[Guest]

Unsatisfied with the lame riddle he previously appeared in, an evil Emperor decided to further toy with his 16 prisoners. He came to them in the night and said:

In the morning you will again arrange yourselves in my courtyard. This time, forget the chess. The 16 of you will form lines of 4 men. I will then count the lines of 4 and that number of you will be released to be permanent ale tasters in the Imperial Brewhaus, whoever is left will meet the hangman. HAR HAR HAR!

The other prisoners again choose you, the nerdy dude to figure out what to do.

How many can you save tomorrow?

Edited September 30, 2009 by Chewbacca

[Guest]

Well, a trivial answer would be 10 as shown in the fig below

[Guest]

Well, a trivial answer would be 10 as shown in the fig below

I was about to post that

If you set them in a straight line, you save 13.

#1-#4 are a line of 4, #2-#5 are a line of 4, etc.

[Guest]

I was about to post that

If you set them in a straight line, you save 13.

#1-#4 are a line of 4, #2-#5 are a line of 4, etc.

nice i didnt think of that.

So far some of the 16 men have been saved. Can more be saved?

Edited September 30, 2009 by Chewbacca

[Guest]

It does not specify a straight line - if you place them all single file in a circle, then all can be saved

[Quantum.Mechanic]

Unsatisfied with the lame riddle he previously appeared in, an evil Emperor decided to further toy with his 16 prisoners. He came to them in the night and said:

“In the morning you will again arrange yourselves in my courtyard. This time, forget the chess. The 16 of you will form lines of 4 men. I will then count the lines of 4 and that number of you will be released to be permanent ale tasters in the Imperial Brewhaus, whoever is left will meet the hangman. HAR HAR HAR!”

The other prisoners again choose you, “the nerdy dude” to figure out what to do.

How many can you save tomorrow?

Bonus points for your self-deprecating style, persistence, and -- the best part -- fresh and interesting puzzles.

[Guest]

It does not specify a straight line - if you place them all single file in a circle, then all can be saved

I think it is pretty obvious that straight lines are what is intended. The primary definition of a line is straight. Would you really like to try the circle for this evil emperor? I think we know what he would do.

Nice guess though!

P.S. no lining up on the equator either! (that was your next guess i bet!)

Edited September 30, 2009 by Chewbacca

[Guest]

Here's an attempt with 11 saved lives

[Guest]

Here's an attempt with 11 saved lives

Nice.

Different than the solution to save 13, but a step toward the solution I am thinking of to save more than 13.

Edited September 30, 2009 by Chewbacca

[Guest]

1820

As it is not required for the four man forming a line to be in the same distance from each other, a single line of 16 prisoners would set 1820 of them free :-) as there is that many possible choises of 4 numbers from 16, for example #1 #3 #10 #16 form a straight line as well as #1 #2 #3 #4! Actually only seven prisoners in a row would do - 35 combinations, the rest nine can just sit and laugh sipping their ales

[Guest]

What if you put the prisioners in a pattern that would make each a corner of a small cube inside another cube. Then, from the purspective of each plane, one cube would save 6, the second another 6, and then by adding a few diagonal planes you would save an additional 16, potentialy saving 28.

Think of it this way.

Looking at a die, there are six faces. Looking flat against each you see that the four corners can be individual points on a line. The two cubes equals 12 faces or planes. One inside the other means that you can use a combination of the large and small edges to make their corners into lines. Four for each edge combo. You will have more than enough lines of four to save them. Of course you have to get the evil guy to move to each ideal position so from his perspective there are four points on the line, and you have to assume that gravity doesn't play a factor, but still, it works!

[Guest]

Unless I've miscounted, I can get 15

Each side of the star has 5 men, thus saving 2, for a total of 10. You also have a line from each point to the opposite valley, for 5 more.

.

[Guest]

What if you put the prisioners in a pattern that would make each a corner of a small cube inside another cube. Then, from the purspective of each plane, one cube would save 6, the second another 6, and then by adding a few diagonal planes you would save an additional 16, potentialy saving 28.

Think of it this way.

Looking at a die, there are six faces. Looking flat against each you see that the four corners can be individual points on a line. The two cubes equals 12 faces or planes. One inside the other means that you can use a combination of the large and small edges to make their corners into lines. Four for each edge combo. You will have more than enough lines of four to save them. Of course you have to get the evil guy to move to each ideal position so from his perspective there are four points on the line, and you have to assume that gravity doesn't play a factor, but still, it works!

I should have mentioned that gravity is a factor. these are prisoners not the cirque de solie!

Nice thinking though, if they were under water it would work.

Edited September 30, 2009 by Chewbacca

[Guest]

1820

As it is not required for the four man forming a line to be in the same distance from each other, a single line of 16 prisoners would set 1820 of them free :-) as there is that many possible choises of 4 numbers from 16, for example #1 #3 #10 #16 form a straight line as well as #1 #2 #3 #4! Actually only seven prisoners in a row would do - 35 combinations, the rest nine can just sit and laugh sipping their ales

Give me a break!

I swear it is hard to make these riddles bulletproof. If I gave all the caveats it would reduce the fun, and make the answer too obvious. The intent is that they be consecutive, and no, not the same distance nesesarily.

Good try thinking out of a stack of boxes though!

Edited September 30, 2009 by Chewbacca

[Guest]

Unless I've miscounted, I can get 15

Each side of the star has 5 men, thus saving 2, for a total of 10. You also have a line from each point to the opposite valley, for 5 more.

.

You got it! nice! I give you credit for solving it. and differently than my solution. (a little)

My original idea was to make there be no possible lines with more than 4, which produces this...

but I wanted to see what people came up with.

this pic was found on the web by the way, so im only taking credit for putting the riddle together. thanks everybody!

Edited September 30, 2009 by Chewbacca

[Guest]

You got it! nice! I give you credit for solving it. and differently than my solution. (a little)

My original idea was to make there be no possible lines with more than 4, which produces this...

but I wanted to see what people came up with.

this pic was found on the web by the way, so im only taking credit for putting the riddle together. thanks everybody!

Only problems I see with the solution involve whether the lines must be considered unique, and whether all 16 prisoners must be in line.

Each star is drawn with only 5 unique lines in your solution, so only 10 prisoners are saved under that circumstance.

There is one prisoner standing in the middle, not connected to a line. The statement was, "The 16 of you will form lines of 4 men." This could be decided that if all 16 men are not in some line of 4 then the Emperor would rule the lines invalid. Of course, it'd be hard to say that they have to arrange themselves in the courtyard any way they want to make as many lines of 4 as possible without giving too much info.

It makes me like Sashtsoh's answer better, if you stick to that line of answer.

Overall I'd say that kwiatek had the best mathematical answer!

Edited September 30, 2009 by Dagda333

[Guest]

You got it! nice! I give you credit for solving it. and differently than my solution. (a little)

My original idea was to make there be no possible lines with more than 4, which produces this...

but I wanted to see what people came up with.

this pic was found on the web by the way, so im only taking credit for putting the riddle together. thanks everybody!

Connecting each outer point through the center creates 5 more lines, each of which include the 16th person.

Edited September 30, 2009 by ogden_tbsa

[Guest]

Only problems I see with the solution involve whether the lines must be considered unique, and whether all 16 prisoners must be in line.

Each star is drawn with only 5 unique lines in your solution, so only 10 prisoners are saved under that circumstance.

There is one prisoner standing in the middle, not connected to a line. The statement was, "The 16 of you will form lines of 4 men." This could be decided that if all 16 men are not in some line of 4 then the Emperor would rule the lines invalid. Of course, it'd be hard to say that they have to arrange themselves in the courtyard any way they want to make as many lines of 4 as possible without giving too much info.

It makes me like Sashtsoh's answer better, if you stick to that line of answer.

Overall I'd say that kwiatek had the best mathematical answer!

You need to take a second look. there are 5 lines of 4 that the center point is a part of! (just noticed ogden answered already)

Edited October 1, 2009 by Chewbacca

[Guest]

Chew, I like your answer and you could have required no 2 prisoners to be on more than 1 line together OR no 2 lines can contain the same 2 prisoners.

Nice job.