Finding a second color knowing the distance
I'm doing a project where im finding the patterns of 1 color duo and trying to find the unkown second color of the second color duo.
If for example i have 2 colors (first duo):
RGB(60, 90, 80)
RGB(70, 50, 120)
By using the simplest algorithim i find that:
distance = sqrt((r2 - r1)^2 + (g2 - g1)^2 + (b2 - b1)^2)
where distance is 57.
Then i have the second color duo:
RGB(80,45,150)
RGB(x,y,z)
Finding the second color here by only knowing first color + distance is a bit unrealistic, any suggestions on how i could find something like this, or any good insight on workarounds.
algorithm colors rgb similarity
asked Nov 14 '18 at 5:57
VemboyVemboy
1
add a comment |
I'm doing a project where im finding the patterns of 1 color duo and trying to find the unkown second color of the second color duo.
If for example i have 2 colors (first duo):
RGB(60, 90, 80)
RGB(70, 50, 120)
By using the simplest algorithim i find that:
distance = sqrt((r2 - r1)^2 + (g2 - g1)^2 + (b2 - b1)^2)
where distance is 57.
Then i have the second color duo:
RGB(80,45,150)
RGB(x,y,z)
Finding the second color here by only knowing first color + distance is a bit unrealistic, any suggestions on how i could find something like this, or any good insight on workarounds.
algorithm colors rgb similarity
asked Nov 14 '18 at 5:57
VemboyVemboy
1
add a comment |
I'm doing a project where im finding the patterns of 1 color duo and trying to find the unkown second color of the second color duo.
If for example i have 2 colors (first duo):
RGB(60, 90, 80)
RGB(70, 50, 120)
By using the simplest algorithim i find that:
distance = sqrt((r2 - r1)^2 + (g2 - g1)^2 + (b2 - b1)^2)
where distance is 57.
Then i have the second color duo:
RGB(80,45,150)
RGB(x,y,z)
Finding the second color here by only knowing first color + distance is a bit unrealistic, any suggestions on how i could find something like this, or any good insight on workarounds.
algorithm colors rgb similarity
asked Nov 14 '18 at 5:57
VemboyVemboy
1
I'm doing a project where im finding the patterns of 1 color duo and trying to find the unkown second color of the second color duo.
If for example i have 2 colors (first duo):
RGB(60, 90, 80)
RGB(70, 50, 120)
By using the simplest algorithim i find that:
distance = sqrt((r2 - r1)^2 + (g2 - g1)^2 + (b2 - b1)^2)
where distance is 57.
Then i have the second color duo:
RGB(80,45,150)
RGB(x,y,z)
Finding the second color here by only knowing first color + distance is a bit unrealistic, any suggestions on how i could find something like this, or any good insight on workarounds.
algorithm colors rgb similarity
algorithm colors rgb similarity
asked Nov 14 '18 at 5:57
VemboyVemboy
1
asked Nov 14 '18 at 5:57
VemboyVemboy
1
asked Nov 14 '18 at 5:57
VemboyVemboy
1
asked Nov 14 '18 at 5:57
VemboyVemboy
1
asked Nov 14 '18 at 5:57
VemboyVemboy
1
1
add a comment |
add a comment |
2 Answers
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Welcome to StackOverflow!
If there is only 1 possible second color, you could calculate the distance simply by the difference of each R, G, and B elements. Therefore, the distance between RGB(60, 90, 80) and RGB(70, 50, 120) is (-10, +40, -40).
If the first color is RGB(80,45,150), then the second color with the same distance is RGB(90, 5, 190)
Another way is to use the distance as radial distance, however this will results in infinitely many possible second colors
answered Nov 14 '18 at 6:12
AndreasAndreas
1,99931220
add a comment |
You have a spherical surface around the first color point in RGB space (cutted sphere if distance is too long). For exact distance you have integer equation
dr^2 + dg^2 + db^2 = distance^2
that might have: no solutions, and symmetrical cases: 4 solutions, 8 solutions, 16/24 solutions, perhaps more. So task is to find triplets giving needed sum (c.f. Pythagorean triples for 2D case - there are 4 neighbors for distance 2, 8+4 neighbors for distance 5, no neighbors for distance 1.5 and so on).
If some tolerance is allowed, then you can find even more possible solutions.
Seems you need some kind of constraints to limit results.
Also it might be useful to consider another color model like HSV (if color perception is important)
edited Nov 14 '18 at 14:33
Shudipta Sharma
1,185413
answered Nov 14 '18 at 6:20
MBoMBo
48.9k23050
add a comment |
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2 Answers
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2 Answers
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Welcome to StackOverflow!
If there is only 1 possible second color, you could calculate the distance simply by the difference of each R, G, and B elements. Therefore, the distance between RGB(60, 90, 80) and RGB(70, 50, 120) is (-10, +40, -40).
If the first color is RGB(80,45,150), then the second color with the same distance is RGB(90, 5, 190)
Another way is to use the distance as radial distance, however this will results in infinitely many possible second colors
answered Nov 14 '18 at 6:12
AndreasAndreas
1,99931220
add a comment |
Welcome to StackOverflow!
If there is only 1 possible second color, you could calculate the distance simply by the difference of each R, G, and B elements. Therefore, the distance between RGB(60, 90, 80) and RGB(70, 50, 120) is (-10, +40, -40).
If the first color is RGB(80,45,150), then the second color with the same distance is RGB(90, 5, 190)
Another way is to use the distance as radial distance, however this will results in infinitely many possible second colors
answered Nov 14 '18 at 6:12
AndreasAndreas
1,99931220
add a comment |
Welcome to StackOverflow!
If there is only 1 possible second color, you could calculate the distance simply by the difference of each R, G, and B elements. Therefore, the distance between RGB(60, 90, 80) and RGB(70, 50, 120) is (-10, +40, -40).
If the first color is RGB(80,45,150), then the second color with the same distance is RGB(90, 5, 190)
Another way is to use the distance as radial distance, however this will results in infinitely many possible second colors
answered Nov 14 '18 at 6:12
AndreasAndreas
1,99931220
Welcome to StackOverflow!
If there is only 1 possible second color, you could calculate the distance simply by the difference of each R, G, and B elements. Therefore, the distance between RGB(60, 90, 80) and RGB(70, 50, 120) is (-10, +40, -40).
If the first color is RGB(80,45,150), then the second color with the same distance is RGB(90, 5, 190)
Another way is to use the distance as radial distance, however this will results in infinitely many possible second colors
answered Nov 14 '18 at 6:12
AndreasAndreas
1,99931220
answered Nov 14 '18 at 6:12
AndreasAndreas
1,99931220
answered Nov 14 '18 at 6:12
AndreasAndreas
1,99931220
answered Nov 14 '18 at 6:12
AndreasAndreas
1,99931220
1,99931220
add a comment |
add a comment |
You have a spherical surface around the first color point in RGB space (cutted sphere if distance is too long). For exact distance you have integer equation
dr^2 + dg^2 + db^2 = distance^2
that might have: no solutions, and symmetrical cases: 4 solutions, 8 solutions, 16/24 solutions, perhaps more. So task is to find triplets giving needed sum (c.f. Pythagorean triples for 2D case - there are 4 neighbors for distance 2, 8+4 neighbors for distance 5, no neighbors for distance 1.5 and so on).
If some tolerance is allowed, then you can find even more possible solutions.
Seems you need some kind of constraints to limit results.
Also it might be useful to consider another color model like HSV (if color perception is important)
edited Nov 14 '18 at 14:33
Shudipta Sharma
1,185413
answered Nov 14 '18 at 6:20
MBoMBo
48.9k23050
add a comment |
You have a spherical surface around the first color point in RGB space (cutted sphere if distance is too long). For exact distance you have integer equation
dr^2 + dg^2 + db^2 = distance^2
that might have: no solutions, and symmetrical cases: 4 solutions, 8 solutions, 16/24 solutions, perhaps more. So task is to find triplets giving needed sum (c.f. Pythagorean triples for 2D case - there are 4 neighbors for distance 2, 8+4 neighbors for distance 5, no neighbors for distance 1.5 and so on).
If some tolerance is allowed, then you can find even more possible solutions.
Seems you need some kind of constraints to limit results.
Also it might be useful to consider another color model like HSV (if color perception is important)
edited Nov 14 '18 at 14:33
Shudipta Sharma
1,185413
answered Nov 14 '18 at 6:20
MBoMBo
48.9k23050
add a comment |
You have a spherical surface around the first color point in RGB space (cutted sphere if distance is too long). For exact distance you have integer equation
dr^2 + dg^2 + db^2 = distance^2
that might have: no solutions, and symmetrical cases: 4 solutions, 8 solutions, 16/24 solutions, perhaps more. So task is to find triplets giving needed sum (c.f. Pythagorean triples for 2D case - there are 4 neighbors for distance 2, 8+4 neighbors for distance 5, no neighbors for distance 1.5 and so on).
If some tolerance is allowed, then you can find even more possible solutions.
Seems you need some kind of constraints to limit results.
Also it might be useful to consider another color model like HSV (if color perception is important)
edited Nov 14 '18 at 14:33
Shudipta Sharma
1,185413
answered Nov 14 '18 at 6:20
MBoMBo
48.9k23050
You have a spherical surface around the first color point in RGB space (cutted sphere if distance is too long). For exact distance you have integer equation
dr^2 + dg^2 + db^2 = distance^2
that might have: no solutions, and symmetrical cases: 4 solutions, 8 solutions, 16/24 solutions, perhaps more. So task is to find triplets giving needed sum (c.f. Pythagorean triples for 2D case - there are 4 neighbors for distance 2, 8+4 neighbors for distance 5, no neighbors for distance 1.5 and so on).
If some tolerance is allowed, then you can find even more possible solutions.
Seems you need some kind of constraints to limit results.
Also it might be useful to consider another color model like HSV (if color perception is important)
edited Nov 14 '18 at 14:33
Shudipta Sharma
1,185413
answered Nov 14 '18 at 6:20
MBoMBo
48.9k23050
edited Nov 14 '18 at 14:33
Shudipta Sharma
1,185413
edited Nov 14 '18 at 14:33
Shudipta Sharma
1,185413
edited Nov 14 '18 at 14:33
Shudipta Sharma
1,185413
1,185413
answered Nov 14 '18 at 6:20
MBoMBo
48.9k23050
answered Nov 14 '18 at 6:20
MBoMBo
48.9k23050
answered Nov 14 '18 at 6:20
MBoMBo
48.9k23050
48.9k23050
add a comment |
add a comment |
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