Lecture 2 - Flow control

Lecture 2 - Flow control#

Now we are going to practice branching and iteration

branching (interpreter steps through from top to bottom)
- if : if condition is true, run this block of code
- elif : else if these conditions are true, run this block of code
- else : if none of the above are true, run this block of code
iteration
- for : known number of iterations, can end early with a break condition
- while : unbounded number of iterations but know a break condition

# Let us start by iterating over strings. 

string = "BB1000_anno2023_rocks"

# write a "for" loop using the "range" function that prints each character in string

for index in range(len(string)):
    print(string[index])

B
B
1
0
0
0
_
a
n
n
o
2
0
2
3
_
r
o
c
k
s

# Let us now instead try to iterate over the characters in the string directly and print each of them
for char in string:
    print(char)

B
B
1
0
0
0
_
a
n
n
o
2
0
2
3
_
r
o
c
k
s

# We can also use loops inside loops to print number sequences like
# 0
# 01
# ...
# 0123456
# and so on

for i in range(10):
    for j in range(i+1):
        print(j, end='')
    print()

# Modify the example to print only the even numbers.
# The mod (%) operator may come in handy!

for i in range(0, 10, 2):
    for j in range(i+1):
        if j % 2 == 0:
            print(j, end='')
    print()

# Now we are interested in knowing the numbers of zeros in the string, 
# so you need to write code that will output (print) that using the variable zero_count. 
# Might be helpful to recall the comparison operator '=='

string = "BB1000_anno2023_rocks"

zero_count = 0
for character in string:
    if character == '0':
        zero_count += 1

assert zero_count == 4, error_message(zero_count, 4)

# Rewrite the following while loop to a for loop

letters = "aefhilmnorsxAEFHILMNORSX"
word = input("I will cheer for you! Enter a word: ")
times = int(input("Enthusiasm level (1-10): "))

# While loop example to reproduce with for loop
i = 0
while i < len(word):
    char = word[i]
    if char in letters:
        print("Give me an " + char + "! " + char)
    else:
        print("Give me a  " + char + "! " + char)
    i += 1
print("What does that spell?")
for i in range(times):
    print(word, "!")

---------------------------------------------------------------------------
StdinNotImplementedError                  Traceback (most recent call last)
Cell In[8], line 4
      1 # Rewrite the following while loop to a for loop
      3 letters = "aefhilmnorsxAEFHILMNORSX"
----> 4 word = input("I will cheer for you! Enter a word: ")
      5 times = int(input("Enthusiasm level (1-10): "))
      7 # While loop example to reproduce with for loop

File ~/opt/miniconda3/envs/bb1000_ht23/lib/python3.9/site-packages/ipykernel/kernelbase.py:1201, in Kernel.raw_input(self, prompt)
   1199 if not self._allow_stdin:
   1200     msg = "raw_input was called, but this frontend does not support input requests."
-> 1201     raise StdinNotImplementedError(msg)
   1202 return self._input_request(
   1203     str(prompt),
   1204     self._parent_ident["shell"],
   1205     self.get_parent("shell"),
   1206     password=False,
   1207 )

StdinNotImplementedError: raw_input was called, but this frontend does not support input requests.

for char in word:
    if char in letters:
        print("Give me an " + char + "! " + char)
    else:
        print("Give me a  " + char + "! " + char)
print("What does that spell?")
for i in range(times):
    print(word, "!")

Give me an N! N
Give me an a! a
Give me an n! n
Give me an n! n
Give me an a! a
What does that spell?
Nanna !
Nanna !
Nanna !
Nanna !
Nanna !
Nanna !
Nanna !
Nanna !
Nanna !
Nanna !

# We can also loop over lists of numbers to find the max number!
# You will learn more about lists in lecture 4. For now, think about them as objects you can iterate over like strings (also called iterables).

numbers = [1, 4, 12, 41, 3, 8, 142, 2, 5]
max_number = 0

# Write a loop that checks every number in the list and see if it is bigger than previous max_number

for number in numbers:
    if number > max_number:
        max_number = number
print(max_number)

assert max_number == 142, error_message(max_number, 142)

# Computers are great at doing calculations. Write a piece of code that can print the 10 times multiplication table.
# Here you want to create the entire line of the table before invoking the print function - because print automatically adds a newline

for i in range(1, 11):
    row = ''
    for j in range(1, 11):
        row += f'{j*i:4d} |'
        if j < 11:
            row += ' '
    row += '\n'
    print('-'*69, end='\n')
    print(row, end='')

---------------------------------------------------------------------
   1 |    2 |    3 |    4 |    5 |    6 |    7 |    8 |    9 |   10 | 
---------------------------------------------------------------------
   2 |    4 |    6 |    8 |   10 |   12 |   14 |   16 |   18 |   20 | 
---------------------------------------------------------------------
   3 |    6 |    9 |   12 |   15 |   18 |   21 |   24 |   27 |   30 | 
---------------------------------------------------------------------
   4 |    8 |   12 |   16 |   20 |   24 |   28 |   32 |   36 |   40 | 
---------------------------------------------------------------------
   5 |   10 |   15 |   20 |   25 |   30 |   35 |   40 |   45 |   50 | 
---------------------------------------------------------------------
   6 |   12 |   18 |   24 |   30 |   36 |   42 |   48 |   54 |   60 | 
---------------------------------------------------------------------
   7 |   14 |   21 |   28 |   35 |   42 |   49 |   56 |   63 |   70 | 
---------------------------------------------------------------------
   8 |   16 |   24 |   32 |   40 |   48 |   56 |   64 |   72 |   80 | 
---------------------------------------------------------------------
   9 |   18 |   27 |   36 |   45 |   54 |   63 |   72 |   81 |   90 | 
---------------------------------------------------------------------
  10 |   20 |   30 |   40 |   50 |   60 |   70 |   80 |   90 |  100 | 

# Should look something like this
"""
---------------------------------------------------------------------
   1 |    2 |    3 |    4 |    5 |    6 |    7 |    8 |    9 |   10 | 
---------------------------------------------------------------------
   2 |    4 |    6 |    8 |   10 |   12 |   14 |   16 |   18 |   20 | 
---------------------------------------------------------------------
   3 |    6 |    9 |   12 |   15 |   18 |   21 |   24 |   27 |   30 | 
---------------------------------------------------------------------
   4 |    8 |   12 |   16 |   20 |   24 |   28 |   32 |   36 |   40 | 
---------------------------------------------------------------------
   5 |   10 |   15 |   20 |   25 |   30 |   35 |   40 |   45 |   50 | 
---------------------------------------------------------------------
   6 |   12 |   18 |   24 |   30 |   36 |   42 |   48 |   54 |   60 | 
---------------------------------------------------------------------
   7 |   14 |   21 |   28 |   35 |   42 |   49 |   56 |   63 |   70 | 
---------------------------------------------------------------------
   8 |   16 |   24 |   32 |   40 |   48 |   56 |   64 |   72 |   80 | 
---------------------------------------------------------------------
   9 |   18 |   27 |   36 |   45 |   54 |   63 |   72 |   81 |   90 | 
---------------------------------------------------------------------
  10 |   20 |   30 |   40 |   50 |   60 |   70 |   80 |   90 |  100 |
 """

'\n---------------------------------------------------------------------\n   1 |    2 |    3 |    4 |    5 |    6 |    7 |    8 |    9 |   10 | \n---------------------------------------------------------------------\n   2 |    4 |    6 |    8 |   10 |   12 |   14 |   16 |   18 |   20 | \n---------------------------------------------------------------------\n   3 |    6 |    9 |   12 |   15 |   18 |   21 |   24 |   27 |   30 | \n---------------------------------------------------------------------\n   4 |    8 |   12 |   16 |   20 |   24 |   28 |   32 |   36 |   40 | \n---------------------------------------------------------------------\n   5 |   10 |   15 |   20 |   25 |   30 |   35 |   40 |   45 |   50 | \n---------------------------------------------------------------------\n   6 |   12 |   18 |   24 |   30 |   36 |   42 |   48 |   54 |   60 | \n---------------------------------------------------------------------\n   7 |   14 |   21 |   28 |   35 |   42 |   49 |   56 |   63 |   70 | \n---------------------------------------------------------------------\n   8 |   16 |   24 |   32 |   40 |   48 |   56 |   64 |   72 |   80 | \n---------------------------------------------------------------------\n   9 |   18 |   27 |   36 |   45 |   54 |   63 |   72 |   81 |   90 | \n---------------------------------------------------------------------\n  10 |   20 |   30 |   40 |   50 |   60 |   70 |   80 |   90 |  100 |\n '

# This time I would like that you insert a "super" between "to" and "hero" in our motto string to create a super_string
# Later on you will learn smarter ways of doing this, but for now let us practice branching and loops.

string = 'From zero to hero'
super_string = ""

not_yet_injected = True

for i in range(len(string)):
    if i > 0 and not_yet_injected:
        if string[i] == "o" and string[i-1] == "t":
            print("We located 'to' and are now ready to inject ' super'")
            super_string += string[i] + ' super'
            not_yet_injected = False
            continue
    super_string += string[i]
print(super_string)

We located 'to' and are now ready to inject ' super'
From zero to super hero

assert super_string == "From zero to super hero", error_message(super_string, "From zero to super hero")

Now to while loops#

Suppose that we want to know the square root of some positive number (value) but do not have the math.sqrt() function available. That is, we want to calculate \(x\), where \(x^2 = value\)

The value of \(x\) can be computed as the roots (=zero crossings) of the function \(f(x) = x^2-value\)

We can use the so-called bisection method to get an approximate value of x.

https://en.wikipedia.org/wiki/Bisection_method

This method approximates the root of a continous function by repeatedly dividing the interval that bounds the root at the midpoint.

Preparation for the algorithm

Define the interval. Here we know that \(x>0\) and \(x\le value\). That is low_bound = 0 and high_bound = value
Set up a guess corresponding to the midpoint of the interval, i.e., guess = (low_bound + high_bound) / 2.

Run the algorithm

Calculate guess * guess
If f(guess)is close enough (use threshold) to zero, stop and say that guess is the approximate x
If not, update the bounding interval (i.e., low_bound and high_bound).
- if guess ** 2. smaller than x, we can update low_bound to the value of `guess``
- else if guess ** 2. is greater than or equal to high_bound, we can update high_bound to the value of guess
Then make a new guess by calculating the midpoint of the new bounding interval
Using the new guess, repeat the cycle

bisection_method

# Implement the bisection method that calculates the square root of 10 to within a threshold of 10e-7.
# Print the value and the number of iterations it took to reach the value within the desired threshold.

threshold = 10e-7
value = 10.
num_iterations = 0

low_bound = 0.0
high_bound = value
guess = (high_bound + low_bound) / 2.0

while abs(guess**2. - value) >= threshold:
    if guess ** 2. < value:
        # Look only in the upper half
        low_bound = guess
    else:
        high_bound = guess
    guess = (high_bound + low_bound) / 2.0
    num_iterations += 1

import math
assert guess == approx(math.sqrt(value)), error_message(guess, math.sqrt(value))

Maze challenge#

In this exercise, your task is to write an algorithm that will take you through a maze.

Several pre-defined functions are at your disposal:
- is_right_clear() - checks whether the field to the right is clear or not (wall)
- is_left_clear() - checks whether the field to the left is clear or not (wall)
- is_front_clear()- checks whether the field in front is clear or not (wall)
- is_running()- checks whether a wall has been hit or the maze was solved
- turn_right() - changes your direction to the right
- turn_left()- changes your direction to the left
- move_forward()- move you forward in the current direction
You need to combine these functions in a control flow
The entrance to the maze starts at the yellow square and the exit is marked by the green square
An already given animation functionality is “wrapped” around the the control flow (for you to write) so that you can see whether you made it

NOTE

Should you end up in an infinite loop, you need to manually stop the relevant code block by pressing the “stop” bottom (same location as the play bottom).

%matplotlib qt
from maze import *

# You can change between MAZE0, MAZE1 and MAZE2 for increasing difficulty
reset_start(MAZE2)
while is_running():
    if is_right_clear():
        turn_right()
        move_forward()
    elif is_front_clear():
        move_forward()
    else:
        turn_left()

Yay you completed the maze!

Lecture 2 - Flow control

Contents

Lecture 2 - Flow control#

Now to while loops#

Maze challenge#