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Chin-Shiuh Shieh Assignment 3.1, 文鶑 1

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Maximize F 1 (x,y) Assignment - 20142 Requirement: Find maximize value of F 1 (x,y) used hill-climbing (gradient-ascent) method.

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Step1: Initialize a value for: (x,y) (Choice one point:-5≤x,y<5) Example: x0 = 1; y0 = 1; Assignment - 20143

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If initialize value (x0,y0) is good, we can find global value. If not we can be trapped in local optima The number of steps are change around: 100 Assignment - 20145

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We initialize many points of (x0,y0) and used For loop to find optima value with each point. And then compare this to choice the best optima. Assignment - 20146

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The optimize when we you many initialize points (For loop) seem better than when we only used one point in some cases. But the number of steps size will be increase linear with number of initialize points. Example: The number of steps size with 10 initialize points are change around: 1000 Assignment - 20147

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But when we used Hill-climbing on this, some time I receive local optimize, some time is a strange value, have something are wrong in here? What happening? I will point to you a case that we can’t find the optimize value if we only used Hill-climbing method. Assignment - 201411

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Now, you are standing in point a. You used Hill- climbing and you receive point a’. But it isn’t a optimize value in this location because the peak of mountains isn’t exit (out of range), and you are strapped on back of the mountains. Assignment - 201412

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We find maximum value of F(x,y) when (x,y) in line: y=-5*x/3 + 5 I replace y=-5*x/3 + 5 in function F(x,y). And then we can used For loop or Hill-Climbing to find this. In here I used For loop with step size: 0.01. You can use Hill-Climbing to replace the For Loop to have better result. Assignment - 201413

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After that we compare maximum value in case 1 and case 2 to find exactly global value And we find out the optimize value is point b: Max = 4.1724 (x0, y0)= 1.8600 1.9000 Steps: 2009 Assignment - 201414

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No. Optimize value in Case1 (-5

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