WHAT, Wolfberg's Helpful Anagramming Tool, is really a toolbox of powerful tools, with many facilities. You can make effective use of WHAT right away by using just a few of its features, and the document entitled "WHAT's First" introduces these features. Many users will find these basic features are all they need to employ to benefit by its use, but WHAT provides a wealth of other facilities you ought to explore when you have a chance. One way you can do this is by studying the examples in this document.
These examples were updated in June, 2015 to be based on the OWL3 lexicon.
If you ask WHAT to perform this command:
aeilnos~
it answers with:
There is at least one word.
and perhaps you are thinking it is
SEALION. If so, you would
be wrong, since there is a space in the two-word spelling of that
seallike animal. In case you were wondering,
SEALLIKE is acceptable in the
OWL3 lexicon.
You can ask to see the answer by requesting to WHAT that
it present the words on the slate with the one character command of an
apostrophe.
You can ask WHAT to report the 10-highest opening
horizontal opening plays with one query with a few subcommands. Here is
the query and its resulting presentation:
WHAT?: aefknot /|2 /@*~ /<$ $ 10'
(first 10 items only)
KENAF__34p@8D FAKE___22p@8E KEF____20p@8F TAKEN__20p@8D OFTEN__18p@8D
KOFTA+_34p@8D KAF____20p@8F OAKEN__20p@8D TOKEN__20p@8D KANE___16p@8E
If you type the above command, before you press the Enter
key, look at the Scoring tab in the upper right corner
of the WHAT window and both the
Query and
Presentation tabs at the left. You will see the meaning of
the command by the yellow-colored areas. Recall that you need not have
had to construct the command from your knowledge; the WHAT
GUI can be used to construct the command for you.
These are the individual subcommands in the above command:
| /|2 | - | filter based on word lengths - they are at least 2 |
| /@*~ | - | score for a play which include the center square |
| /<$ | - | sort the presentation by decreasing scores |
| $ | - | include scores in the slate presentation |
| 10' | - | present the first 10 words only |
WHAT?: 3.r4.,acilmt? METRICAL
What words have both a prefix of
OVER
and UNDER?
under.*\
over[6-|] #
There are 330 such words, and here are the first 16:
OVERACHIEVE OVERACHIEVER OVERACT OVERACTIVITIES
OVERACHIEVED OVERACHIEVERS OVERACTED OVERACTIVITY
OVERACHIEVEMENT OVERACHIEVES OVERACTING OVERACTS
OVERACHIEVEMENTS OVERACHIEVING OVERACTIVE OVERAGE
What words which start with
OVER do not also start with
UNDER?
under* \
over[6-|] /1w\
over* \
/y0=0-1
There are 1440 such words, and here are the first 15:
OVERABLE OVERACCENTUATE OVERACTIONS
OVERABSTRACT OVERACCENTUATED OVERACUTE
OVERABUNDANCE OVERACCENTUATES OVERADJUSTMENT
OVERABUNDANCES OVERACCENTUATING OVERADJUSTMENTS
OVERABUNDANT OVERACTION OVERADVERTISE
What 8-letter words can be formed by adding an
S
to the front of a 7-letter word, such as
SCATTIER?
7.\ // get all the 7's
s[]# // for each one, front hook an S
There are 428 such words.
Now, you should try this for 8-letter words which start with a
G:
g7.\ // get all the 8's starting with G
s[]# // for each one, front hook an S
There are 2 such words.
What are the 5-voweled 8-letter words that start with
Z,
such as
ZOOMANIA?
z7.,5(V)2(C)
There are 5 such words:
ZABAIONE ZOOECIUM ZOOGLEAE ZOOGLOEA ZOOMANIA
What are the 7-letter palindromes, such as
REVIVER?
....(R-2)(R-4)(R-6)
This results in these 8 words:
DEIFIED HALALAH REIFIER REPAPER REVIVER ROTATOR SEMEMES SLAHALS+
What 7-letter words can be spelled backwards, including palindromes and
non-palindrome words, such as
GATEMAN?
7.\ // get all 7-letter words
[~] // reverse each one and retain those which are words
There are the resulting words:
DEIFIED DESSERT NAMETAG REKNITS REVILED SALLETS STELLAS TRESSED
DEIFIER GATEMAN REIFIED REKNOTS REVIVER SEMEMES STINKER
DELIVER HALALAH REIFIER REPAPER ROTATOR SLAHALS+ STONKER+
What are the 35 most-likely 7-letter words?
7. /<% 35'
Here is the answer:
ARENITE ANEROID ANESTRI INERTIA RELIANT RUINATE TRENAIL
RETINAE ATONIES ANTIRED LATRINE RETAINS STAINER URANITE
TRAINEE ELATION ANTSIER NASTIER RETINAL STEARIN URINATE
AILERON ERASION DETRAIN RATINES RETINAS TAURINE AERIEST
ALIENOR TOENAIL GOATIER+ RATLINE RETSINA TRAINED ALIENER
You could ask for the same words with the probabilities showing by suffixing
a percent sign to the above query, and this is the resulting presentation:
ARENITE__1154736w TOENAIL__1119744w RATINES___839808w STEARIN___839808w
RETINAE__1154736w ANESTRI___839808w RATLINE___839808w TAURINE___839808w
TRAINEE__1154736w ANTIRED___839808w RELIANT___839808w TRAINED___839808w
AILERON__1119744w ANTSIER___839808w RETAINS___839808w TRENAIL___839808w
ALIENOR__1119744w DETRAIN___839808w RETINAL___839808w URANITE___839808w
ANEROID__1119744w GOATIER+__839808w RETINAS___839808w URINATE___839808w
ATONIES__1119744w INERTIA___839808w RETSINA___839808w AERIEST___769824w
ELATION__1119744w LATRINE___839808w RUINATE___839808w ALIENER___769824w
ERASION__1119744w NASTIER___839808w STAINER___839808w
What are the 35 most-likely 7-letter racks
that have 7-letter words?
7. /Q;W /<% 35'
Here is the answer:
AEEINRT AEINOST AEINRST AEEILRT ADEILOR AEILOST EINORTU
ADEINOR ADEINRT AEINRTU AEEINST ADEINOS AENORST AEINNOT
AEILNOR AEGIORT AAEEINT AEEIRST ADEIORS AENORTU AEENORS
AEILNOT AEIINRT AEEILNR AADEIOR ADEIOST EILNORT AEEORST
AEINORS AEILNRT AEEILNT AAEILNO AEILNOS EINORST EEILNOR
You could ask for the same "words" with their anagrams showing by
suffixing a right parenthesis to the above query, and this is the
resulting presentation:
AEEINRT (ARENITE,RETINAE,TRAINEE)
ADEINOR (ANEROID)
AEILNOR (AILERON,ALIENOR)
AEILNOT (ELATION,TOENAIL)
AEINORS (ERASION)
AEINOST (ATONIES)
ADEINRT (ANTIRED,DETRAIN,TRAINED)
AEGIORT (GOATIER+)
AEIINRT (INERTIA)
AEILNRT (LATRINE,RATLINE,RELIANT,RETINAL,TRENAIL)
AEINRST (ANESTRI,ANTSIER,NASTIER,RATINES,RETAINS,RETINAS,RETSINA,STAINER,STEARIN)
AEINRTU (RUINATE,TAURINE,URANITE,URINATE)
AAEEINT (TAENIAE)
AEEILNR (ALIENER)
AEEILNT (LINEATE)
AEEILRT (ATELIER)
AEEINST (ETESIAN)
AEEIRST (AERIEST,SERIATE)
AADEIOR (AERADIO+)
AAEILNO (AEOLIAN)
ADEILOR (DARIOLE)
ADEINOS (ANODISE+)
ADEIORS (RADIOES+,ROADIES)
ADEIOST (IODATES,TOADIES)
AEILNOS (ANISOLE)
AEILOST (ISOLATE)
AENORST (ATONERS,SANTERO,SENATOR,TREASON)
AENORTU (OUTEARN)
EILNORT (RETINOL)
EINORST (NORITES,OESTRIN,ORIENTS,STONIER)
EINORTU (ROUTINE)
AEINNOT (ENATION)
AEENORS (ARENOSE)
AEEORST (ROSEATE)
EEILNOR (ELOINER)
*led \ [1-4|]lled \ [1-4|]
What words that end in
INGS
are not also words without the
S,
such as ABLINGS?:
*ings\/1w // put ...ings words into wordlist 1
*ing\ // get all ...ing words
[]s\ // get all ...ings words based on ...ing words
/y0=1-0 // take the difference to produce the answer
There are the 9 resulting words:
ABLINGS BEESTINGS CHITTERLINGS EYESTRINGS HUSTINGS
BEASTINGS BIESTINGS EMPTINGS GAYWINGS
How many bingos are there with
LRAIOU and a blank,
and if any, what letters can the
blank be, and then what is the first letter of each of the words, and
then what are the words?
WHAT?: lraiou?#
Number of words = 1
WHAT?: =
The one blank can be only this letter: X
WHAT?: 1/1
Letter 1 of word 1 is: U
WHAT?: '
UXORIAL
Generate a random rack with a 50% chance of having a bingo, such as
AEIORST.
This particular rack has no bingos, but it is a likely one. First
perform these commands:
/.50P // set likelihood to 50%
\! // set slate presentation to show nothing
/."ABCDEFGHIJKLMNOPQRSTUVWXYZ?" // use alphagram order for random racks
Then, request a random rack by performing the command consisting of
merely one period, and you may be shown:
WHAT?: AEIORST
You can then ask to see the answer by performing the command consisting
of merely one apostrophe, and WHAT will report:
(nothing)
colored red.
What 7-letter words can be played above and parallel to
RETAINS,
also forming 7 two-letter words?
(.r) (.e) (.t) (.a) (.i) (.n) (.s) /p
The answer is:
AMENTIA ANALGIA EPINAOI OPUNTIA OTALGIA
Going one step further, if
OPUNTIA
is placed above, what can then be placed below?
(or.) (pe.) (ut.) (na.) (ti.) (in.) (as.) /p
The answer is:
ARABESK CRAVENS GRABENS STAMENS SWEVENS
Newbies want to know about this subject, and here is how
you can find out using WHAT:
*q(!u)* \ // get all Q words where the next letter is not a U
*q /+s // augment the slate with words ending in Q
WHAT presents these words as a result:
BUQSHA KAMOTIQS+ QABALAHS QAMUTIKS+ QIBLAS+ QIVIUT SHEQEL
BUQSHAS MBAQANGA QABALAS QANAT QIGONG+ QIVIUTS SHEQELS
BURQA MBAQANGAS QADI QANATS QIGONGS+ QOPH SUQ
BURQAS NIQAAB+ QADIS QAT QINDAR QOPHS SUQS
CINQ+ NIQAABS+ QAID QATS QINDARKA QULLIQ+ TRANQ
CINQS+ NIQAB+ QAIDS QAWWALI+ QINDARS QULLIQS+ TRANQS
FAQIR NIQABS+ QAJAQ+ QAWWALIS+ QINTAR QWERTY UMIAQ
FAQIRS QABALA QAJAQS+ QI QINTARS QWERTYS UMIAQS
KAMOTIQ+ QABALAH QAMUTIK+ QIBLA+ QIS SHEQALIM
You might want to know those longer words, such as 7's or longer which
contain a certain string inside (not starting or ending the longer word.
Let's say the word is LWEI
and we are looking for 6's or longer:
*?LWEI?*
The answer is:
EDELWEISS EDELWEISSES
Try the same kind of query for the base word of POLICE. If you have not yet seen this before, you may be surprised at the result.
Someone at the Lexington Scrabble Club picked up a rack with four
I's,
and we considered what are the possible bingos.
WHAT can do this in several ways, the simplest of which is:
4I/S
which means what are the shortest steals where rearrangement is not
required. If rearrangement is required (using the command
4I/~S),
there are no answers, since they all have four
I's, with other letters
inserted.
A second way to do this is to try various numbers of blanks until there is
an answer. You might try
4I4?'s and see if anything
comes out, and then
press the F1 function key
to watch more blanks get added, one for each press. If you want
to see all words with four blanks, you can do it this way:
4I!*/A
The exclamation point forces the
four I's
to be in the answer. The asterisk
means any number of blanks, and since the asterisk implies pattern
matching, you then include
/A
to make the query kind anagramming.
It turns out there are more than 1000 such words in
WHAT's OWL3 lexicon.
SATINE
plus a blank produces many words; and how many is that? With the query:
SATINE?#
The answer is:
Number of words = 73
But how many different letters can the blank be? The easiest way you
can ask this is with this query:
SATINE?#=
and WHAT reports:
The one blank can be any of 24 different letters.
Another way would be to bring back the previous command by pressing the
up-arrow key and then suffix the subcommand
/Q;W
to yield words in canonical order. This query results in:
Number of canonically-ordered words = 24
Although you can type the
=
command to see what letters the blank can
be and by observation find the 2 missing letters, here is how you can
explicit list those letters:
/1U=(A) // define set 1 to be one each of all letters
/2U=(=) // define set 2 to be those letters the blank matched
/3U=(1)-(2) // define set 3 as the difference between these sets
(3) // show set 3
You will see that the letters are
Q and
Y.
Steve Hartsman posted a message to CGP about the rack
ACDEINT.
Steve listed various data relating to that rack.
The following WHAT output
shows how you get the same info using the program:
WHAT?: acdeint
(nothing)
WHAT?: acdeint?<>:
C: ACCIDENTs
I: ACTINIDEs CTENIDIA vINDICATEds
K: ANTICKED
M: MEDICANTs
N: INCANTED
O: ACTIONED+ CATENOIDs
P: PEDANTIC
R: DICENTRAs
S: DANCIEST+ DISTANCEds
T: NICTATED
U: INCUDATE
WHAT?: =
The one blank can be any of these letters: CIKMNOPRSTU
WHAT?: acedintu"
INCUDATE - INCUDATE adj incudal
WHAT?:
WHAT is command language oriented, and there is a graphical interface which can optionally be used to construct commands. In the above excerpt from the WHAT workspace, you see the prompt by the program on several lines, which is "WHAT?: ", and the remainder of the command line is the command you typed.
In the first command, as a result of your typing the 7 letters and pressing the Enter key, WHAT came back with the response of " (nothing)" colored red, which means there are no 7-letter words with those 7 letters.
To form the second command, you could have pressed the up-arrow key to get back the previous command and then suffix these four characters:
| ? | - | meaning a blank |
| < | - | meaning "show front hooks (in lowercase)" |
| > | - | meaning "show back hooks (in lowercase)" |
| : | - | meaning "present the output using colons" |
In the third command, you typed the equal sign followed by pressing the Enter key, which caused WHAT to report on what the blank in the previous query can be.
To form the fourth command, you pressed the up-arrow key thrice to bring back the original command, and then you suffixed a U and quotation mark. When you pressed the Enter key, WHAT presented the definition for INCUDATE. You can then use the mouse to double-click "incudal" to see its definition in the Definitions area above.
The following query causes WHAT to present words with a Q which have anagrams:
Q4?/&>1
To get the 8-letter words with words which have the vowels
A,
E,
I,
O,
U
and 3 consonants, give this query to WHAT:
(@V)3(C)
The answer is:
AEQUORIN CAESIOUS+ EQUATION EUPHORIA JALOUSIE SEQUOIAS
AEROBIUM DIALOGUE EUDAIMON EUSOCIAL OUTRAISE THIOUREA
AGOUTIES EDACIOUS EULOGIAS EXONUMIA SAUTOIRE
From John Van Pelt's Verbalobe web site,
there was a list of those 7- and 8-letter
words which can be extended with TY
There are seven 7-letter words and five 8-letter words
which have this characteristic. Here is how to ask WHAT
to present the 7's:
7.\ // first get all the 7's to the slate
[]ty // when TY is suffixed, what are words?
The 8's can be done in the same way.
From John Van Pelt's Verbalobe web site,
there are 64 words ending in either
ED or
T:
*ed#
says there are 14652 words, and
*t#
says there are 9448 words.
So begin with the smaller second group, and then
ask WHAT to perform:
[1-2|]ed
which yields 366 words.
10(A-G)/|5
Collins is the lexicon which includes the words from North America and Great Britain. This lexicon is used in most countries throughout the world and for world competitions, but it has not yet been accepted for all competitive games in North America.
Perhaps you want to look at 2-letter words from the Collins 2012 Lexicon. You probably would like to be shown which of these are from the British list only, such as with a suffix of a number sign. You must use the WHAT graphical interface to alter the lexicons - there are no commands to change the lexicons for you to present in the workspace.
Select the Lexicons tab in the upper right corner of the WHAT window. Click on the Change Lexicon(s) button. A dialog appears a the lower right corner of the WHAT window. Select "Collins Words 2012 ..." as the primary lexicon and "OWL3 ..." as the secondary lexicon at the bottom of the dialog. In the upper part of the dialog, select the middle option which says to identify words NOT also in the secondary lexicon with some suffix character. Then go the right area of the dialog and select a number sign as the suffix character. Click the OK button and you are now going to see answers to queries coming from Collins, and words not in OWL3 will be suffixed with a number sign. For example, take a look at the 2-letter words with a command such as two periods; you are then shown:AA BA EM HO MA OH QI US AB BE EN ID ME OI RE UT AD BI ER IF MI OM SH WE AE BO ES IN MM ON SI WO AG BY ET IO# MO OO# SO XI AH CH# EX IS MU OP ST# XU AI DA FA IT MY OR TA YA AL DE FE JA# NA OS TE YE AM DI# FY# JO NE OU# TI YO AN DO GI KA NO OW TO YU# AR EA# GO KI NU OX UG# ZA AS ED GU# KO# NY# OY UH ZO# AT EE# HA KY# OB# PA UM AW EF HE LA OD PE UN AX EH HI LI OE PI UP AY EL HM LO OF PO UR#
You can see those 19 2-letter words acceptable only from Collins 2012. If you want to make a list of just these words, save the slate to a wordlist. Change the lexicon settings to produce answers from OWL3, get the OWL3 twos and remove these from the wordlist you saved.
As suggested by Joe Edley, let's say you have produced a random rack, and it has been input as a query. What can you now do to find out its subwords?
For the random rack of UMDNEEO, you are shown the answer of EUDEMON. Now, press the up-arrow key and then hold down the Ctrl key and press the digit 3. You will be shown all those words of at least length using the seven letters in UMDNEEO.
When you hold the Ctrl key and press 3, these command characters are suffixed to the query: " /|3".
Wendy Wolfberg asked how to spell "upgradable",
and I first typed that
in to find it is a word. Then I pressed the up-arrow key,
typed an E,
and pressed the Enter key
to see if it could be UPGRADEABLE,
and indeed it is also a word. I did the same for
GRADABLE and found the
E there is
not allowed. So, I wonder what other words of this form can have the
optional E. Here is
some WHAT input and output:
*eable# // get all EABLE words onto the slate; see how many
The answer is 139 words.
[1-6|]able // for each word on the slate get all but its final
// 5 letters, add ABLE, and which of these are words?
The answer is these 57 words:
ATONABLE FRAMABLE MOVABLE SHAPABLE TYPABLE
BITABLE GRAZABLE NAMABLE SHARABLE UNLIKABLE
BLAMABLE HATABLE PLACABLE SIZABLE UNNAMABLE
CITABLE HIRABLE POSABLE SMOKABLE UNSHAKABLE
CLOSABLE LIKABLE RAISABLE SPARABLE UPGRADABLE
DATABLE LINABLE RATABLE STATABLE USABLE
DIAGNOSABLE LIVABLE REMOVABLE TAKABLE VOTABLE
DISLIKABLE LOVABLE RIDABLE TAMABLE WADABLE
DRAPABLE MACHINABLE SALABLE TASTABLE WRITABLE
DRIVABLE MAKABLE SAVABLE TOTABLE
DYABLE MICROWAVABLE SHAKABLE TRADABLE
FINABLE MINABLE SHAMABLE TUNABLE
Then, just for a little more investigation, I did this:
[]/&2 ) // for each word with at least 2 anagrams, show them
Here is the answer:
DATABLE (ABLATED)
DYABLE (BELADY)
FRAMABLE (FARMABLE)
RIDABLE (BEDRAIL,BRAILED,RAILBED+)
SMOKABLE (ABELMOSK)
SPARABLE (PARABLES,PARSABLE,PREBASAL)
STATABLE (ABETTALS,TASTABLE)
TASTABLE (ABETTALS,STATABLE)
TUNABLE (ABLUENT)
USABLE (SUABLE)
What words have both a
J and a
Q?
The easy command to ask for this is:
j!q!*/a
The suffix exclamation points indicate that
J and
Q
are required to be in the answer. The asterisk represents
any number of blanks, but since it also implies the query is a pattern
match kind, you must also change it to be an anagram kind of query, and that
is what the /a does.
The answer is:
JACQUARD JACQUERIE JEQUIRITIES JONQUIL QAJAQ+ QUILLAJA
JACQUARDS JACQUERIES JEQUIRITY JONQUILS QAJAQS+ QUILLAJAS
Another query (of the same length) to accomplish the same thing is:
*,j!q!*
This is a indeed a pattern match query for all words, and the rack
includes the two required letters of
J and
Q
plus any number of other letters.
Issue this query:
x6?/&>1 // get the 46 7-letter words with anagrams
Now ask WHAT to present these in pairs. One way you can
do this is to ask it to present output in rows with two columns and
sort by canonical order:
/ORC9W /OC2C />; // produce the desired answer
What 7-letter words starting with a Z and ending in S cannot drop the S to form a 6-letter word?
John Chew listed these answers for the TWL98 Lexicon:
zealous zilches zincous zipless zloties zygosis zymoses zymosis
and ZAIDIES
is new in OWL3.
To do this in WHAT:
z5.s\ // get the 7's with start with Z and end with S
/1w // save to wordlist 1
[1-6]\ // get the 6's which are acceptable when dropping the S
[]S\ // put the S back on these 6's
/y0=1-0 // produce the answer by taking the difference
' // present the answer of 9 words
Michael Eldeiry announced he was seeking a list of all words of length 2 through 8 which do not take an S as a back hook. There are several ways to do this in WHAT, and here are two of them. Beware that the answer is rather large, so you may want to do this work in pieces. For example, let us begin with just the 3-letter words.
Get words onto the slate for which you want to ask this question. If
you first just want a list of the threes, issue the command:
3.\
Then make the query using these words as input such that the retained words
are those set of back hooks is a subset of the set with all letters but
S:
[]/><=(!S)
These are the 249 such threes:
AAS CRY FOH IFS NOH POX SOX WHA
ABS CUZ+ FOR INS NOO PRY SPY WHO
ADS DAS+ FOU ITS NOR PST STY WIZ
ADZ DEF FOX IVY NOS PYX SYN WOS
AFF DEX FRO JEU NOT QIS TAE WRY
AFT DID FRY JUN NTH QUA TAJ WUD
AGO DUH GAN JUS NUS RAH TAX WUZ+
AGS DUI GAS KAS OCH+ RAJ TES+ XED+
AHA DUM+ GEY KEX ODS RAN THE XIS
AHS EAU GIS+ LAX OES RAS THO YAH
AIS EDS GOR LED OFT RAX THY YAR
ALS EEK GOS LEU OHO RES TIS YAS+
ANY EEW+ GOT LEX OHS REX TIX+ YEH
APT EFS GOX LEZ OMS REZ+ TIZ+ YER+
ARS ELF GRR+ LIS ONS SAD TOO YES
ASH ELS HAD LOR+ OOF+ SAE TRY YET
ASS EMS HAJ LOX OPS SAT TUX YEZ+
ATT ENS HAO LUX ORA SAU UMM YOM
AVA ERE HAS MAX ORS SAX UMS+ YON
AWA ERS HES MEH+ OVA SEN UNS YUM
AYS ESS HEX MEN OXO SEX UPO ZAS
AZO FAR HEY MET OXY SEZ+ UPS ZAX
BAH FAS HIC MIM PAH SHA UTS ZOA
BES FAX HID MIX PAX SHH VAS ZUZ
BIS FER HMM MMM+ PER SHO+ VEG ZZZ
BIZ FEW HOO+ MOI+ PES SHY VEX
BOX FEY HOS MUX+ PHT SIS VIA
BRR FEZ HUH NAH PIU SIX VIS
BYS FIE HUP NAM PIX SKY VOX
CIS FIX ICY NAW PLY SLY VUM
COX FIZ IDS NEE POH SOC+ WAS
COZ FLY IFF NIX POS+ SOS WAX
Here is a completely different way to get the same answer:
3.\ // Start with the candidate words, such as the threes:
/1w // save these to wordlist 1
[]s\ // create the fours which are S extensions of these threes
[1-1|]\ // remove the S, so these are the threes with take S
/y0=1-0 // take the set difference (all minus those which take S)
If you want to make the full list Michael asked for, start with
the following command to have all words of length 2 through 8:
*/|2-8\
This is 88099 words. Once again the second query to yield the final answer is:
[]/><=(!S)
The result has 55031 words, and here are the first 24 of them:
AAHED AALS AARRGHH ABACK ABAKAS ABAPICAL
AAHING AARDWOLF AAS ABACUS ABALONES ABAS
AAHS AARGH ABACAS ABACUSES ABAMPS ABASED
AALIIS AARRGH ABACI ABAFT ABANDONS ABASEDLY
Steve Root notices BUQSHA and BUGSHA are both words. What other word pairs are there with a G or a Q?
Solve this for words up to length 8:
Q!7?/|3-8# // get all words with a Q of length 3-8
Using OWL3, the answer is 1069 words.
/m"abcdefghijklmnopgrstuvwxyz" // define the mapping to convert Q to G
[m] // map each slate word
The answer consists of these 27 words:
BUGSHA GAT GUESTED GUIDS GUILTS PIROGUES ROGUES
BUGSHAS GATS GUESTING GUILT GURSH PLAGUE TOGUE
GADI GIS+ GUESTS GUILTED+ GURSHES PLAGUES TOGUES
GADIS GUEST GUID GUILTING+ PIROGUE ROGUE
Let's look for 6-letter words which are repetitious in that they
are the same 3-letter word duplicated, such as
POMPOM:
3.\ // get all 3-letter words onto the slate
2[] // get words which consist of 2 of the same slate word
The answer is:
BEEBEE CANCAN COOCOO FURFUR PALPAL POMPOM TESTES WEEWEE
BOOBOO CHICHI DUMDUM MOTMOT PAWPAW TARTAR TSKTSK
These above two commands yield those 6-letter words which consist of a
repeated 3-letter word. But notice a word like
ATLATL is missing,
since ATL
is not a 3-letter word. There are at least two methods to
get such words:
3./qs\ // get all 17576 3-letter strings onto the slate 2[] // get words which consist of 2 of the same stringThere are 33 such words, 15 of which are the ones formed with repeated 3-letter words.
6.\ // get all 6-letter words onto the slate
[1-3]/QS\ // for each 6-letter word, yield a 3-letter string (not
// necessarily a word) from the first 3 letters
2[] // for each of those 3-letter strings, yield words
// which consist of 2 of these strings
ATLATL BOUBOU CUSCUS GRIGRI MUUMUU SIKSIK+ TZETZE BEEBEE BULBUL DIKDIK GRUGRU PALPAL TARTAR VALVAL BERBER+ CANCAN DOODOO KUMKUM+ PAWPAW TESTES WEEWEE BONBON CHICHI DUMDUM MOTMOT POMPOM TSETSE BOOBOO COOCOO FURFUR MURMUR SARSAR TSKTSKAssume the answer of the above 33 words, including ATLATL, is on the slate. Now, let's find which of these have a consonant as the second letter. Here are two different methods of doing this:
/1w // save the slate to wordlist 1 .(C)4. // get all 6-letter words with a consonant as letter 2 /y0=0&1 // yield the intersection of the 2 lists
/0I .(C)4. // get the 6's with a consonant as letter 2
ATLATL CHICHI GRIGRI GRUGRU TSETSE TSKTSK TZETZE
What 6-letter words consist of two consecutive 3-letter words, such as STYRAX?
There are many ways to approach this, and this is just one of them:
3.\ // get all the 3's
[]3./1w\ // for each, suffix 3 letters to form a word; save it
3.\ // get all the 3's again
3.[]\ // for each, prefix 3 letters to form a word
/y0=0&1 // intersect the two lists
28' // present just the first 28 of these
There are 3626 such words. Here are the first 28:
ABAKAS ABATIS ABSORB ACETUM ADDEND ADOBES ADSORB
ABASER ABATOR ABSURD ACTINS ADDERS ADOBOS AFFAIR
ABATED ABAYAS ACETIC ACTION ADDLED ADORED AFFLUX
ABATES ABOMAS ACETIN ACTORS ADDLES ADORES AFFRAY
What 8-letter words can be formed by concatenating two 4-letter words?
WHAT does not have a way to iterate over two lists, but it can be used to produce the answer. Here are two completely different ways:
8.\ // get all 8's [1-4]/#=1 \ // yield those 8's which begin with a 4 [5-8]/#=1 \ // from previous, yield those 8's ending with a 4 /0I z7. // use slate as source; get those starting with Z
4.\/4w // put all the fours into wordlist 4
Z7.\ // get all the 8's starting with Z
/4I[]/2p // using the 4's as the word source, do 2-word pattern matching
// with the words on the slate (the 8's)
[] // combine the pairs of 4's into 8's
ZESTLESS ZINCATES ZIZZLING ZONETIME ZOONOSES ZOOSPERM ZILLIONS ZINCKING ZONELESS ZOOMABLE+ ZOONOTIC ZOOSPORE
You are presented with a quiz to find all the 4-letter words in
OATEN.
You think you have the answer, and you want to check it to see if you
found all the words. Make this probe to find out how many words are in
the answer:
OATEN/|=4#
In the above command, the subcommand
following the letters is a filter that the word length must be equal to
four. But if you are unsure of the query you made, and you want to
reassure yourself the words just counted are indeed of length four,
show the slate with the words hidden and with word lengths using
this command:
]|
You could have had the
WHAT GUI make up this command by going to the
Presentation
tab on the left side of the WHAT window, and clicking the
radio button labeled "Hidden, using [] brackets" and the option button
labeled "Lengths". The presented result is:
[].4c []_4c []_4c []_4c []_4c []_4c []_4c []_4c
which should be reassuring that there are indeed 8 4-letter words on
the slate. If you want to check whether a word you have in mind is
indeed a word, type a command such as:
eton/-?
are you will be told whether
ETON
is an acceptable word. You could at
this point, ask whether a word is on the slate, such as:
nota/?
You will be told just whether
NOTA
is on the slate. If you instead ask:
nota/0?
when the word is on the slate, you are also told at what that item number
the word is; to make that datum have no meaning you could first shuffle
the order of the slate using the command:
/0S
where that second character is the digit zero.
But when you then ask for whether a word is on the slate, you should
turn off alphabetical sorting, so you could issue the command:
nota/0?/='
The final subcommand in the above command could be created for you if
you go to the Sorts area of the
In "The Complete Wordbook for Game Players",
author Mike Baron suggests
there are so many 5-letter words that one should approach these in
groups. After studying those with high-scoring letters, it then makes
sense to study those with a first or fifth letter of score 4 or 5.
Mike calls these "The High Fives". This list of high fives,
which can be made by WHAT,
is rather large, and you may want to break it up into
smaller groups, such as those high fives which begin with
A.
Here is the command to get these words:
A3.($4-5)
There are 48 such words. If you would like to present them with related
info, such as anagrams, WHAT
can do this for you, and here is how this
looks for those high fives starting with the letter
O:
OBEAH__(BOHEA) OLOGY__(GOOLY+) OOMPH OUTBY
OCHRY ONERY ORACH__(ROACH) OVARY
ODDLY ONLAY ORACY+ OXBOW
Joey Mallick mentioned that
GAYWINGS
a perennial herb) must have
the final
S.
We wondered whether there are other wuch words ending
in "WINGS".
This is an example of when to use the "unhooks" feature
of WHAT. Ask
WHAT to show unhooks, and you can see that indeed that
is the only such word. With this query:
*wings`
where that final character is the back quote or accent grave, the
result is:
BACKSWINGS- FOREWINGS- MICROBREWINGS- -SWINGS-
BEESWINGS- GAYWINGS MOWINGS- TELEVIEWINGS+-
BELLOWINGS+- GLASSBLOWINGS- OUTSWINGS- THAWINGS+-
BITEWINGS- GNAWINGS- OVERBLOWINGS+- UNDERDRAWINGS+-
BORROWINGS- GREENWINGS- OVERSWINGS- UNDERWINGS-
BOWINGS- HAIRWINGS+- PARAWINGS- UNKNOWINGS-
BREWINGS- HINDWINGS+- PIGEONWINGS- UPSWINGS-
CLEARWINGS- INFLOWINGS+- REDWINGS- VIEWINGS-
DOWNSWINGS- KNOWINGS- ROWINGS- WAPPENSCHAWINGS-
DRAWINGS- LACEWINGS- SEWINGS- WAXWINGS-
ELBOWINGS+- LAPWINGS- SHOWINGS- WHITEWINGS-
FLEXWINGS+- LAWINGS- SOWINGS+- WINDOWINGS+-
FOLLOWINGS- LOWINGS- SWEPTWINGS- WINGS-
All of these words other than
GAYWINGS have a suffix of
"-",
which in an indicator you can unhook the final letter.
If the list of
...WINGS
words were longer, you might prefer to answer
this kind of question in a different manner. Here is another way to
accomplish this:
*wings\/1w // collect all the words and copy them to wordlist 1.
[1-2|]\ // for each of these words, keep those which can drop S
[]s\ // for each of these, put the S back on
/y0=1-0 // take the set difference, and output to the slate
The output from the last command indicates the slate has one item. To
see it, issue the one-character command of an apostrophe.
Here is a way to find the highest-scoring triple-triples of length 8:
/@1A.! // scoring will be based on starting at square 1A
8./$350/>'/<$$/OC18X // filter based on scores at least 350,
// sort by decreasing scores & alphabetical
The answer is:
MEZQUITE__392p QUIZZING__374p SOVKHOZY__365p MAHZORIM__356p
MEZQUITS__392p WHIZBANG__374p BENZYLIC__356p OXAZINES__356p
OXAZEPAM__392p HUTZPAHS__365p CHAZANIM__356p OXAZOLES+_356p
JACQUARD__383p MITZVAHS__365p EXEQUIAL__356p QUIXOTIC__356p
JAZZLIKE__383p MITZVOTH__365p EXEQUIES__356p RHIZOMIC__356p
KHAZENIM__374p QUIXOTRY__365p ISOZYMIC__356p ZYZZYVAS__356p
QUEZALES__374p QUIZZERS__365p JONQUILS__356p
But that is not enough, since those starting at 1H also deserve to be
considered, and these are the results:
QUINZHEE+_401p PIROZHOK__374p JACUZZIS__365p MACHZORS__356p
BEZIQUES__392p QUEAZIER__374p LYSOZYME__365p POLYZOIC__356p
CAZIQUES__392p QUETZALS__374p PUMPJACK+_365p RIBOZYME__356p
CHUTZPAH__383p QUINZIES+_374p QUIZZERS__365p SOVKHOZY__356p
JANIZARY__383p QUIZZING__374p CHUTZPAS__356p WHEEZING__356p
FROWZILY__374p WHEEZILY__374p COENZYME__356p YOKOZUNA__356p
HIGHJACK__374p BLOWZILY__365p DENAZIFY__356p ZABAJONE__356p
PIROZHKI__374p HYLOZOIC__365p FLAPJACK__356p
Find the top-scoring 21-letter words when the horizontal placement
of the word starts at position 1A on a board for the
21 x 21 board game:
/21B // set the board size to 21 x 21
/@1A.! // score horizontal words which start at the upper left square
21./$>8400/>'/<$ $ /o1 // find all 21-letter words which score more than 8400, sort
// by decreasing scores and show with scores
The answers from the WJ2 Lexicon are:
CARBOBENZOXYDIPEPTIDE_9266p
HYDROXYFORMOBENZOYLIC_9122p
CARBOXYMETHYLPACHYMAN_8978p
CHLORHYDROXYQUINOLINE_8978p
ACETYLPHENYLHYDRAZINE_8834p
IODOCHLORHYDROXYQUINS_8546p
BENZYLISOTHIOCYANATES_8402p
BISDEMETHOXYCURCUMINS_8402p
PHTHALYLSULFATHIAZOLE_8402p
Knowing that
UXORIAL
is the only 7-letter word which can be made from
U?ORIAL,
what other 7's have this characteristic? There are 76 such 7's.
Do any of them have anagrams? The answer is "no".
x6?/Q;W // get the 535 7-letter canonically-ordered words with X
[]-X+?/:=(X)/Q;W# // get the 78 canoncially-ordered words that take only an X
[]/A // show these words
/<% // show these sorted by decreasing probabilities,
// seeing UXORIAL is first; perhaps some of these are
// unexpected, such as PTYAEI+X
[]/&>1 // look for those with more than 1 in the anagram set,
// and there are none - perhaps not surprising
Joev Dubach asked for a list of the 5's which are not contained in any 8's.
Here is how you use WHAT to produce the answer:
5.\/1w // get all the 5's and put then into wordlist 1
8.\ // get all the 8's.
[]/a/|=5\/2w // for each 8 find all 5's and put them into wordlist 2
// (this takes 2-3 minutes to complete)
/y0=1-2 // take the difference, and this is the answer
This yields these 127 words:
ADDAX DYKEY GAWKY JIBED KEXES NAKFA QUOTH WOOZY
BAFFY FAWNY GIZMO JIBER KHAPH NIQAB+ SKYEY WUSHU
BEEFY FEAZE GRRRL+ JIFFS KICKY NUBBY THUJA XYLYL
BEVVY+ FEMME GUPPY JIFFY KOMBU OUZOS TIZZY YECHY
BIFFY FEYLY GYVED JIGGY KOPEK PAPAW TOFFY YEUKY
BUFFY FEZZY HAFIZ JIMPY KOPJE PHIZZ+ TWIXT YOLKY
BUPPY FJELD HAJJI JIVED KUDZU PHPHT UPBOW YUCKY
BUZZY+ FLAXY HAYEY JIVEY KYLIX PONZU+ VAMPY YUKKY
BYWAY FLYBY HOKKU JOTTY LEZZY POOJA+ VROUW YUMMY
CUPPY FUDDY HUFFY JOUKS LUVVY+ PUFFY VUGGY YUZUS+
DADDY FUDGY+ JAGRA JOWLY MAMMY PUJAH VUGHS ZAPPY
DOOZY FUGGY JAMMY JUGUM MAUZY+ PUKKA WAZOO ZIPPO+
DUCHY FUJIS JAWAN JUICY MIFFY QAJAQ+ WICCA ZIPPY
DUCKY FURZY JAZZY JUMBO MOMMY QIBLA+ WIMPY ZONKS
DUDDY FUZZY JEFES JUMPY MUJIK QOPHS WOMBY ZUPPA+
DUSKY GAUZY JERKY KEMPY+ MUZZY QUIPU WONKY
Here is a better way to do the above:
5.\ // get all the 5's
[]3?/#=0 // produce the answer
The final step takes about 30 seconds.
Steve Hartsman asked for a way to take a given list of words and convert each of these into alphagrams. What was unclear is if the order of the list should be retained, and if not should duplicates be removed.
If the order of the list is to be retained and duplicates kept, then
import the given list to the slate, clear the workspace via the GUI or by
using the /CW command,
and then present the slate with canonically-ordered words and requesting
one word per line with no sorting (alphabetically), namely:
/o1/=';
Then output the workspace to a file; you will have to remove the first
two lines (the command and a blank line) and the final line (a blank line)
outside of WHAT. This is then the result.
On the other hand, if the order of the list is not to be retained and you
are removing duplicates, then after importing the input list to the slate,
execute the command:
[]/q;w
Then you can export the slate to a result file; there are no extra
lines which must be removed with this method.
An alternate way to express the notion of words which match their alphagrams is: "What words are spelled in alphabetical order?". This assumes the current canonical ordering is the default of alphagrams. Here are the commands to do this:
7./Q;W\ // get canonically-ordered 7-letter words; don't show them [] // for each, yield only words (pattern matching is implied)The answer is merely BEEFILY and BILLOWY.
In a message posted to CGP, Steve Hartsman asked what alphagrams of words of at least 7 letters spell that word reversed. His example was TRIGGED. This is similar to the previous example. All you need do is define the canonical order to be the alphabet in reverse.
/;"ZYXWVUTSRQPONMLKJIHGFEDCBA" // define this canonical order 7./Q;W\ // get the canonically-ordered 7-letter words, but don't show them [] // for each one, yield only words (pattern matching is implied)The answer is:
SNIFFED SPONGED SPOOKED SPOONED TROLLED WOOLLED SPIFFED SPOOFED SPOOLED TRIGGED VROOMED WRONGEDDo this for 8's, and TROLLIED is the only answer, and that is the longest. There are 54 such words of length 6.
We have noticed the word
ZINGARA
can end in any of
A,
E,
I, or
O.
We wonder what
other 7-letter words have this characteristic, where the only
final letters are those four vowels. Here are the commands to
find out:
6.a\ // get all 7-letter words ending in A
[1-6]./:=(AEIO) // get those 7-letter words ending in A whose last
// letter can be A,E,I,O only
This yields only
STRETTA and
ZINGARA, which answers the question.
Even if we change the = to >= the same answer is produced, which is
not surprising.
Aaron Bader thought it was interesting to see the 11-words which are in
3-anagram groups. Here is a way to get the list of alphagrams for these
words. Then you can use these as the basis for a quiz. Aaron suggests
the most interesting group is ACEHOOPRRST.
11. /Q;W /&=3
Here are the 11 racks:
RPOONMIHGCA TSRNMLIIGEA TSRONMLIIEA TSSSSNLIEEA
SPOONLIHGEE TSRNNIIEECA TSROONIEDCA VTTSONIICAA
TSONNMLIIAA TSRONMIIECA TSRRPOOHECA
At this point, you can see the answers by typing the one-character command
of a right parenthesis. This lone character in a command indicates you
want anagrams presented.
What 4-letter words require the use of a blank?
(B) 3(S) /1w#
4. #
/y 0=0-1
WHAT reports
the first query results in 4185 words, and the next one
results in 4214 words. This is the answer of these 29 words:
BAZZ+ FAFF+ HMMM+ KAKA KICK KLIK MUMM ZIZZ+
BIBB FIZZ JAZZ KAKI KIKE KONK RAZZ
BUZZ FUZZ JIZZ+ KECK KINK KOOK SHHH+
EKKA+ HAJJ JUJU KEEK KIRK KYAK TIZZ+
Another way to get the first set of words is with this query:
4.,(@B) /1w#
One more technique to get the same list is to seek wotds with a
probability of zero, since probabilities are calculated based on
letters without any consideration to blanks. This command does the trick:
4./%=0
What words contain A, E, I, O. and U, where these letters show up in alphabetical order within the word?
First, find which ones begin with
AB
and may have other vowels too:
a*e*i*o*u*
These 5 words are shown:
ABSTEMIOUS ABSTEMIOUSNESS ABSTENTIOUS
ABSTEMIOUSLY ABSTEMIOUSNESSES
Then, restrict the question to
words which start with
A,
where those five vowels are the only vowels:
a0(C)e0(C)i0(C)o0(C)u0(C)
These words result:
ABSTEMIOUS ABSTEMIOUSLY ABSTENTIOUS ARSENIOUS
If you then prefix with any number of consonants:
0(C)a0(C)e0(C)i0(C)o0(C)u0(C)
These are the words:
ABSTEMIOUS ABSTENTIOUS CAESIOUS+ FACETIOUSLY
ABSTEMIOUSLY ARSENIOUS FACETIOUS
You are provided with a rack in which there is a bingo, and you want
some hints. You can ask for specific letters of specific words, but
perhaps you want to just know whether a certain letter is involved.
For example, you may want to know whether the final letter of any of
the words in the answer is a
C,
or you may want to know this for a
specific word, such as the first one. You can do make such a query, but
it may modify the slate, so it would be advisable to copy the slate
to a wordlist before making the query, and then you can restore it
when you want to. The following query asks whether the 7th letter
of any of the 7-letter words on the slate is a
C:
[1-6]C~
If you want to know the same thing for the fourth letter:
[1-3]C[5-7]~
Having said all this, there is still the possibility that when alternate
source is supported more than for subwords, this will make such a query
easier. Let us assume a pattern match query can use source words from a
given wordlist. Then:
/1W // save the slate to wordlist 1
/1I! // use wordlist 1 as the source (long-term)
6.C~ // are there any 7-letter words in wordlist 1 which end in C?
John Morse posted to the CGP Email List some letter sets with one blank such
that the blank can be exactly
C or
G or
P.
WHAT can be use to find such answers:
6?c# // get all 7's with a C
[]-c+g/#>0# // from those words, keep those which produce a word
// when a G replaces the C
[]-c+p/#>0# // from those words, keep those which produce a word
// when a P replaces the C - there are 569 words
[]-c+?/:=(CGP) // from those words, keep those which produce words
// such that the blank is exactly C,G,P
Two words come out: VINCULA and
WACKILY.
Then, if you want to see these in canonical order:
[]/q;w // yield these answers in canonical order
[]-c/qs // yield strings from these answers less one C
Avi Moss asked for the 8's with no 7-letter subword, i.e. a 7-letter word
prefixed or suffixed with a hook letter which leads to an 8-letter
word is not included. Here are WHAT to find these:
8.\/1w // make a list of all 8's and put into wordlist 1.
7.\ // make a list of all 7's
.[]\/2w // make a list of all 8's which are formed from a 7 with
// a front hook, and put into wordlist 2.
7.\ // get the 7's again
[].\/3w // make a list of all 8's which are formed from a 7 with
// a back hook, and put into wordlist 3.
/y4=2|3 // combine wordlists 2 and 3 into wordlist 4
/y0=1-4 // yield the answer onto the slate
The answer to this is rather large.
Seth Lipkin noticed from a word-of-the-day listing that
PANMIXIA's
longest subword (when anagramming) is of length 5. He asks are there
any 8's with smaller maximum length subwords? Here is how this can
be answered in WHAT:
8.\ // get all the 8's to the slate
[]/a/|=7/#=0# // retain only 8's which have no 7's, say how many
[]/a/|=6/#=0# // retain only 8's which have no 6's, say how many
[]/a/|=5/#=0#/5w // same for 5's, and save in wordlist #5
[]/a/|=4/#=0#/4w // same for 5's, and save in wordlist #4
[]/|=3/#=0 // retain only 8's which have no 3's and show them
...
/5r' //and then go back to see wordlist 5
WHAT?: 8.\
WHAT?: []/a/|=7/#=0#
Number of items = 6351
WHAT?: []/a/|=6/#=0#
Number of items = 853
WHAT?: []/a/|=5/#=0#/5w
Number of items = 63
Wordlist 5 has 63 items.
WHAT?: []/a/|=4/#=0#/4w
Number of words = 0
Wordlist 4 has 0 items.
WHAT?: /5r'
The slate has 63 strings.
ABEYANCE CACHUCHA EPOPOEIA HOODMOLD MAHIMAHI QUINCUNX WINDMILL
ABEYANCY CARACARA EXIGUITY HOOKWORM MILKWOOD QUIZZING WOODCOCK
AMADAVAT COMMENCE FOOFARAW JACQUARD PARALLAX SUSURRUS WOODWORK
APAGOGIC CORRIDOR FUCKFACE JIPIJAPA PELLMELL THATAWAY WOODWORM
AQUAFARM CRUCIFIX FUGITIVE KAVAKAVA PICKWICK UNUNBIUM WOOLWORK
AVADAVAT DAHABIAH GANGBANG KAZACHOK PURPURIC UXORIOUS WORKHOUR
BELLPULL DUMMKOPF GUITGUIT KEFFIYEH QUANDANG VIRILITY WORMWOOD
BUZZKILL DYSPEPSY HIGHBROW LINKWORK QUIDDITY VIVARIUM ZOOTOMIC
BUZZWORD EFFICACY HOMEOBOX LOOKDOWN QUIDNUNC WIFELIKE ZUGZWANG
So the above list are those 8's with 4's as the longest subwords.
In the presentation of the 63 strings, you are not shown which are new words in OWL3. To see this, issue the command consisting of an an open and closed square brace. You will see two of these words are new.
Then, Seth followed up by noticing in the above list that all the words
have repeated letters, and there is a small number of these with only
a single repeat, such as
DUMMKOPF.
We could first run a check to be
sure we did not miss something - which words have no repeated letters
(if any)? Here is one way to answer this in WHAT:
.7(-)\/8w // put 8's with no repeated letters into wordlist 8
/y0=5&8 // the slate gets the answer, lists 5 and 8 intersection
WHAT
can be used to find the words in the above list with have exactly
one repeated letter. It is a small variant of the above scenario:
?6(-)(+)/8w
/y0=5&8
The set (-)
represents a non-repeated letter, and the set
(+)
represents a repeated letter.
Wayne Yorra wondered what two letters combine with 5 blanks to
produce the shortest list of 7-letter words. Let's look for those
that produce less than 3:
../q;s# // get all 351 canonically-ordered digrams onto the slate
[]5?/#<3
This yields the answer:
JQ QQ QX
Then you can see the words with:
[]5?
This produces:
EQUINOX JONQUIL QUIXOTE QULLIQS+S
This question arises on a regular basis: What are the longest words with
no repeated letters? This is very easy to answer using
WHAT, since there is a
set which represents a first-time letter of a word. The only problem is to
formulate a query in which you cover multiple word lengths. Let's say you
guess the answer is at least 10. You can ask for words of length to the maximum
length of 26 with this query:
26(+) /|10 /<| |
The subcommands indicate:
/|10 // filter to yield words of at least length 10
/<| // sort by decreasing length
| // present lengths along with words
The above query yields 1526 words, so WHAT brings up the dialog to have
you indicate how many you want to see. You can click on the option to
see the first 25 words, and these are they:
DERMATOGLYPHICS_15c MOTHERFUCKING___13c
UNCOPYRIGHTABLE_15c MULTIBRANCHED___13c
AMBIDEXTROUSLY__14c SUBORDINATELY___13c
DERMATOGLYPHIC__14c TROUBLEMAKING___13c
TROUBLEMAKINGS__14c UNPREDICTABLY___13c
CONSUMPTIVELY___13c UNPROBLEMATIC___13c
COPYRIGHTABLE___13c AMBIDEXTROUS____12c
DOCUMENTARILY___13c AMBLYGONITES____12c
ENDOLYMPHATIC___13c AMYLOPECTINS____12c
FLOWCHARTINGS___13c BANKRUPTCIES____12c
HYDROMAGNETIC___13c BLUESTOCKING____12c
LYCANTHROPIES___13c CABINETWORKS____12c
METALWORKINGS___13c
If you are then curious what is the last word in the list, you can type
the command:
/>| /1 // sorting by increasing length, what is the first word
You will be shown it is
ABDUCTIONS; so this is the alphabetically
first 10-letter word with no repeated letters.
Steve Root asked what is the longest word with no 1-point consonants. Of course, that would be these consonants only: BCDFGHJKMPQVWXZ.
Two methods are included here:
/1U=($2)|(V) // define set 1
(1) // look at set 1 to be sure
0(1)/<||12' // query for any number of set 1 letters, and
// sort by decreasing length, show length, and
// the first 12 words
Here is the answer:
(first 12 words only) DEHUMIDIFIED_12c BODYCHECKED__11c MACADAMIZED__11c BACKCHECKED__11c COMMODIFIED__11c PADDYWACKED__11c BEACHCOMBED__11c DEACIDIFIED__11c PICKABACKED__11c BIOFEEDBACK__11c HEDGEHOPPED__11c PIGGYBACKED__11c
When Steve Root was doing a postmortem on a Clabbers game he played,
he noticed with whatever word tools he had at his disposal that his
rack made 9's with
DW and
UW only.
He wondered how many other 7-letter
racks have this characteristic. WHAT came to the rescue:
dw7?# // get nines with DW - there are 825 of them
[]-dw/q;s\ // remove DW from each; yield canonically-ordered strings
/1w // save 793 strings to wordlist 1
Do the same thing for UW,
and save the result to wordlist 2. Then:
/y3=1&2 // take the intersection of wordlists 1 and 2
The result is five strings:
ABLORST DFIOORT ELLNOPS GHIINNS GHINORT
Now for each of these strings, find those which produce only two words when
combined with two blanks:
/3[ []??/#=2
The result is
DFIOORT,
since the default canonical order is alphabetical.
What 6-letter words have one unique letter, a pair of another letter, and a trio of another letter?
An initial thought it to first ask WHAT this query:
?(+)(R-1)(+)(R-1)(R-1)/QS // first get all strings which conform
However, WHAT answers:
Queries which result in strings are limited to 4 blanks & sets
This restriction does not prevent the same list from being created;
it just takes two steps to do it:
?(+)(R-1)/QS# // get strings of the form ABB
[](+)(R-1)(R-1)/QS# // get strings of the form ABBCCC
There are 15600 of these.
[]/a // for each of the strings, get anagrammed words
The answer consists of these 101 words:
ACACIA BEDDED FALLAL JEERER MANNAN PEPPED SALSAS TERETE
AGGADA BONOBO FESSES JESSES MASSAS PEPPER SEEDED TESTEE
ALLELE BOWWOW GOOGOL KABAKA MESSES PIPPIN SEISES TETTER
ANANDA+ CESSES HALALA KANAKA MISSIS POWWOW SEMEME TITBIT
ANNONA COCOON HALLAL KEEKED NAGANA REARER SENSES TITTIE
ARREAR DEEMED HAMMAM LALLAN NANNAS+ REDDED SEPSES VENENE
ASANAS DESEED+ HEEDED LESSEE NEEDED REEDED SESHES+ WEDDED
ASSAIS DOODAD HOLLOO LOLLOP NESSES REEFER SETTEE WEEDED
ASSAYS EDDIED HORROR MAMMAE PALAPA REEKER STOTTS YESSES
AYAYAS+ EERIER HUBBUB MAMMAL PAPAYA REELER TATTOO ZANANA
BANANA EFFETE IGGING MAMMAS PAZAZZ RESEES TEDDED
BAOBAB EMESES INNING MAMMEE PEEPED REVERE TEETER
BATATA+ EPOPEE ISSEIS MANANA PEEPER ROCOCO TEETHE
You might be tempted to combine the final two steps above, using the
following query:
*,[](+)(R-1)(R-1) // get words of the form ABBCCC anagrammed
but, at least for now,
WHAT does not support query-based sets in racks.
Seth Lipkin mentioned he likes the 6+2 Unistem list of John Chew. A
"unistem" is a set of letters that, together with a certain number of
blanks, anagrams to form only one word. For example,
HOAGIE is a
two-blank unistem using the OWL3 lexicon, because the only 8-letter word in
AGEHIO?? is
ESOPHAGI.
Since HOAGIE happens to be a word itself,
rather than a random collection of letters, it is a 6-letter two-blank
unistem word, or a 6+2 unistem word for short.
John presents these in reverse probability order on his web site, and
here is how you can produce this list using WHAT.
6.\ // get all the 6-letter words onto the slate
[]+??/#=1\ // replace the slate with only those 6's which lead
// to one 8-letter word
/<% 24' // present the first 24 words of the answer
John Chew provides the answers as 6's and in decreasing probability;
WHAT's first 24 such answers are (read down columns first):
HOAGIE TUYERE JILTER ZOARIA AFTOSA ZOSTER VITTLE GADOID
FRAENA FANEGA OGIVAL TAXITE GONEFS NONWAR EXODOI HEJIRA
JAILER JAILED WITNEY EXILER STOGEY RENNIN AWEIGH MAFTIR
John's list is in a slightly different order, since his probability computation
includes taking blanks into account, but WHAT's computation
does not.
Matthew Hodge asked for a list of the highest probability 7-letter words which have no anahooks, such as Since NOTATES. This means that Since NOTATES plus a blank yields no eights.
Begin by making such a list and then decide how many make sense:7. /Q;W # // get all racks for 7-letter wordsAfter about 4 seconds of working, WHAT announces the number of these is 21063. Now prune this down to those which have no anahooks:
[]? /#=0 #After working about 8 seconds, WHAT announces the number of these is 4136.
Set the presentation of the slate for the long-term so that answers
are sorted by decreasing probability and show nothing:
/<%! \
It is worth noting that you do not have to make the commands from your
memory and knowledge of WHAT.
You can use the GUI to formulate these
commands. The first two of the above three commands are constructed
using choices you make from the Query tab,
except for the terminating
number sign, which you can get from the Presentation
tab, but you just
might recall that particular subcommand without using the GUI. The final
command is constructed using the bottom part of the
Presentation tab.
Now see what the 190th rack is, and show the words (technically, the
anagrams of the answer):
/200 )
The presented answer is:
AEINTXY (ANXIETY)
It might make sense to save this list now to a wordlist. For example,
/1W
saves the 4136 racks to wordlist number 1. Now if you want to work with
only those racks which are of the probability of
ANXIETY or greater, one
way to do this is start over and include the filtering subcommand of:
/%{ANXIETY}
in your first query. Alternatively, you can direct
WHAT to use a word list
as the basis of words for a query which includes filtering. Let's detail
both of these sub-scenarios:
/0I 7. /Q;W /%{ANXIETY} #
The second character in the above command is the digit zero, and that
value of zero indicates the slate is the word source for this command.
WHAT reports there are 202 words. 7. /Q;W /%{ANXIETY} #
The answer is 4889.
[]? /#=0 #The answer is 191.
Now, look at the first 50 of these, since that is about how many can be
shown in the workspace at once (My workspace is 30 rows of 89 columns.).
Show both the number of ways to make the rack (the basis of the
probability) and the word(s):
50' % )
This is the result:
(first 50 items only)
AAEILNO_746496w_(AEOLIAN) EFIOORS_145152w_(ROOFIES)
AAEINNO_466560w_(AEONIAN) AEENNOV_142560w_(NOVENAE)
AABEIOR_373248w_(AEROBIA) ADDEITU_139968w_(AUDITED)
AEIINOP_373248w_(EPINAOI) AEIJNRS_139968w_(INJERAS+)
AENOSTT_311040w_(NOTATES) AEIJNST_139968w_(TAJINES+)
ADEIRTY_279936w_(DIETARY) AEINTVW_139968w_(VAWNTIE)
AEINTUV_279936w_(VAUNTIE) AGIIRST_139968w_(ARTIGIS+)
AEHIORR_233280w_(HOARIER) AACDIOR_124416w_(ACAROID)
EFIOORT_217728w_(FOOTIER) AADGIOS_124416w_(ADAGIOS)
EGIOORS_217728w_(GOOSIER) AADIRSU_124416w_(SUDARIA)
EIILORR_207360w_(ROILIER) AAEGSTU_124416w_(GATEAUS)
AADGIOT_186624w_(AGATOID) ABDEIIL_124416w_(ALIBIED)
AAINORV_186624w_(OVARIAN) ADEFGOR_124416w_(FORAGED)
ADDENOT_186624w_(DONATED) ADEIMOW_124416w_(MIAOWED)
ADENRTV_186624w_(VERDANT) ADENTUV_124416w_(VAUNTED)
AEIKLOR_186624w_(OARLIKE) ADERSTV_124416w_(ADVERTS,STARVED)
ABEERTU_171072w_(BEAUTER+) AEGOTUV_124416w_(OUTGAVE)
ADIILOS_165888w_(SIALOID) BDEGIOT_124416w_(BIGOTED)
DEFIOST_165888w_(FOISTED) DEGHINO_124416w_(HONGIED)
DEIIORV_165888w_(IVORIED+) DEGHIOT_124416w_(HOGTIED)
EIILORV_165888w_(RILIEVO) DEGILTU_124416w_(GUILTED+)
AEMNOTT_155520w_(TOMENTA) DEIJNOT_124416w_(JOINTED)
EENOSTW_152064w_(TOWNEES) EGIMNOU_124416w_(MEOUING)
AADNOOT_145152w_(ODONATA+) EIJLORT_124416w_(JOLTIER)
DEGIOOS_145152w_(GOODIES) EINRSUW_124416w_(UNWISER)
You could export the wordlist of these 191 racks to a
file and use it as the basis for flashcarding.
Bob Gillis would like to get
the top 1000 8-letter words which do not have a 7-letter word as a subword
by removing the first or final letter. An example is
SALMONID. Perhaps a
more interesting set of words are those 8's which have no
included 7-letter words when anagramming, but here is how to answer Bob:
8.\ // get all the 8's
[]/a/|=7/#=0 // keep only those 8's with the sought characteristic
There are 6351 of these. Look at the 25 most likely ones:
/<% // and then click on seeing the first 25
You are shown:
AERATION OEDIPEAN NOTARIZE ANTIWEED DANEWORT IRONWARE ANAEROBE
UNEASIER LOITERER PINOTAGE GAIETIES DOWNRATE ROTATIVE
DETAINEE OUTEATEN RATOONER ABOITEAU TEARDOWN THIONATE
ETIOLATE INDAGATE AGENESIA AUDITION FETATION UINTAITE
Notice
DANEWORT,
DOWNRATE and
TEARDOWN
are anagrams. How many of these 8's have at least one other anagram?
Issue this query:
[]/&>1
and WHAT reports there are 308 of these.
Once again, show these in reverse likelihood:
/<%
Then order these by decreasing number of anagrams and show those
with 4 and 3 anagrams in a set:
MOORWORT ROOTWORM TOMORROW WORMROOT
DANEWORT DOWNRATE+ TEARDOWN
ILMENITE MELINITE TIMELINE
DEPORTER PORTERED REPORTED
CEMENTER CEREMENT RECEMENT
CONIDIUM MUCINOID ONCIDIUM
BEGRUDGE BUGGERED DEBUGGER
From this, one could present a quiz: what group of words of length 8
has the most number of anagrams and there are no 7's in the letters?
There is a web site with considerable info about words. One category there was for words with symmetrically-distributed letters, such as IZAR and WIZARD. Notice I and R are symmetric in that I is the ninth letter of the alphabet, and R is the ninth last letters of the alphabet.
WHAT can help seek these words. For example, find the 6-letter words with this characteristic:/M"ZYXWVUTSRQPONMLKJIHGFEDCBA" // set WHAT's mapping to symmetrical letters 3./qs# // get all trigrams [][m]/a // yield the answersWHAT presents these 19 words:
BOLSHY GIRTHS HOLLOS LOVIER+ RIGHTS SHOVEL WIZARD CERVIX GRITHS HOVELS RAZZIA+ SHIRAZ+ SHRIVE EVOLVE GRIVET HYBRIS REVIVE SHIVER VERMIN
From John Van Pelt's Verbalobe web site, a question was posed: "What are the longest words with all imbedded 2's?" That was then extended to ask the same question for other lengths up to 7, and these are the answers:
| 2's | from | 11's | - | DENOMINATOR + 9 others |
| 3's | from | 10's | - | CALABASHES only |
| 4's | from | 8's | - | PARAKEET + 2 others |
| 5's | from | 9's | - | TRAVERSES and 3 others |
| 6's | from | 8's | - | BLENDERS + 12 others |
| 7's | from | 9's | - | CAROUSERS + 2 others |
To find such words using
WHAT, you need to make
individual probes for specific pairs of lengths.
For example, to find the longest 9-letter words with the most
imbedded 5-letter words, use these commands:
9.\
[1-5]/#=1\
[2-6]/#=1\
[3-7]/#=1\
[4-8]/#=1\
[5-9]/#=1\ // yields the answer with the 4 words
From John Van Pelt's Verbalobe web site, there are many 9-letter words which have no imbedded 2-letter words, from ACROLECTS thru WILLFULLY, and there is 1 anagram pair in the list.
WHAT can find these with several steps:
9.\ // get all 9's
[1-2]/#=0\
[2-3]/#=0\
[3-4]/#=0\
[4-5]/#=0\
[5-6]/#=0\
[6-7]/#=0\
[7-8]/#=0\
[8-9]/#=0\ // yields the answer with 71 words
/1w // saves the answer remains in wordlist 1
[]/&=2 // finds those words with anagrams (in the lexicon)
These three words result:
HURTFULLY,
SPRIGHTLY, and
TRIGLYPHS.
By inspection you can see that the sought answer consists of the
2nd and 3rd words.
RUTHFULLY
is not in the list,
since it has
the 2-letter word UT
imbedded within.
In order to check for the number of anagrams in the answer and not in the lexicon, suffix the final command above with /1I to use wordlist 1 as the word source instead of the lexicon. In this case, only the two words are in the answer.
Someone mentioned that
THEREIN
contains many shorter words spelled in order.
WHAT can be used to check whether this word has the
most imbedded words for all 7-letter words:
therein/i# // find out how many words are imbedded in THEREIN
The answer is 13, which includes the word itself.
7.\ // get all the 7's onto the slate
[]/i/#13 // get those with more than 12 imbedded words
[]/i/#14 // get those with more than 13 imbedded words
[]/i/#15 // get those with more than 14 imbedded words
[]/i/#16 // get those with more than 15 imbedded words
[]/i/#17 // get those with more than 16 imbedded words
[]/i/#18 // get those with more than 17 imbedded words
[]/i/#19 // get those with more than 18 imbedded words
One word is left, and that is AMUSERS.
amusers/i# // be sure how many imbedded words in AMUSERS
The answer is 19, which includes the word itself:
AM AMUSE ER MUS MUSERS US USERS
AMU AMUSER ERS MUSE SER USE
AMUS AMUSERS MU MUSER SERS USER
So, there are 18 imbedded words within
AMUSERS.
Get the 8-letter words
which can be formed from 2 shorter words concatenated,
such that the shorter words are are least of length 3.
This takes a few steps using WHAT:
8.\
[1-3]/#=1\
[4-8]/#=1\
/1w // save the 3+5 list to wordlist 1
8.\
[1-4]/#=1\
[5-8]/#=1\
/y1=1|0 // merge the 4+4 list into wordlist 1
8.\
[1-5]/#=1\
[6-8]/#=1\
/y1=1|0 // merge the 5+3 list into wordlist 1
So now the answer of 11970 words is in wordlist 1. These, of course, are
not all truly compound words. Realize the answer has only 8-letter words.
Another way to go about this is to use the query kind of 2-word patterns,
but with this kind of query, there is no control on the length of the
subwords. Commands to do this are:
8. \
[] /2P #
One could take the answer of 19013 strings from
WHAT and remove those lines
which break the longer word other than what is desired. The lengths of
each of the two-word results can be used for sorting. For example:
/>|/8~W // sort by increasing word lengths and save to wordlist 8
At this point, the wordlist has the two-word results sorted by length. In
this wordlist, the first items have 6 letters as the size of the first
word, and the end of the list of 19013 items has 2 letters as the size of
the first word. The stated goal is to eliminate both of these groups from
the entire answer. We have to do is find where those groups start and stop.
A way to probe wordlist 8 is via the dialog which comes up for wordlist
item deletion - menu pick
With these probes I found the border between the length-6 and length-5
first words. So now, those first 3157 words can be deleted from the
slate using the delete wordlist items command:
/=8:1,3157D
Then, the same item deletion dialog can be used to find the trailing
items to delete. At this point, these are items starting at 12906.
So issue a similar command:
/=8:12906,5000D
The count has to be at least how many items are to be deleted, and 5000
is higher than needed. So, at this point, 12905 items are now remaining
in wordlist 8. This number of items is higher than the number of 8-letter
words found by the first method described above, namely 11970. The
difference is that in wordlist 8, an 8-letter word may be showing up more
than once, since it can be split at more than one place. We can prove
this by running a modified wordlist 8 back through the anagrammer with a pattern
query. Export wordlist 8, use a text editor to remove spaces, and import
the result back into wordlist 8. Then, the following query should then yield 11970 words:
[]/8[# // using wordlist 8 as the source, pattern match each word
Yes, indeed it does - there are 11970 words.
Using the 2-letter abbreviations for the 50 states plus
DC, what 8-letter
words consist of four state abbreviations concatenated? Here is a
way to answer this using WHAT:
WHAT?: /FR2W"C:\msw\whatexe\50states.wds"
Wordlist 2 with 51 items has been imported from
file C:\msw\whatexe\50states.wds
WHAT?: 8.#
Number of words = 31475
WHAT?: /2I! // use the states in wordlist 2 as the word source
WHAT?: [1-2]/#=1#
Number of words = 6911
WHAT?: [3-4]/#=1#
Number of words = 1155
WHAT?: [5-6]/#=1#
Number of words = 210
WHAT?: [7-8]/#=1#
Number of words = 20
WHAT?: /-I ! [] /OC10W
ARMORIAL CALAMINE GANYMEDE MAINLAND MANDORLA+ MEMORIAL SCINCOID
CACONYMS CANDIDAL LAVALAVA MALARIAL MEGADEAL MESCALIN+ UTILIDOR
CALAMARI GAMODEME MAHIMAHI MANDARIN MELAMINE MOORLAND
Yield all 7-letter canonical racks which lead to words:
7./q;w#
and 21063 is how many there are. For each of these, add a blank and retain
those words for which there are at least 7 resulting words:
[]?/#7
and 2584 is how many there are.
Export this list to a file and add a
? to each line in the file.
Set the long-term kind presentation to
=!
Clear the workspace. Execute commands from the file with the question
marks. Export the workspace. We are not there yet,
but we are on the way.