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.
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
TWL 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 KAF...20p@8F OAKEN.20p@8D TOKEN.20p@8D KANE..16p@8E
FAKE..22p@8E KEF...20p@8F TAKEN.20p@8D OFTEN.18p@8D KAON..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 248 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 1409 such words, and here are the first 15:
OVERABLE OVERACCENTUATE OVERACTIONS
OVERABSTRACT OVERACCENTUATED OVERADJUSTMENT
OVERABUNDANCE OVERACCENTUATES OVERADJUSTMENTS
OVERABUNDANCES OVERACCENTUATING OVERADVERTISE
OVERABUNDANT OVERACTION OVERADVERTISED
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 383 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 7 words:
DEIFIED HALALAH REIFIER REPAPER REVIVER ROTATOR SEMEMES
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 REVIVER SEMEMES TRESSED
DEIFIER GATEMAN REIFIED REPAPER ROTATOR STELLAS
DELIVER HALALAH REIFIER REVILED SALLETS STINKER
What are the 35 most-likely 7-letter words?
7. /<% 36'
Here is the answer:
ARENITE ANEROID ANESTRI LATRINE RETAINS STAINER URANITE
RETINAE ATONIES ANTIRED NASTIER RETINAL STEARIN URINATE
TRAINEE ELATION ANTSIER RATINES RETINAS TAURINE AERIEST
AILERON ERASION DETRAIN RATLINE RETSINA TRAINED ALIENER
ALIENOR TOENAIL INERTIA RELIANT RUINATE TRENAIL ATELIER
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 RATLINE..839808w TAURINE..839808w
RETINAE.1154736w ANESTRI..839808w RELIANT..839808w TRAINED..839808w
TRAINEE.1154736w ANTIRED..839808w RETAINS..839808w TRENAIL..839808w
AILERON.1119744w ANTSIER..839808w RETINAL..839808w URANITE..839808w
ALIENOR.1119744w DETRAIN..839808w RETINAS..839808w URINATE..839808w
ANEROID.1119744w INERTIA..839808w RETSINA..839808w AERIEST..769824w
ATONIES.1119744w LATRINE..839808w RUINATE..839808w ALIENER..769824w
ELATION.1119744w NASTIER..839808w STAINER..839808w ATELIER..769824w
ERASION.1119744w RATINES..839808w STEARIN..839808w
What are the 35 most-likely 7-letter racks
that have 7-letter words?
7. /Q;W /<% 36'
Here is the answer:
AEEINRT AEINOST AEINRTU AEEINST ADEIOST EILNORT AEEORST
ADEINOR ADEINRT AAEEINT AEEIRST AEILNOS EINORST EEILNOR
AEILNOR AEIINRT AEEILNR AAEILNO AEILOST EINORTU AEGINRT
AEILNOT AEILNRT AEEILNT ADEILOR AENORST AEINNOT AEEGINR
AEINORS AEINRST AEEILRT ADEIORS AENORTU AEENORS AADEINR
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)
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)
AAEILNO (AEOLIAN)
ADEILOR (DARIOLE)
ADEIORS (ROADIES)
ADEIOST (IODATES,TOADIES)
AEILNOS (ANISOLE)
AEILOST (ISOLATE)
AENORST (ATONERS,SENATOR,TREASON)
AENORTU (OUTEARN)
EILNORT (RETINOL)
EINORST (NORITES,OESTRIN,ORIENTS,STONIER)
EINORTU (ROUTINE)
AEINNOT (ENATION)
AEENORS (ARENOSE)
AEEORST (ROSEATE)
EEILNOR (ELOINER)
AEGINRT (GRANITE,GRATINE,INGRATE,TANGIER,TEARING)
AEEGINR (REGINAE)
AADEINR (ARANEID)
*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 8 resulting words:
ABLINGS BEESTINGS EMPTINGS GAYWINGS
BEASTINGS CHITTERLINGS EYESTRINGS HUSTINGS
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 any of these letters: 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 EPINAOI OPUNTIA
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 STAMENS
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 QAIDS QINDAR QIVIUT QWERTYS TRANQ
BUQSHAS QANAT QINDARKA QIVIUTS SHEQALIM TRANQS
FAQIR QANATS QINDARS QOPH SHEQEL UMIAQ
FAQIRS QAT QINTAR QOPHS SUQ UMIAQS
QAID QATS QINTARS QWERTY SUQ
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 TWL lexicon.
SATINE
plus a blank produces many words; and how many is that? With the query:
SATINE?#
The answer is:
Number of words = 67
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 23 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 = 23
Although you can type the
=
command to see what letters the blank can
be and by observation find the 3 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
J,
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
N: INCANTED
O: CATENOIDs
P: PEDANTIC
R: DICENTRAs
S: DISTANCEds
T: NICTATED
U: INCUDATE
WHAT?: =
The one blank can be any of these letters: CIKNOPRSTU
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 AGOUTIES EDACIOUS EULOGIAS EXONUMIA OUTRAISE SEQUOIAS
AEROBIUM DIALOGUE EQUATION EUPHORIA JALOUSIE SAUTOIRE THIOUREA
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 7 7-letter words and 5 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 13287 words, and
*t#
says there are 8449 words.
So begin with the smaller second group, and then
ask WHAT to perform:
[1-2|]ed
which yields 324 words.
10(A-G)/|5
SOWPODS is the lexicon which includes the words from North America (OSPD) and Great Britain (OSW). This lexicon is used in most countries throughout the world, but it has not yet been accepted for North America. Although the abbreviation OSPD is used for the Official Scrabble Players Dictionary, the North American lexicon for club and tournament play is named TWL, for "The Word List".
Perhaps you want to look at 2-letter words from the SOWPODS lexicon. You probably would like to be shown which of these are from the British list only, such 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 "SOWPODS" as the primary lexicon and "TWL98" 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 SOWPODS, and words not in TWL 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 ID MM ON SH WE AB BE EN IF MO OO# SI WO AD BI ER IN MU OP SO XI AE BO ES IO# MY OR ST# XU AG BY ET IS NA OS TA YA AH CH# EX IT NE OU# TE# YE AI DA# FA JO NO OW TI YO AL DE FY# KA NU OX TO YU# AM DI# GI# KO# NY# OY UG# ZO# AN DO GO KY# OB# PA UH AR EA# GU# LA OD PE UM AS ED HA LI OE PH# UN AT EE# HE LO OF PI UP AW EF HI MA OH PO# UR# AX EH HM ME OI# QI# US AY EL HO MI OM RE UT
You can see those 25 2-letter words acceptable only from the British Official Scrabble Words (OSW). 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 TWL, get the TWL twos and remove these from the wordlist you saved.
If you don't mind the long output of 121 lines, you could look at the SOWPODS twos with definitions. If the definitions file you are using is for TWL only, you will find no definitions given for the OSW-only words.
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?
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 117 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 46 words:
BITABLE DYABLE LIVABLE PLACABLE SHARABLE TUNABLE
CITABLE FINABLE LOVABLE RATABLE SIZABLE TYPABLE
CLOSABLE FRAMABLE MACHINABLE REMOVABLE SMOKABLE UPGRADABLE
DATABLE GRAZABLE MAKABLE RIDABLE SPARABLE USABLE
DIAGNOSABLE HATABLE MICROWAVABLE SALABLE STATABLE VOTABLE
DISLIKABLE HIRABLE MINABLE SAVABLE TAKABLE WADABLE
DRAPABLE LIKABLE MOVABLE SHAKABLE TAMABLE
DRIVABLE LINABLE NAMABLE SHAPABLE TRADABLE
Then just for a little more investigation I did this:
[]/&2 ) // for each word with at least 2 anagrams, show anagrams
Here is the answer:
DATABLE (ABLATED)
DYABLE (BELADY)
FRAMABLE (FARMABLE)
RIDABLE (BEDRAIL,BRAILED)
SMOKABLE (ABELMOSK)
SPARABLE (PARABLES,PARSABLE,PREBASAL)
STATABLE (ABETTALS,TASTABLE)
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 JONQUIL QUILLAJA
JACQUARDS JACQUERIES JONQUILS 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 38 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 lists these answers:
zealous zilches zincous zipless zloties zygosis zymoses zymosis
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 8 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 217 such threes:
AAS BYS FEZ HIC LEZ OFT RAJ TAE WIZ
ABS CIS FIE HID LIS OHO RAN TAJ WOS
ADS COR FIX HIM LOX OHS RAS TAX WRY
ADZ COX FIZ HMM LUX OMS RAX THE WUD
AFF COZ FLY HUH MAX ONS RES THO XIS
AFT CRY FOH HUP MED OPS REX THY YAH
AGO CUM FOR ICK MEN ORA SAD TIS YAR
AHA DEX FOU ICY MET ORS SAE TOO YEH
AIS DID FOX IDS MIM OVA SAT TRY YEP
ALS DUI FRO IFF MIX OXO SAU TUX YES
ANY EAU FRY IFS NAE OXY SAX UMM YET
APT EFS GAN INS NAH PAH SEN UNS YOM
ARS ELF GAS ITS NAM PAX SEX UPO YON
ASH ELS GEY IVY NAW PER SHA UPS YOU
ASS EMS GOR JEU NEE PES SHH UTS YUM
ATT ENS GOT JUN NIX PHT SHY VAS ZAX
AVA ERE GOX JUS NOH PIU SIS VEG ZOA
AWA ERS HAD KAS NOO PIX SIX VEX
AYS ESS HAJ KEX NOR PLY SKY VIA
AZO FAR HAO KOI NOS POH SLY VIS
BAH FAS HAS LAX NOT POX SOS VOX
BIS FAX HEP LED NTH PRY SOX WAS
BIZ FER HES LEU NUS PYX SPY WAX
BOX FEW HEX LEV ODS QUA STY WHA
BRR FEY HEY LEX OES RAH SYN WHO
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 the S
/y0=1-0 // take the set difference (all minus those which take the 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 80367 words. Once again the second query to yield the final answer is:
[]/><=(!S)
The result has 50764 words, and here are the first 27 of them:
AAHED AALIIS AARGH AAS ABACK ABAFT ABAMPS ABAS ABASERS
AAHING AALS AARRGH ABACAS ABACUS ABAKAS ABANDONS ABASED ABASES
AAHS AARDWOLF AARRGHH ABACI ABACUSES ABALONES ABAPICAL ABASEDLY ABASH
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:
Q7?/|3-8# // get all words with a Q of length 3-8
The answer is 977 words.
/m"abcdefghijklmnopgrstuvwxyz" // define the mapping to convert Q to G
[m] // map each slate word
The answer consists of these 22 words:
BUGSHA GATS GUESTING GUIDS GURSH PIROGUES ROGUE TOGUES
BUGSHAS GUEST GUESTS GUILT GURSHES PLAGUE ROGUES
GAT GUESTED GUID GUILTS PIROGUE PLAGUES TOGUE
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 MOTMOT PAWPAW TARTAR WEEWEE
BOOBOO CHICHI FURFUR PALPAL POMPOM 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 3-letter strings onto the slate[TBD - this section needs more words, especially here]
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 COOCOO FURFUR MURMUR POMPOM TSETSE WEEWEE BEEBEE BULBUL CUSCUS GRIGRI MUUMUU SARSAR TSKTSK BONBON CANCAN DIKDIK GRUGRU PALPAL TARTAR TZETZE BOOBOO CHICHI DUMDUM MOTMOT PAWPAW TESTES VALVALAssume the answer of the above 29 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 2840 such words. Here are the first 28:
ABAKAS ABATOR ACETIC ACTORS ADOBOS AFFAIR AGAMAS
ABASER ABOMAS ACETIN ADDEND ADORED AFFLUX AGEISM
ABATED ABSORB ACTINS ADDERS ADORES AFFRAY AGORAS
ABATIS ABSURD ACTION ADDLED ADSORB AFTERS AGOROT
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 the previous, yield those 8's which end with a 4 /0I z7. // use the slate as word source and 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 ZONELESS ZOONOSES ZOOSPERM ZILLIONS ZINCKING ZONETIME 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 of 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:
4.($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 ORACH.(ROACH) OXBOW
OCHRY ONERY OUTBY
ODDLY OOMPH OVARY
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- UNDERWINGS-
BEESWINGS- GAYWINGS MOWINGS- UNKNOWINGS-
BITEWINGS- GLASSBLOWINGS- OVERSWINGS- UPSWINGS-
BORROWINGS- GNAWINGS- PARAWINGS- VIEWINGS-
BOWINGS- GREENWINGS- PIGEONWINGS- WAPPENSCHAWINGS-
BREWINGS- KNOWINGS- REDWINGS- WAXWINGS-
CLEARWINGS- LACEWINGS- ROWINGS- WHITEWINGS-
DOWNSWINGS- LAPWINGS- SEWINGS- WINGS-
DRAWINGS- LAWINGS- SHOWINGS-
FOLLOWINGS- LOWINGS- -SWINGS-
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 the 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 1-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/<$$ // filter based on scores at least 350,
// present sorted by decreasing scores
The answer is:
OXAZEPAM.392p QUEZALES.374p HUTZPAHS.365p ISOZYMIC.356p
MEZQUITS.392p KHAZENIM.374p ZYZZYVAS.356p EXEQUIES.356p
MEZQUITE.392p SOVKHOZY.365p RHIZOMIC.356p EXEQUIAL.356p
JAZZLIKE.383p QUIZZERS.365p QUIXOTIC.356p CHAZANIM.356p
JACQUARD.383p QUIXOTRY.365p OXAZINES.356p BENZYLIC.356p
WHIZBANG.374p MITZVOTH.365p MAHZORIM.356p
QUIZZING.374p MITZVAHS.365p JONQUILS.356p
But that is not enough, since those starting at 1H also deserve to be
considered, and these are the results:
CAZIQUES.392p QUEAZIER.374p HYLOZOIC.365p MACHZORS.356p
BEZIQUES.392p PIROZHOK.374p BLOWZILY.365p FLAPJACK.356p
JANIZARY.383p PIROZHKI.374p ZABAJONE.356p DENAZIFY.356p
CHUTZPAH.383p HIGHJACK.374p YOKOZUNA.356p COENZYME.356p
WHEEZILY.374p FROWZILY.374p WHEEZING.356p CHUTZPAS.356p
QUIZZING.374p QUIZZERS.365p SOVKHOZY.356p
QUETZALS.374p LYSOZYME.365p POLYZOIC.356p
Find the top-scoring 21-letter words when the horizontal placement
of the word starts at position 1A on a board for the
21 x 21 board game:
/21B // set the board size to 21 x 21
/@1A.! // score horizontal words which start at the upper left square
21./$>6900/<$$ // find all 21-letter words which score more than 6900, sort
// by decreasing scores and show with scores
The answers are:
PHOSPHOGLYCERALDEHYDE.7106p
PSYCHOPHARMACOLOGISTS.6962p
OVERINTELLECTUALIZING.6962p
ANTHROPOMORPHIZATIONS.6962p
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 483 7-letter canonically-ordered words with at least one X
[]-X+?/:=(X)/Q;W# // get the 76 canoncially-ordered words that take only an X
[]/A // show these words
/<% // show these sorted by decreasing probabilities,
// and see UXORIAL is first; perhaps some of these are
// unexpected: PTYAEI+X, PRYAEI+X, CNPRIO+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 almost 2 minutes to complete)
/y0=1-2 // take the difference, and this is the answer
This yields these 108 words:
ADDAX FAWNY GYVED JIVED KUDZU PUJAH WONKY
BAFFY FEAZE HAFIZ JIVEY KYLIX PUKKA WOOZY
BEEFY FEMME HAJJI JORUM LEZZY QOPHS WUSSY
BIFFY FEYLY HOKKU JOTTY MAMMY QUIPU XYLYL
BUFFY FJELD HUFFY JOUKS MIFFY QUOTH YECHY
BYWAY FLAXY JAGRA JOWLY MIRKY SKYEY YEUKY
CRAZY FLYBY JAMMY JUGUM MOMMY THUJA YOLKY
CUPPY FUGGY JAWAN JUICY MUGGY TIZZY YUCKY
DADDY FUJIS JAZZY JUMBO MUJIK TOFFY YUMMY
DOJOS FURZY JEFES JUMPY MUZZY TWIXT ZAPPY
DOOZY FUZZY JERKY JUPON NUBBY UPBOW ZIPPY
DUCHY GANJA JIBED KEXES OUZOS VROUW ZONKS
DUCKY GAUZY JIBER KHAPH PAPAW VUGGY
DUDDY GAWKY JIFFS KICKY PAPPY VUGHS
DUSKY GIZMO JIFFY KOPEK PHPHT WIMPY
DYKEY GUPPY JIMPY KOPJE PUFFY WOMBY
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,
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 the canonically-ordered 7-letter words, but don't show them [] // for each one, 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 48 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 9 racks:
AACIINOSTTV ACEEIINNRST ACGHIMNOOPR
AAIILMNNOST ACEHOOPRRST AEGIILMNRST
ACDEINOORST ACEIIMNORST AEIILMNORST
At this point, you can see the answers by typing the 1-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 3883 words, and the next one
results in 3903 words. This is the answer of these 20 words:
BIBB FUZZ JUJU KECK KIKE KONK MUMM
BUZZ HAJJ KAKA KEEK KINK KOOK RAZZ
FIZZ JAZZ KAKI KICK KIRK KYAK
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 an
A.
and may have other vowels too:
a*e*i*o*u*
These 13 words are shown:
ABSTEMIOUS ADENOCARCINOMATOUS ANTIRELIGIOUS AUTOECIOUS
ABSTEMIOUSLY ADVENTITIOUS ARENICOLOUS AUTOECIOUSLY
ABSTEMIOUSNESS ADVENTITIOUSLY ARGENTIFEROUS
ABSTEMIOUSNESSES AMENTIFEROUS ARSENIOUS
ABSTENTIOUS ANEMOPHILOUS ARTERIOVENOUS
Then, restrict the question to were those 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 FACETIOUS
ABSTEMIOUSLY ARSENIOUS FACETIOUSLY
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 wordlist1 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 E-mail 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 466 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 lists 2 and 3 into list 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 those 8's which have no 7's, say how many
[]/a/|=6/#=0# // retain only those 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 those 8's which have no 3's and show them
...
/5r' //and then go back to see wordlist 5
WHAT?: []/a/|=7/#=0#
Number of words = 5990
WHAT?: []/a/|=6/#=0#
Number of words = 863
WHAT?: []/a/|=5/#=0#/5w
Number of words = 69
WHAT?: []/a/|=4/#=0#
Number of words = 0
WHAT?: /5r'
ABEYANCE COMMENCE EPOPOEIA HOMEOBOX LOOKDOWN QUIDNUNC WIFELIKE
ABEYANCY COOKBOOK EXEQUIES HOOKWORM MAHIMAHI QUINCUNX WINDMILL
AMADAVAT CORRIDOR EXIGUITY JACQUARD MILKWOOD QUIZZING WOODCOCK
APAGOGIC CRUCIFIX FILIFORM JIPIJAPA NINETEEN SUSURRUS WOODWORK
AUTOMATA DAHABIAH FOOFARAW KAVAKAVA PARALLAX THATAWAY WOODWORM
AVADAVAT DAHABIYA FORENOON KAZACHOK PELLMELL UBIQUITY WOOLWORK
BELLPULL DUMMKOPF FUGITIVE KEFFIYEH PICKWICK UXORIOUS WORMWOOD
BUZZWORD DYSPEPSY GANGBANG KICKBACK PURPURIC VIRILITY ZIZZLING
CACHUCHA ECONOBOX GUITGUIT LARRIKIN QUANDANG VIVARIUM ZOOTOMIC
CARACARA EFFICACY HIGHBROW LINKWORK QUIDDITY VIZCACHA
So the above list are those 8's with 4's as the longest subwords.
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 all 8's with no repeated letters into wordlist 8
/y0=5&8 // the slate gets the answer, the intersection of lists 5 and 8
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 QW QX XZ
Then you can see the words with:
[]5?
This produces:
EQUINOX JONQUIL OXAZINE OXIDIZE QUIXOTE QWERTYS SQUAWKS
Notice that there are no
QQ words.
To be sure that is the only such digram:
../qs# // get all 676 (26 x 26) digrams onto the slate
[]5?/#=0;
Just QQ is shown.
The similar pair of queries:
../q;s# // get all 351 canonically-ordered digrams onto the slate
[]5?/#=1
produces just JQ.
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 DOCUMENTARILY...13c MULTIBRANCHED...13c AMYLOPECTINS....12c
UNCOPYRIGHTABLE.15c ENDOLYMPHATIC...13c SUBORDINATELY...13c BANKRUPTCIES....12c
AMBIDEXTROUSLY..14c FLOWCHARTINGS...13c TROUBLEMAKING...13c BLUESTOCKING....12c
DERMATOGLYPHIC..14c HYDROMAGNETIC...13c UNPREDICTABLY...13c CABINETWORKS....12c
TROUBLEMAKINGS..14c LYCANTHROPIES...13c UNPROBLEMATIC...13c
CONSUMPTIVELY...13c METALWORKINGS...13c AMBIDEXTROUS....12c
COPYRIGHTABLE...13c MOTHERFUCKING...13c AMBLYGONITES....12c
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)/<||10' // query for any number of set 1 letters, and
// sort by decreasing length, show length, and
// the first 10 words
Here is the answer:
(first 10 words only) DEHUMIDIFIED.12c BODYCHECKED..11c HEDGEHOPPED..11c PIGGYBACKED..11c BEACHCOMBED..11c COMMODIFIED..11c MACADAMIZED..11c BIOFEEDBACK..11c DEACIDIFIED..11c PICKABACKED..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 668 of them
[]-dw/q;s // remove dw from each and yield a canonically-ordered string
/w // save 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
DFIOORT,
since the default canonical order is alphabetical.
I was expecting there might be other 7-letter strings at this point and so I would need to do some more work, but since this is it, this answers Steve's question.
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 85 words:
ACACIA COCOON HALALA LALLAN MISSIS POWWOW SEEDED TESTEE
ALLELE DEEMED HEEDED LESSEE NAGANA REARER SEISES TETTER
ARREAR DOODAD HOLLOO LOLLOP NEEDED REDDED SEMEME TITBIT
ASANAS EDDIED HORROR MAMMAE NESSES REEDED SENSES TITTIE
ASSAIS EERIER HUBBUB MAMMAL PAPAYA REEFER SEPSES WEDDED
ASSAYS EFFETE INNING MAMMAS PAZAZZ REEKER SETTEE WEEDED
BANANA EMESES ISSEIS MAMMEE PEEPED REELER TATTOO YESSES
BAOBAB EPOPEE JEERER MANANA PEEPER RESEES TEDDED ZANANA
BEDDED FALLAL JESSES MANNAN PEPPED REVERE TEETER
BOWWOW FESSES KABAKA MASSAS PEPPER ROCOCO TEETHE
CESSES GOOGOL KEEKED MESSES PIPPIN SALSAS TERETE
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 TWL96 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
[]+??/<% // present the answer
John Chew provides the answers as 6's and in decreasing probability;
WHAT's first 25 such answers are (read down columns first):
HOAGIE FANEGA ZOARIA GONEFS NONWAR AWEIGH QINDAR
FRAENA JILTER TAXITE JOLTER VANNED GADOID
JAILER OGIVAL EUREKA STOGEY VITTLE HEJIRA
AWHILE WITNEY EXEDRA ZOSTER EXODOI 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 19424. Now prune this down to those which have no anahooks:
[]? /#=0 #After working about 8 seconds, WHAT announces the number of these is 3992.
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 200th 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 3992 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 4617.
[]? /#=0 #The answer is 202.
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) DEFIOST.165888w.(FOISTED)
AAEINNO.466560w.(AEONIAN) EIILORV.165888w.(RILIEVO)
AABEIOR.373248w.(AEROBIA) AEMNOTT.155520w.(TOMENTA)
AEIINOP.373248w.(EPINAOI) EENOSTW.152064w.(TOWNEES)
AENOSTT.311040w.(NOTATES) ABDEOOT.145152w.(TABOOED)
ADEOORS.290304w.(ROADEOS) DEGIOOS.145152w.(GOODIES)
ADEIRTY.279936w.(DIETARY) EILOOPR.145152w.(LOOPIER)
AEINTUV.279936w.(VAUNTIE) AEENNOV.142560w.(NOVENAE)
EIOORRT.272160w.(ROOTIER) ADDEITU.139968w.(AUDITED)
ADEEHIR.256608w.(HEADIER) ADEILLT.139968w.(TALLIED)
ADEEIRW.256608w.(WEARIED) AEGNRTY.139968w.(AGENTRY)
ADEGOST.248832w.(DOTAGES) AEIMNRV.139968w.(VERMIAN)
ADEIMOU.248832w.(MIAOUED) AEINTVW.139968w.(VAWNTIE)
ADILOTU.248832w.(OUTLAID) AACDIOR.124416w.(ACAROID)
AEHIORR.233280w.(HOARIER) AADGIOS.124416w.(ADAGIOS)
EFIOORT.217728w.(FOOTIER) AADIRSU.124416w.(SUDARIA)
EGIOORS.217728w.(GOOSIER) AAEGNOP.124416w.(APOGEAN)
EIILORR.207360w.(ROILIER) ABDEIIL.124416w.(ALIBIED)
AADGIOT.186624w.(AGATOID) ADEFGOR.124416w.(FORAGED)
AAINORV.186624w.(OVARIAN) ADEILMU.124416w.(MIAULED)
ADDENOT.186624w.(DONATED) ADEIMOW.124416w.(MIAOWED)
ADENRTV.186624w.(VERDANT) ADENTUV.124416w.(VAUNTED)
AEIKLOR.186624w.(OARLIKE) ADERSTV.124416w.(ADVERTS,STARVED)
DDEINOT.186624w.(DENTOID) AEGOTUV.124416w.(OUTGAVE)
ADIILOS.165888w.(SIALOID) BDEGIOT.124416w.(BIGOTED)
You could export the wordlist of these 202 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 5990 of these. Look at the 25 most likely ones:
/<% // and then click on seeing the first 25
You are shown:
AERATION OEDIPEAN OUTEATEN RATOONER TEARDOWN UINTAITE FETATION
UNEASIER TOLERANT LOITERER GAIETIES DANEWORT THIONATE
ETIOLATE INTERIOR NOTARIZE ANTIWEED AUDITION ROTATIVE
DETAINEE TOLERATE INDAGATE AGENESIA ABOITEAU IRONWARE
Notice
TEARDOWN and
DANEWORT
are anagrams. How many of these 8's have at least one other anagram?
Issue this query:
[]/&>1
and WHAT reports there are 277 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:
THEISTIC ITCHIEST ETHICIST CHITTIES
WORMROOT TOMORROW MOORWORT
TIMELINE MELINITE ILMENITE
ONCIDIUM MUCINOID CONIDIUM
OBLATION LOBATION BOLTONIA
DEBUGGER BUGGERED BEGRUDGE
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 16 words:
BOLSHY EVOLVE GRITHS HOLLOS HYBRIS RIGHTS SHOVEL VERMIN CERVIX GIRTHS GRIVET HOVELS REVIVE SHIVER SHRIVE WIZARD
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 | - | |
| 3's | from | 9's | - | RUTABAGAS + 3 others |
| 4's | from | 8's | - | PARAKEET + 1 other |
| 5's | from | 9's | - | TRAVERSES only |
| 6's | from | 8's | - | BLENDERS + 11 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 1 word
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 85 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 9919 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. One could take the answer 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 16112 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 2813 words can be deleted from the
slate using the delete wordlist items command:
/=8:1,2813D
Then, the same item deletion dialog can be used to find the trailing
items to delete. At this point, these are items starting at 10606.
So issue a similar command:
/=8:10606,3000D
The count has to be at least how many items are to be deleted, and 3000
is higher than needed. So, at this point, 10605 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 9919. 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 wordlist 8 back through the anagrammer with a pattern
query. This query ignores the spaces, and it should then yield 9919 words:
[]/8I# // using wordlist 8 as the source, pattern match each word
Yes, indeed it does.
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 = 28419
WHAT?: /2I! // use the states in wordlist 2 as the word source
WHAT?: [1-2]/#=1#
Number of words = 6263
WHAT?: [3-4]/#=1#
Number of words = 1041
WHAT?: [5-6]/#=1#
Number of words = 184
WHAT?: [7-8]/#=1#
Number of words = 16
WHAT?: '
ARMORIAL CALAMINE GANYMEDE MAHIMAHI MALARIAL MEGADEAL MEMORIAL SCINCOID
CALAMARI GAMODEME LAVALAVA MAINLAND MANDARIN MELAMINE MOORLAND UTILIDOR
Yield all 7-letter canonical racks which lead to words:
7./q;w#
and 19424 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 2166 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.