exercises1.txt

Algorithms 4th Ed

Exercises 1

1.1.1 Give the value of the following expressions:

a) (0 + 15) / 2 ?

= 7

because is integer division.  15/2 = 7.5 gets truncated to 7

b) 2.0e-6 * 100000000.1

== 0.000002 * 100000000.1

= 2 x 10^-6 * 1000000001 * 10^-1 == 2000000002 * 10^-7 == 200.0000002

Program prints this value as expected

c) true && false || true && true

evaluation is left associative so first term is true then next term is false BUT there is then an or so 
checks that term - is true so then checks last term - is true so returns true.

1.1.2 Give the type and value of each of the following expressions:

a) (1 + 2.236) / 2

In brackets there is a floating point number so integer 1 is promoted to floating point and hence brackets 
evaluates to 3.236.  floating point divided by integer is promoted to floating point division and result is: 1.618

b) 1 + 2 + 3 + 4.0 = 10.0

addition is left associative so 1 + 2 + 3 results in integer addition resulting in 6

Then you get 6 + 4.0 which is promoted to 6.0 + 4.0 = 10

c) 4.1 >= 4  = true

d) 1 + 2 + "3" = "123"

*** Got that wrong.  I am getting 33.  Why???

Ah got it, addition is left associative so you get 1+2 = 3 by integer addition. Then java seens + used 
with an integer and a string so the tostring conversion is used to concatenate resulting in "33"

1.1.3 Write a program that takes three integer command line arguments and prints equal if all three are equal, 
and not equal otherwise.

// set PATH="C:\Program Files\Java\jdk1.7.0_07\bin";%PATH%
// javac -cp stdlib.jar;algs4.jar;. ex1a.java
// java -cp stdlib.jar;algs4.jar;. ex1a 6 7 8

public class ex1a {

  public static void main(String[] args) {
      if(args.length < 3)
	  StdOut.println("Must pass 3 integer arguments");
      else {
         int n1 = Integer.parseInt(args[0]);
         int n2 = Integer.parseInt(args[1]);
         int n3 = Integer.parseInt(args[2]);
         StdOut.println( n1 == n2 && n2 == n3 ? "equal" : "not equal");
      }
  }
}

1.1.4 What (if anything) is wrong with each of the following statements?

a) if (a > b) then c = 0;  // then is a syntax error
b) if a > b { c = 0; }     //brackets around a>b expression required
c) if (a > b) c = 0;       // fine.
d) if (a > b) c = 0 else b = 0;   //; required after c = 0

1.1.5 Write a code fragment that prints true if the double variables x and y are both strictly between 
0 and 1 and false otherwise.

  public static void main(String[] args) {
      if(args.length < 2)
	  StdOut.println("Must pass 2 double arguments");
      else {
         double d1 = Double.parseDouble(args[0]);
         double d2 = Double.parseDouble(args[1]);
         StdOut.println(d1 > 0 && d1 < 1 &&  d2 > 0 && d2 < 1 );
      }
  }


1.1.6 - prints fibonacci sequence.

1.1.7 Give the value printed by each of the following code fragments:

a.  
    double t = 9.0;
    while(Math.abs(t - 9.0/t) > .001)
        t = (9.0/t + t) / 2.0;
    StdOut.printf("%.5f\n", t);

while loop produces:
5.0
3.4
3.02
converging towards 3.0000000

When run mine produced: 3.00009

b.

int sum = 0;
for (int i = 1; i < 1000; i++)
    for(int j = 0; j < 1000; j++)
        sum++;
StdOut.println(sum);

should produce 1000 * 1000 (1000 being incremented by 1, 1000 times.

Ah, that was wrong.  Produces 999000.

Ah, because i is from 1 to 999 (excludes 1000) you miss out 1000.  and 999000 + 1000 = 1000000.

Edit for loop to be (int i = 1; i <= 1000; i++)  and get expected 1m. Or better:
(int i = 0; i < 1000; i++)

c.

int sum = 0;
for (int i = 1; i < 1000; i++)
    for(int j = 0; j < 1000; j++)
        sum++;
StdOut.println(sum);

The outer loop doubles each time, bounded by 1000.  log2(1000)/log2(2) = 9.97 so will go from 0 - 9

so sum will be 9 * 1000 = 9000

No, because I forgot log2(1) = 0.  So add 1.

sum will be 10 * 1000 = 10000

ie i increments like this:

1    1    2^0   
2    2    2^1
3    4    2^2
4    8    2^3
5    16   2^4
6    32   2^5
7    64   2^6
8    128  2^7
9    256  2^8
10   512  2^9

1.1.8 What do each of the following print?

a. System.out.println('b');

prints b

Correct.

b. System.out.println('b' + 'c');

prints bc

Wrong! Prints 197.

Why?

The + operator does not work like concatenation on char values, instead it works like addition.

ascii value 'b' is integral value 98 (or 0x62). The UTF value is the same.  Ascii 'c' is 99.  98 + 99 =  197

c. System.out.println('a' + 4);

From above we can probably guess correctly.

ascii 'a' = 97  + 4 = 101

Correct!

1.1.9 Write a code fragment that puts the binary representation of a positive integer N into a String s.

Without looking at solution I did this:

public static String int2binary(int n) {
    String s = new String();
    while(n > 0) {
        s += (n % 2);
        n /= 2;
    }
    StringBuilder rev = new StringBuilder(s);
    return rev.reverse().toString();
}

The reversing bit on end is not very elegant.  so I liked the solution.  all it does differently is add the 
character to the BEGINNING of the string each time.

my version converted:

public static String int2binary(int n) {
    String s = new String();
    while(n > 0) {
        s = (n % 2) + s;
        n /= 2;
    }
    return s;
}

1.1.10 - answer provided - just read it.

1.1.11 Write a code fragment that prints the contents of a two dimensional boolean array, using * to to 
represent true and a space to represent false. Include row and column numbers.

public static void printarray(boolean[][] arr) {
    for(int row = 0; row < arr.length; ++row) {
        for(int col = 0; col < arr[row].length; ++col) {
            System.out.print(arr[row][col] ? "*" : " "); 			        }
    	System.out.print('\n');
    }
    System.out.printf("Array contains %d rows and %d columns\n", arr.length, arr[0].length);
}

1.1.12 What does the following print?

		  int[] a = new int[10];
		  for (int i = 0; i < 10; i++)
			  a[i] = 9 - i;
		  for (int i = 0; i < 10; i++)
			  a[i] = a[a[i]];
		  for (int i = 0; i < 10; i++)
			  System.out.println(i);

Surely it just prints 0 to 9.

And it does.  

1.1.13 Write a code fragment to print the transposition (rows and columns changed) of a two dimensional 
array with M rows and N columns.

import java.util.Arrays;

//http://en.wikipedia.org/wiki/Transpose
public static int[][] transposition(int[][] array,int m,int n){
        int[][] transposed=new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                transposed[i][j]=array[j][i];
            }
        }
        return transposed;
}

1.1.14 Write a static method lg() that takes an int value N as argument and returns the largest int not 
larger than the base-2 logarithm of N. Do not use Math.

// log base 2
public static int lg(int n) {
    //take number and keep dividing by 2 until zero
    int count = 0;
    while(n > 0) {
    	n /= 2;
    	++count;
    }
    return count - 1;
}

1.1.15 Write a static method histogram() that takes an array a[] of int values and an integer M as 
arguments and returns an array of length M whose ith entry is the number of times the integer i 
appeared in the argument array. If the values in a[] are all between 0 and M-1, the sum of the values 
in the returned array should be equal to a.length.

   int[] histogram(int[] array, int m) {
        Arrays.sort(array);
        int[] ret=new int[m];
        int i=0;
        while (i<array.length&&array[i]<m){
            ret[array[i]]= ++ret[array[i]];
            i++;
        }
        return ret;
    }

Call it like this:

int[] arr = {5,5,5,4};
int ret[] = histogram(arr, 6));



1.1.16 Give the value of exR1(6):

    public static String exR1(int n)
    {
        if (n <= 0) return "";
        return exR1(n-3) + n + exR1(n-2) + n;
    }

   exR1(6) -> n >= 0 so returns blank string    
      
1.1.17 Criticise the following recursive function:

    public static String exR2(int n)
    {
        String  s = exR2(n-3) + n + exR2(n-2) + n;
        if(n <= 0) return "";
        return s;
    }

   Example exR2(6)

   exR2(6) -> exR2(3) + 6 + exR2(4) + 6;

   exR2(3) -> exR2(0) + 3 + exR2(1) + 3;

   exR2(4) -> exR2(1) + 4 + exR2(2) + 4;

   exR2(0) -> exR2(-3) + 0 + exR2(-2) + 0;

   Problem is that the base case (ie when a condition met, return) is AFTER recursive calls.  So recursion 
   step never ends.  Get stackoverflow as per book.


1.1.18 Consider the following recursive function:

    public static int mystery(int a, int b)
    {
        if (b == 0) return 0;
        if(b % 2 == 0) return mystery(a+a, b/2);
        return mystery(a+a, b/2) + a;
    }

What are the values of mystery(2, 25) and mystery(3, 11)? Given positive integers a and b, describe what 
value mystery(a, b) computes. answer the same question, but replace + with * and replace return 0 with return 1.

mystery(2, 25) 
   1)     -> mystery(4, 12) + 2
   2)     -> ( mystery(8, 6) ) + 2
   3)     -> ( mystery(16, 3) + 16) + 2
   4)     -> (( mystery(32, 1) + 32 ) + 16) + 2
   5)     -> ( mystery(64, 0) + 32 ) + 16) + 2
   6)     -> ( 0 ) + 32 + 16 + 2
   = 50

   1) -> mystery(3, 11)
   2) -> ( mystery(6, 5) + 3
   3) -> ( mystery(12, 2) + 6) + 3
   4) -> ( mystery(24, 1) + 6) + 3
   5) -> ( mystery(48, 0) + 24 ) + 6) + 3
   5) -> ( 0 + 24 + 6 + 3)
   = 33

   mystery(a, b) = ?

Well, I thought best thing to do is try some small numbers and run them and see what it looks like:

example input:
0,  5 =  0
1,  5 =  5
2,  5 = 10
2, 25 = 50

it looks like the answer is a * b.

Lets put in some values just to check that works.  Try even and odd combinations.  You could use code like 
this to test:

  if(args.length < 2)
    System.out.println("Must pass 2 integer arguments");
  else {
    System.out.println(mystery(Integer.parseInt(args[0]), Integer.parseInt(args[1])));
  }

It seems to work.  Hence mystery(a, b) = a * b

If we change any + to * and 0 to 1,  then the function will eventually return a^b.


1.1.19 Run the following program on your computer:

public class fibonacci {

    public static long f(int n)
    {
        if (n == 0) return 0;
        if (n == 1) return 1;
        return f(n-1) + f(n-2);
    }

  public static void main(String[] args) {
      for(int  n = 0; n<100; n++)
          StdOut.println(n + " " + f(n));
  }
}

This program takes a long..... long.... time.

Largest value of n?  About 55?

The question asks for a better implementation which saves computed values in an array.

OK looked this up.  Idea is memoization.  See here:  
http://en.wikipedia.org/wiki/Memoization

And here for Java code examples: 
http://en.wikibooks.org/wiki/Algorithm_Implementation/Mathematics/Fibonacci_Number_Program#Java

Essentially, the idea is to 'chache' values in an array.

Doing this very simply with fixed size array:

public class fib_memo {

    static long[] fib = new long[100];

    public static long f(int n)
    {
        if (n == 0) return 0;
        if (n == 1) return 1;
        if(fib[n] == 0) {
	    fib[n] = f(n-1) + f(n-2);
        }
        return fib[n];
    }


  public static void main(String[] args) {
      for(int  n = 0; n<100; n++)      
	  StdOut.printf("fib[%4d] = %14d\n", n, f(n));
  }
}

To do more professionally, you would need to use a re-sizeable array and keep expanding the array as large n's 
were required.  But this simple implementation makes operation very quick.  You now do not even notice any 
delay in printing results.

One other issue is that even the 64 bit long overflows with large fib numbers.  Looks like from fib(93).  
So better to use Java's bigInteger type for big n.




If you read up on fibonacci and the golden ratio you will see that there is an algebraic method of 
calculating fibonacci.  It is an estmimation but it is accurate for integer values.

Formula can be found here:
http://www.mathsisfun.com/numbers/nature-golden-ratio-fibonacci.html

In java code:

public class fibonacci {

    public static long f(int n)
    {
        if (n == 0) return 0;
        if (n == 1) return 1;
        return f(n-1) + f(n-2);
    }

    // phi = (1 + sqrt(5) / 2)
    public static double phi()
    {
        return (1 + Math.sqrt(5)) / 2.0;
    }

    public static long fib_alg(int  n) {
        return (long)Math.round(Math.pow(phi(), n)  / Math.sqrt(5.0) );
    }

  public static void main(String[] args) {
      for(int  n = 0; n<100; n++)
          StdOut.println(n + " " + fib_alg(n));
  }
}


1.1.20 Write a recursive static method that computes the value of ln(N!)

public class ex_1_1_20 {

    public static long fact(int n)
    {
        if (n < 2) return 1;
        return n * fact(n-1);
    }

  public static void main(String[] args) {
      for(int  n = 0; n<20; n++)
          StdOut.printf("ln(%d!) = %f\n", n, Math.log(fact(n)));
  }
}


1.1.21 Write a program that reads in lines from std input with each line 
containing a name and two integers and then use printf() to print a table 
with a column of the names, the integers, and the result of dividing the 
first by the second, accurate to three decimal places. You could use a 
program like this to tabulate batting averages for baseball players or grades for students.

public class ex_1_1_21 {

public static class Data
{
    public Data() {}
    public Data(double v1, double v2, String s1) { n1 = v1; n2 = v2; s = s1; } 
    public double n1; 
    public double n2;  
    public String s; 
};

  public static void main(String[] args) {
	int i = 0;
	int idx = 0;  // array index  
	double v1, v2;
	v1 = v2 = 0;
	String s;

	Data items[] = new Data[100];  //asumes no more than 100 items

      while(!StdIn.isEmpty()) {
          switch(i % 3) {
	    case 0: v1 = StdIn.readFloat(); break;
            case 1: v2 = StdIn.readInt();   break;
            case 2:  
                s =  StdIn.readString(); 
		items[idx++] = new Data(v1, v2, s);
               break;
          }
          ++i;
      }
      
      StdOut.println("Item\tNum1\tNum2\tdiv");
      for(int j = 0; j < idx; ++j) {
          StdOut.printf(  "%s\t%3.3f\t%3.3f\t%3.3f\n", 
               items[j].s, items[j].n1, items[j].n2, items[j].n1 / items[j].n2);
      }	
      StdOut.printf("%d items\n", idx);           
  }
}

Improved tidy version:

import java.util.Iterator;
import java.util.LinkedList;
import java.util.Scanner;


public class test {

	public class Data
	{
	    public Data() {}
	    public Data(double v1, double v2, String s1) { n1 = v1; n2 = v2; s = s1; } 
	    public double n1; 
	    public double n2;  
	    public String s; 
	};
	
    LinkedList<Data> getData() {
	      int i = 0;
	      double v1, v2;
	      v1 = v2 = 0;
	      String s;
	      LinkedList<Data> list = new LinkedList<Data>();
		  
	      Scanner input = new Scanner(System.in);
		  
		  while(input.hasNext()) {
		     switch(i % 3) {
		 	    case 0: v1 = input.nextFloat(); break;
	            case 1: v2 = input.nextFloat(); break;
	           case 2:  
	                s = input.next();
	        		list.add(new Data(v1, v2, s));
	           break;
	         }
	         ++i;	            
		  }
		  input.close();
		  return list;
    }
    
    void display_table(LinkedList<Data> list) {
		System.out.printf("%d items\n", list.size());	
		System.out.println("Item\tNum1\tNum2\tdiv");
        Iterator<Data> iter = list.iterator(); 
          
        while(iter.hasNext()) {
        	  Data item = (Data) iter.next();
        	  System.out.printf(  "%s\t%3.3f\t%3.3f\t%3.3f\n", item.s, item.n1, item.n2, item.n1 / item.n2);
        }    	
    }
	
	void run() {
		display_table(getData());
	}
	
	  public static void main(String[] args) {
		  new test().run();
	  }
}


1.1.22 Write a version of BinarySearch that uses the recursive rank() given on page 25 and traces the method 
calls.  Each time the recursive method is called, print the argument values lo and hi, indented by the depth 
of the recursion. hint: Add an argument to the recursive method that keeps track of the depth.

import java.util.Arrays;

public class BinarySearchRec
{
    public static int rank(int key, int[] a)
    {
        return rank(key, a, 0, a.length - 1, 0); 
    }

    public static int rank(int key, int[] a, int lo, int hi, int depth)
    {  // Index of key in a[], if present, is not smaller than lo
       // and not larger than hi
	int tmp = depth;
       while(tmp-- > 0)
	   StdOut.print(' '); //padding by recursion call depth

	StdOut.printf("key = %d, lo = %d, hi = %d\n", key, lo, hi);
	if(lo > hi) {
            StdOut.printf("key = %d not found\n", key);
            return -1;
        }
       int mid = lo + (hi - lo) / 2;
       if (key < a[mid]) return rank(key, a, lo, mid - 1, ++depth);
       else if (key > a[mid]) return rank(key, a, mid + 1, hi, ++depth);
       else return mid;
    }

    public static void main(String[] args) {
        int[] whitelist = In.readInts(args[0]);
        Arrays.sort(whitelist);

        while(!StdIn.isEmpty()) 
        {  //read key, print if not in whitelist
            int key = StdIn.readInt();
            if(rank(key, whitelist) < 0)
               StdOut.println(key);
        }
    }
}