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Worksheet

A15.7 Calculation and theory questions

Exam preparation · A level · OCR H446 1.4.1, AQA 7517 4.4.1.2, Eduqas A500QS 1.8 · about 40 min

BugBotLab
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What this lesson is about

Number representation, floating point, Boolean logic, state machines and complexity answered quickly and checked.

Questions 6 marks in all

  1. [1 mark]What is the denary value of the 8-bit two's complement number 11101100?

  2. [1 mark]A normalised floating point number has mantissa 01101000 (point after the first bit) and exponent 0011. What is its denary value?

  3. [1 mark]Which mantissa is normalised?

    1. A10100000
    2. B11010000
    3. C00110000
    4. D11100000

  4. [1 mark]In how many rows of its truth table is NOT (A AND B) OR C true?

  5. [1 mark]An algorithm has a loop over n items inside another loop over n items. What is its time complexity?

    1. AO(n²)
    2. BO(n)
    3. CO(log n)
    4. DO(2ⁿ)

  6. [1 mark]What does this finite state machine print?

    t = {("even", "0"): "even", ("even", "1"): "odd", ("odd", "0"): "odd", ("odd", "1"): "even"}
for text in ["110", "111", ""]:
    state = "even"
    for symbol in text:
        state = t[(state, symbol)]
    print(repr(text), state)

The task: theory at speed

Write each method as a function, then print the answers. - twos(bits): bits is a string of 0s and 1s of any length. Return its value as a two's complement integer: the leftmost bit is worth minus its normal place value. - float_value(mantissa, exponent): both are strings of bits. The mantissa is two's complement with the binary point just after the leftmost bit; the exponent is a two's complement integer. Return the number's value as a float. - run_fsm(transitions, state, accepting, text): transitions is a dictionary from (state, symbol) tuples to the next state, state is the start state, accepting is a set of accepting states, and text is a string of symbols. Return True if the machine ends in an accepting state. Then print exactly these six lines, using the functions for the values: 1. 11101100 = <value> using twos; 2. 01101000 x 2^0011 = <value> and 3. 10100000 x 2^1111 = <value> using float_value; 4. 1101: <result> and 5. 1001: <result>, where the result is accepted or rejected, using run_fsm with the even-number-of-1s machine above (states "even" and "odd", start "even", accepting {"even"}); 6. true rows: <n>, counting the rows where NOT (A AND B) OR C is true with three nested loops over A, B and C. The robot stays still.

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

def twos(bits):
    pass

Plan your program here, then type it in and press Run.

QR code
Do it on the robot
www.bugbotlab.com/learn/a15-7-theory-at-speed/
The simulator checks it and tells you when it passes. Nothing to install, no account.

Challenges

  1. Normalise the mantissa 00011010 with exponent 0101, keeping the value the same.
  2. Add hex_to_denary(text) without using int(text, 16), and test it on 2F and FF.
  3. Design a state machine that accepts binary strings ending in 01, and test it with run_fsm.