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    <p>From your example it seems that the few bits that change are not always the same (e.g. always the lowest 4). So I would suggest a simple run length encoding of the bits on the transposed array. Without having a distribution of your numbers/data, I would suggest to start with 4 bits for the length, but there you could try a little bit with some of your example inputs.</p> <p>The pseudocode (for compression) would look like this:</p> <pre><code> for bitpos = 0 to 31 for datapos = 0 to 255 BitString.append(getbit(data[datapos], bitpos); endfor endfor result=""; pos = 0; while (notEndOfString) # count 1s count = 0; while (pos &lt; 32*256 AND count &lt; 16 AND BitString[pos]==1) count++; pos++; endwhile result.append4BitNumber(count); # count 0s count = 0; while (pos &lt; 32*256 AND count &lt; 16 AND BitString[pos]==0) count++; pos++; endwhile result.append4BitNumber(count); endwhile </code></pre> <p>Maybe one could increase the compression by applying afterwards Lempel-Ziv or a Huffman encoding - but without more information about the distribution of the input data one can not say more (this holds for this problem in general - with a better information of the input data, one could tailor some kind of compression to it).</p> <p><strong>EDIT</strong>: Another easy approach would be to make an encoding of the changing bit positions: You start with your initial 32 bit word, then you store for every data word 3 bits defining how much bits change (i.e. 0..7), and afterwards you store 0..7 times 4 bits where the 4 bits encode the position of the chaning bit. That means when e.g. on average 2 bits change you need for your 32*256 bit packet 32+255*(3+8)=2837 => approx 35% of its original size.</p> <p>If you often have the same numbers of bits changing, some of these 4 bits pattern would appear very often, while others not at all => a Huffman coding on these 4 bits-groups would compress it optimal (if you know that these pattern probabilities would never change, you could even make a static Huffman tree, so you dont have to store it).</p>
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