Archive for the ‘Puzzles’ Category

Deluge – A Sequel?

October 19, 2016

A New Chapter

I am again returning to the Novel ‘Deluge – Agasthya Secrets’ by Dr Ramesh Babu. It is about a year since the Novel has been published. It is an ancient mystery novel set in modern times, in the style of Dan Brown. I thoroughly enjoyed the novel and had recommended the same to many of my friends. I had also written a review of the same, available elsewhere in my blog site. I felt it required a final chapter to tie up a few loose ends and also to give scope for a sequel, if ever, in the future. Initially I thought this added chapter, when read alone, will spoil the suspense element of the original narration and hence, I did not publish the same in my blog site. Now I have made a few changes which will hold the eventual suspense, and at the same time will induce the readers to read the whole novel. To begin with, I am introducing the important characters of the novel before going for the finale proposed by me.

Swathi: A medical intern starting on a career in medical research in micro biology for finding new antibiotics and pro-biotics. She teams up with an American professor in finding herbal antibiotics and takes the help of Ashwin, for locating herbal plants near Rameswaram.

Ashwin: An IIT(M) graduate in Marine Engineering and on internship at Indian Maritime University at Chennai, interested in Marine biology and Archeology.

Ravi: Director of Indian Maritime University at Chennai, a Marine Archeologist, guiding Ashwin in his internship.

MARCOS: Short form for Marine Commandos, a special operation unit of Indian Navy.

Shivani: Joined the team of Ashwin/Swathi, as a member of MARCOS, a marine commando unit, to prevent a possible terrorist attack from the sea, off the coast of Rameswaram.

Lee: Brings along a mini nuclear submarine Triton-1000 from China to help the team (!?!), in under water exploration.

 

THE FINALE

27th March 2016, 6.00 AM

After seeing off Ashwin/Swathi/Shivani trio, leaving for high seas by speed boat, Ravi was lucky to find a modest inn near Uvari itself and had a nap over night. But he got up early at 5 AM, with the village getting busy so early in the next morning. He tried to call Ashwin/Swathi, but their phones were out of range. He packed up his things and left, to take a tour of the shore temple as he had planned earlier. As he approached the Temple he again tried Ashwin/Swathi, as he was getting worried about them. He was looking at the sea intermittently for any sign of them. When he was trying his phone again, he sensed some movement in the sea. When he looked up with hope, he could only see a floating object slowly swaying towards the shore. It was semi-circular in shape and as it came nearer he realized it as a lifebuoy. But why is it floating vertically, as though some thing heavy is hanging from it? He approached the object and pulled the same to the shore. It was a lifebuoy with the inscription, TRITON 1000. Is it not the family of mini submarines, used for under water explorations, he suspected. As he pulled it completely, he saw a heavy object entangled with lifebuoy through some wild sea weeds. It was looking like a wheel guard of a heavy vehicle made of blue PVC material. It had a slit opening in which a black object was remaining stuck. On closer look he recognized it as the stone tablet Ashwin was handling the previous day. Now he was sure something serious has happened to Ashwin trio in the high seas. He again tried to contact them without success. He noticed the tablet now to have the inscriptions erased on both sides.

Sadly Ravi walks towards the shore temple ruins, after safe guarding his finds from the sea in his car. He went inside the temple ruins, to where sanctum sanctorum should have been. There was no Sivalinga – perhaps, it has been moved to the new Swayambu Linga temple nearby. The entrance to the sanctum was somewhat intact. He could see the carvings of Sages Patanjali and Vyagrapadar on either side of the entrance. But he was pleasantly surprised to see carvings of sage Agasthya below Patanjali and of sage Tirumoolar below Vyagrapadar. This carving of Agasthya was exactly the same as what they saw in Tiruchuli, again with the typical signature of Agasthiyar at the bottom, in Harappan script. Oh! What is this big space beneath the Agasthyar’s carving? It looks as though a tile of granite has been removed from this space. However on the other side there was a granite peace in place below Tirumoolar, though slightly damaged. Just as he was wondering about this, Sun’s rays started falling on these carvings. And in the sunlight he could see faintly a similar map in the space below Agasthya, as he has seen earlier on the tablet which he just now retrieved from the sea (with all inscriptions erased on both sides). But the Brahmi scripts on the map were not same. It did not take long for him to realize the map with the scripts was a mirror image of the original. Does it mean that the tablet originally belonged to this temple?

Ravi came hurrying to his car where he has stored the tablet and the lifebuoy. As he was about to return to the ruins with the tablet, he received a call from Swathi and was relieved to know they are safe. He asked them to come over to Uvari for the surprising revelation. As they arrived there in about twenty minutes along with Marcos team, they explained about what happened to them at the Spot-X. When Ravi showed them the tablet and lifebuoy of TRITON 1000, they all guessed that Shivani and Lee might have been involved in the mysterious explosion that occurred below the sea. Could they have ejected themselves and escaped? Or?

Ravi, Ashwin and Swathi approached the stone tablet in the car with all reverence. When they lifted it out into the sun, they could again see the map in the sun light at a particular angle. Ravi immediately reversed the tablet – Yes, they could see the inscription on the reverse also in the early morning sun light. Apparently the ancient chemical used for inscription is able to show up under ultra violet rays of the early morning sun!  They carried the Tablet as though it is a divine idol and tried to place it in the space below the Agasthya carving in the shore temple. Like a strong magnet the frame attracted the tablet with perfect fitting. It may require some force to take the same thing out again. Ravi observed a smile in the face of sage Agasthya. Or was it his imagination? But what is it about a similar tablet below Thirumoolar? Ravi made a mental note to come back soon to the site with ultraviolet lamp and other accessories to do further research.

With the sun still falling on the temple they could still read the inscription shining in the ultraviolet rays.

ஊழி அடைத்த உலக நாதன்

ஆழி வேலெறிந்த அய்யன் தலம்

Uuzhi adaitha Ulaga naathan

Aazhi vel erintha ayyan thalam

Ravi understood. Yes,

“Know this place from where the Swami

  Threw his spear to seal Tsunami”


 

(PS): I shared this proposed final chapter with the author Mr. Ramesh Babu. He liked this so much, he is proposing to include this finale in the next edition of the book. In his own words, “One reader Mr. L. Nagarajan has suggested an entire new chapter to my novel. It fits in so well just after the climax and also follows my style of narration! Hats off!” (https://plus.google.com/101148246018050836170). You may get this book ‘Deluge (Agasthya Secrets)’, at the following sites:

https://notionpress.com/read/deluge

https://store.kobobooks.com/en-US/ebook/deluge-20

http://bookstore.bookcountry.com/Products/SKU-001042017/Deluge.aspx

https://www.amazon.com/Deluge-Agasthya-Secrets-Ramesh-Babu-ebook/dp/B014XUEVR0

 

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Dan Brown’s Inferno, the hell

August 19, 2016

I just finished (Aug 2016) reading the novel Inferno by Dan Brown. When I finished reading The Amber Room by Steve Berry, I wrote a blog about similar mysteries that abound in ancient and medieval India. I invoked English language writers of Indian origin to write such mysteries a-la Dan-Brown-Style, but with Indian artifacts and mysteries. (Please refer to my blog.

https://lvnaga.wordpress.com/2014/01/28/ancient-mystery-thriller/). My thirst for the same was quenched somewhat by a novel ‘Deluge – Agasthya’s Secrets’, by Dr. Ramesh Babu of Chennai. (https://lvnaga.wordpress.com/2016/03/22/dr-ramesh-babu-indian-version-of-dan-brown/).

This new novel ‘Inferno’ by Dan Brown is based on a biological ‘terrorism’ of scientific age placed in the surroundings of medieval mysteries of Ottoman Empire covering present day Florence of Italy and Istanbul of Turkey.

This novel is heavily based on Italian poet Dante Alighieri’s master piece ‘The Devine Comedy’ consisting of three cantos – Inferno, Purgatory and Paradise, describing the path of the dead/soul towards hell, punishment and finally perhaps the heaven. There are many articles available in the net comparing this piece of literature with similar ideas represented in ancient Hindu scriptures by Saint Veda Vyasa in his Garuda Purana. Some of them even suggest that Dante was inspired by this description of Hell (and the travel of soul with its pseudo-body through the hell to the Paradise). I wonder if any other Indian  literature has given annotations of this work in Sanskrit or any other modern languages. I am vaguely familiar with a story in Mahabharat where King Yutishtra, with all his integrity and moral equipoise, is made to undergo a horrible view of the Hell, as a punishment for his abetment of a lie in killing Aswatthama in the war. Any extract of description such a view, is it available, I wonder.

(Those who have not read this novel ‘Inferno’ and are planning to read the same, may please avoid reading further, to retain the mystery and suspense of the Novel.)

In this novel there are some unexplained ambiguities as below

  • How can a single type of vector virus would do equal harm to the fertility of both men and women? Evidently, their reproductive systems are quite different.
  • Neither it is necessary to affect random one-third of both male and female population equally, to reduce the birth rate by a third.
  • Hence it is better to say that the vector-virus modifies the DNA of whole population but switches on at random only in one-third of male population. This will reduce the birth rate by one-third and gradually reduce the population by a third, as this DNA-Virus from even unaffected males gets inherited by the subsequent population. It will be again switched on at random in one third of males in every generation.
  • Though a lot of anxiety is expressed by all the characters in the novel about this biological ‘terror’, it appears to be a very humane way of controlling the population. It is same as vasectomy and tubectomy, which are of course, voluntary. This type of population control is normally adopted in animals and pests.
  • The characters in the novel, opposed to this type of ‘terrorism’ initially, come around and accept the same and think of making it reversible.

However the novel is quite interesting and highly readable. I understand it is also coming as a movie soon with Indian actor Irfan Khan in the role of Provost, the off-shore expediter and the secondary antagonist in the novel.

PS: I remember playing a board game in my village (India) on the nights of Gokul Ashtami and Shivratri known as Parama Pada Shobanam. It is a kind of a snake and ladder game where we start from hell and pass through several evil images and then on to happy images and finally to the heavenly images of Gods.(i.e. Inferno, Purgatory and Paradise).  On the way we encounter many snakes (named after villains of Hindu epics) and ladders of good conduct and behaviors. I wish somebody adds an image of this board to this blog. (LVN)

 

 

Dr. Ramesh Babu, Indian Version of DAN BROWN

March 22, 2016

Deluge

Agasthya Secrets

A novel by Dr. Ramesh Babu

A REVIEW by L V Nagarajan

An Indian version of Dan Brown has risen in the horizon. Many of us might have read Dan Brown’s Novels Da Vinci Code, Angels & Demons, Digital Fortress and the like. These novels are based on ancient mysteries being interpreted and used in the present days for good or bad deeds. These mysteries involve codes, puzzles and secrets that existed in the context of medieval European History. We always felt Indian history and culture, being much older, offers much more scope for such novels.

Please refer to my blog:

(https://lvnaga.wordpress.com/2014/01/28/ancient-mystery-thriller/) wherein I have given some ideas on such themes.

 

Dr. Ramesh Babu’s novel DELUGE is a very successful attempt to create a full length fiction of ancient mystery thriller set in modern India. He has made an excellent mix of several ancient intrigues like, Nadi jyothish, Siddha philosophy, mythology of Lord Siva as Tripurandaka, mystical land of Lemuria, ancient Tamil-Brahmi scripts etc. This novel is about events set to happen in the near future. The parallel story of devas and asuras of yore and merging it with the tsunami of Lemuria keep the reader quite engaged. The synthesis of Vedic Culture and Tamil Culture has been brought out very well in the narration. Extremist views in Politics and Religion has also been brought out convincingly. Narration is very absorbing and it is difficult to believe that it is author’s first full length novel.

I recommend this novel to all my friends for a good read.

Following are the details

Deluge

Agasthya Secrets

By Dr. Ramesh Babu, MS, MCh, FRCS Glas, FRCS Edin, FRCS Paed
Professor of Paediatric Urology,
SRMC, Chennai.

https://notionpress.com/read/deluge

Nagarajan

Baudhayana’s Circles

September 21, 2013

Western biographers credit Archimedes of Syracuse (287-212 B.C) with the analytical evaluation of the factor Pi associated with circle, within a close range of 31/7 to 310/71. However in the process of establishing this, they also recognize him as the first person to realize that the same factor is associated with both perimeter and area of the circle. He is also said to be the first person to propose and prove that “Area of the circle = ½ x Perimeter x Radius”. Is it really so? Let us look at another person, one Baudhayana, from ancient India (800BC) who also worked on the circles earlier to Archimedes, by a few centuries.

It was known to Baudhayana, or even to people earlier to him, that the perimeter of the circle depends only on its radius or diameter and that it is actually proportional to the radius or diameter. Though it has not been stated explicitly, it is clear from various sutras that they were well aware that for similar figures, the ratio of the areas equals the square of the ratio of the lengths of the corresponding sides. It was also known that the area of the circle depends only on its radius or diameter and that it is actually proportional to the square of the radius. That is, for a circle, it was known that: – Area = Ka x r2, and Perimeter P = Kp x (2r). But do they know that, Ka = Kp = Pi ?.

During 800 BC, this Indian high priest, Baudhayana, has formulated, in his Sulvasutra (I-48), the so-called Pythagoras theorem, centuries before Pythagoras (572 BC). In another sutra (I-51) he has given a general rule for finding the square root of any number, both geometrically and arithmetically. In his Sutra (I-61) he found the value of √2 to a great accuracy and has given the procedure for the same. This Indian mathematician could construct a circle almost equal in area to a square and vice versa. He has described such procedures in his sutras (I-58 and I-59). All these were achieved in 800BC!

As Baudhayana was designing a religious altar for performing the Hindu rites, he constructed a square within a square as below:


Normal+

He observed the inner square is exactly half of the bigger square in area.  This led him to formulate, in his Sutra I-48, the so called Pythagoras theorem, which was reinvented by Pythagoras a few centuries later. Please refer to my earlier blog on “Baudhayana’s (Pythagoras) Theorem”.

Subsequently Baudhayana wanted to evolve procedures for constructing circular altars. He constructed two circles circumscribing the two squares shown above. Now, just as the areas of the squares, he realized that the inner circle should be exactly half of the bigger circle in area. Yes, he knows that the area of the circle is proportional to the square of its radius and the above construction proves the same. By the same logic, just as the perimeters of the two squares, the perimeter of outer circle should also be √2 times the perimeter of the inner circle. This proves the known fact, that the perimeter of the circle is proportional to its radius. Now it is known beyond any doubt that for a circle,

Area = Ka x r2, and Perimeter P = Kp x (2r).

But that is not enough. Baudhayana wants to construct circular altars of specific areas. He needs to know the values of Ka and Kp. Baudhayana and his ilk were more interested in the area of the circle than its perimeter.

At this point, Baudhayana would make an important observation. Areas and perimeters of many regular polygons, including the squares above, can be related to each other just as the case of circles. The perimeters and areas of some simple regular polygons are listed below (‘r’ is the distance from the centre of the polygon to its sides):

Equi. Triangle–  Perimeter = K3(2r) & Area = K3(r2); with K3 =  3√3

Square-  Perimeter = K4(2r) & Area = K4(r2); with K4 = 4

Hexagon-  Perimeter = K6(2r) & Area = K6(r2); with K6 =  2√3

Octagon-  Perimeter = K8(2r) & Area = K8 (r2); with K8 =  8(√2-1)

It may also be noticed that the values of the constants, Ki’s, are gradually reducing from about 5 to about 3.3, by the time we reach the octagon. Another fascinating feature with these polygons is: all their areas are ‘½ r’ times their perimeters. Anybody would have been tempted to conclude from above that for circles also, Kp= Ka= K0. This will automatically make Area of the circle = ½ r x (Perimeter). However Baudhayana wanted to prove this.

Let us now consider an N-gon, a regular polygon of N-sides. Let ‘r’ be the distance of the sides from the centre. Let each side be equal to ‘s’. The area of the triangle one side makes along with the centre is (½)sr. Hence the total area of this N-gon is ½Nsr.

So, for N-gon -> Perimeter = Ns & Area = ½Nsr; with Kn = Ns/2r

Here again, Area = (r/2) x Perimeter. In all the above regular polygons, a circle of radius r can be inscribed. Now let us assume that, for this circle, constants Kp and Ka are different. The areas and perimeters of the above polygons are steadily reducing but still remaining more than those of this circle. (i.e.) Kn(2r) > Kp(2r) and Kn(r2) > Ka(r2). Hence any Kn will have to be greater than both K1 and K2. In the N-gon last considered, Ns being the perimeter, it reduces constantly as number of sides N increases. Kn = Ns/2r will also reduce gradually and finally will converge to a finite value, now known as Pi. Baudhayana realized that there is an N-gon with such a perimeter, whose Kn is just more than Kp by any arbitrarily small amount ‘∂’. Similarly Baudhayana concluded that there is an N-gon of such an area, whose Kn is just more than Ka by any small amount ‘ε’.  However both these Kn’s are greater than Kp and Ka. (i.e.) Kp + ∂ > Ka, and,  Ka + ε  > Kp, for any small amounts of ∂’s and ε’s. The above is possible only when Kp=Ka.

Thus Baudhayana concluded, Kp = Ka = Ko, which is now known as Pi. This automatically makes, Area of the circle = ½ r x (Perimeter).

So we may conclude that, the above facts about the circles were already known to Baudhayana and other ancient Indian mathematicians even before Archimedes. Credit is surely due to Archimedes for narrowing down the value of Pi between 31/7  and 310/71. However, Baudhayana on his own has narrowed down the value of Pi to be between 40/12 and 40/13 (i.e. 31/3 and 31/13). It is the value 40/13 he has used in his Sulva Sutra I-58, (for Circling the Square or to find a circle equal in area to a square), as will be demonstrated later. Before going to I-58, let us see how he derived the above values for Pi.

Ancient mathematicians could have already found the value of Pi with limited accuracy, by actual measurements of diameter and perimeter of the circle by using ropes as per the practices existing in those days. It could have given them, at best, a value between 3.11 and 3.17 (i.e. Pi +/- 1%). So, the attempts continued to analytically find the value of Pi.

Even before Baudhayana, the upper limit for the value of Pi was fixed as 4 by considering a circle inscribed in a square. The lower limit was fixed as 3 by considering a regular hexagon along with its circum-circle. But after Baudhayana’s (Pythagoras) theorem, these limits could be narrowed down to be between 3 and 2√3 by considering both circum-circle and in-circle of a hexagon as below.

Hexagon

The circum-radius is ‘a’ and the in-radius is √3/2(a). The perimeter of the hexagon is larger than that of in-circle and less than that of circum-circle. i.e. π(2xa) > 6a and π(2ax√3/2) < 6a. Hence, 2√3 > π > 3.  There is an indirect reference to the value of 3 in an earlier sutra of Baudhayana for constructing altars: “The pits for the sacrificial posts are 1 pada in diameter, 3 padas in circumference.” This gives an approximate value of 3 for Pi. But they knew it is more than 3, as 3 pada is already known as the perimeter for a hexagon inscribed in the above circle. Hence 3 is just the lower limit for value of Pi.

Baudhayana went one step further by considering a regular octagon along with its circum-circle and in-circle. Just as Archimedes found the bounds for value of Pi, by considering 96-gon circum-scribing and in-scribing a given circle, Baudhayana found the bounds, by considering circum-circle and in-circle of an octagon. (yes, a complimentary procedure to what was done by Archimedes, five centuries later). Baudhayana used for √2, the value of 17/12 (normally used in those days) to arrive at these simple fractions, as below. He first considered a square with 12 units as half side. Hence half diagonal will be 17 units considering 17/12 as √2. Referring to the diagram below, the octagon inscribed within this square will have its side as 10.

octagon

The in-radius of this octagon will be 12 units and circum-radius will be 13 units. (‘5,12,13’ , the Pythagoras triple, as known even in Baudhayana’s times defines this octagon!). The perimeter of the octagon is 80 units. As this octagon is sandwiched between the circum-circles and in-circle, (ref Figure above), we may compare their perimeters as below:

Pi(2×13) > 80 > Pi(2×12)

Hence,

Baudhayana’s values of Pi are given by: 40/12 > Pi > 40/13.

Accuracy of Pi was always sought to be improved throughout the history, even after much closer estimates by Archimedes. A later mathematician Manava (650 ~ 300 BC) has stated in his sutra:

Viskambhah pancabhaagasca  viskambhastrigunasca yah.

sa mandalapariksepo na vaalamatiricyate ||      

(Manava  Sulvasutra 10.3.2.13)

(a fifth of the diameter plus three times the diameter, is

the circumference of the circle, not a hair-breadth remains.)

It gives a value of 31/5 as the upper limit for value of Pi, a better value than Baudhayana’s 31/3, and approaching 31/7 of a later day Archimedes: (as per Manava’s statement “not a bit remains” after 3.2 times the diameter).

Baudhayana cleverly used his values of Pi for finding the area of a circle and for drawing a circle of a given area. He found the value of 40/13 to be closer to the eventual value of Pi and has used the same to derive his Sulvasutra I-58 for “Circling the Square”. Baudhayana would have derived this formula as below:

Let ‘a’ be the distance of the sides of the square from its centre. i.e. each side of the square is ‘2a’. To find a circle of radius ‘r’ with an area equivalent to this square, one may write

πr2 = 4a2;        (i.e.), r = (2/√π)a

Square Circle

Baudhayana used 40/13 as value of Pi, and 17/12 as value of √2.

From the picture above we see, a < r < √2a.

Hence he assumed, r = a + (1/x)(√2a-a) = a [1 + (1/x)(√2 – 1)]

For the above square, πr2 = 4a2

r2 = 4a2/π = 4a2 *(13/40) = a2 * 130/100

Hence, r = a * (√130)/10

By Baudhayana’s method (I-51) for finding the square root of any number,

√130 = √(122 – 14)  = 12 – (14/24) = 10 + 17/12

So, r = a(1 + 17/120) = a [1 + (1/3)(51/120)]

≈ a[1 + (1/3)(5/12)] = a[1 + (1/3)(√2 – 1)]

Thus, r = [a + (1/3)(√2a – a)]

So Baudhayana formulates his sulvasūtra I-58 as below:

caturaśraṃ maṇḍalaṃ cikīrṣann

akṣṇayārdhaṃ madhyātprācīmabhyāpātayet |

yadatiśiṣyate tasya saha

tṛtīyena maṇḍalaṃ parilikhet

 

caturaśraṃ = Square, Mandalam = Circle,

akṣṇayārdhaṃ = Half Diagonal,  

madhyātprācīm = From centre towards east,

abhyāpātayet = laiddown, yadatiśiṣyate = portion in excess,

tasya saha tṛtīyena = using only a third of this, 

parilikhet – Draw around

To make a square into a circle, draw half its diagonal from the centre towards the East; then describe a circle using only a third of the portion which is in excess.

i.e. Using the above formula your are able to draw a circle of given area (=4a2), where a is the measure of half the side of the above square. The radius of this circle is given as:

r = [a+1/3(√2a – a)] = [1+1/3(√2 – 1)] a

Let us see with value of Pi as 40/13, how the areas of square and circle compare.

Area of Square = 4a2

Area of circle = 40/13 x [1 + (1/3)(√2-1)]a2

= (40/13) x (41/36)2 a2  = (67240/16848) a2 = 3.9909 a2

Remarkably close.

As per this construction the value of Pi we obtain as per today’s value of √2 is, Pi = 3.088312

However in the reverse process of squaring the circle, he has gone for corrective fractions as will be demonstrated by Baudhayana’s sutra I-59, as given below:

maṇḍalaṃ caturaśraṃ cikīrṣanviṣkambhamaṣṭau

bhāgānkṛtvā bhāgamekonatriṃśadhā

vibhajyāṣṭāviṃśatibhāgānuddharet |

bhāgasya ca ṣaṣṭhamaṣṭamabhāgonam

viṣkambham = Diameter; aṣṭau bhāgānkṛtvā = making eight parts; 

bhāgaekona = take out one part,

triṃśadhā vibhajya = 29 parts of this part;

āṣṭāviṃśatibhāgānuddharet = of these remove 28 parts; 

bhāgasya ca = from this part also,

ṣaṣṭhamaṣṭamabhāgonam = remove (1/6 minus 1/8 of 1/6).

If you wish to turn a circle into a square, divide the diameter into eight parts and one of these parts into twenty-nine parts: of these twenty-nine parts remove twenty-eight and moreover the sixth part (of the one part left) less the eighth part (of the sixth part). 

The above formula is to make a square of area, equal to a given circle. Baudhayana could have really inverted the earlier formula I-58 to obtain this. But this sutra appears quite complicated. It is so only because Baudhayana in this case used a better value for √2. As per his sutra I-61,

√2 = 1 + (1/3) + (1/3*4) – (1/3*4*34) = 577/408

We may be wondering how handicapped the ancient mathematicians were without the present day decimal point system. However ancient Indians were so facile with fractions they never needed the decimal point system. Even as late as 19th century, Indians were using fractional multiplications tables of ½, ¼, 1/8 , 1/16 , 1/32 and even 3/16  in their every day arithmetic calculations. Baudhayana uses this amazing fractional arithmetic to arrive at the above formula.

We know from I.58, r = a + (1/3)(√2a-a) = [1+1/3(√2 – 1)]a

So we may write, a = r/[1+(1/3)(√2-1)]

With √2 = 577/408,

we get, a = r/[1+(1/3)(169/408)]= (1224/1393)r

Now for fractional magic:

1224/1393 = (1224/1392)*(1392/1393) = (51/58)/(1393/1392)

= [1-(7/56)(56/58)]/[1 + (1/1392)]

= [1-(1/8)(28/29)]*[1- (1/1392)]

= [1-(28/8*29)]*[1 – (1/8*29*6)]

(assuming 28/29 ≈ 1, in the last term),

= 1 – [28/(8*29)] – [1/(8*29*6)] + [1/(8*29*8*6)], .

Thus Baudhayana obtains the final formula of I-59 as:

(With Side of square as ‘s’ and the diameter of the circle as ‘d’)

 s = [1- 28/(8*29) – 1/(8*29*6) + 1/(8*29*6*8)] x d

i.e., s = 0.878682 * d

This increases the value of Pi marginally from 3.088312 to 3.088326.

Conclusion:

It was after studying the book “Journey Through Genius” by William Dunham, I got interested in the History of Mathematics. I read several books to know more on this subject. In order to create interest among our genext, I started to write a few blogs on this subject. The present one is on ancient Indian Mathematician Baudhayana (800BC) and his works on Circles. We seem to know him only through his Sulva-sutras. However by reading extensive material on him, I could make out a few narrations of his works. Of course this narration includes a few of my imaginations and intuitions. Ancient Indians in 800BC were well aware of the basic properties of the circle. Baudhayana’s sulvasutras I-58 and I-59 give ample proof of this. Baudhayana was also able to fix the value of Pi to be between 40/12 and 40/13 (i.e. between 3.33 and 3.08).  Baudhayana’s name is still uttered during many Hindu rituals. Even my own family is linked to Baudhayana through his line of disciples as mentioned often by us in our prayers as, Apasthamba, Aangirasa, Baragaspathya and Bharatwaja.

References:

1. A history of Ancient Indian Mathematics – C N Srinivasaiengar, The World Press Private Ltd. Calcutta. (1967)

2. Journey Through Genius – William Dunham, Penguin Books 1990.

3. S.G. Dani, Geometry in Sulvas_sutras, in ‘Studies in the history of Indian mathematics’, Cult. Hist. Math. 5, Hindustan Book Agency, New Delhi, 2010.

Square Root of 2, by Baudayana

June 16, 2013

Square Root of 2, by Baudayana

L V Nagarajan

Baudhayana is a great mathematician of ancient India estimated to have lived during 800BC. He was an expert mathematician, architect, astronomer and a Hindu high priest. He has proposed several mathematical formulas (Sulva Sutras), some of them with proofs. His statement and proof of the so-called Pythagoras theorem is so simple and elegant.

In ancient times, a Square was held as an important geometrical figure. Every area was expressed in so many squares. There was considerable interest in finding an equivalent square for every area, including circle, rectangle, triangle etc.

Baudhāyana, gives the length of the diagonal of a square in terms of its sides, which is equivalent to a formula for the square root of 2:

samasya dvikaraṇī. pramāṇaṃ tṛtīyena vardhayet
tac caturthenātmacatustriṃśonena saviśeṣaḥ

Sama – Square; Dvikarani – Diagonal (dividing the square into two), or Root of Two

Pramanam – Unit measure; tṛtīyena vardhayet – increased by a third

Tat caturtena (vardhayet) – that itself increased by a fourth, atma – itself;

Caturtrimsah savisesah – is in excess by 34th part

In English syntax, it will read as below:

The diagonal of a square of unit measure (is given by) increasing the unit measure by a third and that again by a fourth (of the previous amount). This by itself is in excess by a 34th part (of the previous amount).

That is,

√2 = 1 + 1/3 + ¼ (1/3) = 17/12

But as per the sulba-sutra above, this in excess by a 34th part of the previous amount.

Hence

√2 = 1 + 1/3 + ¼ (1/3) – 1/34[(¼ (1/3)]

= 1 + 1/3 + 1/(3*4) + 1/(3*4*34) = 577/408

The above value is correct to five decimals.

There have been several explanations as to how this formula was evolved. Apparently, initially (even on works of late 6th century AD) an approximate value of 17/12 was used for √2, which is nothing but [1+ 1/3 + 1/(3*4)].

1. One theory is they just used actual measurement by ropes to arrive at these fractions – they first tried unit rope length and then 1/2 of the same length. As it was too long they next tried 1/3. It was just short and hence ¼(1/3) was added to it. It was quite close and hence 17/12 was used initially. However it was found slightly longer (savisesah) and when measured by rope again it was found longer by 1/34[1/4(1/3)]. This was also found minutely longer (savisesah) but accepted as a sufficiently accurate value. – That was a simple explanation.

2. Another explanation for the evolution of this formula was based on geometrical construction. – Two equal squares, each with side of one unit were taken. One of the squares was vertically divided into three rectangles. Two pieces of the above was placed along the two adjacent sides of the square to form an approximate square of side 1+1/3, but for missing a small square 1/3×1/3. This was also made up by using a piece from remaining rectangle. Hence we get the first approximate value of (1+1/3) for √2.

Square Root 2

But a small piece is still remaining of size (1/3 x 2/3). This was made into 4 equal strips of size [1/4(1/3) x 2/3]. Two pieces, end to end, were kept along one side of the above augmented square and the other two pieces on the other side. Now we get a total area which is the sum of the two squares. The above augmented square is of side (1 + 1/3 + 1/(3×4)) = 17/12, which was found good enough initially. But, we still miss a small portion of (1/12 x 1/12), to complete the square. Hence the size has to be reduced by this extra area (savisesah). – To find this extra measure, (1/12 x 1/12) should be divided by (17/12 +17/12) and that gives 1/(3x4x34). Of course still we have some minute extra area (savisesah). The above construction is explained in the above figures.

Yes, this is really an interesting explanation. But this amazing formula (sutra) evolved in 800 BC deserves a better explanation which I will offer now.

3. Way back in 1967, I was in a class room of IIT/Kanpur. The teacher was Professor Dr. V. Rajaraman, the pioneer of computer education in India. He was teaching us the basic algorithms for programming in Fortran, a (then) popular programming language. One of the very early recursive algorithms he taught us was, to find the square root of a number. It goes thus:

Let N be the number for which square root is required

Make first guess of the square root as r0 – Later, we will know, the guess may be as bad as 10 times N; still the method works as smoothly as ever.

The next guess can be made as r1 = ½ (r0 + N/r0)

Keep improving this value using the recursion: r(n+1) = ½ [r(n) + N/r(n)]

Surprisingly the value converges very fast to √N, to the required level accuracy.

As you have seen in my earlier blog of Evolution of Sine Table by Hindu Maths, our ancient mathematicians have always preferred recursive steps to solve any problems. Hence, in this case also, Baudayana preferred to use recursive steps, exactly as above. However like others, he preferred to calculate individual step sizes as below:

r(n+1) – r(n) = ½[N/r(n) – r(n)], which is same as the above recursive statement.

To find square root of 2, Boudayana used (1+1/3) = 4/3, as the first guess, r(0).

Hence next step, r(1) – r(0) = ½(3/2 – 4/3) = (3/4 – 2/3) = 1/(3×4)

And hence, r(1) = 1 + 1/3 + 1/(3×4) = 17/12 = 1.4166

Next step, r(2) – r(1) = ½(24/17 – 17/12) = [12/17 – 17/(3x4x2)] =  -1/(3x4x34)

And hence,  r(2) = 1 + 1/3 + 1/(3×4) – 1/(3x4x34) = 577/408 = 1.414216 (correct up to 5 decimals)

Baudayana would have gone to the next minute step also (as his savisesah, indicates), as below

Next step, r(3) – r(2) = ½(816/577 – 577/408) = 408/577 – 577/(3x4x34x2) = -1/(3x4x34x1154)

Hence, r(4) =  1 + 1/3 + 1/(3×4) – 1/(3x4x34) -1/(3x4x34x1157) = 1.41421356237469

The above value is correct up to 13 places.

The recursive algorithm is always the first approach of ancient Indian mathematicians.

Just to satisfy myself that it is not just an isolated case, I tried this logic for finding square root of three also. The step sizes came out to be as below:

Start = 1

First Step = ½             , Root = 1.5

2nd Step = 1/(2×2), Root = 1.75

3rd Step = – 1/(2x2x14), Root = 1.7321429

4th Step = – 1/(2x2x14x194), Root = 1.7320508

The above value is correct up to 7 decimal places.

Again hats-off to Baudayana!

Baudhayana’s (Pythagoras) Theorem

June 12, 2013

                              Baudhayana’s    Pythagoras    Theorem

“Long long ago, so long ago, nobody knows how long ago” – that is how we used to start our stories in our younger days. But this story starts exactly like this.   Long long ago, so long ago, nobody knows how long ago, there lived one Baudhayana, who was an ancient Hindu master. He is dated to have lived during 800BC. He was an expert mathematician, architect, astronomer and a Hindu high priest. Once he was designing a sacrificial alter in the shape of a square. He inscribed another smaller square inside this square as below:

Normal

Baudhayana contemplated on this shape and realized the area of the inner square is exactly half the area of the outer square. With the cross-wires drawn as above, it is easy for us also to see this fact.

But the genius of Baudhayana went further. He thought of inscribing an off-set square with in the bigger square as below:

Offset

Now he calculated the area of the inner square as:

Area of the inner Square

= Area of the outer square – area of the 4 bordering triangles

= (a + b) – 4 x (ab/2)

(i.e) Area of the inner Square = a2  +  b2

Aaha..! This sounds very familiar. Is this not called Pythagoras Theorem?  But how come, it exists in 800 BC, almost 300 years before Pythagoras (570 -495 BC)? That too found by an ancient Indian? Should we call this then, as Baudhayana’s Theorem. But Baudhayana  proposed  many more such theorems in his Sulva Sutras. His statement of the so called Pythagoras theorem is as below:

 “dīrghasyākṣṇayā rajjuH pārśvamānī, tiryaDaM mānī, 

cha  yatpthagbhUte kurutastadubhayākaroti.”

The above verse can be written again, by separating the combined words and syllables, as below:

“dīrghasya  akṣṇayā  rajjuH – pārśvamānī, tiryaDaM mānī, 

Cha  yat  pthah  bhUte  kurutah – tat ubhayākaroti.”

Below are the meanings of all the words:

Dirgha – Oblong tank or pond

Akshnaya – Diagonally or transversely

Rajjuh – rope

Pārśvamānī = The longer side of the oblong or the side of a square

Tiryak –across, oblique, sideways

Yat (… tat) – Which ( … the same)

Prthah – ( particular) measure

bhūta – become, produce

kurutaha – they (two) do, both do (typical Sanskrit dual verb)

(Yat …) tat –  (Which …) the same

ubhayā – In two ways, two together

Ubhayangkarothi – Produces or effects the two together

Putting the verse in the English language syntax, it reads as below:

In an oblong tank – (what) longer side and (the other) oblique side, the measures (or areas) they produce – (the same) (sum of) both, is effected or produced – by a diagonally held rope.

The natural evolution of this Baudhayana Sutra (Or this Baudhayana Theorem) speaks volumes of its originality. Our salutations to Baudhayana.

In trying to translate this verse into English I was handicapped by two deficiencies – (i) my highly limited knowledge of Sanskrit and, (ii) Non availability of a English-Sanskrit-English technical dictionary.  Such a dictionary is very much a need of the hour, as lot more technical people are now trying to understand and interpret the immense contribution of ancient Indians to Science and Technology. For example in the case of this verse, Deergha, Parsva and Triya may mathematically mean the three sides of a right angled triangle. Experts in this field should take initiative in developing such a technical dictionary for Sanskrit.

Ref : S.G. Dani, On the Pythagorean triples in the ´ Sulvas¯utras, Current Sci. 85(2003), 219-224;

(available at: http://www.ias.ac.in/currsci/jul252003/contents.htm/)

 

L V Nagarajan

12 June 2013

Polyhedrons

November 28, 2010

 

Solid geometry has always fascinated me and has fired up my imaginative power. I was introduced to Solid Geometry by my elder brother (L V Sundaram) when I was barely 10 years old. He had a habit of reading and explaining to himself as a part of his study process. When I was around as a young boy, he would do the explaining to me, not caring how much of it I understood. But he was rather surprised that I understood a lot of what he told me especially in geometry. As I learnt more of plane geometry, I developed a habit of extending the results of the theorems to the third dimension.

Starting from triangle, quadrilateral and pentagon, there are infinite number of polygons. Polygons with all sides and all internal angles equal are called regular polygons and there are infinite numbers of regular polygons with 3 to infinite numbers of sides. In the limit the regular polygon becomes a circle.

Extending to the third dimension, there are infinite numbers of polyhedrons, starting from tetrahedron.

Two non-parallel straight lines in a plane make an angle. Three straight lines in a plane which are non parallel to each other make a triangle, which is a polygon of least order – 3 sides and 3 angles.

Similarly, two non-parallel planes make an angle all along the common edge. Three planes which are non parallel to each other make a solid angle at a unique point of intersection of all the three planes. This solid angle is referred to as trihedral angle, as they are made by three planes. The solid angle at the vertex of a typical polyhedron is known as a polyhedral angle, made by three or more planes. But three intersecting planes still do not make a closed polyhedron. A fourth non-parallel plan is needed to get a polyhedron of least order – 4 faces, 6 edges and 4 vertices; this is known as a Tetrahedron or sometimes as triangular pyramid. We may also think of a polygonal pyramid with the base as a polygon of n-sides. For a regular polygon of n-sides, this polygonal pyramid will become a circular cone in the limit.

Regular polyhedrons are those whose faces are all regular polygons, congruent to each other, whose polyhedral angles are all equal and which has the same number of faces meet at each vertex. An interesting fact is, there are only five regular polyhedrons: the Tetrahedron (four triangular faces), the Cube (six square faces), the Octahedron (eight triangular faces—think of two pyramids placed bottom to bottom), the Dodecahedron (12 pentagonal faces), and the Icosahedron (20 triangular faces). (as below)

Polyhedrons are classified and named according to the number and type of faces. A polyhedron with four sides is a tetrahedron, but is also called a Triangular pyramid. The six-sided cube is also called a hexahedron. A polyhedron with six rectangles as sides also has many names—a rectangular parallelepiped, rectangular prism, or box. However there are infinite numbers of semi-regular and non-regular polyhedrons which closely approximate to a hollow sphere in the limit. Imagine cutting off the corners of a cube to obtain a polyhedron formed of triangles and squares, for example. Other common polyhedrons are best described as the same as one of previously named that has part of it cut off, or truncated, by a plane.

Euler characteristic

The Euler characteristic χ relates the number of vertices V, edges E, and faces F of a polyhedron:

For a simply connected polyhedron, χ = 2.

For regular polyhedrons as below

Cube ;     χ = 8-12+6 = 2

Tetrahedron     χ = 4 -6 +4 = 2

Octahedron     χ = 6 – 12 + 8 = 2

Dodecahedron χ = 20 – 30 + 12 = 2

And so on ….

Duality

File:Dual Cube-Octahedron.svg

For every polyhedron there exists a dual polyhedron having:

  • faces in place of the original’s vertices and vice versa,
  • the same number of edges
  • the same Euler characteristic and orientability.

Dodecahedron and Icosahedron are the dual of each other. Same is true with Cube and Octahedron. The dual of a Tetrahedron is another smaller tetrahedron.

The five regular polyhedra were discovered by the ancient Greeks. The Pythagoreans knew of the tetrahedron, the cube, and the dodecahedron; the mathematician Theaetetus added the octahedron and the icosahedron. These shapes are also called the Platonic solids, after the ancient Greek philosopher Plato; Plato, who greatly respected Theaetetus’ work, speculated that these five solids were the shapes of the five fundamental components of the physical universe.

Are we sure there are only five regular polyhedrons? By definition the faces of regular polyhedrons should all be regular polygons. Here are the possibilities:

  • Triangles. The interior angle of an equilateral triangle is 60 degrees. Thus on a regular polyhedron, only 3, 4, or 5 triangles can meet at a vertex. If there were 6 or more, their angles would add up to 360 degrees or more and hence not admissible. Consider the possibilities:
    • 3 triangles meet at each vertex. This gives rise to a Tetrahedron.
    • 4 triangles meet at each vertex. This gives rise to an Octahedron.
    • 5 triangles meet at each vertex. This gives rise to an Icosahedron
  • Squares. Since the interior angle of a square is 90 degrees, at most three squares can meet at a vertex. This is indeed possible and it gives rise to a hexahedron or cube.
  • Pentagons. As in the case of cubes, the only possibility is that three pentagons meet at a vertex. This gives rise to a Dodecahedron.
  • Hexagons or regular polygons with more than six sides cannot form the faces of a regular polyhedron since their interior angles are at least 120 degrees.

Hence convex polyhedral angles of regular polyhedrons can only be of three types: Trihedral (Tetra-, Hexa- and Dodeca-hedrons), Tetrahedral (Octahedron) and Pentahedral (Icosahedon). Hence we can say there can be only five regular polyhedrons as above.

The following are the patterns that may be cut off from paper or a card board and may be used to make 3-dimensional models of the regular polyhedrons.

              

My next blog on the subject will concentrate only on Tetrahedrons.

Soma Cubes

September 18, 2008

I read about these Soma Cube puzzles way back in 1966. I could not get those puzzle pieces. I wanted to get them done on my own but it never happened. I remembered them recently and saw the details of the puzzle on the net at http://web.inter.nl.net/users/C.Eggermont/Puzzels/Soma/. Atleast I solved the basic one, by visualising the pieces in my mind. I have attached the details of the puzzle for others to see and enjoy.

soma-cube

Hope people like it

L V Nagarajan

18th Sept 2008