2566 – Lip Service by Leon
Posted by Listen With Others on 7 March 2008
Lip Service was first published in The Listener magazine 25th September 1980 and is notorious for having attracted no correct entries. G Guinness of Richmond managed to complete the grid correctly but failed with the following game of solitaire. Last July, Dave Tilley distributed the puzzle as one of several vintage Listeners. It might have ended there had not Xanthippe also set a puzzle that featured a game of peg solitaire. My admission in the LWO blog that I could do Xanthippe’s version but not Leon’s prompted John Reardon to send me a full solution to Lip Service with notes. This is so impressive it deserves a wider audience so is reproduced below.
Knowing that the puzzle is not impossible might inspire others to follow John’s example but they should note that there were a number of errors in Dave’s version:
Errata to version sent by Dave Tilley, July 2007:
Clues Horizontal
D: 14 [(2FE)^{2}1] should read [(2: FE)^{2}1]
Clues Vertical
2: BD Rabble should probably read Rabbie
3: EF should read 2: EF
7: CE y^{2} + 4y^{2} + 8x + 4 where y is square should read y^{3} + 4y^{2} + 8y + 4 where y is square (3)
SOLUTION
Preamble
When the diagram has been properly completed it will contain four digits, which compose a date and 33 letters of the alphabet. No distinction is made between zero and the letter ‘O’, or between unity and the letter ‘I’: solvers please copy. Of the different letters remaining, each has a distinct digital value other than one or unity.
Clues may lead to the numerical or to the literal light: most, however, clue both forms. Suitable comparison should therefore enable the solver to establish the literal / digital correspondences. The lights, in their numerical form all belong to the same notation. In many cases, the literal equivalent is not a proper word.
After completing the diagram, solvers are invited to play the solitaire. They should begin by removing the peg at position A3 and, by jumping orthogonally one peg at a time, remove the remaining pegs, finishing with one peg left at G5. .The pegs in order of their removal spell out a message. Solvers are asked to include this with their solution. (NB: A jumping peg always takes its letter with it. Only the following pegs are removed whilst occupying their original position: A3, B3, B4, C2, C3, C5, C6, D1, D2, D4, D6, E2, E3, E5, F3, F4, F5, G4)
As a check, solvers might like to know that the sum of the three words CHAT MY SERF equals 10216.
Clues Horizontal
A: 35 Something for the baby: added to its reverse for MOM (3)
COT + TOC = MOM; C + T = M i.e. 3 + 6 = 9
COT + TOC = MOM; C + T = M i.e. 3 + 6 = 9
B: 24 x^{4} + 2x^{3} where x is prime. Highlight of history? (3)
TOR; hidden and x = 5 means 875 B10 and so 60[11] B12. This clue gave the first indication that we were to work in base 12.
TOR; hidden and x = 5 means 875 B10 and so 60[11] B12. This clue gave the first indication that we were to work in base 12.
B: 56 One less than a prime! Steamer, do we hear? (2)
MR; sounds like mister and 9[11] B12 = 119 B10 and 5! = 120
MR; sounds like mister and 9[11] B12 = 119 B10 and 5! = 120
C: 12 Briefcases? Twice E: 12 (2)
CA; short for cases and 34 B12 = 40 B10 E: 12 = 20 B10 = 18 B12
CA; short for cases and 34 B12 = 40 B10 E: 12 = 20 B10 = 18 B12
C: 23 Root of 3: FC’s comicality will look after … (2)
AA; SQRT(1694 B12) = 44 B12 This looks like a rare example of a clue with ellipses that must be read with its partner to be solved:
AA; SQRT(1694 B12) = 44 B12 This looks like a rare example of a clue with ellipses that must be read with its partner to be solved:
Root of ITMA’s comicality will look after one for the road; prime
One root of ITMA’s comicality was its catchphrases, notably the abbreviated ones ITMA itself and TTFN so the clue could be read as:
Abbreviation: will look after car = Automobile Association
C: 46 …one for the road; prime (3)
CAR; 34[11] B12 = 491 B10 which is prime
CAR; 34[11] B12 = 491 B10 which is prime
C: 67 Offside = HOT minus B: 24 (2)
RH; righthand side and HOT – TOR = [11]7 B12
RH; righthand side and HOT – TOR = [11]7 B12
D: 14 Someone who’s red has sun to excess: reverse of [(2: FE)^{2}1] (4)
CMMI; COMMUNIST minus SUN TO rev[3991] = 1993 = 3135 B10; SQRT(3136) = 56 = 48 B12 = AS
CMMI; COMMUNIST minus SUN TO rev[3991] = 1993 = 3135 B10; SQRT(3136) = 56 = 48 B12 = AS
D: 57 4 : C –A minus 102 is fifty, except it recurs (3)
YFF; 3[11]0 – 102 B12 = 2[10][10] and rev[fifty minus letters IT] = yff
YFF; 3[11]0 – 102 B12 = 2[10][10] and rev[fifty minus letters IT] = yff
E: 12 Initially Stravinsky’s score (2)
IS; Igor Stravinsky; 18 B12 = 20 B10
IS; Igor Stravinsky; 18 B12 = 20 B10
E: 53 Strain = MA + 5: B – D (3)
FIT; MA + MAY = 94 + 942 B12 = [10]16 B12
FIT; MA + MAY = 94 + 942 B12 = [10]16 B12
E: 67 Sum of two consecutive squares = difference of two consecutive cubes.
EI; 5^2 + 6^2 = 5^3 – 4^3 = 61 and in B12 this is 51
EI; 5^2 + 6^2 = 5^3 – 4^3 = 61 and in B12 this is 51
Set common to Gauss’s 1, 2 and 3 (2)
EIN, ZWEI and DREI
F: 23 Prime squared prime (2)
41; 41 B12 = 49 B10 = 7^2 and 41 itself appears prime in B10
41; 41 B12 = 49 B10 = 7^2 and 41 itself appears prime in B10
F: 46 Palindromic prime and ancient plough (3)
ERE; 545 B12 = 773 B10 which is prime
ERE; 545 B12 = 773 B10 which is prime
G: 35 (E: 67) (1: ED) = – au – (3)
64C; tacautac; (EI)*(IC) = 51*13 B12 = 61*15 B10 = 915 = 643 = TAC
64C; tacautac; (EI)*(IC) = 51*13 B12 = 61*15 B10 = 915 = 643 = TAC
Clues Vertical
1: C – E 502 – 191 = 201? Yes and no (3)
CCI; 502 – 191 = 331 B12 and CCI = 201 in Roman numerals
CCI; 502 – 191 = 331 B12 and CCI = 201 in Roman numerals
2: B – D How to make nineteen three less? Substitute – we hear. Cap that, Rabbie! (3)
TAM; 649 sounds like 6 for 9 so 19 becomes 16; a TAM is a cap that Rabbie Burns might have worn
TAM; 649 sounds like 6 for 9 so 19 becomes 16; a TAM is a cap that Rabbie Burns might have worn
2: E – F Batsman’s triumph in Capetown? (2)
SA; 84 B12 = 100 B10, a century
SA; 84 B12 = 100 B10, a century
3: A – B Square that every army unit has (2)
CO; 36 B10= 30 B12
CO; 36 B10= 30 B12
3: F – C In WWII it made us laugh. Either way it’s square (4)
1TMA; 1694 B12 = 2704 B10 = 52^2; 4961 B12 = 8281 B10 = 91^2
1TMA; 1694 B12 = 2704 B10 = 52^2; 4961 B12 = 8281 B10 = 91^2
4: A – C No seaforce is complete without this killer (3)
ORC; hidden
ORC; hidden
4: D – E Unlucky number (2)
II; 11 B12 = 13 B10
II; 11 B12 = 13 B10
4: D – G (CS)^{2} (4)
IIE4; 38^2 B12 = 44^2 B10 = 1936 = 1154 B12
IIE4; 38^2 B12 = 44^2 B10 = 1936 = 1154 B12
4: F – G Perfect square river (2)
EA; 54 B12 = 64 B10
EA; 54 B12 = 64 B10
5: B – D Permit: anagram of (E: 12)^{2} (3)
MAY; 18^2 B12 = 20^2 B10 = 400 = 294 B12 = YMA
MAY; 18^2 B12 = 20^2 B10 = 400 = 294 B12 = YMA
5: E – G Physician, surgeon or organist? Product of three consecutive primes, first two being squared (3)
FRC; Fellow of the Royal College (of Physicians, Surgeons or Music); 1575 B10 = 3^2*5^2*7 = [10][11]3 B12
FRC; Fellow of the Royal College (of Physicians, Surgeons or Music); 1575 B10 = 3^2*5^2*7 = [10][11]3 B12
6: B – C Norvic’s abbreviated address: 143 Base X (2)
RR; Norvic = Norvicensis of Norwich; [11][11] B12 = 143 B10
RR; Norvic = Norvicensis of Norwich; [11][11] B12 = 143 B10
6: D – F Spent money – au fond du jardin? (3)
FEE; fée = fairy in French (at the bottom of the garden)
FEE; fée = fairy in French (at the bottom of the garden)
7: C – E Wild plant extracted from two gills, as it were, y^{3} + 4y^{2} + 8y + 4 where y is square (3)
HFI: 1129 B10 with y = 9 = 7[10]1 B12; HALF PINT minus PLANT
HFI: 1129 B10 with y = 9 = 7[10]1 B12; HALF PINT minus PLANT
All the letters assigned numbers are therefore:
A

C

E

F

H

I

M

O

R

S

T

Y

4

3

5

10

7

1

9

0

11

8

6

2

The check given in the preamble:
CHAT MY SERF = 3746 + 92 + 85[11][10] = 10216 B12.
Leon did not follow the convention of A = 10 and B = 11 in base 12.
[John has provided an Excel file that shows 37 diagrams – one for each of the moves. There isn’t room to show them here but I would be glad to forward the file to anyone interested.]
The game of solitaire:
[John has provided an Excel file that shows 37 diagrams – one for each of the moves. There isn’t room to show them here but I would be glad to forward the file to anyone interested.]
The game was started by removing peg C at A3. Pegs in a red font could only be removed if on a green background – these were pegs that were removed when in their original position and there were no cases where one of these moved to an alternative green cell. Bold black pegs could not be removed – whenever they moved from their original position they became regular black and then could be removed at any time.
The message revealed is:
CORMAC MACCARTHY FORTIS ME FIERI FECIT 1446
This can be confirmed in Brewers under Blarney although AD (1446) is missing and they spell Macarthy with one C. It is the inscription on a stone some 20ft above the Blarney Stone and it would be the true stone were it not virtually inaccessible.
Solution received with thanks from John Reardon, February 2008.
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So, had Lip Service been published today there might have been at least one correct solution. There were two major stumbling points for me: realising that the notation was in base 12 and playing the game of solitaire. It is bad enough in Excel but spreadsheets were unknown in 1980 and keeping track of the two types of peg with just pen and paper strikes me as being a near impossibility, especially since the quotation was unfamiliar and not in the ODQ.
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