Prime Numbers

520 Chapter 9 FAST ALGORITHMS FOR LARGE-INTEGER ARITHMETIC
9.6. Use the identity (9.4) to write a program that calculates any product
xy for each of x, y having at most 15 binary bits, using only table lookups,
add/subtracts, shifts, and involving no more than 2 21 bits of table storage.
(Hint: The identity of the text can be used after one computes a certain lookup
table.)
9.7. Modify the binary divide algorithm (9.1.3) so that the value x mod N
is also returned. Note that one could just use equation (9.5), but there is a
way to use the local variables of the algorithm itself, and avoid the multiply
by N.
9.8. Prove that Arazi’s prescription (Algorithm 9.1.4) for simple modular
multiplication indeed returns the value (xy) modN.
9.9. Work out an algorithm similar to Algorithm 9.1.3 for bases B =2 k ,for
k>1. Can this be done without explicit multiplies?
9.10. Prove Theorem 9.2.1. Then prove an extension: that the difference
y/R − (xR −1 )modN is one of {0,N,2N,...,(1 + ⌊x/(RN)⌋)N}.
9.11. Prove Theorem 9.2.4. Then develop and prove a corollary for powering,
of which equation (9.8) would be the special case of cubing.
9.12. In using the Montgomery rules, one has to precompute the residue
N ′ =(−N −1 )modR. In the case that R =2 s and N is odd, show that the
Newton iteration (9.10) with a set at −N, with initial value −N mod 8, and
the iteration thought of as a congruence modulo R, quickly converges to N ′ .
In particular, show how the earlier iterates can be performed modulo smaller
powers of 2, so that the total work involved, assuming naive multiplication and
squaring, can be effected with about 4/3 ofans-bit multiply and about 1/3 of
an s-bit square operation. Since part of each product involved is obliterated
by the mod reduction, show how the work involved can be reduced further.
Contrast this method with a traditional inverse calculation.
9.13. We have indicated that Newton iterations, while efficient, involve
adroit choices of initial values. For the reciprocation of real numbers, equation
(9.10), describe rigorously the range of initial guesses for a given positive real
a, such that the Newton iteration indeed causes x to converge to 1/a.
9.14. We have observed that with Newton iteration one may “divide using
multiplication alone.” It turns out that one may also take square roots in the
same spirit. Consider the coupled Newton iteration
x = y =1;
do {
x = x/2+(1+a)y/2;
y =2y − xy 2 ;
y =2y − xy 2 ;
}