Mice--a pain in the brain

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Mice--a pain in the brain

Postby finallyskeptic » Mon Dec 14, 2009 5:42 am

I have 12 mice with me,one of them is special he eats cake faster or slower than others.You have to find out which is the special one without using any clock,knife or object except cake and the mice.Find the minimum number of cake used if:
1.We know that the special mouse eats faster.
2.We know that the special mouse eats slower.
3.We don't know if he eats faster or slower.
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Re: Mice--a pain in the brain

Postby vanderpoel » Mon Dec 14, 2009 7:54 am

:kit:
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Re: Mice--a pain in the brain

Postby finallyskeptic » Mon Dec 14, 2009 10:35 am

Using a cat would test the running speed.I am sorry. ;)
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Re: Mice--a pain in the brain

Postby Gord » Mon Dec 14, 2009 10:31 pm

Uh...well...if I gave one cake each to 11 of the mice, I could see which one was eating faster or slower than the others. And if none of them were, I could deduce it was mouse #12 who was the different one. That would work when I knew if the special one was faster or if I knew it was slower, but if I didn't know (such as in the third case), I would have to give the twelfth mouse a cake too.
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Re: Mice--a pain in the brain

Postby tenthie » Tue Dec 15, 2009 5:30 am

U actually need to find d minimum no. of cakes u can use.....
So..............
U can do it with four cakes i suppose..... :roll:
i dont know about the third part...but yeah...d first two can definitely be done by four.....
Heres how its done...(the 1st part)
U divide d 12 mice into groups of three...so 4 groups would b formed...den u give each of d grup a cake...and see wich 1 finishes faster...so wen dat cake is completely finished, v know dat grup contains d fastest mice..now v just let des 3 mice(of the fastest grup) gorge on d leftovers by d other 3 grups.. nd find out wich one finishes faster.

The second part is a little more tricky.... U divide d mice into 2 grups of 6 each...
Then give each a piece of cake...So lets say grup 1 finishes faster....Now keep d leftover piece as it is....and exchange 3 mice from d two groups ....give a cake each again and again observe wich grup finished faster... If grup 1 finishes faster again...den dat wud mean d three mice in grup 2 contain d slower one. If grup 2 finishes faster..dat wud mean the 3 exchanged mice put in grup 1 contain d slower one.....and dont forget d two leftovers...wich r of d same level..nd so v feed d leftovers separately to ny 2 mice from d three....if one finishes faster its obvious....but if both finish equally...den its d other mousee!!!!!!

I shud think dat is the solution.. 8-)
Im munching upon d third part...but i guess dat shud be 4 too
Hmmmmmmm.......... :?

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Re: Mice--a pain in the brain

Postby Gord » Tue Dec 15, 2009 7:34 am

Well, the ABSOLUTE minimum would be 2 cakes. If I picked the special mouse by random chance and any other mouse and gave them each a cake, I could know which one was eating faster or slower.

But if I didn't know whether the special mouse was eating faster or slower, I'd need a minimum of 3 cakes....

:mrgreen: Since this would depend entirely on luck, it can't be the correct answer.
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Re: Mice--a pain in the brain

Postby finallyskeptic » Tue Dec 15, 2009 7:46 am

Since this would depend entirely on luck, it can't be the correct answer

Yup you are correct this time about the correctness of the answer.
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Re: Mice--a pain in the brain

Postby Gord » Tue Dec 15, 2009 8:09 am

finallyskeptic wrote:
Since this would depend entirely on luck, it can't be the correct answer

Yup you are correct this time about the correctness of the answer.

I was right the first time, too, I just didn't state it in my post. :mrgreen:
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Re: Mice--a pain in the brain

Postby finallyskeptic » Tue Dec 15, 2009 8:17 am

Then I have never been wrong in my Life.
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Re: Mice--a pain in the brain

Postby Gord » Tue Dec 15, 2009 8:32 am

finallyskeptic wrote:Then I have never been wrong in my Life.

Whereas I'm always wrong, because when I'm wrong I think I'm wrong so I'm still wrong, but when I'm right I think I'm wrong so then I am wrong.

Right? :mrgreen:
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Re: Mice--a pain in the brain

Postby finallyskeptic » Tue Dec 15, 2009 8:34 am

Wrong. :D
I have been wrong then.
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Re: Mice--a pain in the brain

Postby Gord » Tue Dec 15, 2009 10:24 am

finallyskeptic wrote:Wrong. :D

I knew it!!
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Re: Mice--a pain in the brain

Postby Gord » Sat Dec 19, 2009 7:30 am

Is tenthie right? Do you need to use the leftover cakes from previous speed tests?
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"You are also taking my words out of context." -- Justin
"Nullius in verba" -- The Royal Society ["take nobody's word for it"]
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Re: Mice--a pain in the brain

Postby jimlad » Sun Feb 14, 2010 10:04 pm

Split the mice into 4 groups of 3 and give a cake to 3 of the groups
- if all three of the groups finish at the same time, then the special mouse is in the group that didn't receive a cake
- if the special mouse is in one of the groups that received a cake, then that group should finish at a different time than the other groups
You can find which group the special mouse is in and then feed each of those 3 mice a cake and find which one finishes at a different time.

This method uses 6 cakes.

A similar way would be to split them into 3 groups of 4 and see which one is different and then feed 3 of the 4 remaining mice a cake and use the same steps above to locate the mouse.


Tenthies' wont work because you cant assume that the special mouse eats faster... if he eats slower then there wont be all these leftovers

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Re: Mice--a pain in the brain

Postby tenthie » Mon Feb 22, 2010 9:16 am

i've stated two methods out there and yeah.. they only work when we know that d special mouse is either faster or slower...


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