Table of Contents
1. Overview
This is a non-standard but still positional numeral system. Its feature is that digits can have one of the values -1, 0 and 1. Nevertheless, its base is still 3 (because there are three possible values). Since it is not convenient to write -1 as a digit, we’ll use letter Z further for this purpose. If you think it is quite a strange system – look at the picture – here is one of the computers utilizing it.
So here are few first numbers written in balanced ternary:
0 0 1 1 2 1Z 3 10 4 11 5 1ZZ 6 1Z0 7 1Z1 8 10Z 9 100
This system allows you to write negative values without leading minus sign: you can simply invert digits in any positive number.
-1 Z -2 Z1 -3 Z0 -4 ZZ -5 Z11
Note that a negative number starts with Z and positive with 1.
2. Conversion algorithm
It is easy to represent a given number in balanced ternary via temporary representing it in normal ternary number system. When value is in standard ternary, its digits are either 0 or 1 or 2. Iterating from the lowest digit we can safely skip any 0s and 1s, however 2 should be turned into Z with adding 1 to the next digit. Digits 3 should be turned into 0 on the same terms – such digits are not present in the number initially but they can be encountered after increasing some 2s.
Example 1: Let us convert 64 to balanced ternary. At first we use normal ternary to rewrite the number:
$$ 64_{10} = 02101_{3} $$
Let us process it from the least significant (rightmost) digit:
- 1,0 and 1 are skipped as it is.( Because 0 and 1 are allowed in balanced ternary )
- 2 is turned into Z increasing the digit to its left, so we get 1Z101.
The final result is 1Z101.
Let us convert it back to the decimal system by adding the weighted positional values: $$ 1Z101 = 81 \cdot 1 + 27 \cdot (-1) + 9 \cdot 1 + 3 \cdot 0 + 1 \cdot 1 = 64_{10} $$
Example 2: Let us convert 237 to balanced ternary. At first we use normal ternary to rewrite the number:
$$ 237_{10} = 22210_{3} $$
Let us process it from the least significant (rightmost) digit:
- 0 and 1 are skipped as it is.( Because 0 and 1 are allowed in balanced ternary )
- 2 is turned into Z increasing the digit to its left, so we get 23Z10.
- 3 is turned into 0 increasing the digit to its left, so we get 30Z10.
- 3 is turned into 0 increasing the digit to its left( which is by default 0 ), and so we get 100Z10.
The final result is 100Z10.
Let us convert it back to the decimal system by adding the weighted positional values: $$ 100Z10 = 243 \cdot 1 + 81 \cdot 0 + 27 \cdot 0 + 9 \cdot (-1) + 3 \cdot 1 + 1 \cdot 0 = 237_{10} $$