Talk:Puzzles/Action sequences/Crossing the River II

Restatement of Situation: Three Farmers- F1, F2, F3 Three Sons- S1, S2, S3 Any "S" cannot stay with an "F" with a different number. Solution: ASSUME that when someone ("X") returns to the origin and is about to make another journey to the final destination, X stays in his boat, but if two other people are to use the boat to reach the final destination, X gets off to make room. -- Lord Emsworth 00:43, 11 Oct 2003 (UTC)
 * 1) F1 and S1 cross the river; S1 disembarks and F1 returns
 * 2) S3 and S2 cross the river; S2 disembarks and S3 returns
 * 3) F2 and F1 cross the river; F1 disembarks and F2 returns w/ S2
 * 4) F3 and F2 cross the river; F2 disembarks and F3 returns
 * 5) F3 and S3 cross the river; both disembark and F2 returns
 * 6) F2 and S2 cross the river; both disembark: The End.


 * That looks pretty good. I think, In step 4 there is a little problem: you return with F3 to the original shore where there are S2 and S3 which means that S2 will be frightened because F3 is there but not his own father. Thomas
 * I actually thought about that. Now my statement "ASSUME..." says that if a person X returns to the origin and in the immediate next trip is to go back to the final destination, then X does not get off at the origin. Step 4:F3 returns but stays in the boat. On Step5:F3 and S3 go across. F3 was in the boat and S2 was on shore! The instructions say "The sons are afraid to be either on the shore or in the boat whith one or two of the other fathers if their own father is not there." There is no requirement that a son on shore will be afraid of someone closeby in the boat.Lord Emsworth 13:16, 11 Oct 2003 (UTC)


 * I've reformulated the problem a bit. Can you please have a look and check that it now rules out the case you have after step 4. Thanks. There is a solution where you don't have to do the "staying in the boat and therefore technically not on the shore" thing Thomas

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Well...you didn't say the kids on the far side would stay put:

Two kids go across, one rows back. Two kids go across, one rows back. He and his pop row across, scare away two kids - but they're already across the river. Pop rows back. Two other dads row across, scare away the kid (might as well be fair). One rows back, last two farmers go across - there, no one left on the far side of the river!

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Ok, here's my reasoning - I'm not sure this thing has a solution. I'm stating the problem with Aa, Bb, and Cc being Farmers with sons.

Step 1: You can't leave anyone on the far side of the river except a kid. (If you try to leave a farmer on the far side, kids will run off). So we have: A * a Bb    Cc

Step 2: We can't send across any adults except A - otherwise a runs off. If we send A with anyone else, a kid runs off. Sending only A is dumb - that takes us back to the beginning. So, we can't send anyone except kids, e.g, b and c. A  *a B   b C    c

Which can only have one real way to go forward: Aa* B   b C    c

Step 3: We can't send A, because that will scare away kids (either a left on this side, or b or c on the far side). Sending a only takes us back a step, so we have to send B. If we send B, we have to send C, otherwise, c runs away. Aa       *Bb Cc

Now, who is coming back? We can't send the kids - A will scare them away. We can't send only 1 adult - his kid will run away. Sending both adults is dumb - that just takes us back a step. So, we send a pair back: Aa Bb* Cc

Step 4: Who can we send over now? We can't send unescorted kids - C will scare them away. Sending a pair back (Aa) just takes us back a step. We can't send only one adult - his kid will run. Send both adults. a    *A b     B        Cc

Who can we send back? One adult will scare away a or b. Sending back A and B takes us back a step...send c. a    A b    B c*   C

Step 5: Awww, crap - it's got a solution - you just move kids around now!

Anybody else find that too easy?