o 8Va @sddlmZddlmZddlmZddlmZmZm Z ddl m Z ddl m Z ddlmZddlmZdd lmZmZmZdd lmZmZdd lmZdd lmZdd lmZmZddlm Z ddl!m"Z"ddl#m$Z$ddl%m&Z&ddl'm(Z)ddl*m+Z+m,Z,m-Z-m.Z.ddl/m0Z0ddl1m2Z2m3Z3ddl4m5Z5ddl6m7Z7ddl8m9Z9m:Z:ddl;mZ>ddl?m@Z@mAZAddlBmCZCddgZDGdd d eEZFGd!d"d"ZGGd#d$d$eGZHGd%d&d&eGZIGd'd(d(eGZJGd)d*d*eGZKGd+d,d,eGZLGd-d.d.eGZMGd/d0d0eGZNGd1d2d2eGZOGd3d4d4eGZPGd5d6d6eGZQGd7d8d8eGZRGd9d:d:eGZSeIeHeJeKeLeMeNeOePeQeReSg ZTd;d<eTDZUd=d>ZVd?d@ZWdAdBZXdCdDZYdEdFZZdGdHZ[dIdJZ\edKdLdMdNdOfdPdZ]dQdRZ^ddSdTZ_edKdLdMfdUdVZ`ddWdZadXdYbeceUea_dedKdLdMfdZd[Zedd\d]Zfd^d_Zgd`daZhedKdLdMfdbdcZidddeZjedKdLdMfdfdgZkdhdiZldjdkZmdldmZnedKdLdMfdndoZodpdqZpdrdsZqdtduZrdvdwZsdxdyZtdzd{Zud|d}Zvdd~dZwddZxddZyddZzddZ{ddZ|dddZ}ddZ~dddZddZddZddZddZddZddZddZddZeddLdMfddZdddZdddZdddZddZddZddZdddZeZddZdddZddZdNS))Add)check_assumptions)Tuple)as_int is_sequenceordered) factor_terms)_mexpand)Mul)Rational)igcdexilcmigcd)integer_nthrootisqrt)Eq)S)Symbolsymbols)_sympifysign)floor)sqrt)MutableDenseMatrix)divisors factorint multiplicity perfect_power) nextprime) is_squareisprime)sqrt_mod)GeneratorsNeeded)Poly factor_list)signsimp) solveset_real)default_sort_keynumbered_symbols) filldedent diophantine classify_diopcsLeZdZdZfddZfddZddZdd Zd d Zd d Z Z S)DiophantineSolutionSeta Container for a set of solutions to a particular diophantine equation. The base representation is a set of tuples representing each of the solutions. Parameters ========== symbols : list List of free symbols in the original equation. parameters: list List of parameters to be used in the solution. Examples ======== Adding solutions: >>> from sympy.solvers.diophantine.diophantine import DiophantineSolutionSet >>> from sympy.abc import x, y, t, u >>> s1 = DiophantineSolutionSet([x, y], [t, u]) >>> s1 set() >>> s1.add((2, 3)) >>> s1.add((-1, u)) >>> s1 {(-1, u), (2, 3)} >>> s2 = DiophantineSolutionSet([x, y], [t, u]) >>> s2.add((3, 4)) >>> s1.update(*s2) >>> s1 {(-1, u), (2, 3), (3, 4)} Conversion of solutions into dicts: >>> list(s1.dict_iterator()) [{x: -1, y: u}, {x: 2, y: 3}, {x: 3, y: 4}] Substituting values: >>> s3 = DiophantineSolutionSet([x, y], [t, u]) >>> s3.add((t**2, t + u)) >>> s3 {(t**2, t + u)} >>> s3.subs({t: 2, u: 3}) {(4, 5)} >>> s3.subs(t, -1) {(1, u - 1)} >>> s3.subs(t, 3) {(9, u + 3)} Evaluation at specific values. Positional arguments are given in the same order as the parameters: >>> s3(-2, 3) {(4, 1)} >>> s3(5) {(25, u + 5)} >>> s3(None, 2) {(t**2, t + 2)} csBtt|s tdt|stdt||_t||_dS)Nz$Symbols must be given as a sequence.z'Parameters must be given as a sequence.)super__init__r ValueErrortupler parameters)selfZ symbols_seqr2 __class__G/usr/lib/python3/dist-packages/sympy/solvers/diophantine/diophantine.pyr/bs  zDiophantineSolutionSet.__init__cs@t|t|jkrtdt|jt|ftt|dS)Nz+Solution should have a length of %s, not %s)lenrr0r.addrr3solutionr4r6r7r9nszDiophantineSolutionSet.addcGs|D]}||qdSN)r9)r3 solutionsr;r6r6r7updatess zDiophantineSolutionSet.updateccs&t|D] }tt|j|VqdSr<)rdictziprr:r6r6r7 dict_iteratorws z$DiophantineSolutionSet.dict_iteratorcOs2t|j|j}|D] }||j|i|q |Sr<)r-rr2r9subs)r3argskwargsresultr;r6r6r7rB{szDiophantineSolutionSet.subscGsBt|t|jkrtdt|jt|f|tt|j|S)Nz0Evaluation should have at most %s values, not %s)r8r2r0rBlistr@)r3rCr6r6r7__call__szDiophantineSolutionSet.__call__) __name__ __module__ __qualname____doc__r/r9r>rArBrG __classcell__r6r6r4r7r-$s = r-c@sXeZdZdZdZdddZddZeddZed d Z dd e fd d Z dddZ dS)DiophantineEquationTypeaf Internal representation of a particular diophantine equation type. Parameters ========== equation : The diophantine equation that is being solved. free_symbols : list (optional) The symbols being solved for. Attributes ========== total_degree : The maximum of the degrees of all terms in the equation homogeneous : Does the equation contain a term of degree 0 homogeneous_order : Does the equation contain any coefficient that is in the symbols being solved for dimension : The number of symbols being solved for NcCst|jdd|_|dur||_nt|jj|_|jjtd|js&td|j|_ t dd|j Dsz3DiophantineEquationType.__init__..zCoefficients should be Integers)rexpandequation free_symbolsrFsortr(r0as_coefficients_dictcoeffallvalues TypeErrorr$ total_degree homogeneoussethomogeneous_orderr8 dimension _parameters)r3rZr[r6r6r7r/s    z DiophantineEquationType.__init__cCsdS)zf Determine whether the given equation can be matched to the particular equation type. Fr6r3r6r6r7matchesszDiophantineEquationType.matchescCs|jSr<rfrhr6r6r7 n_parameterssz$DiophantineEquationType.n_parameterscCs&|jdurtd|jfdd|_|jS)Nzt_:%iTinteger)rgrrkrhr6r6r7r2s z"DiophantineEquationType.parametersreturncCstd|j)N"No solver has been written for %s.)NotImplementedErrorname)r3r2limitr6r6r7solveszDiophantineEquationType.solvecCsL|s td|j|dur!t||jkr!td|jt|f||_dS)Nz2This equation does not match the %s equation type.z#Expected %s parameter(s) but got %s)rir0rqr8rkrg)r3r2r6r6r7 pre_solves  z!DiophantineEquationType.pre_solver<NN) rHrIrJrKrqr/ripropertyrkr2r-rsrtr6r6r6r7rMs   rMc@&eZdZdZdZddZdddZdS) Univariatea Representation of a univariate diophantine equation. A univariate diophantine equation is an equation of the form `a_{0} + a_{1}x + a_{2}x^2 + .. + a_{n}x^n = 0` where `a_{1}, a_{2}, ..a_{n}` are integer constants and `x` is an integer variable. Examples ======== >>> from sympy.solvers.diophantine.diophantine import Univariate >>> from sympy.abc import x >>> Univariate((x - 2)*(x - 3)**2).solve() # solves equation (x - 2)*(x - 3)**2 == 0 {(2,), (3,)} Z univariatecC |jdkSNrXrjrhr6r6r7ri zUnivariate.matchesNcCsJ||t|j|jd}t|j|jdtjD]}| |fq|S)Nr2r) rtr-r[r2r'rZ intersectrIntegersr9)r3r2rrrEir6r6r7rss zUnivariate.solverurHrIrJrKrqrirsr6r6r6r7rxs rxc@rw) Lineara Representation of a linear diophantine equation. A linear diophantine equation is an equation of the form `a_{1}x_{1} + a_{2}x_{2} + .. + a_{n}x_{n} = 0` where `a_{1}, a_{2}, ..a_{n}` are integer constants and `x_{1}, x_{2}, ..x_{n}` are integer variables. Examples ======== >>> from sympy.solvers.diophantine.diophantine import Linear >>> from sympy.abc import x, y, z >>> l1 = Linear(2*x - 3*y - 5) >>> l1.matches() # is this equation linear True >>> l1.solve() # solves equation 2*x - 3*y - 5 == 0 {(3*t_0 - 5, 2*t_0 - 5)} Here x = -3*t_0 - 5 and y = -2*t_0 - 5 >>> Linear(2*x - 3*y - 4*z -3).solve() {(t_0, 2*t_0 + 4*t_1 + 3, -t_0 - 3*t_1 - 3)} ZlinearcCryrz)rbrhr6r6r7rir{zLinear.matchesNcs|||j|j}dvrd }nd}t||jd}|j}t|dkr>t||d\}}|s<||f|S|S fdd|D} g} t|dkr| t | d| d| d| d| d<| d| d| d<t t| d ddD]!} t | d| | } | d| | d<| | | | | <| d| qz| | d g} t t| D]} gg}}t t |D]n\}}|jr|tj}}|d}n|\}}|||d}t|| | | | |}\}}|tjurd|vr|Sn$t|t r |jd||jd}t|t r|jd||jd}||||q| t |t |}q| ||| |S) NrXrr|csg|]}|qSr6r6)rTvr^r6r7 Vz Linear.solve..)rtr^r[r-r2r8divmodr9appendrrangeinsert enumeraterZ make_args is_IntegerrOne as_coeff_Mulindexbase_solution_linear isinstancerC)r3r2rrvarrUrEparamsqrABrZgcdr=Ztot_xZtot_yjargkpZpnewsolZsol_xZsol_yr6rr7rssj    * 4             z Linear.solverurr6r6r6r7rs rc@,eZdZdZdZddZd defddZdS) BinaryQuadratica Representation of a binary quadratic diophantine equation. A binary quadratic diophantine equation is an equation of the form `Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0`, where `A, B, C, D, E, F` are integer constants and `x` and `y` are integer variables. Examples ======== >>> from sympy.abc import x, y >>> from sympy.solvers.diophantine.diophantine import BinaryQuadratic >>> b1 = BinaryQuadratic(x**3 + y**2 + 1) >>> b1.matches() False >>> b2 = BinaryQuadratic(x**2 + y**2 + 2*x + 2*y + 2) >>> b2.matches() True >>> b2.solve() {(-1, -1)} References ========== .. [1] Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, [online], Available: http://www.alpertron.com.ar/METHODS.HTM .. [2] Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online], Available: https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf Zbinary_quadraticcCs|jdko |jdkSNrrbrfrhr6r6r7rizBinaryQuadratic.matchesNrnc9 s|||j}|j}|\}}||d}|||}||d} |||||tjddt||| D\}}} t||j} | j\} |dd|| } |dkr| dkr|dkr׈|dkrt|\} }|s~| | ft|\} }|s| | f| St |}|dd|D}|D].}t||\}}|st||\} }|st| |\}}|s| ||fq| S| dkr|dkrt |j ||gdj | gd}|D] }| |d |dfq| St|t|| |}| }t||t|t|sntd d d }|d|}t||tj}|D]}t|||}| |d|d fqO| St|ṙfd d}fdd}tdtD]:}t|d|rt|d|r| ||||fq| St| r|dkrt| }tdd d\} }td||| |d||||| |||| d|d|| |d||d|}t|} | D]n\}!}"||"||!||"||!}#t|#d||}$t|!|"d|}%t|!ts[t|"tr}tt |$|%d|||dkr{t |$|%d|||}| j!|q'|$j"r|%j"rt#|||$|%r| |$|%fq'| St |j |ddddj | gd}|r| |$ddd|s| St%||\}&}'t&||\}(t'|(})dkr|)D]8\}}| |fD]-}| |fD]$}|&t(||g|'}z | dd|DWqt)y Yqwqq| St*|)})t+|)D]\}*}+|) |* |+ fqt'd }|dd},|dd }-t,dd|&dd|'ddDr|)D]P\}}||t|,|-t}.||t|,|-t}/tt|.|/d}0tt|.|/dt}1|&t(|0|1g|'}| |qQ| St-dd|&dd|'ddD}2d }3|,}4|-}5|4d |2dks|5|2dkr|4|,|5|-|4|-|5|,}4}5|3d 7}3|4d |2dks|5|2dks|)D]y\}*}+t|3D]o}6t,dd|&t(|*|+g|'DrY|*t|+|4t|5}.|*t|+|4t|5}/t|.|/d}7t|.|/dt}8|&t(|7|8g|'}| ||*|,|-|+|*|-|+|,}*}+qq| S)NrcSg|]}t|qSr6rrTrr6r6r7rrz)BinaryQuadratic.solve..rcSsg|]}| qSr6r6)rTtermr6r6r7rr[r|rXzT)realcsb dd||d|Srr6u)EF_cegsqctr6r7/s4*z'BinaryQuadratic.solve..csPdd||d|Srr6r)Drrrsqarr6r7r2s*"zu, vrlrcSrr6rrT_r6r6r7rircsrQr<rRrr6r6r7rVwrWz(BinaryQuadratic.solve..cSsg|]}|jqSr6)rrr6r6r7rrcsrQr<rRrr6r6r7rVrW).rtr[r^rr _remove_gcdr-r2rr9rrrZrsrrrrr'r}r~ diop_solverSrabs divisibler rr rrr8 check_paramr>ris_solution_quadpop_transformation_to_DN_find_DNdiop_DNMatrixr0rdrFr_r )9r3r2rrrr^xyrrCrErZdiscrrrdivdx0y0sZsolnarUreqrootsrootZansZsolve_xZsolve_yZz0rr;Zs0Zt0Znumx_0y_0PQNZ solns_pellXYTUZ_aZ_bZx_nZy_nLrZT_kZU_krZXtZYtr6) rrrrrrrrrr7rss0     (   ~   m (,` >     E$A   4   ( $$& &  $$$ * zBinaryQuadratic.solverurHrIrJrKrqrir-rsr6r6r6r7rs rc@eZdZdZdZddZdS)InhomogeneousTernaryQuadraticz Representation of an inhomogeneous ternary quadratic. No solver is currently implemented for this equation type. Zinhomogeneous_ternary_quadraticcCs*|jdkr |jdks dS|jsdS|j S)NrrF)rbrfrcrerhr6r6r7ris z%InhomogeneousTernaryQuadratic.matchesNrHrIrJrKrqrir6r6r6r7r rc@r) !HomogeneousTernaryQuadraticNormalaQ Representation of a homogeneous ternary quadratic normal diophantine equation. Examples ======== >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine.diophantine import HomogeneousTernaryQuadraticNormal >>> HomogeneousTernaryQuadraticNormal(4*x**2 - 5*y**2 + z**2).solve() {(1, 2, 4)} $homogeneous_ternary_quadratic_normalcsdjdkr jdks dSjsdSjsdSfddjDtdko1tfddjDS)NrrFcg|] }j|r|qSr6rrTrrhr6r7rz=HomogeneousTernaryQuadraticNormal.matches..c3|] }|dvVqdSrNr6rnonzeror6r7rVz.rbrfrcrer^r8r_r[rhr6rr3r7ris$z)HomogeneousTernaryQuadraticNormal.matchesNrncCs,|||j}|j}|\}}}||d}||d} ||d} t|| | dd\\} } } \}}}\}}}| |}| |}t||jd}|dkrQ|dkrQ|St| ||dusot| ||dusot| ||durq|St||\}}}t|t |\}}||9}||9}t |||\}}}t |t | krt |||t |t |t |\}}}n-t |t | krt |||t |t |t |\}}}nt |||t |t |t |\}}}t |||}t |||}t |||}t| | | }t ||| }t ||| }t ||| }|t ||||S)NrT)stepsr|r)rtr[r^ sqf_normalr-r2r"descent _rational_pqrrrholzer reconstructr r9)r3r2rrrr^rrrrbrUZsqf_of_aZsqf_of_bZsqf_of_cZa_1Zb_1Zc_1Za_2Zb_2Zc_2rrrEz_0rrrZsq_lcmr6r6r7rssP       &&$    z'HomogeneousTernaryQuadraticNormal.solverurr6r6r6r7rs   rc@rw) HomogeneousTernaryQuadratica Representation of a homogeneous ternary quadratic diophantine equation. Examples ======== >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine.diophantine import HomogeneousTernaryQuadratic >>> HomogeneousTernaryQuadratic(x**2 + y**2 - 3*z**2 + x*y).solve() {(-1, 2, 1)} >>> HomogeneousTernaryQuadratic(3*x**2 + y**2 - 3*z**2 + 5*x*y + y*z).solve() {(3, 12, 13)} homogeneous_ternary_quadraticcsfjdkr jdks dSjsdSjsdSfddjDtdko1tfddjD S)NrrFcrr6rrrhr6r7r rz7HomogeneousTernaryQuadratic.matches..c3rrr6rrr6r7rV!rz6HomogeneousTernaryQuadratic.matches..rrhr6rr7ris&z#HomogeneousTernaryQuadratic.matchesNcs,|||j}|j|\}}}|||g}t||jd}dd} tfdd|Ds||rt||||||||} | } t| dt| d} | D]} t| dt| d}|| kro| } |} qY| t | d|| | d|S|d|d|d<|d<| t |\}}}|dur| |||f|S|ddkr|ddkr|d|d|d<|d<| t |\}}}n2|d|d|d<|d<| t |\}}}n||s||r|d}||}||}|d}||}|d}t }d |d||d<d |||d||d<d |||d||d<d ||d|||||<d|||<d|||<| t ||\}}}|durr|St ||||d|\}}|||||||}}}nr||dkr|ddkr|ddkrʈ|d}||}t | |\}}|||}}}n:|d|d|d<|d<| t |\}}}n"|d|d|d<|d<| t |\}}}n | t|\}}}|dur |S| t ||||S) Nr|cSst|dkr t|dSdS)NrNNN)r8rFrr6r6r7 unpack_sol7s  z5HomogeneousTernaryQuadratic.solve..unpack_solc3|] }|dVqdSrr6rrr6r7rV<rz4HomogeneousTernaryQuadratic.solve..rrXrr)rtr[r^r-r2anyr+rrr9r_diop_ternary_quadraticr?r_diop_ternary_quadratic_normal)r3r2rr_varrrrrrErsolsrZmin_sumrmrrrrrrrrr_coeffrrrrr6rr7rs#s     ,$          "   z!HomogeneousTernaryQuadratic.solverurr6r6r6r7rs  rc@r)InhomogeneousGeneralQuadraticz Representation of an inhomogeneous general quadratic. No solver is currently implemented for this equation type. Zinhomogeneous_general_quadraticcCsB|jdkr |jdks dS|jsdStdd|jDr|j SdS)NrrFTcs|]}|jVqdSr<Zis_Mulrr6r6r7rVz8InhomogeneousGeneralQuadratic.matches..rbrfrerr^rcrhr6r6r7risz%InhomogeneousGeneralQuadratic.matchesNrr6r6r6r7rrrc@r)HomogeneousGeneralQuadraticz~ Representation of a homogeneous general quadratic. No solver is currently implemented for this equation type. Zhomogeneous_general_quadraticcCs@|jdkr |jdks dS|jsdStdd|jDr|jSdS)NrrFcsrr<rrr6r6r7rVrz6HomogeneousGeneralQuadratic.matches..rrhr6r6r7risz#HomogeneousGeneralQuadratic.matchesNrr6r6r6r7rrrc@&eZdZdZdZddZd ddZdS) GeneralSumOfSquaresa Representation of the diophantine equation `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`. Details ======= When `n = 3` if `k = 4^a(8m + 7)` for some `a, m \in Z` then there will be no solutions. Refer [1]_ for more details. Examples ======== >>> from sympy.solvers.diophantine.diophantine import GeneralSumOfSquares >>> from sympy.abc import a, b, c, d, e >>> GeneralSumOfSquares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345).solve() {(15, 22, 22, 24, 24)} By default only 1 solution is returned. Use the `limit` keyword for more: >>> sorted(GeneralSumOfSquares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345).solve(limit=3)) [(15, 22, 22, 24, 24), (16, 19, 24, 24, 24), (16, 20, 22, 23, 26)] References ========== .. [1] Representing an integer as a sum of three squares, [online], Available: http://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares general_sum_of_squarescsRjdkr jdks dSjsdStddjDrdStfddjDS)NrrFcsrr<rrr6r6r7rVrz.GeneralSumOfSquares.matches..c3&|]}|dkrj|dkVqdSrXNrrrhr6r7rV$)rbrfrerr^r_rhr6rhr7riszGeneralSumOfSquares.matchesNrXc s|||j}t|jd }|j}t||jd}|dks"|dkr$|Sdd|Dddk}d}t||ddD]#} |rN| fd dt | Dn| | |d7}||kr^|Sq;|S) NrXr|rcSg|] }|jr dndqSrrXZis_nonpositiverTrr6r6r7rrz-GeneralSumOfSquares.solve..rTzeroscg|] \}}||qSr6r6rTrrZsignsr6r7r) rtr[intr^rfr-r2countsum_of_squaresr9r) r3r2rrrrnrEnegstookrr6rr7rss(  zGeneralSumOfSquares.solverzrr6r6r6r7r s   r c@s2eZdZdZdZddZeddZd d d ZdS) GeneralPythagoreana Representation of the general pythagorean equation, `a_{1}^2x_{1}^2 + a_{2}^2x_{2}^2 + . . . + a_{n}^2x_{n}^2 - a_{n + 1}^2x_{n + 1}^2 = 0`. Examples ======== >>> from sympy.solvers.diophantine.diophantine import GeneralPythagorean >>> from sympy.abc import a, b, c, d, e, x, y, z, t >>> GeneralPythagorean(a**2 + b**2 + c**2 - d**2).solve() {(t_0**2 + t_1**2 - t_2**2, 2*t_0*t_2, 2*t_1*t_2, t_0**2 + t_1**2 + t_2**2)} >>> GeneralPythagorean(9*a**2 - 4*b**2 + 16*c**2 + 25*d**2 + e**2).solve(parameters=[x, y, z, t]) {(-10*t**2 + 10*x**2 + 10*y**2 + 10*z**2, 15*t**2 + 15*x**2 + 15*y**2 + 15*z**2, 15*t*x, 12*t*y, 60*t*z)} Zgeneral_pythagoreancsjdkr jdks dSjsdStddjDrdStfddjDr+dStfddjDs9dSttfddjDjdkS) NrrFcsrr<rrr6r6r7rV#rz-GeneralPythagorean.matches..c3r r rrrhr6r7rV%rc3s"|] }ttj|VqdSr<)r rr^rrhr6r7rV's c3s|] }tj|VqdSr<)rr^rrhr6r7rV+s)rbrfrerr^r_rsumrhr6rhr7ris&zGeneralPythagorean.matchescCs |jdSrzrjrhr6r6r7rk-s zGeneralPythagorean.n_parametersNrXcs|||j}|j}|jt||ddt||ddt||dddkr<|D] }|| ||<q2t||jd}d}t|D]\}} t|| ddkrY|}qI|jt ddD} | dddg} | fdd t dD| d|| g| |d} d} t|D];\}} ||ks|dkr|dks|dkr|dkrt | t t|| d} qt || d}t | t|r|n|d} qt|D]\}} | | |t t|| d| |<q|| |S) NrrrXr|rcss|]}|dVqdSrr6)rTZm_ir6r6r7rVFrWz+GeneralPythagorean.solve..cs$g|]}d|dqSrr6rrrr6r7rH$z,GeneralPythagorean.solve..)rtr^r[rfrkeysr-r2rr extendrr rr_oddr9)r3r2rrr^rrPrErrrZithrrZlcmrr6r"r7rs1s8 @ "(& zGeneralPythagorean.solverz) rHrIrJrKrqrirvrkrsr6r6r6r7r s rc@r) CubicThuea Representation of a cubic Thue diophantine equation. A cubic Thue diophantine equation is a polynomial of the form `f(x, y) = r` of degree 3, where `x` and `y` are integers and `r` is a rational number. No solver is currently implemented for this equation type. Examples ======== >>> from sympy.abc import x, y >>> from sympy.solvers.diophantine.diophantine import CubicThue >>> c1 = CubicThue(x**3 + y**2 + 1) >>> c1.matches() True Z cubic_thuecCs|jdko |jdkS)Nrrrrhr6r6r7riqrzCubicThue.matchesNrr6r6r6r7r'Zs r'c@r ) GeneralSumOfEvenPowersas Representation of the diophantine equation `x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0` where `e` is an even, integer power. Examples ======== >>> from sympy.solvers.diophantine.diophantine import GeneralSumOfEvenPowers >>> from sympy.abc import a, b >>> GeneralSumOfEvenPowers(a**4 + b**4 - (2**4 + 3**4)).solve() {(2, 3)} Zgeneral_sum_of_even_powerscsTjdksdSjddkrdStfddjDsdStfddjDS)NrFrrc3s*|]}|dkr|jo|jjkVqdSr )is_Powexprbrrhr6r7rVs(z1GeneralSumOfEvenPowers.matches..c3r r rrrhr6r7rVr)rbr_r^rhr6rhr7ris zGeneralSumOfEvenPowers.matchesNrXc s|||j}|j}d}|D] }|jr||r|j}qt|}|d }t||jd} |dks6|dkr8| Sdd|D ddk} d} t |||D]#} | ra| fddt | Dn| | | d7} | |krq| SqN| S)NrXr|rcSrrrrr6r6r7rrz0GeneralSumOfEvenPowers.solve..rcrr6r6rrr6r7rr) rtr[r^r$r)r*r8r-r2rpower_representationr9r) r3r2rrrr^rrrrrErrrr6rr7rss4    zGeneralSumOfEvenPowers.solverzrr6r6r6r7r(us  r(cCsh|]}|jqSr6)rq)rT diop_classr6r6r7 rr-cCs$zt|WdStyYdSw)NT)rr0rr6r6r7rSs  rScGs tt|Sr<)r1sortedr.r6r6r7 _sorted_tuple r0cszt|Wn-ty)ttd|}t|dkr|YStdd|DYn ty3tdwdkr:|Stfdd|DS)NrcSsg|]}|dqSr)as_content_primitiverr6r6r7rz_remove_gcd..z-_remove_gcd(a,b,c) or _remove_gcd(*container)rXcsg|]}|qSr6r6rrr6r7rr)rr0rFfilterr8rar1)rZfxr6r5r7rs    rcCstt||t|Sr<)rrrrrr6r6r7rsrcCs.t||\}}t|t|dkr|S|dS)NrrX)rr)rrwrr6r6r7_nint_or_floorsr9cCs |ddkSNrrr6r.r6r6r7r&r1r&cCs |ddkSr:r6r.r6r6r7_evenr1r;rTrlNFc) sddlm}m}m}t|}t|tr|j|j}zt |j ddj j t d|r[t|s3tdfdd|D}|kr[tt|tt|fd d t|||d DWS|\}|jrhtWSjst}t||} fd d | DWS|}t|}|jrJ|jd did}t|} tdd| jDrJ| }|sJWntt fytt!dwd} d} d} zt"|\}}}|rt|}t#j$t%j$g}t&j$t'j$t(j$g}||vrd} n||vr|dkrdt ||d}d}d}t)dd|}|D]}z||}Wn t*yd}Ynwt+|o&t+|}q t ||d}|D]!}z||d}Wn t*yHd}Ynwt+|oQt+|}q2t||gs^d} ny|scd} ns|dkrt ||d}d}d}t)dd|}|D]}z||}Wn t*yd}Ynwt+|ot+|}q}t ||d}|D]!}z||d}Wn t*yd}Ynwt+|ot+|}qt||gsd} n|sd} |dkr|dfg}ntWn/tt,fyt-|}|dj.r|ddkrt||d|||dYS|d}Ynwt}|D]Y}|\}} t"|dd\}!} }"t/|dd0\} }t1||}#|"t2j$t&j$t'j$t3j$fvrP|4t5|!|#q|"t(j$t#j$t%j$t6j$fvrn|#D] |4t5|!q_qt,d|"d|vr~|7dt8dgt}$|s|9t|$j:r|4|$t}%|D]Xt;ddDr| rt|}&|%<|&q| rt |}'t t=fdd|'}'t|'}&|%<|&q| rt|}(|%<|(q|%4q|%4q|%S) a; Simplify the solution procedure of diophantine equation ``eq`` by converting it into a product of terms which should equal zero. Explanation =========== For example, when solving, `x^2 - y^2 = 0` this is treated as `(x + y)(x - y) = 0` and `x + y = 0` and `x - y = 0` are solved independently and combined. Each term is solved by calling ``diop_solve()``. (Although it is possible to call ``diop_solve()`` directly, one must be careful to pass an equation in the correct form and to interpret the output correctly; ``diophantine()`` is the public-facing function to use in general.) Output of ``diophantine()`` is a set of tuples. The elements of the tuple are the solutions for each variable in the equation and are arranged according to the alphabetic ordering of the variables. e.g. For an equation with two variables, `a` and `b`, the first element of the tuple is the solution for `a` and the second for `b`. Usage ===== ``diophantine(eq, t, syms)``: Solve the diophantine equation ``eq``. ``t`` is the optional parameter to be used by ``diop_solve()``. ``syms`` is an optional list of symbols which determines the order of the elements in the returned tuple. By default, only the base solution is returned. If ``permute`` is set to True then permutations of the base solution and/or permutations of the signs of the values will be returned when applicable. Examples ======== >>> from sympy.solvers.diophantine import diophantine >>> from sympy.abc import a, b >>> eq = a**4 + b**4 - (2**4 + 3**4) >>> diophantine(eq) {(2, 3)} >>> diophantine(eq, permute=True) {(-3, -2), (-3, 2), (-2, -3), (-2, 3), (2, -3), (2, 3), (3, -2), (3, 2)} Details ======= ``eq`` should be an expression which is assumed to be zero. ``t`` is the parameter to be used in the solution. Examples ======== >>> from sympy.abc import x, y, z >>> diophantine(x**2 - y**2) {(t_0, -t_0), (t_0, t_0)} >>> diophantine(x*(2*x + 3*y - z)) {(0, n1, n2), (t_0, t_1, 2*t_0 + 3*t_1)} >>> diophantine(x**2 + 3*x*y + 4*x) {(0, n1), (3*t_0 - 4, -t_0)} See Also ======== diop_solve() sympy.utilities.iterables.permute_signs sympy.utilities.iterables.signed_permutations r)subsets permute_signssigned_permutationsTrNrOz/syms should be given as a sequence, e.g. a listcsg|]}|vr|qSr6r6r)rr6r7rOr4zdiophantine..cs$h|]tfddDqS)csg|]}|qSr6r6r)dict_sym_indexrr6r7rRr4z)diophantine...)r1)rT)r?r)rr7r-Rszdiophantine..)permutecs$h|]}tt|r|qSr6)r rBr@rTr)rrr6r7r-Zr#as_AddFrXcsrr<) is_number)rTrr6r6r7rVarzdiophantine..z@ Equation should be a polynomial with Rational coefficients.rrcS|d|dSNrrXr6)rr6r6r7rzdiophantine..cSrDrEr6rr6r6r7rrFr )paramsymsr@_dict)Zevaluatezunhandled type: %sr6csrQr<rRrAr6r6r7rVrWcs |d|dddkSrEr6rGrr6r7r )>sympy.utilities.iterablesr<r=r>rrrZlhsZrhsrFrYr[r\r(rrar?r@rr8r+Zas_numer_denomrCrdras_independentr$rZgensZas_exprZ is_polynomialr#AssertionErrorr*r,r rqr(rrrmapKeyErrorboolrpr%Z is_Rationalr&rrrrr9merge_solutionrxremover1rBZis_zeror_r>r6))rrHrIr@r<r=r>rZdsolZgoodrZdo_permute_signsZdo_permute_signs_varZpermute_few_signsrrUrZlen_varZpermute_signs_forZpermute_signs_checkZvar_mulZxy_coeffZx_coeffZ var1_mul_var2Z v1_mul_v2r^Zv1termsZflrrbaservar_teq_typer;ZnullZ final_solnZ permuted_signZlstZpermuted_sign_varr6)rr?rrr7r+s:I                         cCsg}d|vrdSt|}tdddd}|D]}||vr#|t|q|t|qt||D]\}}t|fi|jdurDtSq0t|S)a' This is used to construct the full solution from the solutions of sub equations. Explanation =========== For example when solving the equation `(x - y)(x^2 + y^2 - z^2) = 0`, solutions for each of the equations `x - y = 0` and `x^2 + y^2 - z^2` are found independently. Solutions for `x - y = 0` are `(x, y) = (t, t)`. But we should introduce a value for z when we output the solution for the original equation. This function converts `(t, t)` into `(t, t, n_{1})` where `n_{1}` is an integer parameter. Nr6rTrX)rmstartF)iterr)rnextr@rZ assumptions0r1)rrWr;rrrvalZsymbr6r6r7rSs rScCs.tD]}||r||j|dSqdS)Nr|)all_diop_classesrirs)rr diop_typer6r6r7 _diop_solve"s  r_cCst|dd\}}}|tjkrt||S|tjkrt||S|tjkr(t|ddS|tjkr3t |ddS|t jkr=t ||S|t jkrFt |S|tjkrRt|tjdS|tjkr^t|tjdS|durl|tvrlttdtd|) a Solves the diophantine equation ``eq``. Explanation =========== Unlike ``diophantine()``, factoring of ``eq`` is not attempted. Uses ``classify_diop()`` to determine the type of the equation and calls the appropriate solver function. Use of ``diophantine()`` is recommended over other helper functions. ``diop_solve()`` can return either a set or a tuple depending on the nature of the equation. Usage ===== ``diop_solve(eq, t)``: Solve diophantine equation, ``eq`` using ``t`` as a parameter if needed. Details ======= ``eq`` should be an expression which is assumed to be zero. ``t`` is a parameter to be used in the solution. Examples ======== >>> from sympy.solvers.diophantine import diop_solve >>> from sympy.abc import x, y, z, w >>> diop_solve(2*x + 3*y - 5) (3*t_0 - 5, 5 - 2*t_0) >>> diop_solve(4*x + 3*y - 4*z + 5) (t_0, 8*t_0 + 4*t_1 + 5, 7*t_0 + 3*t_1 + 5) >>> diop_solve(x + 3*y - 4*z + w - 6) (t_0, t_0 + t_1, 6*t_0 + 5*t_1 + 4*t_2 - 6, 5*t_0 + 4*t_1 + 3*t_2 - 6) >>> diop_solve(x**2 + y**2 - 5) {(-2, -1), (-2, 1), (-1, -2), (-1, 2), (1, -2), (1, 2), (2, -1), (2, 1)} See Also ======== diophantine() FrJT) parameterizerrNz Alhough this type of equation was identified, it is not yet handled. It should, however, be listed in `diop_known` at the top of this file. Developers should see comments at the end of `classify_diop`. ro)r,rrq diop_linearrdiop_quadraticrdiop_ternary_quadraticrdiop_ternary_quadratic_normalrdiop_general_pythagoreanrxdiop_univariater diop_general_sum_of_squaresrZInfinityr(diop_general_sum_of_even_powers diop_knownr0r*rp)rrHrr^rXr6r6r7r(s,/              rcCs^d}d}tD]}||}|rd}nq|r)|j|r#t|j|jfS|j|jfSttd)NFTz This equation is not yet recognized or else has not been simplified sufficiently to put it in a form recognized by diop_classify().)r]rir[r?r^rqrpr*)rrKZmatchedr^r,r6r6r7r,}s$ a Helper routine used by diop_solve() to find information about ``eq``. Explanation =========== Returns a tuple containing the type of the diophantine equation along with the variables (free symbols) and their coefficients. Variables are returned as a list and coefficients are returned as a dict with the key being the respective term and the constant term is keyed to 1. The type is one of the following: * %s Usage ===== ``classify_diop(eq)``: Return variables, coefficients and type of the ``eq``. Details ======= ``eq`` should be an expression which is assumed to be zero. ``_dict`` is for internal use: when True (default) a dict is returned, otherwise a defaultdict which supplies 0 for missing keys is returned. Examples ======== >>> from sympy.solvers.diophantine import classify_diop >>> from sympy.abc import x, y, z, w, t >>> classify_diop(4*x + 6*y - 4) ([x, y], {1: -4, x: 4, y: 6}, 'linear') >>> classify_diop(x + 3*y -4*z + 5) ([x, y, z], {1: 5, x: 1, y: 3, z: -4}, 'linear') >>> classify_diop(x**2 + y**2 - x*y + x + 5) ([x, y], {1: 5, x: 1, x**2: 1, y**2: 1, x*y: -1}, 'binary_quadratic') z * cCst|dd\}}}|tjkrLd}|dur td|t|fdd}t|j|d}|dur6|dgt|j}t|dkrBt|dStdgt|jSdS) am Solves linear diophantine equations. A linear diophantine equation is an equation of the form `a_{1}x_{1} + a_{2}x_{2} + .. + a_{n}x_{n} = 0` where `a_{1}, a_{2}, ..a_{n}` are integer constants and `x_{1}, x_{2}, ..x_{n}` are integer variables. Usage ===== ``diop_linear(eq)``: Returns a tuple containing solutions to the diophantine equation ``eq``. Values in the tuple is arranged in the same order as the sorted variables. Details ======= ``eq`` is a linear diophantine equation which is assumed to be zero. ``param`` is the parameter to be used in the solution. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_linear >>> from sympy.abc import x, y, z >>> diop_linear(2*x - 3*y - 5) # solves equation 2*x - 3*y - 5 == 0 (3*t_0 - 5, 2*t_0 - 5) Here x = -3*t_0 - 5 and y = -2*t_0 - 5 >>> diop_linear(2*x - 3*y - 4*z -3) (t_0, 2*t_0 + 4*t_1 + 3, -t_0 - 3*t_1 - 3) See Also ======== diop_quadratic(), diop_ternary_quadratic(), diop_general_pythagorean(), diop_general_sum_of_squares() FrJNz%s_0:%iTrlr|r) r,rrqrr8rsr2rFr1)rrHrr^r^r2rEr6r6r7rbs(   rbcCst|||\}}}|dkr#|dur!|dkr| }||| |fSdStt|t|\}}}|t|9}|t|9}t||rc|dur[|dkrK| }||||||||fS||||fSdS)a Return the base solution for the linear equation, `ax + by = c`. Explanation =========== Used by ``diop_linear()`` to find the base solution of a linear Diophantine equation. If ``t`` is given then the parametrized solution is returned. Usage ===== ``base_solution_linear(c, a, b, t)``: ``a``, ``b``, ``c`` are coefficients in `ax + by = c` and ``t`` is the parameter to be used in the solution. Examples ======== >>> from sympy.solvers.diophantine.diophantine import base_solution_linear >>> from sympy.abc import t >>> base_solution_linear(5, 2, 3) # equation 2*x + 3*y = 5 (-5, 5) >>> base_solution_linear(0, 5, 7) # equation 5*x + 7*y = 0 (0, 0) >>> base_solution_linear(5, 2, 3, t) # equation 2*x + 3*y = 5 (3*t - 5, 5 - 2*t) >>> base_solution_linear(0, 5, 7, t) # equation 5*x + 7*y = 0 (7*t, -5*t) rNrrru)rr rrr)rUrrrrrrr6r6r7rs"    rcCs@t|dd\}}}|tjkrddt||dtjDSdS)a Solves a univariate diophantine equations. Explanation =========== A univariate diophantine equation is an equation of the form `a_{0} + a_{1}x + a_{2}x^2 + .. + a_{n}x^n = 0` where `a_{1}, a_{2}, ..a_{n}` are integer constants and `x` is an integer variable. Usage ===== ``diop_univariate(eq)``: Returns a set containing solutions to the diophantine equation ``eq``. Details ======= ``eq`` is a univariate diophantine equation which is assumed to be zero. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_univariate >>> from sympy.abc import x >>> diop_univariate((x - 2)*(x - 3)**2) # solves equation (x - 2)*(x - 3)**2 == 0 {(2,), (3,)} FrJcSsh|]}t|fqSr6rrr6r6r7r-Xz"diop_univariate..rN)r,rxrqr'r}rr~rrr^r^r6r6r7rg6s rgcCs || S)zN Returns `True` if ``a`` is divisible by ``b`` and `False` otherwise. r6r7r6r6r7r\s rcCsRt|dd\}}}|tjkr'|dur|tdddg}nd}tt|j|dSdS)a Solves quadratic diophantine equations. i.e. equations of the form `Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0`. Returns a set containing the tuples `(x, y)` which contains the solutions. If there are no solutions then `(None, None)` is returned. Usage ===== ``diop_quadratic(eq, param)``: ``eq`` is a quadratic binary diophantine equation. ``param`` is used to indicate the parameter to be used in the solution. Details ======= ``eq`` should be an expression which is assumed to be zero. ``param`` is a parameter to be used in the solution. Examples ======== >>> from sympy.abc import x, y, t >>> from sympy.solvers.diophantine.diophantine import diop_quadratic >>> diop_quadratic(x**2 + y**2 + 2*x + 2*y + 2, t) {(-1, -1)} References ========== .. [1] Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, [online], Available: http://www.alpertron.com.ar/METHODS.HTM .. [2] Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online], Available: https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf See Also ======== diop_linear(), diop_ternary_quadratic(), diop_general_sum_of_squares(), diop_general_pythagorean() FrJNrTrlr|)r,rrqrrdrs)rrHrr^r^r2r6r6r7rccs+ rccs8tt|||ftfdd|D}t|dkS)z Check whether `(u, v)` is solution to the quadratic binary diophantine equation with the variable list ``var`` and coefficient dictionary ``coeff``. Not intended for use by normal users. csg|] \}}||qSr6)ZxreplacerZrepsr6r7rsz$is_solution_quad..r)r?r@ritemsr )rr^rrrr6ror7rs rc s|dkrR|dkr dgS|dkrgS|dkrPg}tt|D]0}td| ||d}|rM|D]\}}|||||f|dkrL|||||fq.q|SdS|dkru|dkr\gS|dkred|fgSt|d\}} | rs||fgSgSt|d\} } | r|dkr| ||fgSg}ttt||dd| dD]&}zt||d|d\} } Wn tyd} Ynw| r|| |fq|Sd|dkr|krnnt ||S|dkrdgSt |dkrt dd|} d} g}g}| D]&}|d}||d||d| dkr|d| krn| d} qt | rf|dkr3|| d}|| d}nG| }|d| dkr]t | }||d||d|d7}|d| dks>||}||}n|dkrx|| d}|| d}ngS||fgSg}g}t|}|D]}t||dr||q|D]}||dt|t d d }fd d |D}t dkr|d d |D}|D]}t |t |} d} g}g}| D]}||d||d| dkrUt |ddkrU|| d}|| d}|d||dkr|||||fn6t|dgkrSt|d}||||dd|dd|||||dd||ddfn| d} | t|t |krfnqڐqȐq|S)a3 Solves the equation `x^2 - Dy^2 = N`. Explanation =========== Mainly concerned with the case `D > 0, D` is not a perfect square, which is the same as the generalized Pell equation. The LMM algorithm [1]_ is used to solve this equation. Returns one solution tuple, (`x, y)` for each class of the solutions. Other solutions of the class can be constructed according to the values of ``D`` and ``N``. Usage ===== ``diop_DN(D, N, t)``: D and N are integers as in `x^2 - Dy^2 = N` and ``t`` is the parameter to be used in the solutions. Details ======= ``D`` and ``N`` correspond to D and N in the equation. ``t`` is the parameter to be used in the solutions. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_DN >>> diop_DN(13, -4) # Solves equation x**2 - 13*y**2 = -4 [(3, 1), (393, 109), (36, 10)] The output can be interpreted as follows: There are three fundamental solutions to the equation `x^2 - 13y^2 = -4` given by (3, 1), (393, 109) and (36, 10). Each tuple is in the form (x, y), i.e. solution (3, 1) means that `x = 3` and `y = 1`. >>> diop_DN(986, 1) # Solves equation x**2 - 986*y**2 = 1 [(49299, 1570)] See Also ======== find_DN(), diop_bf_DN() References ========== .. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P. Robertson, July 31, 2004, Pages 16 - 17. [online], Available: https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf rrkrXrrFrTZ all_rootscs g|] }|tdkr|qSr!rrrr6r7rGrLzdiop_DN..cSsg|]}|r| qSr6r6rr6r6r7rJrmN)r square_factor cornacchiarrrrrr0_special_diop_DNrPQar&r[rr"rlength)rrrrrrrrZsN_exactsDsqZpqarGrrrrZfsrfZzsrrrr6rtr7rs6    (                RrcCs6t|}|dfg}d} |d}|t|krnt||\}}|dkr)|||f|d7}q d}d} d\} } d\} } g}d} t||| }|| |}||d| } || | }|| | }|D]\}}|d||d|kr~|||||fqc|d7}| dkr|ddkr |S| |} } | |} } q?)a1 Solves the equation `x^2 - Dy^2 = N` for the special case where `1 < N**2 < D` and `D` is not a perfect square. It is better to call `diop_DN` rather than this function, as the former checks the condition `1 < N**2 < D`, and calls the latter only if appropriate. Usage ===== WARNING: Internal method. Do not call directly! ``_special_diop_DN(D, N)``: D and N are integers as in `x^2 - Dy^2 = N`. Details ======= ``D`` and ``N`` correspond to D and N in the equation. Examples ======== >>> from sympy.solvers.diophantine.diophantine import _special_diop_DN >>> _special_diop_DN(13, -3) # Solves equation x**2 - 13*y**2 = -3 [(7, 2), (137, 38)] The output can be interpreted as follows: There are two fundamental solutions to the equation `x^2 - 13y^2 = -3` given by (7, 2) and (137, 38). Each tuple is in the form (x, y), i.e. solution (7, 2) means that `x = 7` and `y = 2`. >>> _special_diop_DN(2445, -20) # Solves equation x**2 - 2445*y**2 = -20 [(445, 9), (17625560, 356454), (698095554475, 14118073569)] See Also ======== diop_DN() References ========== .. [1] Section 4.4.4 of the following book: Quadratic Diophantine Equations, T. Andreescu and D. Andrica, Springer, 2015. rXrTrrrXrXr)rrrrr)rrZsqrt_Drr~f2rrrrZG0ZG1ZB0ZB1r=rrZG2ZB2r6r6r7rwlsF6        rwc Cst}t||d}t| ||dd}|sdS|D]S}||dkr#q||}} |||}}||d|kr9nq)|||d} | |dkrm| |} t| d\} } | rm||krb|| krb| |}} |t|t| fq|S)a9 Solves `ax^2 + by^2 = m` where `\gcd(a, b) = 1 = gcd(a, m)` and `a, b > 0`. Explanation =========== Uses the algorithm due to Cornacchia. The method only finds primitive solutions, i.e. ones with `\gcd(x, y) = 1`. So this method can't be used to find the solutions of `x^2 + y^2 = 20` since the only solution to former is `(x, y) = (4, 2)` and it is not primitive. When `a = b`, only the solutions with `x \leq y` are found. For more details, see the References. Examples ======== >>> from sympy.solvers.diophantine.diophantine import cornacchia >>> cornacchia(2, 3, 35) # equation 2x**2 + 3y**2 = 35 {(2, 3), (4, 1)} >>> cornacchia(1, 1, 25) # equation x**2 + y**2 = 25 {(4, 3)} References =========== .. [1] A. Nitaj, "L'algorithme de Cornacchia" .. [2] Solving the diophantine equation ax**2 + by**2 = m by Cornacchia's method, [online], Available: http://www.numbertheory.org/php/cornacchia.html See Also ======== sympy.utilities.iterables.signed_permutations rTrrNr)rdr r"rr9r) rrrrZa1rrrrZm1rrzr6r6r7rvs0#    rvccsd}}d}}| }|}|} |} t| t|| } | ||} | ||} | ||}| | | | | |fV| |}}| |}}||}}| | | } || d| } q)a Returns useful information needed to solve the Pell equation. Explanation =========== There are six sequences of integers defined related to the continued fraction representation of `\\frac{P + \sqrt{D}}{Q}`, namely {`P_{i}`}, {`Q_{i}`}, {`a_{i}`},{`A_{i}`}, {`B_{i}`}, {`G_{i}`}. ``PQa()`` Returns these values as a 6-tuple in the same order as mentioned above. Refer [1]_ for more detailed information. Usage ===== ``PQa(P_0, Q_0, D)``: ``P_0``, ``Q_0`` and ``D`` are integers corresponding to `P_{0}`, `Q_{0}` and `D` in the continued fraction `\\frac{P_{0} + \sqrt{D}}{Q_{0}}`. Also it's assumed that `P_{0}^2 == D mod(|Q_{0}|)` and `D` is square free. Examples ======== >>> from sympy.solvers.diophantine.diophantine import PQa >>> pqa = PQa(13, 4, 5) # (13 + sqrt(5))/4 >>> next(pqa) # (P_0, Q_0, a_0, A_0, B_0, G_0) (13, 4, 3, 3, 1, -1) >>> next(pqa) # (P_1, Q_1, a_1, A_1, B_1, G_1) (-1, 1, 1, 4, 1, 3) References ========== .. [1] Solving the generalized Pell equation x^2 - Dy^2 = N, John P. Robertson, July 31, 2004, Pages 4 - 8. https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf rrXTr)rr)ZP_0ZQ_0rZA_i_2ZB_i_1ZA_i_1ZB_i_2ZG_i_2ZG_i_1ZP_iZQ_iZa_iZA_iZB_iZG_ir6r6r7rx s&%       rxc Cst|}t|}g}t|d}|dd}t|dkr t||S|dkr:d}tt||dd|ddd}nT|dkrett|| d\}}|sP|d7}tt||dd| ddd}n)|dkrldgS|dkrud|fgSt|d\} }|r| ||f| ||fgSdgSt||D]8} zt||| dd\} }Wn tyd}Ynw|r|| | ft| | | | ||s|| | fq|S)a Uses brute force to solve the equation, `x^2 - Dy^2 = N`. Explanation =========== Mainly concerned with the generalized Pell equation which is the case when `D > 0, D` is not a perfect square. For more information on the case refer [1]_. Let `(t, u)` be the minimal positive solution of the equation `x^2 - Dy^2 = 1`. Then this method requires `\sqrt{\\frac{\mid N \mid (t \pm 1)}{2D}}` to be small. Usage ===== ``diop_bf_DN(D, N, t)``: ``D`` and ``N`` are coefficients in `x^2 - Dy^2 = N` and ``t`` is the parameter to be used in the solutions. Details ======= ``D`` and ``N`` correspond to D and N in the equation. ``t`` is the parameter to be used in the solutions. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_bf_DN >>> diop_bf_DN(13, -4) [(3, 1), (-3, 1), (36, 10)] >>> diop_bf_DN(986, 1) [(49299, 1570)] See Also ======== diop_DN() References ========== .. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P. Robertson, July 31, 2004, Page 15. https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf rXrrrrkF) rrrrrrr0r equivalent) rrrrrrZL1ZL2rzr{rrr6r6r7 diop_bf_DNL sF-    (*  rcCs0t||||||ot|||||S)aV Returns True if two solutions `(u, v)` and `(r, s)` of `x^2 - Dy^2 = N` belongs to the same equivalence class and False otherwise. Explanation =========== Two solutions `(u, v)` and `(r, s)` to the above equation fall to the same equivalence class iff both `(ur - Dvs)` and `(us - vr)` are divisible by `N`. See reference [1]_. No test is performed to test whether `(u, v)` and `(r, s)` are actually solutions to the equation. User should take care of this. Usage ===== ``equivalent(u, v, r, s, D, N)``: `(u, v)` and `(r, s)` are two solutions of the equation `x^2 - Dy^2 = N` and all parameters involved are integers. Examples ======== >>> from sympy.solvers.diophantine.diophantine import equivalent >>> equivalent(18, 5, -18, -5, 13, -1) True >>> equivalent(3, 1, -18, 393, 109, -4) False References ========== .. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P. Robertson, July 31, 2004, Page 12. https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf )r)rrrrrrr6r6r7r s0$rcCs\ddlm}||||}t|dtur$t|d}t|d}||Sd}t|}||S)aw Returns the (length of aperiodic part + length of periodic part) of continued fraction representation of `\\frac{P + \sqrt{D}}{Q}`. It is important to remember that this does NOT return the length of the periodic part but the sum of the lengths of the two parts as mentioned above. Usage ===== ``length(P, Q, D)``: ``P``, ``Q`` and ``D`` are integers corresponding to the continued fraction `\\frac{P + \sqrt{D}}{Q}`. Details ======= ``P``, ``D`` and ``Q`` corresponds to P, D and Q in the continued fraction, `\\frac{P + \sqrt{D}}{Q}`. Examples ======== >>> from sympy.solvers.diophantine.diophantine import length >>> length(-2 , 4, 5) # (-2 + sqrt(5))/4 3 >>> length(-5, 4, 17) # (-5 + sqrt(17))/4 4 See Also ======== sympy.ntheory.continued_fraction.continued_fraction_periodic r)continued_fraction_periodicrrX)Z sympy.ntheory.continued_fractionrtyperFr8)rrrrrZrptZnonrptr6r6r7ry s "   rycC*t|dd\}}}|tjkrt||SdS)aD This function transforms general quadratic, `ax^2 + bxy + cy^2 + dx + ey + f = 0` to more easy to deal with `X^2 - DY^2 = N` form. Explanation =========== This is used to solve the general quadratic equation by transforming it to the latter form. Refer [1]_ for more detailed information on the transformation. This function returns a tuple (A, B) where A is a 2 X 2 matrix and B is a 2 X 1 matrix such that, Transpose([x y]) = A * Transpose([X Y]) + B Usage ===== ``transformation_to_DN(eq)``: where ``eq`` is the quadratic to be transformed. Examples ======== >>> from sympy.abc import x, y >>> from sympy.solvers.diophantine.diophantine import transformation_to_DN >>> A, B = transformation_to_DN(x**2 - 3*x*y - y**2 - 2*y + 1) >>> A Matrix([ [1/26, 3/26], [ 0, 1/13]]) >>> B Matrix([ [-6/13], [-4/13]]) A, B returned are such that Transpose((x y)) = A * Transpose((X Y)) + B. Substituting these values for `x` and `y` and a bit of simplifying work will give an equation of the form `x^2 - Dy^2 = N`. >>> from sympy.abc import X, Y >>> from sympy import Matrix, simplify >>> u = (A*Matrix([X, Y]) + B)[0] # Transformation for x >>> u X/26 + 3*Y/26 - 6/13 >>> v = (A*Matrix([X, Y]) + B)[1] # Transformation for y >>> v Y/13 - 4/13 Next we will substitute these formulas for `x` and `y` and do ``simplify()``. >>> eq = simplify((x**2 - 3*x*y - y**2 - 2*y + 1).subs(zip((x, y), (u, v)))) >>> eq X**2/676 - Y**2/52 + 17/13 By multiplying the denominator appropriately, we can get a Pell equation in the standard form. >>> eq * 676 X**2 - 13*Y**2 + 884 If only the final equation is needed, ``find_DN()`` can be used. See Also ======== find_DN() References ========== .. [1] Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0, John P.Robertson, May 8, 2003, Page 7 - 11. https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf FrJN)r,rrqrrnr6r6r7transformation_to_DN sN  rcCs$|\}}||d}|||}||d}||}||}|d} ddt|||||| D\}}}}}} tddd\} } |rtd||\} } t|| d\}}| d|| | | d| d| |||| d| ||| | ||||| d| || i}t| | g|\}}tddtj| t|  | ddg|tddtj| t|  | ddg|fS|rtd||\} } t|| d\}}| d|| | d| d||| d| ||d| ||| di}t| | g|\}}tddtj| dddg|tddtj| dddg|tt|  | dgfS|rtd||\} } t|| d\}}| d||| | d| d|| d| dd| ||| di}t| | g|\}}tdddddtj| g|tdddddtj| g|tdt|  | gfStddtj|dddgtddgfS) NrrXcSrr6rrr6r6r7rY rz)_transformation_to_DN..X, YTrlr)rrrrrrr)rr^rrrrrUrrr~rrrrrrZA_0B_0r6r6r7rN s8   (`P@R<R&rcCr)a This function returns a tuple, `(D, N)` of the simplified form, `x^2 - Dy^2 = N`, corresponding to the general quadratic, `ax^2 + bxy + cy^2 + dx + ey + f = 0`. Solving the general quadratic is then equivalent to solving the equation `X^2 - DY^2 = N` and transforming the solutions by using the transformation matrices returned by ``transformation_to_DN()``. Usage ===== ``find_DN(eq)``: where ``eq`` is the quadratic to be transformed. Examples ======== >>> from sympy.abc import x, y >>> from sympy.solvers.diophantine.diophantine import find_DN >>> find_DN(x**2 - 3*x*y - y**2 - 2*y + 1) (13, -884) Interpretation of the output is that we get `X^2 -13Y^2 = -884` after transforming `x^2 - 3xy - y^2 - 2y + 1` using the transformation returned by ``transformation_to_DN()``. See Also ======== transformation_to_DN() References ========== .. [1] Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0, John P.Robertson, May 8, 2003, Page 7 - 11. https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf FrJN)r,rrqrrnr6r6r7find_DN s'  rc Cs|\}}tddd\}}t||\}}|t||g|d}|t||g|d} |d||d||||||d||d|||||||d} t| t||f|| f} | }||d ||d|d ||dfS)NrTrlrrXr)rrrr rBr@r]) rr^rrrrrrrrrZ simplifiedr6r6r7r s\0rc Csddlm}|jr|jst||g|dS|jr"|js"t||g|dStddd\}}||||\}}||jrCt||g|dS|||d|||d} | \} } t| |dS) z If there is a number modulo ``a`` such that ``x`` and ``y`` are both integers, then return a parametric representation for ``x`` and ``y`` else return (None, None). Here ``x`` and ``y`` are functions of ``t``. r)clear_coefficientsr|zm, nTrlrX)r) sympy.simplify.simplifyrrCrr-rr3rr_) rrrrrrrrUrrZjunkr6r6r7r s      rc Cszt|dd\}}}|tjtjfvr;t||}t|dkr&t|d\}}}nd\}}}|r6t|||f||S|||fSdS)a Solves the general quadratic ternary form, `ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0`. Returns a tuple `(x, y, z)` which is a base solution for the above equation. If there are no solutions, `(None, None, None)` is returned. Usage ===== ``diop_ternary_quadratic(eq)``: Return a tuple containing a basic solution to ``eq``. Details ======= ``eq`` should be an homogeneous expression of degree two in three variables and it is assumed to be zero. Examples ======== >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine.diophantine import diop_ternary_quadratic >>> diop_ternary_quadratic(x**2 + 3*y**2 - z**2) (1, 0, 1) >>> diop_ternary_quadratic(4*x**2 + 5*y**2 - z**2) (1, 0, 2) >>> diop_ternary_quadratic(45*x**2 - 7*y**2 - 8*x*y - z**2) (28, 45, 105) >>> diop_ternary_quadratic(x**2 - 49*y**2 - z**2 + 13*z*y -8*x*y) (9, 1, 5) FrJrrN)r,rrqrrr8rF_parametrize_ternary_quadratic rr`rr^r^rrrrr6r6r7rd s"     rdcsRtfddD}t|rt||dSt|r't||dSdS)Ncg|]}||qSr6r6rrr6r7r r4z+_diop_ternary_quadratic..r)r rrirsr)rr^rr6rr7r s   rcCs(t|dd\}}}|dvrt||SdS)a< Returns the transformation Matrix that converts a general ternary quadratic equation ``eq`` (`ax^2 + by^2 + cz^2 + dxy + eyz + fxz`) to a form without cross terms: `ax^2 + by^2 + cz^2 = 0`. This is not used in solving ternary quadratics; it is only implemented for the sake of completeness. FrJrrN)r,_transformation_to_normalrnr6r6r7transformation_to_normal s rc sTt|}|\}}}tfdd|DsT||}||}||}d} |s1d} ||}}tdd| |fdd| |fdf} | rR| dd| dd| S|d dkr|d dkr|d |d|d<|d <t|} | dd | dd | S|d|d|d<|d<t|} | dd| dd| S||dks||dkrB|d } ||} ||} |d }||}|d }t}d | d ||d <d | || d ||d <d | || d ||d <d | |d | | |||<d|||<d|||<t||}td d dt| d | t| d | ddddddg |S||dkr|d dkr|d dkretd d gd S|d |d|d<|d <t|} | dd | dd | S|d|d|d<|d<t|} | dd| dd| Std S) Nc3rrr6rrr6r7rV* rz,_transformation_to_normal..FTrXr)rrrXrrrr) rXrrrrXrXrrXr) rFrrZrow_swapZcol_swaprr?rZeye)rr^rrrrrrrUZswaprrrrrrrrZT_0r6rr7r% st     &        "          >       rcCsHt|dd\}}}|dvr"tt||d\}}}t|||f||SdS)a Returns the parametrized general solution for the ternary quadratic equation ``eq`` which has the form `ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0`. Examples ======== >>> from sympy import Tuple, ordered >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine.diophantine import parametrize_ternary_quadratic The parametrized solution may be returned with three parameters: >>> parametrize_ternary_quadratic(2*x**2 + y**2 - 2*z**2) (p**2 - 2*q**2, -2*p**2 + 4*p*q - 4*p*r - 4*q**2, p**2 - 4*p*q + 2*q**2 - 4*q*r) There might also be only two parameters: >>> parametrize_ternary_quadratic(4*x**2 + 2*y**2 - 3*z**2) (2*p**2 - 3*q**2, -4*p**2 + 12*p*q - 6*q**2, 4*p**2 - 8*p*q + 6*q**2) Notes ===== Consider ``p`` and ``q`` in the previous 2-parameter solution and observe that more than one solution can be represented by a given pair of parameters. If `p` and ``q`` are not coprime, this is trivially true since the common factor will also be a common factor of the solution values. But it may also be true even when ``p`` and ``q`` are coprime: >>> sol = Tuple(*_) >>> p, q = ordered(sol.free_symbols) >>> sol.subs([(p, 3), (q, 2)]) (6, 12, 12) >>> sol.subs([(q, 1), (p, 1)]) (-1, 2, 2) >>> sol.subs([(q, 0), (p, 1)]) (2, -4, 4) >>> sol.subs([(q, 1), (p, 0)]) (-3, -6, 6) Except for sign and a common factor, these are equivalent to the solution of (1, 2, 2). References ========== .. [1] The algorithmic resolution of Diophantine equations, Nigel P. Smart, London Mathematical Society Student Texts 41, Cambridge University Press, Cambridge, 1998. FrJrrN)r,rFrr)rrr^r^rrrr6r6r7parametrize_ternary_quadraticy s7 rc Cs<d|vsJ|\}}}t|}|durdS|ddkrdS|dkr@|d|d|d<|d<t|||f||\}}} ||| fS|\} } } tddd\} }}tdd |D}t|t| | | f| || ||| ||f}|j | dd \}}||} ||t|| |} ||t|| |} t | | | S) NrXrrrzr, p, qTrlcss|] \}}||VqdSr<r6)rTrrr6r6r7rV rz1_parametrize_ternary_quadratic..)rB) rFrrrr rpr rBr@rNr)r;rr^rrrrZy_pZx_pZz_prrrrrrrZeq_1rrr6r6r7r s0     $ rc Cstt|dd\}}}|tjkr8t||}t|dkr#t|d\}}}nd\}}}|r3t|||f||S|||fSdS)a Solves the quadratic ternary diophantine equation, `ax^2 + by^2 + cz^2 = 0`. Explanation =========== Here the coefficients `a`, `b`, and `c` should be non zero. Otherwise the equation will be a quadratic binary or univariate equation. If solvable, returns a tuple `(x, y, z)` that satisfies the given equation. If the equation does not have integer solutions, `(None, None, None)` is returned. Usage ===== ``diop_ternary_quadratic_normal(eq)``: where ``eq`` is an equation of the form `ax^2 + by^2 + cz^2 = 0`. Examples ======== >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine.diophantine import diop_ternary_quadratic_normal >>> diop_ternary_quadratic_normal(x**2 + 3*y**2 - z**2) (1, 0, 1) >>> diop_ternary_quadratic_normal(4*x**2 + 5*y**2 - z**2) (1, 0, 2) >>> diop_ternary_quadratic_normal(34*x**2 - 3*y**2 - 301*z**2) (4, 9, 1) FrJrrN)r,rrqrr8rFrrr6r6r7re s      recs&tfddD}t||dS)Ncrr6r6rrr6r7r r4z2_diop_ternary_quadratic_normal..r)r rrs)rr^rr6rr7r src Cst|||}tdd|D}tddt||D}\}}} t||} || }|| }t|| } || }| | } t|| } || }|| }|| 9}|| 9}| | 9} |r]||||| ffS||| fS)a Return `a', b', c'`, the coefficients of the square-free normal form of `ax^2 + by^2 + cz^2 = 0`, where `a', b', c'` are pairwise prime. If `steps` is True then also return three tuples: `sq`, `sqf`, and `(a', b', c')` where `sq` contains the square factors of `a`, `b` and `c` after removing the `gcd(a, b, c)`; `sqf` contains the values of `a`, `b` and `c` after removing both the `gcd(a, b, c)` and the square factors. The solutions for `ax^2 + by^2 + cz^2 = 0` can be recovered from the solutions of `a'x^2 + b'y^2 + c'z^2 = 0`. Examples ======== >>> from sympy.solvers.diophantine.diophantine import sqf_normal >>> sqf_normal(2 * 3**2 * 5, 2 * 5 * 11, 2 * 7**2 * 11) (11, 1, 5) >>> sqf_normal(2 * 3**2 * 5, 2 * 5 * 11, 2 * 7**2 * 11, True) ((3, 1, 7), (5, 55, 11), (11, 1, 5)) References ========== .. [1] Legendre's Theorem, Legrange's Descent, http://public.csusm.edu/aitken_html/notes/legendre.pdf See Also ======== reconstruct() csrQr<)rurr6r6r7rV6 rWzsqf_normal..cSsg|] \}}||dqSr!r6rr6r6r7r7 rzsqf_normal..)rr1r@r) rrrUrABCr|ZsqfrrrZpcpaZpbr6r6r7r s$ ""    rcCs,t|tr|nt|}tdd|DS)a Returns an integer `c` s.t. `a = c^2k, \ c,k \in Z`. Here `k` is square free. `a` can be given as an integer or a dictionary of factors. Examples ======== >>> from sympy.solvers.diophantine.diophantine import square_factor >>> square_factor(24) 2 >>> square_factor(-36*3) 6 >>> square_factor(1) 1 >>> square_factor({3: 2, 2: 1, -1: 1}) # -18 3 See Also ======== sympy.ntheory.factor_.core cSsg|] \}}||dqSr!r6)rTrrr6r6r7rc rz!square_factor..)rr?rr rp)rr~r6r6r7ruL srucCs<tt||}|D]\}}|dkrtd||9}q |S)a Reconstruct the `z` value of an equivalent solution of `ax^2 + by^2 + cz^2` from the `z` value of a solution of the square-free normal form of the equation, `a'*x^2 + b'*y^2 + c'*z^2`, where `a'`, `b'` and `c'` are square free and `gcd(a', b', c') == 1`. rXza and b should be square-free)rrrpr0)rrrr~rrr6r6r7rf s  rcCst|t|krt||\}}}|||fS|dkrdS|dkr!dS|dkr'dSt||}|d||}|dkr=d}d}n#t|} d}| D]} tt|| d\} } | r_t|| | }}nqE|durt||\} }}t| ||| ||| |||SdS)a! Return a non-trivial solution to `w^2 = Ax^2 + By^2` using Lagrange's method; return None if there is no such solution. . Here, `A \neq 0` and `B \neq 0` and `A` and `B` are square free. Output a tuple `(w_0, x_0, y_0)` which is a solution to the above equation. Examples ======== >>> from sympy.solvers.diophantine.diophantine import ldescent >>> ldescent(1, 1) # w^2 = x^2 + y^2 (1, 1, 0) >>> ldescent(4, -7) # w^2 = 4x^2 - 7y^2 (2, -1, 0) This means that `x = -1, y = 0` and `w = 2` is a solution to the equation `w^2 = 4x^2 - 7y^2` >>> ldescent(5, -1) # w^2 = 5x^2 - y^2 (2, 1, -1) References ========== .. [1] The algorithmic resolution of Diophantine equations, Nigel P. Smart, London Mathematical Society Student Texts 41, Cambridge University Press, Cambridge, 1998. .. [2] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, [online], Available: http://eprints.nottingham.ac.uk/60/1/kvxefz87.pdf rXrXrXrrXrrXrNrr)rldescentr"rrrr)rrr8rrrrrrrrZsQrzWrrr6r6r7ru s4"  *rcCst|t|krt||\}}}|||fS|dkrdS|dkr!dS|| kr(dS||kr;td|\}}}||||fSt||}t|||\}}|d||d|}t|} || d} t|| \} } } t|| ||| || || | | | S)a Returns a non-trivial solution, (x, y, z), to `x^2 = Ay^2 + Bz^2` using Lagrange's descent method with lattice-reduction. `A` and `B` are assumed to be valid for such a solution to exist. This is faster than the normal Lagrange's descent algorithm because the Gaussian reduction is used. Examples ======== >>> from sympy.solvers.diophantine.diophantine import descent >>> descent(3, 1) # x**2 = 3*y**2 + z**2 (1, 0, 1) `(x, y, z) = (1, 0, 1)` is a solution to the above equation. >>> descent(41, -113) (-16, -3, 1) References ========== .. [1] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, Mathematics of Computation, Volume 00, Number 0. rXrr)rrXrXrr)rrr"gaussian_reducerur)rrrrrr8rrrZt_2Zt_1Zx_1Zz_1Zy_1r6r6r7r s&    0rcCsd}d}t|||||dkr|d |d f}t||||t||||kr+||}}t||||t||||krnt|||||t|||||}||d||d|d||df}}t||||t||||ks9||}}t|||||t|||||dkst|d|d|d|df|||t||||kr|}n|d|d|d|df}|d|||d|dfS)a Returns a reduced solution `(x, z)` to the congruence `X^2 - aZ^2 \equiv 0 \ (mod \ b)` so that `x^2 + |a|z^2` is minimal. Details ======= Here ``w`` is a solution of the congruence `x^2 \equiv a \ (mod \ b)` References ========== .. [1] Gaussian lattice Reduction [online]. Available: http://home.ie.cuhk.edu.hk/~wkshum/wordpress/?p=404 .. [2] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, Mathematics of Computation, Volume 00, Number 0. rrrrXr)dotnorm)r8rrrrrrUr6r6r7r s  . \  rc Cs@|\}}|\}}||||||||t|||S)z Returns a special dot product of the vectors `u = (u_{1}, u_{2})` and `v = (v_{1}, v_{2})` which is defined in order to reduce solution of the congruence equation `X^2 - aZ^2 \equiv 0 \ (mod \ b)`. rs) rrr8rru_1u_2Zv_1Zv_2r6r6r7r s0rcCs$|\}}tt||f||f|||S)z Returns the norm of the vector `u = (u_{1}, u_{2})` under the dot product defined by `u \cdot v = (wu_{1} + bu_{2})(w*v_{1} + bv_{2}) + |a|*u_{1}*v_{1}` where `u = (u_{1}, u_{2})` and `v = (v_{1}, v_{2})`. )rr)rr8rrrrr6r6r7r# srcCs$t|r d|}n|d}|||}d} ||d||d||d} } } | | | kr9|dkr8tdn|||} } }t| | | |krJnt|| | }\}}d|vrZn||| |||  ||}}t||}t|rt||}t||tj ksJn!||}t||||||r|d7}t||tj ksJ||d||d||d}||| ||| |||}t| |d|||}t| |d|||}t||d|||}t dd|||fDsJ|d7}qt d d | | |fDS) ae Simplify the solution `(x, y, z)` of the equation `ax^2 + by^2 = cz^2` with `a, b, c > 0` and `z^2 \geq \mid ab \mid` to a new reduced solution `(x', y', z')` such that `z'^2 \leq \mid ab \mid`. The algorithm is an interpretation of Mordell's reduction as described on page 8 of Cremona and Rusin's paper [1]_ and the work of Mordell in reference [2]_. References ========== .. [1] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, Mathematics of Computation, Volume 00, Number 0. .. [2] Diophantine Equations, L. J. Mordell, page 48. rrTzbad starting solutionNrXcsrr<)rrr6r6r7rVg rzholzer..cSrr6rlrr6r6r7rj rzholzer..) r&r0maxrr r;r9rrZHalfrr_r1)rrrrrrUrZsmallstept1t2Zt3rrrZuvrrrrrr8rrr6r6r7r- sF  ( $  $$#rrcCs^t|dd\}}}|tjkr-|durd}n td|t|fdd}tt|j|ddSdS) a Solves the general pythagorean equation, `a_{1}^2x_{1}^2 + a_{2}^2x_{2}^2 + . . . + a_{n}^2x_{n}^2 - a_{n + 1}^2x_{n + 1}^2 = 0`. Returns a tuple which contains a parametrized solution to the equation, sorted in the same order as the input variables. Usage ===== ``diop_general_pythagorean(eq, param)``: where ``eq`` is a general pythagorean equation which is assumed to be zero and ``param`` is the base parameter used to construct other parameters by subscripting. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_general_pythagorean >>> from sympy.abc import a, b, c, d, e >>> diop_general_pythagorean(a**2 + b**2 + c**2 - d**2) (m1**2 + m2**2 - m3**2, 2*m1*m3, 2*m2*m3, m1**2 + m2**2 + m3**2) >>> diop_general_pythagorean(9*a**2 - 4*b**2 + 16*c**2 + 25*d**2 + e**2) (10*m1**2 + 10*m2**2 + 10*m3**2 - 10*m4**2, 15*m1**2 + 15*m2**2 + 15*m3**2 + 15*m4**2, 15*m1*m4, 12*m2*m4, 60*m3*m4) FrJNz%s1:%iTrlr|r)r,rrqrr8rFrs)rrHrr^r^rr6r6r7rfm s rfrXcC4t|dd\}}}|tjkrtt|j|dSdS)a Solves the equation `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`. Returns at most ``limit`` number of solutions. Usage ===== ``general_sum_of_squares(eq, limit)`` : Here ``eq`` is an expression which is assumed to be zero. Also, ``eq`` should be in the form, `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`. Details ======= When `n = 3` if `k = 4^a(8m + 7)` for some `a, m \in Z` then there will be no solutions. Refer [1]_ for more details. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_general_sum_of_squares >>> from sympy.abc import a, b, c, d, e >>> diop_general_sum_of_squares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345) {(15, 22, 22, 24, 24)} Reference ========= .. [1] Representing an integer as a sum of three squares, [online], Available: http://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares FrJraN)r,r rqrdrsrrrrr^r^r6r6r7rh s" rhcCr)a Solves the equation `x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0` where `e` is an even, integer power. Returns at most ``limit`` number of solutions. Usage ===== ``general_sum_of_even_powers(eq, limit)`` : Here ``eq`` is an expression which is assumed to be zero. Also, ``eq`` should be in the form, `x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0`. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_general_sum_of_even_powers >>> from sympy.abc import a, b >>> diop_general_sum_of_even_powers(a**4 + b**4 - (2**4 + 3**4)) {(2, 3)} See Also ======== power_representation FrJraN)r,r(rqrdrsrr6r6r7ri s riccs~ddlm}|r |dur|||D]}t|VqdStd|dD]}|||D]}t|}d|t||Vq*q#dS)aD Returns a generator that can be used to generate partitions of an integer `n`. Explanation =========== A partition of `n` is a set of positive integers which add up to `n`. For example, partitions of 3 are 3, 1 + 2, 1 + 1 + 1. A partition is returned as a tuple. If ``k`` equals None, then all possible partitions are returned irrespective of their size, otherwise only the partitions of size ``k`` are returned. If the ``zero`` parameter is set to True then a suitable number of zeros are added at the end of every partition of size less than ``k``. ``zero`` parameter is considered only if ``k`` is not None. When the partitions are over, the last `next()` call throws the ``StopIteration`` exception, so this function should always be used inside a try - except block. Details ======= ``partition(n, k)``: Here ``n`` is a positive integer and ``k`` is the size of the partition which is also positive integer. Examples ======== >>> from sympy.solvers.diophantine.diophantine import partition >>> f = partition(5) >>> next(f) (1, 1, 1, 1, 1) >>> next(f) (1, 1, 1, 2) >>> g = partition(5, 3) >>> next(g) (1, 1, 3) >>> next(g) (1, 2, 2) >>> g = partition(5, 3, zeros=True) >>> next(g) (0, 0, 5) r)ordered_partitionsNrXr2)rMrr1rr8)rrrrrrr6r6r7 partition s .  rcCs|ddksdS|ddkrd}nd}t||dd|dkr/t|}t||dd|dkst||dd|}|}|d|krN|||}}|d|ksAt||t|fS)a  Represent a prime `p` as a unique sum of two squares; this can only be done if the prime is congruent to 1 mod 4. Examples ======== >>> from sympy.solvers.diophantine.diophantine import prime_as_sum_of_two_squares >>> prime_as_sum_of_two_squares(7) # can't be done >>> prime_as_sum_of_two_squares(5) (1, 2) Reference ========= .. [1] Representing a number as a sum of four squares, [online], Available: http://schorn.ch/lagrange.html See Also ======== sum_of_squares() rrXNrqrr)powrr)rrrr6r6r7prime_as_sum_of_two_squaress    rc Cs>iddddddddd d d d d ddddddddddddddddddd d!d"d#d$i}d%}|d%krAd&Std'|}|d'|}|d(d)krTd*S||vrs||\}}}td||d||d||St|d\}}|rd||d%d%fSd*}|d(dkrt|r|n|d}t|d+d,D].}||dd}t|rt|\}}td||d|||d|t||Sqd*S|d(dks|d(d-krt|r|n|d}n t|r|dn|}t|d+d,D]'}||d}t|rt|\}}td||d||d||Sqd*S).a Returns a 3-tuple `(a, b, c)` such that `a^2 + b^2 + c^2 = n` and `a, b, c \geq 0`. Returns None if `n = 4^a(8m + 7)` for some `a, m \in Z`. See [1]_ for more details. Usage ===== ``sum_of_three_squares(n)``: Here ``n`` is a non-negative integer. Examples ======== >>> from sympy.solvers.diophantine.diophantine import sum_of_three_squares >>> sum_of_three_squares(44542) (18, 37, 207) References ========== .. [1] Representing a number as a sum of three squares, [online], Available: http://schorn.ch/lagrange.html See Also ======== sum_of_squares() rX)rXrrrrr)rXrXrX )rXrr")rrr:)rrU)rr)r r)rr )r rir)rri)rri)rri)rXri)r!ir)rr$i )r 'i%)89rr)rrrrrrNrrr) rr$r0rr&rr!rr) rZspecialrrrrrrzrr6r6r7sum_of_three_squaresBst&    $  4   (rcCs|dkrdStd|}|d|}|ddkrd}|d}n|ddks*|ddkr1d}|d}nd}t|\}}}td||d||d||d||S) a{ Returns a 4-tuple `(a, b, c, d)` such that `a^2 + b^2 + c^2 + d^2 = n`. Here `a, b, c, d \geq 0`. Usage ===== ``sum_of_four_squares(n)``: Here ``n`` is a non-negative integer. Examples ======== >>> from sympy.solvers.diophantine.diophantine import sum_of_four_squares >>> sum_of_four_squares(3456) (8, 8, 32, 48) >>> sum_of_four_squares(1294585930293) (0, 1234, 2161, 1137796) References ========== .. [1] Representing a number as a sum of four squares, [online], Available: http://schorn.ch/lagrange.html See Also ======== sum_of_squares() r)rrrrrrrrrrX)rrr0)rrrrrrr6r6r7sum_of_four_squaress     .rc cs>dd|||fD\}}}|dkr-|dr+t| |||D] }tdd|DVqdS|dks5|dkr?ttd ||f|dkrL|rJd |VdS|dkru|dkrZ|fVdSt|}|rs|\}}t||\}} | ss||fVdS|dkrt|||d D]}|VqdS|dkrt||} | sdS|s|d kr|d kr||kr||dvr dS| durt|VdS|dkr|dkrt|}|r|d|dkrdS||krt ||d|d} t | ||g|D] }tt |Vq|rt ||d} t d|D]} t | | |g|D]}tt |d || Vq qdSdS)ab Returns a generator for finding k-tuples of integers, `(n_{1}, n_{2}, . . . n_{k})`, such that `n = n_{1}^p + n_{2}^p + . . . n_{k}^p`. Usage ===== ``power_representation(n, p, k, zeros)``: Represent non-negative number ``n`` as a sum of ``k`` ``p``\ th powers. If ``zeros`` is true, then the solutions is allowed to contain zeros. Examples ======== >>> from sympy.solvers.diophantine.diophantine import power_representation Represent 1729 as a sum of two cubes: >>> f = power_representation(1729, 3, 2) >>> next(f) (9, 10) >>> next(f) (1, 12) If the flag `zeros` is True, the solution may contain tuples with zeros; any such solutions will be generated after the solutions without zeros: >>> list(power_representation(125, 2, 3, zeros=True)) [(5, 6, 8), (3, 4, 10), (0, 5, 10), (0, 2, 11)] For even `p` the `permute_sign` function can be used to get all signed values: >>> from sympy.utilities.iterables import permute_signs >>> list(permute_signs((1, 12))) [(1, 12), (-1, 12), (1, -12), (-1, -12)] All possible signed permutations can also be obtained: >>> from sympy.utilities.iterables import signed_permutations >>> list(signed_permutations((1, 12))) [(1, 12), (-1, 12), (1, -12), (-1, -12), (12, 1), (-12, 1), (12, -1), (-12, -1)] cSrr6rrr6r6r7rrz(power_representation..rrcss|]}| VqdSr<r6rr6r6r7rVrz'power_representation..NrXzA Expecting positive integers for `(p, k)`, but got `(%s, %s)`r2rrrq)rrrrqrrrXT) r+r1r0r*rrr_can_do_sum_of_squaresrrpow_rep_recursivereversedr) rrrrrberrrrZfeasiblerrr6r6r7r+sp.   ( r+ccs|dkr|dkrt|VdS|dkrB|dkrDt|d||||EdH|t||}|dkrFt||d|||g|EdHdSdSdSdSrE)r1rr)Zn_irZ n_remainingrUrZresidualr6r6r7r7s$rccst|d||EdHdS)ayReturn a generator that yields the k-tuples of nonnegative values, the squares of which sum to n. If zeros is False (default) then the solution will not contain zeros. The nonnegative elements of a tuple are sorted. * If k == 1 and n is square, (n,) is returned. * If k == 2 then n can only be written as a sum of squares if every prime in the factorization of n that has the form 4*k + 3 has an even multiplicity. If n is prime then it can only be written as a sum of two squares if it is in the form 4*k + 1. * if k == 3 then n can be written as a sum of squares if it does not have the form 4**m*(8*k + 7). * all integers can be written as the sum of 4 squares. * if k > 4 then n can be partitioned and each partition can be written as a sum of 4 squares; if n is not evenly divisible by 4 then n can be written as a sum of squares only if the an additional partition can be written as sum of squares. For example, if k = 6 then n is partitioned into two parts, the first being written as a sum of 4 squares and the second being written as a sum of 2 squares -- which can only be done if the condition above for k = 2 can be met, so this will automatically reject certain partitions of n. Examples ======== >>> from sympy.solvers.diophantine.diophantine import sum_of_squares >>> list(sum_of_squares(25, 2)) [(3, 4)] >>> list(sum_of_squares(25, 2, True)) [(3, 4), (0, 5)] >>> list(sum_of_squares(25, 4)) [(1, 2, 2, 4)] See Also ======== sympy.utilities.iterables.signed_permutations rN)r+)rrrr6r6r7rCs-rcCs|dkrdS|dkr dS|dkrdS|dkrt|S|dkrN|dvr$dSt|r2|ddkr0dSdSt|}|D]\}}|ddkrK|drKdSq:dS|dkra|dtd|d d kradSdS) aReturn True if n can be written as the sum of k squares, False if it can't, or 1 if ``k == 2`` and ``n`` is prime (in which case it *can* be written as a sum of two squares). A False is returned only if it can't be written as ``k``-squares, even if 0s are allowed. rXFrTr)rXrrrrr)r r!rrpr)rrr~rrr6r6r7rss2 rr<)T)F)rX)NF)Zsympy.core.addrZsympy.core.assumptionsrZsympy.core.containersrZsympy.core.compatibilityrrrZsympy.core.exprtoolsrZsympy.core.functionr Zsympy.core.mulr Zsympy.core.numbersr r r rZsympy.core.powerrrZsympy.core.relationalrZsympy.core.singletonrZsympy.core.symbolrrZsympy.core.sympifyrZ$sympy.functions.elementary.complexesrZ#sympy.functions.elementary.integersrZ(sympy.functions.elementary.miscellaneousrZsympy.matrices.denserrZsympy.ntheory.factor_rrrrZsympy.ntheory.generaterZsympy.ntheory.primetestr r!Zsympy.ntheory.residue_ntheoryr"Zsympy.polys.polyerrorsr#Zsympy.polys.polytoolsr$r%rr&Zsympy.solvers.solvesetr'Zsympy.utilitiesr(r)Zsympy.utilities.miscr*__all__rdr-rMrxrrrrrrrr rr'r(r]rjrSr0rrr9r&r;r+rSr_rr,joinr/Zfunc_docrbrrgrrcrrrwrvrxrrryrrrrrrdrrrrrrerrrurrrrrrrrfrhrirrrrr+Z sum_of_powersrrrr6r6r6r7s                   cN I`WGNA   # U &* :9&5 H_B?['-S3, 3TA (, 9E5)  @ # ( &9+P 3o 0