Download presentation
Presentation is loading. Please wait.
1
CS100A, Fall 1997. Lecture 8 1 CS100A, Fall 1997 Lecture 8, Thursday, 25 September. More on Iteration We’ll develop some loops of the form: // general picture, or invariant:... while ( B ) <body S, which makes progress toward termination but keeping the invariant true> // B is false, and from this and the invariant // we can determine that the result has been // calculated
2
CS100A, Fall 1997. Lecture 8 2 Write a method with the following heading: // Yield the position of c in s (or s.length() // if c is not in s) public static int find(char c, String s) Examples cs result yielded ‘a’“All’s well that ends” 13 ‘A’“All’s well that ends”0 ‘i’“All’s well that ends”20 ‘w’ “”0 ‘w’ “ “1 ‘ ‘“ “0
3
CS100A, Fall 1997. Lecture 8 3 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”. General picture: c is not in s[0..j-1], i.e. c is not one of the first j characters of s, and 0 <= j <= s.length(). Initialization: j= 0; When can the loop stop? When j= s.length() or c = s[j]. So the loop condition is j != s.length() && s.charAt(j) != c Not Java notation
4
CS100A, Fall 1997. Lecture 8 4 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 s[0..j-1] and // 0 <= j <= s.length(). while (j != s.length() && c != s.charAt(j)) j= j+1; return j; }
5
CS100A, Fall 1997. Lecture 8 5 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 s[0..j-1] and // 0 <= j <= s.length(). while (j != s.length()) {if c = s.charAt(j) return j; j= j+1; } return j; }
6
CS100A, Fall 1997. Lecture 8 6 Example of a loop: logarithmic spiral containing n lines 1: length d 2: length 2*d 3: length 3*d … k: length k*d... 1 2 3 4 Each line turns turn degrees to the left of its predecessor turn degrees (hc,vc) blue red green
7
CS100A, Fall 1997. Lecture 8 7 // The spiral consists of n line segments. // Line segment 1 has starting point (hc, vc). // Line segment k, for 1<=k<=n, has length k*d. // 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 k*d We demonstrate execution of this program on the Macintosh, using also n = 10,000 and different values of turn (95, 90, 180, 150),
8
CS100A, Fall 1997. Lecture 8 8 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 (k<=n) {//Draw line k if (k%3==0) g.setColor(Color.red); if (k%3==1) g.setColor(Color.blue); if (k%3==2) g.setColor(Color.green); int theta= k*turn %360; double L= k*d; // Length of line k // Compute end (h_next,v_next) of line k int h_next= (int) Math.round ( h+L*Math.cos(theta*Math.PI/180)); int v_next= (int) Math.round ( v+L*Math.sin(theta*Math.PI/180)); g.drawLine(h,v,h_next, v_next); h= h_next; v= v_next; k= k+1; }
9
CS100A, Fall 1997. Lecture 8 9 Problem: Given n>0, set s to the largest power of 2 that is at most n. ns 112^0 = 1 222^1 = 2 32 442^2 = 2*2 = 4 54 64 74 882^3 = 2*2*2 = 8 98 … 50322^5 = 2*2*2*2*2 = 32
10
CS100A, Fall 1997. Lecture 8 10 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 s is a power of 2 and s <= n Initialization: s= 1; Stop when: 2*s > n Continue as long as: 2*s <= n Make progress toward termination and keep general picture true: s= 2*s; // Known: n>0. Set s to the largest power of // 2 that is at most n s= 1; //invariant: 2 is a power of 2 and s <= n while (2*s <= n) s= 2*s;
Similar presentations
