Re: [math-fun] Similar permutations problem
Wouldn't that be 20000*19999/2 = 199990000 *pairwise* comparisons ? Not a lot by today's standards.
Solution 1 compared with each of 19,999 others Soluiion 2 compared with each of 19,998 others .... = 20,000! comparisons.
I suspect the factorial notation may not mean what you think it does, or you have an unusual idea about the sum of arithmetic sequences, or something else is going on. In any case, today, (20,000 choose 2) is not such a big number, and if this is a real life problem, you should just go try that first. When you get to, say, 10M permutations, we can discuss how to do that. -tom On Thu, May 7, 2015 at 12:35 PM, Dave Dyer <ddyer@real-me.net> wrote:
Wouldn't that be 20000*19999/2 = 199990000 *pairwise* comparisons ? Not a lot by today's standards.
Solution 1 compared with each of 19,999 others Soluiion 2 compared with each of 19,998 others ....
= 20,000! comparisons.
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Tom Rokicki