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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


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