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.   
This article is copyright ©2024 by Michael Keller.  All rights reserved.