# The Staircase Challenge

Your work is to recreate this piece of art:

                        _____
|
|
|
|
_____|
|
|
|
|
_____|
|
|
|
|
_____|
|
|
|
|
_____|


The answer must recreate this and print this as a result. All languages allowed, no direct printing of art, ofc, some level of manipulation is required. The answer with the least bytes wins.

Closes on Thursday 6:30 AM UTC or so.

The original thing was shown to me by my friend who did this with Java, he refused to show me the source code and now I'll dazzle him with the brilliance of other languages possibly. :D

You cannot use any alternative character (makes it easier?).

1. Pyth - 28 bytes - isaacg
2. CJam - 30 bytes - Runer112
3. CJam - 32 bytes - Martin Büttner

Highest votes: C - 73 bytes - Paul R

isaacg takes the crown for passing the Staircase Challenge with Pyth. Watch out for more challenges like these on PPCG!

• Welcome to Programming Puzzles & Code Golf Stack Exchange! All challenges here require an objective winning criterion, to indisputably decide which solution should win. This looks like a code-golf question, i.e. shortest code wins, but I'll let you edit it in yourself in case you want to make it something different. Thanks! Commented Mar 2, 2015 at 15:08
• Ah, sorry, I missed that. I've edited the proper tag into your question, then. Commented Mar 2, 2015 at 15:11
• Do we have to print that 1 weird trailing space on the 6th line ? Commented Mar 2, 2015 at 15:45
• More generally, is trailing space allowed? Can I pad this to a rectangle of the width of the first line? Commented Mar 2, 2015 at 15:56
• Can we have a trailing newline? Commented Mar 2, 2015 at 16:26

# C, 86807675 73 bytes

c;main(i){for(i=21;i--;c='|')printf("%*s%c\n",i/5*6+5,i%5?"":"_____",c);}

• There always has to be that guy who will post a C solution. Upvote for you.
– qedk
Commented Mar 2, 2015 at 17:56
• You could get it even shorter by changing the loop to for(i=25;i--;) Commented Mar 2, 2015 at 18:02
• @FelixBytow It should be i=26. Besides this, the ' ' could be changed to 32 for an extra character. My solution is 2 characters longer after these optimizations :( Commented Mar 2, 2015 at 18:37
• You seem to have an extra step. there are 21 rows in the required output, not 26. Some more optimizations: main(i){for(i=21;i--;)printf("%*s%c\n",i/5*6+5,i%5?"":"_____",i<20?'|':0);} 1. Simplify the formula for length 2. As @Allbeert says, you can use the ascii code for ' ' but why stop at ASCII 32 when ASCII 0 will do. Also it runs fine for me with "" instead of " " Commented Mar 2, 2015 at 18:54
• How about this to save two bytes in '|' printing? c;main(i){for(i=21;i--;c='|')printf("%*s%c\n",i/5*6+5,i%5?"":"_____",c);} Commented Mar 3, 2015 at 13:56

# Java, 198158156 146 bytes

This can probably be shortened a lot. As usual, suggestions are welcome.

void a(){String a="",b="_____",c=b;for(int i=-1;i<20;b=i%5<1?b.replace(c,"     "):i%5>3?b+" "+c:b)a=b+(++i>19?"":"|")+"\n"+a;System.out.print(a);}


Indented (kinda):

void a(){
String a="",b="_____",c=b;
for(int i=-1;i<20;b=i%5<1?b.replace(c,"     "):i%5>3?b+" "+c:b)
a=b+(++i>19?"":"|")+"\n"+a;
System.out.print(a);
}


Thanks Martin Büttner, Rainbolt, and Geobits.

• Honestly, since you did it with Java, I'm impressed.
– qedk
Commented Mar 2, 2015 at 16:57

Brainfuck (1065 Bytes)

It's not pretty, it's not short...but i'll optimize later on!

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


# CJam, 36 30 bytes

Try it online.

L{{S5*\+}/S'|5*+]'_5*+}5*1>zN*


My initial 36-byte solution generated the result in the output orientation. Despite my attempts to squeeze more bytes out of the algorithm, I couldn't. Then I saw Martin's brilliant strategy of generating columns instead of rows and transposing the result. I realized that was probably a better approach, so I set off to create a transposition-based solution.

However, my approach to implementing that strategy varies quite a bit. Instead of generating full columns, I use an iterative solution that indents any already-generated "steps" and adds adds a new step at each iteration. So the first iteration of the main loop generates this:

 |||||
_
_
_
_
_


The second iteration of the main loop indents the existing step and adds a new one after it:

      |||||
_
_
_
_
_
|||||
_
_
_
_
_


And the full five iterations of the main loop generate this:

                     |||||
_
_
_
_
_
|||||
_
_
_
_
_
|||||
_
_
_
_
_
|||||
_
_
_
_
_
|||||
_
_
_
_
_


After this, all that needs to be done is eliminate the first line, which would otherwise become the unwanted riser for the bottom step, and transpose.

# Python 2, 8077 74 bytes

n=24;exec"print' '*n+'_'*5+'|'*(n<24)+('\\n'+~-n*' '+'|')*4*(n>0);n-=6;"*5


Got rid of the double exec and fit everything into the one print!

# Clip, 46

{:24S:5'_m[z{*4,:+5*6zS"|
":*6zS:5'_"|
"}vR4


## Explanation

{               .- Put everything in a list -.
:24S           .- 24 spaces                -.
:5'_           .- 5 underscores            -.
m[z            .- Map...                   -.
{           .- A list                   -.
*4         .- 4 of the following       -.
,        .- Append                   -.
:+5*6zS .- 5 + 6 * the iteration of spaces  -.
"|      .- A pipe and newline       -.
"
:*6zS      .- 6 * the iteration of spaces      -.
:5'_       .- 5 underscores            -.
"|         .- A pipe and newline       -.
"
           .- End list (per iteration  -.
}vR4         .- The mapping is onto {3,2,1,0}    -.

• Clip doesn't even have a Wikipedia page. I mean, what....
– qedk
Commented Mar 2, 2015 at 17:56
• @therewillbecoffee Many languages on this site don't have a Wikipedia page. That's the fun ;) Commented Mar 2, 2015 at 18:07
• @Ypnypn Did you design it? It looks really interesting! Although I'd be interested in a true quine. ;) (That one on the examples page is a bit cheaty.) Commented Mar 2, 2015 at 18:11

# CJam, 36 32 bytes

{5*S*'_+a5*~_W<S+5'|*+}5/;]W%zN*


Test it here.

I also tried using an explicit formula, but it's longer in CJam... maybe it helps someone else:

21,29,ff{_2$5/)6*(=@@6/5*=_++" |_|"=}W%N*  ## Explanation I found that the staircase can be built much more easily if you a) transpose the grid and b) reverse the lines: _ _ _ _ _ ||||| _ _ _ _ _ ||||| _ _ _ _ _ ||||| _ _ _ _ _ ||||| _ _ _ _ _  So first I'm building that, then reverse, then transpose. { }5/ "For i in [0 .. 4]."; 5*S*'_+ "Get a string of 5*i spaces and append _."; a5*~ "Get five such lines."; _W<S+ "Duplicate the last, remove the _, add a space."; 5'|*+ "Add 5 copies of |."; ; "The above creates a row too many, so discard the last one."; ]W%zN* "Wrap everything in an array, reverse, transpose, riffle with newlines.";  • Okay, someone's did it with 30 bytes. – qedk Commented Mar 3, 2015 at 10:26 # Python 2, 59 n=21 exec"n-=1;print n/5*6*' '+' _'[n%5<1]*5+'|'*(n<20);"*n  The 21 lines are indexed by n in [20,19,...,1,0]. First prints 6 spaces for each "step" we're up (minus 1), computed as n/5*6. Then, prints five spaces, except these are instead underscores for multiples of five. Finally, prints a vertical line, except for the top line n=20. • Nice and straightforward. I like it! Commented Mar 3, 2015 at 2:37 # JavaScript, 11510796948987 83 bytes This is too long to win, but it's the first time I've come up with an answer on PCG.SE, and I'm kind of proud to have made something postable. With some helpful syntactical advice I shortened the code significantly - even below the scrollbar threshold! for(s='',y=22;--y;s+='\n')for(x=0;++x<29;)s+=6*~(~-y/5)-~x?4+5*~(x/6)+y?' ':'_':'|'  • Nice answer, several things you could do to shorten it are take out the alert. If you run it in the console it works just fine without it. Also, the semicolon inside the last curly brace isn't necessary. You can save 1 byte by using (y/5-.2) instead of ((y-1)/5) – qw3n Commented Mar 3, 2015 at 16:33 • @qw3n thanks for those suggestions; I wasn't sure if the alert was required by the rules. – vvye Commented Mar 3, 2015 at 19:31 • You could also move the s+='\n' after the y-- and get rid of the curly braces so it looks like for(s='',y=21;y>0;y--,s+='\n'). I also initialized the s inside the for loop so your code is all one statement – qw3n Commented Mar 3, 2015 at 19:39 • I thought that was my last one but this should be it for me for(s='',y=22;--y;s+='\n')for(x=0;++x<29;)s+=6*~~(y/5-.2)+5-x?5*~~(x/6)+1-y?' ':'_':'|' if you flip your ternary expression you can test for number - x which is 0 if both terms are equal saving you 2 more bytes. – qw3n Commented Mar 3, 2015 at 21:04 • Tilde games: ~n == -n-1, -~n == n+1, ~-n == n-1, for(s='',y=22;--y;s+='\n')for(x=0;++x<29;)s+=6*~(~-y/5)-~x?4+5*~(x/6)+y?' ':'_':'|' is 83 Commented Mar 4, 2015 at 9:37 # ECMAScript 6, 142138129 91 bytes Special thanks to @edc65 for really reworking this. a=o='',[for(x of!0+o)(o=(a+' |\n').repeat(4)+a+'_____|\n'+o,a+=' ')],a+'_____\n'+o  out.value=(a=o='',[for(x of!0+o)(o=(a+' |\n').repeat(4)+a+'_____|\n'+o,a+=' ')],a+'_____\n'+o) textarea{ width:300px; height:400px; } <textarea id="out";></textarea> The logic of the original version check @edc65 comment for how it morphed. ((f,l,p)=> //variables f(24)+l+p[1]+ //add the non pattern line [0,1,2,3].map(b=>f(18-6*b)) //add the right number of spaces in front of the 4 steps .map((a,b)=>f(4,a+f(5)+p) //the four repeating lines of the step +a+l) //the landing line .join(p)+p) //put it all together ((n,s=' ')=>s.repeat(n) //create an variable array of some character ,'_____','|\n') //string literals  • You can safely remove the new constructor before Array to save a few bytes. Commented Mar 3, 2015 at 0:50 • @hsl thanks for some reason I thought the new was necessary. – qw3n Commented Mar 3, 2015 at 2:59 • Array(n).join(s) is so ES5! did you try repeat Commented Mar 3, 2015 at 7:31 • [1,2,3,4].map((a,b) and using just b => [0,1,2,3].map(b (-4) Commented Mar 3, 2015 at 7:36 • I run Firefox and it works, pretty nicely! – qedk Commented Mar 3, 2015 at 10:23 # MATLAB, 68 bytes I have the strong feeling MATLAB should be able to do better, but I can't think of a way. p(1:5,6)='|';p(1,1:5)=95;w=blkdiag(p,p,p,p);w(21,25:29)=95;flipud(w)  Creates the stairs upside-down and flips it. My thaumometer broke because of all the magic constants around. '|' is intentionally left as-is (instead of ascii codepoint) to initialize p and w as a char array. • '|' = 124 anyway, so that doesn't cost any extra chars. Commented Mar 3, 2015 at 6:42 # Ruby, 48 25.times{|i|puts" "*(4-i/6)*5+(i%6==0??_*5:?|)}  ## Old approach, 68 4.times{|i|(?_*5+" |"*5).each_line{|l|puts" "*(4-i)*5+l}} puts"_"*5  • Welcome to PPCG! A couple of Ruby golfing tips: 1. there's some unnecessary whitespace in there. 2. Single character strings like '_' can be written as ?_. 3. Newlines can be embedded directly in strings (so you can actually do "<linebreakhere>|"). 4. The parentheses around that aren't necessary. The final puts can be replaced with $><< (which allows you to get rid of the space, even after using ?_). Keep it up! :) Commented Mar 3, 2015 at 13:16
• Yeah, I just got rid of some whitespace. Thanks! I didn't know about the single character strings. Commented Mar 3, 2015 at 13:19
• You can probably also replace (1..4).map by 4.times and then use 4-i instead of 5-i. Commented Mar 3, 2015 at 13:26
• Ooh, good call. Done. Commented Mar 3, 2015 at 13:29

# Pyth, 29 28

V21++**6/-20N5d*5?d%N5\_<\|N


Try it here.

A pretty straightforward solution, with the "append five spaces or five underscores" trick from @xnor's solution, but with the loop from 0 to 20, not 20 to 0.

• I dub you thee, Sir isaacg for passing the Staircase Challenge.
– qedk
Commented Mar 6, 2015 at 11:24
• ...and now you can enter the staircase Commented Aug 18, 2015 at 20:42

# Julia, 83 bytes

for n=24:-6:0 print(" "^n*"_"^5*"|"^(n<24)*"\n"*(" "^(n>0?n-1:0)*"|\n"^(n>0))^4)end


In Julia, string concatenation is performed using the * operator and string repetition is performed using ^.

# ><>, 108104 100 bytes

cc+::?v~'_____'o\/' 'o  \
?:o'|'\' 'o1-30.o\v!?:-1<}:{oav!?:<4;!
-20.12^?(+cc:ooo/ \~1-'!|'o1. \~ao6


A simple ><> solution, using the same strategy as my Python answer. The main difference is that ><> doesn't have string multiplication (or even strings), so all of that is done with loops.

## Explanation

cc+                  Push 24 (call this "n")

[outer loop]
[loop 1, print n spaces]

:                    Copy n (call this "i")
:?                   If i is not zero...
' 'o1-30.                Print space, decrement i and go to start of loop 1

~'_____'ooooo        Pop i and print five underscores
:cc+(?               If n < 24...
'|'o                     Print a pipe
21.                  Otherwise skip pipe printing

[loop 2: print vertical parts of stairs]

?!;                  If n is zero, halt
4                    Push 4 (call this "j")
?!                   If j is zero...
~ao6-20.                 Pop j, print a newline, minus 6 from n and go to start of outer loop
ao                   Print a newline
}:{                  Copy n (call this "k")

[loop 3: print n-1 spaces]

1-                   Decrement k
:?!                  If k is zero...
~1-'!|'o1.           Pop k, decrement j, print a pipe and go to start of loop 2
' 'o                 Otherwise print a space and go to start of loop 3

• I feel like I will always upvote a ><> answer. Commented Mar 4, 2015 at 17:14

# Perl, 50

#!perl -l
print$"x6x(-$_/5),($_%5?$":_)x5,"|"x$|++for-20..0  Try me. # T-SQL, 276 bytes declare @x int declare @y int declare @w varchar(30) declare @l char(5) set @l='_____' set @x=23 while @x > 4 begin set @w=replicate(' ',@x) set @y=0 if @x=23 print @w+' '+@l else print @w+' '+@l+'|' while @y < 4 begin set @y=1+@y print @w+'|' end set @x=@x-6 end print @l+'|'  # Visual FoxPro 9.0, 261 bytes ### n = Number of Steps total of 175 characters, but had to output to file to display correctly - so minus 43 chars for file operations = 132 chars. n=10 c=CHR(13) f="st.t" ERAS (f) FOR i=n TO 1 STEP -1 p=(i-1)*6 =STRTO(PADL("_____",p+5)+IIF(i<n,"|","")+c+IIF(i>1,REPLI(PADL("|"+c,p+1),4),""),f,.t.) ?PADL("_____",p+5)+IIF(i<n,"|","")+c+IIF(i>1,REPLI(PADL("|"+c,p+1),4),"") ENDFOR MODI COMM (f)  Note to answerer: Byte count is for the absolute working source code, and the byte counter says that it's 261 bytes, so it is. # Bash (+tac from coreutils): 110 bytes This can be pasted directly into the terminal. (n=;set {1..4};for i do echo "${n}_____|";n+='     ';for i do echo "$n|";done;n+=\ ;done;echo "${n}_____")|tac
• Without |tac`, it doesn't quite work. And I ran this on Git Bash, so I should kill myself.
– qedk
Commented Mar 5, 2015 at 4:42