azk

Thanks CalmLogic, You have been a great help. I ll ask my recommenders and see what they think

Thanks for the reply CalmLogic, can you tell me my chance of getting in for the remaining schools, Cornel University University of Illinois--Urbana-Champaign Purdue University--West Lafayette SUNY--Stony Brook Brown University Duke University University of Pennsylvania University of Virginia and can you suggest one or two good schools that i can get in for safety with my current scores? thanks

I meant awaiting for acceptance, Sorry for the confusion, i honestly didnot know the difference. So that doesnt really mean anything ? My school is wartburg college, i guess it is not that well known. My concentration/ area of interest is computer graphics, software engineering. Do you think i can even get into Duke or Northwestern? thanks for the reply

Thank you so much CalmLogic, I was waiting for the response in desperation. I didnot know what to do because my GRE general score is too low. I am a junior now but I will graduate by Summer 09. I am trying to get into a PhD program for Fall 2009. I am currently doing two researchs that are about to be published, but they would not be published by March which is after the application process. I dont really know what to do. I have to apply to some graduate schools now as their deadlines are in December. Do you have any suggestion on the list of schools that i have now? Is that possible for me to get into any of them? Thanks for the previous response, it really makes me feel that i m not alone.

Hi, I am an international student but I am currently studying in US. My GRE score : verbal 590, quant 790 GPA: 4.0 major, 3.9 overall major in computer science, graduating in 3 years. I am hoping to get into a PhD program somewhere in US. Research: +Physics- granularity (not published) +CS- computer graphics ( about to publish) +CS- stereographic ( about to pubish) I have not yet received my score for AGRE until Dec 19. And here are the universities i have in mind: University of California at Berkeley University of Texas at Austin Cornel University University of Washington University of Illinois--Urbana-Champaign Purdue University--West Lafayette SUNY--Stony Brook Brown University Duke University University of Pennsylvania University of Virginia I have no idea how possible this seems. so please help !

with f: N->{0,1}, function f always map any subset of N to a finite set {0,1}. The number of functions f existing is proportionate to number of subsets that we can derive from N. Then the number of subsets of N is the cardinality of a power set. And the power set is strictly greater than the superset that it is derived from( which is N in this case) And the power set is not countable. We can use diagonalization to prove this through contradiction. There is a neat proof here ( it is just diagonalization leading to contradiction) - please read if interested: www.cs.rpi.edu/~drinep/modcomp/class17.ppt ( nice) www.cs.xu.edu/csci250/06s/Theorems/powerSetuncountable.pdf (ok proof) Excellent question n3urodron3 !!! [bounce] bounce!!

What is AGRE? is it the same as GRE?

I guess we are misunderstanding each other, consider this scenario: *P = x *Q=y //x and y are integers now if P=Q ---> *P =*Q = y Now, if we want to decrement the reference count of *P, we decrement the ref count of y ( even though we want to decrement x because there is one less pointer pointing at it). I guess the only thing here is that we are the system that takes charge of decrementing and incrementing the ref count. did i miss anything?

jb123, every data has a reference count which facilitates reference counting garbage-collector. If we assign P := Q, the data pointed by P will still have reference count =1 and there is no way to decrement that count anymore because we lose the pointer P ( now pointing to Q). Hence, the garbage collector cannt clear that data and there is no way to reach it --> memory leak. [bounce]

because the function can be non-continuous so not all points in N are mapped into S={0,1}. This kind of function can either be surjective and injective( injective if cardinality of the function's domain I found the mistake i made though, we know that the function f maps subset of N -> {0,1} Even though the subset of a countable set is countable, the NUMBER of subset is not. The theorem doesnot say anything about the NUMBER of subsets, it only guarantees that the subsets themselves are countable. so answer I is False according to me [bounce] bounce!!

Sure, i have the same answer, Because the mantissa this float is bounded by 2* 2^3 = 2^4 =16 with the one at the sign bit, this float must be > -16. we can eliminate the rest easily. Can you answer many of them at once so it is easier to discuss as a group? just skip ones that you dont want to answer :D [bounce] bounce!!

I think it doesnt reject any string listed. a) abbccc a(bbc)©© b) abcbcbc a(bc)(bc)(bc) c) abbcc a(bbc)© d) acbc a©(bc) [bounce] bounce!!

I post some more questions so people will enjoy. I dont have answer for these, just found them. . 1. For an ambigous grammar, the compiler writer can either (approach 1) use a non-ambigous version of the same grammar, or (approach 2) retain the ambigous grammar, but resolve the ambiguity by introducing precedence and associative of operators. Why would he sometimes prefer approach 2 to approach 1? I. an ambigous grammar can produce at most two derivations II. ambiguous grammars being smaller, they result in shorter code III. ambigous grammars reduce parsing time by avoiding chain reduction 2. Assuming P != NP, which of the following is/are true concerning propositional formulae: I. All satisfiable formulae are in P II. All unsatisfiable formulae are in P III. All satisfiable formulae are in NP 3. construct DFA for ((Number of 0's) + 2 * (Number of 1's)) mod 3 = 0 4. floating point: (-1)^S * 2^(E-127) (1.M) then which of these cannot be represented exactly in this system? a) 100 b) 15.5 c) 0.25 d) 0.1 e) 1.0625 5. static type checking vs dynamic type checking I. code bigger for dynamic typing II. static takes more compilation time, dynamic takes more run time III. static type checking checks all expressions while dynamic checks only those expressions that are evaluated. 6) Which string does a(b*c)* reject a) abbccc b) abcbcbc c) abbcc d) acbc 7) IEEE 747 representation ( an example was given of how to convert from 747 representation to real form. ) N = (-1)^S * 2^(E-127) * (1.M) Give the value of --------------------------------- | 1 | 10000010 | 1101000000000000 --------------------------------- a) -14.5 b) 14.5 c) -3.5 d) -29 e) 3.5 8) One on Garbage collection. When we make an assignment P = Q, where P & Q are pointers to an integer, what happens a) *P reference count is decremented, P is assigned to Q, *Q reference count is incremented b) *P reference count is incremented, P is assigned to Q, *Q reference count is decremented c) P is assigned to Q, *P reference count is decremented, *Q reference count is incremented d) P is assigned to Q, *P reference count is incremented, *Q reference count is decremented 9. There is a relational schema which has k attributes. The domain of each attribute consists of exaclty 2 elements. A table is defined as subset of tuples where in each tuple, a value is defined for each of the k attributes. The minimum value of k needed for the number of distinct tables to exceed 10^9 is (a) 5 (b) 9 © 17 (d) 512 (e) 1024 10. Suppose in a memory if the 8 RAM chips are used to interleave the memory on the lower bits. Suppose the minor/major times of the memories are 50/200. What is the maximum speed up? (major time = time between two successive accesses to the same chip) (minor time = time between two successive accesses to different chips) a) 2 b) 4 c) 6 d) 8

That is true, my bad. The answer must be A. I should have read the question more carefully [bounce] bounce!!

I think your algorithm works perfectly fine, thanks for detailing it. But step 1 (above) seems to say that you compare a[X] to 4? and you have to do that until N/2 ? I thought we are trying to minimize the number of comparisons to theta(1) i didnt see this earlier, never mind my comment then. [bounce] bounce !!

I. N -> {0,1} (countable) let A be a subset of N, A is countable because of theorem 1 let B be a subset of N, B and A are disjoint, B is also countable -consider sets of functions f: A -> {0} and g: B -> {1} -because A is countable and {0} is finite with cardinality of 1 -> the cartesian product between this two sets is countable (theorem 4) -because B is countabl and {1} is finite --> Cartesian product of this two sets is countable.( thereom 4) -Also the union of two countable sets is countable( theorem 2) Therefore, the mapping from N -> {0,1} is countable ( which contradicts the answer, please correct this for me !!!!) II. {0,1} -> N (countable) well, this mapping is injective, 0 can be mapped to at most one integer and so can 1. so {0} -> N is countable ( we can even enumerate this) {1) -N is also countable ( ditto) the union of the two countable sets of functions should be countable. so i am sure about this. III. theorem 3 --> countable. please correct me [bounce] bounce !!!

I am still trying to understand your method, maybe i am too dumb !!! let me try your algo on this string of integers: 1 2 3 | 4 4 4 4 ( the bar splits this string into 2 halves. floor(7/2) = 3 integer division) then the first half doesnt even have the major integer that we are finding. if we take the ceiling(7/2) =4, then the string will be 1 2 3 4 | 4 4 4 ( indexes from 0..6) So we start from the left at int 1 (index 0), add 4 ( k=4) --> see 4 at ( index=0+4) and not equal to 1 (nope) then int 2 (index 1), add k ( which is 4) --> see 4 at index 5 and not equal to 2 (nope) then int 3 (index 2), add 4 --> see 4 at index 6 and not equal to 3(nope) then int 4 (index 3), add 4 --> crashes at index 7 ( out of bound) So i think it will crash in this case, and it also has to make floor(n/2) comparision --> O(n) I might miss something, correct me [bounce] bounce !!

Bandwidth = 2^23 bps N = (.8)*(2^23)/(1.1*300) = 20 336 terminals

Yeah i think Nonevent99 is right, I interpreted the bits as True and False after getting the output from the gates. I should have interpreted them before applying AND and OR logic. Sorry for any confusion. This is trickier than i though :D

According to this algorithm, if X happens to position somewhere at the middle of the chunk, then X + (N/2 +1) will give a position outside the chunk. Also, how can you find X in the first place? I am so sorry, my algorithm has a missing piece. I use a hash table for this. Because we know that biggest and smallest element in the list so we can create a hash table for every possible values in this range. This hash table will contain the number of repetition of a given value (count). Then i go through the input list and hash that number ( there is a hash function to do this) into the table and increment the count there. Because the major integer will appear more than n/2 +1 times, so the count >= n/2 +1. Whenever the count > n/2, we quit. let's parse this: 2 3 3 3 3 4 6 so we have 5 slots for numbers from 2 -> 6 target = n/2 +1 = 4 parse 2 --> [1,0,0,0,0] 2 3 --> [1,1,0,0,0] 2 3 3 --> [1,2,0,0,0] 2 3 3 3 --> [1,3,0,0,0] 2 3 3 3 3 --> [1,4,0,0,0] hura! it is 4 = target so we quit. The current integer is the one we want (i.e. 3). Because no integer other than the major integer can appear more than n/2 so the algorithm is correct. Otherwise, correct me :D [bounce] bounce !!

Well, according to this source, it doesnt say anything about any exact order of bind the values back to the variables. [bounce] bounce !!

Is that true that the subroutine will rewrite all of those variables in that order? y then x then y ? If it doesnt return the variables in that order ? how does it resolve the conflict between two values of y ?

I think this might do S -> Cc | aD | bE C -> Cc | A 3 # expand c for the case k > n+m D -> aD | A # expand a for the case k E -> bE | B # expand b for the case k A -> aAc | bBc | e B -> bBc | e I normally find out all the possible scenarios then use different symbols to represent each case. for example: Cc is the case where k > n+m aD is the case where k bE is the case where k [bounce] bounce !!

Sorry, i didnt post everything. OK now i added the meanings to the answer. Negative logic just inteprets True in Positive logic as False, and False in Positive logic as True. correct me if it is wrong. [bounce] bounce !!

If we know that the array is sorted and we want to check if an integer k appears more than n/2 times in the array. I would just go through the list and count number of consecutive k in the array ( because it is sorted so all the integer k must be in chunk). if the count surpasses n/2 then we are set, otherwise, no. Another branch of this problem can be: The minimum no.of comparisons required to determine AN INTEGER THAT appears more than n/2 times in a RANDOM array of n integers ( precondition: there is such int) For this problem, we can do this: we keep track of current integer and count, if we see the same integer, count++ if we see a different integer count-- if count reaches 0: CURRENT changes to that new integer COUNT =1 and keep doing that. because the integer that appears >= n/2 +1 will remain at the end. The worse case scenario for this algorithm is this 9 8 9 8 9 8 9 curr=9, count =1 see a 8: so count=0 ---> curr=8, count =1 see a 9: so count =0 ---> curr=9, count =1 and so on which makes the count =1 and current =9 at the end. confirm me [bounce] bounce !!