Castawords solutions (back to main article and Puzzles)
(All solutions are given in alphabetical order)
(1) AIJLTW BEFHOZ CNRSXY DGKMPU
(2) ADGTVW BOPRXY CIKLMZ EFHNSU
(3) ALMNWZ BCDIKS EFHOQR GPTUVY
(4) AKNOUW BCDEHI FGMTXY JLPRSV
(5) ABOPSZ CDFKUY EGMNTX IJLRVW
(6) ABJOWX CDELQV GMRSYZ HKNPTU
(7) AFGOUX BCIPWY DMNSVZ HJKLQR
(8) AFKNPT BHSUWX CIMORY DEGLQV
(9) AGJPTV BHLSWZ DENORX FIMQUY
(10) ACSTWZ BIOQXY DFHJNR EGKLMU
(11) ABCIJO DHKRSV EMNPUX FGTWYZ
(12) ANPTVW BFLSUZ CEHKMY DGJORX
(13) ABCSUX DJOPTV EGKLMY FHINRW
(14) ACEGPT BDMOUV FHIJWY KLNQRS
(15) AMOSTV BEFGKY CDHPUZ ILNQRX
(16) ABFJOW CDIRUX EGHKSY LMNPTV
(17) HATE shares two letters in common with each of FACT and HAND,
so none of them can be the red herring (nor can VOTE and WEST, though
we don't need to know that). If NECK is not the red herring, then
we need to arrange the letters as CH / A / NT / DEF to allow for HATE,
HAND, and FACT. Then K must be with A to allow for NECK, making
MARK the red herring. So the red herring is either MARK or
NECK, and we can add other words to two possible frameworks: CH / AKMR
/ NT / DEF (if MARK is the red herring) or FH / A / DT / CEKN (if NECK
is the red herring). The second group of the first
framework (AKMR) already has four letters and needs either W or S from
WEST, V or O from VOTE, and something from GILD, which would force it
to have seven letters, which is impossible. So NECK is the red
herring and the second framework is correct. Adding HYMN and MARK
makes the framework FHR / AY / DMT / CEKN. G must be with
the first group to allow for both GILD and GAZE, giving us FGHR / AY /
DMTZ / CEKN. I must go with the second group to allow for
JOIN, and O with the first to allow for VOTE, giving us FGHOR / AIVY /
DJMTZ / CEKLN. U must be with the third group to allow for BURY
and PLUS, giving us FGHOR / AIVY / DJMTUZ / BCEKLN. We still have
to place W, S, and P in the first two groups to allow for WEST and
PLUS. The first group only has one spot left, so S must go there
and W and P in the second group, giving the full solution:
(17) AIPVWY BCEKLN DJMTUZ FGHORS
(18) AEGJMT BEHOPX COUVWY DKNSTZ FILQRS
(19) To start, assign the letters Y U K O N to five different
blocks. A and K cannot be on the same block (otherwise we could
not spell DURAK); A also cannot be on the U or N blocks (JUNTA) or the
O block (OWARE), so A is on the Y block. R cannot be on the U,
YA, or K block (DURAK), or the O block (PROBE), so R is on the N block,
and we can assign D to the O block so that we can spell DURAK. So
far the blocks are AY / U / K / DO / NR. I cannot be on YA
(GIZA), U (QUBIC), K, or NR (RISK), so I is on DO. E must be on K
(because of CLUE, REALM, PROBE), and W is on U to complete OWARE.
B must be on AY (PROBE and QUBIC), and P on UW completes PROBE.
Now the partially filled blocks are ABY / PUW / EK / DIO /
NR. L is on DIO (REALM and CLUE), and M on PUW completes
REALM. T is on EK (JUNTA and GHOST), and J on DILO completes
JUNTA. C is on NR (QUBIC and CLUE), and Q on EKT completes
QUBIC. CLUE forces one of the four blanks to be on
ABY. G must be on MPUW (FROG, GIZA, and GHOST), S on
ABY (GHOST and RISK), and H on CNR, completing GHOST. The blocks
are now ABSY* / GMPUW / EKQT / DIJLO
/ CHNR (* indicating a blank). There must also be a blank
on UWPMG (RISK), which fills that block, so there is only room for the
V in VIRA on EKQT. Now EKQTV must contain a blank, since there is
not room to put both F and Z to spell FROG and GIZA. This fills
that block, and the Z in GIZA must be on CHNR, and the F in FROG on
ABSY*, filling that block. The only place left for the X in
HEX is on DIJLO, and the last blank is on CHNRZ. The
solution:
(19) ABFSY* CHNRZ* DIJLOX EKQTV* GMPUW*
(20) To start,
assign the letters S H E E P to five different blocks. We also want to
remember which four letters are repeated: A, E, L, and O.
I cannot be on the S block (SQUID) or the H or P block (APHID), so it
is on one of the E blocks. R is not on the S or H block (SHREW), the EI
block (TIGER), or the other E block (which must be the E in TIGER since
the EI block will be used for I), so R is on the P block. C is not on
the H block (CHIRO), the EI or E block (CIVET), or the P block (COYPU),
so it is on the S block. T is not on the C, EI, E, or PR blocks (CIVET,
TIGER), so it is on the H block, and we can finish CIVET and TIGER by
placing V on the PR block and G on the CS block. We can also place one
of the two Os on the E block to finish CHIRO. D goes on the EO block
too (APHID, SQUID), and one of the two As goes on the SCG block to
finish APHID. COYPU and SQUID force U to go on the HT block, and Q goes
on the PRV block to finish SQUID. So far the blocks are ACGS / HTU / EI
/ DEO / PQRV.
Either F or N must go on the PQRV block (FINCH), but if it is F, there
is only room for one more letter and we need two spots for L/A/M from
LLAMA and B/O/N from BISON. So N must go there (with F going on the DEO
block to finish FINCH), and we have only one spot left. Since we need a
letter from both LLAMA and MOOSE, the sixth letter must be M. The
second O then goes on the HTU block to finish MOOSE. Y goes on the EI
block to finish COYPU. Since the ACGS block has one of the As, it
cannot have an L (LLAMA), and thus must have the J from JUREL. The A
there must be the A in QUAIL, so one of the Ls is on the DEFO block to
finish QUAIL. The DEFLO block cannot have an A (LLAMA), so it has the X
from HYRAX, which fills the block. Since there is no room on DEFLOX,
the W in SHREW must be on the EIY block. Since the DEFLOX block is
full, the O there must be the O in both BISON and OKAPI, so the HOTU
block must have the B for BISON and either K or the second A for OKAPI.
It cannot have K, since the block would then be full and have no room
for any letters from LLAMA. So the block must be ABHOTU. The EIWY block
must have the second L to finish LLAMA, as well as Z to finish ZEBRA
(since the E in ZEBRA must be on the DEFLOX block). The K goes on the
ACGJS block to finish OKAPI, and the puzzle is solved. The full
solution:
(20) ABHOTU ACGJKS DEFLOX EILWYZ MNPQRV
(21)
First we need to list all of the five letter words (including LOGIC
which appears in the title -- sorry!). There are thirteen:
BREAK CHORD LOGIC
LYRIC MAJOR MOTIF POLKA ROUND
SIXTH SUITE TWICE VOICE WALTZ
[I confess that the puzzle contained a mistake when it was published in
2000 on the Games Cafe: it had the erroneous five-letter word which, written (by me) by mistake instead of that.]
We're going to use the second solving method described on the main page (as used to solve puzzle 2): instead
of starting with one letter on each block and adding letters one by
one, we build one entire block at a time. Start with the block
containing O (it's usually best to begin with one of the letters which
appears most often in the word list). What other letters are on
this block? One of the letters must be from TWICE, but not from VOICE
or MOTIF, so it is W. One must be from BREAK, but not from MAJOR,
POLKA, or VOICE, so it is B. One is from SUITE, but not from
MOTIF, ROUND, or VOICE, so it is S. One is from LYRIC, but not from
CHORD, MOTIF, or POLKA, so the fifth letter is Y. Already we have
an entire block: BOSWY. Next, the block containing I cannot
contain any of the letters BOSWY, nor any of the letters from other
words containing I. One of the letters must be from CHORD, but
not from LYRIC or SIXTH, so it is D. From MAJOR we can use either
A or J, but if we use A, we will find that there are too few letters
left, since every word contains A, D, or I. So we eliminate A and
use J from MAJOR, and subsequently Z from WALTZ and K from BREAK.
That completes a second block: DIJKZ. One advantage of this
method is that it usually gets easier with each completed block.
The block containing R must also contain T from WALTZ, P from POLKA, V
from VOICE, and G from LOGIC. The last two blocks can be
done at the same time, by starting with any word (e.g. BREAK) and
putting the two remaining letters on different blocks:
BOSWY/DIJKZ/GPRTV/E????/A????. Then pick a word with either E or
A (e.g. MAJOR), and add the fifth letter to the other block:
BOSWY/DIJKZ/GPRTV/EM???/A????. Using MOTIF, TWICE, and
CHORD in succession gives us
BOSWY/DIJKZ/GPRTV/EHM??/ACF??. The remaining letters
come from POLKA, SIXTH, SUITE, and finally ROUND. The full solution is:
(21) ACFUX BOSWY DIJKZ EHLMN GPRTV
(22) AFKOPY BENSUX CDJLMT HIRVWZ
(23) AUXZ** BHIKR* CDEFJY GMNQT* LOPSVW
(24)
STEM has six possible arrangements fitting 211: ST/M/E/-, SE/T/M/-,
SM/E/T/-, TE/S/M/-, TM/S/E/-, and EM/S/T/-. Five of the six have
only one cube which can contain A so that MAST and TAME both fit; the
sixth has three, giving eight possible arrangements of AEMST, but SAME
eliminates five of these, leaving ST/EA/M/-, SE/TA/M/-, and
TE/SA/M/-. Adding SWAM allows the W in eight positions, but
WENT makes one of those impossible (mark these seven with lower case w
on the three main arrangements). N can occur on two or three
cubes for each of the seven remaining arrangements (it's easier now to
write the two arrangements with MW on the third cube separately).
CANE, GNAT, NOSE, and SAND make four placements of N impossible,
yielding 12 arrangements so far (we won't try to place CDGO yet).
We want to add I next, so it will be easier to write each N position
separately, leaving two possible w positions (lower case) on some:
STw STw SEw SEw TE STN ST SE
EAw EAw TAw TAw SA EA EA TA
MN
M MN
M MN MW MWN MWN
-
N -
N W -
- -
Now we can add
I (using MAIN and SITE), which has two possible cubes (third and
fourth) for two arrangements, and one for the others. Using MINK
and TAKE to add K reduces to one arrangement, and places K definitely
on the fourth cube:
STN
EA
MWI
K
NOSE and WHOM
force O onto the fourth cube as well, and yields two possible cubes for
H (first and second). CANE and LOCK put C on the third cube, with
L also on first or second. GNAT and GRIM put G on the fourth
cube, and R once again first or second; FARE forces R onto the first
and F onto third or fourth. SAND and DOGS put D
third. PICK puts P first or second:
STNR hlp
EA hlp
MWICD f
KOG f
OKAY and STAY
put Y third, filling that cube and forcing F to the fourth. Since
the third cube is full, SOUR puts U second and JOKE puts J first and
JAMB puts B first or second. L and P cannot be on the same cube
because of FLOP, so one is on the first cube, filling it, and B and H
are forced to the second, filling it too. CHIP puts P on the
first and L on the second. Finally QUIZ puts Q and Z on the
fourth cube, completing the solution:
(24) ABEHLU CDIMWY FGKOQZ JNPRST
(25) Using the second method with letter E, we have
the following: E(BASK)(CS)(GY)(HU)(JUY)(MUSK)(PA)(VAY)(WK). Since
there are no overlaps with CS/GY/HU/PA/WK, they must consitute the five
letters grouped with E, and we can eliminate any other letters (BJMV),
leaving E(ASK)(CS)(GY)(HU)(UY)(USK)(PA)(AY)(WK). (ASK)(USK)
indicate that either both of AU, or neither of AU, are on the E
cube. If neither, then HPY must be on the E cube, leaving
EHPY(SK)(CS)(WK), which gives either CEHKPY or EHPSWY. If
both, then HKPSY are eliminated, leaving ACEGUW.
Put the E cube aside for the moment and use the same method on A.
We get A(COT)(DE)(FET)(GO)(HUNT)(JU)(MU)(WN)(ZONE). We look at three
cases:
(1) If O is on the A cube we get ADFOW(HU)(JU)(MU), which must be
ADFOUW, otherwise there would be too many letters. This can only
pair up with CEHKPY among the three E possibilities.
(2) If O is absent, we get
AG(CT)(DE)(FET)(HUNT)(JU)(MU)(WN)(ZNE). If E is present, we
get ACEGW(HU)(JU)(MU), which must be ACEGUW, which is one of the three
possibilities we found for E.
(3) If neither O nor E is present, we have
ADG(CT)(FT)(HUNT)(JU)(MU)(WN)(ZN). There are no overlaps
with FT/MU/ZN, which must be the other three letters, eliminating C
(forcing T and also eliminating F) and J (forcing U and eliminating M);
but this gives both U and T, contradicting HUNT. So this case is
impossible, and we have either ADFOUW/CEHKPY, or ACEGUW
alone. Trying ADFOUW/CEHKPY with T on the third cube, and
using the third/fourth cube method fails, giving
ADFOUW/CEHKPY/BGIJMTVZ/LNRS, with too many letters on the third cube
and too few on the fourth. So we know that ACEGUW is one
cube, and we need another letter for the second cube.
Trying I by the same method, we get
I(BS)(COST)(FT)(GOY)(HUT)(JUY)(MUS)(VY)(ZO), but we can eliminate U and
W because they are on the AE cube, leaving
I(BS)(COST)(FT)(OY)(HT)(JY)(MS)(VY)(ZO). There are no overlaps
with BS/FT/VY/ZO, so four of the remaining five letters on the I cube
must be among these eight. The sixth letter can be any of the
remaining letters CHJM. (BS)(MS) indicate that B and M must
either both be on the I cube (eliminating CHJS and leading to BIMTYZ),
or neither (forcing S and leading to FHISYZ). Finally we try the
third/fourth cube method with both ACEGUW/BIMTYZ and ACEGUW/FHISYZ,
putting N on the third cube and doing both cases at the same
time. FHISYZ fails quickly, as HUNT and ZONE put TO on the fourth
cube, contradicted by COST. BIMTYZ works, however, giving
ACEGUW/BIMTYZ/NRS???/DHKOV? with FELT, JULY, and PAIL left to
place. All three share L, which puts L on the fourth cube, and
FJP on the third, giving the solution:
(25) ACEGUW BIMTYZ DHKLOV FJNPRS
(26) COIL, FISH, and GOSH put H/I/O/S on four separate
blocks. GOSH and FISH then place G and F:
H/GI/FO/S. COIL puts C and L on blocks 1 and 4 in some
order; CLAY puts A and Y on 2 and 3 in some order, but OVAL puts A on 2
and thus Y on 3. So far we have: H[c/l]/AGI/FOY/S[c/l]. OVAL puts
V on the same block as C: H[cv/l]/AGI/FOY/S[cv/l]. SIZE and
DOPE put E on block 1, and SIZE then puts Z on block 3.
MONK and WALK put K on block 1 or 4, but not with L, so it joins CV,
and W goes on block 3. Block 3 has room for only one letter,
which must be in both BRED and TURN, so the sixth letter on block 3 is
R. So far we have:
EH[ckv/l]/AGI/FORWYZ/S[ckv/l]. BRED puts B and D on blocks
2 and 4, while DOPE puts D and P also on 2 and 4, so one has D and the
other B and P. We'll use curly brackets for the second
group: EH[ckv/l]/AGI{bp/d}/FORWYZ/S[ckv/l]{bp/d}.
There are five unassigned letters from JOLT MONK TURN: J,M,N,T,U, which
all must go on blocks 1, 3, and 4. If block 1 contains L,
it would need three more letters, but it would already have L from
JOLT, so it could not add three more letters from two
words. So block 1 contains CKV and block 4 contains L:
CEHKV/AGI{bp/d}/FORWYZ/LS{bp/d}. In the same way, block 4 cannot
contain D, as it would need three more letters from MONK and TURN, so
it contains BP and block 2 contains D: CEHKV/ADGI/FORWYZ/BLPS.
Now block 1 needs only one letter, which must be shared by JOLT and
TURN, so T goes on block 1. JOLT puts J on block 2, which
needs one letter shared by MONK and TURN. So N goes on
block 2, and both U and M finish block 4. What is unusual about
this solution? Each block can be anagrammed to form a word:
(26) ADGIJN BLMPSU CEHKTV FORWYZ (JADING PLUMBS KVETCH FROWZY)
(27) Using the second method, building a block with A
gives only one combination with six letters (you might already guess
the joke). A second block with O gives MNOPRS.
Combining them and adding the remaining letters in pairs gives
UX[gil/tvwy]/HJK[gil/tvwy]; the four-letter group goes with UX and the
three-letter group with HJK. The solution is four groups of six
letters, sequential except for the missing Q:
(27) ABCDEF GHIJKL MNOPRS TUVWXY
(28) ADOWYZ BCLNRS EFKMPU GHIJTV
(28) In
the ten-word non-ideal puzzle, since every word contains A, the five missing letters FHOQW must be on the block with
A. The rest of the problem is not very hard:
(29) AFHOQW BIJKNS CEPRUX DGLMTV
Most recently edited on April 16, 2024.
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