Let's say I've got some securely-generated random data I want to encode as words, 32 bits of it. My wordlist is 2048 words, so each word encodes 11 bits of information. I write a routine to pack these 32 bits into the lower 11 bits of 3 16-bit types and zero the upper 5 bits of each. For the last one, I pack in the remaining 10 bits of random data and zero the top 6 instead. I then look each of these up in my wordlist array to get my words. I'd like to use these words.

Here's my question: does what I'm doing reduce the amount of secure randomness I'm encoding in these words? I'd be looking the final word up in a list that's effectively half the size, because the 11th bit is always zero, so is there some "loss of randomness" here? I know this is a little confusing, but I'd appreciate some guidance. I suspect there might be some existing guidance on this, but I'm not sure what to search for to find it and none of my googling has yielded results.

Edit: To clarify, my intuitive guess is that this won't pose a problem, but I'm not comfortable going with that. The closest I've come to rationalizing this is that $2^{11} \times 2^{11} \times 2^{10} = 2^{32}$.


1 Answer 1


One way to think about this is that you are defining an encoding, and that this is just a change in representation. You are representing a 32-bit value as 3 words, and importantly this is a reversible process. Since it's a bijective mapping, there's as much entropy in the output as there is in the input.

Really, what you're doing is not fundamentally different from representing the 32-bit value in 9 base 10 digits. It's true that there is notable bias in the leading digit, but the important thing is that the full range is covered and it's invertible.


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