Cody

# Sven

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 on 17 Jan 2017 at 20:10 Sven received Speed Demon badge for Solution 45288 on 16 Jun 2015 on 9 Dec 2013 on 9 Dec 2013 on 9 Dec 2013 on 9 Dec 2013 on 9 Dec 2013 on 9 Dec 2013 on 9 Dec 2013 on 9 Dec 2013 on 9 Dec 2013 on 9 Dec 2013 on 26 Feb 2013 on 14 Feb 2013 on 14 Feb 2013 on 14 Feb 2013 on 14 Feb 2013 on 21 Jan 2013 on 7 Oct 2012 on 27 Sep 2012 on 2 Sep 2012 on 2 Sep 2012 on 2 Sep 2012 on 30 Aug 2012 Sven submitted Solution 132555 to Problem 325. 2 b | ~ 2 b on 30 Aug 2012 Sven submitted Solution 132547 to Problem 431. Sorting on 30 Aug 2012 on 30 Aug 2012 on 30 Aug 2012 on 28 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 27 Aug 2012 on 13 Jul 2012 on 13 Jul 2012 Sven submitted a Comment to Solution 111016 Hehe... yep I'm sure that would send us scrambling to add a tolerance to whatever-the-current-best-solution-was. But if this were intended for a bwboundaries(BW,'noholes','minimal') type option to bwboundaries, then all coordinates could be assumed integers and even the normalisation would be an unnecessary step since bwboundaries never skips a pixel. on 13 Jul 2012 Sven submitted a Comment to Solution 110965 Well then, I bet that the following would run faster and work on any dimensioned data (such as a [10,2,50] sized matrix representing 50 sets of 10 XY vectors): normr = @(v)bsxfun(@rdivide, v, sqrt(sum(v.^2, 2))); on 13 Jul 2012 on 13 Jul 2012
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