module exptf; % Functions for raising canonical forms to a power.
% Author: Anthony C. Hearn.
% Copyright (c) 1990 The RAND Corporation. All rights reserved.
fluid '(!*exp);
symbolic procedure exptsq(u,n);
begin scalar x;
if n=1 then return u
else if n=0
then return if null numr u then rerror(poly,4," 0**0 formed")
else 1 ./ 1
else if null numr u then return u
else if n<0 then return simpexpt list(mk!*sq u,n)
else if null !*exp
then return mksfpf(numr u,n) ./ mksfpf(denr u,n)
else if kernp u then return mksq(mvar numr u,n)
else if denr u=1 then return exptf(numr u,n) ./ 1
else if domainp numr u
then x := multsq(!:expt(numr u,n) ./ 1,1 ./ exptf(denr u,n))
else <<x := u;
% Since U is in lowest terms, then so is U^N.
while (n := n-1)>0
do x := multf(numr u,numr x) ./ multf(denr u,denr x);
% We need canonsq for a:=1+x/2; let x^2=0; a^2;
x := canonsq x>>;
if null cdr x then rerror(poly,101,"Zero divisor");
return x
end;
symbolic procedure exptf(u,n);
if domainp u then !:expt(u,n)
else if !*exp or kernlp u then exptf1(u,n)
else mksfpf(u,n);
symbolic procedure exptf1(u,n);
% Iterative multiplication seems to be faster than a binary sub-
% division algorithm, probably because multiplying a small polynomial
% by a large one is cheaper than multiplying two medium sized ones.
if n=0 then 1
else begin scalar x;
x := u; while (n := n-1)>0 do x := multf(u,x); return x
end;
symbolic procedure exptf2(u,n);
% Binary version of EXPTF1, Used with EXP off, since expressions
% formed in that case tend to be smaller than with EXP on.
if n=0 then 1
else begin scalar x; integer m;
x := 1;
a: m := n;
if m-2*(n := n/2) neq 0 then x := multf(u,x);
if n=0 then return x;
u := multf(u,u);
go to a
end;
endmodule;
end;