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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 k-sets 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.
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 "Anti-Ramsey
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), 61-68.

(2) J. Korner, K. Marton, "New bounds ...", European Journal of
Combinatorics 9 (1988), 523-530.

(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, 54-60.

(5) D. Deng, D. Stinson, R. Wei, "Lovasz's local lemma ...", Designs,
Codes, and Cryptography 32 (2004), #1-3, 121-134.

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
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 well-defined and should have been studied. Write to you later. Fenghui
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" e-mail 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
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.
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 3^3-1, 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...)

Cross-posted to alt.sci.math.combinatorics ...

(I thought I had an answer, but I realized it doesn't work.)
--- Christopher Heckman
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/(k-1))/ln(k/(k-1))=(ln(n)-ln(k-1))/(ln(k)-ln(k-1))
Guest

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

Your analysis is fine, but I am afraid this lower bound is not tight.
Because

ln(n/(k-1))/ln(k/(k-1))=(ln(n)-ln(k-1))/(ln(k)-ln(k-1))<ln(n)/ln(k)=log(n,k)

seems a little bit too good.

Consider a simple case when n=6 and k=4, I don't think we can find two
colorings that will do the job.

Anyway your analysis is very helpful, I now know how to attach this
problem and where to find the knowledge I need.

I really appreciate your help. Nice weekend.
Guest

 Posted: Tue Jul 05, 2005 9:32 pm    Post subject: Re: An interesting coloring problem, can anyone help? Great, thank you very much Mr. Heckman. I will look into your proof carefully. Can I write to you later? Fenghui
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?

Proginoskes wrote:
 Quote: fhzh...@gmail.com wrote: Sorry I didn't address the problem very clear. We have only k colors available, Oh, sure! I missed that ... which means all coloring schemes should have colored the set of elements into exact k colors. In that case, only one scheme is not enough. It's similar to hashing in the sense of that in both cases we want to find a group of maps(functions) from [1..n] to [1..k]. I will go get a book on Hypergraph and read it. There aren't any real restrictions on what elements can be colored what colors, so hypergraphs may not be as useful as I originally thought. Look up books on combinatorics instead.

Actually, I think I've solved it, or at least found a non-trivial lower
bound on the number of colorings.

Here's all the background, for those who haven't read the complete
thread. (Or those who are reading alt.sci.math.combinatorics but not
sci.math.)

DEFINITION. A k-coloring [scheme] is a mapping C from {1,2,...,n} to
{1,2,...,k}.

DEFINITION. If A is a subset of {1,2,...,n} with |A| = k, then a
coloring C is perfect for A if the elements of A all receive different
colors; i.e.,
| {c(i) : i in A} | = k.

PROBLEM. Find a set D of k-colorings with minimum size, so that every
subset of {1,2,...n} with size k is perfect with respect to some
coloring in D.

And now for the lower bound ...

THEOREM. Let S = {1, 2, ..., n}, and let C_1, C_2, ..., C_c be
k-colorings of S. Then there is a subset T of S of size at least n *
((k-1)/k)^c such that ANY subset of T with k elements is not perfect
with respect to any of C_1, C_2, ..., C_c.

Proof. Induction on c. If c = 0, we can let T = S.

Now suppose that T_(c-1) is a subset of size n * ((k-1)/k)^(c-1) such
that any subset of T_(c-1) with k elements is not perfect with respect
to any of C_1, C-2, ..., or C_(c-1).

Now let U_i be the set of all j in T_(c-1) such that C_c(j) = i. Let
T_c be union of the (k-1) largest U_i's. This guarantees that T_c has
at least
(k-1)/k * | T_(c-1)| = n * ((k-1)/k)^c elements.

Any subset A of T_c which has k elements is not perfect with respect to
C_1, C_2, ..., C_(c-1), since A is a subset of T_c, which is a subset
of T_(c-1), and any subset of T_(c-1) with k elements is not perfect
with respect to C_1, C_2, ..., C_(c-1).

A can't be perfect with respect to C_c, because {C_c(j) : j is in A} is
a subset of {C_c(j) : j is in T_c}, and the latter has k-1 elements, by
construction.

This proves the theorem.

COROLLARY. If A is perfect with respect to C_1, C_2, ..., C_c, then
c >= ceiling (ln (n/(k-1)) / ln (k/(k-1))).

Proof. Let B be T_c, constructed as in the proof. If B contains at
least k elements, then some subset of B with k elements is not perfect
with respect to any of C_1, C_2, ..., C_c, contrary to assumption.
Hence
k-1 >= |B| >= n * ((k-1)/k)^c.
Solving for c finishes the proof.

I suspect the answer to your problem is the ceiling of
(ln (n/(k-1))) / (ln (k/(k-1))), or not too much above it. To construct
the colorings, follow the proof, and make the U_i's as close the same
size as each other as possible.

For instance, if k = 2 and n = 2^p, then the answer is p. (You can use
the binary expansions of 0 ... n-1 to get the colorings, as well.)
Here's an example with k = 2 and n = 7:

For coloring C1, break the set {1,2,3,4,5,6,7} into 2 sets of about the
same size: {1,2,3,4} and {5,6,7}, for instance. Make C1 color 1,2,3,4
with 1 and 5,6,7 with 2.

We now have the sets {1,2,3,4}, {5,6,7}. If x and y are both elements
of one of these sets, then C1(x) = C1(y); i.e., C1(5) = C1(6) = C1(7).

For coloring C2, break the set {1,2,3,4} (which are all colored 1) into
2 sets of about the same size: {1,3} and {2,4}, for instance. (Any
partition will do.) Make C2 color 1,3 with 1 and 2,4 with 2. You also
need to deal with {5,6,7}, which are all colored color 2 by C1; color
5,6 with 1, and 7 with 2, for instance.

coloring c(1) c(2) c(3) c(4) c(5) c(6) c(7)
C1 1 1 1 1 2 2 2
C2 1 2 1 2 1 1 2

Now, we have the sets {1,3}, {2,4}, {5,6}, and {7}. Note that C1(x) =
C1(y) and C2(x) = C2(y) for any x,y in {1,3}, or in {2,4}, or {5,6}, or
{7}.

The coloring C3 will make this last distinction. Divide {1,3} into 2
sets of about the same size, {1} and {3}, make C3 color everything in
{1} with 1, everything in {3} with 2; divide {2,4} into {2} and {4},
and make C3 color everything in {2} with 1, everything in {4} with 2,
and divide {5,6} into {5} and {6}, and make C3 color 5 with 1 and 6
with 2.

It doesn't matter how C3 colors 7. Now our colorings look like:

coloring c(1) c(2) c(3) c(4) c(5) c(6) c(7)
C1 1 1 1 1 2 2 2
C2 1 2 1 2 1 1 2
C3 1 1 2 2 1 2 x

Now we've found a set of 3 colorings with the property that for any
distinct x and y, there is a coloring C in {C1, C2, C3} such that |
{C(x), C(y)} | = 2 = k. Also, this set of colorings is minimal, since
ceiling (ln (7/(2-1)) / ln (2/(2-1))) = 3.

--- Christopher Heckman
Guest

 Posted: Tue Jul 05, 2005 9:32 pm    Post subject: Re: graph theory 11. sketch pf: A well-known 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.
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 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 NON-NEGATIVE 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 NP-hard. (By a reduction from subset sum.) --Gerhard ___________________________________________________________ Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/
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 well-known 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 11-gon (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
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
 Quote: Path: grapevine.lcs.mit.edu!newsswitch.lcs.mit.edu! bloom-beacon.mit.edu!newsfeed.stanford.edu! postnews.google.com!news4.google.com!news.glorb.com! sn-xit-04!sn-xit-12!sn-xit-01!sn-post-01!supernews.com! corp.supernews.com!not-for-mail From: "PaulKing"
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" 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 84-ticket 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/t-designs.

--- Christopher Heckman

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