quater-imaginary base

The quater-imaginary numeral system is a numeral system, first proposed by Donald Knuth in 1960. Unlike standard numeral systems, which use an integer (such as 10 in decimal, or 2 in binary) as their bases, it uses the imaginary number 2''i'' (equivalent to $\sqrt$) as its base. It is able to (almost) uniquely represent every complex number using only the digits 0, 1, 2, and 3. Numbers less than zero, which are ordinarily represented with a minus sign, are representable as digit strings in quater-imaginary; for example, the number −1 is represented as "103" in quater-imaginary notation.

Decomposing the quater-imaginary

In a positional system with base $b$,

$\ldots d_3d_2d_1d_0.d_d_d_\ldots$

represents$\dots + d_3\cdot b^3+d_2\cdot b^2+d_1\cdot b+d_0+d_\cdot b^+d_\cdot b^+d_\cdot b^\dots$

In this numeral system, $b = 2i$,

and because $(2i)^2=-4$, the entire series of powers can be separated into two different series, so that it simplifies to \begin
& dots+d_4\cdot(-4)^2 +d_2\cdot(-4)^1+d_0+d_\cdot(-4)^+\dots \end for even-numbered digits (digits that simplify to the value of the digit times a power of -4), and \begin
2i\cdot dots+d_3\cdot(-4)^1+d_1+d_\cdot (-4)^+d_\cdot (-4)^ + \dots\end for those digits that still have an imaginary factor. Adding these two series together then gives the total value of the number.

Because of the separation of these two series, the real and imaginary parts of complex numbers are readily expressed in base −4 as $\ldots d_4d_2d_0.d_\ldots$ and $2\cdot(\ldots d_5d_3d_1.d_d_\ldots)$ respectively.

Converting from quater-imaginary

To convert a digit string from the quater-imaginary system to the decimal system, the standard formula for positional number systems can be used. This says that a digit string $\ldots d_3d_2d_1d_0$ in base ''b'' can be converted to a decimal number using the formula
:$\cdots + d_3\cdot b^3+d_2\cdot b^2+d_1\cdot b+d_0$
For the quater-imaginary system, $b = 2i$.

Additionally, for a given string $d$ in the form $d_, d_,\dots d_0$, the formula below can be used for a given string length $w$ in base $b$
:$Q2D_w \vec d \equiv \sum_^ d_k \cdot b^k$

Example

To convert the string $1101_$ to a decimal number, fill in the formula above:
:$1\cdot(2i)^3 + 1\cdot(2i)^2 + 0\cdot(2i)^1 + 1\cdot(2i)^0 = -8i - 4 + 0 + 1 = -3 - 8i$

Another, longer example: $1030003_$ in base 10 is
:$1\cdot(2i)^6 + 3\cdot(2i)^4 + 3\cdot(2i)^0 = -64 + 3\cdot 16 + 3 = -13$

Converting into quater-imaginary

It is also possible to convert a decimal number to a number in the quater-imaginary system. Every complex number (every number of the form ''a''+''bi'') has a quater-imaginary representation. Most numbers have a unique quater-imaginary representation, but just as 1 has the two representations 1 = 0. in decimal notation, so, because of 0._2''i'' = , the number has the two quater-imaginary representations 0._2''i'' = 3· = = 1 + 3· = 1._2''i''.

To convert an arbitrary complex number to quater-imaginary, it is sufficient to split the number into its real and imaginary components, convert each of those separately, and then add the results by interleaving the digits. For example, since −1+4''i'' is equal to −1 plus 4''i'', the quater-imaginary representation of −1+4''i'' is the quater-imaginary representation of −1 (namely, 103) plus the quater-imaginary representation of 4''i'' (namely, 20), which gives a final result of −1+4''i'' = 123_2''i''.

To find the quater-imaginary representation of the imaginary component, it suffices to multiply that component by 2''i'', which gives a real number; then find the quater-imaginary representation of that real number, and finally shift the representation by one place to the right (thus dividing by 2''i''). For example, the quater-imaginary representation of 6''i'' is calculated by multiplying 6''i'' × 2''i'' = −12, which is expressed as 300_2''i'', and then shifting by one place to the right, yielding: 6''i'' = 30_2''i''.

Finding the quater-imaginary representation of an arbitrary real integer number can be done manually by solving a system of simultaneous equations, as shown below, but there are faster methods for both real and imaginary integers, as shown in the negative base article.

Example: Real number

As an example of an integer number we can try to find the quater-imaginary counterpart of the decimal number 7 (or 7₁₀ since the base of the decimal system is 10). Since it is hard to predict exactly how long the digit string will be for a given decimal number, it is safe to assume a fairly large string. In this case, a string of six digits can be chosen. When an initial guess at the size of the string eventually turns out to be insufficient, a larger string can be used.

To find the representation, first write out the general formula, and group terms:
:$\begin 7_& = d_+d_\cdot b+d_\cdot b^+d_\cdot b^+d_\cdot b^+d_\cdot b^ \\ & = d_+2id_-4d_-8id_+16d_+32id_ \\ & = d_-4d_+16d_+i(2d_-8d_+32d_) \\ \end$

Since 7 is a real number, it is allowed to conclude that ''d₁'', ''d₃'' and ''d₅'' should be zero. Now the value of the coefficients ''d₀'', ''d₂'' and ''d₄'', must be found. Because d₀ − 4 d₂ + 16 d₄ = 7 and because—by the nature of the quater-imaginary system—the coefficients can only be 0, 1, 2 or 3 the value of the coefficients can be found. A possible configuration could be: ''d₀'' = 3, ''d₂'' = 3 and ''d₄'' = 1. This configuration gives the resulting digit string for 7₁₀.

:$7_ = 010303_ = 10303_.$

Example: Imaginary number

Finding a quater-imaginary representation of a purely imaginary integer number is analogous to the method described above for a real number. For example, to find the representation of 6''i'', it is possible to use the general formula. Then all coefficients of the real part have to be zero and the complex part should make 6. However, for 6''i'' it is easily seen by looking at the formula that if ''d₁'' = 3 and all other coefficients are zero, we get the desired string for 6''i''. That is:

:$\begin6i_ = 30_\end$

Another conversion method

For real numbers the quater-imaginary representation is the same as negative quaternary (base −4). A complex number ''x''+''iy'' can be converted to quater-imaginary by converting ''x'' and ''y''/2 separately to negative quaternary. If both ''x'' and ''y'' are finite binary fractions we can use the following algorithm using repeated Euclidean division:

For example: 35+23i=121003.2_2i

35 23i/2i=11.5 11=12−0.5
35÷(−4)=−8, remainder 3 12/(−4)=−3, remainder 0 (−0.5)×(−4)=2
−8÷(−4)= 2, remainder 0 −3/(−4)= 1, remainder 1
2÷(−4)= 0, remainder 2 1/(−4)= 0, remainder 1
20003 + 101000 + 0.2 = 121003.2
32i+16×2−8i−4×0+2i×0+1×3−2×i/2=35+23i

Radix point "."

A radix point in the decimal system is the usual . (dot) which marks the separation between the integer part and the fractional part of the number.
In the quater-imaginary system a radix point can also be used. For a digit string $\dots d_d_d_d_d_d_ . d_d_d_ \dots$ the radix point marks the separation between non-negative and negative powers of ''b''. Using the radix point the general formula becomes:

:$d_5 b^5 + d_4 b^4 + d_3 b^3 + d_2 b^2 + d_1 b + d_0 + d_ b^ + d_ b^ + d_ b^$
or
:$\begin 32id_+16d_-8id_-4d_+2id_+d_+\fracd_+\fracd_+\fracd_\\ =32id_+16d_-8id_-4d_+2id_+d_-\fracd_-\fracd_+\fracd_ \end$

Example

If the quater-imaginary representation of the complex unit ''i'' has to be found, the formula without radix point will not suffice. Therefore, the above formula should be used. Hence:

:$\begin i & = 32id_+16d_-8id_-4d_+2id_+d_+\fracd_+\fracd_+\fracd_\\ & = i(32d_-8d_+2d_-\fracd_+\fracd_)+16d_-4d_+d_-\fracd_\\ \end$

for certain coefficients ''d_k''. Then because the real part has to be zero: ''d''₄ = ''d''₂ = ''d''₀ = ''d''₋₂ = 0.
For the imaginary part, if ''d''₅ = ''d''₃ = ''d''₋₃ = 0 and when ''d''₁ = 1 and ''d''₋₁ = 2 the digit string can be found. Using the above coefficients in the digit string the result is:

:$i = 10.2_.$

Addition and subtraction

It is possible to add and subtract numbers in the quater-imaginary system. In doing this, there are two basic rules that have to be kept in mind:

# Whenever a number exceeds 3, ''subtract'' 4 and "carry" −1 two places to the left.
# Whenever a number drops below 0, ''add'' 4 and "carry" +1 two places to the left.

Or for short: "If you add four, carry +1. If you subtract four, carry −1". This is the opposite of normal long addition, in which a "carry" in the current column requires ''adding'' 1 to the next column to the left, and a "borrow" requires subtracting. In quater-imaginary arithmetic, a "carry" ''subtracts'' from the next-but-one column, and a "borrow" ''adds''.

Example: Addition

Below are two examples of adding in the quater-imaginary system:

   1 − 2i                1031             3 − 4i                 1023
   1 − 2i                1031             1 − 8i                 1001
   ------- +     <=>     ----- +          ------- +      <=>   ------- +
   2 − 4i                1022             4 − 12i               12320

In the first example we start by adding the two 1s in the first column (the "ones' column"), giving 2. Then we add the two 3s in the second column (the "2''i''s column"), giving 6; 6 is greater than 3, so we subtract 4 (giving 2 as the result in the second column) and carry −1 into the fourth column. Adding the 0s in the third column gives 0; and finally adding the two 1s and the carried −1 in the fourth column gives 1.

In the second example we first add 3+1, giving 4; 4 is greater than 3, so we subtract 4 (giving 0) and carry −1 into the third column (the "−4s column"). Then we add 2+0 in the second column, giving 2. In the third column, we have 0+0+(−1), because of the carry; −1 is less than 0, so we add 4 (giving 3 as the result in the third column) and "borrow" +1 into the fifth column. In the fourth column, 1+1 is 2; and the carry in the fifth column gives 1, for a result of $12320_$.

Example: Subtraction

Subtraction is analogous to addition in that it uses the same two rules described above. Below is an example:

         − 2 − 8i                       1102
           1 − 6i                       1011
           -------           <=>        -----
         − 3 − 2i                       1131

In this example we have to subtract $1011_$ from $1102_$. The rightmost digit is 2−1 = 1. The second digit from the right would become −1, so add 4 to give 3 and then carry +1 two places to the left. The third digit from the right is 1−0 = 1. Then the leftmost digit is 1−1 plus 1 from the carry, giving 1. This gives a final answer of $1131_$.

Multiplication

For long multiplication in the quater-imaginary system, the two rules stated above are used as well. When multiplying numbers, multiply the first string by each digit in the second string consecutively and add the resulting strings. With every multiplication, a digit in the second string is multiplied with the first string. The multiplication starts with the rightmost digit in the second string and then moves leftward by one digit, multiplying each digit with the first string.
Then the resulting partial products are added where each is shifted to the left by one digit. An example:

              11201
              20121  ×
        ---------------
              11201      ←––– 1 × 11201
             12002       ←––– 2 × 11201
            11201        ←––– 1 × 11201
           00000         ←––– 0 × 11201
          12002      +   ←––– 2 × 11201
        ---------------
          120231321

This corresponds to a multiplication of $(9-8i)\cdot(29+4i) = 293-196i$.

Tabulated conversions

Below is a table of some decimal and complex numbers and their quater-imaginary counterparts.

Examples

Below are some other examples of conversions from decimal numbers to quater-imaginary numbers.

:$5 = 16 + (3\cdot-4) + 1 = 10301_$

:$i = 2i + 2\left(-\fraci\right) = 10.2_$

:$7 \frac - 7 \fraci = 1(16) + 1(-8i) + 2(-4) + 1(2i) + 3\left(-\fraci\right) + 1\left(-\frac\right) = 11210.31_$

Z-order curve

The representation
:$z = \sum_ z_k \cdot (2i)^$
of an arbitrary complex number $z\in \Complex$ with $z_k\in \$ gives rise to an injective mapping
:$\textstyle \begin \varphi \colon & \Complex & \to & \R \\ & \sum_ z_k \cdot (2i)^ & \mapsto & \sum_ z_k \cdot r^\\ \end$
with some suitable $r \in \Z$. Here $r = 4$ cannot be taken as base because of
:$\textstyle \sum_ 3\cdot (2i)^ = \tfrac5 \; \; \; \; \ne \; \; \; \; 1 = \sum_ 3 \cdot 4^.$
The image $\varphi(\Complex) \subset \R$ is a Cantor set which allows to linearly order $\Complex$ similar to a Z-order curve. Since the image is disconnected, $\varphi$ is not continuous.

See also

* Quaternary numeral system
* Complex-base system
* Negative base

References

Further reading

*
*

{{Donald Knuth navbox

Non-standard positional numeral systems
Donald Knuth
Complex numbers