+ iUjRt^RIt^RIHt^RIHt^RIHt^RIH t !RR]4t Rt R t R t R#) z The Schur number S(k) is the largest integer n for which the interval [1,n] can be partitioned into k sum-free sets.(https://mathworld.wolfram.com/SchurNumber.html) N)S)Basic)Function)Integerc:a]tRt^ toRt]R4tRtRtVt R#) SchurNumbera This function creates a SchurNumber object which is evaluated for `k \le 5` otherwise only the lower bound information can be retrieved. Examples ======== >>> from sympy.combinatorics.schur_number import SchurNumber Since S(3) = 13, hence the output is a number >>> SchurNumber(3) 13 We do not know the Schur number for values greater than 5, hence only the object is returned >>> SchurNumber(6) SchurNumber(6) Now, the lower bound information can be retrieved using lower_bound() method >>> SchurNumber(6).lower_bound() 536 c `VP'dV\PJd\P#VP'd\P#VP 'dVP 'd \R4h^^^^^^ ^^,^^/pV^8:d\W!,4#R#R#)zk should be a positive integerN) is_NumberrInfinityis_zeroZero is_integer is_negative ValueErrorr)clskfirst_known_schur_numberss&& ^/var/www/html/photoedit/myenv/lib/python3.14/site-packages/sympy/combinatorics/schur_number.pyevalSchurNumber.eval's ;;;AJJzz!yyyvv <<<1=== !ABB)*Aq!QAr1c(J %Av8;<< c*VP^,pV^8Xd \R4#V^8Xd \R4#VP'd5^VPV^, 4P 4,^, #^V,^, ^, #)ii)argsr is_Integerfunc lower_bound)selff_s& rrSchurNumber.lower_bound4sr YYq\ 73<  74= ===TYYrAv&2244q8 82 1}rN) __name__ __module__ __qualname____firstlineno____doc__ classmethodrr__static_attributes____classdictcell__) __classdict__s@rrr s(4 = =  rrcV\PJd \R4hV^8:d \R4hV^8:d^pM9\P!\P !^V,^,^44p\ V4#)zInput must be finitez&n must be a non-zero positive integer.)rr rmathceillogr)nmin_ks& r_schur_subsets_numberr0AscAJJ/00AvABB a $((1Q37A./ 5>rc\V\4'dVP'g \R4h\ V4pV^8Xd^..pM"V^8Xd^^..pMV^8Xd.RO.pM^^.^^..p\ V4V8dj\ W 4p\\ V4V^, ^,^,4Uu.uFp^V,^,NK ppVR;;,V, uu&KyV#uupi)a This function returns the partition in the minimum number of sum-free subsets according to the lower bound given by the Schur Number. Parameters ========== n: a number n is the upper limit of the range [1, n] for which we need to find and return the minimum number of free subsets according to the lower bound of schur number Returns ======= List of lists List of the minimum number of sum-free subsets Notes ===== It is possible for some n to make the partition into less subsets since the only known Schur numbers are: S(1) = 1, S(2) = 4, S(3) = 13, S(4) = 44. e.g for n = 44 the lower bound from the function above is 5 subsets but it has been proven that can be done with 4 subsets. Examples ======== For n = 1, 2, 3 the answer is the set itself >>> from sympy.combinatorics.schur_number import schur_partition >>> schur_partition(2) [[1, 2]] For n > 3, the answer is the minimum number of sum-free subsets: >>> schur_partition(5) [[3, 2], [5], [1, 4]] >>> schur_partition(8) [[3, 2], [6, 5, 8], [1, 4, 7]] zInput value must be a number)) isinstancerr rr0len_generate_next_listrange)r.number_of_subsetssum_free_subsetsrmissed_elementss& rschur_partitionr=Os^!UAKKK788-a0AvC5 aF8 a%;FQF+  "3 3./?C,1#6F2G!A#PQTU,VW,Vq1Q377,VW/ Xs4C(c.pVFxpVUu.uFqD^,V8:gKV^,NK ppVUu.uF)qD^,^, V8:gKV^,^, NK+ ppWV,pVPV4Kz \\V4^,4Uu.uF*p^V,^,V8:gK^V,^,NK, p pVPV 4TpV#uupiuupiuupi)r4)appendr9r7) current_listr.new_listitemnumbertemp_1temp_2new_itemr last_lists && rr8r8sH)-?vQ(&((?-1GT6AX\Q5F,&(Q,,TG?!  #(L(9!(;"<M"rLs9 "(&2(2j AH r