74 lines
2.2 KiB
Markdown
74 lines
2.2 KiB
Markdown
# HW3
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## Problem 1
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- 1.
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| Deque | Left test | Left test |
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| ----------------- | ------------- | ------------- |
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| <2,1,0,2> | | |
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| <3,2,1,0,3> | Left(2,1,3)T | Left(0,2,3)F |
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| <3,2,1,0,3> | Left(3,2,4)T | Left(0,3,4)T |
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| <5,3,2,1,0,5> | Left(3,2,5)T | Left(0,3,5)F |
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| <6,3,2,1,0,5,6> | Left(5,3,6)F | Left(0,5,6)T |
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| <6,3,2,1,0,5,6> | Left(6,3,7)T | Left(5,6,7)T |
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| <6,3,2,1,0,5,6> | Left(6,3,8)T | Left(5,6,8)T |
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| <9,2,1,0,5,6,9> | Left(6,3,9)F | Left(5,6,9)T |
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| <10,1,0,5,6,9,10> | Left(9,2,10)F | Left(6,9,10)T |
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the first and the last one of each deque are the bottom and the top.
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- 2.
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- 1. Impossible, Edge $v_4v_5$ would be forced to cross the path $v_0v_1v_2$.
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- 2. Possible, 
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- 3. Impossible, <2, 1, 0, 2> means 0 is on the left of $v_2v_1$. However, <4, 1, 2, 4> contains only right of $v_2v_1$, which does not include 0.
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## Problem 2
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``` C
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#include <stdio.h>
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#define true 1
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#define false 0
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typedef int tPointi[2];
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int Area2(tPointi a, tPointi b, tPointi c) {
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return (b[0] - a[0]) * (c[1] - a[1]) -
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(c[0] - a[0]) * (b[1] - a[1]);
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}
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int Collinear(tPointi a, tPointi b, tPointi c) {
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return Area2(a, b, c) == 0;
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}
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int Between(tPointi a, tPointi b, tPointi c) {
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if (!Collinear(a, b, c))
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return false;
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if (a[0] != b[0])
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return ((a[0] < c[0] && b[0] > c[0]) ||
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(a[0] > c[0] && b[0] < c[0]));
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else
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return ((a[1] < c[1] && b[1] > c[1]) ||
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(a[1] > c[1] && b[1] < c[1]));
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}
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int TIntersect(tPointi a, tPointi b, tPointi c, tPointi d) {
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if (Between(a, b, c) && !Collinear(a, b, d))
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return true;
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if (Between(a, b, d) && !Collinear(a, b, c))
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return true;
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if (Between(c, d, a) && !Collinear(c, d, b))
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return true;
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if (Between(c, d, b) && !Collinear(c, d, a))
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return true;
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return false;
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}
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int main() {
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tPointi a = {0, 0};
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tPointi b = {0, 2};
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tPointi c = {0, 1};
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tPointi d = {1, 1};
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printf("TIntersect is %s", TIntersect(a, b, c, d)?"true":"false");
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}
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``` |