Print the American Flag!

Question

Special Independence Day (USA) themed challenge for you today. You must write a program that prints this ascii-art representation of The American Flag.

0
|---------------------------------------------------------
| *   *   *   *   *   * #################################|
|   *   *   *   *   *                                    |
| *   *   *   *   *   *                                  |
|   *   *   *   *   *   #################################|
| *   *   *   *   *   *                                  |
|   *   *   *   *   *                                    |
| *   *   *   *   *   * #################################|
|   *   *   *   *   *                                    |
| *   *   *   *   *   *                                  |
|########################################################|
|                                                        |
|                                                        |
|########################################################|
|                                                        |
|                                                        |
|########################################################|
|                                                        |
|                                                        |
|########################################################|
|---------------------------------------------------------
|
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Trailing spaces on each line, as well as one trailing newline, are allowed.

Note that this isn't quite the way the flag should look, but it's the closest I could get with ASCII.

As usual, this is code-golf so standard loopholes apply and shortest answer in bytes wins!

Answer 1

CJam, 184 120 109 101 76 74 69 67 64 62 58 bytes

0'-57*"  #"56f*'|f+7*2>" *  "50*22/W<Sf+..e&~J$]N'|+a37*.+

Try it online in the CJam interpreter.

Idea

The most interesting part of the flag is the stars and stripes pattern.

If we repeat two spaces and a number sign 56 times and append a vertical bar to each, we get

                                                         |
                                                         |
#########################################################|

Repeating this pattern 7 times and discarding the first two lines, we obtain the stripes:

#########################################################|
                                                         |
                                                         |
#########################################################|
                                                         |
                                                         |
#########################################################|
                                                         |
                                                         |
#########################################################|
                                                         |
                                                         |
#########################################################|
                                                         |
                                                         |
#########################################################|
                                                         |
                                                         |
#########################################################|

Now, if we repeat the string " * " 50 times and split the result into chunks of length 22, we obtain the stars:

 *   *   *   *   *   *
   *   *   *   *   *  
 *   *   *   *   *   *
   *   *   *   *   *  
 *   *   *   *   *   *
   *   *   *   *   *  
 *   *   *   *   *   *
   *   *   *   *   *  
 *   *   *   *   *   *
   

The whitespace is a little off, but we can fix that by eliminating the last chunk and appending a space to the remaining ones.

Now, if we superimpose stripes and stars, we get

 *   *   *   *   *   * #################################|
   *   *   *   *   *                                    |
 *   *   *   *   *   *                                  |
   *   *   *   *   *   #################################|
 *   *   *   *   *   *                                  |
   *   *   *   *   *                                    |
 *   *   *   *   *   * #################################|
   *   *   *   *   *                                    |
 *   *   *   *   *   *                                  |
########################################################|
                                                        |
                                                        |
########################################################|
                                                        |
                                                        |
########################################################|
                                                        |
                                                        |
########################################################|

All that's left to do is adding two lines of 57 dashes, adding a column of 37 vertical bars and putting the cherry on top.

Code

0         e# Push a zero.
'-57*     e# Push a string of 57 dashes.
"  #"56f* e# Repeat each character in the string 56 times.
'|f+      e# Append a vertical bar to each resulting string.
7*        e# Repeat the resulting array of strings 7 times.
2>        e# Discard the first two strings.
" *  "50* e# Repeat the string 50 times.
22/       e# Split the result into chunks of length 22.
W<        e# Discard the last, partial chunk.
Sf*       e# Append a space to each chunk.
..e&      e# Twofold vectorized logical AND.
          e# Since all characters in the strings are truthy, this always selects
          e# the second character, painting the stars over the stripes.
~         e# Dump all resulting strings on the stack.
J$        e# Copy the string of dashes.

]         e# Wrap the entire stack in an array.
N'|+a37*  e# Repeat ["\n|"] 37 times.
.+        e# Perform vectorized concatenation.

Answer 2

Python 2, 113 bytes

for i in range(38):print i and"|"+["-"*57,(" *  "*7)[i%2*2:][:(i<11)*23].ljust(56,"  #"[i%3])+"|"][1<i<21]*(i<22)

String slicing and modulo checks galore.

Answer 3

Brainf**k, 3355 3113 1598 1178 782 bytes

What is this language?

Here is the hand optimized version featuring 28 loops. I think I've taken this about as far as it will go.

Here's the run at ideone.com:

+++[>++++<-]>[>+++>+++>+++>++++++++++>+>++++<<<<<<-]>++++++>---->->>>.<--.
<++++.>>---.>+++++++[<........>-]<<.
<.<<<<+++++[>>.<.>..<<-]>>.<.>.<<++++[>>>........<<<-]>>>.>.>.
<.<<<<+++++[>>...<.<-]+++++[>>.......<<-]>>.>>.>.
<.<<<<++++++[>>.<.>..<<-]++++[>>........<<-]>>>>.>.
<.<<...<<+++++[>.>...<<-]++++[>>>........<<<-]>>>.>.>.
<.<<<<++++++[>>.<.>..<<-]++++[>>........<<-]>>>>.>.
<.<<<<+++++[>>...<.<-]+++++[>>.......<<-]>>.>>.>.
<.<<<<+++++[>>.<.>..<<-]>>.<.>.<<++++[>>>........<<<-]>>>.>.>.
<.<<<<+++++[>>...<.<-]+++++[>>.......<<-]>>.>>.>.
<.<<<<++++++[>>.<.>..<<-]++++[>>........<<-]>>>>.>.
>>>+++[<<<
<.>>>+++++++[<<<<........>>>>-]<<<.>.
>>++[<<
<.<<<<+++++++[>>........<<-]>>>>.>.
>>-]<<
>>>-]<<<
<.>>>+++++++[<<<<........>>>>-]<<<.>.
<.>>.>+++++++[<........>-]<<.
>>++++++++[<<<.>.<.>.>>-]

---

How does this work?

 1: +++[>++++<-]>[>+++>+++>+++>++++++++++>+>++++<<<<<<-]>++++++>---->->>>.<--.
 2: <++++.>>---.>+++++++[<........>-]<<.
 3: <.<<<<+++++[>>.<.>..<<-]>>.<.>.<<++++[>>>........<<<-]>>>.>.>.
 4: <.<<<<+++++[>>...<.<-]+++++[>>.......<<-]>>.>>.>.
 5: <.<<<<++++++[>>.<.>..<<-]++++[>>........<<-]>>>>.>.
 6: <.<<...<<+++++[>.>...<<-]++++[>>>........<<<-]>>>.>.>.
 7: <.<<<<++++++[>>.<.>..<<-]++++[>>........<<-]>>>>.>.
 8: <.<<<<+++++[>>...<.<-]+++++[>>.......<<-]>>.>>.>.
 9: <.<<<<+++++[>>.<.>..<<-]>>.<.>.<<++++[>>>........<<<-]>>>.>.>.
10: <.<<<<+++++[>>...<.<-]+++++[>>.......<<-]>>.>>.>.
11: <.<<<<++++++[>>.<.>..<<-]++++[>>........<<-]>>>>.>.
12: >>>+++[<<<
13: <.>>>+++++++[<<<<........>>>>-]<<<.>.
14: >>++[<<
15: <.<<<<+++++++[>>........<<-]>>>>.>.
16: >>-]<<
17: >>>-]<<<
18: <.>>>+++++++[<<<<........>>>>-]<<<.>.
19: <.>>.>+++++++[<........>-]<<.
20: >>++++++++[<<<.>.<.>.>>-]

This program uses 10 memory locations:

0: loop counter #1
1: loop counter #2
2: "*"  ASCII 42
3: spc  ASCII 32
4: "#"  ASCII 35
5: "|"  ASCII 124
6: "\n" ASCII 10
7: "0"  ASCII 48, "-"  ASCII 45
8: loop counter #3
9: loop counter #4

Line 1

• This line sets up the ASCII characters in registers 2 through 7 (mostly). Some tweaking is done later.
• This code first puts 3 in register 0, and then loops 3 times incrementing register 1 four times each loop: +++[>++++<-]. Then end result is that register 0 is 0, and register 1 is 12.
• The 12 is used as the loop counter for the next loop. For 12 times through the loop, registers 2, 3, and 4 are incremented 3 times, register 5 is incremented 10 times, register 6 is incremented 1 time, and register 7 is incremented 4 times. At the end of this loop, they contain: R2(36), R3(36), R4(36), R5(120), R6(12), R7(48). After the loop register 2 is incremented 6 times, register 3 is decremented 4 times, and register 4 is decremented once. At this point, the values are: R2(42), R3(32), R4(35), R5(120), R6(12), R7(48). All but registers 5 and 6 contain their initial ASCII values.
• Next register 7 is output, the "0" at the top of the flag!
• Next register 6 is decremented twice to 10 (ASCII newline) and output. Done with the first line of the flag!

Line 2

• First it increments register 5 by 4 which makes it "|" (ASCII 124) and outputs it.
• Then it decrements register 7 by three changing it from "0" (ASCII 48) into "-" (ASCII 45) and outputs it.
• Next it puts 7 into loop counter 3 (register 8) and loops 7 times, writing out 8 dashes each time for a total of 7*8 = 56 dashes.
• Finally it ends by outputting a newline.

Line 3

• This line contains two loops.
• The first loop writes " * " 5 times.
• Then " * " is written
• The second loop loops 4 times writing 8 "#" for a total of 32.
• Then "#", "|", and "\n" are written.

Lines 4 - 11

• These lines use the same technique as line 3 to write out the stars and stripes of the flag.

Line 12

• This line starts a loop that runs 3 times.
• The loop ends at line 17.

Line 13

• Writes a strip that goes across the flag.
• Uses a loop that runs 7 times writing "#" 8 times each time through the loop.

Line 14

• The start of a loop that runs 2 times.

Line 15

• Writes a strip that goes across the flag.
• Uses a loop that runs 7 times writing " " 8 times each time through the loop.

Line 16

• End of inner loop that started at line 14.

Line 17

• End of outer loop that started at line 13.

Line 18

• Draws bottom stripe of flag.

Line 19

• Draws bottom border of flag.

Line 20

• Draws the flagpole.
• Loops 8 times, writing "|" and newline twice each time through the loop.

Answer 4

///: 225 characters

/D/ddd//d/--------//H/hhh//h/########//S/sss//s/        //A/aaaaa//a/ *  //b/|HHh|
|SSs|
|SSs|
//p/|
|
|
|
/0
|DDd-
|A * Hh#|
|  A Ss |
|A * Ss |
|  A Hh#|
|A * Ss |
|  A Ss |
|A * Hh#|
|  A Ss |
|A * Ss |
bbb|HHh|
|DDd-
pppp

Answer 5

JavaScript (ES6), 153 156

Using template string, there is 1 newline that is significant and counted

Test running the snippet below (being EcmaScript 6, Firefox only)

// TEST - Just for testing purpose,redefine console.log

console.log = (...x) => O.innerHTML += x+'\n'

// SOLUTION

o=[0];for(o[r=1]=o[21]='-'[R='repeat'](57);++r<21;o[r]=" *  "[R](7).substr(r%2*2,r<11&&23)+'  #'[r%3][R](r<11?33:56)+'|')o[37]='';console.log(o.join`
|`)

<pre id=O></pre>

To be even more patriotic, here is the EcmaScript 5 version

// TEST - Just for testing purpose,redfine console.log

console.log = function(x){ O.innerHTML += x+'\n' }

// SOLUTION - 175 bytes

for(o=(A=Array)(38),o[0]=0,r=2;r<21;r++)o[r]=A(8)[J='join'](" *  ").substr((r&1)*2,r<11?23:0)+A(r<11?34:57)[J]('  #'[r%3])+'|';
o[1]=o[r]=A(58)[J]('-'),console.log(o[J]('\n|'))

<pre id=O></pre>

Answer 6

Ruby, 104 102 bytes

Using ideas from ManAtWork's Ruby answer with permission.

puts 0,s=?|+?-*57,(0..18).map{|i|?|+("#  "[i%3]*(i>8?56:33)).rjust(56," *   *"[i%2*2,4])+?|},s,'|
'*16

Ruby, 127 121 112 bytes

Changed quotes to ? used array instead of conditional for stripe colour. used conditional instead of formula for stripe length.

puts 0,s=?|+?-*57
19.times{|i|puts ?|+("#  "[i%3]*(i>8?56:33)).rjust(56,i%2>0?"   *":" *  ")+?|}
puts s,"|\n"*16

The trick here is to draw the stripes (both red/# and white/space) to the correct length, then right justify them, padding with stars. Ruby's rjust allows us to specify the padding string, which alternates between " * " and " *".

Original version, 127 bytes

puts 0,s="|"+"-"*57
19.times{|i|puts("|"+((i%3>0?" ":"#")*((i+1)/10*23+33)).rjust(56,i%2>0?"   *":" *  ")+"|")}
puts s,"|\n"*16

Answer 7

SWI-Prolog, 275 bytes

In a language of French origin, which is kind of fitting

a:-put(48),nl,b,c(0).
b:-z,w(-,57).
c(I):-nl,I=36;J is I+1,(I=19,b,c(J);I>19,z,c(J);I>8,z,(I mod 3=:=0,w(#,56);tab(56)),z,c(J);z,(I mod 2=:=0,tab(1),w('*   ',5),put(42),tab(1);w('   *',5),tab(3)),(0=:=I mod 3,w(#,33);tab(33)),z,c(J)).
z:-put(124).
w(A,B):-writef('%r',[A,B]).

See the result here

Answer 8

C, 235 211 208 205 203 198 197 186 bytes

i;x(){for(puts("0");i<37;i++){char b[58]="";i<21?memset(b,i%20?i%3&1?35:32:45,56),i&&i<10?memcpy(b," *   *   *   *   *   *   "+(i%2?0:2),23):0,b[56]=i%20?124:45:0;printf("|%.57s\n",b);}}

edit: added some of Cool Guy's suggestions and made use of ?: to replace some if statements.

edit: removed overflow \0 prevention and used string length limiter in printf instead.

edit: reworked both memset conditionals.

edit: moved puts("0") inside the for header to remove its semicolon.

edit: slight refactoring to get 11 more bytes.