From Wikipedia:

In number theory, the Calkin–Wilf tree is a tree in which the vertices correspond one-to-one to the positive rational numbers. The tree is rooted at the number \$1\$, and any rational number expressed in simplest terms as the fraction \$\frac{a}{b}\$ has as its two children the numbers \$\frac{a}{a+b}\$ and \$\frac{a+b}{b}\$.

Calkin-Wilf tree

The Calkin–Wilf sequence is the sequence of rational numbers generated by a breadth-first traversal of the Calkin–Wilf tree, $$\frac11, \frac12, \frac21, \frac13, \frac32, \frac23, \frac31, \frac14, \frac43, \frac35, \frac52, \ldots$$

For this challenge, you are given a fraction found in the \$i\$th position of the Calkin-Wilf sequence, and must output \$i\$. You can start from either 0 or 1.

Test cases

(Starting from 1.)

\$a_i\$ \$i\$
\$\frac11\$ \$1\$
\$\frac13\$ \$4\$
\$\frac43\$ \$9\$
\$\frac34\$ \$14\$
\$\frac{53}{37}\$ \$1081\$
\$\frac{37}{53}\$ \$1990\$

Standard loopholes are forbidden. Since this is , the shortest code wins.


15 Answers 15


JavaScript (ES6), 33 bytes

Expects a pair of integers (p,q) representing the fraction \$p/q\$.


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This version is based on the nice recurrence formula found by alephalpha:


It was modified to work with two integers \$p/q=x\$ instead of a native fraction and without an explicit ceiling function:

$$\lceil q/p\rceil=\begin{cases} q/p,&q\equiv 0\pmod p\\ (q+p-(q \bmod p))/p,&q\not\equiv 0\pmod p\end{cases}$$

We have:

$$f(2\lceil1/x\rceil-1-1/x)=f(2\lceil q/p\rceil-p/p-q/p)$$

which is turned into the following JS code:

f(q % p ? q + p - q % p * 2 : q - p, p)

which can be further simplified to:

f(q + p - (q % p || p) * 2, p)

JavaScript (ES6), 40 bytes

Expects a pair of integers (p,q) representing the fraction \$p/q\$.


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f = (p, q) =>   // (p, q) = input pair
p - q ?         // if p is not equal to q:
  p > q ?       //   if p is greater than q:
                //     we're located on the right branch
    f(p - q, p) //     turn it into the corresponding left branch
    + 1         //     and add 1
  :             //   else:
                //     we're located on the left branch
    2 *         //     double the result of
    f(p, q - p) //     a recursive call with the parent fraction
:               // else:
                //   we've reached the root
  1             //   stop and return 1

PARI/GP, 29 bytes


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  • \$\begingroup\$ How the hell does this even work?? Is it based on the recursive formula on Wikipedia to find the \$i\$'th term in the sequence \$q_{i+1}=\frac1{2\lfloor q_i\rfloor-q_i+1}\$? \$\endgroup\$
    – Aiden Chow
    Apr 27 at 2:29
  • 2
    \$\begingroup\$ @AidenChow Yes. \$q_{i+1}=1/(2\lfloor{q_i}\rfloor-q_i+1)\Rightarrow 1/q_{i+1}-1=\lfloor{q_i}\rfloor-(q_i-\lfloor{q_i}\rfloor)\$. \$\lfloor{q_i}\rfloor\$ is an integer and \$0\le q_i-\lfloor{q_i}\rfloor<1\$, so \$\lceil{1/q_{i+1}-1}\rceil=\lfloor{q_i}\rfloor\$, \$q_i-\lfloor{q_i}\rfloor=\lceil{1/q_{i+1}-1}\rceil-(1/q_{i+1}-1)\$. Thus \$q_i=\lfloor{q_i}\rfloor+(q_i-\lfloor{q_i}\rfloor)=\lceil{1/q_{i+1}-1}\rceil+(\lceil{1/q_{i+1}-1}\rceil-(1/q_{i+1}-1))=2\lceil{1/q_{i+1}}\rceil-1-1/q_{i+1}\$. \$\endgroup\$
    – alephalpha
    Apr 27 at 2:38
  • \$\begingroup\$ what the hell.... how you thinking of these things? i try convert that equation into something useful but came up short..... \$\endgroup\$
    – Aiden Chow
    Apr 27 at 2:46

Vyxal, 12 bytes


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Takes input as a pair of numbers.


             # function taking a pair and returning its parent:
g-           #   subtract minimum
  ¨£∨        #   zip and return the last truthy element from each pair
     )       # end function
      ↔      # apply until fixed point and collect intermediate results
       vÞṘ   # vectorizing is sorted in reverse?
          Ṙ  # reverse
           B # convert from binary

Python, 34 bytes (@Neil)

f=lambda Q:Q and 1+f(1/Q-1-2/Q%-2)

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Port of @Arnauld's latest and greatest which ports @alephalpha's.

Expects a Python Fraction object.

Previous Python, 44 bytes (@xnor)

f=lambda Q:Q==1or(Q>1)+2*f(max(Q/(1-Q),Q-1))

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Original Python, 46 bytes

f=lambda Q:Q==1or(Q>1)+2*f([Q/(1-Q),Q-1][Q>1])

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Expects a Python Fraction object.

Similar but as far as I can tell not identical to @Arnauld's approach.

  • \$\begingroup\$ Selecting with max also seems to work. \$\endgroup\$
    – xnor
    Apr 27 at 5:09
  • \$\begingroup\$ A port of @Arnauld's latest formula is at most 35 bytes: f=lambda p,q:p and-~f(q-q%-p*2-p,p). \$\endgroup\$
    – Neil
    Apr 27 at 19:07
  • \$\begingroup\$ Thanks, @Neil. Using fractions I managed to shave one off. \$\endgroup\$
    – loopy walt
    Apr 27 at 21:02

Raku, 25 bytes


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Port of @loopy walt's answer, which is a port of @Arnauld's, which is a port of @alephalpha's.


J, 24 bytes


-4, port of @Jo King's answer, which is a port of @loopy walt's, which is a port of @Arnauld's, which is a port of @alephalpha's.

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                       *  NB. signum, 0 for 0, 1 for everything else
                     @.   NB. index into the gerund list and execute
*                         NB. if 0, use signum again to return 0
  (                 )     NB. if 1, execute the recursive function
                          NB. uses the formula in alephalpha's PARI/GP answer
      $:@                 NB. recurse here

Original 28 bytes


In short, returns the length of the sequence up to the input. Expects a rational number.

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                           *  NB. signum, returns 1 for all inputs to give initial value
   [                          NB. input unchanged
    (             )^:~:^:a:   NB. form for a do-while that uses a: to collect results into an array
                              NB. u^:v^:_ v is a boolean function and u modifies the initial value
                         a:   NB. empty box
                     ~:       NB. continue if input is not equal to current value
              <.@]            NB. floor the right arg (qᵢ)
            2*                NB. double
         ]-~                  NB. subtract by qᵢ
       1+                     NB. increment
     1%                       NB. reciprocal
[:                            NB. then
  #                           NB. take the length

Arturo, 49 bytes

f:$[p q][(p=q)?->1->(p>q)?->1+f p-q p->2*f p q-p]

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Port of Arnauld's JavaScript answer.


Jelly, 11 bytes

Uses AndrovT's method from their Vyxal answer, go upvote!


A monadic Link that accepts a pair of co-prime positive integers, [numerator, denominator] and yields the 1-indexed index of that fraction in

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_ṂoƊƬ</€¬ṚḄ - Link: pair of positive, coprime integers, [N,D]
    Ƭ       - start with C=[N,D] and collect until a fixed-point applying:
   Ɗ        -   last three links as a monad - f([N,D]):
 Ṃ          -     minimum ([N,D])
_           -     ([N,D]) subtract (that) (vectorises)   -> [0,D-N] or [N-D,0]
  o         -     (that) logical OR ([N,D]) (vectorises) -> [N,D-N] or [N-D,D]
       €    - for each ([numerator, denominator] pair in the collected results):
      /     -   reduce by:
     <      -     is (the numerator) less than (the denominator)?
        ¬   - logical NOT (vectorises)
         Ṛ  - reverse
          Ḅ - convert from binary

Loads of \$11\$s, is there a \$10\$?



Rust, 91 84 81 77 bytes

Based on the recursive relation:

$$ a_n = b_{n-1} $$ $$ b_n = 2\cdot b_{n-1} \cdot \left \lfloor \frac{a_{n-1}}{b_{n-1}} \right \rfloor - a_{n-1} + b_{n-1}$$

0-indexed. Recursion is very verbose in rust so I hope this is shorter than the recursive version most other answers use.

|m|{let mut d=(1,1,1);while(d.0,d.1)!=m{d=(d.1,d.0+d.1-d.0%d.1*2,d.2+1)};d.2}

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  • 1
    \$\begingroup\$ if*n==m{None}else{...;Some(0)} => (*n!=m).then(||{...;0}) \$\endgroup\$
    – alephalpha
    Apr 27 at 10:24
  • \$\begingroup\$ d.0+d.1-d.0%d.1*2 is a shorter formula (a(n) = a(n-2) + a(n-1) - 2*(a(n-2) mod a(n-1)) from OEIS A002487). \$\endgroup\$
    – Neil
    Apr 27 at 19:12

Prolog (SWI), 59 53 bytes

Thanks @Jo King for -6 bytes and fixing the floating point issue

N+X:-A is 2*ceiling(1/N)-1-1rdiv N,A+B,X is B+1.

Took the recursive formula from @alephalpha's answer, so make sure to upvote his answer as well!

Try it online!

  • \$\begingroup\$ You can avoid the float inaccuracies by passing in rational numbers, e.g. 53 bytes. In updated versions of Prolog SWI, you can do 1r2 to represent a rational number \$\endgroup\$
    – Jo King
    Apr 27 at 6:25
  • \$\begingroup\$ @JoKing Huh, thanks for the info! I didn't realize rdiv was a thing... \$\endgroup\$
    – Aiden Chow
    Apr 27 at 7:05
  • \$\begingroup\$ Huh, also didn't realize that putting the 0+0. case on its own would work... usually when i do that the N isn't evaluated or smth and the code doesn't work, so I now just put everything in one line so that the N<(number) would evaluate the N without me having to worry about anything. I guess it just works in this case... \$\endgroup\$
    – Aiden Chow
    Apr 27 at 7:09

Scala, 72 bytes

Golfed version. Try it online!

def f(p:Int,q:Int):Int=if(p-q!=0)if(p>q)f(p-q,p)+1 else 2*f(p,q-p)else 1

Ungolfed version. Try it online!

object Main {
  def main(args: Array[String]): Unit = {
    def f(p: Int, q: Int): Int = {
      if (p - q != 0) {
        if (p > q) {
          f(p - q, p) + 1
        } else {
          2 * f(p, q - p)
      } else {


Charcoal, 34 24 bytes


Try it online! Link is to verbose version of code. 1-indexed. Takes input as two integers. Explanation: Port of @Arnauld's port of @alephalpha's answer.


Input the numerator and denominator.


Repeat until the numerator is zero.


Calculate the next numerator. This uses true modulus, so instead of writing q + p - (q % p || p) I can write q - q % -p; the final expression is then q - q % -p * 2 - p.


Save the copy of the numerator that conveniently happens to be in the loop variable to the denominator.

Increment the count.


Output the final count.

Previous 34-byte version based on @AndrovT's answer:


Try it online! Link is to verbose version of code. 1-indexed. Takes input as two integers. Explanation:


Input the fraction.


Repeat until the numerator and denominator are equal.


If the numerator was larger than push a 1 bit otherwise push a 0 bit.


Subtract the smaller element from the larger.


Push a final 1 bit.


Convert from base 2.

I could have done better by generating the sequence (36 33 bytes):


Try it online! Link is to verbose version of code. 0-indexed. Takes input as two integers. Explanation:


Input the fraction in reverse.


Start with 1 as the fraction.


Repeat until the fraction is found.


Find the next fraction. (And yes, that is legitimately 4 υs in a row.)

Increment the loop count.


Output the final count.


><>, 36 bytes


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Port of Arnauld's answer. Takes input as codepoints, e.g. 53/37 is 5%


Ruby, 38 28 bytes


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Using @alephalpha/@Arnauld formula like most of the answers.


><> (Fish), 59 bytes


Try it

Despite not having a easy way to check tuple equality still shorter than rust. Uses the formula Neil suggested under my Rust answer.


1ii11 pushes the initial state.

$4[:{:{:{:@=r@={+2(?vr{] Checks if the 2 tuples are equal

$:@$:@$:@$2*-+ Performs the basic calculation of the next value for B

{1+} Adds one to the counter

]{n; Prints the result and exits, if the tuples are equal.

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