Together Everyone Achieves More

Question

(related: one, two, three)

An acrostic is a style of poem/writing where the beginning character of each line, when read vertically, also produces a word or message. For example,

Together
Everyone
Achieves
More

also spells out the word TEAM when the first column is read vertically.

Acrostics are a subset of mesostics, where the vertical word can be anywhere in the horizontal words. For example, the TEAM one above could also be written as a mesostic as follows

   togeTher
everyonE
       Achieves
       More

along with several other variations.

The challenge here will be to produce an acrostic or mesostic from a given list of input words.

Input

• A list of words in any any suitable format.
• The list will only contain words made from lowercase [a-z].
• The list is guaranteed to form an acrostic or a mesostic (no need to handle bogus input).
• One of the words in the input will form the vertical word, while the rest make the horizontal words - part of the challenge here is to find the appropriate vertical word, so it cannot be taken separately.

Output

• The ASCII-art acrostic or mesostic formed from the input words, written to STDOUT or returned, in any reasonable format.
• The corresponding vertical word must be capitalized (as in the examples).
• Leading spaces to get the vertical word to line up appropriately are required. Trailing spaces, and leading/trailing newlines are optional. Extra leading spaces are fine as well, so long as the words align correctly.
• If both an acrostic and mesostic are possible, output only the acrostic.
• If more than one acrostic/mesostic is possible, your code can output any or all of them.

Rules

• Either a full program or a function are acceptable.
• Standard loopholes are forbidden.
• This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

Examples

['together', 'team', 'everyone', 'achieves', 'more']
Together
Everyone
Achieves
More

['aaa', 'aaa', 'aaa', 'aaa']
Aaa
Aaa
Aaa
# One output, or multiple (of the same) output is allowed

['aaa', 'aaa', 'aab', 'baa']
Aaa
Aaa
Baa
# This is the only allowed output, since others would be mesostic, which are lower priority

['live', 'every', 'love', 'very', 'ohio']
Live
Ohio
Very
Every
# Note that 'live' couldn't be the vertical word since then it would be a mesostic, which is lower priority output

['cow', 'of', 'fox']
cOw
 Fox
# A shorter mesostic test case

['late', 'ballroom', 'anvil', 'to', 'head']
anviL
   bAllroom
    To
   hEad

Answer 1

Pyth, 52 49 47 46 bytes

jm++*;-lsQhAd<HGr>HG4hhMDfhhShMTm.b,xNYNtdhd.p

Try it online.

This is probably very golfable. It prints a bunch of leading spaces.

Answer 2

Brachylog, 145 bytes

p~c[A:B:C],Bl1,A:CcP@pz:cahgB,P:1a~@nw|lL,?:laot:" "rjg:Ljb:sa:?z:cap~c[A:B:C],Bl1,A:Cc@pz:caZ:LmgB,Zl-yM,Z:M:Lz:2az:ca~@nw
bB,?h@u:Bc
b#=,?h@u|h

Try it online!

(Takes half a minute, so be patient.)

Answer 3

JavaScript (ES6), 255 263 269 286

Edit 17 bytes saved as arbitrary number of leading spaces is allowed
Edit2 some shuffling, 6 byte saved
Edit3 return a list of string instead of a single string with newlines (OP comment to feersum's answer), 8 more bytes saved

For each word in the input list, I use a recursive DFS to find all possibile mesostics/acrostics. Each one is stored as an array with word and target letter position inside the word. Every result found is saved in global result array at position 1 (if it's an acrostic) or 0 if it's a mesostic.

After the complete scan for all words, I get the result at last position in array and build and return its ascii art rapresentation.

l=>(l.map((w,i)=>(r=(p,v,i,a)=>(l[i]='.',w[p]?l.map((v,i)=>~(j=v.search(w[p]))&&r(p+1,v,i,a|j,m[p]=[j,v])):l[p+1]?0:s[+!a]=[...m],l[i]=v))(0,w,i,m=[]),s=[]),m=s.pop(),m.map(([j,v])=>' '.repeat((l+0).length-j)+v.slice(0,j)+v[j].toUpperCase()+v.slice(j+1)))

Less golfed

f=l=>(
  l.map((w,i)=>
    // r: recursive DFS function
    // defined here as it uses local w variable
    (r = (p,v,i,a) => (
     l[i] = '.'
     , w[p] 
     ? l.map(
       (v,i) => ~(j=v.search(w[p])) && 
                r(p+1, v, i, a|j, m[p] = [j,v])

     )
     : l[p+1] ? 0 // invalid if there are still unused words
              : s[+!a]=[...m] // a is 0 if acrostic
     , l[i] = v) 
    )(0, w, i, m=[])
  , s=[]),
  m = s.pop(), // get last result
  // m.map(([j]) => o = o<j ? j : o, o=0), // find offset for alignment
  // no need to find the optimal offset as leading blanks are allowed
  m.map(([j,v]) => ' '.repeat((l+0).length-j) 
                   + v.slice(0,j) 
                   + v[j].toUpperCase()
                   + v.slice(j+1)
  )
)

Test

f=l=>(l.map((w,i)=>(r=(p,v,i,a)=>(l[i]='.',w[p]?l.map((v,i)=>~(j=v.search(w[p]))&&r(p+1,v,i,a|j,m[p]=[j,v])):l[p+1]?0:s[+!a]=[...m],l[i]=v))(0,w,i,m=[]),s=[]),m=s.pop(),m.map(([j,v])=>' '.repeat((l+0).length-j)+v.slice(0,j)+v[j].toUpperCase()+v.slice(j+1)))

console.log=x=>O.textContent+=x+'\n\n'

;[
 ['together', 'team', 'everyone', 'achieves', 'more']
,['aaa', 'aaa', 'aaa', 'aaa']
,['aaa', 'aaa', 'aab', 'baa']
,['live', 'every', 'love', 'very', 'ohio']
,['cow', 'of', 'fox']
,['late', 'ballroom', 'anvil', 'to', 'head']
].forEach(l=>console.log(f(l).join`\n`))

<pre id=O></pre>

Answer 4

Mathematica 10.0, 139 bytes

An unnamed function returning a list of lines:

Sort[{q=Max[p=Min/@Position@@@({z={##2},#})],Array[" "&,q-#2]<>ToUpperCase~MapAt~##&@@@({z,p})}&@@@Permutations@Characters@#][[1,2]]&

Example usage:

In[144]:= f = Sort[{q=Max[p=Min/@Position@@@({z={##2},#})],Array[" "&,q-#2]ToUpperCase~MapAt~##&@@@({z,p})}&@@@Permutations@Characters@#][[1,2]]&;

In[145]:= f @ {"late","ballroom","anvil","to","head"} // Column

 ... several pages of warnings ... 

Out[145]= baLlroom
            Anvil
            To
           hEad

I'm looking for suggestions on better ways to do the capitalization. I found a very nice function MapAt for capitalizing the letter in the string.

Answer 5

Perl6, 287 277 269 bytes

my @w=$*IN.words;my ($q,$r)=gather for ^@w {my @v=@w.rotate($_);my \d=@v.shift;for @v.permutations {my @o=flat($_ Z d.comb).map:{$^a.index: $^b};take $_,@o if @o>>.defined.all}}.sort(*[1].sum)[0];for @$q Z @$r ->(\a,\b){say " "x($r.max -b)~a.substr(0,b)~a.substr(b).tc}

Answer 6

Haskell, 214 206 204 202 bytes

import Data.List
z=zipWith
h i j t|(u,v:w)<-splitAt j t=([1..sum i-j]>>" ")++u++toEnum(fromEnum v-32):w
f x=uncurry(z=<<h)$sort[(head<$>z elemIndices w l,l)|w:l<-permutations x,(True<$w)==z elem w l]!!0

Returns a list of space padded strings, e.g. f ["late","ballroom","anvil","to","head"] -> [" baLlroom"," Anvil"," To"," hEad"] or more display friendly:

*Main> mapM_ putStrLn $ f ["late", "ballroom", "anvil", "to", "head"]
 baLlroom
   Anvil
   To
  hEad

f selects the words that are written horizontally together with a list of offsets. h pads each word according to the corresponding offset and inserts the uppercase letter. In detail:

                permutations x       -- for each permutation of the input list x
         w:l<-                       -- bind w to the first word and l to the rest
             (True<$w)==z elem w l   -- keep it if the list of other words
                                     -- containing the next letter of w
                                     -- equals (length w) times True, i.e. we have
                                     -- as many matching letters as letters in w.
                                     -- This rules out combinations shortcut by zipWith

                                     -- for all the remaining w and l make a pair
         head<$>z elemIndices w l    -- the first element of the list of list of
                                     -- indices where the letter appears in the word 
                                l    -- and l itself
   sort                              -- sort the pairs (all 0 indices, i.e. acrostics
                                     -- go first)
                               !!0   -- pick the first
                                     -- now we have a pair like
                                     -- ([2,0,0,1],["ballroom","anvil","to","head"])
 uncurry(z=<<h)                      -- loop over (index,word) and 
                                     -- provide the third parameter for h 



 h i j t                             -- h takes the list of indices and
                                     -- an index j and a word t
       (u,v:w)<-splitAt j t          -- split the word at the index and bind
                                     --   u: part before the split
                                     --   v: letter at the split
                                     --   w: part after the split
         [1..sum i-j]>>" "           -- the spaces to pad
           ++ u                      -- followed by u
           ++ toEnum(fromEnum v-32)  -- and uppercase v
           :                         -- and w 

Answer 7

Python, 249 bytes

Probably still very golfable

from itertools import*;e=enumerate;lambda l:[[[' ']*(max(j for k,(j,c)in o[1:])-i)+l[k][:i]+[c.upper()]+l[k][i+1:]for k,(i,c)in o[1:]]for p in product(*[list(e(w))for w in l])for o in permutations(list(e(p)))if[c for k,(i,c)in o[1:]]==l[o[0][0]]][0]

Takes and returns a list of list of characters.
- e.g. " bAllroom" is [' ',' ',' ','b','A','l','l','r','o','o','m']

Only ever returns the first result and checks in an order such that all the acrostics are checked first.

See all test cases printed in the display format on ideone

---

Here is a more readable functional form that does the same (except it returns the first result immediately rather than evaluating and then returning the first result):

from itertools import*
def f(l):
    for p in product(*[list(enumerate(w)) for w in l]):
        for o in permutations(list(enumerate(p))):
            if [c for k,(i,c) in o[1:]] == l[o[0][0]]:
                return [ [' '] * (max(j for k,(j,c) in o[1:]) - i)
                       + l[k][:i]
                       + [c.upper()]
                       + l[k][i+1:]
                       for k, (i, c) in o[1:]
                       ]

Answer 8

Perl 6, 177 bytes

->\a{for first({.[0] eq[~] .[1]»[1]},map |*,[Z] map {.[0]X [X] map {.comb[(^$_,$_,$_^..* for ^.chars)]},.[1..*]},a.permutations)[1] ->$/ {say [~] " "x a.comb-$0,|$0,$1.uc,|$2}}

Brute-force solution.

How it works

-> \a {
    for first({.[0] eq[~] .[1]»[1]},          # For the first valid candidate
            map |*, [Z]                       # among the transposed
            map {                             # lists of candidates
                .[0] X [X] map {
                    .comb[(^$_,$_,$_^..* for ^.chars)]
                }, .[1..*]
            },
            a.permutations                    # for all permutations of the input:
        )[1] ->$/ {
        say [~] " "x a.comb-$0,|$0,$1.uc,|$2  # Print the candidate as ASCII art.
    }
}

Each candidate looks like:

"of", (("c"),"o",("w")), ((),"f",("o","x"))

Transposing the list of lists of candidates is necessary to ensure that an acrostic, if it exists, is found before any mesostic.