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Loydova osmička je hlavolam s osmi čverečky v matici 3x3 a jedním prázdným místem. Cílem je uspořádat čtverečky do základní konfigurace (1, 2, 3; 4, 5, 6; 7, 8, prázdno). Normálně to je Loydova patnáctka, ale princip je stejný. Mimochodem, nevymyslel to Loyd, více třeba na Wikipedii, ten na tom jen vydělal peníze.
Následuje Adamova implementace podle návodu ze cvičení, více v komentářích. Testováno jen ve FreePascalu?. Zkoušel jsem to ve starém TurboPascalu?, ale na ten jsou tam moc velká čísla a nelíbí se mu konstantní parametry, tak jsem se s tím raději dál nepatlal. Neměl by ale být problém to upravit. Možná se na vás bude Borland ještě zlobit za to, že ignorujete návratové hodnoty funkce (CQAddItem() vrací boolean, ale já na to kašlu, protože vím, že je to vždy TRUE), v tom případě to přiřaďte do nějaké odpadní booleovské proměnné.
program LoydovaOsmicka(input, output); { nebo trojka:), 15ka je na nas moc, FPC si moc nerozumi s 64 bity:( } { Pouzivam kruhovou frontu (viz CodeQueue), staci pak radove nizsi pamet (resp. nemusim porad dokola sesypavat). Pole indexuju od nuly a prazdne misto mam take oznacene nulou. Jinak je to presne podle navodu ze cviceni. Komentare odtud dal a identifikatory jsou anglicky. Pisu tak vzdycky, prijde mi to lepsi nez mixovat jazyky a prznit cestinu. Vsechna cisla mimo 1 a 0 najdete mezi konstantami na zacatku, Side muzete zmenit na 2 a odpovidajicim zpusobem dalsi cisla, viz nize. Procedura PermMove() by mohla byt rychlejsi, to by ale bylo zbytecne, protoze ji pouzivam jen pro rekonstrukci tahu pri vypisu reseni. Takhle je aspon srozumitelne (snad) a kratce zapsana. Program umi jednak resit resitelne konfigurace, jednak vypsat pocet kroku nutnych pro reseni vsech konfiguraci. Za povsimnuti stoji, ze algoritmus je v podstate totozny a ze (tim padem) vyreseni netrivialni konfigurace trva radove stejne dlouho jako nalezeni poctu kroku pro vsechny konfigurace. Pokud zvolite druhou moznost, vypisou se take dve konkretni konfigurace, pro ktere je reseni nejdelsi (StepsMax=31 kroku), a vypise se i pocet konfiguraci, pro ktere reseni neexistuje, muzete si overit, ze je to presne polovina z celkoveho poctu (PermCodeMax+1), coz je spravne. Kazda konfigurace totiz patri do prave jedne poloviny konfiguraci, ktere ze sebe navzajem lze ziskat soupanim ctverecku. } const { You change Size to 2 (four is too much:(), if you also change StepsMax to 6 and PermCodeMax to 23. } Side = 3; Count = Side*Side; Max = Count-1; NghbrMax = 3; { There are up to four neighbors per square. } NghbrOffsets : array[0..NghbrMax] of integer = (-Side, 1, Side, -1); PermCodeMax = 362879 {23}; { = fact(Side*Side)-1 } CQMax = 24058; { Max queue size-1. Empirically determined:) (for 3x3). } StepsMax = 31 {6}; { How many steps needed to solve a config at most. } SqEmpty = 0; type SqNumber = SqEmpty..Max; { square number } Perm = array[0..Max] of SqNumber; PermCode = longint; PermCodeNghbrs = record count : integer; codes : array[0..NghbrMax] of PermCode; moves : array[0..NghbrMax] of SqNumber; end; CodeQueue = record buf : array[0..CQMax] of PermCode; head : integer; { place for newly added items } tail : integer; { the "oldest" item } isEmpty : boolean; end; function PermToCode(const p : Perm): PermCode; forward; procedure CodeToPerm(c: PermCode; var p: Perm); forward; procedure CodeGetNeighbors(c: PermCode; var neighbors : PermCodeNghbrs); forward; function PermEmptyIndex(const p: Perm) : integer; forward; procedure PermMove(var p: Perm; num : SqNumber); forward; procedure WriteSquarePerm(p: Perm); forward; procedure ReadSquarePerm(var p: Perm); forward; procedure CountSteps; forward; procedure SolveConfig; forward; { CodeQueue } procedure CQMakeEmpty(var buf : CodeQueue); forward; function CQIsFull(const buf : CodeQueue) : boolean; forward; function CQAddItem(var buf : CodeQueue; item : PermCode) : boolean; forward; function CQGetItem(var buf : CodeQueue; var item : PermCode) : boolean; forward; function PermToCode(const p : Perm): PermCode; var i : integer; j : integer; used : array[0..Max] of boolean; repeatCount : longint; { How many times in a row the same number appears at a certain position } c : PermCode; begin for i := 0 to Max do used[i] := FALSE; c := 0; repeatCount := 1; for i :=2 to Max do repeatCount := repeatCount * i; for i := 0 to Max-1 do begin for j := 0 to p[i]-1 do begin if not used[j] then c := c + repeatCount; end; used[p[i]] := TRUE; repeatCount := repeatCount div (Max-i); end; PermToCode := c; end; procedure CodeToPerm(c: PermCode; var p: Perm); var i : integer; j : integer; k : integer; n : integer; used : array[0..Max] of boolean; repeatCount : longint; begin for i := 0 to Max do used[i] := FALSE; repeatCount := 1; for i :=2 to Max do repeatCount := repeatCount * i; for i := 0 to Max do begin n := c div repeatCount; c := c mod repeatCount; j := 0; k := 0; while j <= n do begin if not used[k] then inc(j); inc(k); end; dec(k); p[i] := k; if i <> Max then begin used[k] := TRUE; repeatCount := repeatCount div (Max-i); end; end; end; function PermEmptyIndex(const p: Perm) : integer; var i : integer; begin i := 0; while p[i] <> SqEmpty do inc(i); PermEmptyIndex := i; end; procedure CodeGetNeighbors(c: PermCode; var neighbors : PermCodeNghbrs); var p : Perm; emptyIdx : integer; nghbrIdx : integer; i : integer; begin neighbors.count := 0; CodeToPerm(c, p); emptyIdx := PermEmptyIndex(p); for i := 0 to NghbrMax do begin nghbrIdx := emptyIdx + NghbrOffsets[i]; if (nghbrIdx>=0) and (nghbrIdx<=Max) and { within bounds } (((nghbrIdx div Side) = (emptyIdx div Side)) or ((nghbrIdx mod Side) = (emptyIdx mod Side)) { in the same row or column } ) then begin p[emptyIdx] := p[nghbrIdx]; p[nghbrIdx] := SqEmpty; neighbors.codes[neighbors.count] := PermToCode(p); neighbors.moves[neighbors.count] := p[emptyIdx]; inc( neighbors.count ); p[nghbrIdx] := p[emptyIdx]; p[emptyIdx] := SqEmpty; end; end; end; procedure PermMove(var p: Perm; num : SqNumber); var emptyIdx : integer; i : integer; n : PermCodeNghbrs; begin emptyIdx := PermEmptyIndex(p); CodeGetNeighbors(PermToCode(p), n); i := 0; repeat CodeToPerm(n.codes[i], p); inc(i); until (p[emptyIdx] = num) ; { or (i = n.count) - just for safety, not needed for valid args } end; procedure WriteSquarePerm(p: Perm); var i : integer; j : integer; begin for i := 0 to Side-1 do begin for j := 0 to Side-1 do begin if p[ i*Side + j ] = SqEmpty then write(' _') else write(' ', p[ i*Side + j ]); { works well for 1-digit numbers } end; writeln; end; end; procedure ReadSquarePerm(var p: Perm); var i : integer; j : integer; begin for i := 0 to Side-1 do begin write(' row ',i, '> '); for j := 0 to Side-1 do read(p[ i*Side + j ]); readln; end; end; {CodeQueue} procedure CQMakeEmpty(var buf : CodeQueue); begin buf.head := 0; buf.tail := 0; buf.isEmpty := TRUE; end; function CQIsFull(const buf : CodeQueue) : boolean; begin CQIsFull := (buf.head = buf.tail) and not buf.isEmpty; end; function CQAddItem(var buf : CodeQueue; item : PermCode) : boolean; begin if not CQIsFull(buf) then begin buf.buf[buf.head] := item; buf.head := (buf.head+1) mod CQMax; if buf.isEmpty then buf.isEmpty := FALSE; CQAddItem := TRUE; end else CQAddItem := FALSE; end; function CQGetItem(var buf : CodeQueue; var item : PermCode) : boolean; begin if not buf.isEmpty then begin item := buf.buf[buf.tail]; buf.tail := (buf.tail+1) mod CQMax; if buf.tail = buf.head then buf.isEmpty := TRUE; CQGetItem := TRUE; end else CQGetItem := FALSE; end; procedure CountSteps; var neighbors : PermCodeNghbrs; i : longint; p : Perm; c1 : PermCode; c : PermCode; queue : CodeQueue; codeSteps : array[0..PermCodeMax] of -1..StepsMax; stepStats : array[-1..StepsMax] of longint; begin for i := 0 to PermCodeMax do codeSteps[i] := -1; stepStats[-1] := PermCodeMax; { target config (0 steps) not included } for i := 0 to StepsMax do stepStats[i] := 0; { Target config } for i:=1 to Max do p[i-1] := i; p[Max] := SqEmpty; c1 := PermToCode( p ); CQMakeEmpty(queue); CQAddItem(queue, c1); codeSteps[c1] := 0; inc(stepStats[0]); while CQGetItem(queue, c) do begin CodeGetNeighbors(c, neighbors); for i:=0 to neighbors.count-1 do begin if codeSteps[ neighbors.codes[i] ] = -1 then begin codeSteps[ neighbors.codes[i] ] := codeSteps[c]+1; inc( stepStats[ codeSteps[ neighbors.codes[i] ] ] ); dec( stepStats[-1] ); if codeSteps[ neighbors.codes[i] ] = StepsMax then begin writeln(codeSteps[ neighbors.codes[i] ], ' steps:'); CodeToPerm(neighbors.codes[i], p); WriteSquarePerm(p); end; CQAddItem(queue, neighbors.codes[i]); end end; end; writeln(stepStats[-1], ' configurations are unsolvable.'); for i := 0 to StepsMax do writeln(stepStats[i], ' configurations are solvable in ', i, ' steps.'); end; procedure SolveConfig; var neighbors : PermCodeNghbrs; i : longint; p : Perm; c0 : PermCode; c1 : PermCode; c : PermCode; queue : CodeQueue; codeMoves : array[0..PermCodeMax] of SqNumber; begin for i := 0 to PermCodeMax do codeMoves[i] := SqEmpty; { Initial config } writeln('Enter the initial configuration:'); ReadSquarePerm( p ); c0 := PermToCode( p ); writeln; { Target config } for i:=1 to Max do p[i-1] := i; p[Max] := SqEmpty; c1 := PermToCode( p ); CQMakeEmpty(queue); CQAddItem(queue, c1); while CQGetItem(queue, c) and (codeMoves[c0] = SqEmpty) do begin CodeGetNeighbors(c, neighbors); for i:=0 to neighbors.count-1 do begin if codeMoves[ neighbors.codes[i] ] = SqEmpty then begin codeMoves[ neighbors.codes[i] ] := neighbors.moves[i]; CQAddItem(queue, neighbors.codes[i]); end end; end; if codeMoves[c0] = SqEmpty then writeln('Unsolvable.') else begin writeln('Solution:'); c := c0; CodeToPerm(c, p); i := 1; repeat writeln('#', i, ' Move ', codeMoves[c], ':'); PermMove(p, codeMoves[c] ); WriteSquarePerm(p); c := PermToCode(p); inc(i); until c = c1; end; end; var opt : integer; begin writeln('Choose an option (enter one digit):'); writeln(' [1] Solve a configuration.'); writeln(' [2] Compute steps needed to solve all configurations.'); writeln; readln(opt); if opt = 1 then SolveConfig else if opt = 2 then CountSteps else writeln('Unknown option.'); end.
len par oprav:
1.) je to Lloydova osmicka, nie Loydova
2.) sused po anglicky je neighboUr
ale inak je to fakt dobre, takze nechcel som ti to neijak prepisovat ;)
ad 1) Taky mě to zarazilo.. už jsem chtěl Adamovi napsat, ale pak jsem si to šel raději ověřit, a třeba wikipedia na několika místech shodně tvrdí Sam Lyod (viz třeba Adamův odkaz nahoře v článku nebo rovnou http://en.wikipedia.org/wiki/Sam_Loyd), tak nevím. Ale já ho taky znal jako Lloyda.