Recursion L

1

• Recursion (LO, Chapter 12)
• Recursion is just another way to implement
  repetition. Compare
• while (n gt 0)
• System.out.println(n)
• n--
• and
• void countDown(int n)
• if (n gt 0)
• System.out.println(n)
• countDown(n-1)
• But it is a very powerful way to implement
  repetition!

2

• Recursion (cont.)
• Examples
• Factorial
• Fibonacci
• Towers of Hanoi Recursively move piles of disks
• Quicksort Recursively sort subarrays
• Nested pictures Recursively draw smaller
  pictures
• (Binary) tree traversal TreeSet iteration
• Expression evaluation Recursively evaluate
  subexpressions
• Region growing (MineSweeper, Bubblet and AtariGo)
• Maze searching (backtrack search)
• Shortest path problems
• Program compilation
• ...

3

• Factorial
• The factorial of a nonnegative integer n, denoted
  n!, is the product of the positive integers less
  than or equal to n. E.g., 5! 5 4 3 2 1
  120. Equivalently, 0! 1, n! n(n-1)! (for
  n gt 0). The factorial numbers are thus 1, 1, 2,
  6, 24, 120, 720, 5040, ...
• int factorial(int n) // recursive definition
• if (n 0)
• return 1
• else
• return n factorial(n-1)
• Abbreviate factorial by f
• f(4) 4 f(3) 4 (3 f(2)) 4 (3 (2
  f(1)))
• 4 (3 (2 (1 f(0))))
• 4 (3 (2 (1 1))) ... 24
• (Here the recursion is unnecessary.)

4

• Fibonacci
• The Fibonacci numbers are formed by starting with
  0 and 1 and then adding the latest two numbers to
  get the next one. Equivalently, f(0) 0, f(1)
  1, f(n) f(n-2) f(n-1) (for n gt 1). The
  Fibonacci numbers are thus 0, 1, 1, 2, 3, 5, 8,
  13, 21, 34, 55, 89, ...
• int fibonacci(int n) // naive recursive
  definition
• if (n lt 1)
• return n
• else
• return fibonacci(n-2) fibonacci(n-1)
• f(5) f(3) f(4) (f(1) f(2)) (f(2)
  f(3))
• (1 (f(0) f(1))) ((f(0) f(1))
  (f(1) f(2)))
• (1 (0 1)) ((0 1) (1 (f(0)
  f(1))))
• 2 (1 (1 (0 1))) ...
• 5
• (Here the recursion is unacceptably inefficient.
  Consider the tree for f(10). This is an
  exponential computation.)

5

• Fibonacci (cont.)
• int fibonacci(int n)
• return f(0, 1, n)
• int f(int prev, int curr, int n) // efficient
  recursion
• if (n 0)
• return prev
• else
• return f(curr, prevcurr, n-1)
• Now there is no repeated computation, this is a
  linear computation
• f(0, 1, 6) f(1, 1, 5) f(1, 2, 4) f(2, 3, 3)
• f(3, 5, 2) f(5, 8, 1) f(8, 13,
  0) 8

6

• Recursion (cont.)
• Every recursive method must have the following
  properties
• A termination condition (e.g., n 0, n lt 1 ),
  cf. basis for induction.
• Every recursive call must strictly approach the
  termination condition (e.g., factorial(n-1),
  fibonacci(n-2), fibonacci(n-1) ), f(curr, prev,
  curr, n-1), cf. induction step in proof
  by induction.
• Recursion is analogous to induction.
• Recursive methods can be proved correct by
  induction.

7

• Towers of Hanoi
• (Ref. E. Rouse Ball, Mathematical Recreations )
• How to move the pile of n disks from peg 1 to
  peg 3, one disk at a time, so that a larger disk
  is never above a smaller disk?

1
2
3
8

• Towers of Hanoi (cont.)
• If n 0, we are finished, and don't need to do
  anything at all.
• Otherwise, suppose we are moving n disks from
  peg i to peg k
•  Move n-1 disks from peg i to peg j
•  Move 1 disk (the largest of the n) from peg i
  to peg k
•  Move (the top, smaller) n-1 disks from peg j
  to peg k
• Successive states
• ABC on peg 1
• BC on peg 1
  A on peg 3
• C on peg 1 B on peg 2 A
  on peg 3
• C on peg 1 AB on peg 2
• AB on peg 2
  C on peg 3
• A on peg 1 B on peg 2 C
  on peg 3
• A on peg 1
  BC on peg 3
• 
  ABC on peg 3

9

• Towers of Hanoi (cont.)
• / Moves n disks from from to to using other /
• void move(int n, int from, int other, int to)
• if (n gt 0)
• move(n-1, from, to, other)
• System.out.println("Move disk from "
  from
• " to " to)
• move(n-1, other, from, to)
• The initial call is, say
• move(3, 1, 2, 3)
• Note
• Termination condition n 0
• Each recursive call strictly approaches the
  termination condition move(n-1, from, to,
  other) (n-1 lt n)

10

• Towers of Hanoi (cont.)
• Output
• Move disk (A) from 1 to 3
• Move disk (B) from 1 to 2
• Move disk (A) from 3 to 2
• Move disk (C) from 1 to 3
• Move disk (A) from 2 to 1
• Move disk (B) from 2 to 3
• Move disk (A) from 1 to 3
• Exercise Write a recursive method to move n disks
  from one peg to another under these rules when
  there are 4 pegs available in total. Try to find
  a method that takes as few moves as possible.
• (Clearly this can be done in fewer moves than
  when there are only 3 pegs available. But no
  algorithm has been proved to always take the
  minimum possible number of moves for 4 or more
  pegs. Indeed, the minimum possible number of
  moves is unknown for arbitrary n and 4 or more
  pegs.)

11

• Quicksort (repeated)
• public void quicksort(String a, int m, int n)
• if (m lt n)
• // Partition am..n about a "pivot"
  element x
• // (Omitted here...)
• // Recursively sort the "small" and "large"
  subarrays of a
• // independently
• quicksort(a, m, j) // sort am..j
• quicksort(a, i, n) // sort ai..n
• Note
• Termination condition m gt n
• Each call strictly approaches the termination
  condition  quicksort(a, m, j) (where j lt n),
   quicksort(a, i, n), (where m lt i )

12

• Nested images I

13

• Nested images I (cont.)
• class ImagePane extends JPanel
• public void paintComponent(Graphics g)
• Image image
• Toolkit.getDefaultToolkit().getImage("g
  avarnie.jpg")
• drawPicture(g, 6, size, image)
• private void drawPicture(Graphics g, int n,
  int size,
• Image image)
• if (n gt 0)
• g.drawImage(image, 0, 0, size, size,
  this)
• drawPicture(g, n-1, size/2, image)

14

• Nested images II (Lewis Loftus, pp. 451453)
• (See LL, Figure 12.5, p.452)
• void drawPicture(int size, Graphics g)
• if (size gt STOP)
• // Separate picture into four quadrants
• g.drawLine(size/2, 0, size/2, size)
• g.drawLine(0, size/2, size, size/2)
• // Draw three images in different
  quadrants
• g.drawImage(world, size/2OFFSET,
  0OFFSET,
• size/2-(OFFSET2), size/2-(OFFSET2),
  this)
• g.drawImage(everest, 0OFFSET,
  size/2OFFSET,
• size/2-(OFFSET2), size/2-(OFFSET2),
  this)
• g.drawImage(goat, size/2OFFSET,
  size/2OFFSET,
• size/2-(OFFSET2), size/2-(OFFSET2),
  this)
• // Draw the entire picture again in other
  quadrant
• drawPicture(size/2, page)

15

• Nested images II (cont.)
• import java.applet.Applet
• import java.awt.
• public class NestedImages extends Applet
• private final int SIZE 300 // Size of
  applet
• private final int STOP 20 // Smallest
  picture size
• private final int OFFSET 2 // Picture
  offset from border
• private Image world, everest, goat
• public void init()
• world getImage(getDocumentBase(),
  "world.gif")
• / Similarly for images everest and goat
  /
• setSize(SIZE, SIZE)
• public void paint(Graphics g)
• drawPicture(SIZE, g)
• private void drawPicture(int size, Graphics
  g) ...

16

• Nested images III (Abelson Sussman, SICP,
  Section 2.2.4)
• (See SICP, Figure 2.14)

Image
Top
Top
Right
Image
Image
Image
Right
Top
Right
17

• Nested images III (cont.)
• void drawMain(Image image, int x, int y, int
  size, Graphics g)
• if (size gt STOP)
• // Draw image in bottom left quadrant
• g.drawImage(Image, x, ysize/2, size/2,
  size/2, this)
• // Draw recursive pictures in other three
  quadrants
• drawTop(image, x, y, size/2, g)
• drawMain(image, xsize/2, y, size/2, g)
• drawRight(image, xsize/2, ysize/2,
  size/2, g)
• Note that the code exactly mirrors the pattern.

18

• Nested images III (cont.)
• void drawTop(Image image, int x, int y, int size,
  Graphics g)
• if (size gt STOP)
• // Draw images in bottom quadrants
• g.drawImage(Image, x, ysize/2, size/2,
  size/2, this)
• g.drawImage(Image, xsize/2, ysize/2,
  size/2, size/2,
• this)
• // Draw recursive pictures in top
  quadrants
• drawTop(image, x, y, size/2, g)
• drawTop(image, xsize/2, y, size/2, g)
• Exercise 1 Write method drawRight.
• Exercise 2 Complete and test an application with
  a suitable image.

19

• Binary tree traversal
• The easiest way to visit the nodes of a binary
  search tree in their natural order is
  recursively
• void traverse(Entry t)
• if (t ! null)
• // Traverse left subtree
• traverse(t.left)
• // Visit node
• System.out.print(t.key)
• // Visit right subtree
• traverse(t.right)
• Note
• Termination condition t null
• Recursive calls strictly approach termination
  condition, as subtrees have strictly fewer nodes
  than the whole tree.

20

• Expression evaluation
• (To be completed...)

21

• Region growing
• A bitmap
• 00000000011001111000000001111
• 01110011011100111110000011100
• 01111010000001111111110011000
• 01111010001111111100110011000
• 01111010000001111100011011000
• 01111010000001111100111011100
• 01111010000000011110000011110
• 01111110011000000000100011110
• What is the area of the "island" containing a
  given point, (row, col)? I.e., write a method
  which takes a bitmap and a point as arguments,
  and returns the number of ones connected to the
  given point.
• (We interpret ones as "land" and zeros as "sea".)
• (This problem arises in MineHunt, Bubblet,
  AtariGo, ...)

22

• Region growing (cont.)
• We need to look outward from the given point to
  each neighbour that we have not yet seen.
• Initially, assign all elements of an auxiliary
  boolean array, seen, to false (unseen). (Array
  seen is just a representation of a set of
  points.)
• Main method
• // Returns the number of unvisited points in the
  island
• // containing point (row, col).
• int area (int row, int col)
• int area 0
• if (gridrowcol 1 ! seenrowcol)
  
• area
• seenrowcol true
• for each neighbour (r, c) of (row, col)
• area area(r, c)
• return area

23

• Region growing (cont.)
• Note An alternative solution to this problem was
  given in the lecture notes on the collection
  framework. The problem can be solved equally
  easily iteratively or recursively. (Though the
  recursive solution is arguably simpler.)
• Exercise Modify either solution to return the set
  of points in the region.

24

• Maze Solving (LL, pp.455459)
• Maze input
• 4 8 // number of rows and columns
• 1 1 0 1 1 1 1 0 // 0 blocked, 1 free
• 0 1 1 1 0 1 0 1
• 0 1 0 1 0 1 1 1
• 1 1 1 0 0 1 0 1
• Goal To get from the top left corner (0, 0) to
  the bottom right corner (3, 7).
• Solution At each unvisited position, try going
  down, right, up, left, and repeating, in turn.
• 7 7 0 7 7 7 1 0 // 1 still free
• 0 7 7 7 0 7 0 1 // 3 visited
• 0 3 0 3 0 7 7 7 // 7 visited and on path
• 1 3 3 0 0 3 0 7

25

• Maze Solving (cont.)
• Maze representation
• final int NROWS 4, NCOLS 8
• final int grid new intNROWSNCOLS
• Main algorithm
• // Finds a path (if any) from (row, col) through
  unvisited
• // squares to the destination (NROWS-1,NCOLS-1).
• boolean solve (int row, int col)
• boolean done false
• if (valid(row, col)) // Unvisited position
• gridrowcol 3 // Mark position visited
• if (goal(row, col)) // Destination reached?
• done true
• else

26

• Maze Solving (cont.)
• // ... continuation
• if (! done)
• done solve(row, col1) // Try
  right
• if (! done)
• done solve(row-1, col) // Try
  up
• if (! done)
• done solve(row, col-1) // Try
  left
• if (done)
• gridrowcol 7 // Mark
  position on path
• return done
• Initial call

27

• Maze Solving (cont.)
• Auxiliary methods
• boolean valid (int row, int col)
• return (0 lt row row lt NROWS
• 0 lt col col lt NCOLS
• gridrowcol 1) // position is
  FREE
• // and
  UNVISITED
• boolean goal (int row, int col)
• return (row NROWS-1 col NCOLS-1)
• Exercise Write a method to print the coordinates
  of the elements of the path from the start, (0,
  0), to the goal, (nrows-1, ncols-1).
• See sample programs on Web site. Program OnePath
  completes this implementation. Program AllPaths
  makes minor changes to count the total number of
  paths.

28

• General path finding algorithm
• (To be completed...)

29

• Counting rook paths on a chess board
• Find the number of non-intersecting rook paths
  from the top left square of an mxn chess board to
  the bottom right square of the board. E.g., for
  mn2, there are 2 paths, for m2, n3 there are
  4 paths, for mn3 there are 12 paths.
• This can be solved by a simple generalisation of
  the all-paths version of the maze solving
  program.
• See the example program RookPaths on the Web
  site.
• See the following Web site for extensive
  information on this recreational problem
• http//pauillac.inria.fr/algo/bsolve/constant/cnnt
  v/gammel/gammel.html
• Exercise (difficult) Modify the program RookPaths
  so that it can count the number of solutions on
  an 8x8 board in an acceptable time.

30

• Graph coloring
• A map
• The corresponding graph

1
2
0
3
6
4
1
2
0
3
6
4
5
31

• Graph coloring (cont.)
• Problem Given n colors, can we assign a different
  color to each region (vertex) so that adjacent
  regions (vertices) are assigned different colors?
• Answer (Haken and Appel, ca. 1970) For a planar
  map, with n 4, yes!
• But how?
• Graph representation
• 0 1 6 // nbrs0 lt1, 6gt
• 1 0 2 6 // nbrs1 lt0, 2, 6gt
• 2 1 3 6 // ...
• 3 2 4 6
• 4 3 6
• 5
• 6 0 1 2 3 4 // nvertices 7

32

• Graph coloring (cont.)
• Graph representation (cont.)
• private int ncolors // Maximum number of colors
  available
• private int nvertices // Number of vertices in
  graph
• private int nbrs
• // List of neighbours of vertices 0, 1, ...,
  nvertices-1
• private int colors
• // Color of vertices 0, 1, ..., nvertices-1
• // For all v, 0 lt colorsv lt ncolors
• // Initially, colorsv 0, for 0 lt v lt
  nvertices

33

• Graph coloring (cont.)
• Main algorithm
• // Assuming vertices 0,...,v-1 are already
  consistently colored,
• // tests whether there is a consistent assignment
  of colors
• // to vertices v,...,nvertices-1.
• public boolean solve (int v)
• if (goal(v)) // All vertices consistently
  colored?
• return true
• else
• // Try to color vertex v with each color c in
  turn
• for (int c 0 c lt ncolors c)
• if (valid(v, c))
• colorsv c
• if (solve(v1))
• return true

34

• Graph coloring (cont.)
• Auxiliary methods
• private boolean goal (int v)
• return (v nvertices)
• // Tests whether c is different from the colors
  of all preceding
• // neigbours of v.
• private boolean valid (int v, int c)
• for (int i 0 i lt nbrsv.length i)
• int u nbrsvi
• if (u lt v colorsu c)
• return false
• return true

35

• Backtrack Search Single solution (not
  examinable)
• The maze solving and graph colouring problems are
  examples of backtrack search for a single
  solution. There are many other examples.
• Backtrack search arises whenever there is a need
  to make repeated choices (or assignments) subject
  to some consistency condition, possibly requiring
  undoing previous choices (or assignments). This
  "undoing", followed by a different choice (or
  assignment) is called backtracking.
• Backtrack search is most easily implemented
  recursively, as in the previous examples.
• It can be recognised whenever there is an
  apparent need for arbitrarily deeply nested
  loops.
• for each choice for position 0
• for each valid choice for position 1
• for each valid choice for position 2
• ...
• for each valid choice for position n
• Print the choices for all positions,
  and exit

36

• Backtrack Search Single solution (cont.)
• Corresponding recursive schema
• // Assuming valid choices for positions
  0,...,k-1, finds a
• // consistent assignment of choices for positions
  k,...,n-1.
• boolean solve (int k)
• if (choice0..k-1 is a solution) // e.g.,
  k n
• return true
• else
• for each valid choice c for position k,
• given choice0..k-1
• choicek c
• if (solve(k1))
• return true
• return false

37

• Backtrack Search All solutions (not examinable)
• Backtrack search is often required to generate
  all solutions to some combinatorial problem, as
  in the rook paths problem (or the nested loops
  example). In this case we need a slight variant
  to the general schema.
• // Assuming valid choices for positions
  0,...,k-1, finds all
• // consistent assignment of choices for positions
  k,...,n-1.
• void solve (int k)
• if (choice0..k-1 is a solution) // e.g.,
  k n
• print, process or count choice0..k-1
• else
• for each valid choice c for position k,
• given choice0..k-1
• choicek c
• solve(k1)
• Again, initially, call solve(0).

38

• Backtrack Search Iterative version (not
  examinable)
• Backtrack search can also be implemented
  iteratively, e.g., to generate all solutions
• k 0
• S0 set of possibilities for choice0
• while (k gt 0)
• while (Sk is not empty) // advance
• Choose c in Sk and remove it
• choicek c
• if (choice0..k is a solution) // e.g.,
  k n-1
• Print, process or count choice0..k
• else
• k k 1
• Sk set of possibilities for
  choicek,
• given choice0..k-1
• k k - 1 // backtrack

39

• Backtrack Search Iterative version (cont.)
• This iterative schema may run faster than the
  corresponding recursive schema, but only by a
  small constant factor, and is more complex since
  the array S0..n-1 of sets of possible choices
  needs to be explicitly maintained.
• Exercise Write an iterative schema for the
  iterative search to find a single solution to
  such a combinatorial problem.
• Exercise Re-implement either solution to the Maze
  problem iteratively.

40

• Backtrack Search Least solution (not examinable)
• Often we wish to find the least or cheapest
  solution. This requires a variation on the
  all-solutions schema
• float minCost infinity // cost of current
  min-cost solution
• Value solution // current min-cost
  solution
• Value choice new ValueMAX // current
  partial solution
• void solve (int k, float cost)
• if choice0..k-1 is a solution // new
  min-cost solution
• minCost cost
• solution copy(choice)
• else
• // Try each choice c for position k in
  turn
• for each valid choice c for position k
• given choice0..k-1
• newCost f(cost, k, c) // cost of
  new partial soln
• if (newCost lt minCost)
• choicek c

41

• Backtrack Search Shortest solution (not
  examinable)
• We may adapt this schema to find the shortest
  path to a solution (cost is the number of moves)
  in a state-space exploration problem such as
  Traffic.
• int minMoves infinity // length of current
  shortest solution
• Move solution // current shortest
  solution
• Move choice new MoveMAX // current partial
  solution
• void solve (int k, Position pos, int nMoves)
• if p is the goal position // new shortest
  solution
• minMoves nMoves
• solution copy(choice)
• else
• for each valid move m from position pos
• newPos apply move m to position pos
• if we have not been to newPos before
  (on this path)
• nMoves
• if nMoves lt minMoves
• choicek m // new partial
  solution

42

• Backtrack Search Shortest solution (cont.)
• Initially, call solve(0, initialPos, 0). The
  solution is array solution, length minMoves.
• Basically, this is a classical AI problem, and
  efficient solutions are still discussed in AI
  courses.
• Warning There could be errors in the above
  schema.
• Note It may be simpler to solve this problem by
  breadth-first search, using an explicit queue of
  positions to be expanded, as with the first
  solution to the island area problem.