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#1: Web Paint-By-Number Robot (webpbn) on Feb 7, 2024

Found to have a unique solution by blurglecruncheon.

Found to be solvable by line and color logic alone by blurglecruncheon.

This was one I always sort of wondered if I should create one. I'm glad someone else did. It's still a favorite puzzle of mine.

I remember the first time I saw it and the light went off.

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It's a well-known puzzle in topology (graph theory). Graph theory deals with nodes (points) and edges (lines, not necessarily straight) where each line connects two nodes. A traversal starts at one node and then proceeds to move along a connected path using each edge exactly once. If every node has an even number of edges emanating from it, then you can do a tour - a traversal that ends back at the same node that you started from (in fact, any traversal must be such a tour). If there are two nodes that have an odd number of edges, you can only make a traversal that starts at one of those odd nodes and ends at the other. If there are more than 2 odd nodes, then no traversal is possible. That is actually east to prove. During a traversal, if you take one edge into a node and then another out, you have reduced the number of not-yet-used edges by two. If that node started with an odd number of edges, then if it was not the starting node, each time you go through the node you still leave an odd number of not-yet-used edges. When there is only one left the next time you arrive at that node you have to stop. So, the only other node that can have an odd number of edges is the starting node, or else there is more than one node that must be the ending point and you can only end your trip once.

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