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For example, in negadecimal (base -10): how to take the square root of $185_{-10}=25_{10}$?

Or in negabinary (base -2): the square root of $1100100_{-2}=36_{10}$?

Converting to another base, taking the square root and converting back to the negative base is not an option. I'm looking for an algorithm fully done in the negative base.

I take the square root of a binary number by dividing a number like $11001_2=25_{10}$ as follows: $01|10|01$

And then conditionally subtracting $01$ from each segment from left to right, adding a $1$ to the answer and appending the answer to the $01$ subtraction. enter image description here

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  • $\begingroup$ Why would anybody want to do that?! $\endgroup$ – Professor Vector Aug 3 '17 at 21:09
  • $\begingroup$ I'm interested in designing negabinary arithmetic circuits for mathematical interest. To design a square-root circuit, I first need a digit-by-digit algorithm. I don't want to use things like Newton's method. $\endgroup$ – gilianzz Aug 3 '17 at 21:13
  • $\begingroup$ What's the range of numbers your circuit will support? How accurate do you want square root to be? Do you have memory to devote to a lookup table or a partial lookup table? BTW, doing a square root digit by digit essentially means taking the square root of a polynomial, which in general requires taking a square root. For example, in base 10, $\sqrt{256}\approx 10\sqrt{2}+\sqrt{6}$ or $\sqrt{256}\approx 10\sqrt{2}+\frac{5\sqrt{2}}{4}$. These are both attempts at taking $\sqrt {256}$ digit by digit. The exact answer is $10\sqrt{2}+16-10\sqrt{2}$, which you might as well had obtained "normally". $\endgroup$ – Χpẘ Aug 3 '17 at 22:16
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    $\begingroup$ Do you know the classical pencil and paper algorithm for calculating square roots? If you do, then you should tell us what is going wrong with your attempt to adapt the algorithm to the negative bases. If you don't, then you should learn about them. $\endgroup$ – Rob Arthan Aug 3 '17 at 22:22
  • $\begingroup$ @Xpw: extracting square roots using pencil and paper, digit by digit, is one of the classical algorithms. Google will find you lots of information about this. It does not involve any calculation in quadratic number rings. $\endgroup$ – Rob Arthan Aug 3 '17 at 22:29
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The standard positive base $b$ square root algorithm is:

  1. Split the input digits into groups of two (such that the decimal point is one of the split points).
  2. Initialize $\mathit{remainder}$ to $0$ and $\mathit{output}$ to $0$.
  3. Append the next digit group to the end of $\mathit{remainder}$.
  4. Find the largest digit $d$ such that $(2 \cdot b \cdot \mathit{output} + d) \cdot d \le \mathit{remainder}$.
  5. Subtract $(2 \cdot b \cdot \mathit{output} + d) \cdot d$ from $\mathit{remainder}$.
  6. Append $d$ to $\mathit{output}$.
  7. While more digits are desired, repeat from step 3.

The reason this fails in a negative base is that step 4 is really looking for the first digit of the real-number solution to

$$\begin{align*} (b \cdot \mathit{output} + x)^2 &= b^{2n} \cdot \mathit{input} \\ \iff (2 \cdot b \cdot \mathit{output} + x) \cdot x &= b^{2n} \cdot \mathit{input} - (b \cdot \mathit{output})^2, \end{align*}$$

where $\mathit{remainder}$ is maintained as the integer part of the right side. A negative base presents four difficulties:

  • Because we’re outputting one digit at a time, on any given iteration, output may be of either sign, which means that $(2 \cdot b \cdot \mathit{output} + x) \cdot x$ may be either increasing or decreasing in $x$.
  • The first digit of $x$ may now be greater than $x$. The set of numbers starting with a digit $d$ is no longer $[d.000000\cdots, d.999999\cdots) = [d, d + 1)$, but rather $[d.909090\cdots, d.090909\cdots) = \left[d + \frac{b}{1 - b}, d + \frac{1}{1 - b}\right)$.
  • For the same reason, we might sometimes need more than two input digits to produce one output digit: $\sqrt{02.9921_{-10}} = 1.09_{-10}$ starts with $1$, but $\sqrt{02.81_{-10}} = 2.9_{-10}$ starts with $2$.
  • Without taking into account whether we’re outputting a digit of positive or negative place value, it’s easy to inadvertently produce the negative square root instead of the positive square root.

One strategy for addressing these difficulties involves building $\mathit{remainder}$ from the digits of $\mathit{input} \cdot (1 - b)$ instead of $\mathit{input}$. The points at which the first digit of $x$ changes are described by $x = d + \frac{1}{1 - b}$:

$$\left(2 \cdot b \cdot \mathit{output} + d + \frac{1}{1 - b}\right) \cdot \left(d + \frac{1}{1 - b}\right) = b^{2n} \cdot \mathit{input} - (b \cdot \mathit{output})^2 \\ \begin{multline} \iff (2 \cdot b \cdot \mathit{output} + d) \cdot d \cdot (1 - b) + 2 \cdot b \cdot \mathit{output} + 2 \cdot d + \frac{1}{1 - b} \\ = b^{2n} \cdot \mathit{input} \cdot (1 - b) - (b \cdot \mathit{output})^2 \cdot (1 - b). \end{multline}$$

The pre-decimal part of the right side also changes here (since the right side equals an integer plus $\frac{1}{1 - b}$), so maintaining that pre-decimal part as $\mathit{remainder}$ guarantees sufficient precision to pin down the first digit of $x$ correctly.

Taking this all into account, we can update the algorithm for negative base $b$ as follows:

  1. Multiply the input by $1 - b$.
  2. Split the resulting digits into groups of two (such that the decimal point is one of the split points).
  3. Initialize $\mathit{remainder}$ to $0$ and $\mathit{output}$ to $0$.
  4. Append the next digit group to the end of $\mathit{remainder}$.
    • If $\mathit{output} = 0$ and we’re outputting a digit of negative place value, set $d = 0$.
    • Otherwise if $\mathit{output} \le 0$, find the smallest digit $d$ such that $$(2 \cdot b \cdot \mathit{output} + d) \cdot d \cdot (1 - b) + 2 \cdot b \cdot \mathit{output} + 2 \cdot d \ge \mathit{remainder}.$$
    • Otherwise, find the smallest digit d such that $$(2 \cdot b \cdot \mathit{output} + d) \cdot d \cdot (1 - b) + 2 \cdot b \cdot \mathit{output} + 2 \cdot d < \mathit{remainder}.$$
  5. Subtract $(2 \cdot b \cdot \mathit{output} + d) \cdot d \cdot (1 - b)$ from $\mathit{remainder}$.
  6. Append $d$ to $\mathit{output}$.
  7. While more digits are desired, repeat from step 4.
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  • $\begingroup$ Could you explain how you found the expressions in step 5? $\endgroup$ – gilianzz Jan 12 at 21:26
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    $\begingroup$ @gilianzz Yeah, I’ve added an expanded explanation. $\endgroup$ – Anders Kaseorg Jan 13 at 0:18

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