Ackermann Function - Expansion

Expansion

To see how the Ackermann function grows so quickly, it helps to expand out some simple expressions using the rules in the original definition. For example, we can fully evaluate in the following way:

\begin{align}
A(1,2) & = A(0, A(1, 1)) \\
& = A(0, A(0, A(1, 0))) \\
& = A(0, A(0, A(0, 1))) \\
& = A(0, A(0, 2)) \\
& = A(0, 3) \\
& = 4.
\end{align}

To demonstrate how 's computation results in many steps and in a large number:

\begin{align}
A(4, 3) & = A(3, A(4, 2)) \\
& = A(3, A(3, A(4, 1))) \\
& = A(3, A(3, A(3, A(4, 0)))) \\
& = A(3, A(3, A(3, A(3, 1)))) \\
& = A(3, A(3, A(3, A(2, A(3, 0))))) \\
& = A(3, A(3, A(3, A(2, A(2, 1))))) \\
& = A(3, A(3, A(3, A(2, A(1, A(2, 0)))))) \\
& = A(3, A(3, A(3, A(2, A(1, A(1, 1)))))) \\
& = A(3, A(3, A(3, A(2, A(1, A(0, A(1, 0))))))) \\
& = A(3, A(3, A(3, A(2, A(1, A(0, A(0, 1))))))) \\
& = A(3, A(3, A(3, A(2, A(1, A(0, 2)))))) \\
& = A(3, A(3, A(3, A(2, A(1, 3))))) \\
& = A(3, A(3, A(3, A(2, A(0, A(1, 2)))))) \\
& = A(3, A(3, A(3, A(2, A(0, A(0, A(1, 1))))))) \\
& = A(3, A(3, A(3, A(2, A(0, A(0, A(0, A(1, 0)))))))) \\
& = A(3, A(3, A(3, A(2, A(0, A(0, A(0, A(0, 1)))))))) \\
& = A(3, A(3, A(3, A(2, A(0, A(0, A(0, 2)))))) \\
& = A(3, A(3, A(3, A(2, A(0, A(0, 3))))) \\
& = A(3, A(3, A(3, A(2, A(0, 4))))) \\
& = A(3, A(3, A(3, A(2, 5)))) \\
& = ... \\
& = A(3, A(3, A(3, 13))) \\
& = ... \\
& = A(3, A(3, 65533)) \\
& = ... \\
& = A(3, 2^{65536} - 3) \\
& = ... \\
& = 2^{2^{ \overset{65536}{} }} - 3. \\
\end{align}

Written as a power of 10, this is roughly equivalent to 106.031×1019,727.

Read more about this topic:  Ackermann Function

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