Merge branch 'pr'

Author: abhatia1205

1_beginner/chapter3/solutions/temperature.py

@@ -1,4 +1,4 @@
-# Temperature
+  # Temperature
 # Write a program that converts Celsius to Fahrenheit.
 # It should ask the user for the temperature in
 # degrees Celsius and then print the temperature in degrees Fahrenheit.

1_beginner/chapter5/practice/add_all_the_way.py

@@ -5,3 +5,8 @@
 # You can use a for or while loop.
 
 # write code here
+
+
+
+
+#Try using the other loop and do the same p[roblem again

1_beginner/chapter5/practice/alternating.py

@@ -0,0 +1,17 @@
+""" 
+
+Ask the user for an integer. The print the numbers from 1 to that number,
+but alternating in sign. For example, if the input was 5, what would be printed
+is 1, -1, 2, -2, 3, -3, 4, -4, 5. (Note, DO NOT include the last negative
+number).
+
+Do this with a for loop
+
+"""
+
+#Write code here.
+
+number = int(input("Enter Number Here: "))
+
+
+#Now try it with a while loop

1_beginner/chapter5/practice/even.py

@@ -1,5 +1,5 @@
 # Even
 # Print every even number greater than 10
-# and less than 101
+# and less than 101 (10 ia included)
 
 # write code here

1_beginner/chapter5/practice/fibonnaci.py

@@ -0,0 +1,20 @@
+''' CHALLENGE PROBLEM!! NOT FOR THE FAINT OF HEART!
+
+The Fibonacci numbers, discovered by Leonardo di Fibonacci,
+is a sequence of numbers that often shows up in mathematics and,
+interestingly, nature. The sequence goes as such:
+
+1,1,2,3,5,8,13,21,34,55,...
+
+where the sequence starts with 1 and 1, and then each number is the sum of the
+previous 2. For example, 8 comes after 5 because 5+3 = 8, and 55 comes after 34
+because 34+21 = 55.
+
+The challenge is to use a for loop (not recursion, if you know what that is),
+to find the 100th Fibonnaci number.
+'''
+
+#write code here
+
+
+#Can you do it with a while loop?

1_beginner/chapter5/practice/prime.py

@@ -0,0 +1,28 @@
+"""
+
+Write a program to check if a number is prime or not. A prime
+number is one that is not divisible by any number other than
+1 and itself. For example, 11 is prime because it is not divisible
+by 2,3,4,5,...10 (i.e. 11/10, for example, is not an integer).
+
+Write a for loop to check if a number is prime or not.
+
+"""
+
+#write code here
+
+numer = int(input("Enter number here: "))
+
+"""
+Given a number less than or equal to 10 billion, see if you can check if it is
+prime in UNDER 2 SECONDS. The code you wrote above probably wont do that, so
+you will have to figure out a clever solution.
+
+NOTE: THE ABOVE IS A CHALLENGE PROBLEM, AND IS EXTRREEMMLY HARD TO DO.
+I COULDN'T DO IT OPTIMALY UNTIL I LEARNED HOW TO. If you can't
+figure out the solution, don't feel discouraged, its seriously a really
+hard problem (I have seen this asked to college students :O)
+
+"""
+
+#write code here

1_beginner/chapter5/practice/series.py

@@ -0,0 +1,16 @@
+"""
+
+Here are some interesting results from mathematics that we can
+model with computer science.
+
+In class we added up 1 + 1/2 + 1/4 + 1/8 ...
+See if you can add up 1 + 1/3 + 1/9 + 1/27 + 1/81 ... in a similar fashion.
+What is the answer? What if, instead of 3, we use 5 (1 + 1/5 + 1/25 + 1/125...)
+7? Do you see a patter? (Note: make sure not to end up with an infinite loop).
+
+
+
+Also, try adding up 1 -1/3 + 1/5 - 1/7 + 1/9 - 1/11 ... 10 million times. then
+multiply this result by 4. What number is this close to?
+
+"""

2_intermediate/chapter10/practice/img_avg.py

@@ -0,0 +1,46 @@
+""" Here is the challenge problem for 2d loops>
+Images are often represented as 3d arrays,
+where the rows and columns are the pixels in the image,
+and each pixel has an r, g, and b value.
+
+The interesting thing is that we can iterate over images.
+The challenge is, given an image, create a program that
+will return a different image where each pixel is the average
+of the pixels surrounding it in the original image.
+
+The neighbors of an image are all the pixels that surroun it,
+1 on each side, and 4 on the diagonals, for 8 in total. 
+Each pixel doesn't necessarily have 8 neighbors, though (think about why)
+
+The code to grab an image from the internet and make it
+into an array is given to you. The code also displays the new image
+you create in the end.
+
+NOTE: The image is 3 dimensional because each pixel has rgb values.
+To find the average value of all of a pixels neighbors, you must
+change the average of the red value to the red value, blue to blue, etc.
+For example, if the neighbors of a pixel with value [1,2,3]
+were [20,30,40] and [10,120,30], the new pixel that would replace the original one would be
+[15,75,35]
+"""
+
+from PIL import Image
+import requests
+import numpy
+import matplotlib.pyplot as plt
+
+url = "https://images.dog.ceo/breeds/waterdog-spanish/20180723_185544.jpg"
+img = numpy.array(Image.open(requests.get(url, stream=True).raw)).tolist()
+newimg = img
+transpose = numpy.transpose(img)
+
+plt.imshow(img)
+plt.show()
+
+#write code to create newimg here
+
+plt.imshow(newimg)
+plt.show()
+
+plt.imshow(transpose)
+plt.show()

2_intermediate/chapter10/solutions/img_avg.py

@@ -0,0 +1,90 @@
+""" Here is the challenge problem for 2d loops>
+Images are often represented as 3d arrays,
+where the rows and columns are the pixels in the image,
+and each pixel has an r, g, and b value.
+
+The interesting thing is that we can iterate over images.
+The challenge is, given an image, create a program that
+will return a different image where each pixel is the average
+of the pixels surrounding it in the original image.
+
+The neighbors of an image are all the pixels that surroun it,
+1 on each side, and 4 on the diagonals, for 8 in total. 
+Each pixel doesn't necessarily have 8 neighbors, though (think about why)
+
+The code to grab an image from the internet and make it
+into an array is given to you. The code also displays the new image
+you create in the end.
+
+NOTE: The image is 3 dimensional because each pixel has rgb values.
+To find the average value of all of a pixels neighbors, you must
+change the average of the red value to the red value, blue to blue, etc.
+For example, if the neighbors of a pixel with value [1,2,3]
+were [20,30,40] and [10,120,30], the new pixel that would replace the original one would be
+[15,75,35]
+"""
+
+from PIL import Image
+import requests
+import numpy
+import matplotlib.pyplot as plt
+
+url = "https://images.dog.ceo/breeds/waterdog-spanish/20180723_185544.jpg"
+img = numpy.array(Image.open(requests.get(url, stream=True).raw))
+newimg = img
+transpose = numpy.transpose(img)
+
+
+#write code to create newimg here
+def solution1():
+	"""Iterating over the image here. i is a variable from 0 to the width of the image.
+	j is a variable that ranges from 0 to the height of the image. i is associated with
+	values"""
+	for i in range(len(img)):
+		for j in range(len(img[0])):
+			x_n = [0]
+			y_n = [0]
+
+			if(i == 0):
+				x_n.append(1)
+			elif(i ==  len(img)-1):
+				x_n.append(-1)
+			else:
+				x_n.append(1)
+				x_n.append(-1)
+
+			if(j == 0):
+				y_n.append(1)
+			elif(j ==  len(img[0])-1):
+				y_n.append(-1)
+			else:
+				y_n.append(1)
+				y_n.append(-1)
+
+			r_avg = -1*img[i][j][0]
+			g_avg = -1*img[i][j][1]
+			b_avg = -1*img[i][j][2]
+			c = -1
+
+			for x in x_n:
+				for y in y_n:
+					r_avg += img[i+x][j+y][0]
+					g_avg += img[i+x][j+y][1]
+					b_avg += img[i+x][j+y][2]
+					c+=1
+			r_avg = r_avg/c
+			g_avg = g_avg/c
+			b_avg = b_avg/c
+
+			newimg[i][j] = [r_avg, g_avg, b_avg]
+
+
+
+
+solution1()
+
+plt.imshow(newimg)
+plt.show()
+
+plt.imshow(transpose)
+plt.show()