Castawords solutions (back to main article and Puzzles)
(All solutions are given in alphabetical order)
The parallel solving method
Let's try solving problem 1, which is quite easy. We will try to
build up all four blocks at the same time, gradually assigning letters
to the various blocks, calling each
block by the letters which we know are on it, in alphabetical order. Ideally
we start with a word that has mostly common letters, preferably no
singles. In this problem we find that SODA has four letters
which appear three times, so we assign the letters to the four blocks:
A / D / O / S. We compare DOWN with
SODA, and see that N and W must combine with A and S in some order. But
N cannot go with A because of MANE, so we have AW / D / O / NS. Next, M
and E (from MANE) must combine with D and O in some order: we'll write
this as AW / Dme / Ome / NS, the lower-case letters indicating
possibilities. Consider the words BACK and BIRD; B cannot go with
A or D. If B combines with NS, then C and K must go with D and O in
some order -- but this is impossible, since either the D or O group
must include E, which can combine with neither C (CUTE) nor K (JERK).
So B must combine with O, and since C or K must combine with D, the D
group must include M (not E, which we already know cannot combine with
C or K). So far we have: NSck / BEO / AW / DMck. Now add I
and R from BIRD: I cannot go with S (FIGS), so I goes with A and R with
N. This forces C to go with N, and K to go with D (JERK), and we can add the J
to A. So far we have: CNRS / BEO / AIJW / DKM. Next we add U and T from CUTE (T cannot go with M), H and Y from
MYTH (Y cannot go with O), and P and L from PLOY (L cannot go with U).
Now we have: CNRSY / BEHO / AIJLTW / DKMPU. Only four letters
remain. Look at FIGS and FLUX; F goes with B, X with C, and G with D.
Finally Z from ZIPS goes with B and we have the solution:
(1) AIJLTW BEFHOZ CNRSXY DGKMPU
Few problems will be as easy as this, and it might be
necessary to split the array of blocks into two or more separate cases
(see (28) for a very extreme case). We'll see some other
possibilities in walkthroughs of later solutions.
The sequential solving method
A more complex, but very powerful, method is to determine the six
letters of each cube one by one, starting with one of the letters of
highest frequency. In problem 2, we start with the letter A, and try to
find the five letters which go with it. List the letters of each word
not containing A as possibilities, each group in parentheses:
A(COVE)(CURD)(EXIT)(MOWS)(PITH)(TEMP)(UGLY). Now remove each letter
found in words containing A (i.e., LESRMPIY). We get:
A(COV)(CUD)(XT)(OW)(TH)(T)(UG). Since T is the only possibility from
TEMP, we add it to A, outside the parentheses, and eliminate H and X.
Now we have only four words left (not containing A or T), and each must
provide a different letter. Thus we can eliminate the repeated letters
O, C, and U. This produces ADGTVW as the only possibility for the cube
containing A. Now follow the same procedure with another high-frequency
letter not already placed -- let's try E, eliminating all other letters
in words containing E, as well as the established letters ADGTVW. We
get E(SK)(UR)(FR)(S)(N)(H)(UY)(ZNY), and can immediately add HNS to E,
eliminating K, Y, and Z. This also leaves us U, eliminates R, and adds
F, yielding EFHNSU.
Neither of the first two cubes contain M, so we can
start a third cube starting with M, and do the third and fourth cubes together:
you can add OPR to the fourth cube with MOWS, TEMP, and FARM, giving
ADGTVW/EFHNSU/M/OPR, then use OPR to add CI to the third with COVE and
PAIN, and use I to add X to the fourth with EXIT, giving
ADGTVW/EFHNSU/CIM/OPRX. The remaining words BALE BASK UGLY ZANY
divide the last five letters BY/LKZ, so LKZ must go on the third cube
and BY on the last, completing the solution. [The third and fourth cube method
is usually faster than constructing the third cube in the same way as
the first two, but that can be done also: We get
M(BL)(BK)(C)(XI)(I)(LY)(ZY), leading to CIM(BL)(BK)(LY)(ZY). If Y is
right, L is wrong and B is right, giving us only five letters. So Y
must be wrong and L right, giving us CIKLMZ. The fourth cube can be
found by listing the unused letters from each word (BOPRXY), completing
the solution.]
(2) ADGTVW BOPRXY CIKLMZ EFHNSU
Sometimes it takes trying several letters to find one which
gives
only one or two possibilities. If a letter produces a single set
for
one block, and no others seem to work well for a second, it might be
reasonable to use that as a base and return to the parallel
method. In the solution to (8) we will do the opposite,
placing some letters with the parallel method and using them to seed
the sequential method.
(3) Using the sequential method to build the first block with the most
common letter I quickly gives BCDIKS. Trying A for the
second block gives four possibilities, but E works much better and
gives EFHOQR. Putting A on the third block leads to the
solution easily:
(3) ALMNWZ BCDIKS EFHOQR GPTUVY
(4) We'll use the parallel method with two other important themes:
[a] Putting too many common letters on the same block will leave too few words to fill the remaining faces;
[b] A block with only one or two available faces must take letters
which satisfy all of the words it has not used. If
there is only one face available, it must be filled with a letter
common to all the words it has not used. If there are two
faces available and two words left, it cannot use any letters common to
both.
We find that PATH has no singles and the two most common letters AT,
plus a third common letter H. AXIS and SHOT put S on the block
with P, and O goes on the A block to finish SHOT: PS / AO / T /
H. E must be on the AO or H block (KEYS VENT), but if it
were on the AO block, there would only be two words left (PING WIRY) to
fill three faces, so it must be on the H block. KEYS and VENT
also put N on the AO block, and putting V on the PS block finishes
VENT: PSV / ANO / T / EH. JAMB and JUTE put J
on the PSV block, and JUTE puts U on the ANO block. Only
two words (KEYS WIRY) are available to fill the last two faces on the
ANOU block: Y would use up both, so it must be K from KEYS and W from
WIRY (not I because of AXIS, or R because of CRAG). Now KEYS
puts Y with T, and AXIS and WIRY put I with EH (putting R with JPRSV): JPRSV
/ AKNOUW / TY / EHI. Now the first block is almost full, and
needs to finish both HALF and MOLD, so its sixth letter must be
L. AXIS, HALF, and PING put XFG on the third block, and CRAG puts
C on the fourth. We have JAMB and MOLD left with one spot
on the third (M) and two on the fourth (BD):
(4) AKNOUW BCDEHI FGMTXY JLPRSV
(5) Using the sequential method, we try the most
common letters: A, L, N, and U. The first three all give
multiple possibilities for the first block, but U gives
U(FAWN)(FAX)(GOD)(GAY)(JAD)(NAVY)(INK)(AC)(ZINC). If A is
correct, there would only be three words left, not enough to fill the
block. So A is eliminated, putting C on the first block and
eliminating ZIN, leaving CU(FW)(FX)(GOD)(GY)(JD)(VY)(K).
K is added, leaving space for three letters from six
words. If J is correct, there must be one each from
(FW)(GO)(VY), which is impossible since there would only be two faces
left. So D is correct, eliminating GJO, leaving CDKU(FW)(FX)(Y)(VY).
Y is correct and V is wrong, leaving one spot for (FW)(FX), which must
be F, giving the first block as CDFKUY. Trying A again now gives
ABOS(PI)(PM)(ZI), giving either ABOPSZ or ABIMOS.
Putting L on the third block and filling in words on both arrays at
once, we get JLR on the third block and EGTX on the fourth with either
M or P. There is only one spot left on the fourth
block. In the case with P on the fourth block, N must
go on the third (PINK), but then nothing else can: VWZ are forced to
the fourth block, leaving one block of foour and one of
eight. So EGMTX is correct, which corresponds
to ABOPSZ. EGMTX has only one face left for FAWN NAVY PINK
ZINC, so N goes on the fourth and IVW on the third:
(5) ABOPSZ CDFKUY EGMNTX IJLRVW
(6) Using the sequential method, we build a block with A: A(CB)(J)(LOG)(LX)(OCK)(W)(ZB) puts J and W with A, but leaves
three solutions, so we put it aside and try E, which gives
E(C)(DMP)(D)(LONG)(LNX)(MOCK)(Q)(VN). We add CDQ and remove KMOP,
giving CDEQ(LNG)(LNX)(VN). N cannot be correct, or
there is no sixth letter, so N is wrong and V is correct, leaving
(LG)(LX), which gives L as the sixth letter. CDELQV allows
us to reduce the first block to AJW(B)(OG)(X)(OK)(ZB),
which adds B and X and removes Z; the sixth letter come from (OG)(OX),
so it must be O, giving ABJOWX. Starting the last two blocks with
U and Y from QUAY allows us to easily enter the remaining letters
(using CURB and ZEBU with U, LYNX and WHEY with Y, TEAR using R, LONG
using N, etc.) Eventually we have DAMP and MOCK with
two letters left on one block and one on the last. The common M
goes on the last and PK on the third:
(6) ABJOWX CDELQV GMRSYZ HKNPTU
(7) Using the sequential method, we first note that A and I cannot be
on the same block, since there would be only three words left to fill
four faces. So building a block with A but not I gives
A(BO)(DO)(F)(G)(X)(PU)(QUD), or AFGX(BO)(DO)(PU)(QUD). If U is
wrong, we add P and (QD), and there is no face left for (BO). So
U is correct, and eliminating D gives O, so the first block is
AFGOUX. Building a second block with I gives
I(B)(C)(CK)(HZY)(WS)(PSH)(VY)(YK); adding BC and removing K gives BCI(HZY)(WS)(PSH)(VY)(Y); adding Y and eliminating HVZ gives BCIY(WS)(PS);
S must be wrong and PW correct to fill the other two faces.
Building the last two blocks is straightforward:
(7) AFGOUX BCIPWY DMNSVZ HJKLQR
(8) We will combine parallel and sequential methods to solve this one,
which is harder than usual. PURL and QUIP have two letters in
common, so R and L have to be on the same two blocks as Q and I, in
some order. But WAIL prevents LI, so L is with Q and R with I,
and we add those to WAIL to get W / A / IR / LQ. E cannot
be on the first three blocks (EXAM WREN), so it is with LQ, and N goes
with A to complete WREN, giving W / AN / IR / ELQ. It is
hard to find an easy way forward, so we switch to the sequential method
starting with ELQ,
which has only three unused words to fill the three remaining
faces. We get ELQ(OD)(OGS)(VS), and the repeated letters
are wrong, yielding DELGQV. Building
a second block starting with AN, and removing DEGLQV, gives
AN(CK)(CHF)(UT)(PU)(UP)(THY)(T). We add T and remove HUY to
get ANT(CK)(CF)(P), then add P; with two faces left C is wrong and FK
are correct. The last two block start with IR and W, and the rest
follows:
(8) AFKNPT BHSUWX CIMORY DEGLQV
(9) This is one of the harder problems on the page. NAIL compared
to HINT puts either H or T with A. We can try constructing
the first block with both AH and AT. Starting with AH gives
AH(DUS)(F)(GU)(J)(V)(XS), which adds FJV, leaving AFHJV(DUS)(GU)(XS),
and there is no letter to all three remaining words.
Starting with AT gives AT(FOP)(G)(J)(QOP)(VO), which adds GJ to AT and
gives AGJT(FOP)(QOP)(VO). O cannot be right, or there will only
be five letters, so O is wrong, V is added, and we have AGJTV(FP)(QP),
where P must be the last letter. Since T is with A, H must
be with L, and we have the parallel start HL / I / N /
AGJPTV. Building a second complete block from HL will speed
up the solution. We get HL(BRY)(DS)(B)(MZ)(WRM)(XYS), and add B
and remove RY to get BHL(DS)(MZ)(WM)(XS). This
has two solutions: BLHSWZ and BDHLMX. Adding letters to the
other two blocks using both possibilities for BHL eventually leads to a
contradiction on one array, while the other one yields the solution:
(9) AGJPTV BHLSWZ DENORX FIMQUY
(10) Using the sequential method with I gives
I(MY)(BUN)(COE)(OZ)(O)(MNX)(QU). Adding O and removing CEZ
gives IO(MY)(BUN)(MNX)(QU). The four remaining words
must provide four different letters, so the repeated letters MNU are
removed, giving BIOQXY. Building a second block with A gives A(T)(CE)(ZE)(FW)(GFT)(SK)(SGH)(WE); adding T and removing FG gives AT(CE)(ZE)(W)(SK)(SH)(WE); adding W and removing E gives ATW(C)(Z)(SK)(SH);
adding CZ leaves one face, which must be filled with the common
S. Put D and E on the two remaining blocks and add all but
four ambiguous words, giving DNR / EKMU; the remaining four letters
(from FLOW GIFT JAIL SIGH) divide as FHJ / LG, so LG goes with EKMU and
FHJ with DNR:
(10) ACSTWZ BIOQXY DFHJNR EGKLMU
(11) Using the sequential method with S gives S(CH)(D)(FV)(FOR)(K)(HOM)(XR)(ZRO), which adds DK to S. If O is correct, we have DKOS(C)(V)(X), which is seven letters. So O is wrong, and we have DKS(CH)(FV)(FR)(HM)(XR)(ZR).
If R is wrong, we have DFKSXZ(CH)(CM), too many letters again. So
R is also correct, and we have DKRS(CH)(V)(HM),
which adds V and must have H as the sixth letter. So the
first block is DHKRSV. Building a second block with E
gives E(BUY)(NG)(U)(PG)(AN)(AN)(M)(XAY), which adds MU and reduces to
EMU(NG)(PG)(AN)(XA). G cannot be right, as it would reduce
to five letters (AEGMU), so N is correct, G and A are wrong, and we get
EMNPUX. Putting A and W on the other two blocks leads
easily to:
(11) ABCIJO DHKRSV EMNPUX FGTWYZ
(12) There are three equally common letters: A, O, and U; each of the
12 words contains exactly one of the three. Using the
sequential method with A gives
A(BON)(UCT)(UNK)(JUP)(POH)(COW)(THUG)(VO). U is impossible, as
it eliminates HP and forces O, giving only three letters. O is
also impossible, as it reduces to AO(T)(K)(J)(TG), which eliminates G and leaves only five
letters. So we know that A, O, and U are all on different
blocks. A now reduces to A(BN)(CT)(NK)(JP)(PH)(CW)(THG)(V),
which adds V. Looking at G, H, and T, we see that G
eliminates T and H and gives CP; H eliminates G and P and gives CJ; T
eliminates C and H and gives PW. In all three cases we have five
letters with (BN)(NK) left, so we know that N goes with AV, but leaves
three possible sets, so we will put it aside for the moment. Now
that we know that it cannot contain A or O, U gives
U(XLE)(BNY)(YS)(FR)(S)(RZE)(SW)(VLE), which adds S and removes WY,
giving SU(XLE)(BN)(FR)(RZE)(VLE). We also know that NV are with
A, leaving SU(XLE)(B)(FR)(RZE)(LE), and adding B. E is not possible, as it would give F also, but leave nothing for the last face. So we have BSU(XL)(FR)(RZ)(L), which adds L and removes X, leaving BLSU(FR)(RZ),
which makes R impossible and gives us BFLSUZ. This doesn't
help with the partial A-block, so we build a third block with O,
greatly diminished by knowing one and a half blocks, and the letters in
the four O words: O(X)(D)(DT)(DK)(RM)(JM)(R)(TG) gives D, R, and X, and eliminates KMT, leaving
(J)(G), which finishes another block: DGJORX. G and J
eliminate two of the three cases for A, leaving ANPTVW, and all of the
unused letters go on the last block:
(12) ANPTVW BFLSUZ CEHKMY DGJORX
(13) We'll use both methods to solve a fairly difficult
problem. A is the most common letter, and cannot be on the
same block as I: there are only four words left, and no way to choose
four distinct letters from (OD)(DOV)(JY)(ODY). If we start
with CHAT, where can the letter I go? It must go with C because
of STIB and WHIM. If we compare TODY and COLY, C must go with T
or D, but CHAT prevents T, so C is with D (and L with T) and we have
CDI / H / A / LT. CDI is enough to make building a full block
easy: CDI(PU)(GU)(JX)(R)(X) adds RX, and (PU)(GU) makes U the sixth
letter. Returning to the parallel method, we
have CDIRUX / H / A / LT. LARK puts K with H, FINK and GUAN put N
with LT; now FINK puts F with A and GUAN puts G with HK, giving CDIRUX
/ GHK / AF / LNT. APUS and STIB put S with GHK, and then put B
with AF and P with LNT. WHIM and XEMA put M with
LNPT, and then put W with ABF and E with GHKS, giving CDIRUX / EGHKS / ABFW / LMNPT.
COLY and DOVE put O with ABFW and then put Y with EGHKS (finishing that
block and also TODY) and V with LMNPT (also finishing a block).
JINX puts J on the unfinished block:
(13) ABFJOW CDIRUX EGHKSY LMNPTV
(14a) The sequential method works well here: building the
first cube with the most comon letter U gives
U(BOWL)(CLI)(A)(GAT)(IST)(OX)(WAV), and adding A to U simplifies to AU(BOL)(CLI)(IS)(OX). The last four cubes must have no repeated letters, eliminating ILO, and giving ABCSUX. Building a
second cube with N gives six combinations, but using E instead gives
E(WL)(LIP)(RK)(GT)(HMP)(JK)(JRY)(MIT)(WVY). J would
eliminate R and K, and R would eliminate J and K, so neither J nor R is
possible, leaving EKLY(GT)(HM)(MT), giving EGKLMY or
EHKLTY. Building the last two cubes with N on the third cube
gives two arrays until the last three words are reached: CLIP, HUMP,
and MIST cause a contradiction on the EHKLTY array (HUMP and MIST put M
on one and IP on the other, but CLIP fails). The other works perfectly, giving the solution:
(14a) ABCSUX DJOPTV EGKLMY FHINRW
(14b) Building the first cube
with I quickly leads to FHIJWY. Finding a good letter for a second cube is difficult; eventually A gives four possibilities: A(VUK/ETC)(PG/RB). Building
the last two cubes with E on the third cube for the two VUK cases, and
U for the two ETC cases, eventually eliminates three arrays and gives
the solution:
(14b) ACEGPT BDMOUV FHIJWY KLNQRS
(14c) The first cube
with E gives FHIJWY. The second cube with N has three possible arrays. Put O on the third cube, and eventually the solution is:
(14c) AMOSTV BEFGKY CDHPUZ ILNQRX
(14d) The two most common letters, E and I, cannot be
on the same block, since there are only three other words to fill the
other four faces. Using the sequential method with I then gives
I(BVY)(HZ)(UG)(HS)(JUY)(TBU)(WH). If H is correct, we have
HI(BVY)(UG)(JUY)(TBU), and all of the repeated letters can be
eliminated, giving GHIJTV. If H is wrong, we have
ISWZ(BVY)(UG)(JUY)(TBU). U must then be correct, otherwise we
have to add G, and have only one face and no letter shared by
(BVY)(JY)(TB). With U correct we have ISUVWZ. Building a
second block with E (but not I) we have
EF(DRUG)(JUL)(KL)(MD)(PAR)(TAU). But H must be on the same block
as either F or I (because of CHEZ FOCI HOSE), and this is impossible if
ISUVWZ is on one block and EF is on another (CHEZ prevents H from being
on a block with E or Z). So ISUVWZ is impossible and the first
block is GHIJTV, reducing the second block to
EF(DRU)(UL)(KL)(MD)(PAR)(AU). If U is wrong, we get ADEFL, not
enough letters. So U is correct and we eventually get EFKMPU for
the second block. The last two blocks follow easily:
(14d) ADOWYZ BCLNRS EFKMPU GHIJTV
(14e) The three most common letters, A, I, and O, are
on different blocks (QUIA and TRIO prevent A or O from being on the
same block as I; A and O on the same block leaves only three unused
words to fill the other four faces). Building a sequential
block with O gives five possibilities; I gives four
possibilities. These match up in four ways: BCIJNP with
either DFOQSV or EGHOQS; or either FHIJKY or FHILMS with
BCDEOQ. Putting A on the third block, and adding
letters from the four A words, will eliminate two pairings (SU on the
same block makes BUSY impossible; KN makes MONK
impossible). BCIJNP is now established as
correct. BUSY MONK POLY can be entered on the two
remaining pairings, then JOWL. One pairing already has the
letters from VIEW; adding V to the other pairing leaves no face
available for TRIO, establishing the pairing with DFOQSV as
correct. This has only one face available on the fourth
block for GRID and TRIO, so R must be on the fourth block and GT on the
third:
(14e) AGMTWY BCIJNP DFOQSV EHKLRU
(14f) The first block with I gives DIPUWZ. The
second block with A has two solutions: AFHLOV and AHJLSV.
Putting E on the third block allows most of the words to be entered,
but JOIN POSY ZANY can only be entered correctly on one
pairing. [This was an attempt at another red herring, but
turned into a regular 11-word puzzle by accident.]
(14f) AFHLOV BCGJTY DIPUWZ EKMNRS
(14g) The two most common letters are A and O. The first
block with A gives two solutions: AGIVWX and AGLRWX. The
second block with O gives BKOPQY.
Putting D on the third block and entering DRIP and QUAD makes ROUX
impossible with AGIVWX, so AGLRXW is correct. The rest is
straightforward:
(14g) AGLRVW BKOPQY CEHIMU DFNSTX
(15) 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:
(15) ADGIJN BLMPSU CEHKTV FORWYZ (JADING PLUMBS KVETCH FROWZY)
(16) Using the sequential 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:
(16) ABCDEF GHIJKL MNOPRS TUVWXY
(17) 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:
(17) AFHOQW BIJKNS CEPRUX DGLMTV
(18a) Using the sequential 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:
(18a) ACEGUW BIMTYZ DHKLOV FJNPRS
(18b) Using the sequential method with O gives three possibilities:
DFOPUX, FHIOVY, and HJOPTY. A yields four possibilities:
ACEJLU, ACPUWX, AEGHNQ, and ANPUVW. This looks awful,
but only three of the 12 pairings are possible: all of the others
contain at least one repeated letter (ANPUVW cannot pair with any of
the three O possibilities, nor can HJOPTY pair with any of the four A
groups). Setting up the three pairings with S and K on the
last two blocks, and filling in BOAS GUYS KADY, we find that one
pairing is eliminated by K and Y on the same block, and FIHOVY is
definitely correct. Adding CONK JINX LYNX gives one of the
pairings five letters on the last block, with letters still to be
placed from five words, with none in common with all five.
This reduces to one pairing and establishes ACEJLU as correct
also. The last five words can be placed (VEXT QUIT FRAT
PIER WORE) to complete the solution:
(18b) ACEJLU BGKQRX DNPSTW FHIOVY
(18c) Only one pair of words shares two letters in common: comparing GASH and HALF puts either F or L with S.
Starting a block with SL gives one solution; SF gives two.
A second block with A gives three possibilities. The
three possibilities for each block combine in five ways.
Building the last two blocks simultaneously with all five groups
eventually eliminates all but one pairing:
(18c) ADKOQR BGILNW CEHTUZ FMPSVY
(18d) Comparing SLIM and KILN puts either K or N with M.
[1] Starting a block with KM gives only one solution: DKMPTY. A
second block with NS has two solutions, but one conflicts with the MK
block, leaving BJNRSX. Putting I and L on the other two
blocks, and adding letters from JULY DEUX OPUS, puts EO on the same
block and contradicts OBEY.
[2] MN gives two solutions: AFBMNU and
BCMNRU. KS also gives two solutions: EJKRSW and
KSTWXY. The four solutions pair up in three ways; putting O
on the third block quickly eliminates two (DOWN and OBEY put DE on the
same block, contradicting DEUX). The third pairing leads to
the solution:
(18d) ABFMNU CGLOTX DHIPVY EJKRSW
(19a) 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:
(19a) AIPVWY BCEKLN DJMTUZ FGHORS
(19b) COLD shares two letters with each of CHOW, GODS, and PLOY, so
none can be the red herring, nor can BEND and VINE for the same
reason. BEND, FIRM, SACK, and PLOY have no letters in common, so
FIRM and SACK also cannot be the red herring (e.g. FIRM?? would have
three words to fill two faces). If JURY were the red herring,
BEND CHOW MATH would force H to be on the same block, impossible
because of HERS. If MATH were the red herring, BEND JURY
PLOY would force Y to be on the same block and eliminate JLOPRU; then
COLD and BEND would make the sixth letter D, but there would be no
faces to allow for VINE. By elimination, HERS is the red
herring. The last two letters of the block must come from
COLD PLOY QUIT; QUIT has no letters in common with the other two, so
there must be a common letter from them, which must be L (not O because
of CHOW). FIRM JURY MATH eliminate three letters from QUIT,
leaving Q and making the first block EHLQRS. A second block built
from O eventually gives three possibilities, so we try I instead, which
starts I(BD)(COW)(COD)(GOD)(JY)(A)(POY)(ACK). Adding A and
eliminating CK gives AI(BD)(OW)(OD)(GOD)(JY)(POY). O is not
possible, as it would produce BOJ, not enough letters. So instead
we get AI(BD)(W)(D)(GD)(JY)(PY), and add DW, leaving ADIW(JY)(PY),
which must be ADIJPW. Put I on the third block, and you can
add COLD GODS PLOY, then SACK JURY QUIT, then MATH FIRM, giving KMOU
and CFGTY on the last two blocks, with BEND and VINE left. N
must go on the last block, and BV on the third:
(19b) ADIJPW BKMOUV CFGNTY EHLQRS
(20) 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:
(20) ABFSY* CHNRZ* DIJLOX EKQTV* GMPUW*
(21) 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:
(21) ABHOTU ACGJKS DEFLOX EILWYZ MNPQRV
(22)
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. This is one of the easiest puzzles on the page. It can also be easily solved by the parallel method.]
We're going to use the sequential method, starting 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:
(22) ACFUX BOSWY DIJKZ EHLMN GPRTV
(23) A occurs in half of the words, so use the sequential method with
A, giving A(BLOW)(FIST)(HELP)(HULK)(RUDY)(TRON), reducing to A(BLO)(F)(LP)(ULK)(UY)(O). F and O join A, eliminating B and L and leaving AFO(P)(UK)(UY).
U cannot be correct, as there would be only five letters, so the first
block is AFKOPY. Building a second block with S leads to
BENSUX. Put T on the third block, adding letters to
give AFKOPY / BENSUX / CDT / IRVZ. The last five letters
pair up as JLM / HW, so the three letters join the group of three, and
two join four:
(23) AFKOPY BENSUX CDJLMT HIRVWZ
(24) This is one of the easiest puzzles on the page: ALIEN has the four
commonest letters and one of the fifth commonest. Use
the parallel method, and consider the words in the following order:
EVITA KLUTE (place T and V); ZELIG GILDA (place G, V, and D); KLUTE
PIQUE (place U and V); PANIC (place P, C, and Q); JANIS SHANE (place S,
J, and H); FARGO TOXIC (place O and X); WATER (place R, F, and W);
MARCO (place M); MELBA (place B); BUGSY (place Y). The four
blanks fill the unused faces:
(24) AUXZ** BHIKR* CDEFJY GMNQT* LOPSVW
(25)
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:
(25) ABEHLU CDIMWY FGKOQZ JNPRST
Most recently edited on May 22, 2024.
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