( ||X|| ||Y|| ) , where . (dot) indicates dot product, ||X|| indicates length of vector X

Here, X.Y= (1.0 + 0.1 + 0.0 + 1.1 + 1.1 + 0.0 + 1.0 + 1.0) = 2

||X|| = ( 1*1 + 0*0 + 0*0 + 1*1 + 1*1 + 0*0 + 1*1 + 1*1)^1/2 = (5 )^1/2 = 2.24

||Y|| = (0*0 + 1*1 + 0*0 + 1*1 + 1*1 + 0*0 + 0*0 + 0*0)^1/2 = (3)^1/2 = 1.73

Cosine Similarity = 2/ ( 2.24 X 1.73) = 0.516

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4.

L2 (Euclidean)