Hello everyone, I am mac Jiang, first of all, congratulations to everyone Happy Ching Ming Festival! As a bitter programmer, Bo Master can only nest in the laboratory to play games, by the way in the early morning no one sent a microblog. But I still wish you all the brothers to play happy! Today we share the coursera-ntu-machine learning Cornerstone (Machines learning foundations)-Exercise solution for job two. I encountered a lot of difficulties in doing these topics, when I find the answer on the Internet but can not find, and Lin teacher does not provide answers, so I would like to do their own questions on how to think about the writing down, for everyone to provide some ideas. Of course, my understanding of the topic is not necessarily correct, if you bo friends found errors please contact, thank you! Again: Please do not use this blog as a way to pass the exam, or better to learn and understand the course! Hope my blog is helpful to your study! The source of this article: http://blog.csdn.net/a1015553840/article/details/51043019
1. The first question (1) test instructions: The error rate of the objective function f is MU, the sample is added to the noise so that P (x/y) as above formula (here 1-lambda is the noise ratio, because the output y is two yuan, so add noise means the original f (x) variable number, to become Y), The error rate of a noisy y of a hyphothesis h is obtained. (2) Analysis: Here is the solution is the two-dollar classification problem, that is, y={-1,+1}. So the noise here means that the 1-lamba ratio (x, y) becomes (x,-y). When Y=f (x), by test instructions we know that the error rate for H to F is Mu, and the y=f (x) probability is lambda, so this part of the error rate is LAMBDA*MU. When y is not equal to f (x), because we just flip the sign of Y, so the original correct to become wrong, the original error becomes correct, so the error rate of H to F is 1-mu, this part error rate is (1-LAMBDA) (1-MU) (3) Answer: the third
2. The second question
(1) Test instructions: When the lambda takes what value, the performance of H (here refers to the error rate) is independent of Mu? (2) Analysis: We in the above to the error rate calculation formula, our aim is to dissolve the MU, very simple answer, lambda=0.5 when the error rate and mu irrelevant, Total error Rate is 0.5 (3) Answer: 0.5
3. Question three (1) test instructions: When n is greater than or equal to 2,d_vc greater than or equal to 2 o'clock, we can replace MH (n) with N^D_VC. When D_VC = 10 o'clock, you want a 95% confidence rate, epsilon is less than or equal to 0.05, then we need more samples (N)? (2) Analysis: This problem is very simple, bring into the teacher class to speak formula calculation on it. (3) Answer: 460,000
4. Question fourth (1) test instructions: d_vc=50, the confidence rate is 1-delta,delta = 0.05, the number of training samples is n. When n = 10000, the value of the epsilon is calculated using the following five formulas, which is the smallest (most compact)? (2) Analysis: No way, calculate with calculator. I calculated the results otiginal VC bound=0.63,radenmacher Penalty bound = 0.33,parrondo and Van den Broek=0.224,devroye = 0.215,variant VC bound=0.86 (3) Answer: Decoroye
5. Question Fifth (1) test instructions: when n = 5 o'clock, calculate respectively, which minimum (2) Analysis: Original VC bound=13.83,radenmacher penalty bound=7.04,parrondo and Van den broek=4.7, Devroye=4.9,variant VC bound=16.2 (3) Answer: Parrando and Van den Broek
6. Question Sixth (1) test instructions: What is the MH (N) of positive and negatibe intervals on R? (2) Analysis: (3) Answer: N^2-n +2
7. Question Seventh (1) test instructions: vc-dimension (2) analysis of Positive and negative: We have obtained MH (N) = n^2-n+2. When N=1,MH (1) =2=2^ (1); when N=2,MH (2) =4=2^2, when N=3,MH (3) =8=2^3; when N=4,MH (4) =16! =2^4 (3) Answer: 3
8. Title (1) test instructions: The use of x1^2+x2^2 as a classification line, the formation of two concentric circles, within the concentric circle of +1 classes, in the concentric circle outside the classified-1 class. Suppose there are n samples, MH (n) (2) Analysis: Think carefully, this problem and a positive interval is a problem, but the dimension from one dimension to the two-dimensional (3) answer: the Last
9. The ninth question (1) test instructions: polynomial discriminant hyphothesis as shown above, for H D_VC (2) Analysis: From H can be seen, the problem is perceptrons, that is, the PLA algorithm implementation. And we have proved in class that his D_VC = d+1 (3) Answer: d+1
10. The tenth blogger is not! Ask the great God for answers.
11. Question 11th (1) test instructions: h is defined as, for H D_VC, (2) Analysis: In fact, can be plotted to find H is a square wave, considering the situation of n=1,n=2,n=3 can be obtained 2^n classification method, extended to N-dimensional can be (3) Answer: Infinity
12. Question 12th (1) test instructions: Which of the following formulas is the upper bound function of the growth function MH (n), when N is greater than or equal to d_vc greater than or equal to 2 (2) Analysis: In class when the teacher said, when the D_VC is greater than or equal to 2,n greater than or equal to 2 o'clock, MH (N) the upper limit function So the second mistake. For the third item, it is correct before N is less than D_VC, but when N is greater than D_VC, then his MH (n) must be greater than his, error. The first one is inexplicably wrong. Fourth, expand the I=1 section, 2*MH (N-1), in fact, this part is bigger than MH (N). Imagine, the D_VC is infinite, the classification number is 2^n,n each increases 1, the classification number increases one times. And here d_vc! = Infinity, then he has a multiplier of less than 2 each time, therefore less than 2*MH (N-1). ----In fact, you can find ideas from which table B (n,k)! (3) Answer: item Fourth
13. Question 13th (1) test instructions: which MH (n) below is not possible (2) Analysis: MH (N) is monocytogenes, and the fourth is not. In fact, this reason Bo Master did not understand thoroughly, but it seems because the fourth item is not strictly monocytogenes (3) Answer: Fourth
14. Question 14th (1) test instructions: Some of the options below are correct and some are wrong. Find the smallest range in the correct options. (2) Analysis: This problem is to ask for and set of D_VC. First, we look at the left border. If a HK is an empty set, then d_vc=0, so the minimum value is 0, we look at the right border. Since it is the intersection, we retain the smallest D_VC (3) Answer: item Fourth
15. Question 15th (1) test instructions: In consideration of the intersection situation, the discussion and the situation (2) Analysis: A. Left: H1 ... HK can do the same set can do, so for Max B. The question on the right is there is no K-1. Imagine that there is a H1 that divides the plane all points into +1,h2 to divide the plane all points into-1. H1 and H2 words for VC dimension for 1, and each D_VC add up to 0 (3) Answer: Fourth
第16-18 title: http://blog.csdn.net/a1015553840/article/details/51023193 第19-20 Title: http://blog.csdn.net/a1015553840/ article/details/51029765
The source of this article: http://blog.csdn.net/a1015553840/article/details/51043019
Robotic Learning Cornerstone (Machine learning foundations) Learn Cornerstone job Two after class exercise solution
