20150504, 16:05  #23 
"Forget I exist"
Jul 2009
Dartmouth NS
20507_{8} Posts 
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 (vmiddle 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.
Last fiddled with by science_man_88 on 20150504 at 16:27 
20150504, 16:42  #24 
"Serge"
Mar 2008
San Diego, Calif.
2^{5}·5^{2}·13 Posts 
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.

20150504, 16:43  #25  
"Forget I exist"
Jul 2009
Dartmouth NS
7·1,217 Posts 
Quote:


20150504, 23:24  #26 
May 2013
East. Always East.
11·157 Posts 
axn's pseudocode is exactly what I went for using Excel but 32bit 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. 
20150505, 00:02  #27  
"Forget I exist"
Jul 2009
Dartmouth NS
7·1,217 Posts 
Quote:
Last fiddled with by science_man_88 on 20150505 at 00:19 

20150505, 01:00  #28 
May 2013
East. Always East.
11·157 Posts 
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] Last fiddled with by TheMawn on 20150505 at 01:01 
20150505, 01:18  #29 
"Forget I exist"
Jul 2009
Dartmouth NS
7×1,217 Posts 
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.

20150505, 03:39  #30  
Jun 2003
2·7·17·23 Posts 
Quote:


20150505, 04:06  #31 
"Serge"
Mar 2008
San Diego, Calif.
2^{5}·5^{2}·13 Posts 
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. } 
20150505, 09:40  #32 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2876_{16} Posts 
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 )

20150505, 14:50  #33 
"Robert Gerbicz"
Oct 2005
Hungary
2^{2}×7×59 Posts 
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. 
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