MC: Ping-Pong Cubes
05-27-2018, 05:18 PM
Post: #21
 Maximilian Hohmann Senior Member Posts: 575 Joined: Dec 2013
RE: MC: Ping-Pong Cubes
Hello!

(05-27-2018 04:48 PM)David Hayden Wrote:  In another case, we deliberately wrote a piece of thow-away code. The throw-away couldn't handle the rapidly growing load that we anticipated, but it could be written and deployed quickly. It gave us time to write a more robust implementation that could handle the load which increased four orders of magnitude.

Even more off-topic, but in my experience (from my code-writing-and-maintaining-and-extending-legacy-code days) "throw-away-code" is the worst of nightmares. Because in many cases it doesn't get thrown away - either because the team changes and no one seems to remember or because new challenges leave to time to clean up the old mess - but creeps onward and onward... In the end result a large software is built on top of a rotten foundation. Ask Volkswagen!

And back on topic: I personally do not have objections about brute force solutions to mathematical problems. These days computer time is a lot cheaper than developer time and if a job needs to be done on a limited budget (as it is always in engineering and the industrial environment) "elegance" of the solution is the least of worries.
05-27-2018, 06:50 PM (This post was last modified: 05-27-2018 06:51 PM by ijabbott.)
Post: #22
 ijabbott Senior Member Posts: 593 Joined: Jul 2015
RE: MC: Ping-Pong Cubes
(05-27-2018 05:18 PM)Maximilian Hohmann Wrote:  And back on topic: I personally do not have objections about brute force solutions to mathematical problems. These days computer time is a lot cheaper than developer time and if a job needs to be done on a limited budget (as it is always in engineering and the industrial environment) "elegance" of the solution is the least of worries.

I think that depends on the scale of the problem to be solved though. If you need to sort (say) a million records, you probably shouldn't use the nearest $$\mathcal{O}(n^2\,)$$ sorting algorithm that comes to hand. You should use an $$\mathcal{O}(n \log n)$$ algorithm, and preferably one that makes use of auxiliary storage to minimize copying.
05-27-2018, 08:46 PM
Post: #23
 Maximilian Hohmann Senior Member Posts: 575 Joined: Dec 2013
RE: MC: Ping-Pong Cubes
(05-27-2018 06:50 PM)ijabbott Wrote:  I think that depends on the scale of the problem to be solved though. If you need to sort (say) a million records, you probably shouldn't use the nearest $$\mathcal{O}(n^2\,)$$ sorting algorithm that comes to hand. You should use an $$\mathcal{O}(n \log n)$$ algorithm, and preferably one that makes use of auxiliary storage to minimize copying.

As you say, it depends. If you have to sort that list one time only, copy-paste the first sorting algorithm Google finds for you and let it run over night. If you need to sort it again every 10 minutes it might be worth thinking about a clever sorting routine first!
05-27-2018, 09:19 PM
Post: #24
 pier4r Senior Member Posts: 1,989 Joined: Nov 2014
RE: MC: Ping-Pong Cubes
(05-27-2018 05:18 PM)Maximilian Hohmann Wrote:  These days computer time is a lot cheaper than developer time

I had some hard time to crack this sentence (that is very common on subreddits like /r/sysadmin , /r/cscareerquestions and so on). I mean I know that a computing device is cheaper than an employee, but I often though it was nonetheless a waste of computing power.

Then I realized that also the brain is quite some computing device - although more for problem solving. In general the solution of a problem should be computed from the start and not only when the computing device runs. So if I use more refined techniques that takes more time to bring me the solution, would be faster - as described by David Hayden - to use a less refined technique that let me reach the solution faster.

That is: "running time" is only a part of the "solution time". So if I solve something with a state of the art algorithm in assembly and I take 3 months, of which 4 minutes for the computation, while someone else does it in gw-basic with a almost brute force algorithm and takes 3 days, of which 2.5 days for the computation, the latter is better.

Of course it depends on the context. For a function often used, say in a framework, in embedded devices, in devices/services sold/used by hundreds of people or more, then it is better to optimize it a little bit.

Wikis are great, Contribute :)
05-27-2018, 10:33 PM
Post: #25
 brickviking Senior Member Posts: 330 Joined: Dec 2014
RE: MC: Ping-Pong Cubes
Merely thinking helps get the job done
(05-27-2018 09:19 PM)pier4r Wrote:  ...
Then I realized that also the brain is quite some computing device - although more for problem solving. In general the solution of a problem should be computed from the start and not only when the computing device runs. So if I use more refined techniques that takes more time to bring me the solution, would be faster - as described by David Hayden - to use a less refined technique that let me reach the solution faster.

That is: "running time" is only a part of the "solution time". So if I solve something with a state of the art algorithm in assembly and I take 3 months, of which 4 minutes for the computation, while someone else does it in gw-basic with a almost brute force algorithm and takes 3 days, of which 2.5 days for the computation, the latter is better.

Want to know what the laugh is? I was trying to find a way to say exactly this in my last post, but I think you've covered it better than I could. Nicely put.

In reply to this, I'll post something that Fred Brooks wrote about. Requoted from the wikiquote page for Fred Brooks):
Quote:…well over half of the time you spend working on a project (on the order of 70 percent) is spent thinking, and no tool, no matter how advanced, can think for you. Consequently, even if a tool did everything except the thinking for you – if it wrote 100 percent of the code, wrote 100 percent of the documentation, did 100 percent of the testing, burned the CD-ROMs, put them in boxes, and mailed them to your customers – the best you could hope for would be a 30 percent improvement in productivity. In order to do better than that, you have to change the way you think.

As quoted [paraphrased] from Toolbox, Java World, July 1999

(Post 233)

Regards, BrickViking
HP-50g |Casio fx-9750G+ |Casio fx-9750GII (SH4a)
05-27-2018, 11:15 PM
Post: #26
 pier4r Senior Member Posts: 1,989 Joined: Nov 2014
RE: MC: Ping-Pong Cubes
Nice quote. Remembers me one point in this list https://www.hpcalc.org/hp48/docs/columns/aphorism.html

Wikis are great, Contribute :)
05-29-2018, 08:07 PM (This post was last modified: 05-29-2018 08:22 PM by ijabbott.)
Post: #27
 ijabbott Senior Member Posts: 593 Joined: Jul 2015
RE: MC: Ping-Pong Cubes
I have a list of the first 200 or so cubes that are Ping-Pong numbers if anyone is interested in seeing them. The first 10 Ping-Pong Cubes as defined by Joe are contained therein. I was curious to see if the sequence of cubes that are Ping-Pong Numbers hits a wall (or at least a very large gap) in the same way that the sequence of Ping-Pong Cubes does after the first few. It doesn't.
06-01-2018, 01:16 AM
Post: #28
 Valentin Albillo Senior Member Posts: 386 Joined: Feb 2015
RE: MC: Ping-Pong Cubes

Hi, all:

This is my original research for this interesting challenge (you won't find any of this on the internet prior to this very post, AFAIK). First of all, a couple of

Definitions:

- In what follows I'll call APN (Alternate-Parity Number) to any positive integer N (not necessarily a cube) which has all its digits alternating in parity.

- In what follows I'll call PPN (Ping-Pong Number, as per Mr. Horn's definition) to any positive integer N such than both N and N$$^3$$ are APNs. Notice that there are no single-digit PPNs (they would be just "Ping" or "Pong" but not "Ping-Pong").

So far I've worked out the following:

For D ranging from 2 to 20 digits, the following table lists: the number of APNs having D digits, having up to D digits, and the probability that an arbitrary positive integer having up to D digits is an APN:

Digits  #Numbers up to D digits    # of D-Digit APNs  # APNs up to D dig.   Prob. N is APN
-------------------------------------------------------------------------------------------
2                          90                   45                   45   0.500000000000
3                         990                  225                  270   0.272727272727
4                       9,990                1,125                1,395   0.139639639640
5                      99,990                5,625                7,020   0.070207020702
6                     999,990               28,125               35,145   0.035145351454
7                   9,999,990              140,625              175,770   0.017577017577
8                  99,999,990              703,125              878,895   0.008788950879
9                 999,999,990            3,515,625            4,394,520   0.004394520044
10               9,999,999,990           17,578,125           21,972,645   0.002197264502
11              99,999,999,990           87,890,625          109,863,270   0.001098632700
12             999,999,999,990          439,453,125          549,316,395   0.000549316395

13           9,999,999,999,990        2,197,265,625        2,746,582,020   0.000274658202
14          99,999,999,999,990       10,986,328,125       13,732,910,145   0.000137329101
15         999,999,999,999,990       54,931,640,625       68,664,550,770   0.000068664551
16       9,999,999,999,999,990      274,658,203,125      343,322,753,895   0.000034332275
17      99,999,999,999,999,990    1,373,291,015,625    1,716,613,769,520   0.000017166138
18     999,999,999,999,999,990    6,866,455,078,125    8,583,068,847,645   0.000008583069
19   9,999,999,999,999,999,990   34,332,275,390,625   42,915,344,238,270   0.000004291534
20  99,999,999,999,999,999,990  171,661,376,953,125  214,576,721,191,395   0.000002145767

The following code for the 12-digit HP-71B will produce the above table for D ranging from 2 up to 12 digits.

1  DEF FNC1(D)=D
2  DEF FNC2(D)=10^D-10
3  DEF FNC3(D)=9*5^(D-1)
4  DEF FNC4(D)=45/4*(5^(D-1)-1)
5  DEF FNC5(D)=45/4*(5^(D-1)-1)/(10^D-10)
6  !
7  DESTROY ALL
8  FOR D=2 TO 12 @ DISP FNC1(D);FNC2(D);FNC3(D);FNC4(D);FNC5(D) @ NEXT D

(I computed the results for D from 13 up to 20 digits using multiprecision software.)

Inspecting the table reveals that by the time we're considering 20-digit numbers (and so 60-digit cubes) we're dealing with 99,999,999,999,999,999,990 (almost 100 quintillion, 10$$^{20}$$) candidates if we're conducting a pure brute-force search where we check both N and N$$^3$$ for APN-ness.

A much faster strategy consists in directly generating each APN in turn and then checking only its cube for APN-ness. This reduces the search order from O(10$$^N$$) to O(5$$^N$$), i.e., a mere 214,576,721,191,395 (214 trillion) candidates instead of 100 quintillion. This reduces the number of candidates by a factor of 466,033x.

Obviously, checking the cubes of 214 trillion APNs to try and find any that also happen to be PPN is beyond the computation capabilites of a typical PC, at least in a reasonable amount of time. For instance, even using hardware capable or generating one million/sec. APNs up to 20-digits long while also computing and checking their up to 60-digit cubes, it would take almost 2,500 days to complete the search so it's clear that a more refined idea is sorely needed.

After thinking about it for a while I came up with a workable idea to speed the search greatly by trading extending checking time (increasing it 5x) for reducing the number of candidates to check (decreasing it to O(2.5$$^N$$)). This provides a tremendous, exponential reduction in search time, to the point that for 20-digit candidates we only need to check the cubes of 751,937,115 candidates. Compared to the previous 214 trillion candidates, this further reduces the number of candidates by a factor of 285,365x, for a total reduction over pure brute-force of almost 133 billion times. I'm still using brute force but less and less brute and more and more refined.

The technique involves generating and checking what I call "root" candidates, which are not so easy to describe (and doing so here would make this already very long post about twice or thrice longer) but which, as stated above, take 5x more time to check but their number is reduced to O(2.5$$^N$$).

I wrote code for and tested it first in the 12-digit HP-71B, which reproduces just the first 3 rows of the following table, as 4-digit candidates already generate 12-digit cubes. The remaining rows up to 20 digits were generated by converting the 71B code and then running it using multiprecision software:

Digits  #Numbers up to D digits   # APNs up to D dig. #Roots up to D dig.
-------------------------------------------------------------------------
2                          90                   45             35
3                         990                  270            110
4                       9,990                1,395            315

5                      99,990                7,020            810
6                     999,990               35,145          2,035
7                   9,999,990              175,770          5,055
8                  99,999,990              878,895         12,630
9                 999,999,990            4,394,520         31,605
10               9,999,999,990           21,972,645         79,190
11              99,999,999,990          109,863,270        197,650
12             999,999,999,990          549,316,395        493,380
13           9,999,999,999,990        2,746,582,020      1,232,630
14          99,999,999,999,990       13,732,910,145      3,080,445
15         999,999,999,999,990       68,664,550,770      7,701,600
16       9,999,999,999,999,990      343,322,753,895     19,253,260
17      99,999,999,999,999,990    1,716,613,769,520     48,128,440
18     999,999,999,999,999,990    8,583,068,847,645    120,313,780
19   9,999,999,999,999,999,990   42,915,344,238,270    300,777,310
20  99,999,999,999,999,999,990  214,576,721,191,395    751,937,115

I've found no simple formula or recurrence relation for the 4th column (# of roots up to D digits long) but the simple formula   #roots(D) = 8.2675*2.5$$^D$$   gives a good approximation from D=3 onwards. Notice that the ratio between the number of roots up to 20 and 19 digits respectively is already 751937115/300777310 = 2.49998 so the limit ratio seems clearly to be 2.5 and thus there are O(2.5$$^N$$) root candidates to check.

The reduction in the number of candidates to generate and check is most dramatic when going from checking all the number to generating and checking only APNs to generating and checking only roots. Matter of fact, the search and check procedure for the 751,937,115 up-to-20-digit candidates takes just 5 hours to complete in pretty old hardware & software.

Conclusions:

Regrettably, after all this work, except for the 10 small ones already known, no further solutions were found for numbers from 10 up to 10$$^{20}$$ so if an 11th PPN does exist it has more than 20 digits (and its cube more than 60 digits). Thus, any further search should begin with numbers 21-digit long and up, i.e.: N >= 10$$^{20}$$, and a fast modern PC coupled with suitably fast, compiled software will probably be capable to extend the search up to 27 or 28 digits in the same timet took me to reach 20 digits.

Regards.
V.
.
06-03-2018, 04:26 AM
Post: #29
 Warbucks Junior Member Posts: 15 Joined: Mar 2018
RE: MC: Ping-Pong Cubes
For me the interest is in the series underpinning the n^3 alt parity.

As I would normally enter this in Maple or equivalent, we are moving and I have no access to hardware, so forgive my symbolism.

This seq. is a stair step graph that is 5 wide and 9 groups per decimal radix digit. So if you wanted to know the number of terms < 10^n, you would use:

9*(5^n-1)/4 - 9 n>=2

So if you wanted to now the number of terms<100 (10^2) n=2, there would be 45 terms.

If you wanted to know the number of terms 10^n < #terms < 10^(n+1) we have:

9*5^n

There is indeed a solution. The general case is 2^a(5^b)(j). There are three cases. If anyone is interested, I could do a proof, but not on this phone!

It was said that the next term in the sequence is 21 digits in length, you would start at a number of this form, n is number of digits (21):

101010101010101... or for each digit n >= 1

a sub n = 1/2(1+(-1)^(n+1))

Also if there is interest, for completeness, I can generate the 21 digit cube answer.
06-03-2018, 05:42 AM
Post: #30
 ijabbott Senior Member Posts: 593 Joined: Jul 2015
RE: MC: Ping-Pong Cubes
Interesting stuff from Valentin Albillo and Warbucks. I'd be interested in more details.

A note on confusing terminology: Valentin's APN (Alternate Parity Number) is what Joe called a Ping-Pong Number, and his PPN (Ping-Pong Number) is what Joe called a Ping-Pong Cube.
06-03-2018, 12:29 PM
Post: #31
 pier4r Senior Member Posts: 1,989 Joined: Nov 2014
RE: MC: Ping-Pong Cubes
(06-03-2018 05:42 AM)ijabbott Wrote:  .

I have to say that from the initial message the tables have turned. You fostered quite some contributions. Kudos!

Wikis are great, Contribute :)
06-03-2018, 03:24 PM
Post: #32
 Joe Horn Senior Member Posts: 1,459 Joined: Dec 2013
RE: MC: Ping-Pong Cubes
(06-03-2018 04:26 AM)Warbucks Wrote:  Also if there is interest, for completeness, I can generate the 21 digit cube answer.

That would be greatly appreciated.

<0|ɸ|0>
-Joe-
06-14-2018, 03:07 PM
Post: #33
 Valentin Albillo Senior Member Posts: 386 Joined: Feb 2015
RE: MC: Ping-Pong Cubes
(06-03-2018 03:24 PM)Joe Horn Wrote:
(06-03-2018 04:26 AM)Warbucks Wrote:  Also if there is interest, for completeness, I can generate the 21 digit cube answer.

That would be greatly appreciated.

You're barking up the wrong tree, Mr. Horn.

V.
.
06-15-2018, 02:12 AM
Post: #34
 Joe Horn Senior Member Posts: 1,459 Joined: Dec 2013
RE: MC: Ping-Pong Cubes
(06-14-2018 03:07 PM)Valentin Albillo Wrote:
(06-03-2018 03:24 PM)Joe Horn Wrote:  That would be greatly appreciated.

You're barking up the wrong tree, Mr. Horn.

Are you implying that Warbucks cannot produce a 21-digit Ping-Pong Number whose cube is also a Ping-Pong Number? (If your previous reply explains why, please accept my apology for not understanding it).

<0|ɸ|0>
-Joe-
06-15-2018, 04:52 AM
Post: #35
 Werner Senior Member Posts: 349 Joined: Dec 2013
RE: MC: Ping-Pong Cubes
My understanding is that he offered to produce the next 21-digit Ping Pong Number that is itself a cube, but not of another Ping Pong Number.

Cheers, Werner
06-15-2018, 03:37 PM (This post was last modified: 06-15-2018 03:38 PM by Joe Horn.)
Post: #36
 Joe Horn Senior Member Posts: 1,459 Joined: Dec 2013
RE: MC: Ping-Pong Cubes
(06-15-2018 04:52 AM)Werner Wrote:  My understanding is that he offered to produce the next 21-digit Ping Pong Number that is itself a cube, but not of another Ping Pong Number.

Aha, yes, quite right. Valentin handily demonstrated that no Ping-Pong Cube (which he calls a PPN) exists between 381078125 and 10^60.

<0|ɸ|0>
-Joe-
06-17-2018, 10:33 AM
Post: #37
 ijabbott Senior Member Posts: 593 Joined: Jul 2015
RE: MC: Ping-Pong Cubes
(06-15-2018 03:37 PM)Joe Horn Wrote:
(06-15-2018 04:52 AM)Werner Wrote:  My understanding is that he offered to produce the next 21-digit Ping Pong Number that is itself a cube, but not of another Ping Pong Number.

Aha, yes, quite right. Valentin handily demonstrated that no Ping-Pong Cube (which he calls a PPN) exists between 381078125 and 10^60.

Even so, the possibility that there are infinitely many of them hasn't been ruled out yet!
06-20-2018, 09:30 PM
Post: #38
 Valentin Albillo Senior Member Posts: 386 Joined: Feb 2015
RE: MC: Ping-Pong Cubes
.
Hi Mr. Horn, pier4r, ijabbott and everyone else interested:

After waiting a considerable number of days for Warbucks to post anything on this thread either replying to Mr. Horn or substantiating his assertions, I'm posting here my final considerations on it all. If you're interested, read on (Warning: Long Read):

(06-15-2018 02:12 AM)Joe Horn Wrote:
(06-14-2018 03:07 PM)Valentin Albillo Wrote:  You're barking up the wrong tree, Mr. Horn.

Are you implying that Warbucks cannot produce a 21-digit Ping-Pong Number whose cube is also a Ping-Pong Number?

Of course I'm "implying" that Warbucks cannot produce a 21-digit solution to Mr. Horn's Mini-challenge (MC henceforth). Furthermore, I'm explicitly stating here that not only can't he do that but also he doesn't even seem to understand the sources he used to make the assertions he posted.

I'll explain in detail but first the relevant timeline:

- Mr. Horn posted an interesting MC, "Ping-Pong Cubes" (PPC henceforth), which asks for an HP calculator program that finds the first ten PPC as defined by him.

- Within a few hours, Didier posted a program for the HP Prime, which Mr. Horn immediately commented enthusiastically ("Wow ! [...] Very cool !")

- A few hours later I saw the MC and posted a 5-line program for the HP-71B which finds all 10 small solutions in 0.49 sec. I naively expected some comment by Mr. Horn, as my code includes a novel technique to check non-iteratively whether a number is a PPN or not, and also because Mr. Horn hosted in the very distant past a column called "The Titan File" dedicated to the HP-71B so I thought he would appreciate my solution all the more. Alas, Mr. Horn didn't see it fit to comment anything and he didn't post even simply to acknowledge he had read my solution either.

- Nevertheless (and as this is not the first time Mr. Horn ignores my solutions to one of his challenges) I still was interested in the last part of his MC (finding the 11th PPC) and did some original research (i.e., entirely done by myself, without searching the net or using external references) in order to either find the coveted 11th PPC, or else prove that there's none, or at least improve the lower bounds for future searches.

- Thus, two weeks after posting my 71B solution, I posted a 100+ line message with tables, the formulas which generated them (in HP-71B BASIC), an explanation of the techniques used, comments and conclusions, the most important being that except for the 10 small ones already known there is no PPN less than 10^20 that generates a PPC when cubed, so any further search should necessarily begin with PPN candidates at least 21-digit long.

- Again, I naively thought that Mr. Horn would appreciate my original research and so would post some comment or at least would acknowledge this time that he had read it, but it was not to be, he completely ignored it. Again.

- Then someone identified as "Warbucks" posted about a dozen lines of what on examination turned out to be nothing but ramblings where he included such utter bluffs as "If anyone is interested I could do a proof [...]" and "If there is interest, for completeness, I can generate the 21 digit cube answer".

- Somewhat surprised at those bold statements, I analyzed what he said in his post and discovered that it's all a bunch of assorted lines directly copy-pasted from the Internet, taken from publicly available sources which have nothing to do with Mr. Horn's MC (or even with alternating cubes) and mixed with unsupported assertions a.k.a. bluffing. Yet Warbucks, who seems unaware of the irrelevancy of the sources he's parroting, had nevertheless the guts to offer "a proof" (of what !?) and a 21-digit solution right after copying whatever mumbo-jumbo he thought looked relevant.

- To my utter surprise, Mr. Horn then went on and while he didn't ever reply to either my 71B solution or my extensive original research, he did immediately reply to Warbucks post, telling him that "that" (i.e., posting a 21-digit solution) "would be greatly appreciated."

Facepalm ... Needless to say, Warbucks never replied to Mr. Horn's plea (of course!), nor did he post any "proof", nor did he "generate the 21 digit cube answer" he offered, despite being encouraged by Mr. Horn to do so. To help readers of this post to understand why, let's analyze Warbucks post in detail for evidence supporting what I've just said about it:

Warbucks Wrote:"So if you wanted to know the number of terms < 10^n, you would use:
9*(5^n-1)/4 - 9 n>=2
So if you wanted to now the number of terms<100 (10^2) n=2, there would be 45 terms. If you wanted to know the number of terms 10^n < #terms < 10^(n+1) we have:
9*5^n"

All these formulas and values apply only to PPNs, not PPCs or even just alternating cubes, and he took them verbatim either from internet sources (IMO, OEIS, PDF documents, specialized webs/books, etc.) or even adapted them from my own original research post where I gave them as user-defined functions for the HP-71B (shifted versions of my FNC4 and FNC3 respectively), so absolutely nothing new or original here, just mere googling and pasting to sound authoritative.

Warbucks Wrote:There is indeed a solution. The general case is 2^a(5^b)(j). There are three cases. If anyone is interested, I could do a proof [...]

This is sheer bluffing, he doesn't seem to understand or care what is being sought (Ping-Pong Cubes generated by cubing a Ping-Pong Number) or, in case he's just dealing with mere alternating cubes, he doesn't seem to understand either that his internet sources have nothing to do with such cubes. Matter of fact, he simply copy-pasted what is said in this PDF document:

45th International Mathematical Olympiad - Solutions

(or one of its several versions available in the net and even in some specialized books), which gives the solutions to six problems of the 45th International Mathematical Olympiad which took place at Athens, Greece in July 2004. At the end of this PDF document we find Problem 6, which reads:

"6. We call a positive integer alternating if every two consecutive digits in its decimal representation are of different parity. Find all positive integers n such that n has a multiple which is alternating."

so this Problem 6 deals with alternating integers alright, i.e., Ping-Pong Numbers, but has nothing do to with their cubes or alternating cubes or cubes at all, much less with finding PPNs which have PPN cubes as well. Nothing-to-do. At all.

It seems Warbucks didn't understand or care whether the info was relevant or not, he possibly just found the PDF document in the net after some googling and then proceeded to regurgitate and copy-paste lines from it. In particular, quoting once more what Warbucks posted:

Warbucks Wrote:There is indeed a solution. The general case is 2^a(5^b)(j). There are three cases. If anyone is interested, I could do a proof[...]

which was copy-pasted from the Official Solution in the aforementioned document, namely the lines in the document that read:

"For the general case n=2^a*5^b*k, where k is not a multiple of 5 or 2 and a = 1"

and also from the second solution by Joel Tay (in the document as well) which includes the Cases 1,2a,2b and 3 which are the "There are three cases" which Warbucks mentions. Then Warbucks continues with:

Warbucks Wrote:It was said that the next term in the sequence is 21 digits [...] Also if there is interest, for completeness, I can generate the 21 digit cube answer.

where the "It was said" obviously refers to my post (because nothing is said in the PDF document or anywhere else on the net about a 21-digit PPN or PPC) but actually I never said that there's a 21-digit solution, I said that 21-digit PPNs are the lower limit to search for the 11th PPC. Also, an "answer" to what ? Not certainly to what Mr. Horn is looking for, the 11th PPC.

In conclusion, it seems that Warbucks either has no idea of what Mr. Horn's MC is about or else he does not care, nor does he seem to be capable of producing any relevant proof (of what?) or even understand that the PDF document or source he's allegedly parroting has nothing to do with alternating cubes.

He seems to be utterly clueless and certainly can't produce a valid 11th solution in the form of a 21-digit PPN which remains PPN when cubed, which is the main interest of Mr. Horn, or even simply produce a non-PPC 21-digit alternating cube, which is way way easier. For instance, these are the eight 21-digit non-PPC alternating cubes, found by a simpler variant of my code in next to no time:

105858789694589210761 = 4730521^3
123274525814321496103 = 4976887^3
307810347014567834983 = 6751927^3
325874149438941050527 = 6881503^3
361858767434383618729 = 7126009^3
656181898125276107096 = 8689766^3
905292149232581638541 = 9673781^3
943014381450761214161 = 9806321^3

and just for show, these are the smallest 25-digit and 30-digit ones, found very quickly using this same simple variant (essentially just removing the PPN constraint for the base number):

1270525474365816309892523 = 108308147^3
101652769272709496147418783432 = 4667020818^3

Why Warbucks did something like this, parroting a document he googled out and extracting lines from it and then offering "proof" and a "solution" while seemingly understanding nothing of it, is beyond me. I can't fathom what he's trying to achieve with this kind of behavior.

To end on a realistic note and contingent on there being some interest, in a few days I'll post updated details extracted from my Original Research, establishing new, improved lower bounds and commenting on several heuristics I've thoroughly researched and the conclusions that can be extracted from them. Oh, and last-but-not-least a quite solvable PPN/PPC-based counter-MC for Mr. Horn and interested readers (including Warbucks).

P.S.: By the way, the final conclusion to Problem 6 of that 45th IMO is:

"Concluding all, we have shown that a positive integer n is alternating if and only if it is not a multiple of 20.".

Great. Now someone please ask Warbucks what this has to do with Mr. Horns's MC or with alternating cubes in general, even if not PPC.

Regards.
V.
.
06-21-2018, 02:50 AM (This post was last modified: 06-21-2018 02:53 AM by Joe Horn.)
Post: #39
 Joe Horn Senior Member Posts: 1,459 Joined: Dec 2013
RE: MC: Ping-Pong Cubes
(06-20-2018 09:30 PM)Valentin Albillo Wrote:  ...
- A few hours later I saw the MC and posted a 5-line program for the HP-71B which finds all 10 small solutions in 0.49 sec. I naively expected some comment by Mr. Horn, as my code includes a novel technique to check non-iteratively whether a number is a PPN or not, and also because Mr. Horn hosted in the very distant past a column called "The Titan File" dedicated to the HP-71B so I thought he would appreciate my solution all the more. Alas, Mr. Horn didn't see it fit to comment anything and he didn't post even simply to acknowledge he had read my solution either.

- Nevertheless (and as this is not the first time Mr. Horn ignores my solutions to one of his challenges)...

...
- Thus, two weeks after posting my 71B solution, I posted a 100+ line message with tables, the formulas which generated them (in HP-71B BASIC), an explanation of the techniques used, comments and conclusions, the most important being that except for the 10 small ones already known there is no PPN less than 10^20 that generates a PPC when cubed, so any further search should necessarily begin with PPN candidates at least 21-digit long.

- Again, I naively thought that Mr. Horn would appreciate my original research and so would post some comment or at least would acknowledge this time that he had read it, but it was not to be, he completely ignored it. Again.

...
- To my utter surprise, Mr. Horn then went on and while he didn't ever reply to either my 71B solution or my extensive original research, he did immediately reply to Warbucks post, telling him that "that" (i.e., posting a 21-digit solution) "would be greatly appreciated."

YIKES! Sorry! YES, I didn't reply, but NO, I was NOT ignoring your posts! You were clearly WAY ahead of me regarding this MC, so I thought it prudent to stay out of your way. I'm very grateful for your postings (ALL of them, going back many many years!), and I'm very sorry if saying this NOW is too little too late to prevent you from feeling offended by my silence.

(06-20-2018 09:30 PM)Valentin Albillo Wrote:  To end on a realistic note and contingent on there being some interest, in a few days I'll post updated details extracted from my Original Research, establishing new, improved lower bounds and commenting on several heuristics I've thoroughly researched and the conclusions that can be extracted from them. Oh, and last-but-not-least a quite solvable PPN/PPC-based counter-MC for Mr. Horn and interested readers (including Warbucks).

Yes, I'm interested. Thanks in advance.

<0|ɸ|0>
-Joe-
06-21-2018, 03:22 AM
Post: #40
 brickviking Senior Member Posts: 330 Joined: Dec 2014
RE: MC: Ping-Pong Cubes
Any time someone poses important information, while expecting a reply, and is offended when one is not offered, is someone who will be disappointed in life often. I generally stay out of the way of people like these, as it doesn't generally end well for either me or the other person. This discussion was already above my head when the MC was posed originally, and talk of solutions on the venerable HP-71B has not improved my understanding. So I stayed out of the discussion. I feel Joe might have felt the same way.

I think I'll just step into the other room for a while...

(Post 247)

Regards, BrickViking
HP-50g |Casio fx-9750G+ |Casio fx-9750GII (SH4a)
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