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Whenever you sit back and smile proudly to yourself about how clever the block of code you just wrote is, your next move should be to delete and rewrite it.
This is a clever block of code! Great job, now rewrite it to be sane 😂
I think it depends; some smart code is good actually, think 0x5f3759df. As long as you properly document it and leave plenty of comments. This one is not smart though, at best it’s what I would call witty.
I’d accept that “smart code” and “clever code” are 2 different things
Fast inverse square root eh?
Still linear time at least, could always be much MUCH worse
True. Lost opportunity to blow things up with useless recursivity
This isn’t sufficiently enterprisey for Java. There should be a Roman numeral factory followed by relevant fromString and toInteger methods.
Ugh. Literally refactored multiple factories into straightforward functions in the most recent sprint where I work.
Someone saw a public factory method which was a factory for a reason and just cargo culted multiple private methods using the same pattern.
Why don’t you just ask Chat-GPT o3 every time? Works like a charm!
Because there are better random generators
My first thought was something along the lines of a “zip bomb”. For every “M” in the input string, it’d use more than a KiB of memory. But still, it’d take a string of millions of "M"s to exhaust memory on even a low-end modern server. Still probably not a good idea to expose to untrusted input on a public networked server, though. And it could easily peg a CPU core for a good while. Very good leveraged target for DDOSing.
It also works the other way round: wanna convert Arabic n to Roman? Just write n times ‘I’ and revert these replacement in inverse order.
I don’t know what happens when the substring overlaps. Like for the number 6, will it replace the first 5 I’s with V and end up correctly with VI or the last ones and come to IV? I would guess the former and maybe you know but I never thought about it before
Also does not handle ‘IIIIIIIII’ -> ‘IX’ properly
If the substitution went right to left it might work.
It does.
I’ve written it that way and it works, as in it will replace left to right and you replace iiii to iv after iiiii to v
Makes sense but it will fail at 9 (VIV) it would only work for 9 if the replace went from right to left or the V and IV statements were exchanged but in both cases, 6 would fail
9 is IX though, and that works.
6 works fine, as it replaces the first set of 5 I with V and then there’s nothing to replace.
I’d written it in typescript for all it’s worth; go ahead and try it yourself :)
Does 9 really work? Wouldn’t it be:
IIIIIIIII -> VIIII -> VIVI like your questions about this and they all seem fair but I kinda wanna encourage you to go ahead and write it yourself; it’s a fun way to convert into Roman numerals that both is and isn’t intuitive at the same time.
No, cause you do the replacement from large to small, I.e. you’d first check for 10 I to replace with X (none found); then replace 9 with IX (check), then check for 5, 4 and so on.
The original doesn’t have an extra check for 9 and it works for Roman->Indioarabic because it’s:
IX ->IVV ->IIIIV ->IIIIIIIIIBut the other way around, you need an extra step for 9. That’s where our misunderstanding comes from.
I noticed my “and so on” is literally a noop here so yeah.
They forgot “CM” so this doesn’t work for any number that ends in 900s
No, M will be replaced by DD and then CD will be picked up, so it will go
- CM
- CDD
- CCCCD
- CCCCCCCCC
- …
It’s not too bad, it’s readable and easily optimised by adding intermediate sums and removing whatever power of 10 you’re working on.
My first thought was that it’d be a great oracle for randomized testing.
Should do a regex find all then iterate over each chunk recursively until unchanged.
there was no regex in ancient rome
Actually there were seven kings prior to the establishment of the republic, at which point they expelled the rulers… a reg-ex if you will.
IIV becomes IIIII, hm?
IIV would never be used. In Roman numerals at most one smaller unit can come in front of a larger one. The code doesn’t do any validation though.
Yes, that does demonstrate my point.
It’s got some code duplication. Who can code
gulfgolf this?public static int convertRomanNumeral(String numeral) { return 4; // todo }Code gulf, you say?
public static String convertRomanNumeral(String numeral) { numeral = numeral.replace("America", "Mexico"); return numeral; }Gulf clap.
public static int convertRomanNumeral(String numeral) { numeral = numeral.replace("M", "DD") .replace("CD", "CCCC") .replace("D", "CCCCC") .replace("C", "LL") .replace("XL", "XXXX") .replace("L", "XXXXX") .replace("X", "VV") .replace("IV", "IIII") .replace("V", "IIIII"); return numeral.length(); }public static int convertRomanNumeral(String numeral) { return numeral.replace("M", "DD") .replace("CD", "CCCC") .replace("D", "CCCCC") .replace("C", "LL") .replace("XL", "XXXX") .replace("L", "XXXXX") .replace("X", "VV") .replace("IV", "IIII") .replace("V", "IIIII") .length(); }
until(original=new) { run convertOriginal }
Alternatively pip install roman.
unary!
I just wrote something similar for decoding binary asm instructions.
If you have the time it’s a good solution!
