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r38/packages/poly/exptf.red
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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 n < 0 then errach {"exptf",u,n} else 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;