Trong ml;lenciy chung ta se neu len thu~t giai t6ng quat cho va'n d§ 1. Y
tu'dng cd ban Ia thvc hi~n IDQtqua trinh guy di€n tie'n ke't h<;5pvoi IDQtso' qui tae
heuristic nh~m tang cuC1ngt6c dQ giai quye't bai roan va d~t dU<1emQt IC1igiai t6t nhanh hdn. B~ thie't ke' thu~t giai nciy ta ccin dinh nghla mQt so' khai ni~m lien
quail baa g6m cac khai ni~m: "Sf! h(lp nh(ft" eua cae s1;1'ki~n, mQt "bade gidi", mQt "lili gidi" va "sf! gidi dLt(le".
- Dinh nghla 3.3: Ta noi 2 s1;1'ki~nfaetl va faet2 la h(lp nha:t khi co cac di§u
ki~n sail day:
(l)faetl vafaet2 la cling lo~i k, va (2)faetl =faet2 ne'u k =1, 2, 6.
[faetl[l], {faetl[2..nops(faetl)]}] =[faet2[1], {faet2[2..nops(faet2)]}]
ne'u k =6 va quail h~ trong s1;1'ki~nfaetl co tinh "d6i xung".
lhs(faetl) =Ihs(faet2) va eompute(rhs(faetl)) =compute(rhs(faet2)) ne'u k
=3.
( lhs(faetl) = Ihs(fact2) va rhs(faetl) =rhs(faet2)) hay
( lhs(faetl) =rhs(faet2) va rhs(faetl) =Ihs(faet2)) ne'u k =4.
evalb(simplify( expand(lhs(faetl)-rhs(factl)- Ihs(faet2)+rhs(faet2)))= 0)
hay
evaIb(simplify( expand(lhs(factl)-rhs(factl)+ Ihs(fact2)-rhs(fact2))) =0) ne'uk=5.
Trang do cae ky hi~u co y nghla nhu'sau:
{. . .} : mQt t~p h<;fp. [. . .] : mQt danh saeh (list).
L[ 1] : thanh phfin thu nh§t eua danh saeh L.
L[Lj] : day cae thanh phfin tu vi tri i de'n vi tri j trong danh saeh L. nops(L) : s6 thanh phfin eua danh saeh L.
lhs(eqn) : vfi treEeua d~ng thue <eqn>, rhs(eqn) : vfi phai eua d~ng thue <eqn>,
compute(expr) : kfit qua tinh roan eua bi6u thuc <expr>, expand(expr) : khai tri6n bi€u thue <expr>,
simplify(expr) : don gian bi€u thue <expr>,
evalb(exprl = expr2) : danh gia vi~e so sanh b~ng nhau gifi'abi€u thue <exprl> va bi6u thuc <expr2>.
Cae vi d\l v~ cae s1,1'ki~n h<;fpnh§t voi nhau:
[ObI, "TA1tLGIAC"] va [ObI, "TAM_GIAC"]. DOAN[A,B] va DOAN[A,B].
TAM_GIAC[A,B,C]. a va DOAN[B,C]. Ob.a =(m+I)"'2 va Ob.a =mA2 + 2*m + 1. a =b va a=b.
ObI =Ob2 va Ob2 =ObI.
aA2 =bA2 + cA2 va bA2 =aA2 - cA2.