Worthwhile task shaving
Recreate https://xkcd.com/1205/ which shows a (humorous) table of how long you can work on making a routine task more efficient before spending more time than saved, for various s(h)avings against how often the task is run.
There are of course several possible interpretations of "day" and "week" in this context. The Phix implementation assumes 8 hour days and 5 day weeks might be more realistic, whereas it seems the original author worked with 24 hour days and 7 day weeks, and, tbh, my interest is piqued to see what built-in facilities other languages might have for handling such non-standard terms, if any. Extra kudos awarded for getting into the mind of the original author and reproducing their results exactly, or drumming up non-trivial (but still elegant) and potentially actually useful routines. This task can be made as trivial or as convoluted as you please, and should aim more for a little playfulness than rigid scientific accuracy.
Phix
One outlier here is 1hr 5/day ==> 3 years vs original 10 months: as per notes above for 5/8ths the cutoff is indeed 3 years.
Note that the standard builtins such as elapsed() have no facilities for non-standard terms such as 8 hour working days.
with javascript_semantics constant SEC = 1, MIN = 60, HOUR = 60*MIN, DAY = 8*HOUR, -- (allow some sleepage) WEEK = 5*DAY, -- (omit weekends) MONTH = 4*WEEK, YEAR = 12*MONTH, -- (as 48 weeks/omit holidays) shavings = {1,5,30,MIN,5*MIN,30*MIN,HOUR,6*HOUR,DAY}, frequencies = {{50,DAY},{5,DAY},{1,DAY},{1,WEEK},{1,MONTH},{1,YEAR}}, roundto = {SEC, MIN, HOUR, DAY, WEEK, MONTH, YEAR}, ts = {"sec", "min", "hour", "day", "week", "month", "year"} function duration(atom a) string es for rdx=1 to length(roundto) do atom t = trunc(a/roundto[rdx]) if rdx>1 and t<1 then exit end if es = sprintf("%d %s%s",{t,ts[rdx],iff(t=1?"":"s")}) end for return es end function printf(1," 50/day 5/day daily weekly monthly yearly\n") for s=1 to length(shavings) do integer si = shavings[s] string line = sprintf("%10s ",duration(si)) for f=1 to length(frequencies) do integer {per,slot} = frequencies[f] if si*per > slot then line &= sprintf("%10s ","n/a") else atom shaving = (5*YEAR/slot * per) * si line &= sprintf("%10s ",duration(shaving)) end if end for printf(1,"%s\n",line) end for
- Output:
50/day 5/day daily weekly monthly yearly 1 sec 2 days 1 hour 20 mins 4 mins 1 min 5 secs 5 secs 2 weeks 1 day 1 hour 20 mins 5 mins 25 secs 30 secs 3 months 1 week 1 day 2 hours 30 mins 2 mins 1 min 6 months 2 weeks 2 days 4 hours 1 hour 5 mins 5 mins 2 years 3 months 2 weeks 2 days 5 hours 25 mins 30 mins n/a 1 year 3 months 3 weeks 3 days 2 hours 1 hour n/a 3 years 7 months 1 month 1 week 5 hours 6 hours n/a n/a 3 years 9 months 2 months 3 days 1 day n/a n/a 5 years 1 year 3 months 1 week