MC: PingPong Cubes

05272018, 05:18 PM
Post: #21




RE: MC: PingPong Cubes
Hello!
(05272018 04:48 PM)David Hayden Wrote: In another case, we deliberately wrote a piece of thowaway code. The throwaway 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 offtopic, but in my experience (from my codewritingandmaintainingandextendinglegacycode days) "throwawaycode" 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. 

05272018, 06:50 PM
(This post was last modified: 05272018 06:51 PM by ijabbott.)
Post: #22




RE: MC: PingPong Cubes
(05272018 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. 

05272018, 08:46 PM
Post: #23




RE: MC: PingPong Cubes
(05272018 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, copypaste 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! 

05272018, 09:19 PM
Post: #24




RE: MC: PingPong Cubes
(05272018 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 gwbasic 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 :) 

05272018, 10:33 PM
Post: #25




RE: MC: PingPong Cubes
Merely thinking helps get the job done
(05272018 09:19 PM)pier4r Wrote: ... 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 CDROMs, 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. (Post 233) Regards, BrickViking HP50g Casio fx9750G+ Casio fx9750GII (SH4a) 

05272018, 11:15 PM
Post: #26




RE: MC: PingPong Cubes
Nice quote. Remembers me one point in this list https://www.hpcalc.org/hp48/docs/columns/aphorism.html
Wikis are great, Contribute :) 

05292018, 08:07 PM
(This post was last modified: 05292018 08:22 PM by ijabbott.)
Post: #27




RE: MC: PingPong Cubes
I have a list of the first 200 or so cubes that are PingPong numbers if anyone is interested in seeing them. The first 10 PingPong Cubes as defined by Joe are contained therein. I was curious to see if the sequence of cubes that are PingPong Numbers hits a wall (or at least a very large gap) in the same way that the sequence of PingPong Cubes does after the first few. It doesn't.


06012018, 01:16 AM
Post: #28




RE: MC: PingPong 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 (AlternateParity 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 (PingPong 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 singledigit PPNs (they would be just "Ping" or "Pong" but not "PingPong"). 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 DDigit 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 12digit HP71B will produce the above table for D ranging from 2 up to 12 digits. 1 DEF FNC1(D)=D 2 DEF FNC2(D)=10^D10 3 DEF FNC3(D)=9*5^(D1) 4 DEF FNC4(D)=45/4*(5^(D1)1) 5 DEF FNC5(D)=45/4*(5^(D1)1)/(10^D10) 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 20digit numbers (and so 60digit cubes) we're dealing with 99,999,999,999,999,999,990 (almost 100 quintillion, 10\(^{20}\)) candidates if we're conducting a pure bruteforce search where we check both N and N\(^3\) for APNness. A much faster strategy consists in directly generating each APN in turn and then checking only its cube for APNness. 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 20digits long while also computing and checking their up to 60digit 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 20digit 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 bruteforce 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 12digit HP71B, which reproduces just the first 3 rows of the following table, as 4digit candidates already generate 12digit 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 upto20digit 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 21digit 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. . 

06032018, 04:26 AM
Post: #29




RE: MC: PingPong 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^n1)/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. 

06032018, 05:42 AM
Post: #30




RE: MC: PingPong 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 PingPong Number, and his PPN (PingPong Number) is what Joe called a PingPong Cube. 

06032018, 12:29 PM
Post: #31




RE: MC: PingPong Cubes
(06032018 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 :) 

06032018, 03:24 PM
Post: #32




RE: MC: PingPong Cubes
(06032018 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. <0ɸ0> Joe 

06142018, 03:07 PM
Post: #33




RE: MC: PingPong Cubes  
06152018, 02:12 AM
Post: #34




RE: MC: PingPong Cubes
(06142018 03:07 PM)Valentin Albillo Wrote:(06032018 03:24 PM)Joe Horn Wrote: That would be greatly appreciated. Are you implying that Warbucks cannot produce a 21digit PingPong Number whose cube is also a PingPong Number? (If your previous reply explains why, please accept my apology for not understanding it). <0ɸ0> Joe 

06152018, 04:52 AM
Post: #35




RE: MC: PingPong Cubes
My understanding is that he offered to produce the next 21digit Ping Pong Number that is itself a cube, but not of another Ping Pong Number.
Cheers, Werner 

06152018, 03:37 PM
(This post was last modified: 06152018 03:38 PM by Joe Horn.)
Post: #36




RE: MC: PingPong Cubes
(06152018 04:52 AM)Werner Wrote: My understanding is that he offered to produce the next 21digit Ping Pong Number that is itself a cube, but not of another Ping Pong Number. Aha, yes, quite right. Valentin handily demonstrated that no PingPong Cube (which he calls a PPN) exists between 381078125 and 10^60. <0ɸ0> Joe 

06172018, 10:33 AM
Post: #37




RE: MC: PingPong Cubes
(06152018 03:37 PM)Joe Horn Wrote:(06152018 04:52 AM)Werner Wrote: My understanding is that he offered to produce the next 21digit Ping Pong Number that is itself a cube, but not of another Ping Pong Number. Even so, the possibility that there are infinitely many of them hasn't been ruled out yet! 

06202018, 09:30 PM
Post: #38




RE: MC: PingPong 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): (06152018 02:12 AM)Joe Horn Wrote:(06142018 03:07 PM)Valentin Albillo Wrote: You're barking up the wrong tree, Mr. Horn. Of course I'm "implying" that Warbucks cannot produce a 21digit solution to Mr. Horn's Minichallenge (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, "PingPong 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 5line program for the HP71B 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 noniteratively 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 HP71B 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 HP71B 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 21digit 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 copypasted 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 21digit solution right after copying whatever mumbojumbo 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 21digit 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: 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 userdefined functions for the HP71B (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 (PingPong Cubes generated by cubing a PingPong 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 copypasted 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., PingPong 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. Nothingtodo. 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 copypaste 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 copypasted 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 21digit PPN or PPC) but actually I never said that there's a 21digit solution, I said that 21digit 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 21digit PPN which remains PPN when cubed, which is the main interest of Mr. Horn, or even simply produce a nonPPC 21digit alternating cube, which is way way easier. For instance, these are the eight 21digit nonPPC 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 25digit and 30digit 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 lastbutnotleast a quite solvable PPN/PPCbased counterMC 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. . 

06212018, 02:50 AM
(This post was last modified: 06212018 02:53 AM by Joe Horn.)
Post: #39




RE: MC: PingPong Cubes
(06202018 09:30 PM)Valentin Albillo Wrote: ... 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. (06202018 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 lastbutnotleast a quite solvable PPN/PPCbased counterMC for Mr. Horn and interested readers (including Warbucks). Yes, I'm interested. Thanks in advance. <0ɸ0> Joe 

06212018, 03:22 AM
Post: #40




RE: MC: PingPong 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 HP71B 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 HP50g Casio fx9750G+ Casio fx9750GII (SH4a) 

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