Solver’s blog – Listener 4036 – Base Jog by Brimstone 

In the elusive pursuit of an all-correct tally for the year, the numerical puzzles offer me a particular challenge of their own. On the one hand I am one of the solvers who genuinely seems to enjoy them – reminiscences of Maths A level classes and sultry summers in the Upper Sixth – but on the other hand there is more of a possibility of being stymied by the language, of just not understanding what is going on. The last numerical, the Pentomino Factory, was a gentle piece of work, though I certainly didn’t manage the 10-minute solve that some of the other bloggers made claim to. It would seem likely that this month’s offering will be somewhat stiffer.

So first problem is a printer malfunction. Access the Times website OK, check the solution to Beat It (think I was OK on that one), and click print on Base Jog. Nothing. My printer has got it into its head that it’s on holiday, feet up, all jobs stopped. One restart later, one printer utility disinterred from the depths of the Hard Disc folder, one new black ink cartridge for good measure (make your printer feel really special) and a rather streaky copy of Brimstone’s puzzle chugs out of my printer. At first sight I guess this has to do with baseball about which I know nothing whatever. Still a bit to do first. Takes me a while to figure this concatenation lark and try various permutation of 2, 7 and 21 before I figure out that it’s actually a rather simple way of assembling the digits of the answer. That’s ok. Otherwise nothing too fancy in the jargon – no modulos and stuff like that. I think a numerical is a good place to make my blogger’s debut: on past experience I know that most numericals involve finding a way in and then using tight logic to derive successive answers. In most cases I wind up making weird scribblings on the back of the copy, assuming some parameter somewhere that I shouldn’t and having to make a second run at it. This way I can keep tracks on what I do.

So this uses the numbers from 0 to 25 and letters A to Z. Soon spot that the clue numbers are indicated by letters and thus by brilliant deduction figure out that Y and Z, not used as clue indications, must be 0 and 25. H across – B^(UR+Y) – suggests that Y is 0 otherwise the answer could be 1 or 0 (if B was 1 or 0) or impossibly long. No single digit answers so Y = 0 and Z = 25.

Next notice that only four clue letters are both across and down clues so A,B, H and U correspond to 2, 4, 7 and 9. Comparing B ac and B dn we are going to have very similar answers (both have AS in them) with B ac the slightly smaller of the two by (A+B+H), so by somewhere between 13 and 22. So B must = 2 as none of the other grid entries would work out (eg 9 ac is 6 digits and 9 dn is 3 digits). Again comparing H ac and H dn it’s clear that the down answer is the longer of the two so H must be 4. U down = HUH = 16 x 7 (112) or 16 x 9 (144) = 3 digits so U = 9 and thus A = 7 and 9 dn is 144. Take pause to pay for very nice but hideously over-priced take-away curry and wander home to be assailed by kids cheering a very motley bunch of entrants in Britain’s Got Talent. All mathematical skills completely obliterated for the night. Maybe I should go on the box and demonstrate solving a Listener crossword – happy to do it in a squeaky falsetto or with a ventriloquist dummy if it would get me the audience vote. Six numbers sussed with a minimum of work, no spreadsheet in sight. It’s going ok.

Saturday. Decide to write the blog. Reassemble the now rather hazy logic of last night’s deductions. Decide that they look quite compelling in type. Try to erase disturbing recollection of sado-masochistic dream about Simon Cowell and decide not to even mention this is in the blog.

Meanwhile H ac (4 ac) = B^(UR+Y) = 2^(9xR + 0) = 512 x R and is a 3 digit answer so R=1. B dn (2 dn) = AS = Sx7 and has 1 as a middle digit. S is max 24 so answer lust be 110 ish (210 ish would require S to be 30+) so S = 16 (AS = 112) or 17 (119). Thus B(2) ac starts with a 1 and = AS – (A+B+H) = 112 or 119 – 13. Cunningly 112 is thus too small so S = 17. Can now 4 ac and 2 ac and dn. Going frightfully well. Meanwhile Q ac (ASH) = 28 x 17 (dust off calculator) = 476 and V ac therefore = 477. Note that first digit of 16 ac is a 4 but not making any guesses just yet. Also M dn = 36 – 8 + 18 = 46. As there only two down answers of two digits and 7 dn is A dn, M is presumably 19. Memory of Simon Cowell in peaked leather cap thankfully begins to fade…

A(7) dn = M+A+Y+S = 19 + 7 + 0 + 17 = 43. Stick it in. Think is going to turn nasty pretty soon. Uneasy sense of lull. Toy with the idea of finding out what kind of multiple of 17(S) would create a six digit answer starting 13 for 9 ac but decide that that is a bit too much of a slog. Have a little look at A ac whose final digits include U^E. Reckon E can’t be that big otherwise we’ll run out of digits. Still have 3, 5, 6, 8 etc up for grabs which would give 729, 59049, 531441 and so on. As there are three concatenations in the clue conclude that E must be 3 otherwise it’ll squeeze the other bits of the clue. Slightly dodgy but seems a reasonable assumption. Further reckon that the first concatenation – R + ED – gives the first two digits = 4? = 3xD + 1, so D = somewhere between 13 and 16. As D is an across clue letter that rules out 14. First concatenation of D ac is B^U + R = 513 so D can’t be 13 ac (already have a 4 as the second digit) or 16 ac (too short and starts with a 4) so reckon D is 15. Enter first concatenation of 15ac and 7ac.

Peer at 3(E) dn. 6???5?. Reckon each concatenation generates two digits so RUN = 9 x N = 5?. So N = 6. Have a nasty feeling that I am beginning to construct a house of cards here but let’s plough on for the moment. Complete 15ac. Last concatenation (time to copy/paste that word) of 17(S) ac is OSE = 51 x O and either = ?4? or 4?4?. Latter would require value of O to be 80 odd so settle on first alternative. Hmm. Doesn’t work. Sound of tumbling cards. Entertain fantasy of misprint in crossword. Apparently not.

Change of tack. Investigate Wikipedia article on baseball in 1974 and read about Hank Aaron surpassing Babe Ruth’s home run record. Seems a probable theme and therefore presume that their names will appear in the final grid and thus that the letters of their names must be single digits. So far have 1/R, 2/B, 3/E, 4/H, 7/A and 9/U so looking hopeful. Still puzzled about 17 ac.

Aah, feel very stupid. Loyal readers I can hear you shouting from the back of the auditorium – 19 DOWN, 19 DOWN, you subtracted 2 x 4 instead of 2 x 2. Yup. Big dunce hat. So let’s try that again. HU – B – B + UB = 36 – 2 – 2 + 18 does of course = 50 so last concatenation (hey the paste worked!!!) = ?5? so O = odd number so must be 5, and K and N must be the missing 6 and 8 unless 0 is used as O for the lettering so maybe leave O = 5 out for the moment. There again 51 x something = ?5? so can only be 3, 5, 7 or 9 so must be 5. As we were. 3(E) dn starts 6? so GO = 5 x G must be 60 or 65. 60 is out (gives a zero at the start of 5 ac) so G = 13 and 3 dn is 657554.

JOE (F dn) is going to end in 5 or 0. 12 dn looks a likely candidate. ?65 = 15 x J giving unique value of 11 for J. Putting that value into J(11) ac for JADE/D = JAE = 11 x 21 = 231 which would marry up 1 as the first digit for 12 dn. So will run with J = 11 and F = 12.

Starting to brave up on the guessing front. With theme identified and half the numbers sorted it would be easy to lose interest so will go for the reckless fast track route. So 8 dn is ?33. Reckon only candidates from remaining down clues are T, W and X as the others have concatenations (good old paste). From known numbers T dn = 9 x (27 + T). Only 333 would work giving T = 10 so no good if T is 8 dn. Humph.

Meanwhile U ac = JOG(G – E + R)S = 715 x 11 x 17 = 133705 which thankfully fits. R dn ends 2?0 so second concatenation, which = C x 23 means that C must = 10. C(10) dn starts TU and 7? so guess that T is 8 (U = 9). SO check T dn again. Maybe I made another boob. (B + A) = 7 + 2 = 9. (T + HE + S) would = 8 + 12 + 17 = 37. Whoops. Back to the back of the class again. 9 x 37 = 333. Last concatenation of 10 dn = S + T + E + N = 34 and middle concatenation = 6 x 13 = 78.

Fill in 4 dn – 5632:9:28. So K = 22 looking at the last 2 digits of 13 ac. 5 ac = 53. 22 ac comes out to 83 so confirming K. I ac = 39 x 11 = 429 so must be 16 ac: I = 16. P must be 20, it’s the only remaining across entry that’s long enough for the clue. So – 32:846:X – 4 which equals ?0 so X = 14 or 24. X is a down clue letter so has to be 14.

Q ac (476) and V ac (477) must therefore be 23 and 24 or vice versa so both of those entries start 47?.

For 7 ac I now have 46937?3. Looking at the concatenations R + ED = 46 so W + HI + T presumably = 93 = W + 64 + 8, so W = 21. EB + L + U^E = 6 + L + 729 = L + 735 = 7?3 so L = 18 as 8 is already taken. 13 ac – WEA = 441 which gives value of 123 for X dn which works against the individual values. S ac, 1^st concatenation is (W + I + N)L = 774. Confirm L dn: LE + F + T = 54 + 12 + 8 = 74 which is good; (-A + N + DR + I + G + H) = -7 + 6 + 15 + 16 + 13 + 4 = 47 which again is good; final digit is T = 8.

N dn = N + E + W + Y + O = 6 + 3 + 21 + 0 + 5 = 35 which is good; -R + K = 22 – 1 = 21; Y + A – N + KEE + S = 0 + 7 – 6 + 198 + 17 = 216 which makes 23 ac 476 so = Q ac and thus V = 24.

Confirm W dn: QU + IZ + Z + ED = 207 + 400 + 25 + 45 = 677 which is good 

Finally calculate first concatenation of 1 dn = PI/T = 20 x 16/8 = 40 so first digit of 1 dn is 4.

Reckon I’m looking for BABE RUTH = 2723 1984 which is easily spotted in two downward diagonals. AARON translates as 77156 which lies along upward diagonal adjoining the end of BABE and RUTH. Numbers on symmetrical diagonal give 43610 = HENRY (not Hank) and the Wikipedia article talks about Aarons’ 715^th home run beating Ruth’s previous record so reckon missing central digit is 1 allowing for intersecting diagonals of 714 and 715.

Reckon that’s done. Slightly before midnight on Saturday including an afternoon drive to Oxford and back to hear my son singing evensong at New College with his school choir whilst fragile students lean sweetly against each other, weary from their graduation parties. So no prizes for speed but another surprisingly gentle Listener done and dusted.

Just before turning in wonder about the title. Slightly awkward if obviously descriptive title. Idly convert letters to numbers. 714 and 715. Very nice.

For this puzzle, one letter had to be omitted from every clue (not restricted to the definitions) which needed to be restored, and then added somewhere into the answer as well. Sounds straightforward; yes, that’s what I thought.

The first clue ends “animal exhibiting loppy udders”. Well that’s going to be “floppy udders” when the missing F is replaced. A shame I can’t think of such a beast, but it indicates this puzzle is going to be a breeze. FLW (Famous Last Words)! It was downhill from then on. After the first pass of clues, I had 6 answers: EYES, STAIR, LILY, RUT, SPRING and SHRI. But they had to be entered in pencil, since the missing letter in the clue had to be fitted in somewhere. True, I had been able to identify a number of potential (but probably incorrect) missing letters in clues, and then some more answers were solved, such as EMIGRATE, WILLOW and ELUENT, but progress was quite slow.

Along the way, there were some lovely misdirections: “One socially acceptable” where the definition is just the “One”; “extremely” relating to “vibrate” rather than to “liable” in the clue to ACTIVE; and who would have guessed that “Allah” was the word that had lost a letter (N for NALLAH meaning a ravine)? Then there was the clue to GISM which was almost the last I solved: GORMANDISM without a jumble of RANDOM.

After some time, the moment of revelation: the missing letters I had started ?O?E?HH?Y?? and there he is, JOSEPH bloody HAYDN again! Now I know it’s the 200th anniversary of his death, but anyone would think he was the greatest British composer. Yes, I know he was good, but that’s not the point! And lo and behold, there are his symphonies again: SURPRISE, MIRACLE, LONDON, etc. Previous puzzles in the “series” are MynoT’s EV840 “Late” last December, Phi’s IQ122 “Papa” in April and Llig’s “Exit Pursued by a Bear” in last month’s Crossword. In my opinion, however, and not because I’m writing this blog, this was the best implementation, not to mention the most difficult. It also appeared as near as possible to the anniversary date, bearing in mind that the 30th May was Listener numerical puzzle day (about which I don’t want to speak yet!).

All in all, though, an enjoyable and testing puzzle, so many thanks to Nibor.

Oh, and no animal with floppy udders I’m afraid; it was of course POLYP, being LOPPY juddering.

Oh what a struggle for the junior easy clues 8 X 8 team. We gazed in shock, as usual, at the empty grid on Friday evening and got nowhere at all. Why? As usual, we had not fully understood the preamble and we were searching for miraculous letters that could make the clues into completely sound sense and also be added to whatever word we slotted into the grid – and, we fondly imagined, make sense there too. Take ‘Maybe Ripon and Exeter, for instance? Yes!’ GREE seemed to be a fair anagram (‘maybe’) of R, E and EG, (it’s Shakespearian and Scottish for ‘agree’) and if we added an A we had a lovely word to slot in. Of course, this sort of thinking didn’t get us far when we worked out PROSCRIBE, FRONTAGERS, EMIGRATE, WILLOW, ON THE NOD, ACRONYMS and so on, and we soon realized that Exeter and Ripon, as SEES, made a fine additional letter on EYES.

Horror-struck, we recognised that those additional letters could go in anywhere, so we had to split our lights into mini-lights and even minier lights when they crossed each other, with minute letters. We chipped away at it and struggled, producing over half the solutions and an obscure phrase about ?OSE and THIRTY-FIRST and EIGHTEEN.

If only we had solved 1ac. sooner! We saw the anagram in LOPPY but had ‘pudders’, ‘judders’ and ‘dudders’, all as potential anagram indicators, (‘pudder’ implies a pother and ‘dudder’ is confusion!) with that curious word ‘exhibiting’ in the middle of the clue (that we still don’t understand – help please Denis?) Amateurs that we are, of course we decided the most obvious ‘dudder’ had to be the one. It was after a night’s sleep that we made the break-through and Joseph Haydn gave us our first pdm. From there it was an easy step to ‘JOSEPH HAYDN DIED THE THIRTY-FIRST MAY EIGHTEEN-O-NINE. Why ever did it take so long? It isn’t a month since we completed Llig’s Crossword Club crossword on the same theme!

With the letters we had, plus all those additional extra letters, it didn’t take us long to identify the jumbled symphonies, after all, Haydn et al. didn’t give names to many of the 106 and word-length helped. Surprise, Military, Drumroll, London and Clock were soon in place. Drumroll was one word in our book so that led us to the two words of LE MATIN for the one with the M of ACRONYMS crossing it – and then to MERCURY as another red herring. We hadn’t recognised that these were only ‘half a dozen’ of the twelve London symphonies and that MIRACLE was required. (Kick ourselves again and make a resolution to pay attention to the title next time!)

This is where we originally failed. The symphonies helped us to fill some of our gaps but not all of them. We couldn’t work out ‘God saves beloved hero(i)n in sea, acceptable not base’ (3d), and were especially mystified by ‘Excessive f(e)asting gets rid of random explosive energy’ (35ac). What a long detour these caused! The first clue was for JOSHUA, and the second for GISM, and it needed explaining to us by a wise friend that GORMANDISM had to lose ‘explosive’, or ‘jumbled’ RANDOM to leave GISM meaning energy. (Isn’t that one of those indirect anagrams that get me into trouble?)

Energy was needed to struggle through the last small words – the little ones were the most difficult for us – TOLE, ASS, YOLK and even TIDE! Thank you, anyway, Nibor for showing us just how far we still have to go as solvers – this was a fearsome challenge.

Shirley Curran.

Sheer joy, this one, for the junior 8 X 8 coffee break team! We were away from home and son, James, lent me his computer to download Schadenfreude’s Overhead Reduction. ‘Oh, no!’ I groaned, when the bar-less grid appeared. He leant over and commented, ‘Looks as though the crunch has hit the crossword world. The editor is probably cutting down on ink!’ Out of the mouths …

However, even with no Chambers and no Bradford, we cold-solved almost half the clues, sitting with our knees up by our noses on Easyjet. These were wonderfully clear clues, EASIER, RIFTED, TIERCES, VANISH, MANTRA, TERSEST, ARCTIC, FENNEC, IGAPO, MAORI and so on.

Fortunately, these clues intersected and allowed us to construct a putative grid (sadly beginning at the bottom, which, of course, renders things more difficult!) We had a bit of a squabble about what 180-degree symmetry is and fiddled with top/bottom, mirror, left/right, attempting to suss out the difference, but it soon became clear that we were almost creating two separate half crosswords. The endgame looked manageable – we just had to complete the solving to get there.

Back home, we got to work with Crossword Compiler and everything fell into place. Well – of course we had the junior 8 X 8 team habitual red herrings. How easily one letter can cause trouble! Thoughtlessly, I put AREOLAS for AREOLAE (overlooking the Scots ‘ae’ for ‘very’). That led to SPONSORS in the place of ENSURERS – defensible as a solution with lots of wriggling, which clashed with a tentative ACHEBE for the writer. I can hear the experts wondering how we could go so wrong! STERN (S for sea and TERN for ‘Unusual back on a seabird’) seemed to be confirmed by Chambers definition of STERN as an obsolete word for ‘back’ – and so on. Our top, right-hand corner was a totally different crossword from the official solution – so it goes! Luckily MURRE and MACE put us back on track.

We were somewhat nonplussed by DONNEE as ‘Laureate’s last letter’ might have been an s (I need Denis, here – could the apostrophe s have counted as a last letter in this case?) but the intersection with WHETS obliged us to accept the Chambers definition of DONNEE as ‘basic assumptions’. Not quite French, but, as Eeyore would say with bonhommie, ‘It’s an English word for ‘donnee”.

So there we were – a delightful filled grid. I printed a few in case we had to attach top to bottom, left to right or do some zig-zag snipping and turn half upside down. You must understand, we are new to this and never sure what sort of hoop we might have to leap through. However, the dénouement was wonderful. THE MIDDLEMAN quickly obliged by filling the strip down the middle, and all that was left to do was find the glue stick.

What a fine puzzle! The clues were so clear (even if I did manage to bungle a few) and the endgame was within reach for relative novices. There was none of the extreme complexity that sometimes keeps us struggling all week. This is one to suggest to newcomers who are terrified of attempting the Listener challenge. Thank you, thank you, Schadenfreude!

Shirley Curran.