How efficient is this program? How could I make it more efficient?

package main;

import java.math.BigInteger;

public class Ackermann {

    public static void main(String[] args) {


    public static BigInteger ack(BigInteger a, BigInteger b) {
        BigInteger ans;
        if (a.equals(BigInteger.ZERO)) ans = b.add(BigInteger.ONE);
        else if (b.equals(BigInteger.ZERO)) ans = ack(a.subtract(BigInteger.ONE),BigInteger.valueOf(1));
        else ans = ack(a.subtract(BigInteger.ONE), ack(a,b.subtract(BigInteger.ONE)));
        return (ans);

  • 4
    \$\begingroup\$ As far as I remember, ack(4,2) is 2^65533-3. Since the only way that ack() can return a specific value is to recurse that deep, there is no way that a machine will ever execute that algorithm. \$\endgroup\$ Commented Oct 10, 2015 at 19:33

2 Answers 2


The Ackermann function is not designed to be efficient... ;-) It is designed to be a counter-proof to a theoretical construct.

But, the reality is that the BigInteger is required. The conditions are required.

About the only things I can suggest are:

  • use better names
  • return-early logic
  • the ans variable is unnecessary
  • always use braces for conditional blocks.
  • BigInteger.ONE is already defined, you have used it in some places already, no need for BigInteger.valueOf(1)
  • call it ackermann not ack

I would have:

public static BigInteger ackermann(BigInteger a, BigInteger b) {
    if (a.equals(BigInteger.ZERO)) {
        return b.add(BigInteger.ONE);
    if (b.equals(BigInteger.ZERO)) {
        return ackermann(a.subtract(BigInteger.ONE),BigInteger.ONE);
    return ackermann(a.subtract(BigInteger.ONE), ackermann(a, b.subtract(BigInteger.ONE)));

This guy Jon Hanna in the linked Stackeroverflow question uses a non-recursive but iterative strategy with some extra Ackermann specific optimizations. It is in c# but I hope it gives you some interesting insights.


He gets an answer for Ackerman(4,2) in a little over a second which I find impressive. My implementation will take a couple of lifecycles of the universe to complete.

  • \$\begingroup\$ This doesn't seem to be a review of the code in the question. \$\endgroup\$ Commented Sep 12, 2023 at 16:01
  • \$\begingroup\$ True. However it does provide some pointers in answering the question in the question: "How could I make it more efficient?" Since I am new to this platform I am open to removing my answer if it is not in line with expectations and habits. \$\endgroup\$ Commented Sep 14, 2023 at 12:57

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