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Fri, 20 Oct 2023
The discrete logarithm, shorter and simpler
I recently discussed the “discrete logarithm” method for multiplying integers, and I feel like I took too long and made it seem more complicated and mysterious than it should have been. I think I'm going to try again. Suppose for some reason you found yourself needing to multiply a lot of powers of !!2!!. What's !!4096·512!!? You could use the conventional algorithm: $$ \begin{array}{cccccccc} & & & & 4 & 0 & 9 & 6 \\ × & & & & & 5 & 1 & 2 \\ \hline % & & & & 8 & 1 & 9 & 2 \\ & & & 4 & 0 & 9 & 6 & \\ & 2 & 0 & 4 & 8 & 0 & & \\ \hline % & 2 & 0 & 9 & 7 & 1 & 5 & 2 \end{array} $$ but that's a lot of trouble, and a simpler method is available. You know that $$2^i\cdot 2^j = 2^{i+j}$$ so if you had an easy way to convert $$2^i\leftrightarrow i$$ you could just convert the factors to exponents, add the exponents, and convert back. And all that's needed is a simple table: \begin{array}{rr} 0 & 1\\ 1 & 2\\ 2 & 4\\ 3 & 8\\ 4 & 16\\ 5 & 32\\ 6 & 64\\ 7 & 128\\ 8 & 256\\ 9 & 512\\ 10 & 1\,024\\ 11 & 2\,048\\ 12 & 4\,096\\ 13 & 8\,192\\ 14 & 16\,384\\ 15 & 32\,768\\ 16 & 65\,536\\ 17 & 131\,072\\ 18 & 262\,144\\ 19 & 524\,288\\ 20 & 1\,048\,576\\ 21 & 2\,097\,152\\ \vdots & \vdots \\ \end{array} We check the table, and find that $$4096\cdot512 = 2^{12}\cdot 2^9 = 2^{12+9} = 2^{21} = 2097152.$$ Easy-peasy. That is all very well but how often do you find yourself having to multiply a lot of powers of !!2!!? This was a lovely algorithm but with very limited application. What Napier (the inventor of logarithms) realized was that while not every number is an integer power of !!2!!, every number is an integer power of !!1.00001!!, or nearly so. For example, !!23!! is very close to !!1.00001^{313\,551}!!. Napier made up a table, just like the one above, except with powers of !!1.00001!! instead of powers of !!2!!. Then to multiply !!x\cdot y!! you would just find numbers close to !!x!! and !!y!! in Napier's table and use the same algorithm. (Napier's original table used powers of !!0.9999!!, but it works the same way for the same reason.) There's another way to make it work. Consider the integers mod !!101!!, called !!\Bbb Z_{101}!!. In !!\Bbb Z_{101}!!, every number is an integer power of !!2!!! For example, !!27!! is a power of !!2!!. It's simply !!2^7!!, because if you multiply out !!2^7!! you get !!128!!, and !!128\equiv 27\pmod{101}!!. Or: $$\begin{array}{rcll} 14 & \stackrel{\pmod{101}}{\equiv} & 10\cdot 101 & + 14 \\ & = & 1010 & + 14 \\ & = & 1024 \\ & = & 2^{10} \end{array} $$ Or: $$\begin{array}{rcll} 3 & \stackrel{\pmod{101}}{\equiv} & 5844512973848570809\cdot 101 & + 3 \\ & = & 590295810358705651709 & + 3 \\ & = & 590295810358705651712 \\ & = & 2^{69} \end{array} $$ Anyway that's the secret. In !!\Bbb Z_{101}!! the silly algorithm that quickly multiplies powers of !!2!! becomes more practical, because in !!\Bbb Z_{101}!!, every number is a power of !!2!!. What works for !!101!! works in other cases larger and more interesting. It doesn't work to replace !!101!! with !!7!! (try it and see what goes wrong), but we can replace it with !!107, 797!!, or !!297779!!. The key is that if we want to replace !!101!! with !!n!! and !!2!! with !!a!!, we need to be sure that there is a solution to !!a^i=b\pmod n!! for every possible !!b!!. (The jargon term here is that !!a!! must be a “primitive root mod !!n!!”. !!2!! is a primitive root mod !!101!!, but not mod !!7!!.) Is this actually useful for multiplication? Perhaps not, but it does have cryptographic applications. Similar to how multiplying is easy but factoring seems difficult, computing !!a^i\pmod n!! for given !!a, i, n!! is easy, but nobody knows a quick way in general to reverse the calculation and compute the !!i!! for which !!a^i\pmod n = m!! for a given !!m!!. When !!n!! is small we can simply construct a lookup table with !!n-1!! entries. But if !!n!! is a !!600!!-digit number, the table method is impractical. Because of this, Alice and Bob can find a way to compute a number !!2^i!! that they both know, but someone else, seeing !!2^i!! can't easily figure out what the original !!i!! was. See Diffie-Hellman key exchange for more details. [ Also previously: Percy Ludgate's weird variation on this ] [Other articles in category /math] permanent link |