Code Golf: Letter-Ception

Question

The Challenge:

For an input of one letter X (upper or lower case from A to Z) and one digit N (0-9) print the corresponding letter X made of N * X.

The letter has to be from this list:

 AAA      BBBB       CCCC     DDDD      EEEEE     FFFFF      GGG      H   H
A   A     B   B     C         D   D     E         F         G         H   H
AAAAA     BBBB      C         D   D     EEEE      FFFF      G  GG     HHHHH
A   A     B   B     C         D   D     E         F         G   G     H   H
A   A     BBBB       CCCC     DDDD      EEEEE     F          GGG      H   H


IIIII         J     K   K     L         M   M     N   N      OOO 
  I           J     K  K      L         MM MM     NN  N     O   O
  I           J     KKK       L         M M M     N N N     O   O
  I       J   J     K  K      L         M   M     N  NN     O   O
IIIII      JJJ      K   K     LLLLL     M   M     N   N      OOO 


PPPP       QQQ      RRRR       SSSS     TTTTT     U   U     V   V     W   W
P   P     Q   Q     R   R     S           T       U   U     V   V     W   W
PPPP      Q   Q     RRRR       SSS        T       U   U     V   V     W   W
P         Q  QQ     R  R          S       T       U   U      V V      W W W
P          QQQQ     R   R     SSSS        T        UUU        V        W W 


X   X     Y   Y     ZZZZZ
 X X       Y Y         Z 
  X         Y         Z  
 X X        Y        Z   
X   X       Y       ZZZZZ

Examples:

input: a 1

output:

 AAA
A   A
AAAAA
A   A
A   A

---

input: A 0

output: A

---

input: A 2

output:

      AAA  AAA  AAA
     A   AA   AA   A
     AAAAAAAAAAAAAAA
     A   AA   AA   A
     A   AA   AA   A
 AAA                 AAA
A   A               A   A
AAAAA               AAAAA
A   A               A   A
A   A               A   A
 AAA  AAA  AAA  AAA  AAA
A   AA   AA   AA   AA   A
AAAAAAAAAAAAAAAAAAAAAAAAA
A   AA   AA   AA   AA   A
A   AA   AA   AA   AA   A
 AAA                 AAA
A   A               A   A
AAAAA               AAAAA
A   A               A   A
A   A               A   A
 AAA                 AAA
A   A               A   A
AAAAA               AAAAA
A   A               A   A
A   A               A   A

---

input: A -1

output: what ever: it doesn't matter

---

Additional Rules:

• The input parameters can be separated by what ever character you want.
• Each letter must use the capital of itself as the ascii-character to draw it.
• Trailing spaces, new lines etc. are allowed
• Instead of a program, you may write a function that takes the digit string as an argument. The output should be printed normally.
• Stdout / Stderr doesn't matter, just pick one. If stuff gots printed on the other doesn't matter either.
• Possible output formats can be printed to STDOUT, returned as a list of strings, returned as a character matrix, etc. as long as the result can simply be printed using the languages default print method.*

*: like the function f(a,1) returns the string and one can simply say print(f(a,1)) dont make the print() call part of the answer. (This was pointed out by Kevin Cruijssen and Arnauld).

Winning:

This is code-golf, lowest byte-count wins. Have fun!

---

Edit: this question seems very identical to this however I would say it is not, as it should not only work for H but for each letter from the alphabet.. Guess you decide rather or not it is a duplicate.

Answer 1

JavaScript (ES8), 281 bytes

Takes input as (letter)(N). Returns a string.

c=>n=>(g=y=>y--?''.padEnd(w).replace(/./g,(_,x)=>(h=d=>~~(d/=5)?(P=parseInt)('hhvhefhfhfu111ufhhhfv1f1v11f1vehp1ehhvhhv444vehgggh979hv1111hhlrhhpljhehhhe11fhfuphheh9fhffge1u4444vehhhh4ahhhalhhhha4ah444ahv248v'[y/d%5+5*P(c,36)-50|0],36)>>x/d%5&1?h(d):' ':c)(w))+`
`+g(y):'')(w=5**n)

Try it online!

How?

Font encoding

The letters are \$5\times5\$, which means that each row can be encoded as a 5-digit binary value, i.e. \$0\$ to \$31\$ in decimal. These values can conveniently be stored as a single digit in Base36.

The pattern that is stored is mirrored both horizontally and vertically.

Example for 'F':

#####     ....#     00001      1     '1'
#....     ....#     00001      1     '1'
####. --> .#### --> 01111 --> 15 --> 'f' --> '11f1v'
#....     ....#     00001      1     '1'
#....     #####     11111     31     'v'

The whole font is stored as a single string of \$26\times5 = 130\$ characters.

To test the 'pixel' at \$(x,y)\$ for the \$n^{\text{th}}\$ letter, we do:

parseInt('hhvhefhfh...'[y + 5 * n], 36) >> x & 1

Main algorithm

Given \$n\$, we define \$w=5^n\$, which is the width of the final output.

For \$0\le x<w\$ and \$0\le y<w\$, we want to know if the output pixel at \$(x,y)\$ is set or not. We use the recursive function \$h\$, which tests the letter pixel located at:

$$\left(\left\lfloor\frac{x}{5^k}\right\rfloor\bmod 5,\left\lfloor\frac{y}{5^k}\right\rfloor\bmod 5\right)$$

for all \$k\$ in \$[0 \dots n-1]\$.

The function returns a space as soon as a blank pixel is detected at some depth, or the character corresponding to the input letter if all iterations are successful.

Answer 2

R, 348 bytes

function(K,N){if(N)for(i in 1:N)T=T%x%matrix(c(15269425,32045630,16269839,32032318,33061407,33061392,15224366,18415153,32641183,1082926,18444881,17318431,18732593,18667121,15255086,32045584,15255151,32045649,16267326,32641156,18400814,18400580,18400938,18157905,18157700,32575775)[utf8ToInt(K)-64]%/%2^(24:0)%%2,5,5)
write(c(" ",K)[T+1],1,5^N,,"")}

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Uses an encoding nearly identical to Ouros'; however, it does not reverse the bits, instead opting to use them directly.

It then creates a 5x5 matrix of bits and builds the Kronecker Power matrix to generate the necessary pattern, writing the results to stdout.

Answer 3

Clean, 436 372 bytes

Significantly shorter with new IO format.

import StdEnv,StdLib
t=transpose
f=flatten
$0c=[[c]]
$n c=f[t(f[t($(n-1)if(isOdd({#18415150,16301615,31491134,16303663,32554047,1096767,15262766,18415153,32641183,15254032,18128177,32539681,18405233,18667121,15255086,1097263,32294446,18136623,16267326,4329631,15255089,4539953,11191857,18157905,4329809,32575775}.[toInt(max'A'c)-65]>>p))c' ')\\p<-[i..i+4]])\\i<-[0,5..20]]

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Compresses the letter patterns into the bits of integer literals to save ~700 bytes. For example, A:

1. Flatten [[' AAA '],['A A'],['AAAAA'],['A A'],['A A']]
2. Reverse [' AAA A AAAAAAA AA A']
3. Turn ['A AA AAAAAAA A AAA '] into binary ('A' = 1, ' ' = 0)
4. Turn 0b1000110001111111000101110 into decimal
5. Get 18415150

Answer 4

R, 259 bytes

function(K,N,`!`=utf8ToInt){if(N)for(i in 1:N)T=T%x%(sapply(!"








		





	





",intToBits)[1:5,5*(-64+!K)-4:0]>0)
write(c(" ",K)[T+1],1,5^N,,"")}

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Disclaimer :
this solution has been obtained by taking @Giuseppe's answer and replacing the matrix compression with another approach very very similar to the one used in @Arnauld's answer, so first of all go upvote them :)

The idea is the following :

Given this 5 x 26*5 matrix of 0/1 :

(1 replaced by '#', 0 replaced by '.' and '|' added for readability)

.####|#####|.###.|#####|#####|#####|.###.|#####|#...#|...#.|#####|#####|#####|
#.#..|#.#.#|#...#|#...#|#.#.#|#.#..|#...#|..#..|#...#|....#|..#..|....#|.#...|
#.#..|#.#.#|#...#|#...#|#.#.#|#.#..|#...#|..#..|#####|....#|..#..|....#|..#..|
#.#..|#.#.#|#...#|#...#|#.#.#|#.#..|#.#.#|..#..|#...#|....#|.#.#.|....#|.#...|
.####|.#.#.|#...#|.###.|#...#|#....|..##.|#####|#...#|####.|#...#|....#|#####| ...
  ^     ^     ^     ^     ^     ^     ^     ^     ^     ^     ^     ^     ^   
  |     |     |     |     |     |     |     |     |     |     |     |     |   
  A     B     C     D     E     F     G     H     I     J     K     L     M   

each column is considered as binary number and converted to an integer. These integer are then converted to unprintable ASCII in the range 1...31 :

e.g. for the columns of "B" the final string will be "\017\021\017\021\017" (unprintable chars written in octal representation):

#####                   ####.     11110          15            '\017'
#.#.#                   #...#     10001          17            '\021'
#.#.#      ------->     ####. --> 11110  ------> 15   ------>  '\017'
#.#.#                   #...#     10001          17            '\021'
.#.#.                   ####.     11110          15            '\017'

          (transposed                  bin to int   int to ASCII
         for reability)    

Hence, given the final string of 5*26 = 130 characters, we convert that string back to the matrix of 0/1 using :

sapply(utf8ToInt(STRING),intToBits)

then we simply subsect the matrix selecting only the first 5 rows (intToBits returns 32 bits) and only the columns corresponding to the letter passed as input and finally we apply kronecker as explained in @Giuseppe's answer.