Re: Curve crossing point algorithm
Bahram HOUCHMANDZADEH
Tue, 24 Jun 1997 08:24:50 +0200
Mike Brewer wrote:
>
> I have two (for the sake of argument) *closed* loops defined by a set of
> points for each loop. What I want to do is to form the union of the two
> regions enclosed by the two loops, such that if the loops do not cross at
> all, then the union will just be the two loops. If however they DO cross,
> then I want a single loop enclosing the union.
>
> Is the best way to go about it to find the crossing points of the two loops
> and to cut out the bits of the loop in between? If so, what is the best way
> of doing this? If not, what method should I use?
Here one not very satisfactory method :
Suppose you have two loops S1 (A,B,C,D) and S2(P,Q,R,S) :
S1 S2
A B P Q
D C S R
To know if P belongs to the interior of S1, you have to check that P is
always at the right side of AB,BC, CD and DA. doing that simultaneously
for all points of S2 gives you the points of S2 which belongs to the
interior of
S1. This method is fast, but assumes that some points of the
intersection
belong two both loops. All situations are not covered.
S2
New book about Exponential distribution
isexiem@leonis.nus.sg (Xie Min)
25 Jun 1997 03:53:07 GMT
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Re: Scrabble game
wpilib+@pitt.edu (Richard F Ulrich)
25 Jun 1997 14:38:22 GMT
Carolyn Longworth (clongwor@ultranet.com) wrote:
: Hi,
: This is probably not the place to ask this question, but I am
: desperate. (I want to prevent a riot at the Senior Center).
: A group of Scrabble players had their year-end fete. They determined
: the winners by taking the total of each person's game scores for the
: year and dividing this by the number of games the person played. Some
: people only showed up for 13 games and so their total was divided by 13,
: while others showed up for all 34 games and their total was divided by
: 34. The people who showed up for all the games claim that this method
: of determining the average scores for the year is more advantagious for
: people who play the fewest games. Is there any truth to this?
: Thanks!
-- As others have pointed out, it is easier to be extreme with a
shorter series.
-- Nobody has mentioned that the quality of competition has an
effect on scores, so that who-plays-whom could make a difference:
some opponents are tough; some players play defensively and make
it hard to score high (so, winning-margin might be a reasonable
criterion, rather than highest score...). I was going to suggest
that some players might take it easy, with an easy victory in sight,
and not run up the highest score they could, but that never happened
in any Scrabble game that I lost.
If you were trying to set up a formal ranking, you might start with
the averages, and then look at narrower competition which features
only the top-scoring people. But if you are trying to make people
feel good, why not announce the BEST SCORE of the year, and also
HIGHEST-FIVE-SCORE-AVERAGE ? etcetera.
Rich Ulrich, wpilib+@pitt.edu
HELP with exponential smoothing, PLEASE!
arosa@mail.telepac.pt (Antonio Rosa)
Wed, 25 Jun 1997 15:30:30 GMT
Hi !, i'm using an programm called STATISTICA to forecast
telephones demand per month. I've acheived good results using an
linear trend in exponential smoothing, but now i'm trying to develop
an program to make the same calculations.....but i'm getting different
results. Can you help me? here are the results from the program
STATISTICA:
month: JANUARY
S0 (estimated from data) = 341884
T0 (estimated from data) = 102675
alpha=0.55 ; beta=0.1 (constants)
case Real Values Smoothed Series Resids
1 393221 444559 -51337.5
2 562774 614586 2380.3
3 611754 682960 -71206.1
4 803921 736751 67169.7
5 870344 (* forecasted)
I'm using the following formulas:
T0=(Xn-X1)/(N-1) , where N is the length of series
S0=X1-(T0/2)
Formula of trend: T(t)=beta* ( F(t) - F(t-1) ) + (1-beta)*T(t-1) ,
where T is trend and F is forecast
F(t+1)=alpha*( A(t) ) + (1-alpha)*( F(t-1) - T(t-1) ) , where A is
the real value
I think that this last formula is wrong, can anyone tell me the
correct formula(s) so that i can get the same values of STATISTICA ?
Thanks in Advance!
Antonio Rosa
----
Antonio Rosa
Portugal Telecom
Special Systems and Services
Coimbra - Portugal
mail: arosa@mail.telepac.pt
Re: Curve crossing point algorithm
Robert Smith
Wed, 25 Jun 1997 22:11:10 GMT
Mike Brewer wrote:
>
> I have two (for the sake of argument) *closed* loops defined by a set of
> points for each loop. What I want to do is to form the union of the two
> regions enclosed by the two loops, such that if the loops do not cross at
> all, then the union will just be the two loops. If however they DO cross,
> then I want a single loop enclosing the union.
>
> Is the best way to go about it to find the crossing points of the two loops
> and to cut out the bits of the loop in between? If so, what is the best way
> of doing this? If not, what method should I use?
>
>
>
> Mike: These problems are fun, but they're never as easy as they seem! I
would use an "edge-walker" technique, somewhat like the following.
1. Repeatedly move horizontally from one edge of the picture until you
intersect a curve.
2. start moving along the curve in a clockwise direction.
3. When you get to some sort of a node in the curve, take the most
COUNTERclockwise path.
4. When you return to the starting point, you will have traversed the
periphery of the union of any intersecting curves.
5. Continue as in 1. to see if you encounter any curves you haven't
traversed before. The secret of this is to use a curve which is marked
by one bit of, say, and 8-bit pixel. You can now set various other bits
to indicate the starting point, pixels you have already traversed, etc
etc.
The main trouble you may encounter is odd-ball cases where the curves
overlap for awhile, for example. Also remember that unless you make
some assumptions about shape, you can never know if the crossing of two
lines is really two straight lines crossing or two jagged lines touching
at one point which is also an elbow. Also do you allow
self-intersecting curves like a figure 8?
Have fun.
Bob
--
. Robert A. Smith, Ph.D.
_____ . Vision Systems' Analyst
| |<. Current Technology, Inc.
|_____| . (603) 868-2270
^ . mailto:ras@curtech.com
/ \
/ \
Re: Scrabble game
"Bob Wheeler"
Wed, 25 Jun 1997 20:08:42 GMT
My reading of this was that the problem was only partially
statistical, as most real problems are. I suspect the "regular"
players object to an "occasional" player who by chance obtains
a very high score and thus wins a top spot. In such a case, dividing
by the range helps. Of course, without talking to Carolyn Longwoth,
the original poster, I cannot be sure. My experience, however,
is that taking a questioner's statement about a problem at
face value often leads to a lot of work in solving the wrong
problem -- although it sometimes leads to a publication.
--
Bob Wheeler, ECHIP, Inc.
Reply to bwheeler@echip.com)
Richard F Ulrich wrote in article
<5oragu$k0l@usenet.srv.cis.pitt.edu>...
> Carolyn Longworth (clongwor@ultranet.com) wrote:
> : Hi,
> : This is probably not the place to ask this question, but I am
> : desperate. (I want to prevent a riot at the Senior Center).
> : A group of Scrabble players had their year-end fete. They determined
> : the winners by taking the total of each person's game scores for the
> : year and dividing this by the number of games the person played. Some
> : people only showed up for 13 games and so their total was divided by
13,
> : while others showed up for all 34 games and their total was divided by
> : 34. The people who showed up for all the games claim that this method
> : of determining the average scores for the year is more advantagious for
> : people who play the fewest games. Is there any truth to this?
> : Thanks!
>
> -- As others have pointed out, it is easier to be extreme with a
> shorter series.
>
> -- Nobody has mentioned that the quality of competition has an
> effect on scores, so that who-plays-whom could make a difference:
> some opponents are tough; some players play defensively and make
> it hard to score high (so, winning-margin might be a reasonable
> criterion, rather than highest score...). I was going to suggest
> that some players might take it easy, with an easy victory in sight,
> and not run up the highest score they could, but that never happened
> in any Scrabble game that I lost.
>
> If you were trying to set up a formal ranking, you might start with
> the averages, and then look at narrower competition which features
> only the top-scoring people. But if you are trying to make people
> feel good, why not announce the BEST SCORE of the year, and also
> HIGHEST-FIVE-SCORE-AVERAGE ? etcetera.
>
>
> Rich Ulrich, wpilib+@pitt.edu
>
>
Re: Do random events really exist?
Dan Bonnick
Mon, 23 Jun 1997 23:32:18 +0100
In article <33AA08EB.75F61B58@eos.ncsu.edu>, "Pavel E. Guarisma"
writes
>ModTollens wrote:
>>
>> Such an interesting thought. I think this allows me to ask the question,
>> "Is life fair?"
>> I am very interested in any response.
>
>Sure!
>Life, in the long run, is fair!!
[apologies in advance for the near-crosspost]
What about the second law of thermodynamics?
>
>The central limit theorem allows us to assume a normal distribution for
>large number of observations of any random variable. So this means that,
>in the long run, the bad stuff that happens should even out with the
>good stuff that happens. Furthermore, since bad experiences and good
>experiences should cancel each other out in the long run then the mean
>of this Gaussian should be zero.
I don't know about you, but I think my life is finite - there won't be
enough time to get to the central limit...
Dan Bonnick
