CS100A,%20Lect.%209,%2029%20Sept.%201997

1
CS100A, Lect. 9, 29 Sept. 1997

• More on Iteration
• Well develop some loops of the form
• lt initialization to make invariant truegt
• // general picture, or invariant P ...
• while ( B )
• ltbody S, which makes progress toward
• termination but keeping P truegt
• // B is false, and from this and invariant P
• // we can see that the result has been
• // achieved

2

• Problem Given ngt0,
• set s to the largest power of 2 that is at most
  n.
• n s
• 1 1 20 1
• 2 2 21 2
• 3 2
• 4 4 22 22 4
• 5 4
• 6 4
• 7 4
• 8 8 23 222 8
• 9 8
• 50 32 25 22222 32
• 63 32
• 64 64 26 222222 64
• 65 64

3

• Strategy Start s at 1 and continue to multiply
  it by 2 until the next multiplication by 2 would
  make it too big.
• General picture (invariant)
• s is a power of 2 and s lt n
• Initialization s 1
• Stop when 2s gt n
• Continue as long as 2s lt n
• Make progress toward termination and keep
• general picture true s 2s
• // Known ngt0. Set s to the largest power of
• // 2 that is at most n
• s 1
• //invariant 2 is a power of 2 and s lt n
• while (2s lt n)
• s 2s

4

• Example of a loop logarithmic spiral
• containing n lines

2
green
turn degrees
red
3
(hc,vc)
blue
1
4
blue
1 length d 2 length 2d 3 length 3d k
length kd ...
Each line turns turn degrees to the left of its
predecessor
5

• // The spiral consists of n line segments.
• // Line segment 1 has starting point (hc, vc).
• // Line segment k, for 1ltkltn, has length kd.
• // Each line segment makes an angle of turn
  degrees
• // with the previous line segment.
• // Line colors alternate between blue, green, red
• final static int hc 300 // Center of spiral is
  (hc,vc)
• final static int vc 250
• final static int n 200 // Number of sides
  to draw
• final static int turn 121 // The turn factor
• final static double d .2 // Length of leg k is
  kd
• We demonstrate execution of this program on the
  Macintosh, using also n 10,000 and different
  values of turn (85, 89, 90, 85, 135, 180, 150),

6

• public void paint(Graphics g)
• int h hc int v vc
• int k 1
• //Invariant lines 1..k-1 have been drawn, and
  line k
• // is to be drawn with start
  point (h,v)
• while (kltn)
• //Draw line k
• if (k30) g.setColor(Color.red)
• if (k31) g.setColor(Color.blue)
• if (k32) g.setColor(Color.green)
• int theta kturn 360
• double L kd // Length of line k
• // Compute end (h_next,v_next) of line k
• int h_next (int) Math.round (
• hLMath.cos(thetaMath.PI/180))
• int v_next (int) Math.round (
• vLMath.sin(thetaMath.PI/180))
• g.drawLine(h,v,h_next, v_next)

7

• Write a method with the following heading
• // Return the position of c in s (or s.length()
• // if c is not in s)
• public static int find(char c, String s)
• Examples
• c s result
• a Alls well that ends 13
• A Alls well that ends 0
• i Alls well that ends 20
• w 0
• w 1
• 0

8

• Strategy look at the characters of s one by one,
  starting from the beginning. Return as soon as c
  is found. Use a variable j to tell how much of s
  has been scanned (looked at).
• General picture c is not in s0..j-1, i.e. c is
  not one of the first j characters of s, and
• 0 lt j lt s.length().
• Initialization j 0
• When can the loop stop?
• When j s.length() or c sj.
• So the loop condition is
• j ! s.length() s.charAt(j) ! c

Not Java notation
9

• What should be done in the body to get closer to
  termination?
• Add one to j.
• What should be done in the body to keep the
  general picture true? Nothing.
• // Yield the position of c in s (yield s.length()
• // if c is not in s)
• public static int find(char c, String s)
• int j 0
• // Invariant c is not in s0..j-1 and
• // 0 lt j lt s.length().
• while (j ! s.length( ) c ! s.charAt(j))
• j j1
• return j

10

• Another version
• // Yield the position of c in s (yield s.length()
• // if c is not in s)
• public static int find(char c, String s)
• int j 0
• // Invariant c is not in s0..j-1 and
• // 0 lt j lt s.length().
• while (j ! s.length( ))
• if c s.charAt(j)
• return j
• j j1
• return j