2.8. Pseudocode

Sometimes it can be useful to write algorithms in code form without the syntax of a specific language. We can do this using pseudocode. This is a way of writing code in a format that is easy for a human to read. The exact symbols and keywords may vary depending on the convention you use.

The way in which we represent if-elif-else statements as follows. This is an example with 3 conditions.

IF condition A THEN
    process 1
ELSEIF condition B THEN
    process 2
ELSEIF condition C THEN
    process 3
ELSE
    process 4
ENDIF

Here’s an example!

Python code

x = 1

if x < 0:
    print('x is a negative number')
elif x > 0:
    print('x is a positive number')
else:
    print('x is 0!')

Pseudocode

IF x < 0 THEN
    Display 'x is a negative number'
ELSEIF x > 0 THEN
    Display 'x is a positive number'
ELSE
    Display 'x is 0!'
ENDIF

Question 1

Which of the following provides the pseudocode that corresponds to the following Python program?

weather = 'rainy'

if weather == 'rainy':
    print('Take an umbrella!')

1. IF weather == rainy:
       Display 'Take an umbrella'
   

2. IF 'Take an umbrella'
       weather == 'rainy'
   ENDIF
   

3. ELSEIF weather == 'rainy'
       Display 'Take an umbrella'
   ENDELSE
   

4. IF weather == 'rainy' THEN
       Display 'Take an umbrella'
   END IF
   

Solution

IF weather == rainy:
    Display 'Take an umbrella'

Invalid. Uses : instead of THEN and missing ENDIF at the end.

IF 'Take an umbrella'
    weather == 'rainy'
ENDIF

Invalid and Incorrect. Here the condition and the process have been swapped. The IF is also missing the THEN.

ELSEIF weather == 'rainy'
    Display 'Take an umbrella'
ENDELSE

Invalid. The first condition should be an IF not an ELSEIF. The closing tag should be ENDIF instead of ENDELSE

IF weather == 'rainy' THEN
    Display 'Take an umbrella'
END IF

Correct.

Question 2

Which of the following provides the pseudocode that corresponds to the algorithm illustrated below?

1. IF light == 'green' THEN
       Display 'GO!'
   ELSEIF light == 'yellow' THEN
       Display 'Slow down!'
   ELSEIF light == 'red' THEN
       Display 'Stop!'
   ENDIF
   

2. IF light == 'green' THEN
       Display 'GO!'
   ELSEIF light == 'yellow' THEN
       Display 'Slow down!'
   ELSE
       Display 'Stop!'
   ENDIF
   

3. IF light == 'green' THEN
       Display 'GO!'
   ELSE
   IF light == 'yellow' THEN
       Display 'Slow down!'
   ELSE
   IF light == 'red' THEN
       Display 'Stop!'
   ELSE
   

4. IF light == 'green' THEN
       Display 'GO!'
   ENDIF
   IF light == 'yellow' THEN
       Display 'Slow down!'
   ENDIF
   IF light == 'red' THEN
       Display 'Stop!'
   ENDIF
   

Solution

Solution is locked

Code challenge: Starting Player

The following algorithm is used to pick a starting player. Both players roll a die and whoever rolls the largest number gets to start. If it’s a tie, then the younger player starts. If it’s a tie again then player 2 starts.

IF roll1 > roll2 THEN
    Display 'Player 1 starts'
ELIF roll2 > roll1 THEN
    Display 'Player 2 starts'
ELSE
    IF age1 < age2 THEN
        Display 'Player 1 starts'
    ELSE
        Display 'Player 2 starts'

Write the Python code that corresponds to the given pseudocode.

Solution

Solution is locked