CS195H Homework 1 Grid homotopies and free groups. Due: February 5, 2015, before class

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1 CS195H Homework 1 Grid homotopies and free groups

2 This second homework is almost all about grid homotopies and grid curves, but with a little math in the middle. This homework, last year, took people about ten hours on average. It had a few additional problems, but lacked the problems on free groups at the start. Please again work on this problem set in pairs, but pick a new partner for this week. I recommend that each group have someone whose main strength is math, and someone else whose main strength is CS. (These could, I suppose, be the same person!) To hand in the homework, you ll need to do two things: (a) hand in the math parts as a handwritten or (better) pdf document. (b) hand in the code. Handwritten work should be given in during class and code should be submitted by in the form of a zip file, which should contain: A README.txt document saying whose work this is, and clarifying anything unusual you did. A bunch of.m files, one for each procedure you wrote. The must have the names and arguments specified, because we ll be using an automated program to test them. If I invoke pseq2lseq and you ve written psequencetolsequence, you ll fail the tests we ve written and get no credit. Make our lives easy so that you can make your grade better! The PDF for your written portion of the assignment, if you ve chosen to hand it in this way. You should name your zip file with your login and the 2-digit assignment number 01. I, for instance, would call mine jfh01.zip. You might call yours Robert_Smith01.zip, if you like using your brown.edu login. Math Stuff Problem 1 [15 pts] In class we talked about loops in the punctured plane, concentrating on loops that start and end at E = (0, 0), with the puncture at P = ( 1 2, 1 2 ). When we consider homotopy classes of loops (two loops 1

3 γ 1 and γ 2 are the same if there s a homotopy from one to the other, leaving the basepoint E fixed throughout, i.e., a continuous function with the properties that H : [0, 1] [0, 1] R 2 H(0, t) = γ 0 (t) for all t (1) H(1, t) = γ 1 (t) for all t, and (2) H(s, 0) = H(s, 1) = E for all s. (3) This set of homotopy classes of loops has a standard name. Letting R 2 0 denote R 2 {(0, 0)}, the set s called π 1 (R 2 0, E). (In general, for any space, X, and any point x 0 X, π 1 (X, x 0 ) denotes the group of homotopy classes of paths in X, based at x 0.) And as we saw in class, there s a group structure on the set, given by concatenating curves. The constant curve at the origin serves as an identity, etc. And we also saw that this particular group corresponded to Z, the integers, which can also be considered the free group on one letter. In general by a free group on a set like {a, b} we mean all finite sequences of elements of the set, or those elements with a superscript of 1. The simplest such sequence is the empty sequence, which is usually denoted e just to make it visible, but other instances of legal sequences are aaaba 1 b 1 babb, and aba. Multiplication in the group is done by concatenation, and there is one rule: any sequence like aa 1 or a 1 a should be replaced with the empty sequence. (That means that the second example sequence given above should be rewritten as aaaba 1 b 1 babb = aaaba 1 abb = aaabbb. It s traditional to write a 3 instead of aaa, just to keep things simpler, but that s just a notational convenience. That means that the second example can be written a 3 b Warmup. In the free group on two letters a and b (which we ll denote F ({a, b}), write down the left-inverses of ab, a 2 ba, and b 1 a, i.e., for the first problem, find a string u such that uab = e, once you simplify. 2

4 Explain why, in each case, the left-inverse is also a rightinverse, and why this will generally be true for any element of a free group. (In doing this, you re confirming another of the group axioms, i.e., showing that a free group really is a group, which we glossed over pretty fast in class.) One great thing about free groups is that they serve as the foundation for the idea of groups described by generators and relations. I m going to ask you to just play with this idea without trying too hard to formalize it just yet. An example of generators-and-relations description of a group is a a 3 = e or, more commonly, a a 3. That notation means Take the free group with one generator, a, and add a new rule: whenever you have three as in a row, you can replace them with an empty string. Perhaps a better version of the rule is you can, after any set of algebraic manipulations, replace three as by an empty string. So if we have a 1 a 2, we can change it to a 1 a 1 a 3, and hence to a 1 a 1 = a 2. In the notation, the symbols to the left of the vertical bar are called generators and those to the right are relations. Sometimes relations are written in the form a 6 = e, where e denotes the identity, i.e., the empty string, or even ab = ba, but more often they re written without an equals sign; in our two examples, we d write a 6 and aba 1 b 1 ; in each case, the presence of a relation on the right hand side without an equals sign means this sequence can be replaced by an empty sequence. A little algebra will show you that in the example we ve given, we have a 2 = a 1, and various similar things. In fact, there are really only three elements: e, a, and a 2. By the time you get to a 3, it s just another name for the identity e. So this description is just another way to denote Z/3Z, the integers mod 3. With that in mind, here are a couple of other generators-and-relations descriptions of some small groups. Your job is, for each one, to tell me the order of the group (i.e., the number of elements) and whether you recognize it (i.e., know some standard group that s isomorphic to it). 2. a, b a 2, b 3, a 2 ba 1 b 1 3. a, b a 4, b 2, abab 3

5 4. a a 7, a 4 5. a, b, c, d a 2, abab 1, acac 1, adad 1, ab 2, ac 2, ad 2, bcd 1, cdb 1, dbc 1. (This last one is difficult... feel free to give up. After all, it s possible that I missed some relation.) Problem 2 [10 pts] We ve studied grid homotopy with a rectangular grid. What would happen if instead we d started with a triangular grid, in which each vertex has six edges coming out of it? (It might be good to label these from 0 to 5, where edge i comes out in direction 60i degrees from the positive x-axis, measured counterclockwise.) How would you do work analogous to what we ve done so far? For instance, what would you choose as your basic moves? Would every grid-loop in the plane still be grid-homotopic to the constant grid-loop (the one with no edges)? Would grid-homotopy classes of grid-loops in the punctured plane still be in 1-1 correspondence with the integers? How would you choose to represent your triangle-grid curves? Problem 3 [10 pts] In the programming section below, you ll be writing code to interconvert among three representations of grid-loops. There s one more representation namely an edge followed by a sequence of turns, where a turn of 0 means reverse direction, 1 means turn left, 2 means go straight, and 3 means turn right. One disadvantage of this representation is that you can t represent the empty loop, since it starts with an edge. (An alternative choice would be to say that a loop always starts from the origin, having arrived there going north; then a 1 would mean that the first step was to the west, a 2 would mean north, and so on.) Describe what a hairpin insertion/deletion looks like in this labeling scheme; then described what a slide-move looks like. You may assume, for simplicity, that your hairpin or slide move is applied to a pair of edges in the middle of a long sequence of edges, i.e., there are turn numbers to both the left and right of the part you re fiddling with you don t have to worry about the corner case where the hairpin comes on the first two moves of the loop, etc. 4

6 Programming In class we discussed grid curves polygonal paths in the plane whose edges were all length-one, and whose vertices were all at integergrid points. A grid-loop is a grid path whose starting and ending points are identical. We declared two loops grid-homotopic if one could be obtained from the other by a sequence of moves of two types: 1. Deleting or inserting a hairpin, i.e., replacing a segment of the path like (i, j), (i, j + 1), (i, j) with just (i, j) [the hairpin can be in either x or y, and the +1 can be replaced by a 1, of course. 2. Sliding across a square: replacing (i, j), (i + 1, j), (i + 1, j + 1) with (i, j), (i, j + 1), (i + 1, j + 1). Note that the examples of hairpins and slides above are just examples. There are also vertically oriented harpins, etc., and you need to deal with all of them. Problem 1 [15 pts] There are three easy representations for a grid-loop: 1. a sequence of points, 2. a starting point and a sequence of edges (an edge being one of the vectors [0, 1], [1, 0], [0, 1], and [ 1, 0]), 3. a starting point and a sequence of edge-labels (like E, N, W, S for east, north, etc.) To make it easier, use 1, 2, 3, and 4 for east, north, west, and south respectively. Write a simple program for converting from each of these to each other one, in Matlab. Your points should be represented as 1 2 matrices, so a sequence of n points is an n 2 matrix. Your vectors should be represented as 1 2 matrices as well. Use structures (a Matlab data type) to represent an edge-sequence or a label-sequence, but a simple matrix to represent a point-sequence. We chose the names pseq, eseq, and lseq, so that your functions should be called pseq2lseq, etc. Each should take a single argument and return a single value, i.e., should start with function ls = pseq2lseq(ps), for example. 5

7 Hint: once you ve written three of these, the other three should be very easy to write. Details: A pseq should contain an odd number of points, with the starting point (typically (0,0)) being the same as the last point. So the constant pseq at the origin looks like 0 0 while a loop that goes around a single square looks like An eseq should have two fields, es.start, a 1 2 array containing the starting point, and es.dirs, an n 2 array containing the directions of the edges in order. An lseq should have two fields, ls.start, a 1 2 array containing the starting point, and ls.dirs, an n 1 array containing numbers 1 through 4, indicating the directions of the edges. Be very careful to give your functions the signatures we ve described, and your structures the fields we ve described; we ll be grading your code with an automatic grading script, and if your function s not there because you gave it the wrong name, you lose. Problem 2 [10 pts] A grid-loop has to be closed, i.e., have the same starting and ending point. That s an invariant of the structure, and it s easy to check in representation 1. Write programs checkeseqclosed, etc., to check that this invariant holds in representations 2 and 3. Your program should take a single argument, and return either true (if the invariant holds) or false otherwise. Problem 3 [10 pts] There are other invariants of the data structure: the points must all be on the grid, all edges must have length one, etc. For each representation, write an isvalid program to check that everything 6

8 about a particular instance of this data structure is valid. (names: isvalidpseq, isvalideseq, isvalidlseq). Once again: single argument, return single boolean result. Problem 4 [10 pts] Write a program that takes a grid loop (in pseq form!) as input and produces as output a sequence of grid-loops (in pseq form), each differing from the last by a single grid-homotopy step, and with the first being the input loop, and the last being a constant loop at the input-loop s starting point. Your input should be a pseq, and your output should be a cell array (another matlab data type) containing pseqs. Call your program findnullhomotopy(pseq). Problem 5 [10 pts] Write a program that displays the resulting homotopy, by showing the initial loop, and then repeatedly re-drawing: each subsequent drawing should show the previous loop in light grey, and the current loop in black. Use figure(gcf); pause(0.5); to pause a halfsecond between drawings. Call your program drawhomotopy(hom); the input, hom, should be a cell array of the kind produced by findnullhomotopy, so that drawhomotopy(findnullhomotopy(pseq)) is a legal invocation. Problem 6 [10 pts] Suppose that your input loop has n edges. Since the final (constant) loop will have no edges, you ll need to apply at least ceiling(n/2) hairpin-deletions during the homotopy. That means that producing the homotopy takes Ω(n) steps (where a step is a slide or a hairpin move, not an elementary operation like addition or assignment). Let s let H(n) be the greatest time required to reduce any length-n loop to a constant loop. Show that H(n) is Ω(n 2 ). Also show that it s O(n 3 ) by exhibiting an n 3 algorithm. Can you do better than O(n 3 )? 7

9 Problem 7 [0 pts] Tell us how long you spent on this problem set, and who you worked with. (Please hand in only one copy for the two of you.) 8