Number spiral diagonals
Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?
You will have to analyze the question and find the pattern to solve this question. It took me a few hours to solve this problem but I was not able to post the solution these many days because I had some other important work which I had to complete.
My approach to solve this problem was to find the pattern and write a program which can generate the solution. We are not going to do any mathematics deriving any new equations to get the solution.
Please make sure that you will read all of this post so that you will understand the underlying concept perfectly. Remember that these patterns are only applicable for any nxn square matrix where n is odd number.
Consider the following matrices:
You can see that, if we draw lines connecting the outer elements then there are four numbers on the corner of each squares and also these numbers are part of diagonal elements. One more observation is that the numbers are all odd numbers.
Consider the 3x3 matrix in the first picture, we can see that the largest number in the matrix is 9(3x3), see the 5x5 matrix, the largest number in the matrix is 24(5x5) and similarly if you draw a nxn matrix then the largest number in the matrix is n2.
Keeping aside above two matrices see the following 7x7 matrix. I have highlighted all the diagonal elements.
21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
It can be verified that the sum of the numbers on the diagonals is 101.20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?
You will have to analyze the question and find the pattern to solve this question. It took me a few hours to solve this problem but I was not able to post the solution these many days because I had some other important work which I had to complete.
My approach to solve this problem was to find the pattern and write a program which can generate the solution. We are not going to do any mathematics deriving any new equations to get the solution.
Please make sure that you will read all of this post so that you will understand the underlying concept perfectly. Remember that these patterns are only applicable for any nxn square matrix where n is odd number.
Consider the following matrices:
 |
| 3x3 number spiral matrix |
 |
| 5x5 number spiral matrix |
Consider the 3x3 matrix in the first picture, we can see that the largest number in the matrix is 9(3x3), see the 5x5 matrix, the largest number in the matrix is 24(5x5) and similarly if you draw a nxn matrix then the largest number in the matrix is n2.
Keeping aside above two matrices see the following 7x7 matrix. I have highlighted all the diagonal elements.
 |
| 7x7 number spiral matrix |
I have chosen a larger matrix because I wanted to show you the pattern I found, which is very hard to show in case of a smaller matrix.
As of now according to our analysis we know that only odd numbers get the chance to be diagonal elements in the number spiral, for each and every square formed we have only four diagonal elements. So in the above matrix(7x7 matrix) we have 3 squares and 12 diagonal elements excluding 1.
Writing the odd numbers for 7x7 matrix:
1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49.
Highlighting the numbers which are diagonals in different squares in 7x7 matrix:
1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49.
Now if you have observed first square the values doesn't have a gap between the numbers. For the second square the numbers have a gap of 1 number between each. For the third square the numbers have a gap of 2 numbers between each. This continues upto infinity.
So if we can understand the pattern then we can write the diagonal elements for any size of number spiral matrix. Now for 9x9 matrix the diagonal elements will be(writing the values without actually creating the diagram):
1,3,5,7,9,11,13,15,17,19,21,23,,25,27,29,31,33,35,37,39,41,43,45,47,49,51, 53,55,57,59,61,63 ,65,67,69,71,73,75,77,79,81.
I think you have understood the pattern. To be simple the gap between the odd numbers increases for every four iterations. Four numbers because we can only have four corners for a square.
Now see the program and you will understand everything easily.
Program
According to the question, we know that the last number for our iteration will be 1002001. Also the number will be the fourth iteration in the outer square of the 1001x1001 matrix. So we can stop looping when we hit the number 1002001. The same thing has been accomplished in the program.
In solution we have added 1 already because it is an exception for our pattern😋.
You can download the source code form Github Gist pep28.py
Output
Summary
It took me some time to find the pattern in the question given, but I am really satisfied with the performance of the program. Few others have solved the problem by creating a formula but as I wanted to solve the problem writing a program I have not created any formula. Hope you have liked the solution.
If you haven't understood anything or have any doubt feel free to comment and I will reply to you as soon as possible.
Please feel free to comment if you have found a typo or have a better program or have a different solution or have any suggestion.
You can also contact me.
Thank you. Have a nice day😃.