String Patterns: Searching for Interesting Words and Numbers

8/3/2019 String Patterns: Searching for Interesting Words and Numbers

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String Patterns: Searching for InterestingWords and Numbers

Roger Bilisoly, PhD

Associate Professor of Statistics

Central Connecticut State University

Department of MathematicsAmherst College, Amherst, Massachusetts

Thursday, October 6, 2011


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Overview of Talk

String Patterns and Examples

Unusual words, squares, and primes

Anagrams of Words and Numbers Including square anagrams

Birthday Problem and Pangrams

Analyzing Dickens A Christmas Carol.


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1. String Patterns

Regular expressions(also called regexes) are used tofind string patterns. A variety of software packages hasthem implemented, e.g., Mathematica, Perl, SAS, Emacs,

and so forth. Well use them to find interesting words (from wordlists

available on the web) and interesting numbers (e.g.,squares with unusual digit patterns).


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Perl Regexes

For a wordlist, one word per line:

/cat/ would match cat cats scatter but NOT Cat

/[cC]at/ would match cat Cat or Catcher

/cat/i would match cat CaT or sCaTtEr i stands for case insensitive

/cat|dog/ would match either cat or dog

Well see examples of more complex string patterns.

See Chapter 2 of Bilisoly (2008b).


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Example: Some Unusual Words

Are there other words like bookkeeper with threedouble letters in a row?

Essentially no: {bookkeeper, bookkeepers, bookkeeping,bookkeepings}

Are there words containing mile? {besmiled, besmiles, camomiles, facsimiles, homiletic,

outsmiled, outsmiles, similes, smiled, smiler, smilers, smiles}

wordlist = Import["c:\CROSSWD.TXT","Lines"];

threepair = Pick[wordlist, StringMatchQ[wordlist,

RegularExpression[".*(.)\\1(.)\\2(.)\\3.*"]]]

milewords = Pick[wordlist, StringMatchQ[wordlist,

RegularExpression[".+mile.+"]]]


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before state within preparedeagleb-e-f s-t-a w-i-t p-r e-a| | | |\| || | |r-o e n h a e-d l-g

concern decency rather prosperse-c d-e-c r-a-t p-r| |\ |/| \ | ||\r-n-o n y e-h e s-o

Each node is a distinct letter, and each edge connects letters that areadjacent in the word. Graphs are directed: the arrows are understood.

From Section 24 of Eckler (1996).

Word Graphs


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Searching for word graphs

Linear words have no branches, e.g., A-M-H-E-R-S-T.o What is longest such word?o Answer: ambidextrouslyhas 14 letterso lycanthropies, metalworkings, multibranched, unpredictablyare only

examples with 13 letters

How many square cyclic words are there? (like EAGLE)o Need to match regular expression /(.)...\1/ and have 4 distinct

letters. Latter can be checked by taking intersection.o 417 such words, including dazed(and other 4 letter weak verbs starting with

d in the past tense), sails(and other 4 letter nouns starting with s, etc.

Longest cyclic words are 12 letters long: spaceflights, speculations, subharmonics, subordinates,switchblades, switchboards, sympathizers.


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C(3) through C(8) Alphabets

aroma

blurb

comic

dread

eagle

-----

going

hatch

iambi-----

knock

local

maxim

nylon

outdo

plump

-----

razorstars

theft

-----

-----

widow

xerox

yolky

-----

asthma

benumb

cosmic

demand

excuse

------

gaming

health

incubi------

kopeck

lawful

medium

napkin

overdo

pickup

------

rathershirts

throat

------

------

window

------

yearly

------

area

bomb

chic

dead

ease

fief

gong

high

impi----

kick

leal

maim

noun

ouzo

pump

----

roarsaws

text

unau

----

whew

----

----

----

amnesia

brewpub

chronic

dogsled

eclipse

-------

glowing

hawkish

intagli-------

kinfolk

logical

midterm

newborn

oregano

parsnip

-------

regularsailors

tourist

-------

-------

whipsaw

-------

-------

-------

asphyxia

--------

catholic

disabled

earphone

--------

gambling

hyacinth

----------------

kinsfolk

lightful

mealworm

nitrogen

obligato

pawnshop

--------

roadstersardines

tolerant

--------

--------

withdraw

--------

yeastily

--------

angostura

---------

chromatic

diagnosed

enjoyable

---------

gathering

haircloth

------------------

---------

---------

mechanism

---------

---------

playgroup

---------

regulatorseafronts

therapist

---------

---------

worldview

---------

---------

---------


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Cyclic Word Mathematica Code


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Applying String Patterns toSquare Integers (Squares)

Numbers are strings, too, so amenable to regexes.

Well apply regexes to find some unusual squares.

We will also investigate the randomness of the digits of

squares.


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Squares without Doubled Adjacent Digits

Suppose the digits in a square (in base 10) are random.

Then P(adjacent digits are unequal) = 9/10.

So P(100 digit square has no adjacent digits equal) = 0.9^99= 0.0000295127, or about 30 in a million.

We can check this estimate by stochastic computer searches.

51737187749414391248343906418265954222307660356158 ^2 =

2676736596218154562635394063470527614352180564931651707698543214192825076469507142492903267408520964

92247034353159048872216907430912506658722672391337 ^2 =

8509515346952905702167423230637367421917540367430367521058492897148214941569637843428692738072647569

46684912798986257941402247358809860358456884854826 ^2 =

2179481083048950920786370528301507294579140187693485325858417852926259858127813068545985375095490276

72400570183600937943662908692744405525154232178778 ^2 =

5241842562910525153020949368218058035430952649615191319089815789036984216473806049461870608953573284

etc.


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Code to Compute Counts of Such Squares

Do[

total = 0;

nsteps = 1000000;

ndigits = 100;

start = Ceiling[Sqrt[10^(ndigits-1)]];

stop = Floor[Sqrt[10^(ndigits)]];

Do[square = Random[Integer, {start,stop}]^2;

match = StringCases[ToString[square], RegularExpression["(.)\\1"]];

If[Length[match] == 0, ++total, Null],

{i,1,nsteps}

]

Print[total],

{nreps,1,50}

]50 repetitions of searching

a million squares.


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Are the Digits of Squares Random?The initial digits are not.

Lower Upper Lower^2 Upper^2

1 100 141 10000 19881

2 142 173 20164 29929

3 174 200 30276 40000

4 200 223 40000 49729

5 224 244 50176 59536

6 245 264 60025 69696

7 265 282 70225 79524

8 283 300 80089 90000

9 300 316 90000 99856

10 317 447 100489 199809

20 448 547 200704 299209

30 548 632 300304 399424

40 633 707 400689 499849

50 708 774 501264 599076

60 775 836 600625 698896

70 837 894 700569 799236

80 895 948 801025 898704

90 949 999 900601 998001

The limiting proportion of digits 1 k 9 is given by:

(Sqrt[k+1] - Sqrt[k] + Sqrt[10(k+1)] Sqrt[10k])/9.

Digit Prob.1 19.16%

2 14.70%

3 12.39%

4 10.92%

5 9.87%

6 9.08%7 8.45%

8 7.93%

9 7.50%

Total = 100.00%


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Aside: Benfords Law

Benfords Law says that initial digits oftenfollow the following probability distribution:

Log[10, (k + 1)/k] for 1 k 9.

{an} satisfies Benfords law iff Log[10, an]

(mod 1) is uniformly distributed. Seehttp://en.wikipedia.org/wiki/Benford's_law.

Benfords Law does not fit the distributionof initial digits of squares.

Digit Prob.

1 30.10%

2 17.61%

3 12.49%4 9.69%

5 7.92%

6 6.69%

7 5.80%

8 5.12%

9 4.58%

Total 100.00%
http://en.wikipedia.org/wiki/Benford's_lawhttp://en.wikipedia.org/wiki/Benford's_law


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The final digits of squaresare not random.

Final digit can only be: {0,1,4,5,6,9} (only 60% of possibleone digit endings are allowed)

Final two digits can only be: {00,01,04,09,16,21,24,25,29,36,41,44,49,56,61,64,69,76,

81,84,89,96} (only 22% of possible two digit endings allowed)

Final three digits can only be: {000,001,004,009,016,024,025,036,041,044,049,056,064,076

,081,084,089,096,100,104,116,121,124,129,136,144,156,161,164,169,176,184,196,201,204,209,216,224,225,236,241,244,249,256,264,276,281,284,289,296,304,316,321,324,329,336,344,356,361,364,369,376,384,396,400,401,404,409,416,424,436,441,444,449,456,464,476,481,484,489,496,500,504,516,521,524,529,536,544,556,561,564,569,576,584,596,600,601

,604,609,616,624,625,636,641,644,649,656,664,676,681,684,689,696,704,716,721,724,729,736,744,756,761,764,769,776,784,796,801,804,809,816,824,836,841,844,849,856,864,876,881,884,889,896,900,904,916,921,924,929,936,944,956,961,964,969,976,984,996} (only 15.9% allowed)

The percentages are decreasing, but to what value?


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Walter Penneys Theorem

Pick ndigits at random, and let P(n) = Probability that these ndigits are the final digits of a square.

Theorem Penney (1960): As n, P(n) 5/72 6.94%.

Penney shows that P(n) = (2n-1 + 4)(5n+1 + 7)/(36*10n) for neven

and P(n) = (2n-1 + 5)(5n+1 + 11)/(36*10n) for nodd.

For a proof see Walter Penney (1960)

On the Final Digits of Squares.

Also see Walter Stangl (1996)

Counting Squares in Zn

n P(n) P(n)/P(n-1)

1 .6000000 .6000

2 .2200000 .3667

3 .1590000 .7227

4 .1044000 .6566

5 .0912100 .8736

6 .0781320 .8564

7 .0748719 .9590


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Equi-Pandigital Primes

An equi-pandigitalnumberin base b contain each digit from 0 through (b-1)

exactly the same number of times.

Theorem. For b > 3, there are no equi-pandigital primes.

Proof. Let n be an equi-pandigital number in base b. Then mod (b 1) n is

congruent to the sum of its digitsbecause bn

1n

= 1. Let rbe the # ofrepetitions of 0, 1, 2, , b 1, which sum to b(b 1)/2. So we have:

Ifb is even, then n 0 (mod b 1) since b/2 is an integer, so (b 1) divides n.

Ifb is odd, then (b 1)/2 is an integer, so either n 0 or (b 1)/2. In both cases,

(b 1)/2 divides n.For b > 3, (b 1) and (b 1)/2 are nontrivial, so n is not prime. QED

Remark 1: Finding a base 10 equi-pandigital prime will take some trickery.

Remark 2: 102, 1001012, 1010012, 100010112, etc. are prime, as are

1023, 2013, 1000122123, 1000221123, etc.


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Equi-Pandigital Gaussian Primes

Theorem 9.15 of Deskins (1964): Letp be a prime in Z, then all Gaussian primes

in Z[i] fall into one of the following three cases, up to units.

(a) Ifp 3 (mod 4), thenp is a Gaussian prime.

(b) Ifp 1 (mod 4), thenp = (a + b i)(a b i) is the Gaussian prime factorization,

where a, b is the unique (up to order) solution top = a2 + b2.

(c) Ifp = 2, then 2 = (1 + i)(1i) is the factorization into Gaussian primes.

Proof: See Deskins (1964).

Remark: Brillharts algorithm can find a and b forp 1 (mod 4). See Williams (1995).

Lets make an estimate of the number of (5,5)-digit pandigital primes.

Since P(mZis prime) 1/log(m) (just differentiate the logarithmic integral), we need tomultiply this by the number of (a + b i)s satisfying a > b and condition (b) above. Hence

105

> a > b > 104.5

, which implies that 2(1010

) > a2

+ b2

. Pick two digits from 1 through 9 touse as a and b, and the rest of the digits can be put in any order, so the total number of

expected Gaussian primes is approximately:


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Search Results

By computer search, there are 69774 equi-pandigitalGaussian primes of the form a + b i, a > b > 0. Here aresome interesting ones:

Pandigital Gaussian prime Distinguishing property

96530 + 87421i Max norm

20468 + 13597i Min norm

98765 + 10234i Max realimaginary parts

60143 + 59872i Min realimaginary parts

86420 + 79513i Largest real part with all even digits

20864 + 13579i Smallest imaginary part with all odd digits

97531 + 82604i Largest real part with all odd digits


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All the Equi-Pandigital Gaussian Primes


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2. Anagrams of Words and Numbers

An anagramof a word is a non-identity permutation ofthat words letters.

E.g., Amherstis an anagram of hamster.

One word anagrams are sometimes called transpositionsin

wordplay. In wordplay, some require an anagram to have a related

meaning to the original word.

Well also consider anagrams of numbers. In what

follows, initial zeros are forbidden. E.g., 132 = 169, 142 = 196, and 312 = 961 are anagrams of eachother (in base 10).


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How to find English word anagramsStep 1: Obtain a wordlist.

There are now a variety of sources available:

American Cryptogram Association athttp://cryptogram.org/cdb/words/words.html

National Puzzlers League at

http://www.puzzlers.org/dokuwiki/doku.php?id=solving:wordlists:about:start

Grady Wards Moby word lists (in public domain)

http://icon.shef.ac.uk/Moby/

The above wordlists include all the inflected forms of

words: nouns with both singular and plural forms,adjectives with comparative forms, verbs with allconjugated forms, etc.
http://cryptogram.org/cdb/words/words.htmlhttp://www.puzzlers.org/dokuwiki/doku.php?id=solving:wordlists:about:starthttp://www.puzzlers.org/dokuwiki/doku.php?id=solving:wordlists:about:starthttp://icon.shef.ac.uk/Moby/http://icon.shef.ac.uk/Moby/http://www.puzzlers.org/dokuwiki/doku.php?id=solving:wordlists:about:starthttp://www.puzzlers.org/dokuwiki/doku.php?id=solving:wordlists:about:starthttp://cryptogram.org/cdb/words/words.html


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Step 2: Read in the wordlist andstore it in a hash.

aa

aahaahed

aahing

aahs

aal

aalii

aaliis

aals

aardvarkaardvarks

aardwolf

aardwolves

aas

aasvogel

aasvogels

abaabaca

abacas

abaci

aback

abacus

abacuses

open(WORDS, "CROSSWD.TXT") or die;

while () {

chomp;

@letters = split(//);$key = join('',sort(@letters));

if ( exists($dictionary{$key}) ) {

$dictionary{$key} .= ",$_";

} else {

$dictionary{$key} = $_;

}

}

foreach $key (sort keys %dictionary) {

print "$key, $dictionary{$key}\n";

}

Perl program from Section 3.7.2 of Bilisoly (2008b):The hash key equals the

letters of the word sorted

in alphabetical order.

Examples:

aah -> aah

aahed -> aadeh

aahing -> aaghin

aardvark -> aaadkrrv

If key already exists, then

an anagram has been

discovered. Example:evil, live, vile, veilall have

the key eilv.


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Step 3: Print out the hash withthe keys sorted in alphabetical order.

The result (see right) is ananagram dictionary.

Invaluable for word gamessuch as Scrabble and

Jumble: just sort the lettersat hand and check if theyform a word.

Looking for entries with two

or more commas revealsword anagrams. Most words do not have

anagrams.

aa, aa

aaaaabbcdrr, abracadabra

aaaabcceelrstu, baccalaureates

aaaabcceelrtu, baccalaureate

aaaabdilmorss, ambassadorial

aaaabenn, anabaenaaaaabenns, anabaenas

aaaaccdiiklllsy, lackadaisically

aaaaccdiiklls, lackadaisical

aaaaccrr, caracara

aaaaccrrs, caracaras

aaaacgnr, caraganaaaaacgnrs, caraganas

aaaacmnrst, catamarans

aaaacmnrt, catamaran


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Numbers are formed from thealphabet {0,1,2,3,4,5,6,7,8,9}.

The program above can easily be modified to find anagrams of a set ofnumbers.

In recreational mathematics, it is well known that 122 = 144, 212 = 441; and132 = 169, 312 = 961, 142 = 196.

Unlike words, it turns out that it is easy to find two or more squares that areanagrams. For example, the following 87 squares are anagrams of each other:

1026753849, 1042385796, 1098524736, 1237069584, 1248703569, 1278563049,1285437609, 1382054976, 1436789025, 1503267984, 1532487609, 1547320896,1643897025, 1827049536, 1927385604, 1937408256, 2076351489, 2081549376,2170348569, 2386517904, 2431870596, 2435718609, 2571098436, 2913408576,3015986724, 3074258916, 3082914576, 3089247561, 3094251876, 3195867024,3285697041, 3412078569, 3416987025, 3428570916, 3528716409, 3719048256,3791480625, 3827401956, 3928657041, 3964087521, 3975428601, 3985270641,4307821956, 4308215769, 4369871025, 4392508176, 4580176329, 4728350169,4730825961, 4832057169, 5102673489, 5273809641, 5739426081, 5783146209,

5803697124, 5982403716, 6095237184, 6154873209, 6457890321, 6471398025,6597013284, 6714983025, 7042398561, 7165283904, 7285134609, 7351862049,7362154809, 7408561329, 7680594321, 7854036129, 7935068241, 7946831025,7984316025, 8014367529, 8125940736, 8127563409, 8135679204, 8326197504,8391476025, 8503421796, 8967143025, 9054283716, 9351276804, 9560732841,9614783025, 9761835204, 9814072356.


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A Pattern Emerges

# Digits # Squares # Anasquares Proportion

1 3 0 0.00%

2 6 0 0.00%

3 22 7 31.82%

4 68 13 19.12%

5 217 86 39.63%

6 683 293 42.90%

7 2163 1212 56.03%

8 6837 4699 68.73%

9 21623 17380 80.38%

10 68377 60623 88.66%

In fact, looking at n-digit squares, it seems that as n increases, the proportion of

squares with square anagrams (lets call these anasquares) keeps increasing.

What is the limit?

The above table is Table 1 from Bilisoly (2008a).

Also see http://oeis.org/A177952.
http://oeis.org/A177952http://oeis.org/A177952


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The limit is 100%!

Let Sd,b be the set of squares with exactly ddigits when written in base b.

Define apattern of a number n to be the digits ofn in base b sorted

from least to greatest. Note that a pattern is a hash key.

Theorem (Bilisoly, 2008a): The proportion of anasquares in Sd,b1 as

dand for b fixed.

Proof: A lower bound to the number of anasquares occurs when as many as

possible patterns correspond to exactly 1 square. To find this lower bound,

we count the number of patterns and d-digit squares.

First, thinking back to the Perl program, the hash key of a number is obtainedby sorting its digits. Let dibe the number of times the digit iappears in a

square. Then this hash key can be represented by:

110

**...*|...|**...*|**...*

bddd


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End of Proof

The number of distinct hash keys is:

Second, the number ofddigit squares is:

Hence the number ofd-digit squares is exponential (in d), but the number of

patterns is a polynomial (in d), so the proportion of anasquares is bounded

below by the following, which 1 as d (and b is fixed.)

QED

)./11(11 2/2/)1(2/1 bbbbbbddddd

0,)/11(

1)!1(

)1)...(2)(1()/11(

max2/

2/

bb

b

dbdbdbb

d

d

)!1(

)1)...(2)(1(1

b

dbdbd

d

bd


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3. Birthday Problem and Pangrams

The basic birthday problem is famous: For npeople,assuming all days are equally likely, what is theprobability that at least two people share the samebirthday?

The following are related: Let Nshared = number of people such that 1 birthday appears at

least 2 times.

Let Nall = number of people such that all 365 birthdays appear atleast once.

Note E(Nshared) = 24.6166 > the usual # of peoplequoted. Why? P(22 people, 2 share) = 0.475695

P(23 people, 2 share) = 0.507297


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Results from:Flajolet, Gardy, and Thimonier (1992)

Suppose all days are not equally likely, then let pi = P(day i is a birthday).

Corollary (The Birthday Problem) We need j = 1 day to appear at least k = 2 times.

0

3651shared

)exp()1()( dtttpNEi i

0

3651all

))exp(1(1)( dttpNEi i

Corollary (The Coupon Problem) We need all letters to appear at least once.


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Application to Birthdays

What is the expected number of people needed so that 2 people share a birthday?

Mathematica gives E(Nshared) =24.6166, which assumes each day is equallylikely. Note E(Nall) =2364.65.

What is the expected number of people born in 1978 needed so that 2 peopleshare a birthday?

Mathematica gives E(Nshared) =24.5262 and note E(Nall) =2435.14.

Plot of Julian Day vs. Proportion

of births on that day for 1978.

Which days does the lower

band represent?

Data Source: Todd Swansons Home Page:

http://www.math.hope.edu/swanson/da

ta/birthdays.txt
http://www.math.hope.edu/swanson/data/birthdays.txthttp://www.math.hope.edu/swanson/data/birthdays.txthttp://www.math.hope.edu/swanson/data/birthdays.txthttp://www.math.hope.edu/swanson/data/birthdays.txt


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Pangrammatic Windows

The Spirit dropped beneath it, so that the extinguisher

covered its whole form; but though Scrooge pressed it down

with all his force, he could not hide the light: which streamed

from under it, in an unbroken flood upon the ground.

He was conscious of being exhausted, and overcome by anirresistible drowsiness; and, further, of being in his own

bedroom. He gave the cap a parting squeeze, in which his hand

relaxed; and had barely time to reel to bed, before he sank

into a heavy sleep.

AWAKING in the middle of a prodigiously tough snore, and

sitting up in bed to get his thoughts together, Scrooge hadno occasion to be told that the bell was again upon the

stroke of One. He felt that he was restored to consciousness

in the right nick of time, for the especial purpose of holding

a conference with the second messenger dispatched to him

through Jacob Marley's intervention.

This text is from Charles

DickensA Christmas Carol.

The blue portion is a

pangrammatic window, i.e.,

it contains each letter of thealphabet at least once.

There are 679 letters in

color.

The search started with

The Spirit and thewindow could be shortened

by dropping letters from the

beginning.


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Pangrams in A Christmas Carol

a 9308 0.076892

b 1943 0.016051

c 3035 0.025072

d 5674 0.046872

e 14850 0.122674

f 2433 0.020099

g 2979 0.024609

h 8368 0.069127

i 8294 0.068515

j 113 0.00093

k 1031 0.008517

l 4553 0.037612

m 2840 0.023461

n 7960 0.065756

o 9690 0.080048

p 2119 0.017505

q 97 0.000801

r 7031 0.058082

s 7900 0.065261

t 10869 0.089787

u 3335 0.02755

v 1022 0.008443

w 3096 0.025576

x 131 0.001082

y 2298 0.018983

z 84 0.000694

Well search A Christmas Carolfor pangrams byselecting random starting positions. Then we

compare this to independently generated lettersusing the letter frequencies of this novel. Thecounts and the proportions are listed to the right.

Of course, letters are not independent, but thequestion is this: How does the actual pangram

lengths differ from the simulated independentpangram lengths?

Letter Frequencies

in Christmas Carol


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Pangram Lengths

The left histogram shows lengths of

pangrams found inA Christmas Carol

using random starting points.

The right histogram shows lengths of

pangrams found in a simulated string of

independent letters using the

proportions found inA Christmas Carol.

Note the long right tail

N = 1000N = 1000

Theoretical mean = 2473.8

Figures from Bilisoly (2009)


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Concordancing words with the letter zreveals

"Why, it's old Fezziwig! Bless his heart; i

ess his heart; it's Fezziwig alive again!"

Old Fezziwig laid down his pen,

"Yo ho, there! Ebenezer! Dick!"

ho, my boys!" said Fezziwig. "No more work to-n

e, Dick. Christmas, Ebenezer! Let's have the shuters up," cried old Fezziwig, with a sharp clap

illi-ho!" cried old Fezziwig, skipping down from

-ho, Dick! Chirrup, Ebenezer!"

ared away, with old Fezziwig looking on. It was

aches. In came Mrs. Fezziwig, one vast substanti

came the three Miss Fezziwigs, beaming and lovabl

brought about, old Fezziwig, clapping his hands

overley." Then old Fezziwig stood out to dance

to dance with Mrs. Fezziwig. Top couple, too; wah, four times--old Fezziwig would have been a m

, and so would Mrs. Fezziwig. As to her, she was

eared to issue from Fezziwig's calves. They shon

next. And when old Fezziwig and Mrs. Fezziwig hd Fezziwig and Mrs. Fezziwig had gone all throug

gain to your place; Fezziwig "cut"--cut so deftl

ke up. Mr. and Mrs. Fezziwig took their stations

hearts in praise of Fezziwig: and when he had do

luence over him, he seized the extinguisher-ca

e the cap a parting squeeze, in which his hand

ore and centre of a blaze of ruddy light, whi

ore alarming than a dozen ghosts, as he was p

; and such a mighty blaze went roaring up the

, half thawed, half frozen, whose heavier part

ught fire, and were blazing away to their dearanding his gigantic size, he could accommoda

chit, kissing her a dozen times, and taking o

erness and flavour, size and cheapness, were

, so hard and firm, blazing in half of half-a-q

e flickering of the blaze showed preparations

g grew but moss and furze, and coarse rank gr

of endeavouring to seize you, which would ha

relents," she said, amazed, "there is! Nothing

grave his own name, EBENEZER SCROOGE.

er they've sold the prize Turkey that was han

re?--Not the little prize Turkey: the big one

it. It's twice the size of Tiny Tim. Joe Mi

e passed the door a dozen times, before he ha

out after dark in a breezy spot--say Saint Paud-stone, Scrooge! a squeezing, wrenching, graspinThe cold within him froze his old features, n

e court outside, go wheezing up and down, beatin

'em through a round dozen of months presented

e chattering in its frozen head up there. The

d a great fire in a brazier, round which a part

eir eyes before the blaze in rapture. The wat

Scrooge seized the ruler with such

n the gloom. Half-a-dozen gas-lamps out of th

her-beds, Abrahams, Belshazzars, Apostles putting o

ring at those fixed glazed eyes, in silence fothe vision's stony gaze from himself.

from other regions, Ebenezer Scrooge, and is conpe of my procuring, Ebenezer."

Exchange pay to Mr. Ebenezer Scrooge or his orde

st have sunk into a doze unconsciously, and

er a long way below freezing; that he was clad b

The Spirit gazed upon him mildly. It

Middle section has

43 of the 84 zs, but

represents only 3 of 83

pages of Dickens (1986).


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Pangram Lengths: Fezziwig Effect

N = 1000

Simulated pangrams.

Theoretical Mean = 3620.5

N = 1000

Actual pangrams.

Endpoint with Fezziwig


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100,000 Simulated Pangram Lengths

Best fit lognormal distribution shown.


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References

Roger Bilisoly (2008a). Anasquares: Square Anagrams of Squares. Mathematical Gazette, 92,58-63.

Roger Bilisoly (2008b). Practical Text Mining with Perl, Wiley.

Roger Bilisoly (2009). Two Language-based Examples for Use in the Statistics Classroom.American Statistical Association Proceedings of the Joint Statistical Meetings, Section onStatistical Education.

Gunnar Blom, Lars Holst, and Dennis Sandell (1993). Problems and Snapshots from the World ofProbability, Springer.

W. E. Deskins (1964). Abstract Algebra, MacMillan.

Charles Dickens (1986). A Christmas Carol, Bantam.

Philippe Flajolet, Daniele Gardy, and Loys Thimonier (1992). Birthday Paradox, CouponCollectors, Caching Algorithms and Self-Organinzing Search. Discrete Applied Mathematics, 39,207-229.

Walter Penney (1960). On the Final Digits of Squares. The American Mathematical Monthly, Vol.67, No. 10, pp. 1000-1002.

Walter Stangl (1996). Counting Squares in Zn. Mathematics Magazine, Vol. 69, No. 4, pp. 285-189.

Kenneth Williams (1995). "Some Refinements of an Algorithm of Brillhart," CanadianMathematical Society Conference Proceedings, Volume 15, 409-416. Available athttp://www.math.carleton.ca/~williams/papers/pdf/202.pdf .
http://www.math.carleton.ca/~williams/papers/pdf/202.pdfhttp://www.math.carleton.ca/~williams/papers/pdf/202.pdf


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Web References

Benfords Law http://mathworld.wolfram.com/BenfordsLaw.html http://en.wikipedia.org/wiki/Benford's_law

Squares with 3 distinct digits http://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htm

Counterexample to Ed Pegg http://mathworld.wolfram.com/Baxter-HickersonFunction.html

American Cryptogram Association http://cryptogram.org/

National Puzzlers Association http://www.puzzlers.org/

Moby Word Lists http://icon.shef.ac.uk/Moby/

Anasquare counts http://oeis.org/A177952. 1978 birthday data

http://www.math.hope.edu/swanson/data/birthdays.txt

Word Ways http://wordways.com/
http://mathworld.wolfram.com/BenfordsLaw.htmlhttp://en.wikipedia.org/wiki/Benford's_lawhttp://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htmhttp://mathworld.wolfram.com/Baxter-HickersonFunction.htmlhttp://cryptogram.org/cdb/words/words.htmlhttp://www.puzzlers.org/dokuwiki/doku.php?id=solving:wordlists:about:starthttp://icon.shef.ac.uk/Moby/http://oeis.org/A177952http://www.math.hope.edu/swanson/data/birthdays.txthttp://wordways.com/http://wordways.com/http://www.math.hope.edu/swanson/data/birthdays.txthttp://oeis.org/A177952http://icon.shef.ac.uk/Moby/http://www.puzzlers.org/dokuwiki/doku.php?id=solving:wordlists:about:starthttp://cryptogram.org/cdb/words/words.htmlhttp://mathworld.wolfram.com/Baxter-HickersonFunction.htmlhttp://mathworld.wolfram.com/Baxter-HickersonFunction.htmlhttp://mathworld.wolfram.com/Baxter-HickersonFunction.htmlhttp://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htmhttp://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htmhttp://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htmhttp://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htmhttp://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htmhttp://en.wikipedia.org/wiki/Benford's_lawhttp://mathworld.wolfram.com/BenfordsLaw.html


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Wordplay References

Tony Augarde (1994). The Oxford A to Z of Word Games, Oxford. Tony Augarde (2003). The Oxford Guide to Word Games, Oxford.

o Has historical information. Dmitri Borgmann (1967). Beyond Language,Scribners. Ross Eckler (1979). Word Recreations, Dover.

o Most examples originally appeared in Word Ways. Ross Eckler (1996). Making the Alphabet Dance, St. Martin's.o Most examples originally appeared in Word Ways.

Dave Morice (1997). Alphabet Avenue, Chicago Review Press. Dave Morice (2001). The Dictionary of Word Play, Teachers and Writers

Collaborative. Warren F. Motte, Jr. (1998). Oulipo: A Primer of Potential Literature, Dalkey

Archive.o Oulipostands for Ouvroir de Litterature Potentielle, which is a group of

writers, mathematicians, and other people interested in literarystructures.


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The Key Wordplay Resource:Word Ways: A Journal of Recreational Linguistics

Established by Dmitri Borgmann in 1968o He is author of Language on Vacationand Beyond Language

Bought by A. Ross Eckler, Jr. in 1968. He waseditor and publisher from 1968-2006.

o PhD in mathematics from Princeton, 1954o Worked at Bell Labs, 1954-84o Published Word Recreations(1979), Names and Games:

Onomastics and Recreational Linguistics(1986), Making theAlphabet Dance(1996)

Current editor is Jeremiah Farrell, professor

emeritus of mathematics at Butler University

Online at http://wordways.com/
http://wordways.com/http://wordways.com/


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Open question:

What are the upper

and lower bounds of

this plot? Points aresquares in base 10

with 12 or less

digits. This is Figure

2 of Bilisoly (2008a).


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Brillhart Alogithm (See Slide 18)


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Let us generalize the birthday problem.

Let represent an alphabet of size na.

For birthdays let = {d1, d2,, d365}, so naequals 365.

Let pi= P(dioccurs), so that each day need not be equally likely.

Define Njk= number of letters drawn from (with replacement) so

that there arejdistinct letters that each appear at least ktimes.

Let ek(t) = kth order Taylor series expansion of exp(t). Theorem 1 of Flajolet, Gardy and Thimonier (1992) states:

1

00

n

1 11)exp()))()(exp()((][)(

a

j

l

i ikiik

l

jk dtttpetpxtpexNE

Corollary (The Birthday Problem) We need j = 1 day to appear at least k = 2 times.Note that the sum has only one term, and N12 = Nshared.

0

365

112)exp()1()( dtttpNE

i i

See Corollary 1 of Flajolet et al. (1992)

Product of 1st degree

polynomials in x


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Generalized Coupon Collectors Problem

Theorem 2 of Flajolet, Gardy and Thimonier (1992)The expected number of letters drawn to get thecomplete alphabet, , is given below. Their proof

follows fairly easily from Theorem 1.

0 1all ))exp(1(1 dttpN

an

i i

For uniformly likely birthdays, 2364.65 people are neededon average to get all 365 days to appear. For 1978, weexpect to need 2435.14 people.

Pangrams have = {a, b, c, , z}, na= 26, and pi

determined by frequencies found in a text sample.


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Example of Mathematica 8 code

to find 14 letter words with nomultiple edges and diameter = 2.


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Squares Having only Three Distinct Digits:Investigated by Hisanori Mishima

Largest known sporadic example:81401637345465395512991484^2 =6626226562522666562566262626266252566552622656522256

However, there are an infinite number of patterned 3-digit squares.

97 9409

997 994009

9997 99940009

99997 9999400009

999997 999994000009

1235 1525225

12335 152152225

123335 15211522225

1233335 1521115222225

12333335 152111152222225

See http://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htm
http://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htmhttp://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htmhttp://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htmhttp://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htmhttp://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htmhttp://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0203.htm


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How many sporadic solutions?

Assume that digits of squares are independent.

Binomial[10,3] (3/10)^n P(n-digit square with 3 distinct digits)

10^(n/2)(1 - 1/Sqrt[10]) number of n-digit squares

Expected # n-digit squares with 3 distinct digits Binomial[10,3] (3/10)^n * 10^(n/2)(1 - 1/Sqrt[10]) =constant * (3/10)n

But n (3/10)n converges, which suggests a finite # of solutions.

Sum = 360(1-1/10)(3+10) 1517.

However, the analogous argument for squares with

4 distinct digits results in n (4/10)n, which diverges.


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Three Examples of 150-digit Squareswith exactly 9 distinct digits.

This square has no 1s.

590286760507408218847058025821601275020644462041449539546951025992081988403 ^2 =

348438459630330307258664735742002640854590982059075240585330803623545707923

270640840935682702648690975443382535993405344806539300344597863650226490409

This square has no 8s.705635480731670264258949343062158505097813112879657762505544377338770578675 ^2 =

497921431667415396779522503575462537630442531656110964327039001763590201921

651000116765556629356041206321334237073176796236102099130225925794364755625

This square has no 7s.

624228579548317386188320909320329013264935555613585034560249848356296289668 ^2 =

389661319524910006943311531238425925016482192128500480080280569984852381981

266389000443336616268805696010466681530390083280245809868986959183363550224


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Squares with 9 Distinct Digits

How common is an n-digit square with only 9 distinct digits?

Binomial[10,9] (9/10)^n probability of square with 9 digits

For n = 150, this gives 1.36891 E-6 1 in a million.

Hence a computer program checking 5,000,000 random 150-digit

squares should find 5,000,000*1.36891 E-6 = 6.84 such squares. This was done 30 times, and the counts are given in the histogram.

Mean = 7.27

SD = 2.65


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Ed Peggs Failed Conjecture

In April of 1999, on sci.math, Ed Pegg conjectured thatthere are only finitely many cubes without the digit 0.

D. Hickerson found a counterexample and a few dayslater Lew Baxter found the example given below.

baxter[n_] := (2 10^(5 n)-10^(4 n)+2 10^(3 n)+10^(2 n)+10^n+1)/3

Do[Print[{baxter[i],baxter[i]^3}],{i,1,5}]

{64037, 262598918898653}

{6634003367, 291962492648791178822648631863}

{666334000333667, 295852962482593148779111778815593148629851963}

{66663334000033336667,296251862962481592598148777911117778814892598148629651852963}

{6666633334000003333366667,

296291851962962481492592648148777791111177778814822592648148629631851862963}

Function given at http://mathworld.wolfram.com/Baxter-HickersonFunction.html
http://mathworld.wolfram.com/Baxter-HickersonFunction.htmlhttp://mathworld.wolfram.com/Baxter-HickersonFunction.htmlhttp://mathworld.wolfram.com/Baxter-HickersonFunction.htmlhttp://mathworld.wolfram.com/Baxter-HickersonFunction.html


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Let A = 1, B = 2, C = 3, ..., Z = 26. Let s(word) = Sum of its alphabetic values

o Example: s(bad) = 2 + 1 + 4 = 7 Let nn(number) = its number name

o Example: nn(3) = three

Consider the dynamical system of composing sand nno That is, iterate n-> nn(n) -> f(nn(n)) -> nn(f(nn(n)), etc.o Example: 1, 34, 160, 205, 174, 278, 291, 253, 254, 258,

247, 281, 240, 216, 228, 288, 255, 240 1 becomes a 5-cycle, so what else can happen?

o Answer first published by Dmitri Borgmann in 1967 inBeyond Language.

Miscellaneous Example:Number Words and Numbers Graph


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Documents

String Patterns: Searching for Interesting Words and Numbers