Eniac

Dr. Halo · Post by **Dr. Halo** » Sat Feb 05, 2011 6:01 pm

Don't stick to the name too much

Sat Jun 25, 2011 11:00 am

So what do we know from the challenge? It is a 10*10 abacus, which means it has 100 beads. This plus the name of the challenge seems to indicate that we should use base 10 for this challenge (ENIAC did computations with base 10 numbers, not base 2 like most computers). Now the question is, which value does a bead have? Are they the same for each row, or maybe different?

If all beads have the same value of 10^0 = 1, then the highest number which can be represented is 100*10^0 = 100. But that is not accepted as the solution.

The solution is also not that each row's value is 10 times the value of the previous row. That would give us 10 * (10^0 + 10^1 + 10^2 + 10^3 + 10^4 + 10^5 + 10^6 + 10^7 + 10^8 + 10^9) = 11111111110. Which is also not accepted as the solution.

Is there an error in my conclusions (which is entirely possible), or is there an error in the challenge itself, as some posts in this thread seem to indicate?

Karian · Post by **Karian** » Tue Jul 12, 2011 9:25 am

Once you know where the catch is in this puzzle, it is pretty straight to solve. Although I solved this puzzle in march 2009, I could pretty easily find back the solution in a few seconds (and a calculator at hand).

In itself, the puzzle is flawed because it is possile to represent a lot more numbers with it than the correct answer. As long as you utilize a good coding scheme, you can represent very big numbers.
Maybe to claify things a bit, imagine the bars of the abacus are shaped like this /\, so that beads will always be on the left against the wood, or to the right. I don't think that with this restriction, you can get a higher number then the real answer (for others, feel free to disaprove me)

Imagine using (or if possible use) an abacus like this. Maybe you will find a solution to this answer.

rmplpmpl · Post by **rmplpmpl** » Tue Jul 12, 2011 11:07 am

I have been working on this for more than 2,5 years, just solved it... after many clever and not-so-clever ways to represent huge numbers with it.

Perhaps a hints: as said above, the abacus is used the most simple way you could think of as you might have learned in your first year in school... but you have to omit redundancies, than you'll be fine.

Tue Jul 12, 2011 11:11 am

Karian wrote: Maybe to claify things a bit, imagine the bars of the abacus are shaped like this /\, so that beads will always be on the left against the wood, or to the right.

Well... I'm not sure if I am really much wiser from your explanation

But I will follow your advice given in the other thread, and let this challenge rest for some time, and after a while I am going to try again.

Thank you for helping me, and for confirming that this challenge is erroneous.

rmplpmpl wrote: Perhaps a hints: as said above, the abacus is used the most simple way you could think of as you might have learned in your first year in school... but you have to omit redundancies, than you'll be fine.

Hm, in my opinion the most simple way to use it would be to let each bead have a value of 1, but since that is not the solution... okay, like I said, I will let it rest for now.

Karian · Post by **Karian** » Tue Jul 12, 2011 2:45 pm

If you check the previous posts, you can have a lot of ways of interpreting the position of the beads. based on this, you can get pretty large numbers. My indication about the shape is just to only allow to groups of beads per row, one touching the left side and one touching the right. With this limitation, you should be able to get to the correct solution.

Giving all the value of 1 isn't the solution, since you already stated a better way of interpreting a configuration as a number with your second example. But this still isn't the most optimal way.

Tue Jul 12, 2011 5:10 pm

Karian wrote:My indication about the shape is just to only allow to groups of beads per row, one touching the left side and one touching the right.

Actually I thought that this is how it's always done with an abacus - there are never three or more groups per row, just one (touching either the left, or the right side) or two (with one group touching the left side, and the other touching the right side). The videos which one can search for on the Internet seem to show people moving beads all the way to either side; not stopping in the middle of a row, thus not letting one or more beads stay in-between. I could be mistaken but that is how I perceive it.

moose · Post by **moose** » Tue Jul 12, 2011 5:48 pm

The colors confused me ... now I've solved it

(now I think I'll Rank < 1000

)

Mon Jun 18, 2012 11:05 pm

Having solved it now, I think that the accepted solution is seriously flawed. I'll write a mail to adum and ask him to do something about it. See also my post in the "Challenges Solved" section.

Hippo · Post by **Hippo** » Thu Dec 03, 2015 10:26 pm

Hmm, I have tried 10((10^10-1)/9), 6^20-1, 30((10^10-1) /9), 30((30^10-1)/29) with no success

.

I have tried some more stupid ones like, 30^10-1.

May be I would be more lucky next time.

Fri Dec 04, 2015 8:28 pm

This challenge is baseically flawed.

Hippo · Post by **Hippo** » Mon Dec 07, 2015 8:07 am

OK, I was close (next try was OK). The most important info I have missed is ... the color split of the beads is not important!!! (I have revealed too much, but the challenge deserves it).

Deamhan · Post by **Deamhan** » Sat May 23, 2020 12:46 pm

Question for this challenge is definitely incorrect, number range could be extremely large. Hint: do not split beads on groups, and solution is not related to eniac.