Author 
Message 
Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Sun Jul 16, 2006 4:34 am Post subject:
Re: Combinatorics



Thomas Mautsch wrote:
Quote:  In news:<1152934283.850075.249680@b28g2000cwb.googlegroups.com
schrieb Proginoskes <CCHeckman@gmail.com>:
Whatever5k@web.de wrote:
Suppose you have n fields, where n is at least 10. Each field is to be
filled with a ball. There are infinitely many balls, but they can be
distincted by their colour. In total there are 10 different colours. So
now, you want to find out how many distinct possibilities there are to
fill the n fields with those balls. There are, however, two conditions:
a) at the end, each colour has to appear at least once
b) the first field cannot contain the colour blue
Any idea how to tackle this?
I am an absolute beginner in combinatorics.
Please help.
The answer should be "infinitely many possibilities". You should check
the wording of the problem.
From what can be gathered from the OP's postings in de.sci.mathematik,
the problem is to determine the number of ndigit decimal numbers
with first digit different from zero
that contain each of the ten digits 0,1,2,3,4,5,6,7,8,9 at least once.

I don't see this at all. _IF_ there was a condition that stated each
field had to have at most one ball, then that would be the case. But as
I read the question, the following are all solutions to the problem,
for any nonnegative integer K: (where 0 = "blue")
Field 1: Infinitely many balls of colors 1 ... 9,
Field 2: Infinitely many balls of color 0, K balls of color 1,
Fields 3n: Empty.
This satisfies conditions (a) and (b). If fields aren't allowed to be
empty, put an infinite number of balls of color 1 in fields 3n.
That was what I was asking clarification about.
 Christopher Heckman 

Back to top 


Rich Holme science forum beginner
Joined: 06 Jun 2005
Posts: 45

Posted: Mon Jul 10, 2006 2:24 pm Post subject:
Re: arithmetic Robak 0,(9)+{1+}0=1 Time Theory



dbd@gatekeeper.vic.com (David DeLaney) writes:
Quote:  ksrobak <ksrobak@o2.pl> wrote:
Rich Holmes <rsholmes+usenet@mailbox.syr.edu
And why, WHY, has this information been withheld from a.r.k until now?
[Rich Holmes]
Manley Hubbel had a kid?
~~~~~~~~~~~~~~~~~~~~~~~~
X + oo = oo <== religion
X + oo = X + oo <== LOGIC
~~~~~~~~~~~~~~~~~~~~~~~~
X + oo = oo ; oo + X = oo + X <== ordinal transfinite arithmetic
* <== PERTH
A=A
Dave "time=inertia?" DeLaney

Ah yes, I remember that post. Good times, good times.

 Doctroid Doctroid Holmes <http://www.richholmes.net/doctroid/>
Ancient use of incendiary pigs as an antielephant measure is
disqualified on grounds of pigs not being cows, even when on fire.
 John D Salt 

Back to top 


David DeLaney science forum beginner
Joined: 08 May 2005
Posts: 20

Posted: Sat Jul 08, 2006 4:28 pm Post subject:
Re: arithmetic Robak 0,(9)+{1+}0=1 Time Theory



ksrobak <ksrobak@o2.pl> wrote:
Quote:  Rich Holmes <rsholmes+usenet@mailbox.syr.edu
And why, WHY, has this information been withheld from a.r.k until now?
[Rich Holmes]

Manley Hubbel had a kid?
Quote:  ~~~~~~~~~~~~~~~~~~~~~~~~
X + oo = oo <== religion
X + oo = X + oo <== LOGIC
~~~~~~~~~~~~~~~~~~~~~~~~

X + oo = oo ; oo + X = oo + X <== ordinal transfinite arithmetic
* <== PERTH
Dave "time=inertia?" DeLaney

\/David DeLaney posting from dbd@vic.com "It's not the pot that grows the flower
It's not the clock that slows the hour The definition's plain for anyone to see
Love is all it takes to make a family"  R&P. VISUALIZE HAPPYNET VRbeable<BLINK>
http://www.vic.com/~dbd/  net.legends FAQ & Magic / I WUV you in all CAPS! K. 

Back to top 


BDH science forum beginner
Joined: 25 Oct 2005
Posts: 11

Posted: Thu Jun 29, 2006 10:06 am Post subject:
Re: Combinations satisfying linear equations?



Exactly. 

Back to top 


Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Mon Apr 03, 2006 6:08 am Post subject:
Re: Help in solving this puzzle



Michael Harrington wrote:
Quote:  "James Waldby" <jwaldby@pat7.com> wrote in message
news:442CDE7C.89CBBFB4@pat7.com...
Michael Harrington wrote:
"john@x" ... wrote ...
There is a lottery drawn on 1 through 20 numbers and six distinct balls
are randomly picked by the lottery every day. To play you need to buy a
ticket and select six numbers of your choice. An example ticket might
be [1,20,6,19,5,12] You can buy any number of tickets. If any three of
the numbers in one of your tickets are among the six winning numbers
picked by the lottery, in any order then you win.
What is the minimum number of tickets you have to buy to be certain of
winning with atlease one?
...
Now for your stated 20 numbers we have 10 blocks of 2
{1,2} {3,4}.............{17,18}{19,20}
This will equate to 10C3 = 120 tickets.
This is unproven as the minimum, but it is some partial solution.
Would be interested to see any improvement on this.
This is not the simplistic problem others are making it out to be.
You need to cover at least 17 of the numbers to force a win,
not 20 of them. Rather than 10C3 tickets your scheme should
need somewhere between 8C3 and 9C3, ie, between 56 and 84.
Return on investment might be better or worse with your 120
ticket scheme vs. an 84ticket one, depending on how many of
the 20 (6C3) winning triples occur on distinct tickets.
Having multiple winning triples on some tickets will decrease
the total payoff if each winning ticket pays only once.
jiw
Good point on only covering 17 numbers. Would be good to
know if there is a mathematical method of finding the minimum
number of covering tickets.

The problem clearly belongs in the field of combinatorics. If you want
every triple to be in exactly K tickets, then you're in the area of
design theory/tdesigns.
 Christopher Heckman 

Back to top 


Richard Reddy science forum beginner
Joined: 16 Nov 2005
Posts: 1

Posted: Wed Nov 16, 2005 5:33 am Post subject:
Re: > Challenge



The statistics on child mortality for AIDS are fairly accurate, but the
"third world" depends on who you talk to. We have underdeveloped nations
in Africa, Asia and Latin America. Many developed nations also have
AIDS epidemics, including the USA. There is substantial margin for error in
the business of counting deaths for epidemics of all kinds. We also find
politics in official numbers, which are frequently wrong.
The issue of quality of life for people infected with AIDS is even
more tragic, given the high cost of drugs proven to be effective. The
poorest
nations are simply helpless in the face of this epidemic, lacking medical
infrastructure, public health education, or the ability to assist sick and
disabled people.
In my opinion, the age of AIDS victims is irrelevant. We should endeavor
to assist any nation or state unable to cope with this pandemic. It's the
human thing to do, whatever we think of particular statistics. The
suffering
is real and the AIDS epidemic is enormous. So too are millions of deaths
that might have been delayed by years or decades with appropriate medical
intervention.
I offer another challenge. Find a reputable organization and try to
be of help in some way. An organization can be in error with statistics,
but still be highly effective in delivering assistance, a process that
frequently
involves danger and hardship. There are many wonderful people stretched
out beyond reckoning in combating this global menace. You will find many
charities who hire "pros" to raise money, sometimes with questionable
tactics
or poor percentages reaching the intended recipients. By law, this
statistics
are public information.
Apathy is certainly a dangerous condition. I know little about this
organization,
but there are many putting their whole heart into the effort to fight this
epidemic.
If you want to hang your hat on some real statistical offenders, try the
antismoking crew. Prohibitionists love to exaggerate. For example, public
adds that say each cigarette shortens your life by 5 minutes.
<dsaklad@zurich.csail.mit.edu> wrote in message
news:1117871877.716280.18410@o13g2000cwo.googlegroups.com...
Quote:  Path: grapevine.lcs.mit.edu!newsswitch.lcs.mit.edu!
bloombeacon.mit.edu!newsfeed.stanford.edu!
postnews.google.com!news4.google.com!news.glorb.com!
snxit04!snxit12!snxit01!snpost01!supernews.com!
corp.supernews.com!notformail
From: "PaulKing" <aimulti at aimultimedia.com
Newsgroups: misc.health.aids
Subject: Challenge
Date: Fri, 03 Jun 2005 19:51:13 0400
Organization: www.talkabouthealthnetwork.com
MessageID: <431b337a0657600e87b4f777796371e2 at
localhost.talkabouthealthnetwork.com
XNewsreader: www.talkabouthealthnetwork.com
XProblemsTo: info at talkaboutnetwork.com
XPostedBy: USERID7496
ContentType: text/plain;
XComplaintsTo: abuse at supernews.com
Lines: 28
Xref: grapevine.lcs.mit.edu misc.health.aids:98998
CHALLENGE TO BELIEVERS
If anyone still believes 'AIDS' statistics, let them answer
just this one question.
The high budget apathykills.org TV commercial claims 25 million
children in the Third World have died of 'AIDS'.
The UN figures (see graph from them I posted above) says that
ONLY 3% of child mortalities in the Third World are from
'AIDS'.
If 3% = 25 million then 100% must = 832.5 million.
If nearly ONE BILLION children in the Third World have died
then how come the population growth is at an all time high.
ONE BILLION DEAD CHILDREN would almost mean every child in
Africa and India is dead.
Explain this and you can still believe 'AIDS' figures.
Fail to do so and you must admit 'AIDS' figures are TOTAL
GARBAGE.



Back to top 


Matthew Roozee science forum beginner
Joined: 22 Oct 2005
Posts: 7

Posted: Sat Oct 22, 2005 5:56 pm Post subject:
Re: graph theory



amanda.pascoe@gmail.com wrote:
Quote:  11.
sketch pf:
A wellknown fact in graph theory is the theorem that the sum of the
degrees of the vertices of G is equal to twice the number of edges
(imagine adding the number of edges leaving each vertex. When all
vertices are accounted for, each edge has been counted twice). Then by
assumption the sum of the degrees of V(G) is greater than or equal to
3p, so p is less than or equal to 2*E(G)/3.
p<=2*(17)/3<=11.333...
p is an integer, so the maximum value for p is 11.

This establishes the upper bound for p.
To complete the proof, all we have to show is that there exists a
solution for p = 11.
Start with an 11gon (which uses 11 of our edges), and label the
vertices 0,...,10 in order. Ignoring the points 0,5,10 for a moment,
draw the edges that are equivalent mod 5... so 1 connects to 6, 2
connects to 7, etc. There are 4 of these so we have used 15 edges total
and each of these 8 vertices has degree 3. We then connect 0 to both 5
and 10 (17 edges total) to give 5 and 10 degree 3 and 0 degree 4.
 Matt 

Back to top 


Woeginger Gerhard science forum beginner
Joined: 30 Jul 2005
Posts: 5

Posted: Tue Aug 16, 2005 8:30 am Post subject:
Re: Integer Optimization Problem P+p(j)/C+c(j)



unix68 <unix68@networld.at> wrote:
# Hi,
# I need help for solving the following problem (references to
# algorithms and papers).
#
# Given:
# P, p1,p2,...,pn all integers
# C, c1,c2,...,cn all integers
# P = constant
# C = constant
#
# Goal:
# maximize [P+sum(pj)]/[C+sum(cj)] for any j=1..n
#
# This means we have to find the optimal subset of elements
# out of a given set (1...n) such that the ratio
# [P+sum(pj)]/[C+sum(cj)] gets maximal.
In case all c1,...,cn are NONNEGATIVE integers,
your problem is polynomially solvable. (There is
a direct solution via sorting, you do not even
need fractional programming.)
In case c1,...,cn are ARBITRARY integers,
your problem is NPhard. (By a reduction from
subset sum.)
Gerhard
___________________________________________________________
Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/ 

Back to top 


Guest

Posted: Tue Jul 05, 2005 9:32 pm Post subject:
Re: An interesting coloring problem, can anyone help?



Well, yeah, actually a "trivial" lower bound can be derived easily.
Consider all ksets of [1..n], we have totally
{n choose k} such sets.
and each coloring is going to color at most
(n/k)^k
of them. Therefore another lower bound should be
{n choose k}/(n/k)^k
with stirling's approximation, we can easily get a lower bound of
e^k, when n/k is large enough.
Nice day. 

Back to top 


Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Tue Jul 05, 2005 9:32 pm Post subject:
Re: An interesting coloring problem, can anyone help?



fhzhang@gmail.com wrote:
Quote:  I tried some simple cases, your result seems correct.
I really appreciate your help. Nice weekend.

This problem looks like it's going to be a difficult one in general.
I did some reading through the literature earlier today (Monday), and I
found some papers related to the problem, which evidently goes by the
name of Perfect Hashing. (Colorings which color all the vertices in a
set of size k with different colors are sometimes called "AntiRamsey
Colorings." You may also want to search for related papers on this
subject.)
Some of the titles are incomplete, but once you find the appropriate
journal, you'll see what the rest of them are.
(1) M. Fredman, J. Komlos, "On the size of separating ...", SIAM
Journal of Algebraic and Discrete Methods 5 (1984), 6168.
(2) J. Korner, K. Marton, "New bounds ...", European Journal of
Combinatorics 9 (1988), 523530.
(3) J. Korner, G. Simonyi, "Trifference", Studia Sci. Math Hungar 30
(1995), 95ff.
(4) S. Blackburn, "Perfect Hash Families ...", Journal of Combinatorial
Theory Series A 92 (2000), #1, 5460.
(5) D. Deng, D. Stinson, R. Wei, "Lovasz's local lemma ...", Designs,
Codes, and Cryptography 32 (2004), #13, 121134.
Unfortunately, (3) and (5) weren't in the ASU library. Paper (4) gives
the best bounds, as well as some constructions, and (2) is also worth
checking into.
 Christopher Heckman 

Back to top 


Guest

Posted: Tue Jul 05, 2005 9:32 pm Post subject:
Re: An interesting coloring problem, can anyone help?



That's what I was thinking too. This problem seems welldefined and
should have been studied.
Write to you later.
Fenghui 

Back to top 


Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Tue Jul 05, 2005 9:32 pm Post subject:
Re: An interesting coloring problem, can anyone help?



fhzh...@gmail.com wrote:
Quote:  Great, thank you very much Mr. Heckman.
I will look into your proof carefully. Can I write to you
later?

Yes. My "professional" email address is checkman@math.usa.edu.
(Actually, you need to reverse the orders of the letters in USA; this
is to throw off spambots.)
I can't shake the feeling that someone has answered this problem
before, though, and written a paper about it.
 Christopher Heckman 

Back to top 


Guest

Posted: Tue Jul 05, 2005 9:32 pm Post subject:
Re: An interesting coloring problem, can anyone help?



I tried some simple cases, your result seems correct.
I really appreciate your help. Nice weekend. 

Back to top 


Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Tue Jul 05, 2005 9:32 pm Post subject:
Re: Combinatorics question.



Binesh Bannerjee wrote:
Quote: 
But, say I wanted the number of ways to pick _3_ items
from there?
How would I get that number, with 2A's, 3B's and 4C's?
(Or more basically, for a group consisting of any number
of items, with any number of those items possibly repeated,
how do I find the number of ways of picking n items, with
order being significant.)
I can see that AABBBCCCC empirically can have 26 possible
ways of picking 3 items. I can see that's because it's
3^31, with the 1 being the possibility of picking 3
A's... But, I'm not sure how to genericize the calculation,
which is what I'm after. (for instance, it's less easy for
me to see how to calculate that to pick _4_ items from
there, would have 71 possible ways, even tho, I can
enumerate those...)

Crossposted to alt.sci.math.combinatorics ...
(I thought I had an answer, but I realized it doesn't work.)
 Christopher Heckman 

Back to top 


Guest

Posted: Tue Jul 05, 2005 9:32 pm Post subject:
Re: An interesting coloring problem, can anyone help?



Sorry I made a mistake,
ln(n/(k1))/ln(k/(k1))=(ln(n)ln(k1))/(ln(k)ln(k1))<ln(n)/ln(k)=log(n,k)
is not correct. But still when n=6,k=4, we have
ceil(ln(n/(k1))/ln(k/(k1)))=3
Three colorings are not enough do distinguish all 4subsets of a 6set. 

Back to top 


Google


Back to top 



The time now is Tue Apr 23, 2019 12:04 pm  All times are GMT

