*hfSrSSKrSSKJrJrJr SSKJr SSKJ r SSK J r J r J r Jr SSKJrJr SSKJrJrJrJr SS KJr SS KJr SS KJr SS KJr SS KJ r J!r!J"r" SSK#J$r$ SSK%J&r&J'r' SSK(J)r)J*r* SSK+J,r, Sr-Sr.Sr/SSjr0Sr1Sr2S\RfS4Sjr4Sr5S Sjr6Sr7/4Sjr8g)!z>> from sympy import solve_poly_inequality, Poly >>> from sympy.abc import x >>> solve_poly_inequality(Poly(x, x, domain='ZZ'), '==') [{0}] >>> solve_poly_inequality(Poly(x**2 - 1, x, domain='ZZ'), '!=') [Interval.open(-oo, -1), Interval.open(-1, 1), Interval.open(1, oo)] >>> solve_poly_inequality(Poly(x**2 - 1, x, domain='ZZ'), '==') [{-1}, {1}] See Also ======== solve_poly_inequalities z8For efficiency reasons, `poly` should be a Poly instancer%could not determine truth value of %sF)multiple==!=T)NF><>=)r$T<=)r%Tz'%s' is not a valid relation) isinstancer ValueErroras_expr is_numberrrtrueRealsfalseEmptySetNotImplementedError real_rootsrappendNegativeInfinityInfinityLCreversedinsert)polyreltreals intervalsroot_intervalleftrightsigneq_signequal right_open multiplicitys W/home/mande/repo/quber/.venv/lib/python3.13/site-packages/sympy/solvers/inequalities.pysolve_poly_inequalityrKsl, dD ! ! FH H ||~ t||~q# . ;GG9  !''\JJ< %7!;= =69 d{GD+H   X &d _ !!!**a 11HET48H   X &D2X O 779q=DD$ #:G CZG D[%NGU D[%NGU;cAB BJJz"*5/ Da?$$8DUJGI,0%yZZ?5$$8DzBD(,d:_$$Q(<=#2 ?   8A..tZH J c b[UVVs/sHn[U6Ho"PM M snn6$s snnf)auSolve polynomial inequalities with rational coefficients. Examples ======== >>> from sympy import Poly >>> from sympy.solvers.inequalities import solve_poly_inequalities >>> from sympy.abc import x >>> solve_poly_inequalities((( ... Poly(x**2 - 3), ">"), ( ... Poly(-x**2 + 1), ">"))) Union(Interval.open(-oo, -sqrt(3)), Interval.open(-1, 1), Interval.open(sqrt(3), oo)) )rrK)polyspss rJsolve_poly_inequalitiesrQqs0 eGe-BA-F1-F1eG HHGs+ cj[RnUGHnU(dM [[R[R5/nUHuupEn[ XE-U5n[ US5n/n [ R"Xs5H<upU RU 5n U [RLdM+U RU 5 M> U n/n UH6n UHn X-n M U [RLdM%U RU 5 M8 U nU(aM O UHn URU 5nM GM U$)aSolve a system of rational inequalities with rational coefficients. Examples ======== >>> from sympy.abc import x >>> from sympy import solve_rational_inequalities, Poly >>> solve_rational_inequalities([[ ... ((Poly(-x + 1), Poly(1, x)), '>='), ... ((Poly(-x + 1), Poly(1, x)), '<=')]]) {1} >>> solve_rational_inequalities([[ ... ((Poly(x), Poly(1, x)), '!='), ... ((Poly(-x + 1), Poly(1, x)), '>=')]]) Union(Interval.open(-oo, 0), Interval.Lopen(0, 1)) See Also ======== solve_poly_inequality r") rr2rr6r7rK itertoolsproduct intersectr5union)eqsresult_eqsglobal_intervalsnumerdenomr<numer_intervalsdenom_intervalsr?numer_intervalglobal_intervalrBdenom_intervals rJsolve_rational_inequalitiesrbs%.ZZF $Q%7%7DE#' NUC3EKEO3E4@OI3<3D3D#47/)33OD1::-$$X. 47 ) I#3&5N#5O'6#!**4$$_5 $4 ) ##7$(:)H\\(+F)GL MrLTc vSn/n[RnUGH=nU(dM /n[RnUGHn [U [5(aU upO8U R (a%U R U R- U RpOSn U [RLa"[R[RSpn OUU [RLa"[R[RSpn O U R5R5up[X4U5uupn U R$R&(d!U R)5U R)5Sp>> from sympy import Symbol >>> from sympy.solvers.inequalities import reduce_rational_inequalities >>> x = Symbol('x', real=True) >>> reduce_rational_inequalities([[x**2 <= 0]], x) Eq(x, 0) >>> reduce_rational_inequalities([[x + 2 > 0]], x) -2 < x >>> reduce_rational_inequalities([[(x + 2, ">")]], x) -2 < x >>> reduce_rational_inequalities([[x + 2]], x) Eq(x, -2) This function find the non-infinite solution set so if the unknown symbol is declared as extended real rather than real then the result may include finiteness conditions: >>> y = Symbol('y', extended_real=True) >>> reduce_rational_inequalities([[y + 2 > 0]], y) (-2 < y) & (y < oo) Tr"z only polynomials and rational functions are supported in this context. Fr) relational) rr2r0r+tuple is_Relationallhsrhsrel_opr/ZeroOner1togetheras_numer_denomrrrdomainis_Exactto_exact get_exactis_ZZis_QQrsolve_univariate_inequalityr5rbhasoneevalf as_relational)exprsgenrdexactrWsolution_exprsrY_solexprr<r[r\optrnindrAexcludes rJreduce_rational_inequalitiesrsS: E CzzH wwD$&& c%% $488 3T[[#Cqvv~$%FFAEE4cc$%EE155$cc#}}==?  &=NC')#::&&&+nn&68H%eZZ))+FLLFLL{!$3/3D%PP e^S12GJ  /7 7D14AA&1a!QUU3Z5GaZ4F015G4A3BCG GODad X>>#))#. OA# %j2' (4AsJ?%J3(J3J0c(^URSLa[[S55eU4SjmSSS.n/nT"U5HKupXR5;a[ USU5nO[ U*SX15nUR U/U-5 MM [ XB5$)a]Reduce an inequality with nested absolute values. Examples ======== >>> from sympy import reduce_abs_inequality, Abs, Symbol >>> x = Symbol('x', real=True) >>> reduce_abs_inequality(Abs(x - 5) - 3, '<', x) (2 < x) & (x < 8) >>> reduce_abs_inequality(Abs(x + 2)*3 - 13, '<', x) (-19/3 < x) & (x < 7/3) See Also ======== reduce_abs_inequalities Fzs Cannot solve inequalities with absolute values containing non-real variables. c  >^/nUR(dUR(aqURnURHSnT "U5nU(dUnM[R "X5VVVVs/sHuupupgU"X5XW-4PM nnnnnMU U$UR (aUURmTR(d [S5eURU4SjT "UR555 U$[U[5(adT "URS5nUHGupURX[US5/-45 URU*U[!US5/-45 MI U$U/4/nU$s snnnnf)Nz'Only Integer Powers are allowed on Abs.c36># UHupUT-U4v M g7fN).0rcondsrs rJ Areduce_abs_inequality.._bottom_up_scan..LsX=Wkd$'5)=Wsr)is_Addis_MulfuncargsrSrTis_Powexp is_Integerr,extendbaser+rr5r r ) rryopargr}r_expr_condsr_bottom_up_scans @rJr.reduce_abs_inequality.._bottom_up_scan9sY ;;$++Byy(-"E&--e<><Db=DRaSXbou~><>E !. [[A<< !JKK LLX_TYY=WX X c " "$TYYq\2F%  tbqk]%:;< teUbqk]%:;< &  BZLE #>s0E> r&r(r'r)r)is_extended_real TypeErrorrkeysrr5r)rr<rzmapping inequalitiesrrs @rJreduce_abs_inequalityrs( u$ $  >t$GL&t,  lln $tQ,DteQ 5DTFUN+ - ( ::rLc \[UVVs/sHup#[X#U5PM snn6$s snnf)aReduce a system of inequalities with nested absolute values. Examples ======== >>> from sympy import reduce_abs_inequalities, Abs, Symbol >>> x = Symbol('x', extended_real=True) >>> reduce_abs_inequalities([(Abs(3*x - 5) - 7, '<'), ... (Abs(x + 25) - 13, '>')], x) (-2/3 < x) & (x < 4) & (((-oo < x) & (x < -38)) | ((-12 < x) & (x < oo))) >>> reduce_abs_inequalities([(Abs(x - 4) + Abs(3*x - 5) - 7, '<')], x) (1/2 < x) & (x < 4) See Also ======== reduce_abs_inequality )rr)ryrzrr<s rJreduce_abs_inequalitiesrfs:* !ID(37! ""!s( Fc ^^^)SSKJn UR[R5SLa[ [ S55eU[RLa5[TTSUS9RU5nU(aURT5nU$TnUnTRSLa*[RnU(dU$URU5$TRc[SSS 9mTRUT05mSnT[RLaUnGOT[R La[RnGOiTR"TR$- n ['U T5n U [R(:XaY[+U 5n TR-U S5n U [RLaUnGODU [R La[RnGOU Gb[/U TU5n TR0n U S ;aVTR-U R2S5(aUnOTR-U R4S5(d[RnO[U S ;aUTR-U R4S5(aUnO1TR-U R2S5(d[RnUR4UR2pX- [R6La[9SU SS5R;U5nUnUGcU R=5unnTUR>;a[AU R>5S :a[Be[EU TU5nUc[Be[+U 5m)U)UU4Sjn/nU"TT5HnURK[EUTU55 M! U(d [MT)TU5nSTR0;=(a TR0S:gn[OURP[SUR4UR25- 5n[SUU-[UU5-6R[9UR4UR2UR4U;UR2U;55n[WSU55(a[YUSS9SnO=[[US5nUS(a[ eUSn[AU5S :a []U5n[RnT)R_[R`5=n[R(:wGaSn[S5n[cUTU5n[eU[85(dCUH;nUU;dM U"U5(dMUR(dM-U[SU5- nM= GOUR4UR2n!n [YU[SU!5-5HnU"U 5n"U U!:waU"U5n#[gU U5n$U$U;aU$R(aU"U$5(a|U"(aU#(aU[9U U5- nO^U"(aU[8Rh"U U5- nO>> from sympy import solve_univariate_inequality, Symbol, sin, Interval, S >>> x = Symbol('x') >>> solve_univariate_inequality(x**2 >= 4, x) ((2 <= x) & (x < oo)) | ((-oo < x) & (x <= -2)) >>> solve_univariate_inequality(x**2 >= 4, x, relational=False) Union(Interval(-oo, -2), Interval(2, oo)) >>> domain = Interval(0, S.Infinity) >>> solve_univariate_inequality(x**2 >= 4, x, False, domain) Interval(2, oo) >>> solve_univariate_inequality(sin(x) > 0, x, relational=False) Interval.open(0, pi) rdenomsFz| Inequalities in the complex domain are not supported. Try the real domain by setting domain=S.Reals)rd continuousNrzT extended_realz When gen is real, the relational has a complex part which leads to an invalid comparison like I < 0. r)r&r(r$z The inequality, %s, cannot be solved using solve_univariate_inequality. xc>TRT[U55nTRUS5nU[R [R 4;aU$URSLa[R $URS5nUR(aTRUS5$[SU-5e![a [R nNf=f)NrFr*z!relationship did not evaluate: %s) subsrrrrr1r/rr is_comparabler3)rvr expanded_errzs rJvalid*solve_univariate_inequality..validsOOCA7  !QA))H%%.77NAA#yyA.-;a?AA! A sB77CC=r#c38# UHoRv M g7fr)r.)rrs rJr.solve_univariate_inequality..@s-solve_univariate_inequality..Cs Q=O=OrLz'sorting of these roots is not supportedz zZ contains imaginary parts which cannot be made 0 for any value of zm satisfying the inequality, leading to relations like I < 0. )<sympy.solvers.solversr is_subsetrr0r3rrt intersectionrxrr2r xreplacerr/r1rgrhrrjrrrrisupinfr7rrUrm free_symbolslenr,rrr rrsetboundaryrlistallrrsortedcoeff ImaginaryUnitrr+_ptRopenLopenopen is_finiter5removerr)*rrzrdrnrrrv_gen_domaineperiodconstfranger<rrrrsolnsr singularities include_xdiscontinuitiescritical_pointsr>sifted make_realcoeffIcheckim_solazstartend valid_startvalid_zptrPsol_setsr_validrs*`` @rJrtrts r-  E)!*."##$ $ qww  ( ce <$*5MEI$**1??;;F"" f5A%a22!"A 5%((qGYGYGY &)A, 6"#&'UUAEEs!')C.(H!IA*/,K$|*/(%(]#%]#:r?R?RW\]_W`W`'2w(.(5!2D(D)4(.(..2J(J)0(.(..2J(J(.(--q2I(I$%E"J"/A"il2F"/ QZZ'$Z!% #t 4d 1<&=>> &//7  |HJJE5<U [ R LaX:*OX:nU [ R La['U*U:U5nGNU[U5nGNcf=f) aReturn the inequality with s isolated on the left, if possible. If the relationship is non-linear, a solution involving And or Or may be returned. False or True are returned if the relationship is never True or always True, respectively. If `linear` is True (default is False) an `s`-dependent expression will be isolated on the left, if possible but it will not be solved for `s` unless the expression is linear in `s`. Furthermore, only "safe" operations which do not change the sense of the relationship are applied: no division by an unsigned value is attempted unless the relationship involves Eq or Ne and no division by a value not known to be nonzero is ever attempted. Examples ======== >>> from sympy import Eq, Symbol >>> from sympy.solvers.inequalities import _solve_inequality as f >>> from sympy.abc import x, y For linear expressions, the symbol can be isolated: >>> f(x - 2 < 0, x) x < 2 >>> f(-x - 6 < x, x) x > -3 Sometimes nonlinear relationships will be False >>> f(x**2 + 4 < 0, x) False Or they may involve more than one region of values: >>> f(x**2 - 4 < 0, x) (-2 < x) & (x < 2) To restrict the solution to a relational, set linear=True and only the x-dependent portion will be isolated on the left: >>> f(x**2 - 4 < 0, x, linear=True) x**2 < 4 Division of only nonzero quantities is allowed, so x cannot be isolated by dividing by y: >>> y.is_nonzero is None # it is unknown whether it is 0 or not True >>> f(x*y < 1, x) x*y < 1 And while an equality (or inequality) still holds after dividing by a non-zero quantity >>> nz = Symbol('nz', nonzero=True) >>> f(Eq(x*nz, 1), x) Eq(x, 1/nz) the sign must be known for other inequalities involving > or <: >>> f(x*nz <= 1, x) nz*x <= 1 >>> p = Symbol('p', positive=True) >>> f(x*p <= 1, x) x <= 1/p When there are denominators in the original expression that are removed by expansion, conditions for them will be returned as part of the result: >>> f(x < x*(2/x - 1), x) (x < 1) & Ne(x, 0) rrcURX5nU[RLaU$US;agU$![a [Rs$f=f)NTF)rrNaNr)ierPrrs rJclassify#_solve_inequality..classifysL  AAEEz-'H 55L s%111A ANr$T)as_AddF)r#r")linear)rrrrhr9rgrr7rdegreerr-r3rrrtr/r1rras_independentris_zero is_negative is_positiverirk_solve_inequalityr r+r5)rrPrrrroorrOokoooknoorrrhbaxefrbeginning_denomscurrent_denomsrcrs rJrrsT-  vv{ [[ vv{q 3 33   B B 66BFF?D M 88:?a(BAHHJN% %0 E z IIK   4 0   !_  5 11 II   &!%& -AA q ==B!!!)B""&&>F266N:!N2A!"Q(A=A!R  QUUaZB155)QVV3LL!$ 3 #rA"aff,RA&aff4 a2gQU1q59  LL ;A 0 1 81B4&!<" 807 8B2&Dqvv~(2""5"@WWQVT*RRC(ERRC(AGG3WWbS1Wd+WWQWd+QVV|"&!&&.agqv&bS1Wb)BT A+sC?K?!K??QL('Q(M=Q?MDQ QQc ^ 00p2/nUGHrnURURpvUR[5n[ U5S:XaUR 5m OjUR U-n [ U 5S:Xa8U R 5m UR[[USU5T 55 M[[S55eURT 5(a$URT /5RXg45 MURU 4Sj5n U (a<[SU 55(a%URT /5RXg45 GMLUR[[USU5T 55 GMu UR!5V V s/sHup[#U /U 5PM n n n UR!5V V s/sHup[%X5PM nn n ['X-U-6$s sn n fs sn n f)Nr$rzZ inequality has more than one symbol of interest. c>URT5=(aA UR=(d. UR=(a URR(+$r)ru is_Functionrrr)urzs rJr&_reduce_inequalities..s<c D B!B!%%2B2B.BDrLc3B# UHn[U[5v M g7fr)r+rrrs rJr'_reduce_inequalities..s!Ij*Q"4"4js)rgriatomsr rpoprr5rrr3r is_polynomial setdefaultfindritemsrrr)rsymbols poly_partabs_partother inequalityrr<genscommon componentsrzry poly_reduced abs_reduceds ` rJ_reduce_inequalitiesr"tsbx E" NNJ$5$5c zz&! t9>((*C&&0F6{ajjl .z$3/GMN)*6+   c " "  b ) 0 0$ =$DEJc!Ij!III##C,33TK@ .z$3/GMN?#BR[Q`Q`QbcQb:30%#>QbLcIQIYZIY:3*56IYKZ +e3 55dZs G)G/c [U5(dU/nUVs/sHn[U5PM nn[5R"UVs/sHo"RPM sn6n[U5(dU/n[U5=(d UU-n[ SU55(a[ [S55eUVs0sH&o"RbMU[URSS9_M( nnUVs/sHo"RU5PM nnUVs1sHo"RU5iM nn/nUHn[U[5(aFURURR!5UR"R!5- S5nOUS;a [%US5nUS:XaMxUS:Xa[&R(s $URR*(a[-SU-5eUR/U5 M UnA[1X5nURUR35VVs0sHupxX_M snn5$s snfs snfs snfs snfs snfs snnf) a!Reduce a system of inequalities with rational coefficients. Examples ======== >>> from sympy.abc import x, y >>> from sympy import reduce_inequalities >>> reduce_inequalities(0 <= x + 3, []) (-3 <= x) & (x < oo) >>> reduce_inequalities(0 <= x + y*2 - 1, [x]) (x < oo) & (x >= 1 - 2*y) c3<# UHoRSLv M g7f)FNrrs rJr&reduce_inequalities..s 81   &szP inequalities cannot contain symbols that are not real. TrrrFr )rrrrVranyrrrr namerr+rrrgr-rhr rr1r.r3r5r"r) rrrrrecastkeeprkrs rJreduce_inequalitiesr+s L ! !$~ (45 1GAJ L5 5;;>A> ?D G  )7|#tt+G 8 888 $  5A++3aqvvT22 50<= 1JJv& L=+237azz&!7G3 D  a $ $quu}}8!5=309s)H<I9I I,I  I&I )T)F)9__doc__rSsympy.calculus.utilrrr sympy.corersympy.core.exprtoolsrsympy.core.relationalrr r r sympy.core.symbolr r sympy.sets.setsrrrrsympy.core.singletonrsympy.core.functionr$sympy.functions.elementary.complexesr sympy.logicr sympy.polysrrrsympy.polys.polyutilsrsympy.solvers.solvesetrrsympy.utilities.iterablesrrsympy.utilities.miscrrKrQrbrrrr0rtrrr"r+rrLrJr<sB-88+DD"*4FF(44+XvI"?DXvD;N"27;177W\d