+ jhRt^RIt^RIHtHtHt^RIHt^RIH t ^RI H t H t H t Ht^RIHtHt^RIHtHtHtHt^RIHt^R IHt^R IHt^R IHt^R IH t H!t!H"t"^R I#H$t$^RI%H&t&H't'^RI(H)t)H*t*^RI+H,t,Rt-Rt.Rt/RRlt0Rt1Rt2R]PfR3Rlt4Rt5RRlt6Rt7.3Rlt8R#) 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 instance%could not determine truth value of %sF)multiple==!=TN><>=<=z'%s' is not a valid relation) isinstancer ValueErroras_expr is_numberrrtrueRealsfalseEmptySetNotImplementedError real_rootsr appendNegativeInfinityInfinityLCreversedinsert)polyreltreals intervalsroot_intervalleftrightsigneq_signequal right_open multiplicitys&& [/home/luca/screen-checking/.venv/lib/python3.14/site-packages/sympy/solvers/inequalities.pysolve_poly_inequalityrHsq, dD ! ! FH H ||~ t||~q# . ;GG9  !''\JJ< %7!;= =69 d{GD+H   X &d _ !!!**a 111HET48H   X &D2X O 779q=DDu #:G CZG D[U D[U;cAB BJJz"*5/ Da?$$8DUJGI,0%yZZ?5$$8DzBD(,d:_$$Q(<=#2 ?   8A..tZH J c b\VUUu.uFp\V!Fq"NK K upp!#uuppi)auSolve polynomial inequalities with rational coefficients. Examples ======== >>> from sympy import Poly >>> from sympy.solvers.inequalities import solve_poly_inequalities >>> from sympy.abc import x >>> solve_poly_inequalities((( ... Poly(x**2 - 3), ">"), ( ... Poly(-x**2 + 1), ">"))) Union(Interval.open(-oo, -sqrt(3)), Interval.open(-1, 1), Interval.open(sqrt(3), oo)) )rrH)polyspss& rGsolve_poly_inequalitiesrNqs0 eGe-BA-F1-F1eG HHGs+ c\PpVEF'pV'gK\\P\P4.pVFwwrEp\ WE,V4p\ VR4p.p \ P!Ws4F=wrV PV 4p V \PJgK,V PV 4K? T p.p VF;p VF p W,p K V \PJgK*V PV 4K= T pV'dKM VFp VPV 4pK EK* V#)aSolve a system of rational inequalities with rational coefficients. Examples ======== >>> from sympy.abc import x >>> from sympy import solve_rational_inequalities, Poly >>> solve_rational_inequalities([[ ... ((Poly(-x + 1), Poly(1, x)), '>='), ... ((Poly(-x + 1), Poly(1, x)), '<=')]]) {1} >>> solve_rational_inequalities([[ ... ((Poly(x), Poly(1, x)), '!='), ... ((Poly(-x + 1), Poly(1, x)), '>=')]]) Union(Interval.open(-oo, 0), Interval.Lopen(0, 1)) See Also ======== solve_poly_inequality r!) rr/r r3r4rH itertoolsproduct intersectr2union)eqsresult_eqsglobal_intervalsnumerdenomr9numer_intervalsdenom_intervalsr<numer_intervalglobal_intervalr?denom_intervals& rGsolve_rational_inequalitiesr_s%.ZZF $Q%7%7DE#' NUC3EKEO3E4@OI3<3D3D#47/)33OD1::-$$X. 47 ) I#3&5N#5O'6#!**4$$_5 $4 ) ##7$(:)H\\(+F)GL MrITc Rp.p\PpVEF`pV'gK.p\PpVEFp \V \4'dV wrM=V P 'd*V P V P, V PrMRp V \PJd#\P\PRrp MVV \PJd#\P\PRrp M V P4P4wr\W3V4wwrp T P$P&'g"T P)4T P)4Rr>> from sympy import Symbol >>> from sympy.solvers.inequalities import reduce_rational_inequalities >>> x = Symbol('x', real=True) >>> reduce_rational_inequalities([[x**2 <= 0]], x) Eq(x, 0) >>> reduce_rational_inequalities([[x + 2 > 0]], x) -2 < x >>> reduce_rational_inequalities([[(x + 2, ">")]], x) -2 < x >>> reduce_rational_inequalities([[x + 2]], x) Eq(x, -2) This function find the non-infinite solution set so if the unknown symbol is declared as extended real rather than real then the result may include finiteness conditions: >>> y = Symbol('y', extended_real=True) >>> reduce_rational_inequalities([[y + 2 > 0]], y) (-2 < y) & (y < oo) Tr!z only polynomials and rational functions are supported in this context. F) relational) rr/r-r(tuple is_Relationallhsrhsrel_opr,ZeroOner.togetheras_numer_denomrrrdomainis_Exactto_exact get_exactis_ZZis_QQrsolve_univariate_inequalityr2r_hasoneevalf as_relational)exprsgenraexactrTsolution_exprsrV_solexprr9rXrYoptrkindr>excludes&&& rGreduce_rational_inequalitiesrs]: E CzzH wwD$&& c%%% $488 3T[[#Cqvv~$%FFAEE4cc$%EE155$cc#}}==?  &=NC')#::&&&&+nn&68H%eZZ))+FLLLFLLL{!$3/3D%PP e^S12GJ  /7 7D14AA&1a!QUU3Z5GaZ4F015G4A3BCG GODad X>>#))#. OA# %j2' (4As"J7#KK7 Kc>aVPRJd\\R44hV3RloRRRR/p.pS!V4FTwrWP49d\ V^V4pM\ V)^W1,4pVP V.V,4KV \ WB4#)a]Reduce an inequality with nested absolute values. Examples ======== >>> from sympy import reduce_abs_inequality, Abs, Symbol >>> x = Symbol('x', real=True) >>> reduce_abs_inequality(Abs(x - 5) - 3, '<', x) (2 < x) & (x < 8) >>> reduce_abs_inequality(Abs(x + 2)*3 - 13, '<', x) (-19/3 < x) & (x < 7/3) See Also ======== reduce_abs_inequalities Fzs Cannot solve inequalities with absolute values containing non-real variables. c 6<a.pVP'gVP'dvVPpVPFWpS !V4pV'gTpK\P !W4UUUUu.uFwwrwrgV!W4WW,3NK pppppKY V#VP 'dWVPoSP'g \R4hVPV3RlS !VP444V#\V\4'dqS !VP^,4pVFOwrVPW\V^4.,34VPV)V\!V^4.,34KQ V#V.3.pV#uuppppi)z'Only Integer Powers are allowed on Abs.c3><"TFwrVS,V3xK R#5iN).0r|condsrs& rG Areduce_abs_inequality.._bottom_up_scan..LsX=Wkd$'5)=Ws)is_Addis_MulfuncargsrPrQis_Powexp is_Integerr)extendbaser(rr2r r) r|rvopargrzr_expr_condsr_bottom_up_scans & @rGr.reduce_abs_inequality.._bottom_up_scan9sa ;;;$+++Byy(-"E&--e<><Db=DRaSXbou~><>E !. [[[A<<< !JKK LLX_TYY=WX X c " "$TYYq\2F%  tbqk]%:;< teUbqk]%:;< &  BZLE #>s3 F r$r#r&r%)is_extended_real TypeErrorrkeysrr2r)r|r9rwmapping inequalitiesrrs&&& @rGreduce_abs_inequalityrs( u$ $  >Ct$GL&t,  lln $tQ,DteQ 5DTFUN+ - ( ::rIc \\VUUu.uFwr#\W#V4NK upp!#uuppi)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)rvrwr|r9s&& rGreduce_abs_inequalitiesrfs:* !ID(37! ""!s( Fc $aaa)^RIHpVP\P4RJd\ \ R44hV\PJd:\SSRVR7PV4pV'dVPS4pV#SpTpSPRJd,\PpV'gV#VPV4#SPf"\RRR7oSPVS/4oRpS\PJdTpEM!S\P Jd\PpEMSP"SP$, p \'V S4p V \P(8Xd\\+V 4p SP-V ^4p V \PJdTpEMQV \P Jd\PpEM+V Ee'\/V SV4p SP0p V R9dYSP-V P2^4'dTpMSP-V P4^4'g\PpM^V R9dXSP-V P4^4'dTpM2SP-V P2^4'g\PpVP4VP2rW, \P6Jd \9^V RR4P;V4pTpVEf3V P=4wppSVP>9d!\AV P>4^8d\Bh\EV SV4pVf\Bh\+T 4o)T)TT3R lp.pT!SS4FpTPK\ETST44K! T'g\MS)ST4pR SP09;'dSP0R8gp\OTPP\STP4TP24, 4p\STT,\UT4,!P\9TP4TP2TP4T9TP2T944p\V;QJdRT4F 'dK RM RM !RT44'd\YTRR7^,pMG\[TR4pTR,'d\ hTR,p\AT4^8d \]T4p\PpS)P_\P`4;p\P(8wEdRp\S4p\cTST4p\eT\84'gKTFBpTT9gK T!T4'gKTP'gK0T\ST4, pKD EM2TP4TP2p!p \YT\ST!4,4FpT!T 4p"T T!8wdT!T4p#\gT T4p$T$T9dT$P'dT!T$4'dT"'dT#'dT\9T T4, pMlT"'d T\8Ph!T T4, pMET#'d T\8Pj!T T4, pMT\8Pl!T T4, pTp K TFp%T\ST%4,pK T\PJd.\C\ RSPGST4: RT: R244hTP;T4p\P.p&TP4p T T9d;T!T 4'd-T Pn'dT&Pq\ST 44TFp'T'p!T!\gT T!44'dT&Pq\9T T!RR44T'T9dTPsT'4MET'T9dTPsT'4T!T'4p(MTp(T('dT&Pq\ST'44T!p K TP2p!T!T9d;T!T!4'd-T!Pn'dT&Pq\ST!44T!\gT T!44'd'T&Pq\8Pl!T T!44T\P(8wdX'dTP;T4pM$\u\wT&!TT4PGST4pV'gV#VPV4# \d\\ R 44hi;i \B\ 3d6\ \ R SPGS\IR 44,44hi;i \d\ hi;i \ d \ R4hi;i \d\PpRpELi;i)aSolves a real univariate inequality. Parameters ========== expr : Relational The target inequality gen : Symbol The variable for which the inequality is solved relational : bool A Relational type output is expected or not domain : Set The domain over which the equation is solved continuous: bool True if expr is known to be continuous over the given domain (and so continuous_domain() does not need to be called on it) Raises ====== NotImplementedError The solution of the inequality cannot be determined due to limitation in :func:`sympy.solvers.solveset.solvify`. Notes ===== Currently, we cannot solve all the inequalities due to limitations in :func:`sympy.solvers.solveset.solvify`. Also, the solution returned for trigonometric inequalities are restricted in its periodic interval. See Also ======== sympy.solvers.solveset.solvify: solver returning solveset solutions with solve's output API Examples ======== >>> 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) denomsFz| Inequalities in the complex domain are not supported. Try the real domain by setting domain=S.Reals)ra continuousNrwT extended_realz When gen is real, the relational has a complex part which leads to an invalid comparison like I < 0. z The inequality, %s, cannot be solved using solve_univariate_inequality. xc<SPS\V44pSPV^4pV\P \P 39dV#VPRJd\P #VP^4pVP'dSPV^4#\RV,4h \d\P pLi;i)Fz!relationship did not evaluate: %s) subsrrrrr.r,rr is_comparabler0)rvr expanded_er|rws& rGvalid*solve_univariate_inequality..validsOOCA7  !QA))H%%.77NAA#yyA.-;a?AA! A sB>>CC=r"c38"TFqPxK R#5ir)r+)rrs& rGr.solve_univariate_inequality..@s-solve_univariate_inequality..Cs Q=O=OrIz'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. )r$r&)r#r%)<sympy.solvers.solversr is_subsetrr-r0rrq intersectionrurr/r xreplacerr,r.rdrerrgrrrrfsupinfr4r rRrj free_symbolslenr)rrr rrsetboundaryrlistallrrsortedcoeff ImaginaryUnitrr(_ptRopenLopenopen is_finiter2removerr)*r|rwrarkrrrv_gen_domaineperiodconstfranger9rrrrsolnsr singularities include_xdiscontinuitiescritical_pointsr;sifted make_realcoeffIcheckim_solazstartend valid_startvalid_zptrMsol_setsr_validrs*ff&&& @rGrqrqs)r-  E)!*."##$ $ qww  ( ce <$*5MEI$**1??;;F"" f5A%a22!"A 5%((qGYGYGY &)A, 6"#&'UUAEEs!')C.(H!IA*/,K$|*/(%(]#%]#:r?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<>> 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) rcVPW4pV\PJdV#VR9dR#V# \d\Pu#i;i)TNTF)rrNaNr)ierMr~rs&&& rGclassify#_solve_inequality..classifysL  AAEEz-'H 55L s&444AANT)as_AddF)linear)r"r!)rrrrer6rdrr4rdegreerr*r0rrrqr,r.rras_independentris_zero is_negative is_positiverfrh_solve_inequalityr r(r2)rrMrrrroor|rLokoooknoorrrebaxefrbeginning_denomscurrent_denomsrcr~s&&& rGrrsT-  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:!N22A!"Q(A=A!R  QUUaZB155)QVV3LL!$ 3 #rA"aff,RA&aff4 a2gQU1q59  LL ;A 0 1v 81B4&!<" 807 8B2&Dqvv~(2""5"@WWQVT*RRC(ERRC(AGG3WWbS1Wd+WWQWd+QVV|"&!&&.agqv&bS1Wb)BT A+sJAL4L4L44RMRM85R7M88DR RRc Na //r2.pVEFpVPVPrvVP\4p\ V4^8XdVP 4o MoVP V,p \ V 4^8Xd9V P 4o VP\\V^V4S 44K\\R44hVPS 4'd%VPS .4PWg34KVPV 3Rl4p V 'd`\;QJdRV 4F 'dK RM RM !RV 44'd&VPS .4PWg34EKwVP\\V^V4S 44EK VP!4U U u.uFwr\#V .V 4NK p p p VP!4U U u.uFwr\%W4NK pp p \'W,V,!#uup p iuup p i)zZ inequality has more than one symbol of interest. c<VPS4;'dDVP;'g0VP;'dVPP'*#r)rr is_Functionrrr)urws&rGr&_reduce_inequalities..sHc DD BB!B!B!%%2B2B.BDrIc3B"TFp\V\4xK R#5ir)r(rrr~s& rGr'_reduce_inequalities..s!Ij*Q"4"4jsFT)rdrfatomsr rpoprr2rrr0r is_polynomial setdefaultfindritemsrrr)rsymbols poly_partabs_partother inequalityr|r9genscommon componentsrwrv poly_reduced abs_reduceds&& ` rG_reduce_inequalitiesr tsbx E" NNJ$5$5c zz&! t9>((*C&&0F6{ajjl .z$3/GMN)*6+   c " "  b ) 0 0$ =$DEJcc!Ij!Iccc!Ij!III##C,33TK@ .z$3/GMN?#BR[Q`Q`QbcQb:30%#>QbLcIQIYZIY:3*56IYKZ +e3 55dZs >H,H!c "\V4'gV.pVUu.uFp\V4NK pp\4P!VUu.uFq"PNK up!p\V4'gV.p\V4;'gTV,p\ ;QJdRV4F 'gK RM RM !RV44'd\ \R44hVUu/uF*q"PeKV\VPRR7bK, ppVUu.uFq"PV4NK ppVUu0uFq"PV4kK pp.pVFp\V\4'dKVPVPP!4VP"P!4, ^4pMVR9d \%V^4pVR8XdKVR8Xd\&P(u#VPP*'d\-RV,4hVP/V4K Tp?\1W4pTPVP34UUu/uFwrxWbK upp4#uupiuupiuupiuupiuupiuuppi)a!Reduce a system of inequalities with rational coefficients. Examples ======== >>> from sympy.abc import x, y >>> from sympy import reduce_inequalities >>> reduce_inequalities(0 <= x + 3, []) (-3 <= x) & (x < oo) >>> reduce_inequalities(0 <= x + y*2 - 1, [x]) (x < oo) & (x >= 1 - 2*y) c3<"TFqPRJxK R#5i)FNrrs& rGr&reduce_inequalities..s 81   &sTFzP inequalities cannot contain symbols that are not real. rrr)rrrrSranyrrrr namerr(rrrdr*rer rr.r+r0r2r r) rrr~rrecastkeeprkrs && rGreduce_inequalitiesr)s L ! !$~ (45 1GAJ L5 5;;>A> ?D G  )7|##tt+G s 8 8sss 8 888 $  5A++3aqvvT22 50<= 1JJv& L=+237azz&!7G3 D  a $ $quu}}8!5=309s)I2I7"I<6I<J7J J )T)F)9__doc__rPsympy.calculus.utilrrr sympy.corersympy.core.exprtoolsrsympy.core.relationalrrr r sympy.core.symbolr r sympy.sets.setsr rrrsympy.core.singletonrsympy.core.functionr$sympy.functions.elementary.complexesr sympy.logicr sympy.polysrrrsympy.polys.polyutilsrsympy.solvers.solvesetrrsympy.utilities.iterablesrrsympy.utilities.miscrrHrNr_rrrr-rqrrr r)rrIrGr:sB-88+DD"*4FF(44+XvI"?DXvD;N"27;177W\d