# Combinatorics

What is your favourite combinatorial object, or mathematical object you find the most fascinating?
I always am curious of what types of combinatorial or mathematical objects (in general) interest other researchers. For myself I have a pretty keen interest in types of restricted weak integer compositions (weak implies 0 is applied in integer sequence whereas the lack of this description means you can't use 0 in integer sequences, and restricted means finding subsets of these objects).

What combinatorial or mathematical object (preferably more elusive ones) fascinate you the most, or are your favourite

Not sure if the following qualifies but this phi spiral?

Which operations can be constructed using only xy and x+y+z?

What method can I use to charcterize which operations repeated multiplication xy and 3-input addition x+y+z are able to construct mod 4, or mod 8, or .. or mod 232?

The reason I ask is because one can prove that it is impossible to construct a known constant from unknown inputs using these operators (no formula f(x,y,..)=K for all x,y,..), and one can also characterise those other operators that one can add into the mix with them that preserve the property.

But I don't know what the totality of operations one can construct with xy and x+y+z is. Indeed, I haven't figured it out for plain old xy and x+y ! (yes, I can describe those: "all polynomials"; but how many are there and what operations cannot be formed as polynomials?)

Any clever ways to go about this? I am trying symbolic computation to generate the orbits of  "x" and "y" under substitution in the binary operators xy and x+y+z mod 4.

Peter T Breuer · Birmingham City University

Strangely, I see 1436 operator tables constructed by xy and x+y mod 4. That's multiplication and the ordinary addition, not the more rarified three-input addition.

Shouldn't these plain old operations construct all polynomials in x,y with 0 constant term? Going by the coefficients, there should be 415 of them (15 coefficients taking the values 0,1,2,3). But I suppose there are polynomials with different coefficients that are functionally equal, since there are divisors of zero mod 4? The difference might be zero mod 4, but have nonzero coefficients?

Can anyone count the number of functionally distinct multinomials in x,y mod 4?

How many polynomials are there? 64, according to the computer:

[0,1]x3+[0,1]x2+[0,1,2,3]x+[0,1,2,3]

That's 16 that are zero at zero.

That was due to reductions from 2x3=2x2=2x, and x4=x2. I'll try counting the multinomials. There are 4 possible coefficients of x, y and xy. There are 2 possible coefficients of x2, y2, x2y,xy2,x2y2. There are 2 possible coefficients of x3,y3,x3y,xy3,x3y3, x3y2,x2y3. That is all. That's 43 212, or 218. It's no good .. I'll have to count them computationally.

What if a polynomial (zero at zero) is zero on all odd numbers? The coefficients satisfy a3+a2+a1=0=a3-a2+a1, so 2a2=0 and a2=0 or a2=2. In one case a3=-a1, and the other a3=2-a1. so one has -a1x3+a1x and (2-a1)x3+2x2+a1x. If the polynomial is equal at even numbers too, then 2a1=0 so a1=0 or a1=2. The polynomial is then 0, 2x3+2x, 2x3+2x2, 2x2+2x.

That gives four versions of each polynomial with the same functionality. This is the kernel of the embedding of polynomials into functions. It's a simple count .. there are 44/4=43=64 polynomials by functionality, 16 with zero constant term.

What if a multinomial x3p3(y)+x2p2(y)+xp1(y)+p0(y) is functionally zero?

p3(y)+p2(y)+p1(y)+p0(y)=0

-p3(y)+p2(y)-p1(y)+p0(y)=0

2p1(y)+p0(y)=0

p0(y)=0

So 2p3(y)=2p2(y)=2p1(y)=p0(y)=0.

So p3,p2,p1  take only even values (0 or 2) and p0(y)=0.

So p0 is one of the four functionally zero polynomials (that's 4 options).

That p takes only even values means it is b3x3+b2x2+b1x+[0,2], where one or three of b1,b2,b3 are even: [0,2]x3+[1,3]x2+[1,3]x+[0,2], [1,3]x3+[0,2]x2+[1,3]x+[0,2], [1,3]x3+[1,3]x2+[0,2]x+[0,2], [0,2]x3+[0,2]x2+[0,2]x+[0,2], for 244=26=64 options.

That is a total of 6434=220 functions.

That would bring the number of functionally distinct multinomials down to about 416/220 =212=4096! I don't seem to be able to get a stable estimate for this quantity :-(. My computer says the answer is 3686, 906 of which are zero at zero. That can't be right.

What is an interesting and hot research topic in combinatorial group theory?

I need an interesting topic about combinatorics(graph theory) in relation with group theory, which I want to have application in other sciences.

What are the values of the special Bell polynomials of the second kind?

Bell polynomials of the second kind Bn,k(x1,x2,...,xn-k+1) are also called the partial Bell polynomials, where n and k are positive integers. It is known that Bn,k(1,1,...,1) equals Stirling numbers of the second kind S(n,k).

What are the values of the special Bell polynomials of the second kind Bn,k(0,1,0,1,0,1,0,...) and Bn,k(1,0,1,0,1,0,...)? Where can I find answers to Bn,k(0,1,0,1,0,1,0,...) and Bn,k(1,0,1,0,1,0,...)? Do they exist somewhere?

Feng Qi · Tianjin Polytechnic University

Bai-Ni Guo and Feng Qi, Explicit formulas for special values of Bell polynomials of the second kind and Euler numbers, ResearchGate Technical Report, available online at http://dx.doi.org/10.13140/2.1.3794.8808.

Can anyone give me an interpretation or link about improper uniform prior as a prior distribution in Bayesian estimation?

I need a good reason why improper uniform prior could be use as a prior in Serial Numbered Population (SNP) problem. Or maybe someone can tell me about improper uniform prior itself. An pdf link might be very helpful for me.

I attached the journal.

Brian S. Blais · Bryant University

I'm not sure if Jochen's answer is entirely correct; "An improper prior can be used when the resulting posterior is proper."  I'd modify it to say that an improper prior can be used when the resulting posterior can be achieved with a *proper* prior and the limits of the distribution, after the posterior is calculated, is taken to infinity and the result is shown to be the same.  For example, you can use a proper, bounded uniform prior and then, after you calculate the posterior, take the limits as the bounds grow to infinity to show that if you used the improper prior in the first place it would give the same answer.

Seen this way, the improper prior is a short-cut to make the analysis cleaner.  For example, I personally find E T Jaynes' analyses very clear because he uses improper priors (when he can!), and warns that one has to be careful to do it properly if there is even a hint of trouble.  I find a paper like Bretthorst's "Difference of Means" paper to be a much harder read because he uses proper priors throughout, even in cases where I think the improper prior would work.

Can anyone help me with the inequality for the ratio of two Bernoulli numbers?

Dear Colleagues

I need an inequality for the ratio of two Bernoulli numbers, see attached picture. Could you please help me to find it? Thank you very much.

Best regards

Feng Qi (F. Qi)

Feng Qi · Tianjin Polytechnic University

[1] Feng Qi, A double inequality for ratios of Bernoulli numbers, ResearchGate Dataset, available online at http://dx.doi.org/10.13140/2.1.2367.2962.

[2] Feng Qi, A double inequality for ratios of Bernoulli numbers, RGMIA Research Report Collection 17 (2014), Article 103, 4 pages; Available online at http://rgmia.org/v17.php.

What is the power series expansion at zero of the secant to the power of three?

It is well known that the secant $\sec z$ may be expanded at $z=0$ into the power series
\label{secant-Series}
\sec z=\sum_{n=0}^\infty(-1)^nE_{2n}\frac{z^{2n}}{(2n)!}

for $|z|<\frac\pi2$, where $E_n$ for $n\ge0$ stand for the Euler numbers which are integers and may be defined by

What is the power series expansion at $0$ of the secant to the power of $3$? In other words, what are coefficients in the following power series?

It is clear that the secant to the third power $\sec^3z$ is even on the interval $\bigl(-\frac\pi2,\frac\pi2\bigr)$.

Herbert H H Homeier · Universität Regensburg

Assume that secant numbers are finitely expressed in closed form. Then coefficients of m-th powers of the secant powers series are finitely expressed by using (m-1)-fold convolutions of the sequences of secant numbers (cf. http://en.wikipedia.org/wiki/Power_series#Multiplication_and_division), e.g. the k-th coefficient of the third power is given by

(E*E*E)_k = \sum_{m=0}^k E_{k-m} \sum_{n=0]^m  E_{m-n} E_n

Beware that here, the odd secant numbers are simply 0.

Can anyone suggest references for Iterated Function Systems and combinatorics: uniqueness of addresses?

Could anyone suggest me some references about the uniqueness of the addresses of an Iterated Function System? E.g. points of Cantor set can be coded by a unique address, how general is this property?

More precisely, I am interested in the sufficient conditions to have a one-to-one relation between the shift space of the Iterated Function Systems and  its attractor.

Miroslav Rypka · Palacký University of Olomouc

Dear Anna,

There are three kinds of attractors of iterated function systems, totally disconnected, just touching and overlapping. Totally disconnected attractors have metrically equivalent structure to the Cantor set. This can be found in Barnley's book Fractals Everywhere. The remaining cases may be treated with the help of lifted IFS, which is also explained in the book.

Best regards

Miroslav

• Let A & B be two square matrices such that A^2 is not equal to B^2, A is not equal to B but A^3=B^3 & A^2B=AB^2. What is determinant of A^2-B^2?
If A and B Are non-singular then we have determinant of (A^2-B^2)=-determinant of AB , Am I correct? If so, then the problem is If A and B are singular, how can we prove?
Prasanth G. Narasimha-Shenoi · Government College Chittur

@ Samuli  Yeah the facts are correct.  Sorry that I am not able to answer

What is the formula to find the number of simple cycles in a graph? Is this problem NP Complete?
I would like to list out all simple cycles in a connected graph.
Sergey Perepechko · Petrozavodsk State University

You need to give additional details about your problem. If you know adjacency matrix of your graph and want to count number of simple cycles of fixed length look at slides and papers in my profile.

Sergey Perepechko

Which Software Package is the best for computations in Codes over Rings?
What is the best software package available for carrying out computations related to algebraic coding theory, especially codes over rings.
Mostafa Eidiani · Khorasan Institute of Higher Education

Hi

Can anyone help me find amortized splay tree operation derivation?
Please help me with a derivation of "amortized cost of a splay tree operation" . I am waiting for that please consider it fast and let me know.
Albert Manfredi · The Boeing Company
Both of these articles include proofs. Not sure whether this is what you're asking. Hope they help.

http://www.cs.cornell.edu/Courses/cs312/2006fa/recitations/rec20.html

http://www.bowdoin.edu/~ltoma/teaching/cs231/fall10/Lectures/13-amortized/splay.pdf
• Cheng Tianren asked a question:
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)
In integral theory, how can you integrate all four perspectives in human development into a model?
According to Wikipedia, http://en.wikipedia.org/wiki/Permutation, if you have 3 objects [a,b,c] combinatorics there are 6 possible permutations. But positioning in space and time are not considered. What if a 'c' is placed 0,5xx behind [a,b] or 'a' and 'b' separately either horizontal/vertical, or 'c' and 'b' are isolated back to back and 'a' spins around and around for 30 seconds more until reaching a stable position?

I am looking at Integral theory and trying to integrate all four perspectives in human development (internal-individual; internal-collective; external-individual; external-collective) into a model. I am not satisfied with those similar to the above ''max 6 possible permutation of the three elements". I believe that for 3 objects there are more than 1+2+3 permutations; and for 4 objects (my case) more than 10.
Alina Abraham · ICL Business School
Adapting Ken Wilber/Terri O'Fallon model, using a spiraling trajectory
In terms of combinatorics, could (6677,333,166) be cyclical on the torus?
Would like to know if (6677,333,166) could be cyclical on the torus.
Christopher Landauer · The Aerospace Corporation
The notation is ambiguous - what combinatorial object are these the parameters for? Is it some kind of design? If so, what kind, and what parameters are indicated? The problem is that there are hundreds of different parameterized combinatorial designs that use the same kind of notation
Is there a database available on the net of symmetric designs?
Symmetric designs=combinatorial 2-designs with as many points as blocks.
Mohan Shrikhande · Central Michigan University
Nice to hear from you, Patrick!
Spin - could you provide some clarification?
I always thought that the sum of angular momentum should be zero. I am wondering if this applies on a sub-atomic scale as well as a macroscopic scale.
Daniel Crespin · Central University of Venezuela
Hello Donald,
A reasoning similar to yours implies that the total linear momentum of the Universe is zero. And furthermore, there should be a motionless center of mass of the Universe.
But how many technical details regarding these arguments can be worked out is not so clear. It may be worth a trial.

The term "spin" in Physics usually refers to angular momentum of atomic scale objects. Technically, for n >= 3 the special orthogonal group SO(n) is connected and has first homotopy group (also called fundamental group) isomorphic to Z_2, the integers modulo 2. Therefore its universal covering space, denoted Spin(n), is a well defined connected topological group (and a Lie group as well) with fibers consisting of two points, equivalently, it is a connected double cover.

If you consider instead the orthogonal group O(n), this is non-connected, has two connected components and the connected component of the identity is SO(n). The universal cover of O(n) is a topological space known as Pin(n), necessarily a non-connected double cover of O(n). But the group structure of Pin(n) is not unique. This subtlety is mentioned in http://en.wikipedia.org/wiki/Pin_group See references there.

In Classical Physics, the natural formalism for rotations in ordinary three dimensional space is based on SO(3), its tangent bundle, and the Lie algebra so(3) = tangent space to SO(3) at the identity. A good reference is the book Classical Mechanics by V. Arnold.

Then comes Quantum Mechanics and descendants. Schrödinger time dependent equation involves complex numbers. This forces the introduction of complex valued wave functions \psi. To go beyond the mathematical formalism of QM, the mathematical object \psi requires physical interpretation. The wave "amplitude" |\psi|^2 is interpreted "physically" as a probability distribution. This makes the "phase" disappear physically, because |\psi|^2 and |\exp(-i n t) \psi|^2 are one and the same physical state. But the phase is the natural way to consider rotations. Thus, rotations disappear in QM.

The quantistic way to recover something resembling rotation is to use Pauli matrices, equivalently, to use spin(3). I have been unable to make sense of these as rotations. Here the quantum dictum "Shut up and calculate" is acutely present. A nearby neon sign says "Abandon hope all ye who enter here".

In my opinion the unfortunate and mistaken choice of a unitary evolution equation ---that is, of Schrödinger evolution equation--- for the hydrogen atom made it impossible to understand microscopic rotational phenomena. Even worse, transitions themselves are contradicted by this equation.

On the other hand, Schrödinger eigenvalue equation is one of the most impressive wonders of science.

Most cordially and with best regards,
Daniel Crespin
Can anyone help me with a combinatorial interpretation?
I am asking for a combinatorial interpretation of a formula for Bell numbers in terms of Kummer confluent hypergeometric functions and Stirling numbers of the second kind. See the formula (8) and Theorem 1 in the attached PDF file or http://arxiv.org/abs/1402.2361．Could you please help me? Thank a lot.
Feng Qi · Tianjin Polytechnic University
Yes, th formula (8) is not the only such formula. Some mathematician asked me to provide a combinatorial interpretation, but I do not know the combinatorial meanings of the formula (8). I think that "the formula (8) just represents one more expression for those numbers, using Kummer confluent hypergeometric function" may be not a combinatorial interpretation of the formula (8).
Does the discrete n-circle (n even) admit a partition into n/2 pairs, all with a distinct diameter?
A (discrete) n-circle is the set of complex n-th roots of unity, or: the vertices of a regular n-gon. The above question arose as part of a (nearly finished) research project on a method to produce unpredictable number sequences. Although my partial answers are no longer needed for the project, the simple-looking and still unsettled problem keeps intriguing me. I proved that if a partition exists into pairs of distinct diameters, then n must be of type 8k or 8k+2 (k>0 integer). Computer generated examples confirm that for n <= 112, these types are *exactly* the sizes that work. The computer was stopped after running for two days on the case n=114 (having inspected nearly 0.000...001% (about 300 zeros) of the total search space). The only hope on further information must come from construction methods other than brute-force search with back-tracking and from proofs. Specifically, the problem becomes this: Design an algorithm that is guaranteed to produce a partition (as desired) whenever there exists one and reports failure otherwise. Unlike the current backtracking brute-force search, the algorithm should provide answers in a reasonable time. [Added 09-12-2013: solved] The problem is certainly NP (Nondeterministic Polynomial), but chances are that it is NP-complete. [Added 09-12-2013: not NP complete] A weaker problem is to find a number b <= n/2 such that *any* b vertex pairs with different diameters can be rotated apart in the n-circle for *any* (even) n. It might be "(n/2)-1", I haven't checked on this yet. Ultimately, one should be able to determine the best b for each individual n (including the odd case). [Added 09-12-2013: this is still wide open. Exhaustive computer search is getting quite demanding, even for fairly low n]
Marcel Van de Vel · VU University Amsterdam
I have finished my paper containing the questions that I collected in my original posts. There are several more questions (and partial answers) in it, which all arose with the development of one major result on a theoretical method to produce unpredictable numbers.
As this paper is intended for publication in a regular journal, I cannot place it it here in public. If anyone is interested in receiving a copy (25 pp), please let me know. I'll send a pdf file by e-mail. The paper classifies mainly as combinatorial mathematics.
How close is spectral partitioning to the solution of the min-cut problem?
There are many approaches using different matrices and eigenvectors to solve the min-cut problem. What is the best theoretical result providing a good approximation from a spectral cut to the solution?
• Donald beverly Giles asked a question:
Is there any interest in a very different spreadsheet algorithm for generating incidence matrices of projective planes?
I have devised an algorithm for generating the incidence matrices of projective geometries not shown in traditional texts. Will include as an attachment. This design is self dual and relies on the 4 mols of order 5.Rather than use M*Mt can use M^2.
Square Root of a Symmetric Matrices
The square root of a 31 by 31 matrix with 6"s down the main diagonal and 1"s elsewhere is a symmetric binary matrix with six 1's in each row and column. If someone has an algorithm for this square root then perhaps they can apply it to a larger matix that Iam presently working on. This larger matrix is an 81 by 81 matrix with 16's down the main diagonal and 3's eveywhere else. The square root of this will be a binary symmetric martix with sixteen 1's in each row and column. For me this is not a simple problem. My knowledge of matrices does not extend to taking square roots of symmetric matrices and getting symmetric binary matrices as the answer.
Christopher Landauer · The Aerospace Corporation
since the given matrices are linear combinations of the identify and the matrix commonly denoted J (the all 1 matrix), the eigenvalues are simple to compute (the eigenvalues of the k x k matrix J are one k and the rest 0), but only one of them is a square (36 in the first 31 x 31 example, 256 in the 81 x 81 example), and the rest are all equal (5 in the first example and 13 in the second) - also, since J . J = kJ, a square root of J is J/sqrt(k) - therefore, it is easily shown that a square root of rI + sJ in k x k matrices can be found having the same form aI+bJ, with

r = a^2, s = 2*a+b^2*k, which is easily solved for a and b (need r >= 0 for a to be real, and s-2*a >= 0 for b to be real) - in the two given examples, r=5, s=1 and r=13, s=3, then second condition does not hold and b will be imaginary

there are, of course, other square roots, as other people pointed out
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
That is very instructive. Are there some references for further readings?
What do the three parameters represent i.e. (44,22,10)?
Was thinking these were parameters of some type of combinatorial design
Bahattin Yildiz · Fatih University
Actually it is [44,22,10] and it describes the parameters of a binary linear code, of length 44, dimension 22 and minimum Hamming weight 10.
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.
Daniel Page · University of Manitoba
I can't recommend a book better than Combinatorial Designs by Douglas Stinson. It's how I learnt Design Theory. It is a little more expensive, but it is full of handy results.

http://www.amazon.com/Combinatorial-Designs-Constructions-Douglas-Stinson/dp/0387954872

Another good book is the Handbook of Combinatorial Designs. My previous supervisor and my current one wrote a pretty interesting bit in there on Lotto Designs.
• Donald beverly Giles asked a question:
Can anyone recommend ways to find a skew starter for a room square of side 667?
Skew room squares exist for all odd values greater than 5. If n is prime it is a simple matter to generate a skew starter. But 667 is not prime. 667=23*29 which means a computer search has to be done in order to generate one. I would be satisfied with the skew room square of side 667, even though we can show it exists we can't seem to construct it. Any suggestions on this particular problem are appreciated.
Why do mathematicians think that the four colour problem cannot be solved theoretically?
I published a proof. If there is any error in my proof, kindly inform me.
Donald beverly Giles · Board of Governors of the Federal Reserve System
This is because Bill Tutte, the man who broke the German Tunny code on his own, spent a great deal of time trying to to prove the four colour theorem. He had a nice short proof for the five colour theorem which was not his proof. He also constructed the Tutte Fragment which ended Heawood's conjecture of trivalent graphs. Any graph can be triangualized.The dual of the triangualized graph is a trivalent graph. Now a Hamiltonian circuit with an even number of vertices can then be constructed and 2 coloured. The remaining third of the vertices can then be 3 coloured This implies that any map can be 4 coloured. Well the Tutte fragment shut this conjecture down flatlly. A strange proof he made was that the averaqe number of colours for all graphs was pi. This meant some were more than pi and some were fewer than pi. I was with Tutte as a student in 74 when he was pursuing this problem. It was a wonderful class on graph theory. Most of the students were professors. Only a few of us were there for a credit.
Why does the four colorability of planar graphs not ensure the non-biplanarity of K_9?
I hope the proof of the four color theorem is sufficient to explain the answer.
Sanjib Kuila · Panskura Banamali College
I have already studied, that paper of Beineke. I also gone through the original works of J Battle et. al., and that of Tutte. In this topic, Harary wrote in his book, "No elegant or even reasonable proof is known."
Proof of the existence of balanced tournament designs?
I'm writing my senior research paper on balanced tournament designs, and I am looking for the proof of the existence of them which is in this paper.
Donald beverly Giles · Board of Governors of the Federal Reserve System
Jennarose: Ron Mullin, Waterloo's first graduate student, proved the existence of room squares for all odd v greater than 5. The last one that needed proving was 257 which was constructed in the orient in the mid 70's. Further to this skew room squares exist for all odd v greater than 5. A skew room square is a room square where exactly one the cells (i,j) or (j,i) is occupied and the other cell is empty.
Don Giles is converting skew room squares of order 4n+3 into symmetric block designs with parameters (4n+3,2n+1,n) which in turn are being converted into Hadamard matrices of order (4n+4). All of these combinatorial structures are interrelated.
Paul Schellenberg developed the room square of order 25 as part of his PhD. He too was a student and then professor at Waterloo in the late 60's and early 70's
Addition and multiplication theorems were used after a collection of smaller room squares had been established by a host of researchers.
Hope this helps you with BALANCED TOURNAMENT DESIGNS>
I can elaborate more on their usage if you wish.