Ngamanye amaxesha umbuzo omnye kwinkqubo yangokuhlwa (kule meko evela kumsasazi ohloniphekileyo uKai Pflaume) wanele ukuguqula isiphelo somboniso wemibuzo esingenabungozi sibe yingxaki encinci yokulungisa. Yiloo nto kanye eyenzekayo kwi-"Who Knows What?" Umbuzo oyingcali: Olu didi luyaziwa, impendulo ayikabonakali - kodwa imiba sele imisela ukuba zeziphi iziphumo ezisalungileyo.
Masithathe amaqela amabini \(A\) kunye \(B\) . Ngaphambi kombuzo wokugqibela, iqela \(A\) laliphumelele isixa \(x_a\) , kwaye iqela \(B\) laliphumelele isixa \(x_b\) . Siqwalasela ityala
$$
x_a > x_b > 0.
$$
Amaqela ngoku abheja amanani apheleleyo.
$$
1 \leq y_a \leq x_a,\qquad 1 \leq y_b \leq x_b.
$$
Ukuba impendulo ichanekile, isixa esibekwe kwi-stake siyongezwa; ukuba impendulo ayichanekanga, iyasuswa. Kwiziphumo ezine ezinokwenzeka, iziphumo zamanqaku okugqibela alandelayo.:
$$
\begin{array}{c|c|c}
\text{Fall} & A & B\\
\hline
A \text{ richtig}, B \text{ richtig} & x_a+y_a & x_b+y_b\\
A \text{ richtig}, B \text{ falsch} & x_a+y_a & x_b-y_b\\
A \text{ falsch}, B \text{ richtig} & x_a-y_a & x_b+y_b\\
A \text{ falsch}, B \text{ falsch} & x_a-y_a & x_b-y_b
\end{array}
$$
Kulo mboniso, i-tie ikhokelela kumbuzo woqikelelo. Ke ngoko, ihlala ibonakala kwi-matrix njengetyala elahlukileyo "=". Kwixabiso lepesenti le-optimal probability, sibala i-true wins kwi-sub-cases ezine ezinokwenzeka ngokulinganayo. Ipateni efana ne \(|A|B|A|A|\) ivelisa i-true wins ezintathu kwiQela \(A\) kunye nenye kwiQela \(B\) . I-tie ayibalwa njengoloyiso oluthe ngqo kwiQela ngalinye. Olu bala luchaneke ngakumbi lubalulekile; akwanelanga ukubala nje iseli yonke njenge-"blue" okanye "red".
Le modeli inengcamango: Siphatha iindibaniselwano zeempendulo ezine ngokulinganayo. Ke ngoko, ayikuko ukuba iQela \(A\) okanye iQela \(B\) liyazi ngcono na icandelo, kodwa kuxhomekeke kuphela kwindlela esetyenziswayo ngaphambi kokuba kuphendulwe.
Masi
$$
d=x_a-x_b.
$$
Emva koko \(d>0\) lilungelo leqela \(A\) . Umbuzo ngoku ngulo: Yeyiphi intlawulo efanelekileyo?
I-matrix epheleleyo yokubheja okunokwenzeka ingabalwa ngokuguquguqukayo.:
Umbono weQela A
Siqala ngokujonga ukuba iqela \(A\) liphumelela nini ngee-fixed stakes \(y_a\) kunye \(y_b\) .
Ityala
$$
A \text{ richtig}, B \text{ falsch}
$$
Ihlala isiya kwiQela \(A\) , kuba
$$
x_a+y_a > x_b-y_b
$$
Oku kusebenza ngokuzenzekelayo kuba \(x_a>x_b\) kunye \(y_a,y_b>0\) .
Kwezinye iimeko ezintathu, sifumana:
$$
\begin{array}{c|c}
\text{Fall} & A \text{ gewinnt genau dann}\\
\hline
A \text{ richtig}, B \text{ richtig} & x_a+y_a>x_b+y_b\\
A \text{ falsch}, B \text{ richtig} & x_a-y_a>x_b+y_b\\
A \text{ falsch}, B \text{ falsch} & x_a-y_a>x_b-y_b
\end{array}
$$
Nge \(x_a=x_b+d\) oku kuba:
$$
\begin{array}{c|c}
\text{Fall} & A \text{ gewinnt genau dann}\\
\hline
A \text{ richtig}, B \text{ richtig} & d+y_a>y_b\\
A \text{ falsch}, B \text{ richtig} & d-y_a>y_b\\
A \text{ falsch}, B \text{ falsch} & d-y_a>-y_b
\end{array}
$$
Ngoko ke:
$$
\begin{array}{c|c}
\text{Fall} & A \text{ gewinnt genau dann}\\
\hline
A \text{ richtig}, B \text{ richtig} & y_b<y_a+d\\
A \text{ falsch}, B \text{ richtig} & y_b<d-y_a\\
A \text{ falsch}, B \text{ falsch} & y_b>y_a-d
\end{array}
$$
Indlela yokubala ngoku ibalulekile. Ngaphambili, umntu unokulingeka ukuba ahlole iseli nganye ngokusekelwe ekubeni iqulethe iimeko ezingaphezulu kune \(A\) kune \(B\) . Nangona kunjalo, oku kulula kakhulu ekubaleni amathuba okuphumelela. Ezi meko zine zizinto ezinokwenzeka ngokulinganayo. Ke ngoko, \(|A|B|A|A|\) ayibalwa njengokuphumelela okukodwa kwi \(A\) , kodwa ibalwa njengeemeko ezintathu eziphumeleleyo kwi- \(A\) .
Kwi-stake esisigxina \(y_a\) yeqela \(A\) ngoko ke sidibanisa \(A\) iingeniso zomntu ngamnye kwiimeko ezine kuzo zonke ii-stake ezinokwenzeka \(y_b=1,2,\ldots,x_b\) .
Ityala elithi " \(A\) correct, \(B\) correct" lihlala lisiya kwiqela \(A\) . Oku sele kuvelisa \(x_b\) ii-subcases eziphumeleleyo.
La manani alandelayo aphumela kwezinye iimeko ezintathu.:
$$
\begin{aligned}
N_1(y_a)&=\min(x_b,d+y_a-1),\\
N_3(y_a)&=\min(x_b,\max(0,d-y_a-1)),\\
N_4(y_a)&=\begin{cases}
x_b, & y_a\leq d,\\
\max(0,x_b-y_a+d), & y_a>d.
\end{cases}
\end{aligned}
$$
Oku kuthetha ukuba inani lee-subcases eziphunyelelwe yiQela \(A\) li
$$
N_A(y_a)=x_b+N_1(y_a)+N_3(y_a)+N_4(y_a).
$$
Amathuba okuphumelela ahambelanayo ngu
$$
P_A(y_a)=\frac{N_A(y_a)}{4x_b}.
$$
Ekubeni i-tie ingafakwanga apha, \(P_A\) lithuba lokuphumelela umbuzo oyintloko ngokuthe ngqo (ngaphandle kombuzo woqikelelo).
Olu balo luchanekileyo lutshintsha kancinci i-optimal xa luthelekiswa ne-implementation elula yeeseli. KwiQela \(A\) iziphumo zezi ndawo zilandelayo ze-application ezifanelekileyo.:
$$
\boxed{
\begin{cases}
1\leq y_a\leq2, & x_b=1,\ d=2,\\
d\leq y_a\leq x_b-d+1, & 2d\leq x_b+1,\\
1\leq y_a\leq d, & 2d=x_b+2,\\
1\leq y_a\leq \max(1,x_b-d+1,d-x_b-1), & 2d>x_b+2.
\end{cases}
}
$$
Zonke iibhethi kule ndawo zinyusa amathuba okuphumelela kweQela \(A\) Ukuba ufuna ukubheja elona nani likhulu phakathi kweebhethi ezilungileyo ngokulinganayo, kufuneka usebenzise umda ochanekileyo wendawo.
Umzekelo:
$$
x_a=30,\qquad x_b=22.
$$
Emva koko
$$
d=x_a-x_b=8.
$$
Pha
$$
2d=16\leq 23=x_b+1
$$
Uluhlu olufanelekileyo luyasebenza.
$$
8\leq y_a\leq 15.
$$
Eyona ndlela ilungileyo yokusebenzisa ke ngoko
$$
\boxed{y_a=15}.
$$
Indlela endala yokuqwalasela iiseli ezipheleleyo ibingayi kucebisa uluhlu \(9\leq y_a\leq 14\) . Inani lamatyala angaphelelanga libonisa ngokuchanekileyo ukuba amaxabiso amabini emida \(8\) kunye \(15\) nawo alungile.
Umbono weQela B
Ngoku sijonga imeko efanayo ngokwembono yeqela elilandela emva \(B\) . Nalapha, asisabali kuphela iiseli ezipheleleyo, kodwa sibala iingeniso ezizimeleyo \(B\) kwii-subcases ezine.
Ityala
$$
A \text{ richtig}, B \text{ falsch}
$$
Iqela \(B\) lisoloko lilahlekelwa. Kwiimeko ezintathu ezisele, Iqela \(B\) lifumana inani elilandelayo lee-sub-cases eziphumeleleyo kwi-stake esisigxina \(y_b\) xa lihlanganisa yonke \(y_a=1,2,\ldots,x_a\):
$$
\begin{aligned}
M_1(y_b)&=\max(0,y_b-d-1),\\
M_3(y_b)&=x_a-\max(0,d-y_b),\\
M_4(y_b)&=\max(0,x_b-y_b).
\end{aligned}
$$
Kunjalo ke
$$
N_B(y_b)=M_1(y_b)+M_3(y_b)+M_4(y_b)
$$
kunye namathuba anxulumene nokuphumelela
$$
P_B(y_b)=\frac{N_B(y_b)}{4x_a}.
$$
\(y_b\leq d\) kwenza kube lula
$$
N_B(y_b)=2x_b.
$$
Kwi \(y_b>d\) umntu ufumana kuphela
$$
N_B(y_b)=2x_b-1.
$$
Le yindawo efanelekileyo yeQela \(B\)
$$
\boxed{
1\leq y_b\leq \min(d,x_b).
}
$$
Olu lulungiso olubalulekileyo xa luthelekiswa nendlela ye-cruder cell majority: Ubhejo olufanelekileyo lweQela \(B\) alusoloko luyi \(1\) kuphela. Umzekelo, ukuba iQela \(A\) liphambili ngo- \(d=8\) kwaye iQela \(B\) linokubheja ubuninzi bama \(22\) , ngoko ke zonke iibhejo ze- \(B\) zilungile malunga namathuba okuphumelela.:
$$
1\leq y_b\leq 8.
$$
Ingqiqo ihlala ifana: Iqela elilandela emva akufuneki libheje ngokungeyomfuneko. Ngelixa ukubheja okuphezulu kakhulu kuphucula iimeko zomntu ngamnye, kwenza abanye babe mandundu. Kwakamsinya nje ukuba \(y_b\) udlule kwi-deficit \(d\) , iqela \(B\) lilahlekelwa yimeko iyonke. Ke ngoko, iqela elikhokelayo libheja ngendlela yokuba, kuzo zonke iibhejo ezinokwenzeka ngumchasi, liphumelele iimeko zomntu ngamnye kangangoko kunokwenzeka.
Umlandeli akabheji nge-euro enye kuphela, kodwa ubuninzi bufana ne-deficit. Umbuzo oyintloko ke ngoko ngumzekelo omhle wokuba ingakanani ithiyori yomdlalo equlethwe kumgaqo wemibuzo obonakala ulula: okubalulekileyo asikuko kuphela ukuba yeyiphi iseli egqibela iluhlaza okwesibhakabhaka okanye ibomvu, kodwa nokuba zingaphi kwezine ii-sub-cases ezikwiseli eziphunyeziweyo.