Suppose I have $P$, which is the base-$p$ representation of an integer $n$ and I want to calculate it base-$q$ representation $Q$. The obvious algorithm is: interpret $P$ to obtain $n$, then calculate $Q$ from $n$. This is easy.

But if $n$ is very large this could require a lot of memory, and it could require a large amount of processing before any of the digits of $Q$ are emitted. In certain special cases we can do better; there is an online algorithm which can start producing output before all the input is read. For example, if $p^i = q^j$ then the algorithm can read $i$ digits of $P$ and immediately produce $j$ digits of $Q$ without any further input and with only a constant amount of memory. This method is used, for example, in converting generic computer data ($p=256$) to “Base64” format ($q=64$).

Is there a general online algorithm for radix conversion? I expect that if there is such an algorithm, in the worst case it cannot emit any digits of $Q$ without first reading all of $P$, but are there algorithms that can often do better than this, in cases more general than the very unusual $p^i=q^j$ situation? Or are there any other special cases of interest?

[ Related thread: The math behind converting from any base to any base without going through base 10? but all of the suggestions there ingest the entire input before producing any output. ]

  • $\begingroup$ There seems to be no hope of producing the units digit of $Q$ before reading all of $P$ unless the sequence $p^i\bmod q$ is eventually zero, which in general it isn't; exceptions include the $p^i = q^j$ case. And to produce the initial digit of $Q$ it seems that we would need to get at least a rough approximation of $p^aq^{-b}$ where $a$ is the length of $P$, which we can't know without reading $P$. The idea of producing the middle digits when we can't get the initial or final ones is risible. But I hope someone has thought of something I haven't. $\endgroup$ Oct 23, 2017 at 16:20
  • $\begingroup$ In an odd coincidence, there is a post on this topic today on Jeremy Gibbons' blog. $\endgroup$ Nov 10, 2017 at 17:42

1 Answer 1


This answer is not complete, but it is the start of a complete answer, and nothing else has been offered. I hope to finish it another time.

I have found that an online right-to-left algorithm exists in a more general case than I mentioned in my original question.

I suspect that no such algorithm exists in general, for reasons I outlined in my comment. What would the online algorithm produce first?

Units digit first

Suppose the algorithm wants to produce the units digit of $Q$ without reading all of $P$. Represent $n$ in base-$p$ as $n = \sum_i d_ip^i$ where each base-$p$ digit $d_i$ is between $0$ and $p-1$ inclusive. The units digit of $Q$ is equal to $n\bmod q$ which is $$\Bigl(\sum_i d_ip^i\Bigr)\bmod q.\tag{$\star$}$$ The result depends on every one of the $d_i$ unless the corresponding term $d_ip^i$ has the same value $\bmod q$ for each possible value of $d_i$. In particular we must have $$0\cdot p^i \equiv 1\cdot p^i\pmod q$$ so $p^i$ must be a multiple of $q$ in order for us to ignore the $i$th term of the sum. Say $p^{k_0}$ is a multiple of $q$ for some $k_0$. Then in computing the units digit of $Q$ we can disregard all the terms of the sum with $i\ge k_0$, and we need only examine the first $k_0$ digits of $P$.

To compute the next digit of $Q$ we can disregard all the terms of the sum with $i\ge k_1$ where $k_1$ is the smallest number for which $p^{k_1}$ is a multiple of $q^2$, and so on.

For example, suppose $p=6$ and $q=8$ and $n=3021213_6$. $6^3$ is a multiple of $8$ so we can generate the units digit of $Q$ just by examining the last three digits of $P$. $2\cdot6^2 + 1\cdot 6 +3 \equiv 1\pmod 8$ so the units digit of $Q$ is 1. Reading the next three digits of $P$ will allow us to compute another digit of $Q$, and so on.

This situation includes the case of $p^i=q^j$ that I discussed in the original question, but it is more general.

Note that although the $k_i = O(i)$ it may not increase smoothly. We actually have $k_i = \lfloor c(i+1)\rfloor$ for some rational constant $c$ that depends on $P$ and $Q$. For example, $24$ is a multiple of $4$, and so $24^2$ must be a multiple of $4^2$, but it is also a multiple of $4^3$. Next $24^3$ is a multiple of $4^4$ but not of $4^5$. So when converting from base $24$ to base $4$, reading the first digit of $P$ gets you one digit of $Q$, but reading the next digit of $P$ gets you not one but two more digits of $Q$. Reading the third digit of $P$ gets you only one more digit of $Q$. Here $c = \frac32$.

(To come: most significant digit first; intermediate digits first.)


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