Recursion - cs.cornell.edu...Recursion vs Iteration •Recursionis provably equivalentto iteration...

49
Recursion Lecture 14

Transcript of Recursion - cs.cornell.edu...Recursion vs Iteration •Recursionis provably equivalentto iteration...

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Recursion

Lecture 14

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Announcements for Today

Prelim 1

• Tonight at 7:30 pm§ A–F in Uris G01§ G-H in Malott 228§ I–L in Ives 305§ M-Z in Statler Aud.

• Graded by noon on Sun§ Scores will be in CMS§ In time for drop date

Other Announcements

• Reading: 5.8 – 5.10• Assignment 3 now graded

§ Mean 93.6, Median 98§ Time: 7.1 hr, StdDev: 3.3 hr§ But only 552 responses

• Assignment 4 posted Friday§ Parts 1-3: Can do already§ Part 4: material from today§ Due 2 weeks from yesterday

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So You Want to be a Consultant

• Applications are now open for Spring semester§ https://www.cs.cornell.edu/undergrad/ugradcoursestaff§ Applications close October 31

• CS 1110 will not do much hiring this year§ Already have a large course staff§ There will be less students in the Spring

• But CS 1133 (Short Course) is hiring!§ Decided to give it dedicated staff§ If you can get an A on prelim 1, good enough

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Recursion

• Recursive Definition: A definition that is defined in terms of itself

• Recursive Function: A function that calls itself (directly or indirectly)

PIP stands for “PIP Installs Packages”

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A Mathematical Example: Factorial

• Non-recursive definition:n! = n× n-1 ×… × 2 × 1

= n (n-1 ×… × 2 × 1)

• Recursive definition:n! = n (n-1)!0! = 1

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for n > 0 Recursive caseBase case

What happens if there is no base case?

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Factorial as a Recursive Function

def factorial(n):"""Returns: factorial of n.Pre: n ≥ 0 an int"""if n == 0:

return 1

return n*factorial(n-1)

• n! = n (n-1)!• 0! = 1

10/17/19 Recursion 6

What happens if there is no base case?

Recursive case

Base case(s)

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Example: Fibonnaci Sequence

• Sequence of numbers: 1, 1, 2, 3, 5, 8, 13, ...a0 a1 a2 a3 a4 a5 a6

§ Get the next number by adding previous two§ What is a8?

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A: a8 = 21B: a8 = 29C: a8 = 34D: None of these.

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Example: Fibonnaci Sequence

• Sequence of numbers: 1, 1, 2, 3, 5, 8, 13, ...a0 a1 a2 a3 a4 a5 a6

§ Get the next number by adding previous two§ What is a8?

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A: a8 = 21B: a8 = 29C: a8 = 34D: None of these.

correct

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Example: Fibonnaci Sequence

• Sequence of numbers: 1, 1, 2, 3, 5, 8, 13, ...a0 a1 a2 a3 a4 a5 a6

§ Get the next number by adding previous two§ What is a8?

• Recursive definition:§ an = an-1 + an-2 Recursive Case§ a0 = 1 Base Case§ a1 = 1 (another) Base Case

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Why did we need two base cases this time?

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Fibonacci as a Recursive Function

def fibonacci(n):"""Returns: Fibonacci no. an

Precondition: n ≥ 0 an int"""if n <= 1:

return 1

return (fibonacci(n-1)+fibonacci(n-2))

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Recursive case

Base case(s)

Note difference with base case conditional.

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Fibonacci as a Recursive Function

def fibonacci(n):"""Returns: Fibonacci no. an

Precondition: n ≥ 0 an int"""if n <= 1:

return 1

return (fibonacci(n-1)+fibonacci(n-2))

• Function that calls itself§ Each call is new frame§ Frames require memory§ ∞ calls = ∞ memory

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n

fibonacci 3

5

n

fibonacci 1

4 n

fibonacci 1

3

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Fibonacci: # of Frames vs. # of Calls

• Fibonacci is very inefficient.§ fib(n) has a stack that is always ≤ n§ But fib(n) makes a lot of redundant calls

fib(5)

fib(4)

fib(3) fib(2)

fib(2) fib(1) fib(0)

fib(0)

fib(1)

fib(1)12Recursion

fib(3)

fib(2) fib(1)

fib(0)fib(1)

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Fibonacci: # of Frames vs. # of Calls

• Fibonacci is very inefficient.§ fib(n) has a stack that is always ≤ n§ But fib(n) makes a lot of redundant calls

fib(5)

fib(4)

fib(3) fib(2)

fib(2) fib(1) fib(0)

fib(0)

fib(1)

fib(1)13Recursion

fib(3)

fib(2) fib(1)

fib(0)fib(1)

Path to end = the call stack

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Recursion vs Iteration

• Recursion is provably equivalent to iteration§ Iteration includes for-loop and while-loop (later)§ Anything can do in one, can do in the other

• But some things are easier with recursion§ And some things are easier with iteration

• Will not teach you when to choose recursion§ This is a topic for more advanced classes

• We just want you to understand the technique10/17/19 Recursion 14

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Recursion is best for Divide and Conquer

Goal: Solve problem P on a piece of data

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data

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Recursion is best for Divide and Conquer

Goal: Solve problem P on a piece of data

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dataIdea: Split data into two parts and solve problem

data 1 data 2

Solve Problem P Solve Problem P

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Recursion is best for Divide and Conquer

Goal: Solve problem P on a piece of data

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dataIdea: Split data into two parts and solve problem

data 1 data 2

Solve Problem P Solve Problem P

Combine Answer!

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Divide and Conquer Example

Count the number of 'e's in a string:

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p e nn e

Two 'e's

p e nn e

One 'e' One 'e'

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Divide and Conquer Example

Count the number of 'e's in a string:

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p e nn e

Two 'e's

p e nn e

Zero 'e's Two 'e's

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Divide and Conquer Example

Count the number of 'e's in a string:

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p e nn e

Two 'e's

p e nn e

Zero 'e's Two 'e's

Will talk about how to break-up later

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Three Steps for Divide and Conquer

1. Decide what to do on “small” data§ Some data cannot be broken up§ Have to compute this answer directly

2. Decide how to break up your data§ Both “halves” should be smaller than whole§ Often no wrong way to do this (next lecture)

3. Decide how to combine your answers§ Assume the smaller answers are correct§ Combining them should give bigger answer

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Divide and Conquer Exampledef num_es(s):

"""Returns: # of 'e's in s"""# 1. Handle small dataif s == '':

return 0elif len(s) == 1:

return 1 if s[0] == 'e' else 0

# 2. Break into two partsleft = num_es(s[0])right = num_es(s[1:])

# 3. Combine the resultreturn left+right

“Short-cut” forif s[0] == 'e’:

return 1else:

return 0

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p e nn e

0 2+

s[0] s[1:]

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Divide and Conquer Exampledef num_es(s):

"""Returns: # of 'e's in s"""# 1. Handle small dataif s == '':

return 0elif len(s) == 1:

return 1 if s[0] == 'e' else 0

# 2. Break into two partsleft = num_es(s[0])right = num_es(s[1:])

# 3. Combine the resultreturn left+right

“Short-cut” forif s[0] == 'e’:

return 1else:

return 0

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p e nn e

0 2+

s[0] s[1:]

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Divide and Conquer Exampledef num_es(s):

"""Returns: # of 'e's in s"""# 1. Handle small dataif s == '':

return 0elif len(s) == 1:

return 1 if s[0] == 'e' else 0

# 2. Break into two partsleft = num_es(s[0])right = num_es(s[1:])

# 3. Combine the resultreturn left+right

“Short-cut” forif s[0] == 'e’:

return 1else:

return 0

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p e nn e

0 2+

s[0] s[1:]

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Divide and Conquer Exampledef num_es(s):

"""Returns: # of 'e's in s"""# 1. Handle small dataif s == '':

return 0elif len(s) == 1:

return 1 if s[0] == 'e' else 0

# 2. Break into two partsleft = num_es(s[0])right = num_es(s[1:])

# 3. Combine the resultreturn left+right

“Short-cut” forif s[0] == 'e’:

return 1else:

return 0

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p e nn e

0 2+

s[0] s[1:]

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Divide and Conquer Exampledef num_es(s):

"""Returns: # of 'e's in s"""# 1. Handle small dataif s == '':

return 0elif len(s) == 1:

return 1 if s[0] == 'e' else 0

# 2. Break into two partsleft = num_es(s[0])right = num_es(s[1:])

# 3. Combine the resultreturn left+right

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Base Case

RecursiveCase

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Exercise: Remove Blanks from a String

def deblank(s):"""Returns: s but with its blanks removed"""

1. Decide what to do on “small” data§ If it is the empty string, nothing to do

if s == '':return s

§ If it is a single character, delete it if a blankif s == ' ': # There is a space here

return '' # Empty stringelse:

return s10/17/19 Recursion 27

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Exercise: Remove Blanks from a String

def deblank(s):"""Returns: s but with its blanks removed"""

2. Decide how to break it upleft = deblank(s[0]) # A string with no blanksright = deblank(s[1:]) # A string with no blanks

3. Decide how to combine the answerreturn left+right # String concatenation

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Putting it All Together

def deblank(s):"""Returns: s w/o blanks"""if s == '':

return selif len(s) == 1:

return '' if s[0] == ' ' else s

left = deblank(s[0])right = deblank(s[1:])

return left+right

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Handle small data

Break up the data

Combine answers

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Putting it All Together

def deblank(s):"""Returns: s w/o blanks"""if s == '':

return selif len(s) == 1:

return '' if s[0] == ' ' else s

left = deblank(s[0])right = deblank(s[1:])

return left+right

10/17/19 Recursion 30

Base Case

RecursiveCase

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Minor Optimization

def deblank(s):"""Returns: s w/o blanks"""if s == '':

return selif len(s) == 1:

return '' if s[0] == ' ' else s

left = deblank(s[0])right = deblank(s[1:])

return left+right

10/17/19 Recursion 31

Needed secondbase case to handle s[0]

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Minor Optimization

def deblank(s):"""Returns: s w/o blanks"""if s == '':

return s

left = s[0]if s[0] == ' ':

left = ''right = deblank(s[1:])

return left+right

10/17/19 Recursion 32

Eliminate the second base

by combining

Less recursive calls

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Following the Recursion

a b cdeblank

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Following the Recursion

a b cdeblank

a b cdeblank

10/17/19 Recursion 34

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Following the Recursion

a b c

a

deblank

a b cdeblank

b cdeblank

10/17/19 Recursion 35

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Following the Recursion

a b c

a

deblank

a b cdeblank

b cdeblank

b cdeblank

10/17/19 Recursion 36

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Following the Recursion

a b c

a

b

deblank

a b cdeblank

b cdeblank

b cdeblank

cdeblank

10/17/19 Recursion 37

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Following the Recursion

a b c

a

b

deblank

a b cdeblank

b cdeblank

b cdeblank

cdeblank

cdeblank

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Following the Recursion

a b c

a

b

deblank

a b cdeblank

b cdeblank

b cdeblank

cdeblank

cdeblank

c10/17/19 Recursion 39

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Following the Recursion

a b c

a

b

deblank

a b cdeblank

b cdeblank

b cdeblank

cdeblank

cdeblank

c c10/17/19 Recursion 40

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Following the Recursion

a b c

a

b

deblank

a b cdeblank

b cdeblank

b cdeblank

cdeblank

cdeblank

c c

c✗10/17/19 Recursion 41

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Following the Recursion

a b c

a

b

deblank

a b cdeblank

b cdeblank

b cdeblank

cdeblank

cdeblank

c c

c✗cb

10/17/19 Recursion 42

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Following the Recursion

a b c

a

b

deblank

a b cdeblank

b cdeblank

b cdeblank

cdeblank

cdeblank

c c

c✗cb

cb✗

10/17/19 Recursion 43

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Following the Recursion

a b c

a

b

deblank

a b cdeblank

b cdeblank

b cdeblank

cdeblank

cdeblank

c c

c✗cb

cb✗cba

10/17/19 Recursion 44

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Following the Recursion

a b c

a

b

deblank

a b cdeblank

b cdeblank

b cdeblank

cdeblank

cdeblank

c c

c✗cb

cb✗cba

cba✗

10/17/19 Recursion 45

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Following the Recursion

a b c

a

b

c c

c

cb

cb

cba

cba

cba

deblank

a b cdeblank

b cdeblank

b cdeblank

cdeblank

cdeblank

10/17/19 Recursion 46

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Final Modification

def deblank(s):"""Returns: s w/o blanks"""if s == '':

return s

left = s[0]if s[0] == ' ':

left = ''right = deblank(s[1:])

return left+right

10/17/19 Recursion 47

Real work done here

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Final Modification

def deblank(s):"""Returns: s w/o blanks"""if s == '':

return s

left = sif s[0] in string.whitespace

left = ''right = deblank(s[1:])

return left+right

10/17/19 Recursion 48

Module string has special constants to simplify

detection of whitespace and other characters.

Real work done here

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Next Time: Breaking Up Recursion

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