Minor changes to scripts of 3rd module.

3 files changed, 97 insertions, 94 deletions

diff --git a/accessing_parts_of_arrays/script.rst b/accessing_parts_of_arrays/script.rst
index c4a74fe..aeb1938 100644
--- a/accessing_parts_of_arrays/script.rst
+++ b/accessing_parts_of_arrays/script.rst
@@ -317,7 +317,7 @@ We can see the contents of the image, using the command
 
 We do not see white and black because, ``pylab`` has mapped 
 white and black to different colors. 
-This can be changed by using a different colormap. 
+This can be changed by using a different color map. 
 
 To see that ``I`` is really, just an array, we say, I, at the prompt 
 
@@ -460,6 +460,10 @@ Pause the video here, try out the following exercise and resume the video.
 
  Obtain the elements [[23, 24], [33, -34]] from C. 
 
+.. R40
+
+Switch to the terminal for solution.
+
 .. L40
 
 {{{continue from paused state}}}
@@ -468,10 +472,8 @@ Pause the video here, try out the following exercise and resume the video.
 
     C[1:3, 2:4] 
 
-.. R40
+.. R41
 
-Switch to the terminal for solution.
-<Type the command>
 C[1:3, 2:4] will give us the required elements. 
 
 Now, we wish to obtain the top left quarter of the image. How do
@@ -484,35 +486,39 @@ columns.
 
     I[:150, :150]
 
-.. R41
+.. R42
 
 I[:150, :150] gives us the top-left corner of the image. 
 
-.. R42
+.. R43
 
 We use the ``imshow`` command to see the slice we obtained in the
 form of an image and confirm. 
 
-.. L42
+.. L43
 ::
 
     imshow(I[:150, :150])
 
-.. R43
+.. R44
 
 Pause the video here, try out the following exercise and resume the video.
 
-.. L43
-
 .. L44
 
+.. L45
+
 {{{ show slide containing exercise 5 }}} 
 
-.. R44
+.. R45
 
  Obtain the square in the center of the image.
 
-.. L45
+.. R46
+
+Switch to the terminal for solution.
+
+.. L46
 
 {{{continue from paused state}}}
 {{{ Switch to the terminal }}}
@@ -520,10 +526,8 @@ Pause the video here, try out the following exercise and resume the video.
 
     imshow(I[75:225, 75:225])
 
-.. R45
+.. R47
 
-Switch to the terminal for solution.
-<Type the command>
 Hence, we get the center of the image.
 
 Our next goal is to compress the image, using a very simple
@@ -537,78 +541,78 @@ We shall first learn the idea of striding using the smaller array
 C. Suppose we wish to access only the odd rows and columns (first,
 third, fifth). We do this by, 
 
-.. L46
+.. L47
 ::
 
     C[0:5:2, 0:5:2]
 
-.. R46
-
-.. R47
+.. R48
 
 if we wish to be explicit, we say, 
 
-.. L47
+.. L48
 ::
 
     C[::2, ::2]
 
-.. R48
+.. R49
 
 This is very similar to the step specified to the ``range``
 function. It specifies, the jump or step in which to move, while
 accessing the elements. If no step is specified, a default value
 of 1 is assumed. 
 
-.. L48
+.. L49
 ::
 
     C[1::2, ::2] 
 
-.. R49
+.. R50
 
 we get the elements, [[21, 23, 0], [41, 43, 0]]
 Pause the video here, try out the following exercise and resume the video.
 
-.. L49
-
 .. L50
 
+.. L51
+
 {{{ show slide containing exercise 6 }}} 
 
-.. R50
+.. R51
 
  Obtain the following. 
 [[12, 0], [42, 0]]
 [[12, 13, 14], [0, 0, 0]]
 
-.. L51
+.. R52
+
+The solution is on your screen.
+
+.. L52
 
 {{{continue from paused state}}}
 {{{ show slide containing Solution 6 }}} 
 
-.. R51
-
-The solution is on your screen.
+.. R53
 
 Now, that we know how to stride over an array, we can drop
 alternate rows and columns out of the image in I. 
 
-.. L52
+.. L53
 ::
 
     I[::2, ::2]
 
-.. R52
+.. R54
 
 To see this image, we say, 
 
-.. L53
+.. L54
 ::
 
     imshow(I[::2, ::2])
 
-.. R53
+.. R55
 
 This does not have much data to notice any real difference, but
 notice that the scale has reduced to show that we have dropped
@@ -616,18 +620,16 @@ alternate rows and columns. If you notice carefully, you will be
 able to observe some blurring near the edges. To notice this
 effect more clearly, increase the step to 4. 
 
-.. L54
+.. L55
 ::
 
     imshow(I[::4, ::4])
 
-.. R54
-
-.. L55
+.. L56
 
 {{{ show summary slide }}}
 
-.. R55
+.. R56
 
 This brings us to the end of this tutorial. In this tutorial, we
 have learnt to, 
@@ -639,11 +641,11 @@ have learnt to,
  #. Slice and stride on arrays.
  #. Read images into arrays and manipulate them.
 
-.. L56
+.. L57
 
 {{{Show self assessment questions slide}}}
 
-.. R56
+.. R57
 
 Here are some self assessment questions for you to solve
 
@@ -675,11 +677,11 @@ Change the array to
     B = array([[10, 11, 10, 11],
                [20, 21, 20, 21]])
 
-.. L57
+.. L58
   
 {{{solution of self assessment questions on slide}}}
 
-.. R57
+.. R58
 
 And the answers,
 
@@ -699,11 +701,11 @@ And the answers,
 
     B[:2, 2:] = B[:2, :2]
 
-.. L58
+.. L59
 
 {{{ Show the Thank you slide }}}
 
-.. R58
+.. R59
 
 Hope you have enjoyed this tutorial and found it useful.
 Thank you!
diff --git a/getting_started_with_arrays/script.rst b/getting_started_with_arrays/script.rst
index 242ac8e..84c1f6b 100644
--- a/getting_started_with_arrays/script.rst
+++ b/getting_started_with_arrays/script.rst
@@ -192,15 +192,14 @@ Pause the video here, try out the following exercise and resume the video.
 Find out the shape of the other arrays i.e. a1, a3, ar that we have 
 created.
 
-.. L15
-
-{{{ Continue from paused state }}}
-
 .. R15
 
-It can be done as,
+Switch to the terminal for solution
 
-.. L16
+.. L15
+
+{{{ Continue from paused state }}}
+{{{ Switch to the terminal }}}
 ::
 
     a1.shape
@@ -212,7 +211,7 @@ It can be done as,
 Now let us try to create a new array with a mix of elements and see what
 will happen,
 
-.. L17
+.. L16
 ::
 
     a4 = array([1,2,3,'a string'])
@@ -224,7 +223,7 @@ arrays handle elements with the same datatype, but it didn't raise an
 error. Let us check the values in the new array created. 
 Type a4 in the terminal,
 
-.. L18
+.. L17
 ::
 
     a4
@@ -240,6 +239,8 @@ Also,if you have noticed,we got something like 'dtype S8' in the output.
 dtype is nothing but the datatype which is the minimum type required
 to hold the objects in the sequence.
 
+.. L18
+
 .. L19
 
 {{{ switch to the next slide, identity & zeros methods }}}
@@ -257,6 +258,8 @@ The function ``identity()`` takes an integer argument which specifies the
 size of the desired matrix,
 
 .. L20
+
+{{{ Switch to the terminal }}}
 ::
 
     identity(3)
@@ -289,19 +292,18 @@ Pause the video here, try out the following exercise and resume the video.
 
 .. R22
 
-We learned two functions ``identity()`` and ``zeros()``, find out more
-about the functions ``zeros_like()``, ``ones()``, ``ones_like()``.
+Find out about the functions 
+    - zeros_like()
+    - ones()
+    - ones_like()
 
-.. L23
-
-{{{ continue from paused state }}}
-{{{ Switch to the terminal }}}
+< pause for some time and then continue >
 
 .. R23
 
 Try the following, first check the value of a1,
 
-.. L24
+.. L23
 ::
 
     a1
@@ -311,16 +313,17 @@ Try the following, first check the value of a1,
 We see that ``a1`` is a single dimensional array, 
 Let us now try a1*2
 
-.. L25
+.. L24
 ::
 
     a1 * 2
 
 .. R25
+
 It returned a new array with all the elements multiplied by 2.
 Now let us again check the contents of a1
 
-.. L26
+.. L25
 ::
 
     a1
@@ -329,17 +332,15 @@ Now let us again check the contents of a1
 
 note that the value of a1 still remains the same.
 
-.. R27
-
 Similarly with addition,
 
-.. L27
+.. L26
 ::
 
     a1 + 2
     a1
 
-.. R28
+.. R27
 
 it returns a new array, with all the elements summed with two. But
 again notice that the value of a1 has not been changed.
@@ -347,53 +348,53 @@ again notice that the value of a1 has not been changed.
 You may change the value of a1 by simply assigning the newly returned
 array as,
 
-.. L28
+.. L27
 ::
 
     a1 += 2
 
-.. R29
+.. R28
 
-Notice the change in elements of a,
+Notice the change in elements of a by typing 'a'
 
-.. L29
+.. L28
 ::
 
     a
 
-.. R30
+.. R29
 
 We can use all the mathematical operations with arrays, Now let us try 
 this
 
-.. L30
+.. L29
 ::
 
    a1 = array([1,2,3,4])
    a2 = array([1,2,3,4])
    a1 + a2
 
-.. R31
+.. R30
 
 This returns an array with element by element addition
 
-.. L31
+.. L30
 ::
 
     a1 * a2
 
-.. R32
+.. R31
 
 a1*a2 returns an array with element by element multiplication, notice 
 that it does not perform matrix multiplication.
 
-.. L32
+.. L31
 
-.. L33
+.. L32
 
 {{{ switch to summary slide }}}
 
-.. R33
+.. R32
 
 This brings us to the end of the end of this tutorial.In this tutorial, 
 we have learnt to, 
@@ -408,13 +409,13 @@ we have learnt to,
     - zeros() & zeros_like()
     - ones() & ones_like()
 
-.. L34
+.. L33
 
 {{{Show self assessment questions slide}}}
 
-.. R34
+.. R33
 
-Here are some self assessment questionss for you to solve
+Here are some self assessment questions for you to solve
 
 1. ``x = array([1, 2, 3], [5, 6, 7])`` is a valid statement
 
@@ -435,11 +436,11 @@ Here are some self assessment questionss for you to solve
    - Both statement A and B are correct.
    - Both statement A and B are incorrect.
 
-.. L35
+.. L34
 
 {{{solution of self assessment questions on slide}}}
 
-.. R35
+.. R34
 
 And the answers,
 
@@ -453,11 +454,11 @@ And the answers,
 2. The function ``ones_like()`` returns an array of ones with the same 
    shape and type as a given array.
 
-.. L36
+.. L35
     
 {{{ switch to thank you slide }}}
 
-.. R36
+.. R35
 
 Hope you have enjoyed this tutorial and found it useful.
 Thank you!
diff --git a/matrices/script.rst b/matrices/script.rst
index 68eb709..0979882 100644
--- a/matrices/script.rst
+++ b/matrices/script.rst
@@ -144,10 +144,10 @@ m3 can be created as,
 
 .. R11
 
-Let us now move to matrix matrix operations.
+Let us now move to matrix operations.
 We can do matrix addition and subtraction easily.
 m3+m2 does element by element addition, that is matrix addition.
-Note that both the matrices are of the same order.
+Note that both the matrices should be of the same order.
 
 .. L11
 ::
@@ -187,8 +187,8 @@ Matrix multiplication in matrices are done using the function ``dot()``
 
 .. R15
 
-Due to size mismatch the multiplication could not be done and it
-returned an error,
+Due to size mismatch, the multiplication could not be done and it
+returned an error.
 
 Now let us see an example for matrix multiplication. For doing matrix
 multiplication we need to have two matrices of the order n by m and m
@@ -306,7 +306,7 @@ And the Frobenius norm of the matrix ``im5`` can be found out as,
 
 .. R25
 
-Thus we have successfully obtained the frobenius norm of the matrix m5
+Thus we have successfully obtained the Frobenius norm of the matrix m5
 
 Pause the video here, try out the following exercise and resume the video.
 
@@ -355,7 +355,7 @@ The norm of a matrix can be found out using the method
 
 .. R30
 
-Inorder to find out the Frobenius norm of the matrix im5,
+In order to find out the Frobenius norm of the matrix im5,
 we do,
 
 .. L30
@@ -377,7 +377,7 @@ And to find out the Infinity norm of the matrix im5, we do,
 .. R32
 
 This is easier when compared to the code we wrote. Read the documentation 
-of ``norm`` to read up more about ord and the possible type of norms
+of ``norm`` to read up more about ``ord`` and the possible type of norms
 the norm function produces.
 
 Now let us find out the determinant of a the matrix m5. 
@@ -545,10 +545,10 @@ And the answers,
 
 2. False.
    ``eig(A)[0]`` and ``eigvals(A)`` are same, that is both will give the 
-   eigen values of matrrix A.
+   eigen values of matrix A.
 
 3. ``norm(A,ord='fro')`` and ``norm(A)`` are same, since the order='fro' 
-   stands for frobenius norm. Hence true.
+   stands for Frobenius norm. Hence true.
 
 .. L45