Re^5: Speeding permutation counting

BrowserUk,
I am likely misunderstanding the problem. For the sake of simplicity, I am using strings of 1 character. Do I have the correct output for the provided input?

string 1: 0
string 2: 1
string 3: 0
string 4: 1
string 5: 0

1 & 2 -> 01 10
1 & 3 -> 00 00
1 & 4 -> 01 10
1 & 5 -> 00 00
2 & 3 -> 10 01
2 & 4 -> 11 11
2 & 5 -> 10 01
3 & 4 -> 01 10
3 & 5 -> 00 00
4 & 5 -> 10 01

Totals

01 = 6
10 = 6
11 = 2
00 = 6
[download]

If that is correct, then I think my math approach will work.

Cheers - L~R

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Re^6: Speeding permutation counting by BrowserUk (Patriarch) on Jul 20, 2007 at 00:19 UTC
What the OP is calculating is the number of bit pairing (4 values) for each combination of 2 strings. Eg: Assuming 32-bit strings, typical values might be: `strings: '00' '01' '10' '11' 1 & 2 10 5 12 15 1 & 3 2 17 7 6 1 & 4 1 & 5 2 & 3 ....` [download] So the OP wants 4 numbers per pairing of strings, regardless of the number of bits. What you would be calculating is 4 numbers per bit position, regardless of the number of strings. Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error. "Science is about questioning the status quo. Questioning authority". In the absence of evidence, opinion is indistinguishable from prejudice. "Too many [] have been sedated by an oppressive environment of political correctness and risk aversion."	[reply] [d/l]
Re^7: Speeding permutation counting by Limbic~Region (Chancellor) on Jul 20, 2007 at 00:58 UTC
BrowserUk, See, I told you I was misunderstanding the problem ;-) Cheers - L~R	[reply]