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megabreit



Joined: 03 Jan 2009
Posts: 141

Post Eniac Reply with quote
Is this challenge related to the normal use of that abacus?
E.g. is this a base 10 abacus? Or do you expect people
to "redefine" the meaning of the beads?
Fri Jan 30, 2009 7:56 pm View user's profile Send private message
rmplpmpl



Joined: 26 Oct 2008
Posts: 113
Location: Germany

Post Reply with quote
I am asking myself the same question, "normal" use obviously not, cause that answer is not accepted.

I don't get the connection to good old Eniac, neither.
Fri Jan 30, 2009 8:39 pm View user's profile Send private message
MerickOWA



Joined: 07 Apr 2008
Posts: 182
Location: HkRkoz al KuwaiT 2019 HaCkEr 101

Post Reply with quote
I didn't get the connection to Eniac either. I don't think its meant as a hint to this challenge. I think its possibly just a historical reference.
Sat Jan 31, 2009 4:10 am View user's profile Send private message
Belriel



Joined: 20 Dec 2008
Posts: 16

Post Reply with quote
The biggest number I could think of to be displayed with this 10x10 abacus is a 16 digit number starting with 3 and ending with 5 ... but that's not the correct solution Sad. Is there really a way to display even bigger numbers?
Thu Feb 05, 2009 9:49 pm View user's profile Send private message
MerickOWA



Joined: 07 Apr 2008
Posts: 182
Location: HkRkoz al KuwaiT 2019 HaCkEr 101

Post Reply with quote
thats too big, so your method of using the abacus is too complicated.

Its been suggested that, if you use the abacus in a very complicated way, you could represent very very large numbers. The answer to this problem doesn't require any terribly complicated method of using, however its not exactly the easiest method either.
Fri Feb 06, 2009 1:39 am View user's profile Send private message
Belriel



Joined: 20 Dec 2008
Posts: 16

Post Reply with quote
But the question is
Quote:
What is the biggest number that can be displayed using this abacus (without destroying or rearranging it)?

You can use the Chinese abacus with base 16, the two upper beads stand for 5 each, the lower 5 represent 1 each, so 5+5+5=16-1. This one could be used with the left five as the ones and the right five as 6 each, so 5*6+5*1=36-1, then you can represent 36^10-1 as the biggest number. Complicated, true, but without destroying or rearranging.
Fri Feb 06, 2009 6:47 pm View user's profile Send private message
fridolin



Joined: 30 Nov 2008
Posts: 16

Post Reply with quote
I also applied various versions of the abacus, also the chinese one and some variants - but no right solution. Considering the heading "Eniac" didn't effect anything as eniac had also a ten decimal digit computing unit.
Sun Feb 08, 2009 3:28 pm View user's profile Send private message
MerickOWA



Joined: 07 Apr 2008
Posts: 182
Location: HkRkoz al KuwaiT 2019 HaCkEr 101

Post Reply with quote
The Chinese system is not being used for this challenge. Perhaps the challenge is flawed in that way, but the answer to this challenge doesn't use a complicated system for the beads. All beads have the same "meaning". Hopefully that isn't too much of a giveaway.
Mon Feb 09, 2009 4:28 pm View user's profile Send private message
higgs



Joined: 03 Jan 2009
Posts: 4

Post Reply with quote
Ugh I could represent 2^91 -1 with that... Rather big number I'd say... roughly 27 digits in decimal. But it won't take it Sad
If I use the beads at bits, where (as with a Compact Disk) two beads touching represents a 0 and a space represents a 1 (or you could take chance vs. not change or a different interpretation...), you'd have 9 binary digits per row. There being 10 rows, you'd have 9*10 digits which makes the highest number the one where all digits are 1's, 2^91 -1.
But it doesn't like my idea or something
Tue Feb 17, 2009 5:07 pm View user's profile Send private message
CoreEvil



Joined: 27 Mar 2008
Posts: 18

Post Reply with quote
Ok, here is my calculation, I believe it's mathematically accurate, and that it represents the biggest possible number. Please let me know if there is anything wrong with my reasoning:

1) Every row is a 15 slot, base 3 string, which gives you 3^15 possibilities. Each row starts with the string AAAAA-AAAAA-AAAAA (where 'A' represents an empty slot), and keeps increasing until it reaches CCCCC-CCCCC-CCCCC.The default setup when you load the challenge is AAAAA-BBBBB-CCCCC
2) Since you need to count only the combinations where you start (from the right) with 5 Cs followed by 5 Bs (with As allowed anywhere in between) You have a total of 3003 possible combination for each row, this can be verified programmatically by iterating over the 3^15 combinations and counting the ones that follow this pattern.
3) Since you have 10 of these rows, the biggest number should be 3003^10 which equals 59642154303295182755378537795549049

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Tue Oct 13, 2009 3:39 pm View user's profile Send private message
laz0r



Joined: 04 Feb 2010
Posts: 290
Location: Within the depths of Unix

Post Reply with quote
I think this challenge should be reworded or removed due to the evident incorrectness of the accepted solution!

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Mon Feb 15, 2010 9:46 pm View user's profile Send private message
nighthalk



Joined: 31 Jul 2009
Posts: 41

Post Reply with quote
do the colors matter? what if all the beads were blue? with the "simplest form possible" im assuming it meens theres no fancy space measuring or trinary state stuff going on (meaning 11 states per row)
Fri Mar 05, 2010 8:21 am View user's profile Send private message
Dr. Halo



Joined: 27 Oct 2008
Posts: 6
Location: Munich

Post Reply with quote
As MerickOWA stated a year ago, the colors have no meaning in this challenge.
Sun Mar 07, 2010 10:20 pm View user's profile Send private message
polarlemniscate



Joined: 03 Mar 2010
Posts: 6

Post ENIAC Reply with quote
Think about number bases - it's really very simple. I suspect the ENIAC title is something to do with this. What number base did ENIAC work in?
Thu May 13, 2010 9:51 pm View user's profile Send private message
nask00s



Joined: 13 Jan 2011
Posts: 1

Post Reply with quote
I think that eniac ran in base 10 but I'm not pretty sure
Sat Feb 05, 2011 4:24 pm View user's profile Send private message
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