Re: 15! permutations
- To: mathgroup at smc.vnet.net
- Subject: [mg108676] Re: 15! permutations
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Sat, 27 Mar 2010 05:09:25 -0500 (EST)
10^4 seconds is less than three hours. 10^4/60/60. 2.77778 In case you had your numbers switched, 10^9 seconds is less than 32 years: 10^9/60/60/24/364.25 31.7751 However... you shouldn't have used 9 orders of magnitude for one number and 4 for the other, so maybe you meant: (15!)^2/2/60/60/24/364.25/10^9 // Round 27167891 About 2.7 million years. Bobby On Fri, 26 Mar 2010 05:35:23 -0500, Nicola Mingotti <n-no-mingotti at g-spam-mail.com> wrote: > On 2010-03-25 12:11:52 +0100, bn77 said: > >> Hi, >> >> I'm trying to write a program in mathematica that compares roughly >> (15!)^2 / 2 pairs of permutations of length 15. Can mathematica do this >> in a reasonable time? Any experience of this sort? >> >> TIA, >> bn77 > > The number of permutations is then : > N[((15!)^2)/2] => 8.55006*10^23 > > Supposing you can compare 10^9 objects per second > you would need 10^4 seconds that is 3.171 milion years > according to Wolfram Alpha conversion. > > So, no ! If you really need to cycle through all these objects > it's impossible. Brute force here fails, you need to find a smarter > way to solve it. > > bye > > Nicola. > -- DrMajorBob at yahoo.com