... your code will also match for rgb(2,2,4) etc.
I do not believe so. Given,
- picture 1 contains r=1, g=2, b=3
- picture 2 contains r=2, g=3, b=4
And assuming for simplicity, 3-bit/color images, the data would be encoded as
my %lookup = {
## 0 1 2 3 4 5 6 7
red => [ 0b00, 0b10, 0b01*, 0b00, 0b00, 0b00, 0b00, 0b00, ],
green => [ 0b00, 0b00, 0b10*, 0b01, 0b00, 0b00, 0b00, 0b00, ],
blue => [ 0b00, 0b00, 0b00, 0b10, 0b01*, 0b00, 0b00, 0b00, ],
};
Anding the 3 bitstrings selected by rgb( 2,2,4 ) (* above) together
0b01 & 0b10 & 0b01 == 00 ==> No hits.
So, for the simple case of looking up one rgb value at a time, there is no overlap.
For the more complex case of locating images that contain either rgb( 1,2,3 ) or rgb( 2,3,4 ), there is the problem that oring all six selected bitstrings together would also select rgb( 1,2,4 ) and rgb( 1,3,4 ) and rgb( 2,2,4 ) etc., but that is different from the problem description, which is more akin to r = 1 or 2 and g = 2 or 3, and b = 3 or 4, for which my code algorithm description would work correctly.
For the case you describe, that of only matching those images that contain either rgb( 1,2,3 ) or rgb( 2,3,4 ), you have to do it in two separate operations. And the individual strings of the first 3 values together and extract the image identities from the result. Then do the second set of three and extract the image identities.
It's and versus or.
Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.
| [reply] [Watch: Dir/Any] [d/l] [select] |