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Oh Congrats cjlax26! 720 is not bad. :) ..BTW can you let us know your raw score?

Hey jb123, great to hear that! I could not attempt so many questions and had to be contended with around high 30s. And I think this implies I already missed the 700 mark! :( But thats just one isolated case of mine. So lets hear what others have to say! :)

Okay, this helps to clear it up in a better manner. The argument which krkq35 used was the exact same one my friend used against the sum, which even I had thought to be true (till azk's post). Thanks for clarifying it with those links azk! and thanks krkq35 for participating in this! :)

Yes, even I am arriving at the same above scenario. Could there be any other way to do the first part? Like, we know that for any function f:A -> B to be injective, cardinality(A) S ? (assuming S = {0,1})

Is the answer to 8 - (A) ? (As we assign Q into P and not vice-versa).

I think I did not explain it quite clearly. What I meant with the addition part was that we add (N/2) when we find the first K and not all the numbers. That is, 1. We scan the complete array from start till max (N/2) places (oh yes, we have to find CEIL(N/2) or else the complete algo will be pointless :)). 2. If we find no K within that range, then clearly no such number exists in the array. 3. Else, AFTER we find K at location X, we add FLOOR(N/2) to X, keeping in mind the overflow, and check if the value of current location is still K. If yes, we have found the answer. So, taking your cases in consideration, please check the inline comments in the quote. Please let me know if there still are any loose ends you can help me with. Thanks! :) (I hope I am not getting the question completely wrong.) Oh dear! I guess I did get the question wrong. The above explanation is seeming pointless now. Yes azk, hashing is the best method to do this thing. 4 days away and I commit such an unforgivable blunder. I hope things do not get more messier. :(

azk sure has a wonderful knack of solving grammer problems! Thanks for the solution and explanation azk! :)

Thats a fantastic approach azk! Thanks for detailing it! However, if I am not wrong, I see that everytime you have to scan the atleast N/2 + 1 elements by the hashing approach while the way I suggested scans atmost ( N/2 - 1 ) elements (from start - sequentially). I do not know if I am missing something here. Or I think I will have to recheck this again. Btw let me restate it properly now what I had intended the algo to act as earlier. I still might be missing some small bits and pieces but I think this might be a bit faster than the approach with hash tables (as we already know K in our case). Please let me know if I am leaving out something. Thanks! :)

70. Let N be the set of all natural numbers. Which of the following sets are countable? I. The set of all functions from N to { 0,1 } II. The set of all functions from { 0,1 } to N III. The largest subset of N (A) None (B) I and II only © I and III only (D) II and III only (E) I, II, and III Ans: Can anyone please explain me this. I had an opinion which was contradicted by my friend, so I wanted to know how exactly to proceed while working on such type of questions (i.e. countable sets). Thanks! :)

Thanks for clarifying that, azk. But, I still have a small doubt in here. Aren't the outputs exactly opposite for both AND and NAND gates shown above? Although, our interpretation of Truth logic leads both of them to have the same Truth results (true/false), the actual output with respect to the bits, is still acting as a differentiating factor between the two gates. Would the gates be called one and the same then? Thanks!

Oh Thanks. I think that explains quite a lot. So, let me re-summarize what you just said. Integer to be checked for (N/2 + 1) times appearance: K Array of N length: a[N] Algo: 1. Start searching for the 1st K. If K exists we will find it within first N/2 places. 2. Let 1st K be found at Xth place i.e. a[X] = K ..(X 3. Now as the array is already sorted, there has to exist atleast ( N/2 + 1 ) instances of K in the form of a continuous chuck. So, we directly compute the supposed end of this chunk by adding ( N/2 + 1 ) places to X and see if its value is still K. i.e. Check if a[X + (N/2 + 1)] = K and we are done. So, the number of comparisons would include just, - Computing the answer through above formula and verifying it after we have found out the first K and which amounts to theta(1) time. Did I get it right? Incase of an unknown integer within a random array, your method seems to be perfect as it gets an outcome within theta(n). But, I would like to experiment with it too and see if it can be done through any other way just for sake of it. :) Thanks again!

Hey azk, can you please elucidate on the algorithm used to do that? Thanks!

Hey azk, did you check out nonevent99's explanation (Post #4) for Q4 ? I think the way they have mentioned the "positive" and "negative" terms it sure has to do something with the implication of input of bit values. Else, they could have expressed it directly as "An AND" gate. Plus, this being an AGRE question, and not a undergrad quiz question, I would be in dual minds to tick (a) as an answer, as that seems to be an automatic choice at first glance, which in turn spells it as a trap. Or maybe I am just too paranoid. :)

I think we have a really good explanation to that by nonevent in the same thread. (Although he acts a bit ambiguous about it in the later part, this does serve as a basic explanation in a way. If you try experimenting with it, you will surely hit a more correct version.) http://www.www.urch.com/forums/30197-post16.html Do check the rest of the posts on that thread too. And if you still are stuck with any subtleties, feel free to ask about them specifically.

It would be real great if it works that way. Thanks for your insight CL. You have helped a lot. :)