Build a staircase for my kid

Question

The other day, my son asked me to build him a staircase using Lego-ish blocks. And I came up with something like this:

Then my kid asked me for a program using the least number of bytes that generated a similar staircase in a computer screen. I am not that good at code-golf, so I need your help. I need a program that:

• Receives a positive integer with the number of levels the staircase needs to have.
• Outputs a drawing of a staircase, with the pattern you see in the image.

The output will be in text format but the bricks can be distinguished one from another. For instance, you may use the '█' character as half a block and paint it in any color you want, or just choose any character of your choice.

Restrictions:

• Blocks need to be of three different colors, which will be used as long as possible (if input is 1 or 2 there are not enough blocks to use all three colors). If you want, you may use the '░▒▓' characters, for instance, or just select three different characters.
• No two blocks of the same color or pattern can be side to side in a single row.

My son does not really care about trailing spaces or new lines as long as a staircase is drawn.

Examples (sorry for the bad choice of characters):

Input: 1
Output:
██

Input: 2
Output:
██
 ▓▓

Input: 3
Output:
██
 ▓▓
██░░

Input: 5
Output:
██
 ██
██░░
 ██░░
██░░▓▓

Answer 1

Python 2, 55 bytes

i=2
exec"print(i%2*' '+`2%i*1122`*i)[:i];i+=1;"*input()

Try it online!

Cycles between blocks of 22, 44, except the top row is 00. For example, on input 10, prints

00
 22
2244
 2244
224422
 224422
22442244
 22442244
2244224422
 2244224422

Prints rows of increasing length i=2,3,.. by creating a space for odd lengths, repeating the pattern i times, and trunctating to length i. The pattern is 2244 for all rows except the first one i=2 for which it's 0. This is achieved with the arithmetic expression 2%i*1122.

Answer 2

Jelly,  21 19  16 bytes

d2SR+%3x2⁶;ṙḂµ€Y

A full program printing the result.

Uses 00, 11, and 22 as the blocks.

Try it online!

How?

d2SR+%3x2⁶;ṙḂµ€Y - Main link: number n
             µ€  - for €ach "row" in n (i.e. for row, r = 1 to n inclusive):
d2               -   divmod 2   -> [integer-divide by 2, remainder] i.e. [r/2, r%2]
  S              -   sum        -> r/2 + r%2
   R             -   range      -> [1, 2, 3, ..., r/2 + r%2]
    +            -   add r      -> [r+1, r+2, r+3, ..., r + r/2 + r%2]
     %3          -   modulo 3   -> [r%3+1, r%3+2, r%3+0, ..., (r + r/2 + r%2)%3]
                 -   e.g.: r: 1  , 2  , 3    , 4    , 5      , 6      , 7       , ...
                             [2], [0], [1,2], [2,0], [0,1,2], [1,2,0], [2,0,1,2], ...
       x2        -   times 2 - repeat each twice (e.g. [2,0,1,2] -> [2,2,0,0,1,1,2,2]
         ⁶       -   literal space character
          ;      -   concatenate (add a space character to the left)
            Ḃ    -   r mod 2 (1 if the row is odd, 0 if it is even (1 at the top))
           ṙ     -   rotate (the list) left by (r mod 2)
               Y - join with newlines
                 - implicit print (no brackets printed due to the presence of characters)

Answer 3

JavaScript (ES6), 80 bytes

n=>eval(`for(s=11,i=1;i++<n;)s+='\\n'+(' '+'2233'.repeat(n)).substr(i%2,i+1);s`)

f=
n=>eval(`for(s=11,i=1;i++<n;)s+='\\n'+(' '+'2233'.repeat(n)).substr(i%2,i+1);s`)
console.log(f(+prompt()))

---

JavaScript (ES6), 87 bytes

Previous solution.

n=>[11,...Array(n).fill(' '+'2233'.repeat(n)).map((r,n)=>r.slice(n%2,n+3+n%2))].join`
`

Answer 4

05AB1E, 22 21 20 18 17 bytes

Uses the interesting fact that 4^(N+2)/5 = [3,12,51,204,...] = b[11,1100,110011,11001100,...]

F4NÌm5÷bDðì}r·IF,

Try it online!

Explanation

F                     # for N in 0...input-1 do
 4                    # push 4
  NÌ                  # push N+2
    m                 # push 4^(N+2)
     5÷               # integer division by 5
       b              # convert to binary
        D             # duplicate
         ðì           # prepend a space to the copy
           }          # end loop
            r         # reverse stack
             ·        # multiply top of stack by 2
              IF      # input times do
                ,     # print with newline

Answer 5

SOGL, 31 28 27 25 bytes

∫³2\@*O"²b“2⁵I%*r*;I»«nKp

Explanation:

∫                          iterate input times, pushing 1-indexed counter
 ³                         get 3 total copies of it on stack
  2\                       1 if divides by 2, else 0
    @*                     get that many spaces
      O                    output in a new line
       "²b“                push 1122
           2⁵I%*           multiply 1122 by 2%(iteration+1)
                r          convert to string
                 *         multiply by iteration
                  ;I»«     get one iteration variable ontop of stack
                      n    increase, floor divide by 2, multiply by 2 (gets the amount of bricks in a line)
                       Kp  split that multiplied string in pieces of that length

using this technique
Example output for 9:

00
 22
2244
 2244
224422
 224422
22442244
 22442244
2244224422

22 bytes

∫³2\@*O"²b“2⁵I%*;I»«mp

The command m has been documentated on the 1st SOGL commit, just not implemented.

Answer 6

PHP, 61 59

aa<?for(;++$i<$argn;)echo"
",str_pad(" "[~$i&1],2+$i,bbcc);

works pretty much like the python versions but uses all three colors if possible. No trailing new line.

-2 bytes by @user63956. Thanks !

Answer 7

Pyth, 29 bytes

VQI!%hN2pd)Vh/N2p*2@G%+NH3)pb

Test it online!

Explanations

VQI!%hN2pd)Vh/N2p*2@G%+NH3)pb

VQ                               For N in range(0, input)
  I!%hN2pd)                      If N is odd, print a leading space
           Vh/N2          )      For H in range(0, N / 2 + 1)
                   @G%+NH3       Select the letter at position (N + H) % 3 in the alphabet
                 *2              Then make it a two letters string ("aa" or "bb" or "cc")
                p                Print it
                           pb    End the line by printing a new line

I am sure there are a lot of ways to shorten that code, but I am king of tired right now... Will try later.

Answer 8

Batch, 125 bytes

@set s=█
@for /l %%i in (2,1,%1)do @call:c
:c
@set s= %s:█= %
@set s=%s:▓=█%
@set s=%s:░=▓%
@set s=%s:  =░░%
@echo %s%

Note: Save this in CP437 or CP850 or some such. Works by rotating the colours each time. Since I can't map over the string to perform the rotation, I use four replacements, using spaces as a temporary stage. This then also allows me to prefix a space every line, so that two spaces turn into a new block. Sample output:

░░
 ▓▓
░░██
 ▓▓░░
░░██▓▓