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Transcript
7. RECURSIONS
Rocky K. C. Chang
October 12, 2015
Objectives
• To be able to design recursive solutions to solve "simple"
problems.
• To understand the relationship between recursion and
mathematical induction.
• To understand that recursion is a postponement of work.
• To know how to design recursive algorithms
systematically.
WHAT IS RECURSION?
Example 1: Computation of n!
• Standard method:
• fact(n) = 1 * 2 * … * n
• Recursive method:
• fact(n) = n * fact(n-1)
• fact(n-1) = (n-1) * fact(n-2)
• …
• fact(2) = 2 * fact(1)
• fact(1) = 1 * fact(0)
• fact(0) = 1
EXERCISE 7.1
Implement a recursive function to compute n!.
Example 2: Birthday cake cutting
• Question: how many pieces of a circular cake as a result
of n cuts through the center?
• Write down your answer here.
Example 2
• Recursive approach:
Example 3: Fibonacci numbers
• The rabbit population problem:
• Start out with a male-female pair of rabbits.
• It takes two months for each pair to be reproductive.
• Each pair reproduces another male-female pair in each subsequent
month.
• How many pairs of male-female rabbits after n months if no rabbits
die during this period?
• Standard method:
• 1st month: 1
• 2nd month: 1
• 3rd month: 2
• 4th month: 3
• …
Example 3
• Recursive method:
• Just prior to the start of nth month, there are fib(n-1) pairs.
• However, only fib(n-2) pairs can reproduce new pairs.
• Therefore, fib(n) = ?
RECURSION AND
MATHEMATICAL
INDUCTION
Parallel w. mathematical induction
• Mathematical induction is a very powerful theorem-
proving technique.
• Proving that a statement T(n) is true for n ≥ 1 requires the
proofs for
• (Base) T(1) is true.
• (Induction) For every n > 1, if T(n-1) is true, then T(n) is true.
Mathematical induction, e.g.,
• Prove that for all natural numbers x and n, xn – 1 is
divisible by x – 1.
• Base: when n = 1, xn – 1 is divisible by x – 1.
• Induction: For n > 1, assume that xn-1 – 1 is divisible by x – 1.
• Write xn - 1 = x(xn-1 – 1) + (x – 1).
• Therefore, xn – 1 is also divisible by x – 1.
Mathematical induction, e.g.,
• Prove that if n is a natural number and 1 + x > 0, then (1 +
x)n ≥ 1 + nx.
• Base: for n = 1, LHS = RHS = 1 + x.
• Induction: For n ≥ 1, assume that (1 + x)n ≥ 1 + nx.
• (1 + x)n+1 = (1 + x)(1 + x)n
• (1 + x)n+1 ≥ (1 + x)(1 + nx) = 1 + (n + 1)x + nx2
• (1 + x)n+1 ≥ 1 + (n + 1)x
The parallels
• A recursive process consists of two parts:
• A smallest case that is processed without recursion (base case)
• A general method that reduces a particular case into one or more of
the smaller cases (induction step)
• Factorials
• Base:
• Induction:
• The birthday cake cutting problem
• Base:
• Induction:
• Fibonacci numbers
• Base:
• Induction:
RECURSION AS A
POSTPONEMENT OF
WORK
Tracing the recursive calls
fact(6) =
=
=
=
=
=
=
=
=
=
=
=
=
6 * fact(5)
6 * (5 * fact(4))
6 * (5 * (4 * fact(3)))
6 * (5 * (4 * (3 * fact(2))))
6 * (5 * (4 * (3 * (2 * fact(1)))))
6 * (5 * (4 * (3 * (2 * (1 * fact(0))))))
6 * (5 * (4 * (3 * (2 * (1 * 1)))))
6 * (5 * (4 * (3 * (2 * 1))))
6 * (5 * (4 * (3 * 2)))
6 * (5 * (4 * 6))
6 * (5 * 24)
6 * 120
720
A recursive tree for fact(n)
fact(n)
:
:
fact(0)
Returning the values
Invoking the calls
fact(n-1)
How does recursion work?
• When a program is run, each execution of a method is
called an activation.
• The data objects associated with each activation are
stored in the memory as an activation record.
• Data objects include parameters, return values, return address and
any variables that are local to the method.
• A stack is a data structure to store all activation records in
a correct order.
A recursive tree for fib(n)
Memory requirement for recursions
HOW TO DESIGN
RECURSIVE
ALGORITHMS?
Several important steps
• Find the key step.
• How can this problem be divided into parts?
• Find a stopping rule.
• It is usually the small, special case that is trivial or easy to handle
without recursion.
• Outline your algorithm.
• Combine the stopping rule and the key step, using an if statement
to select between them.
• Check termination.
• Verify that the recursion will always terminate.
• Draw a recursion tree.
MORE COMPLEX
EXAMPLES
Tower of Hanoi
• Problem: To move a stack of n disks from pole A to pole B,
subjecting to 2 rules:
• Only one disk can be moved at a time.
• A larger disk can never go on top of a smaller one.
• n = 1, trivial
• n = 2, trivial again
• n = 3: trivial to some people?
• n = 4: trivial to fewer people?
• n = 64?
Think recursively …
Think recursively …
A recursive solution
• solveTowers(n, A, B, C)
• n: the number of disks
• moving the disks from A to B via C
• no larger disks put on top of smaller disks
• A recursive approach:
• Base: trivial and solved for n = 1
• Induction; for n > 1, solveTowers(n, A, B, C) can be solved by
• solveTowers(n - 1, A, C, B) // the top n-1 disks
• solveTowers(1, A, B, C)
// the largest disk
• solveTowers(n - 1, C, B, A) // the top n-1 disks
A recursion tree for n = 3
(1)
solveTowers
(3,A,B,C)
(6)
(2)
solveTowers
(2,A,C,B)
(4)
(3)
solveTowers
(1,A,B,C)
(7)
solveTowers
(2,C,B,A)
(9)
(5)
solveTowers
(1,B,C,A)
(8)
solveTowers
(1,C,A,B)
(10)
solveTowers
(1,A,B,C)
Computing 
• Write R(n) = Q(1) × Q(2) × Q(3) × … × Q(n), where
• Q(1) = sqrt[½+ ½ sqrt(½)],
• Q(2) = sqrt[½+ ½ Q(1)], …,
• Q(n) = sqrt[½+ ½ Q(n – 1)], n > 1.
• Therefore,
• Base: R(1) = Q(1)
• Induction: Given R(n – 1) and Q(n – 1),
Q(n) = sqrt[½+ ½ Q(n – 1)] and R(n) = R(n – 1) × Q(n).
The recursion tree
rhs(n)
term(n)
rhs(n-1)
term(n-1)
:
:
:
:
rhs(1)
term(1)
Summary
• Recursion is a divide-and-conquer approach to solving a
complex problem.
• Divide, conquer, and glue
• Similar to mathematical induction, recursion has a “base”
case and an “induction” step.
• A recursion implementation essentially postpones the
work until hitting the base case.
• Recursions appear in many, many, …, different forms.
Acknowledgments
• The figures on pp. 19-20 are taken from
• Robert L. Kruse and Alexander J. Ryba, Data Structures and
Program Design in C++, Prentice-Hall, Inc., 1999.
• The figures on pp. 25-26 are taken from
• F. Carrano and J. Prichard, Data Abstraction and Problem Solving
with JAVA Wall and Mirrors, First Edition, Addison Wesley, 2003.
END