CS 112 Introduction to Programming

1
CS 112 Introduction to Programming

• Lecture 29
• Recursive Programming
• Richard Yang

2
Outline

• Admin. and review
• Recursive thinking defining concept recursively
• Recursive method calls
• Places of recursion matter

3
Admin.

• Assignment 6 is returned
• All the unpicked up assignments are outside of my
  office at Watson 202
• Assignment 7
• How to convert a char to a String?char
  wordString myStr new String(word)

4
Review Java GUI Programming

• GUI components
• Components e.g. Button, Label
• Containers, e.g. Frame, Panel
• A container has a layout manager, which manages
  how components in it are laid out
• e.g., Flow, Border, Grid, Box
• Often you will need to use multiple containers to
  achieve desired effects
• Components can generate events to which listener
  objects can respond
• Add a listener to a component when an event
  happens, the method of the listener object is
  automatically invoked
• We often use one listener object to serve
  multiple components, if the components are
  similar
• e.g., use getSource() to determine the component
  generating the event.

5
Recursive Programming

• Recursion a method calls itself
• Recursion is often used to search for a solution
  to a problem, e.g.,
• Hanoi Towers how to move the disks?
• Maze search how to find a way out of a maze?
• Change maker how many ways of making changes to
  a one-dollar bill?
• Recursion is also useful for writing elegant
  programs, e.g.,
• KochSnowflake

6
Outline

• Admin. and review
• Recursive thinking defining concept recursively
• Recursive method calls
• Places of recursion matter

7
Recursive Thinking

• A recursive definition is one which uses the word
  or concept being defined in the definition itself
• When defining an English word, a recursive
  definition is often not helpful
• Good is not bad, bad is no good.
• But in other situations, a recursive definition
  can be an appropriate way to express a concept
• Before applying recursion to programming, it is
  best to practice thinking recursively

8
Recursive Definition LIST

• Consider the following list of numbers
• 24, 88, 40, 37
• Such a list can be defined as
• A LIST is a number
• or a number comma LIST
• That is, a LIST is defined to be a single number,
  or a number followed by a comma followed by a
  LIST
• The concept of a LIST is used to define itself

9
Recursive Definitions

• The recursive part of the LIST definition is used
  several times, terminating with the non-recursive
  part
• number comma LIST
• 24 , 88, 40, 37
• number comma LIST
• 88 , 40, 37
• number comma LIST
• 40 , 37
• number
• 37

10
Recursive Definitions Factorial

• N!, for any positive integer N, is defined to be
  the product of all integers between 1 and N
  inclusive
• This definition can be expressed recursively as
• 1! 1
• N! N (N-1)!
• The concept of the factorial is defined in terms
  of another factorial

11
Recursive Definitions

• 5!
• 5 4!
• 4 3!
• 3 2!
• 2 1!
• 1

12
Outline

• Admin. and review
• Recursive thinking defining concept recursively
• Recursive method calls
• Places of recursion matter

13
Traditional Method Calls

• Flow of control flows from one method to another
  and returns as specified

foo
bar
main


bar()
foo()




14
Recursive Method Calls

• What if a method calls itself?

void LotsOfPrints() System.out.println(i
nside LotsOfPrints) LotsOfPrints()
15
Termination

• So we have another way to write infinite loops!
• To terminate, we have to stop making recursive
  calls
• Use conditionals to determine termination

void LotsOfPrints() if
(some-condition) do
non-recursive computation else
System.out.println(inside
LotsOfPrints) LotsOfPrints()

16
Infinite Recursion

• All recursive definitions have to have a
  non-recursive part
• If they didn't, there would be no way to
  terminate the recursive path
• Such a definition would cause infinite recursion
• This problem is similar to an infinite loop, but
  the non-terminating "loop" is part of the
  definition itself
• The non-recursive part is often called the base
  case

17
Termination (contd.)

• Introduce an input variable
• Value of input variable determines termination

void LotsOfPrints(int num) if (num 0)
System.out.println(the
bug stops here!) else
System.out.println(inside LotsOfPrints)
LotsOfPrints(num - 1)
18
Control Flow
LotsOfPrints(2)
LotsOfPrints(1)
LotsOfPrints(0)






LotsOfPrints(1)
LotsOfPrints(0)












19
Place of Recursion Ordering
void LotsOfPrints(int num) if (num
0) else
System.out.println(num)
LotsOfPrints(num - 1)
LotsOfPrints(5) 5 4
3 2 1
20
Place of Recursion Ordering
void LotsOfPrints(int num) if (num
0) else
LotsOfPrints(num - 1)
System.out.println(num)
LotsOfPrints(5) 1 2
3 4 5
21
Local and Formal Variables

• Each invocation of a method has its own set of
  method variables formal parameters and local
  variables

void LotsOfPrints(int num) int local
2num if (num 0)
else
LotsOfPrints(num - 1)
System.out.println(local)
LotsOfPrints(3) 2 4
6
22
Control Flow






LotsOfPrints(2)
LotsOfPrints(1)
LotsOfPrints(0)
print(local)
print(local)
print(local)




23
Exercise 1

• Without using any loop statement, write a sum
  method that calculates the sum of the numbers
  from 0 to n, where parameter n is a non-negative
  integer?

24
Exercise 2

• Without using any loop statement, write method
  power(int x, int y) that raises integer x to the
  power of y, where y is a non-negative integer?
• Examples
• power(2, 1) 2
• power(2, 2) 4
• power(2, 10) 1024

25
CS 112 Introduction to Programming

• Lecture 30
• Recursive Programming
• Richard Yang

26
Outline

• Review and admin.
• Palindrome
• Change maker
• Towers of Hanoi

27
Review What is Recursion?

• Follows the mathematical notion of induction
• There is a base case
• Inductive step provides solution for a larger
  problem
• solve a smaller problem
• extend it a bit
• LotsOfPrints(0) do nothing
• LotsOfPrints(n) LotsOfPrints(n-1), Print

28
Recursive Programming

• A method in Java can invoke itself if set up
  that way, it is called a recursive method
• Each call to the method sets up a new execution
  environment, with new parameters and local
  variables
• As always, when the method completes, control
  returns to the method that invoked it (which may
  be an earlier invocation of itself)
• The code of a recursive method must include two
  cases
• the base case
• the recursive case

29
Recursive Sum
int sum(int n) if (n 0) //
base case return 0 else
// recursive case return
sum(n-1) n
sum(4)
sum(0)
4
0
10
0
30
Local and Formal Variables

• Each invocation of a method has its own set of
  method variables formal parameters and local
  variables

void LotsOfPrints(int num) int local
2num if (num 0)
else
LotsOfPrints(num - 1)
System.out.println(local)
LotsOfPrints(3) 2 4
6
31
Control Flow






LotsOfPrints(2)
LotsOfPrints(1)
LotsOfPrints(0)
print(local)
print(local)
print(local)




32
Recursive Exponentiation
int power(int x, int n) if (n 0)
// base case return 1
else // recursive case
temp power(x, n/2)
temp temp if (n 2 1)
temp x return temp

33
Writing Recursive Programs

• Determine the structure of your method
• Think of the sub-problems that you may want to
  solve, then make the definition of your method
  general enough to handle them so that you can use
  recursion
• Determine the recursive case
• Determine the base case

34
Outline

• Review and admin.
• Palindrome
• Change maker
• Towers of Hanoi

35
Palindrome Problem

• A palindrome is a string that reads the same
  backward or forward
• Question how do you check a String to see
  whether or not it is a Palindrome?
• What is the signature of the method to allow
  recursion?
• What is the base case?
• What is the recursive case?

x
y
z
a
z
y
x
36
Palindrome

• checkPalindrome(string, left, right)
• Base case
• if right lt left, then true
• if string.charAt(left) ! string.charAt(right),
  then false
• Recursive case
• check palindrome (string, left 1, right - 1)

37
Palindrome Check
boolean checkPalindrome(String s, int left, int
right) if (right lt left) // base
return true else if (s.charAt(left) !
s.charAt(right)) // base return
false else return checkPalindrome(s,
left 1, right - 1)
38
Outline

• Review and admin.
• Palindrome
• Change maker
• Towers of Hanoi

39
Change Maker

• Recursion is useful for trying all possibilities
• Consider the change making problem
• Count number of ways to make change for a given
  amount with pennies, nickels, dimes, quarters
• Questions
• What is the signature of the method to allow
  recursion?
• What is the base case?
• What is the recursive case?

40
Change Maker Structure

• Order coin types in increasing order and number
  them
• count(amount, num_types)
• number of ways to make amount cents using
  num_types coins where num_types could be 0, 1,
  2, 3, 4

41
Change Maker Recursive Case

• Question count(11, 2) ?
• count(11, 2) sum of
• count(6, 2)
• count(11, 1)
• Two paths
• Use one more coin of the highest denomination,
  reduce amount and try to make change for
  remaining money
• Or, discard the highest denomination and try to
  make money with remaining denominations

42
Change Maker Recursive Case
3
11, 2
2
1
1
1
1
0
43
Change Maker Recursive Case
4
15, 2
3
1
2
1
1
1
44
Change Maker Base Case

• amount lt 0 num_types 0 0
• amount 0 num_types 1 1

45
Change Maker Code
int coins 1, 5, 10, 25 int count(int
amount, int num_types) if (amount lt 0
num_types 0) return 0 else if
(amount 0 num_types 1) return
1 else return count(amount -
coinsnum_types - 1, num_types)
count(amount, num_types - 1)
46
Outline

• Review and admin.
• Palindrome
• Change maker
• Towers of Hanoi

47
Towers of Hanoi

• The Towers of Hanoi is a puzzle made up of three
  vertical pegs and several disks that slide on
  the pegs
• The disks are of varying size, initially placed
  on one peg with the largest disk on the bottom
  with increasingly smaller ones on top
• The goal is to move all of the disks from peg 1
  to peg 3 under the following rules
• We can move only one disk at a time
• We cannot move a larger disk on top of a smaller
  one

48
Tower of Hanoi code
void moveTower (int numDisks, int start, int end,
int temp) if (numDisks 1)
System.out.println(move disk from start
to end) else
moveTower (numDisks-1, start, temp, end)
System.out.println(move disk from start
to end) moveTower (numDisks-1,
temp, end, start) public static void
main(String args) moveTower(4, 1, 3,
2)
49
Towers of Hanoi

• An iterative solution to the Towers of Hanoi is
  quite complex
• A recursive solution is much shorter and more
  elegant
• See SolveTowers.java
• See TowersOfHanoi.java

50
CS 112 Introduction to Programming

• Lecture 31
• Recursive Programming
• Richard Yang

51
Outline

• Admin. and review
• Examples

52
Admin.

• Assignment 8
• Recursive programming
• Assignment 9
• Enhance a web browser
• Replace your lowest assignment
• Date of second exam?

53
Writing Recursive Programs

• Determine the structure of your method
• Think of the sub-problems that you may want to
  solve, then make the definition of your method
  general enough to handle them so that you can use
  recursion
• Determine the recursive case
• Determine the base case

54
Examples

• int sum(int n)
• int power(int x, int n)

55
Palindrome Check
boolean checkPalindrome(String s, int left, int
right) if (right lt left) // base
return true else if (s.charAt(left) !
s.charAt(right)) // base return
false else return checkPalindrome(s,
left 1, right - 1)
void main(String args) String str
xyzazyx boolean result
checkPalindrome(str, 0, str.length())
56
Change Maker
int coins 1, 5, 10, 25 int count(int
amount, int nCoins) if (amount lt 0
nCoins 0) // base return 0 else
if (amount 0 nCoins 1) // base
return 1 else return
count(amount - coinsnCoins - 1, nCoins) //
use the largest coin
count(amount, nCoins - 1) // not use
void main(String args) int n count (99,
coins.length)
57
Exercise Find Maximum

• Find the maximum of an array of numbers
• int findMax(int array)
• Questions
• What is the structure?
• What is the base case?
• What is the recursive case?

58
Recursive findMax
int findMax(int items, int numElems) if
(numElems 1) return items0
int partResult findMax(items, numElems - 1)
if (partResult lt itemsnumElems - 1)
return itemsnumElems - 1 else
return partResult public static void
main(String args) int items 2,
4, 8, 7, 5, 9, 21, 11 int max
findMax(items, items.length)
59
Binary Search

• Searching for a number in a sorted array (say 7)
• void search(int items, int targetElem)
• Linear scan would work, but could do better
• Check in the middle and decide which half to
  search in
• Questions
• What is the structure of the method?
• What is the base case?
• What is the recursive case?

60
Binary Search Structure and Base

• boolean find(items, targElem, rangeBeg, rangeEnd)
• - items is an array of values
• - targElem is element we are searching for
• - rangeBeg is beginning of current array section
• - rangeEnd is end of current array section 1
• Base case array section is size one
• return true if targElem is contained in the
  single entry remaining
• return false if targElem is not contained in that
  entry
• initiate recursion find(items, targElem, 0,
  items.length)

61
Binary Search Recursive Case
static boolean find(int array, int elem, int
rangeBeg, int rangeEnd) if (rangeBeg
rangeEnd - 1) // base return
(arrayrangeBeg elem) int mid
(rangeBeg rangeEnd)/2 if (arraymid
elem) // base return true else
if (elem lt arraymid) return find(array,
elem, rangeBeg, mid) else return
find(array, elem, mid 1, rangeEnd)
62
Exercise Tiled Pictures

• Questions
• What is the structure of the draw method
• What is the recursive case?
• What is the base case?
• See TiledPictures.java

63
Exercise Fractals

• A fractal is a geometric shape made up of the
  same pattern repeated in different sizes and
  orientations
• The Koch Snowflake is a particular fractal that
  begins with an equilateral triangle
• To get a higher order of the fractal, the sides
  of the triangle are replaced with angled line
  segments
• See KochSnowflake.java
• See KochPanel.java

64
Towers of Hanoi

• The Towers of Hanoi is a puzzle made up of three
  vertical pegs and several disks that slide on
  the pegs
• The disks are of varying size, initially placed
  on one peg with the largest disk on the bottom
  with increasingly smaller ones on top
• The goal is to move all of the disks from peg 1
  to peg 3 under the following rules
• Move only one disk at a time
• Cannot move a larger disk on top of a smaller one

65
move(3 1 -gt 3 2)
move(2 1 -gt 2 3)
Disk 3 1 -gt 3
move(2 2 -gt 3 1)
66
Tower of Hanoi code
void moveTower (int numDisks, int start, int end,
int temp) if (numDisks 1)
System.out.println(move disk from start
to end) else
moveTower (numDisks-1, start, temp, end)
System.out.println(move disk from start
to end) moveTower (numDisks-1,
temp, end, start) public static void
main(String args) moveTower(4, 1, 3,
2)
67
Towers of Hanoi

• An iterative solution to the Towers of Hanoi is
  quite complex
• A recursive solution is much shorter and more
  elegant
• See SolveTowers.java
• See TowersOfHanoi.java

68
CS 112 Introduction to Programming

• Lecture 32
• Recursive Programming
• Richard Yang

69
Outline

• Admin. and review
• Recursive search
• Indirect recursion
• Recursion vs. iteration

70
Admin.

• Assignment 8
• Recursive programming
• Assignment 9
• Enhance a web browser
• Replace your lowest assignment
• Second exam 7 to 9pm on Dec. 6

71
Writing Recursive Programs

• Determine the structure of your method
• Think of the sub-problems that you may want to
  solve, then make the definition of your method
  general enough to handle them so that you can use
  recursion
• Determine the recursive case
• Determine the base case

72
move(3 1 -gt 3 2)
move(2 1 -gt 2 3)
Disk 3 1 -gt 3
move(2 2 -gt 3 1)
73
Tower of Hanoi Code
void moveTower (int numDisks, int start, int end,
int temp) if (numDisks 1)
System.out.println(move disk from start
to end) else
moveTower (numDisks-1, start, temp, end)
System.out.println(move disk from start
to end) moveTower (numDisks-1,
temp, end, start) public static void
main(String args) moveTower(4, 1, 3,
2)
74
Summary of Examples

• int sum(int n)
• int power(int x, int n)
• int findMax(int items, int numElems)
• boolean checkPalindrome(String s,
  int left, int right)
• int count(int amount, int nCoins)
• boolean find(int items, int targElem,
  int rangeBeg, int rangeEnd)
• - void moveTower (int numDisks,
  int start, int end, int temp)

75
Outline

• Admin. and review
• Recursive search
• Indirect recursion
• Recursion vs. iteration

76
Search

• Often we need to search for a solution among a
  large number of possibilities, e.g.,
• When playing Chess, we want to check what the
  best next step is
• We can use recursion to search the space
• One major issue is that we do not fall into
  infinite loop
• In the previous examples, we always go from a
  larger problem to a smaller problem, therefore no
  such possibility
• We may need to remember history
• We may need to revise history so that we will
  search the whole space
• We study two examples
• Search for a route in a Maze
• The Knights tour
• Your homework will consider anther one
• Eight queens

77
Maze

• Notation
• 1 allowed
• 0 blocked
• Objective search a path through a maze?
• Questions
• What is the structure
• What is the recursive case
• What is the base case

1
0
1
1
1
0
0
1
1
1
1
1
0
1
1
1
78
Maze Trace
1
0
1
1
1
0
0
1
1
1
1
1
0
1
1
1
79
Maze Dead End
1
0
1
1
1
0
0
1
1
1
1
1
0
1
1
1
1
4
2
3
80
Maze Keep State
1
0
1
1
1
0
0
1
1
1
1
1
0
1
1
1
1
4
2
3
81
Maze Traversal

• From each location, we use recursion to search in
  each direction
• Recursion keeps track of the path through the
  maze
• To avoid returning to a place that has already
  been visited, mark any place that has already
  been visited
• Otherwise, may have infinite loop
• Lesson to avoid infinite loop, recursion may
  need to keep track of state
• The base cases
• Reaching the final destination
• An invalid move
• See MazeSearch.java
• See Maze.java

82
Knight's Tour Problem

• Knight's Tour Problem On an n x n Chess Board,
  starting from a square, can a knight visit all
  the squares in exactly n2-1 moves

83
Knights Tour

• Questions
• What is the structure
• What is the recursive case
• What is the base case

84
Knights Tour

• From each location, we use recursion to search in
  each possible new position
• Recursion keeps track of the path through the
  board
• To avoid returning to a location that has already
  been visited, mark any location that has already
  been visited
• If a search from a location is not successful,
  unmark the previous mark at that location!
• The base case
• The whole board is visited!
• See KnightMain.java
• See ChessBoard.java

85
Outline

• Admin. and review
• Backtrack and search
• Indirect recursion
• Recursion vs. iteration

86
Indirect Recursion

• A method invoking itself is considered to be
  direct recursion
• A method could invoke another method, which
  invokes another, etc., until eventually the
  original method is invoked again
• For example, method m1 could invoke m2, which
  invokes m3, which in turn invokes m1 again
• This is called indirect recursion, and requires
  all the same care as direct recursion
• It is often more difficult to trace and debug

87
Indirect Recursion
88
Outline

• Admin. and review
• Backtrack and search
• Indirect recursion
• Recursion vs. iteration

89
Recursion vs. Iteration

• For some problems, recursion provides an elegant
  solution, often cleaner than an iterative version
• However, iterative solutions are generally faster
• Note that just because we can use recursion to
  solve a problem, doesn't mean we should
• For instance, we usually would not use recursion
  to solve the sum of 1 to N problem, because the
  iterative version is easier to understand
• You must carefully decide whether recursion is
  the correct technique for any problem

90
Recursive Fibonacci

• Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21
• A fibonacci number is sum of previous two numbers
• Fibonacci(1) 1
• Fibonacci(2) 1
• Fibonacci(n) Fibonacci(n-1) Fibonacci(n-2)

int Fibonacci(int n) if (n 1 n
2) return 1 else
return Fibonacci(n-1) Fibonacci(n-2)
91
Iterative Fibonacci
int Fibonacci(int n) if (n1 n2)
return 1 int prev_fib 1
int curr_fib 1 for (int count3 count
lt n count) temp
curr_fib prev_fib prev_fib
curr_fib curr_fib temp
return curr_fib
92
Backup Slides
93
Recursive Sorting

• Goal sort a jumbled set of numbers
• recursively sort all but one element
• Insert last item at the right place

94
Sort Code
void sort(int items) if (items.length
1) return int
newItems allButLast(items)
sort(newItems) insert(newItems,
itemsitems.length - 1, items)
95
Sort Code allButLast
int allButLast(int items) int
result new int items.length - 1 for
(int j 0 jltitems.length - 1 j)
resultj itemsj return result
96
Sort Code insert
void insert(int partResult, int value, int
result) int j for (j0
jltpartResult.length
partResultj lt value j)
resultj partResultj resultj
value for ( j lt partResult.length
j) resultj1 partResultj
97
Recursive Merge-Sorting

• Sort a jumbled set of numbers
• Break into two halves, recursively sort
• Merge

98
Recursive Sorting Merge

• Merge(a1,a2,an b1,b2,bn)
• if a1 lt b1 result a1, Merge(a2an
  b1,b2,bn)
• if b1 lt a1 result b1,
  Merge(a1,a2an b2bn)
• Merge(nothing, b1,b2,bn) b1, b2,bn
• Merge(aArray, apos, bArray, bpos, resArray,
  resPos)
• if aArrayapos lt bArraybpos then
• resArrayresPos aArrayapos
• Merge(aArray, apos 1, bArray, bpos, resArray,
  resPos 1)
• if aArrayapos gt bArraybpos then
• resArrayresPos bArraybpos
• Merge(aArray, apos , bArray, bpos 1, resArray,
  resPos 1)

99
Recursive Sort code
void Sort(int items) // base case
could have considered a base case of 1 if
(items.length 2) if (items0 gt
items1) int temp
items1 items1 items0
items0 temp
return
100
Sort (contd.)
// recursive case int half1
FirstHalf(items) int half2
SecondHalf(items) Sort(half1)
Sort(half2) Merge(half1, 0, half2, 0,
items, 0)
101
Recursive Merge
void Merge(int a, int aPos, int b,int bPos,
int res, int resPos) if (aPos
a.length bPos b.length)
return if (aPos a.length)
for (int jbPos, k resPos jltb.length j,
k) resk bj if
(bPos b.length) for (int jaPos, k
resPos jlta.length j, k)
resk aj
102
Merge (contd.)
if (aaPos lt bbPos)
resresPos aaPos
Merge(a, aPos 1, b, bPos, res, resPos 1)
else resresPos
bbPos Merge(a, aPos, b, bPos
1, res, resPos 1)
103
Helper Methods
int FirstHalf(int items) int
result new int items.length/2 for
(int j0 jltresult.length j)
resultj itemsj return result
int SecondHalf(int items) int
result new int items.length -
items.length/2 for (int j0, k
items.length/2 jltresult.length j, k)
resultj itemsk return
result