1. 6/12/2016 Prepared by Dr.Saad Alabbad1 CS100 : Discrete Structures Proof Techniques(2) Mathematical Induction & Recursion Dr.Saad Alabbad Department of Computer Science E-mail: salabbad@gmail.com Tel # 2581888

2. 6/12/2016 Prepared by Dr.Saad Alabbad2 Mathematical Induction  Principle of Mathematical Induction To prove that P(n) is true for all positive integers n where P(n) is a propositional function we complete two steps: Basis Step: prove that P(1) is true. Inductive Step: show that if P(k) is true then P(k+1) is true for all positive integers k (i.e P(k)  P(k+1) ) This can be expressed as [P(1)  k P(k)  P(k+1)]  n P(n)

3. 6/12/2016 Prepared by Dr.Saad Alabbad3 Mathematical Induction  Example(1): Show that if n is a postive integer then 1+2+…+n=[n(n+1)]/2  Basis Step: P(1) is true since 1=[1(1+1)]/2=2/2=1  Inductive Step:  Assume that P(k) is true i.e 1+2+…+k=[k(k+1)]/2 (*)  We need to show that P(k+1) is true  By adding k+1 to (*) we obtain 1+2+…+k+(k+1)= [k(k+1)]/2+(k+1) = [k(k+1)+2(k+1)]/2 =(k+1)(k+2)/2 by taking k+1 as a common factor =P(k+1) So P(k+1) is true under the assumption that P(k) is true which completes the proof.

4. 6/12/2016 Prepared by Dr.Saad Alabbad4 Mathematical Induction  Example(2): Show that n<2 n for all positive integers n  Basis Step: P(1) is true since 1<2 1  Inductive Step:  Assume that P(k) is true i.e k<2 k (*)  We need to show that P(k+1) is true i.e k+1<2 k+1  By adding 1 to (*) we obtain  K+1<2 k +1.  But since 1  2 k for all positive integers k. We have  K+1<2 k +1   2 k +2 k  =2.2 k  =2 k+1  K+1<2 k+1 So P(k+1) is true under the assumption that P(k) is true which completes the proof.

5. 6/12/2016 Prepared by Dr.Saad Alabbad5 Mathematical Induction  Example(3): Show that the number of subsets of a finite set S is 2 n where n is the number of elements of the set S.  Basis Step: P(0) is true since the set with zero elements (empty set) has exactly one subset. Namely itself 2 0 =1  Inductive Step:  Assume that P(k) is true i.e every set with k elements has 2 k subsets  We need to show that P(k+1) is true i.e every set with k+1 elements has 2 k+1 subsets  Let T be a set with k+1 elements then we can write T=S  {a} where a is an element of T and S=T-{a}  For each subset X of S there are two subsets of T: subset that contains a i.e X  {a} and the subset that does not contain a i.e X (see next slide)  Since S has k elements, there are 2 k subsets of S and hence there are 2. 2 k = 2 k+1 subsets of T (see next slide)  So P(k+1) is true under the assumption that P(k) is true which completes the proof.

6. 6/12/2016 Prepared by Dr.Saad Alabbad6 Mathematical Induction { } { } { } {1} { } {1} { } {2} {1} {1,2} { } {2} {1} {1,2} { } {3} {2} {2,3} {1} {1,3} {1,2} {1,2,3} { } {3} {2} {2,3} {1} {1,3} {1,2} {1,2,3} S={1,2} T={1,2,3} a=3 Every subset of T can be generated by adding or omitting element a from every subset of S Example(3): Illustration

7. 6/12/2016 Prepared by Dr.Saad Alabbad7 Mathematical Induction  Example(4): Show that 2 n  n! for all positive integers n  4  Basis Step: P(4) is true since 2 4 =16  4!=24 (note: basis step is 4)  Inductive Step:  Assume that P(k) is true i.e 2 k  k! (*) for all k  4  We need to show that P(k+1) is true i.e 2 k+1  (k+1)!  Multiply (*) by 2, we obtain  2.2 k =2 k+1  2.K!   (k+1)k! Since 2<k+1 =(k+1)! So 2 k+1  (k+1)! So P(k+1) is true under the assumption that P(k) is true which completes the proof.

8. 6/12/2016 Prepared by Dr.Saad Alabbad8 Strong Mathematical Induction  Principle of Strong Mathematical Induction To prove that P(n) is true for all positive integers n where P(n) is a propositional function we complete two steps: Basis Step: prove that P(1) is true. Inductive Step: show that if P(1)  P(2) ..  P(k) are true then P(k+1) is true for all positive integers k (i.e P(k)  P(k+1) ) This can be expressed as [P(1)  P(2)…  P(k)  P(k+1)]  n P(n)  Sometimes it is called second principle of induction  It is equivalent to the mathematical induction we have seen.However, in many cases it is much easier to prove using the strong induction.  Note that the difference lies in the inductive step. We assume that P(1),P(2) up to P(k) are true and we show that P(k+1) is true where in the normal induction we assume that only P(k) is true and show that P(k+1) is true.

9. 6/12/2016 Prepared by Dr.Saad Alabbad9 Strong Mathematical Induction  Example: Show that if n is an integer greater than 1, then n can be written as the product of primes  Basis Step: P(2) is true since 2 can be written as the product of one prime (itself!)  Inductive Step:  Assume that P(2),…P(k) are true i.e P(j) is true for 2  j  k  We need to show that P(k+1) is true i.e k+1 can be written as product of primes  Now we have two cases:  K+1 is prime which means that P(k+1) is true.  K+1 is composite and can be written as the product of two positive integers 2  a  b  k+1  By the inductive step we see that both a and b can be written as the product of primes. Thus if K+1 is composite it can hbe written as the product of primes (primes in the factorization of a and b) So P(k+1) is true under the assumption that P(j) is true for 2  j  k which completes the proof. Note: this theorem is called the Fundamental Theorem of Arithmetic

10. 6/12/2016 Prepared by Dr.Saad Alabbad10 Recursive Definitions: Recursively Defined Functions Recursion See: "Recursion" if you still don't get it,  Recursion: Defining something (process,function,set,concept,etc..) in terms of itself  Example(1): One’s ancestors : his father and his father’s ancestors  Example(2): Factorial of n is defined as n!=n*n-1*….*1 Can be defined recursively as: n!=n* (n-1)!

11. 6/12/2016 Prepared by Dr.Saad Alabbad11 Recursive Definitions: Recursion See: "Recursion" if you still don't get it,  Recursion: Defining something (process,function,set,concept,etc..) in terms of itself  Example(1): One’s ancestors : his father and his father’s ancestors  Example(2): Factorial of n is defined as n!=n*n-1*….*1 Can be defined recursively as: n!=n* (n-1)!

12. 6/12/2016 Prepared by Dr.Saad Alabbad12 Recursive Definitions: Recursively Defined Functions Recursive functions have two parts 1)Basis step: gives the value of the function at some initial value (usually zero). It is needed to terminate the recursive step. 2)Recursive Step: gives a rule of finding the value of the function from previous values  Example(1): Factorial of n is defined as n!=n*n-1*….*1 Can be defined recursively as: 0!=1 [basis step] n!=n* (n-1)! [recursive step]  Example(2): Let F(n)=0+1+2+…+n. Define F(n) recursively: F(0)=0 [basis step] F(n)=n+F(n-1) [recursive step] we can also write F(n+1)=(n+1)+F(n)  Question : Define F(n)=2 n recursively.

13. 6/12/2016 Prepared by Dr.Saad Alabbad13 Recursive Definitions: Recursively Defined Functions  Example(3):Febonacci numbers f n are defined as: f 0 =0 [basis step] f 1 =1 f n = f n-1 + f n-2 [inductive step] Compute f 5. f 5 = f 4 +f 3 f 2 =1+0=1 f 3 =f 2 +f 1 =1+1=2 = f 3 +f 2 +f 2 +f 1 f 4 =f 3 +f 2 =2+1=3 f 5 =f 4 +f 3 =3+2=5 =f 2 +f 1 +f 1 +f 0 +f 1 +f 0 +1 =f 1 +f 0 + 1+ 1+ 0+1+ 0+1=5 More efficient way

14. 6/12/2016 Prepared by Dr.Saad Alabbad14 Recursive Definitions: Recursively Defined Sets Recursive definition of sets have two parts 1)Basis step: initial collection of elements in the set. 2)Recursive Step: gives a rule for forming new elements in the set from those already in the set.  Example(1): Define the set S as follow 2  S [basis step] If x  S and y  S then x+y  S [ Recursive step] What is S ?? 2 in S 2+2=4 in S [Recursive step] 2+4=6 in S [Recursive step] Etc… Can you figure it out?

15. 6/12/2016 Prepared by Dr.Saad Alabbad15 Recursive Definitions: Recursive Algorithm Recursive algorithms: solve a problem by reducing it to an instance of the same problem with smaller input. Factorial Algorithm int fac(int n) { if (n == 0) return 1; return n * fac(n-1); } Exponentiation Algorithm int power(int a, int n) { if (n == 0) return 1; return a * power (a,n-1); }

16. 6/12/2016 Prepared by Dr.Saad Alabbad16 Recursive Definitions: Fibonacci Recursive Algorithm Fibonacci Algorithm int fib(int n) { if (n == 0) return 0; if (n == 1) return 1; return fib(n-1)+ fib(n-2); } f5f5 f4f4 f3f3 f3f3 f2f2 f2f2 f1f1 f0f0 f1f1 f1f1 f0f0 f2f2 f1f1 f1f1 f0f0

17. 6/12/2016 Prepared by Dr.Saad Alabbad17 Recursive Definitions: Fibonacci Recursive Algorithm Fibonacci Algorithm int fib(int n) { if (n == 0) return 0; if (n == 1) return 1; return fib(n-1)+ fib(n-2); } f5f5 f4f4 f3f3 f3f3 f2f2 f2f2 f1f1 f0f0 f1f1 f1f1 f0f0 f2f2 f1f1 f1f1 f0f0