Mathematical Miracle of God in Quran’s H.M. initialed Suras based on # 19

[2:23] If you have any doubt regarding what we revealed to our servant,* then produce one sura like these, and call upon your own witnesses against GOD, if you are truthful. *2:23-24 The Quran’s miraculous mathematical code provides numerous proofs as it spells out the name “Rashad Khalifa” as God’s servant mentioned here. Some literary giants, including Al-Mutanabby and Taha Hussein, have answered the literary challenge, but they had no awareness of the Quran’s mathematical composition. It is the Quran’s mathematical code, revealed through God’s Messenger of the Covenant, Rashad Khalifa, that is the real challenge ― for it can never be imitated. See Appendices 1, 2, 24, & 26 for the detailed proofs.

[11:1] A.L.R. This is a scripture whose verses have been perfected, then elucidated.* It comes from a Most Wise, Most Cognizant. *11:1 Our generation is fortunate to witness two awesome phenomena in the Quran: (1) an extraordinary mathematical code (Appendix 1), and (2) a literary miracle of incredible dimensions. If humans attempt to write a mathematically structured work, the numerical manipulations will adversely affect the literary quality. The Quran sets the standard for literary excellence.

The verse translations are from Quran: The Final Testament, Authorized English Version by Rashad Khalifa, Ph.D.. This is available for free down load at: http://www.masjidtucson.org/downloads/qe/index.html or http://www.submission.info/downloads/qe/index.html . Or for online viewing at: http://www.masjidtucson.org/quran/

To summarize, a grid based super-computer operating at petaflop speeds (quadrillion operations per second), would take trillions of years (many times longer than the universe’s age: http://en.wikipedia.org/wiki/Age_of_the_universe) to replicate the mathematical structure of one set of Quranic chapters, prefixed by the initials H and M. This does not factor in the need for literary excellence, nor the words and chapters that contain meaningful passages with religious enlightenment, to go along with the mathematical structure.

Chapters 40 – 46 of the Quran are prefixed with the initials H and M (in Arabic Hã and Mîm). Other initials prefix various chapters (suras) of the Quran (please see Appendix 1 of the Quran at http://www.masjidtucson.org/quran/appendices/appendix1verify.html for details). The role of the Quranic initials is summarized in Quran 10:1 (and other verses as well).

[10:1] A. L. R.* These (letters) are the proofs of this book of wisdom. *10:1 These letters constitute a major portion of the Quran’s awesome mathematical code and proof of divine authorship. See Appendix 1 for details.

In sura 11 verse 13 of the Quran, we see a mathematical challenge that proves that the Quran’s authorship is far beyond human capability. Please note also that at the time of revelation of the Quran (AD 610), and establishment of all its parameters, there were no computers. Yet this challenge is open to all generations.

The Quran: Impossible to Imitate

[11:13] If they say, “He fabricated (the Quran),” tell them, “Then produce ten suras like these, fabricated, and invite whomever you can, other than GOD, if you are truthful.”* *11:13 The Quran’s mathematical miracle is inimitable (See Appendix 1).

Let us consider just the contiguous block of chapters from 40-46 (seven chapters) having initials H and M as the first verse. In aggregate, the total frequency of these initials in the seven chapters is 2147 or 19×113. The counts of these initials in the seven chapters along with the challenge of coming up with the specific values of H and M that appear in each of the chapters is summarized in Table 1. The part of this Table in italics is a sample of the challenge from the Arabic Quran.

 

Table 1: The H.M. Counts

	Challenge		An illustration of the inimitability inthe Arabic Quran
Col	Col 1	Col 2	Col 3	Col 4	Col 5	Col 6	Col 7	Col 8
Row				Sura	Letter Count ”H”	LetterCount “M”	Digits addition of H+M	Total count of H+M
R-A	A1	A2	Digit Sum (A1, A2)	40	64	380	21	444
R-B	B1	B2	Digit Sum (B1, B2)	41	48	276	27	324
R-C	C1	C2	Digit Sum (C1, C2)	42	53	300	11	353
R-D	D1	D2	Digit Sum (D1, D2)	43	44	324	17	368
R-E	E1	E2	Digit Sum (E1, E2)	44	16	150	13	166
R-F	F1	F2	Digit Sum (F1, F2)	45	31	200	6	231
R-G	G1	G2	Digit Sum (G1,G2)	46	36	225	18	261
					292	1855	113	2147(19x113)

In this table, the columns are numbered C-1 through C-8, and the rows R-A through R-G.

The specific challenge is to come up with values for the table elements A1, A2, B1, B2, … G1, G2 (shaded yellow) so that the combination of existing numerical patterns is preserved. The existing counts of the initials in 40-46 are seen in columns 5-6, rows A-G (shaded green). Column C-7 is obtained by adding the digits of the H and M occurrence for that corresponding sura. So for sura 40, we see 64 Hs and 380 Ms, and (6 + 4) + (3 + 8 + 0) = 21. The corresponding challenge column is C-3.

(II) The Requirements for variables A1-G2:

1. The sum of values in column C-3 (which is derived by adding the digits of the corresponding solution in C-1 and C-2) for rows A,B,C =21+27+11=59. 21, 27 and 11 are the specific solutions found in the Quran, however the challenge simply requires the numbers to add to 59.

2. The sum of values in column C-3 for rows D,E,F,G = 54 (17+13+6+18).

3. The sum of the values in C-3 for rows B,C,D = 55 (27+11+17).

4. The sum of the values in C-3 for rows A,E,F,G = 58 ( 21+13+6+18).

(III) Conditions for Conforming with the 19-based Quranic structure:

1. The sum of the numbers in C8 for rows A,B,C are a multiple of 19. In the Arabic Quran, the sum comes to 1121=19x59 – compare with (1) above.

2. The sum of the numbers in C8 for rows D,E,F,G are also a multiple of 19. In the Arabic Quran, the sum comes to 1026=19x54 – compare with (2) above.

3. The sum of the numbers in C8 for rows B,C,D are a multiple of 19. In the Arabic Quran, the sum comes to 1045=19x55 – compare with (3) above.

4. The sum of the numbers in C8 for rows A,E,F,G are also a multiple of 19. In the Arabic Quran, the sum comes to 1102=19x58 – compare with (4) above.

 (IV) What is involved in solving this 19-linked challenge?

As any computer scientist will tell you, the following steps are involved in developing an algorithm to solve the challenge.

Since no single-digit combinations would satisfy the criteria listed in the “Requirements” or “Conditions” section, let us assume that the solution values involve 2 digit numbers and work on that basis (i.e., assuming values under 100). When it does not work out and no matching result found, we could switch to a 3-digit solution (limiting number: 1000).

Since the total count of the 7 H’s & 7 M’s (7+7=14) across the chapters is 2147, we can place an estimate that on average each cell value is (2147/14) = approximately 150.

What is the cost of such a search?
Suppose there we just 2 cells labeled A and B (instead of 14 cells). Since we don’t know exactly what number is in each cell, we’ll write our program to try all numbers from 1 – 150 in each cell. So it would start with the number 1 in A, and the number 1 in cell B. Since that won’t work, we’ll keep the 1 in A, and increment to a 2 in cell B. If that doesn’t work, then we’ll try a 3 in cell B. Until we’ve tried 150 numbers in cell B. Next we’ll change cell A to the number 2, and try 150 combinations in cell B. In total our program tries 150 combinations in A, and for each of those 150 numbers, we’ll try 150 different numbers in B. In total we’ll be trying 150 x 150 = 150² combinations. Similarly, if there were 3 cells, we’d be trying 150 x 150 x 150 = 150³ combinations.

When we have 14 cells, we will write 14 nested loops in our program. This gives us a total of 150¹⁴ numerical combinations (i.e., 150^14) = (2.9 x 10³⁰)[where ^ denotes to the power of, i.e., 150^14 means 150¹⁴].

Next we must add some factor for the cost of each combination. Each time we come up with a new combination in the 14 cells, we perform the 84 digit additions (see requirements 1-4 above) and 28 number additions (see conditions 1-4 above). This gives us (84+28) = 112. Assume that the cost of these operations is fixed at one unit of computation time, in programming terms: one floating point operation. On average let us say we only perform 25% of the operations because if any of the checks fail we will probably not try the next.
So, let us say in each iteration, or trial, the  cost of the algorithm is: 0.25 * 112 = 28.
For the total run of the program, the cost is: 28 * 150¹⁴   = 8.17 * 10³¹ operations [where * denotes multiplication].

How long will it take?
We have the number of operations from the previous paragraphs. Now we need to divide by number of operations a computer performs per second. And then by the seconds in a year.

Assume a computer can perform 1 trillion operations /second. To find out how long that computer would take to solve our problem, we divide the number of operations (8.17 x 10³¹) by 1 trillion
(10¹²). So we get:
= (8.17 * 10³¹)/ ( 10¹²)
= 8.17 * 10¹⁹

The above number is in seconds. To find how many years those number of seconds take, we further divide the result above by the number of seconds per year. The number of seconds in a year is:
(60 seconds * 60 min * 24 hour * 365.25 days) = 3.16 x 10⁷

= 8.17 * 10¹⁹ / 3.16 * 10⁷
= 2.59 * 10¹² years

We know however, that solutions going only up to 150 will not work. We also know that with grid computing, we now have speeds of petaflops instead of teraflops. So, a more accurate estimation of the cost would be as follows. Let us assume our 14 nested loop counters go from 1-500 instead of 1-150.

This means, our operation cost estimate is:
28 x 500¹⁴ = 1.71 * 10³⁹

Strictly speaking, the last iteration we will not go up to 500 for cell A1, instead only to 64. So that the time reduces slightly to:
28 * ( 64 * 500¹³)  = 2.19 * 10³⁸

Let us assume we have a super computer operating at 10 petaflops, i.e., 10 quadrillion operations per second (more on this in the next section). In other words, the computer performs 10¹⁶ operations per second.

This means, the time taken is:
(2.19 x 10³⁸ ) / (10¹⁶) = 2.19 x 10²² seconds

As before, let us divide by number of seconds in a year:
(2.19 x 10²²) / (3.16 x 10⁷) = 6.93 x 10¹⁴ years

(V) How will this change with improving computing speeds?