Assignment 0 (Written): Combinatorial Surfaces — due 2/8

For the written part of your first homework you will do some exercises that will help familiarize you with basic descriptions and representations of combinatorial surfaces (simplicial surfaces, adjacency matrices, halfedge meshes), which will help prepare you to work with such surfaces as we continue through the course. (If any of this stuff seems abstract right now, don’t worry: we’ll use it over and over again to implement “real” algorithms starting in just a couple weeks!)

You must complete 8 out of 15 exercises in the Written Exercises section of Chapter 2 of the course notes. You may choose any set of 8 exercises you like, but if you do more than 8, please mark clearly on your submission which ones you would like us to grade.

The assignment is due on February 8, 2020 at 5:59:59pm Eastern (not at midnight!). Remember to turn in the whole assignment via a single email including both the written exercises (as a PDF file) and the code (in a ZIP file). Further hand-in instructions can be found on this page.

14 thoughts on “Assignment 0 (Written): Combinatorial Surfaces — due 2/8”

  1. For definition of next map $\rho$ in half edge, the note says
    $\rho(ij) = jk \ \forall ijk \in K$.
    Is it required that $i\neq j, j \neq k, i \neq k$?
    If not, then in the cone example given on page 19, why $\rho(h_1) = h_2$, and not $\rho(h_1) = h_0, h_2$?

    1. I now see below this it says “oriented triangle ijk”, implying ijk are different vertices. However, it still leaves me confused about exercise 2.10, “surface as permutation”, as there are many possible “next” map.
      For example, one way to obtain the surface in 2.10 is by connecting two triangles with two edges, and gluing the two connecting edge together. Then the top triangle could contain edges 0->1->2, or 1->2->4.

      1. You are correct. With special (almost degenerate) cases like these, the combinatorics does not completely fix the geometry. If you give a valid permutation, you will be marked as correct!

    1. A polygonal disc is a closed disc, meaning it has $\chi = 1$. It is compact with boundary. A polygonal sphere on the other hand has $\chi = 2$ and is compact without boundary. You may think of a polygonal disc as a polygonal sphere with a face removed.

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