Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” Thailand – Japan Students Science Fair 2023 “Seeding Innovations Through Fostering Thailand-Japan Youth Friendship” 20 – 23 December 2023
Thailand – Japan Student Science Fair 2023 Princess Chulabhorn Science High School are proud to organize the Thailand - Japan Student Science Fair 2023 (TJ-SSF 2023) on December 19 to 23, 2023 at Princess Chulabhorn Science High School Loei. This collaborative international event is under the motto of TJ- SSF 2023 “ Seeding Innovations Through Fostering Thailand - Japan Youth Friendship” which will be graciously preside by her Royal Highness Maha Chakri Sirindhorn who has been the driving force in the promotion of education to all Thai youths, especially in learning of science, mathematics, and technology. The goal of this Thailand - Japan Student Science Fair 2023 is to bring together groups of talented students in science and mathematics, from Thailand and Japan to share and exchange their research finding and to build closer and stronger collaboration between the two countries. There will be 36 schools from Thailand 18 schools from Super Science High School and 12 college from KOSEN participating in TJ-SSF 2023. The Thailand Japan - Student Science Fair 2023 will be a cordial bilateral religious relation between Japan and Thailand. The event will not only provide a platform for the exchange in scientific knowledge between these like-minded young scientists but also will act as springboard for sustainable development and promotion of many more 21st century skills which range from communication, collaboration to long lasting friendships among youths. Thank you all for your participating in on this auspicious cooperation. We hope you will have pleasant stay in Loei, Thailand
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 1 Table of Contents Congratulatory Message from Minister of Education of Thailand 2 Congratulatory Message from the Ambassador of Japan to Thailand 3 Congratulatory Message from Deputy Minister of Education of Thailand 4 Congratulatory Message from Secretary General of Office of The Basic Education Commission 5 Keynote Speaker 6 Contents 10 Contributors for TJ-SSF 2023 79 Contact Persons of TJ-SSF 2023 80
2 Congratulatory Message from Minister of Education of Thailand I would like to extend my heartfelt congratulations to everyone concerned for their efforts in organising the Thailand - Japan student science fair 2023 (TJ-SSF 2023 ). Building on the successes of previous years, the 2023 event will not only foster creativity, innovation and science awareness but bring out two countries even closer through future-oriented collaborative projects. Since there is establishment in 1993, the Princess Chulabhorn Science High Schools, which nature outstanding scientific achievement in Thailand, have been proactive in developing cooperation and building relationships with many Super Science High Schools, and National Institutes of Technology or KOSEN in Japan, with support from other academic agencies. The shared goal is to enhance the teaching and learning of science, mathematics and technology among gifted students, and to gear curricula towards addressing the challenges of the 21st century. Participating students and teachers to the Science Fair benefit from having opportunities to showcase their talents and express their potential in science through project presentations. Since science and technology are the key driving forces for economic and social development, cooperation among new generations from our respective countries can only enhance the prosperity and well-being of our societies. On behalf of the Thai government, I would like to reaffirm our continuing commitment and support for inspirational events of this kind. I would like to offer my sincere thanks to everyone who has contributed to making this year’s Science Fair a success. My special thanks go to the Embassy of Japan, Japan’s Ministry of Education, Culture, Sports, Science and Technology (MEXT), Japan Science and Technology Agency (JST), participating Super Science High Schools (SSHs) and KOSEN Institutes, as well as the Japan International Cooperation Agency (JICA), and the Japan Foundation. I am confident that the continued collaboration between our two countries will lead to educational sustainability for generations to come. Police General Permpoon Chidchob Minister of Education of Thailand
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 3 Congratulatory Message from the Ambassador of Japan to Thailand On behalf of the Government of Japan, it is a great pleasure to see the Thailand-Japan StudentScience Fair 2023 (TJ-SSF 2023) again that can confirm our close and strong bond between Thailand and Japan. This is a little pace of science and technology education. It is obvious that we have already made significant progress within the students and teachers gathering from Super Science High Schools of Japan, KOSEN Institutes of Japan and Princess Chulabhorn Science Schools of Thailand in this event, Thailand –Japan Student Science Fair 2023. I would like to congratulate all students from both Japan and Thailand for their creative capacities with their science projects. I sincerely believe that these new generation students will play a vital role to develop their countries in the future. I hope this TJ-SSF 2023, which is based on cooperative work and new ways of diffusing knowledge by using science and technology methods, will further motivate both Japanese and Thai students to expand their golden opportunities in their higher education and contribute to the development of relevant fields. Finally, may I express my appreciation to an excellent arrangement and the hospitality of the Thailand - Japan Student Science Fair 2023 (TJ-SSF 2023) at Princess Chulabhorn Science High School Loei. Last but not least, I would like to express my great respect to all concerned with this great event. H.E Mr. NASHIDA Kazuya Ambassador of Japan to Thailand
4 Congratulatory Message from Deputy Minister of Education of Thailand It’s my great pleasure to praise the event of Thailand – Japan Student Science Fair 2023 (TJ-SSF 2023) between Japan and the Princess Chulabhorn Science High Schools. I would like to express my deepest admiration for the contribution to the collaboration between Thailand and Japan which promotes the fostering of the next young generation of students through various activities. The cooperation programs between the Princess Chulabhorn Science High Schools, the Super Science High Schools and KOSEN in Japan show how much of an achievement they are and how much they empower students of both countries. Science is one of the most important channels of knowledge. It has a specific role, as well as a variety of functions for the benefit of our society: creating new knowledge, improving education, and increasing the quality of our lives. Thailand – Japan Student Science Fair 2023 ( TJ- SSF 2023) ensures collaborative teaching and learning with emphasis on students’ science projects. The event has helped strengthen student’s scientific and technological competences. The activities included in the event encourage students to be creative, analytical and critical thinkers who will ultimately contribute significantly to both countries. The students were fortunate enough to participate in various activities supported by MEXT, JST, SSHs, EOJ, KOSEN Institutes, JICA, the Japan Foundation and other Japanese agencies. I whole heartily wish Thailand –Japan Student Science Fair 2023 will continue to flourish for many more years to come and make our two nations more progressively advanced in the future. Mr. Surasak Phancharoenworakul Deputy Minister of Education of Thailand
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 5 Congratulatory Message from Secretary General of Office of The Basic Education Commission On behalf of the General of Office of Basic Education commission, it is an honor to extend my sincere congratulations for Thailand –Japan Student Science Fair 2023 (TJ-SSF 2023). I am pleased to see the cooperation mature into something above and beyond our expectations. The number of students and teachers in both our countries often comment on the unique experiences and scientific knowledge they gained. It was not easy for all Princess Chulabhorn Science High Schools to maintain and sustain their vision. However, with the perseverance of the students, who represent the younger generation, I am confident they will improve their creative, analytical, and critical thinking skills for the 21st century which include achieving a sustainable future in line with the Sustainable Development Goals. It is that potential future and realization of the SDGs that makes me really appreciate all the cooperation of Thailand – Japan Student Science Fair over three times of the collaboration between the Princess Chulabhorn Science High Schools and the Super Science High Schools of Japan and National Institute of Technology (KOSEN). Students gained valuable experiences related to their projects on science. The event also provides opportunities for students and teachers to perform their potential on science; therefore, students have the chance to present their projects on science. I believe that this memorable science fair will greatly inspire our students’ passion for science, provide valuable learning opportunities for both our students, promote cross-cultural exchange, enhance language skills, and cultivate lifelong friendships between our students. Besides, “ The 2nd Thailand- Japan Educational Leaders Symposium: Science Education for Sustainability (TJ-ELS 2023)” is a stage for teachers to exchange their knowledge, experiences and share their best practice that is beneficial for students. All of these activities have not only led to the goal of the collaboration program but also created a deep relationship and network amongst students, teachers, schools and our two nations. All the support and opportunities are created by academic agencies in Japan, providing a great chance for our students to explore the scientific world and learn Japanese culture. Once again, I would like to take the opportunity to wish the academic cooperation continued success in the future for many more brilliant milestones to come. Acting Sub Lt. Thanu Vongjinda Secretary-General of the Vocational Education Commission Acting for Secretary-General of the Basic Education commission
6 Keynote Speaker Dr. TOMOYUKI Naito Vice President of Kobe Institute of Computing (KIC) Tomoyuki (Tomo) Naito is Vice President and Professor at Graduate School of Information Technology, Kobe Institute of Computing, Japan. In his over 25 years of professional career, he has been working with clients on digital economy acceleration policy and strategy formulation as well as its implementation for effective development; in particular ICT use leapfrogging practice in developing countries. His professional interests include digital economy, distance learning, ICT innovation ecosystem, Internet of Things, FabLabs, Mobile Big Data solution and other related areas. Prior to assuming his current position as a graduate school professor, he was Senior Advisor in charge of ICT and Innovation for the development field at Japan International Cooperation Agency ( JICA) . Previously, he was Program Manager at the World Bank in charge of the Tokyo Development Learning Center, Director of Planning as well as Director of Transportation and ICT at JICA headquarters. He is serving for several public advisory committees as designated member including Global Steering Committee of “Internet for All” project at the World Economic Forum (2016- 2019) , Regional Governing Committee of the Global Development Learning Network AsiaPacific (2011–2021), Global Strategy Working Group under the Minister for Internal Affairs and Communications of the Government of Japan (2018–2019), and others. His recent paper “Redefining the Smart City for Sustainable Development” is contained in the Brookings Institution’s book “Breakthrough (2021) . ”The paper “Role of ICT in education redefined by COVID-19” is contained in the book “SDGs and International Contribution under the Pandemic Era (2021: in Japanese). ” Another paper “Indispensable ICT for achieving SDGs” is contained in the book “International Contribution and Realization of SDGs (2019: in Japanese).” He is also a registered 1st class architect in Japan since 1997. He holds a Master of Arts in international relations degree from Graduate School of Asia-Pacific Studies, Waseda University, Japan
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 7 Keynote Speaker Prof. Dr. Nonglak Meethong Khon Kaen University Contact information : E-mail : [email protected] Phone : 043-009700 Ext. 50660 Education: • B Sc. (Honors) in Ceramic Engineering, Alfred University, New York, USA • Ph D. in Structural and Environmental Materials, Massachusetts Institute of Technology, Massachusetts, USA Work Experiences: • Director, Battery and New Energy Science and Technology Factory, Khon Kaen University, Thailand (2022-present) • Chair of the Battery and New Energy Science Program, Department of Physics, Faculty of Science, Khon Kaen University, Thailand (2022-present) • Vice President & Committee of the Thailand Energy Storage Technology Association (TESTA) https://www.testa.or.th/ (2021- present) • Professor, Department of Physics, Faculty of Science, Khon Kaen University, Thailand (since March 2023) • Associate Professor, Department of Physics, Faculty of Science, Khon Kaen University, Thailand (2020-present) • Assistant Professor, Department of Physics, Faculty of Science, Khon Kaen University, Thailand (2013-2019) • Chair of the Materials Science and Nanotechnology Program, Department of Physics, Faculty of Science, Khon Kaen University, Thailand (2014-2017) • Member of several national level committees for the Thai government and advisory boards of several leading energy related companies in Thailand. Selected Honors and Awards: • National Innovation Award 2023 in the field of Society and Environment (2023) • WIIPA Grand Prize for Commercial Potential Award, Kaohsiung International Invention & Design Expo (2022) • Special Recognition award, National Research Council of Thailand (2020) • Thailand New Gen Innovators Award (2020) • Thailand Young Scientist Award from Foundation for the Promotion of Science and Technology Under the Patronage of His Majesty the King (2015) • Excellent Ph.D. Thesis Award from National Research Council of Thailand (2014) • Winner, Recipient of the H.R.H Princess Maha Chakri Sirindhorn Cup from the 4th National Competition on Innovative Nanotechnology (2012) • Special Recognition award, Thai Rice Foundation Under the Patronage of His Majesty the King and National Innovation Association of Thailand (2014) Research Expertise: • Synthesis of energy storage materials • Characterizations of nanomaterials by advanced techniques Pilot scale production of materials for batteries
8 Keynote Speaker Dr. Hattapark Dejakaisaya Princess Srisavangavadhana College of Medicine Contact information : E-mail : [email protected] Phone : 061-115-5991 • Work Experiences February 2022 – Present: Princess SrisavangavadhanaCollege of Medicine Position:Neuroscientist,Lecturer May - August 2017: Chulabhorn Research Institute Position:Short-termResearcher Projecttitle: Sesame and cancer Supervisors: A/Prof. Jutamaad Satayavivad & Dr. Nuchanart Rangkadilok July - August 2014: Chulabhorn Research Institute Position:TraineeResearcherin Laboratory ofPharmacology • Qualifications 2018 – 2021: PhDinNeuroscienceMonash University, Melbourne, Australia Supervisors:Prof.PatrickKwan&A/Prof.NigelJones Thesistitle:The role of glutamate in the pathogenesis of epilepsy in Alzheimer’s disease Main techniques: • Advancedanimal handling and euthanisation skills(mouse) • Mouse brainsurgery, electrode implantation, kindling-induced seizure&EEGrecording • Experience inmetabolomic analysisusingliquidchromatography-massspectrometry • Trainedinvariousmolecular biologytechniques(westernblot,immunohistochemistry) 2015 - 2016: Masters byResearchinBiologicalSciences Universityof Leeds, Leeds, United Kingdom Thesistitle: MolecularPharmacology ofthe Slo2.2PotassiumChannel Supervisors:Dr.Jonathan Lippiat&Dr.StephenMuench *Graduated with Distinction/FirstClassHonour* Main techniques: • Trainedintwo-electrodevoltage clamps electrophysiology techniques • Plasmidconstruction andpointmutation insertion • Polymerase chain reaction 2011-2015: Bachelor of Science in PharmacologyUniversity ofLeeds,Leeds, UK *Graduated with 2:1/Second Class Honour* 2007-2011: HarrowInternationalSchool,Bangkok
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 9 • Achievements 2021 • Data-Blitz oralpresentation at‘EpilepsyMelbourneSymposium2021’, Melbourne 2020 • Poster presented at‘EpilepsySociety ofAustraliaAnnualScientificMeeting 2020’, Hobart • Poster presented at‘Epilepsy MelbourneSymposium 2020’, Melbourne 2019 • Won‘People’sChoiceaward’fromMonash UniversityCentralClinicalSchool‘3 Minute Thesis’ competition, Melbourne. • Poster presented at‘EpilepsySociety of AustraliaAnnualScientificMeeting 2019’, Sydney • Oral presentation at‘Student ofBrain ResearchSymposium2019’, Melbourne • Poster presentation at‘CNSDiseases 2019’, Melbourne • Competed in the ‘ 2019 Translational Research Symposium Poster Competition’ , Melbourne 2018 • Recipient of‘TheScholarshipinCommemorationofHMKingBhumibol Adulyadej’s 90th BirthdayAnniversary’fromChulabhornRoyalAcademy,Bangkok,Thailand • Position of Responsibility 2022– Present: 1. Course coordinator of “Mechanism of drug action” for year 3 medicine 2. Course coordinator of “Neuroscience & Behaviour” for year 3 medicine 3. Student affairs committee member
10 Contents Section Title page MA-01T Mathematical model between pH and PO4 3- 11 MA-02T Comparison of Durian Prices Forecast Model by Statistical Forecasting Methods and create a website for durian prices. 15 MA-03T Approximation of Euler’s Number by using Geometry and Probability 21 MA-04T Tautology calculates association by using IVA Diagram applied from Venn-Euler Diagram with Set Theory 26 MA-05T Studying the relationship between the remainder of integer divided by 13 and the digits of integer 30 MA-06T Finding the probability of winning Tic Tac Toe 3x3 grid 34 MA-07T A study of cryptography using mathematical method 38 MA-08T A Forecasting Model for the Number of Establishments Registered with The Social Security Fund 42 MA-09T A New Design of Auxetic Structure Using the Centroid of Linear and Non-Linear Shapes and the Golden Spiral for the New Design of the Helmet to Decrease the Chance of Injury from an Accident 45 MA-10T A study of geometric transformations in conjunction with the GSP program for the design of Pa Daeng weaving patterns 49 MA-11T The number of polygons in an equilateral triangle n units 53 MA-12T The Parallel Parking Trajectory Equation 59 MA-13T The Linear Programming for Produce Charcoal Solid Fuel Stick 64 MA-14J The Pi in Origami 68 MA-15J Solving 3D Puzzles Using Tensors Calculation 70 MA-16J Flow lines for housework efficiently 74 MA-18T Subset with Counting Rules 78
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 11 Nattapat Singsuphan1 SuphakonChiangkew1 Tanongkiat Polnchaiya2 MontriTorvattanabun3 1,2Princess Chulabhorn Science HighSchool Loei, That, ChiangKhan District, Loei Province 42110, Thailand Abstract The purpose of this AppliedMathematics Project on Mathematical Modeling of the Relationship Between Phosphate pH is to 1) utilize the information obtained of degree 2-4 polynomials used to determine the pH 2) to develop a mathematical model of the relationship between pH and phosphate 3) to serve as a guideline for future research on soil pH adjustment.The charge balance equation produced the result. We begin with Ferrari's fourth power equation. to determine the worth of [3 +] by providing [3 +] = x In summary, the findings of the phosphate pH relationship investigation produce the following equations: pH = −log + √ 2 − 4(1)(− − 1 ) 2(1) when + 21 = , = , + 1 2 = + 1 [34 ] = 12 [34 ] = 123 [34 ] = Mathematical model between pH and − The pH equation above is merely one value that the authors have summarized because it is a simple format. If we can deduce the link between [34 ] [4 3−] from the reaction equation The stoichiometry allows us to compute the amount of phosphate in grams. from the study of contents on acids, bases, chemical equilibrium, geology, and finding roots of degree 2-4 polynomials, to the study of mathematical modeling between pH and phosphate levels. This model will serve as a reference for future soil pH modification research. Key word: Phosphoric acid, phosphates, acids, bases, chemical equilibrium, geology, and polynomials of the second, third, and fourth degrees. Introduction At the moment, there is a problem with soil deterioration caused by long-term land use and a lack of soil repair. The soil in the center region is very acidic, with a pH less than 4.6. the northeast, the south, and the east The pH level of the topsoil was determined using a map. It revealed that the majority of the severely acidic areas were discovered in acid sulfate soils, showing that the acidic conditions in acidic soils maintained. It is a major contributor to decreasing soil fertility, lower output potential, and adverse agricultural production conditionssoil deterioration since they have physical, chemical, and biological properties that are unsuitable for plant growth, and pH is a crucial element since it controls many soil parameters that influence plant growth but do not effect plant growth. demonstrating the soil's fertility The optimal pH range for most plants is between 5.5 and 7.0, so the soil must be modified by fertilizing to boost plant nutrients in the soil. By fertilizing, the major nutrient is nitrogen, phosphorus (P), or potassium, which increases plant nutrients in the soil. Nitrogen, phosphorus (P), or potassium are the most important nutrients. Put the highlight picture of your project in this area.
12 Phosphoric acid is a type of acid that is employed in a variety of industries. Whether it's in laundry detergent, soap, or even soft drinks and other beverages. Phosphates A phosphate is a chemical anion, salt, functional group, or ester produced from phosphoric acid. In most cases, it refers to orthophosphate. Among numerous phosphates, the synthesis of orthophosphoric acid (34) phosphate is highly essential. Because of its significance in biochemistry, biogeochemistry, ecology, and agriculture and industry. As a result, the organizing team questioned if there was any way to apply understanding of acid-base balance, chemistry, and geology to solve the problem of acidic soil conditions. We used the previously provided knowledge to maths. in the mathematical modeling of pH and phosphate Materials and Methods modeling steps 1. Write the dissociation constant equation using the water-phosphoric acid reaction equation. 2. Enter the charge balance equation with the dissociation constant. 3.Construct the pH equation using the quadratic, third, and fourth equations. 4. Analyze the performance and document the results of the analysis.c 5. Compile the material from the summary and deliver the findings. Results and Discussion mathematical modeling results The charge balance equation can be written as follows: [3 +] = [−] + [24 −] + 2[4 2−] + 3[4 3−] Let [34 −] = 1[34] [3+] ___(1) [4 2−] = 2[24 −] [3+] ___(2) [4 3−] = 3 [24 2−] [3+] ___(3) [−] = K [3+] ___(4) put (1) in (2) [4 2−] = 12[34] [3+] 2 ___(5) put (5) in (3) [4 3−] = 123[34] [3+] 3 ___(6) Will get [3 +] = [−] + [24 −] + 2[4 2−] + 3[4 3−] ___(7) Put (1), (4), (5), (6) in (7) Get [3 +] = [3+] + 1 [34 ] [3+] + 12 [34 ] [32] 2 + 123 [34 ] [3+] 3 multiply [3 +] 3 [3 +] 4 = [3 +] 2 + 1 [34 ][3 +] 2 + 12 [34 ][3 +] + 123 [34 ] [3 +] 4 = [ + 1 [34 ]][3 +] 2 + 12 [34 ][3 +] + 123 [34 ] Using the fourth power polynomial equation chemical equation equilibrium constan Equation 34 + 2 ↔ 24 − + 3 + 1 = [24 −][3 +] [34 ] (1) 24 − + 2 ↔ 4 2− + 3 + 2 = [4 2−][3 +] [24 −] (2) 4 2− + 2 ↔ 4 3− + 3 + 3 = [4 3−][3 +] [24 2−] (3) 22 ↔ 3 + + − K= [3 +][−] (4)
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 13 4 + 2 2 + 2 = [ + 1 [34 ] + 2] 2 + 12 [34 ] + 123 [34 ] + 2 instead of coefficients + 1 [34 ] = 12 [34 ] = 123 [34 ] = Form equation ( 2 + ) 2 [12 [34 ]] 2 = 4[ + 1 [34 ] + 2][123 [34 ] + 2 ] Will get 2 = 4( + 2)( + 2 ) 2 = 4( + 2 = 2 + 2 3 ) 0 = 8 3 + 4 2 + 8 + (4 − 2 ) from Cardan's theorem common cube 3 + 2 + + = 0 = − 2 32 = 192−16 2 192 and = 2 3 27 3 + 27 3 + = 64 4 216 + 32 1536 + 4 − 2 8 1 = √− 2 ± √ 3 27 + 2 4 3 + √− 2 ± √ 3 27 + 2 4 3 − − 3 Will get equation 4 + 2 21 + 1 2 = ( + 21 ) 2 + + + 1 2 ( 2 + 1 ) 2 = ( + ) 2 when + 21 = , = , + 1 2 = Divided into 2 cases as follows the case that 2 + 1 ≥ 0 2 + 1 = + 2 − − − 1 = 0 and the case that 2 + 1 ≤ 0 2 + 1 = − − 2 + + + 1 = 0 From finding the square root of a quadratic polynomial equation .1 = n+√2−4(1)(−−1) 2(1) 2 n−√2−4(1)(−−1) 2(1) 3 = −n+√−2−4(1)(+1) 2(1) 4 −n−√−2−4(1)(+1) 2(1) Getting the pH of phosphoric acid (34 ) pH = −log + √ 2 − 4(1)(− − 1 ) 2(1) Conclusions Applied Mathematics Project on Mathematical Modeling of the Relationship between Phosphate pH The following is an overview and discussion of the findings and recommendations. In summary, the following equations provide the results of the investigation of the link between the pH of phosphate: pH = −log +√2−4(1)(−−1) 2(1) when + 21 = , = , + 1 2 = + 1 [34 ] =
14 12 [34 ] = 123 [34 ] = The pH equation above is just one value that the authors have summarized because it is a format that is easy to understand. If we can find the relation of [34 ] with [4 3−] from the reaction equation. The stoichiometry allows us to compute the amount of phosphate in grams. From learning about acids and bases, chemical equilibrium, geology, to finding roots of degree 2 - 4 polynomials. and a quantitative modeling analysis of the relationship between pH and phosphate. This model can be investigated further in the future to alter soil pH. Acknowledgments The organizing committee would like to thank Teacher Tanongkiat Phonchaiya, Teacher Nampueng Supromin, and Assistant Professor Dr. Montri Torwattanaboon for their assistance in creating a mathematical project on the research of a mathematical model of the link between Phosphate pH. Thank you to the teachers who help students study mathematics and science. Everyone who offers advise and contributes to the project's improvement. Thank you to the school's and department's directors. All executives who support and support the budget in the project's preparation. Thank you, father and mother, for your encouragement and support in completing this project. Thank you to my school brothers and sisters for their cooperation, assistance, and inspiration. until the project is completed References กรมพัฒนาที่ดิน. 2558. สถานภาพทรัพยากรดินและที่ดิน ของประเทศไทย. กรมพัฒนาที่ดิน กระทรวงเกษตรและ สหกรณ์. 303 น. คณะกรรมการจัดท าพจนานุกรมปฐพีวิทยา, 2551. พจนานุกรมปฐพีวิทยา, ส านักพิมพ์มหาวิทยาลัยเกษตรศาสตร์ กรุงเทพฯ. 206น. คณาจารย์ภาควิชาปฐพีวิทยา, 2536. คู่มือปฏิบัติการ ปฐพีวิทยาเบ้อืงตน้ , ภาควิชาปฐพีวิทยา คณะเกษตร มหาวิทยาลัยเกษตรศาสตร์. 119 น. รักบ้านเกิดทีม. (2555). ดินเค็ม(Sanlinty) . ทุ่งรวงทอง. สืบค้น 18 มิถุนายน 2565, จาก https://www.nupress.grad.nu.ac.th บริษัท นีโอนิคส์จ ากัด. (2564). ค่า pH ของดินที่เหมาสม ต่อการปลูกพืช. สืบค้น 18 มิถุนายน 2565, จาก https://www.neonics.co.th ไพศาล นาคมหาชลาสินธุ์. (2559). ภาควิชาคณิตศาสตร์ และวิทยาการคอมพิวเตอร์. คณะวิทยาศาสตร์. จุฬาลงกรณ์ มหาวิทยาลัย. ถ.พญาไท เขตปทุมวัน กรุงเทพฯ 10330 สารานุกรมไทยส าหรับเยาวชนฯ. ธาตุอาหารในดิน. สืบค้น 18 มิถุนายน 2565, จาก https://www.Saranukromthai.or.th วิกิพีเดีย. (2561). กรดฟอสฟอริก. สืบค้น 18 มิถุนายน 2565, จาก https://th.wikipedia.org/wiki วิกิพีเดีย. (2565). กรดฟอสฟอริก. สืบค้น 18 กุมพาพันธ์ 2566, จาก https://th.wikipedia.org/wiki
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 15 Suthira Srichai1 and Thunyaporn Phonchan1 Advisor : Rattanate Wichaiphin1 Special Advisor : Waratchaya Tongsan2 and Adisak Moumeesri3 1Princess Chulabhorn Science High School Buriram, Satuek, Buriram, 31150, Thailand 2Burirum Rajabhat University, Naimuang, Muang, Buriram,31000, Thailand 3Silpakorn University, Nakhon Pathom, 73000, Thailand Abstract This project aims to compare durian price forecasting models. That is the most accurate and suitable for forecasting the price of durian. The data set was divided into 2 sets. The first set since 1999-2019, 19 2 values for predictive model construction by simple linear regression, Polynomial regression, and Using transformations to make curves straight. Data set 2 from AprilSeptember 2020 and April-June 2021, 9 values were used for comparison of the accuracy of the forecasting model. The performance of the forecasting model was measured using the Mean Absolute Percentage Error (MAPE) and the lowest root mean squared error (Root Mean Squared Error: RMSE) by Excel program. Choose the method with the most minor discrepancy to create a website using the svelte language. Use the most accurate and suitable forecasting model to analyze durian price data during the study period. to predict future durian prices The study found that with the Polynomial regression method, The Comparison of Durian Prices Forecast Model by Statistical Forecasting Methods and create a website for durian prices forecast value is the closest to the actual data and the lowest MAPE and RMSE values are 9.36% and 11.47, respectively it is the most accurate and suitable for forecasting prices. which has a forecast model y = 0.002x 2 − 0.386x + 31.909 Defind represents the predicted value represents the interval Keywords: forecasting model, svelte language Introduction Durian, also known as king of fruits. It is a seasonal fruit that is considered important to the Thai agricultural sector. and is also the number 1 export fruit in the group of Thai fruits exported to markets around the world Although in the present era there is an epidemic of COVID 19 that began to affect since the end of 2019. But durian exports can continue to expand well according to consumer demand, especially in China. Which is considered the main market, which in 2020 has durian export volume accounting for 77.33 percent of the world's fresh durian export value but from the problem that farmers face every year is overproduction in the market during the same production period. causing the price of produce to drop. Farmers therefore met with loss. Therefore, the authors are interested in studying and creating a predictive model. By bringing data during the years 1999 - 2019, the results of the forecasts obtained It will be the starting point for making decisions and managing various risks. In addition, the organizer has created a website for durian price forecasting. This is another channel that allows consumers and manufacturers to contact each other more conveniently. And it is information for making a decision to buy products. Put the highlight picture of your project in this area.
16 Methods Preparation of a durian price forecasting model comparison project by statistical forecasting methods and creating a website for durian prices the preparation team divided the operation method into 2 parts: comparison of the model for forecasting the price of golden durian. and create a website for forecasting Monthong Durian The details are as follows. Part 1: Comparison of Monthong durian price forecasting models Data from 1999-2019, 192 values for predictive modeling using a simple linear regression, polynomial regression, and using transformations to make curves straight. the second set of data from April-September 2020 and April-June 2021 for 9 values were used for comparison of the accuracy of the forecast model. The performance of the forecasting model was measured using the Mean Absolute Percentage Error (MAPE) and the lowest root mean squared error (Root Mean Squared Error: RMSE) criteria by using the Excel program for processing. Graph 1: Show Data Set 1 Graph showing durian price data from 1999-2019. 1. Simple linear regression Simple linear regression and correlation analysis is a statistical study method used in the case of The variables we want to study are even variables. where one variable, called the "dependent variable", is represented by Y, and another variable called Independent variables or independent variables is represented by X. In regression analysis, the relationship between the two variables is searched. through prediction in the form of linear equations That is the ability of another variable to be able to predict another variable Correlation analysis is used to determine the magnitude and direction of the relationship between two variables = + (1) Σy = aΣ + bn (2) Σxy = aΣx 2 + bΣx (3) Define represents the time period represents durian price , represent constants 2. Polynomial regression Polynomial regression analysis (Jiangqing, 1996) is a study of the relationship between variables that There is a non-linear relationship. (The relationship is a curve), which the analysis process is difficult and more complicated y = ax 2 + bx + c (1) Σy = aΣx 2 + bΣx + cn (2) Σxy = aΣx 3 + bΣx 2 + cΣx (3) Σx 2y = aΣx 4 + bΣx 3 + cΣx 2 (4) Define represents the time period represents durian price , represent constants 3. Using transformations to make curves straight. Can use simple linear regression in reverse to estimate coefficients and observe a system. Transform the data to make a linear relationship, Use simple linear regression to find the best-fit line for the transformed data, and use the best-fit line to estimate parameters. 3.1 Exponential Model Since the data fits the curve, if we transform the data appropriately, the transformed data should fit the linear relationship. Here the linear relationship is between and , so we transform the data by taking the natural log of all the values We find the best-fit line for the transformed data using simple linear regression. Suppose the regression line is defined by = 0 + 1 We compare the regression line with the linearized equation = 1 + 1 i.e. 1 + 1 = 0 + 1 We find estimates for 1 and 1 by equating coefficients on either side of the equation. = 1 1 (1) ln = ln(1 1 ) (2) ln = ln 1 + ln 1 (3)
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 17 ln Σ = ln 1 + 1Σ (4) ln Σ = ln 1 + 1Σ (5) Define ln 1 represents 0 1 represents 1 3.2 Power Equation We apply the transformation to the observed data (take log base 10 of the and values). We find the best-fit line for the transformed data using simple linear regression (least squares). Suppose this line is defined by log = 0 + 1 log We compare equations and equate coefficients. = 2 2 (1) log = log(2 2) (2) log = log 2 + log 2 (3) log Σ = log 2 + 2log Σ (4) 3.3 Saturation-Growth-Rate equation We see from the previous slide that 1 has a linear relationship with 1 . We apply the transformation to the observed data (take inverse of the and values). We find the best fit line for the transformed data using simple linear regression (least squares). Suppose this line is defined by 1 = 0 + 1 1 Comparing equations we get 1 3 + 3 3 1 = 0 + 1 1 Define y = ax + b (1) Σy = aΣx + bn (2) Σxy = aΣx 2 + bΣx (3) Part 2: Creating a Mon Thong Durian Forecasting Website The website was built using Svelte (https://svelte.dev/), a web programming language similar to JavaScript, which uses SvelteKit (kit.svelte.dev) as the Meta Frameworks. Figure 1: Code used to create the project: https://github.com/ptsgrn/durian-priceprediction Figure 2: Display graphs using Chart.js Figure 3: Use tailwind for web page decoration. Figure 4: Website for forecasting durian prices: https://durian-price-prediction.app.ptsgrn.dev
18 Results and Discussion 1. Forecasting Model After substituting the equations using x and y variables, the predictive model can be obtained. After that, the predictive model can be used to predict the next time. Polynomial, Exponential model, Power equation and Saturation-Growth-Rate equation are detailed below. 1.1 Forecasting results by simple linear regression Substitute the values of the x and y variables into the general form equation. After that, a suitable forecasting model will be obtained. y = ax + b Σy = aΣx + bn Σxy = aΣx 2 + bΣx = 0.2411 + 5.2345 1.2 Forecasting results by polynomial regression Substitute the values of the x and y variables into the general form equation. After that, a suitable forecasting model will be obtained. y = ax 2 + bx + c Σy = aΣx 2 + bΣx + cn Σxy = aΣx 3 + bΣx 2 + cΣx Σx 2y = aΣx 4 + bΣx 3 + cΣx 2 = 0.0024671 2 − 0.38554 + 31.909 1.3 Forecasting results by Exponential model Substitute the values of the x and y variables into the general form equation. After that, a suitable forecasting model will be obtained. = 1 1 = 1 1 ln = ln 1 + ln 1 ln Σ = ln 1 + 1Σ ln Σ = ln 1 + 1Σ = 13.7714 0.0061 1.4 Forecasting results by Power equation Substitute the values of the x and y variables into the general form equation. After that, a suitable forecasting model will be obtained. = 2 2 log = log(2 2) log = log 2 + log 2 log Σ = log 2 + 2log Σ = 7.5078 0.3025 1.5 Forecasting results by Saturation-GrowthRate equation Substitute the values of the x and y variables into the general form equation. After that, a suitable forecasting model will be obtained. = 3 3 + 1 = 3+ 3 1 = 3 3 + 3 1 Σ = 1 3 + 3 3 ( 1 Σ ) = 25.5102 0.3887+ 2. Comparison of the accuracy of forecasting models When obtaining a predictive model from the simple linear regression method and polynomial regression method, then import into excel for processing and comparing the accuracy of the Monthong durian price forecasting model as follows: Graph 2: Show Comparison of durian price forecasting models by statistical forecasting method From the comparison graph of the durian price forecasting model using and the statistical forecasting method, it was found that the durian price obtained from the model using the linear regression, the Exponential model, the Power equation, and the Saturation-Growth-Rate equation were not consistent with the durian price curves that were used. actual value And the durian price from the polynomial regression forecasting model is consistent and follows the actual durian price curve.
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 19 3. Analysis of results Analyze and compare the accuracy of the forecast values. In this experiment, a forecasting model suitable for the durian price time series in each month was selected. By comparing MAPE and RMSE values from both forecasting methods, namely the simple linear regression method and polynomial regression method, the predictive model with the lowest MAPE and RMSE values is considered the model with the least difference from the actual value. MAPE and RMSE criteria are shown as follows. = [ ∑ | – forecast value| × 100] = √ 1 ∑( – forecast value) 2 =1 Table 1: Show Actual value and forecast value of durian price. The results from using simple linear regression, polynomial regression, exponential model, power equation, and Saturation-GrowthRate equation were used to obtain the price forecast of Monthong durian price of the second data set from April-September 2020 to April-June 2021, 9 values are shown in Table 1 Results of comparison of accuracy and actual values found that the polynomial regression method is the most accurate method. Because it gives the value that is the least different from the actual data. And the percentage error (MAPE) was 9.36% and the root mean square error (RMSE) was 11.47, which was the lowest value. Discussion From the use of predictive models, simple linear regression, polynomial regression, exponential model, power equation and Saturation-Growth-Rate equation, it was found that the MAPE values were 36.36%, 9.36%, 36.72%, 62.31%, 76.25% and the RMSE values were 41.19, 11.47, 41.47, 69.13, 84.12 respectively. From the MAPE and RMSE values, it can be seen that simple linear regression, Exponential model, Power equation and Saturation-Growth-Rate equation There is a higher tolerance than the polynomial regression, because the durian price graph does not correspond to the prediction equation. Conclusions The purpose of this study was to compare various durian price forecasting models that were most accurate and suitable for durian price forecasting by simple linear regression, polynomial regression, exponential model, power equation and saturation-Growth-Rate equation. polynomial regression the input prediction value is closest to the actual data, and there are MAPE and RMSE values, which are percent error and square of error. The lowest was 9.36% and 11.47 respectively, so it was the most accurate and suitable for forecasting durian prices. It is appropriate to forecast future durian prices. Suggestion 1. Add statistical forecasting methods and develop the website to update the information. 2. The resulting model may be limited to the data to be analyzed (overfitting) in terms of being able to predict future prices. 3. Processes and data lack flexibility in actual use. The study framework should be defined, such as the prediction period, in making a clear forecast. 4. Data analysis methods are still not suitable for forecasting durian prices. Should study more about the variables that affect the price Year Month Durian prices Forecast Model Simple Linear Regression Polynomial regression Exponential Model Power Equation Saturation- Growth-Rate Equation 2563 Apr 115.92 66.41 93.41 65.64 40.18 25.472 May 92.52 66.92 95.21 66.04 40.23 25.472 Jun 97.63 67.43 97.02 66.45 40.27 25.472 Jul 90.95 67.94 98.85 66.85 40.32 25.472 Aug 100.30 68.46 100.70 67.26 40.37 25.472 Sep 100.00 68.97 102.57 67.67 40.42 25.472 2564 Apr 130.09 69.48 104.47 70.63 40.74 25.473 May 117.95 69.99 106.38 71.06 40.79 25.473 Jun 140.98 70.50 108.31 71.49 40.83 25.474 MAPE (%) 36.36 9.36 36.72 62.31 76.25 RMSE 41.19 11.47 41.47 69.13 84.12
20 of durian and use advanced statistics to forecast for reliable results and that can be used for real Acknowledgments Comparison of durian price forecasting by statistical forecasting methods and construction of durian price websites it is another project that helps solve economic problems in the Thai agricultural sector. The durian is considered an important fruit and is the country’s number 1 exported fruit. Website users can use the knowledge of durian price trends as a guideline for making a purchase decision at the time of purchase. Appropriate and also makes the seller able to sell products according to product quantity causing market equilibrium. The preparation of this project was successfully accomplished. With the courtesy and support of Mrs. Rattanate Wichaiphin, Dr. Waratchaya Tongsan and Dr. Adisak Moumeesri, who gave advice on forecasting the price of durians. And Teacher Suriyan Nandoon, who helped verify the information and provide knowledge on the overview of the project. Lastly, I would like to thank Mr. Patsakorn Yuenyong, who gave me advice and knowledge on website construction and was able to successfully complete the project. References [1] Worawit Chansuwan. 2021. Regression and correlation analysis using MS Excel program. Bangkok : Rajamangala University of Technology Phra Nakhon. [2] Krittiya Chainork. 2014. Durian, the King of Fruits. Bangkok: Faculty of Pharmacy Mahidol University [3] Mukda Mannamin. 2006. Time series and forecasting. Bangkok: Four Printing
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 21 Thanaphum Saiwongin 1 and Praphawit Wongyai 1 Advisors: Manrak Premtong 2 1,2 Princess Chulabhorn Science High School Chiang Rai, Rob Wiang, Mueng, Chiang Rai, 57100, Thailand Abstract Euler’s number or e is the importance irrational number in mathematic which has many applications in many fields. Nowadays, there are various algorithms to approximate Euler’s number. Most of them required advance mathematics knowledge and not suited for high school students. We propose a new algorithm to approximate Euler’s Number based on only the mathematical knowledge in high school level which is Geometry and Probability. We divided the process into two parts, a Theoretical part and an Experimental part. The Theoretical part aims to creating the equation and determining the areas in graph. The Experimental part aims to random points in the graph, calculating the areas, The results show that the average Euler’s number obtained from 100 approximations is in range of 2.5 - 2.7 and the deviation from the general e value is more than 7%. Keywords Euler’s number Approximation of Euler’s Number by using Geometry and Probability Introduction The mathematical constant or Euler’s number [1] approximately equal to 2.71828. Euler’s number is an irrational number [2] which is having shown up since the mid-16th century. It was first mentioned by Swiss mathematician Leonhard Euler. Euler’s number is of paramount importance across various branches of mathematics and finds numerous applications in Finance, Physic, Computer Science and other fields. Throughout history, mathematicians have tirelessly sought to improve the accuracy and efficiency of methods to approximate Euler’s number. Various approaches have been devised. Euler's number can be expressed in various mathematical forms, often introduced as definitions in mathematical analysis courses. These representations include infinite series (see Fig. 1), continuous ratios, limit sequences (see Fig. 2) [3] and gamma function (see Fig. 3) [4] = ∑ 1 ! ∞ =0 Fig. 1. Infinite series = lim→∞ (1 + 1 ) Fig. 2. Limit sequence ≈ ∑ 1 ( − 1)! + 2 , ℎ ≥ 2 −2 =0 Fig. 3. Gamma function In our recent study, our primary focus was on approximately the value of π using a combination of probability and geometry. We employed a method that involved generating random points within a predefined area and then counting the number of points that fell within specific regions. After completing this project, we delved deeper into the function and its integral to gain further insights and broaden our understanding. From our researching, many general approximations of Euler’s number are quite hard to understand especially for the normal high school student because those approximations require the high-level math knowledge so by
22 combining our knowledge of probability theory, geometric concepts, we aim to develop a new and understandable method for approximating "e". Not only provide the new approximation of Euler’s number but we hope this research offers novel insights into approximation techniques, making it a pivotal contribution for mathematicians worldwide, inspiring fresh ideas and advancements in this field. This work aims to create the new way to approximate Euler’s number by using Geometry and Probability. The rest of the paper is organized as follows. Section 2 is a literature review. Section 3 explains the theory. Section 4 provides the experimental setting, result, and discussion. Last, Section 5 gives a summary of the paper and a future plan for the research. Materials and Methods Methods Part 1 Theory Assigning 3 areas in graph = 1 in range 1 to x (1) specifying the area of a small rectangular area has a width of 1 unit and a length of − 1 unit as shown. Calculating a small rectangular area by using ℎ × ℎ equation The area = ( − 1)( 1 ) = 1 − 1 (2) specifying the area of a big rectangular area has a width of 1 unit and a length of − 1 unit as shown. Calculating a big rectangular area by using ℎ × ℎ equation The area = ( − 1)(1) = − 1 (3) specifying the area of an area under the curve has a length of − 1 unit as shown Calculating an area by using Integration The area = ∫ 1 1 = ln | 1 = ln − ln 1 = ln If substitute x with e value, the area will equal to ln or 1 Considering that the entire area is a point so that the number of points will be equal to that area. Assign the variables = area ∆ = area of the point = the number of points Create the equation for proving = ∫ = ∑ ∆ =1 = ∆ + ∆ + ∆+. . . +∆ = ∆ Therefore, 1 2 = 1 ∆ 2 ∆ = 1 2 (2) Creating the equations Assign the variables = the number of points in the area under the curve = the number of points in the big rectangular area = the number of points in the small rectangular area U = the number of points in the graph When define x = e Creating an equation from the ratio of the area under the graph and the area of the large rectangle. = 1 − 1 = 1 + (1) Creating a second equation from the ratio of the area under the graph and the area of the large
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 23 square and the ratio of the area of the small square and the area under the graph. × = 1 −1 × 1− 1 1 = 1 = (2) (1) = (2); = 1 + 1 = 1 − 1 (3) make the linear equation [8] from equation (3) = + substitute = 1 and = 1 − 1 1 = ( 1 − 1 ) when m equal to 1, this equation will be true. Therefore, we will find a value of x that makes this linear equation have a slope equal to 1, that value of x will be an approximation of the value of e. Part 2 Experiment In this process, the theory will be proved by performing approximation tests in Microsoft Excel. The image in this part only used for visualization. (1) 1500 Random points in the graph y = 1 x to find Ne ,Nuand Ns by number of points is 1500 and our determined areas is in range 1 to 2 The determined areas in range 1 to 2 (2) Create an ordered pair from the equations = 1 , = 1 − 1 and save the ordered pair. (3) Change the number of points by changing to 4000, 6000, 8000, … ,20000 and do (1) – (2) so we will get 10 ordered pairs (4) Use the ordered pair to create a linear equation using the Linear Regression method and store the m or slope value of linear equation (5) changing the x value to different values, starting with changing x to 2.5 and doing steps (1)-(4). The obtained m value has less deviation from Experiments using an x value of 2 and then changing the x value to 3 found more discrepancies. Therefore, reduce the value of x by 0.1 with the above method continuously.
24 It is found that the value of m obtained is close to 1 in the area 2.5 – 3 and save the ordered pairs (x, m) (6) Use the ordered pairs from the previous step to create a linear equation using the Linear Regression method (7) Substituting y with 1 in the equation will give you a value of x that causes m = 1, which is the value of Euler’s number obtained from our approximation. = 1.5138 − 3.1718 1 = 1.5138 − 3.1718 = 1+3.1718 1.5138 = 2.7558 (8) Repeat the experiment many times to observe the results Results and Discussion The results show that the Euler’s numbers from our method in each approximation will not equal because our method use randomization and if we adjust the parameter in our approximation by set the starting number of points has set to be greater than or equal to 1,500 and the number of subsequent points has set to increase by more than or equal to 1,000 and using a total of 10 numbers of points for approximation. With these 3 conditions, the average Euler’s number obtained from 100 approximations is in range of 2.5 - 2.7 and the deviation from the general e value is about 7%. Conclusions Euler’s number from our method is not stable but with the fixed parameter, the average Euler’s number obtained from 100 approximations is in range of 2.5 - 2.7 and the deviation from the general e value is not more than 7%. Acknowledgments We would like to express my heartfelt thanks to our advisor for their unwavering support and encouragement throughout this endeavor. His guidance were indispensable in bringing this project to fruition. References [1] Aghaeiboorkheili, M. and Kawagle, J.G. (2022) The History of the Derivation of Euler’s Number. Journal of Applied Mathematics and Physics, 10, 2780-2795. https://doi.org/10.4236/jamp.2022.109185 [2] Sandifer, Ed (Feb 2006). "How Euler Did It: Who proved e is Irrational?" (PDF). MAA Online. [3] Formulas 2–7: H. J. Brothers, Improving the convergence of Newton's series approximation for e, The College Mathematics Journal, Vol. 35, No. 1, (2004), pp. 34–39. [4] Shekh, Zahid & Ray, Prasanta Kumar. (2015). Approximation of Euler Number Using Gamma Function. American Journal of Undergraduate Research. 12. 10.33697/ajur.2015.014. [5] J. Guillera and J. Sondow, (2008), Double integrals and infinite products for some classical constants via analytic continuations of Lerch's transcendent, Ramanujan Journal ,16 247–270. [6] H. J. Brothers and J. A. Knox, New closedform approximations to the Logarithmic Constant e, The Mathematical Intelligencer, Vol. 20, No. 4, (1998), pp. 25–29. [7] Di Paola, G., Bertani, A., De Monte, L., & Tuzzolino, F. (2018). A brief introduction to probability. Journal of thoracic disease, 10(2),
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 25 1129–1132. https://doi.org/10.21037/jtd.2018.01.28 [8] Pierce, Robyn & Stacey, Kaye & Bardini, Caroline. (2010). Linear functions: Teaching strategies and students' conceptions associated with y = mx + c. Pedagogies: An International Journal. 5. 202-215. 10.1080/1554480X.2010.486151. [9] Jeffrey M. Stanton (2001) Galton, Pearson, and the Peas: A Brief History of Linear Regression for Statistics Instructors, Journal of Statistics Education, 9:3, DOI: 10.1080/10691898.2001.11910537 [10] Mazurek, Leszek. (2020). Division by zero.
26 Apisada Promdeesarn1 , and Ilada Rungprathom1 Advisor: Naeem Binibroheng 2 1Princess Chulaborn Science High School Pathumthani, Bogoen, Ladlumkaeo, Pathumthani,12140, Thailand Abstract A tautology is a statement that is always true, no matter the circumstances. Tautologies are often used in logical arguments to prove a point or to emphasize a statement. The relationship between logic and set theory is one of mutual support and reinforcement, with each field relying on the other to help define and clarify key concepts and ideas. This project aims to solve tautology using Venn – Euler diagram. In order to simplify tautology, therefore, the Venn – Euler diagram has been redrawn (Iva diagram) which can be used in 1-6 sets. Then find the tautology. Any form of expression is a tautology. There are the following situations: If the statement is connected with ∧ or ∨ the answer must be Universe. If connected with ⇒ The answer to the previous proposition must be a subset of the following proposition. And if the statement is connected with ⇔ the answers to the two statements must be equal. Therefore, it will be regarded as a tautology. Keywords: tautology, logic, set theory, Venn diagram Tautology calculates association by using IVA Diagram applied from Venn-Euler Diagram with Set Theory Introduction In logic, a tautology refers to a proposition or statement that holds true under all possible truth value assignments to its constituent variables or components. It represents a statement that is intrinsically true, independent of any specific interpretation or truth assignment. (Joseph, 2018) Tautologies are often used in logical arguments to prove a point or as a way to emphasize a statement. There are several ways to determine whether a statement is a tautology. Truth table: A truth table lists all the possible combinations of truth values for the variables in a statement and shows the resulting truth value of the statement for each combination proof: A statement can also be shown to be a tautology through a formal proof, using rules of logic to demonstrate that it is always true. But this method has its limitations: it takes a lot of time, and if there are more than 6 propositions, it's easy to get confused and make mistakes. From studying information about set theory and logic, it was found that they are connected. Therefore, with the purpose to verify this perpetuity using a diagram created by the authors "IVA diagram" is based on the original Venn-Euler diagram. Materials and Methods Data collection search for uncomplicated ways to find tautology in logic and then find the relationship between logic and set theory in terms of operation. The logical operations ¬, ∧, ∨ translate into the theory of sets in a natural way using truth sets. If A is a set, define Ac = {x: x ∉ A} called the complement of A, If B is a second set, define, A ∩ B = { x: x ∈ A ∧ x ∈ B } called the intersection of A and B, A ∪ B = { x: x ∈ A ∨ x ∈ B }, called the union of A and B, A is a subset of B if ∀ { x ∈ A ⇒ x ∈ B }, the sets A and B are equal if and only if A ⊆ B and B ⊆ A, that is, ∀ { x ∈ A ⇔ x ∈ B } Invent the IVA diagram. Put the highlight picture of your project in this area
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 27 When dealing with sets, the commonly used Venn-Euler diagram is limited in its application to problems involving a maximum of three sets. Consequently, an alternative representation known as Universe with Rectangles and set with circles. In the IVA diagram, sets in the universe are denoted by circles, and sets in shaped as rectangles. Results and Discussion IVA diagram represented 1 proposition (p) and Truth value location (L) The result shown in figure 1 and table 1 Figure 1 Table 1 L p 1 T IVA diagram represented 2 propositions (p and q) and Truth value location (L) The result shown in figure 2 and table 2 Figure 2 Table 2 L p q 3 T T 2 T F 1 F T 0 F F IVA diagram represented 3 propositions (p, q, and r) and Truth value location (L) The result shown in figure 3 and table 3 Figure 3 Table 3 L p q r 7 T T T 6 T T F 5 T F T 4 T F F 3 F T T 2 F T F 1 F F T 0 F F F IVA diagram represented 4 propositions (p, q, r, and s) and Truth value location (L) The result shown in figure 4 and table 4 Figure 4
28 Table 4 L p q r s 15 T T T T 14 T T T F 13 T T F T 12 T T F F 11 T F T T 10 T F T F 9 T F F T 8 T F F F 7 F T T T 6 F T T F 5 F T F T 4 F T F F 3 F F T T 2 F F T F 1 F F F T 0 F F F F IVA diagram represented 5 propositions (p, q, r, s, and t) and Truth value location (L) The result shown in figure 5 and table 5 Figure 5 Table 5 L p q r s t 31 T T T T T 30 T T T T F 29 T T T F T 28 T T T F F 27 T T F T T 26 T T F T F 25 T T F F T 24 T T F F F 23 T F T T T 22 T F T T F 21 T F T F T 20 T F T F F 19 T F F T T 18 T F F T F 17 T F F F T 16 T F F F F 15 F T T T T 14 F T T T F 13 F T T F T 12 F T T F F 11 F T F T T 10 F T F T F 9 F T F F T 8 F T F F F 7 F F T T T 6 F F T T F 5 F F T F T 4 F F T F F 3 F F F T T 2 F F F T F 1 F F F F T 0 F F F F F IVA diagram represented 6 propositions (p, q, r, s, t and u) and Truth value location (L) The result shown in figure 6 and table 6 Figure 6 Table 6 L p q r s t u 63 T T T T T T 62 T T T T T F 61 T T T T F T 60 T T T T F F 59 T T T F T T 58 T T T F T F 57 T T T F F T 56 T T T F F F 55 T T F T T T 54 T T F T T F 53 T T F T F T 52 T T F T F F 51 T T F F T T 50 T T F F T F 49 T T F F F T 48 T T F F F F 47 T F T T T T
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 29 46 T F T T T F 45 T F T T F T 44 T F T T F F 43 T F T F T T 42 T F T F T F 41 T F T F F T 40 T F T F F F 39 T F F T T T 38 T F F T T F 37 T F F T F T 36 T F F T F F 35 T F F F T T 34 T F F F T F 33 T F F F F T 32 T F F F F F 31 F T T T T T 30 F T T T T F 29 F T T T F T 28 F T T T F F 27 F T T F T T 26 F T T F T F 25 F T T F F T 24 F T T F F F 23 F T F T T T 22 F T F T T F 21 F T F T F T 20 F T F T F F 19 F T F F T T 18 F T F F T F 17 F T F F F T 16 F T F F F F 15 F F T T T T 14 F F T T T F 13 F F T T F T 12 F F T T F F 11 F F T F T T 10 F F T F T F 9 F F T F F T 8 F F T F F F 7 F F F T T T 6 F F F T T F 5 F F F T F T 4 F F F T F F 3 F F F F T T 2 F F F F T F 1 F F F F F T 0 F F F F F F Conclusions According to the research on the relationship between sets and logic, the definitions of operation marks in these two theories are related. Then we redrew Venn – Euler diagram to find tautology. Then find the tautology. Any form of expression is a tautology. There are the following situations: If the statement is connected with ∧ or ∨ the answer must be Universe. If connected with ⇒ The answer to the previous proposition must be a subset of the following proposition. And if the statement is connected with ⇔ the answers to the two statements must be equal. Acknowledgments Thank you to Princess Chulabhorn Science High School Pathum Thani School, Pathum Thani for his advice and facilitating research. References [1] Samai Laovanit, and Asst. Prof. Puaphan Traders. (2011). Set, Mathematics Mathayomsuksa 4-6 Volume 1 (p. 1-70). Bangkok: High-Ed Publishing. [2] Pairoj Yearayong. (2016). Proof, Logic and Set Theory (pp. 27-40). Bangkok: Chulalongkorn University. [3] Patrick Keef, & David Guichard. (2018). Introduction to higher mathematics. Department of Mathematics Whitman College. Retrieved Jan 3, 2023, from https://www.whitman.edu/mathematics/h igher_mama_online/section01.05.html [4] Moshe Machover (2020). Set Theory, Logic and their Limitation. Kind collage London [5] Ali Muhammad Rushdii, 1Mohamed Zarouan, and et al (2015). A Modern Syllogistic Method in Intuitionistic Fuzzy Logic with Realistic Tautology. Retrieved Jan 3, 2023, from https://www.hindawi.com/journals/tswj/ 2015/327390/
30 Anchaleeporn Prathumsaen1 and Anchitta Sakulma1 Pirapat Pumkong2 and Jittinart Rattanamoong3 1,2Princess Chulabhorn Science High School Chonburi, Chonburi, 20170, Thailand 3Department of Mathematics, Srinakharinwirot University, Bangkok, 10110, Thailand Abstract The objective of this project is to study the relationship between the remainder of integer divided by 13 and the digits of integer by studying the remainder from dividing positive integer by 13 and the remainder from dividing a negative integer by 13. It has been discovered that we can find the remainder from dividing positive integer by 13 by calculating the sum of the products of the digits in the units, tens, hundreds, thousands, ten thousands, hundred thousands, millions, ten millions, hundred millions, etc., of that number in sequence with 1, -3, -4, -1, 3, 4,... until all place values are accounted for and repeating when the resulting sum is not greater than zero and less than 13. We can also calculate the remainder from dividing negative integer by 13 by subtracting the remainder from dividing the absolute value of that number by 13 from 13. Through this project, it is recommended to conduct further studies on observations with other numbers and expand the findings in the field of data encryption and decryption. Studying the relationship between the reminder of integer divided by 13 and the digits of integer Keywords: Dividing, Positive integer, Negative integer, Relationship, Remainder Introduction In mathematics, division is considered a significant operation due to its fundamental role in various processes. Divisors are concepts that have been known and studied since at least the time of Euclid. The fundamental ideas are developed in this. (Ivan Niven, Herbert S. Zuckerman, and Hugh L. Montgomery, 1991, p. 4). In the context of division, each number has distinct observations. For example, when an integer is divided by 8 and the division is exact, it is observed that the last digit of the number must be divisible by 8. However, when division is not exact, we can find the remainder. There are various methods to determine the remainder, one of which involves studying the relationship between the digits of integer using modular arithmetic. This leads to finding the remainder of integer divided by another integer. Therefore, we are interested in exploring the relationship between remainders of integer divided by 13 and the digits of the integer. By setting conditions to establish patterns, utilizing number theory concepts, and conducting proofs to determine the conditions for exact division and consequently finding the remainder when the number is not divisible by 13. Materials and Methods Part 1 Study the relationship between the remainder of positive integer divided by 13 and the digits of integer. Let A be an n + 1 -digit positive integer, n Î È ¥ {0} . And r is the remainder from dividing A by 13. There is A r º (mod 13) , where 0 13 £ < r 1. Dividing positive integer in the form of 10t by 13. Because 100 ≡ 1(mod13) 101 ≡ -3(mod13) 102 ≡ -4(mod13) 103 ≡ -1(mod13)
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 31 104 ≡ 3(mod13) 105 ≡ 4(mod13) 106 ≡ 1(mod13) Conjecture 1.1 Let t Î È ¥ {0} . Based on the basic properties of congruence, we can deduce that: 3 10 ( 1) (mod13) t t º - 3 1 10 ( 3)( 1) (mod13) t t + º - - 3 2 10 ( 4)( 1) (mod13) t t + º - - 2. The remainder obtained from dividing the n + 1-digit positive integer by 13. It is easy to consider the remainder in the case of positive integers with 1 digit ( n = 0 ). Therefore, we will consider the cases where n = 1 and n > 1. Case 1: n = 1 , let 1 0 A a a = while 0 1 a a, {0,1,2,..., 9} Î and 1 a ¹ 0 Case 2: n > 1 , let 1 2 2 1 0 ... n n n A a a a a a a - - = while {1,2,..., 9} n a Î and {0,1,2,..., 9} j a Î ; j n = - 0,1, 2,..., 1 . From conjecture 1.1, we can divide it into 3 sub-cases as follows. Sub-case 1: n º 0(mod 3) Sub-case 2: n º 1(mod 3) Sub-case 3: n º 2(mod 3) 3. The remainder obtained from dividing the n + 1-digit positive integer, where each digit of the integer is repeated, by 13. Let A aaa a = ... be the n + 1 -digit positive integer while n ³ 1 and a Î {1, 2,..., 9} If n = 1 , then r A a a a º º - + º - 3 2 (mod 13) If n > 1 , then Case 1: n º 0(mod 3) 1 3 0 [( 6)( 1) ] ( 1) (mod 13) n i n i r A a a - = º º - - + - å Because of the parity of an integer, we can get: Sub-case 1: n º 0(mod 6) Sub-case 2: n º 3(mod 6) Case 2: n º 1(mod 3) 1 1 3 1 0 [( 6)( 1) ] ( 2)( 1) (mod 13) n i n i r A a a - - - = º º - - + - - å Because of the parity of an integer, we can get: Sub-case 1: n º 1(mod 6) Sub-case 2: n º 4(mod 6) Case 3: n º 2(mod 3) 2 3 0 [( 6)( 1) ](mod 13) n i i r A a - = º º - - å Because of the parity of an integer, we can get: Sub-case 1: n º 2(mod 6) Sub-case 2: n º 5(mod 6) Part 2 Study the relationship between the remainder of negative integer divided by 13 and the digits of integer. Let 1 A be an negative integer, 1 A A = - while A Î ¥ . And r is the remainder from dividing A by 13. There is A r º (mod 13) , where 0 13 £ < r . Because - º - 1 1(mod 13) , so - º - A r(mod 13) Case r = 0 : 1 A A º - º 0(mod13) Case 0 13 < < r : From 0 13(mod 13) º , So 1 A A r º - º - 13 (mod13) Because 0 13 < < r If and only if - < - < 13 0 r Therefore 0 13 13 < - < r Results and Discussion Part 1 The relationship between the remainder of n + 1 -digit positive integer divided by 13 and the digits of integer. Case 1: n = 1, 1 0 A a a = 1 0 r A a a º º - + 3 (mod13) Case 2: n > 1, 1 2 2 1 0 ... n n n A a a a a a a - - =
32 Sub-case 1: n º 0(mod 3) 1 3 3 3 1 3 2 0 [( 1) ( 3)( 1) ( 4)( 1) ] n i i i i i i i r A a a a - + + = º º - + - - + - - å ( 1) (mod 13) n n + - a Sub-case 2: n º 1(mod 3) 1 1 3 3 3 1 3 2 0 [( 1) ( 3)( 1) ( 4)( 1) ] n i i i i i i i r A a a a - - + + = º º - + - - + - - å 1 1 1 ( 1) ( 3)( 1) (mod 13) n n n n a a - - - + - + - - Sub-case 3: n º 2(mod 3) 2 3 3 3 1 3 2 0 [( 1) ( 3)( 1) ( 4)( 1) ](mod 13) n i i i i i i i r A a a a - + + = º º - + - - + - - å Part 2 The relationship between the remainder of n + 1 -digit positive integer, where each digit of the integer is repeated, divided by 13 and the digits of integers. Case 1: n = 1, A aa = r A a º º - 2 (mod 13) Case 2: n > 1, A aaa a = ... Sub-case 1: n º 0(mod 6) r A a º º (mod 13) Sub-case 2: n º 1(mod 6) r A a º º - 2 (mod 13) Sub-case 3: n º 2(mod 6) r A a º º - 6 (mod 13) Sub-case 4: n º 3(mod 6) r A a º º - 7 (mod 13) Sub-case 5: n º 4(mod 6) r A a º º - 4 (mod 13) Sub-case 6: n º 5(mod 6) r A º º 0(mod 13) Table 1: Shows the relationship between the remainder of n + 1-digit positive integer, where each digit of the integer is repeated, divided by 13 and the digits of integer. In the case where n > 1 and n ≡ 0, 1, 2(mod6). n≡0(mod6) n≡1(mod6) n≡2(mod6) a = 1 1 -2 -6 a = 2 2 -4 -12 a = 3 3 -6 -18 a = 4 4 -8 -24 a = 5 5 -10 -30 a = 6 6 -12 -36 a = 7 7 -14 -42 a = 8 8 -16 -48 a = 9 9 -18 -54 Table 2: Shows the relationship between the remainder of n + 1-digit positive integer, where each digit of the integer is repeated, divided by 13 and the digits of integer. In the case where n > 1 and n ≡ 3, 4, 5(mod6). n≡3(mod6) n≡4(mod6) n≡5(mod6) a = 1 -7 -4 0 a = 2 -14 -8 0 a = 3 -21 -12 0 a = 4 -28 -16 0 a = 5 -35 -20 0 a = 6 -42 -24 0 a = 7 -49 -28 0 a = 8 -56 -32 0 a = 9 -63 -36 0 Part 3 The relationship between the remainder of negative integer divided by 13 and the digits of integer. 1 1 2 2 1 0 ... n n n A A a a a a a a - - = - = - and 1 A A r º - º - 13 (mod13) , where 0 13 £ < r Conclusions In this project, we have studied the relationship between the remainder of integer divided by 13 and the digits of integer. This investigation is divided into two cases. In case 1, the study is conducted on positive integers. In case 2, the study is carried out on negative integers. From the study of both cases, we can conclude that we can find the remainder from dividing positive integer by 13 by calculating the sum of the products of the digits in the units, tens, hundreds, thousands, ten thousands, hundred thousands, millions, ten millions, hundred millions, etc., of that number in sequence with 1, -3, -4, -1, 3, 4,... until all place values are accounted for and repeating when the resulting sum is not greater than zero and less than 13. We can also calculate the remainder from dividing negative integer by 13 by subtracting the remainder from dividing the absolute value of that number by 13 from 13.
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 33 Acknowledgments We would like to extend our sincere thanks to my project advisor and special advisor for providing suggestions, correcting mistakes, and overseeing the project's progress from start to finish. We are immensely grateful and extend our highest appreciation. Thanks are also extended to everyone involved who contributed to the completion of this project, as well as to Princess Chulabhorn Science High School Chonburi for fostering a learning environment that allowed us to acquire knowledge and develop an awareness of its value. Lastly, we would also like to thank our parents. Their belief in us has kept our spirits and motivation high during this process. References [1] David M Burton. (2010). The Theory of Congruences. In Martin Lange (Eds.), ELEMENTARY NUMBER THEORY (7th ed.). (pp. 61-84). https://jgcsr.org/wpcontent/uploads/David_M._Burton_ Elementary_Number_Theoryzlib.org_.pdf [2] Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery. (1991). An Introduction to the Theory of Numbers. (5th ed.). https://editorialdinosaurio.files.word press.com/2012/03/itn-niven.pdf [3] Paul Dawkins. (2018). Calculus I – Summation notation. Paul’s Online Notes. https://tutorial.math.lamar.edu/classes/calci/ summationnotation.aspx [4] Summation notation. (n.d.). Colombia University. http://www.columbia.edu/itc/sipa/math/sum mation.html [5] TechTarget Contributor. (n.d.). What is an integer and what are examples of integers. https://www.techtarget.com/whatis /definition/integer
34 Sothita Chiso1 and Thanisorn Tubtim1 Advisor: Amornrut Aountri2 Special Advisor: Jenjira Thipcha3 1,2Princess Chulabhorn Science High School Lopburi, Huaypong, Khoksamrong, Lopburi, 15120, Thailand 3Department of Mathematics, Maejo University, Chiang Mai, 50290, Thailand Abstract The mathematical project “Finding the probability of winning Tic Tac Toe 3x3 grid” came about because the organizers noticed something interesting in Tic Tac Toe or XO games. The first player will have a higher chance of winning than the second player. The organizers, therefore, brought the Tic Tac Toe to study, so this project aims to find reasons that make the first player more likely to win than the second player. Using probabilities to prove, find reliable reasons, and convert the value into a number will make the reasoning clear. In addition, it can be further developed into a fair game and applied to writing code to find probabilities more easily. From doing this project, the organizing team had a systematic thinking process. Gain skills and understanding of probability estimation and techniques for playing the Tic Tac Toe. Keywords : Probability , Tic Tac Toe , fair game Finding the probability of winning Tic Tac Toe 3x3 grid Introduction Because the organizers have foreseen some interesting things in the TIC TAC TOE game or the XO game that we know. Is that the first player will have a higher chance of winning than the second player. The organizers, therefore, brought the TIC TAC TOE game or the XO game to study the probability to prove and find out why the first player has a higher chance of winning the XO game than the second player. The organizers will find the probability of both landing in different positions and finding the probability of winning 5-9 times and ending the game. Methods From the Tic Tac Toe game table, which has 9 squares, we will put letters in each square to determine the position of X and O. The letters in each square of the grid will be assigned as in Table 1. A B C D E F G H I Table 1 Take the order of descending X and O to write in a row. Part 1 Finding the Probability of Entering Different Positions and Winning Let the first player start from the corner of the table. If the second player is positioned on the side of the table (position B D F H), then the first player is positioned on the other side of the table in line. (vertical or horizontal) that a win can occur This will force the second player into the space between the X in defense and then place it in the other corner to win a two-way win. For example, see Table 2.
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 35 X O O X X Table 2 (ABGDI) The winning rate is 7/8 because if the second player placed in the middle is the first landing, there will be a draw as in Table 3. X win win win draw win win win win Table 3 The case starts from the middle. If the second player lays on the side of the grid (B D F H), they place in the corner next time. to enforce protection Then put another corner can win two ways. Example X X O X O Table 4 (EDAIC) But in the case of the second player laying in a corner (A C G I), in this case, it lays in the remaining corner diagonally. It can be divided into 2 cases as follows. Case 1 If the second player is placed in a side position There will be a defense that can be won in two ways. O O X X X O X Table 5 Case 2 If the second player is placed in the next corner position It will be the defense that makes this game a draw. O X X X X O O O X Table 6 (EAIGDBHC) Figured out to get a win rate of 5/6 as in Table 7. 2 3 win 2 3 win X win 2 3 win 2 3 Table 7 Part 2 Find the probability that you can win in 5-9 turn. For example, the character set ABDCG will be written X and O as in Table 8. X O O X X Table 8 (ABDCG) At the same time, Table 8 can also be caused by other character sets, such as GBDCA, etc. Therefore, we are looking for a winning Tic Tac Toe game through all possible letter combinations. and bring to probabilities Rules of Tic Tac Toe game The game of Tic Tac Toe is won when one of the players can arrange their letters (X or O) to complete three letters before winning. In the
36 game of Tic Tac Toe, the 3x3 table has 8 types of arrangements to win, which when detached into the letters are as follows: ABC DGF GHI ADG BEH CFI AEI CEG, which within the 3 letters of each type can be switched to which one will be placed first, for example, ABC can enter ABC BCA BAC CBA ACB CAB. Tic Tac Toe game can end with 5-9 landings, we will find each one ending with 5, 6, 7, 8 and 9 landings. Winning a tic tac toe game in 5th turn The winner of a game with 5 downs must be X because it ends the game with X down rounds. _ _ _ _ _ Therefore, the letter combination must have 5 characters. The landing position of the X must have all 8 letter combinations in a win that can be swapped. One example is ABC. A_B_C Which, as mentioned at the beginning, that these 3 letters can switch positions. So the alternation of these 3 letters will give you a total of 3! or 6 types The next step we have to put the remaining 6 letters in the empty position, which is the round of O. From the formula P(n,r)=n!/((n-r)!) where n = 6 and r = 2, when substituted in the formula, 6!/4!= 30. When all the letters in the descending order of X and O are added together, all possible patterns are 6x30 = 180 patterns. And here is an example of 1 out of 8 ways to win the game of Tic Tac Toe in a 3x3 grid, so you have to multiply 8 to know all the ways that can win the game of Tic Tac Toe in 5 downs, total 180 x 8 = 1440 ways. Do the same thing to find the probability that you can win in 5-9 turn. Equations , = ! (−)! (1) () = (2) Equations (1) to find the way to win in 5-9 turn, Equations (2) to fine the probability that you can win in 5-9 turn. Results and Discussion Tables 9 : Shows the probability that X will win when X start at the middle and O play in each place 2 3 win 2 3 win X win 2 3 win 2 3 Table 9 Tables 10 : Shows the mean probability that X will win = 0.4826927 The mean probability that O will win = 0.232237365 Win rate in corner and center position = 21/24 and 20/24 Turn Probability Winner 5th 0.0952381 X 6th 0.07317073 O 7th 0.552840 X 8th 0.391304 O 9th 0.8 X Table 10
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 37 Conclusions X has more probability than O so X or the first player is more likely to win. In the future, we have an idea to find probability in even number grid such as 4x4 , 6x6. Acknowledgments This mathematical project integrates mathematical knowledge to apply Finding the probability of winning the TIC TAC TOE 3x3 grid project. We would like to thank Teacher Amornrat Ouantrai for advising on the preparation of the project and teachers from Princess Chulabhorn science high school Lopburi everyone who encouraged me to prepare this project as well as for various suggestions. References [1] Timeanddate. (1995). Moonrise, Moonset, and Phase Calendar for Bangkok. [website]. Retrieved February 7, 2023, Available https://www.timeanddate.com/moon/thai land/bangkok. [2] Wikipedia. (1996). Jasmine. [website]. Retrieved July 1, 2022, Available https://shorturl.asia/vJklE ,Wikipedia. (2004). Moon. [wedsite]. Retrieved July 1, 2022, Available https://citly.me/ZhiH7
38 Thanarat Prachaiyo1 Poonyapat Pattale1 Sarawuth Klongdee1 1Princess Chulabhorn Science High School Mukdahan, Mukdahan, 49000, Thailand Abstract The purpose of this project is to study how to crypt using mathematical methods and camouflage latent information in common images. In terms of matrix and equations, it consists of Thai letters, Thai vowels. The first is RSA encoding to encode the characters and the second encoding is the matrix encoding to encode a group of characters. The message set is in the form of (a group of letters , letters converted to numbers) and has a table showing the relationship between numbers and letters into 3 groups: the first group is the English alphabet, the second group is the Thai alphabet, and the third group is the Thai vowel and tonal. Sending data with encrypted messages and sending information in encrypted form can be easily verified. Therefore, the hidden information in the image is disguised to conceal the information and secret code that is a secret message. The results of the experiment showed that the two stages of encryption (RSA and matrix) It can be kept confidential and the relationship table is constantly changing. Every time you enter and decrypt Keywords : cryptography, encryption, decryption A study of cryptography using mathematical method Introduction Ever since humans began to communicate. Data security, or the confidentiality of that type of information, is to expect that information will not be disclosed to the outside world or to the extent that they want to disclose it. At present Modern technology has been developed. As a result, data security levels are becoming more and more high. There are different formats and methods of data retention. Increase complexity to prevent threats from bad actors who want to steal data for their own benefit. Prevent hacker attacks and prevent computer criminals In 2010, Komson Singh Hiranusorn gave examples of applying the theory to solve various problems as follows: RSA Encryption is a good method of data encryption, which is commonly used to encrypt sensitive data through electronic networks and is highly reliable in data security. The steps are as follows: 1. Generate public key is n,e and private key is n,d 2. Create encryption equation (C = M e (mod n) or C = Md (mod n) 3. Create a decryption equation: M = C d (mod n) or M = C e (mod n), where M is Plain Text and C is Cipher Text. M.Jayachandran uses Steganography technique in combination with SAR Image (Synthetic Aperture Radar) to transmit confidential information while sending SAR Image back to the waiting unit using LZW to assist in encoding and decoding. The author wants to apply matrix and equity knowledge to encoding and decoding in cryptography. A table showing the relationship between letters and numbers is used using knowledge of linear equations to determine the desired encoding and decoding.
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 39 Materials and Methods RSA cryptography, the first public-key encryption algorithm known to be ideal for digital signatures as well as encryption, is one of the first major advances in public-key encryption. RSA is still used in protocols for electronic commerce and is believed to be secure if the key is long enough. RSA encryption process 1.Define the number of primes p and q. 2.Let n = p*q and z = (p-1)*(q-1) 3.e<n and e and n must not be co-factorial. 4.e*d mod z =1 5.Let m be the value obtained from the hash function. Encryption Public Key: (e,n) ≡ Decryption Private Key: (d,n) Image based Steganography Steganography refers to the process of hiding a secret message within a larger message in such a way that someone cannot know the presence or content of the hidden message. The purpose of steganography is to maintain confidential communication between two parties. This is different from encryption, in which we can conceal the contents of secret messages. Although Steganography differs from Cryptography, there are many similarities between the two. Some classify steganography as a form of cryptography, since hidden communication seems like a secret message. Image from https://www.javatpoint.com/imagesteganography Image from https://www.javatpoint.com/imagesteganography Infiltration in images 1.Select the image you want to embed data into. This is a normal image, which involves the process of embedding data into the image without only minor changes in the size of the image. The size of the image is proportional to the size of the text entered. 2. The process of embedding data into images can be done using LSB Steganography, which is a very popular method. 3. Can be used in conjunction with cryptography The encrypted message is embedded in the image and forwarded to the recipient. Extraction of latent information in images 1. Images with embedded information Data extraction can be done using the same method as embedding data. 2. The data obtained from the extraction It must be deciphered before the real message can be known.
40 Results and Discussion RSA encoding example 1.Select primes 17 and 19, so n = 323 and r = 288. 2.Consider the integer T greater than r and divide by r to the remainder 1, which will be 289, 577, 865, ... 3.Select a non-prime number, i.e. 865, factored into 173 x 5. 4.Select x as 5 and y as 173 (often y is greater than x for harder guessing). 5.Define the HELLO text by converting letters to numbers first, which can be used in the table above as (2,8),(2,5),(2,12),(2,12),(2,15). (2,8) 5 ≡ (32,145) (mod 323) where (a,b)n = (a n,bn ) and (a, b) n ≡ ((a n mod z), (b n mod z)) (mod z) (2,5) 5 ≡ (32,218) (mod 323) (2,12) 5 ≡ (32,122) (mod 323) 2 times (2,15) 5 ≡ (32,2) (mod 323) The encoded text is (32,145), (32,218), (32,122), (32,122), (32,2). Example of encoding by matrix method 1. Consider the text (2,8),(2,5),(2,12),(2,12),(2,15) (-7,(1,-2,2)) , (- 3,(-1,1,3)) , (5,(1,-1,-4)) will get the encryption key in the form of a matrix size 3*3 and 1 −2 2 −1 1 3 1 −1 −4 det = 1 , which is not 0 . −1 −10 −8 −1 −6 −5 0 −1 −1 2.Encode text, define (a,b)*(c,d) = (a*c,b*d) Each set of text in a 1*n matrix can be said to be the first set of texts, which in this matrix contains the letters (32,145) (32,218) (32,122)H, E, L, respectively. The second set of text is (32,122) (32,2) (2,0) , where 0 represents a space or space in the text. When all messages are encrypted, the encrypted message is (32,49) (-64,-194) (32,456) (32,120) (-64,-242) (32,250) It should be noted that RSA encryption converts text into a cryptographic code (a,x) , (a,y) , (a,z). The same ahead is a, but when encoded in a matrix way, it makes The A value that defines the block of text changes. This results in more unpredictability. For decoding by means of a matrix Define the decryption key as is −1 −10 −8 −1 −6 −5 0 −1 −1 , and the encrypted message is: (32,49) (-64, -194) (32,456) (32,120) (-64, -242) (32,250) The decoded text is (32,145), (32,218), (32,122), (32,122), (32,2). RSA method decoding The decoded text from the matrix method is arranged in order: (32,145), (32,218), (32,122), (32,122), (32,2). Because (2,0) is a space or a space, it is not decoded by the RSA method. Decoding can be done by where M ≡ C y (mod n)M is RSA-encoded text:M ≡ C y (mod n) (32,145) 173 ≡ (2,8) mod 323 (32,218) 173 ≡ (2,5) mod 323 (32,122) 173 ≡ (2,12) mod 323 (32,122) 173 ≡ (2,12) mod 323 (32,2) 173 ≡ (2,15) mod 323 Therefore, the text is decoded in both formats: (2,8),(2,5),(2,12), (2,12),(2,15) converted using a table showing the relationship between numbers and letters to get the word HELLO as before. Working principle of LSB Steganography Each pixel consists of three values - red, green, blue. These values range from 0 to 255, which means they are 8-bit values. Let us understand how this technique works using an example. Suppose we want to hide the text "hi" into a 4 x4 image which has pixel values as shown below. [(225, 12, 99), (155, 2, 50), (99, 51, 15), (15, 55, 22), (155, 61, 87), (63, 30, 17), (1, 55, 19 ), (99, 81, 66), (219, 77, 91), (69, 39, 50), (18, 20 0, 33), (25, 54, 190)] From the ASCII table, we can convert the secret text to decimal value and then change it to binary format: 0110100 0110101 Now we can iterate through the pixel values one by one. Once we have converted it to binary. We can replace each least significant bit with that text bit sequentially (for example, a binary of 225 is 11100001, and then we can replace the last bit, which is the right bit (1), with the default data bit (0) to get 11100000, that is, a binary of 224, etc.), which will allow us to modify the value of the pixel. +1 or -1 only, which cannot be recognized at all. The output value of the pixels after performing LSBS operations is as follows: [(224, 13, 99), (154, 3, 50), (98, 50, 15), (15, 54, 23), (154, 61, 87), (63, 30, 17), (1, 55, 19
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 41 ), (99, 81, 66), (219, 77, 91), (69, 39, 50), (18, 20 0, 33), (25, 54, 190)] means each pixel, such as (225,12,99), where a binary of 225 is 11100001, a binary of 13 is 00001100, and a binary of 99 is 01100011. The binary of h is 01101000, and the binary of I is 0110101, so each byte change in pixels changes accordingly. The last bit in each byte changes to the binary values of h and i, respectively. Conclusions This project is a study of cryptography using knowledge of equations and matrices. In regard to the proportion. RSA is the 1st encoding method using the knowledge of the equation and matrix encoding is the 2nd encoding to increase data security. Each set of letters is in the form (a, b), where a is a group of letters. In the 1st encoding, a group of letters is changed into the same set of codes, which can adversely affect the data. For the second encoding, the group of letters is replaced by an increasingly diverse code. This results in higher safety. Acknowledgments This project was successfully completed due to the advisor, Prof. Sarawuth who gave advice as well as guidelines for correcting various shortcomings with care. The author would like to thank you very much. The authors would like to thank all the researchers who have used their research as references. The authors are happy to listen to suggestions from everyone who comes to study to be useful for the further development of this project. References Dima Grigoriev. (2005). Constructions in public-key cryptography over matrix groups. Retrieved January 1st 2022. From https://www.researchgate.net/publication/296108 70_Constructions_in_publickey_cryptography_over_matrix_groups. Hoffstein J. (2008). An introduction to mathematical cryptography. Retrieved January 1 st 2022. From. https://link.springer.com/book/10.1007/97 8-0-387-77993-5. Thapanapong Rukkanchanunt. (2020). RSA cryptography. Access 1 January 2022. From www2.cs.science.cmu.ac.th.
42 Pawit Akarawanichakun1 Piyasak Kongthong1 Advisors:Supamit Wiriyakulopast2 Special Advisor: Dr.Warangkhana Riansut3 1,2Princess Chulabhorn Science High School Nakhon Si Thammarat,Bangchak, Muang Nakhon Si Thammarat,Nakhon Si Thammarat,80330 3Department of Mathematics and Statistics, Faculty of Science, Thaksin University, Phatthalung Campus, Ban Phrao, Pa Phayom, Phatthalung 93210 Abstract The objective of this study is to construct an appropriate forecasting model for the number of establishments registered with the social security fund. Four statistical methods were used, including Box-Jenkins method, Holt’s exponential smoothing method, Brown’s exponential smoothing method and the damped trend exponential smoothing method. Monthly data obtained from the website of the Social Security Office from January 2015 to June 2023 (138 months) were used to analyze. The forecasting model’s accuracy was checked using the mean absolute percentage error and root mean square error. The result showed that Brown’s exponential smoothing method was t m 1 Y 505,375.77083 1,162.44335 (m 1) ˆ 0.53514 + = + − + where m = 1 represented of January 2023 Keywords: Establishment registered with the social security fund , Forecast , Accuracy , Statistical method, Social security fund A Forecasting Model for the Number of Establishments Registered with The Social Security Fund Introduction Currently, during the economic recovery period from the impact of the COVID-19 pandemic that occurred in the recent past, there has been an increase in the number of businesses due to the emergence of new ventures. Establishing a new business has become easier nowadays, and we can create businesses even in our teenage years, with a tendency to further develop them in the future. In the 5th year of secondary education, there's a subject called "Math Project" which includes practical teaching. During the semester break after completing the 4th year of secondary education, the team participated in practical training on time series data analysis and forecasting by a professor. This led the team to be interested in analyzing the dataset of the number of registered establishment with the Social Security Fund and creating the most accurate forecasting model. The primary aim is to assess the economic situation in the future for decision-making in establishing businesses or investing in ventures. Furthermore, the formular gained from this project can be applied to analyze and forecast similar time series datasets of a differnt genre. Materials and Methods Materials 1.Computer 2. Statistical Package for the Social Sciences (SPSS) 3.Microsoft Excel Methods Part 1 : Data Analysis Conduct a comprehensive examination of the complete dataset through the construction of graphical representations Examine trends and seasonal patterns from the graph. Put the highlight picture of your project in this area.
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 43 Part 2 : Model Selection 1.Utilize the trend and seasonal values obtained from part 1 to select the model that best fits the data. 2.Select a model that is considered suitable for the data to be used in formulating the model for consideration. Table 1:Forcasting Models Part 3 Create a forcasting models 1.Using the partial data set to create all of the forecasting models using Statistical Package for the Social Sciences(SPSS). 2.Use the forecasting model to predict a value from the start date of the data set till the end. Part 4 Create a new dataset 1.Take the obtained model and generate a forecast dataset. 2.From the acquired dataset, calculate the error values to conduct a accuracy comparison. Part 5 Accuracy comparison 1.Use the error value from each model to calculate MAPE and RMSE. 2.Compare to find models that have the lowest MAPE and RMSE values. n2 t 2 t t 1 100 e MAPE n Y = = (1) 2 n 2 t 2 t 1 1 RMSE e n = = (2) Results and Discussion Table 2: Characteristics of the movement of the time series of the number of establishments registered with the Social Security Fund. From January 2012 to December 2022 From table 2 we can conclude that this data set has a trend but no seasonal. So we have chosen 4 models which are Holt's exponential smoothing method, Brown's smoothing method, Damped trend smoothing method, and Box-Jenkins method. Table 3: The obtained forcasting models After using SPSS or Statistical Package for the Social Sciences to calculate all the parameters we can create all the models as seen in the table 3. Method Methods Models 1 Box-Jenkins ARIMA (1, 1, 1) ARIMA(p, d, q): ( )( ) ( ) d p t q t f d q ˆ ˆ ˆ ˆ B 1 - B Y = + B e 2 Holt’s exponential smoothing method t+ m t t Y = a + b (m) ˆ where a = Y + (1 - )(a + b ), b = a - a + (1 - )b t t t -1 t -1 t t t -1 t -1 a a g g ( ) 3 Brown’s exponential smoothing method t+ m t t ( ) 1 Y = a + b m - 1 + ˆ é ù ê ú ê ú a ë û 4 Damped trend exponential smoothing method n i t+ m t t i= 1 Y = a + b ˆ where a Y 1 a b ,b a a 1 b t t t 1 t 1 t t t 1 t 1 = + − + = − + − ( )( − − − − ) ( ) ( )
44 Table 4: Comparison of MAPE and RMSE from each method Predict Method MAPE RMSE Holt’s 0.11 662.803 Brown’s 0.12 693.808 Damped 0.26 1366.69 BoxJenkins 0.27 1437.79 Then we used the models to create a data set of predicted values to calculate the error values in table 4. Using error values from table 4 we can comparison accuracy by using MAPE and RMSE. Conclusions In this project after consider about forecasting method including the Box-Jenkins method, Holt’s exponential smoothing method, Brown’s exponential smoothing method, and the damped trend exponential smoothing. We have found that the Brown’s exponential smoothing method have the lowest RMSE and MAPE so we consider this method to be most propriate t m 1 Y 505,375.77083 1,162.44335 (m 1) ˆ 0.53514 + = + − + (3) As we used to predict value of Establishments Registered with The Social Security Fund from July 2023 to December 2023. Table 5 : Pridicted Value of Establishments Registered with The Social Security Fund from July 2023 to December 2023 Time Predicted Value July 2023 514,597 August 2023 515,755 September 2023 516,953 October 2023 518,131 November 2023 519,310 December 2023 520,488 Acknowledgments This research paper can not be finish without my advisors. I would like to express my sincere thanks to my two research advisors, Mr. Supamit Wiriyakulopast and Assoc. Prof. Dr. Warangkhana Riansut for invaluable help and constant encouragement throughout the course of this research. References [1] Social Security Fund. [Internet]. [Accessed May 14, 2565]. Available at: https://www.hrm.chula.ac.th/newhrm/soc ial-security-fund/ [2] Warangkhana Riansut. Coffee Seed Price Forecasting Model. Journal of Ratchamongkol Bangkok Research 2019; 13(1): 141-155. [3] Social Security Insurance. [Internet]. [Accessed May 14, 2565]. Available at: https://www.sso.go.th/wpr/
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 45 Peeraphat Priaram1 And Tanawin Chinpattanawanich1 Advisors: Saknarin Channark2 And Jirakoon Erbim2 1,2 Princess Chulabhorn Science High School Phetchaburi, Khao-Yai, Cha-Am, Phetchaburi, 76120, Thailand Abstract This project designed the new auxetic structure using a centroid of linear and non-linear shapes and the golden spiral. Focus on the two types of auxetic structures: the re-entrant structure class and the rotating rigid structure class. In re-entrant structure, set the centroid inside of linear and non-linear shapes and used the new method to invent the unit cell of auxetic structure in 5x5 units. In the new procedure, use the square for a linear shape to create the unit cell of a re-entrant honeycomb and the Reuleaux triangle for a non-linear shape inspired by specific characteristics of three-leave clover to invent the unit cell of a Reuleaux triangle honeycomb. Prove that re-entrant honeycomb and Reuleaux triangle honeycomb are auxetic structures. Then, create a new auxetic structure in 4x4 units of rotating rigid structure inspired by specific characteristics of the Nautilus for the new unit cell of the new auxetic structure, named the golden spiral. In addition, create four models using Solidworks, including re-entrant honeycomb structures, Reuleaux triangle honeycomb structures, S-shape structures, and golden spiral A New Design of Auxetic Structure Using the Centroid of Linear and Non-Linear Shapes and the Golden Spiral for the New Design of the Helmet to Decrease the Chance of Injury from an Accident structures. Next, simulate a force of 5 newtons at the front and top views of all models for the stress simulation test. The results showed that the Reuleuax triangle honeycomb structure has the lowest relative standard deviation (RSD) at the front view and top views. Therefore, the Reuleaux triangle honeycomb structure has the best force distribution. Then, use that to design the new helmet design using Solidworks, simulate a force of 30,000 newtons on top view, and compare the results with the original helmet design. The results showed that the RSD of the new helmet design is 20.904%, while that of the original helmet design is 24.674%. Therefore, the new helmet design can reduce the force of the original helmet design by 15.280% and decrease the chance of injury from an accident. Keywords: Centroid, Reuleaux triangle, Golden spiral, Relative standard deviation, Re-entrant structure, Rotating rigid structure, Re-entrant honeycomb structures Introduction Helmets worn by motorcycle riders often suffer from shocks resulting from severe accidents, this problem puts the driver at greater risk of serious injury [8], whereas the Auxetic structure has good force distribution properties [2,4]. Then developer would bring the Auxetic structure to design the new helmet that can reduce impacts better. Focus on the two types of Auxetic structures: re-entrant structure and rotating rigid structure class [3]. In the re-entrant structure class, define the Centroid [10] within linear and non-linear shapes, using the new method to create a new design of linear and non-linear shapes to invent the Auxetic structure. Then, prove that the new design is a re-entrant structure. The linear shape that was interesting is the square shape of the re-entrant honeycomb and the non-linear Put the highlight picture of your project in this area.
46 shape is the Reuleuax triangle [5] which is inspired by the characteristic of three-leave clover [6]. About the rotating rigid structure class, the S-shape structure based on S-shape unit cells [1] which was derived by research is strong because of the characteristic of the structure, then interested in designing a newly rotating rigid structure from the golden spiral inspired by both the Nautilus [7] and the structural characteristics of the Burj Khalifa. Create both types of auxetic structures, including re-entrant honeycomb, Reuleaux triangle honeycomb, s-shape, and golden spiral, to select the structure that is the best force distribution model from the stress concentration of the experimental. Therefore, this project designed a new Auxetic structure and procedure to select the Auxetic structure that best has the best force distribution and to design a new helmet that reduces impacts and injury from an Figure 1. Shows the unit cells of the Auxetic structures in two-classes: Re-entrant structure class, and Rotating rigid structure class. In first class, the unit cells have re-entrant and Reuleaux triangle respectively. And another class, the unit cells have S-shape and golden spiral respectively. Methods Part 1 Designing the new Auxetic structure. 1. Designing the new method to design the unit cell of linear and non-linear shapes using a centroid. 2. Designing the unit cells of linear shape for re-entrant honeycomb structure in re-entrant structure class. Figure 1. Shows the procedure to designing the unit cells for the re-entrant honeycomb structure. 3. Designing the unit cells of non-linear shape for the Reuleaux honeycomb structure in re-entrant structure class. Figure 2. Shows the procedure to designing the unit cells for the Reuleaux triangle honeycomb structure. 4. Designing the unit cells of golden spiral for the golden spiral structure in the rotating rigid structure class using geometric procedure. Figure 3. Shows the procedure to designing the unit cells for the golden spiral structure. 5. Creating an Auxetic structure model in Solidworks for 4 models: re-entrant honeycomb, Reuleaux triangle honeycomb, S-shape, and golden spiral. Figure 2. Shows the Auxetic structures: re-entrant honeycomb () a ,Reuleaux triangle honeycomb () b , S-shape () c , and golden spiral ( ) d . ( ) a ( ) b ( ) c ( ) d acciden t.
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 47 Part 2 Defining the structure which is the best force distribution. 1. Simulating the force into models on top and front views for 5 newtons. 2. Save the images that showed the stress concentration of all models on top and front views. 3. Calculating the relative standard deviation (RSD) from the images using both the ImageJ and Python coding. 4. Comparing which is the best force distribution model. Part 3 Designing the new helmet. 1. Designing a new helmet which applied by the best force distribution model. 2. Creating both the original helmet and the new helmet models in Solidworks. 3. Simulating the force on top of both models for 30,000 newtons, when the average human weight is 70 kilograms, and the average speed is 80 kilometers per hour. 4. Calculating the RSD. 5. Comparing the force distribution of both models. Results and Discussion Results 1. Results of a stress concentration of Auxetic structures in Solidworks Figure 4. Shows the stress concentration from stress simulation in Solidworks on top view: re-entrant honeycomb () e , Reuleaux triangle honeycomb ( ) f , S-shape ( ) g , and golden spiral () h . And on front view: re-entrant honeycomb ()i , Reuleaux triangle honeycomb ()j , S-shape () k , and golden spiral ()l . 2. Results of a stress concentration of Helmet models in Solidworks Figure 5. Shows the stress concentration from stress simulation in Solidworks on top view: original helmet ( ) m , and new helmet () n . And on the side view: original helmet () o , and new helmet ( ) p . Table 1: Shows the RSD of stress concentration of all auxetic structures. Types of Auxetic Structures RSD (%) Front Top Re-entrant honeycomb 101.042 31.640 Reuleaux triangle honeycomb 49.848 25.531 S-shape 97.739 155.409 Golden spiral 81.135 33.285 Table showed that the Reuleaux triangle honeycomb structure has lowest RSD on top and front views: 49.848% and 25.531% respectively. Table 2: Shows the RSD of stress concentration of all helmet models. Types of helmet models RSD (%) Original design 24.674 Auxetic design 20.904 Table showed that the Auxetic design and the original design have RSD are 20.904% and ( ) e ( ) g ( ) h ()i ( )j ( ) k ()l ( ) f ( ) m ( ) n ( ) o ( ) p Calculating the relative s t a n d a r d Comparing which is the b e s t f o r c e Comparing the force d i s t r i b u t i o n o f