Balanced based representation

Question

I'm interested in balanced base $B$ representation for fixed point arithmetic. The paper Fixed-point arithmetic in SHE schemes (Costache, Smart, Vivek and Waller, in Proceedings of 23rd International Conference on Selected Areas in Cryptography (SAC 2016), Springer Lecture Notes in Computer Science, vol. 10532, pp. 401–422, 2016; PDF) provides two examples using balanced based $B=3$. However, I don't understand the 2nd second representation, i.e. $$\frac{8}{3}=10.\bar{1} $$ where $\bar{1}=-1$. More generally speaking, I'm also interested in balanced form representation $B=5, 7$. I believe this should be a well-known problem, though I was not able to google it. Do you know some relevant literature to this?

Answer 1

The expression $10.\overline{1}$ should be interpreted as follows: $$ 1 \cdot 3^1 + 0 \cdot 3^0 + (-1) \cdot 3^{-1} = 3 - \frac{1}{3} = \frac{8}{3}. $$ The interpretation of balanced base $b=2d+1$ is the same, with the digits ranging from $-d$ to $d$.

Answer 2

$$\frac{8}{3}=3 - \frac13=1\cdot3^1 + 0\cdot3^0 +(-1)\cdot3^{-1}=(10.\bar{1})_3$$

Since $\frac83$ is not equal to $\dfrac p{5^i}$ for any integer $p$ and $i$, it cannot be expressed as a fixed-point balanced base-5 number. Instead, infinite periodical fraction part has to be used, where $\bar2=-2$.

$$\frac{8}{3}=5-2+\frac{-2}5+\frac2{25}+\frac{-2}{125}+\frac2{625}+\frac{-2}{3125}+\frac2{15625}+\cdots=(1\bar2.\bar{2}2\bar{2}2\bar{2}2\cdots)_5$$

Let us see it step by step.
$\frac{8}{3}\approx5$
$\frac{8}{3}-5=\frac{-7}3\approx -2$
$\frac{8}{3}-(5-2)=\frac{-1}3\approx \frac {-2}5$
$\frac{8}{3}-(5-2+\frac{-2}5)=\frac15\frac{1}3\approx \frac15\frac {2}5$
$\frac{8}{3}-(5-2+\frac{-2}5+\frac15\frac25)=\frac15\frac15\frac{-1}3\approx \frac15\frac15\frac {-2}5$
$\frac{8}{3}-(5-2+\frac{-2}5+\frac15\frac25+\frac15\frac15\frac{-2}5)=\frac15\frac15\frac15\frac{1}3\approx \frac15\frac15\frac15\frac {2}5$
$\frac{8}{3}-(5-2+\frac{-2}5+\frac15\frac25+\frac15\frac15\frac{-2}5+\frac15\frac15\frac15\frac25)=\frac15\frac15\frac15\frac15\frac{-1}3\approx \frac15\frac15\frac15\frac15\frac {-2}5$
$\vdots$

You can work out the the balanced base-7 number with periodical fraction for $\dfrac 83$.

For the case of base 3, you can take a look at balanced ternary at Wikipedia.