Sections 9-11
- Determine the outcome of algorithms
- Determine the outcome of similar algorithms
- Editing Algorithms
- Developing Algorithms
- Homework
- Challenges and Homework
Algorithms can be written in different ways and still accomplish the same tasks. Algorithms that look similar often yield differnet outputs. To solve the same problem, many different algorithms can be used.
Therefore, algorithms are very important for programmers, and today we're going to explore how to determine the outcome of algorithms, how to deteremine the output of similar algorithms, how to edit existing algorithms, and how to develop our own algorithms.
Consider the following algorithm.
def mystery(num, num2):
if (num % num2 == 0):
print("True")
else:
print("False")
mystery(20, 4)
- What does the algorithm do? Please explain in words.
The algorithm checks if num is divisble my num2. If it is and there is no remainder, the output is True, and if not, the output is False. In this case, the output is true becasue 20 is divisible by 4.
- What if I put in 30 as num and 4 as num2. What would be the output?
The output would be False. 30 is not divisble by 4, as it is 7 with a remainder 2.
What is the output of this algorithm?
temp = 95
if (temp >= 90):
print("it is too hot outside")
else:
if (temp >= 65):
print("I will go outside")
else:
print("it is too cold outside")
What is the output of this algorithm? it looks similar but the output is different!
Edited code:
temp = 95
if (temp >= 90):
print("it is too hot outside")
elif (temp >= 65):
print("i will go outside")
else:
print("it is too cold outside")
To develop algorithms, we first need to understand what the question is asking. Then, think about how you would approach it as a human and then try to find what pattern you went through to arrive at the answer. Apply this to code, and there you have it! An algorithm!
Let's say you wanted to sum up the first five integers. How would you do this in real life? Your thought process would probably be:
- The sum of the first integer is 1.
- Then, increase that integer by 1. I now add 2 to my existing sum (1). My new sum is 3.
- Repeat until I add 5 to my sum. The resulting sum is 15.
Now let's translate this into code.
sum = 0 # start with a sum of 0
for i in range (1, 6): # you will repeat the process five times for integers 1-5
sum = sum + i # add the number to your sum
print(sum) # print the result
Task: Write an algorithm in python that sums up the first 5 odd integers. You can use the following code as a starter.
Done with step in range function
sum = 0
counter = 2
for i in range (1, 10, counter):
sum = sum + i
print(sum)
Homework
Create an algorithm that will start with any positive integer n and display the full sequence of numbers that result from the Collatz Conjecture. The COllatz Conjecture is as follows:
- start with any positive integer
- if the number is even, divide by 2
- if the number is odd, multiply by 3 and add 1
- repeat steps 2 and 3 until you reach 1
Example: if the starting number was 6, the output would be 6, 3, 10, 5, 16, 8, 4, 2, 1
x = int(input("Enter any positive integer:"))
print(x)
def collatz(x):
while x != 1:
if x % 2 == 0:
x = int(x/2)
else:
x = x * 3 + 1
print(x)
collatz(x)
Challenges and Homework
You have one homework problem.
Yes just one.
Don't get excited though.
Problem: Given a specific integer N, return the square root of N (R) if N is a perfect square, otherwise, return the square root of N rounded down to the nearest integer
Input: N (Integer)
Output: R (Integer)
Constraints: Do not use any built-in math operations such as sqrt(x) or x**(0.5), Try complete the problem in logarithmic time.
Hint 1: Maybe you can use Binary Search to try and reduce the number of checks you have to perform?
Hint 2: Is there a mathematical pattern amongst numbers and their square roots that could help you reduce the number of searches or iterations you must execute? Is there some value or rule you can set before applying binary search to narrow the range of possible values?
def findSqrt(num):
if num >= 1:
lower_bound=0
upper_bound=num
else:
lower_bound=num
upper_bound=1
temp=0;
err = 0.000001
while(abs(num - (temp * temp)) > err):
temp = (lower_bound+upper_bound)/2
temp0 = round(temp)
if temp0*temp0 == num:
return temp0
if temp*temp >= num:
upper_bound = temp
elif temp*temp < num:
lower_bound = temp
return int(temp) # Rounding down to nearest int
print("Sqrt: " + str(findSqrt(49)))
print("Sqrt: " + str(findSqrt(8)))
# Extra for exact value, not rounded
def exactSqrt(num):
if num >= 1:
lower_bound=0
upper_bound=num
else:
lower_bound=num
upper_bound=1
temp=0;
err = 0.000001
while(abs(num - (temp * temp)) > err):
temp = (lower_bound+upper_bound)/2
temp0 = round(temp)
if temp0*temp0 == num:
return temp0
if temp*temp >= num:
upper_bound = temp
elif temp*temp < num:
lower_bound = temp
return temp
print("Sqrt: " + str(exactSqrt(8)))
from math import sqrt as sq
test_cases = [0,1,4,85248289,22297284,18939904,91107025,69122596,9721924,37810201,1893294144,8722812816,644398225]
answers = [int(sq(x)) for x in test_cases]
def checkValid():
for i in range(len(test_cases)):
if findSqrt(test_cases[i]) == answers[i] and exactSqrt(test_cases[i]) == answers[i]:
print("Check number {} passed".format(i+1))
else:
print("Check number {} failed".format(i+1))
checkValid()