Wörter in einem Array finden.

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	Pdd Betreff: Wörter in einem Array finden.	Fr, Dez 08, 2006 15:32 Antworten mit Zitat
Hallo. Ich habe ein Problem dabei, ein Wort in einem Array zu finden. Mein Array sieht so aus: Code: [AUSKLAPPEN] [EINKLAPPEN] AChars$(20, 14, 5) d.h. 19 hoch, (ich rechne die 0 nicht mit.), 13 breit. Immer in (x, y, 1) ist ein Buchstabe gespeichert. Das gesamte Feld wird von unten Links angewendet: Code: [AUSKLAPPEN] [EINKLAPPEN] ^ AChars$(20, 0, 1) \| \| \| ... \| 1---------> AChars$(1, 14, 1) Jetzt habe ich dazu eine Liste mit Wörtern. Code: [AUSKLAPPEN] [EINKLAPPEN] Dim AWords$(10) Nun will ich herausfinden, ob in einer Reihe des AChars$ ein Wort aus AWords$ vorhanden ist. Das ganze sollte später auch RÜCKWÄRTS, DIAGONAL und SENKRECHT sein. Um nun zu überprüfen, ob das Wort enthalten ist, folgender Code: Code: [AUSKLAPPEN] [EINKLAPPEN] For i = 1 To 20 ; Alle Zeilen durchgehen For g = 1 To (14 - Len(Word$)) wort_neu$ = "" For r = g To (g + Len(Word$)) wort_neu$ = wort_neu$ + AChars$(i, r, 1) Next DebugLog(wort_neu$) Next Next Somit sollte das Game sich beenden, wenn er das Wort in einer Zeile findet, oder nicht? Das blöde is nur, das das game jetzt noch 1FPS hat, wenn nicht sogar weniger :/ Hat jemand ne Idee, wie man das weniger Rechen-INtensiv machen kann?
Lieber Apfel als Birne.

Pdd

Betreff: Wörter in einem Array finden.

Fr, Dez 08, 2006 15:32
Antworten mit Zitat

Hallo.
Ich habe ein Problem dabei, ein Wort in einem Array zu finden.
Mein Array sieht so aus:
Code: [AUSKLAPPEN]

AChars$(20, 14, 5)

d.h. 19 hoch, (ich rechne die 0 nicht mit.), 13 breit.
Immer in (x, y, 1) ist ein Buchstabe gespeichert.
Das gesamte Feld wird von unten Links angewendet:
Code: [AUSKLAPPEN]

^ AChars$(20, 0, 1)
|
|
| ...
|
1---------> AChars$(1, 14, 1)

Jetzt habe ich dazu eine Liste mit Wörtern.

Code: [AUSKLAPPEN]

Dim AWords$(10)

Nun will ich herausfinden, ob in einer Reihe des AChars$ ein Wort aus AWords$ vorhanden ist. Das ganze sollte später auch RÜCKWÄRTS, DIAGONAL und SENKRECHT sein.

Um nun zu überprüfen, ob das Wort enthalten ist, folgender Code:
Code: [AUSKLAPPEN]

         For i = 1 To 20 ; Alle Zeilen durchgehen
               For g = 1 To (14 - Len(Word$))
                  wort_neu$ = ""
                  For r = g To (g + Len(Word$))
                     wort_neu$ = wort_neu$ + AChars$(i, r, 1)
                  Next
                  DebugLog(wort_neu$)
               Next
         Next

Somit sollte das Game sich beenden, wenn er das Wort in einer Zeile findet, oder nicht? Das blöde is nur, das das game jetzt noch 1FPS hat, wenn nicht sogar weniger :/ Hat jemand ne Idee, wie man das weniger Rechen-INtensiv machen kann?

Lieber Apfel als Birne.

	MhhF	Mo, Dez 11, 2006 16:59 Antworten mit Zitat
Probier mal hiermit Code: [AUSKLAPPEN] [EINKLAPPEN] Dim AChars$(20, 14, 5) Dim AWords$(10) Awords$(1)="blitz" Awords$(2)="super" Awords$(3)="basic" ;Einlesen des Feldes Restore chardata For x=1 To 20 Read aChars(x,1,1),aChars(x,2,1),aChars(x,3,1),aChars(x,4,1),aChars(x,5,1),aChars(x,6,1),aChars(x,7,1),aChars(x,8,1),aChars(x,9,1),aChars(x,10,1),aChars(x,11,1),aChars(x,12,1),aChars(x,13,1),aChars(x,14,1) Next ;Spalten For y=1 To 14 s$="" For x=1 To 20 s$=s+AChars(x,y,1) Next For i=1 To 3 If Instr(s,AWords(i)) Then Print AWords(i)+" in Spalte: "+y+" Zeile: "+Instr(s,Awords(i))+" bis "+(Instr(s,Awords(i))+Len(AWords(i)))+" gefunden!" Next Next ;Zeilen For x=1 To 20 s$="" For y=1 To 14 s$=s+AChars(x,y,1) Next For i=1 To 3 If Instr(s,AWords(i)) Then Print AWords(i)+" in Zeile: "+x+" Spalte: "+Instr(s,Awords(i))+" bis "+(Instr(s,Awords(i))+Len(AWords(i)))+" gefunden!" Next Next ;Diagonal For x=-21 To 20 s$="" For y=1 To 14 If x+y>0 And x+y<21 Then s$=s+AChars(x+y,y,1) Next For i=1 To 3 If Instr(s,AWords(i)) Then Print AWords(i)+" in Z.: "+(x+(Instr(s,Awords(i))))+" S.: "+Instr(s,Awords(i))+" bis Z.: "+(x+Instr(s,Awords(i))+Len(AWords(i)))+"S.: "+(Instr(s,Awords(i))+Len(AWords(i)))+" gefunden!" Next Next .chardata Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z" Data "g","e","g","b","l","i","t","z","a","g","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z" Data "g","e","g","j","a","s","h","d","a","s","h","a","z","z" Data "g","e","g","j","a","r","i","d","a","u","h","a","z","z" Data "g","e","g","j","a","r","h","c","a","p","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","e","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","r","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z" Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"

MhhF

Mo, Dez 11, 2006 16:59
Antworten mit Zitat

Probier mal hiermit
Code: [AUSKLAPPEN]

Dim AChars$(20, 14, 5)
Dim AWords$(10)
Awords$(1)="blitz"
Awords$(2)="super"
Awords$(3)="basic"

;Einlesen des Feldes
Restore chardata
For x=1 To 20
Read aChars(x,1,1),aChars(x,2,1),aChars(x,3,1),aChars(x,4,1),aChars(x,5,1),aChars(x,6,1),aChars(x,7,1),aChars(x,8,1),aChars(x,9,1),aChars(x,10,1),aChars(x,11,1),aChars(x,12,1),aChars(x,13,1),aChars(x,14,1)
Next

;Spalten
For y=1 To 14
s$=""
For x=1 To 20
   s$=s+AChars(x,y,1)
Next

For i=1 To 3
   If Instr(s,AWords(i)) Then Print AWords(i)+" in Spalte: "+y+" Zeile: "+Instr(s,Awords(i))+" bis "+(Instr(s,Awords(i))+Len(AWords(i)))+" gefunden!"
Next
Next

;Zeilen
For x=1 To 20
s$=""
For y=1 To 14
   s$=s+AChars(x,y,1)
Next

For i=1 To 3
   If Instr(s,AWords(i)) Then Print AWords(i)+" in Zeile: "+x+" Spalte: "+Instr(s,Awords(i))+" bis "+(Instr(s,Awords(i))+Len(AWords(i)))+" gefunden!"
Next
Next

;Diagonal
For x=-21 To 20
s$=""
For y=1 To 14
   If x+y>0 And x+y<21 Then s$=s+AChars(x+y,y,1)
Next

For i=1 To 3
   If Instr(s,AWords(i)) Then Print AWords(i)+" in Z.: "+(x+(Instr(s,Awords(i))))+" S.: "+Instr(s,Awords(i))+" bis Z.: "+(x+Instr(s,Awords(i))+Len(AWords(i)))+"S.: "+(Instr(s,Awords(i))+Len(AWords(i)))+" gefunden!"
Next
Next

.chardata
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"
Data "g","e","g","b","l","i","t","z","a","g","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"
Data "g","e","g","j","a","s","h","d","a","s","h","a","z","z"
Data "g","e","g","j","a","r","i","d","a","u","h","a","z","z"
Data "g","e","g","j","a","r","h","c","a","p","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","e","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","r","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"
Data "g","e","g","j","a","r","h","d","a","g","h","a","z","z"

	hectic Sieger des IS Talentwettbewerb 2006	Mo, Dez 11, 2006 17:09 Antworten mit Zitat
zB: Suchwort hat eine bestimmte ql:len. Wurde ein Buchstabe anhand ql:Mid (Suchwort,Position,1) des Suchwortes im Feld gefunden, so wird diese Position=Position+1 gerechnet. Ist Position= ql:len (Suchwort) so wurde das Wort gefunden. Das geht in alle Richtungen. Wird eine neue Zeile/Spalte/Diagonale eingelesen, sollte die Position wieder auf 1 gesetzt werden.

hectic

Sieger des IS Talentwettbewerb 2006

Mo, Dez 11, 2006 17:09
Antworten mit Zitat

zB: Suchwort hat eine bestimmte ql:len. Wurde ein Buchstabe anhand ql:Mid (Suchwort,Position,1) des Suchwortes im Feld gefunden, so wird diese Position=Position+1 gerechnet. Ist Position= ql:len (Suchwort) so wurde das Wort gefunden. Das geht in alle Richtungen. Wird eine neue Zeile/Spalte/Diagonale eingelesen, sollte die Position wieder auf 1 gesetzt werden.