Code Golf - Banner Generation

Answer 1

Python, 233 231 225 224 205 chars

Candidate for the shortest python solution here :-)
It is a two-liner - last line was broken in 3 for urrrr readability

s=raw_input()
for i in range(5):print' '.join(''.join((' ',c)[ord(
'W_E__U__QQ_QN_UQ_EA_Q]_D_Q_QYQ__D[_PP_B__F__Q__EG_Y__EZWU]A_A_P_OPO_\\_QNQWT_YUS'
[max(0,3*ord(c)-192-k)])>>i&1]for k in(2,1,0))for c in s)

ps. Thanks for comments, corrected issues and now using 7-bit ASCII only!

Test banner:

TTT H H EEE     QQQ U U III CCC K K     BBB RR  OOO W W N N     FFF OOO X X     JJJ U U M M PPP SSS     OOO V V EEE RR      TTT H H EEE     L   AAA ZZZ Y Y     DD  OOO GGG !!!
 T  H H E       Q Q U U  I  C   K K     B B R R O O W W NNN     F   O O  X        J U U MMM P P S       O O V V E   R R      T  H H E       L   A A   Z Y Y     D D O O G   !!!
 T  HHH EE      Q Q U U  I  C   KK      BBB RR  O O WWW NNN     FF  O O  X        J U U M M PPP SSS     O O V V EE  RR       T  HHH EE      L   AAA  Z  YYY     D D O O G G !!!
 T  H H E       QQQ U U  I  C   K K     B B R R O O WWW N N     F   O O  X      J J U U M M P     S     O O V V E   R R      T  H H E       L   A A Z     Y     D D O O G G    
 T  H H EEE     QQQ UUU III CCC K K     BBB R R OOO WWW N N     F   OOO X X     JJJ UUU M M P   SSS     OOO  V  EEE R R      T  H H EEE     LLL A A ZZZ YYY     DD  OOO GGG !!!

Answer 2

Ruby : 207 215 252 345 characters

i=gets.chomp;5.times{|t|p i.gsub(/./){|c|j=3*(c>?@?c.ord-64:0);(3*t..3*t+2).map{|d|"mini5mbmzjf2bqjmof3prl72i5pn138iuhylmkpi65i278kq3qjfaihyjb66787odp8ktiy5hwt78tmnb"[j..j+2].to_i(36)[d]==1?c:" "}.join+" "}}

Answer 3

Python 2.6, 251 - 243 - 227 characters

I tried a slightly different approach (bitpacking the parts the letters are made of) ...

• handles uppercase letters, spaces, exclamation mark.
• 2 unnecessary linebreaks added here for readability (the for loop could be one line)
• be sure to save this as UTF-8 with BOM!

As always, all comments and suggestions welcome! Contender for shortest Python solution (again a few characters behind at the moment)...

w=raw_input()
for l in range(5):print''.join("1111 11  11  1   1   "
[int(("%05d"%ord(u"<ϳϲࢬ禉ऐऒ࠾⬃ᅘᖆⰯ囌❿✛іϾь穏ࠂᅜ⭦⭪⫸㡩⬪㰼"
[max(0,ord(c)-64)]))[l])*3:][:3].replace("1",c)+" "for c in w)

Answer 4

Python, 340 characters

d=dict((i,[23535,31727,29263,15211,29391,4815,31567,23533,29847,31527,23277,29257,23421,23549,31599,5103,32623,23275,31183,9367,31597,11117,32749,21653,31213,29351][i-65])for i in range(65,91))
d[33]=29183
d[32]=0
s=raw_input()
for l in range(5):
 p=""
 for c in s:
  for n in range(3):
   if d[ord(c)]&2**(3*l+n):p+=c
   else:p+=" "
  p+=" "
 print p

sample output

>>> 
ABCDEFGHIJKLMNOPQRSTUVWXYZ !
aaa bbb ccc dd  eee fff ggg h h iii jjj k k l   m m n n ooo ppp qqq rr  sss ttt u u v v w w x x y y zzz     !!! 
a a b b c   d d e   f   g   h h  i    j k k l   mmm nnn o o p p q q r r s    t  u u v v w w  x  y y   z     !!! 
aaa bbb c   d d ee  ff  g g hhh  i    j kk  l   m m nnn o o ppp q q rr  sss  t  u u v v www  x  yyy  z      !!! 
a a b b c   d d e   f   g g h h  i  j j k k l   m m n n o o p   qqq r r   s  t  u u v v www  x    y z           
a a bbb ccc dd  eee f   ggg h h iii jjj k k lll m m n n ooo p   qqq r r sss  t  uuu  v  www x x yyy zzz     !!! 
>>> 

not too great, but it was fun writing it

edit whoops, I made the input be lowercase. fixed now, saved me one character too :)

Answer 5

C#, 239 231 229 chars (292 bytes)

I'm a bit late but this just looked like fun.

using C=System.Console;class P{static void Main(){var t=C.ReadLine();for(int
b=15,s;b>0;b-=3){foreach(var c in t)for(s=0;s++<4;)C.Write(s>3||c<33?' ':((
"翇篭篯礧歮禧禤祯寭璗牯宭䤧彭忭筯篤筿殭秏璒孯孪寿咕寏犧"[c<34?0:c-64])&1<<b-s)>0
?c:' ');C.WriteLine();}}}

Answer 6

C89, 345 characters

Newlines added for your sanity (they are not included in the character count and can/should be deleted):

char o[5][99];
d[]={0x2df7fbef,0x3927bb6b,0x396792cf,0x3da7dbed,0x3a4bfb27,0x2d76f249,0x2dbedbfd,0x3db793ef,0x3fb7daeb,0x3ce7a497,0x3db6ab6d,0x3ff6d495,0x3cf6f2a7,0x38ff8000};
c,i,j;
main(){memset(o,32,495);
while((c=getchar())>0){
for(j=0;j<15;j++)
o[j/3][i+j%3]=d[c-33?(c-65)/2:13]>>((c&1)*15+j)&1?c:32;i+=4;}
for(j=0;j<5;j++)printf("%.*s\n",i,o[j]);}