How to calculate distance to point when there is an obstacle in the way?
I'm writing a genetic algorithm program where "cells" have to make their way to a point that is located behind an obstacle that they have to get to. The structure looks something like this
target
---------------------
cell pool
where the cells come from the pool and have to get around the obstacle and make their way to the target. My question is how I should go about calculating the distance from a cell to the target? I need this because I am going to dynamically color the closest cell each frame. My guess would be to essentially find the closest edge of the obstacle (left or right side) and draw a line to that, then bee line it to the target, and add the lengths of those two lines together, but I was wondering if there was a more optimal way.
Thank you for your help!
geometry trigonometry
asked Nov 28 at 17:41
Tristin Baker
61
|
show 6 more comments
I'm writing a genetic algorithm program where "cells" have to make their way to a point that is located behind an obstacle that they have to get to. The structure looks something like this
target
---------------------
cell pool
where the cells come from the pool and have to get around the obstacle and make their way to the target. My question is how I should go about calculating the distance from a cell to the target? I need this because I am going to dynamically color the closest cell each frame. My guess would be to essentially find the closest edge of the obstacle (left or right side) and draw a line to that, then bee line it to the target, and add the lengths of those two lines together, but I was wondering if there was a more optimal way.
Thank you for your help!
geometry trigonometry
asked Nov 28 at 17:41
Tristin Baker
61
If you know the coordinates of the target, the cell, and the obstacle, you can stitch together two straight lines to get the distance, right? Or do you have more complicated obstacles in the way?
– Adrian Keister
Nov 28 at 17:43
Right now it's static, but the obstacle will move left to right eventually. Probably should have stated that in the OP, my apologies.
– Tristin Baker
Nov 28 at 17:45
I assume that that doesn't change the calculation now that I think of it though, because I'll still have all that info frame-by-frame.
– Tristin Baker
Nov 28 at 17:46
Ok, so I would just make two calculations. 1. $d_1=$ distance from cell to LHS of obstacle + distance from LHS of obstacle to target. 2. $d_2=$ distance from cell to RHS of obstacle + distance from RHS of obstacle to target. Then you would write $d=min(d_1, d_2).$
– Adrian Keister
Nov 28 at 17:51
Appreciate the help @AdrianKeister!
– Tristin Baker
Nov 28 at 17:53
|
show 6 more comments
I'm writing a genetic algorithm program where "cells" have to make their way to a point that is located behind an obstacle that they have to get to. The structure looks something like this
target
---------------------
cell pool
where the cells come from the pool and have to get around the obstacle and make their way to the target. My question is how I should go about calculating the distance from a cell to the target? I need this because I am going to dynamically color the closest cell each frame. My guess would be to essentially find the closest edge of the obstacle (left or right side) and draw a line to that, then bee line it to the target, and add the lengths of those two lines together, but I was wondering if there was a more optimal way.
Thank you for your help!
geometry trigonometry
asked Nov 28 at 17:41
Tristin Baker
61
I'm writing a genetic algorithm program where "cells" have to make their way to a point that is located behind an obstacle that they have to get to. The structure looks something like this
target
---------------------
cell pool
where the cells come from the pool and have to get around the obstacle and make their way to the target. My question is how I should go about calculating the distance from a cell to the target? I need this because I am going to dynamically color the closest cell each frame. My guess would be to essentially find the closest edge of the obstacle (left or right side) and draw a line to that, then bee line it to the target, and add the lengths of those two lines together, but I was wondering if there was a more optimal way.
Thank you for your help!
geometry trigonometry
geometry trigonometry
asked Nov 28 at 17:41
Tristin Baker
61
asked Nov 28 at 17:41
Tristin Baker
61
asked Nov 28 at 17:41
Tristin Baker
61
asked Nov 28 at 17:41
Tristin Baker
61
asked Nov 28 at 17:41
Tristin Baker
61
61
If you know the coordinates of the target, the cell, and the obstacle, you can stitch together two straight lines to get the distance, right? Or do you have more complicated obstacles in the way?
– Adrian Keister
Nov 28 at 17:43
Right now it's static, but the obstacle will move left to right eventually. Probably should have stated that in the OP, my apologies.
– Tristin Baker
Nov 28 at 17:45
I assume that that doesn't change the calculation now that I think of it though, because I'll still have all that info frame-by-frame.
– Tristin Baker
Nov 28 at 17:46
Ok, so I would just make two calculations. 1. $d_1=$ distance from cell to LHS of obstacle + distance from LHS of obstacle to target. 2. $d_2=$ distance from cell to RHS of obstacle + distance from RHS of obstacle to target. Then you would write $d=min(d_1, d_2).$
– Adrian Keister
Nov 28 at 17:51
Appreciate the help @AdrianKeister!
– Tristin Baker
Nov 28 at 17:53
|
show 6 more comments
If you know the coordinates of the target, the cell, and the obstacle, you can stitch together two straight lines to get the distance, right? Or do you have more complicated obstacles in the way?
– Adrian Keister
Nov 28 at 17:43
Right now it's static, but the obstacle will move left to right eventually. Probably should have stated that in the OP, my apologies.
– Tristin Baker
Nov 28 at 17:45
I assume that that doesn't change the calculation now that I think of it though, because I'll still have all that info frame-by-frame.
– Tristin Baker
Nov 28 at 17:46
Ok, so I would just make two calculations. 1. $d_1=$ distance from cell to LHS of obstacle + distance from LHS of obstacle to target. 2. $d_2=$ distance from cell to RHS of obstacle + distance from RHS of obstacle to target. Then you would write $d=min(d_1, d_2).$
– Adrian Keister
Nov 28 at 17:51
Appreciate the help @AdrianKeister!
– Tristin Baker
Nov 28 at 17:53
If you know the coordinates of the target, the cell, and the obstacle, you can stitch together two straight lines to get the distance, right? Or do you have more complicated obstacles in the way?
– Adrian Keister
Nov 28 at 17:43
If you know the coordinates of the target, the cell, and the obstacle, you can stitch together two straight lines to get the distance, right? Or do you have more complicated obstacles in the way?
– Adrian Keister
Nov 28 at 17:43
Right now it's static, but the obstacle will move left to right eventually. Probably should have stated that in the OP, my apologies.
– Tristin Baker
Nov 28 at 17:45
Right now it's static, but the obstacle will move left to right eventually. Probably should have stated that in the OP, my apologies.
– Tristin Baker
Nov 28 at 17:45
I assume that that doesn't change the calculation now that I think of it though, because I'll still have all that info frame-by-frame.
– Tristin Baker
Nov 28 at 17:46
I assume that that doesn't change the calculation now that I think of it though, because I'll still have all that info frame-by-frame.
– Tristin Baker
Nov 28 at 17:46
Ok, so I would just make two calculations. 1. $d_1=$ distance from cell to LHS of obstacle + distance from LHS of obstacle to target. 2. $d_2=$ distance from cell to RHS of obstacle + distance from RHS of obstacle to target. Then you would write $d=min(d_1, d_2).$
– Adrian Keister
Nov 28 at 17:51
Ok, so I would just make two calculations. 1. $d_1=$ distance from cell to LHS of obstacle + distance from LHS of obstacle to target. 2. $d_2=$ distance from cell to RHS of obstacle + distance from RHS of obstacle to target. Then you would write $d=min(d_1, d_2).$
– Adrian Keister
Nov 28 at 17:51
Appreciate the help @AdrianKeister!
– Tristin Baker
Nov 28 at 17:53
Appreciate the help @AdrianKeister!
– Tristin Baker
Nov 28 at 17:53
|
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If you know the coordinates of the target, the cell, and the obstacle, you can stitch together two straight lines to get the distance, right? Or do you have more complicated obstacles in the way?
– Adrian Keister
Nov 28 at 17:43
Right now it's static, but the obstacle will move left to right eventually. Probably should have stated that in the OP, my apologies.
– Tristin Baker
Nov 28 at 17:45
I assume that that doesn't change the calculation now that I think of it though, because I'll still have all that info frame-by-frame.
– Tristin Baker
Nov 28 at 17:46
Ok, so I would just make two calculations. 1. $d_1=$ distance from cell to LHS of obstacle + distance from LHS of obstacle to target. 2. $d_2=$ distance from cell to RHS of obstacle + distance from RHS of obstacle to target. Then you would write $d=min(d_1, d_2).$
– Adrian Keister
Nov 28 at 17:51
Appreciate the help @AdrianKeister!
– Tristin Baker
Nov 28 at 17:53