Staredit Network

Staredit Network -> Miscellaneous -> Enigma
Report, edit, etc...Posted by DT_Battlekruser on 2006-11-15 at 23:04:39
QUOTE(DT_Battlekruser @ Nov 15 2006, 05:54 PM)
The simple way to disprove it is this:

If you can travel through each door only once, then the ONLY possible way to travel through each door of a room with an odd number of doors is for the path which goes through the odd final door to end within the room.  With one continuous path, the maximum number of times this can happen, and therefore the maximum number of odd-number-door rooms you can have, is two.

The given puzzle has three rooms with an odd number of doors (5), therefore the puzzle is impossible.  Q.E.D.
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Report, edit, etc...Posted by Oo.Insane.oO on 2006-11-16 at 18:58:18
QUOTE(DT_Battlekruser @ Nov 15 2006, 11:04 PM)
QUOTE(DT_Battlekruser @ Nov 15 2006, 11:04 PM)

The simple way to disprove it is this:

If you can travel through each door only once, then the ONLY possible way to travel through each door of a room with an odd number of doors is for the path which goes through the odd final door to end within the room.  With one continuous path, the maximum number of times this can happen, and therefore the maximum number of odd-number-door rooms you can have, is two.

The given puzzle has three rooms with an odd number of doors (5), therefore the puzzle is impossible.  Q.E.D.
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Report, edit, etc...Posted by green_meklar on 2006-11-16 at 19:50:40
QUOTE
If you can travel through each door only once, then the ONLY possible way to travel through each door of a room with an odd number of doors is for the path which goes through the odd final door to end within the room. With one continuous path, the maximum number of times this can happen, and therefore the maximum number of odd-number-door rooms you can have, is two.

The given puzzle has three rooms with an odd number of doors (5), therefore the puzzle is impossible. Q.E.D.

But what if you start inside one of the odd-doored rooms?

If it still doesn't work for exactly the same reason, then excuse my ignorance, I don't know much about network theory.
Report, edit, etc...Posted by DT_Battlekruser on 2006-11-16 at 20:29:41
The "path ending within the room" is either the start or end. Hence, you can have a path terminating in a room twice, one for the start and once for the end.
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