SR = {s 2 S
I1(s) + I2(s,R) 2 {0, 2}} (2.3)
1
3 Some results
Lemma 1 The following two properties hold whatever the ranking function R 2 R
(1) Xs2S
na(s) = 6236
(2) Xs2S
R(s) = 6555 8R 2 R
Property (1) in lemma 1 gives the total number of ayats in the Koran (6236 ayats) and property (2) gives
the sum of ranks of sourats in the Koran (6555). This sum is independent of the used ranking function.
The following theorem shows a very interesting and surprising property of Rm, the current ranking
function of the Koran also called the MASHAF ranking. More specifically:
Theorem 1 Let Rm be the MASHAF ranking function, SRm the set of all Rm-homogenous sourats and
ScR
m
its complementary. Then we have the following two properties:
card(SRm) = N
2
(3.4)
X {s2SRm}
Rm(s) = X {s2Sc
Rm}
na(s) (3.5)
where card(SRm) stands for the cardinal of SRm (i.e. the number of sourats in SRm) and N = 114 is
the total number of sourats in the Koran. Theorem 1 shows the following: The actual Koran’s ranking
function Rm splits the Koran into two equal parts SRm and ScR
m
. Furthermore, the sum of the ranks of
the sourats in the first part is exactely equal to the total number of ayats in the second part!
The question afterwards is whether these two properties are unique for the current ranking function
Rm. In order to answer this question we need to check these two properties for all the N! = 114! possible
functions R 2 R. This is definitely not possible. An estimation of the proportion of R 2 R satisfying
properties (3.4) and (3.5) in theorem 1 is therefore needed. A random sample of size n = 264000 was
drawn from the population R. The estimated proportion is 0.0003. This means that only 0.03% of ranking
functions satisfy the two properties in theorem 1. This result is estonishing and is only a begining to a
deeper one. Adding some other criterias will surely make Rm unique. Another option is to study the
relation between Rm and the 0.03% other functions and see whether they have something in common.
2