commit 2870d40f6b888cb1d91c1d8ed33fbb8287a3aa04 Author: Renge Date: Sat Sep 3 21:05:06 2022 -0400 add: hw1 diff --git a/HW1/2_1_1.png b/HW1/2_1_1.png new file mode 100644 index 0000000..b45647f Binary files /dev/null and b/HW1/2_1_1.png differ diff --git a/HW1/2_1_2.png b/HW1/2_1_2.png new file mode 100644 index 0000000..20b6458 Binary files /dev/null and b/HW1/2_1_2.png differ diff --git a/HW1/2_1_3.png b/HW1/2_1_3.png new file mode 100644 index 0000000..67d6804 Binary files /dev/null and b/HW1/2_1_3.png differ diff --git a/HW1/2_2_1.png b/HW1/2_2_1.png new file mode 100644 index 0000000..90c9d35 Binary files /dev/null and b/HW1/2_2_1.png differ diff --git a/HW1/2_2_2.png b/HW1/2_2_2.png new file mode 100644 index 0000000..c0bd6f1 Binary files /dev/null and b/HW1/2_2_2.png differ diff --git a/HW1/HW1.pdf b/HW1/HW1.pdf new file mode 100644 index 0000000..62d907c Binary files /dev/null and b/HW1/HW1.pdf differ diff --git a/HW1/hw1-polygon1-guard.ipe b/HW1/hw1-polygon1-guard.ipe new file 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+ diff --git a/HW1/hw1.html b/HW1/hw1.html new file mode 100644 index 0000000..cdb126e --- /dev/null +++ b/HW1/hw1.html @@ -0,0 +1,853 @@ + + + + + HW1 + + + + + + + + + + +

HW1

Problem 1

+
1. +
  In this graph, there are 3 forced diagonals (7-12, 6-14 and 5-17) which divide this graph in to 4 parts. For the right most part it is a blunt nose fox so it has 4 ways of triangulations; for the second one there is only one way because it is a fox; the third one is again, a blunt nose fox with 4 ways; as for the last part we have to do case analysis:
  +
  - +
    1. use 4-17 but not 2-5: there are two convex polygon with 3 and 6 vertices, so there are 14*1=14 ways to triangulation
    +
  - +
    1. use 2-5 but not 4-17: for the graph under 2-5, it is a convex polygon with 4 vertices so there are 2 ways; for the upper part, 2-17 is a forced diagonal so it can be treated as a 4 vertices polygon so overall it has 2*2=4 ways
    +
  - +
    1. not use both 4-17 and 2-5: for the last situation, 3-17 is a forced diagonal so it has 5*1*1=5 ways
    +
  +
  Overall, there are 4*1*4*(14+4+5)=368 distinct ways to triangulation this graph
  +
+
+
1. +
  In this graph, there are a pair of diagonals that are mutual exclusive: 1-6 and 0-2, if we do case analysis based on that:
  +
  - +
    1. use 1-6: the left side is a convex polygon with 8 vertices so it has 6 ways; the right side is a blunt nose fox so it has 4 ways
    +
  - +
    1. use 0-2: then there are two forced diagonals (0-6 and 2-6). If we connect them, then there is a convex polygon with 7 vertices on the left and a fox on the right. In total, there should be 1*1*1*42 ways
    +
  +
  Over all, there are 132*4+42=570 distinct ways of triangulations.
  +
+

Problem 2

+
1. +
  - +
    +
  - +
    1. As the graph above, there are 6 reds, 6 blues and 7 yellows, then we put guards on either reds or blues and we can get 6 guards.
    +
  - +
    1. +
      +If we put witness points near 18, 2, 7 and 10, and in between 4 and 5, we got 5 areas that does not overlap on any vertex, $w(P) \geq 5$ . Also, we can find 5 guards A, B, C, D and E that cover P, thus $g(P) \leq 5$
      +
      $5 \leq w(p) \leq g(p) \leq 5$ , and we can get $g(p) = 5$
      +
    +
  - +
    1. +
      +If we put witness points near 18, 2 and 7, and in between 4 and 5, we got 4 areas that does not overlap, so $w(P) \geq 4$ . Also, we can find 4 guards A, B, C and D that cover P, thus $g(P) \leq 4$
      +
      $4 \leq w(p) \leq g(p) \leq 4$ , and we can get $g(p) = 4$
      +
    +
  +
+
+
1. +
  - +
    +
  - +
    1. As the graph above, there are 8 reds, 8 blues and 8 yellows, then we put guards on any of reds, yellows or blues and we can get 8 guards.
    +
  - +
    1. +
      +If we put witness points near 3, 10, 17 and 23, and in between 13 and 14, we got 5 areas that does not overlap on any vertex, $w(P) \geq 5$ . Also, we can find 5 guards A, B, C, D and E that cover P, thus $g(P) \leq 5$
      +
      $5 \leq w(p) \leq g(p) \leq 5$ , and we can get $g(p) = 5$
      +
    +
  - +
    1. +
      +If we put witness points near 3, 10, 17 and 23, and in between 13 and 14, we got 5 areas that does not overlap, $w(P) \geq 5$ . Also, we can find 5 guards A, B, C, D and E that cover P, thus $g(P) \leq 5$
      +
      $5 \leq w(p) \leq g(p) \leq 5$ , and we can get $g(p) = 5$
      +
    +
  +
+

+ + + + + \ No newline at end of file diff --git a/HW1/hw1.md b/HW1/hw1.md new file mode 100644 index 0000000..9166fb2 --- /dev/null +++ b/HW1/hw1.md @@ -0,0 +1,38 @@ +# HW1 +## Problem 1 + - 1. In this graph, there are 3 forced diagonals (7-12, 6-14 and 5-17) which divide this graph in to 4 parts. For the right most part it is a blunt nose fox so it has 4 ways of triangulations; for the second one there is only one way because it is a fox; the third one is again, a blunt nose fox with 4 ways; as for the last part we have to do case analysis: + - 1. use 4-17 but not 2-5: there are two convex polygon with 3 and 6 vertices, so there are 14*1=14 ways to triangulation + - 2. use 2-5 but not 4-17: for the graph under 2-5, it is a convex polygon with 4 vertices so there are 2 ways; for the upper part, 2-17 is a forced diagonal so it can be treated as a 4 vertices polygon so overall it has 2*2=4 ways + - 3. not use both 4-17 and 2-5: for the last situation, 3-17 is a forced diagonal so it has 5\*1\*1=5 ways + + Overall, there are 4\*1\*4\*(14+4+5)=368 distinct ways to triangulation this graph + + - 2. In this graph, there are a pair of diagonals that are mutual exclusive: 1-6 and 0-2, if we do case analysis based on that: + - 1. use 1-6: the left side is a convex polygon with 8 vertices so it has 6 ways; the right side is a blunt nose fox so it has 4 ways + - 2. use 0-2: then there are two forced diagonals (0-6 and 2-6). If we connect them, then there is a convex polygon with 7 vertices on the left and a fox on the right. In total, there should be 1\*1\*1\*42 ways + + Over all, there are 132*4+42=570 distinct ways of triangulations. + +## Problem 2 + - 1. + - 1. ![2_1_1](2_1_1.png) + - 2. As the graph above, there are 6 reds, 6 blues and 7 yellows, then we put guards on either reds or blues and we can get 6 guards. + - 3. ![2_1_2](2_1_2.png) + If we put witness points near 18, 2, 7 and 10, and in between 4 and 5, we got 5 areas that does not overlap on any vertex, $w(P) \geq 5$. Also, we can find 5 guards A, B, C, D and E that cover P, thus $g(P) \leq 5$ + + $5 \leq w(p) \leq g(p) \leq 5$, and we can get $g(p) = 5$ + - 4. ![2_1_3](2_1_3.png) + If we put witness points near 18, 2 and 7, and in between 4 and 5, we got 4 areas that does not overlap, so $w(P) \geq 4$. Also, we can find 4 guards A, B, C and D that cover P, thus $g(P) \leq 4$ + + $4 \leq w(p) \leq g(p) \leq 4$, and we can get $g(p) = 4$ + - 2. + - 1. ![2_2_1](2_2_1.png) + - 2. As the graph above, there are 8 reds, 8 blues and 8 yellows, then we put guards on any of reds, yellows or blues and we can get 8 guards. + - 3. ![2_2_2](2_2_2.png) + If we put witness points near 3, 10, 17 and 23, and in between 13 and 14, we got 5 areas that does not overlap on any vertex, $w(P) \geq 5$. Also, we can find 5 guards A, B, C, D and E that cover P, thus $g(P) \leq 5$ + + $5 \leq w(p) \leq g(p) \leq 5$, and we can get $g(p) = 5$ + - 4. ![2_2_2](2_2_2.png) + If we put witness points near 3, 10, 17 and 23, and in between 13 and 14, we got 5 areas that does not overlap, $w(P) \geq 5$. Also, we can find 5 guards A, B, C, D and E that cover P, thus $g(P) \leq 5$ + + $5 \leq w(p) \leq g(p) \leq 5$, and we can get $g(p) = 5$ \ No newline at end of file