Can You Dij It? #2

It’s very simple, but I’m fascinated by it. I’m talking about something I call the digit-line, or the stream of digits you get when you split numbers in a particular base into individual digits. For example, here are the numbers one to ten in bases 2 and 3:

Base = 2: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010…
Base = 3: 1, 2, 10, 11, 12, 20, 21, 22, 100, 101…


If you turn them into digit-lines, they look like this:

Base = 2: 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0… (A030190 in the Online Encyclopedia of Integer Sequences)
Base = 3: 1, 2, 1, 0, 1, 1, 1, 2, 2, 0, 2, 1, 2, 2, 1, 0, 0, 1, 0, 1… (A003137 in the OEIS)


At the tenth digit of the two digit-lines, both digits equal zero for the first time:

Base = 2: 1, 1, 0, 1, 1, 1, 0, 0, 1, 0
Base = 3: 1, 2, 1, 0, 1, 1, 1, 2, 2, 0


When the binary and ternary digits are represented together, the digit-lines look like this:

(1,1), (1,2), (0,1), (1,0), (1,1), (1,1), (0,1), (0,2), (1,2), (0,0)


But in base 4, the tenth digit of the digit-line is 1. So when do all the digits of the digit-line first equal zero for bases 2 to 4? Here the early integers in those bases:

Base 2: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101…

Base 3: 1, 2, 10, 11, 12, 20, 21, 22, 100, 101, 102, 110, 111, 112, 120, 121, 122, 200, 201, 202, 210, 211, 212, 220, 221, 222, 1000, 1001, 1002…

Base 4: 1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23, 30, 31, 32, 33, 100, 101, 102, 103, 110, 111, 112, 113, 120, 121, 122, 123, 130, 131, 132, 133, 200…


And here are the digits of the digit-line in bases 2 to 4 represented together:

(1,1,1), (1,2,2), (0,1,3), (1,0,1), (1,1,0), (1,1,1), (0,1,1), (0,2,1), (1,2,2), (0,0,1), (1,2,3), (1,1,2), (1,2,0), (0,2,2), (1,1,1), (1,0,2), (1,0,2), (1,1,2), (0,0,3), (0,1,3), (0,1,0), (1,0,3), (0,2,1), (0,1,3), (1,1,2), (1,0,3), (0,1,3), (1,1,1), (0,1,0), (1,1,0), (0,1,1), (1,2,0), (1,1,1), (1,2,1), (1,0,0), (0,1,2), (0,2,1), (1,1,0), (1,1,3), (0,2,1), (1,2,1), (1,2,0), (1,0,1), (1,0,1), (0,2,1), (1,0,1), (1,1,1), (1,2,2), (1,0,1), (1,2,1), (0,2,3), (0,1,1), (0,0,2), (0,2,0), (1,1,1), (0,1,2), (0,2,1), (0,1,1), (1,2,2), (1,2,2), (0,2,1), (0,0,2), (1,2,3), (0,2,1), (1,1,3), (0,2,0), (0,2,1), (1,2,3), (1,1,1), (1,0,1), (0,0,3), (1,0,2), (0,1,1), (0,0,3), (1,0,3), (0,1,2), (1,1,0), (0,0,0)

At the 78th digit, all three digits equal zero. But the 78th digit of the digit-line in base 5 is 1. So when are the digits first equal to zero in bases 2 to 5? It’s not difficult to find out, but the difficulty of the search increases fast as the bases get bigger. Here are the results up to base 13 (note that bases 11 and 12 both have zeroes at digit 103721663):

dig=0 in bases 2 to 3 at the 10th digit of the digit-line
dig=0 in bases 2 to 4 at the 78th digit of the digit-line
dig=0 in bases 2 to 5 at the 182nd digit of the digit-line
dig=0 in bases 2 to 6 at the 302nd digit of the digit-line
dig=0 in bases 2 to 7 at the 12149th digit of the digit-line
dig=0 in bases 2 to 8 at the 45243rd digit of the digit-line
dig=0 in bases 2 to 9 at the 255261st digit of the digit-line
dig=0 in bases 2 to 10 at the 8850623rd digit of the digit-line
dig=0 in bases 2 to 12 at the 103721663rd digit of the digit-line
dig=0 in bases 2 to 13 at the 807778264th digit of the digit-line


I assume that, for any base b > 2, you can find some point in the digit-line at which d = 0 for all bases 2 to b. Indeed, I assume that this happens infinitely often. But I don’t know any short-cut for finding the first digit at which this occurs.


Previously pre-posted:

Can You Dij It? #1

He Say, He Sigh, He Sow #39

— Croyez-vous aux idées dangereuses ?
— Qu’entendez-vous par là ?
— Croyez-vous que certaines idées soient aussi dangereuses pour certains esprits que le poison pour le corps ?
— Mais, oui, peut-être.

  Guy de Maupassant, « Divorce » (1888)


“Do you believe in dangerous ideas?”
“What do you mean by that?”
“Do you believe that certain ideas are as dangerous for some minds as poison is for the body?”
“Well, yes, perhaps.”

Dice in the Witch House

“Who could associate mathematics with horror?”

John Buchan answered that question in “Space” (1911), long before H.P. Lovecraft wrote masterpieces like “The Call of Cthulhu” (1926) and “Dreams in the Witchhouse” (1933). But Lovecraft’s use of mathematics is central to his genius. So is his recognition of both the importance and the strangeness of mathematics. Weird fiction and maths go together very well.

But weird fiction is about the intrusion or eruption of the Other into the everyday. Maths can teach you that the everyday is already Other. In short, reality is weird — the World is a Witch House. Let’s start with a situation that isn’t obviously weird. Suppose you had three six-sided dice, A, B and C, each with different set of numbers, like this:

Die A = (1, 2, 3, 6, 6, 6)
Die B = (1, 2, 3, 4, 6, 6)
Die C = (1, 2, 3, 4, 5, 6)

If the dice are fair, i.e. each face has an equal chance of appearing, then it’s clear that, on average, die A will beat both die B and die C, while die B will beat die C. The reasoning is simple: if die A beats die B and die B beats die C, then surely die A will beat die C. It’s a transitive relationship: If Jack is taller than Jim and Jim is taller than John, then Jack is taller than John.

Now try another set of dice with different arrangements of digits:

Die A = (1, 2, 2, 5, 6, 6)
Die B = (1, 1, 4, 5, 5, 5)
Die C = (3, 3, 3, 3, 4, 6)

If you roll the dice, on average die A beats die B and die B beats die C. Clearly, then, die A will also beat die C. Or will it? In fact, it doesn’t: the dice are what is called non-transitive. Die A beats die B and die B beats die C, but die C beats die A.

But how does that work? To see a simpler example of non-transitivity, try a simpler set of random-number generators. Suppose you have a triangle with a short rod passing through its centre at right angles to the plane of the triangle. Now imagine numbering the edges of the triangles (1, 2, 3) and throwing it repeatedly so that it spins in the air before landing on a flat surface. It should be obvious that it will come to rest with one edge facing downward and that each edge has a 1/3 chance of landing like that.

In other words, you could use such a spiked triangle as a random-number generator — you could call it a “trie”, plural “trice”. Examine the set of three trice below. You’ll find that they have the same paradoxical property as the second set of six-sided dice above. Trie A beats trie B, trie B beats trie C, but trie C beats trie A:

Trie A = (1, 5, 8)
Trie B = (3, 4, 7)
Trie C = (2, 3, 9)

When you throw two of the trice, there are nine possible outcomes, because each of three edges on one trie can be matched with three possible edges on the other. The results look like this:

Trie A beats Trie B 5/9ths of the time.
Trie B beats Trie C 5/9ths of the time.
Trie C beats Trie A 5/9ths of the time.

To see how this works, here are the results throw-by-throw:

Trie A = (1, 5, 8)
Trie B = (3, 4, 7)

When Trie A rolls 1…

…and Trie B rolls 3, Trie B wins (Trie A has won 0 out of 1)
…and Trie B rolls 4, Trie B wins (0 out of 2)
…and Trie B rolls 7, Trie B wins (0 out of 3)

When Trie A rolls 5…

…and Trie B rolls 3, Trie A wins (1/4)
…and Trie B rolls 4, Trie A wins (2/5)
…and Trie B rolls 7, Trie B wins (2/6)

When Trie A rolls 8…

…and Trie B rolls 3, Trie A wins (3/7)
…and Trie B rolls 4, Trie A wins (4/8)
…and Trie B rolls 7, Trie A wins (5/9)


Trie B = (3, 4, 7)
Trie C = (2, 3, 9)

When Trie B rolls 3…

…and Trie C rolls 2, Trie B wins (Trie B has won 1 out of 1)
…and Trie C rolls 3, it’s a draw (1 out of 2)
…and Trie C rolls 9, Trie C wins (1 out of 3)

When Trie B rolls 4…

…and Trie C rolls 2, Trie B wins (2/4)
…and Trie C rolls 3, Trie B wins (3/5)
…and Trie C rolls 9, Trie C wins (3/6)

When Trie B rolls 7…

…and Trie C rolls 2, Trie B wins (4/7)
…and Trie C rolls 3, Trie B wins (5/8)
…and Trie C rolls 9, Trie C wins (5/9)


Trie C = (2, 3, 9)
Trie A = (1, 5, 8)

When Trie C rolls 2…

…and Trie A rolls 1, Trie C wins (Trie C has won 1 out of 1)
…and Trie A rolls 5, Trie A wins (1 out of 2)
…and Trie A rolls 8, Trie A wins (1 out of 3)

When Trie C rolls 3…

…and Trie A rolls 1, Trie C wins (2/4)
…and Trie A rolls 5, Trie A wins (2/5)
…and Trie A rolls 8, Trie A wins (2/6)

When Trie C rolls 9…

…and Trie A rolls 1, Trie C wins (3/7)
…and Trie A rolls 5, Trie C wins (4/8)
…and Trie A rolls 8, Trie C wins (5/9)


The same reasoning can be applied to the six-sided non-transitive dice, but there are 36 possible outcomes when two of the dice are thrown against each other, so I won’t list them.

Die A = (1, 2, 2, 5, 6, 6)
Die B = (1, 1, 4, 5, 5, 5)
Die C = (3, 3, 3, 3, 4, 6)


Elsewhere other-posted:

At the Mountains of Mathness
Simpson’s Paradox — a simple situation with a very weird outcome

Straight to Ell

Another new review of Jesús Ignacio Aldapuerta’s The Eyes:

Γκροτεσκο, ώμο, βίαιο… Σε υπερθετικο βαθμό. Δοκιμαστηκαν οι αντοχές μου και η ανοχή μου. Δεν ξέρω εάν είναι αυτοβιογραφικο, εάν είναι κριτική στον σύγχρονο τρόπο ζωής ή απλά το πόνημα ενός σαδιστη ψυχοπαθους.

Είναι η ματιά του συγγραφέα στα τεκταινομενα ή όσα είδαν τα μάτια του;

Οποια και εάν είναι η απάντηση, η ανάγνωση αυτού του βιβλίου χρειάζεται κότσια, γερό στομάχι και αποστασιοποιηση.

2.5 τα αστεράκια, χάριν στρογγυλοποιησης έγιναν 3. — Review


Translation: Grotesque, shoulder, violent … in the superlative degree. They tested my strength and my tolerance. I do not know if it’s autobiographical, if criticism of modern lifestyle or simply the essay a sadistic psychopath.

It is the look of the author in foursomes or what they saw his eyes?

Whatever the answer, read this book takes guts, a strong stomach and detachment.

2.5 the stars, thanks Rounding became three.


Previously pre-posted:

Eyeway to Ell

De Pluribus Unum

A beautifully subtle puzzle:

Scrambled Box Tops

Imagine you have three boxes, one containing two black marbles, one containing two white marbles, and the third, one black and one white marble. The boxes are labelled according to their contents — BB, WW, and BW — but someone has switched the labels so that every box is now incorrectly labelled. You are allowed to take one marble at a time out of any box, without looking inside, and by this process of sampling you are to determine the contents of all three boxes. What is the smallest number of drawings needed to do this? — Martin Gardner, Mathematical Puzzles and Diversions (1959), chapter 3, “Nine Problems”, #5.

Bald eagle, Haliaeetus leucocephalus (Linnaeus 1776)

Bald eagle, Haliaeetus leucocephalus (Linnaeus 1776)

Answer: You can learn the contents of all three boxes by drawing just one marble. The key to the solution is your knowledge that the labels on all three boxes are incorrect. You must draw a marble from the box labelled “black-white”. Assume that the marble drawn is black. You know then that the other marble in the box must be black also, otherwise the label on the box would be correct. Since you have now identified the box containing two black marbles, you can tell at once the contents of the box labelled “white-white”: you know it cannot contain two white marbles, because its label has to be wrong; it cannot contain two black marbles, because you have identified that box; therefore it must contain one black and one white marble. The third box, of course, must then be the one containing two white marbles. You can solve the puzzle by the same reasoning if the marble you draw from the “black-white” box happens to be white instead of black.

Oh My Guardian #2

“Instead, Mr Comey has rocket-fuelled a venomous contest just when Mr Trump was desperate for a lifeline…” — The Guardian view on the FBI’s Clinton probe: exactly the wrong thing to do


Previously pre-posted…

Oh My Guardian #1
Reds under the Thread

Toxic Turntable #9

Currently listening…

• Talbot Rich, Intangible EP (1983)
• Pimkka, Neptunienne (1976)
• Theoxiphos, Autochthulhu (1997)
• ζ Draconis, გველეშაპის ვარსკვლავი (1993)
• Sindroma-83, Cera Vera (1987)
• Zazara Xokh, Chemicated (1995)
• The Hems, Deinotherium (2003)
• Jethro Tull, The Broadsword and the Beast (1982)
• Veilchen, Seismophil (1996)


Previously pre-posted:

Toxic Turntable #1
Toxic Turntable #2
Toxic Turntable #3
Toxic Turntable #4
Toxic Turntable #5
Toxic Turntable #6
Toxic Turntable #7
Toxic Turntable #8

Eyeway to Ell

Two new reviews of Jesús Ignacio Aldapuerta’s The Eyes:

Δεν είχα άλλο βιβλίο στα ελληνικά να διαβάσω, αγγλικά βαριόμουν λίγο, οπότε διάβασα αυτό το ανοσιούργημα. Κάποιες ελάχιστες ιστορίες ήταν εντάξει, σκληρές, αλλά τις άντεξα μια χαρά (καλό είναι αυτό τώρα;), αλλά κάποιες ήταν εντελώς εμετικές (π.χ. Αποδεικτικά στοιχεία), και με το ζόρι τις τελείωσα. Όπως και να το κάνουμε το βιβλίο είναι εμετικό, νοσηρό, άρρωστο, πείτε ό,τι άλλο θέλετε. Φυσικά η γραφή είναι πολύ καλή. Αν δεν ήταν τόσο άρρωστος ο συγγραφέας (από αυτά που διάβασα στον πρόλογο, δεν τον ήξερα και από πριν τον τυπά), σίγουρα θα μπορούσε να γράψει κάποια καλογραμμένα μυθιστορήματα τρόμου (έστω και σκληρά). Αλλά φευ… — Review

Translation: I had another book to read in Greek, English a little bored, so I read this outrage. Some few stories were okay, tough, but withstood fine (good is it now?), But some were completely emetic (eg Evidence), and by force of finished. Whatever you do book is emetic, unhealthy, sick, say whatever else you want. Of course the writing is very good. If it was not so sick the author (from what I read in the preface, I did not know from before standards) certainly could write a well written horror novels (even hard). But alas …


Αν το πάρει κανείς κυριολεκτικά είναι πραγματικά… απαίσιο! Αυτό βέβαια είναι, κατά την γνώμη μου, η επιφανειακή κριτική.

Λογοτεχνικά, ο μυστηριώδης αυτός συγγραφέας είχε να δώσει πράγματα καθώς τόσο οι περιγραφές όσο και οι εικόνες που δημιουργεί είναι δείγμα λογοτεχνικής δεξιοτεχνίας. Φυσικά, κανείς δε προσπερνά το άρρωστο του νοήματος.

Αξίζει κάποιος να το διαβάσει (αν διαθέτει γερό στομάχι) και θα διαπιστώσετε πως λογοτεχνικά είναι καλύτερο από την “Φιλοσοφία στο Μπουντουάρ” του μαρκήσιου και προσεγγίζει την “Ιστορία του Ματιού” του Ζ.Μπατάιγ. Αυστηρά για ενήλικο και ώρμο κοινό! — Review

Translation: If you get literally is really … horrible! This of course is, in my opinion, the superficial criticism.

Literature, the mysterious writer he had to give things as well as descriptions and images created are literary virtuosity sample. Of course, no one overtakes the sick of meaning.

It is worth to read (if you have a strong stomach) and you will see that literature is better than the “Philosophy in the Boudoir” Marquis and approaches the “Story of the Eye” of Z.Mpataig. Strictly for adult and ormo public!


Jesús say:

M… P… U….. G… G….. E… R…. | M….. P… A… T….. A… I….. G…. | M… P…. U… R… R… O….. U… G… H… S….. | …(…. | M… P… U….. T… | M… P….. A… L… L….. A… R… D… | M… P….. R… I… L….. L… | …)…