Guest Posted October 19, 2007 Report Share Posted October 19, 2007 C= Cannibal M= Missionary /x/ = River /x/CCC MMM /CC/MMMC < C/C/MMMC > C/CM/MMC < CM/C/MMC > CM/CM/CM < CMM/C/CM > CMM/CM/C < CMMM/C/C > CMMM/CC/ < CCCMMM/X/ FINISHED THIS IS THE ANSWER. READ CAREFULLY AND YOU WILL UNDERSTAND NOBODY ON ONE SIDE, NOBODY IN RIVER, ALL ON ONE SIDE TWO CANNIBALS GO ONE CANNIBAL DROPPED OFF, ONE GOES BACK CANNIBAL ON ONE SIDE.CANNIBAL AND MISSIONARY IN RIVER, TWO MISSIONARY AND ONE CANNIBAL OTHER SIDE C&M ONE SIDE, C IN R, TWO M AND ONE CANNIBAL OTHER SIDE C&M ONE SIDE, C & M IN R, C & M OTHER SIDE C & TWO M ONE SIDE, C IN R, C & M OTHER SIDE C & TWO M ONE SIDE, C & M IN R, C OTHER SIDE C& THREE M ONE SIDE, C IN R, C OTHER SIDE C & 3 M ONE SIDE, TWO C IN R, NOBODY OTHER SIDE 3 C & 3 M ONE SIDE, NOBODY IN R, NOBODY OTHER SIDE WAMMY! THAT'S IT! Link to comment Share on other sites More sharing options...
Guest Posted November 14, 2007 Report Share Posted November 14, 2007 Cannible = ME Missionary = YOU You and i cross the river leaving two of me and two of you. I drop you off on the other side of the river. I go back and pick up a another me, leaving two of you and one of me. I drop myself off. Leaving only one of each of us. I go back and pick up you. We start crossing, leaving one me and two you. I drop you off and go back. i pick myself up and cross the river dropping myself off. there are two me and two you. I go back and pick you up the last time. no one is left. We get across and get off. leaving the boat... everyone there...! Link to comment Share on other sites More sharing options...
Guest Posted November 30, 2007 Report Share Posted November 30, 2007 /x/CCC MMM /CC/MMMC < C/C/MMMC > C/CM/MMC < once the CM arives <- you have 2 CC to 1 M = dead/eaten M CM/C/MMC > CM/CM/CM < CMM/C/CM > CMM/CM/C < CMMM/C/C > CMMM/CC/ < CCCMMM/X/ FINISHED Link to comment Share on other sites More sharing options...
Guest Posted November 30, 2007 Report Share Posted November 30, 2007 MMMCCC | | MMMC | |CC - CC cross the river MMMCC | |C - C goes back MMM | |CCC - CC cross the river MMMC | |CC - C goes back M C | |MMCC - MM cross the river MM CC | |M C - MC goes back CC | |MMMC - MM cross the river CCC | |MMM - C goes back C | |MMMCC - CC cross the river CC | |MMMC - C goes back | |MMMCCC - CC cross the river Link to comment Share on other sites More sharing options...
Guest Posted December 7, 2007 Report Share Posted December 7, 2007 Convert the Cannibals!! Just kidding! I like the first solution. As for the question of wording, the riddle states that the point of not having the cannibals outnumber the missionaries is so the cannibals don't eat the missionaries. Therefore it's obvious that having 1 cannibal and 0 missionaries is safe! Link to comment Share on other sites More sharing options...
Guest Posted December 8, 2007 Report Share Posted December 8, 2007 A waaaaay easier way to do it would be to take 1 cannibal and one missionary to the other side, drop them both off and then go back, and then do the same thing 2 more times (send the boat back by pushing it) Link to comment Share on other sites More sharing options...
Guest Posted January 30, 2008 Report Share Posted January 30, 2008 Most of your answers are terribly wrong... The admin's is correct. ablissfulgal said: 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 The bold part is where you would have a missionary eaten. What you’ve got is a cannibal on the left side as well as a cannibal and missionary in the boat. Once the boat gets to the left side to drop the missionary off you’ve got 2 cannibals and 1 missionary on the left side. (When the boat is on the left, you count all in it towards the left sides totals. Same if it’s on the right.) Graphically, missionary about to be dropped off: (can)_[(can)(mis)]………….._(can)(mis)(mis) ^ Missionary is outnumbered in the dropoff. Gonna get eaten. The admin’s solution never has that scenario. This is the admin’s…. ____.................................<-[(can)(mis)]_(can)(can)(mis)(mis) (can)_.[(mis) ]->……….…...……..............……_ (can)(can)(mis)(mis) (can)_................................<-[(can)(can)]_(mis)(mis)(mis) (can)(can)_[(can)]->…….…………...............…_(mis)(mis)(mis) (can)(can)_........................<-[(mis)(mis)]_(can)(mis) (can)(mis)_[(can)(mis)]->…….…….........…..._(can)(mis) (can)(mis)_........................<-[(mis)(mis)]_(can)(can) (mis)(mis)(mis)_[(can)]->…….…….........…..._(can)(can) Now all the missionaries are safe and across so the cannibal left with the boat can ferry the rest of the cannibals. At no point in time is a missionary with a greater number of cannibals (even including those sitting in the boat!) artune, Thor7-10, stephcorbin, brownester, coolkid101, liz5000, seqee girl, theirish2121, and MangaMeggie have all made this mistake. Link to comment Share on other sites More sharing options...
Guest Posted February 5, 2008 Report Share Posted February 5, 2008 (edited) They both start on the same side of the island right!! ok so --- MM ----(MC)>->- MISSIONARY CARRIES CANIBAL ACROSS CC ------------------------- MM -<-<-(M)----- AND RETURNS CC.......................C ------------------------- MM ----(MC)->->- MISSIONARY CARRIES SECOND CANIBAL ACROSS C.........................C ------------------------ MM -<-<-(M)----- AND RETURNS C.........................CC -------------------------- M -----(MM)->->- MISSIONARY CARRIES MISSIONARY ACROSS C........................CC -------------------------- M....................... M -<-<-(MC)----- AND RETURNS WITH A CANIBAL C........................C -------------------------- ..........................M -----(MM)->->- MISSIONARY PICKS UP THIRD MISSINARY AND CARRIES ACROSS CC......................C ------------------------- ...........................MM -<-<-(M)----- AND RETURNS CC......................C ------------------------- .........................MM -----(MC)->->- MISSIONARY CARRIES CANNIBAL ACROSS C.......................C ------------------------- ........................MM -<-<-(M)----- AND RETURNS C.......................CC -------------------------- ..........................MM -----(MC)->->- MISSIONARY CARRIES THIRD AND FINAL CANIBAL ACROSS .........................CC -------------------------- ..........................MMM ..........................CCC YAY!!!! Edited February 5, 2008 by Emapher Link to comment Share on other sites More sharing options...
Guest Posted February 8, 2008 Report Share Posted February 8, 2008 (edited) C=Cannibal M=missionary. One of the cannibals drives everyone over...first a M, then C, M, C then M again Edited February 8, 2008 by Dana Link to comment Share on other sites More sharing options...
Guest Posted March 7, 2008 Report Share Posted March 7, 2008 This one is pretty simple too. Firts, two cannibles go to the other side, and one returns. Then the cannible and one missionarie go to the other side the missionarie stays on that side, and the cannible returns. Next the cannible and the last missionarie go to the other side, the missionarie again stays on that side, and the cannible goes back and brings the third cannible to the other side. Link to comment Share on other sites More sharing options...
Guest Posted March 14, 2008 Report Share Posted March 14, 2008 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. Okay, now we're just getting picky... we can also read the part of the question that state "to avoid a potential tragedy" which inherently implies that 0 missionaries means 0% of a tragedy. Unless you want to consider that a missionary has to be present to keep the canibals from eating one another, but that eliminates all possible solutions. Ovecomplicating things hardly ever makes them easier. Link to comment Share on other sites More sharing options...
Guest Posted March 20, 2008 Report Share Posted March 20, 2008 In all actuality the solution is quite simple but is still outside of the box. They tie to each end of the boat a rope, then you send one cannibal and one missionary over to the other side of the river. When they get out of the boat the others pull the boat back to them and once again one cannibal and one missionary get on board. they continue doing this until they are all on the other side of the river. No breaks in numbers and no tragedies. Link to comment Share on other sites More sharing options...
Guest Posted March 21, 2008 Report Share Posted March 21, 2008 two cannibals there no one back two missionaries there no one back one cannibal and one missionary there the end! At no point did the cannibals outnumber the missionaries. Link to comment Share on other sites More sharing options...
Guest Posted March 21, 2008 Report Share Posted March 21, 2008 Okay, now we're just getting picky... we can also read the part of the question that state "to avoid a potential tragedy" which inherently implies that 0 missionaries means 0% of a tragedy. Unless you want to consider that a missionary has to be present to keep the canibals from eating one another, but that eliminates all possible solutions. Ovecomplicating things hardly ever makes them easier. 1 missionary and 1 cannibal over no one back 1 missionary and 1 cannibal over no one back 1 missionary and 1 cannibal over the end! At no point did the cannibals outnumber the missionaries. Nor were the cannibals left to their own devices and allowed to eat each other. Link to comment Share on other sites More sharing options...
Guest Posted March 21, 2008 Report Share Posted March 21, 2008 forget the two i previously mentioned, I was under the impression someone other than the missionaries and cannibals would be opperating the boat. the only way my other two would work would be to tie the boat to a rope and pull it back and forth between the sides of a river... so disregard my other posts on the issue... I just woke up Link to comment Share on other sites More sharing options...
Guest Posted April 9, 2008 Report Share Posted April 9, 2008 one of each on each boat trip across Link to comment Share on other sites More sharing options...
Guest Posted April 27, 2008 Report Share Posted April 27, 2008 another thing just to point out if a CM goes over the first time, and M comes back then another CM goes over, momentarily there would be 2 Cs and 1 M so would that not be game over too? Link to comment Share on other sites More sharing options...
Guest Posted May 16, 2008 Report Share Posted May 16, 2008 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. Thor7-10 by picking up 2 "M" you would leave 1 "M" with 3 "C" which fails the question Link to comment Share on other sites More sharing options...
Guest Posted May 20, 2008 Report Share Posted May 20, 2008 this is easy. the question states that the should never be more cannibals than missionary but says nothing about having more missionarys!! it is also assumed that there is always one in boat. m = missionary c = cannibal one m and one c go across to left one m comes back (1 c on left and 2 c and 2 m on right, 1 m in boat) one m and one c go across to left one c comes back (1 c and 1m on left and 2 m and 1 c on right, 1 c in boat) one c and one m go across one c comes back (1 c and 2 m on the left and 1 c and 1 m on right, 1 c in boat) one c and one m go across one m comes back (2 c and 2m on left and 1 c on right, 1 m in boat) one c and one m go across all c and m on left one c and one m go across to left Link to comment Share on other sites More sharing options...
Guest Posted June 10, 2008 Report Share Posted June 10, 2008 yeah Link to comment Share on other sites More sharing options...
Guest Posted June 10, 2008 Report Share Posted June 10, 2008 Cannibals and Missionaries - Back to the River Crossing Puzzles 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. 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. I think got the idea of the question but is it worded wrong? It starts out with an even number of cannibals and missionaries, so why cant it be that way on the other side. It says that there cant be more cannibals so this is the easy thing I cane up with... what do you think? CCC 1C and 1M go to side two, the boat returns to side one to grab another C and M, and so on. The numbers are always equal. MMM Link to comment Share on other sites More sharing options...
Guest Posted June 10, 2008 Report Share Posted June 10, 2008 Cannibals and Missionaries - Back to the River Crossing Puzzles 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. 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. I'm sorry but the question says there can never be more cannibals than missionaries on one side so if that means that these possible outcomes fit the question...then all you would have to do is take MC every time simple. MM M CC C Link to comment Share on other sites More sharing options...
Guest Posted June 11, 2008 Report Share Posted June 11, 2008 let cannibals be A and the missinaries be M so first one A comes with one M than A leaves M and comes back Where A gets down n two of M cross the river where one M stays and the other M goes back and than bring one A with him where the A goes back aned Brings another A with him than drops A and goes and picks the last A and they all cross the river Link to comment Share on other sites More sharing options...
Guest Posted June 11, 2008 Report Share Posted June 11, 2008 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. sorry the OP says clearly side indicating that in the boat there could be 1 cannibal and 0 missionaries. otherwise I think it's unsolveable Link to comment Share on other sites More sharing options...
Guest Posted June 11, 2008 Report Share Posted June 11, 2008 one of each on each boat trip across which is this: after the first batch of missionary/cannibal gets to the other side you have to find a way to get the boat back to the first side, remember: there are only 2 seats in boat, so there cannot be a third person manning the boat!!! It's important to read the OP carefully before you answer (me included) Link to comment Share on other sites More sharing options...
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