For math puzzles.

122 responses total.

I came  across this puzzle. Some of you may know the answer, so I
won't ask for that. I will ask for an *algorithm* for finding the
answer that is not trial and error:

Let the decimal numbers 1 through 9 be represented by *  . Use the numbers
1 through 9 only once each to make the following equation true:

      *        *       *
    -----    -----   -----  =  1
     * *      * *     * *

What is the operator between the fractions? It can't be multiplication,
because even the largest possible value for each (and this breaks the rules)
results in a number less that 1: 9/10^3 = 0.729

Oh,  sorry, it is plus (+). I must have drifted off while composing
it.  Here is the correct equation:

       *       *       *
     ----- + ----- + -----  =  1
      * *     * *     * *

Algorithm:  Try each permutation of the digits 1..9, in lexicographic
order, until you find one that works.  Success guaranteed!

(In no more than 362,880 steps, I might add.)

That's not "trial and error"? Well, it's not *random* trial and error. 

I found, with trials, no errors, that one can get very *close* to the
sum of unity quite easily. There should therefore be a search algorithm
in the sample space that converges on unity. That would be a little
better than either a random or exhaustive search. 

Are you aware:
1) In possesion of such an algorithm, and asking us to replicate it?
2) Aware of such an algorithm's existence, and oping that one of us will
construct it?
or 3) Postulating the existence of such an algorithm?

It would help me to know how much time to really spend on this. If (3),
fuggetaboutit. ;}

Ok, to find the answer, I did an exhaustive search. It took me about five
minutes to write it in VB (it was a series of 9 for..next and if..thens,
appropriately nested)and maybe 2 seconds to get an answer. That was the
exhaustive search.

Given the answer, I notice that some of my assumptions were wrong, but all
the same, I could have made an educated guess to greatly reduce John's
options. For instance, we know that the numerators will have to be largish
and the denominators smallish, since the highest POSSIBLE value is 9/12 (or
3/4), and once the 1,2, and 9 have been taken out of the pool, it's hard to
make 1/4 out of the six remaining digits (8/34, the next highest value after
those digits have been removed, is less than 1/4). 

Since I find it's sometimes easier to solve a maze by starting at the exit,
the solution is:
9/12 + 7/68 + 5/34
(5/34 = 10/68, so 5/34 + 7/68 = 17/68 = 1/4)

An exhaustive search demonstrates that this is the only answer (barring the
other five permutations of the same fractions, of course).

Well, that wasn't the question, but thanks for the answer anyway. 8^}

What is neat about this is that there IS an answer. Can this be proven,
that there is an answer, without the exhaustive search? What is the
set of answers for an arbitrary sum? (Many would be impossible). 

You haven't answered the question in #7. I note that my "Are you aware:"
should have been "Are you:"

If you're just guessing that an algorithm exists, then I'm not going to rack
my brain to find it. ;}

My answer to #7 is yes, I am aware.  

I rephrased the question.
Are you aware that such an algorithm exists?
If so, are you in possession of such an algorithm?

No^2.  That's why I asked the question originally. 

Well, this *is* the puzzle item; "puzzle" tends to imply that the solution
is known by the querent, or at least known to exist. ;}

Given the solution, I'm not sure if there's a precise algorithm that would
lead directly to it, or if  there are merely ways to minimizing the search
process by ruling out certain possibilities (for instance, the numerators
would have to be large-ish, but an early prediction of mine that they would
all be in (6, 7, 8, 9) turned out to be false).

I looked in science for a mathematical puzzle item, but there wasn't
one, so I thought this puzzle would interest the puzzle people. It
did. It was solved (by the exhaustive technique), which was nice, but
the other properties of the puzzle interest me more. I saw no need
to separate the specific from the general. So, what are you trying
to do? Analyze exhaustingly the motivations of posting a puzzle in
the puzzle cf, and find fault with them?

In any case, I knew the numerical solution to the puzzle existed, and
there *must* exist a most efficient algorithm for finding it. So the
puzzle meets at least two of your criteria. 

Consider, if you will, that I am posting a meta-puzzle - and puzzle
about a puzzle. Shall we start a meta-puzzle cf?

#15 para 1> Sheesh, Rane, don't get so defensive. I was just pointing out that
my own strategies for approaching apuzzle where a solution is known to exist
is different from approaching a puzzle where that isn't known. I'm not trying
to bitch you out or nuffin. ;}

Since we've got A + B + C = 1, where A, B, and C are all fractions, it would
seem to me that the algorithm would start by identifying, first, all fractions
a/(10b+c), such that a, b, and c are all different, and a/(10b+c) can be
reduced to a simple ratio, like 1/2 or 3/4.

If that fails to produce three fractions with nine different digits, the next
step would be to identify all fraction PAIRS a/(10b+c), d/(10e+f), where a..f
are all different, and the sum a/(10b+c) + d/(10e+f) can be reduced to a
simple ratio.

In the case of this problem, the algorithm would succeed there, since 9/12
= 3/4 and 7/68 + 5/34 = 17/68 = 1/4. Something we notice that also simplifies
the algoithm is that 68 is a multiple of 34. While it's possible that D/E +
F/G = (DG + EF)/EG could be reduced even if E were not a multiple of G (or
vice versa), it's less likely (intuitively, at least) that such a fraction
pair that satisfies the solution could be found.

This isn't what I'd call an algorithm, though, as much as a way to hone wild
guesses into educated ones. For instance, I settled nearly immediately on 9/12
as a likely component, but got caught up in trying to find two other reducible
fractions using the remaining 6 digits.

(I'm trying to figure out how my strategy for solving a puzzle
when a solution is known to exist would differ from one that in
which it isn't known.  Only thing I could come up with is that
in the latter case, I might spend some effort seeing if I could
show that there is no solution.

I might expend that effort even if some source asserted that
there is a solution, if I didn't fully trust the source.)

I'm not sure my specific *stretegy* would differ, but my diligence would, and
I think that's what I meant. If I *knew* there was an answer, I'd put more
effort into finding it, assuming I were working on what is, after all, a
distraction. If I had committed to finding a solution to a puzzle of some
relevance, on the other hand, I'd want to know what solutions had already been
obtained so that I built on them instead of reconstructing them.

I'm more intrigued by the nature of the puzzle. My attention is caught for
example, by the fact there is just one solution. Are there solutions in,
say, base 7 or base 13? (There is none in base 4.) I'm not a number
theorist, but this puzzle seem to offer a wide range of intriguing
relationships. Maybe, an opportunity for one of those mathematics prizes? 

(Re #5: there are only 60,480 fraction combinations - should be some way to
take advantage of that in an exhaustive search.)

(Yes, you're right.  My upper bound of 9! = 362,880 didn't take
commutativity of addition into account.)

Re 17: I have proved that which has been said unprovable.
See http://www.geocities.com/joshudon/bignumbers/proofs.htm

Um, ok. I don't see anything which has been said to be unprovable, nor do I
see any formal proofs.

Re  #22:"The web page you are trying to access doesn't exist on Yahoo!
GeoCities."
?????

there's an s missing... it should be ...hudson... not ...hudon...
Otherwise, it's correct

How'd  you figure that out?

Re #22: I don't see the relevance of the cited page to #17.

On cited page the first proof is someone's (a Greek I think) paradox,
isn't,  and  called ???'s Arrow?

The second proof on the cited page is confusing. First, it is
Pythagorean and theorem (proper name, spelling). The given explanation
is unclear as the word "segment" has a definition for a circle, but
no segment is considered. 

A much simpler proof is to consider a line that intersects the circle
at two points close together and calculate the angle of intersection
between radius to the points and the straight line. Then let the
two points come together.

The name "Zeno's Paradox" comes to mind, although Zeno's version was: If
Achilles runs twice as fast as a turle, and the turtle starts x feet away from
the goal while Achilles starts 2x feet away from the goal, then while common
algebra suggests that the two will tie, circular logic suggests that the
turtle will always be closer to the goal than Achilles will be (because when
Achilles is x feet away, the turtle is x/2 feet away; when Achilles is x/2
fgeet away, the turtle is x/4 feet away, and so on).

The Arrow is also from Zeno (http://www.mathpages.com/rr/s3-07/3-07.htm)
but refers to a different "paradox", which is in this case really a very
primitive version of relativity. Consider an arrow at time t: it occupies space
s, and only space s, and space s is identifcal in volume to the arrow. That is,
at time t, the arrow is motionless, for it can occupy no other space than its
own. But if the arrow is motionless at time t1 and at time t2, then at what
time does the arrow have motion?There is no specific moment at which we can
state that the arrow has motion, and therefore the arrow is always motionless.
But, clearly, if we throw the arrow, it will move: So either time is an
illusion, or motion is.

the easiest solution to Zeno's entire set of paradoxes is to keep Zeno away
from the marijuana.

(I guesses that ...hudson... was correct based on the last name of the
poster)(er, guessed)

Limits and formal proofs don't work together.
I could have described it as a limit explicitly,
but this problem is encountered at the geometric
math level.  When I learned this topic, this was
postulated, and a postulate, is by definition,
unprovable.

Then it was improperly stated to be a postulate, as it can be proven.  I
am unsure about your proof, however, as it is not stated clearly enough to
follow easily. I agree that a formal "geometric" proof is different in
kind from the limit proof I suggest, but they equally establish the
theorem. 

There is another kind of proof. With the line and radius to the mutual
common point of contact in place, either the figure is symmetric, or
it is not. If it is not symmetric, the angle will not be a right angle,
and there will be two points of contact of the line with the circle. Hence
the figure must be symmetric, and hence at right angles. 

Maybe this is a clearer way to state Joshua's argument:  Let PA
denote a line segment from point P to line L of shortest length.
Let PB denote a line segment from P to L that insects L at right
angles.  If these segments do not coincide, then PAB is a right
triangle with hypoteneuse PA, and hence by the Pythagorean
theorem, the length of PA is greater than the length of PB,
contradicting the definition of PA as a line segment from P
to L of shortest length.

... but that doesnit say anyhting about tangents.

Hypothesis: The tangent of a circle is always at a right angle to the
diameter.
Proof:
Let T be the tangent of circle C at any randomly selected point on its
perimeter. Let P be the center of the circle. PA is the segment from the
center to the point at which T tangents C. PB is the segment which forms a
right angle to T.  The length of PB must be equal to or less than the length
of PA. However, if B<>A, then PB crosses the perimeter of T, and therefore
PB>PA for all B<>A. Therefore B=A, and PA forms a right angle with T. QED.

But why are we bothering with this, anyway?

I think you mean "line", not "segment". A segment is an area. And I think
you meant to write "perimeter of C" [not T], and in the hypothesis,
"radius to the point of tangency" and not "diameter" (a "diameter" is not
a defined line). The proof also depends upon proving that the shortest
line from a point to a line makes a ight angle. But otherwise....I agree.

Is B<>A an agreed upon expression for "not coincident" for points?

Why bother? Well, one can't settle for just saying it is obvious or
a definition. What does Euclid say about this?

I don't know why you bothered either.  My point in
mentioning it is that it is difficult to know when
something is not provable.

Some things have been proven to be unprovable (Godel's theorem), but this
isn't one of those. 

#33> My terminology is rusty, I'll admit. I think my gist was correct. Also,
I didn't include the element of demonstrating that the shortest line from a
point to a line make a right angle, because I believe John demonstrated that.

I don't know if there's a proper shorthand for stating that (x1, y1, z1) and
(x2, y2, z2) are not coincident; the long version, of course, would be x1<>x2
V y1<>y2 V z1<>z2. =} But that's a pain. I suppose z is also unnecessary since
we're talking about points on a plane (although we would further need to
demosntrate that the shortest line from A to B lies on a single plane >=} ...
if we wanted to get really sily).

Re #34 and #35:  True or false:  Something is unprovable if it is
the negation of something that you've proved.  :)

False.

this sounds like a semantic game, now.

On the one hand, if x then ~(~x), so if x is proven, then ~x cannot be proven,
because we've already proven that ~(~x). That would be the theory, at least,
but how can we prove the *theory*? Sure, one approach to a proof (as above)
is to prove that ~(~x), from which we can conclude x, but that presumes that
x v ~x = the universe and x & ~x = the null set. If x & ~x > null, then
proving x doesn't disprove ~x, and if x v ~x < the universe, then disproving
~x doesn't prove x.

Also, simple negation of x doesn't always return ~x. Most often, simple
negation does create a situation where x ^ ~x = null, while x v ~x <= the
universe, though:
a. I think you should do that
b. I don't think you should do that
(b), most often, really means:
b.' I think you shouldn't do that
But a v b' doesn't include the case where I have no opinion about whether you
should do that or not.

But, because English is sloppy, it seems like there are also cases where x
^ ~x > null, such as:
a. I like my friends so much, in fact, I love them.
b. I don't like my friends, I *love* them.
In (a), x ["I like my friends"] is implied to be a presupposition to "I love
them," while in (b), ~x is used to support "I love them." But if x and ~x both
support "I love my friends," then x & ~x > null.

I'll leave it to Rane to explain what HE meant by his answer, if he wants to.

(Erg, I meant to use & throughout for the AND operator, but slipped a few
times and used ^, which is visually closer to the symbol I use in writing,
the inverted v. v is the OR operator, while ~ is the NOT operator.)

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