... 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, ],
};
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Anding the 3 bitstrings selected by rgb( 2,2,4 ) (* above) together

0b01 & 0b10 & 0b01 == 00 ==> No hits.
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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.