Subject: Re: Combining Neural/Fuzzy Models with Statistical Models
From: saswss@hotellng.unx.sas.com (Warren Sarle)
Date: Sat, 18 Jan 1997 00:57:53 GMT
In article <32DA5910.2AA2@colorado.edu>, Robert Dodier writes:
|> Andrew Gray wrote:
|>
|> > I'm working on combining neural networks and fuzzy logic models
|> > with statistical techniques (regression and data reduction) for
|> > software metrics (for example, predicting development time based on
|> > the type and size of system). While there has been a lot of work on
|> > neural-fuzzy, neural-genetic, fuzzy-genetic, etc. type systems I've
|> > only ever found a small number of researchers using AI/statistical
|> > techniques (presumably at least partially an indication of how few AI
|> > researchers follow the statistical side of things, and vice versa).
|>
|> At the risk of reviving age-old threads about fuzzy logic vs.
|> probability, let me advise you not to bother with fuzzy logic.
|> There are two parts to fuzzy logic, one defensible and the other
|> not. The ``fuzzy'' part is one solution to the problem of
|> representing uncertain knowledge -- this is the defensible part.
|> The ``logic'' part is an attempt to reason with uncertain knowledge
|> -- this is an undefensible hack.
The trouble with fuzzy logic as usually presented is that you can't
reason about two uncertain propositions without knowing how the
uncertainties are related--it's like trying to work with marginal
probability distributions without knowing the joint distribution.
But some fuzzy logicians have noticed this problem and developed
fuzzy logics involving "correlations" between fuzzy propositions.
See the post included below.
|> ... As discussed at length in
|> Warren Sarle's neural networks FAQ (sorry, I don't have pointer)
|> it's useful to consider neural networks as extensions of or variations
|> on conventional regression and classification schemes.
ftp://ftp.sas.com/pub/neural/FAQ.html
______________________________________________________________________
From: William Siler
Newsgroups: comp.ai.fuzzy
Subject: New Fuzzy Logic
Date: Fri, 20 Sep 96 20:25:09 -0500
Organization: Delphi (info@delphi.com email, 800-695-4005 voice)
Lines: 107
Message-ID:
Reposting article removed by rogue canceller.
DIGEST OF BUCKLEY, JJ AND SILER, W: A NEW T-NORM. SUBMITTED TO
FUZZY SETS AND SYSTEMS, 1996.
Fuzzy systems theory has been criticized for not obeying all
the laws of classical set theory and classical logic. The
t-norm and t-conorm here presented obey all the laws of the
corresponding classical theory. A somewhat similar theory has
been proposed by Thomas (1994), except that he does not claim
that the distributive property is maintained.
We first propose a source of fuzziness. We suppose that the
truth value > 0 and < 1 of a fuzzy logical statement A is drawn
from a number of underlying (probably implicit) correlated
random variables whose values alpha[i] are binary, i.e. 0 or 1
with a Bernoulli distribution, and that the truth value of A is
a simple average of these binary values. (George Klir (1994)
proposed a similar process where the random values are binary
opinions of experts as to truth or falsehood of a statement.)
If this is so, then
a = truth(A) = sum(alpha[i] / n
b = truth(B) = sum(beta[i] / n)
r = correlation coefficient(alpha[i], beta[i])
aANDb = truth value(A AND B)
aORb = truth value(A OR B)
sa = standard deviation(alpha) = sqrt(p(alpha)*(1 - p(alpha))
sb = standard deviation(beta) = sqrt(p(beta)*(1 - p(beta))
aANDb = a*b + r*sa*sb
aORb = a + b - a*b - r*sa*sb
rmax = (min(a,b) - a*b) / (sa*sb)
for r = rmax, min(a,b) = a*b + r*sa*sb
rmin = (a*b - max(a+b-1, 0)) / (sa*sb)
for r = rmin, max(a+b-1, 0) = a*b + rmin*sa*sb
Proofs of the following theorems are in the appendices of our
paper.
Theorem 1:
1. maxr = ru, ru <= 1
2. minr = rl, rl >= -1
3. rl <= r <= ru
Theorem 2:
1. aANDb = a*b + r*sa*sb = a*b + cov(a,b)
2. aORb = a + b - a*b - r*sa*sb = a + b - a*b - cov(a,b)
Theorem 3:
1. If r = ru, aANDb = min(a,b) and aORb = max(a,b)
2. If r = 0, aANDb = a*b and aORb = a+b-ab
3. If r = rl, aANDb = max(a+b-1, 0) and aORb = max(a+b, 1)
We now suppose that this basic process is inaccessible to us,
but that we do have a history of a number of instances of the
truths of statement A and statement B. Now, given a value of r,
the correlation coefficient between a, the truth values of A,
and b, the truth values of B, the t-norm and t-conorm
appropriate to this history, T (t-norm) and C (t-conorm) are
defined for [a, b] on S, a restricted subset of [0,1]x[0,1].
Theorem 4.
(The 5 parts of this theorem define the subset S of
[0,1]x[0,1]
possible for r = 1, 0 < r < 1, r = 0, -1 < r < 0 and r = -1.)
Given a value of r, it may be that not all (a,b) combinations
are possible; e.g. a = .25, b = .75 is not possible for r = 1
in the binary process described above.)
Theorem 5.
1. (Shows that for 0 < r < 1 and (a,b) in S,
ab < T(a,b) <= min(a,b).)
2. (Shows that for -1 < r < 0 and (a,b) in S,
max(a+b-1) <= T(A,B) < ab)
3. (Shows that for -1 <= r <= 1 and (a,b) in S,
max(a+b-1, 0) <= T(a,b) <= min(a,b) and
max(a,b) <= C(a,b) <= min(a+b, 1).
Theorem 6. Shows that T is a t-norm and C is a t-conorm on S.
Theorem 7.
1. A AND A = A, r is 1.
2. A OR 0 = 0, any r.
3. A OR X = A, any r.
4. A AND NOT-A = 0, r is -1.
5. A OR A = A, r is 1.
6. A OR X = A, any r.
7. A OR 0 = A, any r.
8. A OR NOT-A = X, r is -1.
9. NOT-(A AND B) = NOT-A OR NOT-B, any r.
10. NOT-(A AND B) = NOT-A AND NOT-B, any r.
11. A OR (A AND B) = A, any appropriate r.
12. A AND (A OR B) = A, any appropriate r.
13. A AND (B OR C) = (A AND B) OR (A AND C), any appropriate
r.
14. A OR (B AND C) = (A OR B) AND (A OR C), any appropriate
r.
References:
Klir, GJ (1994). Multivalued logics vesus model logics:
alternate frameworks for uncertainty modelling. In: Advances in
Fuzzy Theory and Technology, Vol II: 3-47. Duke University
Press, Durham, NC.
Thomas, SF (1994). Fuzzy Logic and Probability. ACG Press,
Wichita, KS.
______________________________________________________________________
--
Warren S. Sarle SAS Institute Inc. The opinions expressed here
saswss@unx.sas.com SAS Campus Drive are mine and not necessarily
(919) 677-8000 Cary, NC 27513, USA those of SAS Institute.
*** Do not send me unsolicited commercial or political email! ***
Subject: Correlations involving ratios
From: chris@agri.upm.edu.my
Date: Fri, 17 Jan 1997 21:41:06 -0600
I have another "puzzle" that's been on my mind. It is concerning correlating a ratio with another variable. To illustrate my point, consider this simple example (by the way, the correlations are actual not hypothetical). Let's have two indepedent variables: soil carbon (C), and soil nitrogen (N). I'm interested to know what's their linear relationship with another soil property, that is soil stability (S). Correlations (r) with S are as follow:
r between C & S = .63**
r between N & S = .00
r between C:N &S; = .51*
Here, the C:N is a ratio obtained by dividing C with N. As shown, the r
between C:N and S is 0.51*. Something peculiar happens when I start to
play around with the data. Now, suppose I were to change the scale of C
by adding the value 10 to every value in the variable C; that is, tC = C
+ 10. Using this new variable tC, I now form a new ratio between carbon
and nitrogen, that is tC:N. Correlations with S are now as
follow:
r between C & S = .63**
r between tC & S = .63** (ok)
r between N & S = .00
r between C:N & S = .51*
r between tC:N & S = .07 (!!!)
As shown, changing the scale of carbon did not affect its r with S --
this is correctly so. If you were to correlate tC with C you would obtain
a perfect 1.00 relationship. In simple regression, changing the scale of
variable only shifts the line upwards or downwards - the slope remains
equal.
However, the r between the tC:N with S is clearly different from
the r between C:N and S! If I were to use tC to form the ratio between
carbon and nitrogen, my interpretation of results would be different - I
would conclude that carbon:nitrogen ratio is insignificantly related to
S. But had I used C, I would instead conclude that carbon:nitrogen ratio
has a significant relationship! Why is this so???? What's happening
here???
Note: this pecularity also happens with other data, so there is
no data entry or calculation error
here.
Anyone?
P.S. By the way, in soil science using such ratios is quite common. This
is why this is quite shock
finding.
-------------------==== Posted via Deja News ====-----------------------
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Subject: Performance Index for Covering Designs, Wheels
From: bm373592@muenchen.org (Uenal Mutlu)
Date: Sat, 18 Jan 1997 19:37:35 GMT
(I've moved this thread to a meaningful subject and also included sci.stat.consult)
Here are some relevant excerpts from the previous postings. The task
is twofold: creating wheels (covering designs) which cover the most of
the possible cases in less blocks as possible (ie. a combinatorial
optimization problem), and a formula which allows the comparison of
designs with different nbr of blocks etc. (ie. a practical performance
index for such designs):
#####
>>A few years ago, I wrote a program to improve partial wheels. I have
>>ran this program overnight with the improved wheel and it has improved
>>it further to 44.45% And it's not done yet...
>>
>I think 2 conditions have to be fulfilled to find an optimal solution for
>that problem:
>a) all tickets should be as different as possible (any two tickets should
>share as less as possible numbers)
>b) each number should appear as often as any other number
>
>Here is a solution based on these conditions:
>1 9 17 25 33 41
>2 10 18 26 34 42
>3 11 19 27 35 43
>4 12 20 28 36 44
>5 13 21 29 37 45
>6 14 22 30 38 46
>7 15 23 31 39 47
>8 16 24 32 40 48
>1 16 23 30 37 44
>2 9 24 31 38 45
>3 10 17 32 39 46
>4 11 18 25 40 47
>5 12 19 26 33 48
>6 13 20 27 34 41
>7 14 21 28 35 42
>8 15 22 29 36 43
>1 15 21 27 38 48
>2 16 22 28 39 41
>3 9 23 29 40 42
>4 10 24 30 33 43
>5 11 17 31 34 44
>6 12 18 32 35 45
>7 13 19 25 36 46
>8 14 20 26 37 47
>1 14 19 32 34 49
>2 15 20 25 35 49
>3 16 21 26 36 49
>4 9 22 27 37 49
>5 10 23 28 38 49
>6 11 24 29 39 49
>7 12 17 30 40 49
>8 13 18 31 33 49
>
>Using the first 27 tickets your percentage for at least a 3 win is
>44.383%
>Using all 32 tickets percentage is at 50.83%
>
>I would be interested to see the 44.45% wheel.
#####
>>Using the first 27 tickets your percentage for at least a 3 win is
>>44.383% Using all 32 tickets percentage is at 50.83%
>
>We should use a general formula for comparing of such wheels with
>differing nbr of tickets:
>
> WCD = Percent / Tickets
>
>where
> - WCD = "Wheel Covering Degree". The higher, the better.
> - Percent is in the range 0.0 to 100.0 for minWin (ie. for
> "at least x-win")
> - Tickets is the nbr of single tickets the wheel consists of
>
>Warning:
> it can be used only for the same type of wheels (ie. k must
> be equal), and percent should be "for at least minWin"
>
>Example for minWin = 3+ (ie. for at least 3):
>
> Percent Tickets -> WCD
> ---------------------------
> 44.383 27 1.64381
> 50.83 32 1.58844
> 70.0 54 1.29630
> 100.0 168 0.59524
>
>The higher the WCD, the better the wheel. It simply says
>"each ticket covers WCD points of the Percent value"
#####
>I had a similar discussion on this with Normand and currently in 1st
>place is John Rawson with 44.43945% (27 lines - 230,160 combs covered
>per line avg = 1.6459%).
>
>But if one comes up with 28 lines would it be fair to expect them to
>have at least 1.6459% per line ?
#####
>>But if one comes up with 28 lines would it be fair to expect them to
>>have at least 1.6459% per line ?
>
>The WCD formula already covers this case. You're mixing up the
>different kinds of percent values and your calculation doesn't
>take into account the nbr of tickets, but the WCD formula does.
>
>(Your 1.6459% is simply the result of 230160 / 13983816 * 100
>ie. simple percent calculation, but the actual number of tickets
>isn't included anywhere, so this won't help in case of differing
>nbr of tickets)
>
>Cf. for example the first 2 entries in the table above, then you
>should see that the formula indeed allows the comparison of your
>example of 28 tickets too, but one needs both the winPercent (or the
>number of winning combinations to get the winPercent) and the nTickets
>(here 28 and 27) values of both wheels for the WCD formula.
>
>And, your percent value (1,6459%) has nothing todo with the WCD value
>(ie. they are very different things; the WCD is NOT a percent value,
>it simply is a number). Both can't be compared together. Apple and Orange :-)
>
>Anyone know of a better formula for comparing of such wheels of similar
>type but with different nbr of tickets?
#####
>>I think 2 conditions have to be fulfilled to find an optimal solution for
>>that problem:
>>a) all tickets should be as different as possible (any two tickets should
>>share as less as possible numbers)
>>b) each number should appear as often as any other number
>
>Right. But, it's not that simple. If we keep adding tickets, at one
>point it will become possible to cover more combinations by having
>more redundancy between tickets. As an example of this, when the 174
>ticket cover was discovered for the Lotto 6/49 it had 18 tickets that
>had 5 identical numbers, yet we couldn't see how to improve it further
>with tickets that had less redundancy.
>
>Is 27 tickets below or above that point?
>
>Also, if you randomly generate several sets of 27 tickets that have no
>duplicating pairs of numbers you will notice that they do not cover
>the same number of combinations! So, what is the best cover possible
>using only such tickets?
>
>This may or may not be the cover we are looking for.
#####
>>Also, if you randomly generate several sets of 27 tickets that have no
>>duplicating pairs of numbers you will notice that they do not cover
>>the same number of combinations! So, what is the best cover possible
>>using only such tickets?
>>
>>This may or may not be the cover we are looking for.
>
>It is the right direction to go I would say. Of course the best is,
>and always has been, designing wheels which have the highest number
>of winning combinations in less tickets as possible. If we take this
>as the only major criteria then we end up with this kind of wheels,
>which IMHO are much better, and also cheaper, for the player than the
>other 100% guaranteed wheels. And, IMHO exactly here should be the
>true difference between wheels and covering designs.
>
>But the designing of such wheels isn't easy either (this maybe because
>it's a relatively new field for me; no simple combine formula and so on... :-)
>But it's a good challenge for wheel designers, group and design theoretists.
>
>So, for this type of wheels we need a different way to measure their
>strength. In an other posting I introduced the WCD formula, which seems
>suitable for this task; ie.
>
> WCD = winPercent / nTickets
>
>where
> winPercent = NbrOfWinningCombinations / C(v,k) * 100
> nTickets = nbr of tickets the wheel has
>
>(One also should always state, besides the other usual params of the
>wheel, the minWin all these values are for. Ie. "for 3 or more winning
>numbers" etc.)
>
>The WCD is a number only (ie. no percent value). The higher the WCD,
>the better the wheel. It should be used for comparison of 2 or more
>wheels of same type with possibly differing number of tickets.
>
>(cf. also the other postings on this which also include some examples)
#####
>|> The WCD is a number only (ie. no percent value). The higher the WCD,
>|> the better the wheel. It should be used for comparison of 2 or more
>|> wheels of same type with possibly differing number of tickets.
>|>
>|> (cf. also the other postings on this which also include some examples)
>|>
>
> I'm not sure it's that easy to compare wheels of differing number of
>tickets by this method. Let's take a very simple example to make this
>clear. Consider a 6/14 lottery with the winning criteria being a
>3+ match.
> The number of combinations is 3003 and the best wheels I can think of
>are, for 1 to 4 tickets:
>
>
> Number of tickets = 1: Numbers 1, 2, 3, 4, 5, 6
>
> 1589 100
> Coverage is 1589, Therefore WCD = ---- * --- = 52.91 %
> 3003 1
>
>
> Number of tickets = 2: Numbers 1, 2, 3, 4, 5, 6
> 7, 8, 9, 10, 11, 12
>
> 2778 100
> Coverage is 2778, Therefore WCD = ---- * --- = 46.25 %
> 3003 2
>
> Number of tickets = 3: Numbers 1, 2, 3, 4, 5, 6
> 7, 8, 9, 10, 11, 12
> 1, 2, 3, 4, 13, 14
>
> 2988 100
> Coverage is 2988, Therefore WCD = ---- * --- = 33.17 %
> 3003 3
>
> Number of tickets = 4: Numbers 1, 2, 3, 4, 5, 6
> 7, 8, 9, 10, 11, 12
> 1, 2, 3, 4, 13, 14
> 5, 6, 7, 8, 13, 14
>
> 3003 100
> Coverage is 3003, Therefore WCD = ---- * --- = 25.00 %
> 3003 4
>
> Which would you say is the best wheel, and how does your
>judgement compare to the WCD rating?
>
> The point I'm trying to make is that the WCD is always going to go
>down with increasing number of tickets, because you can't avoid
>coverage redundancy as you introduce more tickets. The same applies
>to the various wheels in 6/49. Consider again a 'wheel' of 1 ticket.
>Its coverage is 260524 out of 13983816, giving a WCD of 1.863%. This
maybe a typo: should be 260624; it is the maxCover value, ie. the
nbr of combinations 1 block ideally should cover.
I'm currently working on an new method, but the relations for the
given examples remain similar I fear. The advantage of the new formula
is that it gives a true p-value (ie. percent value if multiplied by 100).
>is to be compared with the best current 27 ticket wheel with a
>coverage of about 44.5%, ie a WCD of 1.648%. Then the 168 wheel has
>a WCD of just 100/168 or 0.595%. Again, does the WCD give a good
>guide to which is the best wheel? Essentially the WCD will _always_
>give the best result for a wheel of 1 ticket, and will favour wheels
>of lower number of tickets. Now I'm not a great expert or user of
>wheels, but I'm fairly sure that most people who are keen on wheels
>do not consider a wheel of one ticket to be a good wheel!
>
> So what I'm really trying to say is that I don't think the WCD
>is a good means to compare two wheels of different numbers of tickets..
>However, I can't say I've got any better suggestion.
#####
>Hi Stephen and all the others,
>
>I've moved the discussion about the Performance Index for Wheels (WCD etc.)
>to a new subject called "Performance Index for Covering Designs, Wheels".
>The first posting under the new subject contains IMHO all the relevant
>excerpts from the previous postings.
Ie. it's exactly this posting :-)
(BTW, the old subject was "Dimitris Challenge" in r.g.l.)
Subject: Performance Index for Covering Designs, Wheels
From: bm373592@muenchen.org (Uenal Mutlu)
Date: Sat, 18 Jan 1997 19:37:35 GMT
(I've moved this thread to a meaningful subject and also included sci.stat.consult)
Here are some relevant excerpts from the previous postings. The task
is twofold: creating wheels (covering designs) which cover the most of
the possible cases in less blocks as possible (ie. a combinatorial
optimization problem), and a formula which allows the comparison of
designs with different nbr of blocks etc. (ie. a practical performance
index for such designs):
#####
>>A few years ago, I wrote a program to improve partial wheels. I have
>>ran this program overnight with the improved wheel and it has improved
>>it further to 44.45% And it's not done yet...
>>
>I think 2 conditions have to be fulfilled to find an optimal solution for
>that problem:
>a) all tickets should be as different as possible (any two tickets should
>share as less as possible numbers)
>b) each number should appear as often as any other number
>
>Here is a solution based on these conditions:
>1 9 17 25 33 41
>2 10 18 26 34 42
>3 11 19 27 35 43
>4 12 20 28 36 44
>5 13 21 29 37 45
>6 14 22 30 38 46
>7 15 23 31 39 47
>8 16 24 32 40 48
>1 16 23 30 37 44
>2 9 24 31 38 45
>3 10 17 32 39 46
>4 11 18 25 40 47
>5 12 19 26 33 48
>6 13 20 27 34 41
>7 14 21 28 35 42
>8 15 22 29 36 43
>1 15 21 27 38 48
>2 16 22 28 39 41
>3 9 23 29 40 42
>4 10 24 30 33 43
>5 11 17 31 34 44
>6 12 18 32 35 45
>7 13 19 25 36 46
>8 14 20 26 37 47
>1 14 19 32 34 49
>2 15 20 25 35 49
>3 16 21 26 36 49
>4 9 22 27 37 49
>5 10 23 28 38 49
>6 11 24 29 39 49
>7 12 17 30 40 49
>8 13 18 31 33 49
>
>Using the first 27 tickets your percentage for at least a 3 win is
>44.383%
>Using all 32 tickets percentage is at 50.83%
>
>I would be interested to see the 44.45% wheel.
#####
>>Using the first 27 tickets your percentage for at least a 3 win is
>>44.383% Using all 32 tickets percentage is at 50.83%
>
>We should use a general formula for comparing of such wheels with
>differing nbr of tickets:
>
> WCD = Percent / Tickets
>
>where
> - WCD = "Wheel Covering Degree". The higher, the better.
> - Percent is in the range 0.0 to 100.0 for minWin (ie. for
> "at least x-win")
> - Tickets is the nbr of single tickets the wheel consists of
>
>Warning:
> it can be used only for the same type of wheels (ie. k must
> be equal), and percent should be "for at least minWin"
>
>Example for minWin = 3+ (ie. for at least 3):
>
> Percent Tickets -> WCD
> ---------------------------
> 44.383 27 1.64381
> 50.83 32 1.58844
> 70.0 54 1.29630
> 100.0 168 0.59524
>
>The higher the WCD, the better the wheel. It simply says
>"each ticket covers WCD points of the Percent value"
#####
>I had a similar discussion on this with Normand and currently in 1st
>place is John Rawson with 44.43945% (27 lines - 230,160 combs covered
>per line avg = 1.6459%).
>
>But if one comes up with 28 lines would it be fair to expect them to
>have at least 1.6459% per line ?
#####
>>But if one comes up with 28 lines would it be fair to expect them to
>>have at least 1.6459% per line ?
>
>The WCD formula already covers this case. You're mixing up the
>different kinds of percent values and your calculation doesn't
>take into account the nbr of tickets, but the WCD formula does.
>
>(Your 1.6459% is simply the result of 230160 / 13983816 * 100
>ie. simple percent calculation, but the actual number of tickets
>isn't included anywhere, so this won't help in case of differing
>nbr of tickets)
>
>Cf. for example the first 2 entries in the table above, then you
>should see that the formula indeed allows the comparison of your
>example of 28 tickets too, but one needs both the winPercent (or the
>number of winning combinations to get the winPercent) and the nTickets
>(here 28 and 27) values of both wheels for the WCD formula.
>
>And, your percent value (1,6459%) has nothing todo with the WCD value
>(ie. they are very different things; the WCD is NOT a percent value,
>it simply is a number). Both can't be compared together. Apple and Orange :-)
>
>Anyone know of a better formula for comparing of such wheels of similar
>type but with different nbr of tickets?
#####
>>I think 2 conditions have to be fulfilled to find an optimal solution for
>>that problem:
>>a) all tickets should be as different as possible (any two tickets should
>>share as less as possible numbers)
>>b) each number should appear as often as any other number
>
>Right. But, it's not that simple. If we keep adding tickets, at one
>point it will become possible to cover more combinations by having
>more redundancy between tickets. As an example of this, when the 174
>ticket cover was discovered for the Lotto 6/49 it had 18 tickets that
>had 5 identical numbers, yet we couldn't see how to improve it further
>with tickets that had less redundancy.
>
>Is 27 tickets below or above that point?
>
>Also, if you randomly generate several sets of 27 tickets that have no
>duplicating pairs of numbers you will notice that they do not cover
>the same number of combinations! So, what is the best cover possible
>using only such tickets?
>
>This may or may not be the cover we are looking for.
#####
>>Also, if you randomly generate several sets of 27 tickets that have no
>>duplicating pairs of numbers you will notice that they do not cover
>>the same number of combinations! So, what is the best cover possible
>>using only such tickets?
>>
>>This may or may not be the cover we are looking for.
>
>It is the right direction to go I would say. Of course the best is,
>and always has been, designing wheels which have the highest number
>of winning combinations in less tickets as possible. If we take this
>as the only major criteria then we end up with this kind of wheels,
>which IMHO are much better, and also cheaper, for the player than the
>other 100% guaranteed wheels. And, IMHO exactly here should be the
>true difference between wheels and covering designs.
>
>But the designing of such wheels isn't easy either (this maybe because
>it's a relatively new field for me; no simple combine formula and so on... :-)
>But it's a good challenge for wheel designers, group and design theoretists.
>
>So, for this type of wheels we need a different way to measure their
>strength. In an other posting I introduced the WCD formula, which seems
>suitable for this task; ie.
>
> WCD = winPercent / nTickets
>
>where
> winPercent = NbrOfWinningCombinations / C(v,k) * 100
> nTickets = nbr of tickets the wheel has
>
>(One also should always state, besides the other usual params of the
>wheel, the minWin all these values are for. Ie. "for 3 or more winning
>numbers" etc.)
>
>The WCD is a number only (ie. no percent value). The higher the WCD,
>the better the wheel. It should be used for comparison of 2 or more
>wheels of same type with possibly differing number of tickets.
>
>(cf. also the other postings on this which also include some examples)
#####
>|> The WCD is a number only (ie. no percent value). The higher the WCD,
>|> the better the wheel. It should be used for comparison of 2 or more
>|> wheels of same type with possibly differing number of tickets.
>|>
>|> (cf. also the other postings on this which also include some examples)
>|>
>
> I'm not sure it's that easy to compare wheels of differing number of
>tickets by this method. Let's take a very simple example to make this
>clear. Consider a 6/14 lottery with the winning criteria being a
>3+ match.
> The number of combinations is 3003 and the best wheels I can think of
>are, for 1 to 4 tickets:
>
>
> Number of tickets = 1: Numbers 1, 2, 3, 4, 5, 6
>
> 1589 100
> Coverage is 1589, Therefore WCD = ---- * --- = 52.91 %
> 3003 1
>
>
> Number of tickets = 2: Numbers 1, 2, 3, 4, 5, 6
> 7, 8, 9, 10, 11, 12
>
> 2778 100
> Coverage is 2778, Therefore WCD = ---- * --- = 46.25 %
> 3003 2
>
> Number of tickets = 3: Numbers 1, 2, 3, 4, 5, 6
> 7, 8, 9, 10, 11, 12
> 1, 2, 3, 4, 13, 14
>
> 2988 100
> Coverage is 2988, Therefore WCD = ---- * --- = 33.17 %
> 3003 3
>
> Number of tickets = 4: Numbers 1, 2, 3, 4, 5, 6
> 7, 8, 9, 10, 11, 12
> 1, 2, 3, 4, 13, 14
> 5, 6, 7, 8, 13, 14
>
> 3003 100
> Coverage is 3003, Therefore WCD = ---- * --- = 25.00 %
> 3003 4
>
> Which would you say is the best wheel, and how does your
>judgement compare to the WCD rating?
>
> The point I'm trying to make is that the WCD is always going to go
>down with increasing number of tickets, because you can't avoid
>coverage redundancy as you introduce more tickets. The same applies
>to the various wheels in 6/49. Consider again a 'wheel' of 1 ticket.
>Its coverage is 260524 out of 13983816, giving a WCD of 1.863%. This
maybe a typo: should be 260624; it is the maxCover value, ie. the
nbr of combinations 1 block ideally should cover.
I'm currently working on an new method, but the relations for the
given examples remain similar I fear. The advantage of the new formula
is that it gives a true p-value (ie. percent value if multiplied by 100).
>is to be compared with the best current 27 ticket wheel with a
>coverage of about 44.5%, ie a WCD of 1.648%. Then the 168 wheel has
>a WCD of just 100/168 or 0.595%. Again, does the WCD give a good
>guide to which is the best wheel? Essentially the WCD will _always_
>give the best result for a wheel of 1 ticket, and will favour wheels
>of lower number of tickets. Now I'm not a great expert or user of
>wheels, but I'm fairly sure that most people who are keen on wheels
>do not consider a wheel of one ticket to be a good wheel!
>
> So what I'm really trying to say is that I don't think the WCD
>is a good means to compare two wheels of different numbers of tickets..
>However, I can't say I've got any better suggestion.
#####
>Hi Stephen and all the others,
>
>I've moved the discussion about the Performance Index for Wheels (WCD etc.)
>to a new subject called "Performance Index for Covering Designs, Wheels".
>The first posting under the new subject contains IMHO all the relevant
>excerpts from the previous postings.
Ie. it's exactly this posting :-)
(BTW, the old subject was "Dimitris Challenge" in r.g.l.)
Subject: Re: PC & CCA
From: Pablo del Monte Luna
Date: Sat, 18 Jan 1997 12:41:40 -0800
Richard F Ulrich wrote:
> If you used ALL of the principal components from each analysis, then
> the Canonical Correlation could have been run on all the inital
> variables, and it would have accounted for the same variance. Is
> that reasonable? "80% of the variability" is enough redundancy
> that I would be cautious about a) overfitting, if there were not a
> large number of cases per variable; or b) correlated error, if
> your data sources are not independent.
>
> Maybe that will help you consider what you have, but your question
> was not specific enough for me to say much more.
>
> Rich Ulrich, biostatistician wpilib+@pitt.edu
> http://www.pitt.edu/~wpilib/index.html Univ. of Pittsburgh
One one hand, I have 8 climatic indices (20 years of monthly standarized anomalies, say
N=192) that represents the oceanic and atmospheric conditions in the Tropical Pacific.
On the other, I have 6 areas (extratropics) each subdivided in 6 2X2 boxes . Each
subdivision is an historical record of sea surface temperature, sea level pressure and
scalar wind, separately, in form of anomalies too. The variables are naturally related.
From the 8 original indices, Principal Components Analysis defined 3 factors accounting
89% of the original variability; and for the 6 series of each box (for each variable)
the first three modes explain 80% of the variance 90% of the time, only 4 modes the last
10%.
Then, I made Canonical Correlation Analysis between the factor scores of the indices
with every set of factor scores of each area (for each variable) to obtain 9 canonical
R's per variable.
I apologize for my poor vocabulary and syntax.
In advance, thak you very much for your time and help.
Pablo del Monte
Marine Biologist
Fisheries Group
Center for Biological Research
La Paz, B.C.S., Mexico
http://www.cibnor.mx
delmonte@cibnor.mx
Subject: CFP: 1997 Fall Technical Conference
From: "Randall D. Tobias"
Date: Thu, 16 Jan 1997 16:02:47 GMT
41st Annual
Fall Technical Conference
1997
Call for Papers
"Mining Data for Quality Improvement"
Omni Inner Harbor Hotel
Baltimore, Maryland
October 16-17, 1997
Co-sponsored by:
American Society for Quality Control
- Chemical and Process Industries Division
- Statistics Division
American Statistical Association
- Section on Physical and Engineering Sciences
Applied and expository papers are needed for parallel sessions in
Statistics, Quality Control, and Tutorial / Case Study.
Detailed submission instructions are available on the Web
http://www.sas.com/ftc97/
or you can request them from one of the following members of the
program committee:
Susan L. Albin
Department of Industrial Engineering
Rutgers University
PO Box 909
Piscataway, NJ 08855-0909
phone: 908-445-2238
email: salbin@rci.rutgers.edu
FAX: 908-445-5467
Sharon Fronheiser (to whom paper correspondance should be addressed)
Eastman Kodak Company
151 Mill Hollow Crossing
Rochester, NY 14626
phone: 716-588-2014
email: sharonf@kodak.com
FAX: 716-722-4415
Randy Tobias (to whom electronic correspondance should be addressed)
SAS Institute Inc.
SAS Campus
Cary, NC 27513-2414
tel: 919-677-8000 x7933
email: sasrdt@unx.sas.com
FAX: 919-677-8123
The submission process will start on August 1, 1996 and conclude on
January 17, 1997. Papers should be strongly justified by application
to a problem in quality control, or the chemical, physical, or
engineering sciences. The mathematical level of papers may range from
none, to that of the Journal of Quality Technology, or that of
Technometrics.
--
Randy Tobias SAS Institute Inc. sasrdt@unx.sas.com
(919) 677-8000 x7933 SAS Campus Dr. us024621@interramp.com
(919) 677-8123 (Fax) Cary, NC 27513-2414
Faith, faith is an island in the setting sun.
But proof, yes: proof is the bottom line for everyone.
-- Paul Simon
