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Oct 14, 2015 at 23:18 history edited J. M.'s missing motivation
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Jul 9, 2013 at 6:49 vote accept Zibadawa
Jul 8, 2013 at 7:49 answer added Zibadawa timeline score: 2
Jul 8, 2013 at 2:20 answer added halirutan timeline score: 14
Jul 7, 2013 at 21:13 comment added Zibadawa @Michael Ah, I see. I've edited the question to use bracket notation, instead. Hopefully that is more readable. I'm not familiar with using that notation in notebooks, or currently understand what the upshot of easier downvalue assignments are. Would I be correct in assuming that notation change is nothing more than replacing all instances of $d_i$ with d[i] in my notebook?
Jul 7, 2013 at 21:08 history edited Zibadawa CC BY-SA 3.0
give better mathematica notation for my variables
Jul 7, 2013 at 20:48 comment added Michael E2 @Zibadawa Preferably, one should be able to copy and paste code blocks into Mathematica and test out the code. TeX code is not functional when pasted into Mma. For big or long problems, a simplified example that illustrates the problem is acceptable (and preferred). You can also use d[i] as your variable and have it formatted as a subscript. See this and perhaps point 4 in this.
Jul 7, 2013 at 20:34 history edited Zibadawa CC BY-SA 3.0
formatting explanation on the d_i
Jul 7, 2013 at 20:32 comment added Sjoerd C. de Vries @halirutan That loop isn't that big. It takes only 22 s on my laptop.
Jul 7, 2013 at 20:29 comment added Zibadawa @Michael The underscore stuff was not part of the original Mathematica code. They should be properly typeset subscripts; so type them in as d ctrl+_ i, so it will look like $d_i$ (similary, $d_i^2$). Sorry about that, I couldn't find an easy way to transfer that into the original post at the time. Is mixing latex into the code block acceptable and functional?
Jul 7, 2013 at 19:58 comment added Michael E2 The real problem aside, I don't think code you posted can be right. I get syntax errors (missing brackets and so forth). And d_i is short for Pattern[d, Blank[i]], which probably ought not to be used as a variable. (For instance Solve[(d_i)^2 == d_i, d_i] returns no solutions.)
Jul 7, 2013 at 18:08 comment added Pankaj Sejwal Sjoerd's suggestion will work good in case your tuples are less but the increment in tuples is 2^n and algorithms become tortoise. In case of using lattice you will always work closely with relevant information.I would suggest you to find info on "lattice of subsets" and first page of google will take you places. No matter how much you break your tuples, you will not get desired results unless you have few days to spare for one run.
Jul 7, 2013 at 16:57 comment added Zibadawa @Blackbird I'm not sure what you mean. It sounds in the spirit of Sjoerd's suggestion of breaking the problem up into pieces. Do you have any further details or links for what you are suggesting?
Jul 7, 2013 at 16:48 comment added Zibadawa @SjoerdC.deVries That idea would seem to be a limited fix. It might work for some examples, but would not scale particularly well to larger ones. I'm not sure right now how to generate a code block that would split the tuples up into a dynamically determined number of manageable subsets. At least not in Mathematica.
Jul 7, 2013 at 16:41 comment added Zibadawa @halirutan n=30 is fairly easily achieved in 'small' examples. A particular example I've tried to compute that ran into a memory error had about n=24 to n=28 on each of 5 sets of equations.
Jul 7, 2013 at 13:01 comment added halirutan How large is n typically in your cases? I mean with n=30 you have 2^30 tuples which is a number too large to do even the simplest loop Do[Null, {i, 2^30}];.
Jul 7, 2013 at 11:10 history tweeted twitter.com/#!/StackMma/status/353833407727411200
Jul 7, 2013 at 9:11 comment added Pankaj Sejwal Refer the algorithm to find subsets using "lattice of subsets" which actually starts for small number of promising subsets and than using them together to generate bigger subsets automatically ignoring irrelevant subsets.
Jul 7, 2013 at 8:26 comment added Sjoerd C. de Vries Would it help if you split your Tuples[{1,0},n] into two Tuples[{1,0},n/2] lists and work through each combination of the two lists, joining elements as you go?
Jul 7, 2013 at 5:26 review First posts
Jul 7, 2013 at 8:21
Jul 7, 2013 at 5:09 history asked Zibadawa CC BY-SA 3.0