The sigma polynomial of a graph counts the number of distinct sigma k-bracket colorings, where adjacent vertices satisfy specific color sum equations. This counting problem can be solved by translating it into a geometric problem involving inside-out polytopes and hyperplane arrangements, then applying generating functions and the constant term method to compute the number of integer points in the intersection of a box polytope with flats (intersections of hyperplanes). The sigma polynomial is a quasi-polynomial of degree n with quasi-period n, and its generating function can be expressed as a ratio of sums involving binomial coefficients.
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Southeast Asian Series on Mathematics Research (SEASMR) Current Trends in Discrete Mathematics
Added:5 6 7 8 9. So we have the nine registered participants. No. 1 2 3 4 5 6 7. So seven registered participants from India, two registered participants from Indonesia. 1 2 3 4. Four registered participants from Kuwait and uh the rest from the Philippines. So how many all in all registered? So uh 248 uh participants registered for the event. So to those who are not here, so maybe they are uh uh joining uh this event uh through live stream in uh YouTube. Okay. So uh let's go to the universities. So where are uh our participants coming from based on the universities. So let me uh sort it ascending. So uh we have from uh department of education Alejandro E Honda National High School. So we have from Adam Son University. Sorry I cannot uh mention all of the institutions. So only those who are uh many. Okay. Uh so we have from Batanga State University.
So uh we have from Bulakan State University. Welcome. So, Katand Katanduani State University, Kavit State University, Cebu Normal uh University main campus of course from Central Lson State University, Central Bo State University of Agriculture, uh Deasal University, uh Department of Education, uh Gordon College, Holy uh name University, uh Isabella State University, Islamic Institute of the Philippines, Okay. So we have from uh Manuel S. In Verga University Foundation, Mapua University, Mariano Marcos State University, Mindanao State University, uh Montisori at the uh Sagadra Familia National University. Uh is Ortigas uh National University. Clark Das Marinas National University Philippines. Nova is University of Science in Technology, Pampanga State University, Philippine High School for the Arts. So uh welcome Risala National High School. So we have from uh Southern Lon State University. So we have from Tarlak State University, University of Stomas, University of uh Da Gupan, University of Kalokan City, University of Mindanao. So for to our brothers and sisters from Midan. So hello and welcome. Uh University of Northern Philippines, University of Southern Philippines, University of the Philippines Baguio, University of the Philippines, Silan, University of the Philippines, Losanos. Okay. So uh there so we have so many so most are faculty members. Then there are uh graduate students and I think uh undergraduate students. So that's uh uh where uh the participants or the characteristic of uh the participants in this event. So uh okay. So let me go to the next one.
So let's now uh call in uh Dr. uh John Rafael Antalan so the president of mathematical society of the Philippines region 3 to introduce to us uh the speaker. Uh so Dr. Dr. Andalan will also serve as the moderator uh for uh uh the talk. So uh Dr. Andalan so you now have the floor.
>> Uh thank you very much uh sir Melure for welcoming our uh participants. Uh dear esteemed speakers, brilliant researchers, passionate educators, students and fellow mathematics enthusiasts. A very warm welcome to all of you. Good morning to those joining us from here in the Philippines from Indonesia and Vietnam and a very pleasant day to our participants joining from India, Kuwait and other parts of the globe. On behalf of the Mathematical Society of the Philippines region 3 chapter, it is my absolute honor to welcome you to the official launching and inaugural session of the Southeast Asian Series on Mathematics Research or CSMR.
Today marks a beautiful milestone. What started as a vision to foster regional collaboration has brought together over 200 registered minds. Today, united by a shared love for numbers, structures, and discovery. This morning, we dive into the fascinating world of current trends in discrete mathematics.
What makes the inaugural session deeply special is the span of perspectives we are about to witness. We are privileged to feature the work of both a brilliant beginning mathematics researcher and a highly distinguished seasoned mathematics researcher showing us that the future of mathematical discovery relies equally on fresh energy and deep expertise. Before we begin our journey into graph theory and matrix design, I would like to express our deepest gratitude to the incredible team who made today possible. First to our brilliant speakers, Mr. Josh Pashon from Central Lon Pashon Sebastian from Central Lson State University and Dr. Priita Putri from Institute Technology Bandong in Indonesia. Thank you in advance for generously sharing your expertise, your time and your research with us today. Second to our visionary initiator. A special heartfelt thanks to Dr. Jonathan Kalin from the University of the Philippines Diman whose spark of inspiration, leadership, and dedication initiated this entire collaborative activity. And third, to our tireless organizing committee, the MSP Region 3 Board of Trustees, Mom Yoli, Mom Nancy, Sir Mel, Mom Jervy, Sir Joe, Sir Aos, Sir Edward, M Glenda, and again, Sir Jonathan, who have worked tirelessly behind the scenes, upgrading our capacities and ensuring that our community has a seamless space to connect, whether here in our Zoom room or live on YouTube.
Mathematics has a unique way of transcending borders. Looking at our screen today, we see a vibrant tapestry of cultures from Southeast Asia, South Asia to the Middle East, all speaking the same universal language, the language of mathematics. May today's session spark new questions, inspire new collaborations, and remind us all of the elegant beauty of discrete structures.
Once again, welcome to CSMR.
Thank you all for being here and let us have a wonderful mathematically enriching morning. So, thank you very much.
Uh now uh for the second part of my talk uh allow me to introduce our first speaker for today's uh session. It is my distinct pleasure and a matter of great personal pride to introduce our first speaker for this morning's inaugural session. Our speaker is a true homegrown talent of region 3. Hailing from numpan mea and a proud member of our very own mathematical society of the Philippines region 3 chapter. I have had the unique privilege of watching his mathematical journey unfold firsthand. First as his teacher and now I am incredibly proud to call him my colleague. Just two years ago, he graduated from Central Asan State University with a degree Bachelor of Science in Mathematics, earning distinction kumlaude. Driven by a deep passion for the discipline, he pursued advanced studies and just a few weeks ago, less than three weeks to be exact, he successfully finished his master of science in mathematics at the University of the Philippines, Baguio. Today he continues to give back to his alma matter serving our community as an inspiring instructor at Central Lon State University.
His research interest thrive at the beautiful intersection of graph theory, combinatorics and algebra. In his graduate work, he tackled complex structures, exploring inside out polytopes and utilizing advanced counting techniques like generating functions through arrangements of hyper plates. Today he joins us to share a piece of his brilliant work in discrete mathematics with talk titled solving a graph coloring problem using geometry and combinotaurics the sigma polomial of graphs. Ladies and gentlemen, please join me in giving a warm welcome our colleague, friend and speaker, Mr. Josh Pashon Sebastian. Sir Josh, uh, take it away. Thank you, sir.
Yeah. Hello.
So, good morning everyone. So, by the way, let me share my screen.
An Yes sir Joshy. So good morning everyone.
I am Jose and today I'm going to present part of my thesis solving a counting problem using geometry specifically relating to the sigma coloring.
of eight.
So let me introduce myself first. So by the way I am joined.
I graduated yes mathematics in the central sun state university batch 2084 and also I graduated MS mathematics in the university of the Philippines batch 2036.
Yeah. So my field of interest I currently research interest primarily in geometry combinatorics and graph theory. So there are problems in in mathematics that I would like to solve using these three branches of mathematics. So hopefully by the end of this discussion connections and if you want to contact me use my personal email as well as my CLS2 email as well.
So for the contents of my discussion we'll proceed first with the motivation of my work. So here we solve a particular geometric problem and we translate this into a coloring problem theory. We also will be needing the following preliminaries regarding in graph coloring inside poloops and as well as generating functions. Yeah. And later we discuss the constant term of this particular generated function and we apply this in solving the number of integer interior points inside and then we have an analogous result here. We have the theorem one and two and then we apply this in solving sigma polomial problem by the conclusion and the references.
So let me start with let me start with this certation.
Every branch of mathematics enriches the others and together it reveal the beauty of your required discipline. So by the end of this discussion hopefully collections different branches of mathematics and solving a problem in one branch another branch connected.
So let me start with so consider this question. So given a polygon how can we determine the number of its integer inferior points. So when we say integer interior points is uh integer interior points having a integer coefficients.
Let's have a particular sample have this robus with the vertices 0 0 2 42 and 66 respectively.
The green points represent integer boundary points while the red points represent integer integer points of the polygon.
Yeah. So actually we can solve this general problem using a fixed which states that the area of a polygon is equal to the number of interior points plus half of the boundary points minus one.
So ano ba area.
So the area of the lumbus is the product of its diagonal lengths divided by two.
And when solving this we have area of 12.
And the number of boundary point is actually as you can see points and using the fix theorem we can conclude that there are exactly 11 integer integer. you have nine such forms. Yeah. So let's consider another problem. So here we have a polygon B which is a cartician product of the closed interval 1 to 4 + 1 to 4. And then we define the line X - Y = Z.
And here we want to determine the number of integer interior points in Z but not in the line. Yeah. So we have these 10 points.
So and as we can see there are exactly six such points and we claim that each of these points actually correspond to a proper vertx coloring using one two and plus colors. Yeah. So from a geometric perspective we have a graph coloring perspective.
Yeah. So hopefully we are all motivated in my work. So let's proceed to the preliminary section of my study.
So we're going to discuss some concepts in graph coloring polytops as well as in the genative function.
Let's start. So a vertx coloring is an assignment of colors wherein you can have same color to adjacent vertices of the graph. And if vertx coloring uses existing color we say that it is a kvertex coloring. Okay, coloring of graph G. Now we have an analogous for this. So a bracket coloring. So we define a new terminology. So this a bracket coloring uses the colors one two up to K.
So between these two coloring.
So when we say t coloring it uses kic colors and it does not need to be ordered. So for example if k is equal to 3 like for example 1 2 and 3 4 5 1 7 colors while in the bracket coloring it uses the colors using the subset of this. So 1 2 up to a. So if t is equal to 3 Another is either one, two and three. So one of two one or two and also it does not need to use all colors in one two up to three. Yeah. Also we will use the concept of the thickness of two colorings.
So we say that okay bracket coloring C and C prime are distinct if you can find a vertx U such that C of U is not equal to C prime of U. Yeah. So let's have an example for this. So take a look in this graph. So it is a star graph with five and we assign a color one here 1 5 3 and two. And as we can see we uses four distinct colors 1 three four and five.
So this is a four coloring of this particular graph. And since we are using a subset of one two up to five colors.
So it is actually a five bracket coloring of y. So you can see the distinction between these two coloring.
And then next a sigma coloring assigns a colors to which every adjacent vertices satisfy this particular equation. So the color sum of you is not equal to the color sum of B. And for the discussion regarding on the color sum, sigma coloring and sigma continuity, sigma range and sigma values. So please refer to the article of chart rank and aomodo titled sigma chromatic number of. So I will give this link for that later.
Yeah. And now we define a new concept sigma polomial of graph. So this particular counting function counts the number of distinct sigma k bracket coloring of graph g. So take note k bracket coloring. Yeah. And at the same time a sigma following.
So we have these two condition.
So actually our main goal in this discussion is to determine the number of stick sigma k bracket coloring or the sigma polomial of some families of graphs including the fat graphs cycle graphs even the multipartite graphs and circulent graphs as well.
So in doing this we need some terminologies on the polytopes.
Yeah. So a polytope in Rn of dimension n is given by this set. So it is a collection of points in Rn satisfying this inequality. So here a is the coefficient matrix and then b is the constant vector. So A is an element is an M dimension M byN dimensional matrix with entries while B here is an M dimensional vector with the autoates and we discussed this particular polytope the box polytope BKM in Rn is a set of points in Rn whose components are between 1 and three and so let's take a look of in different dimension. So in dimension one the box polk one is actually a closed interval.
In dimension two the box pol of bk2 is a partition product of two closed intervals and in general the box pol of bkn is actually an n dimensional hyper cube inside length of n of k minus one.
So moving on we have this the hyper plane in Rn is an n minus one dimensional subspace of the following form. So it is a collection of points satisfying this equality where a here is an n dimensional vector with real entries. So take let's take a look what is a hyper plane in different dimension.
So by the way in dimension one hyper plane actually is just a zero zero satisfy ax equals zero in R1. But in R2, an igopane actually describes a line passing through the point zero commute zero. And in RT, a hyper plane is actually a plane passing through the point 0 0 and zero.
Yeah.
Next, an arrangement of hyper plane is just a collection of hyper plane. So take note that finite finite collection of hyper plane. And to form the intersection latis we simply take the subset of the arrangement and then we take the intersection of it. So elements is a intersection of hyperase which we call as flat. So the elements of the intersection latis flats.
Next to form the intersection semilattis. So take note it is a subset of the intersection latis is a collection of dots whose intersection with the polytope is an empty and then this is the concept of instead of polytop. So here we just create a pair a pair of an dimensional polytope and an arrangement of hyper.
So also we'll be using this particular counting function relating to the inside out polytope. So given an inside out polytope this circle and strip A. So here to I is the found the interior points is the interior of the polytop.
So the open enumerator of this set alpha actually counts the number of elements in the disc set intersection with the integer of P minus the union of P sweep.
So this particular counting function actually counts the number of interior of P intersection B that does not lie in any higher plate.
Yeah. So if you go back to our example of B and hypoten in R2 is actually st. So this is hyperane in R2.
Now we can actually write this problem using the concept of open numerators. So here we have this inside out polytop the interior of B and strip A where strip A contains the hypotain.
And then if we cut all interior points of B excluding those lying in any hypothes.
So now from now on if you see this type of problem we can use the open emulator to simplify and to use symbols.
Yeah. So actually we can apply or use the concept of mo function to better solve this problem. So actually we can write the open enumerator of instead of look like this. So it is a sum of the product of the mod function of zero hat f times the number of elements in this set the intersection of a discrete set the integer of b and this f for the sum is taken over all elements of the intersection similar and for the concept of mo function and as diagram so kindly refer to the article So slabs the convex sorry the inside outs diagram for the mo function of every and also we'll be using this theorem the constant term of aative function. So this this actually counts the number of integer points and to use this.
So first from a polytope. So we have a rational polytope meaning is a polytope with rational vertices and then we write it in this manner.
It is a set of points satisfying this matrix inequality.
So we have this coefficient matrix and this is a uh constant vector. Now from this coefficient matrix the column vectors actually is the C subi. So C1, C2 and CD are the column vectors of this matrix and B here is just the um constant vector. And once you do that you take the constant term of this generative function and it will give you the number of integer points in P.
So since we are dealing with generative function so let me have some recall or some concepts in generative function.
So by the way a generative function encodes sequence into a formal power series. So from a sequence you have a particular object with formal power series wherein for every term in the sequence actually corresponds to a coefficient in this sequence in this formal power series. So the formal power series of this sequence is actually the sum of a subn x to the n where nges from 0 to infinity and also the constant term of the generating function is just the coefficient a sub0 and we will use the notation co of g of x. So let's have some well-known genetic function. So for this sequence the sequence of one one one one. So using generating function we can write it in this manner. So it is a formal power series whose coefficients are all one. And using some property of the geometric series you can write it in this way 1 / 1 /x.
So for this particular sequence 1 2 and 3. So you can write it as a generating function 1 all over 1 - x^2 and for this sequence of triangular numbers we have this generating function 1 all over 1 - x and in general we can observe this particular sequence of n + k - 1 k or k is equal to 0 1 to up to infinity corresponds to this generating function 1 all over 1 - x to n.
So later we will be using this.
So moving on we present the main result of this discussion. So we have three main theorems of four data. So the first two one is the analog of the oilless generating function. Next in theorem two we apply this theorem in solving the sigma polomial blocks and theorem three and four we show the generating or s general form of the sigma polomial and as well as the generating function.
So for this first student it's actually the analog of the ear generating function. So what we what does what actually is this counts the number of integer points in this set the intersection of the box polytope and the flat s by taking the constant term of this expression.
So we know that a flat f is intersection of hypotenics. So you can write it in this manner.
And once you get the coefficient matrix and then the constant vector.
So the column vectors of this matrix A is actually this piece of C1 CB and then the column vector B.
So once you get the constant term of this we will get the integer codes in this particular set. And take note the column vectors here must not be must not be identical to the zero vector.
So next in theorem two sorry practical applications of the theorem one in solving the sigma polomial problem. Nice. So what we have here? So given a graph G simple graph with the following vertx set and edge set and we consider an arbitrary a bracket color C and we will define the arrangement of hydroplane as follows. So for every hydro plate we define at it as the color sum of vi1 minus the color sum of vi2 equals zero. For every adjacent vertices v i1 and vi.
So if you have let's say m edges you have m hyper as well. Yeah. And after that you can actually count the number of st sigma k bracket coloring of any graph using this formula.
So take note theorem one and we just take the product of the mo function and this constant and we take the sum over all flats in the intersection semi.
So actually there is a general method on computing single polar. This utilizes the one and the two. So let's have this.
So in step one for a graph G, we like to determine the vertx set edge set and the set of neighbors in this graph. So by the way, what is a set of neighbors? So actually it is just a collection of adjacent vertices to vertx P.
Then after that in step two we define this inside out polytope the box polytop BKB and the script A.
And for every adjacent vertices adjacent vertices in the graph we define a hyper plane sigma of U minus sigma of B equals zero.
So by the way a color sum is just a sum of the colors in the neighborhood of a vertex C.
Next in step three we construct the intersection semianis it's as diagram and for every flat f we compute for this two the cardinality of the box polytope intersection this split set z to intersection with n and then we compute for the function of z hat and f and then in this chapter we actually compute sigma polomial using theorem two. Yeah. So let's have a particular example for this.
So here we have a graph before with the following vertx set and edge set. So from here actually we can actually conclude the set of neighbors and set of the following vertices. So for example v_sub_1 is adjacent to v_sub2. V2 is adjacent to both v1 and v3.
V3 is adjacent to V2 and V4 and V4 is adjacent to V3. So we have the following set of neighborhood for every vertices.
And then in step two we form this inside out polytops and we define the following hyperane depending on the adjacency of the graph.
So we have the following and then we let x sub is equal to the color of the vertex bi and then we have p this hyperase. So actually these are all equivalent hyperase. So we just use another way to write it. Yeah. And then after that we form the intersection semilotis.
So by definition an intersection semilattis is a collection of flats such that its intersection with the polotope is not empty. And if once we do that we have the following plots of hate H1 H2 H3 intersection of H1 and H2 H1 and H3 h2 and H3 and intersection of H1 H2 and H3.
We have the following with the minimum element is equal to R4. So question so why is the minimum element is actually ambient space. So actually in the intersection semilatis it is partially ordered by the reverse inclusion.
So meaning the minimum element is actually the largest among them. It is the object case and for the h diagram. So we have the following actually as diagram intersection. So as we can see the minimum element is in the bottom and the maximum element is to the top.
So this function one one and one.
So then after that we compute for the following. We want to determine the number of elements in the intersection of the box poly the split set n plot for every flat in the intersection semi. So let's consider let's say f is equal to the hyper plane h1.
So you need to write this plot as a matrix equation and once you do that the column vector of the coefficient matrix will be written like this. So me one and one. So actually we apply only the one here and if you simplify this you will get the constant term of this is actually k^ 2 - k a cub - k^ 2 all over 2.
So actually I already solve it for every plot in the intersection.
setting our values and and finally we apply theorem two. So we substitute the values of this to solve the sigma polomial of t4.
So therefore we have this. So we have k to the 4 - 5 * k + k ^ 2 - 1/3 k which is the single polomial of the foil. But question how can we use this? So actually if you want to determine the number of stick sigma for bracket coloring or key just plug in K is equal to 4 and the value.
So this actually works for every values of K. So even 100 1,00 poly.
So I actually have some results. So for the sigma polomial of P3, P4, P5 and P6 are the following. I already calculated the generating function.
So by the way for solving the generating function, we uses a technique using change of basis. So for basis of 1 k^ squ up to k cube here we use different bases to come up with this generating function.
Yeah. Also we have the following results for the sigma polomial of cytographs C3 C4 C5 and C6. And as we can see the sigma polomial for P5 is actually a pace function.
So the polomial depends on the parity of k. So if k is odd it's the sigma polomial for p5 and for k is even this is the sigma polomial.
Yeah with the following generative function. So actually if k if p is equal to 6 and we have a top graph with six vertices we observe this sigma polomials.
So we have a different single polinomial depending on the parity.
So this actually demonstrate the uh polomial behavior of our sigma polomial.
Yes. So we have this theorem number three. So this is the general form of our sigma polomial. is actually a quasi polomial of dbn with a quasi period of n. So we have n the parity.
Yeah. And take note that each of the components are periodic functions in. So the value of this coefficients depends on a.
Yeah. So we have this fourth theorem the general form of the generative function of sigma polomial which is actually equal to this ratio the ratio of the sum of b sub i x n + n - 1 - i where the sum is taken from i= 0 to n + n - 1 / 1 - x n + 1 So we have other solves in sigma polinomial. So here we connect the sigma polinomial to other existing problems such as the how can we determine the nigma continuous graphs sometimes.
So by the way here is the graphs graph invariance property of our single polinomial that is for every if two graphs for every isomeorphic graphs G1 and G2 we observe that they will have a same sigma polomial and also have this characterization of the sigma range of the graph which is actually equal to the minimum value of K such that the change of p is greater than zero for every k= 1 2 and so on.
Also if a graph g satisfy this particular inequality we say that g is a is not a sigma continuous graph.
So it's actually a way to determine which graphs are non continuous.
And finally for a simple graph G1 and G2 the sigma polomial of the graph union of G1 and G2 is actually this product of their individual sigma polomials.
Yeah. So we apply also a graph operation here.
So for the conclusion establish in this discussion.
So first we show theorem one and two again one is the way to count the number of integer points in the box polytope intersection with the plot and we apply the one in solving the sigma polomial of a graph.
Also we present a method on solving single polinomial groups using theorem one and two.
And also in theorem three and four we show the general form of the single polomials and its generating function.
And finally we show the relationship between sigma polomial sigma continuity and also we show the graph inverance property of the sigma polomial.
and also apply operation to graph.
So actually problem in this topic. So first we can actually count or find the sigma polinomial of wig graph and plot. So actually if I desert in my thesis I actually counted or computed for the sigma polomial of the complete graph as well the some families of complete multipartite graphs as well as other families of circular drops. So open wheel and graphs types of graphs also you can apply other operations as well such as the graph join graphs fitting graph as well as the partition product of drops.
So third partial I already created a program to find the sigma polinomial only for fat graph and cycle graph. So for any general graph so please state this for any graph and also using airhart theory make another method on computing sigma polinomial and finally so let's extend the idea of sigma polomial to the sigma symmetric polomials of so for symmetric polomials. So available internet.
So for the references here are the following references I mean they use this introduction to algebraic graph theory and this computing in continuous discretely as well as this the inside out polytops and the the sigma chromatic number of graph.
Yeah. So I hope you learned something from this lecture. So thank you for listening.
>> So uh thank you very much uh sir Josh for the very insightful uh presentation.
Uh thank you sir. Now uh at this point uh the virtual floor is open for uh your questions. uh due to time constraint as sers and moms are dear participants we may cater only to uh questions and if you have further questions you may email uh sir Josh via the email uh he presented earlier.
So while waiting for uh question you may uh type your question now. Uh dear participants uh while waiting for uh other question I prepared one question for you uh sir Josh >> uh >> in your presentation sir you use constant term of a certain specific multi- variable generating function and the concept of inside out polytope in determining the sigma polomial of p and cycle but take note for a certain uh order three up to six for pat and three up to five for the cycle.
However in your uh abstract you use another method in determining uh the sigma polomial of pads and cycle for higher order. My main question is I am just wondering uh the motivation for deviating from the first method to the second method with regard to uh higher order pots and cycles.
>> So actually about that I already created a method for higher order of graphs. So for graph let's say P6 P7 up to P12. So for that I created a new method. So actually it involves the concept of the neighborhood or the difference the difference matrix of a graph. So using that we can actually compute for the quasi periodicity quasi period of a and once you do that you can actually compute for higher order.
So due to time constraint so discuss thank you very much sir Josh. Uh another question sir uh is uh being raised actually by a certain uh individual that is present here in this zoom meeting. Uh the question is can the method be expanded into other uh type of graph coloring?
Uh about that sir I think sir is just how can you describe this type of problem using this sigma or the inside out poly so translation depend >> okay thank you very much uh sir gosh >> possible possible sir >> thank you sir uh uh having uh no more questions in the chat chat box. Uh again, uh thank you very much, sir Josh.
And I'll give floor back. Thank you, sir Josh. Floor back to our moderator. Uh to our uh uh to our host, uh Sir Melor Kupatan.
>> Okay. Uh hello. Uh thank you very much uh sir Jr. and uh thank you very much for our speaker uh uh uh Mr. Jo Sebastian. our part-time faculty member here at the department of mathematics and physics. Thank you for uh your effort and time. So uh based on the program uh we're going to have a kind of community break for 10 minutes. So in this community break I'll be presenting the MSP region 3 membership promotion.
So I'll start sharing my screen. So I watched earlier uh a crash course on how to share a slide uh in uh in Zoom via uh in Zoom uh via YouTube. So let me uh try doing it because you know I'm not uh uh used to sharing my slides uh here in uh Zoom. So but I think mom Jeremy already have it. Mom Jeremy, can you present my slides or Yeah, let me uh try presenting my slides.
So, okay, it's better. Mom Jeremy, could you present my slides because again I am having difficulty there. Okay, thank you very much. So I'm not used to uh press uh sharing my uh slides via Zoom. So what I'm used to is sharing my slides through Google Meet. So I I apologize.
So uh uh I'll do this for uh 5 minutes only. So uh so my short talk is about MSP region 3 uh membership promotion. So next slide please.
Okay. Thank you. So uh uh first what is mathematical society of the Philippines for uh uh a few students who are not aware of uh mathematical society of the Philippines. Mathematical society of the Philippines is the Philippines uh premier professional organization uh dedicated to the promotion of mathematics research in education in the country. Founded in 1973, it has grown from a small Manila based group of math educators to a nationwide network of individuals with chapters in the three island groups, Luson, Visayas, Mindanao.
So, uh, region 3 is the one of the youngest chapter. The MSP has represented the country in the international mathematical union since uh 1978. Next slide, please.
So, what are the objectives of the society? So, there are three objectives.
So promote interest and awareness of mathematics and its applications. Uh propagate knowledge in mathematics and mathematics education and third promote mathematical research. Next slide please. So what are the benefits of becoming a member of the society? So eligibility for a possible uh research grant of 100,000. I think this one is for graduate students taking up master or PhD. So discount in the registration fee in the national and regional convention of course of the mathematical society of the Philippines. Automatic membership in the Southeast Asian mathematical society. Okay. Possible source of points for promotion and inclusion of the memberships in the curriculum v. So yeah I think uh this would make your curriculum BT look better.
So if you have this item so example so in the uh right hand so you have research grant there. So under eligibility what's the number one applicant must be a regular member of MSP of good standing. So there so it's a being a member will make you eligible uh for a possible research grant. So on the let's on the left hand side so on the right hand side if you see there uh so the one in the red uh box so MSP member 5,600 non-member so 6,600. So there's a discount if you are member of MSP. if you joined the national and regional uh national convention. Next slide please.
So uh that one is a photo of uh the resident 2026 MSP national convention.
So there so we are a big happy family.
Okay, next slide please. So uh this one is uh our photo of uh the recent uh convention and general assembly of MSP region 3. So held here in Central State University. So that that one is our first ever uh face-to-face uh national convention. Next slide please.
Okay. So uh what is the eligibility for the membership of the society? So how could you be eligible? So uh those who have taken and passed calculus uh in college and taught at least a senior high school mathematics subject or engage in research on mathematics its applications or its teaching may apply for regular membership in the organization. So there so I think it's clear and self-explanatory. So how about for undergraduate students?
Undergraduate students in their senior year or graduate students enrolled in graduate programs of mathematics or allied fields may apply for associate membership. So there are two types of membership regular and associate. Okay.
So next slide please. So when applying for MSP membership applicants may choose to be affiliated to the MSP uh regional chapter in the geographical region of their residence. If there is none, uh they may apply at a neighboring MSP original chapter of their preference.
Okay, next slide, please. So, how do you become an MSP member?
So, uh yeah, three steps. So, go to the Facebook page of the mathematical society of the Philippines. Look for a feature post about the MSP uh membership application. And third, uh click the link and fill out the application form. So, very briefly, so let me walk you through how to do it. So next slide please.
So there so that is the MSP uh mathematical site of the Philippines uh Facebook page. So uh at the okay uh the third uh the third one. So inside the red uh box. So you have there the featured post on uh membership. So you have to click it once you click it. So next slide please.
Uh yeah you will get uh into uh this one.
So important announcement. So MSP uh uh membership application. So there are two links there. So uh new member and renewal. Of course if you are renewing I think you already know the process.
That's uh for new member you have to uh click the link. So uh the link it's safe. It's not scam like it will get uh your bank information. Not like that.
Okay. It's safe. Okay. Next slide please. So uh if you click the link so uh it will lead you into uh this one. So MSP member application form. So there and uh you fill out uh some of the information. So next slide please. So for more information so visit the for for more information about the medical society of the Philippines uh you may visit the homepage of mathematical society of the Philippines. So that's the link. Then Facebook page of the mathematical society of the Philippines.
That's the link. Next slide please.
Okay. Uh we have also Facebook page of the mathematical society of the Philippines region 3. So that's our uh chapter. Then uh yeah uh YouTube channel mathematical society of the Philippines region 3 uh where uh this uh event via Zoom meeting is currently being live streamed. So next slide please. So that's all I have to say and uh thank you very much for your time.
Yeah, that's six minutes. Yeah.
So uh thank you very much for your time.
So uh sir uh sir Jr. uh are do you want to go to the next uh talk?
>> Yes sir we may proceed now sir. Thank you sir.
>> Okay. So uh so with the permission of our president. So let's uh now proceed uh to the next talk. So uh our speaker will be introduced by uh Dr. Jonathan Khalim. So uh he Dr. Jonathan Khalen is an associate professor of mathematics at the University of the Philippines Dimman. So he will be introducing our speaker uh yeah Dr. Priita Putri. So uh sir Jonathan uh I yeah I you have the floor. Thank you very much.
>> Okay. Uh good morning everyone. It is my great privilege to introduce our next speaker Dr. Pria Putri.
Dr. Putri is an associate professor at the faculty of mathematics and natural sciences of the institute technology band in Indonesia. Dr. Putri specializes in combinatorial mathematics with research interests spanning design theory specially had matrices.
She earned her doctoral degree in mathematics from To<unk>hoku University in Japan. Also, Dr. Putri holds double master's degrees from Kazawa University in Japan and from Institute Technology Band in Indonesia. She has presented her research at international conferences in Japan and a and Australia and actively contributes to the Indonesian Mathematical Society and the Indonesian Combinatoric Society. Beyond research, she has coordinated academic programs, summer schools, and student development initiatives at ITB and was recognized with the Hitachi scholarship award in 2013 and the Die ITB teaching award in 2025.
Friends, let us all welcome Dr. Putri.
>> Okay. Hello everyone.
Thank you very much Dr. Nathan for the very nice introduction for my CV. Uh actually uh firstly I would like to uh give uh uh yeah gratitude to this the committee to uh invite me for this uh nice uh oper nice event and actually uh today I will uh like Dr. Natan uh introduced uh I will share actually some uh in some information or some topics related with the hadart matrices. So actually this hadart matrices is one of my uh topic of research from master and until now but um but uh there's uh less people uh knowing about this hadart matrices. So in this opportunity I would like to share uh about matrices uh not in the detail but actually a brief history. So hopefully this uh topic can be a nice uh topic to be interesting for uh the researcher actually for the uh elementary level. So uh I will I will uh share my uh slides.
Is it uh okay?
Is it the slides clear? Okay. Thank you very much uh Dr. Natan. So I will begin uh this uh talk with the title classical and modern methods for constructing had matrices. Classical means it's uh around it the hadamat matrices constructed in around 80 uh 18th century but the modern methods here means recently. So not uh modern uh in computition way uh with the uh you using computer or quantum computing but in theoretical way but uh for the recent work. So what is hadamart matrices? So madamat matrices is actually just a plus minus uh plus minus one matrix of order n satisfying uh this condition h * hrpose is equal to n * i.
So this hard matrix uh always have the square mat always square matrix and uh pair wise it means that always pairwise orthogonal. This is one of the or uh one of the had matrix of order for one of example. So you may see that uh they are all only plus minus one uh only have the element of plus minus one.
So actually uh we have the hyamart conjecttor. This is uh still an open problem. So theamart matrix of order 4k exist for every positive integer k. So this is actually equivalent to attaining the maximum possible determinant for uh plus minus matrix you may see. So this equality hold.
Now I will introduce uh the uh who is the first time uh at uh obtaining the first matrix of the construction of hadamart matrix. So actually had not the first one. It was uh Joseph Sylvester in 1867 who first infest who first invent this construction. So to extend the the order of the hadamart matrices they use uh the method called uh chroner product.
Hopefully yeah everyone know about the tensor or chronicer uh constructions. So the first uh you have uh for example uh one only one element and then for the next one next order you may just uh do chronic product. So you may have two power tok matrix. So this is called silfester had matrix and uh it always uh always uh the order always the power of two. So this one is the first one the sylvester but then uh had uh studied general problem. So he uh he uh observed how large can the this one uh the absolute uh of the determinant m before a matrix m with the elements is in plus minus one. So actually uh there is there is uamart inequality the famous hadamart inequality it said that uh this uh equation this inequation hold and uh with the equality uh hold if and only if uh every entry is plus minus one and the rows are orthogonal.
Uh so he exhibit example of order 12 and 20. So 12 and 20 is actually not the powers of two. So this uh showing new order is actually possible.
So after uh Sylvester and also Adam uh there uh there was a pelle who uh also tried to construct the hadamart matrices. Well uh because the hadamat is still uh still a conjecture. So some of our researcher trying to approach to obtain this had the order of hard matrices by using many methods. So uh Sylvester using the chronicer product and now Pelle used the quadratic residue in a uh the infinite field fields uh GFQ with the Q is an odd power prime. So this is uh actually always be an example in the class of number theory. So uh this uh using the quadratic character kai and also Jacob style matrix. So if you interest in the number theory actually these hadamart matrices also can be approached using some of the uh number theory.
Uh there are the first pal and also the second pal. So they uh construct using uh using the quadratic residue. But the second pallet actually using the uh another class of matrix called conference matrices. I will introduce it later.
So uh small order that not reachable by silfester which is which is the chroner one and the pal is 92. So uh you may see that this pal is 1933 and let's see uh how long it will be take until we found uh n until they found the 92.
So this is the one of example of the pal type one. So you may see if Q is uh each module three Q is equivalent to three module 4 is a prime power then uh there exist the matrix of order Q + 1. So this is how to construct the matrix. So you may uh make a Q + one square matrix order of Q + one using quadratic residue model Q. For example, uh in uh like in the slide, you may have q is equal to three because three is uh is uh 3 modulo 4 and then the order will be q + 1. So this is one of the example of the hadamat matrix by using pal construction.
For the second construction uh you may use you may uh use a conference matrix.
So let's see the definition first. Uh uh sorry uh I mean the theorem this is not a definition a theorem mention that if Q is equivalent to one module four is a prime power then there exist a hadamar matrix of order t 2 * q + 1 but uh to do this construction you need to know the conference matrix. So simply say that converse matrix is actually similar with the uh with the hadamart matrix the property of the hadamart matrix which is pairwise orthogonal. But in this case uh we have the diagonal entries are all zero and then uh then if you multiply that matrix with its transpose then it is uh it will be uh equal to n minus one time identity matrix. In hadart case we have n * i but in conference matrix because the diagonal entries are all zero then we will have n minus one * i.
So by using uh conference matrix we can have the pal construction of type two.
Uh so this is one of the example of uh pale construction. So first we can uh make the conference matrix of order q + one using quadratic residue and then we can extend it to hadamar matrix of order 2 * q + 1. For example, if q is equal to 5 then uh because five is congrent to one module four and then we can have the order is uh 2 * q + 1. So 2 * 5 + 1 is equal to 12.
So this is uh the so then we can have the construction of the pal construction of the hadamart matrix uh will be like this. So you can just plug the conference matrix and also the identity matrix as follows. Then you may have the hard matrix of order 12 where the C is just a 6 * 6 poly conference matrix and I is the identity matrix of order six.
This uh this kind of construction uh may uh may known as the plug-in method because we just pluck a matrix uh another type of matrix to uh to this uh much bigger matrix. So we uh also uh say also call this method like the pluck in plug in uh method.
So uh but uh like I mentioned before Sylvester and Pale cannot obtain the smallest order right uh in that moment which is uh n is equal to 92. So uh some of the researcher uh begin to uh searching a way to searching the way to find this order 92 and then uh Bert Gum uh Hall use a computer search by using Williamson uh Williamson construction.
So actually uh the help of the computation uh was the construction of the Williamson uh which is uh John Williamson uh in 1944.
So let's see first uh what Williamson uh make the first uh so he find a way uh so uh we can have four symmet four symmetric circular and plus minus one matrix. So he f they so he found found that this a b c d matrix of order m satisfying this equation.
So there exist uh this class of matrices and then uh combine with the four * 4 block array uh to build this hadamart matrices of order for m. So this 4 m * i i means identity matrix. So then we can have had matrix of order for m. So this construction actually uh this construction uh actually uh give uh give the contribution to uh to find the hard matrix of order 92.
So this is the nice construction of the Williamson matrices. So like like I said before a b c d is just four matrix four plus minus one square matrices they circular and also symmetric uh they uh the one that is important that this uh this equality should be hold. So by using the computation uh we can uh we can find this uh kind of uh matrices and then uh just plug this a b c d to this matrix h then you may have the hadamart matrix of order form. Once again this is uh this is uh plug-in method and by using this Williamson and also a hard computation they found this four * matrices but uh this is actually uh not cover uh any of uh order why because uh this uh matrices a b c d is limited not all of m exist in this in in this uh way.
So I will give this is one of the example of Williamson um Williamson method. For example, we have A B C D as follows. We can see that uh so the the um the excite the exciting matrices is the order of prime number. Right? So we begin with the m is equal to 3 and then we can find uh hadamart matrix with williamson method of order uh 12.
uh so uh Williamson mentions I will be back to the uh slides symmetric and circulan uh was one uh of the important role in this case but let's see uh bal in 1965 so from uh so to find 92 it's have the gap around 30 years and now uh in 1965 Balmer Hal uh find the uh actually find a generalization of the Williamson method.
Sorry.
So uh they mentioned that just uh just use A B C D satisfying the same orthogonality condition. So only this condition and also uh plug the these matrices with this uh this uh kind of uh Williamson Williamson construction then we can have hard matrix. So they uh they make a generalization of the Williamson uh construction.
So uh this is one of the method uh they find a b c d in uh such way and they produce adamart matrix of order eight.
So we can see they are not I'm sorry uh they are not uh circle and matrix but actually they can u produce the hadamart matrix.
So uh so what I have introduced is all related with the plug-in method. So plug in means that you have a possibility to find hammer matrices by uh obtaining some small matrices with such a property and then you plug into the bigger matrix. But actually there are another way uh for those who like uh algebraic approach. Tin construct the uh construction from the bal units of four symbol sequence and also complex had matrix. So complex had matrix is actually have the same property. If you have a a complex had matrix and you multiply it with its transpose then it will be equal to n * i but the difference is that the elements of the hadam h sorry for this matrix is complex numbers. So with the root of unity it will be equal to one. So this will be uh there is a way to find uh find a headart matrix from this kind of uh head uh this kind of complex uh matrix but I will not further discuss in this session uh but uh you may find in the some of the research and also if you interest in different sets and cycltomy and this will be also uh one of the one of the approach approach that you may uh obtain the hadamart matrices. So construction from different sets and also pallet type difference family uh also uh have the possibility to find the hadamart matrices orthogonal design products uh they use the generalized tensor construction combining the hadamart matrices of order k and m into one of order like k time uh k * m. So the general tensor uh actually you may begin with the chroner product and then you may uh do some a generalization to find another kind of a chroner product to find the uh other had mattresses.
And then uh Seiri and Yamada in 1992 they also uh they make a survey actually a comprehensive survey uh using Williamson turin gretel sidel construction.
So uh closing uh closing the largest uh historical gap is actually the order is uh not so big. It is actually 428.
So uh for decades uh n is equal to 428 was the smallest multiple of four uh for which no had matrix known.
In 2005 uh ty rei uh finally construct a hadamar matrix of order 428 using complimentary sequence combined with ex uh extensive computer search. uh I need to mention that uh this uh uh in this talk uh this method is actually need the help of the computer search. So uh we can uh we can reduce the time of the computation by using some of property in uh in the theoretical way.
So one one uh way uh that obtained by Caragani and Typheri was using the complimentary sequences.
So uh actually this complimentary sequences uh uh is one uh one topic that I really interest. So uh I will uh begin to uh introduce how to construct the hard matrices from these complimentary sequences. So complimentary sequences actually have a lot of class uh you may see like in graph theory there are many of class of graph this also the similar one so the complimentary sequences uh has the class of go sequences it's called glay sequences b sequence and etc. I will uh I will introduce only some of them that is related with uh my research.
So first uh I will introduce what is actually the complimentary sequence. uh before uh we go to the uh go deep into the definition of a complimentary sequence I will introduce you to the whole polomial let's say f of the factor a let r be a commitative ring with identity and involative automorphism star and a be a factor with the length l and we define the half polomial uh as follows so we define as the summation of the ai x power to i. Uh so this is just a usual polomial actually and let this notation be a ring of lam polomials over r. In this case we define infolive automorphism as follows. Uh so they just mapping a i to ai star where this the star means the infolium automorphism and also x to xar to minus i. So we will use this lang polomials uh and this is the definition of complimentary sequences. So we begin with the n sequences a1 up to a n uh be a sequence with all entries in r. Then we can say that this uh a1 up to a n. So this is the set of the vectors uh will be uh uh sorry set of the sequence will be a comp uh is called complimentary sequence if this condition hold. So if we uh do the summation of the of f ai time f ai star then uh we will have uh they will be just a scalar uh in r. So there is no other uh poloms uh with xar to something. So they all are in r. If this hold then we can have the complimentary uh sequences.
So note that this a1 up to a n do not necessarily have the same length. For example this uh we have four sequences and you may see that they are not uh they are not have the same length.
So now how to construct the complimentary sequences uh how to construct matrix from complimentary sequence let's begin with the smallest uh only just two sequence uh so let's say a and b and then uh we say uh if this a and b are complmentary sequences then uh we call it uh go sequences uh so this aab will be uh goal is called Gallay sequences and the notation of goal sequences is G S K which is K means the length of the sequences.
So let A and uh so for example let A and B be circular matrices with the first row a and b uh respectively then so you first you uh you construct A and B. So the first uh the first um the first uh row is the goal uh is obtained from the goal sequence. You make a circular matrix then you may have the hard matrix of order two or two 2 * n sorry 2 * n with this construction but please note that go sequences is limited so just only the order only have this order. So of course they not uh they not cover all of the alamar matrix orders.
So we will go to the next class. So the next one we have four uh sequences.
Now is uh now with the element 0 and plus minus one. Of course you may uh give 0 and plus - one not only plus minus one sequence then we can say that these uh four sequences will be uh complimentary sequences if uh in this case t sequences of length n if uh they are mutually disjoint complimentary sequences. So uh in this case we notate this set of sequences uh be with tn t means the uh means the t uh t sequences.
So for example you may see this uh t sequences of order seven. So they have the length seven 03 means 0 0. So you may on and 04 means zero four times. So you may have t sequence order seven we are we interest with the uh prime number. So t7 is a prime number. So this is good. And then from these sequences we can obtain much bigger uh matrix with this kind of black in where this r means uh one in diagonal diagonal entries one the reverse diagonal is have the entries one and the others are zero and then by using this green matrix you may have four * n. So now n uh t sequence of order n uh will be interesting because they can obtain 4 * n matrix of hard matrix directly. So this uh construction is actually a nice construction.
If we begin with the uh complimentary sequences, it called goal sadels arrays.
Uh for the uh I think this one is the last uh the last uh class that I will introduce uh today. It's called base sequences. So we have sequence, we have t sequence and also base sequences. Base sequences uh also four sequences but they have uh pair different order. So the first order have the length m and the second sorry the first pair have the length m and the second pair have the length n but they should plus minus one sequences which is different with the t sequence and then uh together they uh they have a complimentary property then uh it's called base a class of base sequences we can u write is as B S M comma N. So we have T N and also B S M comma N. This is an example of the base sequence of order 3 comma 4 because they have the length three and also uh length four. It's uh usually we write the smallest one and then the larger one.
So like I uh mentioned before 428 is uh the important order right now because this one is uh the last order that they find the smallest one is 428.
It was find found by Caragani and also Ty resi in 2005. how they uh how they find these uh sequences actually begin with the tin type sequences. I need to mention that this touring type sequences is actually complimementaryary sequences and then from uh you may see there are four four sequences and they make a base sequences and then they construct the t sequence from base sequences and then from uh t sequences we can obtain a halamart matrix. So this uh this uh this uh method actually uh found by number uh by theoretical way. So not by using a computer they using in the theoretical way and then uh they proved that there exist such construction.
So uh because so recently uh this method uh to find the find the hardark matrix from complimentary sequence actually still on research until now. So this is actually uh one of the I may say modern method because it's just recently until now still on research. So first I will uh introduce uh the method it's called young multiplication for complimentary sequence. Basically uh the idea is to combine two class of uh complimentary sequences and then you may have the target is you find the t sequence because if you have t of order n then you may have the hard matrix of four * n right so now the target is to find the t sequence not the hadart matrix so to find this t sequence uh there are many method uh young uh young in 1983 It's also uh actually this one is uh yeah uh much older but Young found it by using uh direct construction. So by using direct construction so he find the exact exact sequence of order 2 s + 1 * m + n and then uh the next one he find direct construction of 13 * m + n where this mn is our order of base sequences. So then we try to generalize the generalize the uh construction of young by using uh uh so we find we try to find so how it work then we begin uh with this uh definition.
So we so uh not using f but uh not using lan polinomial but we define this c of a like as follows. So what is uh the different with the f? Well, the different is now we have xar to minus something. If we can easily see that there are x to minus something while the uh the usual f lan polomial there is no x to minus thing. Well actually this seems uh obvious but actually for the computation it is uh much uh reduce the computation way. So we define this psy a as follows and uh so we compute the complimentariness using s rather than using f.
So actually uh why uh why this s is important because we want to use uh rather than sequence we want to uh extend it to a matrix.
So for the for so for in the matrix actually the computation is much more easier. So now let's uh let's a be a r * r power to m * n matrix. So it's matrix of order m * n and then uh we define pi of a x comma y as follows where this pi uh of a i means uh from the last definition. So there is a pi for sequence and pi for the uh matrix. So we define in this way and then we recall the famous identity is called LRA's identity uh LRA identity uh mentioned that if we have A B C D like uh like in a slide is in R and then we construct QR ST as follows then we can have uh this identity. Why this identity is important? because uh it is very nice if we have uh Q Q uh QB uh complimentary sequences and then A B C D also complimentary sequences E F G H also complimentary sequences then we can have uh for example if we have this one uh is base sequences then we can find another sequences so we can have like four times something or or m time something. So this is uh nice if we have primes times primes then we can have primes times prime prime prime prime time prime time four which is uh the existence of the hadamart matrix. So our strategy goes this way. So first I uh we try to uh so this is a joint work with professor masa fromoku university uh in that moment. So we try to construct the qrst they are matrix and then uh we construct in this way and finally we can uh extend this lrang identity uh I mean I use uh lrang identity to find this thing. So uh this uh this equity holds means that this matrix QRST uh can be uh can uh be equal to this this uh two two uh term of the uh multiply things uh multiplication things.
So we may focus on uh on uh four or maybe prime complimentary sequence but a prime numbers and also complimentary sequence with a prime numbers.
Uh but uh recently uh actually uh uh around 2010 Yokovic uh find uh define uh this kind of thing. So uh first they uh he uh he mentioned normal sequence. Normal sequences is actually three uh sequences of order uh three sequences with the element zero and plus minus one and we have such property. So bc are disjoint and a is plus is plus - one. So it means a is plus -1 and bc is 0 and plus -1 sequence and they have together a complimentariness then it's called normal sequence and also a near normal sequence is actually a base sequences but with uh some kind of property that mentioned in here.
So uh well basically near normal is a base sequences and then uh Jookovic mentioned that uh defined a young number because it's actually developed from the young result young number is an odd integer 2 s + one such that uh the near sequence or near normal sequence is not empty. So they find uh so in 2010 u if y be an odd uh if y is an odd positive integer and y is smaller than 73 then y is a young number if uh y is not equal to this number. So 35 is actually the the smallest in that moment. Uh I think until now uh this smallest uh number uh which is near sequence and near normal sequence is not exist is empty. Why this number is important?
Uh because uh we have this one is 2 s + one. So this 2 s + one can be attached in here in here. So if you have 2 s+ one so 2 s + one is odd numbers and then you may uh you may uh apply base sequence in here.
So you apply base sequence in here. Then you may have 2 s + 1's * m + n which is uh the order is quite interesting from uh to obtain 2 s + 1 * m + n * 4 which is the order of hadamar matrices.
So until now uh actually the this is the summary of the timeline. Uh we uh uh I have just introduced Sylvester Hardam.
Uh had introduced the determinant of the bound pal uh Williamson Balmer Gollum Hal Bart Hal which which is using a Williamson type matrices and also uh so in 1970 they find the way to obtain had matrix from the complex uh had complex had like I said before means the matrix have the uh element in complex number in 2005 they find uh 428 the order of hadamart matric is 428 so what is what the next order the smallest order is 668 668 remains open so we still uh don't know know how to use it and how to find it and uh hopefully uh from from the complimentary sequence. Uh there is a way to find and uh actually uh not uh not uh focus on the on the construction we can also uh do the research uh to find the as asytotic result. So uh two in 2012 uh Ivan Levinsky proved an asytoic existence theorem for hadamart matrix.
So he find that uh for any order any odd integer p there exist had matrix of order 2 * t * p where p is odd. uh and uh and this is uh quite interesting for all t with uh is which is bounded by this number.
So the uh interest interesting uh things is that his construction actually using uh lot of algebraic number theory. He use a sign groups the so the sign groups uh is uh one of the class of the group and have the uh have the same properties and it combine with a zero autocorrelation sequence. we call it comp uh I mean this is complimentary sequences. So he uh he combined the sign groups and also complimentary sequences.
Then uh he find the uh the this uh bound boundary this. So uh yeah this com uh this method is actually uh quite nice and this extend the classical research by Pelle and also Sylvester uh providing new infinite uh families of hard matrices. I see uh uh I see in his uh thesis the open problems uh of uh to extend this result actually using pure number theory. So I think if you're interested in number theory it will be nice to see deeper in this uh research.
So uh this is inequality that I mentioned in the first slide. So this uh what the contribution of this lefinsky result. So Lefinsky actually confirms that had matrix exist for infinitely infinitely many orders of n. Therefore uh the bound of n power to n / two is achie uh achie achievable infinitely often. So this is actually strength the asytoicidence for the hadamat conjecttor. So rather than find the construction if you uh interest in the uh asytoic result then it also can be possible because until now this uh lefinsky result is the newest one until now and because this is using the sign group which is also uh introduced by uh Craig uh I think if you also uh interest in the algebraic number theory. Uh this one is also one topic that may you see that you may see.
So last uh what is the application for hadamart matrix? So if you're interest to use the hadamart matrix actually uh coding theory use a lot of hadamart matrix. So for example had codes error correcting codes you if you know a rit mer in inside of the rit mer you may have had matrix recently my student have a research on compression compression also using hadamar matrices so there are a lot of uh application in coding theory in statistics we use uh optimal design for experiments for examp uh for example y weighing design uh in sino processing we have walsh had transform uh and also cdmma sequences in quantum information uh we use a mutually unbiased basis mutually unbiased basis can be approached uh in in theoretical way also for complex had matrix and uh and many more and had gate so for uh as I think this is the my last slide Right.
I will update that uh yesterday actually there is a talk from the uh quantum uh computing professor which mentioned that uh the order of mathematics of 668 actually uh is now on the way to search by using quantum computing and it might be a pair in 2029.
So let's see if it is exist or not.
Okay, this will end uh my uh talk. This is my reference. Thank you very much for your time.
Uh thank you very much uh professor Pria for your wonderful uh presentation. So we need to wait until 2029 to know the answer to yeah 600.
>> Let's see. Let's see. Let's see. Let's see if the computer can search or it is possibly say yes. know or they don't know, right?
>> So, let's see.
>> So, there's a question in the chat uh about application, but you already mentioned uh some applications. Uh so the participants are invited to ask questions about the talk and maybe you can also ask uh because there are many students you can ask professor pria uh like things about studying abroad or >> okay >> doing research in bandong etc. So let me be the uh let me ask the first question.
So >> you explained this uh process of starting with a tin type uh sequence and then it became uh a base sequence and then finally a t sequence. Will this all way to to obtain a hadamar matrix? Will this always work if you start with a tin type uh sequence?
>> Okay, thank you very much for the questions.
uh uh let me share the slides that related with uh professor Natan's uh questions. So from here >> yes >> yes yes so if there exist this uh the first tin type sequence uh so uh the only way is you need just to find the existence of the order of these twoin type sequences. If there exist then this always work to this will lead always to hard matrices. But the problem is that is there exist for any order of this tin type sequence then it's also a conjecttor.
So tin type sequence means you have four sequences uh of order 36 three sequences of order 36 and also the other is 35.
And then with this uh with this uh way we can have had matrix. Why uh this why they using terin type? Well actually in that moment 428 was the open problem. So many people try to find a way well actually uh some using another method but this uh sequence tin type sequence uh finally works to obtain the hard matrix. Yes.
And actually still a conjectctor for all if you find all uh all order of type sequence then you you finally uh answer the conjectctor.
>> And of course for people doing combinatorics probably the most combinatorial approach will be to study complimentary sequences.
Um there are many ways. So >> yes. So uh for example uh this one is uh the this uh these are methods to obtain had matrix. So for example if um yes for example uh professor Anatan uh like to do a number theory then number theory can be a cont contribute in the different sets in cyclottomy. So they also uh trying to find not by using the uh con to make a direct construction but using a different sets and also uh pale the pale was the one uh using quadratic residue.
So this one also can be uh one that you may approach using uh number three theory.
Okay. Thank you. Are there questions from the audience maybe from students or researchers?
>> Um well if I may ask another question.
So it's actually not only um um quantum computers that are helping uh mathematical research. Now there's also AI.
>> So >> I think yes I don't know but maybe let's see let's see 2029 if they publish a paper. Well actually for the order of 92 is uh approached by using similar way. Yes we know that in that moment there is no AI but uh using the hard computer uh search and they obtain 92. So it so if you want to uh try to find the 668 or another next uh smallest order maybe uh we need to you know do a collaboration. So we can approach by using in mathematical uh theory approach and uh simply simplify the properties and uh then using the computing quantum computing but maybe because there is an AI well I I have no comment on that one.
>> Okay.
>> Thank you. Thank you very much.
>> Do we have other questions?
Uh yes uh professor manure please go ahead.
>> Uh yeah I think uh what uh the participants are interested maybe partly or a little is that uh uh Dr. Dr. Pa are you available for uh research collaboration like uh could we request you to become a thesis co-advisor of uh uh uh research of our undergraduate students or master students something like that >> is that okay are are you open okay >> yes very okay very welcome actually I'm very happy to to be collaborated if there are any because uh like I uh mentioned before this uh yeah this topics is actually not much known. So just only a few people that really uh want to re do the research in this hadart matrix. So very very welcome to uh be one of the collaborator of this research >> and actually I have this background is uh ITB. So ITB is uh in Bandung Indonesia in west and west Java. Uh this is the one of the maybe it's near uh nearly like in the Philippines we have uh buildings with the green much lot of green area. So please don't worry if you want to visit or for maybe exchange to uh Indonesia to ITB the uh environment is similar so but uh maybe new friends uh can be uh find in this way. So this is the campus my campus in IPB maybe uh this that I want to promote that one.
>> Okay. Maybe one more question before we go.
>> Thank you very much for the question.
Yes.
>> Okay. Uh well uh it seems there are no questions. Uh so maybe they can um message or email you later. So um uh we will now be moving to the next part of the program but again we would like to thank uh professor priita for a very wonderful presentation.
>> Thank you very much for the invitation.
Thank you very much everyone.
>> Okay uh thank you very much uh uh uh Dr. Pria uh Putri for uh your hello.
So, uh sorry, let me uh get my Hello. Okay. So, uh my uh AirPods is better in picking up my voice. Okay. So, uh thank you very much uh Dr. Priita Putri for sharing your expertise on had matrices and for Dr. Jonathan Kalim for uh introducing Dr. Priita and moderating the question and answer. So we're now down to uh the last part of our program which is uh closing ceremonies. So let's uh award the certificates uh to our guest. So I was told that someone will uh show the certificates uh share the certificates on the screen.
So there so uh let me uh read the citation. So mathematical society of the Philippines region 3 certificate of appreciation. This certificate is proudly presented to uh Jos uh P.
Sebastian, Central State University, Philippines. In grateful recognition and sincere appreciation for serving as a research person in the Southeast Asian series on mathematical research with the theme current trends in discrete mathematics organized by the mathematical site of the Philippines MSP region 3 chapter held on uh July 18, 2020 uh no 2026 via Zoom. So signed Nancy Mati and uh secretary MSP regentry and Dr. Jaf John Rafael Antalan president MSP regentry signed. So uh we're going to send uh the certificate with actual signature. So we haven't included it there for data privacy purposes. So uh next uh certificate.
So uh so with the same citation this certificate of appreciation is presented to uh Pitta uh Utri. So Institute Technology Bandong Indonesia. So in grateful uh recognition and sincere appreciation for serving as a resource person uh in the Southeast Asian uh series on mathematical research with the theme current trends and discrete mathematics organized by the mathematical site of the Philippines MSP region 3 chapter held on July 18, 2026 via Zoom signed Nancy Elmati secretary MSP region 3 chapter and uh signed uh John Rafael M. Antalan so president MSP region 3 chapter. So uh thank you very much uh for our speakers for uh your effort and preparing your slides for uh the uh today's uh event.
So uh uh John uh no Josh P. Sebastian and Dr. Petra uh Putri. Okay. Thank you very much uh for your uh uh time and effort in sharing your expertise.
Okay. Yeah. Uh photo op. So uh if it's uh possible on your end, so let's have a photo op of photo opportunity.
Yeah. By the way, I wasn't able to mention earlier. So my my my poll I I I I bought this from Indonesia. Good.
Okay. So uh let's do it. So how many slides do we have? So uh by the way, update on the number of participants uh at the start. So we had 50. So we now have 114. Yeah. Thank you very much uh for coming and joining uh that uh the event for today >> sir. Uh you for the evaluation as well sir kindly remind everyone sir. Thank you sir.
>> Yeah for the evaluation.
Yeah. So uh mom Yolanda Roberto posted the evaluation form link in the uh group chat. So kindly uh fill out the evaluation form link. So once you fill it out uh I think you will get the uh you will get your certificate. If not automatically I think it's automatic.
So if not uh just wait for a few days but I think it's automatic. So there uh you have also the QR code of the evaluation form and the MSP membership uh on your screen. So you could uh take a photo of it or snapshot or something like that. Okay.
To uh uh fill out the evaluation form and possibly join uh MSP for those who are in the Philippines. I think abroad also. Yeah, it's possible.
Okay.
So uh can can uh can we now have the photo opportunity?
>> Yes, sir. Thank you, sir.
Oh yeah.
So I'll be the one uh taking the screenshot. So how many pages is this one?
Okay.
So uh first uh first slide or window.
Yeah. Again, if it's possible, you can open your camera on your head and smile.
Okay, that's first window. Second window.
Okay, next window.
Okay, the next one. There are so many.
Okay, next one.
Okay, next one.
Okay, one more.
Okay.
So, uh there uh thank you very much uh for uh attending uh this uh this day's event. So, uh for the closing remarks and reminder, so may I give the floor to uh uh Dr. uh John Rafael M. Antalan uh the president of MSP region 3 sir Jr.
for the closing remarks.
>> Thank you very much uh sir mur. So again to our invited speakers Dr. Putri and uh sir Josh thank you very much and to everyone present here today and with that po uh stay tuned po for the upcoming events of uh the MSP region 3. Uh we will we will be posting our future events in our Facebook page. With that, take care always.
Integral.
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