May 2015

[science_man_88]

my hardest part is implementing conclusions I have about the starting space trying not to use the || because not all of the statements are exclusive some can happen at the same time but other times might be exclusive I just came up with another possible idea to do this while posting and I have more I can add to eliminate things from being a starting position if the middle value is a power of two multiple of the lowest value then it can only work with highest values v such that (v-middle value)<(middle value - lowest value) because for example in the [1,2,4] example I gave earlier the reason 2 doesn't work is because the amount 4 can supply to the 1 would get it up to the power of two multiple of 1 that is 2. this also destroys things like [1,4,v>6] and [3,6,v>8] and [ 5,10,v>14]. I haven't implemented this yet but I'm hoping it slims it down. edit: yeah that did it down to just over 1.1 million potential starters ( from over 2 million even with proper checks form( cut the time ( including printing) from over 5 minutes to under 3 minutes as well). edit2: now I'm so happy I forgot the new setup possibility and will have to figure it out again. edit3: I implemented the check I just made wrong at first now I'm down to 13427 starters. edit4: I had to change both parts not just part of it fallen to 11734 starters now.

[Batalov]

Originally Posted by axn

I was hoping that I had covered everything ...

I am fairly sure that you are right and I am wrong. I'll run some more debugging in the evening. One extra case was fishy; also with other hash parameters, I found the fifth one, but I am pretty sure that something is leaking in my code. The three core solutions come up every time, though.

[science_man_88]

Originally Posted by science_man_88

my hardest part is implementing conclusions I have about the starting space trying not to use the || because not all of the statements are exclusive some can happen at the same time but other times might be exclusive I just came up with another possible idea to do this while posting and I have more I can add to eliminate things from being a starting position if the middle value is a power of two multiple of the lowest value then it can only work with highest values v such that (v-middle value)<(middle value - lowest value) because for example in the [1,2,4] example I gave earlier the reason 2 doesn't work is because the amount 4 can supply to the 1 would get it up to the power of two multiple of 1 that is 2. this also destroys things like [1,4,v>6] and [3,6,v>8] and [ 5,10,v>14]. I haven't implemented this yet but I'm hoping it slims it down. edit: yeah that did it down to just over 1.1 million potential starters ( from over 2 million even with proper checks form( cut the time ( including printing) from over 5 minutes to under 3 minutes as well). edit2: now I'm so happy I forgot the new setup possibility and will have to figure it out again. edit3: I implemented the check I just made wrong at first now I'm down to 13427 starters. edit4: I had to change both parts not just part of it fallen to 11734 starters now.

I'm not even sure if I'm coding this right so maybe not.

[TheMawn]

axn's pseudocode is exactly what I went for using Excel but 32-bit is not robust enough to handle it.

Batalov's suggestion of splitting by "family" is very smart and I am disappointed in myself for not noticing. I may be able to salvage something from my spreadsheets though.

[science_man_88]

Originally Posted by TheMawn

axn's pseudocode is exactly what I went for using Excel but 32-bit is not robust enough to handle it.

Batalov's suggestion of splitting by "family" is very smart and I am disappointed in myself for not noticing. I may be able to salvage something from my spreadsheets though.

I worked out the problems with my checks I think, and added you simple suggestion of not all being odd numbers ( since three odd numbers add to a odd number I only had to check if the number I was partitioning was odd to cut the number of that simple check potentially in half) and it cut the number left down to 467139 which isn't as close as I originally thought I was but it prints some parts so quick I can't even tell what number it's a partition of. guess I still have a ways to go though. edit: doh I'm semi-stupid I could of ran a parallel version of the game potentially parapply(v->[v[1],v[2],v[3]-v[1]],a) or something like that, I guess I haven't whittled it down enough though.

[TheMawn]

You can have [odd, even, even] which adds to an odd total but is still a very legal configuration.


[4, 12, 5] -> [4,7,10]

[science_man_88]

Originally Posted by TheMawn

You can have [odd, even, even] which adds to an odd total but is still a very legal configuration.


[4, 12, 5] -> [4,7,10]

what I did as a test was I tested the number I was breaking was odd then I checked if the numbers I was getting it broken down to were all odd. I know for example that any configuration that can't last at least 5 moves (because a 255+ could add 8 turns so failure happens on the 13th move or later) has no chance at making the hour. For some reason I think modular restrictions are going to play a role, that or known cycling ratios ( I thought I had an answer out of the blue but when I tested the cycles formed they ended up violating the mersenne ratio rule) because the end of a potential chain in one direction can determine how far in another direction it can go.

[axn]

Originally Posted by Batalov

I am fairly sure that you are right and I am wrong. I'll run some more debugging in the evening. One extra case was fishy; also with other hash parameters, I found the fifth one, but I am pretty sure that something is leaking in my code. The three core solutions come up every time, though.

If you can give the extra solutions (here or by PM), I can get my proggy to print a move tree.

[Batalov]

Debugged already. They have min.length=11; they are a bug ;-)
{ I have coded my own verifier, and for sure, they are "ghosts" that leaked from somewhere. Needless to say, the three real ones are the same as yours. }

[LaurV]

I have a very nice, heuristic, solution to this problem. I am going to send it after a double check tonight (need to write a small program for that )

[R. Gerbicz]

Oh, I'm the first solver of the problem.

I don't see why is it good for you to give away (all) solutions and methods for the challenge problem. In some cases the text of the problem is a little cryptic and hard to understand, so interpretation of the problem (to understand it) is acceptable. What is good that they write that "We will post the names of those who submit a correct, original solution! " so they don't accept copy paste solutions, a little different from ProjectEuler. But if you are the first solver you can't be a cheater.