Sudan function: Difference between revisions

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

=={{header|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>
</lang>

Revision as of 23:33, 10 July 2022

Task
Sudan function
You are encouraged to solve this task according to the task description, using any language you may know.

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)

 x

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