Math Trick: Calculating the Square Root by Hand | Hacker News Reader

Math Trick: Calculating the Square Root by Hand

itl.nist.gov

48 pointsaneesh20 comments

Much faster algorithm, calling your input value X:

1. Move the decimal point 2N digits to the left, so that your value X is of the form a.bcdef... or ab.cdef... Remember the value N.

2. If X now has one digit before the decimal place, set Y = 5. Otherwise, set Y = 2.

3. Compute Y = Y * (300 - X * Y * Y) / 200 to as many decimal places as you want. Feel free to start by computing only a few decimal places.

4. Repeat step 2 until you're computing Y to the number of decimal places you want in your answer and it's not changing.

5. Compute Y = X * Y.

6. Move the decimal point in Y to the right by N-1 places.

I'm sorry, but this is totally unclear (to me at least).

For example what are you computing (in part 3) if you know what X and Y are?

What is N?

What is "the number of decimal places you want"?

Can you explain in another way? Do you have an example? Why does this work?

Part 3 makes perfect sense to me, it's an "x = x+1" style line.

N is (((the number of digits of X), divided by 2), rounded down.)

Edit: hmm, maybe I've got it wrong, the convergence is very slow. Anyways, here's how it works out for me:

  X = 105362
  X = 10.5362 so N = 2 (half the number of digits to the right of the decimal point...)
  Y = 2
  Y = 2*(300-10.5362*2*2) = 2.578552
  Y = 2.578552*(300-10.5362*2.578552*2.578552) = 2.9646326
  Y = 2.9646326*(300-10.5362*2.9646326*2.9646326) = 3.0742773

It should converge to around 3.24, at which point you would shift the decimal point back 2 digits to get 324.

I guess I'm confused by the fact that you are repeating step 2 which reassigns Y...

Clear on N thanks.

An example would make this easier...

Yeah, it should read "repeat step three". Frequently, if you want to find x satisfying f(x) = x you can just let x be the limit of the sequence t, f(t), f(f(t)), f(f(f(t))) where t is some arbitrary number. In this case, we have:

  f(Y) = Y*(300-XYY)/200

If f(Y) = Y, then:

  Y = Y*(300-XYY)/200
  1 = (300-XYY)/200
  200 = 300 - XYY
  100 = XYY

So it looks like the algorithm calculates an inverse square root instead, and then multiplies it by the original number (step 5.)

Yeah, it should read "repeat step three".

Oops, quite right.

So it looks like the algorithm calculates an inverse square root instead, and then multiplies it by the original number (step 5.)

Exactly. The nice thing about this method is that it converges nicely even if you make errors (arithmetical or rounding) along the way -- so you can start by working with only a few digits and only use full precision right at the end.

You could compute square roots using a direct NR iteration (Y := (Y + X/Y) / 2), but that requires that you perform a division at each step -- most people find multiplications to be easier and faster.

Nice, but it is hardly fit for computing a root by hand.

I was taught this method of calculating a square root by hand when I was a child in public school. I have since been surprised by the number of people my own age who were never taught.

Since many Canadian high schools only teach "pre-Calculus" now, it makes me wonder if anyone is learning the "long hand" method for square roots these days.

Same with me, learned this in 8th grade (although I forgot most of the details since). 1996, Romania.

For those intrigued by this kind of thing, this book (paper only, but the web page has more resources) may be interesting:

Dead Reckoning: Calculating Without Instruments, by Ronald W. Doerfler (http://www.myreckonings.com/Dead_Reckoning/Dead_Reckoning.ht... )

More good stuff at http://myreckonings.com/wordpress/ . The nomography posts had caught my eye a while ago, and there may be more of those coming soon.

(Disclosure: I recently exchanged a couple of emails with the author, after leaving a comment on the Heaviside post which was briefly featured here.)

Thinking in powers of two, aids to get faster estimates of square roots. When you are calculating just by hand, you can tolerate errors associated with this method.

For example, the number of the explication is near to 2^16, so the square root is 2^8 = 256 approximately (with relative error of 0.7% and in less of a second :)

Of course, it won't work when numbers are farther of powers of two.

You just have to be familiar with how to scale numbers logarithmically. (That is, half-way between two units is 70% of the value, 1/10th of the way is 30% value, and so on.)

"What is the largest number that we can put in the units and multiply times the divisor and still be less than or equal to what we have?"

Oh my goodness.

"Algebraically, what is d such that d × 4d ≤ 265?" made it much clearer. :)

Just read on, the process becomes obvious after the next iteration. It is very simple actually.

Oh yes. I just thought that sentence was fantastic!

We found longhand square root instructions with worked examples in my Grandma's recipe collection. :)

dare I ask, why?

Among other things, knowing how to solve a problem with limited means helps you understand more precisely how hard it is.

My dad and I were on a backpacking trip and while hiking we calculated how thin the atmosphere was compared to sea level.

It feels more elegant sometimes to solve a problem simply (in paper, in your head) without bringing out the computers.

See Dijkstra on sum/product puzzle: http://www.cs.utexas.edu/users/EWD/transcriptions/EWD06xx/EW...

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