Three missionaries and three cannibals are on one side of the river, wanting to get across.

Unfortunately, the only boat available can hold a maximum of two people. The missionaries, wanting to stay safe, can never be on a side with more cannibals than missionaries (even for a moment!). The boat cannot travel under its own power, so there must be at least one person on board for it to cross.

How can the missionaries get safely across?

(In reply to No Subject by armando)

There only one solution and it doesn't involve handcuffs or cord.

Left side = Boat = Right side
M is a missionary
C is a cannibal
* can be either

MMMCCC = =
MM*C = *C =
MM*C = = *C
MM*C = * = C
MMMCC = = C
MMM = CC = C
MMM = = CCC
MMM = C = CC
MMMC = = CC
MC = MM = CC
MC = = MMCC
MC = MC = MC
MMCC = = MC
CC = MM = MC
CC = = MMMC
CC = C = MMM
CCC = = MMM
C = CC = MMM
C = = MMMCC
C = * = MM*C
*C = = MM*C
 = *C = MM*C
 = = MMMCCC

Note that the solution is a mirror image. Once you find yourself at MC = MC = MC, you've solved it. Also note that the question asks only how all the missionaries can get across safely; if this is the case, ignore the last four boat trips.

Edited on May 29, 2005, 3:54 pm

Posted by Charley on 2005-05-29 15:29:55