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The Chakravala Algorithm in R

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A class of analysis that has piqued the interest of mathematicians across millennia are Diophantine equations. Diophantine equations are polynomials with multiple variables and seek integer solutions. A special case of Diophantine equations is the Pell's equation. The name is a bit of a misnomer as Euler mistakenly attributed it to the mathematician John Pell. The problem seeks integer solutions to the polynomial
x^{2} - Dy^{2} = 1
Several ancient mathematicians have attempted to study and find generic solutions to Pell's equation. The best known algorithm is the Chakravala algorithm discovered by Bhaskara circa 1114 AD. Bhaskara implicitly credits Brahmagupta (circa 598 AD) for it initial discovery, though some credit it to Jayadeva too. Several Sanskrit words used to describe the algorithm appear to have changed in the 500 years between the two implying other contributors. The Chakravala technique is simple and implementing it in any programming language should be a breeze (credit citation)

Diophantine Equations (Pure & Applied Mathematics)

The method works as follows. Find a trivial solution to the equation. \(x=1,y=0\) can be used all the time. Next, initialize two parameters \([p_i,k_i]\) where \(i\) is an iteration count. \(p_i\) is updated to \(p_{i+1}\) if the following two criteria are satisfied.
  • \(p_i + p_{i+1} mod  k_{i} = 0\) i.e. \(p_i + p_{i+1}\) is divisible by \(k_i\)
  • \(| p_{i+1} - d |^{2}\) is minimized
After updating \(p_{i+1}\) \(k_{i+1}\) is found by evaluating
k_{i+1} = \frac{p_{i+1}^{2} - d}{k_{i}}
and the next pair of values for \([x,y]\) is computed as
x_{i+1} = \frac{p_{i+1}x_i + dy_{i}}{|k_{i}|}
y_{i+1} = \frac{p_{i+1}y_i + x_{i}}{|k_{i}|}
The algorithm also has an easy way to check if the found solution is a solution. It does so by only accepting values where \(k_{i} = 1\).

A screen grab of the entire algorithm done in R is shown below.

A Related Puzzle:
A drawer has \(x\) black socks and \(y\) white socks. You draw two socks consecutively and they are both black. You repeat this several times (by replacing the socks) and find that you get a pair of blacks with probability \(\frac{1}{2}\). You know that there are no more than 30 socks in the draw in total. How many black and white socks are there?

The probability that you would draw two black socks in a row is
P = \frac{x}{x+y}\times\frac{x - 1}{x+y - 1} = \frac{1}{2}
Simplifying and solving for \(x\) yields
x^{2} - (2y + 1)x + y - y^2 = 0
which on further simplification gives
x = \frac{2y + 1 \pm \sqrt{(2y+1)^2 +4(y^2 - y)}}{2}
We can ignore the root with the negative sign as it would yield a negative value for \(x\) which is impossible. The positive root of the quadratic equation yields
x = \frac{2y + 1 + \sqrt{8y^2 + 1}}{2}
For \(x\) to be an integer, the term \(\sqrt{8y^2 + 1}\) has to be an odd integer number \(z\) (say). We can now write it out as
z = \sqrt{8y^2 + 1}
z^{2} - 8y^2 = 1
This is Pell's equation (or Vargaprakriti in Sanskrit).
As we know that there are no more than 30 socks in the draw, we can quickly work our way to two admissible solutions to the problem \(\{3,1\}, \{15,6\}\).
If you are looking to buy books on probability theory here is a good list to own.
If you are looking to buy books on time series analysis here is an excellent list to own.


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