Pythagorean theorem to check whether a cell is inside the radius

Question

0.00/5 (No votes)

See more:

I have the dimensions of a matrix and a given cell with radius.

So let's take a look at this example:
https://i.imgur.com/51hTJpv.png[Matrix]
The matrix has 5 rows and 6 columns. We are given the cell (2, 3) with radius 2. It has an impact, and it destroys all of the items in a certain radius (the impact cell is shaded black and the other cells within the radius are shaded grey). I found that I could use the Pythagorean Theorem to check whether a cell is inside the radius:

if (Math.Pow(targetRow - row, 2) + Math.Pow(targetColumn - col, 2) <= radius * radius)
{
    matrix[row, col] = 1; 
}

I don't understand why it works, and I would be very grateful if you could explain it to me. I tried to debug, but I still don't get it.

What I have tried:

if (Math.Pow(targetRow - row, 2) + Math.Pow(targetColumn - col, 2) <= radius * radius)
{
    matrix[row, col] = 1; 
}

Posted 22-May-19 7:48am

Member 14422952

Updated 22-May-19 8:07am

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

OriginalGriff · Accepted Answer · 2019-05-22T08:08:00

Solution 1

This is basic maths.
targetRow - row is the horizontal distance from the target to the centre of your circle with radius radius, and targetColumn - col is the vertical distance. The radius by definition never shorter than the longest of the absolute values of those two, so it is always the hypotenuse. Since the square of any number is always zero or positive, your squares of the other two sides are always positive, and Pythagoras tells us that root(x² + y²) is the distance from the centre to your point. If that is smaller than the radius, your point is inside the circle.

Posted 22-May-19 8:08am

OriginalGriff

Updated 22-May-19 9:45am

v2

Comments

Member 14422952 22-May-19 14:15pm

Thank you for your response! I really appreciate that. But where there are circles? This is what confuses me.

OriginalGriff 22-May-19 14:20pm

You have one circle, with radius radius, yes? And that the center where the "bomb" goes off, yes? Draw that, on squared background. Now add your point (x, y) to the diagram, and draw a circle with the same centre as your original one, and which intersects (x, y). That gives you the other radius, which if it's smaller than the original radius means (x, y) is inside your "blast radius".

Draw it on paper, and you'll see what I mean!

Member 14422952 22-May-19 15:09pm

But can't I calculate the distance between the points using this formula and compare it with the given radius? https://www.mathplanet.com/education/algebra-2/conic-sections/distance-between-two-points-and-the-midpoint

OriginalGriff 23-May-19 2:13am

That is what you are doing!

[no name] 22-May-19 14:46pm

Looks like <sup>xyz</sup> conflicts inside a <code> tag

a^xyz
vs.
a<sup>xyz</sup>

OriginalGriff 22-May-19 15:45pm

Fixed! Thanks!