[VA] Short & Sweet Math Challenges #23: "May the 4th Be With You !" Special

06022018, 06:01 PM
Post: #34




RE: [VA] Short & Sweet Math Challenges #23: "May the 4th Be With You !" ...
(05242018 09:13 PM)Valentin Albillo Wrote:Jeff O. Wrote:In any case, a 10fold or more increase in speed is really needed to make this practical. [...] I’ll keep thinking about the problem to see if I can develop a method that will be much quicker  hopefully it won’t keep me awake at night. So as to not use anyone else's concepts for the time being, I have not reviewed Peter's work. Should anything I say or present below be a repetition of concepts presented by Peter, I fully acknowledge his priority. In an effort to reduce execution time, I developed a method which basically checks 10 numbers at a time. I guess a better way to describe it would be to say that it determines if there is a solution between successive numbers that end in zero. Since it is still basically just brute force, i.e., does not break any ground in attacking the problem in a superefficient manner, I won't explain the procedure in detail. If anyone wants an explanation, I can provide it. Since it effectively checks 10 numbers at a time, it seemed like a program based on the procedure should theoretically be up to 10 times faster than the previous program, as it checks all N+1 digit numbers by counting up to N. In practice, it appears that the various manipulations required to implement it eliminate some of the time saving. The speedup seems to be more like a factor of 4 to 5. But with that improvement, I went ahead and turned it loose to find the 11digit selfies. After (only) several days (again, on Free42 running on a couple of PCs), it found the seven 11digit selfies: 15 694 046 123 52 249 382 004 30 609 287 624 97 653 680 744 60 605 588 394 87 561 939 628 41 919 540 249 The new program listing is presented below. Unfortunately, this program does not technically meet the original challenge. It cannot determine the single digit selfies from 1 through 9 since if you start counting at 1, it is checking candidates starting at 10. I could add code to just print out 1 through 9 at the start, but that seems unnecessary. As noted, this is still essentially just brute force, and would be totally impractical on a DM42. Running on a physical machine in minutes or hours would require a massive speedup factor. I'll keep trying to think of some other method to attack the problem that would be faster, but there's not much time until Sunday. Then I'll have to decide if I want to admit defeat and review Valentin's solution, or let it haunt me the rest of my days... 00 { 112Byte Prgm } 01▸LBL "SELF" 02 CLA 03 CF 29 04 FIX 00 05 RCL 02 06 ARCL ST X 07 10 08 × 09 ALENG 10 1 11 + 12 STO 00 13 R↓ 14▸LBL 04 15 ATOX 16 X=0? 17 GTO 05 18 48 19  20 RCL 00 21 Y↑X 22  23 X<0? 24 GTO 08 25 GTO 04 26▸LBL 05 27 R↓ 28 RCL ST X 29 RCL 00 30 1/X 31 Y↑X 32 1 33 + 34 IP 35 STO 11 36 RCL 00 37 Y↑X 38 RCL 11 39 X≠Y? 40 GTO 08 41 10 42 RCL× 02 43 RCL+ 11 44 ARCL ST X 45 0 46 STO 01 47▸LBL 06 48 ATOX 49 X=0? 50 GTO 07 51 48 52  53 RCL 01 54 10↑X 55 ISG 01 56 DEG 57 × 58 + 59 GTO 06 60▸LBL 07 61 R↓ 62 PRX 63▸LBL 08 64 ISG 02 65 DEG 66 GTO "SELF" 67 END Dave  My mind is going  I can feel it. 

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