# Thinkpython

My answers for the Think Python excersizes (Book by Allen B. Downey)

"""

Created on Tue Mar 5 19:08:56 2019

@author: Usman Kotwal

def countdown(n):

if n <= 0:

print('Blastoff!')

else:

print(n)

countdown(n-1)

def do_n(a, n):

for i in range(n):

countdown(a)

do_n(5, 3)

#Chapter 5 Exercise 5.1

#The time module provides a function, also named time, that returns

#the current Greenwich Mean Time in "the epoch", which is an arbitrary

#time used as a reference point. On UNIX systems, the epoc is 1 January 1970

#Write a script that reads the current time and converts it to a time of day

#in hours, minutes and seconds, plus the number of days since the epoch

import time

def timeofday():

now = time.time()

days = now//(3600*24)

hours = (now//3600) % (days*24)

minutes = (now//60) % (days*24*60) - hours*60

seconds = int(now) % (days*24*60) - hours*3600 - minutes*60

print(days, hours, minutes, seconds)

timeofday()

#Chapter 5 Exercise 5.2

#Fermat's Last Theorem says that there are no positive integers a, b and c such

#that

#a^n + b^n = c^n

#for any values of n greater than 2

#------------------------------------------------------------

#1. Write a function named check_format that takes four parameters

#-a, b, c and n- and checks to see if Fermat's theorem holds. If n is greater

#than 2 and

#a^n + b^n = c^n

#the program shoud print "Holy smokes, Fermat was wrong!" Otherwise

#the program should pirnt, "No, that doesn't work."

def check_fermat(a,b,c,n):

a = int(a)

b = int(b)

c = int(c)

n = int(n)

if a^n + b^n == c^n and n > 2:

print("Holy smokes, Fermat was wrong!")

else:

print("No, that doesn't work")

check_fermat(3,4,5,2)

#----------------------------------------------------------------

#2. Write a function that promts the user to input values for

#a, b, c and n, convertest them to integers, and uses check_fermat to check whether they

#violate Fermat's theorem

def check_fermat():

promta = input("an integer for a \n")

promtb = input("an integer for b \n")

promtc = input("an integer for c \n")

promtn = input("an integer for n \n")

a = int(promta)

b = int(promtb)

c = int(promtc)

n = int(promtn)

if n <= 2:

print("wow, this has nothing to do with Fermat")

elif a^n + b^n == c^n:

print("Holy smokes, Fermat was wrong!")

else:

print("No, that doesn't work")

check_fermat()

#5.3 If you are give three sticks, you may or may not be able

#to arrange them in a triangle. For example, if one of the sticks

#is 12 inches long and the other two are one inch long, you will

#not be able to get the short stick to meet in the middle.

#For any three lenghts, there is a simple test to see

#if it is possible to form a triangle:

# If any of the three lengths is greater than the sum of

# the other two, then you cannot form a triangle. Otherwise

# you can.

#1. Write a function named is_triangle that takes threee integers

#as an argument, and that prints either "Yes" or "No" depending

# on whether you can or cannot form a triangle from the sticks

# with the given lenghts.

def is_triangle(a,b,c):

if c > a+b or a > b+c or b > c+a:

print("No")

else:

print("Yes")

is_triangle(2,8,5)

#--------------------------------------------

#2. Write a function that prompts the user to input

#three stick lengths, converts them to integers, and uses

#is_triangle to check whether sticks with the given

#lenght can form a triangle

"""

def is_triangle(a,b,c):

if c > a+b or a > b+c or b > c+a:

print("No")

else:

print("Yes")

def check_triangle():

a = int(input("an integer for a \n"))

b = int(input("an integer for b \n"))

c = int(input("an integer for c \n"))

is_triangle(a,b,c)

check_triangle()

#-----------------------------------------------

#Ex. 5.5 Read the following function and see if

#you can figure out what it does. Then run it and see

#if you got it right

import turtle

t = turtle.Turtle()

def draw(t, length, n):

if n == 0:

return

angle = 50

t.fd(length*n)

t.lt(angle)

draw(t,length,n-1)

t.rt(2*angle)

draw(t, length, n-1)

t.lt(angle)

t.bk(length*n)

draw(t, 20, 8)

#I did not get it right at all :(

#----------------------------------------------

#Exercise 5.6 The Koch curve is a fractal that looks

#something like Figure 5.2. To draw a Koch curve with

#length x, all you have do do is

#Step1. Draw a Koch curve with length x/3

#Step2. Turn left 60 degrees

#Step3. Draw a Koch curve with length x/3

#Step4. Turn right 120 degrees

#Step5. Draw a Koch curve with length x/3

#Step6. Turn left 60 degrees

#Step7. Draw a Koch curve with length x/3

#The exception is if x is less than 3 --> straight line length x

#1. Write a function called koch that takes a turtle

#and a length as parameters, and that uses the turtle

#to draw a Koch curve with the given length

import turtle

t = turtle.Turtle()

def koch(t = "turtle", x = "length"):

if x < 3:

t.fd(x)

else:

koch(t, x/3)

t.lt(60)

koch(t, x/3)

t.rt(120)

koch(t, x/3)

t.lt(60)

koch(t, x/3)

#2. Write a function called snowflake that draws three

#Koch curves to make the outline of a snowflake

def snowflake(t,x):

for i in range(3):

koch(t,x)

t.rt(120)

t.speed(0)

t.pu()

t.goto(-400, 300)

t.pd()

snowflake(t, 300)

turtle.mainloop()