Your goal is to create an alphabet song as text in the following form (in order):

A is for <word starting with A>
B is for <word starting with B>
C is for <word starting with C>
Z is for <word starting with Z>

Example output:

A is for Apple
B is for Banana
C is for Carrot
D is for Door
E is for Elephant
F is for Frog
G is for Goat
H is for Hat
I is for Icicle
J is for Jelly
K is for Kangaroo
L is for Lovely
M is for Mom
N is for Never
O is for Open
P is for Paste
Q is for Queen
R is for Rice
S is for Star
T is for Test
U is for Underneath
V is for Very
W is for Water
X is for X-ray
Y is for Yellow
Z is for Zipper


  • Each "letter" of the song has its own line, so there are 26 lines, and a possible trailing linefeed.

  • The output is case-sensitive:

    • The letter at the start of each line must be capitalized.
    • is for is lowercase.
    • The chosen word does not need to be capitalized, but may be. All lines should be consistent.
  • The chosen word for each line is up to you, but must be a valid English word with at least 3 letters, and it cannot be a conjunction (like and or but), interjection/exclamation (like hey or yay), abbreviation (like XLS), or a name (like Jon).

  • Though I doubt anyone would find it shorter, I'd find it acceptable to use a phrase instead of a single word. So if for some reason S is for Something smells fishy... is shorter, go for it.

  • Put your program's output in your answer, or at least the list of words you used (if there's a link to run your code online, we don't need to see the entire output).

  • Shortest code wins

This challenge was inspired by this video.

  • 2
    \$\begingroup\$ Given some of the answers, this song by Barenaked Ladies seems relevant. \$\endgroup\$ Feb 8, 2017 at 19:21
  • 1
    \$\begingroup\$ @JonathanAllan No slang. Dictionaries contain a lot of things that aren't technically words. Abbreviations is one, slang is another. \$\endgroup\$
    – mbomb007
    Feb 8, 2017 at 19:34
  • 4
    \$\begingroup\$ It's too bad that this devolved into finding 3 letter words that end in the same letter. \$\endgroup\$
    – 12Me21
    Feb 9, 2017 at 0:33
  • 1
    \$\begingroup\$ @12Me21 how is that devolution, is this not code-golf?! If you can find a way to access longer words with fewer bytes just do it! Edit - Find a list of 3-letter words using a set of two middle letters and a set of two end letters and you can beat the single enders (which use a set of seven middles). \$\endgroup\$ Feb 9, 2017 at 12:32
  • 1
    \$\begingroup\$ There's a couple of answers using an external dictionary. Shouldn't they have to add the size of that file to their code? \$\endgroup\$
    – pipe
    Feb 9, 2017 at 15:30

37 Answers 37


PHP, 91 87 86 76 bytes

based on Jonathan Allan´s word list:

for($a=A;$i<26;)echo"$a is for ",$a++,neaoaaeiseioeuuoaaiineiaei[$i++],"t

older versions, 86 bytes:

for($a=A;a&$c=reaoaaei0eioeuuoaaii0eiaei[$i++];)echo"$a is for ",$a++,$c?$c.t:nto,"


for($a=A;$c=reaoaaeiHeioeuuoaaiiHeiaei[$i++];)echo"$a is for ",$a++,$c^x?$c.t:nto,"

Run with -nr.


A is for Art
B is for Bet
C is for Cat
D is for Dot
E is for Eat
F is for Fat
G is for Get
H is for Hit
I is for Into
J is for Jet
K is for Kit
L is for Lot
M is for Met
N is for Nut
O is for Out
P is for Pot
Q is for Qat
R is for Rat
S is for Sit
T is for Tit
U is for Unto
V is for Vet
W is for Wit
X is for Xat
Y is for Yet
Z is for Zit

For the weird words, see Ismael´s answer


Vyxal j, 38 bytes


Try it Online!

Port of Jelly answer.


»hrZ⌈∪≠ẏ¡ɽ»                             # Push compressed number 1849391794619481940361
           `…„ṙµ`                       # Push dictionary compressed string "anisole"
                 τ                      # Convert to custom base (this will result in "niooaaoasoioaiaaaoiineeaei")
                  kAZ                   # Zip with the uppercase alphabet
                     ƛ÷:                # For each item, push both to the stack and duplicate the top item (e.g. stack: "n", "A", "A")
                        ðJ              # Add a space to the end of the alphabet letter
                          «⁼Ė⁼|«J       # Concatenate compressed string "is for " to it
                                 pp     # Concatenate the alphabet letter and the other letter to the end (e.g. stack: "A is for An")
                                   \tJ  # Append the letter "t" to the end
                                        # j flag joins the top of the stack on newlines

Python 2.X, 147 bytes

Yet another try in python using nltk -

from nltk.corpus import*;k=lambda y:filter(lambda x:y in x[0],words.words());print'\n'.join([chr(i)+' is for '+k(chr(i))[64]for i in range(65,91)])

Output -

A is for Abranchiata
B is for Baconic
C is for Cahokia
D is for Damara
E is for Echium
F is for Fascista
G is for Galenic
H is for Halawi
I is for Ichthyornithidae
J is for Jamaica
K is for Kanauji
L is for Lahontan
M is for Macropodinae
N is for Nankingese
O is for Odelsthing
P is for Palaeeudyptes
Q is for Quintius
R is for Rambo
S is for Sac
T is for Tagula
U is for Ulyssean
V is for Vanguardist
W is for Wallon
X is for Xiphodontidae
Y is for Yazoo
Z is for Zea


from nltk.corpus import*
k=lambda y:filter(lambda x:y in x[0],words.words())
print'\n'.join([chr(i)+' is for '+k(chr(i))[27]for i in range(65,91)])

Python 2, 112 bytes

for x in range(26):y=chr(x+65);print y+' is for '+y+'naaoaaeaneieaiaaaaairaeeei'[x:x+1]+('t','n')[x in(8,20,23)]

Try it online!

All of the words end with 't' except for 3 which end with 'n' so we just store the middle letters of the three letter words and indexes of the 3 'n' words and sort out the 'misfits' at the end.


Befunge-93, 73 bytes

This just uses three-letter words, where the last letter is always t.

naaoaauanoioaefeauoosaeaei:0\55+\"t"\:0g\"A"+:," rof si ">:#,_$1+:55*`#@_

Try it online!

But with an additional two bytes, we can vary the last letter between t and e and get a more interesting selection.

teuowaeocoaooudouaeosaoaei:"A"+:,0" rof si ">:#,_$,:0g,1+:2%0g,55+,:55*`#@_

Try it online!


Jelly, 153 bytes (hardcoded words)


Try it online!

A is for Apple
B is for Bald
C is for Cabin
D is for Dairy
E is for Eagle
F is for Fact
G is for Gauge
H is for Humid
I is for Iamb
J is for Jar
K is for Keg
L is for Lad
M is for Music
N is for Nose
O is for Oak
P is for Page
Q is for Quart
R is for Rag
S is for Sock
T is for Taco
U is for Udon
V is for Vault
W is for Week
X is for Xylophone
Y is for Yard
Z is for Zoo

Scala, 85 bytes

Based on @JonathanAllen's answer

Golfed version. Try it online!

var c=65
for(x<-"niooaauusoioaiuaaoiineeaei"){println(f"$c%c is for $c%c$x%st");c+=1}

Ungolfed version. Try it online!

object Main {
  def main(args: Array[String]): Unit = {
    var c = 65
    val input = "niooaauusoioaiuaaoiineeaei"
    for (x <- input) {
      val output = f"$c%c is for $c%c$x%st"
      c += 1


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