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.
Work out the combinations necessary to get them all safely to the other side. The solution is going to be something like:
1 missionary and 1 cannibal there, 1 missionary back;
1 missionary and 1 cannibal there .....
And so on. Solution will be given tomorrow morning.
Tomorrow morning: Answer is here.
Surely you mean the boat can take 3.
ReplyDeleteBest regards
You always catch me out, Nigel.
ReplyDeleteI'm sure I've done this before, years ago. Anyway, I have a newly calculated solution; that is with 3 people allowed in the boat.
ReplyDeleteI don't think it can be done with just a 2-person boat, but I will look a little while longer for a solution, or proof that there is no path through the graph.
Best regards
2 cs cross
ReplyDelete1 returns.
1c+1m crosses
1m returns
1m+1c crosses
1m returns
2ms cross
1 returns
2ms cross.
Everybody is cross in this tale.
STB.
Yes 2 cs cross
ReplyDeleteYes 1 returns.
Yes 1c+1m crosses
Yes 1m returns
Yes 1m+1c crosses
Yes 1m returns
Yes 2ms cross
No 1 returns
2ms cross.
That would leave the missionary outnumbered.
Nigel - the boat's for two.
ReplyDeleteScotsToryB said...
ReplyDelete2 cs cross
1 returns.
1c+1m crosses
Delivery service to 2c's of 1m at far side. Yummy.
btw: Why is it a tragedy that the poor misunderstood black cannibals are fed by the oppressive missionaries.
Trip 1 MC
Return M
Trip 2 CC
Return C
Trip 3 MM
Return MC
Trip 4 MM
Return C
Trip 5 CC
Return C
Trip 6 CC
All done. Cannibals hungry.
http://en.wikipedia.org/wiki/Missionaries_and_cannibals_problem
ReplyDeleteAnd congrats to Lord T.
Best regards
Thanks all - the "official answer" has been up all along here.
ReplyDeleteCongrats!
[No women took part - hmmmm.]