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Author Topic: LUT: Slope between 2 points -  (Read 548 times) Bookmark and Share
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Hot_Dog
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« on: 21 December, 2011, 01:50:17 »
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For Elimination, let's say I have a playerX and a playerY, and an EnemyX and an EnemyY.  I'm wanting to do something similar to a Bresenham line, but to do that I need to know the slope between the two locations, of the player and the enemy.  I'm assuming a LUT is my best bet, where the LUT will have values for rise and run.  That is, go up/down this many times, and left/right this many times. 

How do I go about implementing this?  What I need to know is given (PlayerX, PlayerY) and (EnemyX, EnemyY), what do I need to do to find the correct value in the LUT?  I want my LUT to have no more than 64 sets of rise over run.
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« Reply #1 on: 21 December, 2011, 02:30:05 »
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I'm not sure I exactly understand what exactly you are trying to do, but why couldn't you just divide your Y's by your X's. That's what you do in math and division is relatively quick. If you need decimal places, you can just tack on an extra byte to your Y values and your final result will have 1 extra byte of decimals.
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« Reply #2 on: 21 December, 2011, 02:36:15 »
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Quote from: thepenguin77 on 21 December, 2011, 02:30:05
I'm not sure I exactly understand what exactly you are trying to do, but why couldn't you just divide your Y's by your X's. That's what you do in math and division is relatively quick. If you need decimal places, you can just tack on an extra byte to your Y values and your final result will have 1 extra byte of decimals.

It is relatively quick, but I was hoping for something slightly faster.  Still, I guess I don't have to be 100% optimized Grin
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« Reply #3 on: 21 December, 2011, 03:22:47 »
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What exactly do you need the slope for? Because line drawing algorithms, for instance, don't actually need a numerical slope value. They just use dy and dx as-is without having to divide one by the other.
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« Reply #4 on: 21 December, 2011, 03:27:14 »
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Quote from: Runer112 on 21 December, 2011, 03:22:47
What exactly do you need the slope for? Because line drawing algorithms, for instance, don't actually need a numerical slope value. They just use dy and dx as-is without having to divide one by the other.

I'm not really dividing dy and dx, I need to find dy and dx.  (y2-y1) / (x2-x1) The slope is used to start at the enemy, and travel little by little to the player to see if there's a wall in the way.
« Last Edit: 21 December, 2011, 03:27:55 by Hot_Dog » Logged
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« Reply #5 on: 21 December, 2011, 03:30:35 »
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So you are looking to see if the player can see the enemy or vica verse?
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« Reply #6 on: 21 December, 2011, 03:36:40 »
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Quote from: Builderboy on 21 December, 2011, 03:30:35
So you are looking to see if the player can see the enemy or vica verse?

Yes, or more importantly, if an enemy can fire a shot without it being blocked by a wall. 

I apologize for not being too clear on this topic.  Let's say that the slope between the enemy and the player is 2/1.  The enemy is at (100, 100), and the player is at (103, 106).  So the calculator checks (101, 102).  Is there a wall?  No.  Check (102, 104).  Is there a wall?  YES!  So the enemy cannot harm the player.
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« Reply #7 on: 21 December, 2011, 04:23:33 »
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What, finding slope? Doesn't (y2-y1)/(x2-x1) find the slope?
« Last Edit: 21 December, 2011, 04:24:13 by epic7 » Logged


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« Reply #8 on: 21 December, 2011, 04:46:38 »
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So just use a line-drawing algorithm with pixel-test instead of pixel-on?
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« Reply #9 on: 21 December, 2011, 06:18:59 »
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Quote from: calc84maniac on 21 December, 2011, 04:46:38
So just use a line-drawing algorithm with pixel-test instead of pixel-on?

The game engine is not built that way.  I'm really sorry that I have been confusing.  I would start over and explain from the beginning, except that I like thepenguin77's suggestion for division, so I'm going to use that.
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