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krkq35

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  1. One thing maybe worth trying is to just not report your scores on the actual graduate school applications, and maybe the schools will not look for/match up your official score reports later. This may not work (or even be a good idea for that matter), but just wanted to throw it out there.
  2. I attempted just over 60, but was sure of only about 30-40... I'm sure I missed a bunch unfortunately due to guessing. I thought it was harder than the other two practice exams as well, but I think that's because these questions exploited my weaknesses more so than the practice tests.
  3. Call-by-value should just copy the values of the parameters and store them essentially as local variables, which will be thrown away after the execution of R. So, I'm almost certain the answer would be x=10, y=15 => ©. Now, if x and y were called explicitly inside of R, then the values would be updated (this only makes sense if x and y are global wrt procedure Q, but in this case they are). Ex: Procedure Q begin var x,y Procedure R(a,b,c) begin //a=15, b=10, c=15 b:=b+10; //b=20 a:=b+c; //a=35 end //no modification to original parameters, x and y y=15; x=10; R(y,x,y) //Passing by value, no changes made write(y,x) end On the other hand, call-by-reference will pass the address of the parameters to the procedure. So, any modification to a variable within the procedure will actually be changing the value at the given address in memory. Once the procedures ends, then the passed parameters will reflect any updates that have occurred. Ex: Procedure Q begin var x,y Procedure R(a,b,c) begin //a=15, b=10, c=15 b:=b+10; //b=20 (and x=b=20) a:=b+c; //a=20+15=35 (and y=a=35) end //x=b=20, y=a=c=35 y=15; x=10; R(y,x,y) //Passing by reference, update parameters write(y,x) //write 35 20 end Since the call was R(y,x,y), then a=c=y and x=b throughout the procedure since these are all just addresses that point to a value.
  4. Wonderful, thanks for digging those threads up - I probably should've looked harder in the first place. Anyway, that will definitely reduce some stress going into Saturday, thanks again.
  5. I'm a domestic student in my final year as an undergraduate hoping to go to a ranked PhD (I'd also settle for MS) program. I've been studying for about a month and half now purely for the CS GRE, but my main problem is that I haven't taken several classes (Automata Theory, Algorithms/Complexity, Operating Systems) necessary for doing well on the GRE. I'm taking all of these classes in my final semester, but unfortunately that's too late. After taking a practice test this weekend, I'm guessing my score on the 8th will probably be in the 50-60th percentile. On the plus side, I have a 3.98/4.00 GPA from a somewhat recognized CS program (University of Missouri), and I have 11 conference publications with a couple journal publications under review. My main concern is that a bad score will overwhelm the rest of my application. Has anyone else had any experience(s) with this? If it helps at all, several of the schools I'm applying to (CS-PhD programs) are: UT-Austin, UIUC, Michigan-Ann Arbor, and CMU (although I know CMU is a long shot). From reading these forums, I know UIUC doesn't require GRE scores, but will that hinder your chances of acceptance by not submitting them anyway? Thanks for any information/advice, these forums have been a huge help while studying!
  6. First off, I came up with the same answer (that all three are true) when I went through this problem. I'm still not quite sure why (I) is false. The number of subsets in any given set S is 2^cardinality(S). So, if cardinality(S) is countable, then won't a set with 2^cardinality(S) elements also be countable? It seems like it would be possible to enumerate the sets mapping N to {0,1}: f_1 maps all of N to {0} f_2 maps 1 to {0}, N\{1} to {1} f_3 maps 2 to {0}, N\{2} to {1} ... f_m maps all of N to {1} which would make the number of functions countable. That's definitely not a proof, but I would think there would be some obvious counterexample if this were false.
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