The power sums of rearrangements of a list of numbers match. f=.[:+/]^/[:i.# (f 5?5)-:f i.5 1
Den 10:07 torsdag den 10. juli 2014 skrev Jon Hough <[email protected]>: > > >The following two 4x4 arrays are rearrangements of each other's rows. > > >3 2 1 0 > > > >1 0 3 2 > > > >2 3 0 1 > > > >0 0 0 0 > > > > > > > > >2 3 0 1 > > > >3 2 1 0 > > > >1 0 3 2 > > > >0 0 0 0 > > >I would like to know a way to acknowledge two arrays as being rearrangements >of each other. Eventually my goal is to compare long lists of such arrays and >nub out duplicates - duplicates being rearrangements. > > >The only way I can think to do this is to cycle through all permutations of >the row of one of the arrays and test for equality with the other array, using >A. . Of course, there are 24 permutations to test for 4x4 arrays, but >obviously for bigger arrays things get worse. > > >Is there a faster way to check two arrays are (ignoring row permutations) >equivalent? > > > > > >---------------------------------------------------------------------- >For information about J forums see http://www.jsoftware.com/forums.htm > > > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
