[Question]
  [
We've had a lot of alphabet challenges. For this challenge, **you are passed the *output* of an alphabet challenge, and you need to output the pattern scaled to size `N`**. 
For example, if `N=5` and you were passed the [L-phabet](https://codegolf.stackexchange.com/questions/87064/print-output-the-l-phabet):
```
ABCDEFGHIJKLMNOPQRSTUVWXYZ
BBCDEFGHIJKLMNOPQRSTUVWXYZ
CCCDEFGHIJKLMNOPQRSTUVWXYZ
DDDDEFGHIJKLMNOPQRSTUVWXYZ
EEEEEFGHIJKLMNOPQRSTUVWXYZ
FFFFFFGHIJKLMNOPQRSTUVWXYZ
GGGGGGGHIJKLMNOPQRSTUVWXYZ
HHHHHHHHIJKLMNOPQRSTUVWXYZ
IIIIIIIIIJKLMNOPQRSTUVWXYZ
JJJJJJJJJJKLMNOPQRSTUVWXYZ
KKKKKKKKKKKLMNOPQRSTUVWXYZ
LLLLLLLLLLLLMNOPQRSTUVWXYZ
MMMMMMMMMMMMMNOPQRSTUVWXYZ
NNNNNNNNNNNNNNOPQRSTUVWXYZ
OOOOOOOOOOOOOOOPQRSTUVWXYZ
PPPPPPPPPPPPPPPPQRSTUVWXYZ
QQQQQQQQQQQQQQQQQRSTUVWXYZ
RRRRRRRRRRRRRRRRRRSTUVWXYZ
SSSSSSSSSSSSSSSSSSSTUVWXYZ
TTTTTTTTTTTTTTTTTTTTUVWXYZ
UUUUUUUUUUUUUUUUUUUUUVWXYZ
VVVVVVVVVVVVVVVVVVVVVVWXYZ
WWWWWWWWWWWWWWWWWWWWWWWXYZ
XXXXXXXXXXXXXXXXXXXXXXXXYZ
YYYYYYYYYYYYYYYYYYYYYYYYYZ
ZZZZZZZZZZZZZZZZZZZZZZZZZZ
```
You would need to output:
```
ABCDE
BBCDE
CCCDE
DDDDE
EEEEE
```
**For the purposes of explanation, I'll be using only `ABCD`, instead of the full alphabet.** You need to be able to match the L-phabet (above), as well as the following patterns:
The single line:
```
ABCD    or     A
               B
               C
               D
```
The single line repeated `N` times
```
ABCD    or    AAAA
ABCD          BBBB
ABCD          CCCC
ABCD          DDDD
```
The [Tabula Recta](https://codegolf.stackexchange.com/questions/86986/print-a-tabula-recta):
```
ABCD
BCDA
CDAB
DABC
```
This alphabet triangle:
```
A      or   AAAAAAA
AB           BBBBB
ABC           CCC
ABCD           D
ABC
AB
A
```
We also have half triangles in lots of varieties:
```
A           AAAA      A        ABCD
BB          BBB       AB       ABC
CCC         CC        ABC      AB
DDDD        D         ABCD     A
```
Finally, the square:
```
AAAAAAA
ABBBBBA
ABCCCBA
ABCDCBA
ABCCCBA
ABBBBBA
AAAAAAA
```
All of the above patterns are of size 4. However, you will be passed a pattern of size 26, as well as an `N` between 1 and 26, and you need to scale the pattern. You do not need to handle any other patterns.
* The output for 1 will always be the single character `A`
* The output for 26 will always be the same, full-sized pattern passed in.
* Trailing spaces are allowed at the end of each line, as well as a trailing newline at the end
* You can find all patterns of size 26 [here](http://pastebin.com/Kt8NX5MF)
This is a [code-golf](/questions/tagged/code-golf "show questions tagged 'code-golf'"), so do it in as few bytes as possible!
  
[Answer]
# PHP, 502 Bytes
```
<?$c=count($x=explode("\n",$_GET[p]));for($t=$u=$o="",$f="substr",$n=0;$n<$g=$_GET["n"];$n++){if(2651==$l=strlen($s))$o.=$f($x[$n],0,$g).$f($x[$n],1-$g)."\n";elseif($l==1026)echo($t=$f($x[$n],0,$g)).$f(strrev($t),1)."\n";elseif($f($s,-1)=="Y")echo$f($x[$n],0,$g-$n).$f($x[$n],-$n,$n)."\n";elseif($l==376&&$f($s,-2,1)=="\n")echo$f($x[$n],0,$g-$n)."\n";elseif($l==726){$t.=$x[$n]."\n";$n+1==$g?:$u=$x[$n]."\n".$u;}else echo$f($x[$n]??"",0,$g)."\n";}if($o)echo$o.substr(strrev($o),2*$g+1);if($t)echo$t.$u;
```
Works with the string length of a pattern. A pattern has these conditions.
Letter at begin an end. CR are removed.
## Expanded
```
foreach($p as$s){ # all patterns in an array
    $c=count($x=explode("\n",$s));
    for($t=$u=$o="",$f="substr",$n=0;$n<$g=$_GET["n"];$n++){
    if(2651==$l=strlen($s))$o.=$f($x[$n],0,$g).$f($x[$n],1-$g)."\n";
# square pattern
    elseif($l==1026)echo($t=$f($x[$n],0,$g)).$f(strrev($t),1)."\n";
#alphabet triangle up down
    elseif($f($s,-1)=="Y")echo$f($x[$n],0,$g-$n).$f($x[$n],-$n,$n)."\n";
# Tabula recta
    elseif($l==376&&$f($s,-2,1)=="\n")echo$f($x[$n],0,$g-$n)."\n";
# two half triangle
    elseif($l==726){$t.=$x[$n]."\n";$n+1==$g?:$u=$x[$n]."\n".$u;}
#alphabet triangle left right
    else echo$f($x[$n]??"",0,$g)."\n";
# all other
    }
    if($o)echo$o.substr(strrev($o),2*$g+1);
    if($t)echo$t.$u;
}
```
[Answer]
# R, 483 412 bytes
```
    x=function(n){d=1:n;a=LETTERS[d];z=rep;f=function(...)cat(...,"\n",sep="");g=c(d,n:2-1);for(i in d)f(a[z(i,i)],if(i!=n)a[(i+1):n]);f(a);for(i in d)f(a[i]);for(i in d)f(a);for(i in d)f(z(a[i],n));for(i in d)f(a[i:n],if(i>1)a[1:i-1]);for(i in g)f(a[1:i]);for(i in d)f(z(" ",i-1),z(a[i],2*(n-i)+1));for(i in d)f(z(a[i],i));for(i in d)f(z(a[i],n-i+1));for(i in g)f(if(i>1)a[2:i-1],z(a[i],2*(n-i)+1),if(i>1)a[i:2-1])}
```
This is my first time posting, I was told I didn't have a recent years experience of R, so just practise a little bit here.
## Expanded
```
    x=function(n){
    # array of 1 to n
    d=1:n
    # first n capital letters 
    a=LETTERS[d]
    # use z to represent the repeat function
    z=rep
    # use f to represent concatenate, with new line and close the gap
    f=function(...)cat(...,"\n",sep="");
    # use g to represent 1 to n then n to 1
    g=c(d,n:2-1)
    
    # L-phabet 
    # start and repeat the first letter to i, combine the i+1 to the end
    for(i in d) f(a[z(i,i)],if(i!=n)a[(i+1):n])
    
    # single line - horizontal
    f(a)
    
    # single line - vertical
    for(i in d)f(a[i])
    
    # single line - repeated by row
    for(i in d)f(a)
    
    # single line - repeated by column
    for(i in d)f(z(a[i],n))
    # Tabula Recta
    # start from i, combine the 1 to i-1
    for(i in d) f(a[i:n],if(i>1)a[2:i-1])
    # alphabet triangle
    for(i in g)f(a[1:i])
    
    # alphabet triangle - upside down
    for(i in d)f(z(" ",i-1),z(a[i],2*(n-i)+1))
    
    # half triangles 
    for(i in d)f(z(a[i],i))
    for(i in d)f(z(a[i],n-i+1))
    # the square
    # combine the first part from i to i-1 letters, repeat the i in the middle, combine the last part from i-1 to 1 letters
    for(i in g) f(if(i>1)a[2:i-1],z(a[i],2*(n-i)+1),if(i>1)a[i:2-1])
    } 
```
[Answer]
# JavaScript (ES6), ~~382~~ ~~380~~ 370 bytes
```
f=(a,N)=>{X=Y=0
if(a[1]){Y=a[1].trim()[0]=='B'
X=z=a[0][1]==a[1][0]?1:!Y
l="ABCDEFGHIJKLMNOPQRSTUVWXYZ"[s='slice'](0,N)
if(z){if(a[0][1]=='A'){for(a=[],i=0,L=l;i<N;i++)a[N-i]=a[N+i]=L[s](0,-1)+l[N-i-1].repeat(i*2)+[...L].reverse().join``
L=L[s](0,-1)
return a}if(a[1][1]=='C')for(i=0;a[i++]=l,i<N;)l=l[s](1)+l[0]}}else X=1
a=X?a.map(l=>l[s](0,N)):a
return Y?a[s](0,N):a}
```
Pass an array of strings to the function `f()`, like this:
```
f(
`AAAAAAA
ABBBBBA
ABCCCBA
ABCDCBA
ABCCCBA
ABBBBBA
AAAAAAA`.split('\n'), 3)
```
---
Less golfed version with comments:
```
f=(a,N)=>{
    // Whether to truncate array horizontally to width N.
    X=0
    // Whether to truncate array vertically to height N.
    Y=0
    // If a second row exists...
    if(I=a[1]){
        // If the first non-whitespace character in the second row == 'B', truncate vertically.
        if(I.trim()[0]=='B')Y=1
        // Truncate horizontally if 2nd character in row 1 == 1st character in row 2; otherwise, if not truncating vertically.
        X=z=a[0][1]==I[0]?1:!Y
        // If 2nd character in row 1 == 1st character in row 2
        if(z){
            // Make an alphabet.
            l="ABCDEFGHIJKLMNOPQRSTUVWXYZ".slice(0,N)
            // If 2nd character in row 1 == 'A', forget everything we just did. Make a new array, generate a Square pattern, then return it.
            if(a[0][1]=='A'){
                for(a=[],i=0,L=l;i<N;i++)
                    a[N-i]=a[N+i]=L.slice(0,-1)+l[N-i-1].repeat(i*2)+[...L].reverse().join``,
                    L=L.slice(0,-1)
                return a
            }
            // If 2nd character in row 2 == 'C', fill array with a Tabula Recta.
            if(I[1]=='C')
                for(i=0;a[i++]=l,i<N;)
                    l=l.slice(1)+l[0]
        }
    }else{
        // If a second row doesn't exist, it's a horizontal line; truncate horizontally.
        X=1
    }
    // Truncate array horizontally if necessary.
    a=X?a.map(l=>l.slice(0,N)):a
    // Truncate array vertically if necessary.
    return Y?a.slice(0,N):a
}
```
]