I will study your 13 coin solution when you give it. You caused me
to study my solution further. The basic solution can be done for 12
coins. We can do it for an additional coin is we leave one coin off
to the side for all three weighings but we won't know is this coin is
heavy or light. We can do an additional coin if we add a genuine
coin on pan 2 for each of the three weighings so that we test 5 coins
on pan 1 and 4 coins on pan 2 each time. This gets us up to 14 coins
but we will not know if the unweighed coin is heavy or light.
Let's say we call the two pans on our balance pan 1 and pan 2.
IDEA: We could have a balance with 3 pans arranged in a triangle.
Group 3 is off to the side and not weighed.
Let's write all the three digit numbers 0-26 in the base 3 number
system:
000
001
002
010
011
012
020
etceteras
Let these 27 numbers be the individual ID numbers of the coins.
Let the first digit be the first weighing, the second digit be the
second weighing and the third digit be the third weighing.
Let the "0" mean the coin is not on the balance.
Let the "1" mean the coin is on pan 1.
Let the "2" mean the coin is on pan 2.
If we know that one of the coins is 25% heavier, we can identify
which of the 27 coins is the heavy coin. If the balance balances,
then we have a zero. If pan 1 goes down, then we have a "1." If pan
2 goes down, then we have a "2."
If we don't know if the bad coin is heavy or light, we will not be
able to distinguish between 001 and 002 or 012 and 021 for examples.
So in this case we must pair up the indistinguishable pairs of
numbers as follows:
000
001, 002
010, 020
011, 022
012, 021
100, 200
101, 202
102, 201
110, 220
111, 222
112, 221
120, 210
121, 212
122, 211
What we need to do next is to choose one of each of the two numbers
so that we have some coins on each pan. There are many solutions so
just try some choices at random and then make a few switches:
000
001, 002 choose 001
010, 020 choose 010
011, 022 choose 011
012, 021 choose 012
100, 200 choose 200
101, 202 choose 101
102, 201 choose 102
110, 220 choose 220
111, 222 choose 111
112, 221 choose 221
120, 210 choose 120
121, 212 choose 212
122, 211 choose 122
If we count the 1's in the first column of the 13 choices, we find we
have 5 coins in pan 1 and 4 coins in pan 2. The same is true for the
second and third weighings.
Options:
If we delete coin 111, then we will have 4 coins in each pan. For 12
coins, we can tell if the bad coin is heavy or light.
We can add a thirteenth coin 000 which will be indicated as the bad
coin if the balance balances all three weighings but we won't know if
it is heavy or light.
We can add a genuine coin so that we have 5 coins on each side
instead of 4 on one side and 5 on the other. Now we can tell whether
the bad coin is heavy or light out of 13 coins.
We can add a fourteenth coin 000 which will be indicated as the bad
coin is the balance balances all three weighings but we won't know if
it is heavy or light.
Regards,
Brian
--- In Math4u@yahoogroups.com, "sinichi7kudo" <sinichi7kudo@...>
wrote:
>
> I just have another question: If you know that the fake coin is
> heavier than other coins. And you are allowed to do 3 weightings.
So
> what is the maximum number of coins?
> And by the way, I found out a way to weight 3 times for 13 coins to
> find out the fake coin (without knowing the fake is heavier or
> lighter)
> Thank you.
> --- In Math4u@yahoogroups.com, "Michael S." <M.Suesserott@> wrote:
> >
> > There is a large body of literature on this coin weighing
problem.
> For
> > starters, you might look at
> > http://www.cut-the-knot.org/blue/OddballProblem1.shtml
> > and follow the links given at the end of the article.
> >
> > This problem can also be attacked via an information theoretical
> > approach, using entropy calculations. IMHO, this is one of the
> best and
> > most elucidating ways to solve the puzzle. For a brief
> introduction, see
> > http://www.av8n.com/physics/twelve-coins.htm
> >
> > For those who want to delve deeper into this, there is the
> excellent
> > book, "Probability and Information" (Theory and Decision Library)
> by
> > A.M. Yaglom and I.M. Yaglom.
> >
> > Best wishes,
> >
> > Michael
> >
> >
> > Mosaad Alabdullatif wrote:
> > > So great
> > > So magnificent
> > >
> > > I really love it
> > >
> > > The question now: if they were 13 coins, we would be forced to
> hit the
> > > 6-6 route. Would this way leads to such a briliant solution?
> > >
> > > Thank you Sinichi7kudo
> > > Thank you Brian
> > >
> > > Mosaad
> > >
> > > */sinichi7kudo <sinichi7kudo@>/* wrote:
> > >
> > > Here is how you do it. It certainly can be solved. I'm now
> trying to
> > > apply the same logic to more than 12 coins, but no
successes
> so far.
> > >
> > > Divide the coins into three groups of 4 coins each. First,
> weigh two
> > > of the groups. If they balance, then the fake coin is in
the
> third
> > > group, so take three of the coins, and weigh them against
> three of
> > > the coins that we know for sure are real (from the first 8
> coins).
> > > If they balance, then the remaining coin is the fake one,
> but we
> > > still don't know if it is lighter or heavier, so weigh it
> against a
> > > real coin to determine that. But if it didn't balance, then
> we know
> > > that the fake coin is one of the three, and we also know if
> it is
> > > lighter or heavier, because we know which side had the real
> coins.
> > > So weigh two of the coins against each other, if they
> balance then
> > > the third coin is fake, if they don't balance we know which
> coin is
> > > fake, because we already know if it is lighter or heavier.
> > > Now let's go back to the first weighing; if the two groups
> didn't
> > > balance, we know that the fake coin is in one of the
groups,
> but we
> > > don't know which group. Let's mark the group that was
> heavier as
> > > group A, and the group that was lighter group B. For the
> second
> > > weighing, weigh two coins of group A and one coin of group
B,
> > > against the other two coins of group A and one coin of
group
> B. If
> > > they balance, the fake coin must be one of the remaining
> two, and a
> > > weigh between them will tell us which one (they are both
> from group
> > > B so we know that it's lighter). If the second weigh didn't
> balance,
> > > look at the side that is heavier: Either the fake coin is
> one of the
> > > two group A coins, or it can the group B coin on the other
> side of
> > > the scale. Take the two group A coins, and weigh them
> against each
> > > other. If they balance, then it is the group B coin that is
> fake,
> > > and if they don't balance, then the heavier one is the fake
> coin.
> > > And that's that. We have covered all possibilities.
> > >
> > > --- In Math4u@yahoogroups.com <mailto:Math4u%
> 40yahoogroups.com>,
> > > BRIAN JENSEN <brianejensen@> wrote:
> > > >
> > > > My sixth grade teacher gave this problem to us around
> 1962. It
> > > took me 38 years to figure out. He didn't remember me or
the
> problem.
> > > > I have found many solutions to this problem. There is no
> regular
> > > pattern that I could find. If you can put the coins on the
> left pan,
> > > right pan, or to the side, then how many different ways are
> there to
> > > position the coins in three weighings? It certainly can be
> solved.
> > > > Regards,
> > > > Brian
> > > >
> > > >
> > > > Mosaad Alabdullatif <sam362yah@> wrote:
> > > > I do not think 3 wieghings will be enough if we do not
> > > know if that defected coin is light or heavy!!
> > > >
> > > > Mosaad
> > > >
> > > > Brian Edward Jensen <brianejensen@> wrote:
> > > > I made a mistake! (Stephen Tavener pointed this out to
me.)
> > > On the
> > > > second problem, I meant to say "sphere," not "cube." I
> started
> > > > writing this problem thinking that the area of a cube
> could be
> > > > obtained from derivative of the volume with respect to
> half a
> > > side.
> > > > But this would be too easy because someone could easily
> calculate
> > > the
> > > > area of a side and multiply by 6 sides for the total
area.
> So then
> > > I
> > > > decided to use a sphere.
> > > >
> > > > Here is a resubmission of the two problems and I am
> deleting the
> > > > original:
> > > >
> > > > Problem 1:
> > > > Here's an old problem, one of my favorites that we
haven't
> had in
> > > a
> > > > couple years:
> > > > A man has 12 coins and one is bad. So eleven coins have
> the same
> > > > weight and one has a different weight. The man puts them
> on a
> > > > balance, some coins on each pan. He can make 3 weighings.
> This is
> > > an
> > > > old-fashioned balance with 2 pans on opposite ends of an
> arm. Of
> > > > course he must have the same number of coins on each side
> but he
> > > > doesn't need to weigh all the coins each time.
> > > > How can he figure out which coin is bad with 3 weighings
> if he
> > > knows
> > > > the bad coin is heavy? (Actually, he can pick the bad one
> out of
> > > 27
> > > > coins if he knows one is 25% heavy)
> > > > Alternately and far more difficult,
> > > > How can he figure out which coin is bad with 3 weighings
> if it is
> > > not
> > > > known if the coin is heavy or light? (Actually Stephen
> Tavener is
> > > > correct in a private communication saying 13 coins would
be
> > > > possible. If a 13th coin is left on the side without
> weighing, and
> > > a
> > > > bad coin is not identified among the twelve, we would
know
> it is
> > > coin
> > > > 13. We would not know if this coin were heavy or light,
> however.)
> > > >
> > > > Problem 2:
> > > > Another problem:
> > > > If the volume of a sphere is C*r^3
> > > > Where
> > > > C is a constant
> > > > r is the radius
> > > > Using simple differentiation instead of looking it up in
a
> book,
> > > what
> > > > is the
> > > > surface area of the sphere? Show your logic.
> > > > Regards,
> > > > Brian
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > ---------------------------------
> > > > Looking for last minute shopping deals? Find them fast
with
> > > Yahoo! Search.
> > > >
> > >
> > >
> > > ----------------------------------------------------------------
-
> -------
> > > Be a better friend, newshound, and know-it-all with Yahoo!
> Mobile. Try
> > > it now.
> > >
>
<http://us.rd.yahoo.com/evt=51733/*http://mobile.yahoo.com/;_ylt=Ahu0
> 6i62sR8HDtDypao8Wcj9tAcJ%20>
> > >
> >
>
Yahoo! Groups Links
<*> To visit your group on the web, go to:
http://groups.yahoo.com/group/Math4u/
<*> Your email settings:
Individual Email | Traditional
<*> To change settings online go to:
http://groups.yahoo.com/group/Math4u/join
(Yahoo! ID required)
<*> To change settings via email:
mailto:Math4u-digest@yahoogroups.com
mailto:Math4u-fullfeatured@yahoogroups.com
<*> To unsubscribe from this group, send an email to:
Math4u-unsubscribe@yahoogroups.com
<*> Your use of Yahoo! Groups is subject to:
No comments:
Post a Comment