Tuesday 27 December 2005

Neuroscience bolsters Agatha Christie promotion

Via the always excellent Language Log, this: The Agatha Christie Code: Stylometry, serotonin and the oscillation overthruster. Mark Liberman expresses a little caution about the alleged and much-publicised neuroscientific proof that Agatha Christie is a great author: the BBC's Scientists study Christie success, and the like. The repeated quotation of a "Dr Roland Kapferer, the project's leader" smacks of appeal to authority: LL notes that he appears to be a film and TV producer, not (as is implied by context) one of the "scientists from universities in London, Birmingham and Warwick" involved in the work.

Only Grantham Today hints at the back-story: Sisters play famous author in seasonal TV special mentions how The Agatha Christie Code "has been made to mark the 75th anniversary of the creation of Miss Marple and will use scientific analysis of Christie's novels to find out why she was such a successful author" - indicating that the documentary, which screens today, is a belated part of the late-2005 Christie relaunch by Chorion plc.

As I said to LL, the scientific exercise appears to have "promotional" written all over it (given that ITV and Chorion plc have a long-standing co-production agreement for Agatha Christie programmes). Even the premise of the research is to be doubted: sales of Christie's fiction were in decline in the late 1990's, until rebadged by Chorion after its acquisition of the rights in 1999.

Chorion furthermore took the unusual step of a moratorium on professional production of Christie's plays, again to clear the field for revisionist West End productions. Agatha Christie's grandson, Matthew Prichard, in what sounds like an admission of the datedness of her work, commented: "If Agatha Christie is to be as popular as the 21st century as she was in the 20th, we have to be open-minded about interpreting stories in modern ways, much as Shakspeare is reinvented for successive generations".

No, Christie is not staying popular due to some intrinsic quality, but by massive and well-managed reinvention.

Addendum:
I watched The Agatha Christie Code and it confirmed all suspicions: promotion backed up by questionable science. It took the form of a dramatised documentary of Agatha Christie's career, intercut with praise by various pundits (no dissenting voices) and a purported investigation into the reason for her success.


Down to the science. The main experts were computational linguist Dr Pernilla Danielsson (billed as "academic champion of communications") and Dr Markus Dahl, a research fellow (a Johnny Depp lookalike, including top hat) at the University of London Institute of English Studies (he has a good track record of work on textual analysis as a tool in tracing attribution and authorship). The camera tracked around them portentously as they sat at glowing laptops in a dimly-lit smoky room and, bit by bit, revealed the purported secret of Christie's success. As LL guessed, Dr Roland Kapferer wasn't among the experts; he was revealed in the credits as the associate producer/writer for the programme.

The science boiled down to a) computerised textual analysis - word frequencies, and so on - and b) decidedly non-mainstream experts interpreting the psychological meaning of those results. Christie's nearly invariable use of "said", rather than said-bookisms, was claimed to enable readers to concentrate on the plot. Her works' narrow range on a 3D scatter plot (axes and variables unknown) indicated a consistent style, assumed to be a Good Thing compared to Arthur Conan Doyle. Sudden coherent sections in her otherwise messy notebook indicated her getting "into the flow" - a trancelike writing state (Darian Leader, a Lacanian psychoanalyst, endorsed the idea that she had been in a similar, deeper trance during her famous 10-day disappearance) and Dr Dahl argued that this trance transferred to the reader.

The trance theory was the central thrust of the programme. David Shephard, a Master Trainer in Neurolinguistic Programming (chosen, says his CV, for "his mastery of linguistic patterns") asserted that the level of repetition of key concepts over small spaces (e.g "life", "living", "live", "death" in a couple of paragraphs) consolidated concepts in the reader's mind. He claimed further that we can only hold nine concepts in the mind at once (I assume a reference to Miller's classic Seven plus or minus two figure for short-term memory capacity) and that Christie's use of more than nine characters overloads the reader's conscious mind, making them literally go into a trance. Dr Dahl cited a further textual result - Christie's "ingenious device" of controlling the reading speed by decreasing the level of detail toward the ends of books, and stage hypnotist Paul McKenna claimed that this invoked the neurotransmitters of craving and release, making the books addictive.

Finally they rolled out the big gun, Dr Richard Bandler, "father of Neurolinguistic Programming", who repeated the assertion of Christie literally hypnotising readers, and said the lack of detail helped maintain that trance. And that was it: "extensive computer analysis", concluded the Joanna Lumley voiceover, has enabled a "quantum leap" in understanding the source of Christie's enduring popularity.

Why, surely only a cynic could remain unconvinced by such rigorous science...

- Ray

Friday 29 July 2005

William Spindler: backdated footnote

Footnote - German text and Google Translate - for 2014 post Spindler's list

Im Herbst 1878 wurde bekanntlich, angeblich aus Anlaß der Attentate von Hödel und Nobiling, die sozialdemokratische Partei unter ein Ausnahmegesetz gestellt. Das am 21. Oktober 1878 vom Reichstage beschlossene „Gesetz gegen die gemeingefährlichen Bestrebungen der Sozialdemokratie" hatte gewissermaßen eine politische Kirchhofsruhe zur Folge. Über Berlin-Potsdam und zweimeiligen Umkreis wurde sofort nach Inkrafttreten des Gesetzes der „Kleine Belagerungszustand" verhängt. Auf Grund dieses Gesetzes wurden sogleich 45 bekannte Sozialdemokraten ausgewiesen. Es herrschte geradezu ein panischer Schrecken, denn die Polizei konnte jeden Menschen ohne Angabe von Gründen ausweisen. Selbstverständlich wagte es in Groß-Berlin und Potsdam niemand mehr, sich als Sozialdemokrat, oder auch nur als Demokrat zu bezeichnen, die sofortige Ausweisung wäre ihm sicher gewesen. Der Ausgewiesene erhielt zumeist den Befehl, innerhalb drei Tagen, bisweilen auch innerhalb drei Stunden, das Gebiet des „Kleinen Belagerungszustandes" zu verlassen. Wer diesem polizeilichen Befehl nicht nachkam, wurde verhaftet, wegen Bannbruchs mit Gefängnis bestraft und nach der Strafverbüßung mittelst polizeilicher Eskorte aus dem Gebiete des „Kleinen Belagerungszustandes" entfernt. Ob die Ausgewiesenen das erforderliche Reisegeld besaßen, ob Frau und Kinder durch die gewaltsame Entfernung des Gatten und Vaters dem Elend und dem Hunger preisgegeben waren, war kein Hinderungsgrund, dem Ausweisungsbefehl den nötigen Nachdruck zu verleihen. Wenn ein Ausgewiesener in heller Verzweiflung die polizeiliche Ausweisungsstelle fragte, weshalb er ausgewiesen werde, da er sich doch seit Inkrafttreten des Sozialistengesetzes jeder agitatorischen Tätigkeit enthalten habe, da wurde ihm geantwortet: „Die Polizei ist gesetzlich nicht verpflichtet, den Ausweisungsbefehl zu begründen; nehmen Sie an, der Polizei gefällt Ihre Nase nicht." Ein Rechtsmittel gegen die polizeiliche Ausweisung gab es nicht. Selbst am Weihnachtsheiligenabend mußten brave, fleißige Arbeiter Frau und Kinder in Hunger und Elend zurücklassen, den Wanderstab ergreifen und das Weichbild Berlins verlassen. Den Lebensunterhalt mußten sich diese wegen ihrer politischen Gesinnung Ausgestoßenen durch Betteln verschaffen.

In the fall of 1878, the Social Democratic Party was known , allegedly on the occasion of the terrorist attacks of Hödel and Nobiling , placed under an emergency law . The adopted by the Reichstag on October 21, 1878 "Law against the growing strength of social democracy " had something of a political graveyard peace result . Via Berlin -Potsdam and zweimeiligen radius was imposed " Small state of siege " immediately after the entry into force of the Law of . On the basis of this Act 45 well-known Social Democrats were immediately reported . There was almost a panic terror , because the police were able to identify every person without giving a reason. Of course, it ventured into Greater Berlin and Potsdam nobody , to call himself a Social Democrat, or even as a Democrat , the immediate expulsion would certainly have been him. The Designated mostly received the order within three days , sometimes within three hours to leave the area of the "small state of siege ." Those who do not respond to that police command , was arrested , punished for violating a prison and after serving the sentence by means of police escort from the field of "small state of siege " away. Whether those expelled had the necessary travel money whether wife and children were abandoned by the forcible removal of the husband and father's misery and starvation , was no obstacle to giving the deportation order the necessary emphasis . If a Marked police expulsion point asked in brighter despair , so he 'll expelled , since he has been in force of the Socialist Law any agitational activities have abstained because he was told : " The police are not legally obliged to give reasons for the expulsion order ; accept you , the police do not like your nose. " An appeal against the police eviction did not exist. Even at Christmas Christmas Eve had to brave , industrious worker's wife and children behind in hunger and misery , take the walking stick and leave the precincts of Berlin. The livelihood had to procure them because of their political beliefs outcast by begging.

Die hinterbliebenen Frauen und Kinder mußten naturgemäß auch betteln gehen, wenn sie nicht verhungern wollten. Jeder, der diese armen Leute unterstützte, geriet in Gefahr, ebenfalls ausgewiesen zu werden. Der damalige Besitzer der großen Spindlerschen Färberei, William Spindler, ein mehrfacher Millionär, war nicht Sozialdemokrat, aber bürgerlicher Demokrat und ein glühender Verehrer Johann Jacobys. Es war bekannt, daß William Spindler, den leider schon sehr lange der kühle Rasen deckt, ein Philanthrop im vollsten Sinne des Wortes war. Es war William Spindler geradezu Herzensbedürfnis, Tränen zu trocknen, Not und Elend zu lindern. Es war daher selbstverständlich, daß die Hinterbliebenen der Ausgewiesenen in großen Scharen den Wohltätigkeitssinn Spindlers in Anspruch nahmen. Meines Wissens nach ist William Spindler, dieser selten edelmütige Mensch als einer der größten Wohltäter der Menschheit zu bezeichnen.

The surviving women and children were naturally go begging, if they did not starve. Everyone who supported these poor people, was in danger of becoming also reported. The owner at the time of the great Spindler's dyeing, William Spindler, a multi-millionaire, was not a social democrat, but bourgeois democrat and an ardent admirer of Johann Jacoby. It was known that William Spindler, the unfortunately very long the cool grass cover, a philanthropist in the fullest sense of the word was. It was William Spindler to dry downright heartfelt need, tears, relieve suffering and misery. It was therefore natural that the families of the deportees took in large numbers the charitable sense in Spindler's claim. To my knowledge, is William Spindler, this noble man rarely be described as one of the greatest benefactors of mankind.

Diese vielen Unterstützungen der Hinterbliebenen Ausgewiesener kamen naturgemäß sehr bald zur Kenntnis der Polizei, zumal das Spitzeltum in üppigster Blüte stand. Nach damaliger Ansicht der Polizei hatte William Spindler die polizeilichen Maßnahmen zur Unterdrückung der Sozialdemokratie durch seine Unterstützungen zu lähmen versucht, er war daher nach polizeilicher Anschauung für die Ausweisung reif. Allein Spindler war Besitzer eines ausgedehnten Fabriketablissements. Er beschäftigte mehrere tausend Arbeiter. Einen solchen Mann auszuweisen, erschien der Polizei als Risiko. Wenn durch die Ausweisung Spindlers mehrere tausend Arbeiter brotlos würden? Es herrschte ohnedies große Arbeitslosigkeit in Berlin und es war Winter. Da bat eines Tages der Vorstand des betreffenden Polizeireviers Herrn Spindler, ihm eine Unterredung zu gewähren. Bereitwilligst empfing William Spindler den Polizeileutnant mit der ihm eigenen Liebenswürdigkeit. Herr Spindler, so etwa äußerte der Polizeioffizier, es wird Ihre Ausweisung in Erwägung gezogen. Ich habe deshalb den Auftrag erhalten, Sie zu fragen, ob, wenn das geschähe, Ihr Etablissement fortbestehen würde. William Spindler antwortete: Ihre Mitteilung kommt mir nicht überraschend. Angesichts der Praktiken der Berliner Polizei habe ich sie längst erwartet. Ich werde, sobald ich den Ausweisungsbefehl erhalten habe, mich vielleicht am Gardasee oder am Golf von Neapel niederlassen und selbstverständlich, da ich fern von Berlin weilen muß, mein Etablissement sofort auflösen. Zu meinem großen Bedauern würden dadurch einige tausend Arbeiter brotlos werden. Die Verantwortung hat aber alsdann die Berliner Polizei und nicht ich. — Sie würden also bestimmt Ihre Etablissements schließen und Ihre Arbeiter entlassen, wenn Sie den Ausweisungsbefehl erhielten, fragte der Polizeileutnant. Mein fester, schon lange gefaßter Entschluß, erwiderte William Spindler. Der Polizeioffizier empfahl sich und — die Ausweisung Spindlers unterblieb. Einige Monate vor Erlaß des Sozialistengesetzes hatte Hofprediger Stöcker zwecks Bekämpfung der Sozialdemokratie die christlich-soziale Partei begründet. Nachdem die Sozialdemokratie von der öffentlichen Bildfläche verschwunden war, wurde es in den christlich-sozialen Versammlungen langweilig. Es ging doch nicht an, fortdauernd eine Partei zu bekämpfen, die in der Öffentlichkeit nicht mehr existierte. Auch in höheren Kreisen schien man die politische Kirchhofsruhe unangenehm zu empfinden. Es mußte ein anderes Angriffsobjekt gefunden werden. Und dies fand sich sehr bald.

This supports many of the bereaved Marked came naturally very soon to the attention of the police , especially since the Spitzeltum stood in luxuriant bloom. According to police, former William Spindler had tried to paralyze the police measures for the suppression of social democracy through his support , he was, therefore, according to police intuition ripe for expulsion. Only Spindler was the owner of a vast Fabriketablissements . He employed several thousand workers . Expel such a man, the police appeared as a risk. If several thousand workers would destitute by the expulsion Spindler ? There was in any case large unemployment in Berlin and it was winter. Because one day asked the board of the concerned police station Mr Spindler, to grant him an interview . Willingly received the William Spindler police lieutenant with the kindness of his own . Mr. Spindler, such as the police officer remarked , it is your designation considered. I have therefore been awarded the contract , to ask whether , if that happened , would continue your establishment. William Spindler said: your message comes as no surprise to me . Given the practices of the police in Berlin I have long been expecting . I will , once I have received the expulsion order , maybe settle down on Lake Garda or the Gulf of Naples and of course, since I have to stay away from Berlin , close my establishment immediately. To my great regret , several thousand workers would be out of work . However, the responsibility has then the Berlin police , not me . - So you would definitely include your establishment and dismiss your workers when you received the expulsion order , asked the lieutenant . My solid, long ago more composed decision said William Spindler . The police officer took his leave , and - the expulsion Spindler was omitted.
- from Interessante Kriminal-Prozesse von kulturhistorischer Bedeutung, Hugo Friedlander, 1913

Thursday 28 July 2005

Tagliente's multiplication

Moved to here.

Mindbender 4: solution


Backdated post to avoid spoiler to Mindbender 4.

This is an interesting one: find a number of indeterminate length (the puzzle doesn't say what kind of phone number it is) that's one less than the 200 x its last four digits.

?????number = 200*(number) - 1

You could fiddle around for ages trying to find analytical solutions for various assumptions of the length of the unknown segment (it'll come down to a Diophantine equation with two variables). You could equally try a deductive method to work out what 4 digits will produce themselves in the rightmost place when 200*x - 1 is applied: for starters, any number of the form 200*x - 1 is going to end in 99.

But this one looked far more readily amenable to an iterative method, finding a stable solution to the iteration x -> f(g(x)) ... where f(x) is a decreasing function and g(x) is an increasing one. The beauty of it is that the functions don't have to be continuous (e.g. polynomials); they can be integer functions, and even bit operations such as removing the leading digits.

In this case, g(x) = 200*x - 1 and f(x) = "take last four digits"

That is, start with any 4-digit number x, find 200*x - 1, select the last four digits to get new x, and repeat.

This is the first one I tried:

1234 ... 1234*200 - 1 = 246799
6799 ... 6799*200 - 1 = 1359799
9799 ... 9799*200 - 1 = 1959799
9799 ... 9799*200 - 1 = 1959799 ... solution!

It doesn't seem to matter what seed you choose to start the iteration; the iteration always rapidly converges to 1959799.

- Ray

Mindbender 3: solution


This proved an interesting one. I shouldn't have done it today, but I needed a break from finishing my tax.

The problem comes down to finding a 3-digit number x such that (x^2)/7 - 4 produces a result of unknown length with x as the rightmost digits.

This didn't have the feel of a problem where iteration would work, and I felt that narrowing down the search space might be a good approach. Firstly, we can say that:

Because (x^2)/7 - 4 is an integer, then x is divisible by 7.

This slightly reduces the range of x from 100-999 to 105-994.

Secondly, we can get the final digit of x, using the 7 times table. For example, if x ends in 1, x^2 ends in 1, therefore (x^2)/7 ends in 3, and (x^2)/7 - 4 ends in 9.

Tabulating these: x, x^2, (x^2)/7, (x^2)/7 - 4
1, 1, 3, 9
2, 4, 2, 8
3, 9, 7, 3
4, 16, 8, 5
5, 25, 5, 1
6, 36, 8, 4
7, 49, 7, 3
8, 64, 2, 8
9, 81, 3, 9

So x must end in 8 or 9.

Mindbender 2: solution


Backdated post, to avoid spoiler to puzzle.

The above puzzle, from the Western Morning News's daily Mindbender series, comes down to this: find a 8-digit number such that stripping the outer two digits off leaves you with that number divided by 83.

?NUMBER? = 83 x NUMBER

If you call the outer digits a and c, and the middle number b, then this can be expressed as the equation:

(10^7)a + 10b + c = 83b ... which reduces to:

(10^7)a - 73b + c = 0

As is very often the case with these Mindbender puzzles, this is a Diophantine equation: an integer relation with multiple variables and constraints. There are formal ways to tackle these, as well as "quick and dirty" number-crunching by computer. But as I mentioned earlier - see Mindbender 1: solution - iterative solutions often give much faster results.

If an equation can be expressed as a relation x = f(g(x)) (where f(x) is an ascending function and g(x) a descending function, or vice versa) then repeatedly applying x -> f(g(x)) can often converge on a solution. This looked a promising route here.

Define g(x) = 83*x
and f(x) = DIGSTRIP(x) ... where DIGSTRIP removes the outer two digits from an 8-digit number.
gives x -> DIGSTRIP(83*x)

For example, start with x = 666666.
x -> DIGSTRIP(83*666666) = DIGSTRIP(55333278) = 533327
DIGSTRIP(83*533327) = DIGSTRIP(44266141) = 426614
DIGSTRIP(83*426614) = DIGSTRIP(35408962) = 540896
DIGSTRIP(83*540896) = DIGSTRIP(44894368) = 489436
and so on ...

This rapidly turned out to be a dead end. Even though it has one of the required properties - it generally stays in range - whatever the seed, it simply refused to converge. I think the problem is that DIGSTRIP is not a well-behaved function; while it reduces an 8-digit number by two orders of magnitude, the result can actually be anywhere in the range of 6-digit numbers. So it isn't actually a function useful for homing in on a result.

So a different approach was needed. Returning to the equation ...

(10^7)a - 73b + c = 0

... another possibility is to try to reduce it to the smallest search space. In this case, the 6-digit number b is not what you want to search through, so I tried expressing b in terms of a and c (since a is in the range 1-9, and c in the range 0-9):

b = ((10^7)a + c)/73 ... or in short ... (10^7)a + c ≡ 0 modulo 73

This has brought the problem to very simple form. All that's needed is to search through 9 cases -  (10^7)a for a=1...9 - and see if any have a multiple of 73 that's less than 9 greater. This is plain calculator territory.


10000000/73 = 136986.3014... 
136986*73 = 9999978, next multiple is 10000051 ... no


20000000/73 = 273972.6027...
273972*73 = 19999956, next multiple is 20000029 ... no


30000000/73 = 410958.9041...
410958*73 = 29999934, next multiple is 30000007 ... yes!


So a=3, c=7, and b = 30000007/73 = 410959


And confirming: 34109597 = 410959*83 ...the solution sought.


- Ray

Mindbender 1: solution

Backdated post to avoid spoilers:

So, the problem was to find a number that is one more than seventeen times the total of its digits.

There's no simple analytical solution, as it's a multi-variable problem. For example, if the solution has three digits a, b and c, then:

17(a + b + c) + 1 = 100a + 10b + c
... which simplifies to the Diophantine equation ...
83a -7b - 16c = 1

... and you could construct similar equations for any assumed number of digits. Generally, linear Diophantine equations have an infinite number of solutions. For example:

if {a, b, c} satisfy 83a -7b - 16c = 1
then so do {a + 7k, b+83k, c} where k is any integer

But this may not be the case when there are further restrictions; in the example above, for instance, the variables all have to be in the range 0-9.

There are methods for tackling equations of this format - see Extended Euclidean Algorithm - but I'm not sufficiently au fait with number theory to try that route. So it looked at first glance as if enumeration of cases would be necessary, trying all numbers of the form 1 modulo 17 (that is 18, 35, 52, ...). It could be a long task if the number is large, but it's easily programmed in a maths package; I still use my old copy of Derive 5. However, number-crunching puzzles is a bit of a last resort.

Then it occurred to me that the solution ...

n = 17 * (digit sum of n) + 1

... might be amenable to an iterative process: seeding the formula with an assumed digit sum, then using the answer to make a new seed, and so on, iterating until a stable result appears (or it goes cyclic). Furthermore the format suggested to me that the whole thing might be cyclic modulo 17, so only seeds up to 16 needed trying.  So:

1: 18, 154, 171, 154 ... gone cyclic
2: 35, 137, 188, 290, 188 ... gone cyclic
3: 52, 120, 52 ... gone cyclic
4: 69, 256, 222, 103, 69 ... gone cyclic
5: 86, 239, 239 ... stable solution.

And there it is, far sooner than I'd expected: 239 = 17*(2+3+9) + 1

From the stated terms of the puzzle, I assume 239 is a unique solution. But how do we know?  It leaves me wondering if there's a more rigorous and/or more efficient way of getting to the answer, and of showing that it's unique.

At the very least, it can be assumed there would be a finite number of solutions, as number size rapidly outstrips the sum of its digits (for a n-digit number, the order of magnitude increases as 10^n, while the maximum digit sum is 9n).

Iteration, by the way, is an immensely powerful tool for some problems virtually impossible to solve in other ways. I knew of it as a standard mathematical procedure for finding roots to equations, but Douglas Hofstadter's book Metamagical Themas introduced me to the idea of using it for problems that don't involve continuous functions. Hofstadter refers to its use for creating "self-enumerating pangrams", like this example by Lee Sallows.

This pangram tallies five As, one B, one C, two Ds, twenty-eight Es, eight Fs, six Gs, eight Hs, thirteen Is, one J, one K, three Ls, two Ms, eighteen Ns, fifteen Os, two Ps, one Q, seven Rs, twenty-five Ss, twenty-two Ts, four Us, four Vs, nine Ws, two Xs, four Ys, and one Z.

The problem is to find the right numbers so that the sentence is correctly self-referential. To produce the one above, Sallows used dedicated combinatorial hardware, and offered a challenge that no-one would be able to find a self-enumerating pangram beginning "This computer-generated pangram contains ..." within ten years.

In fact it was found quite rapidly by Larry Tesler, using an iterative method. It turns out that there is no need to slog through the entire search space for all 26 variables. Remarkably, you can start out with random seed values - "one A, two Bs, three Cs ..." or whatever - and iterate as I did with the Mindbender puzzle. Not all seed sets work; the iteration will get stuck in a loop. But many do converge on a solution.
.
Addendum: Felix Grant at The Growlery just posted a response - The quick'n'dirty approach to puzzles - describing a spreadsheet solution to the Mindbender problem which, whatever its level of elegance may be, highlights the sheer computational power available nowadays. It took a few seconds to get a solution - I  remember using a Sinclair ZX81 to do similar puzzles from PC Plus in the late 1980s. Something like this would have taken hours, even with pre-analysis to reduce the search space.

Addendum 2: Following Felix and Dr C's lead, I also tried computing and graphing a table.

Let f(n) = 17*(digitsum)+1
For three digits, the function in Derive is:
f(n) := 17·(MOD(n, 10) + MOD(FLOOR(n/10), 10) + MOD(FLOOR(n/100), 10)) + 1
Now plot n-f(n) against n.


Interesting pattern. It shows, empirically at least, that the only possible solution(s) can lie between n=100 and n=300. After that, n permanently overtakes f(n).

- Ray

Ancient Greek IOL solution

This is my solution to the Ancient Greek language puzzle mentioned
in the International Linguistics Olympiad post for 28th July 2011. I've backdated it to avoid spoilers.  If you want to try the puzzle, go there first.

Ancient Greek

Consider these phrases in Ancient Greek (in a Roman-based transcription) and their unordered English translations:

(A) ho tōn hyiōn dulos ____ (1) the donkey of the master
(B) hoi tōn dulōn cyrioi ____ (2) the brothers of the merchant
(C) hoi tu emporu adelphoi ____ (3) the merchants of the donkeys
(D) hoi tōn onōn emporoi ____ (4) the sons of the masters
(E) ho tu cyriu onos ____ (5) the slave of the sons
(F) ho tu oicu cyrios ____ (6) the masters of the slaves
(G) ho tōn adelphōn oicos ____ (7) the house of the brothers
(H) hoi tōn cyriōn hyioi ____ (8) the master of the house

1. Match the Ancient Greek phrase (A-H) with the corresponding English translation (1-8).
2. Translate into Ancient Greek:
a) the houses of the merchants
____________________________
b) the donkeys of the slave.
____________________________

Note: the letter ō stands for a long o.

- By Todor Tchervenkov
For the North American Computational Linguistics Olympiad 2007



To start, it helps to list the structures of the translations, whether the nouns of the subject and owners are singular or plural.

(A) ho tōn hyiōn dulos ____ (1) the donkey of the master SS
(B) hoi tōn dulōn cyrioi ____ (2) the brothers of the merchant PS
(C) hoi tu emporu adelphoi ____ (3) the merchants of the donkeys PP
(D) hoi tōn onōn emporoi ____ (4) the sons of the masters PP
(E) ho tu cyriu onos ____ (5) the slave of the sons SP
(F) ho tu oicu cyrios ____ (6) the masters of the slaves PP
(G) ho tōn adelphōn oicos ____ (7) the house of the brothers SP
(H) hoi tōn cyriōn hyioi ____ (8) the master of the house SS

Looking the elements of the phrases, clearly the last two words are the nouns, and the first two are "the ... of the" with variable endings depending on the number of the nouns they attach to.

An immediate observation is that there are only three occurrences of "tu" and three translations with singular owners. Take these aside:

the donkey of the master SS
the brothers of the merchant PS
the master of the house SS

hoi tu emporu adelphoi
ho tu cyriu onos
ho tu oicu cyrios

One in each group has different structure, so we conclude that:

(C) hoi tu emporu adelphoi = (2) the brothers of the merchant

This also shows that the phrase "the B of the A" is structured as "the of-the-A B" in Ancient Greek.

(A) ho tōn hyiōn dulos ____ (1) the donkey of the master SS
(B) hoi tōn dulōn cyrioi ____ (2) the brothers of the merchant PS
(C) hoi tu emporu adelphoi ____ (3) the merchants of the donkeys PP
(D) hoi tōn onōn emporoi ____ (4) the sons of the masters PP
(E) ho tu cyriu onos ____ (5) the slave of the sons SP
(F) ho tu oicu cyrios ____ (6) the masters of the slaves PP
(G) ho tōn adelphōn oicos ____ (7) the house of the brothers SP
(H) hoi tōn cyriōn hyioi ____ (8) the master of the house SS

Now there's only one other "adelph-", so:

(C) hoi tu emporu adelphoi = (2) the brothers of the merchant SS
(G) ho tōn adelphōn oicos = (7) the house of the brothers     SP


(A) ho tōn hyiōn dulos ____ (1) the donkey of the master SS
(B) hoi tōn dulōn cyrioi ____ (2) the brothers of the merchant PS
(C) hoi tu emporu adelphoi ____ (3) the merchants of the donkeys PP
(D) hoi tōn onōn emporoi ____ (4) the sons of the masters PP
(E) ho tu cyriu onos ____ (5) the slave of the sons SP
(F) ho tu oicu cyrios ____ (6) the masters of the slaves PP
(G) ho tōn adelphōn oicos ____ (7) the house of the brothers SP
(H) hoi tōn cyriōn hyioi ____ (8) the master of the house SS

Now there's only one other "oic-"

(C) hoi tu emporu adelphoi = (2) the brothers of the merchant SS
(G) ho tōn adelphōn oicos = (7) the house of the brothers SP
(F) ho tu oicu cyrios = (8) the master of the house SS

(A) ho tōn hyiōn dulos ____ (1) the donkey of the master SS
(B) hoi tōn dulōn cyrioi ____ (2) the brothers of the merchant PS
(C) hoi tu emporu adelphoi ____ (3) the merchants of the donkeys PP
(D) hoi tōn onōn emporoi ____ (4) the sons of the masters PP
(E) ho tu cyriu onos ____ (5) the slave of the sons SP
(F) ho tu oicu cyrios ____ (6) the masters of the slaves PP
(G) ho tōn adelphōn oicos ____ (7) the house of the brothers SP
(H) hoi tōn cyriōn hyioi ____ (8) the master of the house SS

Now there's only one "cyri-" in a position matching "masters"

(C) hoi tu emporu adelphoi = (2) the brothers of the merchant SS
(G) ho tōn adelphōn oicos = (7) the house of the brothers SP
(F) ho tu oicu cyrios = (8) the master of the house SS
(B) hoi tōn dulōn cyrioi = (6) the masters of the slaves PP


(A) ho tōn hyiōn dulos ____ (1) the donkey of the master SS
(B) hoi tōn dulōn cyrioi ____ (2) the brothers of the merchant PS
(C) hoi tu emporu adelphoi ____ (3) the merchants of the donkeys PP
(D) hoi tōn onōn emporoi ____ (4) the sons of the masters PP
(E) ho tu cyriu onos ____ (5) the slave of the sons SP
(F) ho tu oicu cyrios ____ (6) the masters of the slaves PP
(G) ho tōn adelphōn oicos ____ (7) the house of the brothers SP
(H) hoi tōn cyriōn hyioi ____ (8) the master of the house SS

So "dul-" = word stem for "slave"


(C) hoi tu emporu adelphoi = (2) the brothers of the merchant SS
(G) ho tōn adelphōn oicos = (7) the house of the brothers SP
(F) ho tu oicu cyrios = (8) the master of the house SS
(B) hoi tōn dulōn cyrioi = (6) the masters of the slaves PP
( A) ho tōn hyiōn dulos = (5) the slave of the sons SP

(A) ho tōn hyiōn dulos ____ (1) the donkey of the master SS
(B) hoi tōn dulōn cyrioi ____ (2) the brothers of the merchant PS
(C) hoi tu emporu adelphoi ____ (3) the merchants of the donkeys PP
(D) hoi tōn onōn emporoi ____ (4) the sons of the masters PP
(E) ho tu cyriu onos ____ (5) the slave of the sons SP
(F) ho tu oicu cyrios ____ (6) the masters of the slaves PP
(G) ho tōn adelphōn oicos ____ (7) the house of the brothers SP
(H) hoi tōn cyriōn hyioi ____ (8) the master of the house SS

So "hyi-" matches "son"

(C) hoi tu emporu adelphoi = (2) the brothers of the merchant SS
(G) ho tōn adelphōn oicos = (7) the house of the brothers SP
(F) ho tu oicu cyrios = (8) the master of the house SS
(B) hoi tōn dulōn cyrioi = (6) the masters of the slaves PP
(A) ho tōn hyiōn dulos = (5) the slave of the sons SP
(H) hoi tōn cyriōn hyioi = (4) the sons of the masters PP


(A) ho tōn hyiōn dulos ____ (1) the donkey of the master SS
(B) hoi tōn dulōn cyrioi ____ (2) the brothers of the merchant PS
(C) hoi tu emporu adelphoi ____ (3) the merchants of the donkeys PP
(D) hoi tōn onōn emporoi ____ (4) the sons of the masters PP
(E) ho tu cyriu onos ____ (5) the slave of the sons SP
(F) ho tu oicu cyrios ____ (6) the masters of the slaves PP
(G) ho tōn adelphōn oicos ____ (7) the house of the brothers SP
(H) hoi tōn cyriōn hyioi ____ (8) the master of the house SS

And now we know enough to finish:

(C) hoi tu emporu adelphoi = (2) the brothers of the merchant
(G) ho
tōn adelphōn oicos = (7) the house of the brothers
(F) ho tu oicu cyrios = (8) the master of the house
(B) hoi tōn dulōn cyrioi = (6) the masters of the slaves
(A) ho
tōn hyiōn dulos = (5) the slave of the sons
(H) hoi tōn cyriōn hyioi = (4) the sons of the masters
(E) ho tu cyriu onos = (1) the donkey of the master
(D) hoi tōn onōn emporoi = (3) the merchants of the donkeys


We can see that the structure is:

"the B of the A" is written as "the of-the-A B" in Ancient Greek
"ho" = "the" (singular), and "hoi" = "the" (plural)
"tu" = "of the" (singular), "tōn" = "of the" (plural)
"-os" = singular noun, "-oi" = plural noun
"-u" = singular genitive ending, -"ōn" = plural genitive ending

So:

The houses of the merchants = "hoi tōn emporon oicoi"
The donkeys of the slave = "hoi tu dulu onoi"


In this problem, a bit of general knowledge can short-cut the process. Knowing that "Philadelphia" means "brotherly love", and that an emporium is a store or marketplace, immediately gives clues to "adelph-" and "empor-", and I can't pretend that this didn't somewhat steer the direction of my initial guesses. However, the majority of International Linguistics Olympiad puzzles don't have such back doors.

Mindbender 5: solution


Solution backdated to avoid spoiler to Mindbender 5.

At first this looked like a case for trying an iterative solution, but it turned out for once to be simple arithmetic. In terms of strings, the puzzle is:

ABCDEF x 6 = DEFABC

This problem reduces since you can call the two blocks P and Q, so numerically:

(1000P + Q) x 6 = 1000Q + P

so expanding and collecting:

6000P + 6Q = 1000Q + P
5999P = 994Q
Q = (5999/994)P

... which simplifies on inspection to Q = (857/142)P

This means P has to be a multiple of 142. And further, given the condition that the number must lie between 100,000 and 200,000, 142 is the only possibility within range.

So the number is 142857.

142857 x 6 = 857142 QED.