Sure. You have to solve it from inside out:
- not()…See comment below for this one, I was tricked
is a base function that negates what’s inside (turning True to False and vice versa) giving it no parameter returns “True” (because no parameter counts as False) - str(x) turns x into a string, in this case it turns the boolean True into the text string ‘True’
- min(x) returns the minimal element of an iterable. In this case the character ‘T’ because capital letters come before non-capital letters, otherwise it would return ‘e’ (I’m not entirely sure if it uses unicode, ascii or something else to compare characters, but usually capitals have a lower value than non-capitals and otherwise in alphabetical order ascending)
- ord(x) returns the unicode number of x, in this case turning ‘T’ into the integer 84
- range(x) creates an iterable from 0 to x (non-inclusive), in this case you can think of it as the list [0, 1, 2, …82, 83] (it’s technically an object of type range but details…)
- sum(x) sums up all elements of a list, summing all numbers between 0 and 84 (non-inclusive) is 3486
- chr(x) is the inverse of ord(x) and returns the character at position x, which, you guessed it, is ‘ඞ’ at position 3486.
The huge coincidental part is that ඞ lies at a position that can be reached by a cumulative sum of integers between 0 and a given integer. From there on it’s only a question of finding a way to feed that integer into chr(sum(range(x)))
Don’t wanna be pedantic (yes I do) but after a little bit of searching, the originally presented quote seems correct (without the “and” though, it seems). Your version is mixed with a common misquote “Slowly at first, then all at once.” of that quote.