
My blood is tested +ve for Java.
Mani
Quaerendo Invenietis
Originally posted by Mani Ram:
There is a better way with which you can find out the odd bottle (unlike 10 bottles as it happens in your case). Also, you can test it with just 7 prisoners.
My blood is tested +ve for Java.
Originally posted by Chetan Parekh:
Pounding at a thick stone wall won't move it, sometimes, you need to step back to see the way around.
Originally posted by Angela Poynton:
Toom uch hassle for me, I'd just chuck em all in the bin
The Best way to predict your future is to create it  Every great individual common man
Originally posted by Ankur Sharma:
Hey Angela,
......
Hope for Good Next time
My blood is tested +ve for Java.
There is a better way with which you can find out the odd bottle (unlike 10 bottles as it happens in your case). Also, you can test it with just 7 prisoners.
Originally posted by Ankur Sharma:
Hey Angela,
Welcome to ranch ....
This forum is of international readership, please use english while posting your opinion !!!
Otherwise Sheriff's of this forum will delete your account....
Hope for Good Next time
Pounding at a thick stone wall won't move it, sometimes, you need to step back to see the way around.
Originally posted by fred Joly:
Ok, easy with 7 prisonners but I need 70 hours.(instead of 10 alloweded)
Have you got a solution for 7 prisonners in 10 hours ?
I go on searching.
Originally posted by fred Joly:
Yes, with my method I need the results from one taste test to determine what to do next.
Divide 100 bottles in 2 groups of 50 bottles.
Gave them in two prisonners. One dies, one survives.
result : one dead!
Divide 50 bottles (from the group wich contains poison) in 2 groups of 25 bottles.
Gave them in two prisonners. One dies, one survives.
result : two deads.
...with the same method until 1 bottle with poison found
Yes, with my method I need the results from one taste test to determine what to do next.
Divide 100 bottles in 2 groups of 50 bottles.
Gave them in two prisonners. One dies, one survives.
result : one dead!
Divide 50 bottles (from the group wich contains poison) in 2 groups of 25 bottles.
Gave them in two prisonners. One dies, one survives.
result : two deads.
...with the same method until 1 bottle with poison found
Originally posted by Ritika Saxena:
This way won't it take more than 10 hours?
Nothing is impossible; for those who doesnt have to do it themselves.
myjotting.blogspot.com
Originally posted by Nitin Nigam:
Arrange bottles in the formation of 10x10 as shown in the figure.
...
"I'm not back."  Bill Harding, Twister
Originally posted by Jim Yingst:
I don't know that this is optimal, but I get an expected loss of 3.16 prisoners, with a range of 07.
"I'm not back."  Bill Harding, Twister
Originally posted by Greg Charles:
OK, maybe I understand now, but it seems by Nitin's method we'd have to throw out two bottles. Say prisoners B and C died, then we'd have to throw out bottles 13 and 22. (That's with B drinking 11  20, and 2, 12, 22, 32, 42, ... and C drinking 21  30, and 13, 23, 33, 43, ...) Am I missing a detail there? In certain cases, ie., the ones on the diagonal, only one prisoner would die and the offending bottle would be exactly identified. The way I've drawn it out, those would be bottles 1, 12, 23, 34, etc.
Nothing is impossible; for those who doesnt have to do it themselves.
myjotting.blogspot.com
Vikrant Pandit
"I'm not back."  Bill Harding, Twister
Originally posted by Ryan McGuire:
Hmmm... I came up with an expected loss of only 2.32 prisoners and a max of only 3. If bottle 1 is poison, nobody dies; bottles 211, one prisoner; bottles 1256, two prisoners; 57100, three prisoners. (1*0 + 10*1 + 45*2 + 44*3)/100 = 2.32
[ September 08, 2006: Message edited by: Ryan McGuire ]
You think you know me .... You will never know me ... You know only what I let you know ... You are just a puppet ... CMG
Originally posted by Ajay Mathew:
If the selection of prisoners is not done properly a few(couple) of prisoners will face a higher chance of dieing. thats the guys who drink from bottle 57100.
"Sex and drugs and women being set on fire! I've never heard of such a Christmas!"  Christine Baranski in "The Ref"
Originally posted by Ryan McGuire:
If it's more important to save as many prisoners on average as possible, then the mortality rate can be kept to 2.32 but with uneven chances across the prisoners. Eight prisoners have a .16 chance of dieing and two prisoners have a .17 chance of not seeing another day. (Tak Ming Laio, how did you get .09833 and .10166?)
Originally posted by Tak Ming Laio:
Yes, I calculated wrongly. When I think it over the probalilities distribution should be:
8 prisoners have chance of dieing 0.23
2 prisoners have chance of dieing 0.24
Nothing is impossible; for those who doesnt have to do it themselves.
myjotting.blogspot.com
"I'm not back."  Bill Harding, Twister
Originally posted by fred Joly:
Just by curiosity : does this puzzle have any practical application ?
It is sorta covered in the JavaRanch Style Guide. 