R0jkumar

So in the actual exam all the questions are part of the final score? In other words the above raw score is out of 70? And how exactly do you know that more Chinese are taking the exam? And even if they are, how does it mean test takers are improving?

i have a question about the raw score and how it relates to the percentile. One of the guys on this forums has posted his AGRE CS scores as follows : Score / Percent / Raw score/Date Taken 800 77% 50.....2007.04.14 830 88% 46.....2005.11.12 If the look at the sample AGRE CS test booklet, a raw score of 50 which i think is very good is equal to 95%. So, how is it that a raw score of 50 gives only 77%? What am i missing here?

From the very definition of L1, a string x is in L1 if and only if there is a substring of x that is in L. Your language a^nb^n certainly has a substring that belongs to L and hence a^nb^n is a subset of L1 but not the entire set. In other words by using L1 = a^nb^n you are showing that if x belongs to L1, then x contains a substring that belongs to L but you have to show the converse as well. That is you have to show that given a string x belonging to L, if you add any string at the beginning and at the end , you should be able to get a string in L1.But thats wont be true in the case of L1=a^nb^n since if i take any string of a's and b's and add any string at the beginning and the end, i am not necessarily going to get a string of a's followed by b's such that the number of a's and b's are equal. Hope that helps.

If L is (a+b)* then L1 cant be a^nb^n, since L1 = (a+b)*L1(a+b)* = (a+b)*(a+b)*(a+b)* = (a+b)* which is regular.

To answer the original question, if A is known to be NP-Hard, all you need to prove is that A is in NP to prove that A is NP-Complete. gtts : Your statement below isn't really correct. "If A can be polynomially reduced to B then A is also in NP. Combining that with the fact that A is NP-hard we get that A is NP-complete." A can be polynomial time reducible to B and yet be outside of NP. Typically, you don't use polynomial time reducibility to prove the "NP" portion of NP-Completeness proof. You independently prove that a problem is NP by showing that given a solution to the problem, you can verify that the solution is correct in Polynomial time. Polynomial time reducibility is only used to prove the NP-Hard portion of the NP-Completeness proof. rajkumar

Correct answer is (D) as gttts23 mentioned. The language (0 +1)* is recursive and hence contains every language of 0's and 1's . So, II holds as well.

For the 1st question, if the array is 2-ordered, then the elements are such that the following hold : a[0] and a[1] In other words, we have 2 non-decreasing sequences of numbers each of length n. And 1-ordered is basically a sorted array. In a sorted array we have a[0] If you take any number from a 2-ordered list, it is already in its correct position relative to the rest of the (n-1) numbers from the same list but it maybe out of order with respect to the n-numbers from the other list. Hence each number can be a maximum of n-positions from its correct position in the sorted list. I am still thinking about the second part :)

Thats not correct. You are suggesting there are 4 ways to select each of the n-2 positions which isn't correct.Remember, the string in the language is such that it begins with any number of a's followed by any string of b's and c's, followed by string ab and then followed by any number of c's. In other words you don't have a choice of picking an alphabet among a,b,c and c for each of the n-2 positions. That would be true for say a string in the language (a+b+c+d)* since for each position of a string in the language , we have the choice of making 4 choices and hence the number of possible strings would be 4^n.

Got confirmed news from ETS. They don't have standby testing in India. Quite some reputation we Indians have built up over the years !!!!

I agree with taro's answers. By the way, where did you get the questions? rajkumar

I did talk to ETS the other day and this is what they told me. Supposedly, there has to be atleast person taking the same subject test as you at the center that you go to for standby testing to work. Like you found out, there is no point going to a test center where there is no one taking the CS exam. But i was given the impression that ETS does send some extra test booklets for standby testing provided there is atleast one person taking that particular subject test. I am going to call them today and make sure that is the case. But http://www.ets.org/portal/site/ets/m...22f951 90RCRD does mention that standby testing is not available in India which means i am screwed. I wonder why the ETS customer service guy didn't tell me that because she did check for available space in the Bangalore test center and also if any person had registered for the CS test in Bangalore. If that is the case, i have no chance of standby testing in India, right? rajkumar rajkumar

I cant believe i missed the registration deadline. Has anyone here tried standby testing? What are my chances of being allowed to take the exam on exam date if i try standy testing? rajkumar

Answers for (d) and (e) are 3 and 1 respectively since then the pumping lemma holds vacuously. Notice,there is no string of length 3 in {01} and there is no string of length 1 in the empty string language. rajkumar

You are right. I was looking at the wrong columns. rajkumar

My understanding is that there is no test center in India for the November 3rd test for AGRE. Is that correct? rajkumar