1. Iterative statements :  a for loop

  *************************
  A for loop is used when you know in advance that a sequence of statements
  MUST be repeated a GIVEN number of times (eg 10)
  *************************

 - ex0.m
    Print the numbers from 1 to 3  ... or 1 to 100

 - for1.m
    A simple for loop
    for ii=1:10
      disp(ii);
    end
     
    for ii= 1:10  means
   % Generate a row vector 1:10
   % variable ii takes the values of the row vector one by one from the first to the last
   % each time the sequence of statements betweent he for and matching end is executed
   % until control reaches end and then the next value of ii is taken from the rowvector for
   % the loop to continue
   % As soon as the last value of the row vector is assigned to ii (say 10 in our example above)
   % the disp(ii) statement is executed for the last (10th time) and the for loop stops at end
   % and exits the for loop by executing the statement following the end (in our case nothing).

 -   Syntax of a for loop in for2syn.m or sub7p7for.m


 - for3.m
    an if statement in the for loop

 - for4.m

    you and me three times (thrice)
 - Nested for loops nested1.m and nested2.m
   A break     STOPs the execution of the for loop completely and control goes to the
   statement immediately following the matching end
  
   disp() before and after the break!

 - Continue ccontinue.m and ccontinue1.m
   A continues STOP the execution of the CURRENT loop only and control goes to the
   end statement to return to the top so that ii gets the next value from the row vector.
   Code continues with the execution of the next iteration of the loop (if any left).


 - danger.m
   always preallocate arrays never allow them to grow dynamically. It's inefficient and slow.


 - a= ones(1,10);
   for ii=a
    disp(ii)
   end
  
   This forloop is still executed 10 times. Variable ii however takes the same value of 1 in each one
   of the 10 times disp(ii) is executed! Just because you have a for loop it does not mean that
   ii must take (in this example,) 10 different values. The values can be the same i.e. arbitrarily
   chosen in general.

 - Fibonacci numbers   fib.m and sub7p8fibfor.m

 - nested for loops nested1.m nested2.m  nested3.m

    sub7p9break.m sub7p9continue.m

3. Iterative statements :  a while loop

  *************************
  A while loop is used when you know in advance that a sequence of statements
  MUST be repeated some number of times until a condition becomes true or false
  but you don't know the exact number of times
  *************************

  - Syntax of a while    sub7p2awhile.m   
  - Count to 10  : sub7p2bwhileex.m

  - Count to  3  : ex1.m ex2.m. ex3.m  : 3 while loops

  - ex1.m  Details
1.   limit=4;
2.   i=1;
3.   while (i < limit)
4.     fprintf('     %d\n',i);
5.     i=i+1; 
6    end
7

     Line    i   i<limit    output
     1               4
     2       1       4
     3       1   1 < 4 : t  
     4       1                1
     5       2
     6->
     3       2   2 < 4 : t
     4       2                2
     5       3
     6->
     3       3   3 < 4 : t
     4       3                3
     5       4
     6->
     3       4   4 < 4 : f
     7

 How many times was line 3 executed   ?   4
 How many times was line 4/5 executed ?   3

  - Print only odd numbers ex4.m

  - Fibonacci sub7p4fib.m

  - sub7ex1.m : find max and index with max value for vector A

  - sub7ex2.m : for loop
 
  - sub7ex3.m : for loop 

  - sub7ex4.m : while loop


  - sub7extra5.m sub7extra5a.m   : Past Quiz

  - sub7you.m  : Past Quiz

4. More  on for and while loops


   How much is the sum

   S= 1 + 2 +  ... + 100  ?
   S=100+ 99+  ... + 1    ?
  2S=(1+100)+(2+99)+ ... + (100+1) =
    =101    + 101  + ... +  101
    = 100 x 101  = 

 i.e  S= 100 x 101 /2

 In general
  S= 1+ 2 + ... + n = n(n+1)/2  = last term * next-to-last /2

 Write a program that computes S from 1 to 100

  x=1:100;
  sum(x)
  disp(ans)
 mysum=0;
  for ii=1:100
    mysum= mysum+ii;
  end
  disp(mysum);


  mysum=0;
  ii=1;
  while (ii  ?  100)
    mysum = mysum +ii;
  end
  disp(mysum);

Replace ? with the correct relational operator!


5.  b=a>5 ! Selected array operations

   a= [ 10 20 3; 4 50 6;  7 80 9]

  -   What is the maximum of a?   
      Ans: max(max(a))

  -   How many elements greater than 5
      b= a>5
      sum(sum(b))

      or the cuter

      length(a(b))

      a(b) vectorizes (i.e. produces a vector) of the elements of a in column-major
      order that satisfy condition b , i.e. they are greater than 5!

  -  For loop :matrix  form

   total=0
    for ii=1: size(a,1)
     for jj=1: size(a,2)
      if a(ii,jj) > 5
        total=total+1;
      end
     end
    end
    disp (total)

  -  For loop :array form


    total=0
    for ii=1: size(a,1) * size(a,2)
      if a(ii) > 5
        total=total+1;
      end
     end
    disp (total)
  - Change all >5 entries to   half of original value


   b= a>5;
   a(b) = a(b) ./ 2

   Well, i changed my mind. restore original values

   a(b) = a(b) .* 2

   (This is possible because b keeps track of the locations of the original values that were
    greater than 5!)

   b
    1   1  0
    0   1  1
    1   1  1

  FILES
 ******
  FOR LOOPS
  **********
  ex0.m : print 1 to 3;
  for1.m: print 1 to 3 for style
  
  for2syn.m: SYNTAX FOR
  sub7p7for.m

  for3.m : more for loop
  for4.m : more for loop

  danger.m: Resizing arrays !!!


  nested0.m  : nested for loop
  nested1.m  :

  nested2.m  : nested CONTINUE


  ccontinue.m
  ccontinue1.m 
  sub7you.m    : CONTINUE

  bbreak.m:
  brreak1.m:   : BREAK


  nested3.m  :  ... CONTINUE ... BREAK  ?

  
  WHILE LOOPS
  ***********

  sub7p3awhhile.m  : SYNTAX WHILE

  ex1.m : 1 2 3.

ex2.m : smae
  ex3.m : same
  ex4.m : odd numbers

  fib.m : fibonacci
  fib2.m: int32 fibonacci

  sub7ex1.m  : Find MAX
  sub7ex2.m  : for loop find max
  sub7ex3.m  : fancy for loop find max
  sub7ex4.m  : A(10) =27 ?




EXAMPLES (MATLAB DIARY)
***********************
1. Print 1 to 10 (one per row)
a= 1:10

a =

  Columns 1 through 9

     1     2     3     4     5     6     7     8     9

  Column 10

    10
a'
ans =

     1
     2
     3
     4
     5
     6
     7
     8
     9
    10



2. Print hello 4 times

a='hello'

a =

hello

b= [ a ; a ; a ; a ]

b =

hello
hello
hello
hello

for ii = 1:4
     disp('hello');
   end
hello
hello
hello
hello

for ii=1:4 ; disp('hello'); end
hello
hello
hello
hello

 for ii=1:4
     disp('hello');
    end
hello
hello
hello
hello
ii

ii =

     4

for i= 1:2:0 ;  disp('hello'); end
for ii= [ 1 1 1 1 ]
     disp('hello');
   end
hello
hello
hello
hello
ii

ii =

     1

for ii= zeros(1,4) ;
      disp('hello');
   end
hello
hello
hello
hello
for ii= zeros(4,1);
     disp('hello');
   end
hello
for ii=1:4
     disp('hello');
     ii = 10;
   end
hello
hello
hello
hello
ii
ii


% the vector generated on the right of = in the for loop does not need to
% generate different values...

for ii= [ 1 1 1 1 ];
     disp(10);
   end 
    10

    10

    10

    10

% or non zero values

for ii= zeros(1,5) ;
     disp(10);
   end 
    10

    10

    10

    10

    10

% if the vector is empty the lines of the body (between for  and end) are
% not executed.

for ii= 1:2:0
     disp(10);
   end 
1:2:0

ans =

   Empty matrix: 1-by-0


4.% A more complex for loop. the body has an if else statement

type for3.m

for ii=1:10
 if (mod(ii,2) == 0)
   fprintf('ii=%d\n',ii);
 else
   fprintf('ii=%d is odd\n',ii);
 end
end







mod(3,2)

ans =

     1

mod(2,2)

ans =

     0

type for3.m

for ii=1:10
 if (mod(ii,2) == 0)
   fprintf('ii=%d\n',ii);
 else
   fprintf('ii=%d is odd\n',ii);
 end
end


% the body has an if elseif statement

for ii=1:10
 if (ii==10)
   fprintf('\n');
 elseif (mod(ii,3) == 2)
   fprintf('and ');
 elseif (mod(ii,3)==1)
   fprintf('you ');
 elseif (mod(ii,3)==0)
   fprintf('me\n');
 end
end
type for4.m

% the body has an if elseif statement

for ii=1:10
 if (ii==10)
   fprintf('\n');
 elseif (mod(ii,3) == 2)
   fprintf('and ');
 elseif (mod(ii,3)==1)
   fprintf('you ');
 elseif (mod(ii,3)==0)
   fprintf('me\n');
 end
end

for4
you and me
you and me
you and me



5.  % AVOID DYNAMIC ALLOCATION OF ARRAYS

x(1)=1;
for k=2:10000
  x(k)=k;
end

% x(1) =1 creates a 1x1 array
% for k=10000 THE LINE x(k)=k  creates a 1x10000 array from 
% the previous iteration k=9999 that created a 1x9999 array.
% What this means is that you create an array of 10000 elets,
% populate them with zeroes, copy the 1x9999 array there and
% set the x(k) i.e. x(1,k) value to k. This takes 10000 or 2x10000
% steps (SEE QUIZ 2)
% thus the total cost of the array is
%  1+ 2 +3 + 4 + ... + 10000 = 10000 x 10001 /2 ~ 50,000,000

% if one had preallocated the array x=zeros(1,10000);
% the line x(k)=k would have taken 1 step not k and the total cost
% would have been
%  1 + 1 + 1 +1 + ... + 1    = 10000

clear
type danger.m

x(1)=1;
for k=2:10000
  x(k)=k;
end

danger
whos
  Name      Size               Bytes  Class     Attributes

  k         1x1                    8  double              
  x         1x10000            80000  double              

sum(1:10000)

ans =

    50005000
clear;
tic
for k=2:100000000
    x(k)=k;
   end
 toc

Elapsed time is 17.662734 seconds.  <<< EXECUTION TIME


%OR INLINE (I.E IN ONE LINE ...)

clear

tic;x(1)=1; for k=2:100000000 ; x(k)=k; end ; toc

Elapsed time is 15.771215 seconds.
whos
  Name      Size                       Bytes  Class     Attributes

  k         1x1                            8  double              
  x         1x100000000            800000000  double              

% NOTE THAT AFTER THIS RUN ARRAY x has been created
% THERE AFTER THE FOR LOOP IS FAST TO EXECUTE; 1 step per iteration not k!


5. NESTED FOR LOOPS


% NESTED FOR LOOPS
% THE INNER IS COMPLETED FIRST

type nested0.m

for ii= 1:1:3
  for jj= 1:1:4
    fprintf(' ii= %d jj= %d \n',ii,jj);
  end
end


nested0
 ii= 1 jj= 1 
 ii= 1 jj= 2 
 ii= 1 jj= 3 
 ii= 1 jj= 4 
 ii= 2 jj= 1 
 ii= 2 jj= 2 
 ii= 2 jj= 3 
 ii= 2 jj= 4 
 ii= 3 jj= 1 
 ii= 3 jj= 2 
 ii= 3 jj= 3 
 ii= 3 jj= 4 
% A MORE COMPLEX ONE

type nested1.m

% Code Nest1                FILE: sub7p10nested.m
sum =0;
for ii= 1:1:3
  for jj= 1:1:4
    sum = sum + 1;
    fprintf(' ii= %d jj= %d \n',ii,jj);
  end
end
disp(sum);








nested1
 ii= 1 jj= 1 
 ii= 1 jj= 2 
 ii= 1 jj= 3 
 ii= 1 jj= 4 
 ii= 2 jj= 1 
 ii= 2 jj= 2 
 ii= 2 jj= 3 
 ii= 2 jj= 4 
 ii= 3 jj= 1 
 ii= 3 jj= 2 
 ii= 3 jj= 3 
 ii= 3 jj= 4 
    12

% Continue ... stop further execution of the current iteration for
% the innermost loops get a new value for the variable of the innermost
%loop
type nested2.m

for ii=1:3
  for jj=1:4
    if (ii==2) && (jj==3)
      continue;
    end
  end
  fprintf('ii=%3d jj=%3d\n',ii,jj);
end

nested2
ii=  1 jj=  4
ii=  2 jj=  4
ii=  3 jj=  4


% NOTE THAT THE PRINTF ABOVE IS OUTSIDE OF THE JJ loop





type nested2.m

for ii=1:3
  for jj=1:4
    if (ii==2) && (jj==3)
      continue;
    end
  end
  fprintf('ii=%3d jj=%3d\n',ii,jj);
end



nested2
ii=  1 jj=  4
ii=  2 jj=  4
ii=  3 jj=  4


% FORMAL INTRODUCTION OF CONTINUE
type ccontinue.m

for ii=1:10
   if (ii==5)
     continue;
   end
   disp(ii);
end
ccontinue
     1

     2

     3

     4

     6

     7

     8

     9

    10

% ANOTHER EXAMPLE

type ccontinue1.m

for ii=1:10
   disp(ii);
   if (ii==5)
     continue;
   end
end


ccontinue1
     1

     2

     3

     4

     5

     6

     7

     8

     9

    10
type bbreak.m

for ii=1:10
  if (ii==5)
    break;
  end
  disp(ii);
end






bbreak
     1

     2

     3

     4



type  bbreak1.

for ii=1:10
  disp(ii);
  if (ii==5)
    break;
  end
end



bbreak1
     1

     2

     3

     4

     5

type nested3.m

for ii=1:3
  for jj=1:4
    if (ii==2) && (jj==3)
      continue;
    end
    fprintf('ii=%3d jj=%3d\n',ii,jj);
  end
end


type nested1.m

% Code Nest1                FILE: sub7p10nested.m
sum =0;
for ii= 1:1:3
  for jj= 1:1:4
    sum = sum + 1;
    fprintf(' ii= %d jj= %d \n',ii,jj);
  end
end
disp(sum);


type nested0.m

for ii= 1:1:3
  for jj= 1:1:4
    fprintf(' ii= %d jj= %d \n',ii,jj);
  end
end

nested0
 ii= 1 jj= 1
 ii= 1 jj= 2
 ii= 1 jj= 3
 ii= 1 jj= 4
 ii= 2 jj= 1
 ii= 2 jj= 2
 ii= 2 jj= 3
 ii= 2 jj= 4
 ii= 3 jj= 1
 ii= 3 jj= 2
 ii= 3 jj= 3
 ii= 3 jj= 4
type nested0.m


for ii= 1:1:3
  for jj= 1:1:4
    fprintf(' ii= %d jj= %d \n',ii,jj);
  end
end


type nested0.m

for ii= 1:1:3
  for jj= 1:1:4
    fprintf(' ii= %d jj= %d \n',ii,jj);
  end
end


nested0

 ii= 1 jj= 1
 ii= 1 jj= 2
 ii= 1 jj= 3
 ii= 1 jj= 4
 ii= 2 jj= 1
 ii= 2 jj= 2
 ii= 2 jj= 3
 ii= 2 jj= 4
 ii= 3 jj= 1
 ii= 3 jj= 2
 ii= 3 jj= 3
 ii= 3 jj= 4
type nested3.m

for ii=1:3
  for jj=1:4
    if (ii==2) && (jj==3)
      continue;
    end
    fprintf('ii=%3d jj=%3d\n',ii,jj);
  end
end


nested3
ii=  1 jj=  1
ii=  1 jj=  2
ii=  1 jj=  3
ii=  1 jj=  4
ii=  2 jj=  1
ii=  2 jj=  2
ii=  2 jj=  4
ii=  3 jj=  1
ii=  3 jj=  2
ii=  3 jj=  3
ii=  3 jj=  4

for ii=1:3
  for jj=1:4
    if (ii==2) && (jj==3)
      break ;
    end
    fprintf('ii=%3d jj=%3d\n',ii,jj);
  end
end



 nested3a
ii=  1 jj=  1
ii=  1 jj=  2
ii=  1 jj=  3
ii=  1 jj=  4
ii=  2 jj=  1
ii=  2 jj=  2
ii=  3 jj=  1
ii=  3 jj=  2
ii=  3 jj=  3
ii=  3 jj=  4
while (0)
    ;
   end
while (0)
     disp('hello');
   end
while (1)

type ex1.m

limit=4;
i=1;
while (i < limit)
 fprintf('     %d\n',i);
 i=i+1;
end



type ex1.m

limit=4;
i=1;
while (i < limit)
 fprintf('     %d\n',i);
 i=i+1;
end

type ex2.m

limit=3;
i=1;
while (i <= limit)
 fprintf('     %d\n',i);
 i=i+1;
end


ex1
     1
     2
     3
ex2
     1
     2
     3
type ex1.m

limit=4;
i=1;
while (i < limit)
 fprintf('     %d\n',i);
 i=i+1;
end

type ex3.m

type ex3.m

limit=3;
i=0;
while (i < limit)
 i=i+1;
 fprintf('     %d\n',i);
end

ex3
     1
     2
     3
type ex4.m

% print only odd numbers
limit=10;
i=0;
while (i < limit)
 i=i+1;
 if ( mod(i,2) == 1)
    fprintf('     %d\n',i);
 end
end




for ii = 1:100 ; ssum = sum(1:100); end ; disp(ssum);
        5050

for ii = 1:100 ; ssum = ii+1 ; end ; disp(ssum);
   101

for ii = 1:100 ; ssum = ii+1 ; end ; disp(ssum);
   101

for ii = 1:100 ; ssum = ssum+1 ; end ; disp(ssum);
   201

ssum=0;
for ii = 1:100 ; ssum = ssum+1 ; end ; disp(ssum);
   100

for ii = 1:100 ; ssum = ssum+ii ; end ; disp(ssum);
        5150

ssum=0;
for ii = 1:100 ; ssum = ssum+ii ; end ; disp(ssum);
        5050
type ex1.m

limit=4;
i=1;
while (i < limit)
 fprintf('     %d\n',i);
 i=i+1;
end



type for2syn.m

for index = expr             % FILE: sub7p7for.m
   Statement 1
end



3. type for2syn.m

for index = expr             % FILE: sub7p7for.m
   Statement 1
end




% for loop 10:5:30 generates a row vector of 5 values [10 15 20 25 30]
% Each one of them is assigned to variable ii in turn (i.e. 5 assignments).
% After each assignment the body of the for loop is executed i.e. the lines
% between the for and the end
% If there is no any value to be assigned to ii control goes to the statement
% after the end.

>>
for ii=10:5:30
    disp(10);
    disp(ii);
   end
    10
for ii=10:5:30
    disp(10);
    disp(ii);
   end
    10

    10

    10

    15

    10

    20

    10

    25

    10

    30


% AFter the completion of the loop
% ii still retains its last assigned value
ii

ii =

    30

%  Even if ii =ii *10 multiplies ii by 10
%  when ii gets assigned to the new value from [10 15 20 25 30]
%  the previous values gets overwritten
for ii=10:5:30 ;
     disp(10);
     disp(ii);
     ii = ii * 10;
   end
    10

    10

    10

    15

    10

    20

    10

    25

    10

    30
ii

% Because the disp precede the ii=ii*10; the same values get printed
% Note however that the value of ii after the completion of the
% for loop is affected by the ii=ii*10

ii =

   300
% CONCLUSION THE TWO PROGRAMS ARE EQUIVALENT IN WHAT THEY DISPLAY...
% HOWEVER THE TWO PROGRAMS ARE NOT IDENTICAL. THE STATE OF VARIABLES
% IS NOT THE SAME (II is 30 in the former, and 300 in the latter case).
% IE THE TWO PROGRAMS HAVE DIFFERENT SIDE EFFECTS (value of ii)

% Do not forget that i is such that i*i=-1 . That's why i use ii or jj instead
i*i

ans =

    -1