Sudan function

The Sudan function is a classic example of a recursive function, notable especially because it is not a primitive recursive function. This is also true of the better-known Ackermann function. The Sudan function was the first function having this property to be published.

The Sudan function is usually defined as follows (svg):

  <big>
    :<math>\begin{array}{lll}
      F_0 (x, y) & = x+y \\
      F_{n+1} (x, 0) & = x & \text{if } n \ge 0 \\
      F_{n+1} (x, y+1) & = F_n (F_{n+1} (x, y), F_{n+1} (x, y) + y + 1) & \text{if } n\ge 0 \\
      \end{array}
    </math>
  </big>

Task

Write a function which returns the value of F(x, y).

Hoon

<lang Hoon> |= [n=@ x=@ y=@] ^- @ |- ?: =(n 0)

 (add x y)

?: =(y 0)

$(n (dec n), x $(n n, x x, y (dec y)), y (add $(n n, x x, y (dec y)) y)) </lang>

J

Translation of: Javascript

This is, of course, not particularly efficient and some results are too large for a computer to represent. <lang J>F=: {{ 'N X Y'=. y assert. N>:0

 if. 0=N do. X+Y
 elseif. Y=0 do. X
 else. F (N-1),(F N,X,Y-1), Y+F N, X, Y-1
 end.

}}"1</lang>

Examples: <lang J> F 0 0 0 0

  F 1 1 1

3

  F 2 1 1

8

  F 3 1 1

10228

  F 2 2 1

27</lang>

JavaScript

<lang Javascript> /**

* @param {bigint} n
* @param {bigint} x
* @param {bigint} y
* @returns {bigint}
*/

function F(n, x, y) {

 if (n === 0n) {
   return x + y;
 }

 if (y === 0n) {
   return x;
 }

 return F(n - 1n, F(n, x, y - 1n), F(n, x, y - 1n) + y);

} </lang>

Julia

<lang ruby>using Memoize

@memoize function sudan(n, x, y)

  return n == 0 ? x + y : y == 0 ? x : sudan(n - 1, sudan(n, x, y - 1), sudan(n, x, y - 1) + y)

end

foreach(t -> println("sudan($(t[1]), $(t[2]), $(t[3])) = ",

  sudan(t[1], t[2], t[3])), ((0,0,0), (1,1,1), (2,1,1), (3,1,1), (2,2,1)))

</lang>

Output:

sudan(0, 0, 0) = 0
sudan(1, 1, 1) = 3
sudan(2, 1, 1) = 8
sudan(3, 1, 1) = 10228
sudan(2, 2, 1) = 27