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#61




On the modelbuilder:
I'm using a quadcore processor with two threads per core. But the program never seems to use more than one thread (even when nothing else is running, it just jumps between 12 and 13 percent of processor time). Did I fuck up compiling? Or does each thread need to be called on explicitly?
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#62




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So the short answer: yes you have to create each thread explicitly. Depending on the program this doesn't have to be a lot of work. I have made an existing program multithreaded in an hour, once. However, it was quite suitable and well written when I started. 
#63




lol.
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#64




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Computationally difficult, but not the hardest mathematically. The factorial is a pain, but you can work with it. And over any finite set, you can create a lattice basis that includes both integer multiples of Pi and the multiples of the integers. When I was working on it the hardest part seemed to be getting the product of the sin values. And then you have to hope the function never zeros out over the positive real numbers, because then you need to take a cosine integral (which, as we've stated, is NPComplete).
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#65




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#66




The program is taking a ridiculously long time to run. And apparently web surfing cuts efficiency dramatically. One block completed in 28000 seconds while I was sleeping. Another is running now, and we're up to 123000.
Was the guy Von Neumann?
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#67




So it takes over an hour to read in the data after every iteration...should I just learn how to program in pauses and prompts? Or would the speedup of not restarting each iteration not be worth it?
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#68




It might be worthwhile to try to figure out why it takes so long to read. Maybe it will help you come up with a more efficient reading mechanism. If you're really lucky, it could even be something that would be useful in some other cases too.
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I do not anticipate the invention of a working time machine in the foreseeable future. 
#69




The problem may be that it's in a text format (predicting the type and location of the outputs seemed like a pain in the ass). Or that I didn't actually 'compile' anything in the end, I decided instead to use some tool a friend had that let me use either Fortran or Maple's language (which I am most comfortable with). I'll run it by some people who have some specific knowledge of the problem I'm working on. Though he hates being bothered, so it would be nice to have something to show him.
Anyway, on Terezian analysis: So apparently my lattice idea is, and I quote, 'fucking stupid', mostly for reasons I was aware of. The person I ran it by said while it might be the way to go in theory (though we had a debate on the efficiency of regression techniques I would rely on to sort elements into spines), in practice it is too slow. He did like my original idea (encoding music into an image, then using various techniques for finding edges, eigenfaces, and stegonographic information to build the metrics and operators), but mostly because it draws on some fields that are wellunderstood and documented. I'll walk you guys through both if you care, or I can give it another pass first. Also, something crawled across my feeds today that may be worth looking into for guidance:
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#70




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What might be the problem (though I don't know your code, so I'm guessing based on ignorance*) could be the way you add each newly read item to memory. If that involves rearranging the whole rest of what was already read, then you've created an Nsquared mess right there. And if it also means making memory assignments, then it would be a very inefficient Nsquared mess to boot. If this is the case, then the solution would be: first read, then sort in place. * Which, of course, is the best kind of guess to make. It gives one oodles of plausible deniability, and if one is right then that's amazing.
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I do not anticipate the invention of a working time machine in the foreseeable future. 
#71




Okay, I think the problem is the fact that I'm treating some very large lists as sets. A few thousand sets with a few thousand elements in each. So, since I'm using the Maple language for those, I'm guessing that Maple doesn't check for duplicate elements very efficiently.
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#72




As for this:
They don't have anything really novel in terms of audio encoding.
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#73




When I compile it's too slow, and when I use the interpreter it can't read in efficiently. I can't even begin to describe how pissed I am. I'm ready to fucking crucify someone for this. And if it's me, so be it. As long as someone bleeds.
Anyway, on a lighter note, something I've worked on before is reversible computing as SAT. It's closer to what we've done before in this thread, and pretty interesting in its own right. Unless someone wants to take over, or wants to see the Fourier+lattice techniques for music theory (so we'll ignore computability issues), or wants this thread to die.
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#74




How can you crucify yourself? Suppose you manage to nail down one hand, what will you nail the other hand with? The first hand being full at that point. Of bloody nails that is. Not the kind at the ends of your fingers. Man there are a lot of puns in this line of thought.
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Warder of Freya Sedai Firstbrother of Cary Sedai Great Lord of Fire Lord Captain Commander of Singing Chipmunks Master of Nazgul Kitchen 
#75




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No problem, apart perhaps from some engineering stuff which a proper scientist dismisses as "trivial".
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I do not anticipate the invention of a working time machine in the foreseeable future. 
#76




As Gonzo notes, designing a crucifixion machine is basically just an engineering problem, and well within the scope of this thread. Though my Catholic background compels me to note that the nails don't actually go through the hand (you can ripthe hand away). They gobetween the major bones in the forearm.
Why can interpreters run in multiple threads without special coding? Also, I might get to reversible SAT later, but anyone's interested in Terezian analysis... Quote:
Anyway: one of the above, or another pass at Kryptos?
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#77




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Warder of Freya Sedai Firstbrother of Cary Sedai Great Lord of Fire Lord Captain Commander of Singing Chipmunks Master of Nazgul Kitchen 
#78




Okay, a quick rundown on some mathematics of nonstandard computing. True quantum computers (iow, based on entanglement, not the Canadian adiabatic bullshit) rely on quantum bits that each take the form of a probability distribution. You can set each one independently, and when entangled they are understood by the model of a 2^n dimensional circle. From here you perform a number of flips and rotations to the system until it encodes the problem under consideration. At this point, you perform a quantum operation (like the quantum fourier transformation) that collapses the state. A quantum computer is 'complete' if it has a functionally complete set of logical operators and at least one quantum operator. It's important to note that results are only probabalistic, the qubits are not entirely independent, the functional operators need to be reversible, and the quantum operator is irreversible (you can't uncollapse a state).
There are other nonstandard computing models, like the analog computer the Dutch use to manage their unique water issues. I believe it uses voltage and current to directly model the water. Other models use photonic processes, pneumatic processes, mechanical processes (some of these are really cool. I met a local artist who builds some amazing clockwork computers for collectors), or even purely hypothetical ones. And the model I am currently trying to design is another. As we've seen, any operation computable by a Turing machine can be reduced to a circuit of AND/OR/NOT gates. But we can intuitively see that more information per 'unit' (bit, qubit, obit) means greater efficiency. As it happens, my model uses sets of elements in a finite field instead of bits. But, with just a little tweaking, we can see that we stil have gates that function like AND, OR, and NOT; and De Morgan's laws still apply. In addition, you can develop special operators that act directly on the elements of the sets themselves (for instance, if the finite field was integers modulo 7, you could have an affine transformation operator). So: AND becomes the intersection of the sets, OR becomes the union, and NOT becomes the complement (sticking with integers modulo 7, the complement of {0,1,2,3} would be {4,5,6}). The empty/null set is analogous to FALSE, and the full set corresponds to TRUE. Partial sets are used to rigorously manipulate discrete probability distributions of nondetermistic variables. I have to say, I am really fucking proud of the work I've done in the field, and my model will work with certain hardware better than anything else currently out there. Unfortunately the guy in charge of the project just wants to throw computing power at a primitive genetic algorithm to build the model. He has no understanding of what it will look like, what the extra operators can be, and how they behave. So like I said before, I'm on my own.
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#79




Okay, I was bored and already testing some things in Maple when I decided to take another look at the problem I got hung up on last time. Turns out I just misinterpreted a line (and I confused the technique with another I'm more familiar with. Though the authors could have been a bit clearer in their notation). Turns out s is just twice the target minus the summation of the multiset. And since lambda is just there to ensure what we might call 'technical compliance', we'll ignore it, and just look at the straight residue (as opposed to the quadratic diophantine version). But, to show the relationship:
Quadratic Diophantine: A*x^2+B*y=C with A, B, C, x, and y all in N (natural numbers). Since you can assume A, B, and C are all relatively prime (no common factors. And you can assume it because the Euclidean algorithm makes it easy to find gcd(gcd(A,B),C), and you can then just factor it out), you can rewrite it as: A*x^2=CB*y A*x^2 = C mod B x^2 = C*A^(1) mod B Since C*A^(1) mod B is easily worked out, you can show that if the square of a natural number x is congruent to some number Q modulo N (where Q=C*A^(1), and N=B), then there exists a solution to the subsetsum problem used to generate Q and N, provided x is less than some positive number t (also generated from the subsetsum problem. It's essentially a holdout of the restriction of the domain on the Quadratic Diophantine prolem). You find Q using the Chinese Remainder Theorem on the following congruences: Q=t^2 mod P and Q=s^2 mod 2^(m+1) while N is equal to P*2^(m+1) P is the product of a set of distinct prime numbers (other than 2) equal in size to the multiset raised to the power m (the number of bits n the summation of all the elements in the multiset), while s is the summation of a set of values obtained with the Chinese Remainder Theorem on the following congruences: value[i]=multiset[i] mod 2^m and value[i]=0 mod P*primeset[i]^(m) You must also ensure that the values are not congruent to zero modulo the corresponding prime. Now, this problem is NPComplete, which means that it is a deciscion problem. Finding a suitable x only means that a solution to the subsetsum problem exists. To deconstruct the x value into the members of the subsets, you need to test each member (appropriately tweaking the values) and see if the congruence still holds (but these are much easier to do than finding the initial satisfiable x). That being said, here are the Q, N, and t values for the tattoo problem, along with a satisfying x: Q= 13336001699742420686934233547272257409195346594156 82724534116136000988597221417465786711259086166981 18171681770816177404214012655963100828749652830208 54284290255887671935141179833791741577165394255213 75283889443416823485393064744585773608773611552790 75397713471851802327593114552596382780605290453652 07339740598748555180385668281023749958339598750179 20885085371761835725344335119084050637212831886860 15401275935321702864399786025 N= 17637633085770622730167837158713586993955291176605 22975805775356565230649360756421710721188160488822 39650926420928748408886434689636325996966686327607 33996956386638348176583805095951413458974055718811 51510682141417609797054161585105219795825853625704 22912499358432081648005992562325172825869068050796 05252968582528272876272691000467076037340989243503 05805319436099832720320261381112451194704105295364 60897537747220000000000000000 t= 14617619954027659287490340612037252463102826513514 76973197696361287780797932664617294740806305686793 85859046671490428808078940755393128988066259149277 37188540029303521452816955002512704789065694036647 13578929639092785029041463685283867334672994721157 83464940159141940503120547586825551777893350315713 84559579256550517603272266403906272756186227781821 64874364346618291695662594231636567551201082790345 54834031231243103645 x= 14617619633525147492129890288700729114868914534751 46277820262346268891743548951014757483385733388587 36365592106335149370810012590102613115015950247250 45588814771855724069959090134347241495837143808517 32054100734545378712219504178935025638417080045028 60889944097138095731539434077244907991703641901644 39733968042396973272497368511048003935827847781320 03469245351589859856719536806406258223874158777088 79853608867961853645 As you can see, even a relatively small problem blows up pretty quickly under repeated reductions.
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#80




Just for fun, I took a factoring problem as 3 SAT and converted to a Quadratic Residue problem. The number being factored was 6. And look at what I got:
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