Subject: Re: [xsl] How to Do Random "Shuffle"? From: "David Rudel fwqhgads@xxxxxxxxx" <xsllistservice@xxxxxxxxxxxxxxxxxxxxxx> Date: Thu, 18 Sep 2014 12:05:13 0000 
Wolfgang, the same type of weakness will appear in pretty much any version of this algorithm. And I suspect that the above is actually less susceptible to such bizarre cases than other algorithms because only one sequence is calculated rather than many. In any event he random:randomsequence function provides values to 64bit precision, so the likelihood that a problem arises is practically 0. The probability that two random values in a sequence are equal is less than n^2 / 2 * precision, so you would have to have an N of about a 1,000,000,000 before any significant probability of error (about a 3% chance) You could always do a check on count(distincevalues($rand)) if you were really worried about it, but unless your code is meant to navigate nuclear weapons, I doubt such a step is necessary. On Thu, Sep 18, 2014 at 1:48 PM, Wolfgang Laun wolfgang.laun@xxxxxxxxx <xsllistservice@xxxxxxxxxxxxxxxxxxxxxx> wrote: > What if the unsorted sequence is (0.413,0.192,0.888,0.513,0.522,0.413)? > > Admittedly, given existing implemenations of random generators for doubles > in [0,1.0), this is rather unlikely, but you may have a hard time proving > that it is impossible. > > W > > > On 18 September 2014 13:35, David Rudel fwqhgads@xxxxxxxxx > <xsllistservice@xxxxxxxxxxxxxxxxxxxxxx> wrote: >> >> That is not what random:randomsequence does. It creates a sequence of >> N random numbers between 0 and 1. >> >> But if you then find the index of each of these numbers in the sorted >> version of this sequence, **then** you have created a random >> permutation of the numbers from 1 to N, as the OP requested. >> >> So, by way of example, let's say random:randomsequence(5) spits out >> (0.413,0.192,0.888,0.513,0.522) >> >> Then the sorted version is (0.192,0.413,0.513,0.522,0.888). >> >> Taking each element (in sequence) from the original output of >> random:randomsequence() and finding the index in the sorted sequence >> yields (2,1,5,3,4), a random permutation of the numbers from 1 to 5. >> >> >> On Thu, Sep 18, 2014 at 1:14 PM, Wolfgang Laun wolfgang.laun@xxxxxxxxx >> <xsllistservice@xxxxxxxxxxxxxxxxxxxxxx> wrote: >> > random:randomsequence(N) >> > >> > If this is supposed to produce a sequence of numbers in the range 1..N >> > while >> > expecting it to contain every number of that range exactly once: would >> > this >> > truly be a "random" sequence? I don't think so. >> > >> > W >> > >> > >> > >> > On 18 September 2014 11:05, David Rudel fwqhgads@xxxxxxxxx >> > <xsllistservice@xxxxxxxxxxxxxxxxxxxxxx> wrote: >> >> >> >> When I have to do this (essentially create a permutation of the numbers >> >> from 1 to N), I combine random:randomsequence with saxon:sort >> >> >> >> I'm away right now so I'm not working on a machine with XSLT, so the >> >> following syntax may be off, but I use: >> >> >> >> <xsl:variable name="rand" select="random:randomsequence(N)"/> >> >> >> >> <xsl:variable name="sorted.rand" select="saxon:sort($rand)"/> >> >> >> >> <xsl:variable name="permutation" >> >> select="$rand!indexof($sorted.rand,.)"/> >> >> >> >> The select attribute of the last can also be written as "for $i in >> >> $rand >> >> return indexof($sorted.rand,$i)" . >> >> >> >> >> >> On Saturday, September 13, 2014, Eliot Kimber ekimber@xxxxxxxxxxxx >> >> <xsllistservice@xxxxxxxxxxxxxxxxxxxxxx> wrote: >> >>> >> >>> Using XSLT 2 I need to implement rendering of "match table" questions >> >>> where you have two sets of items, the match item and the thing it >> >>> matches >> >>> to. I want to present this as a literal table, where the first column >> >>> is >> >>> the matchfrom items in source order and the second column is the >> >>> matchto >> >>> items, in random order. >> >>> >> >>> I think this is best characterized as a "shuffle" problem, where you >> >>> want >> >>> to reorder a list randomly but all items in the list must be accounted >> >>> for. >> >>> >> >>> I can think of a recursive algorithm: given a list, generate a random >> >>> integer between 1 and the list length, select that item and add it to >> >>> the >> >>> result list, then call this function on the original list minus the >> >>> node >> >>> you just selected. >> >>> >> >>> Is there an easier or more efficient way to do it? >> >>> >> >>> Thanks, >> >>> >> >>> Eliot >> >>> bbbbb >> >>> Eliot Kimber, Owner >> >>> Contrext, LLC >> >>> http://contrext.com >> >>> >> >>> >> >> >> >> >> >>  >> >> >> >> "A false conclusion, once arrived at and widely accepted is not >> >> dislodged >> >> easily, and the less it is understood, the more tenaciously it is >> >> held."  >> >> Cantor's Law of Preservation of Ignorance. >> >> XSLList info and archive >> >> EasyUnsubscribe (by email) >> > >> > >> > XSLList info and archive >> > EasyUnsubscribe (by email) >> >> >> >>  >> >> "A false conclusion, once arrived at and widely accepted is not >> dislodged easily, and the less it is understood, the more tenaciously >> it is held."  Cantor's Law of Preservation of Ignorance. >> > > XSLList info and archive > EasyUnsubscribe (by email)  "A false conclusion, once arrived at and widely accepted is not dislodged easily, and the less it is understood, the more tenaciously it is held."  Cantor's Law of Preservation of Ignorance.
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