Fake coin detection

advertisement
Fake coin detection
Fake coins
• Suppose we have a number of coins, at most
one is fake (i.e. either one is fake or none is
fake). You have a pair of scales (a balance),
and you want to find the fake coin, if it exists,
with the minimum number of comparisons.
Coins can be placed on the left or right of the
scales or not (i.e. we place the remaining coins
on the table).
problem solving process.
• What is important here is the problem solving
process. We first find that we cannot solve the
problem as it is stated (if we only have one
coin, what does it mean to say if it is genuine
or fake). We then go on to solve a second
problem (which we will call the marked coin
problem), and this will help us to solve the
original problem (which we can call the
unmarked coin problem).
2m+1 possibilities
• Suppose that there are m coins, and one is
heavy (then there are m possibilities). If one
coin is heavier or lighter, then there are 2m
possibilities. And if there is not a fake coin
then there are 2m+1 possibilities.
1+2m≥3^{n}.
• When we make a comparison with the scales
there are three possibilities, tip left, right or
balance. With n comparisons there are 3^{n}
different possibilities, giving an upper bound
on the number of situations we can
distinguish. Therefore 1+2m≥3^{n}.
c(n)
• The most coins we can distinguish between is
where there is equality. By rearranging with a
little algebra m=1/2 (3^{n}-1). Let us call this
value c(n) = 1/2 (3^{n}-1).
how this function grows
• We can see how this function grows for the
first few values.
• c(0)=0, c(1)=1, c(2)=4, c=(13).
Two coins???
• The base case is trivial, with no coins we make
no comparisons, and the problem is solved.
But with one coin we cannot say if it is
genuine or fake as we have nothing to
compare it with. Therefore the question does
not make sense, so let us assume that we
have a reference coin which is genuine.
The induction step
• The induction step is as follows. We assume
that if the fake coin exists among a set of c(n)
coins, it can be found with at most n
comparisons. We have to show (build on this
assumption) how to find the fake coin among
c(n+1) coins, with at most n+1 comparisons.
• If the scales balance, then the fake coin must
be on the table (i.e. the coin is not on the
scales). Therefore we can examine the coins
on the table. We have made one comparison,
therefore we must be left with no more than
c(n) coins on the table. As we know how
many coins are not on the scales, we know
how many coins are on the scales.
c(n+1)
• c(n+1) = 3.c(n)+1. (by basic algebra)
• so c(n+1)-c(n) = 2c(n)+1=3^{n}.
the additional reference
• This number is odd (2c(n) is even), 3^{n} is
odd. Therefore we cannot split this number in
two and put half on left and right of the
scales. But we have the additional reference
coin, therefore we put c(n)+1 coins on one
side and c(n)+reference coin on the other side
of the scales.
we cannot apply the inductive
hypothesis.
• If the scales tip, we can eliminate the coins on
the table (i.e. not on the scales), but then we
are left with 2c(n)+1 coins!!! And this is
greater than c(n), so we cannot apply the
inductive hypothesis.
possibly heavier and possibly lighter.
• But this comparison not only tells us that the
fake coin is among the coins on the scales, but
it also tells us which coins are possibly heavier
and possibly lighter. This is important
information and should be utilized. Let us now
move onto a new problem where the coins
are marked either possibly heavier or lighter.
marked coin problem
• Here is the marked coin problem. We are
supplied with a number of coins, each marked
possible heavy (H) or possibly light (L), and
exactly one coin is fake (note that this is
different to the original problem above, and is
a sub-problem).
L1+H1=L2+H2
• Suppose L1 coins are placed on the left and H1
coins are placed on the left.
• Suppose L2 coins are placed on the right and
H2 coins are placed on the right.
• The number of coins on the left and right
must be equal (in order to make any
meaningful comparison). L1+H1=L2+H2
(=3^{n}).
L1+H2=L2+H1
• If the scales tip (e.g. to the left), then the L1
coins and H2 coins must be genuine (and vice
versa). In order to apply the inductive
hypothesis, this number must be equal to
3^{n}. This means L1+H2=L2+H1.
L1=L2 and H1=H2
• These two equations imply that L1=L2 and
H1=H2, in other words, each side must
contain the same number of coins of the same
type. This is another example of symmetry.
This can always be achieved as, given three
coins at least two must be of the same type.
We can now describe the solution to the
marked coin problem and then the solution to
the unmarked coin problem.
marked coin problem
• The solution to the marked coin problem is as
follows. Given 3^{n+1} marked coins, place
3^{n} on each side of the scale so that there
are the same number of each type of coin on
each side. If the scales balance proceed with
the coins on the table (i.e. not on the scale).
Else, if the scales tip to the left proceed with
the possibly heavy coins on the left and
possibly light coins n the right (vice versa if tip
right).
unmarked coins problem
• The solution to the unmarked coins problem is
as follows. Divide the coins into 3 piles (3^{n}1)/2 + reference coin (put on the left), (3^{n}1)/2 +1 (put on the right), and (3^{n}-1)/2 (to
leave on the table). If the scales balance,
proceed with the coins on the table with the
unmarked solution. If the scales tip, mark all
of the coins as either possibly heavy or
possibly light, and proceed with the marked
coin solution.
Download
Related flashcards

Iranian peoples

32 cards

Cryptography

26 cards

Create Flashcards