generated from mwc/lab_encoding
94 lines
2.3 KiB
Markdown
94 lines
2.3 KiB
Markdown
# Boolean questions
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Create the following variables.
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```
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a = Bits("11110000")
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b = Bits("10101010")
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```
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For each of the following bytes, give an equivalent
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expression which uses only `a`, `b`, and bit operators.
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The answers to the first two questions are given.
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1. 01010101
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~b
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2. 00000101
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~a & ~b
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3. 00000001
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4. 10000000
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5. 01010000
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6. 00001010
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7. 01010000
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8. 10101011
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## Integer questions
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These questions are difficult! Try exploring ideas with `Bits`
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in Terminal, a paper and pencil, and a whiteboard. And definitely
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talk with others.
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9. If `a` represents a positive integer, and `one = Bits(1, length=len(a))`,
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give an expression equivalent to `-a`, but which does not use negation.
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10. It is extremely easy to double a binary number: just shift all the bits
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to the left. (`a << 1` is twice `a`.) Explain why this trick works.
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11. Consider the following:
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```
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>>> hundred = Bits(100, 8)
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>>> hundred
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01100100
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>>> (hundred + hundred)
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11001000
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>>> (hundred + hundred).int
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-56
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```
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Apparently 100 + 100 = -56. What's going on here?
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12. What is the bit representation of negative zero? Explain your answer.
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13. What's the largest integer that can be represented in a single byte?
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Explain your reasoning.
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14. What's the smallest integer that can be represented in a single byte?
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Explain your reasoning.
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15. What's the largest integer that can be represented in `n` bits?
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Explain your reasoning.
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## Text questions
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16. Look at the bits for a few different characters using the `utf8` encoding.
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You will notice they have different bit lengths:
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```
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>>> Bits('a', encoding='utf8')
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01100001
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>>> Bits('ñ', encoding='utf8')
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1100001110110001
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>>> Bits('♣', encoding='utf8')
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111000101001100110100011
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>>> Bits('😍', encoding='utf8')
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11110000100111111001100010001101
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```
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When it's time to decode a sequence of utf8-encoded bits, the decoder
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somehow needs to decide when it has read enough bits to decode a character,
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and when it needs to keep reading. For example, the decoder will produce
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'a' after reading 8 bits but after reading the first 8 bits of 'ñ', the
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decoder realizes it needs to read 8 more bits.
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Make a hypothesis about how this could work.
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