PS6-DCT-Soln-correction

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Joan Prudence Fox
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1 PS6-DCT-Soln-correction Unknown Author March 18, 2014 Part I DCT: Discrete Cosine Transform DCT is a linear map A R N N such that the N real numbers x 0,..., x N 1 are transformed into the N real numbers X 0,..., X N 1 according to the formula N 1 ( π X k = x x i cos N (k + 1 ) 2 )i. (1) i=1 1 Question 1: Consider the case where N = 2. Write Python function using (1) to compute X 0, X 1 given x 0, x 1. Specifically your function should take in x 0, x 1 as arguments and return X 0, X 1. In [1]: import numpy as np """ Your code goes here """ def f(x0,x1): N=2 X0=x0+2*x1*np.cos(np.pi/N*(1+1/2)*0) X1=x0+2*x1*np.cos(np.pi/N*(1+1/2)*1) return X0,X1 def vald(i,k): N=8 return 2*np.cos((np.pi/N) * i * (k + 0.5) ) #x[n]*cos(pi*(k+0.5)*n/n) Verify that your function works by computing X 0, X 1 given x 0 = 2, x 1 = 4. In [2]: np.set_printoptions(precision=3) """ Check your function here """ y=f(2.0,4.0) print "%.2f "*len(y) % y

2 2 Question 2: JPEG compression uses the DCT with N = 8. Create a matrix called D that is the matrix representation of the discrete cosine transform, i.e. when you multiply D by the vector x 0. x N 1 you should get the vector X 0. X N 1 In [3]: """ your code goes here""" N = 8 D = np.zeros((n, N)) for i in range(n): D[0,i] = 1 for i in xrange(1,n): for k in xrange(n): D[i,k] = 2*np.cos((np.pi/N) * i * (k + 0.5) ) D=D.T print D [[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]] It is interesting to note that the rows of the matrix D are basis vectors which oscillate with successively higher spatial frequencies. Below is code to plot them. In [4]: import matplotlib.pyplot as plt %matplotlib inline # Graphing helper function from our last assignment def setup_graph(title=, x_label=, y_label=, fig_size=none): fig = plt.figure() if fig_size!= None: fig.set_size_inches(fig_size[0], fig_size[1]) ax = fig.add_subplot(111) ax.set_title(title) ax.set_xlabel(x_label) ax.set_ylabel(y_label) fig=plt.figure(figsize=(9,12)) for u in xrange(n): setup_graph(title= u= +str(u), x_label=, y_label=, fig_size=(6,3)) _=plt.plot(d[u, :]) _=plt.plot(d[u, :], ro )

3 <matplotlib.figure.figure at 0x10f068490>

6 3 Question 3: Consider the vector x = Compute the DCT using the the map D you defined above and compare it to the result you get when you use the function dct from scipy.fftpack. Use type=3 in the dct function.

7 In [6]: from scipy.fftpack import dct, idct x=np.array([[1.0, 2, 3, 4, 5, 6, 7, 8]]) """ Your Code Goes Here""" print np.dot(d,x.t) print (dct(x,type=3)).t [[ ] [ ] [ ] [ ] [ 6.549] [ ] [ 2.184] [ ]] [[ ] [ ] [ ] [ ] [ 6.549] [ ] [ 2.184] [ ]] 4 Question 4: Now, we will take the image we have provided you with and apply the 2D DCT to it. In [7]: import matplotlib.image as mpimg img = mpimg.imread( montypython.png ) p=plt.imshow(img, origin= upper )

8 The above picture is of Monty Python (Python s namesake). Below we select a portion of the image to which we will apply the 2-D DCT. Let s just look at values from one tiny 8 x 8 block (which is what s used JPEG compression). Then, we use a false color spectrum to visualize the pixel intensity. In [8]: img.shape b = img[50:58, 50:58, 0] def show_image(img): plt.imshow(img) plt.colorbar() show_image(b)

9 Now, we display the actual values in b. In [9]: Out [9]: b array([[ 0.157, 0.161, 0.165, 0.165, 0.157, 0.153, 0.157, 0.145], [ 0.165, 0.161, 0.165, 0.165, 0.161, 0.161, 0.157, 0.133], [ 0.161, 0.153, 0.149, 0.149, 0.149, 0.153, 0.161, 0.133], [ 0.165, 0.157, 0.149, 0.145, 0.145, 0.153, 0.169, 0.145], [ 0.169, 0.161, 0.149, 0.149, 0.149, 0.157, 0.173, 0.157], [ 0.157, 0.149, 0.141, 0.141, 0.141, 0.149, 0.165, 0.157], [ 0.165, 0.157, 0.141, 0.137, 0.133, 0.141, 0.153, 0.161], [ 0.153, 0.153, 0.153, 0.153, 0.145, 0.145, 0.153, 0.161]], dtype=float32) The 2-D DCT is just the 1-D DCT applied to every column first and then applied to every row. Write a function to compute the 2-D DCT. Compute the 2-D dct of b and display it as an array of numbers as well as using the false color spectrum to visualize as above. In [15]: """ your Code for computing 2-D DCT goes here""" def dodct(grid): return np.dot(np.dot(d, grid), D.T) """ your Code for viualizing""" print dodct(b) show_image(dodct(b))

[[ 1.589e+01-4.998e+00 3.182e+00-1.717e+00 1.068e+00-4.531e-01 2.959e-01-7.638e-02] [ -4.719e+00 1.632e+00-1.529e+00 7.532e-01-5.843e-01 2.733e-01-1.605e-01 8.717e-02] [ 2.862e+00-7.852e-01 4.

10 [[ 1.589e e e e e e e e-02] [ e e e e e e e e-02] [ 2.862e e e e e e e e-02] [ e e e e e e e e-02] [ 1.485e e e e e e e e-02] [ e e e e e e e e-03] [ 3.353e e e e e e e e-02] [ e e e e e e e e-02]] Now, write a function to undo the 2-D DCT, that is write a function to compute the inverse of the 2-D DCT. Print out the values of the inverse of the 2-D dct applied to b and use the false color spectrum visualization technique to view the result. Compare to show_image(b).

11 In [16]: """ Your code goes here""" def undodct(grid): Di=np.linalg.inv(D) return np.dot(np.dot(di, grid), Di.T) print undodct(dodct(b)) tiny_do_undo = undodct(dodct(b)) show_image(tiny_do_undo) # Yup, looks the same. [[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]] 5 Question 5: Apply the built in scipy functions dct and idct to the image b and compare what you have done above using the show_image function. You should get the same images. Note that the scipy function dct does the 1-D DCT. So, you must figure out how to use it to do the 2-D DCT. In [17]: """Your code for computing the 2-D dct using dct() and for plotting it using show_image g=dct(dct(b, type=3).t, type=3) show_image(g)

12 In [18]: """Your code for computing the inverse 2-D dct using idct() and for plotting it using s ig=idct(idct(g, type=3).t, type=3) show_image(ig) In []:

MATPLOTLIB. Python for computational science November 2012 CINECA.

MATPLOTLIB. Python for computational science November 2012 CINECA. MATPLOTLIB Python for computational science 19 21 November 2012 CINECA m.cestari@cineca.it Introduction (1) plotting the data gives us visual feedback in the working process Typical workflow: write a python