Combinatorics

Combinatorics

  • Jean Moulin-Ollagnier added an answer:
    Is there a bijective proof of : the rational number q_{m,n} = \frac {(2m)! (2n)!} {m! n! (m+n)!}, where m,n are positive integers, is an integer ?

    I found this problem without comments in a french exercise book of 1982, now out of print : {\em 1932 exercices de mathématiques} by Luc Moisotte, ISBN 2-04-015483-3.

    Jean Moulin-Ollagnier · École Polytechnique

    Thank you, Anna,

    but I look for a possible "bijective" proof, I mean :
    this number is an integer because it counts something than we can describe.

    Sincerly, Jean

  • O.V.Shanmuga Sundaram added an answer:
    Re: research
    dear friends. Presently I am doing research in automata theory. What is the latest research going on automata theory. friends, I am now in confusion. how to start and how it grow because I am basically mathematician. I want to know more ideas. also Is there any relavant materials available net. actually I started with buchi automaton in automata theory. but I do not know proceed that I have seen many materials. please help me & suggest me good ideas.
    thanks & regards
    O.V.SHANMUGA SUNDARAM
    ovs3662@gmail.com
    O.V.Shanmuga Sundaram · Sri Shakthi Institute of Engineering and Technology

    we completed ph.d thesis work. now preparing synopsis. for this we need external foreign examiner for evaluation of my ph.d thesis. probably we submit thesis before December 2015.. Our topic is related to automata theory using graph theory concept in the core chapter. we request you kindly know or interested please send your latest profile... thank you for considering my obligation....to my email-id is ovs3662@gmail.com

  • SIMON RAJ F added an answer:
    Is there any Barnette Graph with 2k (even number greater than 84) vertices?

    In 1969 , David Barnette conjectured that 3 regular , 3-connected , bipartite , planar graph is Hamiltonian. . I am interested  to generate Barnette  graph for given  even number of vertices. There are countably finite number of Barnette graphs available in the literature.

    Consider a grid graph with 4 vertices (cycle C 4) which is Hamiltonian. Increasing dimension in one direction , we see that the resulting graph is always Hamiltonian but not Barnette. Can one generate  countably infinite number of  Barnette graphs from one small Barnette graph?     

    SIMON RAJ F · Hindustan University

    Increasing dimension of the graph or network. like Hexagonal network.

  • Krishnan Umachandran added an answer:
    What are the applications of additive combinatorics in the field of engineering and architecture?

    I would like to know some practical applications of additive combinatorics in the field of engineering and architecture. Please suggest some useful reference.

    Krishnan Umachandran · NELCAST

    To enable agility by additive manufacturing (complexity, agility, efficiency) implementing concurrent, hybrid processes, considering the design space with topology organization,
    production constraints, and optimization of variables and materials. Innovation is better through additive manufacturing, such as using Additive Topology Optimized Manufacturing (ATOM) which helped place brackets on the Black Hawk helicopter, and drastically reduces wasted material.

  • Maimaiti Wulayimu added an answer:
    Baker Hausdorff decoupling formula
    I have a two mode squeezed state ,
    exp(ab-a^+b^+)|0>-a|0>_b . I would like to show entangled form which it will give. In this case [A,[A,B]] !=0 & [B,[A,B]] !=0 (!= means not equal) .therefore I can not use well known Glauber (Baker Hausdorff) decoupling formula directly. Does someone has experience on this problem or good suggestion for me. I think, I can do Taylor expansion then try to find some way to decouple this, I hope it will work ...
    Maimaiti Wulayimu · University of Camerino

    I think I misunderstood your question. 

  • Miguel A. Pizaña added an answer:
    What are the major real life or practical applications of intersection graphs?

    What are the major real life or practical applications of intersection graphs? Please suggest some good reference materials.

    Miguel A. Pizaña · Metropolitan Autonomous University

    Clique Graphs, a special case of intersection graphs, have been used in Loop Quantum Gravity: 


    M. Requardt.
    (Quantum) spacetime as a statistical geometry of lumps in random networks.
    Classical Quantum Gravity 17 (2000) 2029--2057.


    M. Requardt.
    Space-time as an order-parameter manifold in random networks and the emergence of physical points.
    In Quantum theory and symmetries (Goslar, 1999), pages
    555--561. World Sci. Publ., River Edge, NJ, 2000.


    M. Requardt.
    A geometric renormalization group in discrete quantum space-time.
    J. Math. Phys. {\bf 44} (2003) 5588--5615.

  • Sam W. Murphy added an answer:
    Knapsack Packing Problem. Anyone have experience with this?

    I am trying to calculate the most compact way of grouping a set of pixels together. Does anyone have a readable guide on how to do this?

    My initial results are given below for clusters of up to 10 pixels. Results are expressed in terms of the sum of unique interpixel distances for a given cluster (e.g. for a 3 pixel cluster it is the sum of the the distances ab, ac, bc).

    1 = 0

    2 = 1

    3 = 3.4

    4 = 6.8

    5 = 13.5

    6 = 21.1

    7 = 31.4

    8 = 44.1

    9 = 58.9

    10 = 78.5

    Sam W. Murphy · University of Campinas

    Thank you Artiom you have been very helpful! (I have only just seen your previous post now, I must have posted mine at the same time!).. I will keep working on this problem

  • Daniel Wright added an answer:
    Just where do we draw the distinctive line between traditional data analysis and present day (Big) data analytics?

    Some authorities like Davenport have already explained that traditional (Small?) data relates to corporate operations while Big data relates to corporate products and services.

    Daniel Wright · ACT, Inc.

    The size of what people mean when they say big data changes (and varies by discipline), so 20 years ago it would have been a lot smaller than now. It might be what you can't store (or do simple analysis of) on a new desktop. Here is a quote from a White House report:

    There are many definitions of “big data” which may differ depend ing on whether you are a computer scientist, a financial analyst, or an entrepreneur pitch
    ing an idea to a venture capitalist. Most definitions reflect the growing technological ability to capture, aggregate, and process an ever-greater volume, velocity, and variety of data.

    https://www.whitehouse.gov/sites/default/files/docs/big_data_privacy_report_may_1_2014.pdf

    So, imo, there is no line (and if someone drew one today, its wrong by tomorrow).

  • Sudev Naduvath added an answer:
    Where can I find useful literature on graph theoretical applications to biological networks?

    Please provide me information regarding the recent developments in the mathematical, especially graph theoretical, studies on biological networks. Please give some good reference too..

    Thanking you in advance,

    Sudev

    Sudev Naduvath · Vidya Academy of Science & Technology

    Thank you so much Professor Ljubomir Jacić, I will go through that.

  • Marzieh Eini Keleshteri added an answer:
    What can be said about the $q$-analogue of the Sheffer identity?

    In \cite{Roman} page 25 we read that,  a sequence $s_n(x)$ is Sheffer for $(g(t), f(t))$, for some invertible $g(t)$, if and only if

    $$ s_n(x+y)=\sum\limits_{k=0}^{\infty}\binom nk p_k(y) s_{n-k}(x),$$

    for all $y$ in complex numbers,  where $p_n(x)$ is associated to $f(t)$. 

    Noting to the fact that $e_q(x+y) \neq e_q(x)e_q(y)$, leads to conclude that $ s_{n,q}(x+y) \neq e_q(yt)s_{n,q}(x)$, and ,therefore, we do not have the $q$-analogue of the identity above directly. Is it possible to express the $q$-analogue of the above mentioned identity in any other way, or should we neglect such an identity for $q$-Sheffer sequences at all?

    Any contribution is appreciated in advance. 

    \bibitem{Roman}Roman S., Rota G. The umbral calculus. Advances Math. 1978;27:95–188.

    Marzieh Eini Keleshteri · Eastern Mediterranean University

    Dear Dr. Waldemar W Koczkodaj

    Thanks for your response. I faced with this question, for the first time, while I was studying the sequence of $q$-Appell polynomials. Since that time, this question has been in my mind for a long time and so far I have not been able to make myself convinced by a perfect answer to it. The reason to ask this question here, actually, is to consult with the experts of this field and read about their different ideas which are originated, clearly, from different points of view. Although I am enthusiastic for learning more and more and go forward through the science, in case that I found a good answer which helps me to publish my studies, I make you sure that I will definitely obey the publication rules and humanity. 

  • Joe Mccollum added an answer:
    What is the number of subsets of a finite set of non-negative integers which are neither the sumsets or the summands of other subsets of $X$?

    Suppose that X is a finite set of positive integers. The sumset of two subsets A and B of X is defined as A+B={a+b:a\in A, b\in B}. Then, what is the number of subsets of X which are neither the non-trivial sumsets of any two other subsets of X nor the non-trivial summands of any other subsets of X?  Also, please suggest some useful references in this area.

    Thanking you in advance,

    Sudev

    Joe Mccollum · US Forest Service

    I'm still trying to understand the question.

    Suppose X = {1,2,3,5}.

    Then X has 2^4 possible subsets.  But {2,3} is out because it is the sumset of {1,2} and {1}.  Then are {1,2} and {1] out because they are summands of {2,3}? 

    If so, we're left with just two subsets - {1,3,5} and {1, 2, 3, 5}.

  • George Stoica added an answer:
    Can we use discrete mathematics in modeling of control and dynamical systems?

    We usually use differential equations, ordinary and partial, difference and delayed. But, could dynamics be captured using discrete mathematics structures, or combinatorics?

    George Stoica · Canada

    Dear Sonam,

    It works in stochastic set-up as well, using appropriate modifications of the discrete approximations. The convergence schemes of the stochastic approximations are very well established; the rates of convergence therein are however slightly inferior to the deterministic ones.

    George 

  • Natig Aliyev added an answer:
    What are the graphs whose total graphs are complete graphs?

    The total graph T(G) of a given graph G is a graph such that the vertex set of T corresponds to the vertices and edges of G and two vertices are adjacent in T if their corresponding elements are either adjacent or incident in G. Can we have non-trivial graphs whose total graphs are complete?

    Natig Aliyev · Auburn University

    I checked  response of Prof. Singaram Dharmalingam now, quick proof :)

    Thank you for the question though.

    Good Luck!

  • Qefsere Doko Gjonbalaj added an answer:
    How can we relate set theory with networks theory?

    Would you please help me to identify some applications of set theory, in particular sumset theory, in networks such as  in social and biological networks? Please suggest some useful reading too/

    Qefsere Doko Gjonbalaj · University of Prishtina

    Dear Sudev Naduvath

    I will suggest you this link

    Connected Dominating Set: Theory and Applications - Springer
    http://www.springer.com/mathematics/applications/book/978-1-4614-5241-6

  • Wiwat Wanicharpichat added an answer:
    Can anybody suggest any references for combinatorial studies on perturbation theory?

    Are there any combinatorial studies on perturbation theory? Can we relate graph theory to that area? If so, please suggest some useful references for a beginner like me.

    Wiwat Wanicharpichat · Naresuan University

    Thank you Prof. Sudev.

  • Tapas Chatterjee added an answer:
    Is it right to take the sumset A + ∅ = ∅?

    Is the concept A + ∅ = ∅ correct? A is any set of non-negative integers. I
    think that if it is so, it contradicts the condition on the cardinality of sum sets
    that |A| + |B| − 1 ≤ |A + B| ≤ |A| |B|. Please give your expert opinions.

    Tapas Chatterjee · Indian Institute of Technology Ropar

    I guess, Kemperman's theorem is for non-empty sets. 

  • Attila Por added an answer:
    Is there any possibility for the sumset of two sets of integers, which are not arithmetic progressions, to be an arithmetic progression?

    The notion of sumset of two sets is defined as A+B={a+b:a in A, b in B}. If the elements of A and B are not in AP,  then can the elements of A+B be an arithmetic progression? If so, what are the conditions required for that? Please can you recommend me some good references.

    Thanking you in advance,

    Sudev Naduvath

    Attila Por · Western Kentucky University

    Don't know if that was already pointed out, but the example by Alex Ravsky is as "bad" as that neither A nor B contain a an arithmetic progression of length 3.

  • Sanjib Kuila added an answer:
    Why does the four colorability of planar graphs not ensure the non-biplanarity of K_9?

    My earlier expectation regarding the sufficiency of 4CT to adress this issue, is not correct.

    Sanjib Kuila · Panskura Banamali College

    My latest publication in http://www.ijeijournal.com/pages/v4i6.html may help to give some insight on this issue.

  • Sudev Naduvath added an answer:
    Compute the number of path from a node to a different one in a weighted undirected connected 3-regular graph.
    The graph is a weighted undirected connected 3-regular graph. The number of nodes is N.
    For each node there is one loop with weight $= \frac{1}{N}, and two other edges which goes from the node to its two ``nearest neighburs with weight $= \frac{\epsilon}{N}, where \^epsilon is a small parameter.
    Therefore, we would like to know for a given value of $N$ how many different paths of lenght $2T$ are possible to go from one node to another passing through $k$ edges with weight $= \frac{\epsilon}{N}$ ?
    Sudev Naduvath · Vidya Academy of Science & Technology

    I agree with Prof. Patrick Solé.  Maple is a very useful software for any kind of mathematical operations.

    The number of paths of length n on the pair of given vertices (vi, vj) is the (i, j)th entry of An where A is the adjacency matrix of the given graph G.

    You may refer refer the following link.

    B Roberts, D P Kroese, Estimating the Number of s-t Paths in a Graph, http://jgaa.info/accepted/2007/RobertsKroese2007.11.1.pdf

    Also visit the following page for a recursive algorithm to find all paths between two given vertices of a given graph.  http://www.technical-recipes.com/2011/a-recursive-algorithm-to-find-all-paths-between-two-given-nodes/

  • Sudev Naduvath added an answer:
    Are there any specific studies on the maximal bipartite sub-graphs of different products of two regular graphs?

    Could you suggest me some of the articles on the maximal bipartite sub-graphs of different products of graphs, especially two regular graphs?

    Sudev Naduvath · Vidya Academy of Science & Technology

    Dear Prof. Vitaly Voloshin, Thank you for the links...

  • Mihai Prunescu added an answer:
    Where can I find some real life or scientific applications of the sumsets of two sets of real numbers?

    Please provide me some practical/real life applications for the sumsets of the sets of integers or real numbers. Please suggest me some useful references too..

    Mihai Prunescu · Institute of Mathematics of the Romanian Academy

    The sum is sometimes a direct sum, like in coordinate axes. R2 = (1,0) R + (0,1) R and this fact has a lot of applications (!) Another interesting fact is that R = Z + [0,1] and the decomposition is again unique. This leads to the functions integer part and fractional part. Also interesting, the sum Z + sqrt(2) Z is dense in R. These are the most easy (trivial) examples, but there are a lot of nice sumsets of reals!

  • Patrick Solé added an answer:
    Is it possible to find an infinite arithmetic progression of value of k such that we can get primes of the form 6k+1 successively?

    for k=1, 2, 3 we get 7, 13, 19 as primes.

    for k=5,  6, 7 we get  31, 37,  43,  as primes.

    Patrick Solé · Institut Mines-Télécom

    Even if true this is likely beyond current technology. It would imply Green Tao theorem:

    http://en.wikipedia.org/wiki/Primes_in_arithmetic_progression

  • Christopher Stover added an answer:
    How to decrease the bounded gaps between primes?

    Last year, Dr. Yitang Zhang has published a paper for the upper bound of twin primes, which is 7*10^7. Does anyone has idea to achieve lower bounded gap?

    Christopher Stover · Florida State University

    Terence Tao and his collaborators did tons of work on this in Polymath 8 (mentioned ambiguously in the above comment). You can read part of what was done here:

    http://terrytao.wordpress.com/2014/09/30/the-bounded-gaps-between-primes-polymath-project-a-retrospective/

    Tao's complete corpus of work on that project is:

    http://terrytao.wordpress.com/tag/polymath8/

  • Issofa Moyouwou added an answer:
    Up to now, what is known on (the maximal) domains that guarantee the transitivity of the majority rule?
    It is well known that the majority rule may not be transitive for some configurations of individual preferences. Domain restrictions are possible ways out. But what is known about maximal such domains (i) with respect to the cardinality? (ii) via set inclusion?
    Issofa Moyouwou · University of Yaounde I

    Thank you very much. I also like the job done on mutiple issues.

  • Patrick Solé added an answer:
    How can we express the recurrence formula about the labeled tree in Maple16?

    In Maple 16,  how can we with the software combstruct,  to give the sentence about the recurrence formula,

    A(x)=1+x[A(x)3+3A(x)A(x2)+2A(x3)]/6

    Patrick Solé · Institut Mines-Télécom

    Very strange, first the gf then the object to count!

  • Peter T Breuer added an answer:
    Are there any statisticians to collaborate on a combinatorics challenge for a paper?

    I and another three colleagues have an ongoing paper about the application of sociometry to multiple human resource allocation to multiple projects. The problem is mathematically quite challenging but we are on a dead end concerning one of the outcomes we are dealing with.

    Particularly, we have been applying meta-heuristics and evolutionary algorithms to solve the problem of allocating groups of people so as to maximize cohesion among them. The research is quite interesting and we think it is going to open multiple and very interesting research and industry application in the near future.

    However, we are stuck in one part. We are trying to calculate the number of viable combinations of people who can work either full-time, part-time or not work at all in several simultaneous projects and we need a person with advanced knowledge in combinatorics to give us a hand. We are willing to pay or to put his/her name as co-author on the paper.

    Getting straight to the point, the problem statement is as follows:

    There are N people (i=1...N) who can be selected to work in P simultaneous projects (j=1...P). Each person can have a dedication of work full-time (1), half-time (0.5) or not work (0), that is, three possible allocations (0, 0.5, 1).

    Now, we know that each project j requires Rj people. How many different and viable combinations are there?

    Numerical Example:

    People available  Project j=1     Project j=2      Project j=3=P

    i=1                       0 or 0.5 or 1   0 or 0.5 or 1    0 or 0.5 or 1     Row sum =<1

    i=2                       0 or 0.5 or 1   0 or 0.5 or 1    0 or 0.5 or 1     Row sum =<1

    i=3                       0 or 0.5 or 1   0 or 0.5 or 1    0 or 0.5 or 1     Row sum =<1

    i=4                       0 or 0.5 or 1   0 or 0.5 or 1    0 or 0.5 or 1     Row sum =<1

    i=5                       0 or 0.5 or 1   0 or 0.5 or 1    0 or 0.5 or 1     Row sum =<1

    Requirements      R_1=2            R_2=1             R_3=1              Row sum=4

    ,that is, we need 2+1+1 people working in these 3 projects and 1 out of the five available people is not used. We can use each person totally (1), half-time (0.5) or not use him/her, but the combinations must be feasible, i.e., each person cannot be assigned over 1 (full-time) and the requirements have to be fulfilled.

    This problem is easier when the people can only work full-time or not work, but with half-times is far more complicated.

    If anyone think that he/she is able to solve it, or just want further details, please  contact me.

    Thanks

    Peter T Breuer · Birmingham City University

    Yes, notice that I added a few words in my answer saying effectively "well, I think you meant to be generous, and allow work to be shared among too many workers by letting many do only part-work, but ended up accidentally saying that you wanted parsimonious allocations, where as many workers as possible are fully employed; still never mind because the one reduces to the other". The reduction is

    [generous]  fg(p;n1,n2)=fp(p;n1,n2,p-n1-n2) [parsimonious]

    There's an extra dummy project to absorb the efforts of under-employed workers.

    So one can stick to the easier (parsimonious) calculation, which in the simplest picture is basically a 3-way recursion from p;n1,n2 to the sum of three smaller values in n1,n2 at p-1. The three ways are how to to distribute two half-person efforts among two projects. Those are 1+0, 0.5+0.5, 0+1. That leads to a recursion from p;n1,n2 to p-1;n1-1,n2-0 and p-1;n1-0.5;n2-0.5 and p-1;n1-0;n2-1. One sums those contributions.

    If you are going to consider 1/3 allocations, and 1/4 allocations, etc, the same principle applies. For each person you need to consider how many ways there are of distributing his m/m of available effort among exactly n projects.

    That's the number of partitions of a set of m objects into up to n labeled partitions (some of the partitions may be empty). Each partition represents the allocation of effort among the n projects by that person. That is the sum of Stirling numbers of the second kind  mS0+mS1+mS2+...+mSn. Unfortunately you don't so much as want that number as actually the partitions themselves. Each partition x1+...+xn=m is a point in the parameter space (p-1;n1-x1/m,...nn-xn/m) to recurse to and the sum of the Stirling numbers that I mentioned is merely a count of the number of terms in the recursion!

    I should say that when n=m, i.e. you allow 1/nths of effort to be distributed among exactlu n projects, the count of terms above that I expressed as a sum of Stirling numbers is a well-known number, the Bell number Bn. It's big, as you might imagine. There are horrible recursions for it.

    All that indicates that you don't actually want to calculate the numbers you say you want to calculate, because they're just nasty and meaningless. They're also unscaled, in that the number goes up if you allow each person to divide their efforts more finely. That should not happen! You should probably be scaling by 1/mn, where n is the number of  projects, and 1/m is the minimum unit of effort. Then your numbers would maybe stay on-scale.

    I know that you want numbers, but I'm saying they aren't meaningful.

    -----------

    I'd rather look at approximations. In particular imagine that each person can give away a bit dp of their effort to any of n projects:

    f(p;n1,n2,..)=f(p-dp;n1-dp,n2,..)+f(p-dp;n1,n2-dp,...)+...                  (*)

    [This is an overestimate. In reality not every assignment of a dp to a project comprises a different overall assignment in the end, since if I assign 2dp to project 1 that is one choice, not two].That's not going to make sense unless I scale by n, the number of projects:

    f(p;n1,n2,..)=(f(p-dp;n1-dp,n2,..)+f(p-dp;n1,n2-dp,...)+...)/n

    in which case one gets

    f(p;n1,n2,...)=f(p;n1,n2,...) -df/dp- 1/n df/dn1- 1/n df/dn2-...

    which says

    df/dp = -1/n(df/dn1+...+df/dnn)

    This is a partial differential equation. My PDEs are rusty, but I believe that is saying that if I were to move by k dp and k ndn1, or k dp and k ndn2, etc, the nett change in f would be zero. That means that this f is an arbitrary function of p-(n1+n2+...)/n . I.e. f(p;n1,n2,...)=F(p-(n1+n2+...)/n) for some F.

    At least my scaling does "remove the infinities", but what am I counting? I scaled by 1/n when allocating a dp of a person, so if I had been allocating 1/mths of a person I would have scaled by 1/nm when allocating all of one person. Therefore my number is 1/nmp times too small for what you intend to count.

    If you start with p=n1+...+nn exactly (supply equals demand), I conclude that your numbers grow as F(p-p/n) nmp  in the limit when you can allocate small amounts 1/m of each of p persons and distribute them among n projects.

    The work ni needed by each particular project does not seem to matter.

    I don't know what the constant of proportionality F is, but the prediction is that it depends closely only on the number p of persons available (p-p/n, exactly speaking). I presume it is monotonic increasing, so you would be ok in taking it as a bound not an asymptote and writing F(p) instead of F(p-p/n).  If one thinks about the one-project case (that needs p workers), the implication is that it is exactly 1.

    [Note: this is an overestimate, deriving from the overestimate at (*) at the beginning of this approximation].

    -----------

    There's a more accurate way of expressing the intention of yours that I expressed as (*) above, namely:

    f(p;n1,n2,..)= Integral f(p-1;n1-x1,n2-x2,..) dA

    where the integral is the sum that appeared in (*) extended to take account of the whole possible set of assignments of a unit of work from p to the projects 1 to n, so the integral is over the sums x1+x2+...=1 for xi in the range [0,1]. That's the "diagonal" plane of the unit hypercube, so I suppose the area is something like 2n/2, but you can tell me!

    This would make sense as a stochastic expectation if one were to divide by that 2n/2. I suggest the xi are uniformly distributed over the range [0,1] taking into account the x1+x2+...=1 stipulation, which is to say that each equally sized subunit of the area of that hyperplane through the unit cube is equally probable. That's why dividing by the total area A=2n/2 make sense. It makes the distribution stochastic (total probability = 1).

    The expectation is a kind of average count of the weighted number of ways of dividing up the work, counting a tiny fraction weight for each possible way.

    If one extends this integral to an assignment of all of p, then

    f(p;n1,n2,..)= Integral 1 dA

    where the integral is over all xi in the range [0,ni] such that x1+x2+...=p. I believe one should weight all these possibiities equally again to get an expectation, which is to say that each subunit of equal area of the hyperplane x1+x2+...=p=n1+n2+... in the scaled cube [0,n1]x[0,n2]x... gets equal weight [but I'm willing to take suggestions!]. I imagine the total area of that plane segment is 2n/2 n1n2... or something very like that, so I have divided by that in getting an expectation, which will be E[1]=1 and the count represented by the unvarnished integral above is that multiplied by what I divided by, which is 2n/2 n1 n2 ....

    However, the same scaling arguments as I applied for my earlier estimation applies again. Whereas I have been applying an integral in which each "way" of dividing up the effort counts for a tiny increment in the integral above, you have been counting it as "1" for each different way.... so I need to scale up by something more again to get to the kind of count you were doing. I'll work it out later, but I guess the extra scaling number is mn-1. So the estimate here is something like 2n/2 n1 n2 ...nn mn-1 for your numbers. This replaces the mnp over-estimate I had before [or will when I check my working and perfect the working .. watch this space].

  • James F Peters added an answer:
    How do I use the Ptolemy inequality to study the geodesic angle?
    here, we try to use the ptolemy inequality to study the geodesic angle. and we introduce 2 preliminary results :

    1. the ptolemy theorem in euclid spherical geometry
    by studying the matrix of 4 quadruple points on the euclid spherical, we find the fact that, since there are 6 lines which connect these 4 points, so if 5 of the 6 lines are equal. then we will get that the sixth line's length is double as the other equally 5 lines;this result imply that the angle between the lines is included in $\frac{\sqrt{a}}{2r}$$\in(2k\pi+\pi/5,2k\pi+5\pi/6)$ ; then we substitute this result into the discriminant, and we get an inequality about the radius in $n+1$ dimension:
    $op_{n+1}^2$$\ge$$\frac{r^2}{t^2}(1+\frac{r^2}{n})-\frac{r^2}{n}$

    2. ptolemy inequality in minkowski geometry
    here ,we use the centroid method to study the n-polygon problem in minkowski geometry. firstly, we introduce a well known problem such that:
    if every angle of a polygon is equal, and the sidelines are :
    $1^2,2^2,......,N^2$, then $\sum{n_{s}}e^{isa}=0$
    by this theorem, we can factorize the mass on the vertex into pairs and the number of pairs is primes $N/2=\Pi{p_{i}^{a_{i}}}$ ,the weight of each pair is $\sum{4k-1}$,then we can divide these pairs into groups and every group has prime points too(page 21 in[1]); consequently we can divide the sidelines into 2 parts: $1^2,3^2,......$ and $2^2,4^2,......$;

    the next step is to construct regular n-polygon and use the ptolemy inequality to make a regulation for the average of the sum of the mass in different group, and we can rearrange these groups of mass to ensure the first part $1^2,3^2,......$ is larger than the average and the second part $2^2,4^2,......$ is less than it.

    therefore we can apply this average of sum to the distance formula in minkowski geometry in polar coordinate (page 24 in [1]).
    here we also use combinatorics method (result we get in step 2) to study the natural logorithm in distance formula of minkowski geometry(page 25).

    our goal is to represent the polar angle in minkowski geometry as the product of the mass lie on different vertex (page 26 in [1]).
    so our question is how to apply the 2 results above to the geodesic angle?

    by the inequality for the radius in $n+1$ dimension euclid spherical, we can ensure $v_{n+1}\ge0$; consequently we can substitute the representation of the polar angle in step 2 to spherical equation , which imply that we can also restrict the range of
    $cos^2{\varphi}$ that is $[\sqrt{33}-4l,3]$

    lastly,we apply the property of ptolemy space to get our estimate for geodesic angle,the bounde is :
    $1-4e^2d/3+\frac{2}{3-4/3e^{2d}}$

    is this method feasible? for more detail, you can refer to :

    application of the ptolemy theorem (3) (page 15-27)
    the analysis techniques for convexity: CAT-spaces (3)
    James F Peters · University of Manitoba

    This is a good question with lots of interesting paths to consider in looking for answers.    This question leads to a consideration of metric spaces that are Ptolemaic and Buseman convex spaces, stemming from

    H. Busemann, The Geometry of Geodesics, 1955

    A good place to start is

    T. Foertsch, A. Lytchak, V. Schroeder, Non-positive curvature and the Ptolemy inequality:

    http://carma.newcastle.edu.au/jon/Preprints/Papers/CAT(0)/Papers/fls-ptolemy.pdf

    These authors prove that a metric space is CAT(0) if and only it is Ptolemy and Busemann convex, p. 11.  Ultrarays assocated with geodesics and their enclosed angles are considered, starting on page 8.

  • Lie Zhu added an answer:
    Can anyone suggest some good reference papers for beginners in the field of Combinatorial Design?
    It can either be related to Key Distribution or any other application, just an overview needed.
    Lie Zhu · Suzhou University

    Design Theory, Zhe-Xian Wan, World Scientific Publishing Co Pte Ltd, 2009

  • Ricky X. F. Chen added an answer:
    The maximum number of cycles of the product of a given permutation and a n-cycle?

    Given a permutation A on the set [n], is there a way to determine the maximum number of disjoint cycles of AC where C ranges over all n-cycles on [n]? For which class of permutations A, this problem has been studied before?  Thanks

    Ricky X. F. Chen · University of Southern Denmark

    The titles of the papers I mentioned are something like:

    Zagier, on the distribution of cycles....

    Stanley, two enumerative results on product of cycles...

About Combinatorics

Combinatorics is a branch of mathematics concerning the study of finite or countable discrete structures. Aspects of combinatorics include counting the structures of a given kind and size (enumerative combinatorics), deciding when certain criteria can be met, and constructing and analyzing objects meeting the criteria (as in combinatorial designs and matroid theory), finding "largest", "smallest", or "optimal" objects (extremal combinatorics and combinatorial optimization), and studying combinatorial structures arising in an algebraic context, or applying algebraic techniques to combinatorial problems (algebraic combinatorics).

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