Chapter 5 Iteration

Chapter 5 Iteration

while

while (expression)
   statement;

The expression is evaluated before each repetition of the loop and so long as it is nonzero the statement is executed.

Examples of while statement.

  i=1;
  while(i<4) {
     cout << i << endl;
     i++;
  }

  i=1;
  while(i++ < 4)
     cout << i << endl;

  i=1;
  while(++i < 4)
    cout << i << endl;

do .. while

do
 statement;
while (expression);

Expression is evaluated after the the statement is executed. If the expression is nonzero the statement is repeated.

do .. while examples.

  i=1;
  do {
      cout << i << endl;
      i++;
  } while (i < 4);

  i=5;
  do {
      cout << i << endl;
      i++;
  } while (i < 4);

for
```
for(expression1;expresssion2;expression3)
   statement;

 equivalent to

  expression1;
  while(expression2) {
       statement;
       expression3;
  }
```
1. expression1
  initializes the loop. executed only once.
2. expression2
  same as while test. It checks the termination of the loop before statement is executed.
3. expression3
  is evaluated after statement is executed.
4. Sample for statement:
```
   for(i=1;i<4;i++) cout << i << endl;
```
comma operator ,
evaluates the expression to the left then removes the result from the stack. The primary use the the comma (sequential operator) is in for statements.
```
   for(i=1,j=4;i<4;i++,j--) cout << i << " " << j << endl;
```

break The break statement may be used to terminate switch, while, do..while and for loops.

   while (fabs(a-rp(x,r)) > 0.001) {
         if(k++ > 40) break;
         x = x-(rp(x,r)-a)/(r*rp(x,r-1));
         cout << x << endl;
   }

continue The continue statement is only used with loops. It skips to evaluate the loop again.

While loop

  i=1;
  while(i++ < 10) {
      if(i%2) continue;
      cout << i << endl;
  }

For Loop.

  for(i=1;i<10;i++) {
      if(i%2) continue;
      cout << i << endl;
  }

goto
The got statement transfers control unconditionally to another statement. The format of the goto is :
```
   goto label;
  

   if(i == 2) goto ok;
   cout << i << endl;
 ok: i=2;
```
The use of go to is frowned upon in any structured program. However, the statement may prove useful to break out of nested loops.

. Examples.

The lock problem.

#include <iostream>
using namespace std;
/*
  A safe has five locks 1,2,3,4 and 5, all of which must be unlocked
  for the safe to open. The keys to the locks are distributed among
  five employees in the following manner:
    Mr. Alden has keys for locks 1 and 3
    Ms  Baker has keys for locks 1 and 4
    Ms. Cable has keys for locks 2 and 4
    Ms  Deere has keys for locks 3 and 5
    Mr. Evans has keys for locks 1 and 5
*/

int here(char *);    // Function prototype

void main()
{ 
   int Lock1,Lock2,Lock3,Lock4,Lock5,OPEN;
   int test = 0;
   char tryagain;


   Lock1=020;   //  010 000 or 16
   Lock2=010;   //  001 000 or 8
   Lock3=004;   //  000 100 or 4
   Lock4=002;   //  000 010 or 2
   Lock5=001;   //  000 001 or 1
   OPEN=  Lock1 | Lock2 | Lock3 | Lock4 | Lock5;   //  011 111 or 31
   
   do {
	cout  << " Who is Here? " << endl;
	if(here("Mr. Alden"))  test |= Lock1 | Lock3;
	if(here("Ms. Baker"))  test |= Lock1 | Lock4;
	if(here("Ms. Cable"))  test |= Lock2 | Lock4;
	if(here("Ms. Deere"))  test |= Lock3 | Lock5;
	if(here("Mr. Evans"))  test |= Lock1 | Lock5;
  
	if(test == OPEN) cout << "Safe can be Opened" <<endl;
	else             cout << "Safe Cannot be Opened" << endl;
        test=0;
	cout << "Try Again? ";
	cin >> tryagain;
   } while (tryagain == 'y' || tryagain == 'Y');
}
 
/*
 The here function reurns 0 (false) or 1 (true)
 based on the question passed.
*/
int here(char *person)
{
	char response;

	cout << "Is " << person << " here? ";
	cin >> response;
	return ( response == 'y' || response == 'Y');
}

 Who is Here?
Is Mr. Alden here? y
Is Ms. Baker here? y
Is Ms. Cable here? y
Is Ms. Deere here? n
Is Mr. Evans here? n
Safe Cannot be Opened
Try Again? y
 Who is Here?
Is Mr. Alden here? y
Is Ms. Baker here? n
Is Ms. Cable here? y
Is Ms. Deere here? n
Is Mr. Evans here? y
Safe can be Opened
Try Again? n
Press any key to continue

From prior problem print all the combinations that can open the safe.

#include <iostream>
#include <iomanip>
using namespace std;
/*
  A safe has five locks 1,2,3,4 and 5, all of which must be unlocked
  for the safe to open. The keys to the locks are distributed among
  five employees in the following manner:
    Mr. Alden has keys for locks 1 and 3
    Ms  Baker has keys for locks 1 and 4
    Ms. Cable has keys for locks 2 and 4
    Ms  Deere has keys for locks 3 and 5
    Mr. Evans has keys for locks 1 and 5
*/



void main()
{ 
   int Lock1,Lock2,Lock3,Lock4,Lock5,OPEN;
   int test;

   Lock1=020;   //  010 000 or 16
   Lock2=010;   //  001 000 or 8
   Lock3=004;   //  000 100 or 4
   Lock4=002;   //  000 010 or 2
   Lock5=001;   //  000 001 or 1
   OPEN=  Lock1 | Lock2 | Lock3 | Lock4 | Lock5;   //  011 111 or 31

   int
   alden=020,  // Alden Pattern
   baker=010,  // Baker pattern
   cable=004,  // Cable pattern 
   deere=002,  // Deere pattern
   evans=001;  // Evans pattern

   int pattern,count=1;

   cout << "The following sets of Employees can open the safe" << endl << endl;
   cout << "Mr. Alden |"
        << "Ms. Baker |" 
	<< "Ms. Cable |" 
	<< "Ms. Deere |" 
	<< "Mr. Evans |" << endl;
   
   char space[] = "______________________________________________________";
   char here[] ="    xxx   |";
   char gone[] ="          |";
   
   cout << space << endl;
   for(pattern=1;pattern<32;pattern++) {   // Genate patterns
       test=0;                             // Pattern to zero
	   if(pattern&alden)  test |= Lock1 | Lock3;
       if(pattern&baker)  test |= Lock1 | Lock4;
       if(pattern&cable)  test |= Lock2 | Lock4;
       if(pattern&deere)  test |= Lock3 | Lock5;
	   if(pattern&evans)  test |= Lock1 | Lock5;
	   if( test == OPEN) {
		   cout << (pattern&alden ?  here : gone);
		   cout << (pattern&baker ?  here : gone);
		   cout << (pattern&cable ?  here : gone);
        	   cout << (pattern&deere ?  here : gone);
		   cout << (pattern&evans ?  here : gone);
		   cout <<endl<< space << endl;
	   }
   }
   
}

The following sets of Employees can open the safe

Mr. Alden |Ms. Baker |Ms. Cable |Ms. Deere |Mr. Evans |
______________________________________________________
          |          |    xxx   |    xxx   |    xxx   |
______________________________________________________
          |    xxx   |    xxx   |    xxx   |          |
______________________________________________________
          |    xxx   |    xxx   |    xxx   |    xxx   |
______________________________________________________
    xxx   |          |    xxx   |          |    xxx   |
______________________________________________________
    xxx   |          |    xxx   |    xxx   |          |
______________________________________________________
    xxx   |          |    xxx   |    xxx   |    xxx   |
______________________________________________________
    xxx   |    xxx   |    xxx   |          |    xxx   |
______________________________________________________
    xxx   |    xxx   |    xxx   |    xxx   |          |
______________________________________________________
    xxx   |    xxx   |    xxx   |    xxx   |    xxx   |
______________________________________________________
Press any key to continue

The problem from Vos Savant.

#include <iostream.h>
#include <stdlib.h>
#define MAX 1000000
void simulate(int);
int pick(int, int);
void main()
{
/*
   Scenario 1
   a coin is under one of three cups
   you select one.  The program selects and
   turnover one where the coin isn't.
   You don't switch.
*/
 simulate(1); 
/*
   Scenario 2
   a coin is under one of three cups
   you select one.  The program selects and
   turnover one where the coin isn't.
   You always switch.
*/
 simulate(2);
/*
   Scenario 3
   a coin is under one of three cups
   you select one.  The program selects and
   turnover one where the coin isn't.
   New person randomly selects one of the remaining cups.
*/
 simulate(3);

}
void simulate(int policy)
{ 
  int plays,right,wrong,isat,guess,turnover,other;
  int i;


   right=wrong=plays=0;
   while(plays++<MAX) {
         isat=rand()%3+1;    // where  coin is at 1, 2, or 3
         guess=rand()%3+1;   // random guess
         turnover=pick(isat,guess);
         other=pick(guess,turnover);
         switch(policy) {
            case 1 :  break;
            case 2 :  guess=other; break;
            case 3 :  if(rand()%2) guess=other; break;
         }
         if(guess==isat) right++;
         else wrong++;
   }
   cout << "Scenario " << policy;
   cout << "\n right ="  << right;
   cout << "\n wrong =" << wrong << endl;
}
//
//  return a number 1 2 3 that is not x or y
//
int pick(int x, int y)
{
  int value;

  if( x != 1 && y != 1 ) value=1;
  else if( x!=2 && y != 2) value=2;
       else value=3;
  return(value);
}

$ a.out
Scenario 1
 right =333297
 wrong =666703
Scenario 2
 right =666448
 wrong =333552
Scenario 3
 right =500055
 wrong =499945