CS 173: Discrete Structures, Fall 2010 Homework 6 Solutions

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1 CS 173: Discrete Structures, Fall 010 Homework 6 Solutions This homework was worth a total of 5 points. 1. Recursive definition [13 points] Give a simple closed-form definition for each of the following subsets of the real plane. Give both a precise definition using set-builder notation and also an informal geometrical description using a picture and/or words. (a) ( points) The set T defined by: i. (1, 1) T, ii. If (x, y) T, then (x, y ) T, iii. If (x, y) T, then (x +, y) T, iv. If (x, y) T, then (x, y + ) T. [Answer:] T {(x, y) (Z + ) : x, y Z + and x+y is even}. That is, the set T consists of all ordered pairs of positive integers in the plane where the sum of the elements in each pair is even. (b) ( points) The set T R defined by: i. (0, 0) T, ii. If (x, y) T, then (x + r, (x + r) ) T where r is some real number 0, iii. If (x, y) T, then ( x, y) T. [Answer:] T {(x, y) R : y x }. That is theset T contains all points on the parabola y x. The next two problems are related. (c) ( points) The set T R containg all points of the form (3α, 9 (3α) ) T, α is a real number in the range [0, 1], (Assume that x returns the positive square root of x.) [Answer:] T {(x, y) : x, y 0 and x + y 9}. That is set T contains all points (x, y) in the first quadrant which are on the circle of radius 3 centered at the origin. 1

2 (d) (3 points) The set T R defined by: i. For all real numbers α in the range [0, 1], (3α, 9 (3α) ) T, ii. If (x, y) T, then (x, y) T, iii. If (x, y) T, then ( x, y) T. iv. If (x, y) T and (p, q) T then for some real number β in the range [0, 1], β(x, y) + (1 β)(p, q) T. [Answer:] T {(x, y) R : x + y 9}. That is, the set T contains all points on and inside the circle of radius 3 centered at the origin.. Induction [10 points] Use induction to prove that the following equation holds for all non-negative integers n: [Answer:] n j0 ( 1 )j (n+1) + ( 1) n 3 n Proof: We will prove this by using induction on n. Base Case: Consider n 0. Checking the two sides of the equation, we get: 0 j0 ( 1 )j ( 1 )0 1 and 0+1 +( 1) Since these are equal, the claim holds for n Inductive Hypothesis: Assume that the claim holds for n k N i.e. k j0 ( 1 )j k+1 +( 1) k 3 k Inductive Step: We need to show that following claim holds: k+1 j0 ( 1 )j (k+1)+1 +( 1) k+1 3 k+1. We can write the sum of the first k terms as follows Using the inductive hypothesis, we get k+1 j0 ( 1 )j k j0 ( 1 )j + ( 1 )k+1 k+1 ( 1 )j k+1 + ( 1) k + ( 1 3 k )k+1 k+1 + ( 1) k + ( ( 1)k+1 ) 3. k k+1 j0 k+1 + ( 1) k + 3 (( 1)k+1 3 k 3 ) (k+1 + ( 1) k ) + 3( 1) k+1 k 3 k k+ + ( 1) k 3( 1) k k+ 1( 1) k 3 k+1 3 k+1 k+ + ( 1)( 1) k k+ + ( 1) k+1 3 k+1 3 k+1 (k+1)+1 + ( 1) k+1 3 k+1 This is what we needed to show. Hence the claim holds for all n 0.

3 3. Son of induction [10 points] Consider a function f : N N defined by f(0). f(n) f(n 1) + ( 7) n for all non-negative integers n. (a) (5 points) Calculate the next four values of f, i.e. f(1), f(), f(3) and f(). [Answer:] f(1) f(0) + ( 7) 1 (7) 1 f() f(1) + ( 7) 1 + (9) 86 f(3) f() + ( 7) 3 86 (33) 600 f() f(3) + ( 7) (01) 0 (b) (5 points) Use induction to prove that f(n) 1 ( 7)n+1 for every integer n 0. [Answer:] Prove this claim using induction on n. Base Case: Consider n 0 and we have seen from above part that f(0). Show that f(n) 1 ( 7)n+1 when n 0 1 ( 7) 0+1 f(0). This is what we need to show. Inductive Hypothesis: Suppose that claim holds for some k N i.e. f(k) 1 ( 7)k+1 Inductive Step: We need to show that f(k ) 1 ( 7)(k+1)+1. By problem definition, f(k ) is defined as f(k ) f(k) + ( 7) k+1 Using inductive hypothesis f(k ) 1 ( 7)k+1 1 ( 7)k+1 + ( 7) k+1 + ( 7)k+1 1 ( 7)k+1 + 8( 7) k ( 7)k+1 1 ( 7)( 7)k+1 1 ( 7)k+ 1 ( 7)(k+1)+1 This is what we needed to show. Thus claim holds for all n 0. 3

4 . More on Induction [8 points] Define p(n) n 7n 0. For which integers n is p(n) non-negative? Prove your answer using induction on n. Hint 1: You can handle smaller values of n using non-inductive methods. Hint : In the proof, your inductive step needs to use its inductive hypothesis. [Answer:] This problem requires you to find value of n for which p(n) 0. Let s first try handling smaller and larger values of n. If n is negative, then 7n is positive. Since the other two terms (n and 10) are clearly positive, then p(n) is positive. Also notice that for n 7, n 7n n(n 7) is always 0. Thus p(n) 0 when n 7. Finally we are left with values from 0 to 6. At n 3 and n, p(n) < 0 and for rest of values i.e. p(n) 0 for n 0, 1,, 5, 6. Therefore p(n) 0 for all n 5. The problem asked for an inductive proof for the larger numbers. That is, we need to prove that p(n) n 7n 0 0 for all n 5. We ll prove this using induction on n. Base Case: Consider case of n 5. We can verify that p(5) 5 7(5) 0 0. Inductive Hypothesis: Assume that p(k) 0 for some integer k where k 5 i.e. k 7k 0 0. Inductive Step: We need to show that p(k ) (k ) 7(k ) 0 0. p(k ) (k ) 7(k ) 0 k + k 7k 7 0 k + k 7k 7 0 k 7k 0 + k 7 k 7k 0 + k 6 From the inductive hypothesis, we know that k 7k 0 0 and we know k 6 0 whenever k 5. So (k 7k 0) + (k 6) 0. That is p(k 0. This is what we needed to show. Therefore the claim holds for all n Induction and Geometry [13 points] Consider n lines in the plane so that no two lines are parallel and no three lines pass through a common point. (a) ( points) Find the number of regions in which a plane is partitioned using n lines where n 0, 1,, 3,. [Answer:] Let the plane be partitioned into g(n) regions using n lines. g(0) 1 region because there is no line so whole plane is one region. g(1) regions, one line can partition the plane into regions. g() regions, two lines with one intersection points breaks plane into regions. g(3) 7 regions and g() 11 regions, see figure 1.

5 Figure 1: (Left) Seven regions obtained by using 3 lines, (Right) Eleven regions obtained by using lines. (b) (3 points) By examining solutions of part (a), find a formula for the number of regions into which n lines can partition the plane. [Answer:] The formula for the number of regions into which n lines can partition the plane is g(n) n(n+1) See solution of part (c) and (d) for details on why this is correct representation of number of regions obtained using n lines. Solutions for next two problems are related. (c) (3 points) Explain why the general formula you derived in part (b) is the correct representation of number of regions obtained by using n lines. [Answer:] We know that no line in the plane is parallel to other line and no three lines intersect at a common point. Thus if we already have k lines in the plane then (k ) th line will have to cross k lines (i.e. k intersections). In order to have k intersections (k ) th line will always pass through k regions and while passing through each of k +1 regions it will divide this region into two parts. Thus we would have g(k ) g(k) + k i.e. old regions g(k) plus k new regions. Let s look at the first few values produced by the recursive formula g(k ) g(k)+k +1, to see why it matches the closed form g(k +1) k +1((k +1)+1)/+1. Notice that: g(0) 1 g(1) g(0) 1 g() g(1) g(3) g() g(k ) g(k) + k (k ) 1 + (k )((k ) )/ The next part will prove this formally, using induction. 5

6 (d) (5 points) Use induction to prove the formula derived in part (b). [Answer:] From part (c), we know that the function for the number of lines is g defined by g(0) 1 g(n) g(n 1) + n We ll use induction to prove that g(n) n(n )/. Base Case: Consider n 0. We know g(0) 1. and g(0) 0(0 )/ 1, so the two are equal. Inductive Hypothesis: Assume that g(k) k(k )/, for some k N Inductive Step: We need to show that g(k ) (k )((k ) )/. By the definition of g, we know that g(k ) g(k) + (k ) So, by the inductive hypothsis, g(k ) k(k+1) + (k ) So then g(k ) This is what we needed to show. k(k ) + (k ) k(k ) k k(k ) + + k k(k ) + (k ) (k )(k + ) (k )((k ) ) 6