Re: How to make a loop for this problem!

*To*: mathgroup at smc.vnet.net*Subject*: [mg75397] Re: How to make a loop for this problem!*From*: Roger Bagula <rlbagula at sbcglobal.net>*Date*: Sat, 28 Apr 2007 05:56:57 -0400 (EDT)*References*: <f0pkt5$28h$1@smc.vnet.net> <f0sf6s$n2u$1@smc.vnet.net>

Peter Pein wrote: >pskmaths at googlemail.com schrieb: > > > >Hi, > >there are only 512 distinct 3x3 matrices with elements out of the set {0,1}. >Therefore you can shuffle the list {0,1,2,...,511} and convert each number n >of this list into a matrix. To do this, convert n into binary representation >with leading zeroes and partition this list of 9 digits into three parts: > >In[1]:= ><< "DiscreteMath`Combinatorica`" >mats = (Partition[IntegerDigits[#1, 2, 9], 3] & ) /@ > RandomPermutation[Range[0, 511]]; > >the first two matrices: > >In[3]:= Take[mats, 2] >Out[3]= >{{{1, 1, 0}, > {0, 1, 1}, > {1, 1, 0}}, > {{1, 1, 0}, > {0, 1, 0}, > {1, 0, 0}}} > >To check for duplicates (impossible): >In[4]:= Sort[mats] === Union[mats] >Out[4]= True > >I had to sort "mats" because Union gives a sorted output. > > >HTH, >Peter > > > Peter, Nice function! Looking at the determinants of the matrices you get a distribution! Try: << "DiscreteMath`Combinatorica`" mats = (Partition[IntegerDigits[#1, 2, 9], 3] & ) /@ RandomPermutation[Range[0, 511]]; a = Take[mats, 512]; Dimensions[a] b = Table[Det[a[[n]]], {n, 1, 512}] ListPlot[b] c = Table[Count[b, n], {n, Min[b], Max[b]}] ListPlot[c, PlotJoined -> True, PlotRange -> All] Roger Bagula