A square medieval castle on a square island
is under siege. All around the castle there is a
square moat 10 meters wide. Due to a
regrettable miscalculation the raiders have
brought footbridges, which are only 9.5
meters long. The invaders cannot abandon
their campaign and return empty-handed.
How can the assailants resolve their
predicament?

annnikky:

Yeah baby correct

annnikky:
Just that u didn't differentiate them.... Anyway..

NO NO NO. That's not what you do to champions.

annnikky:
A square medieval castle on a square island
is under siege. All around the castle there is a
square moat 10 meters wide. Due to a
regrettable miscalculation the raiders have
brought footbridges, which are only 9.5
meters long. The invaders cannot abandon
their campaign and return empty-handed.
How can the assailants resolve their
predicament?

I would just put one footbridge across the corner (forming triangle) and the other footbridge from the corner to the middle of the island. Lobatan!

femi4:
I would just put one footbridge across the corner (forming triangle) and the other footbridge from the corner to the middle of the island. Lobatan!

Yeeeaaaaah correct

emorse:

NO NO NO. That's not what you do to champions.

Ok baby wat do I do then?

A military car carrying an important letter
must cross a desert. There is no petrol
station in the desert, and the car's fuel tank
is just enough to take it half way across.
There are other cars with the same fuel
capacity that can transfer their petrol to one
another. There are no canisters or rope to
tow the cars.
How can the letter be delivered?
You can play a little game with model cars to
simulate this puzzle.

A distant planet "X" has only one airport
located at the planet's North Pole. There are
only 3 airplanes and lots of fuel at the
airport. Each airplane has just enough fuel
capacity to get to the South Pole. The
airplanes can transfer their fuel to one
another.
Your mission is to fly around the globe above
the South Pole with at least one airplane, and
in the end, all the airplanes must return to
the airport.

annnikky:
Nope try again
Wlcm to d club

gracias,

royalvalentine19:
gracias,

Someone pls explain how d parrot z a lady

emyblinkx:
Someone pls explain how d parrot z a lady

That one has been solved already n if u watch closely u'll c d lady, d tail is d leg n d forehead is d eye of d parrot meanwhile there r other ones to solve so am waiting for u

annnikky:
A military car carrying an important letter
must cross a desert. There is no petrol
station in the desert, and the car's fuel tank
is just enough to take it half way across.
There are other cars with the same fuel
capacity that can transfer their petrol to one
another. There are no canisters or rope to
tow the cars.
How can the letter be delivered?
You can play a little game with model cars to
simulate this puzzle.

Amount of cars isn't stated, but let's assume there are 8 cars. All eight cars begin to drive. A quarter through the desert, all cars have half a tank. At this point four of the cars transfer their half tanks to the rest. So now four cats are empty, and four have full tanks.

Those with full tanks proceed, and when their tanks are at half again, 2 transfer all their fuel to the other two, which would give 2 full tanks. Enough to deliver the letter.

HyDef:


Amount of cars isn't stated, but let's assume there are 8 cars. All eight cars begin to drive. A quarter through the desert, all cars have half a tank. At this point four of the cars transfer their half tanks to the rest. So now four cats are empty, and four have full tanks.

Those with full tanks proceed, and when their tanks are at half again, 2 transfer all their fuel to the other two, which would give 2 full tanks. Enough to deliver the letter.

U got it right HyDef buh in my sol. they used 2 cars

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.

annnikky:
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.

First trip: Two cannibals will go

Second Trip: One of the cannibals return the boat

Third Trip: - Two cannibals will go

Forth Trip: One of the cannibas comes back.

Fifth Trip: Two missionaries leaves

Sixth Trip: One cannibal and one missionary return the boat.

Seventh Trip - The last 2 missionaries cross,

Eighth Trip: One cannibal return the boat

Nineth trip: Two cannibals crosses over and

Tenth trip: One of them return for the last trip

Finally, the last two cannibals crosses over

femi4:
First trip: Two cannibals will go

Second Trip: One of the cannibals return the boat

Third Trip: - Two cannibals will go

Forth Trip: One of the cannibas comes back.

Fifth Trip: Two missionaries leaves

Sixth Trip: One cannibal and one missionary return the boat.

Seventh Trip - The last 2 missionaries cross,

Eighth Trip: One cannibal return the boat

Nineth trip: Two cannibals crosses over and

Tenth trip: One of them return for the last trip

Finally, the last two cannibals crosses over

Femi4 u tried buh u didn't get it correct this time

annnikky:
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.

Let's assume they are crossing from A to B; C = cannibals and M = missionaries.

At the beginning,
A=(3C & 3M). B=(NOBODY)

1. Two cannibals cross to B, one returns.
A=(3M & 1C) ------> 2C -------> B=(2C)
A=(3M & 2C) <------ 1C <------- B=(1C)

2. He picks the last cannibal and returns the boat to A.
A=(3M). ------->2C------->B=(3C)
A=(3M & 1C) <-----1C<----- B=(2C)

2. Two missionaries cross to B, one missionary and one cannibal return to A.
A=(1M & 1C) -----> 2M. ----->B=(2C & 2M)
A=(2M & 2C) <-----1M &1C<-----B=(1C & 1M)

3. Two missionaries cross to B and the only cannibal there present returns with the boat to pick his cannibal friends one after the other.
A=(2C) ----->2M-----> B=(1C & 3M)
A=(3C) <-----1C<----- B=(3M)

A=(1C) ----->2C-----> B=(2C & 3M)
A=(2C) <-----1C<----- B=(1C & 3M)

A=(NOBODY) ----->2C-----> B=(3C & 3M)!!!!

PS: The items between each pair of arrows shows the movement of Cs and Ms while the items in brackets shows the number of Cs and Ms at either side of the river after each movement.

PSS: You know what to do baby

emorse:

Let's assume they are crossing from A to B; C = cannibals and M = missionaries.

At the beginning,
A=(3C & 3M). B=(NOBODY)

1. Two cannibals cross to B, one returns.
A=(3M & 1C) ------> 2C -------> B=(2C)
A=(3M & 2C) <------ 1C <------- B=(1C)

2. He picks the last cannibal and returns the boat to A.
A=(3M). ------->2C------->B=(3C)
A=(3M & 1C) <-----1C<----- B=(2C)

2. Two missionaries cross to B, one missionary and one cannibal return to A.
A=(1M & 1C) -----> 2M. ----->B=(2C & 2M)
A=(2M & 2C) <-----1M &1C<-----B=(1C & 1M)

3. Two missionaries cross to B and the only cannibal there present returns with the boat to pick his cannibal friends one after the other.
A=(2C) ----->2M-----> B=(1C & 3M)
A=(3C) <-----1C<----- B=(3M)

A=(1C) ----->2C-----> B=(2C & 3M)
A=(2C) <-----1C<----- B=(1C & 3M)

A=(NOBODY) ----->2C-----> B=(3C & 3M)!!!!

PS: The items between each pair of arrows shows the movement of Cs and Ms while the items in brackets shows the number of Cs and Ms at either side of the river after each movement.

PSS: You know what to do baby

U tried buh not correct yet

annnikky:

U tried buh not correct yet

Why na? There's no time when the cannibals are more than the missionaries. And they all crossed.

emorse:


Why na? There's no time when the cannibals are more than the missionaries. And they all crossed.

Lemme post d sol. R u ok with that

emorse:


Why na? There's no time when the cannibals are more than the missionaries. And they all crossed.

Don't mind her......your answer and mine were the same.

I dey here dey wait her in HD to reveal her solution

annnikky:

Lemme post d sol. R u ok with that

Oya post. But if its the same ehn, I won't take ordinary peck as reward o. Na French kiss I go collect.

femi4:
Don't mind her......your answer and mine were the same.

I dey here dey wait her in HD to reveal her solution

Oya make we dey watch na

emorse:

Oya post. But if its the same ehn, I won't take ordinary peck as reward o. Na French kiss I go collect.

Ok ooo

D sol.


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.


So check this out

annnikky:
D sol.


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.


So check this out

In real life condition: The bolded trips are suicidal. The cannibal will eat the missionary

Also, your trip 3 will be invalid. That is "2 cannibals there" since you stated that there would be tragedy if there are more cannibals than missionaries
together. Remember the boat can only take two people.

Finally, I think the right tragic life situation should be: A missionary and a cannibal cannot be together

femi4:
In real life condition: The bolded trips are suicidal. The cannibal will eat the missionary

Also, your trip 3 will be invalid. That is "2 cannibals there" since you stated that there would be tragedy if there are more cannibals than missionaries
together. Remember the boat can only take two people.

Finally, I think the right tragic life situation should be: A missionary and a cannibal cannot be together

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

annnikky:





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

Interprete this:There can never
be more cannibals than missionaries
together.

Cos we have "Before the Trip" "The Trip" and "Destination"

When does that condition holds?

femi4:
In real life condition: The bolded trips are suicidal. The cannibal will eat the missionary

Also, your trip 3 will be invalid. That is "2 cannibals there" since you stated that there would be tragedy if there are more cannibals than missionaries
together. Remember the boat can only take two people.

Finally, I think the right tragic life situation should be: A missionary and a cannibal cannot be together

Watch this then




https://www.youtube.com/watch?v=W9NEWxabGmg

annnikky:





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

emorse:


A=(3C & 3M). B=(NOBODY)
1. Two cannibals cross to B, one returns.
A=(3M & 1C) ------> 2C -------> B=(2C)
A=(3M & 2C) <------ 1C <------- B=(1C)

2. He picks the last cannibal and returns the boat to A.
A=(3M). ------->2C------->B=(3C)
A=(3M & 1C) <-----1C<----- B=(2C)

2. Two missionaries cross to B, one missionary and one cannibal return to A.
A=(1M & 1C) -----> 2M. ----->B=(2C & 2M)
A=(2M & 2C) <-----1M &1C<-----B=(1C & 1M)

3. Two missionaries cross to B and the only cannibal there present returns with the boat to pick his cannibal friends one after the other.
A=(2C) ----->2M-----> B=(1C & 3M)
A=(3C) <-----1C<----- B=(3M)

A=(1C) ----->2C-----> B=(2C & 3M)
A=(2C) <-----1C<----- B=(1C & 3M)

A=(NOBODY) ----->2C-----> B=(3C & 3M)!!!!

Oya show me the difference. Don't even bother sef, just give me my thing sharp sharp

annnikky:




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

. Correct! In fact, you made a good point.