Generate MathJax for the Golden Ratio Continued Fraction

Question

In anticipation of MathJax being temporarily disabled, the rendered MathJax in this question has been replaced with images. You are still welcome to post answers but you'll have to view the rendered MathJax on another site.

PPCG just got MathJax! This means we can now easily include well formatted mathematical formulas into posts. (Handy MathJax tutorial.)

For example, here is the golden ratio expressed as an infinite continued fraction:

eq0

The MathJax code for this equation is

$$\varphi=1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}}$$

^{You can find this by right clicking the formula and following Show Math As → TeX Commands.

The $$ means it is displayed on its own in the center of the page instead of inline. Use a single $ for inline.}

Challenge

Write a program that takes in a non-negative integer, n, and outputs the MathJax code for that many "steps" of the continued fraction for the golden ratio.

To keep things standard across answers, you must use this exact MathJax syntax:

For n = 0, the output must be $$\varphi=1+\dots$$.
Which is rendered as:
For n = 1, the output must be $$\varphi=1+\cfrac1{1+\ddots}$$.
Which is rendered as:
For n = 2, the output must be $$\varphi=1+\cfrac1{1+\cfrac1{1+\ddots}}$$.
Which is rendered as:
For n = 3, the output must be $$\varphi=1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}$$.
Which is rendered as:

This pattern continues on for larger n. You could say that n represents the number of division lines in the equation.

Notes

\cfrac is used instead of the more common \frac.
\dots is used instead of \ddots for n = 0.
Take input from stdin or the command line.
Output to stdout (with an optional trailing newline).
Alternatively, you may write a function that takes in n as an integer and returns the MathJax code as a string (or still prints it).

Scoring

The smallest submission in bytes wins. Tiebreaker goes to the earlier submission.

Just a note to those wanting to run the stack snippet: Like many (most?) stack snippets, this doesn't work in Safari. — Alex A., Commented Apr 8, 2015 at 0:53
The stack snippet doesn't work when you type stuff... it gives Uncaught ReferenceError: textbox is not defined — soktinpk, Commented Apr 8, 2015 at 2:50
@soktinpk That's strange, I'm having the same problem. But the snippet over here works even though it's the exact same... Here's an external byte counter in case. — Calvin's Hobbies, Commented Apr 8, 2015 at 2:57

Sp3000 · Accepted Answer · 2015-04-08 02:02:45Z

10

Python, 70 68 67 bytes

lambda n:"$$\\varphi=1+\%sdots%s$$"%("cfrac1{1+\\"*n+"d"[:n],"}"*n)

This defines an anonymous function which just uses simple string multiplication and string formatting.

(Thanks to @xnor for pointing out that \\c can just be written as \c, since c can't be escaped. Unfortunately this doesn't hold true for \\v, since \v is ASCII 11.)

Previous attempts:

lambda n:"$$\\varphi="+"1+\\cfrac1{"*n+"1+\\"+"ddots"[n<1:]+"}"*n+"$$"
lambda n:r"$$\varphi=%s1+\%s$$"%("1+\cfrac1{"*n,"ddots"[n<1:]+"}"*n)

edited Apr 8, 2015 at 2:02

answered Apr 8, 2015 at 1:20

Sp3000

61.9k13 gold badges115 silver badges287 bronze badges

\$\begingroup\$ You beat me by 14 seconds! Also 70 chars. \$\endgroup\$
– xnor
Commented Apr 8, 2015 at 1:21
2

\$\begingroup\$ I think this works without using a double \ before cfrac. \$\endgroup\$
– xnor
Commented Apr 8, 2015 at 1:29
\$\begingroup\$ @xnor Seems like it, thanks! And sorry for taking the Python golfs all the time... \$\endgroup\$
– Sp3000
Commented Apr 8, 2015 at 1:33
\$\begingroup\$ Nah, I've stolen my fair share of races to post from you. \$\endgroup\$
– xnor
Commented Apr 8, 2015 at 1:34

Add a comment |

Martin Ender · Accepted Answer · 2015-04-11 22:13:54Z

6

CJam, 51 50 bytes

$$\varphi=1+""\cfrac1{1+"ri:R*'\"ddots"R!>'}R*'$_

Code explanation:

"$$\varphi=1+"             "This is a static string";
  "\cfrac1{1+"ri:R*'\      "Repeat this string input number times. Put a \ at the end";
    "ddots"R!>             "If input is 0, remove 1st characters, else not";
      '}R*                 "Put the closing bracket R times";
        '$_                "The final $$";

Few examples:

N = 0

$$\varphi=1+\dots$$

N = 4

$$\varphi=1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}}$$

N = 15

$$\varphi=1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}}}}}}}}}}}}}$$

UPDATE - 1 byte saved thanks to Sp3000!

Try it online here

edited Apr 11, 2015 at 22:13

Martin Ender

197k67 gold badges447 silver badges975 bronze badges

answered Apr 8, 2015 at 7:26

Optimizer

26.6k7 gold badges66 silver badges142 bronze badges

1

\$\begingroup\$ A little shuffling gives 50: "$$\varphi=1+""\cfrac1{1+"ri:R*'\"ddots"R!>'}R*'$_ \$\endgroup\$
– Sp3000
Commented Apr 8, 2015 at 9:14
\$\begingroup\$ I didn't look at your answer at all - just coincidence. Either way, your 49 byte Pyth solution is 50 bytes in reality, because you have to escape \v to \\v. \$\endgroup\$
– orlp
Commented Apr 8, 2015 at 11:49
\$\begingroup\$ @orlp and that is why I rolled back as there was no potential benefit in that solution being present here as well. \$\endgroup\$
– Optimizer
Commented Apr 8, 2015 at 12:20

Add a comment |

Sp3000 · Accepted Answer · 2015-04-08 06:07:38Z

4

><>, 89 86 + 3 = 89 bytes

:&"$$"{\l?!;o70.
}-1v!?:<{"}"
&:&\~"stod"&:&?:
{1->:?!v}"\+1{1carfc"
rav\$$"\~"\+1=ihp

Run with the -v flag, e.g.

py -3 fish.py program.fish -v 3

Surprisingly ><> doesn't do too badly here, since we can mimic string multiplication by having a counter which we decrement every iteration.

:&"$$"{\                     Put n into the register and push "$$"
}-1v!?:<{"}"                 Push n "}"s
&:&\~"stod"&:&?:             Push "stod", and copy the final "d" if n != 0
{1->:?!v}"\+1{1carfc"        Push n "\+1{1carfc"s
rav\$$"\~"\+1=ihp            Push "\+1=ihprav\$$"
       \l?!;o70.             Keep printing chars until the stack is empty

(-3 bytes thanks to @randomra)

edited Apr 8, 2015 at 6:07

answered Apr 8, 2015 at 5:26

Sp3000

61.9k13 gold badges115 silver badges287 bronze badges

\$\begingroup\$ Two small improvements for 3 bytes (at the end of line 1 and line 3): pastebin.com/wEbKhuUH \$\endgroup\$
– randomra
Commented Apr 8, 2015 at 5:53
\$\begingroup\$ Hah the first and last line mirrors match up, didn't notice :P \$\endgroup\$
– Sp3000
Commented Apr 8, 2015 at 6:08

Add a comment |

Community · Accepted Answer · 2017-03-16 16:03:59Z

Retina, 160 + 7 = 167 bytes

;`.+
$$$$\varphi=1+\dots#$0$$$$

;+(\d*)#(?:(((((((((9)|8)|7)|6)|5)|4)|3)|2)|1)|0) $1$1$1$1$1$1$1$1$1$1$2$3$4$5$6$7$8$9$10# ;#

;+`\\d?dots\d(\d*)
\cfrac1{1+\ddots$1}

Each line goes into a separate source file, so I've added 1 byte for each file after the first. However, for convenience, Retina now also supports the -s command-line flag, which allows you to put all of this into a single file (in which case the newlines are treated as file separators).

The largest part of the code (98 bytes) is used to convert the input from decimal to unary (files 3 to 6). The basic idea of the code is to surround the input in $$\varphi=1+\dots...$$, then convert the input to unary, and then expand \dotsN or \ddotsN to the next level of the continued fraction (while reducing N to N-1).

Alex A. · Accepted Answer · 2015-04-11 21:33:39Z

4

Julia, 76 73 bytes

n->("\$\$\\varphi=1+"*"\\cfrac1{1+"^n*"\\"*"d"^(n>0)*"dots"*"}"^n*"\$\$")

This creates a lambda function that takes a single integer as input and returns the MathJax as a string. To call it, give it a name, e.g. f=n->....

Unfortunately both backslashes and dollar signs have to be escaped in Julia strings because they both have special meaning. String concatenation is performed using * and string repetition with ^.

Examples:

julia> f(0)
"$$\varphi=1+\dots$$"

julia> f(4)
"$$\varphi=1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}}$$"

Suggestions are welcome as always!

Edit: Saved 3 bytes thanks to plannapus!

edited Apr 11, 2015 at 21:33

answered Apr 8, 2015 at 0:26

Alex A.

24.7k5 gold badges38 silver badges119 bronze badges

\$\begingroup\$ "d"^(n>0) instead of (n>0?"d":"") would make it shorter. \$\endgroup\$
– plannapus
Commented Apr 8, 2015 at 8:39
\$\begingroup\$ @plannapus: I don't understand what your name means but you're a genius! I had forgotten that "string"^0 was legit. \$\endgroup\$
– Alex A.
Commented Apr 8, 2015 at 20:24
\$\begingroup\$ you're welcome. My pseudonym is the name of a radiolarian genus, being myself a radiolarian paleontologist. It translates to "flattened turnip" i think :) \$\endgroup\$
– plannapus
Commented Apr 9, 2015 at 6:07
\$\begingroup\$ I am waiting for a BF answer=) \$\endgroup\$
– flawr
Commented Apr 9, 2015 at 9:06
\$\begingroup\$ @flawr: I hope you're not waiting on me for that... \$\endgroup\$
– Alex A.
Commented Apr 10, 2015 at 4:27

Add a comment |

PhiNotPi · Accepted Answer · 2015-04-08 02:36:16Z

Element, 63 Chars

_+2:'\$\$\\varphi\=1\+`[\\cfrac1\{1\+`]?\\[d.]`"dots`[\}`]\$\$`

This is the most straight-forward solution. Unfortunately, the large amount of symbols in the output causes a significant increase in program length (putting the strings in the program directly causes the symbols to perform operations). I'm sure there is room for golfing, but I don't have more time right now.

Since this language is still relatively unknown, here is a link to the interpreter, written in Perl.

_+2:                     take input, add 0 to it to make it a number, and duplicate
'                        put one copy onto the control stack
\$\$\\varphi\=1\+        a "bare" string
`                        output the string
[                        start a for loop, based on the input from earlier
    \\cfrac1\{1\+        a bare string
    `                    output it
]                        end the for loop
?                        test the second copy of the input for non-zero-ness
\\                       a bare \
[d.]                     a "for" loop used as an if block, appends a "d"
`                        output it
dots`                    output dots
"                        get rid of the if condition result so the old result is on top
[                        another for loop, still using the input from earlier
    \}`                  output a }
]                        end for loop
\$\$`                    output $$

+1 for using Element! It's time for Element to become a household name! — Alex A., Commented Apr 8, 2015 at 20:45

Martin Ender · Accepted Answer · 2015-04-11 22:14:12Z

T-SQL, 229 227 138

Been a while since I did an SQL answer and as always it's very verbose. Edit Of course I over complicated it and didn't need a recursive query at all.

CREATE FUNCTION A(@ INT)RETURNS TABLE RETURN SELECT'$$\varphi=1+\'+REPLICATE('cfrac1{1+\',@)+IIF(@>0,'d','')+'dots'+REPLICATE('}',@)+'$$'S

Original

CREATE FUNCTION A(@ INT)RETURNS TABLE RETURN WITH R AS(SELECT CAST('$$\varphi=1+\dots'AS VARCHAR(MAX))S,0N UNION ALL SELECT REPLACE(STUFF(S,14,0,'cfrac1{1+\'),'\do','\ddo')+'}',N+1FROM R WHERE N<=@)SELECT S+'$$'S FROM R WHERE N=@

This creates an inline table function that uses a recursive query to stuff in the additional cfrac1{1+\ per iteration. Changing the dots to ddots was expensive, but saved a couple getting rid of the replace :). Also having to cast the original string as 'VARCHAR(MAX)' cost a bit.

It's used as follows SQLFiddle:

SELECT * 
FROM (SELECT N FROM(VALUES(0),(1),(2),(3),(4),(5))A(N)) N
    CROSS APPLY A(N.N)
N   S
--- ---------------------------------------------------------------------------
0   $$\varphi=1+\dots$$
1   $$\varphi=1+\cfrac1{1+\ddots}$$
2   $$\varphi=1+\cfrac1{1+\cfrac1{1+\ddots}}$$
3   $$\varphi=1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}$$
4   $$\varphi=1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}}$$
5   $$\varphi=1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}}}$$

Martin Ender · Accepted Answer · 2015-04-11 22:14:35Z

3

Ruby, 76 75 71 70 bytes

This feels suspiciously straightforward, so please let me know if I messed up somewhere.

Incidentally, this is the first thing I've ever written in Ruby - I was looking for a language that supported string repetition by multiplying, and Ruby seemed to do the trick.

f=proc{|n|'$$\varphi=1+'+'\cfrac1{1+'*n+'\dd'[0,n+2]+'ots'+'}'*n+'$$'}

To be applied like so:

f.call(0)
$$\varphi=1+\dots$$

f.call(3)
$$\varphi=1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}$$

edited Apr 11, 2015 at 22:14

Martin Ender

197k67 gold badges447 silver badges975 bronze badges

answered Apr 8, 2015 at 12:56

vvye

2611 silver badge6 bronze badges

\$\begingroup\$ @Sp3000 The former doesn't, since Ruby apparently can't convert Booleans to integers. The latter worked though, so thanks for that! \$\endgroup\$
– vvye
Commented Apr 8, 2015 at 14:00

Add a comment |

2 revs · Accepted Answer · 2015-04-08 04:51:42Z

2

J, 60 bytes

(<;._2'$$\varphi=1+\ cfrac1{1+\ d dots } $$ ');@#~1,~5$1,],*

Usage:

   ((<;._2'$$\varphi=1+\ cfrac1{1+\ d dots } $$ ');@#~1,~5$1,],*) 0
$$\varphi=1+\dots$$

   ((<;._2'$$\varphi=1+\ cfrac1{1+\ d dots } $$ ');@#~1,~5$1,],*) 3
$$\varphi=1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}$$

Method:

The string '$$\varphi=1+\ cfrac1{1+\ d dots } $$ ' is cut up at spaces and the parts are repeated 1 n signum(n) 1 n 1 times and then these parts are concatenated.

Try it online here.

edited Apr 8, 2015 at 4:51

community wiki

2 revs
randomra

Add a comment |

Alex A. · Accepted Answer · 2015-04-08 20:42:45Z

2

R, 93 90

Much the same as the other answers. Thanks to @plannapus for the scan tip.

cat('$$\\varphi=1+\\',rep('cfrac1{1+\\',n<-scan()),if(n)'d','dots',rep('}',n),'$$',sep='')

cat used rather than paste0 as the result would end up with \\ rather than \.

In use

> > cat('$$\\varphi=1+\\',rep('cfrac1{1+\\',n<-scan()),if(n)'d','dots',rep('}',n),'$$',sep='')
1: 3
2: 
Read 1 item
$$\varphi=1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}$$

edited Apr 8, 2015 at 20:42

Alex A.

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answered Apr 8, 2015 at 2:25

MickyT

12.2k2 gold badges26 silver badges48 bronze badges

\$\begingroup\$ +1 but instead of making it a function, if you have the user enter n as stdin on its first occurrence, you can save some characters: cat("$$\\varphi=1+\\",rep("cfrac1{1+\\",n<-scan()),if(n)"d","dots",rep("}",n),"$$",sep="") \$\endgroup\$
– plannapus
Commented Apr 8, 2015 at 10:32

Add a comment |

score 2 · Accepted Answer · 2015-07-10 20:48:21Z

2

JavaScript, 114 109 106 85 bytes thanks to George Reith

f=n=>'$$\\varphi=1+\\'+((x='cfrac1{1+\\'.repeat(n))&&x+'d')+'dots'+'}'.repeat(n)+'$$'

This is my first entry in a codegolf contest! Please tell me how to improve.

Previous entry (106 bytes):

w="$$\\varphi=";y=n=>{return a=!n?w+"1+\\dots$$":w+"1+\\cfrac1{".repeat(n)+"1+\\ddots"+"}".repeat(n)+"$$"}

Previous entry (109 bytes):

x="repeat",w="$$\\varphi=";y=n=>{return a=!n?w+"1+\\dots$$":w+"1+\\cfrac1{"[x](n)+"1+\\ddots"+"}"[x](n)+"$$"}

Previous entry (114 bytes):

x="repeat";y=n=>{return a=!n?"$$\\varphi=1+\\dots$$":"$$\\varphi="+"1+\\cfrac1{"[x](n)+"1+\\ddots"+"}"[x](n)+"$$"}

Paste into browser console and call as f(n) where n is the number of 'steps'.

Simplified code:

function y(n) {
   if(n === 0) {
      return "$$\\varphi=1+\\dots$$";
   } else {
      return "$$\\varphi=" + "1+\\cfrac1{".repeat(n) + "1+\\ddots"+"}".repeat(n)+"$$";
   }

edited Jul 10, 2015 at 20:48

answered Apr 8, 2015 at 5:48

user31556

2

\$\begingroup\$ x='repeat' makes it longer, not shorter: use .repeat as is and save 3 chars \$\endgroup\$
– edc65
Commented Apr 8, 2015 at 15:39
\$\begingroup\$ pastebin.com/uU7tgFm9 some more improvements \$\endgroup\$
– edc65
Commented Apr 9, 2015 at 6:17
\$\begingroup\$ Can be made into 87 bytes like so: pastebin.com/0Hkv9uft \$\endgroup\$
– George Reith
Commented Apr 9, 2015 at 12:37
\$\begingroup\$ Or 85 bytes pastebin.com/k90Fyr2m \$\endgroup\$
– George Reith
Commented Apr 9, 2015 at 13:41
\$\begingroup\$ @edc65 I like it how you checked if n is equal to 0, by just n? : ... I didn't realise that if n = 0, n is false. \$\endgroup\$
– user31556
Commented Apr 10, 2015 at 8:53

Add a comment |

Maltysen · Accepted Answer · 2015-04-08 02:38:36Z

Pyth - 52 bytes

The simple approach in Pyth, pretty much stolen from @Sp3000's Python solution. Uses string formatting operator %.

%"$$\\varphi=1+\%sdots%s$$"(+*"cfrac1{1+\\"Q<\dQ*\}Q

Try it online here.

%                  String formatting
 "$$ . . . $$"     String to be formatted
 (                 Tuple (no need to close it)
  +                String concatenation
   *"..."Q         String repetition input times
   <\dQ            If Q>0 then d
  *                String repetition
   \}              The character "}"
   Q               Q times

orlp · Accepted Answer · 2015-04-08 11:45:03Z

1

Pyth, 50 bytes

s["$$\\varphi=1+"*Q"\cfrac1{1+"\\<\dQ"dots"*Q\}"$$

answered Apr 8, 2015 at 11:45

orlp

39.2k7 gold badges81 silver badges195 bronze badges

1

\$\begingroup\$ See my edit history :) \$\endgroup\$
– Optimizer
Commented Apr 8, 2015 at 11:46

Add a comment |

Martin Ender · Accepted Answer · 2015-04-11 22:14:53Z

1

JavaScript (ES6), 76 80

Partly recursive. The single/double d is most annoying part.

F=n=>"$$\\varphi=1+\\"+(R=d=>n--?"cfrac1{1+\\"+R("d")+"}":d+"dots")("")+"$$"

Test In Firefox /FireBug console

> for(i=0;i<5;i++)console.log(F(i))

$$\varphi=1+\dots$$
$$\varphi=1+\cfrac1{1+\ddots}$$
$$\varphi=1+\cfrac1{1+\cfrac1{1+\ddots}}$$
$$\varphi=1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}$$
$$\varphi=1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac1{1+\ddots}}}}$$

edited Apr 11, 2015 at 22:14

Martin Ender

197k67 gold badges447 silver badges975 bronze badges

answered Apr 8, 2015 at 15:42

edc65

32.2k3 gold badges36 silver badges90 bronze badges

Add a comment |

DenDenDo · Accepted Answer · 2015-04-08 16:31:42Z

0

Python, 90 116

since the most efficient solution has already been posted multiple times, i go with string replacement instead

f=lambda n:'$$\\varphi=1+\ddots$$'if n==0 else f(n-1).replace('\ddots','\cfrac{1+\ddots}')
# or, with exactly the same length
x='\ddots';f=lambda n:'$$\\varphi=1+'x+'$$'if n==0 else f(n-1).replace(x,'\cfrac{1+'x+'}')

Edit: damn, overlooked the dots instead of ddots for n=0, now the recursive solution with an extra clause tacked on is too ugly to compete.

x='$$\\varphi=1+\d%sots$$';f=lambda n:x%''if n==0 else x%'d'if n==1 else f(n-1).replace('\ddots','\cfrac{1+\ddots}')

edited Apr 8, 2015 at 16:31

answered Apr 8, 2015 at 14:57

DenDenDo

2,88119 silver badges18 bronze badges

\$\begingroup\$ This is currently missing the n=0 special case (dots instead ddots). \$\endgroup\$
– randomra
Commented Apr 8, 2015 at 15:08

Add a comment |

lynn · Accepted Answer · 2015-07-11 00:13:42Z

0

Haskell, 86

n%x=[1..n]>>x
f n="$$\\varphi=1+"++n%"\\cfrac1{1+"++'\\':drop(0^n)"ddots"++n%"}"++"$$"

Essentially the same approach as all solutions here. drop(0^n)"ddots" is cute, though!

edited Jul 11, 2015 at 0:13

answered Jul 10, 2015 at 23:43

lynn

69.2k11 gold badges133 silver badges283 bronze badges

Add a comment |

Stack Exchange Network

Generate MathJax for the Golden Ratio Continued Fraction

Challenge

Notes

Scoring

16 Answers 16

Python, 70 68 67 bytes

CJam, 51 50 bytes

><>, 89 86 + 3 = 89 bytes

Retina, 160 + 7 = 167 bytes

Julia, 76 73 bytes

Element, 63 Chars

T-SQL, 229 227 138

Ruby, 76 75 71 70 bytes

J, 60 bytes

R, 93 90

JavaScript, 114 109 106 85 bytes thanks to George Reith

Pyth - 52 bytes

Pyth, 50 bytes

JavaScript (ES6), 76 80

Python, 90 116

Haskell, 86

Your Answer

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code-golf
math
kolmogorov-complexity
or ask your own question.

Hot Network Questions

Generate MathJax for the Golden Ratio Continued Fraction

Challenge

Notes

Scoring

16 Answers 16

Python, 70 68 67 bytes

CJam, 51 50 bytes

><>, 89 86 + 3 = 89 bytes

Retina, 160 + 7 = 167 bytes

Julia, 76 73 bytes

Element, 63 Chars

T-SQL, 229 227 138

Ruby, 76 75 71 70 bytes

J, 60 bytes

R, 93 90

JavaScript, 114 109 106 85 bytes thanks to George Reith

Pyth - 52 bytes

Pyth, 50 bytes

JavaScript (ES6), 76 80

Python, 90 116

Haskell, 86

Your Answer

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Hot Network Questions

Not the answer you're looking for? Browse other questions tagged
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math
kolmogorov-complexity
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