Start a timer that will count down the number of seconds from 52! to 0.
Then walk around the Earth’s equator with one step every billion years.
Then, after you make your way around the earth equator (by taking 1 step every billion of years), you take one drop of water out of the Pacific Ocean.
Then, you repeat the process of walking around the equator, and everytime you walk around, you keep draining one singular drop of water.
After the ocean is fully drained, you refill the ocean and put a piece of paper underneath you.
Now, you once again repeat this process of walking, draining, and placing papers.
After your stack of papers has reached the Sun, you repeat another 1000 times.
After all this, you have completed just about a third of the timer.
52 cards is the first thing I think of when I think factorials. It's such a great and relatable way to convey the subject to people, plus it usually ends up blowing their minds like it did mine when I first learned of it. Not from this page, but from a YT vid many moons ago.
The main.css file it imports dates itself to March 9 of 2005, and is housed in an "ancient history" section of the website that covers everything before October 26, 2010, so: "sometime between those two years" =P
To me it feels like a smuggled exponentation. If I take half of 52 and raise it to itself and add a base unit to it like grams or meters, that's an incomprehensibly large amount.
This brings to mind the analysis in Bender & Orszag; they approach this through difference equations (a bit of a lost art in formal mathematics; very 19th-century feel) rather than integration.
Instead of introducing the gamma function, they instead start from the observation that log(F_n) - log(F_n-1) = log(n), so treating this difference as analogous to integration, it says that F_n ~= nlogn + n as the leading asymptotic behavior. This is clear just by substitution and algebra; no calculus necessary (though it helps to "know the answer beforehand").
From there you can treat the error term in this as F_n = n^n * e^n * E_n and plug that into the same relationship (F_n = n * F_n-1) to derive what that error term looks like asymptotically, and end up in the same place that the integration on the OP leads to.
Stirling's approximation is also used a lot in statistical mechanics, because you often have to calculate logs of state space sizes, which means lots of combinatorics and thus lots of factorials. Plus it's continuous so you can do calculus.
A quick and dirty approximation of the number of digits in n! is n lg n, which approximates n! from above, via the inequality
1 * 2 * … * n ≤ n * … * n.
(This approximation should be familiar to many from an algorithmics class.)
For a tighter bound, use n lg n - n/2, or a better approximation of ln 10 in place of 1/2 if you wish. This comes from Stirling's approximation which notes that
> (This approximation should be familiar to many from an algorithmics class.)
You need both sides though :)
What makes it interesting for estimating algorithmic complexity is that \log{n!} \in \Theta(n \log n). One side is obvious as you note, the other less so, but there's a famous trick to do both at once:
lg(n!) grows roughly as (n lg n). Constants matter, of course, but to that's the rough estimate.
As an aside, if you take numbers from 0 to (n-1) in an array, there are n! configurations, so representing each configuration or differentiating each configuration take n lg n bits. So, in some sense, taking a mapping that's able to differentiate the input state to map to the ordered state takes at least O(n lg n) time, the standard runtime of a basic sorting algorithm.
Any additional assumptions (n larger than maximum element, distribution of elements) helps reduce this.
My kids love taking about big numbers. TREE(3) is a family favorite. So, I was going over sequences with them, and I decided to go slow instead. My sequence was: 1 1 1 1 ... 1 ...
They accused me of using just "all 1s" (which is, naturally, cheating). Ai contraire!
The count of the number of digits in the decimal representation of the number of unique primes in the prime factorization of the natural numbers.
The best part is that even pretty young kids can compute this sequence; by the first "2" is at 2*3*5*7*11*13*17*19*23*29!
(Hopefully I got that right; the phone doesn't make it easy to type!)
I came across an interesting feature of factorials while making the puzzle books at https://www.kakurokokoro.com
The widest two rows are nine digits across, but while the first can be any of arrangements of the digits 1-9, the second cannot repeat any digits in the same columns, and so it limits allowable permutations to the number of derangements — which is close to 9!/e (where e is Euler’s Constant 2.718…)
No idea what this has to do with the relationship of i and pi.
There is a algorithm call Prime Swing Factorial that can compute large factorials exactly in arbitrary precision math using prime factorization. Like 10000000! in under second depending of how optimized the math library it. Probably like 100x faster than the normal method.
With Lisp you can use iterative algos and get that under a second too.
SBCL can be ridiculously fast; and if you optimize the compilation for integers... the speed gets really close to your solution.
Worth noting for anyone reaching for this in practice rather than out of curiosity: several standard library implementations (Python's math.factorial is one) already use a divide-and-conquer multiplication scheme instead of naive sequential multiplication for exactly this reason, so you often get most of that speedup for free without implementing prime swing yourself.
I think you mean "using base C without any arbitrary-precision library (e.g. GMP)" . All that illustrates is that Lisp has built-in support for arbitrary-precision arithmetic, whereas C doesn't. Otherwise, how is this surprising, and what is the reason for the performance difference?
Dog slow but the old n270 netbook (32 bit) handles big factorials >20 fine, and OFC it's instant under Common Lisp (SBCL) and Scheme (both S9 and Chicken).
Start a timer that will count down the number of seconds from 52! to 0. Then walk around the Earth’s equator with one step every billion years. Then, after you make your way around the earth equator (by taking 1 step every billion of years), you take one drop of water out of the Pacific Ocean. Then, you repeat the process of walking around the equator, and everytime you walk around, you keep draining one singular drop of water. After the ocean is fully drained, you refill the ocean and put a piece of paper underneath you. Now, you once again repeat this process of walking, draining, and placing papers. After your stack of papers has reached the Sun, you repeat another 1000 times.
After all this, you have completed just about a third of the timer.
https://sites.imsa.edu/hadron/2025/02/26/how-big-is-52/
reply