Riddle and Puzzle Thread

[Lihu]

Quote:

Originally Posted by Golfingdad

Let's say you have 9 marbles of the exact same size and shape, however, one is slightly heavier than the rest.  Using only a (balance) scale, and using it only 2 times, how do you determine which is the heavier ball?

Not sure if it's been answered, but didn't want to scroll through and ruin possible answers to other riddles!

Randomly separate the balls into three groups of three, say groups A, B, and C.  Compare groups A and B on the balance scale.  If they weigh the same, then the heavier ball is in group C.  Otherwise, you the heavier group of three contains the heavier ball.  Say the group with the heavier ball is group A.  Now choose two balls randomly from group A.  Compare them on the balance scale.  Either one is heavier than the other and you've found the heavier ball, or they weigh the same, and the heavier ball is the one ball you left out.

#########################

New Riddle:

You are given 99 coins that are heads up, and an unknown number of coins that are tails up. You are blindfolded and somehow can't feel the difference between the heads and tails sides of the coins. You can count the coins, put them in arbitrary many piles, flip whichever coins you want, but remember, when you flip and when you sort, you DO NOT know which ones are heads up, which are tails up. In the end, you must end up with just two piles, each containing an equal number of heads. How do you do this?

Sorry to be a spoiler, but if you give up here is the answer: Pai's is correct.

http://classic-puzzles.blogspot.com/2006/12/microsoft-puzzle-coins-on-table.html

[mdl]

Sorry to be a spoiler, but if you give up here is the answer: Pai's is correct.

http://classic-puzzles.blogspot.com/2006/12/microsoft-puzzle-coins-on-table.html

That's a different puzzle, and Pai's answer is not the correct one to the version of this puzzle I posted.  Though the correct answer to my version is related.

[Golfingdad]

Sorry to be a spoiler, but if you give up here is the answer: Pai's is correct.

http://classic-puzzles.blogspot.com/2006/12/microsoft-puzzle-coins-on-table.html

@Lihu , what the hell?!?!?!  You aren't supposes to look up the answers to these online, you are supposed to try and figure them out yourself and post the answer (not a link!) and then post a new riddle.

Come on, man!!!

[Lihu]

[QUOTE name="Lihu" url="/t/71500/riddle-thread/0_30#post_929594"]   Sorry to be a spoiler, but if you give up here is the answer: Pai's is correct. [URL=http://classic-puzzles.blogspot.com/2006/12/microsoft-puzzle-coins-on-table.html]http://classic-puzzles.blogspot.com/2006/12/microsoft-puzzle-coins-on-table.html[/URL] [/QUOTE] @Lihu , what the hell?!?!?!  You aren't supposes to look up the answers to these online, you are supposed to try and figure them out yourself and post the answer (not a link!) and then post a new riddle. Come on, man!!!:beer:

Oops, didn't read the rules. Post when I get back from my round.

[iacas]

Sorry to be a spoiler, but if you give up here is the answer: Pai's is correct.

http://classic-puzzles.blogspot.com/2006/12/microsoft-puzzle-coins-on-table.html

You're also not supposed to be using the Internet for these answers.

There are two outstanding riddles right now. They are:

Jack went to the store to sell corn and peas. He only had one bag and needed to keep them separate, so he put the corn in first and tied the bag over the corn, then he put the peas in the bag above the knot.   A buyer for the corn came by with his own bag.  He did not want the peas.

How do we pour the corn into the buyers sack?

Constraints - Pouring the content anywhere else but the sacks is considered soiling.  You can't soil the peas. The buyer wants to keep his own bag.  Jack needs to keep the peas in his own bag. You can't cut a hole in either bag.  (they are very big and loose bags)

You are given 99 coins that are heads up, and an unknown number of coins that are tails up. You are blindfolded and somehow can't feel the difference between the heads and tails sides of the coins. You can count the coins, put them in arbitrary many piles, flip whichever coins you want, but remember, when you flip and when you sort, you DO NOT know which ones are heads up, which are tails up. In the end, you must end up with just two piles, each containing an equal number of heads. How do you do this?

Post your guesses.

[Golfingdad]

\

Jack went to the store to sell corn and peas. He only had one bag and needed to keep them separate, so he put the corn in first and tied the bag over the corn, then he put the peas in the bag above the knot.   A buyer for the corn came by with his own bag.  He did not want the peas.

How do we pour the corn into the buyers sack?

Constraints - Pouring the content anywhere else but the sacks is considered soiling.  You can't soil the peas. The buyer wants to keep his own bag.  Jack needs to keep the peas in his own bag. You can't cut a hole in either bag.  (they are very big and loose bags)

Not sure if this is right but ... You pour the peas into the buyers bag, then untie the knot, then pour the corn on the table, pour the peas back into Jack's bag, and then put the corn into the buyers bag.

I got that from the bold line (nowhere did it say that you couldn't soil the corn), however, as I'm writing, I also see that it says Jack needs to KEEP the peas in his own bag, which he's not technically doing in my solution.  He takes them out and puts them back in.

Am I right?  If so, I don't have another puzzle, but we still have Matt's to solve below:

New Riddle:

You are given 99 coins that are heads up, and an unknown number of coins that are tails up. You are blindfolded and somehow can't feel the difference between the heads and tails sides of the coins. You can count the coins, put them in arbitrary many piles, flip whichever coins you want, but remember, when you flip and when you sort, you DO NOT know which ones are heads up, which are tails up. In the end, you must end up with just two piles, each containing an equal number of heads. How do you do this?

[mdl]

Quote:

Originally Posted by rehmwa

Jack went to the store to sell corn and peas. He only had one bag and needed to keep them separate, so he put the corn in first and tied the bag over the corn, then he put the peas in the bag above the knot.   A buyer for the corn came by with his own bag.  He did not want the peas.

How do we pour the corn into the buyers sack?

Constraints - Pouring the content anywhere else but the sacks is considered soiling. You can't soil the peas. The buyer wants to keep his own bag.  Jack needs to keep the peas in his own bag. You can't cut a hole in either bag.  (they are very big and loose bags)

Well if the bags are really big, Jack can just cinch the top of the bag with one hand, turn the bad sideways, spin the ball of peas part of the bag to close off the peas, hold the spun/twisted close over the peas, untie the corn knot, and dump the corn into the buyer's bag.  This depends on the bag being big enough and the corn ball being small enough relative to the size of the bag that the corn ball can be threaded through the loop in the knot above it to allow untying, or at least significant loosening, where you could sort of funnel squeeze the corn out through the partially untied bag until enough is gone that you can fully untie.  You can maximize the untying room by closing off the peas as close as possible to the mouth of the bag.

[iacas]

Jack went to the store to sell corn and peas. He only had one bag and needed to keep them separate, so he put the corn in first and tied the bag over the corn, then he put the peas in the bag above the knot.   A buyer for the corn came by with his own bag.  He did not want the peas.

How do we pour the corn into the buyers sack?

Constraints - Pouring the content anywhere else but the sacks is considered soiling.  You can't soil the peas. The buyer wants to keep his own bag.  Jack needs to keep the peas in his own bag. You can't cut a hole in either bag.  (they are very big and loose bags)

Will this work?

1. Turn the buyer's bag inside out.
2. Dump the peas into the buyer's bag.
3. Tie the buyer's bag and turn it inside out again. They'll have a "blob" of peas sitting "inside" of the new, right side out bag.
4. Pour the corn from Jack's bag into the buyer's bag.
5. Un-tie the bag and pour the corn peas back into Jack's bag.

I believe that meets the conditions. I'm not sure how you'd tie the bag, but you could use a basic knot and leave one piece of the string long so that you could untie it by pulling on it.

[rehmwa]

\

Not sure if this is right but ... You pour the peas into the buyers bag, then untie the knot, then pour the corn on the table, pour the peas back into Jack's bag, and then put the corn into the buyers bag.

sorry, you can't spoil either corn or peas - they can travel from bag to bag all you like

Erik got it right (don't need a 'knot' really, twist the bag enough to close off the peas prior to reversing the bag will contain the material - and you can open that little sphinctor feature enough to pour back the corn... - I even like the first step to turn the buyer's bag inside out so that when it's finished, it's right side out and full of corn.

though you mixed up the corn with the peas in the 5th step....(I find it's easier to describe peas and diamonds to keep people from mixing up what's the target purchase vs the kept material.)

Still thinking on the pennies.......

[Golfingdad]

Quote:

Originally Posted by rehmwa

Jack went to the store to sell corn and peas. He only had one bag and needed to keep them separate, so he put the corn in first and tied the bag over the corn, then he put the peas in the bag above the knot.   A buyer for the corn came by with his own bag.  He did not want the peas.

How do we pour the corn into the buyers sack?

Constraints - Pouring the content anywhere else but the sacks is considered soiling.  You can't soil the peas. The buyer wants to keep his own bag.  Jack needs to keep the peas in his own bag. You can't cut a hole in either bag.  (they are very big and loose bags)

Well if the bags are really big, Jack can just cinch the top of the bag with one hand, turn the bad sideways, spin the ball of peas part of the bag to close off the peas, hold the spun/twisted close over the peas, untie the corn knot, and dump the corn into the buyer's bag.  This depends on the bag being big enough and the corn ball being small enough relative to the size of the bag that the corn ball can be threaded through the loop in the knot above it to allow untying, or at least significant loosening, where you could sort of funnel squeeze the corn out through the partially untied bag until enough is gone that you can fully untie.  You can maximize the untying room by closing off the peas as close as possible to the mouth of the bag.

sorry, you can't spoil either corn or peas - they can travel from bag to bag all you like

Erik got it right (don't need a 'knot' really, twist the bag enough to close off the peas prior to reversing the bag will contain the material - and you can open that little sphinctor feature enough to pour back the corn... - I even like the first step to turn the buyer's bag inside out so that when it's finished, it's right side out and full of corn.

though you mixed up the corn with the peas in the 5th step....(I find it's easier to describe peas and diamonds to keep people from mixing up what's the target purchase vs the kept material.)

Still thinking on the pennies.......

I agree that Erik's solution is pretty clever ... but I like Matt's as well.  Seems like it would work too, wouldn't it?   After all, you did say that the bags were "very big and loose."

Yeah, I am completely stumped on the coins as well.  I'll keep thinking about it though ...

[iacas]

I think I have the coins one. It's very similar to some questions that Microsoft and other computer companies would ask in the early 2000 as interview questions. I'm watching the Penguins game, and will let others have a chance to answer.

[billchao]

You are given 99 coins that are heads up, and an unknown number of coins that are tails up. You are blindfolded and somehow can't feel the difference between the heads and tails sides of the coins. You can count the coins, put them in arbitrary many piles, flip whichever coins you want, but remember, when you flip and when you sort, you DO NOT know which ones are heads up, which are tails up. In the end, you must end up with just two piles, each containing an equal number of heads. How do you do this?

I know I'm focusing too much on the semantics again, but if you just separate the coins into two even piles, you'd have an equal number of heads in each pile. Well, assuming you don't have any two-headed coins or something like that. It's probably a logic problem though, so I fail.

I think I have the coins one. It's very similar to some questions that Microsoft and other computer companies would ask in the early 2000 as interview questions. I'm watching the Penguins game, and will let others have a chance to answer.

Yea, this is why I'm not a software guy. Also, OT, but the game is awful from my perspective.

[mdl]

I know I'm focusing too much on the semantics again, but if you just separate the coins into two even piles, you'd have an equal number of heads in each pile. Well, assuming you don't have any two-headed coins or something like that. It's probably a logic problem though, so I fail.

But you have no idea which ones are the 99 that are heads up to start.  So split them blindly in half and you have no idea how many heads are in each pile.  Plus, since you start with 99 heads, you obviously need to do some flipping to have two piles each with the same number of heads.

[Golfingdad]

But you have no idea which ones are the 99 that are heads up to start.  So split them blindly in half and you have no idea how many heads are in each pile.  Plus, since you start with 99 heads, you obviously need to do some flipping to have two piles each with the same number of heads.

Yeah, @billchao 's answer was the first thing that came to mind for me too, but I figured that I was just reading too much into your word choice.  You said "how many heads in each pile."  Well, no matter if they are up or down, you still know exactly how many heads are in each pile, know what I mean?  But I didn't bother to answer thinking that I was probably focusing too much on words, as billchao said as well. ;)

Anyways, I'll keep thinking.  Actually, going to pose this one to my nerdy dad right now ...

[billchao]

Yeah, @billchao's answer was the first thing that came to mind for me too, but I figured that I was just reading too much into your word choice.  You said "how many heads in each pile."  Well, no matter if they are up or down, you still know exactly how many heads are in each pile, know what I mean?  But I didn't bother to answer thinking that I was probably focusing too much on words, as billchao said as well. ;)

I used to argue this kind of stuff with my high school teachers. I'm sure they loved having me in their classes.

[rehmwa]

I agree that Erik's solution is pretty clever ... but I like Matt's as well.  Seems like it would work too, wouldn't it?

yeah yeah, actually I like it.  Or the pillow case solution. make piles of each and 'pillow up' each grain.  and then push those pillows around each other.  visualize scooting two piles past each other......

not really in the spirit of the question, obviously.  but points for being clever and not snarky.  In that vein, one could also push the peas off to the side of Jack's bag and sew it into a pocket....untie the know and flow the corn along the edge out of Jack's bag and into the buyer's bag... pick the stiches out afterwards for good measure (is that Matt's?)

'big and loose' was to keep the snarky types from whining that they don't know if they can tie the knots or not (again stupid distractions from the intent - much like the other marble question with the "I'll just assume the bags are clear" type comments).  But, worth trying - sometimes it's hard to tell.  I haven't pulled out any really difficult puzzles yet.  the coin one has me stalled out.

coins......snarky answer is all the coins have heads up or down...again, not the intent.  Something likely to do with sorting into particular sized groups, and flipping certain groups and then recombining......I'm not yet seeing it...one thing is both have the same number of UPS, not the same ups and downs, so a solution of all ups is just as valid as 50 each or 2 each etc etc.

[SloverUT]

Here is my guess on the coins riddle:

I separate the 99 coins with heads up and the unknown with tails up and put them in two separate piles.

I then take all 99 coins in the heads up pile and flip them over to tails up.  I now have two piles of tails up coins.

I now combine all the tails up coins together and tally up how many coins I have.  Once this is done, I divide that total number by two and split them into two even piles.

Once I know exactly how many I have in each pile I then turn an equal number over to heads up in each pile thus giving me an equal amount of heads up.

Note: I am assuming I have the pile of 99 heads up in the beginning and can tell which these are to flip them over.  If not, this is not correct :)

[Golfingdad]

I separate the 99 coins with heads up and the unknown with tails up and put them in two separate piles.

I like your solution, except for one thing.  Matt said a few posts back that ...

you have no idea which ones are the 99 that are heads up to start.

I'm stil completely stumped on this one.  Don't tell us yet Matt!  Erik said 3 days or so of no solution and then you can tell us.