It's time to drag the Island of Knights and Knaves kicking and screaming into the 19th century! We're going to run a train. For these puzzles, you'll have a set of stations with possible connections, and you need to find a route that starts at one station, goes through every other station exactly once, and ends in a station. IE if our map was
A -- B -- C | | | D -- E -- F
Valid routes might be
A B C F E D, or
A D E B C F, but
A B E D C F is invalid. The ordering matters:
A B C is a different route from
C B A.
But also, every station is run by either a Knight or a Knave. The stationmaster has a set of requirements for the route. Knights tell the truth about their requirements, but Knaves always lie. You must plan the route so that every requirement by a Knight is satisfied, and every requirement by a Knave is unsatisfied. IE if we have
A --- B | | -- C --
If C says "I must come first", if C is a Knight, then the route starts at C. But if C is a knave, the route does not start with C.
At least one of the two stationmasters is a Knave.
A -- B
A: "Start at my station."
B: "End at my station."
Two of the stationmasters are knights, two are knaves.
A -- D | | B -- C
A: "The route must connect to me right after a Knave's station."
B: "I should come at some point before C."
C: "Make sure the two knights are adjacent to one another: one has to come right after the other."
D: "A and B should both come at some point before me."
B (again): "Oh, also: I should come at some point after A."
A / \ B --- C
A: "Don't put Raymond right before me."
B: "Don't put Raymond right after me."
C: "Don't put Raymond right before or right after me."
You: "What's wrong with Raymond?"
A: "He's an asshole and a Knave."
B: "He wears sunglasses indoors."
C: "He calls himself a 'singer-songwriter'."
A: "Also he smells bad."
B: "He smells so bad. Like 'rotting socks' bad."
(Assume one of them is Raymond)
The annoying thing with these is that I want people to express requirements like "If X is true, then do Y". That's formally
X => Y, but the negation is
X /\ !Y. What I want is that if the person is a Knave, you have to satisfy
X => !Y.