All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
Back to the Wall (Posted on 2010-04-28) Difficulty: 3 of 5

No! Not a firing squad nor the need for a continuous line to cross all line segments just once!

To each vertex labeled A to L apply a different value from 1 to 12. Let V, W, X, Y and Z be the sums of their respective surrounding vertices.

Provide at least one example where V=W=X=Y=Z, or offer a reason why this, like the continuous line, is impossible.

See The Solution Submitted by brianjn    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Impossibility | Comment 15 of 17 |
(In reply to Impossibility by brianjn)

I see that your program evaluates all the sums within a loop that goes through all 12! = 479,001,600 possible sequences of A - L.

Speed can be gained by doing the P(12,8) = 19,958,400 possible sequences of A, B, C, D, E, F, H, I first to greatly narrow down the possibilities via X = V and D>A before continuing on with the 4! = 24 ways of arranging the remaining 4.

In fact, even though not taking advantage of this in my program due to the initial misunderstanding of the problem, once those 8 vertices are checked you could also check W before proceeding and cut down on the time even more.

Edited on May 1, 2010, 1:28 pm
  Posted by Charlie on 2010-05-01 13:26:15

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information