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* chore: Switch to Node 20 + Vitest * chore: migrate to vitest mock functions * chore: code style (switch to prettier) * test: re-enable long-running test Seems the switch to Node 20 and Vitest has vastly improved the code's and / or the test's runtime! see #1193 * chore: code style * chore: fix failing tests * Updated Documentation in README.md * Update contribution guidelines to state usage of Prettier * fix: set prettier printWidth back to 80 * chore: apply updated code style automatically * fix: set prettier line endings to lf again * chore: apply updated code style automatically --------- Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com> Co-authored-by: Lars Müller <34514239+appgurueu@users.noreply.github.com>
54 lines
1.7 KiB
JavaScript
54 lines
1.7 KiB
JavaScript
/**
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* Problem 28 - Number spiral diagonals
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*
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* @see {@link https://projecteuler.net/problem=28}
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*
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* Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
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*
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* 21 22 23 24 25
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* 20 07 08 09 10
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* 19 06 01 02 11
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* 18 05 04 03 12
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* 17 16 15 14 13
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*
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* It can be verified that the sum of the numbers on the diagonals is 101.
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* What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?
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*
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* @author ddaniel27
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*/
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function problem28(dim) {
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if (dim % 2 === 0) {
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throw new Error('Dimension must be odd')
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}
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if (dim < 1) {
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throw new Error('Dimension must be positive')
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}
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let result = 1
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for (let i = 3; i <= dim; i += 2) {
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/**
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* Adding more dimensions to the matrix, we will find at the top-right corner the follow sequence:
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* 01, 09, 25, 49, 81, 121, 169, ...
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* So this can be expressed as:
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* i^2, where i is all odd numbers
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*
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* Also, we can know which numbers are in each corner dimension
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* Just develop the sequence counter clockwise from top-right corner like this:
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* First corner: i^2
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* Second corner: i^2 - (i - 1) | The "i - 1" is the distance between corners in each dimension
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* Third corner: i^2 - 2 * (i - 1)
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* Fourth corner: i^2 - 3 * (i - 1)
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*
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* Doing the sum of each corner and simplifying, we found that the result for each dimension is:
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* sumDim = 4 * i^2 + 6 * (1 - i)
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*
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* In this case I skip the 1x1 dim matrix because is trivial, that's why I start in a 3x3 matrix
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*/
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result += 4 * i * i + 6 * (1 - i) // Calculate sum of each dimension corner
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}
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return result
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}
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export { problem28 }
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