Humans and Monkeys

[Guest]

im not sure but heres what i think please correct me if im wrong but i thing the big monkey should row the little monkey across and then row back and oh nvm i dont know lol <_< *sigh* complicated much hummmmmm

[Guest]

By saying the monkeys can jump out of the boat does that mean if the big monkey is paddling the boat he does not need to endanger the humans on shore when he drops one of the small monkeys off? That makes sense now that I spell it out. I love it when I answer my own questions.

[Guest]

Actually, if you properly adhere to the puzzle requirements there is no apparent solution. Even the admin's solution violates the requirement "At all times, the number of humans on either side of the river must be greater or equal to the number of monkeys on that side". At any time, any side having only monkeys violates this rule since 0 Humans is not greater or equal to 1, 2 or 3 monkeys.

For the admin solution to be correct, the requirement would need to be revised to state "At all times, the number of humans must be greater or equal to the number of monkeys on any side of the river having both monkeys and humans".

Some would say having only monkeys on a given side would not result in a human being eaten by monkeys. That may sound reasonable, but the rule still exists that # Humans >= Monkeys "At all times ... on either side".

Chris

Actually, 0 >= 0 is true. In the beginning if there are zero humans on one side and zero monkeys, this condition holds.

I will attempt to word my solution as follows:

The large monkey "M" always rows the boat.

In each series of two transfers, the first takes a human, the second takes a small monkey.

*Note* There is no restriction on what can go in the boat

So then:

MH goes over, M comes back

Mm goes over, M comes back

MH goes over, M comes back

Mm goes over, M comes back

MH goes over, both get out

This assumes: M is always in the boat, and HM get in the boat and leave the boat (beginning and end) at the same time.

By keeping one monkey in the boat at all times we can always have one "surplus" human to move around without violating the human-monkey rule.

Comments?

[Guest]

no i guess not!!

[Guest]

Good puzzle

[Guest]

2Humans, 1Big monkey and 1small monkey ride the boat to the other side of the bank - 2 humans and 1small monkey get down. 1Big monkey rides the boat back and hops out of the boat

1Human and 1small monkey take the boat to the other side. 1small monkey hops out of the boat. 1human rides the boat back

big monkey hops in and they ride together to the other side of the coast. Correct?

[Guest]

if there wasnt a limit on how many fit in the boat then they could all just get in!

and as for Jarod's post, the monkey in the boat, when he got to the bank would get out and eat the humans because there would be less humans than monkeys.

[Guest]

NICE AND TRICKY THE ONLY THING I DON'T UNDERSTAND IS WHY IS IT USEFUL TO KNOW THAT THE MONKEYS THROW THEMSELVES OFF THE BANK.. WAS THAT ACTUALLY NEEDED??

[Guest]

Let's consider

__ = boat

..... = water

H = Human

M = big monkey

m = small monkey

>< directions

HHH Mmm __ .....

Hm >

HH Mm .....__ H m

< H

HHH Mm __ ..... m

Mm >

HHH .....__ Mmm

< M

HHH M __ ..... mm

HH >

H M .. ...__ HH mm

< Hm

HH Mm __ ..... H m

HM >

H m .....__ HH Mm

< Hm

HH mm __ ..... H M

HH >

mm .....__ HHH M

< M

Mmm __ ..... HHH

Mm >

m .....__ HHH Mm

< M

Mm __ ..... HHH m

Mm >

.....__ HHH Mmm

[Guest]

I think that this is an easier and quicker solution:

Same aspects as original idea:

H = Human

M = Big Monkey

m = Littel Monkey

< > direction of boat

. no humans or monkeys

HHHMmm . .

HHmm HM> .

HHmm M< H

Mmm HH> H

Mmm H< HH

mm HM> HH

mm M< HHH

m Mm> HHH

m M< HHHm

. Mm> HHHm

. . HHHMmm

Only took nine total trips with the boat.

Enjoy

[Guest]

first one human and the big monkey go in the boat, leaving behind 2 human and 2 small monkey and drop the human on other side

the big monkey goes back to pick a small monkey, leaving behind 2human and a small monkey and they go to the other side to drop the small monkey with one human

big monkey now picks a human leaving behind 1 human and 1 small monkey and drops him making two humans and a small monkey on the other side

now big monkey picks a small monkey, leaving just 1 human and droping him, making it 2 humans and two small monkeys on the other side

then the big monkeyy picks up the last human and they drop to the other side

BUT LETS FACE IT, BIG OR SMALL,Y WOULD D MONKEY COME BACK FOR HUMANS OR MONKEYS AGAIN AND AGAIN EVEN IF HE KNEW HOW T ROW... AND D BEST PART MONKEYS ARE VEGGIES AND DO NAT EAT HUMANS!!!!!!!!!!!!!!!!!!!!

[hugemonkey]

how about this:-

HHHMmm...

HHMm >Hm...

HHMm...Hm

HM >Hm...Hm

HM...HHmm

>HM...HHmm

...HHHMmm

sounds rite ???

I must weigh in on this slanderous puzzle. I assume that by "big monkey" you are referring to me and I am shocked...SHOCKED I SAY at your libelous accusations that I would EAT a human!

but seriously the boat cannot cross the river by itself so it looks like you have gone wrong at the steps marked in blue.

[hugemonkey]

I think that this is an easier and quicker solution:

Same aspects as original idea:

H = Human

M = Big Monkey

m = Littel Monkey

< > direction of boat

. no humans or monkeys

HHHMmm . .

HHmm HM> .

HHmm M< H

Mmm HH> H

Mmm H< HH

mm HM> HH

mm M< HHH

m Mm> HHH

m M< HHHm

. Mm> HHHm

. . HHHMmm

Only took nine total trips with the boat.

Enjoy

A the red (for murder) marked steps you would have more monkeys than humans on one side. The puzzle states that you cannot have more monkeys than humans on one side (note it does not say shore) at a time. That would include if one or the other is in the boat, since the monkeys are able to jump onto (or off of) the boat from the shore.