Ukuphazamiseka kukaSimpson yenye yezinto eziqondakala ngokulula kwaye kwangaxeshanye izinto ezimangalisayo kumanani. Kuyenzeka nanini na xa amaqela edatha ebonisa imeko ethile, kodwa loo meko iyahlehla xa amaqela edityanisiwe. Ngoncedo lomzekelo olula, indida inokuqondwa kwangoko.
Siqwalasela yeeseti ezimbini disjoint \(\#1\) kwaye \(\#2\) kwakunye \(G = \#1 \cup \#2\) kwaye ukuvavanya izinga lempumelelo \(A\) kwaye phakathi kwezi iiseti \(B\):
| \(A\) | \(B\) | \(win\) | |
| \(\#1\) | \(\frac{1}{1}=100\%\) | \(\frac{3}{4}=75\%\) | \(A\) |
| \(\#2\) | \(\frac{2}{5}=40\%\) | \(\frac{1}{3}=33\%\) | \(A\) |
| \(\#1 \cup \#2\) | \(\frac{3}{6}=50\%\) | \(\frac{4}{7}=57\%\) | \(B\) |
Kuyavela ukuba \(A\) uphumelele ngakumbi kune \(B\) kwi \(\#1\) kunye \(\#2\) \(B\) , kodwa ngelishwa kwi \(G\) \(B\) uphumelele ngakumbi \(A\) . Lo mzekelo ukwenye yezo zineseti encinci \(G\) nge \(|G|=13\) . Akukho \(G\) nge \(|G|<13\) (ubungqina ngamandla anyanzelekileyo).
Ngoku sahlulahlula iseti \(G\) endaweni ye \(2\) zibe \(3\) disjoint subsets \(\#1, \, \#2, \, \#3\) with \(\#1 \cup \#2 \cup \#3 = G\) Emva koko sakha imeko enomdla yokuba kuyo yonke into \(e_k \neq \emptyset\) yamandla \(P(G)\) ye \(G\) kulandelayo kuyasebenza: $$\forall e_1, e_2 \in P(G): |e_1| \neq |e_2| \Rightarrow win(e_1) \neq win(e_2) \land |e_1| = |e_2| \Rightarrow win(e_1) = win(e_2)$$ $$\forall e_1, e_2 \in P(G): |e_1| \neq |e_2| \Rightarrow win(e_1) \neq win(e_2) \land |e_1| = |e_2| \Rightarrow win(e_1) = win(e_2)$$
Emva kweeyure ezimbalwa zokunyanzelwa kwi-Core i7 esemgangathweni, lo mzekelo ulandelayo unokufunyanwa:
| \(A\) | \(B\) | \(C\) | \(win\) | |
| \(\#1\) | \(\frac{6}{7}=85,71\%\) | \(\frac{12}{15}=80,00\%\) | \(\frac{22}{37}=59,46\%\) | \(A\) |
| \(\#2\) | \(\frac{95}{167}=56,89\%\) | \(\frac{48}{88}=54,55\%\) | \(\frac{38}{67}=56,72\%\) | \(A\) |
| \(\#3\) | \(\frac{48}{144}=33,33\%\) | \(\frac{16}{50}=32,00\%\) | \(\frac{2}{20}=10,00\%\) | \(A\) |
| \(\#1 \cup \#2\) | \(\frac{101}{174}=58,05\%\) | \(\frac{60}{103}=58,25\%\) | \(\frac{60}{104}=57,69\%\) | \(B\) |
| \(\#1 \cup \#3\) | \(\frac{54}{151}=35,76\%\) | \(\frac{28}{65}=43,08\%\) | \(\frac{24}{57}=42,11\%\) | \(B\) |
| \(\#2 \cup \#3\) | \(\frac{143}{311}=45,98\%\) | \(\frac{64}{138}=46,38\%\) | \(\frac{40}{87}=45,98\%\) | \(B\) |
| \(\#1 \cup \#2\cup \#3\) | \(\frac{149}{318}=46,86\%\) | \(\frac{76}{153}=49,67\%\) | \(\frac{62}{124}=50,00\%\) | \(C\) |
Ngokwenjenje (uthatha ixesha elilinganiselweyo lekhompyuter) imizekelo ye \(n\) ukungadibanisi iiseti kunye nokuziphatha okufanayo kunokufunyanwa. Ukuba iimeko ezinjalo zenzeka ngokwenyani, nasiphi na isigqibo kwisindululo sempumelelo yeqela zombini zisengqiqweni kwaye azinantsingiselo.
Okwangoku, sikwacebisa ukuba kufundwe ngokuBangel ' umdla : Iimodeli, Ukuqiqa kunye nokuchaphazeleka kukaPearl waseJudiya .