You do indeed need to turn the card upside down.
Some of the dots are given, since the left-most column and the lowest row are always filled in completely, while the top row and the right-most column are always alternating.
I can't do an accurate count right now, but it looks like we're missing around 60-70 dots. Each dot has 2 options, so that's at least 2^60=1,152,921,504,606,846,976=~1.15 quintillion combinations, making a brute force check impossible. If we get another part of the code, we might narrow that figure down by quite a bit.
One possibility for the ersult of the code in the picture is that it unlocks the full image. In that case, it will also be likely to require all the codes.
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