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thuật toán AKT
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Thut Ton AKT Bc 1: Mi nh n v hm g,h, f n. M nh u tin S0 . Gn g(S0) =0 S dng tri thc b sung c tnh h(S0) Tnh f(S0) = g(S0) + h(S0)
Bc 2: Chn mt nh m vi hm f l min v gi l nh N. Nu N l ch => dng. ng i t nh ban u n nh N l ngn nht v
bng g(N). Nu khng tn ti N th cy biu din vn khng c ng i ti mc tiu =>
dng (bi ton khng li gii) Nu tn ti nhiu hn mt nh N c cng hm fmin th phi kim tra xem trong s
c nh no l ch hay khng? o Nu c => Dng o Nu khng => chn ngu nhin mt trong cc nh v gi l nh N
Bc 3: ng nh N v m mi nh sau N. Vi mi nh S sau N, tnh g(s)=g(N) + g(N ->S) Dng tri thc b sung c tnh hm h(S) Tnh f(S) = g(S) + h(S)
Bc 4: Quay li bc 2
p Dng: p dng thut ton AKT gii bi ton Taci vi c lng nh sau:
trong l s bc phi dch chuyn t nht theo chiu
ngang v chiu dc y ai a v ng v tr bi b.
Start Goal
1 2 3
4 5 6
7 8
Ta c:
1 2 3
4 8 5
7 6
HoangXuanTypewriternh chnh l trng thi, nguyn c ci
HoangXuanTypewriter
V h = 0 => ch => dng Kt lun: ng i ngn nht g = 5
g = 0
h = 5
f = 5
g = 1
h = 6 f = 7
g = 1
h = 6 f = 7
g = 1
h = 4 f = 5
g = 2 h = 3
f = 5
g = 3
h = 2
f = 5
g = 4
h = 3 f = 7
g = 4 h = 1
f = 5
g = 3
h = 4
f = 7
g = 4 h = 3
f = 7
g = 5
h = 2 f = 7
g = 5
h = 0
f = 5