rookie1ja Posted March 31, 2007 Report Share Posted March 31, 2007 Cannibals and Missionaries - Back to the River Crossing Puzzles Three missionaries and three cannibals want to get to the other side of a river. There is a small boat, which can fit only two. To prevent a tragedy, there can never be more cannibals than missionaries together. This old topic is locked since it was answered many times. You can check solution in the Spoiler below. Pls visit New Puzzles section to see always fresh brain teasers. Cannibals and Missionaries - solution 1 cannibal and 1 missionary there, missionary back. 2 cannibals there, 1 cannibal back. 2 missionaries there, 1 missionary and 1 cannibal back. 2 missionaries there, 1 cannibal back. This one cannibal takes the remaining cannibals to the other side. Three missionaries and three cannibals wanted to get on the other side of a river (Edited: all 6 of them have to get across alive). There was a little boat on which only two of them can fit. There can never be on one side more cannibals than missionaries because of a possible tragedy. Link to comment Share on other sites More sharing options...
Guest Posted May 24, 2007 Report Share Posted May 24, 2007 Alternative Solution:: Cannibals are X's and Missionaries are O's pick up two cannibals: in boat XX leave one cannibal: left side of river X, right side of river X OOO pick up one missionary: in boat XO leave missionary: left side of river XO. right side X OO pick up one missionary: in boat XO leave missionary: left side of river XOO, right side XO pick up one missionary: in boat XO leave missionary: left side of river XOOO, right side of river X pick up cannibal: in boat XX leave both cannibals: left side of the river XXXOOO Link to comment Share on other sites More sharing options...
Guest Posted May 27, 2007 Report Share Posted May 27, 2007 c= canibal m=missionary c+m go across m comes back gets another c c drops m off and comes back to get another m m drops of c and comes back for another c repeat. Link to comment Share on other sites More sharing options...
Guest Posted June 3, 2007 Report Share Posted June 3, 2007 Once again, C=cannibal M=missionary M + M go over. M comes back, picks up another M brings it over, then comes back again to pick up C, brings him over. C comes back over and picks up another C, drops him off and comes back again with the last C. Link to comment Share on other sites More sharing options...
Guest Posted June 5, 2007 Report Share Posted June 5, 2007 all answers are wrong according to question. The question says there should not be more cannibals than missionaries at one place at any time. In all answers, at some time, there is a cannibal (1) and no missionaries (0) failing the condition. To fit the solution, the question should be rephrased as 'There should not be any missionaries present such that they are outnumbered. Link to comment Share on other sites More sharing options...
Guest Posted June 8, 2007 Report Share Posted June 8, 2007 Yes, I've thought about the "outnumbering" issue myself. However, if there are no Missionaries with the Cannibals, then (perhaps, we can consider) there are no Missionaries to be outnumbered. I've converted this puzzle to Algebraic Code, it's an Excel file (not allowed). I'll try again later. When I do, I'd appreciate feedback on this for pleasure or education It's filled with coding and decoding opportunities. Alan Link to comment Share on other sites More sharing options...
Guest Posted June 24, 2007 Report Share Posted June 24, 2007 abliss has the best possible solution for speed although admin is correct also. at no time are there more cannibals than missionaries other than when there are no missionaries at which point it wouldn't matter because they can't be eaten by the cannibals if they are not there Link to comment Share on other sites More sharing options...
Guest Posted June 24, 2007 Report Share Posted June 24, 2007 either i misunderstood the question or the fact that all 6 of them must get across the river should be stated. the question just says all 6 of them "wants" to get across. didn't really specify that all 6 of them has to get across alive. I just took it the different way. though i did kind of eliminate the tragedy part of the question.... cannibal 1 eats cannibal 3 (3 missionary, 2 cannibal). cannibal 2 eats missionary 3 (2 missionary 2 cannibal). cannibal 1 eats cannibal 2 for eating missionary 3 (2 missionary, 1 cannibal) missionary 2 tries to escape and gets eaten by cannibal 1 too. (1 missionary 1 cannibal) cannibal 1 and missionary 1 get on the boat and cross the river happily..... the end Link to comment Share on other sites More sharing options...
Guest Posted June 28, 2007 Report Share Posted June 28, 2007 Visualisation will help understand the solution: At the beginning CCC MMM [boat empty] [other coast empty] "1 cannibal and 1 missionary there, missionary back" CC MM......[CM]->...... CC MM ...... <-[M] ...... C CC MMM ...... [] ...... C "2 cannibals there, 1 cannibal back" MMM ...... [CC]-> ...... C MMM ...... <-[C] ...... CC C MMM ...... [] ...... CC "2 missionaries there, 1 missionary and 1 cannibal back" C M ...... [MM]-> ...... CC C M ...... <-[MC] ...... C M CC MM ...... [] ...... C M "2 missionaries there, 1 cannibal back" CC ...... [MM]-> ...... C M CC ...... <-[C] ...... MMM CCC ...... [] ...... MMM "This one cannibal takes the remaining cannibals to the other side" C ...... [CC]-> ...... MMM C ...... <-[C] ...... C MMM ...... [CC]-> ...... C MMM Link to comment Share on other sites More sharing options...
Guest Posted July 8, 2007 Report Share Posted July 8, 2007 I misread the problem and therefore struggled a bit more to solve it. I was on the impression that neither the missionaries nor the cannibals can outnumber each other. Below the results: Given that missionaries are 123 and cannibals abc. West East 123abc 23bc > 1a a < 1 123 > bc ab < c 23 > 1c abc < 1 3 > 12 12ab < c > 3c 123abc Link to comment Share on other sites More sharing options...
Guest Posted July 10, 2007 Report Share Posted July 10, 2007 I misread the problem and therefore struggled a bit more to solve it. I was on the impression that neither the missionaries nor the cannibals can outnumber each other. Below the results: Given that missionaries are 123 and cannibals abc. West East 123abc 23bc > 1a a < 1 123 > bc ab < c 23 > 1c abc **** < 1 3 > 12 12ab < c > 3c 123abc No, this is wrong. Like many of the others you are assuming that the missionary in the boat is safe, which is not the case. I have noted where the missionary is in trouble with ****. Link to comment Share on other sites More sharing options...
Guest Posted July 13, 2007 Report Share Posted July 13, 2007 I came up with a different solution, that seems to work. There are never more cannibals than missionaries at any one time and there are never 2 cannibals together either on one side of the river, with the assumption that a cannibal by itself can't eat anybody (but him/herself) and they don't eat each other crossing the river. we'll go with m-missionary and c-cannibal: MC MC MC--> -- MC MC <--- M C MC M MC--> MC M <--C MC MC CM --> MC MC <--C MMC C MC --> MMC C <--C MMMC -- CC--> MMMC -- MMMCCC Link to comment Share on other sites More sharing options...
Guest Posted July 14, 2007 Report Share Posted July 14, 2007 i think admin is wrong with taking 2 cannibals over at the same time and expecting on to return, if there is a m on the other side i think that both c's will stay and ,,,,well do what they do. Link to comment Share on other sites More sharing options...
Guest Posted July 19, 2007 Report Share Posted July 19, 2007 take 2 Cannibals to the right side of the river then take 2 missionaries to the right and take 1 cannibal to the left, the right side now has 1 C. and 2 M. and the left side has one cannibal and 1 misionary and there is 1 cannibal one the boat! then take 1 missionary to the right and then go back and bring the last 2 cannibals to the right Link to comment Share on other sites More sharing options...
Guest Posted July 25, 2007 Report Share Posted July 25, 2007 c=cannibals m=missionaries one m or one c always stays in the boat. c+m cross the river leaving c m resturns in boat picking up c and cross the river m returns in boat picking up m and cross the river but takes one c back with him m drops off c picks up last m and cross the river m returns picking c and cross the river m returns picks up last c and return Link to comment Share on other sites More sharing options...
Guest Posted July 27, 2007 Report Share Posted July 27, 2007 1> C+C and one C return 2> C+C and one C return now situation is CMMM with boat and CC at destination end 3> M+M will go and C+M return now CCMM with boat and CM is situation 4> now M+M will go and and C will return 5> C+C go and C return 6> C+C go ...end Link to comment Share on other sites More sharing options...
Guest Posted August 1, 2007 Report Share Posted August 1, 2007 Is it just me or are you guys making this far, far too complicated for yourselves? All six need to get across? There are 3 of each? And just two can fit? Surely the best way is to just put one cannibal and one missionary in the boat. (one of each on one side two of each on the other. go back, and again just take one cannibal and one missionary. Then again return and take the remaining missionary and cannibal. Simple and easy. Link to comment Share on other sites More sharing options...
Guest Posted August 2, 2007 Report Share Posted August 2, 2007 asuming that one canibal has has row back and forth and ferry everyone across. he ids going to get petty mad though Link to comment Share on other sites More sharing options...
Guest Posted August 3, 2007 Report Share Posted August 3, 2007 That makes no sense - no matter what way you start it, there's always got to be one person on the other side and 1>0... Link to comment Share on other sites More sharing options...
Guest Posted August 3, 2007 Report Share Posted August 3, 2007 ah, wait, I've made a fatal error ... I've had in my head that there is someone else rowing the boat. (other than the six) ... Now I feel like a right plum! Link to comment Share on other sites More sharing options...
Guest Posted August 9, 2007 Report Share Posted August 9, 2007 ok you give the cannibals vegetarian lessons then you wount have any cannibal prodlems no fine how says you cant put a rope on the boat so you can send a missionarie and a cannibal to cross. then the Cannibals and Missionaries on the other side pull the boat back and then the next missionarie and cannibal go. then the last missionarie and cannibal but the rope in the boat and cross the river there you go. Link to comment Share on other sites More sharing options...
Guest Posted August 14, 2007 Report Share Posted August 14, 2007 duh!!!! 1m and 1 c m go back 1m and 1 c c go back 1m and 1c c go back wait!!what the Link to comment Share on other sites More sharing options...
Guest Posted August 29, 2007 Report Share Posted August 29, 2007 Once again, C=cannibal M=missionary M + M go over. M comes back, picks up another M brings it over, then comes back again to pick up C, brings him over. C comes back over and picks up another C, drops him off and comes back again with the last C. This one fails right off the bat. As soon as the two missionaries go over, one missionary is left w/ three cannibals. Link to comment Share on other sites More sharing options...
Guest Posted August 29, 2007 Report Share Posted August 29, 2007 i think admin is wrong with taking 2 cannibals over at the same time and expecting on to return, if there is a m on the other side i think that both c's will stay and ,,,,well do what they do. Agreed. The only way this works is assuming that the cannibals are only cannibals while on shore, not in the boat. If they maintained their cannibal tendencies in the boat, here is how the given answer would break down... Cannibals and Missionaries - solution 1 cannibal and 1 missionary there, missionary back. 2 cannibals there, all three cannibals stay and eat lone missionary. Link to comment Share on other sites More sharing options...
Guest Posted August 31, 2007 Report Share Posted August 31, 2007 This is another one of those non-riddle riddles on this site. There was no question posed, just a few statements made. Link to comment Share on other sites More sharing options...
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