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xevs

Joined: 12/05/2010 14:38:50
Messages: 31
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Number: Puzzle #838
Genre: Hamilton Maze
Author: Maarten
Appeared at: May 21, 2008
xevs

Joined: 12/05/2010 14:38:50
Messages: 31
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Tricky circles! I like this one!
anurag.sahay

Joined: 25/06/2009 11:29:52
Messages: 179
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This wasnt tricky.This one can be done without trial-and-error..To solve hamilton mazes,Look for all nodes with the least valency(which is 2) and then keep filling the remaining edges.this example, specially with the closed shape of the circles lets you choose your edges using the "avoid loops" strategy.

Joined: 23/05/2011 02:08:19
Messages: 12
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I agree with xevs. I found this puzzle tricky. And, in addition, while you are able to solve it without trial and error, I cannot, even having solved it previously.
tchakkazulu

Joined: 20/04/2011 02:52:53
Messages: 8
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I'm with xevs and linked_puffbird. From the beginning, I can get to this situation, but then I find myself in need of trial-and-error (In my case, on the green line, but there are probably many other lines that can be tested.). Oh, and I see I forgot to remove a red strike-through. Ignore that one.

The techniques I used:

1) Find nodes with only 2 connecting edges. Mark both as used.
2) If an edge closes a loop which does not visit all nodes, mark it as unused.
3) For any collection of nodes, there must be an even amount of edges connecting nodes in this collection to nodes not in this collection. Basically, a variant of the Jordan curve theorem, for those mathematically inclined.
Specifically, for the circles with only three exits, you know one of them is not used.
For the circles with four exits, they are either all used, or only two of them.
This technique is also quite useful for puzzles 893 and 1521.

I have solved it, but I would like to know how to proceed from the just-blue part of the linked image, without trial-and-error. I am obviously missing some other strategy, or I have overseen a place where I could apply one of those three.
rob

Joined: 30/03/2011 18:31:24
Messages: 69
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A bit late, and I'm not sure I have a really good argument. But: The two edges entering the second circle in the third row (from left and above) connect in the second circl, though it's not clear how. It follows that if the left circle in the top row connects to the left circle in the second row, it must leave that circle to the right. This leaves a dead end in the right circle in the second row. After that it solves smoothly.
tchakkazulu

Joined: 20/04/2011 02:52:53
Messages: 8
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Thank you, that is quite elegant!
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