MIT Undergraduate Math Association Magazine

MIT Undergraduate Math Association Magazine

MIT Undergraduate Math Association Magazine Fall 2013 A Note from the UMA President The MIT Undergraduate Math Association is pleased to bring you our Fall 2013 issue of the UMA Magazine! The UMA has published a magazine for the math community at MIT pe- riodically for over 25 years. Now, after a short hiatus, we are The UMA Magazine writing and editing team resuming the tradition. In the Spring, we held a survey to gauge your opinions about The UMA officers: President Josh Alman, Vice President life as a math major at MIT and interests for our magazine. Mitchell Lee, Treasurer Leon Zhou, and Secretary Felipe This issue contains everything you asked for, and more: Hernandez. • Four articles by current MIT undergraduates about math Holden Lee, last year’s Vice President, who continued to that we find interesting. help us despite having graduated. • An interview with Professor Andrew Sutherland, focus- ing on his work and the recent progress toward the Twin Soohyun Park from USWIM, who helped to organize our Prime Conjecture. math articles. • Interviews with mathematics students who graduated Ping Ngai Chung who contributed an article. from MIT and are pursuing various fields of study. Our staff advisor Prof. Ju-Lee Kim. • The more interesting results from our survey. Special thanks to Prof. Andrew Sutherland, Leonid Chin- • Some mathematical nuggets and jokes. delevitch, Delong Meng, and David B. Rush for being in- We hope that you find the content useful and interesting! terviewed for this magazine. An online version is available at the UMA website http:// web.mit.edu/uma/www/, which will also include unabridged copies of our four interviews. We are already in the process of making the next edition of this magazine! If you are interested in contributing, please contact us at [email protected]. The UMA also holds various events throughout the year. Our first two lectures of the fall semester will be given by Prof. Scott Aaronson on Tuesday, September 10th at 5 pm, and by Prof. Henry Cohn on Tuesday, September 17th at 5 pm, loca- tions TBA. To hear more about these and other UMA events, subscribe to our mailing list [email protected], either directly, or by asking us to add you at [email protected]. Have a great semester! Josh Alman UMA President 1 Table of Contents Combinatorial Applications of Network Flow / Josh Alman 3 Generalized Fibonacci Sequences / Soohyun Park 6 The Young-Frobenius Identity / Mitchell Lee 8 Sieve Methods and the Twin Prime Conjecture / Ping Ngai Chung 12 Interview with Prof. Andrew Sutherland / Felipe Hernandez 16 Interview with Leonid Chindelevitch / Holden Lee 19 Interview with Delong Meng / Holden Lee 22 Interview with David B Rush / Holden Lee 24 Math Major Survey Responses / Holden Lee and Soohyun Park 27 Nuggets and Jokes / Leon Zhou 30 2 Combinatorial Applications of Network Flow Josh Alman Network flow is one of the most important problems in 1.2 The Ford-Fulkerson Algorithm computer science and operations research, and also has ap- A natural question to ask is: given a network, what is the plications to other subjects such as ecology and physics. How- maximum possible value of a flow through it? In computer sci- ever, the problem is inherently combinatorial, and has some ence, one seeks fast algorithms to answer this question. Here, cool applications to combinatorial problems. In this article, I we give one simple algorithm, the Ford-Fulkerson algorithm, will first introduce the theory behind network flows, and then which yields important insights into the problem. present some of my favorite applications. This article is meant The algorithm is based on paths through a graph called aug- to be self-contained, and requires no previous knowledge of menting paths. Given a flow f in our network, an augmenting network flows. A reader familiar with network flows may want 0 path is a list e ,e ,e ,...,e of distinct edges such that: to skip to the section on applications. 0 1 2 k 1. e eR for any 0 i, j k, i 6Æ j · · 1 Introduction 2. e s, and e t 0¡ Æ kÅ Æ 1.1 Definitions 3. eiÅ ei¡ 1 for each 0 i k, Æ Å · Ç A network is a complete directed graph G (V,E), together Æ 4. f0(e j ) c(e j ) for each 0 j k. with an edge capacity function c : E R 0 and two distin- Ç · · ! ¸ guished vertices, the source s and the sink t. In other words, it is a path of edges from s to t which are not If an edge e goes from vertex a to vertex b, we will write e¡ filled to capacity. We can augment f0 by this augmenting path R Æ a and eÅ b. If e is an edge, then e will denote its reverse as follows: we let ag min{c(e j ) f0(e j ) 0 j k} 0, and Æ R R Æ ¡ j · · È edge, namely e e and e e . then we get a new flow f1 defined by: ¡ Æ Å Å Æ ¡ A flow f in the network is a map f : E R with the following ! 8 f (e) ag if e {e ,e ,...,e }, properties: <> 0 Å 2 0 1 k R R R f1(e) f (e) ag if e {e ,e ,...,e }, Æ 0 ¡ 2 0 1 k 1. For each edge e E, f (e) c(e), :> 2 · f0(e) otherwise. R 2. For each edge e E, f (e) f (e ), We can see by definition that f is a new valid flow with f 2 Æ¡ 1 j 1j Æ f ag . From this, it is clear that whenever there is an aug- j 0j Å 3. For each vertex v other than s and t: menting path in our flow, the flow does not have the maximum X possible value. It turns out that the converse is true as well, f (e) 0. Æ and this is how the Ford-Fulkerson algorithm works. eÅ v Æ In the Ford-Fulkerson algorithm, we start with the ‘empty’ flow f , where f (e) 0 for all edges e. Then, while there ex- If e is an edge, and e a and e b, we will often write 0 0 Æ ¡ Æ Å Æ ists an augmenting path, we replace f by augmenting it by f (a,b): f (e). 0 Æ that path. We repeat until there are no augmenting paths left. The value of the flow, denoted f , is the net amount of flow j j Then, the resulting flow is a flow of maximum possible value. into t: X Of course, it is not even clear that this algorithm will ever f f (e). j j Æ terminate. However, it will if we restrict ourselves to rational eÅ t Æ capacities, meaning c(e) Q 0 for all edges e: A network can be thought of as a series of water pipes con- 2 ¸ nected to each other. The capacity of a pipe represents how Lemma 1. If all the edge capacities in a network are rational much water can flow through the pipe in that direction per numbers, then the Ford-Fulkerson algorithm will terminate in unit of time. Then, a flow corresponds to water going through a finite number of steps. the pipes, so that the water is conserved everywhere other Proof. If we multiply all the capacities by a value d 0, ap- than the source and the sink. The value of the the flow is the È ply the Ford-Fulkerson algorithm to get a flow in the resulting amount of water going from the source to the sink per unit of network, then divide the capacities and the resulting flows by time. d, the result is the same as if we had just applied the Ford- From the definitions and this intuition, we can see that: Fulkerson algorithm. Hence, since d can be the LCM of all the X X X X denominators of the edge capacities, we can assume without f f (e) f (e) f (e) f (e). j j Æ Æ¡ Æ Æ¡ loss of generality that all the edge capacities are integers. e t e¡ t e¡ s e s ÅÆ Æ Æ ÅÆ 3 a 3/3 Theorem 1 (Max Flow–Min Cut). For any network G (V,E) b Æ with capacity function c : E Q 0, source s, and sink t: ! ¸ 5/8 2/2 max { f } min {c(C)} flows f j j Æ cuts C s 2/2 2/1 1/3 1/4 t Proof. The fact that − − max { f } min {c(C)} flows f j j · cuts C 2/2 5/7 follows from (1). Now, to complete the proof, we will show that: c 4/6 d max { f } min {c(C)}. flows f j j ¸ cuts C Figure 1: A network with a flow. The notation x/y on an edge Consider any flow f ¤ that results from the Ford-Fulkerson al- means that edge has capacity y, and flow x along it. All edges gorithm. There are no augmenting paths in f ¤. Let C ¤ be the not shown have capacity 0, and have flow equal to negative set of all vertices we can reach from s by a path that uses only the flow in the opposite direction if such a flow is shown, or 0 directed edges that are not filled to capacity. otherwise. The flow shown has value 7. The cut {s,a} also has First, notice that if C ¤ contained t, then there would be an capacity 7.

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