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Old 2009-02-26, 14:11   #3
Raman
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"Mr. Tuch"
Dec 2007
Chennai, India

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Once the matrix has been solved, the history matrix contains the rows which contribute to the dependency. The relations are multiplied to get the value of Q^2\ (mod\ N), which by using square root, Q is resolved.
This is the technique that is being mentioned so, in the paper of Morrison and Brillhart, rather than maintaining a count of each prime power, and then multiplying each prime by half the value of power.

For each relation Q_i, that is examined, the square root is computed as follows, where \overline{Q} is the value of the least positive remainder of Q\ (mod\ N).

i=2,\ \overline{Q}=1,\ R=Q_1
X=GCD(R,\ Q_i)
\overline{Q}=X\overline{Q}\ (mod\ N)
R=\left( \frac{R}{X} \right) \left( \frac{Q_i}{X} \right)
i=i+1
If\ i \le s\ Then\ Goto\ 2
X=\sqrt R
\overline{Q}=X\overline{Q}\ (mod\ N)

Consider the step 7 now. It is given that if R is sufficiently small, then X can be easily computed by taking its square root. I don't understand this fact. Can someone please explain clearly?

If R is greater than N, it will be reduced (mod N). How do we know (by using so the brute force techniques?) what multiple of N should be added up to R, so as to make it a perfect square?

It is somewhat pointed to remark 3.4
Quote:
One method of calculating g is the following modification of the Newton Raphson recursion: When an initial estimate x_0 > \sqrt{kN} (which can be calculated by using the leading point of kN), successively compute x_{n+1} = \lfloor \frac{(x_n^2+kN)}{2x_n} \rfloor for n \ge 0, where the bracket denotes the greatest integer function. When x_{n+1} - x_n \ge 0, then g=x_{n+1}.
How is the initial estimate x_0 being computed so from \sqrt{kN}? Thus, please explain clearly.

Last fiddled with by Raman on 2009-02-26 at 14:17
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