What/Quid:
Algebraic applications in creating imaginary abecedaries
Ex algebrae applicationis creationem abecedārii commentīcia
When/Quando:
June, 2013. VI (Iūnius), MMXIII.
Copyright © 2013 by Alejandro Hector Ochoa-Gonzalez.
Guadalajara, Jalisco, Mexico.
© 2013 by
No part of this e-book may be used, reproduced or transmitted in any manner or by any means whatsoever without written or electronic permission from the publisher, except in the case of brief or long quotations, embodied in critical reviews and articles.
Of course, it is authorized, from now and on, the copying of all or any parts of this article, provided there are no profit purposes; only for study, personal use, reproduction with no monetary interests, duplication to give it as a gift, etcetera.
It is intended that the following writing constitute, in the distant future (due to lack of time, for now), a partial basis for an Introduction of a reverse dictionary in which the words that constitute the entries are written from right to left, but the order or alphabetical indexing will be from left to right according to the order of the letters that make the words resulting from the reversions; such as:
yrreb berry
yrrebkcalb blackberry
yrrebnarc cranberry
yrrebpsar raspberry
yrrebwarts strawberry
yrrehc cherry,
et cetera.
It is not about discovering warm water.
Reverse dictionaries already exist, for example, one by John Walker, The rhyming dictionary of the English language: in which the whole language is arranged according to its terminations..., Routledge & Kegan Paul, 1983.
Imaginary abecedaries
Abecedārii commentīcia
1. Sometimes, grammar and mathematics agree on some points, or may reveal certain relationships, or the human brain can make, or imagine, that certain elements, fragments, procedures, routes, et cetera, harmonize, coincide or tune in some places, subdivisions, times, systems... under certain conditions.
All sciences, disciplines and specialties have certain relations to each other. A truism.
This paper will match certain mechanisms of juxtapositions of letters to form words (or groups of letters), with some values and formulas of algebra.
English Grammar and Babylonian-Persian-Indo-Arab Algebra will come together, but algebra –that very important discipline, a branch or division of the mathematical science– this time will be serving Grammar as an auxiliary.
In 1586, William Bullokar printed the first English grammar, Bref Grammar for English.
By the 1930s American mathematicians Paul K. Rees and Fred W. Sparks released their College Algebra, published by McGraw-Hill Book Company, Incorporated.
This book, expanded, was still printed in the 1990s.
1.1. Note 1: in chapter number 7 there is a summary… Should you want to skip lengthy lucubrations and constructions.
1.2. Note 2: in chapter number 8 there are exposed two theories resulting from an observation exercise* (inductive method). Please see item 7.2.3.
*This one.
2. Symbols used.
In this paper, three main symbols will be used:
B base
p power
R result.
2.1. In this algebraic-grammarian exercise, the number of letters in each imaginary abecedary that we will create will constitute a mathematical base (which is represented by the symbol B).
2.2. The number of letters contained in the words that we will be able to build, will constitute the exponent or power, represented by the symbol p, written as a superscript character, an indicator of the power that the bases cited in 2.1. will have to be raised to; that is, Bp.
2.2.1. Furthermore, the different values that p will be acquiring, depending on the number of letters in each word we will be able to build, will indicate three things:
2.2.1.1. The number of terms each algebraic formula will have, and the highest exponential value of each monomial, binomial, trinomial, quadrinomial, pentanomial, et cetera. The highest exponential value of each nomial will match the number of terms of such nomial, as we will discuss below:
B —this formula has one term, and this monomial is the simplest formula of all; here, p equals one, id est: Bp = B1 = B.
B(B-(p-1)) =
B(B-(2-1)) =
B(B-1) =
B2-B —this formula has two terms, and in this binomial the highest exponential value of p is two.
B(B-(p-2))(B-(p-1)) =
B(B-(3-2))(B-(3-1)) =
B(B-1)(B-2) =
(B2-B)(B-2) =
B3-B2-2B2+2B =
B3-3B2+2B —this formula has three terms, and in this trinomial the highest exponential value of p is three.
B(B-(p-3))(B-(p-2))(B-(p-1)) =
B(B-(4-3))(B-(4-2))(B-(4-1)) =
B(B-1)(B-2)(B-3) =
(B2-B)(B-2)(B-3) =
(B3-B2-2B2+2B)(B-3) =
(B3-3B2+2B)(B-3) =
B4-3B3+2B2-3B3+9B2-6B =
B4-6B3+11B2-6B —this formula has four terms, and in this quadrinomial the highest exponential value of p is four.
B(B-(p-4))(B-(p-3))(B-(p-2))(B-(p-1)) =
B(B-(5-4))(B-(5-3))(B-(5-2))(B-(5-1)) =
B(B-1)(B-2)(B-3)(B-4) =
(B2-B)(B-2)(B-3)(B-4) =
(B3-B2-2B2+2B)(B-3)(B-4) =
(B3-3B2+2B)(B-3)(B-4) =
(B4-3B3+2B2-3B3+9B2-6B)(B-4) =
(B4-6B3+11B2-6B)(B-4) =
B5-6B4+11B3-6B2-4B4+24B3-44B2+24B =
B5-10B4+35B3-50B2+24B —this formula has five terms, and in this pentanomial the highest exponential value of p is five.
You can create more formulas (please see the fifth paragraph of section 6.6.), but I think that with the above examples it will be suffice.
2.2.1.2. In the algebraic “pre-formulas”, within algebraic factors that have two or more elements, the successive values of the symbol p will serve as the bases for the construction of certain subtrahends to be subtracted from B, for example:
(The different nomials, in these five “pre-formulas” are separated by semicolons.)
B; monomial
B(B-(p-1)); binomial
B(B-(p-2))(B-(p-1)); trinomial
B(B-(p-3))(B-(p-2))(B-(p-1)); quadrinomial
B(B-(p-4))(B-(p-3))(B-(p-2))(B-(p-1)); pentanomial.
Note: In the above examples, in the first nomial the symbol p has a value of 1; in the second nomial the symbol p has a value of 2; in the third nomial the symbol p has a value of 3; in the fourth nomial the symbol p has a value of 4; and in the fifth nomial the symbol p has a value of 5.
After being substituted the different values of p, math operations will generate the following “pre-formulas”:
B;
B(B-(2-1));
B(B-(3-2))(B-(3-1));
B(B-(4-3))(B-(4-2))(B-(4-1));
B(B-(5-4))(B-(5-3))(B-(5-2))(B-(5-1));
B;
B(B-1);
B(B-1)(B-2);
B(B-1)(B-2)(B-3);
B(B-1)(B-2)(B-3)(B-4);
According to the algebraic multiplications made in 2.2.1.1., we obtain the following five nomials, which are examples and nothing more, because the possibilities are endless:
B
B2-B
B3-3B2+2B
B4-6B3+11B2-6B
B5-10B4+35B3-50B2+24B
2.2.1.3. In the algebraic formulas resulting from the algebraic multiplications indicated by the “pre-formulas”, the symbol p indicates the highest exponential value of each nomial, and exponents or powers of the terms of each nomial go down 1 by 1, from left to right, up to the value of 1.
The number of terms of each nomial equals the highest exponent or power, p, of such nomial, as it can be seen in the final part of 2.2.1.2.
2.3. The symbol R, which means: result, will be used rarely.
Note: in certain mathematical disciplines, as well as in computer programming and software creation, some authors and engineers indicate or denote raising a number to a power, not by writing a superscript number (exempli gratia, 23 = 8), but through the collation of a caret, hood, or circumflex accent (which in Word 2003 and Word 2007 can be written by pressing the following keys: ALT 94 ^, or by using the Unicode key 005E ^; or another Unicode key, 0302: ^), exempli gratia, 2 ^ 3 = 8. Both forms are correct; here, we will be using predominantly the superscript character, but in tables of chapter 7, we will use the hood or caret (or circumflex accent).
Moreover, the multiplication sign (“by”) can be obtained in Word 2003 and Word 2007 by pressing ALT 158: ×, or ALT 0215, or by using the Unicode key 00D7: ×.
If in some Windows or Unix platform you use, you cannot write some symbols because the software does not support ALT or Unicode keys, or if your keyboard does not respond, I will suggest something: please firstly write in Word the symbols, signs, funny letters, Greek letters, strange characters, rarely used characters, et cetera, and then select, copy and paste them in the Web space in which you are writing.
In another post of one of my other blogs, you can find a number of ALT and Unicode keys to symbols, signs, Latin letters, Greek letters, et cetera:
3. Let’s imagine an ary (abecedary) with a single letter: a.
3.1. Imagine that we could form one-letter words each, only.
3.1.1. According to sections 2.1. and 2.2., and by virtue of that in our ary there is only one letter, let’s raise the base 1 to the first power: Bp = B1 = 11 = 1, which is the number of words that would exist –a single word: a.
3.1.2. Now let’s introduce a restriction: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our ary (1) to a power equal to the number of letters that each and every word could have (1 in this example), under the restriction we apply the formula: B.
Then, as in this case our ary has only one letter: the letter a, we apply the formula B, and substitute: B = 1 (one), which is the number of words we obtain, according to our restriction: a single word, because if our ary has only one letter (a), and we want to form one-letter words each, words in which no letter is repeated, that gives as a result a single word: a.
3.2. Then, imagine that we could form two-letter words each, only.
3.2.1. According to sections 2.1. and 2.2., and as in our ary there is only one letter, we raise the base 1 to the second power (or squared): 12 = 1 × 1 = 1, which is the number of words that would exist –a single word: aa .
3.2.2. Now, let’s introduce a restriction: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our ary (1) to a power equal to the number of letters that each and every word could have (2 in this example), under the restriction we apply the formula: B(B-1) = B2-B.
Then, as in this case our ary has only one letter: the letter a, we apply the formula, B2-B, and substitute: 12-1 = 1-1 = 0 (zero), which is the number of words we obtain, according to our restriction: no word, in view that if our ary has only one letter (a), and we want to form two-letter words each, words in which no letter is repeated, that is impossible.
3.3. Then imagine that we could form three-letter words each, only.
3.3.1. In accordance with sections 2.1. and 2.2., and as in our ary there is only one letter, we raise the base 1 to the third power (or cubed): 13 = 1 × 1 × 1 = 1, which is the number of words that would exist –a single word: aaa.
3.3.2. Now, let’s introduce a restriction equal to the previous one: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our ary (1) to a power equal to the number of letters that each and every word could have (3 in this example), under the restriction we apply the formula: B(B-1)(B-2) = (B2-B)(B-2) = B3-B2-2B2+2B = B3-3B2+2B.
Then, as in this case our ary has only one letter: the letter a, we apply the formula B3-3B2+2B, and substitute: 13-(3)(12)+2(1) = 1-3+2 = 0 (zero), which is the number of words we obtain, according to our restriction: no word, in view that if our ary has only one letter (a), and we want to form three-letter words each, words in which no letter is repeated, that is impossible.
3.4. Further, imagine that we could form four-letter words each, only.
3.4.1. According to sections 2.1. and 2.2., and as in our ary there is only one letter, we raise the base 1 to the fourth power: 14 = 1 × 1 × 1 × 1 = 1, which is the number of words that would exist –a single word: aaaa.
3.4.2. Now, let’s introduce a restriction equal to the above: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our ary (1) to a power equal to the number of letters that each and every word could have (4 in this example), under the restriction we apply the formula: B(B-1)(B-2)(B-3) =
(B2-B)(B-2)(B-3) =
(B3-B2-2B2+2B)(B-3) =
(B3-3B2+2B)(B-3) =
B4-3B3+2B2-3B3+9B2-6B =
B4-6B3+11B2-6B.
Then, as in this case our ary has only one letter: the letter a, we apply the formula B4-6B3+11B2-6B, and substitute: 14-(6)(13)+(11)(12)-(6)(1) = 1-6+11-6 = 0 (zero), which is the number of words we obtain, according to our restriction: no word, because if our ary has only one letter (a), and we want to build words of four letters each, words in which no letter is repeated, that is impossible.
(An explanatory parentheses: as you can see, the formulas for the restrictions come from simpler formulas, such as the ones shown above, in the second paragraph of item 3.2.2., and in the second paragraph of item 3.3.2 .)
3.5. Further, imagine that could form five-letter words each only.
3.5.1. According to sections 2.1. and 2.2., and as in our ary there is only one letter, we raise the base 1 to the fifth power: 15 = 1 × 1 × 1 × 1 × 1 = 1, which is the number of words that would exist –a single word: aaaaa.
3.6. And so on.
3.7. Now, let’s imagine that we could form one-letter and two-letter words each, only.
There would be two words: a, aa (11 + 12 = 1 + 1 = 2). This last arithmetic operation is a result from the algebraic formula: Bp + Bp = R, where B is the base or number of letters in our ary (1), p is the power or exponent that represents the number of letters the words we may form will have, and R is the result, in this case: 11 + 12 = 1 + 1 = 2.
The above statement is according to what was set in sections 2.1. and 2.2. of this algebraic-grammarian text.
3.8. Also, imagine that we could form one-letter, two-letter, and three-letter words each, only.
There would be three words: a, aa, aaa (11 + 12 + 13 = 1 + 1 + 1 = 3).
3.9. In addition, let’s assume that we could form one-letter, two-letter, three-letter, and four-letter words each, only.
There would be four words: a, aa, aaa, aaaa (11 + 12 + 13 + 14 = 1 + 1 + 1 + 1 = 4).
3.9.1. And so on.
4. Now, let’s imagine an abeary (abecedary) with two letters: a, b.
4.1. Imagine that we could form one-letter words each, only.
4.1.1. According to sections 2.1. and 2.2., and as in our abeary there are only two letters, let’s raise the base 2 to the first power Bp = B1 = 21 = 2, which is the number of words that would exist –two words: a, b.
4.1.2. Now let’s introduce a restriction: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our abeary (2) to a power equal to the number of letters that each and every word could have (1 in this example), under the restriction we apply the formula: B.
Then, as in this case our abeary has only two letters (a, b), and there is a restriction, we apply the formula B, and substitute: B = 2 (two), which is the number of words we obtain, according to our restriction: two words, because if our abeary has only two letters (a, b), and we want to form one-letter words each, words in which no letter is repeated, that gives as a result two words: a, b.
4.2. Then imagine that we could form two-letter words each, only.
4.2.1. According to sections 2.1. and 2.2., and as in our abeary there are only two letters, let’s raise the base 2 to the second power (or squared) 22 = 2 × 2 = 4, which is the number of words that would exist –four words: aa, ab, ba, bb.
4.2.2. Now, let’s introduce a restriction, as above: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our abeary (2 in this example) to a power equal to the number of letters that each and every word could have (2 in this example), under the restriction we apply the formula: B(B -1) = B2-B.
This formula is the same as the one that appears in the second paragraph of item 3.2.2.
Then, as in this case our abeary has only two letters (a, b), we apply the formula B2-B, and substitute: 22-2 = 4-2 = 2, which is the number of words that would exist, according to our restriction –two words: ab, ba.
4.3. Imagine that we could form three-letter words each only.
4.3.1. In accordance with sections 2.1. and 2.2., and as in our abeary there are only two letters, let’s raise the base 2 to the third power (or cubed): 23 = 2 × 2 × 2 = 8, which is the number of words that would exist –eight words:
aaa
aba aab baa
bba bab abb
bbb.
4.3.2. Now, let’s introduce a restriction as above, in the item 4.2.2.: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our abeary (2) to a power equal to the number of letters that each and every word could have (3 in this example), under the restriction we apply the formula: B(B-1)(B-2) = (B2-B)(B-2) = B3-B2-2B2+2B = B3-3B2+2B.
This formula is the same as the one that appears in the second paragraph of item 3.3.2.
Then, as in this case our abeary has only two letters (a, b), we apply the formula B3-3B2+2B, and substitute: 23-(3)(22)+2(2) = 8-12+4 = 0 (zero), which is the number of words we obtain, according to our restriction: no word, in view that if our abeary has only two letters (a, b), and we want to form three-letter words each, words in which no letter is repeated, that is impossible.
4.4. Imagine that we could form four-letter words each only.
4.4.1. In accordance with sections 2.1. and 2.2., and as in our abeary there are only two letters let’s raise the base 2 to the fourth power: 24 = 2 × 2 × 2 × 2 = 16, which is the number of words that would exist –sixteen words:
aaaa aaab aaba abaa baaa
aabb abab abba baab baba bbaa
abbb babb bbab bbba bbbb.
4.4.2. Now, let’s introduce a restriction as above (in the item 4.3.2.): that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our abeary (2) to a power equal to the number of letters that each and every word could have (4 in this example), under our restriction we apply the formula:
B(B-1)(B-2)(B-3) =
(B2-B)(B-2)(B-3) =
(B3-B2-2B2+2B)(B-3) =
(B3-3B2+2B)(B-3) =
B4-3B3+2B2-3B3+9B2-6B =
B4-6B3+11B2-6B.
This formula is the same as the one that appears in the second paragraph of item 3.4.2.
Then, we apply the formula B4-6B3+11B2-6B, and substitute: 24-6(23)+11(22)-6B = 16-(6)(8)+(11)(4)-(6)(2) = 16-48+44-12 = 0 (zero), which is the number of words we obtain, according to our restriction: no word, in view that if our abeary has only two letters (a, b), and we want to form four-letter words each, words in which no letter is repeated, that is impossible.
4.5. Imagine that we could form five-letter words each only.
4.5.1. According to sections 2.1. and 2.2., and as in our abeary there are only two letters, let’s raise the base 2 to the fifth power: 25 = 2 × 2 × 2 × 2 × 2 = 32, which is the number of words that would exist –thirty two words:
(16 with initial “a”):
aaaaa (5 a’s)
aaaab aaaba aabaa abaaa (4 a’s and 1 b)
aaabb aabab aabba abaab ababa abbaa (3 a’s and 2 b’s)
aabbb ababb abbab abbba (2 a’s and 3 b’s)
abbbb (1 a and 4 b’s)
(16 with initial “b”):
baaaa (1 b and 4 a’s)
baaab baaba babaa bbaaa (2 b’s and 3 a’s)
baabb babab babba bbaab bbaba bbbaa (3 b’s and 2 a’s)
babbb bbabb bbbab bbbba (4 b’s and 1 a)
bbbbb (5 b’s)
4.6. And so on.
(In mathematics and computer programming, when it is not possible to write a power or exponent in the form of a superscript character [as in: Bp] many individuals use the hood or caret or circumflex accent, ^ –please read the Note under section 2.3.
(Thus, it is true that: 25 = 2 ^ 5 = 32.
(Bp = B ^ p)
4.7. Next, imagine that we could form one-letter, and two-letter words each, only.
There would be six words: a, b, aa, ab, ba, bb (21 + 22 = 2 + 4 = 6).
4.8. Then, imagine that we could form one-letter, two-letter, and three-letter words each, only.
There would be fourteen words:
a b
aa ab ba bb
aaa
aab aba baa
abb bab bba
bbb
(21 + 22 + 23 = 2 + 4 + 8 = 14).
4.9. Then, imagine that we could form one-letter, two-letter, three-letter, and four-letter words each, only.
There would be 30 words:
a b
aa ab ba bb
aaa
aab aba baa
abb bab bba
bbb
aaaa aaab aaba abaa baaa
aabb abab abba baab baba bbaa
abbb babb bbab bbba bbbb.
(21 + 22 + 23 + 24 = 2 + 4 + 8 + 16 = 30).
4.10. Further, imagine that we could form one-letter, two-letter, three-letter, four-letter, and five-letter words each, only.
There would be 62 words:
a b
aa ab ba bb
aaa
aab aba baa
abb bab bba
bbb
aaaa aaab aaba abaa baaa
aabb abab abba baab baba bbaa
abbb babb bbab bbba bbbb
aaaaa
aaaab aaaba aabaa abaaa
aaabb aabab aabba abaab ababa abbaa
aabbb ababb abbab abbba
abbbb
baaaa
baaab baaba babaa bbaaa
baabb babab babba bbaab bbaba bbbaa
babbb bbabb bbbab bbbba
bbbbb
(21 + 22 + 23 + 24 + 25 = 2 + 4 + 8 + 16 + 32 = 62).
4.11. And so on.
5. Then, imagine an abeceary (abecedary) with three letters: a, b, c.
5.1. Imagine that we could form one-letter words each, only.
5.1.1. According to sections 2.1. and 2.2., and as in our abeceary there are only three letters, let’s raise the base 3 to the first power Bp = B1 = 31 = 3, which is the number of words that would exist –three words: a, b, c.
5.1.2. Now, let’s introduce a restriction: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our abeceary (3) to a power equal to the number of letters that each and every word could have (1 in this example), under the restriction we apply the formula: B.
Then, as in this case our abeceary has only three letters (a, b, c), and there is a restriction, we apply the formula B, and substitute: B = 3 (three), which is the number of words we obtain, according to our restriction: three words, because if our abeceary has only three letters (a, b, c), and we want to form one-letter words each, words in which no letter is repeated, that gives as a result three words: a, b, c.
Then, as in this case our abeceary has three letters (a, b, c), we apply the formula, B, and substitute: B = 3, which is the number of words that we agree with our restriction, which requires not repeat any letters in words: a, b, c.
5.2. Imagine that we could form two-letter words each only.
5.2.1. According to sections 2.1. and 2.2., and as in our abeceary there are only three letters, let’s raise the base 3 to the second power (or squared): 32 = 9, which is the number of words that would exist –nine words: aa, ab, ac, ba, bb, bc, ca, cb, cc.
5.2.2. Now, let’s introduce a restriction: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our abeceary (3 in this example) to a power equal to the number of letters that each and every word could have (2 in this example), under the restriction we apply the formula: B(B -1) = B2-B.
This formula is the same as the one that appears in the second paragraph of item 4.2.2.
Then, as in this case our abeceary has only three letters (a, b, c), we apply the formula B2-B, and substitute: 32-3 = 9-3 = 6, which is the number of words that would exist, according our restriction –six words: ab, ac, ba, bc, ca, cb.
5.3. Now imagine that we could form three-letter words each only.
5.3.1. In accordance with sections 2.1. and 2.2., and as in our abeceary there are only three letters, let’s raise the base 3 to the third power (or cubed): 33 = 3 × 3 × 3 = 27, which is the number of words that would exist –twenty-seven words:
By abecedaric order:
aaa, aab, aac, aba, abb, abc, aca, acb, acc,
(9 begin with an “a”)
baa, bab, bac, bba, bbb, bbc, bca, bcb, bcc,
(9 start with a “b”)
caa, cab, cac, cba, cbb, cbc, cca, ccb, ccc
(9 begin with a “c”).
Another way to arrange the same 27 words:
abc, acb, bac, bca, cab, cba, (with no repeated letters)
one letter repeated:
aab, aac, aba, aca, baa, caa, (words with 2 a’s)
abb, bab, bba, bbc, bcb, cbb, (words with 2 b’s)
acc, cac, cca, bcc, cbc, ccb, (words with 2 c’s)
each letter appears three times:
aaa, bbb, ccc (3 a’s, 3 b’s, 3 c’s).
5.3.2. Now, let’s introduce a restriction: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our abeceary (3) to a power equal to the number of letters that each and every word could have (3 in this example), under the restriction we apply the formula:
B(B-1)(B-2) = (B2-B)(B-2) = B3-B2-2B2+2B = B3-3B2+2B.
This formula is the same as the one that appears in the second paragraph of item 4.3.2.
Then, as in this case our abeceary has only three letters (a, b, c), we apply the formula B3-3B2+2B, and substitute: 33-(3)(32)+2(3) = 27-27+6 = 6 (six), which is the number of words we obtain, according to our restriction: –six words: abc, acb, bac, bca, cab, cba.
5.4. Imagine that we could form four-letter words each only.
5.4.1. In accordance with sections 2.1. and 2.2., and as in our abeceary there are only three letters, we raise the base 3 to the fourth power: 34 = 3 × 3 × 3 × 3 = 81, which is the number of words that would exist:
(27 begin with an “a”):
aaaa, aaab, aaac,
aaba, aabb, aabc, (9 begin with “aa”)
aaca, aacb, aacc,
abaa, abab, abac,
abba, abbb, abbc, (9 begin with “ab”)
abca, abcb, abcc,
acaa, acab, acac,
acba, acbb, acbc, (9 begin with “ac”)
acca, accb, accc.
(27 start with a “b”):
baaa, baab, baac,
baba, babb, babc, (9 start with “ba”)
baca, bacb, bacc,
bbaa, bbab, bbac,
bbba, bbbb, bbbc, (9 start with “bb”)
bbca, bbcb, bbcc,
bcaa, bcab, bcac,
bcba, bcbb, bcbc, (9 start with “bc”).
bcca, bccb, bccc.
(27 begin with a “c”)
caaa, caab, caac,
caba, cabb, cabc, (9 begin with “ca”)
caca, cacb, cacc,
cbaa, cbab, cbac,
cbba, cbbb, cbbc, (9 begin with “cb”)
cbca, cbcb, cbcc,
ccaa, ccab, ccac,
ccba, ccbb, ccbc, (9 begin with “cc”)
ccca, cccb, cccc,
5.4.2. Now, let’s introduce a restriction: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our abeceary (3) to a power equal to the number of letters that each and every word could have (4 in this example), under the restriction we apply the formula:
B(B-1)(B-2)(B-3) =
(B2-B)(B-2)(B-3) =
(B3-B2-2B2+2B)(B-3) =
(B3-3B2+2B)(B-3) =
B4-3B3+2B2-3B3+9B2-6B =
B4-6B3+11B2-6B.
This formula is the same as the one that appears in the second paragraph of item 4.4.2.
We apply the formula B4-6B3+11B2-6B, and substitute:
34-(6)(33)+(11)(32)-(6)(3) = 81-(6)(27)+(11)(9)-18 = 81-162+99-18 = 0 (zero), which is the number of words we obtain, according to our restriction: no word, in view that if our abeceary has only three letters (a, b, c), and we want to form four-letter words each, words in which no letter is repeated, that is impossible.
5.5. Now, imagine that we could form one-letter, and two-letter words only, each.
There would be twelve words: a, b, c, aa, ab, ac, ba, bb, bc, ca, cb, cc.
(31 + 32 = 3 + 9 = 12).
5.6. Now imagine that we could form one-letter, two-letter, and three-letter words each, only.
There would be 39 words:
a, b, c,
aa, ab, ac, ba, bb, bc, ca, cb, cc,
aaa, aab, aac, aba, abb, abc, aca, acb, acc,
baa, bab, bac, bba, bbb, bbc, bca, bcb, bcc,
caa, cab, cac, cba, cbb, cbc, cca, ccb, ccc.
(31 + 32 + 33 = 39).
5.7. Now imagine that we could form one-letter, two-letter, three-letter, and four-letter words each, only.
There would be 120 words:
a, b, c,
aa, ab, ac, ba, bb, bc, ca, cb, cc,
aaa, aab, aac, aba, abb, abc, aca, acb, acc,
baa, bab, bac, bba, bbb, bbc, bca, bcb, bcc,
caa, cab, cac, cba, cbb, cbc, cca, ccb, ccc,
aaaa, aaab, aaac,
aaba, aabb, aabc,
aaca, aacb, aacc,
abaa, abab, abac,
abba, abbb, abbc,
abca, abcb, abcc,
acaa, acab, acac,
acba, acbb, acbc,
acca, accb, accc.
baaa, baab, baac,
baba, babb, babc,
baca, bacb, bacc,
bbaa, bbab, bbac,
bbba, bbbb, bbbc,
bbca, bbcb, bbcc,
bcaa, bcab, bcac,
bcba, bcbb, bcbc,
bcca, bccb, bccc,
caaa, caab, caac,
caba, cabb, cabc,
caca, cacb, cacc,
cbaa, cbab, cbac,
cbba, cbbb, cbbc,
cbca, cbcb, cbcc,
ccaa, ccab, ccac,
ccba, ccbb, ccbc,
ccca, cccb, cccc.
(31 + 32 + 33 + 34 = 3 + 9 + 27 + 81 = 120).
5.8. And so on.
6. Now, imagine an abecedary with four letters: a, b, c, d.
6.1. Imagine that could form one-letter words each, only.
6.1.1. According to sections 2.1. and 2.2. and as in our abecedary there are only four letters, we raise the base 4 to the first power 41 = 4, which is the number of words that would exist –four words: a, b, c, d.
6.1.2. Now, let’s introduce a restriction: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our abecedary (4) to a power equal to the number of letters that each and every word could have (1 in this example), under the restriction we apply the formula: B.
Then, as in this case our abecedary has only four letters (a, b, c, d), we apply the formula B, and substitute: B = 4 (four), which is the number of words we obtain, according to our restriction: a, b, c, d.
6.2. Imagine that we could form two-letter words each, only.
6.2.1. According to sections 2.1. and 2.2., and as in our abecedary there are only four letters (a, b, c, d), we raise the base 4 to the second power (or squared): 42 = 16, which is the number of words that would exist: aa, ab, ac, ad, ba, bb, bc, bd, ca, cb, cc, cd, da, db, dc, dd.
6.2.2. Now, let’s introduce a restriction: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our abecedary (4) to a power equal to the number of letters that each and every word could have (2 in this example), under the restriction we apply the formula: B2-B.
Then as in this case our abecedary has only four letters (a, b, c, d) we apply the formula, B2-B, and substitute: 42-4 = 16-4 = 12 (twelve), which is the number of words we obtain, according to our restriction: ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc.
6.3. Now, let’s imagine that we could form three-letter words each, only.
6.3.1. In accordance with sections 2.1. and 2.2., and as in our abecedary there are four letters, we raise the base 4 to the third power (or cubed): 4.3 = 4 × 4 × 4 = 64, which is the number of words that would exist:
By abecedaric order:
aaa, aab, aac, aad, aba, abb, abc, abd, aca, acb, acc, acd, ada, adb, adc, add.
(16 begin with an “a”)
baa, bab, bac, bad, bba, bbb, bbc, bbd, bca, bcb, bcc, bcd, bda, bdb, bdc, bdd.
(16 begin with a “b”)
caa, cab, cac, cad, cba, cbb, cbc, cbd, cca, ccb, ccc, ccd, cda, cdb, cdc, cdd.
(16 begin with a “c”).
daa, dab, dac, dad, dba, dbb, dbc, dbd, dca, dcb, dcc, dcd, dda, ddb, ddc, ddd.
(16 begin with a “d”).
6.3.2. Now, introduce a restriction: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our abecedary (4 in this case) to a power equal to the number of letters that each and every word could have (3 in this example), under the restriction we apply the formula: B3-3B2+2B.
Then as in this case our abecedary has only four letters (a, b, c, d) we apply the formula B3-3B2+2B, and substitute: 43-(3)(42)+(2)(4) = 64-(3)(16)+8 = 64-48+8 = 24 (twenty-four), which is the number of words we obtain, according to our restriction:
abc, abd, acb, acd, adb, adc.
(6 begin with an “a”)
bac, bad, bca, bcd, bda, bdc.
(6 begin with a “b”)
cab, cad, cba, cbd, cda, cdb.
(6 begin with a “c”).
dab, dac, dba, dbc, dca, dcb.
(6 begin with “d”).
6.4. Now imagine that could form four-letter words each, only.
6.4.1. In accordance with sections 2.1. and 2.2., and as in our abecedary there are four letters (a, b, c, d), we raise the base 4 to the fourth power 44 = 4 × 4 × 4 × 4 = 256, which is the number of words that would exist:
aaaa baaa caaa daaa
aaab baab caab daab
aaac baac caac daac
aaad baad caad daad
aaba baba caba daba
aabb babb cabb dabb
aabc babc cabc dabc
aabd babd cabd dabd
aaca baca caca daca
aacb bacb cacb dacb
aacc bacc cacc dacc
aacd bacd cacd dacd
aada bada cada dada
aadb badb cadb dadb
aadc badc cadc dadc
aadd badd cadd dadd
abaa bbaa cbaa dbaa
abab bbab cbab dbab
abac bbac cbac dbac
abad bbad cbad dbad
abba bbba cbba dbba
abbb bbbb cbbb dbbb
abbc bbbc cbbc dbbc
abbd bbbd cbbd dbbd
abca bbca cbca dbca
abcb bbcb cbcb dbcb
abcc bbcc cbcc dbcc
abcd bbcd cbcd dbcd
abda bbda cbda dbda
abdb bbdb cbdb dbdb
abdc bbdc cbdc dbdc
abdd bbdd cbdd dbdd
acaa bcaa ccaa dcaa
acab bcab ccab dcab
acac bcac ccac dcac
acad bcad ccad dcad
acba bcba ccba dcba
acbb bcbb ccbb dcbb
acbc bcbc ccbc dcbc
acbd bcbd ccbd dcbd
acca bcca ccca dcca
accb bccb cccb dccb
accc bccc cccc dccc
accd bccd cccd dccd
acda bcda ccda dcda
acdb bcdb ccdb dcdb
acdc bcdc ccdc dcdc
acdd bcdd ccdd dcdd
adaa bdaa cdaa ddaa
adab bdab cdab ddab
adac bdac cdac ddac
adad bdad cdad ddad
adba bdba cdba ddba
adbb bdbb cdbb ddbb
adbc bdbc cdbc ddbc
adbd bdbd cdbd ddbd
adca bdca cdca ddca
adcb bdcb cdcb ddcb
adcc bdcc cdcc ddcc
adcd bdcd cdcd ddcd
adda bdda cdda ddda
addb bddb cddb dddb
addc bddc cddc dddc
addd bddd cddd dddd.
6.4.2. Now, let’s introduce a restriction: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our abecedary (4 in this case) to a power equal to the number of letters that each and every word could have (4 in this example), under the restriction we apply the formula: B4-6B3+11B2-6B, and substitute: 44-(6)(43)+(11)(42)-(6)(4) = 256-(6)(64)+(11)(16)-(6)(4) = 256-384+176-24 = 24 (twenty-four), which is the number of words we obtain according to hour restriction:
abcd, abdc, acbd, acdb, adbc, adcb,
bacd, badc, bcad, bcda, bdac, bdca,
cabd, cadb, cbad, cbda, cdab, cdba,
dabc, dacb, dbac, dbca, dcab, dcba.
6.5. Now, let’s imagine that we may form five-letter words each, only.
6.5.1. In accordance with sections 2.1. and 2.2., and as in our abecedary there are four letters (a, b, c, d), we raise the base 4 to the fifth power: 45 = 4 × 4 × 4 × 4 × 4 = 1,024, which is the number of words that exist:
aaaaa, aaaab, aaaac, aaaad...
… and 1,020 more words… in addition to the four written above as “samples”…
6.5.2. Now, introduce a restriction: that in each and every of the resulting words, we may not repeat any letter.
Instead of raising the number of letters in our abecedary (4) to a power equal to the number of letters that each and every word could have (5 in this example), under the restriction we apply the formula: B5-10B4+35B3-50B2+24B.
Then, as in our abecedary there are four letters (a, b, c, d), we apply the formula:
B5-10B4+35B3-50B2+24B,
and we substitute:
45-(10)(44)+(35)(43)-(50)(42)+(24)(4) =
1024-(10)(256)+(35)(64)-(50)(16)+(24)(4) =
1024-2560+2240-800+96 = 0 (zero), which is the number of words we obtain, according to our restriction: no word, in view that if our abecedary has only four letters (a, b, c, d), and we want to form five-letter words each, words in which no letter is repeated, that is impossible.
6.5.3. And so on.
6.5.4. Next, imagine that could form one-letter, and two-letter words each, only.
There would be 20 (twenty) words: a, b, c, d, aa, ab, ac, ad, ba, bb, bc, bd, ca, cb, cc, cd, da, db, dc, dd (41 + 42 = 4 + 16 = 20).
6.5.5. Then, imagine that we could form one-letter, two-letter, and three-letter words each, only.
There would be 84 words:
a, b, c, d,
aa, ab, ac, ad, ba, bb, bc, bd, ca, cb, cc, cd, da, db, dc, dd,
aaa, aab, aac, aad, aba, abb, abc, abd, aca, acb, acc, acd, ada, adb, adc, add,
baa, bab, bac, bad, bba, bbb, bbc, bbd, bca, bcb, bcc, bcd, bda, bdb, bdc, bdd,
caa, cab, cac, cad, cba, cbb, cbc, cbd, cca, ccb, ccc, ccd, cda, cdb, cdc, cdd,
daa, dab, dac, dad, dba, dbb, dbc, dbd, dca, dcb, dcc, dcd, dda, ddb, ddc, ddd.
(41 + 42 + 43 = 4 + 16 + 64 = 84).
6.5.6. And so on.
6.6. In order to obtain very many combinations of letters, you could use the formula 26 ^ 26, id est 2626, that is twenty-six raised to the twenty-sixth power, a mathematical operation that would result in a “frightening”, figure balúmbica. Twenty-six is the number of letters in the international Latin order… also, of the English abecedary, et cetera.
a b c d e f g h I j k l m n o p q r s t u v w x y z.
[In a blog “of mine”, algodatos.blogspot.com in some posts of June, 2013, you will find many groups of letters similar to those of item 6.4.1. of this algebraic-grammatical writing, but those groups are much larger, with the inconvenience for you English-speaking readers, those groups of letters include the Spanish letter ñ (sounds always like ny in canyon).
[In some May 2013 posts, there are 27 posts with 729 three-letter groups each, which makes a total of 19,683 three-letter groups.]
Now imagine formulas similar to those written at the end of item 2.2.1.2., Starting with: B26-(an “X” number)B25+(an “X” number)B24... and rather very lengthy formulas that would be “horrific”.
Furthermore, the Hawaiian alphabet/abecedary has only 12 letters: a, e, h, i, k, l, m, n, o, p, u, w.
Here, by applying 12 ^ 12 = 1212 = 8,916,100,448,256, or eight trillion nine hundred and sixteen billion one hundred million four hundred and forty-eight thousand two hundred fifty-six possible words (or groups of letters), a figure also somewhat “scary”, but not as high as the ones obtained when performing operations outlined in previous paragraphs.
7. Summary.
7.1. Table 1.
Table 1.
NL
|
1L
|
1Lw/o
|
2L
|
2Lw/o
|
3L
|
3Lw/o
|
4L
|
4Lw/o
|
5L
|
5Lw/o
|
6L
|
6Lw/o
|
1
|
1
|
1
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
2
|
2
|
2
|
4
|
2
|
8
|
0
|
16
|
0
|
32
|
0
|
64
|
0
|
3
|
3
|
3
|
9
|
6
|
27
|
6
|
81
|
0
|
243
|
0
|
729
|
0
|
4
|
4
|
4
|
16
|
12
|
64
|
24
|
256
|
24
|
1024
|
0
|
4096
|
0
|
5
|
5
|
5
|
25
|
20
|
125
|
60
|
625
|
120
|
3125
|
120
|
15625
|
0
|
6
|
6
|
6
|
36
|
30
|
216
|
120
|
1296
|
360
|
7776
|
720
|
46656
|
720
|
B
|
I
|
B^2
|
II
|
B^3
|
III
|
B^4
|
IV
|
B^5
|
V
|
B^6
|
VI
|
Note: B ^ 2 means “B squared”; B ^ 3, “B cubed”; B ^ 4, “B raised to the fourth power”, et cetera.
Roman numerals indicate in Table 1, which formulas listed below, are applied in the respective columns of Table 1:
I) B.
II) B2-B.
III) B3-3B2+2B.
IV) B4-6B3+11B2-6B.
V) B5-10B4+35B3-50B2+24B.
VI) B6-15B5+85B4-225B3+274B2-120B.
Meanings of the acronyms or abbreviations that appear in the first row of Table 1:
NL Number of letters in each imaginary abecedary.
1L 1-letter words.
1Lw/o 1-letter words without repeating any letters.
2L 2-letter words.
2Lw/o 2-letter words without repeating any letters.
3L 3-letter words.
3Lw/o 3-letter words without repeating any letters.
4L 4-letter words.
4Lw/o 4-letter words without repeating any letters.
5L 5-letter words.
5Lw/o 5-letter words without repeating any letters.
6L 6-letter words.
6Lw/o 6-letter words without repeating any letters.
██████
Addendum, Saturday 21st day of June, 2014:
II) B2-B.
III) B3-3B2+2B.
IV) B4-6B3+11B2-6B.
V) B5-10B4+35B3-50B2+24B.
VI) B6-15B5+85B4-225B3+274B2-120B.
VII) B7-21B6+175B5-735B4+1624B3-1764B2
██████████ ██████████
End of the addendum.
██████
7.2. Now, let's examine some points in Table 1:
7.2.1. First, a curiosity: we can see two parallel “quasi-diagonal broken” lines, from the upper left to the lower right corner of Table 1, marked or constituted by the respective factorials of the following numbers: 1, 2, 3, 4, 5, and 6.
Let's see what is a factorial:
The Merriam-Webster Dictionary defines the factorial: “the product of all the positive integers from 1 to n.” The symbol of the factorial is: ! (the same as an exclamation mark).
So, as an example, the factorial of 4 is 4! = 1 × 2 × 3 × 4 = 24.
Factorial of 1: 1! = 1 × 1 = 1.
Factorial of 2: 2! = 1 × 2 = 2.
Factorial of 3: 3! = 1 × 2 × 3 = 6.
Factorial of 4: 4! = 1 × 2 × 3 × 4 = 24.
Factorial of 5: 5! = 1 × 2 × 3 × 4 × 5 = 120.
Factorial of 6: 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720.
So, in the Table 1, we can see these two parallel quasi-diagonal broken lines, each formed by the numbers 1, 2, 6, 24, 120, and 720.
These factorials are always in columns whose titles/heads end with a lower case “w”, id est, in columns that do not allow the repetition of letters.
7.2.2. Second, something obvious: in the even-number columns (second, fourth, sixth, eighth, tenth and twelfth columns) of Table 1, the exponents or powers (defined as “p” in 2.2. of this text) increase their values 1 by 1 power from left to right: B (or: B ^ 1), B ^ 2, B ^ 3, B ^ 4, B ^ 5, B ^ 6.
7.2.3. Third, and most importantly: it is observed that there is a pattern in each row, from left to right, which justifies, and simultaneously follows the “pre-formula”* B(B-1 ), that is, “B” multiplied by “B-minus-one”.
* Apparently, if step by step each element of the pattern has given rise to a “pre-formula”, then it has been utilized the inductive-synthetic method (from particulars to generals-from parts to wholes); a theoretical formula has been created, it is justified.
If the “pre-formula” is applied in a practical way and it can be applied in equal cases (or very similar cases), always giving equal results (or very similar results, according to the respective instance), it has been used the deductive-analytical method (from generals to particulars-from wholes to parts): the formula is followed by us.
The first “pre-formulas” are simple, and give rise to more complex “pre-formulas” and formulas, such as those developed in item 2.2.1.2.
The explanation is as follows:
If you look at the third column (a column whose title ends with a lower case “w”, which means it does not support repetition of letters, and at the bottom of such column there is a Roman numeral I, which corresponds to the formula B, the simplest), and you represent the value of each number with a letter B, this “B” is the first “pre-formula” or formula, and it is the simplest of all, as noted in the second paragraph of 2.2.1.1. (The symbol B means “base”, as indicated in 2.1.) - You will be able to see that if you apply the “pre-formula” B(B-1), you get certain results, which were written in the fifth column of each respective row, in Table 1.
Let’s verify
–The “pre-formula” B(B-1) is equal to the formula B2-B.
First row: 1(1-1) = (1)(0) = 0.
Or: 1(1-1) = 1-1 = 0.
Second row: 2(2-1) = (2)(1) = 2.
Or: 2(2-1) = 4-2 = 2.
Third row: 3(3-1) = (3)(2) = 6.
Or: 3(3-1) = 9-3 = 6.
Fourth row: 4(4-1) = (4)(3) = 12.
Or: 4(4-1) = 16-4 = 12.
Fifth row: 5(5-1) = (5)(4) = 20.
Or: 5(5-1) = 25-5 = 20.
Sixth row: 6(6-1) = (6)(5) = 30.
Or: 6(6-1) = 36-6 = 30.
So far, we have the “pre-formula” B(B-1), which gives rise to the formula B2-B.
7.2.4. If we consider our results –product of algebraic operations performed in item 7.2.3. when following the “pre-formula” B(B-1)–, values that have also been written in the fifth column of Table 1, as a result of an induced data capture (inductive-synthetic method), and we take those data or values individually, and then multiply each one by (B-2), we obtain some results, which were written in the seventh column of Table 1:
First row: 0(B-2) = 0(1-2) = 0(-1) = 0.
Second row: 2 (B-2) = 2 (2-2) = 2 (0) = 0.
Third row: 6 (B-2) = 6 (3-2) = 6 (1) = 6.
Fourth row: 12 (B-2) = 12 (4-2) = 12 (2) = 24.
Fifth row: 20 (B-2) = 20 (5-2) = 20 (3) = 60.
Sixth row: 30 (B-2) = 30 (6-2) = 30 (4) = 120.
Remember that the six values of the far left in the above operations, id est, 0, 2, 6, 12, 20, and 30, are the product of applying the “pre-formula” B(B-1) in item 7.2.3. and are written in the fifth column of Table 1, and in turn the results of the algebraic operations performed in 7.2.4. (id est, in this item) of the extreme right, id est, 0, 0, 6, 24, 60 and 120, are the result of applying the “pre-formula” B(B-1)(B-2) and have been written in the seventh column of Table 1.
This “pre-formula” B(B-1)(B-2) has been shown in item 2.2.1.1 (please see such item in this writing), and leads to the formula B3-3B2+2B), so that having used the inductive-synthetic method after an empirical or experimental exercise, we can state that the algebraic “pre-formulas” and formulas shown in item 2.2.1.1. have emerged from the pattern observed in Table 1 and mentioned in the opening lines of item 7.2.3., although Table 1 has been placed many paragraphs below item 2.2.1.1., because the intention has been to take the reader first by the empirical road.
So far, we have the “pre-formula” B(B-1)(B-2) which leads to the formula B3-3B2+2B.
7.2.5. Now, let’s consider our results of the algebraic operations in item 7.2.4., product of having followed the “pre-formula” B(B-1)(B-2) –or the formula B3-3B2+2B–, values that have also been written in the seventh column of Table 1, as a result of an induced data capture (inductive-synthetic method), and take those data or values in the individual, and then multiply each by (B-3), to obtain certain results, which were written in the ninth column of Table 1:
First row: 0(B-3) = 0(1-3) = 0(-2) = 0.
Second row: 0(B-3) = 0(2-3) = 0(-1) = 0.
Third row: 6(B-3) = 6(3-3) = 6(0) = 0.
Fourth row: 24(B-3) = 24(4-3) = 24(1) = 24.
Fifth row: 60(B-3) = 60(5-3) = 60(2) = 120.
Sixth row: 120(B-3) = 120(6-3) = 120(3) = 360.
So far, we have the “pre-formula” B(B-1)(B-2)(B-3), which leads to the algebraic formula B4-6B3+11B2-6B.
7.2.6. Now, similarly to what has been performed in the immediately preceding paragraph, let’s consider our results of the algebraic operations in item 7.2.5., product of having followed the “pre-formula” B(B-1)(B-2)(B-3), values that have also been written in the ninth column of Table 1, a result of an induced data capture (inductive-synthetic method), and let’s take the data or values in the individual, and then multiply each by (B-4) to obtain certain results, which have also been written in the eleventh column of Table 1:
First row: 0(B-4) = 0(1-4) = 0(-3) = 0.
Second row: 0(B-4) = 0(2-4) = 0(-2) = 0.
Third row: 0(B-4) = 0(3-4) = 0(-1) = 0.
Fourth row: 24(B-4) = 24(4-4) = 24(0) = 0.
Fifth row: 120(B-4) = 120(5-4) = 120(1) = 120.
Sixth row: 360(B-4) = 360(6-4) = 360(2) = 720.
So far, we have the “pre-formula” B(B-1)(B-2)(B-3)(B-4), which leads to the algebraic formula B5-10B4+35B3-50B2+24B.
7.2.7. Next, and to come to an end in our task of illustrating by examples, a similar maneuver to what was done in the preceding item, let’s consider the results of our algebraic operations in item 7.2.6., product of having followed the “pre-formula” B(B-1)(B-2)(B-3)(B-4) –or the formula B5-10B4+35B3-50B2+24B–, values that have also been written in the eleventh column of Table 1 as a result of an induced data capture (inductive-synthetic method), and take the data or values in the individual, and then multiply each by (B-5) to obtain certain results, which were written in the thirteenth column of Table 1:
First row: 0(B-5) = 0(1-5) = 0(-4) = 0.
Second row: 0(B-5) = 0(2-5) = 0(-3) = 0.
Third row: 0(B-5) = 0(3-5) = 0(-2) = 0.
Fourth row: 0(B-5) = 0(4-5) = 0(-1) = 0.
Fifth row: 120(B-5) = 120(5-5) = 120(0) = 0.
Sixth row: 720(B-5) = 720(6-5) = 720(1) = 720.
So far, we have the “pre-formula” B(B-1)(B-2)(B-3)(B-4)(B-5), which leads to a long formula whose highest exponential value is 6, and that will not be developed in this text.
The odd column titles, from the third column to the right, in Table 1, end with a lower case “w” (“without repetition”), which means that the resulting words, formed from corresponding formulas, do not support any letter repetition, whereby the formulas for the odd columns are much more complex than those of the even columns.
7.3. Table 1, a copy.
Below, Table 1 has been copied, in order to facilitate any possible comparison with Table 2 if the “original” Table 1 has been printed faraway (above) from Table 2, in the text:
Table 1, a copy
NL
|
1L
|
1Lw/o
|
2L
|
2Lw/o
|
3L
|
3Lw/o
|
4L
|
4Lw/o
|
5L
|
5Lw/o
|
6L
|
6Lw/o
|
1
|
1
|
1
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
2
|
2
|
2
|
4
|
2
|
8
|
0
|
16
|
0
|
32
|
0
|
64
|
0
|
3
|
3
|
3
|
9
|
6
|
27
|
6
|
81
|
0
|
243
|
0
|
729
|
0
|
4
|
4
|
4
|
16
|
12
|
64
|
24
|
256
|
24
|
1024
|
0
|
4096
|
0
|
5
|
5
|
5
|
25
|
20
|
125
|
60
|
625
|
120
|
3125
|
120
|
15625
|
0
|
6
|
6
|
6
|
36
|
30
|
216
|
120
|
1296
|
360
|
7776
|
720
|
46656
|
720
|
B
|
I
|
B^2
|
II
|
B^3
|
III
|
B^4
|
IV
|
B^5
|
V
|
B^6
|
VI
|
Note: B ^ 2 means “B squared”; B ^ 3, “B cubed”; B ^ 4, “B raised to the fourth power”, et cetera.
Roman numerals indicate in Table 1 which formulas listed below, are applied in the respective columns of Table 1:
I) B.
II) B2-B.
III) B3-3B2+2B.
IV) B4-6B3+11B2-6B.
V) B5-10B4+35B3-50B2+24B.
VI) B6-15B5+85B4-225B3+274B2-120B.
Meanings of the acronyms or abbreviations that appear in the first row of Table 1:
NL Number of letters in each imaginary abecedary.
1L 1-letter words.
1Lw/o 1-letter words without repeating any letters.
2L 2-letter words.
2Lw/o 2-letter words without repeating any letters.
3L 3-letter words.
3Lw/o 3-letter words without repeating any letters.
4L 4-letter words.
4Lw/o 4-letter words without repeating any letters.
5L 5-letter words.
5Lw/o 5-letter words without repeating any letters.
6L 6-letter words.
6Lw/o 6-letter words without repeating any letters.
7.4. Table 2.
Next, it is shown Table 2, very similar to Table 1, except that in the fourth, sixth, eighth, tenth and twelfth columns, the numbers have been expressed with a base (B) raised to the appropriate power.
Tabla 2.
NL
|
1L
|
1Lw/o
|
2L
|
2Lw/o
|
3L
|
3Lw/o
|
4L
|
4Lw/o
|
5L
|
5Lw/o
|
6L
|
6Lw/o
|
1
|
1
|
1
|
1^2
|
0
|
1^3
|
0
|
1^4
|
0
|
1^5
|
0
|
1^6
|
0
|
2
|
2
|
2
|
2^2
|
2
|
2^3
|
0
|
2^4
|
0
|
2^5
|
0
|
2^6
|
0
|
3
|
3
|
3
|
3^2
|
6
|
3^3
|
6
|
3^4
|
0
|
3^5
|
0
|
3^6
|
0
|
4
|
4
|
4
|
4^2
|
12
|
4^3
|
24
|
4^4
|
24
|
4^5
|
0
|
4^6
|
0
|
5
|
5
|
5
|
5^2
|
20
|
5^3
|
60
|
5^4
|
120
|
5^5
|
120
|
5^6
|
0
|
6
|
6
|
6
|
6^2
|
30
|
6^3
|
120
|
6^4
|
360
|
6^5
|
720
|
6^6
|
720
|
B
|
I
|
B^2
|
II
|
B^3
|
III
|
B^4
|
IV
|
B^5
|
V
|
B^6
|
VI
|
Note: B ^ 2 means “B squared”; B ^ 3, “B cubed”; B ^ 4, “B raised to the fourth power”, et cetera.
Roman numerals indicate in Table 2 which formulas listed below, are applied in the respective columns of Table 2:
I) B.
II) B2-B.
III) B3-3B2+2B.
IV) B4-6B3+11B2-6B.
V) B5-10B4+35B3-50B2+24B.
VI) B6-15B5+85B4-225B3+274B2-120B.
Meanings of the acronyms or abbreviations that appear in the first row of Table 2:
NL Number of letters in each imaginary abecedary.
1L 1-letter words.
1Lw/o 1-letter words without repeating any letters.
2L 2-letter words.
2Lw/o 2-letter words without repeating any letters.
3L 3-letter words.
3Lw/o 3-letter words without repeating any letters.
4L 4-letter words.
4Lw/o 4-letter words without repeating any letters.
5L 5-letter words.
5Lw/o 5-letter words without repeating any letters.
6L 6-letter words.
6Lw/o 6-letter words without repeating any letters.
8. Theories.
In accordance with what was exposed in items 2.1. and 2.2., the following theoretical statements are formulated:
8.1. Given an abecedary consisting of B number of letters, the maximum number of words of p letters that can be formed is obtained by applying the formula: Bp.
(The symbol B represents the number of letters in a given abecedary, while the symbol p represents the number of letters to be included in each word that will be formed.)
The author asks that this theory, although obvious to almost any teenager versed in algebra and grammar, be known as: Theory of abecedaries and maximum number of words.
However, this name is a proposal, an outline, barely. In the future someone may suggest, propose or even impose a better name.
In section 7.1., in Table 1 (there is a copy in section 7.3.), the number of words that can be formed by applying the above formula is, respectively, in the second, fourth, sixth, eighth, tenth, and twelfth columns, according to the number of letters in each abecedary, and the maximum number of letters each word will be able to include.
8.2. Given an abecedary consisting of B number of letters, the maximum number of words of p letters that can be formed without repeating any letter within each word, is obtained by applying the following formulas:
1) B, where p = 1.
2) B(B-1) → B2-B, where p = 2.
3) B(B-1)(B-2) → B3-3B2+2B, where p = 3.
4) B(B-1)(B-2)(B-3) → B4-6B3+11B2-6B, where p = 4.
5) B(B-1)(B-2)(B-3) (B-4) → B5-10B4+35B3-50B2+24B,
where p = 5.
6) B(B-1)(B-2)(B-3)(B-4)(B-5) →
where p = 5.
6) B(B-1)(B-2)(B-3)(B-4)(B-5) →
B6-15B5+85B4-225B3+274B2-120B, where p = 6.
7) B(B-1)(B-2)(B-3)(B-4)(B-5)(B-6) →
B7-21B6+175B5-735B4+1624B3-1764B2+7 20B, where p = 7,
7) B(B-1)(B-2)(B-3)(B-4)(B-5)(B-6) →
B7-21B6+175B5-735B4+1624B3-1764B2
et cetera.
(The symbol B represents the number of letters in a given abecedary, while the symbol p represents the number of letters to be included in each word that will be formed.)
The author asks that this theory be known as: Theory of abecedaries and maximum number of words, with the restriction of not repeating any letters in words.
However, this name is a proposal, an outline, barely. In the future someone may suggest, propose or even impose a better name.
In section 7.1., in Table 1 (there is a copy in item 7.3.), the number of words that can be formed by applying the above formula is, respectively, in the third, fifth, seventh, ninth, eleventh and thirteenth columns, according to the number of letters in each abecedary, and the maximum number of letters each word will be able to include, with their respective restraints.
9. Other possible methods.
Besides of using algebraic formulas of the types outlined above, id est:
1) B.
2) B2-B.
3) B3-3B2+2B.
4) B4-6B3+11B2-6B.
5) B5-10B4+35B3-50B2+24B.
+7 20B...
6) B6-15B5+85B4-225B3+274B2-120B.
7) B7-21B6+175B5-735B4+1624B3-1764B2
et cetera, several formulas, procedures or easier ways can be applied in order to find answers, either by vectors or Cartesian graphs (invented by the father of French skepticism, le père du scepticisme français, philosopher and mathematician René Descartes, 1596-1650) or natural or napierian logarithms* (whose base is the irrational number e,** id est 2.718281828459 ...) or decimal logarithms (whose base is number 10), or calculus, invented by such portentous scientists, German mathematician, logician, and philosopher Gottfried Wilhelm Leibniz (1646-1716) and English mathematician and physicist Sir Isaac Newton (1643-1727).
* It should be noted that the Italian mathematician and Franciscan friar Fra Luca Pacioli (1445-1517), a systematic analyzer of the accounting method called double-entry accounting system and precursor of probability calculus, utilized logarithmic approximations, a century before John Napier did the same (Scottish mathematician, 1550-1617, a Presbyterian and furious anti-Catholic), excellent definer of natural or napierian logarithms.
** This is represented by the symbol e, in honor of Leonhard Euler (1707-1783), a bright Swiss mathematician, whose last name begins with an “e”.
Let us consider slightly the procedure of the infinitesimal calculus in its branch called differential calculus.
(The philosopher, logician and mathematician Gottfried Leibniz German [1646-1716] began his works on calculus in 1674, and in 1677 he had a coherent system, but did not published it until 1684. According to Leibniz's notebooks, on the 11th day of November, 1675, a major event happened: for the first time, he used integral calculus to find the area under the curve of a function y = ƒ (x).
(The English physicist, inventor, and mathematician Sir Isaac Newton [1643-1727] also carried out important works about calculus and functions.)
y = ƒ (x)
y is in function of x
y is a function of x
y is the function
Arbitrarily, we shall replace ƒ (x) by the letter B and/or by certain formulas. This is allowed, because the y, the x, the B, any other letters, and some formulas, are used, in this case, simply as symbols of variables.
y = ƒ (x) = B.
B is the independent variable, while y is the dependent variable: it depends on the value we assign to B.
(1) y = ƒ(x) = B
1L
B y = B
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 14
15 15
16 16
17 17
18 18
19 19
20 20
(2) y = ƒ(x) = B^2
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 14
15 15
16 16
17 17
18 18
19 19
20 20
(2) y = ƒ(x) = B^2
2L
B y = B^2
1 12 4
3 9
4 16
5 25
6 36
7 49
8 64
9 81
10 100
11 121
12 144
13 169
14 196
15 225
16 256
17 289
18 324
19 361
20 400
(3) y = ƒ(x) = B^3
3L
B y = B^3
1 1
3 27
4 64
5 125
6 216
7 343
8 512
9 729
10 1000
11 1331
12 1728
13 2197
14 2744
15 3375
16 4096
17 4913
18 5832
19 6859
20 8000
(4) y = ƒ(x) = B^4
4L
B y = B^4
1 1
2 16
3 81
4 256
5 625
6 1296
7 2401
8 4096
9 6561
10 10000
11 14641
12 20736
13 28561
14 38416
15 50625
16 65536
17 83521
18 104976
19 130321
20 160000
5L
B y = B^5
1 1
2 32
3 243
4 1024
5 3125
6 7776
7 16807
8 32768
9 59049
10 100000
11 161051
12 248832
13 371293
14 537824
15 759375
16 1048576
17 1419857
18 1889568
19 2476099
20 3200000
6L
B y = B^6
1 1
2 64
3 729
4 4096
5 15625
6 46656
7 117649
8 262144
9 531441
10 1000000
11 1771561
12 2985984
13 4826809
14 7529536
15 11390625
16 16777216
17 24137569
18 34012224
19 47045881
20 64000000
7L
B y = B^7
1 1
2 128
3 2187
4 16384
5 78125
6 279936
7 823543
8 2097152
9 4782969
10 10000000
11 19487171
12 35831808
13 62748517
14 105413504
15 170859375
16 268435456
17 410338673
18 612220032
19 893871739
20 1280000000
3 81
4 256
5 625
6 1296
7 2401
8 4096
9 6561
10 10000
11 14641
12 20736
13 28561
14 38416
15 50625
16 65536
17 83521
18 104976
19 130321
20 160000
(5) y = ƒ(x) = B^5
5L
B y = B^5
1 1
3 243
4 1024
5 3125
6 7776
7 16807
8 32768
9 59049
10 100000
11 161051
12 248832
13 371293
14 537824
15 759375
16 1048576
17 1419857
18 1889568
19 2476099
20 3200000
(6) y = ƒ(x) = B^6
6L
B y = B^6
1 1
3 729
4 4096
5 15625
6 46656
7 117649
8 262144
9 531441
10 1000000
11 1771561
12 2985984
13 4826809
14 7529536
15 11390625
16 16777216
17 24137569
18 34012224
19 47045881
20 64000000
(7) y = ƒ(x) = B^7
7L
B y = B^7
1 1
3 2187
4 16384
5 78125
6 279936
7 823543
8 2097152
9 4782969
10 10000000
11 19487171
12 35831808
13 62748517
14 105413504
15 170859375
16 268435456
17 410338673
18 612220032
19 893871739
20 1280000000
█████████████
(1) y = ƒ(x) = B
1Lw/o
B y = B
1 1
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 14
15 15
16 16
17 17
18 18
19 19
20 20
(2) y = ƒ(x) = B^2-B
2Lw/o
B y = B^2-B
1 0
3 6
4 12
5 20
6 30
7 42
8 56
9 72
10 90
11 110
12 132
13 156
14 182
15 210
16 240
17 272
18 306
19 342
20 380
(3) y = ƒ(x) = B^3-3B^2+2B
3Lw/o
B y = B^3-3B^2+2B
1 0
3 6
4 24
5 60
6 120
7 210
8 336
9 504
10 720
11 990
12 1320
13 1716
14 2184
15 2730
16 3360
17 4080
18 4896
19 5814
20 6840
(4) y = ƒ(x) = B^4-6B^3+11B^2-6B
4Lw/o
B y = B^4-6B^3+11B^2-6B
1 0
3 0
4 24
5 120
6 360
7 840
8 1680
9 3024
10 5040
11 7920
12 11880
13 17160
14 24024
15 32760
16 43680
17 57120
18 73440
19 93024
20 116280
(5) y = ƒ(x) = B^5-10B^4+35B^3-50B^2+24B
5Lw/o
B y = B^5-10B^4+35B^3-50B^2+24B
1 0
3 0
4 0
5 120
6 720
7 2520
8 6720
9 15120
10 30240
11 55440
12 95040
13 154440
14 240240
15 360360
16 524160
17 742560
18 1028160
19 1395360
20 1860480
(6) y = ƒ(x) = B^6-15B^5+85B^4-225B^3+274B^2-120B
6Lw/o
B y = B^6-15B^5+85B^4-225B^3+274B^2-120B
1 0
3 0
4 0
5 0
6 720
7 5040
8 20160
9 60480
10 151200
11 332640
12 665280
13 1235520
14 2162160
15 3603600
16 5765760
17 8910720
18 13366080
19 19535040
20 27907200
(7) y = ƒ(x) = B^7-21B^6+175B^5-735B^4+1624B^3-1764B^2+720B
7Lw/o
B y = B^7-21B^6+175B^5-735B^4+1624B^3-1764B^2+720B
1 0
3 0
4 0
5 0
6 0
7 5040
8 40320
9 181440
10 604800
11 1663200
12 3991680
13 8648640
14 17297280
15 32432400
16 57657600
17 98017920
18 160392960
19 253955520
20 390700800
10. Four useless virtual devices.
10.1. A sphere of characters.
A computer file/program, not created yet, simulating a giant hollow sphere (virtual, of course), with many characters (letters, Greek letters, accented vowels, numbers, symbols, punctuation, signs, spaces, et cetera) printed on its inside; of each character, there are at least two copies; of the most used characters, there are more copies, about 50, for example, for the letter e, however, the x could have only four copies, and within the sphere a virtual mosquito flying at the speed of light can stop (and in fact stops) before (or at) numerous letters, numbers, symbols, one by one… (for a millisecond, a microsecond or a nanosecond each) and forms words, text, books, treatises, encyclopedias, et cetera.
It would have to be implemented an algorithm for the creation of words, one after another, and a repository which would keep the words, paragraphs, and pages created, and also, perhaps, an algorithm that prevents the creation of groups of characters such as xaçh^vyñwh*tl~añezdd8ff, that would not make sense in any language. Or maybe it would make sense in the so called machine language, a concept which is widely used by computer engineers and programmers.
The mosquito could write a word in Croatian, then another in German, one in Spanish, one fourth word in English language, and so on, thus, someone has to create one or more intra-algorithms to classify the words by language.
10.2. A virtual circumference of characters.
Imagine a circle, crossed by two diametrical straight lines, one horizontal, and a second vertical line, perpendicular to the first one.
This circle would have 3,600 (three thousand six hundred) degrees (it would be divided into 3,600 parts), and in each grade it would be a character (letter, number, symbol or sign).
For example, in the position of 0 degrees, at the right end of the horizontal line, it could be the value to “space”, in the grade 1, it could be the number zero; in grade 2, it could be the 1; in the grade 3, the number 2; in the grade 10, the number 9; in the grade 11, the lower case letter a; in grade 12, the lowercase letter b, and so on.
The position of 0 degrees (“space”) would be at the right end of the horizontal diametrical straight line; the grade 1 would be one division above, at the degree 1 position (it would have the "printable" value of 0 [zero]); the grade 2 (with the "printable" value of number 1 [one], would be two divisions above the position of 0 degrees, counterclockwise, et cetera.
Each character would be printed or “engraved” on the circumference or immediately outside it, at least twice, the characters that represent the sounds most used in the language (such as vowels e, a, i, o, u, and certain consonants) would have more representation (would be repeated more times in the circumference, and placed not together, but scattered).
A central hand with a virtual rotary axis in the center of the circle defined by the circumference, could move in any of two directions (clockwise, and counterclockwise), in search of the nearest convenient character.
The hand would stop for a millisecond, a microsecond or a nanosecond before each character “chosen” to be marked (selected), and a “virtual printer” would be “printing” (writing) on a virtual page or virtual repository, each character to form words, phrases, sentences, paragraphs, pages articles, treatises, books, and so on.
Similar to section 10.1., it would be needed a programmer or computer engineer, and a linguist or a philologist, to create a suitable program, algorithms, constraints, and so on.
10.3. The lock of characters.
A combination lock, similar to those used to lock the closing of some suitcases (called suitcase combination locks), which usually have three swivel wheels with ten divisions each, numbered 0-9; each wheel has a notch in a certain position. When you match three notches, the suitcase can be opened. The owner of the suitcase can reprogram the opener combination, to be the one he wants, for example, 842.
In our case, our virtual wheels would not have ten “notches” or divisions, but some 1,024 (one thousand twenty-four divisions); each division would have a printed character, and three wheels would not be enough, so our virtual combination lock would have something like 35 gyrating wheels.
Each character (letter, number, symbol, sign, space, formula, et cetera) would be engraved or printed at least twice on each wheel, and the most used characters would be represented (printed or recorded) more times.
No wheel would have printed or engraved characters in the same order. There would be 35 ways to accommodate the characters, one way at each wheel.
Every millisecond, microsecond or nanosecond, the wheels would stop (would rotate at different speeds and in opposite directions alternately, or all swirling at the same speed, but in opposite directions alternately, and according to a special algorithm), and a “reader” or “position sensor” would record or ”would print” (write) in a virtual page or in a virtual space the words or groups of letters that would be forming.
As in previous cases, it is necessary to create one or more algorithms to run this virtual machine.
10.4. “Recombinant DNA language.”
The structure of deoxyribonucleic acid (DNA) was determined and defined in 1953 by English molecular biologist, biophysicist, and neuroscientist Francis Crick (1916-2004), American molecular biologist James D. Watson (1928 -), and British physicist and molecular biologist born in New Zealand, Maurice Wilkins (1916-2004).
Basic structure of DNA, deoxyribonucleic acid: a double helix with plenty of four nitrogenous bases molecules, linked by pairs:
adenine=thymine, A=T; and guanine≡ cytosine, G≡C.
[In RNA, ribonucleic acid, there is no thymine; uracil occupies the place of thymine: adenine=uracil, A=U; and guanine≡ cytosine, G≡C]
Some argue that the three winners of the Nobel Prize in Physiology or Medicine in 1962, Crick, Watson and Wilkins, had disparaged British biophysicist and X-ray crystallographer Rosalind Franklin (1920-1958), and also say that scientifically she deserved more recognition, since she was the one who took the first pictures of X-ray diffraction of the double helix structure of DNA, who first interpreted them, and who showed that the support structure, phosphate, should be on the outside, and nitrogenous bases on the inside.
Although, as the Nobel Prizes are not given posthumously, her relatives could not have received the Nobel Prize on her behalf in 1962.
Well, by now enough of the history of chemistry, biochemistry, biology, medicine, physiology, and so on.
In the fifth column of Table 1, shown in section 7.1. of this writing, it appears the formula II (“two Roman numeral”), which is: B2-B, and in the row corresponding to an abecedary of four letters to form words of two letters with the restriction set in item 6.2.2., in the sense that no letter can be repeated in each word formed, the result of applying the formula was: B2-B = 16-4 = 12, meaning that out of 16 possible combinations of two letters, four should be discarded, in order twelve remain.
The twelve combinations which meet the restriction are:
ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc.
The four combinations discarded (because a letter is repeated) are:
aa, bb, cc, dd.
But here in this section where we are right now, 10.4., we have not an abecedary of four letters, but an acegetary, consisting of four letters: a, c, g, t, which are the initials of four nitrogenous bases: adenine, cytosine, guanine, and thymine.
Also here the restriction is broader than in 6.2.2.: besides not repeating any letter (id est, a nitrogenous base may not bind to another equal to itself, so the combinations aa, cc, gg , tt are impossible ones), each nitrogenous base may bind to only one of the three other bases, to form pairs that are always produced between adenine and thymine: at (or thymine and adenine: ta) on one side; and cytosine and guanine : cg (or guanine and cytosine: gc) on the other.
(The author will repeat here paragraphs 2.1., and 2.2. of this writing:
2.1. In this algebraic-grammarian exercise, the number of letters in each imaginary abecedary that we will create will constitute a mathematical base (which is represented by the symbol B).
2.2. The number of letters contained in the words that we will be able to build, will constitute the exponent or power, represented by the symbol p, written as a superscript character, an indicator of the power that the bases cited in 2.1. will have to be raised to; that is, Bp.
We are faced here, in this case if item 10.4., that the value of base B is 4 (the number of letters in our abecedary, which in this case is an acegetary, which also fulfills the condition of 2.1.) And the power or exponent p is 2 as it will form two-letter words each.
Then, in view of a broader restriction in this case, we will apply a different “pre-formula”: B(B-3), resulting in a formula that also differs from all previous formulas: B2-3B.
The “pre-formula” complies with the condition 2.1. (In this case, B has a value of 4, which is the number of letters in our acegetary).
Furthermore, the “pre-formula” meets the broader restriction that each of the four letters (a, c, g, t) can bind to only one of the remaining three, which may be expressed algebraically as follows: B-3 (operation that results in 1, in this case, and that meets the broader restriction that each nitrogenous base may bind to only one of the other three, and besides that the nitrogenous base it can bind to, may not be a base equal to itself). And if the “pre-formula” meets these conditions, the formula also meets them.
Finally, the formula B2-3B, meets the condition specified in 2.2.: in this case the highest exponential (power) value of p in the formula is 2: B2-3B.
By applying the formula and substituting, we have: B2-3B = 16-12 = 4, which is the number of combinations that we have, and these are: at, ta, cg, gc.
The twelve combinations discarded are: aa, cc, gg, tt, ac, ag, ca, ct, ga, gt, tc, tg.
Now if here our acegetary has only four letters, our super-abecedary could have 256 or 512 or 1,024 characters.
That is, here the architecture or structure of combinations to represent letters would be different from the points 3, 4, 5, and 6 of this paper. Please note that at this item 10.4. we are dealing with one of four useless virtual devices, and there is more freedom to act.
To form chains of pairs of nitrogenous bases we can form random combinations structures almost ad infinitum (remember: the only valid ones are the following four: at, ta, cg, gc), and arbitrarily have been sectioned lengthy chains consisting of billions of pairs of nitrogenous bases, in order to form chains of eight pairs each.
Each of these chains formed by eight pairs or 16 nitrogenous bases represents one character belonging to our super-abecedary of, let’s say, 1,024 characters.
For example, the chain:
at
ta
at
ta
cg
gc
cg
gc
could represent the letter a.
The chain:
at
ta
at
cg
ta
gc
cg
gc
could represent the letter b.
The chain:
at
ta
cg
at
ta
gc
cg
gc
could represent the letter c, and so on.
After a computer program (including certain algorithms, conditions and restrictions) section off chains to form mini-chains of eight pairs, the latter can be arranged so that letters can form words lead to “real” or “reasonably possible to exist “ words (after the program have been trashed all of the mini-chains of eight pairs that hinder or are nonexistent in our range of 1,024 mini-chains of eight pairs of nitrogenous bases each) in any languages that use the Latin abecedary, a copy of the original very, very long (“billionaire”) chain can be considered.
After having formed a considerable number of actual words, let's see a copy of our original chain.
In this copy, the left column moves up, let’s say, “n” number of places, and the right column either remains stationary or moves down “n” or “o” number of places (according to algorithms) to form trillions of new pairs; the impossible pairs (the twelve pairs discarded in the above paragraphs) are trashed, and the possible, “surviving” pairs (at, ta, cg, gc) are grouped into sections of eight pairs each, to form new groups representing different letters and characters of our super-abecedary, which will be forming new words.
It is likely that after the realignment of sequences, the program (software) thrash away more than 75 percent of the new base pairs (for not complying with the broader restriction), so it might be possible to consider the introduction of an additional algorithm or an “under-algorithm” or “intra-algorithm” inside the main algorithm which form groups of eight pairs each, from valid pairs (at, ta, cg, gc) that are not necessarily contiguous, but several or many rows apart, but keeping the order they bring from the original sequence.
There may be millions (or billions?) of billions of possibilities, and the largest computers of today may not have the capacity to handle many rearrangements, as it happens with computers that are intended to simulate the multiple possible recombinations of the real DNA.
The TX5 Cray supercomputer price is about a million dollars, plus a further 50 thousand dollar to maintain it in a controlled, dust-free, low-temperature room, an appropriate place where to locate it. Experts claim that this is an inflated price, that it should cost half, but as Cray has little competition...Too much “box” or continent and less content (really useful hardware).
10.5. In the Wikipedia, you can read about the infinite monkey theorem:
Also, you may want to check something very interesting: the Turing machine:
The virtual Turing machine was created by outstanding British mathematician Alan Turing (1912-1954).
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The author, Alejandro Hector Ochoa-Gonzalez, a guadalajarense or tapatio by birth, inheritance, residency, conviction and convenience –with a Sonoran stamp– dedicates this writing, with much gratitude and appreciation to his teachers of preschool and elementary education at the Colegio Niebla, in Ciudad Obregon, State of Sonora —747* South Sinaloa Street: Cornelia Niebla, Dolores Niebla de García, Isabel Caloca de Velázquez, Teresa García Niebla, Yolanda Duarte Castro, Gloria Yépiz Coronado, Magda Acosta, Evangelina Camacho (English teacher), and the principal, professor Rosario Niebla Lugo, and to my Inductive Spanish teacher (the author of the textbook, Español Inductivo was: Maurilio Barriga Gaona) at high school in the Instituto La Salle, in Ciudad Obregon, Sonora (a school where basketball and baseball are widely practiced), professor Gregorio Patrón, as well to my math teachers at high school, same institute: engineer Bátiz and Lasallian brother Ignacio Navarro Castañeda (who used to smoke cigarettes of a brand named Fiesta [in English: Party]), and at college, same institute: Alfonso Rodríguez García de Alba (Lasallian brother and principal of the school, who told us, his disciples, that we had to bought the book College Algebra [translated into Spanish, of course], by Rees & Sparks, and also taught chemistry and sometimes English), engineer Bórquez, and Victoriano García Angüis (patron of my graduating class, the 10th Graduation Class, 1974-1977), who could write with his back toward the blackboard with either of both hands while verbally exposing the class, and my literature teacher at College, Jorge Herrera Chavarría. Some have died; for them, In memoriam.
* Unforgettable as for any former child your elementary school; in this case, also memorable because the house number matches the number of the “Jumbo” jet, the Boeing 747.
