Assignment - 4.docx - Assignment 4 1 Recitation Problems...

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Assignment – 4 1. Recitation Problems 1.1. Chapter 9 1.1.1. Three computers, A, B, and C, have the numerical features listed below: We may imagine these values as defining a vector for each computer; for instance, A’s vector is [3.06, 500, 6]. We can compute the cosine distance between any two of the vectors, but if we do not scale the components, then the disk size will dominate and make differences in the other components essentially invisible. Let us use 1 as the scale factor for processor speed, α for the disk size, and β for the main memory size. Feature Processor Speed Disk Size Main-Memory Size A 3.06 500α B 2.68 320α C 2.92 640α (a) In terms of α and β, compute the cosines of the angles between the vectors for each pair of the three computers. A=[3.06, 500α, 6β] B=[2.68, 320α, 4β] Cos θ = 8.2008 + 160000α 2 + 24β 2 √(9.36 + 250000α 2 + 36β 2 ) * √(7.18 + 102400α 2 + 16β 2 ) A=[3.06, 500α, 6β] C=[2.92, 640α, 6β] Cos θ = 8.9352 + 320000α 2 + 36β 2 √(9.36 + 250000α 2 + 36β 2 ) * √(8.53 + 409600α 2 + 36β 2 ) B=[2.68, 320α, 4β]
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C=[2.92, 640α, 6β] Cos θ = 7.8256 + 204800α 2 + 24β 2 √(7.18 + 102400α 2 + 16β 2 ) * √(8.53 + 409600α 2 + 36β 2 ) (b) What are the angles between the vectors if α = β = 1? A=[3.06, 500α, 6β] B=[2.68, 320α, 4β] Cos θ = 8.2008 + 160000α 2 + 24β 2 √(9.36 + 250000α 2 + 36β 2 ) * √(7.18 + 102400α 2 + 16β 2 ) = 8.2008 + 160000(1) 2 + 24(1) 2 √(9.36 + 250000(1) 2 + 36(1) 2 ) * √(7.18 + 102400(1) 2 + 16(1) 2 ) = 8.2008 + 160000 + 24 √(9.36 + 250000 + 36) * √(7.18 + 102400 + 16) = 0.99 The angle between vector A and B is approximately 0◦ A=[3.06, 500α, 6β] C=[2.92, 640α, 6β] Cos θ = 8.9352 + 320000α 2 + 36β 2 √(9.36 + 250000α 2 + 36β 2 ) * √(8.53 + 409600α 2 + 36β 2 ) = 8.9352 + 320000(1) 2 + 36(1) 2 √(9.36 + 250000(1) 2 + 36(1) 2 ) * √(8.53 + 409600(1) 2 + 36(1) 2 ) = 8.9352 + 320000 + 36 √(9.36 + 250000 + 36) * √(8.53 + 409600 + 36) = 0.99 The angle between vector A and C is approximately 0◦ B=[2.68, 320α, 4β]
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C=[2.92, 640α, 6β] Cos θ = 7.8256 + 204800α 2 + 24β 2 √(7.18 + 102400α 2 + 16β 2 ) * √(8.53 + 409600α 2 + 36β 2 ) = 7.8256 + 204800(1) 2 + 24(1) 2 √(7.18 + 102400(1) 2 + 16(1) 2 ) * √(8.53 + 409600(1) 2 + 36(1) 2 ) = 7.8256 + 204800 + 24 √(7.18 + 102400 + 16) * √(8.53 + 409600 + 36) = 0.99 The angle between vector B and C is approximately 0◦ (c) What are the angles between the vectors if α = 0.01 and β = 0.5? A=[3.06, 500α, 6β] B=[2.68, 320α, 4β] Cos θ = 8.2008 + 160000α 2 + 24β 2 √(9.36 + 250000α 2 + 36β 2 ) * √(7.18 + 102400α 2 + 16β 2 ) = 8.2008 + 160000(0.01) 2 + 24(0.5) 2 √(9.36 + 250000(0.01) 2 + 36(0.5) 2 ) * √(7.18 + 102400(0.01) 2 + 16(0.5) 2 ) = 8.2008 + 16 + 6 √(9.36 + 25 + 9) * √(7.18 + 10.24 + 4) = 0.99 The angle between vector A and B is approximately 0◦ A=[3.06, 500α, 6β] C=[2.92, 640α, 6β] Cos θ = 8.9352 + 320000α 2 + 36β 2 √(9.36 + 250000α 2 + 36β 2 ) * √(8.53 + 409600α 2 + 36β 2 ) = 8.9352 + 320000(0.01) 2 + 36(0.5) 2 √(9.36 + 250000(0.01) 2 + 36(0.5) 2 ) * √(8.53 + 409600(0.01) 2 + 36(0.5) 2 ) = 8.9352 + 32 + 9 √(9.36 + 25 + 9) * √(8.53 + 40.96 + 9) = 0.99 The angle between vector A and C is approximately 0◦
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B=[2.68, 320α, 4β]
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  • Fall '09
  • Data Mining, Cos, Jaccard index, Jaccard, Cosine similarity

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