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.

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: \[100Z10 = 243 \cdot 1 + 81 \cdot 0 + 27 \cdot 0 + 9 \cdot (-1) + 3 \cdot 1 + 1 \cdot 0 = 237_{10}\]

Practice Problems

• Topcoder SRM 604, Div1-250