CS110_Lesson9_Notes.py

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#-------------------------------------------------------------------------------------
#            Modular Arithmetic
#-------------------------------------------------------------------------------------

# The modulo operator in Python is %
# Use this operator to get the remainder of a/b. 
# If a % b == 0, then a is divisible by b. Otherwise, a is not divisible by b.

# While you can do this with floats, you will most likely only use this for integers. 
# Due to the slight lack of precision of floating-point numbers, some properties of modular arithmetic are not exact.

print("13 % 5 =", 13 % 5)    # 13 % 5 = 3
print()

# prints out the remainder dividing each value by 5 from 0-15 inclusive
for i in range(16):
    print(i, '% 5 =', i % 5)
print()
    
# Notice how the values rotate like a clock?
# A % N is some number from 0 to N-1 for any integer A.

for i in range(31):
    if i % 5 == 0:
        print(i, 'is divisible by 5.')
print('--------------------------------------------------------------\n')

#---------------------------------------------------------------------------------------
#        Binary Numbers
#---------------------------------------------------------------------------------------

# Our regular number system that we use to count is base-10, called 'decimal' and uses digits 0-9.
# When we count in decimal, we go
# 0 1 2 3 4 5 6 7 8 9 
# when we reach the last digit, we add a 1 to the left and start over at zero
# 10 11 12 13 14 15 16 17 18 19 
# and so on
# Notice that if you do any number mod 10, you would get 0-9 as your remainder.

# These numbers are represented by powers of 10.
# 67,342 = 60,000 + 7,000 + 300 + 40 + 2 = (6 * 10000) + (7 * 1000) + (3 * 100) + (4 * 10) + (2 * 1)
#        = (6 * 10^4) + (7 * 10^3) + (3 * 10^2) + (4 * 10^1) + (2 * 10^0)
#----------------------------------------------------------------------------------

# Binary numbers are base-2. They use digits 0 and 1.
# Let's count in Binary
# 0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111

# If we have a number like 110101, this number is represented by powers of 2, if you were to convert back to decimal (base-10).
# 2^0 is on the right, the power increases as you go left.

# 110101 = (1 * 2^5) + (1 * 2^4) + (0 * 2^3) + (1 * 2^2) + (0 * 2^1) + (1 * 2^0)
#        = (1 * 32)  + (1 * 16)  + (0 * 8)   + (1 * 4)   + (0 * 2)   + (1 * 1)
#        = 32 + 16 + 0 + 4 + 0 + 1 = 53
# so 110101 in binary is 53 in decimal

# How do you convert from decimal to binary? There are a few ways to do this:
# METHOD 1: Divide, Mod, and Build backwards (right-to-left) with the remainders.
# 53/2 = 26, 53 % 2 = 1
# 26/2 = 13, 26 % 2 = 0
# 13/2 = 6, 13 % 2 = 1
# 6/2 = 3, 6 % 2 = 0
# 3/2 = 1, 3 % 2 = 1
# 1/2 = 0, 1 % 2 = 1
# since a/b = 0, build backwards with remainders, going back up: 110101

# METHOD 2: Find first power of 2 that is less than the number.
# 64 > 53, but 32 < 53
# Place a 1 for that spot, then subtract it from original number and repeat.
# 100000, 53-32 = 21
# 16 < 21
# 110000, 21-16 = 5
# 8 > 5, place a 0
# 110000
# 4 < 5, place a 1 in that spot
# 110100, 5-4 = 1
# 2 > 1, place a 0
# 110100
# 1 == 1, place a 1
# 110101, 1-1 = 0 --> STOP

# It's up to you to figure out how to program these conversions.

#----------------------------------------------------------------------------------------
#        Hexadecimal numbers
#----------------------------------------------------------------------------------------

# Hexadecimal numbers are base-16. They use digits 0-9 A-F. 10 = A, 11 = B, 12 = C, 13 = D, 14 = E, 15 = F
# Counting with hexadecimal we have this
# 0 = 0
# 1 = 1
# 2 = 2
# 3 = 3
# 4 = 4
# 5 = 5
# 6 = 6
# 7 = 7
# 8 = 8
# 9 = 9
# 10 = A
# 11 = B
# 12 = C
# 13 = D
# 14 = E
# 15 = F
# 16 = 10
# 17 = 11
# 18 = 12
# 19 = 13 ...

# This works the same as binary and decimal, except it uses powers of 16.
# 2A = (2 * 16^1) + (10 * 16^0) = (2 * 16) + (10 * 1) = 32 + 10 = 42
# FF = (15 * 16^1) + (15 * 16^0) = 240 + 15 = 255

# To convert from decimal to hexadecimal, you can do the same methods that were done converting from decimal to binary.
# Though Method 1 is the easiest and I will show you this one specifically.

# Divide, Mod, Build
# NOTE: If the number is less than 255, you can build it easier by just Divide, Mod, Build with those two digits
# 165/16 = 10 = A
# 165 % 16 = 5
# 165 = A5

# If the number is greater than 255, it will look more like the decimal-to-binary conversions with multiple steps.
# You need to BUILD BACKWARDS, so you take the last value from mod that you get and go up.
# 314/16 = 19, 314 % 16 = 10 = A
# 19/16 = 1, 19 % 16 = 3
# 1/16 = 0, 1 % 16 = 1
# 314 = 13A

#----------------------------------------------------------------------------------------------------
#        Built-in Methods for Conversions
#----------------------------------------------------------------------------------------------------

# There are built-in functions you can do that will convert numbers for you. 
# However, you CANNOT use them in the challenges.
# You can use them to double check if your functions are working correctly.

# Use int(x, base) to convert a number x in the given base to base-10.
print("13A in base-16 =", int('13A', 16))
print("110101 in base-2 =", int('110101', 2))
print()

# Use bin() to convert to binary
print("53 in binary =", bin(53))
# You can use the format function to get rid of the '0b' in front
print("53 in binary =", format(53, 'b'))
print()

# Use hex() to convert to hexadecimal
print("314 in hexadecimal =", hex(314))
# Again, you can use the format() function to get rid of the '0x' in front
print("314 in hexadecimal =", format(314, 'x'))

( around 127 lines Python3 code )

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Created on 2020-12-07T15:53:28Z

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