Day of the week for a given date of this year. Do not use any date tools

Question

Please write a program that takes as input two numbers 'day of the month' and 'month number' for any day of this year (2012) and outputs day of the week. Examples:

% ruby ./golf.rb 14 07 # 14th of July 
"Sat"
% ruby ./golf.rb 14 7  # the same
"Sat"
% ruby ./golf.rb 01 12 # Dec 1
"Sat"
% ruby ./golf.rb 31 12 # Last day of the year
"Mon"

• Output format should be like Mon, Tue, Wed, Thu, Fri, Sat, Sun, not less than 3 letters
• Do not use date libs/modules, just math and logic
• Wins the shortest script in any language.

Answer 1

GolfScript (46 44 chars)

~.5*1$8>+2/\3<++7%"WedThuFriSatSunMonTue"3/=

Calculation can be more efficient than lookup if the days are rotated. I doubt that the mapping to day names can get shorter, but it may still be possible to shave a character or two off the hash function.

However, my best idea for how to do it, finding a power which starts with a suitable string, doesn't look like a contender. After about 9 CPU-days of brute forcing I found a couple of solutions, but they're also 44 chars. The faster one is

~16795 33303?`=+7%"WedThuFriSatSunMonTue"3/=

Answer 2

C++ (142) (135) (129 chars)

main(){string s[]={"Mon","Tue","Wen","Thu","Fri","Sat","Sun"};int d,m,t[]={0,5,1,1,4,6,2,4,0,3,5,1,3};cin>>d>>m;cout<<s[(d+t[m]+15-(m<3))%7];}

The requirement that the letters Mon, Tue etc. appear made it quite bad, otherwise it would be only 84 chars. I'll search a golf for the strings...

Hm, trying to golf the string only made it 2 chars longer, but I find it more aesthetically pleasing.

main(){char s[]={"MonTueWenThuFriSatSun"};int d,m,t[]={0,5,1,1,4,6,2,4,0,3,5,1,3};cin>>d>>m;d=((d+t[m]+15-(m<3))%7)*3;s[d+3]=0;printf("%s",s+d);}

Well, this for my first entry at a golf contest, I have still much to learn...

HA! Got it!

main(){char s[]={"MonTueWenThuFriSatSun"},t[]={"0511462403513"};int d,m;cin>>d>>m;d=((d+t[m]-33-(m<3))%7)*3;s[d+3]=0;printf("%s",s+d);}

_{with a well set compiler and some luck we might chop another 3 characters, by using char* instead of char[], but it will crash on most systems. Anyway, it doesn't matter, as I believe soon someone will post a solution in J or K with less than 10 characters, as usual :P}

I'll never give up!

main(){char s[]={"MonTueWenThuFriSatSun0511462403513"};int d,m;cin>>d>>m;d=((d+s[m+21]-33-(m<3))%7)*3;s[d+3]=0;printf("%s",s+d);}

Answer 3

c, 101 chars

d,m;main(){scanf("%d %d",&d,&m);printf("%.3s\n","SunMonTueWedThuFriSat"+(d+".034025036146"[m])%7*3);}

Answer 4

APL (57 56)

D M←⎕⋄3↑'MonTueWedThuFriSatSun'↓⍨3×7|D+⍎M⌷'512503514624'

Answer 5

Python3 (93) (86)

print("MTWTFSSouehraunenuitn"[eval('%s+%s+14'%(input(),"0512573514624"[int(input())]))%7::7])

Thanks to vsz. That is just his idea translated to python (plus comments, plus python-tricks).

print("MTWTFSSouehraunenuitn"[(int(input())+int("0512573514624"[int(input())]))%7::7])

Answer 6

Ruby, 75 74

Ok, my ruby:

a,b=$*.map &:to_i;p:MonTueWedThuFriSatSun[(a-:x365274263153[b].to_i)%7*3,3]

Saved one char:

a,b=$*;p:MonTueWedThuFriSatSun[(a.to_i-:x365274263153[b.to_i].to_i)%7*3,3]

Answer 7

Python (130 chars)

d=raw_input().split()
n=int(d[0])+sum([31,29,31,30,31,30,31,31,30,31,30,31][:int(d[1])-1])-1
print 'SMTWTFSuouehranneduit'[n%7::7]

It's rather longish for Python, but I like it.

Answer 8

Factor 98

: day ( x x -- x )
{ 0 3 6 5 2 7 4 2 6 3 1 5 3 } nth - 7 mod "MonTueWedThuFriSatSun" 3 group nth ;

Output:

( scratchpad ) 23 7 day

--- Data stack:
"Mon"

Answer 9

C, 96 chars

Reads from standard input. Usage: echo 23 7 | ./a.out

main(a,b){
    scanf("%d%d",&a,&b);
    a+=b*30.56-(b>2);
    puts("Thu\0Fri\0Sat\0Sun\0Mon\0Tue\0Wed"+a%7*4);
}

The formula b*30.56, rounded down, gives the accumulated number of days before each month, assuming a 30-day February. -(b>2) corrects for 2012's 29-day February.