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required and cannot have a multiple. That still deviates from the subset sum problem. Allowing for an infinite supply for all the other integers means that I only have to show what the minimum difference is in the set. In your example of {-2, -4, 6}, the minimum difference is 2. So, with that set, I can get any result of n*2. If n must be an integer, n*2 will never equal 7. If it was a problem where the answer was 'yes', I still wouldn't know what the answer might be. I would just know that the destination result is a multiple of the minimum difference and, therefore, there is an answer. I admit that I'm not taking into account the number of times you can go up/down in this part of the discussion. I'm merely trying to explain why I don't see how this reduces properly to the subset sum problem - which doesn't mean that I haven't read what you wrote. I merely disagree with what you wrote. I am happy to continue discussing this with you if you like. You are handling this discussion very well compared to the students (and often professors) that I normally talk to. In reality, this is the time of year that I am flooded with students who just realized that they don't know how to handle a two-dimensional array or somehow skipped every class that discussed how pipelining works. But, I fully understand if you do not want to discuss it. I will only respond if it appears that you want a response.
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variables and constants must be integers. I believe that the entire argument centers on the fact that you do not see it as ax+by+cz... = d. They way I see it, it is a linear
Diophantine equation (I know that it is also listed as a denumerant problem and as an unbounded knapsack problem). Therefore, all previous work with linear Diophantine equations apply. My point, from my very first statement, is that asking "is there a solution" is often easier than "what is the solution". Getting a solution for a linear Diophantine equation is a counting problem and known to be #P. Asking if a solution exists is a very different question. Papadimitriou, Faliszewski, and Hemaspaandra are, to my knowledge, still debating this (and I plan to discuss it with them at conference in July). They all agree that if all variables are positive, it is #P-complete to ask "is there a solution." What if some are negative? I posed that I could convert a linear Diophantine equation with negatives in one that only contains positives and therefore, it would be #P-complete. That was shot down quickly because I would be adding extra constraints on the constants. Until I see proof that it is NP (or more precisely, #P), I will continue to state that it is possible that the "is there a solution" question might not be NP.
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science enthusiasm ... not implying that you must be from
California or that you are worse off if you aren't). To the problem: The minimum difference in {7, 3, 5} is 3 since there is no negative value to compare the positive values to. There is no n such that n3 = 4. So, I can easily say that there is no solution. In the example you previously had and replaced, the minimal difference for {-9, 3, 5} is 4. There is an n such that n4=4. So, there is a solution. You don't compare same signed numbers for difference, only opposite signed numbers. Then, you ignore multiples. This is based on a proof of the Chinese Remainder Theorem that I read decades ago. It is what immediately comes to mind when looking at this problem (just the "can I make a sum", not the entire "balloon" problem). In that proof, it was shown that the subset sum problem was not NP if multiples were allowed. I'm sorry that we are obviously miles apart in our understanding of looking at the same text.
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C and I want them to sum up to D, I am looking for xA+yB+zC=D. The issue is that there are two problems. One is "What are the values for x, y, and z?" The other is "Is there an x, y, and z that can solve the first problem?" The "is there a solution" version, which is what I believe you asked, is much easier. First, you can easily answer no if all of the values of A, B, and C are negative and D is positive. Similarly, if A, B, and C are all positive and D is negative, the answer is no. As long as at least one of A, B, or C is the same sign as D, there is *a* set of values x, y, and z that will solve the problem. But, that doesn't mean that there are integer values of x, y, and z that solve the problem. Further, it doesn't tell you what those values might be. Do you agree with this so far or am I still way off on your problem?
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tropical cyclones which I felt was trivial, when your looking at telling the story of
Meranti since there are 4 typhoons which make the list this year thus far. The other part is that the JTWC and their intensity estimates are unofficial and we have to remember that most of the world is based on 10-min winds. I also noticed that we are currently saying that Winston had the exact same 1-minute winds as Meranti. However, you reverted fairly as you felt that "Meranti was notable for its intensity estimates and this should be reflected here." As a result I think we need to work together to find a solution to both of our concerns.
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polynomial time approximate solutions." The knapsack problem does have a FPTAS. The textbook I use for reference is "Approximate
Solutions" by Vazirani. I admit that it is considered a "dirty" science. Who really wants an approximate solution? But, it answers the question "does it exist" and often there is a fully polynomial time approximation. That goes back to the only point I've intended to make: There *may* be a polynomial time answer to "does it exist" for your original problem.
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won't have it. Please continue to pursue this and you should eventually see that you can turn it into a linear
Diophantine equation. You don't need to solve the equation to get a "yes" or "no" answer. However, solving those equations is very simple. Then, adding the part about going up and down is a twist that I don't feel will change the solution much.
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I'm not claiming to be an expert in this field, but it's clear to me that you aren't reading what I'm writing/wrote. I'm not sure what you're wondering about, but if it is the NP-completeness of the unbounded subset sum problem, make sure you understand precisely what that problem means, and refer to
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Ok I misread both Chaba and
Meranti's estimates and yeah i agree that the MH should be cut down to size while being cleaned up. I have made several seasons GA quality and would love to take a PTS there if we could all pull together on it. However, I still disagree that it should be mentioned that it
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While I seriously doubt we will ever come to any agreement about the problem statement, I have a gut feeling that your problem boils down to taking a set of numbers (I believe they will be integers) and seeing if you can sum them up to a specific value. As an example, if I have the integer A, B, and
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If what you are asking for is a proof sketch, I can happily provide it at a later time (I am about to go off and do something else). But again, I doubt you'd understand. You did not even get the order of the reduction in question correct, and as before, "not NP" is clearly false - this problem most
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I'm sorry, but I still see this as being far too similar to asking if ax+by=c for some value of x, y, and c. You are allowing multiples of x and y, so you have the a and b to contend with. I've tried to nudge you back through time, hoping to eventually get to
Diophantus. It is clear that you simply
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The difference between the subset sum problem and your problem is that you have multiples of each integer. That is a huge difference. As you appear to have already noticed, getting a sum of zero when multiples are allowed for all integers is trivial. Your complication is that one of the integers is
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You are correct that the
Knapsack Problem is NP as an optimization problem. Asking "does a solution exist" is different. What if I can approximate a solution? It isn't optimal, but it is a solution? That is what I work with, approximate solutions to NP problems. Technically, they are called "fully
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Apparently, this still bugs you. I still disagree with the claim that this absolutely must be an NP-complete problem. I see that you have a set of variables (x, y, z...) and you want to come up with a set of constants (a, b, c...) such that ax+by+cz... = d. The constraint is that all values of all
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does not apply here because the numbers are not coprime, and we have negative numbers in the mix as well. So the problem is not equivalent to "is my target number an integer multiple of the minimal difference". Have you read the text I linked for the proof of the NP-completeness? Another source is
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It appears that the gap between what you are writing and what I am reading is far to wide to cross, but I do admire your enthusiasm. I haven't been lucky to have many enthusiastic students or colleagues since I left
California (I think proximity to Silicon Valley is strongly correlated to computer
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Meranti had higher 1-minute winds than
Winston and that is reflected in the article ("surpassing..."). Instead, it would be better to cut down on the excessive information on the storm's meteorological history (which belongs in the storm's own article). And I only recall three typhoons making the
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I don't understand what your issue with the improvements that were made to this template are, but reverting all the changes just because some text is not bold is preposterous. If you would like text to be bold that is not currently bold, there is the sandbox. Or you can post on the talk page and
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By the way, my original gut feeling about my original problem was that it had to have an efficient solution. Results of NP-completeness can be surprising and I spent hours convincing myself my reduction was correct (see also my proof sketch below). And on the subject of approximation algorithms,
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I have since updated my problem. You may claim P = NP if you want, but in general your "minimal difference" method is doomed for failure. For example, it would incorrectly answer "no solution" for the instance A = -3, B = 5, C = 9, D = 27, because 27 is not even, even though there's the obvious
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Hi Jasper, As per usual we are now over 200000b for the PTS article and as a result I am looking at all of the sections in order to try and clean them up. As a part of this earlier, I removed some information relating to Meranti's intensity in comparison to Winston and the list of most intense
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can be solved in polynomial time while 3-SAT is NP-complete, even though both are so similar - so "similar" doesn't cut it. In particular, ax + by = c can be efficiently solved like you said, and in fact, introducing a 3rd variable can be most likely solved efficiently (based on what I've been
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that you'll always be able to find an answer, even if one of A, B, and C have the same sign as D, because I have constrained x, y, and z to be integers. The problem instance with D = 7, A = -2, B = -4, C = 6 has no solution because no integer linear combination of even numbers will make an odd
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NP-complete via reduction from subset sum (the proof of this is left as an exercise); for a single Diophantine equation (again with variables constrained to be nonnegative - that is important!), the unbounded subset sum problem shows that it too is NP-complete in general. Your comment is very
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reputable texts that establish NP-completeness of unbounded subset sum. And in any case, given that you earlier did not understand what I meant by a "reduction", you appear to be unqualified to be talking about complexity theory - I've been replying primarily for your benefit, but I doubt you
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of the numbers in question, but there is the obvious solution 1(-1) + 25(4) + 1(8) + 0(16) among many others. In fact, it would erroneously conclude that whenever the minimal difference is greater than one, and the target number is prime, larger than the minimal difference, and not one of the
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my reduction (and clearly you don't understand what a "reduction" is - I don't care about reducing it to the subset sum problem because like every problem in NP, this problem can be reduced to it - I'm interested in proving it's NP-complete, which requires a reduction
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solution x = y = 0, z = 3, and also the solution x = 6, y = 9, z = 0 or x = 6, y = 0, z = 5. A less trivial example with more than 3 elements in the set is the set {-1, 4, 8, 16} with target value 107. Again, the minimal difference is 3, and 107 is not divided by
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generated by the numbers in the set we have (as you propose): the example with A = 6, B = 3, C = 5, D = 4 has no solution either, even though 1 is the "minimal difference". In fact, I should even correct myself here: the instance where A = B = C = 0 actually has
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is open from Monday, 00:00, 21 November through Sunday, 23:59, 4 December to all unblocked users who have registered an account before Wednesday, 00:00, 28 October 2016 and have made at least 150 mainspace edits before Sunday, 00:00, 1 November 2016.
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The unbounded subset sum problem, or the (equivalent) version of this problem where the target sum is not 0, but some other arbitrary number (the proof of this equivalence is left as an exercise for the reader), is shown to be NP-complete in
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And it is completely in line for me to revert a massive change when I don't agree with it - and cosmetic reasons are very valid reasons to revert. Per BRD, you should not be reinstating your change before you settle the dispute over
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understand what I mean by "NP-complete", nor do you seem to be willing to look at the sources yourself. And in general, solving systems of linear Diophantine equations with the variables constrained to be nonnegative is
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I also saw that it's a linear Diophantine equation - with the variables constrained to be nonnegative, and I happen to have had the pleasure of learning from Papadimitriou himself on this subject. I already told you
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are solvable in polynomial time (as alluded to by other commenters on that thread) while the decision version of integer linear programming is NP-complete. Note that rephrasing the unbounded subset problem to allow
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JMA still clearly states a TD (former Nock-ten) active as of 27/12 18Z. I'm sure we talked about this before and stated that if JMA doesn't show a TD or system (in its Weather maps) then it had already dissipated.
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defined for decision problems, and second, the approximation algorithms we know can't be used to find an exact (i.e. always correct) solution to the decision problem. Otherwise we would not be asking if P = NP.
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linear combinations makes it trivial: if A, B, C... are all zero, then there is no solution unless D = 0 as well, otherwise (without loss of generality) A is nonzero, then we can set x = D/A and y, z, ... =
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before making such a change to an important infobox - this is not a typical infobox either, as it is more intended as a status indicator (akin to the vandalism meter above on my talk page) than a usual
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683:. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose
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Hi, after I looked to the 1970 South-West Indian Ocean best track from Tropical Tidbits, I saw that Iseult formed in there, I also looked that Harriet never formed in the Australian region, here:
578:(beginning with "In fact,"). We can enforce there being only one of each subset item by modifying the proof as follows: Change the dimension of the position vector space to |S| + 1; set
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Once again, you demonstrate that you need more experience with theoretical computer science. If you even bothered to read my reduction, you would have noticed that this doesn't matter.--
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Look at the map there very carefully - Do you see that Iseult supposedly formed right on the border of the Australian region which was 80E for a number of years before it was changed.
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For interested readers, my original problem can also be easily proven to be NP-complete using reduction from ordinary (0-1) subset sum. Refer to the proof I supplied
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for any "yes" answer, and obviously all we need to do is add up all those elements to show the validity of the solution, a polynomial time operation. This is the
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subset sum problem is also NP-complete. Want proof? It's in the paper I linked among others. The unbound subset sum problem still only allows you to have
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list (Nepartak, Meranti, Haima) this year (Chaba fell short at 905). The sub-900 pressure definitely merits mentioning that Meranti made the name.--
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what I wrote, that's a different thing, and I thank you for asking! But if it's that you don't understand my reduction, then you need to
590:= (T, 1, 1, ..., 1) for the target value T. Recall that at odd-numbered heights 2l - 1 (where l ranges from 1 to |S|), I specified that
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for new pages. The BBC together with Wikimedia UK is holding a large 12-hour editathon. Many new articles and drafts are expected. See
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as this problem for it imposes further constraints; having an integer linear combination of wind vectors reaching your destination is
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You have exactly described the NP-complete problem I reduced from - unbounded subset sum, which is a similar yes-or-no problem and is
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I would fix it myself, but I don't understand your code right now, and would rather have you fix it since you designed the change. --
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It's not even just that. Cosmetically, I think your version is not as good as the previous (code-wise, yes it is good to re-use
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are asked to be especially on the look out 08:00-20:00 UTC (that's local London time - check your USA and AUS times) on
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Knowledge talk:WikiProject Tropical cyclones#Using JMA weather maps to extend western Pacific storm tracks needs to stop
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and you are reverting a MASSIVE change because you "think it looks ugly that the text is not bold". That is absurd. --
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made the list, especially when you consider that to the best of my knowledge the list isn't replicated anywhere else.
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can be a simple yes/no question not requiring constructive proof of "yes". The only requirement is that if we have a
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I also spoke with Papadimitriou today and he affirms that the unbounded subset sum problem is indeed NP-complete.--
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as a complexity class, because my problem as formulated is a decision problem. General decision problems in NP are
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which proves that unbounded sum is NP-complete, by reduction from 0-1 subset sum. Linear Diophantine equations
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So that means, there was no Harriet-Iseult, just Iseult, if that information is not true, please tell me.
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the time, even without explicitly finding an answer. The same goes for other NP-complete problems like
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if it's occurring in your area of the world, and thanks for your work to maintain, improve and expand
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Can you leave me a link to "Infobox Musical artist", because i cant find it, even after searching--
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Yes, it's possible that P=NP. That's not related to whether this problem is NP-complete though. --
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have a rigorous understanding of the concepts here to be having a formal discussion about it. --
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the variables constrained to be nonnegative are easy, but that's a very important constraint.
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integers in the set, there is no solution - which is false in general. In particular, the
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If you wish to start a new discussion or revive an old one, please do so on the
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hand-wavy, whereas mine are rigorously grounded in complexity theory - you
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It is also wrong that this problem is not in NP. A solution multiset is a
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describes the Committee's roles and responsibilities in greater detail.
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number. The restriction to integers is important - linear programs over
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BBC 12-hour Editathon - large influx of new pages & drafts expected
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A consensus has LONG been reached to make all Infobox templates use
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If you wish to participate in the 2016 election, please review
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by Hans Kellerer, Ulrich Pferschy, David Pisinger (page 492).
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as problems that ask "constructively find me one solution".
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http://www.tropicaltidbits.com/data/TC/SI/tracks/1970.png
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will need modification, too, but this is just a sketch.
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is the panel of editors responsible for conducting the
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various publications that accept it as NP-complete.
152:indeed usually a simple yes/no question. It's also
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959:I'll just leave a message in there starting now.
398:rigorous proof of any method, while I have cited
797:BBC 100 Women 2016: How to join our edit-a-thon
594:would be the lth set member. Now instead let
1015:Have a prosperous, productive and enjoyable
1058:Knowledge:Notifications#Triggering_events
240:- this is a requirement of the original
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261:solution multiset. Also, a problem in
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668:2016 Arbitration Committee elections
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825:Merry Christmas and happy holidays!
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681:Knowledge arbitration process
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178:for a solution to my problem.
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1079:request a change be made. --
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979:Happy New Year, Jasper Deng!
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1224:Abdullah Almarri - A.W.S.T.
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903:Spread the WikiLove; use {{
702:and submit your choices on
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811:MediaWiki message delivery
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700:the candidates' statements
476:And I'm not interested in
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1175:Jasper Deng
1122:Jasper Deng
1011:Jasper Deng
963:Typhoon2013
949:Jasper Deng
940:Typhoon2013
924:Typhoon2013
880:Doug Weller
755:Jasper Deng
619:Jasper Deng
557:Jasper Deng
532:Jasper Deng
490:Jasper Deng
414:Jasper Deng
354:Jasper Deng
346:certificate
294:Jasper Deng
267:certificate
255:no solution
183:Jasper Deng
135:NP-complete
36:This is an
1230:Jason Rees
1165:Zackmann08
1145:Zackmann08
1120:infobox.--
1103:Zackmann08
1082:Zackmann08
877:. Cheers,
861:Enjoy the
768:Jason Rees
748:Jason Rees
732:Jason Rees
689:topic bans
350:definition
286:understand
219:You again
98:ArchiveĀ 30
90:ArchiveĀ 26
85:ArchiveĀ 25
79:ArchiveĀ 24
73:ArchiveĀ 23
68:ArchiveĀ 22
60:ArchiveĀ 20
875:Knowledge
685:site bans
636:Wyatt2049
463:countless
230:unbounded
1017:New Year
725:2016 PTS
552:a priori
400:multiple
352:of NP.--
259:nonempty
164:rational
1137:Infobox
1117:WT:WPTC
1111:Infobox
1063:Tigraan
478:Sharp-P
471:without
39:archive
1178:(talk)
1125:(talk)
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297:(talk)
186:(talk)
610:and K
598:be (S
576:there
387:2-SAT
250:ideal
154:false
143:3-SAT
16:<
1234:talk
1205:talk
1026:talk
884:talk
867:and
815:talk
805:for
772:talk
736:talk
714:talk
675:The
640:talk
630:LINK
612:down
586:and
515:talk
445:talk
410:must
405:also
371:talk
311:talk
226:from
204:talk
117:talk
1170:it.
947:.--
807:NPR
329:any
290:ask
139:all
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1148:(/
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1134:{{
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582:=
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396:no
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263:NP
206:)
167:0.
150:is
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159:Q
131::
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50:.
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