# Newtons Method by Recursive Quines

Your task is to calculate the square root of 2 using Newton's Method - with a slight twist. Your program is to calculate an iteration using Newton's Method, and output the source code for the following iteration (which must be able to do the same).

Newton's method is fairly exhaustively described on Wikipedia

To calculate square root 2 using Newtons method, you:

• Define f(x) = x^2 - 2
• Define f'(x) = 2x
• Define x[0] (the initial guess) = 1
• Define x[n+1] = x[n] - (f[n] / f'[n])

Each iteration will move x[n] closer to the square root of two. So -

• x[0] = 1
• x[1] = x[0] - f(x[0])/f'(x[0]) = 1 - (1 ^ 2 - 2) / (2 * 1) = 1.5
• x[2] = x[1] - f(x[1])/f'(x[1]) = 1.5 - (1.5 ^ 2 - 2) / (2 * 1.5) = 1.416666667
• x[3] = x[2] - f(x[2])/f'(x[1]) = 1.416666667 - (1.416666667 ^ 2 - 2) / (2 * 1.416666667) = 1.414215686
• and so on

• Calculate x[n] where n is the amount of times the program has been run
• Output the source code to a valid program in the same language which must calculate x[n+1] and satisfy the same criteria of this question.
• The first line of the source code must be the calculate result, properly commented. If the source requires something particular (such as a shebang) on the first line, the result may be put on the second line.

Note that

• Your program must use an initial guess of x[0] = 1
• The Standard Loopholes apply
• Any in-build power, square root or xroot functions are forbidden
• Your program must not accept any input whatsoever. It must be entirely self contained.

Your score is the size of your initial program in UTF-8 bytes. The lowest score wins.

• Do we have to define the functions, or can we simplify by writing x = x-(x*x-2)/(2*x)? – Kyle Kanos Jun 18 '14 at 13:31
• That simplification looks valid to me. As long as it performs the calculation using Newton's Method – lochok Jun 18 '14 at 22:02
• Does the program output the approximation, or just the source code? Can it take as its input the previous solution? – Emily Jun 19 '14 at 16:03
• It has to output the approximation (commented) on the first line, with the source code for the next iteration. The approximation may be preceded by a shebang if the language requires it. The program (nor the program it produces) must not accept any input. – lochok Jun 22 '14 at 7:26

# Common Lisp, 2239568 66

(#1=(lambda(x p)(format t"~S~%~S"p(,x',x,(+(/ p 2)(/ p)))))'#1#1)


Now that I read the problem statement more carefully (thanks, primo!) I noticed that the first line must be the result of calculation, not that it needs to contain the result. Thus, I think my earlier attempts did not quite follow the rules. This one should.

Example use (SBCL 1.1.15):

$sbcl --script nq.lisp | tee nq2.lisp 1 ((LAMBDA (X P) (FORMAT T "~S~%~S" P (,X ',X ,(+ (/ P 2) (/ P))))) '(LAMBDA (X P) (FORMAT T "~S~%~S" P (,X ',X ,(+ (/ P 2) (/ P))))) 3/2)$ sbcl --script nq2.lisp | tee nq3.lisp
3/2
((LAMBDA (X P) (FORMAT T "~S~%~S" P (,X ',X ,(+ (/ P 2) (/ P)))))
'(LAMBDA (X P) (FORMAT T "~S~%~S" P (,X ',X ,(+ (/ P 2) (/ P))))) 17/12)
$sbcl --script nq3.lisp | tee nq4.lisp 17/12 ((LAMBDA (X P) (FORMAT T "~S~%~S" P (,X ',X ,(+ (/ P 2) (/ P))))) '(LAMBDA (X P) (FORMAT T "~S~%~S" P (,X ',X ,(+ (/ P 2) (/ P))))) 577/408)$ sbcl --script nq4.lisp | tee nq5.lisp
577/408
((LAMBDA (X P) (FORMAT T "~S~%~S" P (,X ',X ,(+ (/ P 2) (/ P)))))
'(LAMBDA (X P) (FORMAT T "~S~%~S" P (,X ',X ,(+ (/ P 2) (/ P)))))
665857/470832)
$sbcl --script nq5.lisp | tee nq6.lisp 665857/470832 ((LAMBDA (X P) (FORMAT T "~S~%~S" P (,X ',X ,(+ (/ P 2) (/ P))))) '(LAMBDA (X P) (FORMAT T "~S~%~S" P (,X ',X ,(+ (/ P 2) (/ P))))) 886731088897/627013566048)$

• I've been mostly testing with CCL, but it works similarly with both SBCL and CLISP. – jlahd Jun 19 '14 at 5:42
• This is more like I was expecting. +1 – primo Jun 19 '14 at 13:53

## Python 60 bytes

x=1.
o='x=%s\no=%r;print o%%(x/2+1/x,o)';print o%(x/2+1/x,o)


I've simplified the formula slightly, using the following substitutions:

  x-(x²-2)/(2x)
= (2x²)/(2x)-(x²-2)/(2x)
= (2x²-x²+2)/(2x)
= (x²+2)/(2x)
= (x+2/x)/2
= x/2+1/x


I hope that's not an issue.

The program proceeds in the following manner:

$python newton-quine.py x=1.5 o='x=%s\no=%r;print o%%(x/2+1/x,o)';print o%(x/2+1/x,o)$ python newton-quine.py
x=1.41666666667
o='x=%s\no=%r;print o%%(x/2+1/x,o)';print o%(x/2+1/x,o)

$python newton-quine.py x=1.41421568627 o='x=%s\no=%r;print o%%(x/2+1/x,o)';print o%(x/2+1/x,o)$ python newton-quine.py
x=1.41421356237
o='x=%s\no=%r;print o%%(x/2+1/x,o)';print o%(x/2+1/x,o)


etc.

• I don't know if this is legal or not, but you can shorten your initial code to g="x=%s;o=%r;print o%%(x/2+1/x,o)";print g%(1.5,g) @ 50 chars. – cjfaure Jun 18 '14 at 16:11
• @Trimsty I think it's slightly problematic that 1) it doesn't actually calculate the first iteration, and that 2) the first line doesn't contain the current result. As I understand the problem description, both the original program and later generations should satisfy these criteria. – primo Jun 18 '14 at 16:41

# CJam, 20 bytes

1
{\d_2/1@/+p"_~"}_~


Try it online.

### Output

$cjam <(echo -e '1\n{\d_2/1@/+p"_~"}_~'); echo 1.5 {\d_2/1@/+p"_~"}_~$ cjam <(cjam <(echo -e '1\n{\d_2/1@/+p"_~"}_~')); echo
1.4166666666666665
{\d_2/1@/+p"_~"}_~
$cjam <(cjam <(cjam <(echo -e '1\n{\d_2/1@/+p"_~"}_~'))); echo 1.4142156862745097 {\d_2/1@/+p"_~"}_~$ cjam <(cjam <(cjam <(cjam <(echo -e '1\n{\d_2/1@/+p"_~"}_~')))); echo
1.4142135623746899
{\d_2/1@/+p"_~"}_~


### How it works

1       " Push the initial guess.                                                 ";
{       "                                                                         ";
\d    " Swap the code block with the initial guess and cast to Double.          ";
_2/   " Duplicate the initial guess and divide the copy by 2.                   ";
1@/   " Push 1, rotate the initial guess on top and divide.                     ";
+p    " Add the quotients and print.                                            ";
"_~"  " Push the string '_~'.                                                   ";
}       "                                                                         ";
_~      " Duplicate the code block (to leave a copy on the stack) and execute it. ";

• Well that's impressive. +1 – Kyle Kanos Jun 18 '14 at 18:01

# ECMAScript 6, 38 36

(f=x=>"(f="+f+")("+(x/2+1/x)+")")(1)
(f=x=>"(f="+f+")("+(x/2+1/x)+")")(1.5)
(f=x=>"(f="+f+")("+(x/2+1/x)+")")(1.4166666666666665)
(f=x=>"(f="+f+")("+(x/2+1/x)+")")(1.4142156862745097)
(f=x=>"(f="+f+")("+(x/2+1/x)+")")(1.4142135623746899)


## JavaScript, 51

(function f(x){return "("+f+")("+(x/2+1/x)+")"})(1)


This is the same as the above, for older browsers.

• Sometimes I'm just amazed how simple javascript can make things. +1 – seequ Jun 19 '14 at 8:34
• This seems to be lacking any sort of output (print, putstr, console.log, etc.). – primo Jun 20 '14 at 4:22
• @primo - When JavaScript is run in a console, the returned value is automatically printed. – Derek 朕會功夫 Aug 17 '14 at 1:30
• @Derek朕會功夫 Very many languages can be run as REPL - this is an expression, and not a complete program. See: Standard “loopholes” which are no longer funny. – primo Aug 28 '14 at 10:41
• @Derek朕會功夫 The problem description specifically asks for a program - in several places. The program supplied does nothing. Witness: i.stack.imgur.com/Te7Vf.png The above is an expression that evaluates to an expression. It has merit of its own, but it is not a program. – primo Aug 28 '14 at 15:29

# Lua 129

Probably way too long, but the Lua quine sucks because the nested [[ ]] is a deprecated feature. But it works regardless:

x=1.0;x=x/2.+1./x;l=[[io.write('x=',x,';x=x/2.+1./x;l=[','[',l,']','];',l)]];io.write('x=',x,';x=x/2.+1./x;l=[','[',l,']','];',l)


It's a bit nicer to see if you add newlines instead of colons:

x=1.0
x=x/2.+1./x
l=[[io.write('x=',x,'\nx=x/2.+1./x\nl=[','[',l,']','];',l)]];io.write('x=',x,'\nx=x/2.+1./x\nl=[','[',l,']','];',l)


# J - 102 88 bytes

This is as horrible as I'm at making quines (I will probably revise this when I get better ideas). J's floats are limited to 5 decimal places, but by replacing first line with x=:1x it would be a fraction with infinite precision.

Edit 1: I got better idea. Also added the explanation.


x=:1
((3&{.,[:":(x%2)+1%x"_),:3&}.,],{:,{:)'x=:((3&{.,[:":(x%2)+1%x"_),:3&}.,],{:,{:)'''


First few iterations:

x=:1.5
((3&{.,[:":(x%2)+1%x"_),:3&}.,],{:,{:)'x=:((3&{.,[:":(x%2)+1%x"_),:3&}.,],{:,{:)'''

x=:1.41667
((3&{.,[:":(x%2)+1%x"_),:3&}.,],{:,{:)'x=:((3&{.,[:":(x%2)+1%x"_),:3&}.,],{:,{:)'''

x=:1.41422
((3&{.,[:":(x%2)+1%x"_),:3&}.,],{:,{:)'x=:((3&{.,[:":(x%2)+1%x"_),:3&}.,],{:,{:)'''


### Explanation

((3&{.,[:":(x%2)+1%x"_),:3&}.,],{:,{:)'x=:((3&{.,[:":(x%2)+1%x"_),:3&}.,],{:,{:)'''
((3&{.,[:":(x%2)+1%x"_),:3&}.,],{:,{:) The quine-function
3&}.,],{:,{:  Build the second row
3&}.          Get everything but the first 3 characters from the string
,]        Get the whole string and concat
,{:     Get the last item (') and concat
,{:  -||-
(3&{.,[:":(x%2)+1%x"_)                Build the first row
[:":(x%2)+1%x"_                 Calculate x/2 + 1/x (stolen from Pythoneer) and stringify
3&{.                                 Take the first 3 characters from the string (x=:)
,                                Concatenate 'x=:' and the result
,:              Concatenate the two rows

• I actually love how simple this program is (for serious). – seequ Jun 18 '14 at 19:45
• If I get more time, I'm going to see if I can modify the above for Kona. – Kyle Kanos Jun 19 '14 at 16:10
• @KyleKanos At least the digit-rotation-thingy was similar enough, but I don't know Kona. Good luck! :) – seequ Jun 19 '14 at 16:35
• 1%x is the same as %x. Instead of (x%2)+1%x, you can do (%&2+%)x. – Conor O'Brien Feb 10 '17 at 15:57

# Ruby, 65

x=1.0
puts"x=#{x/2+1/x}",<<'1'*2,1
puts"x=#{x/2+1/x}",<<'1'*2,1
1
`

As too often happens, this is almost a straight port of the Python solution.