Ngezinye izikhathi umbuzo owodwa ohlelweni lwasekuseni (kulokhu kusuka kumethuli ohlonishwayo uKai Pflaume) wanele ukuguqula isiphetho sombukiso wemibuzo esingenangozi sibe inkinga encane yokwenza ngcono. Yilokho kanye okwenzekayo ku-"Who Knows What?" Umbuzo oyingcweti: Isigaba siyaziwa, impendulo ayikakafiki - kodwa izinkinga sezivele zinquma ukuthi yimiphi imiphumela esemihle.
Ake sithathe amaqembu amabili \(A\) kanye \(B\) . Ngaphambi kombuzo wokugcina, iqembu \(A\) laliwine inani \(x_a\) , kanye neqembu \(B\) laliwine inani \(x_b\) . Sibheka icala
$$
x_a > x_b > 0.
$$
Amaqembu manje asebheja izinombolo eziphelele.
$$
1 \leq y_a \leq x_a,\qquad 1 \leq y_b \leq x_b.
$$
Uma impendulo ilungile, inani elibekwe eceleni liyangezwa; uma impendulo ingalungile, liyasuswa. Ngemiphumela emine engaba khona, imiphumela yamaphuzu okugcina elandelayo iyatholakala.:
$$
\begin{array}{c|c|c}
\text{Fall} & A & B\\
\hline
A \text{ richtig}, B \text{ richtig} & x_a+y_a & x_b+y_b\\
A \text{ richtig}, B \text{ falsch} & x_a+y_a & x_b-y_b\\
A \text{ falsch}, B \text{ richtig} & x_a-y_a & x_b+y_b\\
A \text{ falsch}, B \text{ falsch} & x_a-y_a & x_b-y_b
\end{array}
$$
Embukisweni, ukulingana kuholela embuzweni wokulinganisa. Ngakho-ke, kuhlala kubonakala ku-matrix njengecala elihlukile elimakwe ngokuthi "=". Kumanani amaphesenti wamathuba okuwina, sibala ukuwina okuqondile ezimweni ezine ezincane ezinokwenzeka ngokulinganayo. Iphethini efana ne \(|A|B|A|A|\) iveza ukuwina okuqondile okuthathu kweQembu \(A\) kanye neyodwa yeQembu \(B\) . Ukulingana akubalwa njengokunqoba okuqondile kunoma yiliphi iqembu. Lokhu kubalwa okunembe kakhulu kubalulekile; akwanele ukubala nje iseli lonke "njengoluhlaza okwesibhakabhaka" noma "obomvu".
Le modeli iqukethe ukucabangela: Siphatha inhlanganisela yezimpendulo ezine njengokungenzeka ngokulinganayo. Ngakho-ke, akusikho ukuthi iQembu \(A\) noma iQembu \(B\) liyasazi kangcono isigaba, kodwa kuphela mayelana nesu elisetshenziswayo ngaphambi kokuphendula.
Ake
$$
d=x_a-x_b.
$$
\(d>0\) kuba yinzuzo yeqembu \(A\) . Umbuzo manje uthi: Iyini imali engcono kakhulu?
I-matrix ephelele yokubheja okungenzeka ingabalwa ngokuguquguqukayo.:
Umbono weQembu A
Siqala sihlole ukuthi iQembu \(A\) liwina nini ngeziteki ezihleliwe \(y_a\) kanye \(y_b\) .
Icala
$$
A \text{ richtig}, B \text{ falsch}
$$
Kuhlala kuya kuThimba \(A\) , ngoba
$$
x_a+y_a > x_b-y_b
$$
Lokhu kusebenza ngokuzenzakalelayo ngoba \(x_a>x_b\) kanye \(y_a,y_b>0\) .
Kwamanye amacala amathathu, sithola:
$$
\begin{array}{c|c}
\text{Fall} & A \text{ gewinnt genau dann}\\
\hline
A \text{ richtig}, B \text{ richtig} & x_a+y_a>x_b+y_b\\
A \text{ falsch}, B \text{ richtig} & x_a-y_a>x_b+y_b\\
A \text{ falsch}, B \text{ falsch} & x_a-y_a>x_b-y_b
\end{array}
$$
Nge- \(x_a=x_b+d\) lokhu kuba:
$$
\begin{array}{c|c}
\text{Fall} & A \text{ gewinnt genau dann}\\
\hline
A \text{ richtig}, B \text{ richtig} & d+y_a>y_b\\
A \text{ falsch}, B \text{ richtig} & d-y_a>y_b\\
A \text{ falsch}, B \text{ falsch} & d-y_a>-y_b
\end{array}
$$
Ngakho-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 manje ibalulekile. Ngaphambilini, umuntu angase alingeke ukuhlola iseli ngalinye ngokusekelwe kuphela ekutheni liqukethe amacala amaningi e \(A\) kune \(B\) . Kodwa-ke, lokhu kulula kakhulu ekubaleni amathuba okuwina. Amacala amancane amane ngokwawo ayizehlakalo ezingaba khona ngokulinganayo. Ngakho-ke, \(|A|B|A|A|\) ayibalwa njengokuwina okukodwa kwe \(A\) , kodwa kunalokho njenge-sub-cases ezintathu eziwinile ze \(A\) .
Nge-stake eqondile \(y_a\) yeqembu \(A\) ngakho-ke sihlanganisa okufakiwe ngakunye \(A\) ezimweni ezine kuzo zonke izi-stake ezingaba khona \(y_b=1,2,\ldots,x_b\) .
Icala elithi " \(A\) correct, \(B\) awright" lihlala liya eqenjini \(A\) . Lokhu sekuvele kuveza ama-subcase awinile angu- \(x_b\) .
Izinombolo ezilandelayo ziphumela kwamanye amacala amathathu.:
$$
\begin{aligned}
N_1(y_a)&=\min(x_b,d+y_a-1),\\
N_3(y_a)&=\min(x_b,\max(0,d-y_a-1)),\\
N_4(y_a)&=\begin{cases}
x_b, & y_a\leq d,\\
\max(0,x_b-y_a+d), & y_a>d.
\end{cases}
\end{aligned}
$$
Lokhu kusho ukuthi inani lama-subcase anqotshwe yiThimba \(A\) liyi
$$
N_A(y_a)=x_b+N_1(y_a)+N_3(y_a)+N_4(y_a).
$$
Amathuba okuwina ahambisanayo yilawa
$$
P_A(y_a)=\frac{N_A(y_a)}{4x_b}.
$$
Njengoba ukulingana kungafakiwe lapha, \(P_A\) yilona kanye ithuba lokuwina umbuzo oyinhloko ngqo (ngaphandle kombuzo wokulinganisa).
Lokhu kubala okunembe kakhudlwana kushintsha kancane okulungile uma kuqhathaniswa neningi lamaseli alula. KwiQembu \(A\) kuvela izindawo ezilandelayo zohlelo lokusebenza olufanele.:
$$
\boxed{
\begin{cases}
1\leq y_a\leq2, & x_b=1,\ d=2,\\
d\leq y_a\leq x_b-d+1, & 2d\leq x_b+1,\\
1\leq y_a\leq d, & 2d=x_b+2,\\
1\leq y_a\leq \max(1,x_b-d+1,d-x_b-1), & 2d>x_b+2.
\end{cases}
}
$$
Zonke izibhejo kule ndawo zikhulisa amathuba okuwina eQembu \(A\) Uma ufuna ukubheja inani elikhulu ngangokunokwenzeka phakathi kwezibhejo ezinhle ngokulinganayo, kufanele usebenzise umkhawulo ofanele wendawo ngaso sonke isikhathi.
Isibonelo:
$$
x_a=30,\qquad x_b=22.
$$
Ngemuva kwalokho
$$
d=x_a-x_b=8.
$$
Lapho
$$
2d=16\leq 23=x_b+1
$$
Ububanzi obuhle buyasebenza.
$$
8\leq y_a\leq 15.
$$
Ngakho-ke ukusetshenziswa okuhle kakhulu
$$
\boxed{y_a=15}.
$$
Indlela endala yokucabangela amaseli aphelele ngabe isikisele ububanzi \(9\leq y_a\leq 14\) . Inani lamacala angaphelele libonisa ngokunembile ukuthi amanani amabili omngcele \(8\) kanye \(15\) nawo angcono kakhulu.
Umbono weQembu B
Manje sibheka isimo esifanayo ngokombono weqembu elilandelanayo \(B\) . Nalapha futhi, asisabali amaseli aphelele kuphela, kodwa nokufakwa ngakunye \(B\) ezinhlakeni ezine ezincane.
Icala
$$
A \text{ richtig}, B \text{ falsch}
$$
Ithimba \(B\) lihlala lihlulwa. Ezimweni ezintathu ezisele, ithimba \(B\) lithola inani elilandelayo lamacala amancane awinile nge-stake ehleliwe \(y_b\) uma lifingqa yonke \(y_a=1,2,\ldots,x_a\):
$$
\begin{aligned}
M_1(y_b)&=\max(0,y_b-d-1),\\
M_3(y_b)&=x_a-\max(0,d-y_b),\\
M_4(y_b)&=\max(0,x_b-y_b).
\end{aligned}
$$
Kanjalo
$$
N_B(y_b)=M_1(y_b)+M_3(y_b)+M_4(y_b)
$$
kanye namathuba ahambisanayo okuwina
$$
P_B(y_b)=\frac{N_B(y_b)}{4x_a}.
$$
\(y_b\leq d\) kwenza kube lula ku-
$$
N_B(y_b)=2x_b.
$$
Ku- \(y_b>d\) umuntu uthola kuphela
$$
N_B(y_b)=2x_b-1.
$$
Lena yindawo efanele kakhulu yeQembu \(B\)
$$
\boxed{
1\leq y_b\leq \min(d,x_b).
}
$$
Lokhu ukulungiswa okubalulekile uma kuqhathaniswa nendlela ye-cruder cell majority: Ukubheja okuhle kakhulu kweThimba \(B\) akusikho ukuthi kuhlukile ku \(1\) Isibonelo, uma iThimba \(A\) lihamba phambili ngo- \(d=8\) futhi iThimba \(B\) lingabheja okungenani \(22\) , khona-ke konke ukubheja kwe \(B\) kulungile maqondana namathuba okuwina.:
$$
1\leq y_b\leq 8.
$$
Umbono uhlala ufana: Ithimba elilandela ngemuva akufanele libheje phezulu ngokungadingekile. Ngenkathi ukubheja okuphezulu kakhulu kuthuthukisa izimo zomuntu ngamunye, kwenza ezinye zibe zimbi kakhulu. Lapho nje \(y_b\) edlula i-deficit \(d\) , iqembu \(B\) lilahlekelwa yisimo sonke. Ngakho-ke, iqembu eliholayo libheja ngendlela yokuthi, kuzo zonke izibhejo ezingaba khona zomphikisi, liwine izimo eziningi zomuntu ngamunye ngangokunokwenzeka.
Umlandeli akabheji ngempela i-euro eyodwa, kodwa ubheja kakhulu kune-deficit. Ngakho-ke umbuzo oyinhloko uyisibonelo esihle sokuthi ingakanani imfundiso yomdlalo equkethwe emthethweni wemibuzo obonakala ulula: okubalulekile akukhona nje ukuthi yiliphi iseli eligcina liluhlaza okwesibhakabhaka noma libomvu, kodwa nokuthi zingaki izigameko ezine ezincane ezikulelo seli ezinqotshiwe.