Print all 2 letter Scrabble Words

Question

The Challenge:

Print every 2 letter word acceptable in Scrabble using as few bytes as possible. I have created a text file list here. See also below. There are 101 words. No word starts with C or V. Creative, even if nonoptimal, solutions are encouraged.

AA
AB
AD
...
ZA

Rules:

The outputted words must be separated somehow.
Case does not matter, but should be consistent.
Trailing spaces and newlines are allowed. No other characters should be outputted.
The program should not take any input. External resources (dictionaries) cannot be used.
No standard loopholes.

Wordlist:

AA AB AD AE AG AH AI AL AM AN AR AS AT AW AX AY 
BA BE BI BO BY 
DE DO 
ED EF EH EL EM EN ER ES ET EX 
FA FE 
GO 
HA HE HI HM HO 
ID IF IN IS IT 
JO 
KA KI 
LA LI LO 
MA ME MI MM MO MU MY 
NA NE NO NU 
OD OE OF OH OI OM ON OP OR OS OW OX OY 
PA PE PI 
QI 
RE 
SH SI SO 
TA TI TO 
UH UM UN UP US UT 
WE WO 
XI XU 
YA YE YO 
ZA

@Sp3000 I'll say no, if something interesting can be thought up — qwr, Commented Aug 13, 2015 at 3:29
Please clarify what exactly counts as separated somehow. Does it have to be whitespace? If so, would non-breaking spaces be allowed? — Dennis, Commented Aug 13, 2015 at 14:14

VoiceOfFolly · Accepted Answer · 2015-08-14 19:18:06Z

1

Perl, 248 bytes

@v=split(//,"AEIOU");@s=split(' ',"ADEGIRSWX LSTX DFN EFIPRX HPST BHLMNTY DFHMNR ST DHMNSWY MN ZFKP WYPB KQXLHPMB LGJTB X");for(0..14){foreach $c(split(//,@s[$_])){$_<10?print @v[$_%5].$c." ":a;$_>4?print $c.@v[$_%5]." ":a;}}print "MM MY BY HM SH";

First time using perl (and first time golfing), so there's definitely room for improvement. Factored out the vowels and grouped the remaining letters based on how the resulting word was created - adding the vowel first, last, or both vowel first and last create a word on the list.

edited Aug 14, 2015 at 19:18

answered Aug 14, 2015 at 17:51

VoiceOfFolly

112 bronze badges

\$\begingroup\$ I don't know Perl, but I assume you are separating the output by spaces. It seems "MM "."MY "."BY "."HM "."SH " could be shortened to "MM MY BY HM SH". \$\endgroup\$
– Level River St
Commented Aug 14, 2015 at 19:05
\$\begingroup\$ @steveverrill Thanks! I got so caught up in the rest of the code that I overlooked how redundant that was. \$\endgroup\$
– VoiceOfFolly
Commented Aug 14, 2015 at 19:18

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ETHproductions · Accepted Answer · 2015-08-14 23:21:03Z

Javascript (ES6), 214

Perhaps not the shortest way to do it, but definitely interesting.

_=>(i=0,[...'ABDEFGHILMNOPRSTUWXY'].map(z=>[...`ABFHKLMNPTYZ
A
AEIO
ABDFHMNOPRWY
EIO
A
AEOSU
ABHKLMOPQSTX
AE
AEHMOU
AEIOU
BDGHJLMNSTWY
OU
AEO
AEIOU
AEIU
MNX
AO
AEO
ABMO`.split`
`[i++]].map(g=>g+z).join` `).join`
`)

Loops through each letter in the first string, adding it onto each letter in the corresponding line of the second. This returns the words in order of their last letter, like so:

AA BA FA HA KA LA MA NA PA TA YA ZA
AB
AD ED ID OD
AE BE DE FE HE ME NE OE PE RE WE YE
EF IF OF
AG
AH EH OH SH UH
AI BI HI KI LI MI OI PI QI SI TI XI
AL EL
AM EM HM MM OM UM
AN EN IN ON UN
BO DO GO HO JO LO MO NO SO TO WO YO
OP UP
AR ER OR
AS ES IS OS US
AT ET IT UT
MU NU XU
AW OW
AX EX OX
AY BY MY OY

Suggestions welcome!

Aofl · Accepted Answer · 2015-08-15 01:26:04Z

Java, 255 254 bytes

Found a way to squeeze one more byte out of it.

class C{public static void main(String[]a){char c='A';for(char l:"ABDEGHILMNRSTWXY AEIOY  EO DFHLMNRSTX AE O AEIMO DFNST O AI AIO AEIMOUY AEOU DEFHIMNPRSWXY AEI I E HIO AIO HMNPST  EO IU AEO A".toCharArray())if(l<'A')c++;else System.out.print(" "+c+l);}}

Or (though not much clearer):

class C {
    public static void main(String[] a) {
        char c = 'A';
        for (char l : "ABDEGHILMNRSTWXY AEIOY  EO DFHLMNRSTX AE O AEIMO DFNST O AI AIO AEIMOUY AEOU DEFHIMNPRSWXY AEI I E HIO AIO HMNPST  EO IU AEO A"
                .toCharArray())
            if (l < 'A')
                c++;
            else
                System.out.print(" " + c + l);
    }
}

You can save two bytes by changing both occurrences of 'A' into 65. — ProgramFOX, Commented Aug 15, 2015 at 12:31

Leif Willerts · Accepted Answer · 2015-09-19 21:25:40Z

1

Haskell, 333 308 298 bytes

Just for fun!

Evaluating s will print all of the words in a weird order - I used the fact that most combinations are vowel-consonant or vice versa, could probably optimize even more with custom character "classes", shortening the encoded matrix (here, w and k).

Does anybody know a shorter way to print strings without quotes and brackets than my monadic one? Type classes are even longer as far as I can tell.

Also, there might also be a shorter way of doing p's job...

v="AEIOU"
c="BCDFGHJKLMNPQRSTVWXYZ"
w=[976693,321324,50188,945708,52768]
k=[15,0,10,3,8,15,8,5,13,31,27,7,4,2,12,13,0,10,20,11,1]
z a b x=[[c:[e]|(e,f)<-b#p d,f]|(c,d)<-a#x]
(#)=zip
p 0=[]
p n=((n`mod`2)>0):p(n`div`2)
s=mapM_ putStrLn$concat$z v c w++z c v k++[words"AA AE AI BY HM MM MY OE OI SH"]

edited Sep 19, 2015 at 21:25

answered Aug 13, 2015 at 22:21

Leif Willerts

1,0505 silver badges6 bronze badges

\$\begingroup\$ Isn't sequence_ the same as void$sequence? Then you can omit also the import. \$\endgroup\$
– nimi
Commented Aug 13, 2015 at 22:40
\$\begingroup\$ That's oddly packaged, but yes. Thanks! \$\endgroup\$
– Leif Willerts
Commented Aug 13, 2015 at 22:48
\$\begingroup\$ Isn't, void had to be imported. Anyways, gonna/gotta remember this. \$\endgroup\$
– Leif Willerts
Commented Aug 13, 2015 at 22:52
1

\$\begingroup\$ Ah, and sequence_$map putStrLn is mapM_ putStrLn. Replace the (, ) around concat$... with another $. \$\endgroup\$
– nimi
Commented Aug 13, 2015 at 23:07

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Community · Accepted Answer · 2020-06-17 09:04:33Z

PHP, 170 bytes

shortest approach I could find so far ...

for(;$c=ABFHKLMNPTYZ1A2AEIO1ABDFHMNOPRWY1EIO1A1AEOSU1ABHKJMOPQSTX3AE1AEHMOU1AEIOU1BDGHJLMNSTWY1OU11AEO1AEIOU1AEIU1MNX2AO1AEO1ABMO[$i++];)$c<1?print$c.chr($k+65)._:$k+=$c;

breakdown

for(;$c=CHARACTERS[$i++];)  // loop $c through map
    $c<1                            // if $c is a letter (integer value < 1)
        ?print$c.chr($k+65)._           // print first letter, second letter, delimiter
        :$k+=$c                         // else increase second letter
    ;

Note The shortest bit-mapping version with printable ascii costs 190 bytes (113 bytes data+77 bytes decoding) using 6 bits=base 64, 174 bytes (97 data, 77 decoding) using 7 bits (base 128); possibly some more for escaping.

for(;$p<676;)                   // loop through combinations
    ord("MAPPING"[$p/6])-32>>$p%6&1     // test bit
        ?                                       // if set, do nothing
        :print chr($p/26+65).chr($p++%26+65)._; // if not, print combination+delimiter

Base 224 (using ascii 32..255) takes 87 bytes data (+ escaping); but I guess the decoding will cost more than 10 bytes extra.
Excluding C and V from the map would save 16/14/13 bytes on the data but cost a lot in decoding.

Chas Brown · Accepted Answer · 2019-08-26 03:46:51Z

1

Python 2, 202 200 bytes

for s in'AABALIFEN KITOELOPE LAYAINUNEREMI AMANAHI MOHMYEX UPI UTAGOFAWOWETI BY OR BI ZAXI PAS BE AD AEDEHAT QI UMMUS JOIDONOSHOYOD BOMESISOXUHEF KAR'.split():
 for i in range(len(s)-1):print s[i:i+2]

Try it online!

Try to join up overlapping pairs; i.e. ZAXI encodes ZA, AX, XI. Looks like 140 bytes of encoded text is about the best possible with this approach.

edited Aug 26, 2019 at 3:46

answered Aug 25, 2019 at 23:32

Chas Brown

9,7671 gold badge13 silver badges39 bronze badges

1

\$\begingroup\$ 140 bytes is the theoretical limit for this approach. \$\endgroup\$
– Joel
Commented Aug 30, 2019 at 2:43

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emanresu A · Accepted Answer · 2022-09-04 06:22:02Z

1

05AB1E, 143 bytes

•ZÐFý8ù6„=4ÅœÿWì±7Ë5Œ¾`ó/îAnµ%Ñ¥Ëø±/ÀéNzÖ„Ñ§IJ¶D—ÙVEStvÖò…¸¾6F³#EXŠm¯CÄµ±ÓoÊ°}^€Ftåß€ðŒ=«j;F-Â1;xX3i&QZÒ'Ü”>lwìlOs>íÙVÇI)~î‰Hç²?Öd0È^ÝQ°•36B2ô»

Try it online!

edited Sep 4, 2022 at 6:22

emanresu A

41.6k5 gold badges97 silver badges230 bronze badges

answered Jan 23, 2017 at 19:59

Magic Octopus Urn

20.8k5 gold badges62 silver badges136 bronze badges

\$\begingroup\$ 95 bytes by porting @Sp3000's Python 3 answer. \$\endgroup\$
– Kevin Cruijssen
Commented Aug 30, 2019 at 7:16
1

\$\begingroup\$ @KevinCruijssen 91 (and I'm sure this method can get lower, but I hand-picked the dictionary words for humor). \$\endgroup\$
– Grimmy
Commented Aug 30, 2019 at 13:33

Add a comment |

Pandu · Accepted Answer · 2022-09-25 03:11:03Z

Python 3, 182 bytes

[print(f+l)for*F,l in'ABFHKLMNPTYZA AB AEIOD ABDFHMNOPRWYE EIOF AG AEOSUH ABHKLMOPQSTXI AEL AEHMOUM AEIOUN BDGHJLMNSTWYO OUP AEOR AEIOUS AEIUT MNXU AOW AEOX ABMOY'.split()for f in F]

Try it online!

This was briefly the shortest Python 3 solution :)

Each section of the magic string, e.g. AEIOD, encodes e.g. AD ED ID OD. (The last letter of the section should be combined with each other letter.)

Equivalent to:

for *FIRSTS, last in (
    'ABFHKLMNPTYZA AB AEIOD ABDFHMNOPRWYE EIOF AG AEOSUH ABHKLMOPQSTXI AEL AEHMOUM AEIOUN BDGHJLMNSTWYO OUP AEOR AEIOUS AEIUT MNXU AOW AEOX ABMOY'
    .split()
):
    for first in FIRSTS:
        print(first+last)

Community · Accepted Answer · 2020-06-17 09:04:33Z

0

F#, 205 bytes

Seq.iteri(fun i s->Seq.iter(printf"%c%c "((char)i+'A'))s;printfn"")("ABDEGHILMNRSTWXY AEIOY  EO DFHLMNRSTX AE O AEIMO DFNST O AI AIO AEIMOUY AEOU DEFHIMNPRSWXY AEI I E HIO AIO HMNPST  EO IU AEO A".Split())

Live version.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 14, 2015 at 21:53

mattnewport

1,6779 silver badges10 bronze badges

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user2378682 · Accepted Answer · 2015-08-15 16:37:41Z

0

scala, 223 bytes

print(("AABDEGHILMNRSTWXY BAEIOY DEO EDFHLMNRSTX FAE GO HAEIMO IDFNST JO KAI LAIO MAEIMOUY NAEOU ODEFHIMNPRSWXY PAEI QI RE SHIO TAIO UHMNPST WEO XIU YAEO ZA".split(" ").flatMap(a=>a.tail.map(a(0).toString+_))).mkString(" "))

Not especially compact, but vaguely idiomatic .. eval with

scala -e '...'

edited Aug 15, 2015 at 16:37

answered Aug 15, 2015 at 16:31

user2378682

11 bronze badge

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Java, 288

class E{public static void main(String[]a){a="ABDEGHILMNRSTWXY,AEIOY,,EO,DFHLMNRSTX,AE,O,AEIMO,DFNST,O,AI,AIO,AEIMOUY,AEOU,DEFHIMNPRSWXY,AEI,I,E,HIO,AIO,HMNPST,,EO,IU,AEO,A".split(",");for(int i=26,j;i-->0;)for(j=0;j<a[i].length();)System.out.println((char)('A'+i)+""+a[i].charAt(j++));}}

By just stringing together the second letters of each word (plus a delimiter), you just loop through while keeping track of the first. Not as fancy as the base-encoded versions, but much simpler and a good score for Java. Prints in order by descending-first-letter (ZA YA YE YO XI XU), one word per line.

With line breaks:

class E{
    public static void main(String[]a){
        a="ABDEGHILMNRSTWXY,AEIOY,,EO,DFHLMNRSTX,AE,O,AEIMO,DFNST,O,AI,AIO,AEIMOUY,AEOU,DEFHIMNPRSWXY,AEI,I,E,HIO,AIO,HMNPST,,EO,IU,AEO,A".split(",");
        for(int i=26,j;i-->0;)
            for(j=0;j<a[i].length();)
                System.out.println((char)('A'+i)+""+a[i].charAt(j++));
    }
}

handrake · Accepted Answer · 2015-08-18 05:27:30Z

0

Python 239 bytes

import re;f=lambdax:''.join(re.findall('..',x));a,b='ABAHALAMANATAYDEDOEFEHEMENERHOISITMOMUNONUOSOWOY','AAADAEAGAIARASAWAXBEBIBOBYELESETEXFAGOHIHMIDIFINJOKAKILILOMIMMMYOEOFOIOPOROXPAPEPIQISHTOUHUPUSUTWEXIXUYEZA';print f(a),f(a[::-1]),f(b)

This is a bit long, but I've given it a try. I've counted all the patterns that also have the reversed counterparts. Ex) 'AB'/'BA' and so on.

edited Aug 18, 2015 at 5:27

answered Aug 17, 2015 at 15:16

handrake

113 bronze badges

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ovs · Accepted Answer · 2017-01-23 22:04:43Z

0

Python 3, 248 bytes

e=enumerate
[print(*(chr(x+65)+chr(j+65)+" "for j,i in e(bin(k)[2:])if int(i)))for x,k in e([30292443,16793873,0,16400,9320616,17,16384,20753,794664,16384,257,16641,17846545,1064977,29798840,273,256,16,16768,16641, 831616,0,16400,1048832,16401,1])]

The numbers were generated by an additional script
I'm sure that it can be optimized further

answered Jan 23, 2017 at 22:04

ovs

60.1k3 gold badges47 silver badges159 bronze badges

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Innat3 · Accepted Answer · 2019-08-23 08:59:12Z

0

C# (Visual C# Interactive Compiler), 208 bytes

()=>"ABDEGHILMNRSTWXY,AEIOY,EO,DFHLMNRSTX,AE,O,AEIMO,DFNST,O,AI,AIO,AEIMOUY,AEOU,DEFHIMNPRSWXY,AEI,I,E,HIO,AIO,HMNPST,EO,IU,AEO,A".Split(',').SelectMany((x,y)=>x.Select(z=>""+"ABDEFGHIJKLMNOPQRSTUWXYZ"[y]+z))

Try it online!

answered Aug 23, 2019 at 8:59

Innat3

8314 silver badges6 bronze badges

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XMark · Accepted Answer · 2019-08-29 18:32:55Z

PHP, 187 bytes

for($n=A,$l=ABDEGHILMNRSTWXYAEIOY1EODFHLMNRSTXAEGOAEIMODFNS2TOAIAIOAEIMOUYAEOUDEFFHIMNPRSWXYAEI3IE4HIOAIOHMNPST5EOIUAEOA;$q=$l[$x++];){if($q<A)$n++;else{if($q<$p)$n++;$p=$q;echo"$n$q
";}}

Try it online!

Basically, the list is mostly just the second letters in the word list - I noticed in almost all cases, when incrementing the first letter, the second letter is before the one before it. For example going from AY to BA, A is before Y in the alphabet, and going from QI to RE, E is before I in the alphabet. In the very few cases where the second letter was the same or after the previous one, or to skip C and V, I put a number in the string so it knows to increment the first letter.

emanresu A · Accepted Answer · 2022-09-04 06:22:07Z

0

Retina, 167 bytes


¶AABDEGHIcWXY¶BbOY¶DEaEDFHcX¶FAE¶GaHbMaIDFNST¶JaKAI¶LAIaMbMOUY¶NAEOU¶ODEFHIMNPRSWXY¶Pb¶QI¶RE¶SHIaTAIaUHMNPST¶WEaXIU¶YAEaZA
c
LMNRST
b
AEI
a
O¶
+`(¶(.)\S+)(\S)
$1 $2$3

Try it online!

edited Sep 4, 2022 at 6:22

emanresu A

41.6k5 gold badges97 silver badges230 bronze badges

answered Jun 21, 2017 at 6:57

ovs

60.1k3 gold badges47 silver badges159 bronze badges

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Acer · Accepted Answer · 2022-09-20 13:37:44Z

C#, 161 bytes

w=>"aBFHKLMNPTYZeBDFHMNOPRWYiBHKLMOPQSTXmEHMOUhEOSUnEIOUsEIOUdEIOtEIUyBMOrEOxEOlEwObgoBDGHJLMNSTWYfEIOuMNXpOU".Select((x,i)=>x>90&&w!=(w=""+x)?i>80?"":"a"+w:x+w)

The string is every possible second letter (lowercase) followed by each of its valid first letters (uppercase).

Every time a lowercase letter is reached, var w is updated to store that letter, and return "". Each time an uppercase letter is reached, return that letter + w.

Additionally, "a" + w is automatically added every time a new lowercase letter is reached, unless that letter is one of f, o, p or u (this is achieved by putting these 4 at the end of the string and conditionally adding "a"+w based on index).

Try it online!

l4m2 · Accepted Answer · 2022-12-23 10:18:01Z

0

JavaScript (V8), 182 bytes

"A ABFHKLMNPTYZB AD AEIOE ABDFHMNOPRWYG AH AEOSUI ABHKLMOPQSTXL AEM AEHMOUN AEIOUR AEOS AEIOUT AEIUW AOX AEOY ABMOO BDGHJLMNSTWYF EIOU MNXP OU".replace(/. ?/g,c=>c[1]?a=c:print(c+a))

Try it online!

Not that bad

edited Dec 23, 2022 at 10:18

answered Dec 23, 2022 at 10:02

l4m2

26k2 gold badges23 silver badges97 bronze badges

\$\begingroup\$ gzip compress this best (comparing to other compress) to 172:( \$\endgroup\$
– l4m2
Commented Dec 23, 2022 at 10:05

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The Thonnu · Accepted Answer · 2022-12-23 10:50:25Z

Python 2, 23+176 = 199 bytes

print('...'.decode('zip'))

Try it online!

Hexdump of the string in place of ...:

['0x78', '0x9c', '0x0d', '0xc7', '0xd1', '0x8d', '0xc3', '0x30', '0x0c', '0x44',
 '0xc1', '0x56', '0x5e', '0x2b', '0x6b', '0x88', '0x8e', '0x78', '0x31', '0x45',
 '0x9f', '0x25', '0xc1', '0x56', '0xfa', '0x2f', '0x24', '0xf9', '0x19', '0x60',
 '0x24', '0xb4', '0xa1', '0x82', '0x0c', '0xbd', '0x50', '0x45', '0x8e', '0x0e',
 '0x14', '0xa8', '0xa1', '0x0b', '0x75', '0x34', '0xd0', '0x8d', '0x1e', '0xb4',
 '0xd8', '0xc4', '0x66', '0x6c', '0xce', '0x96', '0x6c', '0x8b', '0x62', '0x94',
 '0xc4', '0x0a', '0xb6', '0x63', '0x15', '0x3b', '0xb0', '0xc0', '0x1a', '0x76',
 '0x61', '0x1d', '0x1b', '0xd8', '0xc3', '0x2e', '0x76', '0xe3', '0x95', '0x54',
 '0x51', '0x8d', '0xea', '0xd4', '0xa0', '0x26', '0x5e', '0xf0', '0x1d', '0x6f',
 '0x78', '0xc7', '0x07', '0x7f', '0xc9', '0x5b', '0xbc', '0x9d', '0x43', '0x1c',
 '0x3f', '0x93', '0x10', '0x61', '0x84', '0x13', '0x41', '0xfc', '0x3a', '0x89',
 '0x45', '0x13', '0xcd', '0x68', '0x49', '0x9b', '0x64', '0x21', '0x8d', '0xdc',
 '0xc9', '0x4a', '0x3a', '0x19', '0x64', '0x23', '0x4f', '0xf2', '0x22', '0x3b',
 '0x79', '0x93', '0x0f', '0xb9', '0x38', '0xc5', '0x69', '0x9c', '0xce', '0xbf',
 '0x73', '0x19', '0xbd', '0xd2', '0x9d', '0x9e', '0x0c', '0x31', '0x9c', '0x91',
 '0xcc', '0xca', '0x0c', '0x66', '0x63', '0x9e', '0xcc', '0xce', '0x1c', '0xdc',
 '0xc6', '0x9d', '0x3c', '0xce', '0x33', '0x59', '0x62', '0x19', '0x2b', '0xf9',
 '0xe8', '0x0b', '0x93', '0x2f', '0x47', '0xe5']

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Print all 2 letter Scrabble Words

The Challenge:

Rules:

49 Answers 49

Perl, 248 bytes

Javascript (ES6), 214

Java, 255 254 bytes

Haskell, 333 308 298 bytes

PHP, 170 bytes

Python 2, 202 200 bytes

05AB1E, 143 bytes

Python 3, 182 bytes

F#, 205 bytes

Java, 288

Python 239 bytes

Python 3, 248 bytes

C# (Visual C# Interactive Compiler), 208 bytes

PHP, 187 bytes

Retina, 167 bytes

C#, 161 bytes

JavaScript (V8), 182 bytes

Python 2, 23+176 = 199 bytes

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Print all 2 letter Scrabble Words

The Challenge:

Rules:

49 Answers 49

Perl, 248 bytes

Javascript (ES6), 214

Java, 255 254 bytes

Haskell, 333 308 298 bytes

PHP, 170 bytes

Python 2, 202 200 bytes

05AB1E, 143 bytes

Python 3, 182 bytes

F#, 205 bytes

Java, 288

Python 239 bytes

Python 3, 248 bytes

C# (Visual C# Interactive Compiler), 208 bytes

PHP, 187 bytes

Retina, 167 bytes

C#, 161 bytes

JavaScript (V8), 182 bytes

Python 2, 23+176 = 199 bytes

Your Answer

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