Problem 1: Base-ic Counting

Problem 2: Getting Energized

• The Java compiler has a fun quirk where every binary file it produces begins with
  
  

• What is this in octal? hexadecimal?
• There are hard ways to do this, and very easy ways! I recommend the easy ways :)

Floating Representations

• Python represents floating point (fractional) numbers using two integers
  • One to represent the significant digits
  • One to represent the exponent (where the decimal place is)
• \(1\frac{1}{4}\) Example
  • In decimal: \(\quad\displaystyle 1\frac{1}{4} = \frac{1}{1} + \frac{2}{10} + \frac{5}{100} = 1.25 = (125, -2)\)
  • In binary: \(\quad\displaystyle 1\frac{1}{4} = \frac{1}{1} + \frac{0}{2} + \frac{1}{4} = 1.01 = (101, -10)\)

Problem 3: Floating Problems

• With this in mind, how could you convert the value \(\tfrac{7}{8}\) to a binary floating point representation?
• \[\frac{7}{8} = \frac{0}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = 0.111 = (111, -11)\]
• Now how would we convert \(\frac{1}{10}\) to binary??
  • We run into a problem! An infinitely repeating sequence! \[\frac{1}{10} = \frac{0}{1} + \frac{0}{2} + \frac{0}{4} + \frac{0}{8} + \frac{1}{16} + \frac{1}{32} + \frac{0}{64} + \frac{0}{128} + \frac{1}{256} + \cdots = 0.0001100110011\ldots\]
  • Have to stop the sequence somewhere and approximate it: \[\frac{3}{32} = 0.09375\quad\text{or}\quad\frac{25}{256} = 0.09765625\]

Consequences

• The best we can do within the range of normal integers \[\frac{3602879701896397}{2^{55}} = 0.10000000000000000555111512312578270\]
• When doing operations on these numbers, extra decimals will sometimes get rounded off, suddenly making the number look precise, but you might always have a tiny bit of this rounding error showing up in floating point values.
• So be careful using == for floating numeric comparisons! Rounding might result in unexpected falsehoods
  • 0.1 + 0.1 + 0.1 != 0.3
  • Far better to check if two numbers are within a small margin of one another, or greater or less than the other

Concatenating ASCII

• Suppose we wanted to print out the simple ASCII table as we saw it in the video
• Grid of 8 rows and 16 columns
• We can use chr to get the characters
• Need to “build-up” a string to print for each row

A Solution

row=''
for i in range(32,127):
    if len(row) == 16:
        print(row)
        row = ""
    row = row + chr(i)
print(row)