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Topic: Calculator Maze (Read 535 times) |
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Barukh
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Calculator Maze
« on: May 16th, 2006, 11:55pm » |
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Try this. I've found it challenging.
Can you find a solution without refering to a hint?
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towr
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Re: Calculator Maze
« Reply #1 on: May 17th, 2006, 12:32am » |
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on May 16th, 2006, 11:55pm, Barukh wrote:| Can you find a solution without refering to a hint? |
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I could by programming a bit. But I'm getting nowhere by hand atm..
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Wikipedia, Google, Mathworld, Integer sequence DB
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Icarus
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Re: Calculator Maze
« Reply #2 on: May 18th, 2006, 8:39pm » |
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That was interesting. Here are some hints from my experiences:
Label the boxes as:
A B C
D E F
G H I
By a "path" I mean any sequence of adjacent boxes, regardless of whether it is possible to follow. If there is at least one value for the starting box of a path that allows you to traverse the entire path, the path is "integral".
There are exactly 2 paths from A to I that do not pass through E: abcfi and adghi. bcfi is not integral, so you cannot go that way. adghi is integral, but does not work for the starting value of 1.
Hence any solution is going to pass through E. This breaks down all solutions into 3 parts:
A path from A to E
A series of loops from E to itself.
A path from E to I.
The paths from A to E cannot pass through either I or E (other than the end). There are exactly 4: abe, abcfe, ade, adghe.
The loops also cannot pass through I, or E other than at the ends. There are only 12:
efcbe
ebcfe
ebadghe
edghe
edabe
edabcfe
ehgde
ehgdabe
ehgdabcfe
efcbade
efcbadghe
ebade
Of these 12, the last 3 are not integral.
Finally, there are 8 ways of getting from E to I:
ehk
edghk
ebadghk
efcbadghk
efk
ebcfk
edabcfk
eghdabcfk
Only 3 of these paths are integral: ehk, edghk, efk. In order to traverse ehk, E must be = 4 mod 6.
In order to traverse efk or edghk, E must be divisible by 12 (for edghk, it has to be of the form 12(3k+2)).
From here, I calculated the cumulative effect for each of the e-e loops, and figured out for which values it is legal to traverse the loop (for instance, the ebcfe loop can only be traversed if E = 1 mod 4, and the effect is that the new value of E is (3E+29)/4.
By searching through combinations of loops, I was able to find both of his solutions, plus the additional solution that comes about when you allow reversals at E only.
As an added hint, the largest value I encountered for E was 63. If reversals at E are allowed, then the largest value of E that can be reached is 75.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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