Forming a magic square hackerrank solution in python

Forming a magic square hackerrank solution in python

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Solution — HackerRank: Array Manipulation

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Sign up. Branch: master. Go back. Launching Xcode If nothing happens, download Xcode and try again. Latest commit. Naklecha committed ddbe Oct 12, Git stats 73 commits 1 branch 0 tags. Failed to load latest commit information. View code. Python is a very easy-to-read language, hence, the solution can be easily understood by anyone.

Unsolved Questions None All questions in the implementation category has been solved. Author Email: nishant. Releases No releases published. You signed in with another tab or window. Reload to refresh your session. You signed out in another tab or window.The constant sum in every row, column and diagonal is called the magic constant or magic sum, M. A bit of Theory: Magic squares are divided into three major categories depending upon order of square.

Example : 4,8,12,16. Example : 6,10,14,18. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. Become industry ready at a student-friendly price. Writing code in comment? Please use ide.

Examples: Magic Square of order 3: 2 7 6 9 5 1 4 3 8 Magic Square of order 4: 16 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1 Magic Square of order 8: 64 63 3 4 5 6 58 57 56 55 11 12 13 14 50 49 17 18 46 45 44 43 23 24 25 26 38 37 36 35 31 32 33 34 30 29 28 27 39 40 41 42 22 21 20 19 47 48 16 15 51 52 53 54 10 9 8 7 59 60 61 62 2 1 A bit of Theory: Magic squares are divided into three major categories depending upon order of square. Python program to print magic square of double order. Change value of array elements at fix location.

Top left corner. Top right corner. Bottom Left corner.

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Bottom Right corner. Printing the square. Contributed by Harshit Agrawal. WriteLine. Load Comments. We use cookies to ensure you have the best browsing experience on our website.GitHub is home to over 50 million developers working together to host and review code, manage projects, and build software together.

If nothing happens, download GitHub Desktop and try again. If nothing happens, download Xcode and try again. If nothing happens, download the GitHub extension for Visual Studio and try again. Solutions and explanations to HackerRank problems implemented in Java 8 and Python 3.

Skip to content. Solutions to HackerRank problems implemented in Java 8 and Python 3. MIT License. Dismiss Join GitHub today GitHub is home to over 50 million developers working together to host and review code, manage projects, and build software together. Sign up. Branch: master. Go back. Launching Xcode If nothing happens, download Xcode and try again.

How to perform Magic Square Operation in a Matrix using Python3

Latest commit. Git stats 51 commits 1 branch 0 tags. Failed to load latest commit information. Update solution. Apr 29, Added license. Mar 5, Apr 18, View code. Topics hackerrank hackerrank-solutions. Releases No releases published.I must admit that this problem actually took me a good while to solve. Once you have the right insight on forming a magic Square it is really straight forward. But until that point I was just stuck. Anyway, before rambling on lets get to the actual problem.

Then print this cost on a new line. I started out with the idea that I could go through the square and change the number one by one based on an evaluation of the individual digits.

Back to the drawing board. I thought of checking all combinations but on the outset there was 9! What did I know about forming a magic square? Possibly something I could brute force. However, I knew that not all of those solutions would be valid. As an example if I put 1 in the upper left corner I could not put 2 and 3 in the top row.

forming a magic square hackerrank solution in python

Which meant I had to consult google. It turns out there are only 8 possible solutions to the problem as given on the blog post on mindyourdecisions. That should be rather easy to bruteforce.

Just check the difference between the numbers in the given input and the 8 possible solutions and then take the smallest of those. As you can see I start by reading the input and making it into a 1 dimensional array. After that I hardcoded all 8 solutions. And then I just find the difference of each solution and the input and print the minimum.

It might be possible to do it in less lines of code, but this one worked out nicely. This is probably a dumb question and it has been a while for me with math. How did you even start configuring the number of combinations?GitHub is home to over 50 million developers working together to host and review code, manage projects, and build software together. If nothing happens, download GitHub Desktop and try again.

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Skip to content. Hackerrank Solutions for Python - Total Challenges 3 stars 3 forks.

Forming a Magic Square

Dismiss Join GitHub today GitHub is home to over 50 million developers working together to host and review code, manage projects, and build software together. Sign up. Branch: master. Go back. Launching Xcode If nothing happens, download Xcode and try again. Latest commit. Git stats 63 commits 2 branches 0 tags. Failed to load latest commit information. View code. The Minion Game Merge the Tools itertools. OrderedDict itertools.

CompanyLogo Piling Up! About Hackerrank Solutions for Python - Total Challenges Topics python python2 hackerrank-solutions hackerrank-challenges hackerrank hackerrank-python python3 string strings-manipulation searching-algorithms sorting-algorithms. Releases No releases published. You signed in with another tab or window.

forming a magic square hackerrank solution in python

Reload to refresh your session. You signed out in another tab or window.GitHub is home to over 50 million developers working together to host and review code, manage projects, and build software together. If nothing happens, download GitHub Desktop and try again. If nothing happens, download Xcode and try again. If nothing happens, download the GitHub extension for Visual Studio and try again. Solutions and explanations to HackerRank problems implemented in Java 8 and Python 3.

JAVA Random Magic square matrix

Skip to content. Solutions to HackerRank problems implemented in Java 8 and Python 3. MIT License.

forming a magic square hackerrank solution in python

Dismiss Join GitHub today GitHub is home to over 50 million developers working together to host and review code, manage projects, and build software together. Sign up. Branch: master. Go back. Launching Xcode If nothing happens, download Xcode and try again. Latest commit.

Git stats 51 commits 1 branch 0 tags. Failed to load latest commit information. Update solution. Apr 29, Added license. Mar 5, Apr 18, View code.

Topics hackerrank hackerrank-solutions. Releases No releases published.Please Login in order to post a comment. I am not sure that this particular challenge should be listed under "easy" as written. All the other challenges up to this point have been straightforward "in the head" problems. This one requires quite a bit more set-up. The "easy" portion of this task is to evaluate cost as defined is simply comparing lists of lists in python, anyway.

Generating the matrix of possible magic squares is far more time-consuming than anything up to this point, and might best have been it's own "Generate 3x3 Magic Squares" challenge, with "Magic Square Forming" following directly after.

Were I to see this particular excercise in a "team-coding" evaluation for a job interview, my first suggestion to my partner would be to google for the list of possible 3x3 magic squares and just use that as a look up table, since the boundaries are well understood and fast and lazy will always win!

With Python, a brute force way to generate all magic squares is to generate all possible permutations of 1 to 9 with itertools. I added the sums to a set then checked if the length of the set was 1 to determine if it was a magic square. Sure, you could add some intelligence when generating all legal permutations. For example, once you lock in a certain row, you have knowledge on permissible permutations thereafter. However, this is all besides the point.

You are right in that the scaling is O 9! If the constraints were different, then it would in fact be O N! So generating all of them easily avoids TLE. That is why this problem is marked easy. By checking that the rows equal 15 as they're built out I was able to complete all possible permutations in runs.

I posted that solution elsewhere in this thread. It helps to know from wiki [1] and YouTube video [2] there are 8 possible 3x3 magic squares. Each can be obtained by a rotation or reflection or combination of of each other. If you define a starting magic square, you can perform the following relatively simple transormations to obtain all 8 squares:.

You have to define these functions, but atleast they're quite simple and you don't have to search all possible permutations. Hey Chris, Thank you for posting this, it helped me a lot while I was tackling this problem.

However, I found a less complicated way to solve the problem without diagonal column-wise and diagonal reflection. No need for such brute force for 3x3 squares! It's easy to create one and then get the others by reflection knowing the middle must be a 5, the corners some even numbers, and outer middles odd numbers.

forming a magic square hackerrank solution in python

It takes 5 min of thinking then writing them down on paper :. I generated all possible magic squares by bounding what the sums could be. From here you can narrow down the possible sums since you need at least 6 distinct sets of numbers from that add up to the given sum e. From there I just tried to eliminate sums where the sets couldn't be assembled into a magic square.

Basically it ends up that 15 is the only possible sum. The question never mentions that each digit must only appear once. I hadn't heard of Magic Squares before so I didn't make that assumption. In any magic square, the first number i. Let this position be i,j.


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