Hi Everyone,

Post your solutions to the Nov 28 In-Class exercises to this thread.

Best,

Chris

[0 100]

• d=1: [0 50 100] the optimal is 0
• d=2: [0 25 50] the optimal is 25;
• here we divide the range by (k+1)
• d=3: [12.5 25 37.5] the optimal is also 25

• l = 100
• h1 = [0, 50, 100]
• h2 = [16, 33, 50]
• h3 = [16, 24.5, 33]

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Kunal Deshmukh d = 0 possible values for lambda = 0 50 100 d = 1 possible values for lambda = 0 25 50 d = 2 possible values for lambda = 12 25 37 hence selected value = 25

</nowiki>

range = {0..100} k = 3 , d = 3 (we stop when d = 3) d = 1 : [0 50 100] the optimal is 50 d = 2: [0 25 50] the optimal is 25 because lamda = 21 d = 3: [12 25 37] The optimal is still 25

{0, 50, 100} d = 1 {0, 25, 50} d = 2 {12.5, 25, 37.5} d =3

n= 100 k=3 d=3 lamda=21

(0,50,100} 0 is chosen

{0,25,50} 25 is chosen

{12.5,25,37.5}

25 is chosen at depth 3

Depth 0: [0, 50, 100], Depth 1: [0, 25, 50], Depth 2: [13, 25, 37]

Name - Krishna Vojjila for depth 0, [0,50,100] is taken. 0 is chosen for depth 1, [0,25,50] is taken, 25 is chosen for depth 3, [12.5,25,37.5], 25 is chosen

d = 3, k = 3, n = 100 when d = 0, possible lambda values = {0, 50, 100} when d = 1, possible lambda values = {0, 25, 50} when d = 2 possible lambda values = {12, 25, 37} value 25 is taken.