Mixing
Uniform map of R3 colour space onto a discrete number line
$begingroup$
I'm thinking about theoretical colour palette formation.
I have to start with a locus of $1000^3$ points in $R^3$, with each colour component axis being a discrete variable $0 le r, g, b < 1000$. (The resolution of 1000 is due to software constraints). This represents an RGB colour space - the cubic solid shown in the lower left of this diagram:
I have to decide on a scheme to choose up to $2^8$ of those points in a manner that attempts to get approximate uniform representation from each axis, and then map those points to an $R^1$ discrete space $0 le p < 2^8$. The mapping doesn't need to be linear but that would make things easier.
Uniform method
If we choose $n$ values on each axis, then
$n = lfloor 2^{8/3} rfloor = 6$ values per axis
$6^3 = 216$ values
$ frac {100 cdot 6^3} {2^8} approx 84% $ p space usage
For each channel value $c_i$, for $0 le i < 3$,
$c_i = frac {1000} 6 modleft( lfloor frac p {6^i} rfloor, 6 right) $
That method has perfect channel uniformity but wastes a lot (16%) of the p space.
The following is a generated visual representation of the uniform palette. The x-axis is red, the "short" y-axis is green, and the "long" y-axis is blue.
Bitfield method
Another method is to divide the colour channels into bit fields: for field bit count $f_i, 0 le i < 3$:
$sum_i f_i = 8$
$f = (3, 3, 2)$
$c_i = frac {1000} {f_i} modleft( lfloor frac p {f_{i-1}^i} rfloor, f_i right) $
$f_{i-1} = 1$ for $i = 0$.
This method has perfect p space usage but poor uniformity. Two of the channels have $2^3=8$ values per axis but the last only has $2^2=4$ values.
The following is a generated visual representation of the bitfield palette, with the same axes as the previous one.
I'd like to learn of another method to map p to the $R^3$ space that achieves better uniformity than the bit field method while still using all of the p space.
computer-science closed-map
asked Jan 2 at 22:47
ReinderienReinderien
1135
$endgroup$
add a comment |
$begingroup$
I'm thinking about theoretical colour palette formation.
I have to start with a locus of $1000^3$ points in $R^3$, with each colour component axis being a discrete variable $0 le r, g, b < 1000$. (The resolution of 1000 is due to software constraints). This represents an RGB colour space - the cubic solid shown in the lower left of this diagram:
I have to decide on a scheme to choose up to $2^8$ of those points in a manner that attempts to get approximate uniform representation from each axis, and then map those points to an $R^1$ discrete space $0 le p < 2^8$. The mapping doesn't need to be linear but that would make things easier.
Uniform method
If we choose $n$ values on each axis, then
$n = lfloor 2^{8/3} rfloor = 6$ values per axis
$6^3 = 216$ values
$ frac {100 cdot 6^3} {2^8} approx 84% $ p space usage
For each channel value $c_i$, for $0 le i < 3$,
$c_i = frac {1000} 6 modleft( lfloor frac p {6^i} rfloor, 6 right) $
That method has perfect channel uniformity but wastes a lot (16%) of the p space.
The following is a generated visual representation of the uniform palette. The x-axis is red, the "short" y-axis is green, and the "long" y-axis is blue.
Bitfield method
Another method is to divide the colour channels into bit fields: for field bit count $f_i, 0 le i < 3$:
$sum_i f_i = 8$
$f = (3, 3, 2)$
$c_i = frac {1000} {f_i} modleft( lfloor frac p {f_{i-1}^i} rfloor, f_i right) $
$f_{i-1} = 1$ for $i = 0$.
This method has perfect p space usage but poor uniformity. Two of the channels have $2^3=8$ values per axis but the last only has $2^2=4$ values.
The following is a generated visual representation of the bitfield palette, with the same axes as the previous one.
I'd like to learn of another method to map p to the $R^3$ space that achieves better uniformity than the bit field method while still using all of the p space.
computer-science closed-map
asked Jan 2 at 22:47
ReinderienReinderien
1135
$endgroup$
add a comment |
$begingroup$
I'm thinking about theoretical colour palette formation.
I have to start with a locus of $1000^3$ points in $R^3$, with each colour component axis being a discrete variable $0 le r, g, b < 1000$. (The resolution of 1000 is due to software constraints). This represents an RGB colour space - the cubic solid shown in the lower left of this diagram:
I have to decide on a scheme to choose up to $2^8$ of those points in a manner that attempts to get approximate uniform representation from each axis, and then map those points to an $R^1$ discrete space $0 le p < 2^8$. The mapping doesn't need to be linear but that would make things easier.
Uniform method
If we choose $n$ values on each axis, then
$n = lfloor 2^{8/3} rfloor = 6$ values per axis
$6^3 = 216$ values
$ frac {100 cdot 6^3} {2^8} approx 84% $ p space usage
For each channel value $c_i$, for $0 le i < 3$,
$c_i = frac {1000} 6 modleft( lfloor frac p {6^i} rfloor, 6 right) $
That method has perfect channel uniformity but wastes a lot (16%) of the p space.
The following is a generated visual representation of the uniform palette. The x-axis is red, the "short" y-axis is green, and the "long" y-axis is blue.
Bitfield method
Another method is to divide the colour channels into bit fields: for field bit count $f_i, 0 le i < 3$:
$sum_i f_i = 8$
$f = (3, 3, 2)$
$c_i = frac {1000} {f_i} modleft( lfloor frac p {f_{i-1}^i} rfloor, f_i right) $
$f_{i-1} = 1$ for $i = 0$.
This method has perfect p space usage but poor uniformity. Two of the channels have $2^3=8$ values per axis but the last only has $2^2=4$ values.
The following is a generated visual representation of the bitfield palette, with the same axes as the previous one.
I'd like to learn of another method to map p to the $R^3$ space that achieves better uniformity than the bit field method while still using all of the p space.
computer-science closed-map
asked Jan 2 at 22:47
ReinderienReinderien
1135
$endgroup$
I'm thinking about theoretical colour palette formation.
I have to start with a locus of $1000^3$ points in $R^3$, with each colour component axis being a discrete variable $0 le r, g, b < 1000$. (The resolution of 1000 is due to software constraints). This represents an RGB colour space - the cubic solid shown in the lower left of this diagram:
I have to decide on a scheme to choose up to $2^8$ of those points in a manner that attempts to get approximate uniform representation from each axis, and then map those points to an $R^1$ discrete space $0 le p < 2^8$. The mapping doesn't need to be linear but that would make things easier.
Uniform method
If we choose $n$ values on each axis, then
$n = lfloor 2^{8/3} rfloor = 6$ values per axis
$6^3 = 216$ values
$ frac {100 cdot 6^3} {2^8} approx 84% $ p space usage
For each channel value $c_i$, for $0 le i < 3$,
$c_i = frac {1000} 6 modleft( lfloor frac p {6^i} rfloor, 6 right) $
That method has perfect channel uniformity but wastes a lot (16%) of the p space.
The following is a generated visual representation of the uniform palette. The x-axis is red, the "short" y-axis is green, and the "long" y-axis is blue.
Bitfield method
Another method is to divide the colour channels into bit fields: for field bit count $f_i, 0 le i < 3$:
$sum_i f_i = 8$
$f = (3, 3, 2)$
$c_i = frac {1000} {f_i} modleft( lfloor frac p {f_{i-1}^i} rfloor, f_i right) $
$f_{i-1} = 1$ for $i = 0$.
This method has perfect p space usage but poor uniformity. Two of the channels have $2^3=8$ values per axis but the last only has $2^2=4$ values.
The following is a generated visual representation of the bitfield palette, with the same axes as the previous one.
I'd like to learn of another method to map p to the $R^3$ space that achieves better uniformity than the bit field method while still using all of the p space.
computer-science closed-map
computer-science closed-map
asked Jan 2 at 22:47
ReinderienReinderien
1135
asked Jan 2 at 22:47
ReinderienReinderien
1135
edited Jan 2 at 23:05
Reinderien
asked Jan 2 at 22:47
ReinderienReinderien
1135
asked Jan 2 at 22:47
ReinderienReinderien
1135
asked Jan 2 at 22:47
ReinderienReinderien
1135
1135
add a comment |
add a comment |
1 Answer
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$begingroup$
6*6*7 = 252, which is much closer to 256. If $R in [0, 6)$, $G in [0, 7)$, and $B in [0, 6)$, the 8-bit indexed colour could be
$C = R*7*6 + G*6 + B$
That's pretty good.
I also tried a spherical packing. For a face-centred packing into a cube with points at all corners, I get that a pattern $2x + 1$ layers deep has $4x^3 + 6x^2 + 3x + 1$ points in it. When $x = 3$, the pattern has 172 points. When $x = 4$, the pattern has 365 points. So neither of those is any good... On the other hand, I just found a totally awesome way of visually representing non-leap-years.
answered Jan 3 at 3:58
enigmaticPhysicistenigmaticPhysicist
1985
$endgroup$
$begingroup$
6,7,6 yields about 1.6% waste. Awesome!
$endgroup$
– Reinderien
Jan 3 at 4:17
add a comment |
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1 Answer
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$begingroup$
6*6*7 = 252, which is much closer to 256. If $R in [0, 6)$, $G in [0, 7)$, and $B in [0, 6)$, the 8-bit indexed colour could be
$C = R*7*6 + G*6 + B$
That's pretty good.
I also tried a spherical packing. For a face-centred packing into a cube with points at all corners, I get that a pattern $2x + 1$ layers deep has $4x^3 + 6x^2 + 3x + 1$ points in it. When $x = 3$, the pattern has 172 points. When $x = 4$, the pattern has 365 points. So neither of those is any good... On the other hand, I just found a totally awesome way of visually representing non-leap-years.
answered Jan 3 at 3:58
enigmaticPhysicistenigmaticPhysicist
1985
$endgroup$
$begingroup$
6,7,6 yields about 1.6% waste. Awesome!
$endgroup$
– Reinderien
Jan 3 at 4:17
add a comment |
$begingroup$
6*6*7 = 252, which is much closer to 256. If $R in [0, 6)$, $G in [0, 7)$, and $B in [0, 6)$, the 8-bit indexed colour could be
$C = R*7*6 + G*6 + B$
That's pretty good.
I also tried a spherical packing. For a face-centred packing into a cube with points at all corners, I get that a pattern $2x + 1$ layers deep has $4x^3 + 6x^2 + 3x + 1$ points in it. When $x = 3$, the pattern has 172 points. When $x = 4$, the pattern has 365 points. So neither of those is any good... On the other hand, I just found a totally awesome way of visually representing non-leap-years.
answered Jan 3 at 3:58
enigmaticPhysicistenigmaticPhysicist
1985
$endgroup$
$begingroup$
6,7,6 yields about 1.6% waste. Awesome!
$endgroup$
– Reinderien
Jan 3 at 4:17
add a comment |
$begingroup$
6*6*7 = 252, which is much closer to 256. If $R in [0, 6)$, $G in [0, 7)$, and $B in [0, 6)$, the 8-bit indexed colour could be
$C = R*7*6 + G*6 + B$
That's pretty good.
I also tried a spherical packing. For a face-centred packing into a cube with points at all corners, I get that a pattern $2x + 1$ layers deep has $4x^3 + 6x^2 + 3x + 1$ points in it. When $x = 3$, the pattern has 172 points. When $x = 4$, the pattern has 365 points. So neither of those is any good... On the other hand, I just found a totally awesome way of visually representing non-leap-years.
answered Jan 3 at 3:58
enigmaticPhysicistenigmaticPhysicist
1985
$endgroup$
6*6*7 = 252, which is much closer to 256. If $R in [0, 6)$, $G in [0, 7)$, and $B in [0, 6)$, the 8-bit indexed colour could be
$C = R*7*6 + G*6 + B$
That's pretty good.
I also tried a spherical packing. For a face-centred packing into a cube with points at all corners, I get that a pattern $2x + 1$ layers deep has $4x^3 + 6x^2 + 3x + 1$ points in it. When $x = 3$, the pattern has 172 points. When $x = 4$, the pattern has 365 points. So neither of those is any good... On the other hand, I just found a totally awesome way of visually representing non-leap-years.
answered Jan 3 at 3:58
enigmaticPhysicistenigmaticPhysicist
1985
answered Jan 3 at 3:58
enigmaticPhysicistenigmaticPhysicist
1985
answered Jan 3 at 3:58
enigmaticPhysicistenigmaticPhysicist
1985
answered Jan 3 at 3:58
enigmaticPhysicistenigmaticPhysicist
1985
1985
$begingroup$
6,7,6 yields about 1.6% waste. Awesome!
$endgroup$
– Reinderien
Jan 3 at 4:17
add a comment |
$begingroup$
6,7,6 yields about 1.6% waste. Awesome!
$endgroup$
– Reinderien
Jan 3 at 4:17
$begingroup$
6,7,6 yields about 1.6% waste. Awesome!
$endgroup$
– Reinderien
Jan 3 at 4:17
$begingroup$
6,7,6 yields about 1.6% waste. Awesome!
$endgroup$
– Reinderien
Jan 3 at 4:17
add a comment |
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