balanced ternary

{{See also|Signed-digit representation}}

Let $\backslash mathcal\{D\}\_\{3\}\; :=\; \backslash lbrace\; \backslash operatorname\{T\},\; 0,\; 1\; \backslash rbrace$ denote the set of symbols (also called glyphs or characters), where the symbol $\backslash bar\{1\}$ is sometimes used in place of $\backslash operatorname\{T\}.$

Define an integer-valued function $f\; =\; f\_\{\backslash mathcal\{D\}\_\{3\}\}\; :\; \backslash mathcal\{D\}\_\{3\}\; \backslash to\; \backslash mathbb\{Z\}$ by

:$\backslash begin\{align\}$

f_{}(\operatorname{T}) &= -1, \\

f_{}(0) &= 0, \\

f_{}(1) &= 1,

\end{align}The symbols $0$ and $1$ appear twice in the equalities $f\_\{\}(0)\; =\; 0$ and $f\_\{\}(1)\; =\; 1,$ but these instances do not represent the same thing. The right hand side $0$ and $1$ mean the integers $\backslash in\backslash Z,$ but the instances inside $f$'s parentheses (which belong to $\backslash mathcal\{D\}\_\{3\}$) should be thought of as being nothing more than symbols.

where the right hand sides are integers with their usual values. This function, $f\_\{\},$ is what rigorously and formally establishes how integer values are assigned to the symbols/glyphs in $\backslash mathcal\{D\}\_\{3\}.$ One benefit of this formalism is that the definition of "the integers" (however they may be defined) is not conflated with any particular system for writing/representing them; in this way, these two distinct (albeit closely related) concepts are kept separate.

The set $\backslash mathcal\{D\}\_\{3\}$ together with the function $f\_\{\}$ forms a balanced signed-digit representation called the balanced ternary system.

It can be used to represent integers and real numbers.

Let $\backslash mathcal\{D\}\_\{3\}^\{+\}$ be the Kleene plus of $\backslash mathcal\{D\}\_\{3\}$, which is the set of all finite length concatenated strings $d\_n\; \backslash ldots\; d\_0$ of one or more symbols (called its digits) where $n$ is a non-negative integer and all $n\; +\; 1$ digits $d\_n,\; \backslash ldots,\; d\_0$ are taken from $\backslash mathcal\{D\}\_\{3\}\; =\; \backslash lbrace\; \backslash operatorname\{T\},\; 0,\; 1\; \backslash rbrace.$ The start of $d\_n\; \backslash ldots\; d\_0$ is the symbol $d\_0$ (at the right), its end is $d\_n$ (at the left), and its length is $n\; +\; 1$. The ternary evaluation is the function $v\; =\; v\_\{3\}\; ~:~\; \backslash mathcal\{D\}\_\{3\}^\{+\}\; \backslash to\; \backslash mathbb\{Z\}$ defined by assigning to every string $d\_n\; \backslash ldots\; d\_0\; \backslash in\; \backslash mathcal\{D\}\_\{3\}^\{+\}$ the integer

:$v\backslash left(\; d\_n\; \backslash ldots\; d\_0\; \backslash right)\; ~=~\; \backslash sum\_\{i=0\}^\{n\}\; f\_\{\}\; \backslash left(\; d\_\{i\}\; \backslash right)\; 3^\{i\}.$

The string $d\_n\; \backslash ldots\; d\_0$ represents (with respect to $v$) the integer $v\backslash left(\; d\_n\; \backslash ldots\; d\_0\; \backslash right).$ The value $v\backslash left(\; d\_n\; \backslash ldots\; d\_0\; \backslash right)$ may alternatively be denoted by $\{d\_n\; \backslash ldots\; d\_0\}\_\{\backslash operatorname\{bal\}3\}.$

The map $v\; :\; \backslash mathcal\{D\}\_\{3\}^\{+\}\; \backslash to\; \backslash mathbb\{Z\}$ is surjective but not injective since, for example, $0\; =\; v(0)\; =\; v(00)\; =\; v(0\; 0\; 0)\; =\; \backslash cdots.$ However, every nonzero integer has exactly one representation under $v$ that does not end (on the left) with the symbol $0,$ i.e. $d\_n\; =\; 0\; .$

If $d\_n\; \backslash ldots\; d\_0\; \backslash in\; \backslash mathcal\{D\}\_\{3\}^\{+\}$ and $n\; >\; 0$ then $v$ satisfies:

:$v\backslash left(\; d\_n\; d\_\{n-1\}\; \backslash ldots\; d\_0\; \backslash right)\; ~=~\; f\_\{\}\; \backslash left(\; d\_\{n\}\; \backslash right)\; 3^\{n\}\; +\; v\backslash left(\; d\_\{n-1\}\; \backslash ldots\; d\_0\; \backslash right)$

which shows that $v$ satisfies a sort of recurrence relation. This recurrence relation has the initial condition

$v\backslash left(\; \backslash varepsilon\; \backslash right)\; =\; 0$

where $\backslash varepsilon$ is the empty string.

This implies that for every string $d\_n\; \backslash ldots\; d\_0\; \backslash in\; \backslash mathcal\{D\}\_\{3\}^\{+\},$

:$v\backslash left(\; 0\; d\_n\; \backslash ldots\; d\_0\; \backslash right)\; =\; v\backslash left(\; d\_n\; \backslash ldots\; d\_0\; \backslash right)$

which in words says that leading $0$ symbols (to the left in a string with 2 or more symbols) do not affect the resulting value.

The following examples illustrate how some values of $v$ can be computed, where (as before) all integer are written in decimal (base 10) and all elements of $\backslash mathcal\{D\}\_\{3\}^\{+\}$ are just symbols.

:$\backslash begin\{alignat\}\{10\}$

v\left( \operatorname{T} \operatorname{T} \right)

&= && f_{}\left( \operatorname{T} \right) 3^{1} + && f_{}\left( \operatorname{T} \right) 3^{0}

&&= &&(-1) &&3 &&\,+\, &&(-1) &&1

&&= -4 \\

v\left( \operatorname{T} 1 \right)

&= && f_{}\left( \operatorname{T} \right) 3^{1} + && f_{}\left( 1 \right) 3^{0}

&&= &&(-1) &&3 &&\,+\, &&(1) &&1

&&= -2 \\

v\left( 1 \operatorname{T} \right)

&= && f_{}\left( 1 \right) 3^{1} + && f_{}\left( \operatorname{T} \right) 3^{0}

&&= &&(1) &&3 &&\,+\, &&(-1) &&1

&&= 2 \\

v\left( 1 1 \right)

&= && f_{}\left( 1 \right) 3^{1} + && f_{}\left( 1 \right) 3^{0}

&&= &&(1) &&3 &&\,+\, &&(1) &&1

&&= 4 \\

v\left( 1 \operatorname{T} 0 \right)

&= f_{}\left( 1 \right) 3^{2} + && f_{}\left( \operatorname{T} \right) 3^{1} + && f_{}\left( 0 \right) 3^{0}

&&= (1) 9 \,+\, &&(-1) &&3 &&\,+\, &&(0) &&1

&&= 6 \\

v\left( 1 0 \operatorname{T} \right)

&= f_{}\left( 1 \right) 3^{2} + && f_{}\left( 0 \right) 3^{1} + && f_{}\left( \operatorname{T} \right) 3^{0}

&&= (1) 9 \,+\, &&(0) &&3 &&\,+\, &&(-1) &&1

&&= 8 \\

\end{alignat}

and using the above recurrence relation

:$v\backslash left(\; 1\; 0\; 1\; \backslash operatorname\{T\}\; \backslash right)\; =\; f\_\{\}\backslash left(\; 1\; \backslash right)\; 3^\{3\}\; +\; v\backslash left(\; 0\; 1\; \backslash operatorname\{T\}\; \backslash right)\; =\; (1)\; 27\; +\; v\backslash left(\; 1\; \backslash operatorname\{T\}\; \backslash right)\; =\; 27\; +\; 2\; =\; 29.$

In the balanced ternary system the value of a digit n places left of the radix point is the product of the digit and 3ⁿ. This is useful when converting between decimal and balanced ternary. In the following the strings denoting balanced ternary carry the suffix, bal3. For instance,

: 10_bal3 = 1 × 3¹ + 0 × 3⁰ = 3_dec

: 10𝖳_bal3 = 1 × 3² + 0 × 3¹ + (−1) × 3⁰ = 8_dec

: −9_dec = −1 × 3² + 0 × 3¹ + 0 × 3⁰ = 𝖳00_bal3

: 8_dec = 1 × 3² + 0 × 3¹ + (−1) × 3⁰ = 10𝖳_bal3

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

: −{{sfrac|2|3}}_dec = −1 + {{sfrac|1|3}} = −1 × 3⁰ + 1 × 3⁻¹ = 𝖳.1_bal3.

An integer is divisible by three if and only if 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.{{cite web |url=http://www.abhijit.info/tristate/tristate.html |title=Balanced ternary |last=Bhattacharjee |first=Abhijit |date=24 July 2006 |archiveurl=https://web.archive.org/web/20090919053547/http://www.abhijit.info/tristate/tristate.html |archivedate=2009-09-19}}

:

In decimal or binary, integer values and terminating fractions have multiple representations. For example, {{sfrac|1|10}} = 0.1 = 0.1{{overline|0}} = 0.0{{overline|9}}. And, {{sfrac|1|2}} = 0.1₂ = 0.1{{overline|0}}₂ = 0.0{{overline|1}}₂. Some balanced ternary fractions have multiple representations too. For example, {{sfrac|1|6}} = 0.1{{overline|𝖳}}_bal3 = 0.0{{overline|1}}_bal3. 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{{Cite book

|last=Knuth

|first=Donald

|authorlink=Donald Knuth

|title=The art of Computer Programming

|volume=2

|pages=195–213

|publisher=Addison-Wesley

|year=1997

|isbn=0-201-89684-2}} 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 {{sfrac|1|2}} is not exceptional; it has two equally valid representations, and two equally valid truncations: 0.{{overline|1}} (round to 0, and truncate to 0) and 1.{{overline|𝖳}} (round to 1, and truncate to 1). With an odd radix, double rounding is also equivalent to directly rounding to the final precision, unlike with an even radix.

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.

The conversion of a repeating balanced ternary number to a fraction is analogous to converting a repeating decimal. For example (because of 111111_bal3 = ({{sfrac|3⁶ − 1|3 − 1}})_dec):

: $0.1\backslash overline\{\backslash mathrm\{110TT0\}\; \}\; =\backslash tfrac\{\backslash mathrm\{1110TT0-1\}\; \}\{\backslash mathrm\{111111\backslash times\; 1T\backslash times\; 10\}\}=\backslash tfrac\{\backslash mathrm\{1110TTT\}\; \}\; \{\backslash mathrm\{111111\backslash times\; 1T0\}\}\; =\backslash tfrac\{\backslash mathrm\{111\backslash times\; 1000T\}\; \}\; \{\backslash mathrm\{111\backslash times\; 1001\backslash times\; 1T0\}\}\; =\backslash tfrac\{\backslash mathrm\{1111\backslash times\; 1T\}\}\{\backslash mathrm\{1001\backslash times\; 1T0\}\}\; =\backslash tfrac\{1111\}\{10010\}=\backslash tfrac\{\backslash mathrm\{1T1T\}\}\{\backslash mathrm\{1TTT0\}\}\; =\backslash tfrac\{101\}\{\backslash mathrm\{1T10\}\; \}$

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_bal3 = 7_dec.
• If a 2 is present in ternary, turn it into 1T. For example,
• : 0212₃ = 0010_bal3 + 1T00_bal3 + 001T_bal3 = 10TT_bal3 = 23_dec

If the three values of ternary logic are false, unknown and true, and these are mapped to balanced ternary as T, 0 and 1 and to conventional unsigned ternary values as 0, 1 and 2, then balanced ternary can be viewed as a biased number system analogous to the offset binary system.

If the ternary number has n trits, then the bias b is

:$b=\backslash left\backslash lfloor\; \backslash frac\{3^n\}\{2\}\; \backslash right\backslash rfloor$

which is represented as all ones in either conventional or biased form.Douglas W. Jones, [http://www.cs.uiowa.edu/~jones/ternary/numbers.shtml Ternary Number Systems], October 15, 2013.

As a result, if these two representations are used for balanced and unsigned ternary numbers, an unsigned n-trit positive ternary value can be converted to balanced form by adding the bias b and a positive balanced number can be converted to unsigned form by subtracting the bias b. Furthermore, if x and y are balanced numbers, their balanced sum is {{nowrap|x + y − b}} when computed using conventional unsigned ternary arithmetic. Similarly, if x and y are conventional unsigned ternary numbers, their sum is {{nowrap|x + y + b}} when computed using balanced ternary arithmetic.

We may convert to balanced ternary with the following formula:

:$$

\left(a_na_{n-1}\cdots a_1a_0.c_1 c_2 c_3\cdots\right)_b =

\sum_{k=0}^n a_kb^k + \sum_{k=1}^\infty c_kb^{-k}.

where,

: a_na_n−1...a₁a₀.c₁c₂c₃... 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,

−25.4_dec = −(1T×101¹ + 1TT×101⁰ + 11×101⁻¹)

= −(1T×101 + 1TT + 11÷101)

= −10T1.{{overline|11TT}}

= T01T.{{overline|TT11}}

1010.1₂ = 1T¹⁰ + 1T¹ + 1T⁻¹

= 10T + 1T + 0.{{overline|1}}

= 101.{{overline|1}}

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.

:

|

:

|

:

|

:

|}

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 T1 + T T1

______________ ________________ ________________

1T1110.0TT1 1110TT.TTT1 1110TT.TTT1

+ T + T 1 + T 1

______________ ________________ ________________

1T0110.0TT1 1100T.TTT1 1100T.TTT1

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

1TT1.TT

× T11T.1

_____________

1TT.1TT multiply 1

T11T.11 multiply T

1TT1T.T multiply 1

1TT1TT multiply 1

T11T11 multiply T

_____________

0T0000T.10T

Balanced ternary division is analogous to that of binary and decimal.

However, 0.5_dec = 0.1111..._bal3 or 1.TTTT..._bal3. 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 setting 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 111T > 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...

The process of extracting the square root in balanced ternary is analogous to that in decimal or binary.

:$(10\backslash cdot\; x+y)^\{\backslash mathrm\{1T\}\}-100\backslash cdot\; x^\{\backslash mathrm\{1T\}\}=\backslash mathrm\{1T0\}\backslash cdot\; x\backslash cdot\; y+y^\{\backslash mathrm\{1T\}\}=$

\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

100T0 −T0010

_________

111T×10=111T0 1TTT0T 1TTT0T>111T0, set 1

10T110 −10T110

__________

111T1×10=111T10 TT1TT0T TT1TT0T

100TTT0 −T001110

___________

111T1T×10=111T1T0 T001TT0T T001TT0T

10T11110 −T01TTTT0

____________

111T1TT×10=111T1TT0 T001T0T TTT1T110

− T Return 1

___________

111T1TT0×10=111T1TT00 T001T000T TTT1T1100

− T Return 1

_____________

111T1TT00*10=111T1TT000 T001T00000T

...

Extraction of the cube root in balanced ternary is similarly analogous to extraction in decimal or binary:

:$(10\backslash cdot\; x+y)^\{10\}-1000\backslash cdot\; x^\{10\}=1000\backslash cdot\; x^\{\backslash mathrm\{1T\}\}\backslash cdot\; y+100\backslash cdot\; x\backslash cdot\; y^\{\backslash mathrm\{1T\}\}+y^\{10\}=$

\begin{cases}

\mathrm{T000}\cdot x^{\mathrm{1T}}+100\cdot x+\mathrm{T}, & y=\mathrm{T}\\

0, & y=0\\

1000\cdot x^{\mathrm{1T}}+100\cdot x+1, & y=1

\end{cases}

Like division, we should check the value of half the 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

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 {{radic|2|3}} = 1.259921_dec = 1.1T1 000 111 001 T01 00T 1T1 T10 111_bal3.

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

:

The balanced ternary expansions of $\backslash pi$ is given in OEIS as {{OEIS link|A331313}}, that of $e$ in {{OEIS link|A331990}}.

File:Balanced ternary operation tables.svg

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 traditional binary and ternary. 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. In balanced ternary, the one-digit multiplication table remains one-digit and has no carry and the addition table has only two carries out of nine entries, compared to unbalanced ternary with one and three respectively. Knuth wrote that "Perhaps the symmetric properties and simple arithmetic of this number system will prove to be quite important some day," noting that,

{{quote|The complexity of arithmetic circuitry for balanced ternary arithmetic is not much greater than it is for the binary system, and a given number requires only $\backslash log\_3\; 2\; \backslash approx\; 63\; \backslash \%$ as many digit positions for its representation."}}

More recently, balanced ternary numbers have been proposed for some highly-efficient low-resolution implementations of artificial neural networks. In deep learning, neural nets usually use continuous (floating-point) values, but there are many works investigating quantisation and binarisation to create neural nets that can run with much lower power and/or lower memory requirements. Balanced ternary numbers are proposed to be used for the network parameters, because they are extremely compact, but can naturally represent excitatory/inhibitory/null activation patterns.{{cite arXiv

| last = Li

| first = Fengfu

| title = Ternary Weight Networks

|date = 2022

| class = cs.CV

| eprint = 1605.04711

}}{{cite arxiv

| last = Ma

| first = Shuming

| title = The era of 1-bit LLMs: All large language models are in 1.58 bits

| date = 2024

| arxiv = 2402.17764

}}

Balanced ternary may also provide a more natural representation for the qutrit and quantum computing systems that use it.

The theorem that every integer has a unique representation in balanced ternary was used by Leonhard Euler to justify the identity of formal power series{{cite journal

| last = Andrews | first = George E.

| doi = 10.1090/S0273-0979-07-01180-9

| issue = 4

| journal = Bulletin of the American Mathematical Society

| mr = 2338365

| pages = 561–573

| series = New Series

| title = Euler's "De Partitio numerorum"

| volume = 44

| year = 2007| doi-access = free

}}

:$\backslash prod\_\{n=0\}^\{\backslash infty\}\; \backslash left(x^\{-3^n\}+1+x^\{3^n\}\backslash right)=\backslash sum\_\{n=-\backslash infty\}^\{\backslash infty\}x^n.$

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 (60_dec = 1T1T0_bal3) 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¤ (7_dec = 1T1_bal3), the buyer pays 1¤ + 9¤ and receives 3¤ in change.

• Signed-digit representation
• Methods of computing square roots
• Numeral system
• Qutrit
• Salamis Tablet
• Ternary computer
• Setun, a ternary computer
• Ternary logic
• Generalized balanced ternary

{{reflist}}

{{reflist|group=note}}

{{commons category|Balanced ternary}}

• [http://www.computer-museum.ru/english/setun.htm Development of ternary computers at Moscow State University]
• [https://web.archive.org/web/20090312094241/http://abhijit.info/tristate/tristate.html Representation of Fractional Numbers in Balanced Ternary]
• [https://www.americanscientist.org/article/third-base "Third base"], ternary and balanced ternary number systems
• [https://web.archive.org/web/20110712221950/http://www.hostsrv.com/webmaa/app1/MSP/webm1010/ternary.msp The Balanced Ternary Number System] (includes decimal integer to balanced ternary converter)
• {{OEIS el|1=A182929|2=The binomial triangle reduced to balanced ternary lists|formalname=The rows of the binomial triangle reduced to balanced ternary lists encoded as decimal numbers}}
• [http://userpages.wittenberg.edu/bshelburne/BalancedTernaryTalkSu09.pdf Balanced (Signed) Ternary Notation] {{Webarchive|url=https://web.archive.org/web/20160303182504/http://userpages.wittenberg.edu/bshelburne/BalancedTernaryTalkSu09.pdf |date=2016-03-03 }} by Brian J. Shelburne (PDF file)
• [http://www.mortati.com/glusker/fowler/ The ternary calculating machine of Thomas Fowler] by Mark Glusker

Category:Computer arithmetic

Category:Non-standard positional numeral systems

Category:Ternary computers

Category:Numeral systems

de:Ternärsystem#Balanciert