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    <p>Conceptually you could convert this to a single 1264 bit number using the following algorithm:</p> <ol> <li>Initialize an accumulator variable to zero</li> <li>Iterate over all elements, but skipt those which you know to be zero</li> <li>For the other elements, add 2 to obtain values in the range [0,1,2,3,4]</li> <li>For each such value, multiply the accumulator by 5 then add the corresponding value</li> <li>When you have processed all elements, the accumulator will encode your arrays</li> </ol> <p>To reverse that encoding, youd do this:</p> <ol> <li>Read the encoded value into the accumulator</li> <li>Iterate over all elements, in reverse order, but skipt those which you know to be zero</li> <li>For each element, you obtain the corresponding value as the accumulator modulo 5</li> <li>Subtract 2 from that value</li> <li>Divide the accumulator by 5 using a truncating division</li> </ol> <p>The problem with all of this is the fact that JS doesn't provide 1264 bit numbers out of the box. You might try one of the libraries suggested in <a href="https://stackoverflow.com/q/4288821/1468366">How to deal with big numbers in javascript</a>.</p> <p>But unless you <em>absolutely</em> requre an extremely small representation, I'd suggest an alternative approach: you can encode up to 13 such values in a 32 bit signed integer, since 5<sup>13</sup>=1,220,703,125 &lt; 2,147,483,648=2<sup>31</sup>. So after encoding 13 values I'd write out the result using such a number, then reset the accumulator to zero. This way you'll need ⌈544/13⌉∙32=1376 bits, which is not that much worse in terms of space requirements, but will be <em>a lot</em> faster to implement.</p> <p>Instead of iterating once in forward and once in reverse direction, it might be easier to not multiply the accumulator by 5, but instead multiply the value you add to that by a suitable power of 5. In other words, you maintain a factor which you initialize to 1, and multiply by 5 every time you add a value. So in this case, first data values will have less significant positions than later data values, both for encoding and decoding, which means you can use the same iteration order for both.</p> <p>See the ideone link mentioned in my comment below for an example of this latter approach. It encodes the full 9*9*8 values in 50 integers, each of them using no more than 31 bits. It then decodes the original matrix from that encoded form, to show that all the information is still present. The example does not use any fixed zeros, in your case ⌈544/13⌉=42 integers should be enough.</p>
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    1. POConcatenate nested array to vector?
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    1. COOk, I think I understand this. What I'm not clear on is "you can encode up to 13 such values in a 32 bit signed integer"... I should've mentioned that the array above holds 81 octet sets (each representing a point on a 9x9 grid), so that means I can't use a 32 bit signed integer, right? Also a little unclear on "the accumulator will encode your arrays" (as part of step 5 in your original algorithm)--is there additional logic that happens there, or are you just saying that at that point the value of the accumulator will be the encoded vector?
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    2. CO@neezer: At the end, the accumulator will encode the vector without additional steps required. When you switch to 32 bits for every 13 values, you'd executes steps 1 and 5 once for every such set of 13, not only once for all 544 values together. See http://ideone.com/CgeDDj for an example.
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    3. COGreat, this is very helpful. Thanks!
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