Balanced ternary

Balanced ternary is a non-standard positional numeral system (a balanced form), useful for comparison logic. While it is a ternary (base 3) number system, in the standard (unbalanced) ternary system, digits have values 0, 1 and 2. The digits in the balanced ternary system have values −1, 0, and 1.

Different sources use different glyphs used to represent the three digits in balanced ternary. In this article, 'T' (a ligature of the minus sign and 1) represents −1, while 0 and 1 represent themselves. Other conventions include using '−' and '+' to represent −1 and 1 respectively, or using Greek letter theta (Θ), which resembles a minus sign in a circle, to represent −1.

Computational properties

In the early days of computing, a few experimental Soviet computers were built with balanced ternary instead of binary, the most famous being the Setun, built by Nikolay Brusentsov and Sergei Sobolev. The notation has a number of computational advantages over regular binary. Particularly, the plus-minus consistency cuts down the carry rate in multi-digit multiplication, and the rounding-truncation equivalence cuts down the carry rate in rounding on fractions.

Balanced ternary also has a number of computational advantages over traditional ternary. Particularly, the one-digit multiplication table has no carries in balanced ternary, and the addition table has only two symmetric carries instead of three.

A possible use of balanced ternary is to represent if a list of values in a list is less than, equal to or greater than the corresponding value in a second list. Balanced ternary can also represent all integers using a separate minus sign; the value of the leading non-zero digit of a number has the sign of the number itself.

Conversion from decimal

In the balanced ternary system, the value of a digit n places left of the radix point the product of the digit and 3ⁿ. This is useful when converting between decimal and balanced ternary. For instance,

10_{bal. 3} = 1×3¹ + 0×3⁰ = 3₁₀

10T_{bal. 3} = 1×3² + 0×3¹ + −1×3⁰ = 8₁₀

−9₁₀ = −1×3² + 0×3¹ + 0×3⁰ = T00_{bal. 3}

8₁₀ = 1×3² + 0×3¹ + −1×3⁰ = 10T_{bal. 3}

Similarly, the first place to the right of the radix point holds 3⁻¹ = 1/3, the second place to the right of the decimal place holds 3⁻² = 1/9, and so on. For instance,

−2/3₁₀ = −1 + 1/3 = −1×3⁰ + 1×3⁻¹ = T.1_{bal. 3}.

Dec.	Bal. T.	Expansion	Dec.	Bal. T.	Expansion	Dec.	Bal. T.	Expansion	Dec.	Bal. T.	Expansion	Dec.	Bal. T.	Expansion
−12	TT0	−9−3	−7	T1T	−9+3−1	−2	T1	−3+1	3	10	+3	8	10T	+9−1
−11	TT1	−9−3+1	−6	T10	−9+3	−1	T	−1	4	11	+3+1	9	100	+9
−10	T0T	−9−1	−5	T11	−9+3+1	0	0	0	5	1TT	+9−3−1	10	101	+9+1
−9	T00	−9	−4	TT	−3−1	1	1	+1	6	1T0	+9−3	11	11T	+9+3−1
−8	T01	−9+1	−3	T0	−3	2	1T	+3−1	7	1T1	+9−3+1	12	110	+9+3

An integer is divisible by three the digit in the units place is zero.

We may check the parity of a balanced ternary integer by checking the parity of the sum of all trits. This sum has the same parity as the integer itself.

Balanced ternary can also be extended to fractional numbers similar to how decimal numbers are written to the right of the radix point.^[1]

Decimal	−0.9	−0.8	−0.7	−0.6	−0.5	−0.4	−0.3	−0.2	−0.1	0	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9
Balanced Ternary	T.010T	T.1TT1	T.10T0	T.11TT	0.T or T.1	0.TT11	0.T010	0.T11T	0.0T01	0	0.010T	0.1TT1	0.10T0	0.11TT	0.1 or 1.T	1.TT11	1.T010	1.T11T	1.0T01

In decimal or binary, integer values and terminating fractions have multiple representations. For example, $\textstyle {\frac {1}{10}}$ = 0.1 = 0.10 = 0.09. And, $\textstyle {\frac {1}{2}}$ = 0.1₂ = 0.10₂ = 0.01₂. Some balanced ternary fractions have multiple representations too. For example, $\textstyle {\frac {1}{6}}$ = 0.1T_{Balanced ternary} = 0.01_{Balanced ternary}. Certainly, in the decimal and binary, we may omit the rightmost trailing infinite 0s after the radix point and gain a representations of integer or terminating fraction. But, in balanced ternary, we can't omit the rightmost trailing infinite -1s after the radix point in order to gain a representations of integer or terminating fraction.

Donald Knuth has pointed out that truncation and rounding are the same operation in balanced ternary — they produce exactly the same result (a property shared with other balanced numeral systems). The number 1/2 is not exceptional; it has two equally valid representations, and two equally valid truncations: 0.1 (round to 0, and truncate to 0) and 1.T (round to 1, and truncate to 1).

The basic operations—addition, subtraction, multiplication, and division—are done as in regular ternary. Multiplication by two can be done by adding a number to itself, or subtracting itself after a-trit-left-shifting.

An arithmetic shift left of a balanced ternary number is the equivalent of multiplication by a (positive, integral) power of 3; and an arithmetic shift right of a balanced ternary number is the equivalent of division by a (positive, integral) power of 3.

Fraction	Balanced ternary		Fraction	Balanced ternary
1/1	1		1/11	0.01T11
1/2	0.1	1.T	1/12	0.01T
1/3	0.1		1/13	0.01T
1/4	0.1T		1/14	0.01T0T1
1/5	0.1TT1		1/15	0.01TT1
1/6	0.01	0.1T	1/16	0.01TT
1/7	0.0110TT		1/17	0.01TTT10T0T111T01
1/8	0.01		1/18	0.001	0.01T
1/9	0.01		1/19	0.00111T10100TTT1T0T
1/10	0.010T		1/20	0.0011

Conversion to a fraction

The conversion of a repeating balanced ternary number to a fraction is similar to converting a repeating decimal. For example: $0.1{\overline {\mathrm {110TT0} }}={\tfrac {\mathrm {1110TT0-1} }{\mathrm {100000T0} }}={\tfrac {\mathrm {1110TTT} }{\mathrm {100000T0} }}={\tfrac {\mathrm {111\times 1000T} }{\mathrm {111111\times 1T0} }}={\tfrac {\mathrm {1111\times 1T} }{\mathrm {1001\times 1T0} }}={\tfrac {1111}{10010}}={\tfrac {\mathrm {1T1T} }{\mathrm {1TTT0} }}={\tfrac {101}{\mathrm {1T10} }}$

Irrational and transcendental numbers

As in any other integer base, irrational and transcendental numbers do not terminate or repeat. For example:

Decimal	Balanced ternary
${\sqrt {2}}=1.4142135623731...$	${\sqrt {1T}}=1.11T1TT00T00T01T0T00T00T01TT...$
${\sqrt {3}}=1.7320508075689...$	${\sqrt {10}}=1T.T1TT10T0000TT1100T0TTT011T0...$
${\sqrt {5}}=2.2360679774998...$	${\sqrt {1TT}}=1T.1T0101010TTT1TT11010TTT01T1...$
$\phi ={\frac {{\sqrt {5}}-1}{2}}=0.6180339887499...$	$\phi ={\frac {{\sqrt {1TT}}-1}{1T}}=1.T0TT01TT0T10TT11T0011T10011...$
$\pi =3.14159265358979...$	$\pi =10.011T111T000T011T1101T111111...$
$e=2.71828182845905...$	$e=10.T0111TT0T0T111T0111T000T11T...$

Conversion from ternary

Unbalanced ternary can be converted to balanced ternary notation in two ways:

Add 1 trit-by-trit from the first non-zero trit with carry, and then subtract 1 trit-by-trit from the same trit without borrow. For example,

021₃ + 11₃ = 102₃, 102₃ − 11₃ = 1T1_{bal. 3} = 7₁₀.

If a 2 is present in ternary, turn it into 1T. For example,

0212₃ = 0010_{bal. 3} + 1T00_{bal. 3} + 001T_{bal. 3} = 10TT_{bal. 3} = 23₁₀

Conversion to balanced ternary from any integer base

We may convert to balanced ternary with the following formula: $(a_{n}a_{n-1}\cdots a_{1}a_{0}.c_{1}c_{2}c_{3}\cdots )_{b}=\sum _{k=0}^{n}a_{k}b^{k}+\sum _{k=1}^{\infty }c_{k}b^{-k}.$ where,

a_{n}a_{n-1}\cdots a_{1}a_{0}.c_{1}c_{2}c_{3}\cdots

is the original representation in the original numeral system.

b is the original radix. b is 10 if converting from decimal.

a_{k}

and

c_{k}

are the digits k places to the left and right of the radix point respectively.

For instance,

           -25.4₁₀=-(1T×101¹+1TT×101⁰+11*101^-1)
                  =-(1T×101+1TT+11÷101)
                  =-10T1.11TT
                  =T01T.TT11

            1010.1₂=1T¹⁰⁰+1T¹+1T^-1
                   =10T+1T+0.1
                   =101.1

Addition, subtraction and multiplication and division

The single-trit addition, subtraction, multiplication and division tables are shown below. For subtraction and division, which are not commutative, the first operand is given to the left of the table, while the second is given at the top. For instance, the answer to 1-T=1T is found in the bottom left corner of the subtraction table.

Addition
+	T	0	1
T	T1	T	0
0	T	0	1
1	0	1	1T

Subtraction
−	T	0	1
T	0	T	T1
0	1	0	T
1	1T	1	0

Multiplication
×	T	1
T	1	T
0	0	0
1	T	1

Division
÷	T	0	1
T	1	NaN	T
0	0	NaN	0
1	T	NaN	1

Multi-trit addition and subtraction

Multi-trit addition and subtraction is analogous to that of binary and decimal. Add and subtract trit by trit, and add the carry appropriately. For example:

           1TT1TT.1TT1              1TT1TT.1TT1            1TT1TT.1TT1          1TT1TT.1TT1
         +   11T1.T                -  11T1.T              -  11T1.T    ->      +  TT1T.1
        --------------          --------------                               ---------------
           1T0T10.0TT1              1T1001.TTT1                                 1T1001.TTT1
         +   1T                   +  T  T                                   +   T  T
         --------------         ----------------                             ----------------
           1T1110.0TT1              1110TT.TTT1                                 1110TT.TTT1
         +   T                    + T   1                                     + T   1
         --------------         ----------------                             ----------------
           1T0110.0TT1               1100T.TTT1                                  1100T.TTT1

Multi-trit multiplication

Multi-trit multiplication is analogous to that in decimal and binary.

       1TT1.TT
   ×   T11T.1
   -------------
        1TT.1TT multiply 1
       T11T.11  multiply T
      1TT1T.T   multiply 1
     1TT1TT     multiply 1
    T11T11      multiply T
   -------------
    0T0000T.10T

Multi-trit division

Balanced ternary division is analogous to decimal or binary division.

However, 0.5₁₀ = 0.1111..._{bal. 3} or 1.TTTT..._{bal. 3}. If the dividend over the plus or minus half divisor, the trit of the quotient must be 1 or T. If the dividend is between the plus and minus of half the divisor, the trit of the quotient is 0. The magnitude of the dividend must be compared with that of half the divisor before settinh the quotient trit. For example,

                         1TT1.TT      quotient
0.5 × divisor  T01.0 -------------                
      divisor T11T.1 ) T0000T.10T     dividend             
                      T11T1                        T000<T010, set 1
                     -------
                       1T1T0
                       1TT1T                      1T1T0>10T0, set T 
                      -------
                         111T
                        1TT1T                      1110>10T0, set T
                       -------
                          T00.1
                         T11T.1                    T001<T010, set 1
                        --------
                          1T1.00
                          1TT.1T                  1T100>10T0, set T
                         --------
                           1T.T1T
                           1T.T1T                 1TT1T>10T0, set T
                          --------
                                0

Another example,

                           1TTT
       0.5 × divisor 1T  -------
            Divisor  11  )1T01T                   1T=1T, but 1T.01>1T, set 1
                          11
                         -----
                          T10                    T10<T1, set T
                           TT
                         ------
                           T11                   T11<T1, set T
                            TT
                          ------
                             TT                   TT<T1, set T
                             TT
                            ----
                              0

Another example,

                           101.TTTTTTTTT… 
                        or 100.111111111… 
       0.5 × divisor 1T  -----------------
            divisor  11  )111T                    11>1T, set 1
                          11
                         -----
                            1                     T1<1<1T, set 0
                           ---
                            1T                    1T=1T, trits end, set 1.TTTTTTTTT… or 0.111111111…

Square roots and cube roots

Extractin the sqyare root in balanced ternary is analogous to that in decimal or binary.

(10\cdot x+y)^{\mathrm {1T} }-100\cdot x^{\mathrm {1T} }=\mathrm {1T0} \cdot x\cdot y+y^{\mathrm {1T} }={\begin{cases}\mathrm {T10} \cdot x+1,&y=\mathrm {T} \\0,&y=0\\\mathrm {1T0} \cdot x+1,&y=1\end{cases}}

As in division, we should check the value of half the divisor first. For example,

                             1. 1 1 T 1 T T 0 0 ...
                           -------------------------
                          √ 1T                          1<1T<11, set 1
                           - 1
                            -----
                  1×10=10    1.0T                       1.0T>0.10, set 1
                      1T0   -1.T0
                            --------
                  11×10=110    1T0T                     1T0T>110, set 1
                       10T0   -10T0
                              --------
                 111×10=1110    T1T0T                   T1T0T<TTT0, set T
                       100T0   -T0010
                               ---------
                111T×10=111T0    1TTT0T                 1TTT0T>111T0, set 1
                       10T110   -10T110
                                ----------
               111T1×10=111T10    TT1TT0T               TT1TT0T<TTT1T0, set T
                       100TTT0   -T001110
                                 -----------
              111T1T×10=111T1T0    T001TT0T             T001TT0T<TTT1T10, set T
                       10T11110   -T01TTTT0
                                  ------------
               111T1TT×10=111T1TT0    T001T0T           TTT1T110<T001T0T<111T1TT0, set 0
                                     -      T           Return 1
                                     -----------
             111T1TT0×10=111T1TT00    T001T000T         TTT1T1100<T001T000T<111T1TT00, set 0
                                     -        T         Return 1
                                     -------------
           111T1TT00*10=111T1TT000    T001T00000T
                                              ...

Extracting the cube root in balanced ternary analogous to that in decimal or binary cube root too.

(10\cdot x+y)^{10}-1000\cdot x^{10}=y^{10}+1000\cdot x^{\mathrm {1T} }\cdot y+100\cdot x\cdot y^{\mathrm {1T} }={\begin{cases}\mathrm {T} +\mathrm {T000} \cdot x^{\mathrm {1T} }+100\cdot x,&y=\mathrm {T} \\0,&y=0\\1+1000\cdot x^{\mathrm {1T} }+100\cdot x,&y=1\end{cases}}

Like division, we should check half divisor first too. For example:

                              1.  1   T  1  0...
                           3---------------------
                           √ 1T
                            - 1                 1<1T<10T,set 1
                            -------
                              1.000
              1×100=100      -0.100             borrow 100×, do division
                             -------
                      1TT     1.T00             1T00>1TT, set 1
          1×1×1000+1=1001    -1.001
                             ----------
                                T0T000
            11×100            -   1100           borrow 100×, do division
                              ---------
                     10T000     TT1T00           TT1T00<T01000, set T
       11×11×1000+1=1TT1001   -T11T00T
                              ------------
                                1TTT01000
           11T×100             -    11T00        borrow 100×, do division
                               -----------
                   1T1T01TT     1TTTT0100        1TTTT0100>1T1T01TT, set 1
    11T×11T×1000+1=11111001    - 11111001
                               --------------
                                    1T10T000
          11T1×100                 -  11T100      borrow 100×, do division
                                   ----------
                      10T0T01TT     1T0T0T00      T01010T11<1T0T0T00<10T0T01TT, set 0
    11T1×11T1×1000+1=1TT1T11001    -  TT1T00      return 100×
                                   -------------
                                    1T10T000000
                                         ...

Hence ^³√2=1.259921₁₀=1.1T1'000'111'001'T01'00T'1T1'T10'111_{bal. 3}

Other applications

Balanced ternary has other applications besides computing. For example, a classical two-pan balance, with one weight for each power of 3, can weigh relatively heavy objects accurately with a small number of weights, by moving weights between the two pans and the table. For example, with weights for each power of 3 through 81, a 60-gram object (60₁₀ = 1T1T0) will be balanced perfectly with an 81 gram weight in the other pan, the 27 gram weight in its own pan, the 9 gram weight in the other pan, the 3 gram weight in its own pan, and the 1 gram weight set aside.

Similarly, consider a currency system with coins worth 1¢, 3¢, 9¢, 27¢, 81¢. If the buyer and the seller each have only one of each kind of coin, any transaction up to 121¢ is possible. For example, if the price is 7¢ (1T1), the buyer pays 1¢ + 9¢ and receives 3¢ in change.

Notes

References

Hayes, Brian (2001), "Third base", American Scientist, 89 (6): 490–494, doi:10.1511/2001.40.3268.

External links

Development of ternary computers at Moscow State University
Representation of Fractional Numbers in Balanced Ternary
“Third base”, ternary and balanced ternary number systems
The Balanced Ternary Number System (includes decimal integer to balanced ternary converter)
The binomial triangle reduced to balanced ternary lists. OEIS OEIS: A182929
Balanced (Signed) Ternary Notation by Brian J. Shelburne (PDF file)
The ternary calculating machine of Thomas Fowler by Mark Glusker

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