0
$\begingroup$

Given right associative tetration where:

$^{m}n =$ n^(n^(n^…))

And a situation such as:

$^{m}n = y$

$^{q}p = z$

What is a practical way to calculate which of $y$ and $z$ are larger?

I'm particularly looking at the case where:

$(n, m, p, q)$ are $> 1$

$p > n$

$m > q$

To simplify the question further, assume that $(n, m, p, q)$ are positive integers, they could for example be in the range 10 to 100.

Therefore $(y, z)$ are also positive integers.

|cite|improve this question
$\endgroup$
1
$\begingroup$

Generally, the (exponent?, tetrand?)... the thing in the upper left, has more of an effect on the value than the base. For example the difference between $^43$ and $^37$ is astronomical (about 3.6 trillion orders of magnitude). This only gets more true as $m$ and $n$ increase.

As a rule of thumb, I would say whichever has the larger argument $m$ or $n$ is almost always larger. As for an exact answer, I don't know if there is one. Values resulting from tetration of large integers, say $^{987}4910$, tend do be so large that there is no way to calculate them exactly.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I disagree with this answer. There is no quantification of formalization that a reasonable algorithm is impossible. $\endgroup$ – Brout Nov 26 '18 at 0:19
  • $\begingroup$ The quantification is that the number of digits in $^{987}4910$ exceeds the number of subatomic particles in the observable universe more than $10^{100000}$ times over. Even if you could devise an algorithm, it will still be physically impossible to write each digit of this number down, regardless of how small you make each digit. Furthermore, it would take more time than the universe will physically exist to run the algorithm. Even if you use scientific notation, you'll still be off by orders of magnitude. $\endgroup$ – R. Burton Nov 26 '18 at 4:10
  • $\begingroup$ I do not think you understand the point. What you say is not a proof that two numbers represented in some way with $O(n)$ bits cannot be compared in $O(poly(n))$ time. $\endgroup$ – Brout Nov 26 '18 at 6:10
  • $\begingroup$ I suppose... but it doesn't make much difference if the time it takes just to calculate the time it will take to calculate the difference between two sufficiently large numbers exceeds the lifetime of the human species and $O(n)$ is larger than a universe's worth of storage space. $\endgroup$ – R. Burton Nov 26 '18 at 12:23
1
$\begingroup$

There might be a way to use logarithms to simplify in a structured method. Look $a^{a^a}<b^{b^{b^b}}\iff a^a\log a<b^{b^b}\log b\iff (a\log a)+\log\log a<(b^b\log b)+\log\log b$.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.