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July 7, 2020 08:47
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Für das Lösen deiner kleinen Mathematik Probleme...
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b1 <- c(1,1,1, 0,0,0, 0,0,0) | |
b2 <- c(0,0,0, 1,1,1, 0,0,0) | |
b3 <- c(0,0,0, 0,0,0, 1,1,1) | |
b4 <- c(1,0,0, 1,0,0, 1,0,0) | |
b5 <- c(0,1,0, 0,1,0, 0,1,0) | |
b6 <- c(0,0,1, 0,0,1, 0,0,1) | |
b7 <- c(1,0,0, 0,1,0, 0,0,1) | |
b8 <- c(0,0,1, 0,1,0, 1,0,0) | |
# Die Matrix: | |
A <- rbind(b1,b2,b3,b4,b5,b6,b7,b8) | |
print(A) | |
# Die in Zeilenstufenform umgeformte Matrix: | |
AA <- rbind( b4-b2+(b8-b6)+2*1/3*(b7-b4+b2-2*(b8-b6)), | |
b5-(b8-b6)-1/3*(b7-b4+b2-2*(b8-b6)), | |
b6-1/3*(b7-b4+b2-2*(b8-b6))-b3, | |
b2-(b8-b6)-2*1/3*(b7-b4+b2-2*(b8-b6))-b3, | |
b8-b6+(1/3*(b7-b4+b2-2*(b8-b6))), | |
1/3*(b7-b4+b2-2*(b8-b6))+b3, | |
b3, | |
(b1-b4-b5-b6+b2)+b3) | |
print(AA) | |
# Aufgabe aus dem Beispiel: | |
b1 <- 20 | |
b2 <- 14 | |
b3 <- 11 | |
b4 <- 15 | |
b5 <- 10 | |
b6 <- 20 | |
b7 <- 20 | |
b8 <- 10 | |
# Für die neue "Rechte Seite" gilt dann: | |
b_1 <- b4-b2+(b8-b6)+2*1/3*(b7-b4+b2-2*(b8-b6)) | |
b_2 <- b5-(b8-b6)-1/3*(b7-b4+b2-2*(b8-b6)) | |
b_3 <- b6-1/3*(b7-b4+b2-2*(b8-b6))-b3 | |
b_4 <- b2-(b8-b6)-2*1/3*(b7-b4+b2-2*(b8-b6))-b3 | |
b_5 <- b8-b6+(1/3*(b7-b4+b2-2*(b8-b6))) | |
b_6 <- 1/3*(b7-b4+b2-2*(b8-b6))+b3 | |
b_7 <- b3 | |
b_8 <- b1-b4-b5-b6+b2+b3 | |
for(t in 0:300) { | |
for(s in 0:300) { | |
# Berechne nun mit der neuen "Rechten Seite" und | |
# den beiden freien Variablen t=x8, s=x9 die | |
# Lösungen für x1-x7 | |
x1 <- b_1 - 1 * s | |
x2 <- b_2 - 1 * t | |
x3 <- b_3 + 1 * t + 1 * s | |
x4 <- b_4 + 1 * t + 2 * s | |
x5 <- b_5 | |
x6 <- b_6 - 1 * t - 2 * s | |
x7 <- b_7 - t - s | |
x8 <- t | |
x9 <- s | |
x <- c(x1,x2,x3,x4,x5,x6,x7,x8,x9) | |
# Eine Lösung nur mit nicht-negativen Zahlen? | |
if (all(x >= 0)) { | |
# Dann Ausgabe der Lösung: | |
print(s) | |
print(t) | |
L <- matrix(x, nrow=3, ncol=3) | |
print(L) | |
} | |
} | |
} |
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