## Lesson 1 – Challenge Solutions

Challenge 1.1
``````def border_text(text_var):
# Let's create our function

# Take the length of the text and multiply it by "=" (the border)
border = '='*len(text_var)

# Print the border, the test, and then the border with separating lines

print(border+'\n'+text_var+'\n'+border)

return 0

border_text("Hello World")

border_text("World")``````
Output
``````===========
Hello World
===========
=====
World
=====``````
Challenge 1.2
``````# First, import the math library
import math

# The math library's sin, cos, tan, etc. functions
#  deal in radians so we need to convert degrees to radians

# Let's find side c
# cos(54 degrees)=4 inches/c inches
# c inches*cos(54 degrees)=4 inches
# c inches=4 inches/cos(54 degrees)
side_c = 4/math.cos(angle_b)

# Now let's find side b
# sin(54 degrees)=b inches/side_c
# side_c*sin(54 degrees)=b inches
side_b = side_c*math.sin(angle_b)

# Okay, let's print everything out
print("Side a: 4 inches")
print("Side b: "+str(side_b)+" inches")
print("Side c: "+str(side_c)+" inches")``````
Output
``````Side a: 4 inches
Side b: 5.505527681884694 inches
Side c: 6.805206466816319 inches``````
Challenge 1.3
``````# Used to prevent a link from the last_level and pascals_triangle
import copy

# Let's create our list and start it will 2 levels
pascals_triangle = [, [1, 1]]

# Let's create a while loop that will stop on the tenth level
while len(pascals_triangle) != 10:

# We need a copy otherwise it will be linked to pascals_triangle
last_level = copy.deepcopy(pascals_triangle[-1])

# Okay, for our next loop, it's going to create the next level
new_level = 
while 1:
new_level.append(last_level+last_level)
last_level.pop(0)
if len(last_level) == 1:
new_level.append(1)
break

pascals_triangle.append(new_level)

print(pascals_triangle)
print(pascals_triangle)
print(pascals_triangle)
print('...')
print(pascals_triangle)
print(pascals_triangle)``````
Output
``````
[1, 1]
[1, 2, 1]
...
[1, 8, 28, 56, 70, 56, 28, 8, 1]
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]``````

## Lesson 2 Challenge Solutions

Challenge 1.1
``````

def find_factors(number):
# Functions because why not?

# Initialize out list
factors = []

# Starting from 1, let's test each number
for curr in range(1, number + 1):
# Taking the modulas, if there is a reminder,
#  It's not a factor
if (number % curr == 0):
# We found a factor, add it to our list
factors.append(curr)

# We went through every number from 1 to the user's number
# Return what we have
return factors

# Ask the user for an input (and take the integer of the string that comes in)
usr_inp = int(input("What number would you like to find the factors of? "))
# Print out the factors
print("The factors of", usr_inp, "are:", find_factors(usr_inp))``````
Output
``````What number would you like to find the factors of? 12
The factors of 12 are: [1, 2, 3, 4, 6, 12]``````
Challenge 1.2
``````from random import shuffle
from math import pi

def pi_finder():
# Keeping trying until we find a suitable denominator and numerator
while(1):
# First: Take 9 digits, randomize the order, & create a number from it
denominator_list = ["1","2","3","4","5","6","7","8","9"]
shuffle(denominator_list)

# Multiple ways to do this, this take a list of numbers, and takes an
#  empty string and adds each element from the list - finally taking the
#  numeric value of the string using int()
denominator = int(''.join(denominator_list))

# Second: Take that number times PI (denominator times pi equal numerator)
#		    and then take the integer of the value
numerator = denominator * pi
numerator = int(numerator)

# Third: Check if it's 9 digits, and check if those digits are unique
#		   from each other
# Much easier to deal with strings at this point
str_numerator = str(numerator)

if (len(str_numerator) != 9):
# Continue means reset from the top of the loop
continue

# Create a pool and remove each number as it is found
pool = ["1","2","3","4","5","6","7","8","9"]

for curr in list(str_numerator):
# Take each number out as it is found
if (curr in pool):
pool.remove(curr)

if (pool != []):
# If it is not empty, there are dups in the numerator
continue

# At this point all checks have passed, return the number
return [numerator, denominator]

# Run the function
result = pi_finder()

# Print out the results
print("The numerator is:", result)
print("The denominator is:", result)
print("The approximation of PI is:", result/result)``````
Output
``````The numerator is: 613987452
The denominator is: 195438276
The approximation of PI is: 3.1415926530174674``````

# Challenge 2.1

## Objective

Create a script that will take a user input, find the factors of that number, and print those factors.

## Hints

• It would be sad if you don’t know what factors are, but here you go
• Use a list to store the factors you have found
• Use a loop to check each possible factor
• Use the function `input()` to collect a user’s input
• That input is going to be a string, turn it into an integer

## Solution

Here is the possible output of a function:

``````What number would you like to find the factors of? 12
The factors of 12 are: [1, 2, 3, 4, 6, 12]``````

# Challenge 2.2

## Objective

Create a script that can generate a fraction consisting of 10 unique numbers on the numerator and 10 unique numbers on the denominator that can be a close approximation of PI. So, each number (1,2,3,4,5,6,7,9,0) can only be used once in the numerator and same for the denominator.

## Hints

• You will probably use random numbers (`import random`) for either the denominator or the numerator
• Use a list for creating the random unique numbers
• Also, add the math module, it contains PI `math.pi` (`import math`)
• You will probably use loops
• This is simpler than you think

## Solution

This is one of the many answers that could be generated through a good script:

``````The numerator is: 2385140697
The denominator is: 759213864
The approximation of PI is: 3.141592652739018``````

# Challenge 2.3

## Objective

We have been given some file transfer data from the NOvA experiment. These data files contains a lot of information about file transfers for this experiment. We are going to take how many jobs started in one day. We will then plot this data in a histogram with the bin size being 1 hour.

``````JobID,JobStartTime(unix timestamp),FileID,FileSize(byte),NodeName,FileProcessingTime(second),FileTransferTime(second),NodeType,FileTransferRate(byte/s)