Tutorial 1¶

Question 1¶

A time t in seconds can be converted to days, hours, minutes and seconds using integer division:

Divide t by 60; set t to the quotient and call the remainder s.
Divide t by 60; set t to the quotient and call the remainder m.
Divide t by 24; set d to the quotient and call the remainder h.

Calculate the number of days, hours, minutes and seconds in one million seconds. Display the result as follows:

1000000 seconds is xx days H:M:S

# It should be reasonably easy to follow the instructions in order to produce the correct code to perform the calculation.
# The only tricky part is to maintain the original value of `t` in a new variable `t_orig` so that we can include it in the final print statement.

t = 1000000
t_orig = t
s = t % 60
t = t // 60
m = t % 60
t = t // 60
h = t % 24
d = t // 24

print(t_orig, " seconds is ", d, " days ", h, ":", m, ":", s, sep="")

1000000 seconds is 11 days 13:46:40

Question 2¶

Iodine-131 is a radioisotope with a half-life \(t_\mathrm{half} = 8.02\) days.

Use the equation \(r = \left(\frac{1}{2}\right) ^ {1/t_\mathrm{half}}\) to calculate the daily decay rate (i.e. the proportion of a sample of Iodine 131 that decays in one day).
A sample of radioactive material containing Iodine-131 has an activity of \(1.67 \times 10^{12}\) Becquerels. Write a program which calculates the number of days it takes for the sample to reach an activity of \(10000\) Becquerels.
Print the activity of the sample once every 7 days.

# This is very similar to the bacterial growth example in the notes.
# You shouldn't need to know much about radioactive decay to answer this.
# However, this is the kind of foundational knowledge that I would expect all Natural Scientists to be aware of,
# even if you don't specialise in physics.

t_half = 8.02
r = 0.5 ** (1/t_half)
print("Daily decay rate:", r)

activity = 1.67e12

i = 0
while activity > 10000:
    
    activity = activity * r
    
    i += 1 # or i = i + 1
    
    if i % 7 == 0:
        print(activity)
    
print("Number of days:", i)

Daily decay rate: 0.9172022001375253
911952222126.3414
497998117030.64197
271946400862.7711
148504266207.22452
81095087163.42525
44284338288.751495
24182754914.863205
13205698851.345032
7211357133.063099
3937971953.319482
2150444475.1502457
1174312944.7141201
641267843.954353
350183015.1331379
191227651.97698644
104425438.41462004
57024557.250736624
31139923.17400003
17004863.5541155
9286001.8593598
5070892.233720066
2769108.6471282765
1512152.5653041312
825755.0252234432
450927.6228649774
246242.18424542813
134467.72880470785
73430.02639984165
40098.60823121647
21897.014898581714
11957.503829158799
Number of days: 220

Question 3¶

We can calculate how many digits there are in a number by repeatedly dividing by 10 until the number is less than one.

Write peudo-code for this calculation.
Write a Python program to implement it, displaying the result in the format 1234 has 4 digits.
Test your program with a variety of numbers.
Extend your program so that it works for negative numbers.

Pseudocode.

This is just an example. It is most important that it makes sense to you. Notice that each line corresponds to one line of Python.

set value of x
set y to value of x
set i = 0
repeat until x less than 1:
    divide x 10
    increase i by 1
display value of y and i

# 2.

x = 1234
y = x
i = 0
while x > 1:
    x = x / 10
    i += 1
print(y, "has", i, "digits")

#3.

# Change the value of x in the code above and test that the result is correct.
# It's a good idea to test 'edge cases' such as 1000, 99999

#4.

# The simplest way to do this is to multiply x by -1 if it is negative.
# This is quite a clever 'trick' that allows us to re-use our existing code!

x = -1234.
y = x
if x < 0:
    x = x * -1
    
i = 0
while x > 1:
    x = x / 10
    i += 1
print(y, "has", i, "digits")

1234 has 4 digits
-1234.0 has 4 digits

Question 4 (The Collatz Problem)¶

Given a positive integer n, the Collatz Operation is defined as follows:

If the number is even, divide it by two
If the number is odd, triple it and add one

Repeatedly applying the Collatz Operation results in a sequence of integers which eventually reaches the number 1.

Write a program which generates the Collatz Sequence for a given positive integer \(n\).

For example, if \(n = 5\) then your program should produce the following:

Given a positive integer \(n\), define the Collatz Number to be the length of the Collatz Sequence for \(n\). For example, the Collatz number of 5 is 6.

Write a program which calculates and prints the Collatz number of a given number \(n\). For example, if \(n=5\) your program should print The Collatz number of 5 is 6.

Write a program which finds the smallest number \(n\) whose Collatz Number is greater than 200.

# Despite a completely different problem, we can use a very similar pattern that 
# we used in the previous two questions. 

n = 5
m = n
i = 1
while n > 1:
    if n % 2 == 0:
        n = n // 2
    else:
        n = 3*n + 1
    i += 1 
print("The Collatz Number of", m, "is", i)

The Collatz Number of 5 is 6

# The second part is tricky because we need to repeat the previous
# code for sequential numbers. But we already know how to do this:
# using while loops. Thus, we have a while loop within a while loop.
# The outer one terminates when we reach a collatz number greater than 200.

# Keeping track of the various variables is difficult and it might take some
# perseverence to get this right. Using a function to encapsulate the inner loop
# might make it easier - but we haven't studied functions yet.

n = 1
i = 1
while i <= 200: # while i is less than *or equal to* 200
    i = 1
    m = n
    while m > 1:
        if m % 2 == 0:
            m = m // 2
        else:
            m = 3 * m + 1
        i += 1

    print(n, i)
    n = n + 1
    
print("The Collatz Number of ", n - 1, "is", i) #note that we have to subract 1 from n here

20
165
59
59
59
59
59
72
20
121
20
165
72
72
72
121
20
20
59
59
20
20
72
72
46
165
46
121
46
46
165
165
28
46
134
134
134
134
134
134
41
72
41
46
41
41
134
134
41
134
134
134
41
41
134
134
41
134
41
103
72
72
72
209
The Collatz Number of  2463

 is 209

Scientific Computing Answers

Tutorial 1¶

Question 1¶

Question 2¶

Question 3¶

Question 4 (The Collatz Problem)¶