81
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Challenge

Create a function or program that, when given an integer size, does the following:

If size is equal to 1, output

H H
HHH
H H

If size is greater than 1, output

X X
XXX
X X

where X is the output of the program/function for size - 1

(If you prefer, you may have the base case correspond to 0, so long as you specify in your answer)

Any of the following output formats are acceptable, whichever is more convenient for you:

  • A string of the required structure with any two distinct characters corresponding to H and space

  • A two-dimensional array with the required structure, with any two distinct values corresponding to H and space

  • An array/list of strings, with one line of the output in each string, with any two distinct values corresponding to H and space

Leading spaces are allowed, as long as there is a constant amount of leading spaces on each line. The two distinct output characters can be dependent on anything you choose, as long as they are different.

Specify what output format your code is returning.

Test Cases

1

H H
HHH
H H

2

H H H H
HHH HHH
H H H H
H HH HH H
HHHHHHHHH
H HH HH H
H H H H
HHH HHH
H H H H

3

H H H H H H H H
HHH HHH HHH HHH
H H H H H H H H
H HH HH H H HH HH H
HHHHHHHHH HHHHHHHHH
H HH HH H H HH HH H
H H H H H H H H
HHH HHH HHH HHH
H H H H H H H H
H H H HH H H HH H H H
HHH HHHHHH HHHHHH HHH
H H H HH H H HH H H H
H HH HH HH HH HH HH HH HH H
HHHHHHHHHHHHHHHHHHHHHHHHHHH
H HH HH HH HH HH HH HH HH H
H H H HH H H HH H H H
HHH HHHHHH HHHHHH HHH
H H H HH H H HH H H H
H H H H H H H H
HHH HHH HHH HHH
H H H H H H H H
H HH HH H H HH HH H
HHHHHHHHH HHHHHHHHH
H HH HH H H HH HH H
H H H H H H H H
HHH HHH HHH HHH
H H H H H H H H

This is , so the lowest byte count for each language wins!

asked Mar 9, 2018 at 1:37
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17
  • 5
    \$\begingroup\$ Perfect for Charcoal probably... lol. Also welcome to PPCG! :D \$\endgroup\$ Commented Mar 9, 2018 at 1:54
  • 11
    \$\begingroup\$ Welcome to PPCG. Nice first challenge! \$\endgroup\$ Commented Mar 9, 2018 at 2:06
  • \$\begingroup\$ May we use 0 based sizes? \$\endgroup\$ Commented Mar 9, 2018 at 2:07
  • 3
    \$\begingroup\$ related \$\endgroup\$ Commented Mar 9, 2018 at 2:26
  • 2
    \$\begingroup\$ I'd call this a "Sierpinski H" \$\endgroup\$ Commented Mar 12, 2018 at 21:37

30 Answers 30

29
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answered Mar 9, 2018 at 2:51
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5
  • 18
    \$\begingroup\$ what the heck of course Mathematica has a built-in for recursive nested arrays lol. +1 \$\endgroup\$ Commented Mar 9, 2018 at 3:01
  • 1
    \$\begingroup\$ @HyperNeutrino well obviously \$\endgroup\$ Commented Mar 9, 2018 at 9:29
  • 7
    \$\begingroup\$ @HyperNeutrino How is this considered a built-in? Just Nest (repeatedly) the function multiple times. Like any other submissions (Jelly?) The ArrayFlatten is... well, built-in, but it behaves somewhat just like a Flatten[#,{{1,3},{2,4}}] in this case. (didn't test) \$\endgroup\$ Commented Mar 9, 2018 at 10:34
  • 6
    \$\begingroup\$ There is a built-in for this, but longer. Mathematica has long function names. \$\endgroup\$ Commented Mar 9, 2018 at 10:49
  • 1
    \$\begingroup\$ How could it not, given its triumph at the upgoat challenge? \$\endgroup\$ Commented Mar 14, 2018 at 15:15
21
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Canvas, (削除) 14 (削除ここまで) 12 bytes 

×ばつ∔;3*+

Try it here!

Explanation: 
Code |Instruction |Stack
--------+--------------------------------------------------------------------+-------------------------
 |Push input to stack (implicit) |I
H |Push "H" to stack |I,"H"
; |Swap the top two stack items |"H",I
[ |The following ToS (input) times: |X
 ⌐⌐ |Duplicate ToS (result from last loop ("H" if first loop)) four times|X,X,X,X,X
 ∔ |Join vertically |X,X,X,X\nX
 ×ばつ |Prepend |X,X,XX\nX
 ∔ |Join vertically |X,X\nXX\nX
 ; |Swap top two stack items |X\nXX\nX,X
 3*|Repeat three times vertically |X\nXX\nX,X\nX\nX
 + |Join horizontally |X<space>X\nXXX\nX<space>X
 |End loop (implicit) |X
 |Print ToS (implicit) |

Where I is the input, X is the pattern generated by the previous loop ("H" for the first loop), and <space> is the empty space on the first and third row of the pattern, added implicitly by +.

-2 bytes thanks to dzaima!

answered Mar 9, 2018 at 2:39
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1
  • \$\begingroup\$ Amazingly short answer :O \$\endgroup\$ Commented Mar 9, 2018 at 3:40
19
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MATL, (削除) 12 (削除ここまで) 11 bytes

t:"[ACA]BX*

Given input n, this outputs a matrix containing 0 and n.

Try it online!

To convert this into a character matrix of Hand space add g72*c in the header. Try it online too!

Or add ]1YC to see the matrix displayed graphically. Try it at MATL Online!

Explanation

t % Input (implicit): n. Duplicate
: % Range. Gives the array [ 1 2 ... n]
" % For each (that is, do n times)
 [ACA] % Push the array [5 7 5]
 B % Convert to binary. Gives the ×ばつ3 matrix [1 0 1; 1 1 1; 1 0 1]
 X* % Kronecker product
 % End (implicit). Display (implicit)
Suever
11.2k1 gold badge24 silver badges52 bronze badges
answered Mar 9, 2018 at 9:21
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16
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Stax, (削除) 16 (削除ここまで) 15 bytes 

╛c_mê║6{│◙ÖmπV"

Run and debug it

This is the ascii representation of the program with comments. This program builds up the H sideways, and then transposes once at the end.

'H] ["H"]
 { },* repeat block specified number of times
 c copy the matrix
 {3*m triplicate each row
 |S surround; prepend and append like b + a + b
 |C horizontally center rows with spaces
 M transpose back to original orientation
 m output each row

Bonus 14 byte program - uses its input as the output character. Theoretically, this would not produce the right shape at 10, since it has 2 digits, but attempting to run that crashes my browser.

answered Mar 9, 2018 at 3:05
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11
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Ruby, 72 bytes

Output is a list of strings, one string per line.

f=->n{n<1?[?H]:[*a=(x=f[n-1]).map{|i|i+' '*i.size+i},*x.map{|i|i*3},*a]}

Try it online!

answered Mar 9, 2018 at 2:19
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1
  • \$\begingroup\$ Well done! The output looks wrong on tio at first, but it's fine when zoomed out. \$\endgroup\$ Commented Mar 9, 2018 at 19:07
11
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APL (Dyalog Classic), 14 bytes

×ばつ/ ̈∘.≥⍨2|,⍳⎕⍴3

Try it online!

⎕ evaluated input n

,⍳⎕⍴3 all n-tuples with elements from 0 1 2

2| mod 2

×ばつ/ ̈∘.≥⍨ form a matrix by comparing every pair of tuples a and b - if all elements of a are ≥ the corresponding elements of b, it's a 1, otherwise 0

answered Mar 9, 2018 at 13:33
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0
10
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Haskell, 50 bytes

f 0=[[1]]
f n=[x++map(*c)x++x|c<-[0,1,0],x<-f$n-1]

Try it online!

Makes a grid of 0's and 1's. One character longer for spaces and H's.

Haskell, 51 bytes

f 0=["H"]
f n=[x++map(min c)x++x|c<-" H ",x<-f$n-1]

Try it online!

answered Mar 10, 2018 at 6:29
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9
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Jelly, (削除) 17 (削除ここまで) (削除) 16 (削除ここまで) 15 bytes

381B«€s3Z€ẎF€μ¡

This is a full program that prints a 2D array of 1's and 0's.

Try it online! or see the output with H's and spaces.

answered Mar 9, 2018 at 2:39
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8
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SOGL V0.12, 13 bytes 

┌.{32∙4┼+;3∙┼

Try it Here!

answered Mar 9, 2018 at 10:55
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8
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R, 64 bytes

function(n)Reduce(`%x%`,rep(list(matrix(c(1,1,1,0,1,0),3,3)),n))

Try it online!

Reduces by Kronecker product, as a shameless port of Luis Mendo's answer.

The footer prints the result nicely, but this is an anonymous function which returns a matrix of 1 for H and 0 for space.

answered Mar 9, 2018 at 13:51
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8
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Java (JDK), 126 bytes

n->{int s=1,H[][]=new int[n+=Math.pow(3,n)-n][n],x;for(;s<n;s*=3)for(x=n*n;x-->0;)H[x/n][x%n]|=~(x/n/s%3)&x%n/s%3&1;return H;}

Try it online!

Returns an int[][] with 0 for H and 1 for space. This actually "carves" a wall of H's instead of "piling" H's.

Explanations

n->{ // An int to int[][] lambda function
 int s=1, // size of the carvings.
 H[][]=new int[n+=Math.pow(3,n)-n][n], 
 // change n to 3^n, through +=...-n to avoid an explicit cast
 // and create the 2D array to return, filled with 0s
 x; // counter for the 2D array
 for(;s<n;s*=3) // for each size
 for(x=n*n;x-->0;) // for each cell
 H[x/n][x%n] |= // assign 1 to a cell of the array if...
 ~(x/n/s%3) // it is located in the "holes" of the H
 &x%n/s%3 //
 &1; // 
 return H; // return the array
} // end the lambda

Credits

answered Mar 9, 2018 at 9:53
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2
  • \$\begingroup\$ save 5 bytes by adding a static import for Math.pow \$\endgroup\$ Commented Mar 12, 2018 at 9:16
  • 4
    \$\begingroup\$ @Selim the static import is required in the byte count. So I would lose... 19 bytes. \$\endgroup\$ Commented Mar 12, 2018 at 9:20
7
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V, 22 bytes

éHÀñäLgvr PGï3PkyHGpH

Try it online!

Hexdump:

00000000: e948 c0f1 e416 4c67 7672 2050 47ef 3350 .H....Lgvr PG.3P
00000010: 6b79 4847 7048 kyHGpH

This is basically the exact same approach as the Sierpinski carpet and The Fractal Plus on Anarchy Golf.

answered Mar 9, 2018 at 3:00
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1
  • \$\begingroup\$ Is that French? \$\endgroup\$ Commented May 26, 2018 at 21:43
7
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Python 2, 70 bytes

f=lambda r:-r*'H'or[x+[x,' '*3**r][b]+x for b in 1,0,1for x in f(r-1)]

Try it online!

Function outputs a list of strings.

Python 2, 84 bytes

r=input()
for i in range(3**r):x,s=' H';exec"s+=[x,s][i%3%2]+s;x*=3;i/=3;"*r;print s

Try it online!

Uses the same template as other 3*3 fractal patterns:

answered Mar 9, 2018 at 3:38
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6
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J, (削除) 25 (削除ここまで) 22 bytes

,./^:2@(*/#:@5 7 5)^:]

Try it online! 

 */ multiply by
 #:@5 7 5 the binary matrix shaped like H
,./^:2 assemble the 4-dimensional result into a matrix
 ^:] do it input times
answered Mar 9, 2018 at 10:23
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6
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Haskell, (削除) 73 (削除ここまで) (削除) 67 (削除ここまで) (削除) 64 (削除ここまで) 55 bytes

g#f=g<>f<>g
w=map.(id#)
(iterate(w(>>" ")#w id)["H"]!!)

This works only with the latest version of Prelude, because it exports <> from Data.Semigroup. To run it on TIO, add an import as done here: Try it online! 

g#f= -- function # takes two functions g and f and a list s
 -- and returns
 g <> f <> g -- g(s), followed by f(s) and another g(s)
w= -- w takes a function and a list of lists
 -- (both as unnamed parameters, because of pointfree style,
 -- so let's call them f and l)
 map.(id#) -- return map(id#f)l, i.e. apply (id#f) to every element of l
 w(>>" ")#w id -- this partial application of # is a function that
 -- takes the missing list (here a list of lists)
 -- remember: (>>" ") is the function that replaces every element
 -- of a list with a single space
iterate( )["H"] -- starting with a singleton list of the string "H"
 -- which itself is a singleton list of the char 'H'
 -- repeatedly apply the above function
 !! -- and pick the nth iteration
Example for ["H H", "HHH", "H H"], i.e.
 H H
 HHH
 H H
call the iterated function:
 ( w(>>" ") # w id ) ["H H","HHH","H H"]
expand w: ( map(id#(>>" ")) # map(id#id) ) ["H H","HHH","H H"]
expand outermost #: map(id#(>>" "))["H H","HHH","H H"] ++
 map(id#id) ["H H","HHH","H H"] ++
 map(id#(>>" "))["H H","HHH","H H"]
expand map: [(id#(>>" "))"H H", (id#(>>" "))"HHH", (id#(>>" "))"H H"] ++
 [(id#id) "H H", (id#id) "HHH", (id#id) "H H"] ++
 [(id#(>>" "))"H H", (id#(>>" "))"HHH", (id#(>>" "))"H H"]
expand other #: ["H H"++" "++"H H", "HHH"++" "++"HHH", "H H"++" "++"H H"] ++
 ["H H"++"H H"++"H H", "HHH"++"HHH"++"HHH", "H H"++"H H"++"H H"] ++
 ["H H"++" "++"H H", "HHH"++" "++"HHH", "H H"++" "++"H H"]
collaps ++: ["H H H H", "HHH HHH", "H H H H",
 "H HH HH H", "HHHHHHHHH", "H HH HH H",
 "H H H H", "HHH HHH", "H H H H"]
which is printed line by line: 
 H H H H
 HHH HHH
 H H H H
 H HH HH H
 HHHHHHHHH
 H HH HH H
 H H H H
 HHH HHH
 H H H H

Edit: -9 bytes thanks to @Potato44.

answered Mar 9, 2018 at 15:51
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2
  • 3
    \$\begingroup\$ You should be able to golf (#) down to g#f=g<>f<>g if you use GHC 8.4. This is because Semigroup is now in the prelude. \$\endgroup\$ Commented Mar 11, 2018 at 20:52
  • \$\begingroup\$ @Potato44: I'm pretty sure this will help in a lot of challenges. Thanks! \$\endgroup\$ Commented Mar 12, 2018 at 17:47
6
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Vim - (削除) 66 (削除ここまで) (削除) 56 (削除ここまで) 54 bytes

A @ c H esc " r d ^ q c { ctrl-v } " a y g v r space g v d " a P P " a P V G " b y P g v ctrl-v $ d " a P . . G " b p q @ r

The input is taken as a number in the buffer.

answered Mar 11, 2018 at 23:33
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6
  • \$\begingroup\$ What do I have to type, starting from a bash prompt, assuming I have vim installed, to see the result? \$\endgroup\$ Commented Mar 13, 2018 at 13:03
  • \$\begingroup\$ Type vim, press enter, type the input number (e.g. 3) in the buffer then, from normal mode, press the sequence of keys from the post. \$\endgroup\$ Commented Mar 13, 2018 at 13:09
  • \$\begingroup\$ Make sure to use vanilla vim \$\endgroup\$ Commented Mar 13, 2018 at 13:14
  • \$\begingroup\$ There was a typo in the code. Just fixed it. \$\endgroup\$ Commented Mar 13, 2018 at 13:17
  • 1
    \$\begingroup\$ Works! <kbd>I</kbd> is a capital i, not ell. :set nowrap to see the result, for 4 and more. \$\endgroup\$ Commented Mar 14, 2018 at 12:50
5
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Perl 5, (削除) 46 (削除ここまで) (削除) 44 (削除ここまで) (削除) 43 (削除ここまで) (削除) 41 (削除ここまで) 40 bytes

1 based counting. Uses 0 and 1 for H and space, has a leading 1 (space)

say//,map/$'/^1,@;for@;=glob"{A,.,A}"x<>

Based on a classic idea by mtve.

Try it online!

answered Mar 9, 2018 at 13:27
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4
  • 1
    \$\begingroup\$ Output for n ≥ 3 isn't quite right. \$\endgroup\$ Commented Mar 9, 2018 at 14:12
  • \$\begingroup\$ @primo The program was correct but TIO uses the UTF-8 version of the special characters. I fixed the link to use escapes instead, but the program still works if you use the actual literal characters \$\endgroup\$ Commented Mar 9, 2018 at 14:40
  • \$\begingroup\$ I have no idea why 321円 is necessary, any character seems to work. // and $' can also replace //g and $`, but I'm not sure it leads to an improvement. \$\endgroup\$ Commented Mar 9, 2018 at 15:20
  • 1
    \$\begingroup\$ @primo Thanks! I was still working from code derived from the old mtve solution where 321円 was the bit complement of . (used to generate another fractal pattern). But I dropped the bit-complement so of course I don't need that anymore. I used //g and $` so I can easily test the code from the commandline (// and $' don't lead to a gain I can see, the gained byte is wasted with a space or ! again) \$\endgroup\$ Commented Mar 9, 2018 at 15:30
4
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APL (Dyalog Unicode), (削除) 38 (削除ここまで) 34 bytesSBCS 

({(⍵,(×ばつ⍵),⍵){⍺⍪⍵⍪⍺}⍵,⍵,⍵}⍣⎕)1 1⍴1

Output is a 2-dimensional array with 1 representing H and 0 representing space.

Try it online!

answered Mar 9, 2018 at 2:25
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4
  • 2
    \$\begingroup\$ Welcome to PPCG! You can omit f← and count chars as 1 byte each: codegolf.meta.stackexchange.com/questions/9428/… It's also considered legal to take input from ⎕, i.e. replace ⍣⍵ with ⍣⎕ and drop the outer dfn's braces. \$\endgroup\$ Commented Mar 9, 2018 at 2:43
  • \$\begingroup\$ Thanks! I've never actually formally golfed APL before so these should help. \$\endgroup\$ Commented Mar 9, 2018 at 3:19
  • \$\begingroup\$ 1 1⍴1 can be written as ⍪1 and then the parens around the operator become unnecessary. If you're familiar with trains - they can help a lot here. \$\endgroup\$ Commented Mar 9, 2018 at 14:08
  • \$\begingroup\$ Also, ⍨ is your friend: (⍵,(0×⍵),⍵) => (⍵,⍵,⍨0×⍵) \$\endgroup\$ Commented Mar 9, 2018 at 17:23
4
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Charcoal, (削除) 30 (削除ここまで) 29 bytes

HFENX3ι«J0¦0C0ιCιιT⊗ι⊗ι‖OO→↓ι

Try it online! Link is to verbose version of code. Explanation:

H

Print the original H.

FENX3ι«

Loop over the first size powers of 3.

J0¦0

Move the cursor back to the origin. Trim needs this, as both the original printing of the H and the reflection below move the cursor.

C0ι

Copy the previous iteration downwards, creating a domino.

Cιι

Copy the result down and right, creating a tetromino.

T⊗ι⊗ι

Trim the canvas down to an L shape triomino.

‖OO→↓ι

Reflect the canvas horizontally and vertically with overlap, completing the iteration.

Charcoal is better at some fractals than others. Here's a similar idea, but in almost half the size:

HFN«⟲C26‖OOLX3ι

Try it online! Link is to verbose version of code.

answered Mar 9, 2018 at 10:46
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4
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Python 2, 143 bytes

def g(a,x,y,s):
	if s:s/=3;[g(a,x+k/3*s,y+k%3*s,s)for k in 0,2,3,4,5,6,8]
	else:a[x][y]=1
def f(s):s=3**s;a=eval("s*[0],"*s);g(a,0,0,s);print a

Try it online!

-30 bytes thanks to recursive

wrapper code is for nice formatting. it works fine if you remove it

answered Mar 9, 2018 at 2:17
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1
  • \$\begingroup\$ A few golfs \$\endgroup\$ Commented Mar 9, 2018 at 7:15
4
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PHP 7, (削除) 125 (削除ここまで) 109 bytes

a different approach: Instead of nesting and flattening the result recursively, this just loops through the rows and columns and uses a 3rd loop to find out if to print H or _.

Edit: Saved a lot by combining the row/column loops to one, though it took a bit to get the decrease for the inner loop correct. Requires PHP 7 for the power operator.

Try them online!

for($z=3**$argn;$z*$z>$q=$p;print$c."
"[++$p%$z])for($c=H;$q;$q-=$q/$z%3*$z,$q/=3)if($q%3==1&&$q/$z%3-1)$c=_;

prints the result. Run as pipe with -nR.

qualified function, (削除) 147 (削除ここまで) 130 bytes

function r($n){for($z=3**$n;$z*$z>$q=$p;$r.=$c."
"[++$p%$z])for($c=H;$q;$q-=$q/$z%3*$z,$q/=3)if($q%3==1&&$q/$z%3-1)$c=_;return$r;}

returns a single string. Run with default config (no php.ini).

answered Mar 9, 2018 at 21:31
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1
  • 1
    \$\begingroup\$ %3==1 can be replaced with %3&1. \$\endgroup\$ Commented Mar 13, 2018 at 15:34
3
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Jelly, 25 bytes

×ばつ"3S_4A1e
3*çþ`ị) HY

Try it online!

Although this is longer than the existing Jelly submission, it tries to generate each character independently just from the coordinate.

In particular, if the coordinate is (x,y) (1-indexing), the first link returns 0 and 1 corresponds to H and respectively.


, Pair. Get (x,y)
 ’ Decrement. Get (x,y) (0-indexing)
 b3 Convert to base 3 digits.
 U Upend. So next operations can pair the corresponding digits.
 ×ばつ"3 Multiply the first element (list) by 3.
 S Sum (corresponding digit together). Let the sum be s.
 _4A1e Check if any of abs(s-4) is 1. Equivalently, check
 if there is any 3 or 5 in the list of s.

Also, the 5 bytes ị) HY are used for formatting, so this program (20 bytes) is also valid (but the output doesn't look as nice):

×ばつ"3S_4A1e
3*çþ`
answered Mar 9, 2018 at 10:39
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3
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T-SQL, (削除) 267 (削除ここまで) 261 bytes

DECLARE @N INT=3DECLARE @ TABLE(I INT,H VARCHAR(MAX))INSERT @ VALUES(1,'H H'),(2,'HHH'),(3,'H H');WITH
T AS(SELECT 1 A,3 P,I J,H S FROM @ UNION ALL SELECT A+1,P*3,J*P+I,REPLACE(REPLACE(S,' ',' '),'H',H)FROM @,T
WHERE A<@N)SELECT S FROM T WHERE A=@N ORDER BY J
answered Mar 11, 2018 at 12:44
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5
  • \$\begingroup\$ This is my first answer on Code Golf, so please help me if I made any mistakes. Also, my preferred language is Transact-SQL, which is not very suitable for short code. \$\endgroup\$ Commented Mar 11, 2018 at 12:46
  • 1
    \$\begingroup\$ Welcome to PPCG and nice first post! For tips about golfing in T-SQL, make sure to check out this post! \$\endgroup\$ Commented Mar 11, 2018 at 12:50
  • \$\begingroup\$ I tried adding a sqlfiddle, but it doesn't work well with table variables. If I use normal tables, it's even 1 byte shorter: sqlfiddle.com/#!18/eb14e/2. However, the output is not formatted correctly by sqlfiddle, but it works fine in SSMS. \$\endgroup\$ Commented Mar 11, 2018 at 12:50
  • 1
    \$\begingroup\$ You should be able to get this down to 259 by removing some unnecessary whitespace and linefeeds \$\endgroup\$ Commented Mar 11, 2018 at 18:37
  • \$\begingroup\$ I only got to 261. What am I missing? \$\endgroup\$ Commented Mar 11, 2018 at 19:23
2
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PHP 7, 153 bytes

 function p($n){$r=["H H",HHH,"H H"];if(--$n)foreach(p($n)as$s){$r[+$i]=$r[$i+6*$p=3**$n]=str_pad($s,2*$p).$s;$r[3*$p+$i++]=$s.$s.$s;}ksort($r);return$r;}

Run with default config (no php.ini) or try it online.

answered Mar 9, 2018 at 10:28
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2
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Perl, 64 bytes

//;$_ x=3,$.=s|.+|$&@{[$$_++/$.&1?$&:$"x$.]}$&|g for($_=H.$/)x$'

Requires -p, input is taken from stdin. Output is an H of Hs.

Try it online!

answered Mar 9, 2018 at 11:07
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1
  • \$\begingroup\$ Counting on this site has changed, you don't need to count -p anymore (I think it is too lenient for perl, but that's how it is now) \$\endgroup\$ Commented Mar 9, 2018 at 13:29
2
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PHP (5.6+), 94 bytes

<?for(;$H>$e*=3or$e=($i+=$e&&print"$s
")<${$s=H}=3**$argn;)$s.=str_pad($i/$e%3&1?$s:'',$e).$s;

Used with -F command line option. Assumes interpreter defaults (-n). Will not work on versions previous to 5.6, due to the power operator.

Sample usage

$ echo 3|php -nF h-carpet.php
H H H H H H H H
HHH HHH HHH HHH
H H H H H H H H
H HH HH H H HH HH H
HHHHHHHHH HHHHHHHHH
H HH HH H H HH HH H
H H H H H H H H
HHH HHH HHH HHH
H H H H H H H H
H H H HH H H HH H H H
HHH HHHHHH HHHHHH HHH
H H H HH H H HH H H H
H HH HH HH HH HH HH HH HH H
HHHHHHHHHHHHHHHHHHHHHHHHHHH
H HH HH HH HH HH HH HH HH H
H H H HH H H HH H H H
HHH HHHHHH HHHHHH HHH
H H H HH H H HH H H H
H H H H H H H H
HHH HHH HHH HHH
H H H H H H H H
H HH HH H H HH HH H
HHHHHHHHH HHHHHHHHH
H HH HH H H HH HH H
H H H H H H H H
HHH HHH HHH HHH
H H H H H H H H

Try it online!

answered Mar 10, 2018 at 7:18
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7
  • 1
    \$\begingroup\$ You can save one byte: $s.$s.$s instead of $s.=$s.$s. And you don´t need <? with -R instead of -F. \$\endgroup\$ Commented Mar 10, 2018 at 12:23
  • \$\begingroup\$ Thanks for the byte. Regarding -R, can you show me the complete usage? \$\endgroup\$ Commented Mar 10, 2018 at 14:26
  • \$\begingroup\$ Just like -nF: echo <input> | php -nR '<code>'. -r is almost the same: php -nr '<code>' <arguments>. \$\endgroup\$ Commented Mar 10, 2018 at 14:43
  • \$\begingroup\$ Maybe I'm just too stupid to get it working :/ i.sstatic.net/jqpmk.png \$\endgroup\$ Commented Mar 11, 2018 at 4:24
  • 1
    \$\begingroup\$ preg_filter is to iterate each line while preserving newlines, roughly equivalent to join("\n",array_map(function(){...},split("\n",$s.$s.$s))), but significantly less verbose. I initially had str_pad but changed to sprintf because it's one byte shorter: '"0円".str_pad($$i++/$i&1?"0円":"",$i)."0円"' \$\endgroup\$ Commented Mar 13, 2018 at 5:11
1
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CJam - (削除) 103 (削除ここまで) (削除) 97 (削除ここまで) (削除) 87 (削除ここまで) 76 bytes

{H{ae_,S*}%}:Il~a:A];{A_W={)(a9*+:A;[[HIH][HHH][HIH]]{z~}%}{);:A;"H"}?}:H~N*

This program does a quite verbose "handcoded" recursion. No smart matrix multiplications. Throughout the recursion, on top of the stack there is an array gathering the output gained from the parent calls. Right after each set of recursive calls the output of the recursive calls needs to be zipped together, to make sure the output is correct when the stack is printed linearly at the end of the program. The stack of arguments being passed down the recursion is kept in the variable A.

Try online

answered Mar 11, 2018 at 20:41
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1
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K (ngn/k), 18 bytes

{~|/a<\:'a:1=!x#3}

Try it online!

answered Mar 29, 2018 at 6:52
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1
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Japt, 23 bytes

_·£[X3XX3]Ãy c ·û}gQq)y

Try it online!

Unpacked & How it works

Z{ZqR mXYZ{[Xp3 XXp3]} y c qR û}gQq)y
Z{ Declare a function that accepts a string...
 ZqR Split by newline...
 mXYZ{ and map each row into...
 [Xp3 XXp3] an array of [X.repeat(3), X, X.repeat(3)]
 }
 y Transpose the resulting 2D array
 c Flatten
 qR Join with newline
 û Center-pad each row to the longest
}
gQq) Apply the above function to '"' recursively
y Transpose the resulting 2D string

Using the transposed pattern

III
 I 
III

is far easier to handle than the original H pattern, at least in Japt where the I can be done with string repeat and center-padding.

answered May 23, 2018 at 6:29
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0
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C++11 - 138 bytes

Not sure if this answer has a valid syntax here however.

#define A a?1:0
template<int N>struct H{H<N-1>h[9];H(int a):h{A,0,A,A,A,A,A,0,A}{}};template<>struct H<0>{char h;H(int a):h{a?'H':' '}{}};

Ungolfed with working code

#include <iostream>
#define A a?1:0
template<int N>
struct H
{
 H<N-1> h[9];
 H(int a) : h{A,0,A,A,A,A,A,0,A}
 {}
};
template<>
struct H<0>
{
 char h;
 H(int a) : h{a?'H':' '}
 {}
};
int pow(int a, int b)
{
 int res=1;
 for (int i=1; i<=b; ++i)
 res *= a;
 return res;
}
template<int N>
char getHvalue(int i, int j, H<N> &hn)
{
 int n3=pow(3, N-1);
//std::cout << N << " " << i << " " << j << std::endl;
 return getHvalue(i%n3, j%n3, hn.h[i/n3*3+j/n3]);
}
template<>
char getHvalue<0>(int, int, H<0> &hn)
{
 return hn.h;
}
int main()
{
 H<0> h0(1);
 std::cout << getHvalue(0, 0, h0) << std::endl;
 std::cout << "\n====================\n" << std::endl;
 H<1> h1(1);
 for (int i=0; i<3; ++i) {
 for (int j=0; j<3; ++j)
 std::cout << getHvalue(i, j, h1);
 std::cout << std::endl;
 }
 std::cout << "\n====================\n" << std::endl;
 H<2> h2(1);
 for (int i=0; i<9; ++i) {
 for (int j=0; j<9; ++j)
 std::cout << getHvalue(i, j, h2);
 std::cout << std::endl;
 }
 std::cout << "\n====================\n" << std::endl;
 H<3> h3(1);
 for (int i=0; i<27; ++i) {
 for (int j=0; j<27; ++j)
 std::cout << getHvalue(i, j, h3);
 std::cout << std::endl;
 }
 return 0;
}
\$\endgroup\$

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