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This is my first code golf so please let me know if it's too broad or if I'm missing any information for a good puzzle!

Challenge

In Ontario and possibly other areas of the world, electricity is billed using Time-Of-Use (TOU) pricing, which varies the cost per kilowatt-hour according to when you use power.

Given a date and time, I want to know whether I'm in an on-peak (red), mid-peak (yellow), or off-peak (green) time period.

Input

Assume that input is provided in an acceptable timezone-less ISO 8601 date-time format with the minimum precision of hours: YYYY-MM-DDThh[:mm[:ss]] (the T is literal).

Examples

  • 2014-09-01T14
  • 2014-09-01T17:30
  • 2014-09-01T17:30:02

Output

The output should be a string On, Mid, or Off.

Rules

  • Shortest code wins
  • For the purposes of this challenge, ignore statutory holidays
  • Assume the information found in this post. The actual rules of time-of-use pricing might change in the future by the Ontario Ministry of Energy.

Information

Summer weekdays (May 1st to October 31st)

Time-of-use clock for summer weekdays

  • Off-peak: 19h00 - 07h00
  • Mid-peak: 07h00 - 11h00 and 17h00 - 19h00
  • On-peak: 11h00 - 17h00

Winter weekdays (November 1st to April 30th)

Time-of-use clock for winter weekdays

  • Off-peak: 19h00 - 07h00
  • Mid-peak: 11h00 - 17h00
  • On-peak: 07h00 - 11h00 and 17h00 - 19h00

Weekends

Time-of-use clock on weekends

  • Off-peak: All day
share|improve this question
    
Are you sure the winter weekdays don't have mid-peak / on-peak swapped? –  Jan Dvorak Aug 26 at 8:52
3  
@JanDvorak, in winter people use lights and heating in the morning and evening; in summer they use air-con at midday. –  Peter Taylor Aug 26 at 8:53
4  
This is borderline-duplicate of codegolf.stackexchange.com/q/7008/194 (parse a datetime and do a simple calculation based on whether it's a working day or not). I think the season dependency is just different enough, but others may disagree. –  Peter Taylor Aug 26 at 8:55
    
@PeterTaylor the rules seem much simpler here than in the linked question. This doesn't have to handle leap years, for example. –  Jan Dvorak Aug 26 at 9:00
3  
Should the general date format be YYYY-MM-DDThh[:mm[:ss]] since seconds can only be applied if minutes are applied? –  Cruncher Aug 26 at 14:34

12 Answers 12

up vote 3 down vote accepted

Ruby - 147 144 143 141 137 135

x=->s{y,m,d,h=s.scan(/\d+/).map &:to_i
g=Time.new(y,m,d).wday%6<1?0:[0..11,4..9].count{|r|r===h-7}
%W{Off Mid On}[m<5||m>10?(3-g)%3:g]}

This represents a function which takes a string as a parameter and returns a string.

Here's an online demo with some test cases: http://ideone.com/wyIydw

share|improve this answer

Python 2 - 164

from datetime import*
d,t=input().split('T')
y,m,d=map(int,d.split('-'))
t=int(t[:2])
print'OMOfinfd'[(1+((10<t<17)==(4<m<11)))*(date(y,m,d).weekday()<5<6<t<19)::3]

If needed, below is a explanation of the logic in the final line:

The final line prints a slice of 'OMOfinfd' depending on the evaluation of its conditionals.

  • First, evaluate the operation 1+((10<t<17)==(4<m<11)).

    If the XNOR between the conditions 10<t<17 and 4<m<11 is False, this will evaluate to 1+False => 1+0 => 1. Otherwise, the operation will evaluate to 1+True => 1+1 => 2.

  • Finally, multiply that result of the above operation by whether the day is a weekday and whether the time is between 6am-7pm.

    If this is False, either the day is a weekend or the time is between 7pm-6am, and the result will be (1|2)*0 => 0. Otherwise the result will be (1|2)*1 => 1|2.

A result of 0 will print Off, 1 will print Mid, and 2 will print On.

share|improve this answer
    
Isn't the golfed version actually longer because you use semicolons? Or are newlines counted? –  Beta Decay Aug 26 at 14:59
    
Newlines are typically counted. In Notepad++ (where I typically get my counts) the ungolfed version is 4bytes longer at 168. –  BeetDemGuise Aug 26 at 15:54
3  
@BeetDemGuise Edit → EOL Conversion → UNIX/OSX Format. –  Schism Aug 26 at 16:20

C# - 240 220 chars

string x(string s){var d=DateTime.Parse((s+":00:00").Substring(0,19));int h=d.Hour,i=(int)d.DayOfWeek,x=d.Month;string o="off",m="mid",f="on";return i==6|i==0?o:x>=5&x<11?h>18|h<7?o:h>10&h<17?f:m:h>18|h<7?o:h>10&h<17?m:f;}

Nothing special. Straight forward coding.

Thanks to w0lf :)

share|improve this answer
1  
I think you can shorten s.Length==13?s+":00:00":s.Length==16?s+":00":s to (s+":00:00").Substring(0,19) –  w0lf Aug 26 at 15:41

Ruby - 135

Abuses the Time module. Input by command line argument.

d=Time.new(*$*[0].scan(/\d+/)[0..3])
o,m,f=%w{On Mid Off}
o,m=m,o if (d.mon-5)%10<6
p d.wday%6<1||(12>h=(d.hour+5)%24)?f:15<h&&h<22?m:o

Edit: Thanks w0lf for Time which helped shorten and solve a bug.

share|improve this answer
    
Your program is incorrect. For the input 2014-09-01T17:30 it correctly outputs "Mid", but for 2014-09-01T17 it outputs "Off". –  w0lf Aug 28 at 13:51

Groovy - 621 534 524 491 chars

Some further golfing to do, but pretty simple when leveraging Joda-Time

@Grab(group='joda-time',module='joda-time',version='2.3')
f={p,a->org.joda.time.format.DateTimeFormat.forPattern(p).parseDateTime a}
g={p,a->def x=null;try{x=f p,a}catch(Exception e){}}
a=args[0]
d=["",":mm",":mm:ss"].collect{g "yyyy-MM-dd'T'HH$it",a}.find{it}
r="Off"
m="Mid"
j={d,a,b->d.hourOfDay>a&&d.hourOfDay<b}
k={j(it,6,11)||(j(it,16,19))}
if(d.dayOfWeek<6){x=d.monthOfYear;if(x>4&&x<12){if(j(d,10,17))r="On";if(k(d))r=m}else if(x<5||x>10){if(j(d,10,17))r=m;if(k(d))r="On"}}
println r

sample runs:

bash-3.2$ ./run.peak.sh 
groovy Peak.groovy 2014-08-26T19
Off
groovy Peak.groovy 2014-08-26T07:00
Mid
groovy Peak.groovy 2014-08-26T18:00:00
Mid
groovy Peak.groovy 2014-08-26T12:30:30
On
groovy Peak.groovy 2014-11-01T00
Off
groovy Peak.groovy 2014-02-05T11:11:11
Mid
groovy Peak.groovy 2014-01-05T08:08
Off
groovy Peak.groovy 2014-12-18T18:59:59
On
groovy Peak.groovy 2014-08-31T14
Off

Ungolfed:

@Grab(group='joda-time',module='joda-time',version='2.3')

f = { p,a -> org.joda.time.format.DateTimeFormat.forPattern(p).parseDateTime a}
g = { p,a -> def x=null; try{x=f p,a} catch(Exception e) {} }
a=args[0]

d = ["",":mm",":mm:ss"].collect{g "yyyy-MM-dd'T'HH$it",a}.find{it}
r = "Off"
m = "Mid"

j = {d,a,b -> d.hourOfDay > a && d.hourOfDay < b}
k = { j(it,6,11) || (j(it,16,19)) }

if (d.dayOfWeek<6) {
    x = d.monthOfYear;
    if ( x>4 && x<12 ) {
        if (j(d,10,17)) r="On";
        if (k(d)) r=m
    } else if (x<5||x>10) {
        if (j(d,10,17)) r=m;
        if (k(d)) r="On"
    }
}
println r
share|improve this answer
2  
Since Java 8, there's also java.time.*, which is very similar to Joda Time but is part of the JRE. Perhaps this could shorten the code a bit. –  ntoskrnl Aug 26 at 11:37
    
re: Java 8. excellent point –  Michael Easter Aug 26 at 12:15
    
In your 3rd-to-last example '2014-01-05T08:08', January 5th, 2014 is a Sunday. Thus, it should be 'Off' –  BeetDemGuise Aug 26 at 13:56
    
re: Jan 5, 2014. Quite right. The current code is correct, but the output run is wrong. Will fix. –  Michael Easter Aug 26 at 14:18
    
Tried java.time.*. failed miserably. The parser is too damn strict. Use DateParser from nashorn (also part of the JRE8) which is lenient, and with a bit of abusive hacking, even lenient enough to skip minutes&seconds. –  Mark Jeronimus Aug 27 at 19:40

R, 243 204 characters

b=strptime(scan(,""),"%Y-%m-%dT%H");i=function(x)as.integer(format(b,x));h=i("%H");d=i("%m%d");w=d>1100|d<430;f=ifelse;cat(c("Off","Mid","On")[f(i("%u")%in%5:6|h<7|h>19,1,f(h>11&h<17,f(w,2,3),f(w,3,2)))])

Indented and commented:

b=strptime(scan(,""),"%Y-%m-%dT%H") #Takes stdin and converts into POSIXct
i=function(x)as.integer(format(b,x)) #Format the POSIXct and convert it to integer
h=i("%H")      #Format to hours
d=i("%m%d")    #Format to Month/Day
w=d>1100|d<430 #True if winter time, false if summer
f=ifelse
cat(c("Off","Mid","On")[f(i("%u")%in%5:6|h<7|h>19, #If weekend or night
                          1,                       #Case 1
                          f(h>11&h<17,            #Else if mid-day
                             f(w,3,2),             #Case 2 in winter, case 3 in summer
                             f(w,2,3)))])          #else vice versa

Examples:

> b=strptime(scan(,""),"%Y-%m-%dT%H");i=function(x)as.integer(format(b,x));h=i("%H");d=i("%m%d");w=d>1100|d<430;f=ifelse;cat(c("Off","Mid","On")[f(i("%u")%in%5:6|h<7|h>19,1,f(h>11&h<17,f(w,3,2),f(w,2,3)))])
1: 2014-08-26T15
2: 
Read 1 item
On
> b=strptime(scan(,""),"%Y-%m-%dT%H");i=function(x)as.integer(format(b,x));h=i("%H");d=i("%m%d");w=d>1100|d<430;f=ifelse;cat(c("Off","Mid","On")[f(i("%u")%in%5:6|h<7|h>19,1,f(h>11&h<17,f(w,3,2),f(w,2,3)))])
1: 2014-12-10T15
2: 
Read 1 item
Mid
> b=strptime(scan(,""),"%Y-%m-%dT%H");i=function(x)as.integer(format(b,x));h=i("%H");d=i("%m%d");w=d>1100|d<430;f=ifelse;cat(c("Off","Mid","On")[f(i("%u")%in%5:6|h<7|h>19,1,f(h>11&h<17,f(w,3,2),f(w,2,3)))])
1: 2014-08-26T23
2: 
Read 1 item
Off
share|improve this answer

Bash, 286

this is a simple bash answer using the date program

d(){ date -d $1 +%$2; };D=$(echo $1|sed 's/\(T..\)$/\1:00/');H=$(d $D H);M=$(d $D m);if [ $(d $D u) -gt 5 ]||[ $H -lt 7 ]||[ $H -gt 18 ];then echo Off;exit;fi;if [ $M -gt 4 ]&&[ $M -lt 11 ];then I=On;O=Mid;else I=Mid;O=On;fi;if [ $H -gt 10 ]&&[ $H -lt 17 ];then echo $I;else echo $O;fi
share|improve this answer

Here goes another one!

JavaScript, 175 171

function f(x){d=new Date(x.slice(0,10));t=x.slice(11,13),m=(x.slice(5,7)+1)%12;return(t<8||t>18||!(d.getUTCDay()%6)?'off':((t<11||t>17)?(m<5?'on':'mid'):(m<5?'mid':'on'))}

Unminified:

function f(x) {
  d = new Date(x.slice(0, 10));
  t = x.slice(11, 13), m = (x.slice(5, 7) + 1) % 12;
  return (t < 8 || t > 18 || !(d.getUTCDay() % 6) ? 'off' : ((t < 11 || t > 17) ? (m < 5 ? 'on' : 'mid') : (m < 5 ? 'mid' : 'on'))
}

Only works on interpreters where an ISO8601 date string can be passed into the Date constructor.

CoffeeScript, 192 189

Surprisingly, it's longer in CoffeeScript because there's no ternary operator in that language (which as you can see from my JavaScript, I heavily relied on).

f=(x)->d=new Date(x.slice(0,10));t=x.slice(11,13);m=(x.slice(5,7)+1)%12;return'off'if(t<8||t>18||!(d.getUTCDay()%6));if(t<11||t>17)then(if m<5then'on'else'mid')else(if m<5then'mid'else'on')
share|improve this answer

Python 3 - 352 chars

import datetime as dt
t=input()
i=int
p=print
a='on'
b='mid'
c='off'
m=i(t[5:7])
h=i(t[11:13])
d=dt.date(i(t[0:4]),m,i(t[8:10])).weekday()
if 5==d or 6==d:p(c)
elif h>=19 and h<7:p(c)
elif m<=10 and m>=4:
 if h>=7 and h<11:p(b)
 if h>=11 and h<17:p(a)
 if h>=17 and h<19:p(b)
else:
 if h>=7 and h<11:p(a)
 if h>=11 and h<17:p(b)
 if h>=17 and h<19:p(a)
share|improve this answer
1  
You should change s=['high','mid','off'] to s=['on','mid','off'] - not only does this save 2 chars, but the spec says to output "on". –  Sebastian Lamerichs Aug 26 at 12:27

Nothing stops me from entering my own, and there's other shorter ones here anyways, so…

PHP 5.4+, 194

<?function f($x){$t=substr($x,11,2);$m=(substr($x,5,2)+1)%12;if($t<8||$t>18||(new DateTime(substr($x,0,10)))->format('N')>5)return'off';return($t<11||$t>17)?($m<5?'on':'mid'):($m<5?'mid':'on');}

Unminified and commented:

<?
function f($x) {
  $t = substr($x,11,2); // hour
  $m = (substr($x,5,2) + 1) % 12; // month shifted up by 1

  if ($t < 8 || $t > 18 || (new DateTime(substr($x,0,10)))->format('N') > 5)
    return 'off'; // evenings and weekends

  return ($t < 11 || $t > 17)
    ? ($m < 5 ? 'on' : 'mid') // morning and mid-afternoon
    : ($m < 5 ? 'mid' : 'on'); // afternoon
}

Also note that the date.timezone directive in php.ini must be set, otherwise an Exception will be thrown.

share|improve this answer

Java - 426 309 / 301? (see comments)

String z(String a){
    DateParser b=new DateParser(a);
    boolean c=b.parseEcmaDate();
    Integer[]d=b.getDateFields();
    GregorianCalendar e=new GregorianCalendar(d[0],d[1]-(c?0:1),d[2]);
    e.setTimeZone(new SimpleTimeZone(0,""));
    return(124>>e.get(7)&1)>0&d[3]>6&d[3]<19?
           d[3]>10&d[3]<17^(1008>>e.get(2)&1)>0?"Mid":"On":"Off";
}

Example output:

2014-03-02T00   Off
2014-03-02T06   Off
2014-03-02T07   Off
2014-03-02T10   Off
2014-03-02T11   Off
2014-03-02T16   Off
2014-03-02T17   Off
2014-03-02T18   Off
2014-03-02T19   Off
2014-03-02T23   Off
2014-04-02T00   Off
2014-04-02T06   Off
2014-04-02T07   On
2014-04-02T10   On
2014-04-02T11   Mid
2014-04-02T16   Mid
2014-04-02T17   On
2014-04-02T18   On
2014-04-02T19   Off
2014-04-02T23   Off
2014-05-02T00   Off
2014-05-02T06   Off
2014-05-02T07   Mid
2014-05-02T10   Mid
2014-05-02T11   On
2014-05-02T16   On
2014-05-02T17   Mid
2014-05-02T18   Mid
2014-05-02T19   Off
2014-05-02T23   Off

I used the same EXOR trick as the Python submission. I also used a + as an OR function, for when it's weekend OR night.

My other big trick: bit masks.

For example, to see if a number is between 2 and 6 (Monday to Friday), first create a bit pattern where the interesting values are 1:

  6   2 
0b1111100 = 124

Then, use bit shift to get the interesting bit to the LSB and extract it:

(124 >> day_of_week) & 1

Similarly, I made bit patterns for months and hours:

  9    4
0b1111110000 = 1008

      76          76
0b111110000000000001111111 = 16253055
0b000000011111100000000000 = 129024

Unfortunately, it turns out simply x>y&x<z is shorter in most cases, so I didn't use it in some places.

And finally, a bit of hackery (highly implementation dependent) with jdk.nashorn.internal.parser.DateParser: When parseEcmaDate() doesn't completely parse a date (like when it reads hour and hits end of string), it returns false.

  • When it finishes properly, the date is stored in an Integer[] (auto-unboxing ftw), with the month fixed to be base-0 (like other Java classes).
  • When it returns false, it aborted half-way and didn't do this fixing. It however still put whatever it parsed into the array, which is readily available. Hence the -(c?0:1).
share|improve this answer
    
I think using just .parse() might also work (shaving off 8 characters), but I haven't tested it thoroughly with different inputs. parse internally calls parseEcmaDate and if it fails, calls parseLegacyDate. The latter may screw up the array, but this didn't happen with a few cases I tested. –  Mark Jeronimus Aug 27 at 20:48

ES6 - 146

This is in function form, uses a couple nasty hacks.

let y,m,d,h,c=s=>[y,m,d,h]=s.split(/[T:-]/g).map(i=>+i),new Date(y,m,d).getDay()in{0:1,6:1}||h<20||8>h?'Off':['Mid','On'][(10>h&&h>16)^(m>5||m<8)]

Explained:

// These variables are declared outside of the function scope to save
// characters.
let y, // year
    m, // month
    d, // day
    h, // hour
    c = s => // c for check, called as c(isoString)
      [y, m, d, h] = // Assign to fields
        s.split(/[T:-]/g) // Split at delimiters
         .map(i => +i), // Convert all to numbers
                        // Comma used to keep it as a single statement.
      new Date(y, m, d).getDay() // Create Date to get day of week
        in {0:1, 6:1} // Check if it is good (0 = Sunday, 6 = Saturday). It
                      // is shorter to use an object literal than to
                      // do individual checks.
      ||
      h < 20 || 8 > h ? // h is between 7 and 19 (7pm)
       'Off' : // Return 'Off'
       ['Mid','On'][ // Two other outputs
        (10 > h && h > 16) ^ // Check if it is 11-16 (5pm)
        (m > 5 || m < 8)] // Invert that if the month is in summer/fall. Note
                          // that this returns a number, either 1 or 0. This
                          // is an ugly hack using the bitwise xor operator.
share|improve this answer

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