www.mikescher.com/www/statics/euler/Euler_Problem-079_explanation.md

This is one of the problems i really enjoyed solving.

We make the assumption that our final pass code has no duplicate digits (If we hadn't found a solution we would need to change that part). This is a pretty good assumption because no attempt has a duplicate digit in it.

Side note: This leaves us with a 8-digit code, because only 8 digits are used (4 and 5 are missing).

First we generate a 10x10 grid where we remember the absolute ordering of the numbers from our attempts (eg 3 is before 8 or 9 is after 2). If we inspect this data we can see that every field (for numbers in our pass code) is set, the fifty login attempts generate more than enough data for this.

Side note: In fact there are only 33 unique login attempts

Then we can simply sort an array of the valid digits with this ordering. And - frankly - I find this is a really neat way of doing this.

Because we sort only eight numbers, sorting-performance is not a big factor. So I searched for the simplest (easiest to implement) sorting algorithm I could find. This was surprisingly not Bubble-sort, but Gnome Sort (accordingly to the author "the simplest sort algorithm"). Which was pretty easy to implement in Befunge (and for 8 values is the runtime of O(n^2) not that bad).

A last thing: For a problem where I used an two-dimensional cache and that has input data I was surprised to fit everything in the Befunge-93 80x25 size restrictions.