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Griffin
Daedalian Member

 Posted: Fri Feb 04, 2000 11:42 pm    Post subject: 1 Here's the best answer to Digital Mayhem I have so far: / / / / / / / / / / / / / / / / / 7 8 9 4 5 6 1 2 3 one 7 8 9 4 5 6 2 1 3 two 7 8 9 3 5 6 2 1 4 three 6 8 9 3 5 7 2 1 4 four 6 9 8 3 5 7 2 1 4 five 4 9 8 3 5 7 2 1 6 six 4 9 2 3 5 7 8 1 6
Jester
Guest

 Posted: Sat Feb 05, 2000 6:00 pm    Post subject: 2 4 9 2 3 5 7 8 1 6 you can do it in 5 steps.
Wonko the Sane
Daedalian Member

Richk
Last of the Daedalians

 Posted: Sat Feb 05, 2000 11:52 pm    Post subject: 4 I got 6 too....... 7 8 9 3 5 6 1. swap 3 and 4 1 2 4 7 8 9 6 5 3 2. swap 3 and 6 1 2 4 6 8 9 7 5 3 3. swap 7 and 6 1 2 4 6 9 8 7 5 3 4. swap 8 and 9 1 2 4 6 2 8 7 5 3 5. swap 2 and 9 1 9 4 6 1 8 7 5 3 6. swap 1 and 2 2 9 4 [This message has been edited by Richk (edited 02-05-2000).]
Ghost Post
Icarian Member

 Posted: Sun Feb 06, 2000 11:37 pm    Post subject: 5 I think the solution is 6. but if keys swapped must have a same side, is it possible to find a solution with less than 8 moves ? I found this one: 789 456 123 789 456 213 (1) 789 256 413 (2) 789 256 431 (3) 798 256 431 (4) 978 256 431 (5) 278 956 431 (6) 276 958 431 (7) 276 951 438 (8)
markagain
Guest

 Posted: Tue Feb 08, 2000 3:05 am    Post subject: 6 Everyone seems to like that 5 in the center. How about: 924,573,168 ? The puzzle didn't say diagonals had to be equal as well.
markagain
Guest

 Posted: Tue Feb 08, 2000 3:10 am    Post subject: 7 Oops! sorry. I checked the puzzle and it did include diagonals. Never mind.
Wonko the Sane
Daedalian Member

 Posted: Wed Feb 09, 2000 3:32 am    Post subject: 8 You would be right that it must take 8, but it says nothing about the keys' relationship to each other, it simply says move only 2 keys at a time. It's 6.
Ghost Post
Icarian Member

 Posted: Wed Feb 09, 2000 7:49 am    Post subject: 9 I know it ... It was only to have a bite more to eat ... (and happy to have the same solution...) [This message has been edited by DGA (edited 02-09-2000).]
Matt_Banham
Guest

 Posted: Wed Feb 23, 2000 9:31 pm    Post subject: 10 Hi all, I was browsing the web and stumbled upon this site which I have to say is wicked! Anyway, I solved the first part of this problem really quickly (i.e. I found the magic square very quickly). What I wanted to know is whether a general algorithm exists for solving this problem when the order of the square is n (i.e. for 4X4, 5X5, .... nXn sided square)? I ask this as I solved it by first finding the magic number (which I did by taking the average of the numbers from 1 to 9 or alternatively you could take the average of the lowest possible n-number sum and the highest possible n-number sum). Then I realised that four of the sums had to involve the middle number and so I knew this had to be the magic number divided by n (where n is 3 in this case) i.e. this number is 5. However, a.) I can't prove this last sentence in the general case (when the size of the square is n and b.) if I were to use Gaussian-Jordan elimination, then I will end up with a system of n-1 equations and will have to guess one of the terms by up to n amount of times!!!!! If anyone knows an analytical solution to this problem then I would be grateful. Also, I have a little problem that some of you might want to try:- If you take a regular polygon with the number of sides = n, and you call the distance from the centre of the polygon to any of its corners r, can you prove that if the area of the polygon is equal to its circumference, then as n tends to infinity, r = 2?
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