[Guest]

I got this one from my Dad.

There are 20 people in a room. Each of them has a card with an integer (ANY integer). All the cards are different. Devise a strategy to put the cards in order from least to greatest without anyone KNOWING all 20 cards. This means that if anyone can deduce the final card, its no good. One more rule: someone has to know that they're in order. You can't just make every possible combination of 20 cards without looking at any, you need to know.

There are many solutions, some much more complicated than others, but the real goal is to do it with as few people as possible.

I can do this with 2 people.Can you?

There is another problem that is much more difficult, surprisingly so.

The problem is identical EXCEPT the integers must be positive.

2 people, but it's much trickier.

Enjoy.

[dms172]

guy 1 can know 1-19

guy 2 can know 2-20

just has to say which to cards 1 is between

that way no knoe one no one knows all twenty

thats right if i understood the riddle right

[Guest]

Well, the way the problem is stated, it seems to be pretty open-ended. I'm not sure I am understanding it correctly, but...

You have 1 person know all the positive numbers and 0 and line them up and one person know all the negative numbers and line them up. Then simply line up the negative numbers before the positive numbers.

For the second part, you have 1 person look at a card and ask "does anyone have a card higher than ____" and if more than 0 and less than 19 people raise their hands then you can divide the group into two based on that number. Otherwise, pick another card. Then have 1 person know all the numbers higher and one person know all the numbers lower than that card.

[Guest]

guy 1 can know 1-19

guy 2 can know 2-20

just has to say which to cards 1 is between

that way no knoe one no one knows all twenty

thats right if i understood the riddle right

but what if card 1 and 20 are next to each other in order. You wouldn't know which one to put first

[Guest]

guy 1 can know 1-19

guy 2 can know 2-20

just has to say which to cards 1 is between

that way no knoe one no one knows all twenty

thats right if i understood the riddle right

I don't entirely understand your answer. It looks like it could work, but how do they determine the lowest 19 and highest 19? What if the highest card is 20? Please be more descriptive.

[Guest]

Well, the way the problem is stated, it seems to be pretty open-ended. I'm not sure I am understanding it correctly, but...

You have 1 person know all the positive numbers and 0 and line them up and one person know all the negative numbers and line them up. Then simply line up the negative numbers before the positive numbers.

For the second part, you have 1 person look at a card and ask "does anyone have a card higher than ____" and if more than 0 and less than 19 people raise their hands then you can divide the group into two based on that number. Otherwise, pick another card. Then have 1 person know all the numbers higher and one person know all the numbers lower than that card.

Answer 1: How do they determine which are positive and negative? There are many ways, but how many people look at their cards? What if all 20 cards are either positive or negative?

Answer 2: What if the cards he picks is 20 and nobody has a higher card? This would also involve everyone looking at some card.

[Guest]

Answer 1: How do they determine which are positive and negative? There are many ways, but how many people look at their cards? What if all 20 cards are either positive or negative?

Answer 2: What if the cards he picks is 20 and nobody has a higher card? This would also involve everyone looking at some card.

1: Doesn't everyone look at their own card? Just have, say, all the people with positive numbers go to the right side of the room and all the negative numbers go to the left. If all end up being positive or negative, then it ends up being case 2.

2: If he picks 20, and no one has a higher card, then he picks a different card. Basically the goal is to divide the 20 into two groups, each of which is ordered by a different person and they know what the dividing number for the two groups are, but not the numbers in the other group.

The puzzle only says no one person can know all 20 cards, not that the 20 people cannot all know one of the card values. Or am I misunderstanding?

[Guest]

1: Doesn't everyone look at their own card? Just have, say, all the people with positive numbers go to the right side of the room and all the negative numbers go to the left. If all end up being positive or negative, then it ends up being case 2.

2: If he picks 20, and no one has a higher card, then he picks a different card. Basically the goal is to divide the 20 into two groups, each of which is ordered by a different person and they know what the dividing number for the two groups are, but not the numbers in the other group.

The puzzle only says no one person can know all 20 cards, not that the 20 people cannot all know one of the card values. Or am I misunderstanding?

You can have as many people look at their cards as you want, but the secondary goal is to do it with as few people as possible.

If he picks 20, and nobody has a higher card, then he knows all 20 cards. The same thing would happen if he picked 19 and the only person with a higher card had a 20. Think about it.

[Guest]

I got this one from my Dad.

There are 20 people in a room. Each of them has a card with an integer (ANY integer). All the cards are different. Devise a strategy to put the cards in order from least to greatest without anyone KNOWING all 20 cards. This means that if anyone can deduce the final card, its no good. One more rule: someone has to know that they're in order. You can't just make every possible combination of 20 cards without looking at any, you need to know.

There are many solutions, some much more complicated than others, but the real goal is to do it with as few people as possible.

I can do this with 2 people.Can you?

There is another problem that is much more difficult, surprisingly so.

The problem is identical EXCEPT the integers must be positive.

2 people, but it's much trickier.

Enjoy.

If there are 20 people with cards you need 20 people involved, so the answer is 20

[Guest]

The people are given cards, but assume they haven't looked at them yet.

To make things simpler, collect the cards first. Send everyone out of the room, and call on them as you need them (that is, if you're trying to do it with few people).

I think I'll post my solution to the first problem tomorrow. My solution to the second is fairly similar to my solution for the first, but not quite.

Edited October 1, 2008 by Kioshanta

[Guest]

The people are given cards, but assume they haven't looked at them yet.

To make things simpler, collect the cards first. Send everyone out of the room, and call on them as you need them (that is, if you're trying to do it with few people).

I think I'll post my solution to the first problem tomorrow. My solution to the second is fairly similar to my solution for the first, but not quite.

Do any of them remember the card they were given?

[Guest]

Do any of them remember the card they were given?

If anyone reads a card, they will remember it, but they don't read the card their given unless they are instructed to.

[Guest]

Maybe it's best that this one die. but I said I'd post my solutions.

One person collects the cards and gives one of them to someone else. He sorts the 19 and lays them out. He flips up the middle card and tells the person with the final card to place his card in the cards to the right of that if his is greater than the middle card, and to the left if it's less. The first person does not see where he placed his card. He does so and collects the cards. They are in order.

Do the same thing as the first, just be careful that you don't explain anything to the guy with the last card.

[bonanova]

Maybe it's best that this one die. but I said I'd post my solutions.

One person collects the cards and gives one of them to someone else. He sorts the 19 and lays them out. He flips up the middle card and tells the person with the final card to place his card in the cards to the right of that if his is greater than the middle card, and to the left if it's less. The first person does not see where he placed his card. He does so and collects the cards. They are in order.

Do the same thing as the first, just be careful that you don't explain anything to the guy with the last card.

Explain how the 20th card is in its proper place with respect to the other 19 cards.

You've simply insured its order is correct with respect to card # 10, no?

This problem is similar to the red/blue hat problem where the solution is to

have people step into line between a red and blue hat until everyone is in line.

But that involves more than two people.

Your solution sound similar, but with only one comparison.

How is that be enough?

Suppose you order the 19 cards as follows

2 5 8 13 36 44 47 48 52 56 60 63 64 68 69 74 78 79 82.

You put them face down in that order and turn over the middle card:

2 5 8 13 36 44 47 48 52 56 60 63 64 68 69 74 78 79 82.

The 20^th card is 71.

The 2^nd person is asked to place it [as he understands your instructions] to the right of 56.

How does he know he should place it between the 15^th [69] and 16^th [74] cards?

[Guest]

Explain how the 20th card is in its proper place with respect to the other 19 cards.

You've simply insured its order is correct with respect to card # 10, no?

This problem is similar to the red/blue hat problem where the solution is to

have people step into line between a red and blue hat until everyone is in line.

But that involves more than two people.

Your solution sound similar, but with only one comparison.

How is that be enough?

Suppose you order the 19 cards as follows

2 5 8 13 36 44 47 48 52 56 60 63 64 68 69 74 78 79 82.

You put them face down in that order and turn over the middle card:

2 5 8 13 36 44 47 48 52 56 60 63 64 68 69 74 78 79 82.

The 20^th card is 71.

The 2^nd person is asked to place it [as he understands your instructions] to the right of 56.

How does he know he should place it between the 15^th [69] and 16^th [74] cards?

Sorry about that. It's easy to forget to write things that are clear in your mind.

By "place his card", what I meant was "look at at all the cards on the correct side and then place his card in with them". This way, he will only see half of them.