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Demonstration of Jupyterlab Solutions

Last updated: January 8th, 20192019-01-08Project preview

This is a quick demonstration of RMOTR Jupyterlab Solutions, an open source extension that implements visibility toggle for cells in Jupyter Lab. It's similar to previous Jupyter Notebook attempts like Exercise and Exercise2.

Important: The following demonstration assumes you're inside a JupyterLab, fork this project with the button at the right to see it in action ūüĎČ

Overview of functionality¶

The best way to demonstrate how this extension work, is with a real example of one of the coding activities we propose to our students, so let's get started:

Palindromic Primes¶

A palindromic number is a number that reads the same forwards and backwards. Example: "121" (You read: "one, two, one", forwards and backwards). "1001" ("one, zero, zero, one"), etc.

The objective of this exercise is to write a function next_palindromic_prime that receives a number and returns the closest larger number that's both a prime and palindromic.

You could take a stab at next_palindromic_prime right now, but this is a good example of "divide and conquer". We'll divide our bigger problem into smaller ones and resolve them separately. We'll divide it into 2 major parts:

  • Is palindromic? Figuring out if a number is palindromic
  • Is prime? Figuring out if a number is prime.

With these two parts working, next_palindromic_prime would be greatly simplified. Here's some pseudocode:

def next_palindromic_prime(n):
    repeat to infinite:
        n += 1
        if is_palindromic(n) and is_prime(n):
            return n

Is Palindromic¶

We'll start with a function is_palindromic, that given a number, returns True or False depending if the number is palindromic or not.

Let's start with a couple of examples:

In [2]:
n1 = 1001  # is palindromic
n2 = 1234  # is NOT palindromic
In [23]:
print(f"Our Numbers: {n1}, {n2}")
Our Numbers: 1001, 1234

As we said, a Palindromic number is one that looks the same "forwards and backwards", that means that for example, for n2 we'd have:

Forward:    1234
Backwwards: 4321

So, if we can just compare the original number to its "reversed" version, we can answer if n is palindromic or not. So, with all this said: how can you reverse a string?. Try it out below:

In [ ]:
# reverse n1. If you're not sure about it click on "Reveal Solution"
In [3]:
str(n1)[::-1]
Out[3]:
'1001'
In [ ]:
# reverse n2. If you're not sure about it click on "Reveal Solution"
In [4]:
str(n2)[::-1]
Out[4]:
'4321'

Here's the explanation of it:

First of all, the data type of our n1 and n2 variables is int, which "can't" be reversed. When we talk about "reversing 1234" we're actually talking about reversing the digits, or characters of 1234, so we must first turn 1234 into a string:

str(n1)  # "1234", please note the quotes

Once the number is transformed into a string, we can now "reverse" it with a quick Python slicing hack involving the -1 step:

str(n1)[::-1]  # "4321"

So now we can write our is_palindromic function using this hack, give it a try and check the solution if it doesn't work. I've included a couple of tests for you to verify if it works:

In [5]:
def is_palindromic(a_number):
    # your code here
    pass
In [7]:
assert is_palindromic(1001) is True
assert is_palindromic(1234) is False
assert is_palindromic(121) is True
assert is_palindromic(1) is True
In [6]:
def is_palindromic(a_number):
    return str(a_number) == str(a_number)[::-1]

Is Prime¶

You're probably bored already of solving is_prime functions, so we won't get into much detail; I've added a couple of tests for you to try your code and the solution is also available:

In [10]:
def is_prime(a_number):
    # your code here
    pass
In [16]:
# Primes
assert is_prime(3) is True
assert is_prime(17) is True
assert is_prime(97) is True
assert is_prime(7919) is True

# Not Primes
assert is_prime(4) is False
assert is_prime(8) is False
assert is_prime(15) is False
assert is_prime(7833) is False
In [15]:
def is_prime(a_number):
    if a_number <= 1:
        return False
    if a_number in {2, 3}:
        return True

    for divisor in range(2, a_number - 1):
        if a_number % divisor == 0:
            return False
    return True

Finally, next palindromic prime¶

We're now ready to resolve the next_palindromic_prime, give it a shot!

In [ ]:
def next_palindromic_prime(n):
    # your code here
    pass
In [21]:
assert next_palindromic_prime(100) == 101
assert next_palindromic_prime(150) == 151
assert next_palindromic_prime(160) == 181
assert next_palindromic_prime(200) == 313
assert next_palindromic_prime(10000) == 10301
assert next_palindromic_prime(12000) == 12421

Check the solution if you want to see how we solved it:

In [20]:
import itertools
def next_palindromic_prime(n):
    for next_num in itertools.count(n + 1):
        if is_palindromic(next_num) and is_prime(next_num):
            return next_num

And here's a quick explanation about it:

This is one of those times when you have to repeat "to the infinite". You have to keep repeating and looking for a number that is both palindromic and prime. To do that, we could have used a while loop, something like:

while True:
    n += 1
    if is_palindromic(n) and is_prime(next):
        return n

This is of course less "pythonic" (and even dangerous, if we forget n += 1). So we use itertools.count: this method just keeps incrementing infinitely, which is exactly what we need.

The rest is just what we've discussed so far; once you find a number that satisfices both conditions (prime and palindromic) you have found your answer.

Final Notes about Jupyterlab Solutions¶

Hopefully, this example has demonstrated the capabilities of the Jupyterlab Solutions extension. If you want to install it in your own computer, just follow the steps on the [docs].

Do you need help or have found a problem? Create a new issue.

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