1. The synodic month, that is, the duration after which the moon reverts to the same position with respect to the sun is 29.53 days. The phases of the moon – and thus the Islamic calendar – are linked to the synodic month.
2. We note that the sidereal month, that is, the duration after which the moon reverts to the same position with respect to the celestial sphere (the background of stars) is 27.31 days. The sidereal month is shorter than the synodic month because the earth drags the moon along with it in its orbit around the sun. Thus the moon reverts earlier to its position with respect to the background of fixed stars. The sidereal month is of no importance for the Islamic calendar.
3. The synodic month is a fraction of the number of days. Consequently, the months alternate between 29 and 30 days. It is known that, over a cycle of 30 years of the Islamic calendar, 19 years have 354 days and 11 years have 355 days (years of abundance). The rhythm obeys the moon cycles. The celestial clock is thus fashioned, thanks be to God.
4. Several hours after conjunction – and according to the method of the extension of the visibility zone – the crescent will be visible is some countries and not in others. This situation varies each month.
5. However, since the months can have a maximum of 30 days only, if the crescent is not visible on day D, it will certainly be visible on the day D+1. We will never have to wait until the day D+2. This means that, each month, some countries will finish the month in 29 days, whereas for the others, the month will last 30 days. For a given country, the duration of the month will vary from one month to another, but the rhythm of 29 days/30 days will be valid for each and every country.
6. This means that at the end of each month, one can divide the countries into two groups: the group A of countries where the month has lasted 29 days and the group B of countries where the month has lasted 30 days. This has in fact been verified in all our calculations of the dates of important Islamic festivals. Either the festival is on day D, or it is next day. At the end of each month there can never be more than two groups of countries.
7. Depending on the month, a given country will either be in group A or in group B.
8. At the end of the first month of the year, there are only two groups possible, the group A and the group B.
9. At the end of the second month, the number of groups possible is 22 or 4: the groups AA, BB, AB and BA (the countries in which both months have had 29 days, those in which both months have had 30 days, those in which the first month has been of 29 days and the second 30 days, then the inverse).
10. Each month, the number of possibilities increases by the power of 2: 23, 24…
11. At the end of the year the number of possibilities is theoretically 212 or 4096.
12. The number of theoretical possibilities does not correspond to reality, since, from the 4th month onwards, the possibilities “AAAA” or “BBBB” are to be excluded (4 consecutive months of 29 or 30 days). The months must alternate harmoniously between 29 and 30 days. An year where the twelve months are all “A” or all “B” is impossible.
13. In fact, the intermediate months are of no importance to us. What we want to calculate is the number of possible calendars at the end of the year, that is, at the end of the 12 month cycle. Thus we can directly proceed to the 12th month.
14. At the end of 12 months, we have 4 096 lines of all the possible combinations of A and B: 12 times A, 12 times B, 11 times A and one B, and so on. Amongst all these theoretical combinations, what interests us are the lines where we find 6 times A and 6 times B, keeping in mind that there are only 12 events per line. These are the “normal” years of 354 days, enough for our calculation.
15. Amongst these 4 096 lines, what is the frequency of 6 times A (or 6 times B, this being automatic, because there are only 12 events per line)? The answer is given by combinatory analysis. The frequency of finding “r” times the letter A, in a series of “n” events, is determined by the formula: C(n,r) = n! / (r! * (n – r)!). In our case, n = 12 and r = 6. Thus we have, for the number of lines where we will find 6 times A (and thus also 6 times B) = C(12,6) = 12! / (6! * 6!) = (12 x 11 x 10 x 9 x 8 x 7) / (6 x 5 x 4 x 3 x 2). The result is 924. Thus, amongst all the possible combinations at the end of the year, there are 924 with 6 months of 29 days and 6 of 30 days, or 924 “normal” years.
16. But we can reduce this figure yet more. From time to time, we do have three consequent months of 30 or 29 days. Observing the calendars published on our site, we notice that in 1432 as well as in 1433, the months of Rabi’I, Rabi’II and Jumada I each had 30 days, consecutively. Amongst these two years, 1432 was an “abundant” year of 355 days, whereas 1433 was a “normal” year of 354 days. Whatever that may be, we can admit 3 consequent months of 29 or 30 days, but never 4 consequent months of 29 or 30 days. From a mathematical point of view, we have to eliminate from our 924 lines above all the occurrences of the sequences AAAA and BBBB. All the more, we have to eliminate the series “6 times A”, “6 times B”, “5 times A”, “5 times B”. Let us proceed step by step.
17. Amongst the total of 924 lines of 6 As and 6 Bs, there can only be 2 lines with a series of 6 times A and 6 times B. These are the two lines with AAAAAABBBBBB and BBBBBBAAAAAA. We can thus already eliminate these two lines from the total of 924.
18. Let us now calculate the frequency of “5 times A”. We will only calculate “5 times A”, since the calculation of “5 times B” will be identical. We can thus just multiply by 2 the result obtained for “5 times A”. We must examine all the possible manners in which a series of 5 As can occur in in a line with 6 As and 6 Bs. There are three possible cases:
(1) 1st case: The sequence of 5 As occurs at the head of the line: AAAAA. B must necessary follow the 5 As, since, otherwise we will have a sequence of 6 As, already eliminated in (17) above. A sequence of 5 As at the head of the line will thus be represented by the series AAAAABxxxxxx where the letter x represents 1 more A and 5 Bs. The letter A can be in any position in the series of x. The number of combinations of an A in a series of 6 positions is given by the formula C(6,1) = 6! / (1! * (6 – 1)!) = 6 x 5 x 4 x 3 x 2 / 5 x 4 x 3 x 2 = 6. There are thus 6 ways in which the letter A can occur in a series of 1 A and 5 Bs. Conclusion: there will be 6 lines with 5 As at the head of the line (followed by a B) and 1 more A in the series that follows of 6 more elements. These 6 lines have to be eliminated.
(2) 2nd case: The sequence AAAAA is at the end of the line: xxxxxxBAAAAA. B has necessarily to precede the sequence of 5 As. The sequence of 6 times x at the beginning of the line represents 1 A and 5 Bs. The case is identical to the preceding one. The one A that remains can be in 6 different positions before the B. Again, we will have 6 lines with 5 times A at the end of the line. These 6 lines have to be eliminated.
(3) 3rd case: It is the most complicated. The sequence of 5 times A can be anywhere in the series of 12 elements, necessarily preceded or followed by a B in order to avoid a sequence of 6 times A. The sequence can be written as: xBAAAAABxxxx. The first x can either be an A or a B, just as the four times x at the end of the line. The number of times an A can be placed in a series of 5 elements is C(5,1) = 5. But the case can repeat itself 5 times, since there can be 1, 2, 3, 4 or 5 times x at the head of the line. Thus there will be 5 x 5, or 25 lines, with a sequence of 5 times A, in whatever position in a series of 12 elements.
19. The total of the above three cases with a sequence of 5 times A is consequently: 6 + 6 + 25 = 37. The case of a sequence of 5 times B, either at the head of the line, or at the end of the line, or in any position in the line, is identical to the sequence of five times A. Thus, we will have another 37 lines with a sequence of 5 times B in a row. Altogether, the number of lines to be eliminated because of a sequence of 5 times A or 5 times B in our total of 924 lines is 2 x 37 or 74.
20. There now remain the sequences of 4 times A or 4 times B in a row which we also have to eliminate from our 924 lines. Once again, we have three possible cases:
(1) 1st case: The sequence of 4 As are at the head of the line, followed necessarily by a B, in order to avoid a sequence of 5 As, already treated above. We can write the series as AAAABxxxxxxx. The seven times x at the end represent 2 As and 5 Bs. Now the number of combinations of 2 As in a sequence of 7 elements composed of A and B is C(7,2) = 7! / (2! * (7 – 2)!) = 7 x 6 x 5 x 4 x 3 x 2 / 2 x 5 x 4 x 3 x 2 = 21. There are thus 21 different ways in which a sequel of 4 As in a row can figure at the head of a series composed of 6 A + 6 B.
(2) 2nd case: The sequence AAAA figures at the end of the line, with, necessarily, a B before: xxxxxxxBAAAA. The seven times x at the beginning represent 2 As and 5 Bs. The situation is identical to the preceding one. In a series of 7 positions, the 2 As can figure in 21 different ways. There are thus 21 different ways in which a sequel of 4 As in a row can figure at the end of a series composed of 6 A + 6 B.
(3) 3rd case: There remains the case where a sequence of 4 As in a row can be anywhere in a series of 6 As and 6 Bs. We can write the series as follows: xBAAAABxxxxx. 2 Bs must necessarily frame the sequence of 4 As in order to avoid a sequence of 5 As. The 6 times x represent 2 As and 4 Bs. The number of combinations of 2 As in a series of 6 is given by C(6,2) = 6! / (2! * (6 – 2)!) = 6 x 5 x 4 x 3 x2 / 2 x 4 x 3 x 2 = 15. But the case can repeat itself 6 times, since there can be 1, 2, 3, 4, 5, or 6 times x at the head of the line. Thus there will be 6 x 15, or 90, lines with a sequence of 4 As in a row, in any position in a series of 12 elements.
21. The total of the 3 cases with a sequence of 4 As in a row is consequently 21 + 21 + 90 = 132. The case of a sequence of 4 Bs in a row, either at the head of the line, or at the end, or in any other place is identical. Thus we will have another 132 lines with a sequence of 4 Bs in a row. Altogether, the number of lines to be eliminated because of a sequence of 4 times A or 4 times B in a row, in our total of 924 lines is 2 x 132, or 264 lines.
22. Let us summarise: we have 924 calendars, theoretically possible, with 6 times 29 days and 6 times 30 days. From this number, we have to eliminate 2 cases of a sequence of six times 29 or 30 days in a row, 74 cases of a sequence of 5 times 29 or 5 times 30 days in a row, and, finally, 264 cases of a sequence of 4 times 29 or 4 times 30 days in a row. Consequently, there are 584 legitimate calendars with an alternation of 29 and 30 days, and a maximum of 3 consecutive months of 29 or of 30 days.
In short: Following the exact method of combinatorics, the number of legitimate Islamic calendars in the world is equal to 584. This number is superior to the number of countries in the world. Consequently, we can affirm that each country in the world will have its own calendar, calculated in a precise manner by establishing visibility curves of the young crescent, month after month.