[00:03] MSG: Quit: humhum [00:39] i got some neat results [00:39] even just in mircscript [00:39] hehe [00:43] MSG: Ping timeout: 245 seconds [01:20] i should say that i typoed too much and actually had bad results :( [01:22] huh. [01:22] no results [01:25] i wonder if that is some interesting property of primes [02:13] 3/4 is the best it'll get from 1-65536 [02:13] pretty interesting [02:14] i'm fairly certain this method is testing all possibilities too, even though it's only picking two first and then a third later [02:14] thinking about it, even if you picked three all at once, any two of them would have to all come to correct values on their own [02:19] but some of you have perfect solutions for N=4 [02:19] which means that either i'm missing something [02:19] or 'perfect' is less than the total combinations of (N-1) constants [06:21] perfect is less than the total combinations [06:41] so it would seem [06:41] i wonder why, though [06:41] it seems like there must be some basis in math for the fact that a+b and a-b being in the list seem mutually exclusive [06:42] it's probably obvious too [06:45] lol [06:45] like the fact that [06:45] were there any combinations like that, they would not be prime [06:45] 'cause the constant separation is linear, and you won't have two primes along the same linear scale [06:52] i worked it out in my head ;) [06:52] a-s is the common factor [06:52] er wait [06:52] that's multiplication >:( [06:53] fators, i mean [06:53] n, n+step, and n+2step; these cannot all be prime [06:54] these words are lies [06:54] i don't know what i'm talking about [06:54] it seems obvious but i can't define why [06:55] unless step is a multiple of 3, then either n, n+step or n+2*step will be a multiple of 3 [06:56] that's it [06:56] step cannot be a multiple of 3 then? i wonder why [06:58] nevermind; all the steps are multiples of 3 it seems [06:58] which is sensible [06:59] but then, abstracting it one level fails [06:59] i must have messed up somewhere.. trying to work this out in notepad sucks, i need a pencil and paper [07:01] ah, i left a level out [07:09] can't get it in my head [07:09] too late at night perhaps [07:09] i'll try tomorrow ;P [07:09] I've just settled for the brute force method :-) [07:09] hee. [07:09] i want to know why i can't have a, b, a+b, a-b [07:10] You can, providing b is 2 [07:10] but 2 doesn't have primes on either side of it [07:10] i've got a list of about 4000 numbers that have primes +-105 from them [07:11] a=5, b=2 [07:11] but i can't choose a and b from that list such that a, b, a+b, a-b are all within that list [07:11] 3/4 is all i'll get [07:12] Either a, b, a+b or a-b will be non-prime unless a=5 and b=2 [07:12] that's ok, none of the list are prime [07:12] the list are numbers between pairs of primes [07:13] there will be a multiple of 2 and a multiple of 3 [07:13] Oh okay [07:13] for example: 156 338 494 [07:13] all possibilites are 156 338 494 832 4230 4568 4724 4906 5062 5218 5400 5556 5894 [07:14] any of those numbers +-105 is a prime [07:14] er [07:14] I've finished implementing my brute for program in assembly language :-) [07:14] 7 of them are [07:14] D: [07:14] i can't get at 'any' [07:14] hehe [07:14] oh, smarty pants [07:14] you know i have a thought about the relationship of the numbers too [07:15] i think each number will be > the sum of all the others for best results [07:15] Now I can get the same rubbish results 1000 times as fast :-/ [07:15] err, the sum of the previous numbers, assuming they are ordered low to high [07:15] it's like binary [07:15] with 4 digits you can have up to 16 unique values [07:15] the digits are just worth something other than 2 [07:16] (and they have to have primes an odd number away ;) [07:16] 3? [07:16] ? [07:16] You can either add a number, subtract a number or do neither [07:17] that's three options, so maybe it's base 3? [07:17] i can't fit that in my head [07:17] :P [07:17] i was just thinking about addition when that came to mind [07:17] it seems like it might still hold but it's not accurate [07:18] i don't know any numbering system where the digits represent either -place, 0, or +place [07:18] hehe [07:18] i wrote a script to convert things to negabinary though once! [07:18] base -2 [07:18] shit is crazy [07:18] normal base 3 should work [07:18] but that's [07:18] 0, place value, or 2* place value [07:18] instead of using the digit 2, you'd be using -1 [07:19] at least for it to hold to my imaginary distribution of values hehe [07:19] Imaging the values 1,3,9,27,81 [07:19] that works for the base 2 analogy... there are no overlapping sums [07:19] but with subtraction i just don't know [07:19] i guess actually it makes sense [07:19] If you add, subtract or do nothing with each number, you can make every value up to 121 without any overlap [07:20] you just want the 2nd place in the negative to be greater than the sum of all the ones before it [07:20] Having no overlap is important. If there's any overlap in the combinations, you reduce the number of potential primes [07:20] exactly [07:20] the digits are going to increase more than that example though of course [07:21] Yes [07:31] so i guess then that each digit would be > twice the sum of the previous ones [07:32] 'cause it could subtract all of them to reach that sum [07:32] Yes :-) [07:32] i don't know if it is best to start with a small odd number [07:32] (maybe the best odd number is big) [07:33] but it seems like a simple progression if you start with your odd number, then choose an even number higher than it and proceed from there [10:17] Join: fiveop joined #corewars [10:25] Hi Fiveop [10:26] 3hi [11:26] MSG: Ping timeout: 245 seconds [15:21] MSG: Quit: humhum [15:48] Join: CoreOld joined #corewars [15:55] Join: fiveop joined #corewars [18:06] MSG: Quit: humhum [19:47] Join: waknuk joined #corewars [19:48] Part: waknuk left #corewars [20:09] Join: waknuk joined #corewars [20:09] MSG: Remote host closed the connection [23:34] MSG: Quit: Trillian (http://www.ceruleanstudios.com