Karim Belabas on Sat, 20 Aug 2011 17:05:24 +0200

 Re: Unable to run script

> Hi,
>
> Had a script that can't be run.
> Keep prompting syntax error, unexpected ')', expecting KPARROW or ',': ....
>
> related to it.
>
> Thus would like to seek some advice on this list.

The first

until((bitsize(p)>=x),

in Keygen() is missing a closing ')' somewhere.

A few random remarks

1) you should indent your scripts so that the structure becomes clearer, e.g.

until((bitsize(p)>=x),
p=lookprime(enlar);
\\q=divisors(p-1)[1];
test=1;
until((isprime(q))&&(bitsize(q)>=y),
if((test>numdiv(p-1)),
break;
);
q=divisors(p-1)[test];
test++;
);

===>

until((bitsize(p)>=x),
p=lookprime(enlar);
\\q=divisors(p-1)[1];
test=1;
until((isprime(q))&&(bitsize(q)>=y),
if((test>numdiv(p-1)),
break;
);
q=divisors(p-1)[test];
test++;
);

2) The numdiv(p-1) / divisors(p-1) are fixed and should be moved out of
the second 'until' loop. In fact, even then, this would be highly
inefficient. Use something like

Q = factor(p-1)[,1]            \\ prime divisors of p-1
if (bitsize(Q[#Q]) >= y, ...)  \\ largest prime divisor of p-1

3) It's immaterial here, but you probably want to start using my() rather than
local() [ see manual for the differences between the 2 ]

4) Sequences of print() statements quickly become unreadable; printf()

print("P= ",p);print(" |p|_2= ",bitsize(p));

==>

printf("P = %d, |p|_2 = %d", p, bitsize(p));

5) Sequences of boolean tests are evaluated from left to right:
expressions like

(isprime(q))&&(bitsize(q)>=y)

are highly inefficient (contains 2 tests, the first one being slow, the
second one trivial). The following is much better:

(bitsize(q)>=y) && (isprime(q))

6) No need to enclose tests within parentheses: it is enough to write

bitsize(q)>=y && isprime(q)

7) Sequences of lift / Mod quickly become unreadable: decide once and for all
whether you want to use t_INTs or t_INTMODs (almost certainly the latter).
Once a sensible base ring is thus fixed, you can use lift() for printing
purposes (only). E.g.

w=lift(1/Mod(s,q));
u1=lift(Mod(m*w,q));
u2=lift(Mod(r*w,q));
v=lift(Mod(lift(Mod((g^u1)*(y1^u2),p)),q));

Since g / y1 are t_INTs, the last line is almost certainly not why you want
[ computing g^u1 requires exponential time in log(p) ! ] Compare:

\\ ASSUME that g and y1 are t_INTMOD mod p, requires modifying Keygen()
s = Mod(s, q) \\ paranoia in case s is a t_INT or t_INTMOD mod p*q
w = 1 / s;
u1 = lift(m*w); \\ t_INT exponent
u2 = lift(r*w);
v = Mod(g^u1 * y1^u2, q);