Please see my other posting "HELP WANTED", I would like some help
remembering a puzzle
Sue
One.
Start by drawing from A/O box. Whatever comes out, the box contains
nothing else (since all labels are incorrect). Label this box correctly.
Move the A/O label to the box originally marked the opposite of what you
drew. Now put the last label on the last box.
Example:
Box: 1 2 3
StartLabel: A O A/O
FirstDraw: O
ReLabelStep1: O
ReLabelStep2: A/O O
ReLabelStep3: A/O A O
There are only 2 ways to rearrange three things so none of them stays the
same, which can be described as shifting all labels one to the left or one
to the right. By drawing from the A/O box, you can immediately tell which
direction the labels have been shifted.
--
Kindest regards,
Sylvester McMonkey McBean
Step right in, I'm the Fix-it-Up Chappie.
I'll change all your bellies and then you'll be happy!
Susan Jacques <su...@psc.clara.net> wrote in article
<6jc890$c1g$1...@eros.clara.net>...
> There are three boxes, one containing apples, one containing oranges and one
> containing apples and oranges together. The boxes are labeled "Apples",
> "Oranges","Apples and Oranges" but each of the labels is on the wrong box.
> You can't see into the boxes, but you are allowed to remove items from each
> box.
> Q. What is the minimum number of samples of the contents you must make to
> ensure you can put the labels on the correct boxes.
>
> Please see my other posting "HELP WANTED", I would like some help
> remembering a puzzle
>
> Sue
Welcome to rec.puzzles. You may be interested in this
entry from the rec.puzzles archive:
==> logic/wrong.labels.p <==
Three candle boxes are labeled "white", "red", and "white and red." All
three labels are incorrect. How many candles do I have to pull out to
correctly label the boxes, and how do I do it?
Also:
There are four boxes, labeled:
"Pennies and Nickels"
"Nickels and Dimes"
"Dimes and Quarters"
"Just Quarters"
All the labels are incorrect. How many coins do I have to sample to
correctly label the boxes, and how do I do it?
==> logic/wrong.labels.s <==
One candle suffices. Take one candle from the box labeled "white and
red." Say that it's white. You know that this must be the white box,
therefore the box labeled "red" must contain mixed and the box labeled
"white" must actually be the red candles.
In the four-box case, the answer is three.
Number the boxes from 1 to 4, so that Box 1 is labeled P&N, Box 2 is
labeled N&D, Box 3 is labeled D&Q, Box 4 is labeled Q
First, request Box 3. If the coin from Box 3 is a dime or quarter,
request Box 2, and whatever comes out, you will have enough information
to know which two boxes to choose. Namely:
If ?NQ?, meaning a nickel (or penny) from Box 2 and a quarter from Box 2,
then choose boxes 1 & 4.
If ?QD? then choose boxes 1 & 2.
In all other cases (?ND?, ?DD?, ?DQ?, ?QQ?) choose boxes 1 & 3.
If the coin from Box 3 is a nickel (or penny), request Box 1 next. If the
coin from Box 1 is also a nickel, choose boxes 1 & 4. If the coin from Box
1 is a dime, request Box 4; if it's a quarter, request Box 2. Then choose
as follows:
If D?NN then choose boxes 1 & 2.
If D?ND then choose boxes 1 & 4.
If D?NQ then choose boxes 1 & 4
If QNN? then choose boxes 1 & 3.
If QDN? then choose boxes 1 & 2.
If QQN? then choose boxes 1 & 2.
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SPOILER(Answer below)
I think I could do this in two selections. Since I know all the labels are
wrong I would select from the one labeled (A and O), lets say that I picked
an apple then I know that the entire box has apples and label it accordingly.
I would then draw from the box that used to be labeled apples. If I draw
an apple then I know it's the apples+oranges box and lable it accordingly
and if I draw an orange then I know it's the orange box and label it
thus. The remain tag I place on the last box drawing nothing from it.
Dave.
You don't need the second draw. There are only two combinations in which
all of the labels are screwed up:
label A+O A O O A+O A
box A O A+O A O A+O
So all you need to do is draw one from the box labelled "apples and
oranges." If it's an apple, you're in the configuration on the left;
if it's an orange, you're in the one on the right.
--
Jeff Pack <book...@brown.edu>
Brown University, Class of 1999 I refuse the unpoetic existence.
St. Anthony Hall, K'96 The universe will contain passion and
17th-century digital boy grace, even if I have to make it myself...
[snip second selection]
You don't need the second selection. After selecting the apple, you also
know that the box labeled Oranges must be (A and O), since it couldn't
be Oranges to begin with, and now it can't be Apples. That means the box
labeled Apples must be (A and O).
--
Jack Christenson
jack.chr...@network.com