Yes I see it now!
so you will hear soon from me again if i have the right answer
well I guess I'll skip this one, at least until having some more time - I am more "logic riddles" guy, this river crossing I did in my head almost without using paper, but this one requires some math skills I left in high school
MOU quote, "
I am more "logic riddles" guy, this river crossing I did in my head almost without using paper" Most impressive, if I may say so. Therefore, perhaps you could assist with the following problem, which caused a degree of discord.
"How can three missionaries and three cannibals cross a river two at a time in a canoe. If the cannibals must never outnumber the missionaries left on one side, and two cannibals cannot paddle across together"
Your views would be much appreciated.
Does the use of the characters have any impact if not............1 m 1 C then 1 m 1 c, and lastly 1 m 1 c. Wouldn't that make sense since neither ever outnumbers the other, and 2 cs never paddle across together?
Hi, Locke. If only life were that simple. M/C>>> cross. Who rows back? Who's next? As you can see, it gets rather complicated. Those Cannibals sure are hungry. :wink:
this two at the time part confuses me? it's impossible cause in that case nobody will ever stay on the other side - unless canoe can come back without rower?
So, it is impossible cause:
a) M and M cannot row first time because then 1 M is left with 3 C
b) C and C cannot because of condition that two C cannot row together
c) if C and M go, then somebody has to come back, and from the way task is put nobody can come back alone (no single rowers), and it becomes totally pointless because nobody can ever stay on the other side
If one person CAN row then...then I'll try to do it when I come back from work
But i really think if it's not allowed that one person at the time rows it's impossible because of abc.
Edited: second paragraph has wrong step, correct one is in next post, first and third paragraph are okay
there is possibility that clue is in fact (its not specified) that M's can travel even if cannibals are on one side: e.g. if situation is M,M,M,C - C,C...then M and C can go to the other side with condition that M is not STAYING on it, like, he stays on the boat, so in no time situation is that C>M...if boat is "third territory", you know -
but it's kinda easy then - first one M transfers two C's (in two crosses of course) without leaving the boat, then he goes with another M, and they go out while one C comes back and then he, without leaving boat, takes another M to the other side and goes back, so situation is CCC-MMM, C goes there, stays, while M goes into boat and transfers other C's without leaving, except, obviusly, with the last one...
of course, it works only if it's allowed that one person rows when coming back
actually, WITHOUT having boat as "third territory" it's impossible even if one person rowing is allowed
explanation: only possible first step is MC (MM is not possible because one M stays with 3 C, and CC because of condition), and then it's only possible that M is coming back (because if C comes back then on first side situation is 2 M and 3 C)
So, situation is MMMCC - C
ONLY possible situation
And then - nobody can move.
a) if MM then on first side 1M<2C
b) CC not allowed
c) MC not allowed because on the other side would be 1M<2C
d) M alone is pointless because situation is same as in step one then (MC on other side, only M can come back, so in that case M travels eternally)
e) C alone is also pointless, because someone has to come back and in that case C travels eternally
So, only possible solution is that boat is third territory...it has to be it.
I also have wrong step in solution above, so here it is right way...
With steps:
1) MC go - C left on other side, M still in boat
MMCC - M - C
2) MC go - C left on the other side, M still in boat
MMC - M - CC
3) MM go, they are both left on other side, C goes in boat, rows back but STAYS in boat
MC - C - MMC
4) MC go - C is left on the other side, M stays in boat
C - M - MMCC
5) M takes C and they both go to other side
I am pretty sure that it's the only possible solution, because, as I explaned, situation where one rower is not allowed is absolutely impossible, and as I am explaning in this post, situation where one rower is allowed but boat is not third territory turns to be impossible as well
MOU I think you have it!
At no time is one rower excluded, and the boat is not a bank. This answer will be well and truly checked by others, who may have another opinion.
However in the meantime Have a cigar and gold star.
BTW good luck with joining that EU thing next week.
Mr Math today joined the controversy surrounding the World Math Challenge (WMC), when he released a press statement in which he criticised the committee setting the questions. Stating, "In my opinion they all need immediate medical treatment"
Asked if he was worried that he may be sued over such a remark he replied, "I am not an Aardvark, or anti by nature" It would appear his antiestablishmentism views are over the hill'.
175.015
Five ants are on the corners of an equilateral pentagon with side of length 1. They each crawl directly towards the next ant, all at the same speed and travelling in the same orientation.
How long will each ant travel before they all meet in the centre
It is intuitive that at all times the ants will form the corners of a square, ever decreasing in size and rotating about the centre.
nah, we are not joining, it's complicated process - war stopped us a lot, so at the moment we were only given green light to start process of joining, and first possible date for joining is 2007th
and as I said to mathur, I live in town with four rivers so I am good in crossing them
cannibal problem
i've found this:
assuming that only one cannibal can row and all the missionaries:
()position of the boat.
R=Rower
()CCRMMM - ......
CMMM - CR()
()CRMMM - C
MMM - CCR()
()RMMM - CC
RM - CCMM()
()CRMM - CM
CM - CRMM()
()CCMM - RM
CC - MMMR()
()CCR - MMM
C - MMMCR()
()CR - MMMC
...... - MMMCCR()
if i look at your solution My ownusername, it goes wrong at step 2 there are 2 cannibals and on missioner on the other side so it cannot be the right solution
magnum, it's clear said that in my solution boat is third territory (meaning that if someone is in the boat and is not leaving it, getting out to bank, it's allowed), which is allowed in many "cross the river" puzzles that have other difficulties, therefore, no, not a single time there was situation when m<c.
as for your solution, I am not saying that it's not possibly right, however, I must admit that I don't understand it at all
()CCRMMM - ......
CMMM - CR()
assuming that letters mean what we probably all think that they mean, first step is wrong because it has two cannibal rowers in one time - you simply can't have situation CMMM on first side without having two C rowing in first step - Edit: unless you assume that if two C's are in boat but only one is rowing - now, that's interesting fact, but I doubt that it's solution
However, as I am explaning in one of my posts, step by step, without boat as third territory, it's impossible, see explanation there (step by step) and feel free to say if I made some mistakes in it
step 2 in more details if you don't know what I mean...
2) MMCC on first bank - M in boat - C on other bank
MC boating (meaning other C enters boat, leaving MMC on first bank - MC in boat - C on other bank), then they come to other bank, C going out of boat, so situation is MMC on first bank - M in boat - CC on other bank
maybe it's good idea to put explanation why it's not possible without boat as third territory alone, in single post:
explanation: only possible first step is MC (MM is not possible because one M stays with 3 C, and CC because of condition), and then it's only possible that M is coming back (because if C comes back then on first side situation is 2 M and 3 C)
So, situation is MMMCC - C
ONLY possible situation
And then - nobody can move.
a) if MM then on first side 1M<2C
b) CC not allowed
c) MC not allowed because on the other side would be 1M<2C
d) M alone is pointless because situation is same as in step one then (MC on other side, only M can come back, so in that case M travels eternally)
e) C alone is also pointless, because someone has to come back and in that case C travels eternally
cannibal problem
you are right that the puzzle cannot be solved if the boat isn't third territory if and only if none of the cannibals can row.
so you solved the puzzle assuming the boat is a third territory your solution is therefore correct and i didn't read your post good enough so i mist the fact of the boat being a territory on it's own.
but you also mist one important thing in my post that was the fact that i was assuming that only one cannibal could row the boat. Assuming that i solved the puzzle without seeing the bout as a territory on itself
maybe tryagain can give us a puzzle that has no errors in it :wink:
175.015
the ants will reach each other at center having traveled distance 1,
one problem i can't proof it but my instinct says it's the right answer
