Euclidean Algorithm - Discrete Mathematics - Lecture Slides, Slides of Discrete Mathematics

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CSE115/ENGR160 Discrete Mathematics 04/05/

5.4 Recursive algorithms

• An algorithm is called recursive if it solves a problem by reducing it to an instance of the same problem with smaller input
• Give a recursive algorithm for computing n! where n is a non-negative integer
• We can compute n!=n∙(n-1)! Where n is a positive integer, and that 0!=1 for a particular integer
• We use the recursive step n times

Recursive algorithm for an

• procedure power(a: nonzero real number, n: non-negative integer) if n=0 then power(a,n):= else power(a,n):=a ∙ power(a, n-1)

Example

• Formulate a recursive algorithm for computing bn^ mod m, where b, n and m are integers with m≥2, n≥0, 1≤b<m
• Let m be a positive integer and let a and b be integers, then (a+b) mod m=((a mod m) + (b mod m)) mod m ab mod m = ((a mod m)(b mod m)) mod m
• Thus, b n^ mod m = (b (b n-1^ mod m)) mod m, and b 0 mod m = 1

Example

• procedure mpower(b, n, m: integers with m≥2, n≥0) if n=0 then mpower(b, n, m)= else if n is even then mpower(b, n, m)=mpower(b, n/2, m)^2 mod m else mpower(b, n, m)=mpower(b, ⌞n/2⌟, m)^2 mod m ∙ b mod m) mod m {mpower(b, n, m)=bn^ mod m}
• Trace the above algorithm with b=2, n=5, and m=3 (2^5 mod 3)
• Since n is odd, we have mpower(2, 5, 3)=(mpower(2,2,3)^2 mod 3 ∙ 2 mod 3) mod 3
• Next, mpower(2,2,3)=mpower(2,1,3)^2 mod 3, and mpower(2,1,3)= (mpower(2,0,3)^2 mod 3 ∙ 2 mod 3) mod 3, and mpower(2,0,3)=
• So, mpower(2,1,3)=(1^2 mod 3 ∙ 2 mod 3) mod 3 = 2
• So, power(2, 2, 3)=2 2 mod 3 =1 and finally mpower(2, 5, 3)=(1 2 mod 3 ∙ 2 mod 3) mod 3 =2 7

Example

• Give a recursive algorithm for gcd(a, b) with a < b
• A recursive version of Euclidean algorithm: b=aq+r, gcd(b,a)=gcd(a,r)
• procedure gcd(a, b: non-negative integers with a < b) if a=0 then gcd(a, b):=b else gcd(a, b):=gcd(b mod a, a)
• Let a=5 and b=8. gcd(5, 8)=gcd(8 mod 5, 5)=gcd(3, 5)
• Next, gcd(3, 5)=gcd(5 mod 3, 3)=gcd(2, 3), then gcd(2, 3)=gcd (3 mod 2, 2)=gcd(1, 2). Finally gcd(1,2)=gcd(2 mod 1, 2) = gcd(0, 1)=1. Thus, gcd(5,8)=1 (^) Docsity.com 8

Proving recursive algorithm

• Prove that the recursive algorithm power(a, n) is correct
• procedure power(a: nonzero real number, n: non-negative integer) if n=0 then power(a,n):= else power(a,n):=a ∙ power(a, n-1)
• Basis step: If n=0, power(a,0)=1. This is correct as a 0 =

Proving recursive algorithm

• Show the recursive algorithm for a n
• Basis step: n=0, power(a,0)=
• Inductive step: The inductive hypothesis is power(a,k)=a k^ for a≠0 and non-negative k. To complete the proof, we need to show power(a,k+1)= a k+
• As k+1 is a positive integer, the algorithms sets power(a, k+1)=a∙power(a, k)=a∙a k^ = a k+
• This completes the inductive step

Recursion and iteration

• fibonacci(4)

Iterative algorithm for Fibonacci

Number

• procedure iterative_fibonacci(n: nonzero integer) if n=0 then y:= else begin x:= y:= for i:=1 to n- z:=x+y x:=y y:=z end end {y is the n-th Fibonacci number}

• Example: Sort the list 8, 2, 4, 6, 9, 7, 10, 1, 5, 3
• Merge sort: split the list into individual elements by successively splitting lists into two
• Sorting is done by successively merging pairs of lists

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Merge sort

Merge sort

• In general, a merge sort proceeds by iteratively splitting lists into two sbulists represented by a balanced binary tree
• We can also describe the merge sort recursively
• procedure mergesort(L=a 1 , …, a (^) n ) if n>1 then m:= ⌞n/2⌟ L 1 =a 1 , a 2 , …, a (^) m L 2 =a (^) m+1 , …, a (^) n L=merge(mergesort(L 1 ), mergesort(L 2 )) {L is a sorted list in non-decreasing order}

Merge sort

• procedure merge(L1, L 2 : sorted lists) L:=empty set while L1and L 2 are both non-empty begin remove smaller of first element of L 1 and L 2 from the list it is in and put it at the right end of L if removal of this element makes one list empty then remove all elements from the other list and append them to L end {L is a sorted list in non-decreasing order}

Merge sort

• Two sorted lists with m elements and n elements can be merged into a sorted list using no more than m+n-1 comparisons