Convert to binary from base-10: Difference between revisions
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== The Problem == | == The Problem == | ||
Ask the user to enter a number in base 10. Then convert the number the user entered into binary. Extra points if you ensure the binary always has 8 digits | Ask the user to enter a number in base 10. Then convert the number the user entered into binary. Extra points if you ensure the binary always has 8 digits. You must not use built-in math functions, but rather construct the solution on your own. | ||
== Unit Tests == | == Unit Tests == | ||
Line 61: | Line 61: | ||
Click the expand link to see one possible solution, but NOT before you have tried and failed! | Click the expand link to see one possible solution, but NOT before you have tried and failed! | ||
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
This code was donated to our wiki by an 11th grade student, Jake. I am very grateful to him for the code. | |||
<syntaxhighlight lang="python" > | <syntaxhighlight lang="python" > | ||
not | #first off, we need something to convert (duh), so we are asking the user to feed a base 10 number to the program by using the input function | ||
number = int(input("What base 10 nr would you like to convert to binary? Please enter it below \n > ")) | |||
#setting an empty string. This is where the binary equivalent of the above input will be | |||
binary = "" | |||
#getting to the mathy part. This system was inspired by the tutorial over from https://www.electronics-tutorials.ws/binary/bin_2.html, where it tells you to just keep diving the number by 2 until you get it down to 0. | |||
while number != 0: | |||
#checking if the number has a remainder | |||
if number%2 == 1: | |||
#if the number DOES have a remainder, we need to a 1 in our binary string (cause that's the remainder) | |||
binary += "1" | |||
# rounding the number down because decimals would mess this up | |||
number = int(number/2) | |||
# checking if the number does not have a remainder | |||
elif number %2 == 0: | |||
#if the number DOESN'T have a remainder, we need to a 0 in our binary string (cause there's no remainder) | |||
binary += "0" | |||
# rounding the number down because decimals would mess this up | |||
number = int(number/2) | |||
#this part was grabbed over from https://www.educative.io/edpresso/how-do-you-reverse-a-string-in-python. These are the 2nd and 3rd from the last block of code on the page | |||
#this is needed because according to the tutorial, the result of the first time we divide the number (either a 0 or 1) is the "Least Significant Bit" and the result of the last time we divied is the "Most Significant Bit" | |||
#we gotta swap it because it's backwards right now. It's from LSB to MSB right now, and it needs to be from MSB to LSB | |||
reversed =''.join(reversed(binary)) | |||
#printing the result | |||
print(reversed) | |||
</syntaxhighlight> | </syntaxhighlight> |
Latest revision as of 20:25, 21 September 2020
This is a problem set. Some of these are easy, others are far more difficult. The purpose of these problems sets are:
- to build your skill applying computational thinking to a problem
- to assess your knowledge and skills of different programming practices
What is this problem set trying to do[edit]
The actual solution to this problem is fairly straight-forward, but imagining how to solve the problem is tricky. As a side note, you must understand how to convert base 10 to base 2.
The Problem[edit]
Ask the user to enter a number in base 10. Then convert the number the user entered into binary. Extra points if you ensure the binary always has 8 digits. You must not use built-in math functions, but rather construct the solution on your own.
Unit Tests[edit]
- User Input: 1
- Expected output: 1
- User Input: 4
- Expected output: 100
- User Input: 8
- Expected output: 1000
How you will be assessed[edit]
Your solution will be graded using the following axis:
Scope
- To what extent does your code implement the features required by our specification?
- To what extent is there evidence of effort?
Correctness
- To what extent did your code meet specifications?
- To what extent did your code meet unit tests?
- To what extent is your code free of bugs?
Design
- To what extent is your code written well (i.e. clearly, efficiently, elegantly, and/or logically)?
- To what extent is your code eliminating repetition?
- To what extent is your code using functions appropriately?
Style
- To what extent is your code readable?
- To what extent is your code commented?
- To what extent are your variables well named?
- To what extent do you adhere to style guide?
References[edit]
A possible solution[edit]
Click the expand link to see one possible solution, but NOT before you have tried and failed!
This code was donated to our wiki by an 11th grade student, Jake. I am very grateful to him for the code.
#first off, we need something to convert (duh), so we are asking the user to feed a base 10 number to the program by using the input function
number = int(input("What base 10 nr would you like to convert to binary? Please enter it below \n > "))
#setting an empty string. This is where the binary equivalent of the above input will be
binary = ""
#getting to the mathy part. This system was inspired by the tutorial over from https://www.electronics-tutorials.ws/binary/bin_2.html, where it tells you to just keep diving the number by 2 until you get it down to 0.
while number != 0:
#checking if the number has a remainder
if number%2 == 1:
#if the number DOES have a remainder, we need to a 1 in our binary string (cause that's the remainder)
binary += "1"
# rounding the number down because decimals would mess this up
number = int(number/2)
# checking if the number does not have a remainder
elif number %2 == 0:
#if the number DOESN'T have a remainder, we need to a 0 in our binary string (cause there's no remainder)
binary += "0"
# rounding the number down because decimals would mess this up
number = int(number/2)
#this part was grabbed over from https://www.educative.io/edpresso/how-do-you-reverse-a-string-in-python. These are the 2nd and 3rd from the last block of code on the page
#this is needed because according to the tutorial, the result of the first time we divide the number (either a 0 or 1) is the "Least Significant Bit" and the result of the last time we divied is the "Most Significant Bit"
#we gotta swap it because it's backwards right now. It's from LSB to MSB right now, and it needs to be from MSB to LSB
reversed =''.join(reversed(binary))
#printing the result
print(reversed)