The following problem appeared in an assignment in the Princeton course COS 126 . The problem description is taken from the course itself.

Write a program that plots a Sierpinski triangle, as illustrated below. Then develop a program that plots a recursive patterns of your own design.

https://sandipanweb.files.wordpress.com/2018/01/sierpinski91.png?w=150 150w, https://sandipanweb.files.wordpress.com/2018/01/sierpinski91.png?w=300 300w, https://sandipanweb.files.wordpress.com/2018/01/sierpinski91.png?w=768 768w, https://sandipanweb.files.wordpress.com/2018/01/sierpinski91.png?w=... 1024w, https://sandipanweb.files.wordpress.com/2018/01/sierpinski91.png 1179w" sizes="(max-width: 676px) 100vw, 676px" />

**Part 1. **

The *Sierpinski triangle* is an example of a fractal pattern. The pattern was described by Polish mathematician Waclaw Sierpinski in 1915, but has appeared in Italian art since the 13th century. Though the Sierpinski triangle looks complex, it can be generated with a short recursive program.

**Examples.** Below are the target Sierpinski triangles for different values of order *N*.

https://sandipanweb.files.wordpress.com/2018/01/sierpinski21.png?w=133 133w, https://sandipanweb.files.wordpress.com/2018/01/sierpinski21.png?w=265 265w" sizes="(max-width: 564px) 100vw, 564px" />

Our task is to implement a recursive function `sierpinski()`. We need to think recursively: our function should draw one black triangle (pointed downwards) and then call itself recursively 3 times (with an appropriate stopping condition). When writing our program, we should exercise modular design.

The following code shows an implementation:

class

Sierpinski:# Height of an equilateral triangle whose sides are of the specified length.

defheight(self, length):

return sqrt(3) * length / 2.# Draws a filled equilateral triangle whose bottom vertex is (x, y)

# of the specified side length.

deffilledTriangle(self, x, y, length):

h = self.height(length)

draw(np.array([x, x+length/2., x-length/2.]), np.array([y, y+h, y+h]), alpha=1)# Draws an empty equilateral triangle whose bottom vertex is (x, y)

# of the specified side length.

defemptyTriangle(self, x, y, length):

h = self.height(length)

draw(np.array([x, x+length, x-length]), np.array([y+2*h, y, y]), alpha=0)# Draws a Sierpinski triangle of order n, such that the largest filled

# triangle has bottom vertex (x, y) and sides of the specified length.

defsierpinski(self, n, x, y, length):

self.filledTriangle(x, y, length)

if n > 1:

self.sierpinski(n-1, x-length/2., y, length/2.)

self.sierpinski(n-1, x+length/2., y, length/2.)

self.sierpinski(n-1, x, y+self.height(length), length/2.)

The following animation shows how such a triangle of *order 5* is drawn recursively.

The following animation shows how such a triangle of *order 6* is drawn recursively.

**A diversion: fractal dimension.**

Formally, we can define the *Hausdorff dimension* or *similarity dimension* of a self-similar figure by partitioning the figure into a number of self-similar pieces of smaller size. We define the dimension to be the log (# self similar pieces) / log (scaling factor in each spatial direction). For example, we can decompose the unit square into 4 smaller squares, each of side length 1/2; or we can decompose it into 25 squares, each of side length 1/5. Here, the number of self-similar pieces is 4 (or 25) and the scaling factor is 2 (or 5). Thus, the dimension of a square is 2 since log (4) / log(2) = log (25) / log (5) = 2. We can decompose the unit cube into 8 cubes, each of side length 1/2; or we can decompose it into 125 cubes, each of side length 1/5. Therefore, the dimension of a cube is log(8) / log (2) = log(125) / log(5) = 3.

We can also apply this definition directly to the (set of white points in) Sierpinski triangle. We can decompose the unit Sierpinski triangle into 3 Sierpinski triangles, each of side length 1/2. Thus, the dimension of a Sierpinski triangle is log (3) / log (2) ≈ 1.585. Its dimension is fractional—more than a line segment, but less than a square! With Euclidean geometry, the dimension is always an integer; with fractal geometry, it can be something in between. Fractals are similar to many physical objects; for example, the coastline of Britain resembles a fractal, and its fractal dimension has been measured to be approximately 1.25.

**Part 2.**

Drawing a tree recursively, as described here:

https://sandipanweb.files.wordpress.com/2018/01/treed.png?w=150 150w, https://sandipanweb.files.wordpress.com/2018/01/treed.png?w=300 300w, https://sandipanweb.files.wordpress.com/2018/01/treed.png 708w" sizes="(max-width: 629px) 100vw, 629px" />

The following shows how a tree of *order 10* is drawn:

The next problem appeared in an assignment in the Cornell course CS1114 . The problem description is taken from the course itself.

Let’s consider a 2D matrix of values at integer grid locations (e.g., a grayscale image). To interpolate values on a 2D grid, we can use the 2D analogue of linear interpolation: bilinear interpolation. In this case, there are four neighbors for each possible point we’d like to interpolation, and the intensity values of these four neighbors are all combined to compute the interpolated intensity, as shown in the next figure.

https://sandipanweb.files.wordpress.com/2018/01/f6.png?w=150 150w, https://sandipanweb.files.wordpress.com/2018/01/f6.png?w=300 300w, https://sandipanweb.files.wordpress.com/2018/01/f6.png?w=768 768w, https://sandipanweb.files.wordpress.com/2018/01/f6.png?w=1024 1024w, https://sandipanweb.files.wordpress.com/2018/01/f6.png 1030w" sizes="(max-width: 676px) 100vw, 676px" />

In the ﬁgure, the Q values represent intensities. To combine these intensities, we perform linear interpolation in multiple directions: we ﬁrst interpolate in the x direction (to get the value at the blue points), then in the y direction (to get the value at the green points).

Next, we’ll use the interpolation function to help us implement image transformations.

A 2D afﬁne transformation can be represented with a 3 ×3 matrix T:

https://sandipanweb.files.wordpress.com/2018/01/f4.png?w=150 150w" sizes="(max-width: 194px) 100vw, 194px" />

Recall that the reason why this matrix is 3×3, rather than 2 ×2, is that we operate in homogeneous coordinates; that is, we add an extra 1 on the end of our 2D coordinates (i.e., (x,y) becomes (x,y,1)), in order to represent translations with a matrix multiplication. To apply a transformation T to a pixel, we multiply T by the pixel’s location:

https://sandipanweb.files.wordpress.com/2018/01/f5.png?w=150 150w, https://sandipanweb.files.wordpress.com/2018/01/f5.png?w=300 300w" sizes="(max-width: 486px) 100vw, 486px" />

The following figure shows a few such transformation matrices:

https://sandipanweb.files.wordpress.com/2018/01/f8.png?w=1086&h... 1086w, https://sandipanweb.files.wordpress.com/2018/01/f8.png?w=119&h=150 119w, https://sandipanweb.files.wordpress.com/2018/01/f8.png?w=238&h=300 238w, https://sandipanweb.files.wordpress.com/2018/01/f8.png?w=768&h=966 768w, https://sandipanweb.files.wordpress.com/2018/01/f8.png?w=814&h=... 814w" sizes="(max-width: 543px) 100vw, 543px" />

To apply a transformation **T** to an entire image **I**, we could apply the transformation to each of I’s pixels to map them to the output image. However, this *forward warping*procedure has several problems. Instead, we’ll use *inverse mapping* to warp the pixels of the output image back to the input image. Because this won’t necessarily hit an integer-valued location, we’ll need to use the (bi-linear) interpolation to determine the intensity of the input image at the desired location, as shown in the next figure.

https://sandipanweb.files.wordpress.com/2018/01/f7.png?w=150&h=82 150w, https://sandipanweb.files.wordpress.com/2018/01/f7.png?w=300&h=164 300w, https://sandipanweb.files.wordpress.com/2018/01/f7.png?w=768&h=419 768w, https://sandipanweb.files.wordpress.com/2018/01/f7.png 905w" sizes="(max-width: 516px) 100vw, 516px" />

To demo the transformation function, let’s implement the following on a gray scale bird image:

1. Horizontal *ﬂipping*

https://sandipanweb.files.wordpress.com/2018/01/out_flip1.png?w=150 150w, https://sandipanweb.files.wordpress.com/2018/01/out_flip1.png?w=300 300w" sizes="(max-width: 579px) 100vw, 579px" />

2. *Scaling* by a factor of 0.5

https://sandipanweb.files.wordpress.com/2018/01/out_scale.png?w=150 150w, https://sandipanweb.files.wordpress.com/2018/01/out_scale.png?w=300 300w" sizes="(max-width: 510px) 100vw, 510px" />

3. *Rotation* by 45 degrees around the center of the image

https://sandipanweb.files.wordpress.com/2018/01/out_rot1.png?w=150 150w, https://sandipanweb.files.wordpress.com/2018/01/out_rot1.png?w=300 300w" sizes="(max-width: 567px) 100vw, 567px" />

The next animations show rotation and sheer transformations with the Lena image:

Next, let’s implement a function to transform RGB images. To do this, we need to simply call transform image three times, once for each channel, then put the results together into a single image. Next figures and animations show some results on an RGB image.

Some

The next figure shows the transform functions from here:

https://sandipanweb.files.wordpress.com/2018/01/f9.png?w=150 150w, https://sandipanweb.files.wordpress.com/2018/01/f9.png?w=300 300w, https://sandipanweb.files.wordpress.com/2018/01/f9.png 677w" sizes="(max-width: 676px) 100vw, 676px" />

The next figures and animations show the application of the above non-linear transforms on the *Lena* image.

**Wave1**

https://sandipanweb.files.wordpress.com/2018/01/out_f1.png?w=150 150w, https://sandipanweb.files.wordpress.com/2018/01/out_f1.png?w=300 300w" sizes="(max-width: 568px) 100vw, 568px" />

** Wave2**

https://sandipanweb.files.wordpress.com/2018/01/out_f2.png?w=150 150w, https://sandipanweb.files.wordpress.com/2018/01/out_f2.png?w=300 300w" sizes="(max-width: 640px) 100vw, 640px" />

** **

**Swirl**

https://sandipanweb.files.wordpress.com/2018/01/out_swirl.png?w=150 150w, https://sandipanweb.files.wordpress.com/2018/01/out_swirl.png?w=300 300w" sizes="(max-width: 595px) 100vw, 595px" />

** Warp**

https://sandipanweb.files.wordpress.com/2018/01/out_warp.png?w=150 150w, https://sandipanweb.files.wordpress.com/2018/01/out_warp.png?w=300 300w" sizes="(max-width: 607px) 100vw, 607px" />

Some more **non-linear transforms**:

There is a problem with our interpolation method above: it is not very good at shrinking images, due to aliasing. For instance, if let’s try to *down-sample* the following bricks image by a factor of 0.4, we get the image shown in the following figure: notice the *strange banding effects* in the output image.

**Original Image**

https://sandipanweb.files.wordpress.com/2018/01/bricks_gray.png?w=1... 123w, https://sandipanweb.files.wordpress.com/2018/01/bricks_gray.png?w=2... 247w" sizes="(max-width: 542px) 100vw, 542px" />

**Down-sampled Image with Bilinear Interpolation**

https://sandipanweb.files.wordpress.com/2018/01/out_scale_b1.png?w=... 123w" sizes="(max-width: 542px) 100vw, 542px" />

The problem is that a single pixel in the output image corresponds to about 2.8 pixels in the input image, but we are sampling the value of a single pixel—we should really be averaging over a small area.

To overcome this problem, we will create a data structure that will let us (approximately) average over any possible square regions of pixels in the input image: an *image stack*. An image stack is a 3D matrix that we can think of as, not surprisingly, a stack of images, one on top of the other. The top image in the cube will be the original input image. Images further down the stack will be the input image with progressively larger amounts of blur. The size of the matrix will be rows × cols × num levels, where the original (grayscale) image has size rows×cols and there are num levels images in the stack.

Before we use the stack, we must write a function to create it, which takes as input a (grayscale) image and a number of levels in the stack, and returns a 3D matrix stack corresponding to the stack. Again, the ﬁrst image on the stack will be the original image. Every other image in the stack will be a blurred version of the previous image. A good *blur kernel* to use is:

https://sandipanweb.files.wordpress.com/2018/01/f10.png?w=150 150w" sizes="(max-width: 186px) 100vw, 186px" />

Now, for image k in the stack, we know that every pixel is a (weighted) average of some number of pixels (a k × k patch, roughly speaking) in the input image. Thus, if we down-sample the image by a factor of k, we want to sample pixels from level k of the stack.

⇒ Let’s write the following function to create image stack that takes a grayscale image and a number max levels, and returns an image stack.

from scipy import signal

def

create_image_stack(img, max_level):

K = np.ones((3,3)) / 9.

image_stack = np.zeros((img.shape[0], img.shape[1], max_level))

image_stack[:,:,0] = img

for l in range(1, max_level):

image_stack[:,:,l] = signal.convolve2d(image_stack[:,:,l-1], K,

boundary=’symm’, mode=’same’)

return image_stack

The next animation shows the image stack created from the bricks image.

Now, what happens if we down-sample the image by a fractional factor, such as 3.6? Unfortunately, there is no level 3.6 of the stack. Fortunately, we have a tool to solve this problem: *interpolation*. We now potentially need to sample a value at position (row,col,k) of the image stack, where all three coordinates are *fractional*. We therefore something more powerful than ** bilinear** interpolation:

This sounds complicated, but we can write this in terms of our existing functions. In particular, we now interpolate separately along different dimensions: trilinear interpolation can be implemented with two calls to bilinear interpolation and one call to linear interpolation.

Let’s implement a function *trilerp* like the following that takes an image stack, and a row, column, and stack level k, and returns the interpolated value.

def

trilerp(img_stack, x, y, k):if k < 1: k = 1

if k == int(k):

returnbilerp(img_stack[:,:,k-1], x, y)

else:

f_k, c_k = int(floor(k)), int(ceil(k))

v_f_k =bilerp(img_stack[:,:,f_k-1], x, y)

v_c_k =bilerp(img_stack[:,:,c_k-1], x, y)

returnlinterp(k, f_k, c_k, v_f_k, v_c_k)

Now we can ﬁnally implement a transformation function that does proper * anti-aliasing*. In order to do this, let’s implement a function that will

- First compute the image stack.
- Then compute, for the transformation T, how much T is scaling down the image. If T is deﬁned by the six values a,b,c,d,e,f above, then, to a ﬁrst approximation, the
*downscale factor*is:

https://sandipanweb.files.wordpress.com/2018/01/f11.png?w=150 150w" sizes="(max-width: 251px) 100vw, 251px" />

However, if k < 1 (corresponding to scaling up the image), we still want to sample from level 1. This situation reverts to normal*bilinear*interpolation. - Next call the
*trilerp*function on the image stack, instead of*bilerp*on the input image.

The next figure shows the output image obtained image transformation with proper *anti-aliasing*:

**Down-sampled Image with Anti-aliasing using Trilinear Interpolation**https://sandipanweb.files.wordpress.com/2018/01/out_scale_t.png?w=1... 120w" sizes="(max-width: 505px) 100vw, 505px" />

As we can see from the above output, the aliasing artifact has disappeared.

The same results are obtained on the color image, as shown below, by applying the *trilerp* function on the color channels separately and creating separate image stacks for different color channels.

**Original Image** https://sandipanweb.files.wordpress.com/2018/01/bricks_rgb.png?w=12... 123w, https://sandipanweb.files.wordpress.com/2018/01/bricks_rgb.png?w=24... 247w" sizes="(max-width: 414px) 100vw, 414px" />

**Down-sampled Image with Bilinear Interpolation** https://sandipanweb.files.wordpress.com/2018/01/out_scale_b2.png?w=... 120w" sizes="(max-width: 412px) 100vw, 412px" />

**Down-sampled Image with Anti-aliasing**

https://sandipanweb.files.wordpress.com/2018/01/out_scale_t1.png?w=... 123w" sizes="(max-width: 412px) 100vw, 412px" />

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