CSI 32. Lecture 15. Chapter 11 Recursion Functional Recursion 11.4 Binary Search

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Lecture 15 Chapter 11 Recursion 11.3 Functional Recursion 11.4 Binary Search

11.3 Functional Recursion Functional recursion is a method of defining functions in which the function being defined is applied within its own definition. Example 1: Fibonacci sequence: F(0)=1 (base case) F(1)=1 (base case) F(n)=F(n-1) + F(n-2) for all integers n>1 (recursive def.) Example 2: Factorial Recall that usually we have this definition: n! = 1*2*3*4*5*...*(n-2)*(n-1)*n

11.3 Functional Recursion Functional recursion is a method of defining functions in which the function being defined is applied within its own definition. Example 1: Fibonacci sequence: F(0)=1 (base case) F(1)=1 (base case) F(n)=F(n-1) + F(n-2) for all integers n>1 (recursive def) Example 2: Factorial: Recall that usually we have this definition: n! = 1*2*3*4*5*...*(n-2)*(n-1)*n Note (n-1)!

11.3 Functional Recursion Functional recursion is a method of defining functions in which the function being defined is applied within its own definition. Example 1: Fibonacci sequence: F(0)=1 (base case) F(1)=1 (base case) F(n)=F(n-1) + F(n-2) for all integers n>1 (recursive def.) Example 2: Factorial: Recall that usually we have this definition: n! = 1*2*3*4*5*...*(n-2)*(n-1)*n Therefore we can give a recursive definition: 1! = 1 n! = n*(n-1)!

11.3 Functional Recursion Let's see the program that finds factorial of a number, using recursive definition: 1! = 1 n! = n*(n-1)! Here is a draft of the program: def factorial(n): if n == 1: return 1 else: return n*factorial(n-1) def main(): n=input('please, input n:') main() F=factorial(n) print("%d! = %d"%(n,f))

11.3 Functional Recursion Let's trace the call of factorial(4): def factorial(n): if n == 1: return 1 else: return n*factorial(n-1) factorial(4) n=4 return 4* factorial(3)

11.3 Functional Recursion Let's trace the call of factorial(4): def factorial(n): if n == 1: return 1 else: return n*factorial(n-1) factorial(4) n=4 return factorial(3) 4* n=3 return 3* factorial(2)

11.3 Functional Recursion Let's trace the call of factorial(4): def factorial(n): if n == 1: return 1 else: return n*factorial(n-1) factorial(4) n=4 return factorial(3) 4* n=3 return 3* factorial(2) n=2 return 2* factorial(1)

11.3 Functional Recursion Let's trace the call of factorial(4): def factorial(n): if n == 1: return 1 else: return n*factorial(n-1) factorial(4) n=4 return factorial(3) 4* n=3 return 3* factorial(2) n=2 return 2* factorial(1) n=1 return 1

11.3 Functional Recursion Let's trace the call of factorial(4): def factorial(n): if n == 1: return 1 else: return n*factorial(n-1) factorial(4) n=4 return factorial(3) 4* n=3 return 3* factorial(2) n=2 return 2* factorial(1) n=1 1 return 1

11.3 Functional Recursion Let's trace the call of factorial(4): def factorial(n): if n == 1: return 1 else: return n*factorial(n-1) factorial(4) n=4 return factorial(3) 4* n=3 return 3* factorial(2) n=2 2 return 2* factorial(1) n=1 2*1=2 1 return 1

11.3 Functional Recursion Let's trace the call of factorial(4): def factorial(n): if n == 1: return 1 else: return n*factorial(n-1) factorial(4) n=4 return factorial(3) 4* n=3 6 return 3* factorial(2) n=2 3*2=6 2 return 2* factorial(1) n=1 2*1=2 1 return 1

11.3 Functional Recursion Let's trace the call of factorial(4): def factorial(n): if n == 1: return 1 else: return n*factorial(n-1) factorial(4) 24 n=4 return factorial(3) 4* n=3 4*6=24 6 return 3* factorial(2) n=2 3*2=6 2 return 2* factorial(1) n=1 2*1=2 1 return 1

11.3 Functional Recursion See the program factorial_rec.py

11.3 Functional Recursion Another example of recursive function: Ackermann function or Ackermann Péter function. n+ 1 if m=0 A(m,n) = A(m-1,1) if m>0 and n=0 A(m-1,A(m,n-1)) if m>0 and n>0 This function has only recursive definition, and it grows very fast. Let's see how it works: A(0,4) = 5 A(0,7) = 8 A(1,0) = A(0,1) = 2

11.3 Functional Recursion Another example of recursive function: Ackermann function or Ackermann Péter function. n+ 1 if m=0 A(m,n) = A(m-1,1) if m>0 and n=0 A(m-1,A(m,n-1)) if m>0 and n>0 This function has only recursive definition, and it grows very fast. Let's see how it works: A(0,4) = 5 A(0,7) = 8 A(1,0) = A(0,1) = 2 A(2,0) = A(1,1) = A(0, A(1,0))=A(0,2)=3

11.3 Functional Recursion Another example of recursive function: Ackermann function or Ackermann Péter function. n+ 1 if m=0 A(m,n) = A(m-1,1) if m>0 and n=0 A(m-1,A(m,n-1)) if m>0 and n>0 This function has only recursive definition, and it grows very fast. Let's see how it works: A(0,4) = 5 A(0,7) = 8 A(1,0) = A(0,1) = 2 A(2,0) = A(1,1) = A(0, A(1,0))=A(0,2)=3 A(1,2) = A(0, A(1,1))=A(0,3)=4

11.3 Functional Recursion Another example of recursive function: Ackermann function or Ackermann Péter function. n+ 1 if m=0 A(m,n) = A(m-1,1) if m>0 and n=0 A(m-1,A(m,n-1)) if m>0 and n>0 This function has only recursive definition, and it grows very fast. Let's see how it works: A(0,4) = 5 A(0,7) = 8 A(1,0) = A(0,1) = 2 A(2,0) = A(1,1) = A(0, A(1,0))=A(0,2)=3 A(1,2) = A(0, A(1,1))=A(0,3)=4 A(2,1) = A(1, A(2,0))=A(1,3)=A(0, A(1,2))=A(0,4)=5

11.3 Functional Recursion Another example of recursive function: Ackermann function or Ackermann Péter function. n+ 1 if m=0 A(m,n) = A(m-1,1) if m>0 and n=0 A(m-1,A(m,n-1)) if m>0 and n>0 This function has only recursive definition, and it grows very fast. Let's see how it works: A(0,4) = 5 A(0,7) = 8 A(1,0) = A(0,1) = 2 A(2,0) = A(1,1) = A(0, A(1,0))=A(0,2)=3 A(1,2) = A(0, A(1,1))=A(0,3)=4 A(2,1) = A(1, A(2,0))= A(1,3)=A(0, A(1,2))=A(0,4)=5 A(3,2) = A(2,A(3,1))=A(2,A(2,A(3,0)))=A(2,A(2,A(2,1)))= A(2,A(2,5))=A(2,A(1,A(2,4)))=A(2,A(1,A(1,A(2,3))))= A(2,A(1,A(1,A(1,A(2,2))))=A(2,A(1,A(1,A(1,A(1,A(2,1))))))= A(2,A(1,A(1,A(1,A(1,5)))))=A(2,A(1,A(1,A(1,A(0,A(1,4))))))= A(2,A(1,A(1,A(1,A(0,A(0,A(1,3)...)= A(2,A(1,A(1,A(1,A(0,A(0,A(0,A(1,2))...)=

=A(2,A(1,A(1,A(1,A(0,A(0,A(0,4)...)=A(2,A(1,A(1,A(1,A(0,A(0,5)...)= A(2,A(1,A(1,A(1,A(0,6)...)=A(2,A(1,A(1,A(1,7))))=A(2,A(1,A(1,A(0,A(1,6)) )))=A(2,A(1,A(1,A(0,A(0,A(1,5)...)=A(2,A(1,A(1,A(0,A(0,A(0,A(1,4)...)= A(2,A(1,A(1,A(0,A(0,A(0,A(0,A(1,3)...)= A(2,A(1,A(1,A(0,A(0,A(0,A(0,A(0,A(1,2)...)= A(2,A(1,A(1,A(0,A(0,A(0,A(0,A(0,4)...)=A(2,A(1,A(1,A(0,A(0,A(0,A(0,5)...) =A(2,A(1,A(1,A(0,A(0,A(0,6)...)=A(2,A(1,A(1,A(0,A(0,7)...)= A(2,A(1,A(1,A(0,8)...)=A(2,A(1,A(1,9)...)=A(2,A(1,A(0,A(1,8))))= A(2,A(1,A(0,A(0,A(1,7)...)=A(2,A(1,A(0,A(0,A(0,A(1,6)...)= A(2,A(1,A(0,A(0,A(0,A(0,A(1,5)...)= A(2,A(1,A(0,A(0,A(0,A(0,A(0,A(1,4)...)= A(2,A(1,A(0,A(0,A(0,A(0,A(0,A(0,A(1,3)...)= A(2,A(1,A(0,A(0,A(0,A(0,A(0,A(0,A(0,A(1,2)...)= A(2,A(1,A(0,A(0,A(0,A(0,A(0,A(0,A(0,4)...)= A(2,A(1,A(0,A(0,A(0,A(0,A(0,A(0,5)...)= A(2,A(1,A(0,A(0,A(0,A(0,A(0,6)...)=A(2,A(1,A(0,A(0,A(0,A(0,7)...)= A(2,A(1,A(0,A(0,A(0,8)...)=A(2,A(1,A(0,A(0,9))))=A(2,A(1,A(0,10)))= A(2,A(1,11))=A(2,A(0,A(1,10)))=A(2,A(0,A(0,A(1,9))))= A(2,A(0,A(0,A(0,A(1,8)...)=A(2,A(0,A(0,A(0,A(1,8)...)= A(2,A(0,A(0,A(0,A(0,A(1,7)...)=A(2,A(0,A(0,A(0,A(0,A(0,A(1,6)...)= A(2,A(0,A(0,A(0,A(0,A(0,A(0,A(1,5)...)= A(2,A(0,A(0,A(0,A(0,A(0,A(0,A(0,A(1,4)...)= A(2,A(0,A(0,A(0,A(0,A(0,A(0,A(0,A(0,A(1,3)...)=

A(2,A(0,A(0,A(0,A(0,A(0,A(0,A(0,A(0,5)...)= A(2,A(0,A(0,A(0,A(0,A(0,A(0,A(0,6)...)= A(2,A(0,A(0,A(0,A(0,A(0,A(0,7)...)=A(2,A(0,A(0,A(0,A(0,A(0,8)...)= A(2,A(0,A(0,A(0,A(0,9)...)=A(2,A(0,A(0,A(0,10)...)=A(2,A(0,A(0,11)...)= A(2,A(0,12)...)=A(2,13)...)=A(1,A(2,12))=A(1,A(1,A(2,11)))= A(1,A(1,A(1,A(2,10))))=A(1,A(1,A(1,A(1,A(2,9)))))= A(1,A(1,A(1,A(1,A(1,A(2,8)...)=A(1,A(1,A(1,A(1,A(1,A(1,A(2,7)...)= A(1,A(1,A(1,A(1,A(1,A(1,A(1,A(2,6)...)= = 29

11.4 Binary Search A binary search algorithm locates the position of an element (a number, a letter, a word,...) in a sorted list. It inspects the middle element of the sorted list: if equal to the sought value, then the position has been found; otherwise, the lower(left) half or upper(right) half is chosen for further searching based on: whether the sought value is less than or greater than the middle element. This method reduces the number of elements needed to be checked by a factor of two each time, and finds the sought value if it exists in the list or if not determines "not present". Complexity: logarithmic i.e. if we start with n elements, the algorithm will terminate in at most k log 2 n steps, where k is a constant.

11.4 Binary Search Our book has a nice example with lexicon (please, take a look at it). We will deal with numbers in class. Input: a sorted list of numbers (integers of floating point numbers), a number to find (may be not present in the list) Output: location of the element in the list (if present), 'Not in the list' (if not present)

11.4 Binary Search Example: binary search algorithm on the following input: [1,7,9,12,17,19,23,45,67,123,167] find: 7

11.4 Binary Search Example: binary search algorithm on the following input: [1,7,9,12,17,19,23,45,67,123,167] find: 7 start stop Number of elements if the array: 11, start index = 0, stop index = 10

11.4 Binary Search Example: binary search algorithm on the following input: [1,7,9,12,17,19,23,45,67,123,167] find: 7 start stop Number of elements if the array: 11, start index = 0, stop index = 10 Middle element: (start+stop) 2 = (0+10) 2 =5 or (0+10) // 2 =5 th

11.4 Binary Search Example: binary search algorithm on the following input: [1,7,9,12,17,19,23,45,67,123,167] find: 7 start stop Number of elements if the array: 11, start index = 0, stop index = 10 Middle element: (start+stop) 2 = (0+10) 2 =5 or (0+10) // 2 =5 th 5 th element 19 = 7 the number we are looking for? No

11.4 Binary Search Example: binary search algorithm on the following input: [1,7,9,12,17,19,23,45,67,123,167] find: 7 start stop Number of elements if the array: 11, start index = 0, stop index = 10 Middle element: (start+stop) 2 = (0+10) 2 =5 or (0+10) // 2 =5 th 5 th element 19 = 7 the number we are looking for? No 19 7? True

11.4 Binary Search Example: binary search algorithm on the following input: [1,7,9,12,17,19,23,45,67,123,167] find: 7 start stop Number of elements if the array: 11, start index = 0, stop index = 10 Middle element: (start+stop) 2 = (0+10) 2 =5 or (0+10) // 2 =5 th 5 th element 19 = 7 the number we are looking for? No 19 7? True therefore take the left part (with 19)

11.4 Binary Search [1,7,9,12,17,19] find: 7 start stop Number of elements in the new array (sub-array): 6, start = 0, stop = 5

11.4 Binary Search [1,7,9,12,17,19] find: 7 start stop Number of elements in the new array (sub-array): 6, start = 0, stop = 5 Middle element: (0+5) 2 =2 or (0+5)//2 = 2 nd 2 nd element 9 = 7 the number we are looking for? No

11.4 Binary Search [1,7,9,12,17,19] find: 7 start stop Number of elements in the new array (sub-array): 6, start = 0, stop = 5 Middle element: or (0+5)//2 = 2 nd 2 nd element 9 = 7 the number we are looking for? No 9 7? True (0+5) 2 =2

11.4 Binary Search [1,7,9,12,17,19] find: 7 start stop Number of elements in the new array (sub-array): 6, start = 0, stop = 5 Middle element: (0+5) 2 =2 or (0+5)//2 = 2 nd 2 nd element 9 = 7 the number we are looking for? No 9 7? True therefore take the left part (with 9)

11.4 Binary Search [1,7,9] find: 7 start stop Number of elements in the sub-array: 3, start = 0, stop = 2

11.4 Binary Search [1,7,9] find: 7 start stop Number of elements in the sub-array: 3, start = 0, stop = 2 Middle element: (0+2) =1 or (0+2) // 2 = 1 st 2

11.4 Binary Search [1,7,9] find: 7 start stop Number of elements in the sub-array: 3, start = 0, stop = 2 Middle element: (0+2) =1 or (0+2) // 2 = 1 st 2 1 st element 7 = 7 the number we are looking for? Yes Stop, return index 1.

11.4 Binary Search See the program binary_search.py

Homework Assignment Recall Fibonacci numbers/sequence : F(0)=1 (base case) F(1)=1 (base case) F(n)=F(n-1) + F(n-2) for all integers n>1 (recursive definition) Write a program, that for a given non-negative integer n, displays n first Fibonacci numbers. Print all the Fibonacci function calls. Here is a sketch of the program (Fibonacci calls are not printed): def Fibonacci(n): if n == 0:... elif n == 1:... else:... def main(): n = eval(input("please, input any non-negative integer:")) for i in range(n): print(fibonacci(i)) next slide

Homework Assignment Similar to example with Factorial Function (recursive definition) (slides 6 13 ) draw a figure of Fibonacci Function call on n=5 Suggestion: it is probably worth doing it as a tree: F(5) return F(4) + F(3) F(4) F(3) return... return...

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