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Stacking the chips at Morty's


Best Answer k-man, 07 January 2013 - 04:21 PM

Spoiler for how about...
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9 replies to this topic

#1 bonanova

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Posted 04 January 2013 - 08:50 AM

It was amateur night at Morty's. Alex had finished

his penultimate act of mysterious prestidigitation.

Preparing for the finale, he handed a pile of poker

chips to Jamie and, after turning his back to the

table, asked Jamie to construct three equal stacks

of at least four chips.

 

Back still to the table, Alex next asked Davey to

call out at random a number between 1 and 12.

Suppose Davey calls out the number x.

 

With the chips still out of his sight, and without

knowing the number of chips in each stack, Alex

finally gave instructions to Ian to shift chips among

the piles, in a manner that left x chips in the center

stack.

 

What is the method?

 

To clarify, neither the number x nor the word "all"

were mentioned in the instructions. This rules out

the obvious: move all the chips to stack 3, then

move x of them to the middle.


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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#2 CaptainEd

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Posted 04 January 2013 - 05:17 PM

Spoiler for If I understand it right...

Somehow, I expect you want something smarter...


Edited by CaptainEd, 04 January 2013 - 05:19 PM.

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#3 CaptainEd

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Posted 04 January 2013 - 05:56 PM

Spoiler for Oops! Last instruction was a typo!


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#4 TimeSpaceLightForce

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Posted 04 January 2013 - 05:58 PM

Spoiler for x=3


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#5 jhawk

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Posted 05 January 2013 - 04:03 AM

Seems to me that TimeSpaceLighForce's would work, but moving all but 1 chip from the pile is basically the same as saying move all the chips. BTW, an easier and faster solution would be:

Move 4 chips from #1 to #3
Move all but 1 chip from #2 to #3
Move (12-x) chips from #3 to #2

But again too simpla a solution for a Bonavona puzzle
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#6 jhawk

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Posted 05 January 2013 - 04:20 AM

Ooops, last move should be(x-1) chips to #2.

Another simple solution is for Monty to give Ian 12,13,or 14 chips, making 3 stacks of 4, leading to easy solution
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#7 plasmid

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Posted 06 January 2013 - 04:01 AM

There are a number of trivial solutions that could fit the OP, like saying "move half of the chips in the middle pile into the leftmost stack, and move the other half into the rightmost stack" and then saying "move a chip from the left (or right) stack to the middle stack)" X times.

I think we might need more details about what sorts of instructions are legal in this problem, and what sorts aren't.


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#8 bonanova

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Posted 07 January 2013 - 03:20 AM

I had in mind finding a method that did not seem to depend on

knowing the value of x. There may be very many such ways.

 

So let's amend the trick to make it seem more difficult,

but actually providing a guide to the desired solution.

 

  1. Have Jamie construct three equal stacks.
    You do not know how many chips are in each stack.

     
  2. Instruct Ian to make three adjustments to the stacks.
    The words all, half, etc may not be used; at no point is any stack empty.

     
  3. Ask Davey to call out a number 0 < x < 13.
     
  4. Instruct Ian to make a fourth adjustment to the stacks:
    "Move __ chips from Stack _ to Stack _."
    Where "__" is a specific number different from x.

 

The middle stack should now have x chips.


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#9 k-man

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Posted 07 January 2013 - 04:21 PM   Best Answer

Spoiler for how about...

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#10 bonanova

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Posted 07 January 2013 - 10:33 PM

Several solution have been posted.
Mine is not greatly distinguished from any of them.
It differs from k-man's (above) only in that it works with 12 chips.

Spoiler for consider

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell




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