**Alright guys - here is one of the toughest one that i know. Its a very popular interview question for CS students so you can google the answer but it all depends on you if you want to be honest or just win. I have made some changes to the riddle to make it easy and less complex.
**A group of people with blue eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes and if their eye-color is blue then they can leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people and the Guru (she happens to have green eyes). So any given blue-eyed person can see 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 99 blue and 1 not blue.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
Peace MustafaQuraishi
I think I have it ... not sure
All people are x to themselves and will see lots of people with blue eyes. They will ask "who is the guru talking about?" Could it be one of them or me? Since x does not know his own eye colour he would naturally conclude it is one of the others. Then expect them to leave on that night. However, each one of them is in the same situation. And no one will leave on that first night.
This is where I am unsure ... it seems that any another night would not be useful for the blue eyed people since there would always be too many people to eliminate, but may be not for the Guru. After seeing all of the people in the dilemma of inaction she leaves the next night - but I can't explain why - it's just a feeling.
All people are x to themselves and will see lots of people with blue eyes. They will ask "who is the guru talking about?" Could it be one of them or me? Since x does not know his own eye colour he would naturally conclude it is one of the others. Then expect them to leave on that night. However, each one of them is in the same situation. And no one will leave on that first night.
This is where I am unsure ... it seems that any another night would not be useful for the blue eyed people since there would always be too many people to eliminate, but may be not for the Guru. After seeing all of the people in the dilemma of inaction she leaves the next night - but I can't explain why - it's just a feeling.
Can …I just say an off topic thing… When I read first line… I thought whose chest might they be… And why are they in front of me… Bajrangi bhaijan …akshay Kumar…cz photos took time to load ..and just text was visible…
But…fun was killed by the second line…
**Alright guys - here is one of the toughest one that i know. Its a very popular interview question for CS students so you can google the answer but it all depends on you if you want to be honest or just win. I have made some changes to the riddle to make it easy and less complex.
**A group of people with blue eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes and if their eye-color is blue then they can leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people and the Guru (she happens to have green eyes). So any given blue-eyed person can see 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 99 blue and 1 not blue.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
Is it like applying a filter on a collection problem? Each element in the collection needs to check the filter by him/herself. So at noon, only one islander shows up, gets the answer and leaves the island.
Is it like applying a filter on a collection problem? Each element in the collection needs to check the filter by him/herself. So at noon, only one islander shows up, gets the answer and leaves the island.
**Alright guys - here is one of the toughest one that i know. Its a very popular interview question for CS students so you can google the answer but it all depends on you if you want to be honest or just win. I have made some changes to the riddle to make it easy and less complex.
A group of people with blue eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes and if their eye-color is blue then they can leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people and the Guru (she happens to have green eyes). So any given blue-eyed person can see 99 people with blue eyes (and one with green)**, but that does not tell him his own eye color; as far as he knows the totals could be 99 blue and 1 not blue.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
Maybe this is it. Guy 1 hears the guru say there is at least another blue-eyed one and thinks it has to be one of the others. But after day 1 if nobody leaves, being a logician, he concludes,
1. he has blue eyes and knows it
2. others have blue eyes, but don't know it.
So he leaves the island. It is entirely possible, that all 100 of them can come to the same conclusion on day 2 and leave. Hope the boat is big enough!
so lets see - each person sees 99 blue eyes and 1 green eyes
in the dawn after night 1 everyone see the count is same. so they realise it is inconclusive.
so come to the dawn of 99th day - since there were no desertions so what should the thinking be
if I was any other color than blue then there would be 99 blue and 2 non blue.
since the rest are of the same intellect - they would have figured out if I was non blue then they are blue and left the boat.
but since they did not leave then I must be blue.
And hence on 100th night each one would go.
confusing to explain whats in my mind - but if this is the right answer maybe MQ would narrate better
**Alright guys - here is one of the toughest one that i know. Its a very popular interview question for CS students so you can google the answer but it all depends on you if you want to be honest or just win. I have made some changes to the riddle to make it easy and less complex.
A group of people with blue eye colors live on an island. They are all perfect logicians -- **if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes and if their eye-color is blue then they can leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people and the Guru (she happens to have green eyes). So any given blue-eyed person can see 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 99 blue and 1 not blue.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
**Alright guys - here is one of the toughest one that i know. Its a very popular interview question for CS students so you can google the answer but it all depends on you if you want to be honest or just win. I have made some changes to the riddle to make it easy and less complex.
**A group of people with blue eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes and if their eye-color is blue then they can leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people and the Guru (she happens to have green eyes). So any given blue-eyed person can see 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 99 blue and 1 not blue.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
its one wierd quiz. No wonder CSS guyz are sar-phiray.
Anywaz, my guesstimate is no one leaves. Reason being wut the guru has said and the way they see others. Who'll decide to go? Nobody.
And i know its more probability to be wrong but since i'm already marked dumb, why do i care :)
Maybe this is it. Guy 1 hears the guru say there is at least another blue-eyed one and thinks it has to be one of the others. But after day 1 if nobody leaves, being a logician, he concludes,
1. he has blue eyes and knows it
2. others have blue eyes, but don't know it.
So he leaves the island. It is entirely possible, that all 100 of them can come to the same conclusion on day 2 and leave. Hope the boat is big enough!
That sounds right.
But feedback played a part.
But before she spoke how people were finding out their eye color and leaving.
I know MQ said its not a trick question, But where new people come from if ferry picks up some one?
1st night: My guess would be the first night no one would leave because each assumes the guru spoke of the other 99. The second day they make a count and all 100 + 1 guru are still on the island. This continues for 99 nights.
100th night: By the 100th night and realizing no one left the island the assumption falls on each of the 100 (unaware) blue-eyed persons and they all leave.
BTW if they were such pessimists that for 100 nights they did not think it was them.
I would guess they would all commit suicide one after another.
Ok MQ its a bad question.
There too many degrees of freedom. All can think its them. one can think its him.
99 can think and be wrong, making the last one sure its him.
So out come depends on what restrain 100 act under.
Dil par ley li…
