EBavaria, ibakala eliqhelekileyo eliyi- \(2{,}33\) liqinisekisa ukufaneleka kweGymnasium \(2{,}34\) ingavumi. Kule nqanaba libukhali, ulandelelwano lwemisebenzi emibini yokujikelezisa lunokumisela ukuba umntwana uwela kweliphi icala. UGünther Felbinger uxoxa ngemiphumo enokubakho yale mpembelelo yokujikelezisa kwisithuba sebhlog . Ngenxa yomdla nje, ndizibale ngokwam iziphumo kuzo zonke iindibaniselwano zamanqanaba ezivunyelweyo kwimodeli.
Umbuzo wam ophambili ngulo: Kukangaphi apho ukujikelezisa kwasekuqaleni ukuya kumabanga esifundo esipheleleyo kukhokelela kuhlobo olwahlukileyo lokubala kwesikolo kunokubala kusetyenziswa i-avareji yesifundo esingajikeleziswanga? Ngaphezu koko, kuya kuba nomdla ukubona ukuba yintoni etshintshayo ukuba i-avareji yesifundo ngasinye (njengoko kucetyisiwe apho) ekuqaleni ijikeleziswa kwindawo enye yedesimali.
NgokweCandelo 6 leMimiselo yeSikolo samaBavaria esiPhakamileyo, ibakala eliphakathi libalwa ukusuka kumanqaku esiJamani, kwiMathematika, nakwiZifundo zaseKhaya nakwiZifundo eziQhelekileyo. Imithetho yokuphumeza ichaza imida efana nokufikelela kwi \(2{,}33\) kunye nokufikelela kwi \(2{,}66\) . Kwimodeli yokubala elandelayo, umndilili ngamnye unqunyulwa kwiindawo ezimbini zedesimali kwaye awuzuliswa kwakhona. Ngoko ke, umzekelo...
\[
\frac{2{,}33+2{,}34+2{,}34}{3}=2{,}3366\ldots \longrightarrow 2{,}33.
\]
Oku kubalulekile ekuhleleni: \(2{,}3366\ldots\) ayibi \(2{,}34\) , kodwa \(2{,}33\) kwaye ke ngoko isanele kwisikolo segrama.
Ungeniso lokuqala lubonisa ukujikeleziswa kwangoko ngexabiso eliqikelelweyo le-mean values \(1{,}6\) , \(2{,}6\) , kunye \(2{,}6\) . Kwezi, amabanga apheleleyo \(2\) , \(3\) , kunye \(3\) aqala ukuvela. I-mean yabo ngu
\[
\frac{2+3+3}{3}=2{,}6666\ldots \longrightarrow 2{,}66,
\]
oko kuthetha ukuba umfundi akafanelekanga ukuya kwisikolo segrama. Nangona kunjalo, ngaphandle kokusondeza amanqaku apheleleyo, iziphumo kumzekelo ocingelwayo ziya kuba...
\[
\frac{1{,}6+2{,}6+2{,}6}{3}=2{,}2666\ldots \longrightarrow 2{,}26.
\]
Kumzekelo wesibini, i-rounding isebenza kwicala elichaseneyo: \(2{,}4\) , \(3{,}4\) kunye \(2{,}4\) ziba ngamaxabiso enqaku apheleleyo \(2\) , \(3\) kunye \(2\) , ngokusemthethweni \(2{,}33\) . Nangona kunjalo, amaxabiso angajikelezwanga aphumela kwi \(2{,}73\) . Imizekelo ibonisa ngokucacileyo isiphumo. Nangona kunjalo, ayibonisi ukuba kwenzeka kangaphi okanye ukuba iziphumo zendlela eyahlukileyo yokujikelezisa ziya kuba yintoni.
Ngoko ke, xa ndibala ndisebenzisa isithuba esicacileyo sesiphumo. I-mean yesifundo ngasinye inokuba yenye yexabiso \(501\) elingaphakathi kuyo.
\[
1{,}00,\, 1{,}01,\, 1{,}02,\, \ldots, \, 5{,}99, \, 6{,}00
\]
Masithi. Ndilinganisa zonke iindibaniselwano zesiJamani, izibalo, kunye nezifundo zasekuhlaleni kunye nezifundo ngokubanzi ngokulinganayo. Ngoko ke, kukho
\[
501^3=125\,751\,501
\]
Amatyala. Ngoko ke le modeli ayicacanga, kodwa yenziwe ngokungekho mthethweni: Ithatha ukuba imilinganiselo yesifundo isasazwe ngokulinganayo, ukuzimela kwezifundo ezithathu, kunye namanyathelo e-100. Ayibonisi ukusasazwa kwangempela kwabantwana besikolo samabanga aphantsi saseBavaria.
\(d,m,h\) ezintathu sibhala
\[
A(d,m,h)=\frac{\left\lfloor 100\cdot\frac{d+m+h}{3}\right\rfloor}{100}
\]
kwi-avareji iyonke enqunyulwe kwiindawo ezimbini zedesimali. Ngoku sithelekisa iindlela ezintathu zokubala.:
\[
\begin{aligned}
A_0 kunye= A(\operatorname{round}(d),\operatorname{round}(m),\operatorname{round}(h)),\\
A_1 kunye= A(\operatorname{round}_1(d),\operatorname{round}_1(m),\operatorname{round}_1(h)),\\
A_\mathrm{ref} &= A(d,m,h).
\end{aligned}
\]
\(A_0\) ihambelana nokubala kusetyenziswa amabakala apheleleyo, \(A_1\) kwisiphakamiso esinendawo enye yedesimali, kunye \(A_\mathrm{ref}\) kwisalathiso esingajikelezwanga ngaphakathi kwimodeli. Ukubala kungenziwa ngokuthe ngqo kwiRust , umzekelo.:
const SCHOOL_GYMNASIUM: i32 = 0;
const SCHOOL_REALSCHULE: i32 = 1;
const SCHOOL_MITTELSCHULE: i32 = 2;
const FLOAT_TOLERANCE: f64 = 1e-9;
#[derive(Clone, Copy, Default)]
struct Statistic {
conflict_count: u64,
total_count: u64,
}
fn determine_school(average: f64) -> i32 {
if average <= 2.33 {
return SCHOOL_GYMNASIUM;
}
if average <= 2.66 {
return SCHOOL_REALSCHULE;
}
SCHOOL_MITTELSCHULE
}
fn calculate_average(note_deutsch: f64, note_mathe: f64, note_hsu: f64) -> f64 {
let average = (note_deutsch + note_mathe + note_hsu) / 3.0;
((average + FLOAT_TOLERANCE) * 100.0).floor() / 100.0
}
fn round_to_whole_grade(grade: f64) -> f64 {
grade.round()
}
fn round_to_one_decimal(grade: f64) -> f64 {
(grade * 10.0).round() / 10.0
}
fn normalize_grade(grade: f64) -> f64 {
(grade * 100.0).round() / 100.0
}
fn percentage(count: u64, total: u64) -> f64 {
count as f64 / total as f64 * 100.0
}
fn main() {
let mut combination_count = 0;
let mut official_reference_conflicts = 0;
let mut official_proposal_conflicts = 0;
let mut proposal_reference_conflicts = 0;
let mut statistics = [Statistic::default(); 4];
let mut note_deutsch = 1.0;
while note_deutsch <= 6.0 {
let mut note_mathe = 1.0;
while note_mathe <= 6.0 {
let mut note_hsu = 1.0;
while note_hsu <= 6.0 {
combination_count += 1;
let official_average = calculate_average(
round_to_whole_grade(note_deutsch),
round_to_whole_grade(note_mathe),
round_to_whole_grade(note_hsu),
);
let proposed_average = calculate_average(
round_to_one_decimal(note_deutsch),
round_to_one_decimal(note_mathe),
round_to_one_decimal(note_hsu),
);
let reference_average =
calculate_average(note_deutsch, note_mathe, note_hsu);
let official_school = determine_school(official_average);
let proposed_school = determine_school(proposed_average);
let reference_school = determine_school(reference_average);
if official_average == 3.0 {
statistics[0].total_count += 1;
if reference_school != SCHOOL_MITTELSCHULE {
statistics[0].conflict_count += 1;
}
}
if official_average == 2.66 {
statistics[1].total_count += 1;
if reference_school == SCHOOL_GYMNASIUM {
statistics[1].conflict_count += 1;
}
}
if official_average == 2.33 {
statistics[2].total_count += 1;
if reference_school != SCHOOL_GYMNASIUM {
statistics[2].conflict_count += 1;
}
}
if official_average == 2.66 {
statistics[3].total_count += 1;
if reference_school == SCHOOL_MITTELSCHULE {
statistics[3].conflict_count += 1;
}
}
if official_school != reference_school {
official_reference_conflicts += 1;
}
if official_school != proposed_school {
official_proposal_conflicts += 1;
}
if proposed_school != reference_school {
proposal_reference_conflicts += 1;
}
note_hsu = normalize_grade(note_hsu + 0.01);
}
note_mathe = normalize_grade(note_mathe + 0.01);
}
note_deutsch = normalize_grade(note_deutsch + 0.01);
}
println!(
"official/reference: {}%",
percentage(official_reference_conflicts, combination_count)
);
println!(
"official/proposal: {}%",
percentage(official_proposal_conflicts, combination_count)
);
println!(
"proposal/reference: {}%",
percentage(proposal_reference_conflicts, combination_count)
);
for (index, statistic) in statistics.iter().enumerate() {
println!(
"#{}: {}% ({}/{})",
index + 1,
percentage(statistic.conflict_count, statistic.total_count),
statistic.conflict_count,
statistic.total_count
);
}
}Ukunyamezelana okuncinci ku calculate_average ithintela ixabiso elifana neli \(2{,}67\) ngenxa yokumelwa kwayo kokudada okubini, yathathwa ngempazamo \(2{,}669999\ldots\) inqunyulwe. normalize_grade Ngaphezu koko, inciphisa ixabiso ngalinye le-loop libuyele kwigridi ye-100th-order. Ubuncinane kukho amaxabiso amabini ahlukeneyo aphakathi anokubakho. \(1/300\) ukunyamezelana \(10^{-9}\) ngoko ke incinci ngokwaneleyo ukuba ingatshintshi naluphi na uhlobo lokwenyani lwetyala.
Inkqubo iqulunqwe kwaye iqalwe nge
rustc -O calc.rs && ./calcLe nkqubo iphinda-phinda kuzo zonke iindibaniselwano ze \(125\,751\,501\) kunye neembuyekezo:
| Uthelekiso | Inxalenye |
| Amanqaku apheleleyo xa kuthelekiswa nomndilili wesifundo esingajikelezwanga | \(10{,}4411\,\%\) |
| Amanqaku apheleleyo xa ethelekiswa nesiphakamiso esinendawo enye yedesimali | \(10{,}1521\,\%\) |
| Isindululo esinendawo enye yedesimali xa kuthelekiswa nomndilili wengcali ongajikelezwanga | \(0{,}6615\,\%\) |
Ixabiso eliphambili kum leli \(10{,}4411\,\%\): Kwiindibaniselwano ezingaphezu kwesinye kwishumi, ukujikelezisa kwangoko ukuya kumabanga apheleleyo kukhokelela kuhlobo lwesikolo olubaliweyo olwahlukileyo kunereferensi engajikelezwanga. Eli nani lichaza kuphela isiphumo sokujikelezisa ngaphakathi kwimodeli ekhethiweyo.
Ukutshintshela kwi-avareji yeengcali enendawo enye yedesimali kuya kukhokelela kuhlu olwahlukileyo kwi \(10{,}1521\,\%\) yazo zonke iimeko zemodeli xa kuthelekiswa nendlela yangaphambili. Nangona kunjalo, uthelekiso lwesithathu lubalulekile: xa kuthelekiswa nereferensi engajikelezwanga, indlela enendawo enye yedesimali ishiya kuphela \(0{,}6615\,\%\) yamatyala aguqukayo. Ngoko ke ukuphambuka kwireferensi kuncipha nge
\[
1-\frac{0{,}6615}{10{,}4411}\approx 93{,}7\,\%.
\]
Oku, kum, sisiphumo esingathandabuzekiyo sendlela yokubala yangoku: impumelelo ecingelwayo yesifundo esithile ayitshintshi; lixesha kuphela apho ukujikeleziswa kwenzeka khona elinokutshintsha uhlobo lwesikolo olubaliweyo. Nangona ukubala ngendawo enye yedesimali kungawususi ngokupheleleyo lo mphumo, kuyinciphisa kakhulu ngaphakathi kwemodeli.
Nangona kunjalo, la manani awasixeleli ukuba bangaphi abantwana abachaphazelekileyo ngokwenene. Amanqaku eengxelo zekhadi awenziwanga ngokuyimfuneko njenge-avareji ejikeleziweyo ngoomatshini, ukusebenza kwezemfundo akusasazwanga ngokulinganayo kwaye akuzimeleyo, kwaye ukhetho lokwenyani lohlobo lwesikolo aluxhomekekanga kuphela kula manani mathathu. Ubalo lwam luphendula umbuzo othe ngqo ngakumbi: Kukangaphi ukujikelezisa kutshintsha ibakala elibaliweyo xa zonke iindibaniselwano zexabiso lekhulu zinokwenzeka ngokulinganayo?
Okokugqibela, uthelekiso neposi yokuqala: Izinga lilonke elikhankanyiweyo apho, malunga ne \(10\,\%\) lisondele kwi \(10{,}4411\,\%\) yam. Nangona kunjalo, izibalo zam aziqinisekisi iimeko ezine ngazinye, nganye ixelwe njenge \(1/6 \approx 17\,\%\) Kwi "isikolo samabanga aphakamileyo esiphoswe nge- \(3{,}00\) ", "isikolo samabanga aphakamileyo esiphoswe nge \(2{,}33\) ", "isikolo samabanga aphakamileyo esiphumelele nge \(2{,}66\) ", kunye "nesikolo samabanga aphakamileyo esiphumelele nge \(2{,}66\) \(1{,}4615\,\%\) \(59{,}6096\,\%\) \(1{,}2023\,\%\) \(62{,}9395\,\%\) .
Amaxabiso amane angama-probabilities anee-reference groups ezahlukeneyo ngobukhulu. Ngaphandle kwendawo yesampulu echazwe ngokwahlukileyo \(1/6\) ayinakufunyanwa kuzo. Ukubala kwam ke ngoko kuqinisekisa isiphumo sokujikelezisa, kodwa hayi la ma-probabilities ngabanye. Nokuba kunjalo, iingxelo malunga nenani langempela labantwana abachaphazelekayo ziya kufuna ulwabiwo lwamanqaku okwenyani kunye nokuxhomekeka kwabo.