Problem 710

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Ted
Posts: 21
Joined: Sun Apr 02, 2006 10:46 pm

Problem 710

Post by Ted »

Problem 710 (View Problem)
If palindrome elements have more than one digit, can they occupy the center of a tuple, for example, (2,16,2) could be a twopal for 20? What about (4,12,4)? Would that be a twopal? What about (20)?
wrongrook
Posts: 515
Joined: Sat Oct 17, 2009 10:39 pm

Re: Problem 710

Post by wrongrook »

(2,16,2) is a twopal for 20 (tuple is palindromic and 2 is present).

(4,12,4) is not (no element equal to 2).

(20) is not (no element equal to 2).

SNIP - code removed
Last edited by bruce_love on Thu Jul 16, 2020 12:46 am, edited 1 time in total.
Reason: Removed code
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kenbrooker
Posts: 187
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Location: Northern California, USA

Re: Problem 710

Post by kenbrooker »

Problem 710 celebrates PE membership first exceeding one million, earlier this year...

When I add up the Statistics/Levels memberships on PE.net
I get about 120,000 members...

Does that mean there are about 900,000 members who
Have solved 0 to 24 problems?

(And regarding the Post immediately above,
What happened to --
In particular don't post any code fragments or results.)
"Good Judgment comes from Experience;
Experience comes from Bad Judgment
..."
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bruce_love
Administrator
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Re: Problem 710

Post by bruce_love »

Thanks for that Ken - I have removed the code.
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kenbrooker
Posts: 187
Joined: Mon Feb 19, 2018 3:05 am
Location: Northern California, USA

Re: Problem 710

Post by kenbrooker »

(Thanks Bruce and)
Still looking for an answer to my
Non-parenthetical
Question...
"Good Judgment comes from Experience;
Experience comes from Bad Judgment
..."
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mdean
Posts: 171
Joined: Tue Aug 02, 2011 2:05 am

Re: Problem 710

Post by mdean »

kenbrooker wrote: Wed Jul 15, 2020 1:33 am Problem 710 celebrates PE membership first exceeding one million, earlier this year...

When I add up the Statistics/Levels memberships on PE.net
I get about 120,000 members...

Does that mean there are about 900,000 members who
Have solved 0 to 24 problems?
From the statistics main page I see
There are a total of 1030100 registered members who have solved at least one problem.
So far 118179 registered members (11.47%) have solved 25 problems or more.
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kenbrooker
Posts: 187
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Location: Northern California, USA

Re: Problem 710

Post by kenbrooker »

Well, Color me Blind
mdean...

I missed the Main Page and went
on(ward and downward) to
Add up All those Levels!!

ThankYouSIR...
"Good Judgment comes from Experience;
Experience comes from Bad Judgment
..."
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nargo7
Posts: 7
Joined: Thu Aug 27, 2020 5:23 pm

Re: Problem 710

Post by nargo7 »

The statement for this problem indicates that there are 824 symmetrical sequences that add to 20, and that contain at least one time the element 2. That number of permutations seemed too high to me, after all 20 is a low number, but there are many unexpected results in mathematics, which is one of the features that makes it so interesting, and combinatorics is fertile ground for counter-intuitive surprises (I am aware this is redundant, but couldn’t find a better way to finish the sentence). So, I set up to find those 824 sequences, and organized them in Excel, but to my disappointment I could find only exactly 400, not even half. I am sure that I am wrong, but I am tired of looking for those that I am missing. I am sure I am making some silly overlook, so I would appreciate very much if somebody can point me to at least a few examples of the general forms of the sequences I have not been able to find.
I am attaching here the Excel file, but I tried to save it macro-free to avoid any suspicion of me trying to send a virus, but the software I am using is Office 365 (from the college where my wife is taking classes), and I couldn’t see that option to save the file. So, to give a doubt-free option, I am attaching also a PDF of the 400-column table. I marked the elements different from 1 in colors, and marked the axes of symmetry of the sequences when they are not a row of numbers, all these to help me find the patterns and more sequences, but I also hope this helps to review my findings.
I would be grateful for any hints I receive.
Thanks and best regards.
Attachments
Twopal Sums for 20.pdf
10 pages, 40 sequences per page
(524.78 KiB) Downloaded 8 times
Twopal Sums for 20.xlsx
Trust is scarce these days, but there is no virus or macro here
(69.48 KiB) Downloaded 5 times
mdean
Posts: 171
Joined: Tue Aug 02, 2011 2:05 am

Re: Problem 710

Post by mdean »

I assume you used some sort of program to find the sequences and didn't just do 400 by hand. Did you also try a similar program for t(6)? If that fails, you might be able to see where it's failing.
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philiplu
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Location: Redmond, WA, USA

Re: Problem 710

Post by philiplu »

Took a quick look at your PDF, and looking at the palindrome lengths in descending order, you've got all the palindromes of lengths 19 down to 16, but you're missing a bunch at length 15. I count 35 such length-15 twopals, but I only see 28 in your PDF. There are 7 twopals of length 15 that use 1, 2, and 3, but I don't see those in your PDF. So maybe start there.
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nargo7
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Joined: Thu Aug 27, 2020 5:23 pm

Re: Problem 710

Post by nargo7 »

Hello mdean!
Thank you for the quick reply. Your advice is reasonable, but my OCD kicked in and, because my programming skills are very limited, I had to find the 842 sequences manually in Excel. It is hard for me to quit and leave things incomplete. But I thought this was going to be faster than what it actually is, I thought that the automatic dragging and copying in Excel would make this easier. But, after the help from philiplu removed my blockage, now I am a little over 600 sequences, a great improvement from the 400 I had when I posted my question.
Thanks again, and best regards!

Hello philiplu!
I found the seven sequences you told me, and then many more, your post was an excellent help, it made me realize new strategies to find qualifying sequences. Please see above my reply to mdean. I think there is a conservative reluctance to open my files, so I am attaching a snapshot of the Excel screen with a note to you. It is incredible that so many permutations exist with two restricting conditions (symmetrical and containing a 2) for a small number like 20, but that is the beauty of math. I think that now I will be able to find all 842, but I will post another update if I get stuck again.
Thanks again, and best regards!
Attachments
Picture showing progress after I received the replies
Picture showing progress after I received the replies
Twopal Table Picture.PNG (65.23 KiB) Viewed 3842 times
nargo7
Posts: 7
Joined: Thu Aug 27, 2020 5:23 pm

Re: Problem 710

Post by nargo7 »

Hello,

I am partially happy to report that my spreadsheet grew to 822 permutations, so now I am only missing two! (I mean of the 824 sums to get a total of 20). But I will be happy when I find the other two. Any hints, even of the type "you are getting warmer" game, will be very much appreciated.

If anybody is interested in this file, with color-coded numbers making beautiful patterns, it is attached, you can easily see by the extension that it is NOT a macro-enabled Excel file, so for sure that it does not have any virus.
Attachments
Twopal Sums for 20.xlsx
Safe file, extension indicates not-macro enabled
(186.79 KiB) Downloaded 3 times
pjt33
Posts: 70
Joined: Mon Oct 06, 2008 6:14 pm

Re: Problem 710

Post by pjt33 »

nargo7 wrote: Sat Sep 19, 2020 6:41 pm I am partially happy to report that my spreadsheet grew to 822 permutations, so now I am only missing two!
Sorry, not so. E.g. compare columns RY and ACN, which by my reckoning are both (7, 2, 2, 2, 7).

Also, the reason people aren't looking at your spreadsheet might be because it's a nuisance to turn it into something which we can easily compare with output from our own programs. If my CJam transposition code is correct then you have 804 distinct solutions and one non-palindrome, but I wouldn't swear to it being bug-free.

It's possible to tackle problems of this size sufficiently systematically by hand without making mistakes, but there are also plenty of mistakes in the mathematical literature due to manual calculation errors. Although I have solved a small number of problems on this site by hand, most of them are intended to be solved by a combination of mathematical insight and programming, and I think you would be well advised to start translating your generation technique into code. It may be that the technique is perfect and the shortfall is due to losing track of what you did 200 columns to the left.
nargo7
Posts: 7
Joined: Thu Aug 27, 2020 5:23 pm

Re: Problem 710

Post by nargo7 »

Hi pjt33, thank you very much for your reply.

And you are correct, those two columns you indicated are duplicates, and then I realized that their neighbors column RZ and ACO are also duplicated. So, I am missing at least four sums instead of two as I had thought, but after your post now I think that most likely I have more duplicates. I tried to detect them, and found a few by myself some days ago, but this is difficult to do in my spreadsheet.

I agree with you that my file is cumbersome to work with, but it has the advantage of displaying the patterns in colors, which I find beautiful, and Excel allows us to easily change the formatting as we like. If I am understanding your post correctly, you have generated all 824 sums, so I would be grateful if you can send me that list at nargo777@gmail.com, and I will do the job of completing my spreadsheet with no duplicates, and could even explore and present different criteria to sort them.

By the way, I think it is evident from my posts, but just in case let me clarify that I have not even tried to solve Problem 710, it is way too difficult for my current programming skills, maybe some time in the future I can tackle it, but at this moment I only want to see all the patterns in colors of the 824 sums. I started this file just because I was amazed that there are so many different ways to sum to a number as low as 20, even more with two strict restrictions as the symmetry and containing a “2”, so I just wanted to see them by myself.
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gaufowl
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Location: MD,USA

Re: Problem 710

Post by gaufowl »

About 10 minutes ago I was sure I had found the key to this problem but it seems there is probably another key I'm missing. Could someone verify for me that t(100) ends in 6767 and t(1000) ends in 0496?
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NASU41
Posts: 2
Joined: Sun Apr 12, 2020 3:23 pm

Re: Problem 710

Post by NASU41 »

gaufowl wrote: Thu Jan 14, 2021 9:03 pm About 10 minutes ago I was sure I had found the key to this problem but it seems there is probably another key I'm missing. Could someone verify for me that t(100) ends in 6767 and t(1000) ends in 0496?
My program got the correct answer, and it says t(1000) mod 10k equals 496.
Perhaps your program shows the correct t(n) iff n is even.
Let's check for some odd numbers: for example, you can count t(9)=7 by hand.
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