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Timeline for Multiply with restricted operations

Current License: CC BY-SA 4.0

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Apr 2, 2020 at 6:59 answer added Bubbler timeline score: 4
Mar 16, 2019 at 17:27 history edited xnor CC BY-SA 4.0
deleted 564 characters in body
Apr 13, 2017 at 12:39 history edited CommunityBot
replaced http://codegolf.stackexchange.com/ with https://codegolf.stackexchange.com/
May 18, 2015 at 3:17 vote accept xnor
May 18, 2015 at 3:16 history edited xnor CC BY-SA 3.0
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May 7, 2015 at 20:30 comment added xnor @proudhaskeller No, I'm looking for a proof of optimality. (Sorry, I had missed your comment in my feed.)
May 2, 2015 at 17:58 comment added proud haskeller @xnor Currently, I have the answer which works best without assignment allowed (assignment used only once, and will only double it (approximately)). Does that count for the optimal result without assignment, or does it need to be proven to be the optimal result?
Nov 8, 2014 at 21:23 answer added xnor timeline score: 37
Oct 27, 2014 at 18:03 history edited xnor CC BY-SA 3.0
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Oct 27, 2014 at 17:59 comment added Optimizer I see. So the existing 23 operations won't get 300 if no one else posts, right ? Also, are you really sure an answer less than 23 exists ?
Oct 27, 2014 at 17:58 comment added xnor @Optimizer Yes, operations, thanks. Not sure what you mean about a 24 op solution -- it can't be optimal because there's a 23. If you mean for the no-assignment half-prize, yes, I'd count an optimal solution for that that's more than 23 ops.
Oct 27, 2014 at 17:56 comment added Optimizer You mean 22 operations (instead of chars), right ? Also, if someone posts a 24 operations solution, will that count ?
Oct 27, 2014 at 17:52 comment added xnor @Optimizer No, the best theoretically possible. It might be 23, it might be less. A solution of 22 operations with no proof of optimality would not suffice. (Clarified in the post, thanks.)
Oct 27, 2014 at 17:50 comment added Optimizer @xnor What do you mean by optimal here ? less than 23 operations ?
Oct 27, 2014 at 17:49 history edited xnor CC BY-SA 3.0
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Oct 27, 2014 at 17:45 comment added xnor I will give a bounty of 300 rep to anyone who posts an optimal solution with proof. A brute force search can be a proof if you explain how it checks everything needed. An optimal result restricted not to use variable assignment will be given 150 rep. Other partial bounties may be given for partial results.
S Sep 11, 2014 at 4:04 history bounty ended xnor
S Sep 11, 2014 at 4:04 history notice removed xnor
Sep 11, 2014 at 4:03 vote accept xnor
May 18, 2015 at 3:17
Sep 10, 2014 at 23:11 vote accept xnor
Sep 10, 2014 at 23:12
Sep 4, 2014 at 23:57 comment added rationalis I don't know that there's a better choice for an algorithmic search other than brute force... And if the actual best solution is, say, 22 operations, that's not really feasible for a brute force search.
Sep 4, 2014 at 5:23 comment added xnor @proudhaskeller 33 :-P It was basically a worse version of algorithmshark's answer.
Sep 4, 2014 at 5:17 comment added proud haskeller @xnor How much operations you got when you solved this question?
S Sep 4, 2014 at 2:20 history bounty started xnor
S Sep 4, 2014 at 2:20 history notice added xnor Improve details
Sep 2, 2014 at 21:20 answer added proud haskeller timeline score: 27
Sep 2, 2014 at 14:16 answer added user31301 timeline score: 10
Sep 1, 2014 at 21:53 comment added xnor OK, thanks for the feedback, I guess I should have realize that coders would interpret numbers as stored in bits. I edited it to try to make clear I'm requiring arithmetically correct results for all but a measure-zero set of pairs of real numbers.
Sep 1, 2014 at 21:51 comment added Eric Tressler @flawr I was typing up the same stuff when I saw your comments. Exactly, the set of IEEE floating point numbers is measure zero, and so is all pairs of them. And so is the set of all numbers representable on any machine.
Sep 1, 2014 at 21:48 history edited xnor CC BY-SA 3.0
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Sep 1, 2014 at 21:47 comment added flawr Of course it is clear what you mean but I think it does not help using precisely defined technical terminology for something that does not meet the definition. Sorry for being a stickler! In my opinion you can also write I know what you mean or (most as in common sense).
Sep 1, 2014 at 21:45 comment added xnor @flawr I was imagining the fake language to express arithmetic formulas for real numbers, not act on bits or any sort of finite representation, so I think the measure requirement over the real plane (a,b) ∈ R^2 works fine. But if you actually think it's unclear, I'll edit it.
Sep 1, 2014 at 21:43 comment added flawr Ok since not everyone thought my anwer was as funny as I did, I deleted it and comment here: The rule about the measure zero set is not very wisely chosen since rational numbers are a measure zero set regarding the lebesgue measure, I'd suggest using a certain percentage instead. (Or another kind) But I totally like the idea of this challenge!
Sep 1, 2014 at 21:12 answer added algorithmshark timeline score: 14
Sep 1, 2014 at 21:07 history edited xnor
Added tags.
Sep 1, 2014 at 20:31 history edited xnor CC BY-SA 3.0
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Sep 1, 2014 at 20:30 comment added xnor @MartinBüttner Tiebreak is earliest posting in that case. I think there's a good amount of room for optimizations, so I don't think it will just be a race to find one that works and write it cleanly. At least, that's what I found in trying it; maybe someone will find a clearly minimal solution.
Sep 1, 2014 at 20:27 comment added Martin Ender This feels like a challenge where an optimal solution is likely to be found (once any solution has been found). So what's the tie breaker in that case?
Sep 1, 2014 at 20:24 comment added xnor @Ypnypn Yes, and I've written an example to make sure.
Sep 1, 2014 at 20:19 comment added Ypnypn Is this even possible?
Sep 1, 2014 at 20:19 history tweeted twitter.com/#!/StackCodeGolf/status/506536900753375232
Sep 1, 2014 at 20:14 history edited xnor CC BY-SA 3.0
Fixed grammar.
Sep 1, 2014 at 20:03 history asked xnor CC BY-SA 3.0