Recursion

1
Recursion

• n General Form of Recursive methods
• n Examples of Simple Recursion
• n Hand Simulating Recursive methods
• n Proving Recursive methods Correct
• n Modeling Recursion via Onions
• n Recursive methods on Naturals/Integers

2
Recursion

• Recursion is a programming technique in which a
  call to a method appears in that methods body
  (i.e., a method calls itself this is called
  direct recursion)
• Once you understand recursive methods, they are
  often simpler to write than their iterative
  equivalents
• In modern programming languages, recursive
  methods may run a bit slower (maybe 5) than
  equivalent iterative methods in a typical
  application, this time is insignificant (most of
  the time will be taken up elsewhere anyway)
• We will begin by studying the form of general
  recursive methods then apply this form to
  methods operating on int values and finally
  apply that form to methods operating on linked
  lists (where they are most useful). In both cases
  we will discuss how values of these types are
  recursively defined.

3
Fund Raising Iteration vs. Recursion

• Problem Collect 1,000.00 for charity
• Assumption Everyone is willing to donate a penny
• Iterative Solution
• Visit 100,000 people, asking each for a penny
• Recursive Solution
• If you are asked to collect a penny, give a penny
  to the person who asked you for it
• Otherwise
• Visit 10 people and ask them to each raise 1/10th
  of the amount of money that you have been asked
  to raise
• Collect the money that they give you and combine
  it into one bag
• Give it to the person who asked you to collect
  the money

4
General Form of Recursive methods

• Solve(Problem)
• if (Problem is minimal/not decomposable a base
  case)
• solve Problem directly i.e., without
  recursion
• else
• (1) Decompose Problem into one or more
  similar, strictly smaller subproblems
  SP1, SP2, ... , SPN
• (2) Recursively call Solve (this method) on
  each subproblem Solve(SP1),
  Solve(SP2),..., Solve(SPN)
• (3) Combine the solutions to these
  subproblems into a solution that solves
  the original Problem

5
factorial In Mathematics and Java
Example using Definition 4! 4 3! 4 3
2! 4 3 2 1! 4 3 2 1 0!
4 3 2 1 1

• 1 if N 0
• N!
• N(N-1)! if N gt 0
• int factorial (int n)
• if (n 0) //Non-decomposable
• return 1
• else
• int subN n-1 //Decompose
• int subFact factorial(subN) //Solve
  Subproblem
• return n subFact //Combine

6
Simplified and Iterative factorial

• int factorial (int n)
• if (n 0)
• return 1
• else
• return n factorial(n-1)
• int factorial (int n)
• int answer 1
• for (int i1 iltn i)
• answer i
• return answer

Here we combine the three steps in the previous
else block into a single return statement (one
expression including the decompose, solve, and
combine steps). Compare this method to the one
below it, which operates by iteration. The
iterative version requires the declaration of two
variables and the use of state change operators
(which are always difficult to reason about)
7
pow Raising a Number to a Power

• 1 if N 0
• AN
• AAN-1 if N gt 0
• double pow(double a, int n)
• if (n 0)
• return 1.
• else
• return apow(a,n-1)

Example using Definition A4 A A3 A A
A2 A A A A1 A A A A A0
A A A A 1
Calling pow(a,n)requires exactly n multiplications
The pow in the Java Math library actually
computes its answer using logarithms and
exponentials
8
Simulation of Recursive Methods

• Assume that a method is computed according to its
  definition by a person in an apartment complex
• That person can be called to compute an answer
  once he/she computes the answer (possibly helped
  by calling another person in the apartment
  complex), he/she places a return call, and
  reports back the answer to the person who called
  him/her
• Assume that each person knows the phone number of
  the person living in the apartment underneath
  them, who also has the same set of instructions
  so any person can call the one underneath to
  solve a simpler problem
• The first method call initiates a sequence of
  calls to people living in lower level apartments
  (each person solving a simpler problem), whose
  answers percolate back to the top, finally
  solving the original problem

9
Hand Simulation of Factorial
n return
n return
n return
n return
...
n return
10
Proof Rules for Recursive Methods

• To prove that a recursive method, Solve, is
  correct
• Prove that Solve computes (without recursion) the
  correct answer to any minimal (base case) Problem
• Base cases are simple, so this should be easy
• Prove that the argument to any recursive call of
  Solve, e.g. SPI, is strictly smaller (closer to
  the minimal case) than Problem
• The notion of strictly smaller should be easy to
  understand for the recursive argument, so this
  should be easy
• Prove that these answers are combined to compute
  the correct answer for Problem
• In this proof, you may assume that each recursive
  call, Solve(SPI), correctly computes its answer
• The assumption makes this proof easy
• Recursion is computable induction (a math concept)

11
Applying the Proof Rules

• factorial (recurring on the parameter n)
• factorial correctly returns 1 for the minimal
  argument 0.
• In factorial(n-1), n-1 is always closer to 0 than
  n is.
• Assume factorial(n-1) computes the correct
  answer.
• n times this value is, by definition, the correct
  value of factorial(n)
• pow (recurring on the parameter n)
• pow correctly returns 1 for the minimal argument
  0.
• In pow(a,n-1), n-1 is always closer to 0 than n
  is.
• Assume pow(a,n-1) computes the correct answer,
  an-1.
• apow(a,n-1) is aan-1 is an, the correct value
  of pow(a,n)

12
Proof Rules and Bad Factorials

• int factorial (int n)
• if (n 0)
• return 0 //0! is not 0
• else
• return nfactorial(n-1)
• int factorial (int n)
• if (n 0)
• return 1
• else
• return factorial(n1)/(n1) //n1 not closer
  to 0
• int factorial (int n)
• if (n 0)

13
Proof Rules and Bad Factorials (cont)

• In the first method, the wrong answer (0) is
  returned for the base case since everything
  depends on the base case, ultimately this method
  always returns 0
• In the second method, n1 is farther away from
  the base case this method will continue calling
  factorial with ever larger arguments, until the
  maximum int value is exceeded a runaway (or
  infinite) recursion (actually, each recursive
  call can take up some space, so eventually memory
  is exhausted).
• In the third method, the wrong answer is returned
  by incorrectly combining n and the solved
  subproblem this method returns one more than the
  sum of all the integers from 1 to n (an
  interesting method in its own right) not the
  product of these values
• In the first method, it always returns the wrong
  answer the second method never returns an
  answer, the third method returns the correct
  answer only in the base case

14
fpow a faster version of pow
Note if N is odd N/2 truncates so 7/2 3

• 1 if N 0
• AN B2 where B AN/2 if N gt 0 and N is Even
• AB2 where B AN/2 if N gt 0 and N is Odd
• int fpow(double a, int n)
• if (n 0)
• return 1.
• else
• double b fpow(a,n/2)
• if (n2 0)
• return bb
• else
• return abb

Calling fpow(a,n)requires between at least log2n
and at most 2log2n multiplications Compare this
complexity to calling pow(a,n)requiring n
multiplications Contemporary cryptography raises
large numbers to huge (hundred digit) powers it
needs a fast method
15
Recursive Definition of Naturals

• 0 is the smallest Natural
• It is minimal, and cannot be decomposed
• z(x) is true if x is zero and false if x is
  non-zero
• The successor of a Natural is a Natural
• If x is a Natural, s(x) is a Natural
• The successor of x is x1
• If x is a non-0 Natural, p(x) is a Natural
• The predecessor of x is x-1
• p(0) throws an exception
• Note
• z(s(x)) is always false
• p(s(x)) x
• s(p(x)) x only if X ? 0

16
Onion Model of Recursive Naturals
Every Onion is either Base Case The
core Recursive Case A layer of skin around a
smaller onion (which may be the core, or some
deeper layer) The value of an onion is the
number of layers of skin beyond the core
17
Computation with Naturals

• Lets Define 3 methods that operate on Naturals
  in Java
• int s(int x) return x1
• int p(int x)
• if (x0)
• throw new IllegalArgumentException("p with
  x0")
• return x-1
• bool z(int x) return x 0
• We can define all common operators (both
  arithmetic and relational) on Naturals by writing
  simple recursive methods that use these three
  methods

18
Recursive methods on Naturals

• int sum(int a, int b)
• if (z(a))
• return b
• else
• return s(sum(p(a),b)) //1 (a-1 b)
• int sum(int a, int b)
• if (z(a))
• return b
• else
• return sum(p(a), s(b)) //(a-1) (b1)
• In both sum methods, a gets closer to 0 in
  recursive calls

19
Recursive methods on Naturals II

• int product(int a, int b)
• if (z(a))
• return 0
• else
• return sum(b,product(p(a),b)) //b (a-1)b
• int power(int a, int b)
• if (z(b))
• return 1
• else
• return product(a, power(a,p(b)))//a ab-1

20
Recursive methods on Naturals III

• bool equal(int a, int b) //Is a b
• if (z(a) z(b)) //If either is 0...
• return z(a) z(b) //return whether both
  are 0
• else
• return equal(p(a), p(b)) //Is (a-1) (b-1)
• bool less(int a, int b) //Is a lt b
• if (z(b)) //Nothing is lt 0
• return false
• else if (z(a)) //If b!0, a0, then altb
• return true
• else
• return less(p(a),p(b)) //Is (a-1) lt (b-1)

21
Recursive methods on Naturals IV

• bool even(int a)
• if (z(a)) //True if 0
• return true
• else //Opposite of even(a-1)
• return !even(p(a))
• bool odd(int a)
• if (z(a)) //False if 0
• return false
• else //Opposite of odd(a-1)
• return !odd(p(a))

22
Printing an Integers Digits

• void print (int i)
• if (i lt 10) //if (i gt 10)
• System.out.print(i) // print(i/10)
• else //System.out.print(i10)
• print(i/10)
• System.out.println(i10)
• print correctly prints all 1-digit values 0 - 9.
• In print(i/10), i/10 is one digit closer to 0 - 9
  than i is.
• Assume print (i/10) correctly prints all the
  digits in i/10 in the correct order
• Printing i/10 (all digits but the last) and then
  printing i10 (the last digit), prints all the
  digits in i in order

The code above is simplified by bottom factoring
and test reversal and noting for ilt10, i is i10
23
Printing an Integers Digits in Reverse

• void printReversed (int i)
• if (i lt 10) //System.out.print(i10)
• System.out.print(i) //if (i gt 10)
• else // System.out.print(i/10)
• System.out.print(i10)
• printReversed(i/10)
• printReversed correctly prints all 1-digit
  values 0 - 9.
• In printReversed(i/10), i/10 is closer to 0-9
  than i is.
• Assume printReversed(i/10) correctly prints all
  the digits in i/10 reversed.
• Printing the last digit, i10, and then printing
  the i/10 reversed prints all the digits in i
  reversed.

24
Converting an int to a string

• String toString (int i)
• if (i lt 10)
• return ""(char)(i0)
• else
• return toString(i/10) (char)(i100)
• toString correctly returns all 1-digit values 0
  - 9.
• In toString(i/10), i/10 is one digit closer to
  0-9 than i is.
• Assume toString(i/10) correctly returns a String
  of all the digits in I/10.
• Then, after appending the char equivalent of i10
  at the end of this String correctly stores the
  String equivalent of i.

25
Synthesizing Recursive Methods

• Find the base (non-decomposable) case(s)
• Write the code that detects the base case(s) and
  returns the correct answer for them without using
  recursion
• Assume that you can decompose all non
  base-case(s) and then solve these smaller
  subproblems via recursion
• Choose the decomposition
• There should be some natural decomposition
• Write code that combines these solved subproblems
  (often there is just one) to solve the original
  problem
• Typical decompositions
• Integers i - something or i/something
  (digit at a time)
• Linked Lists l.next

26
Problems

• Write the following method public static
  String reverse (String s) recursively. Choose an
  appropriate base case, recursive call (hint
  substring method using one parameter), and a way
  to combine (hint ) solved smaller problems to
  solve the original problem. reverse("abcd")
  should return "dcba"
• Write the following method public static int
  toInt (String s) recursively. It is like
  parseInt, but only for non-negative
  valuestoInt("138") should return 138

27
Problems (continued)

• Write the following method public static
  boolean equals (String s1, String s2)
  recursively. Its recursive structure is like that
  of the equals methods for ints.
• Write the following method public static int
  compare (String s1, String s2) recursively.
  Its recursive structure is similar to equals,
  above.