Homework 13
Muneeb
11/23/2021
Question 8.2
Suppose that in Problem 6.15, only a one-half fraction of the 2^4 design could be run. Construct the design and perform the analysis, using the data from replicate I.
Since in this case we have to run a one-half fraction of the design so we’ll be running a 2^(4-1) design where K=4 and P=1. Our defining relationship is I = ABCD in this case.
library(FrF2)
design <- FrF2(nfactors=4,resolution=3,randomize=FALSE)
design## A B C D
## 1 -1 -1 -1 -1
## 2 1 -1 -1 1
## 3 -1 1 -1 1
## 4 1 1 -1 -1
## 5 -1 -1 1 1
## 6 1 -1 1 -1
## 7 -1 1 1 -1
## 8 1 1 1 1
## class=design, type= FrF2aliasprint(design)## $legend
## [1] A=A B=B C=C D=D
##
## $main
## character(0)
##
## $fi2
## [1] AB=CD AC=BD AD=BCresponse<-c(7.037, 16.867, 13.876, 17.273, 11.846, 4.368, 9.36, 15.653)
design.resp <- add.response(design,response)
summary(design.resp)## Call:
## FrF2(nfactors = 4, resolution = 3, randomize = FALSE)
##
## Experimental design of type FrF2
## 8 runs
##
## Factor settings (scale ends):
## A B C D
## 1 -1 -1 -1 -1
## 2 1 1 1 1
##
## Responses:
## [1] response
##
## Design generating information:
## $legend
## [1] A=A B=B C=C D=D
##
## $generators
## [1] D=ABC
##
##
## Alias structure:
## $fi2
## [1] AB=CD AC=BD AD=BC
##
##
## The design itself:
## A B C D response
## 1 -1 -1 -1 -1 7.037
## 2 1 -1 -1 1 16.867
## 3 -1 1 -1 1 13.876
## 4 1 1 -1 -1 17.273
## 5 -1 -1 1 1 11.846
## 6 1 -1 1 -1 4.368
## 7 -1 1 1 -1 9.360
## 8 1 1 1 1 15.653
## class=design, type= FrF2DanielPlot(design.resp,half=TRUE)
Our Daniel plot shows that none of the factors significantly affect the crack length, since all the data points are approximately normally distributed.
Question 8.24
Construct a 2^(5-1) design. Show how the design may be run in two blocks of eight observations each. Are any main effects or two-factor interactions confounded with blocks?
Our generated design is as under:
des.res4<-FrF2(nfactors=5,resolution=4,randomize=FALSE)
des.res4## A B C D E
## 1 -1 -1 -1 -1 1
## 2 1 -1 -1 -1 -1
## 3 -1 1 -1 -1 -1
## 4 1 1 -1 -1 1
## 5 -1 -1 1 -1 -1
## 6 1 -1 1 -1 1
## 7 -1 1 1 -1 1
## 8 1 1 1 -1 -1
## 9 -1 -1 -1 1 -1
## 10 1 -1 -1 1 1
## 11 -1 1 -1 1 1
## 12 1 1 -1 1 -1
## 13 -1 -1 1 1 1
## 14 1 -1 1 1 -1
## 15 -1 1 1 1 -1
## 16 1 1 1 1 1
## class=design, type= FrF2aliasprint(des.res4)## $legend
## [1] A=A B=B C=C D=D E=E
##
## [[2]]
## [1] no aliasing among main effects and 2fisIn this case we’ll block as per factor combination AB. Based on that we’ll divide our factor combinations into two blocks.
AB <- c("+","-","-","+","+","-","-","+","+","-","-","+","+","-","-","+")
Block <- c(1,2,2,1,1,2,2,1,1,2,2,1,1,2,2,1)
Data <- data.frame(des.res4,AB,Block)
Data## A B C D E AB Block
## 1 -1 -1 -1 -1 1 + 1
## 2 1 -1 -1 -1 -1 - 2
## 3 -1 1 -1 -1 -1 - 2
## 4 1 1 -1 -1 1 + 1
## 5 -1 -1 1 -1 -1 + 1
## 6 1 -1 1 -1 1 - 2
## 7 -1 1 1 -1 1 - 2
## 8 1 1 1 -1 -1 + 1
## 9 -1 -1 -1 1 -1 + 1
## 10 1 -1 -1 1 1 - 2
## 11 -1 1 -1 1 1 - 2
## 12 1 1 -1 1 -1 + 1
## 13 -1 -1 1 1 1 + 1
## 14 1 -1 1 1 -1 - 2
## 15 -1 1 1 1 -1 - 2
## 16 1 1 1 1 1 + 1Analysing the data we can see that AB & CDE factor combinations are confounded with the blocks.
Question 8.25
Construct a 2^(7-2) design. Show how the design may be run in four blocks of eight observations each. Are any main effects or two-factor interactions confounded with blocks?
library(FrF2)
design <- FrF2(nruns = 32,nfactors=7,blocks = 4,randomize=TRUE)
design## run.no run.no.std.rp Blocks A B C D E F G
## 1 1 27.1.7 1 1 1 -1 1 -1 -1 1
## 2 2 29.1.8 1 1 1 1 -1 -1 1 -1
## 3 3 7.1.2 1 -1 -1 1 1 -1 1 1
## 4 4 20.1.5 1 1 -1 -1 1 1 1 -1
## 5 5 1.1.1 1 -1 -1 -1 -1 -1 -1 -1
## 6 6 22.1.6 1 1 -1 1 -1 1 -1 1
## 7 7 16.1.4 1 -1 1 1 1 1 -1 -1
## 8 8 10.1.3 1 -1 1 -1 -1 1 1 1
## run.no run.no.std.rp Blocks A B C D E F G
## 9 9 21.2.6 2 1 -1 1 -1 -1 -1 1
## 10 10 2.2.1 2 -1 -1 -1 -1 1 -1 -1
## 11 11 9.2.3 2 -1 1 -1 -1 -1 1 1
## 12 12 15.2.4 2 -1 1 1 1 -1 -1 -1
## 13 13 30.2.8 2 1 1 1 -1 1 1 -1
## 14 14 28.2.7 2 1 1 -1 1 1 -1 1
## 15 15 8.2.2 2 -1 -1 1 1 1 1 1
## 16 16 19.2.5 2 1 -1 -1 1 -1 1 -1
## run.no run.no.std.rp Blocks A B C D E F G
## 17 17 24.3.6 3 1 -1 1 1 1 -1 -1
## 18 18 31.3.8 3 1 1 1 1 -1 1 1
## 19 19 25.3.7 3 1 1 -1 -1 -1 -1 -1
## 20 20 5.3.2 3 -1 -1 1 -1 -1 1 -1
## 21 21 14.3.4 3 -1 1 1 -1 1 -1 1
## 22 22 3.3.1 3 -1 -1 -1 1 -1 -1 1
## 23 23 12.3.3 3 -1 1 -1 1 1 1 -1
## 24 24 18.3.5 3 1 -1 -1 -1 1 1 1
## run.no run.no.std.rp Blocks A B C D E F G
## 25 25 6.4.2 4 -1 -1 1 -1 1 1 -1
## 26 26 11.4.3 4 -1 1 -1 1 -1 1 -1
## 27 27 17.4.5 4 1 -1 -1 -1 -1 1 1
## 28 28 23.4.6 4 1 -1 1 1 -1 -1 -1
## 29 29 26.4.7 4 1 1 -1 -1 1 -1 -1
## 30 30 4.4.1 4 -1 -1 -1 1 1 -1 1
## 31 31 13.4.4 4 -1 1 1 -1 -1 -1 1
## 32 32 32.4.8 4 1 1 1 1 1 1 1
## class=design, type= FrF2.blocked
## NOTE: columns run.no and run.no.std.rp are annotation,
## not part of the data frameNo main effects or two factor interactions are confounded with the blocks.
Question 8.28
A 16-run experiment was performed in a semiconductor manufacturing plant to study the effects of six factors on the curvature or camber of the substrate devices produced. The six variables and their levels are shown in Table P8.2.
Part A
What type of design did the experimenters use?
K = 6 & P = 2 in this case, and it’s a 2^(6-2), 16 run design with resolution 4.
Part B
What are the alias relationships in this design?
library(FrF2)
des.res4<-FrF2(nfactors=6,resolution=4,randomize=FALSE)
des.res4$F <- c(-1,1,-1,1,1,-1,1,-1,1,-1,1,-1,-1,1,-1,1)
des.res4## A B C D E F
## 1 -1 -1 -1 -1 -1 -1
## 2 1 -1 -1 -1 1 1
## 3 -1 1 -1 -1 1 -1
## 4 1 1 -1 -1 -1 1
## 5 -1 -1 1 -1 1 1
## 6 1 -1 1 -1 -1 -1
## 7 -1 1 1 -1 -1 1
## 8 1 1 1 -1 1 -1
## 9 -1 -1 -1 1 -1 1
## 10 1 -1 -1 1 1 -1
## 11 -1 1 -1 1 1 1
## 12 1 1 -1 1 -1 -1
## 13 -1 -1 1 1 1 -1
## 14 1 -1 1 1 -1 1
## 15 -1 1 1 1 -1 -1
## 16 1 1 1 1 1 1
## class=design, type= FrF2aliasprint(des.res4)## $legend
## [1] A=A B=B C=C D=D E=E F=F
##
## $main
## character(0)
##
## $fi2
## [1] AB=CE=DF AC=BE AD=BF AE=BC AF=BD CD=EF CF=DEAlias relationships are shown above.
Part C
Do any of the process variables affect average camber?
A <- c(-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1)
B <- c(-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1)
C <- c(-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1)
D <- c(-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1)
E <- c(-1,1,1,-1,1,-1,-1,1,-1,1,1,-1,1,-1,-1,1)
F <- c(-1,1,-1,1,1,-1,1,-1,1,-1,1,-1,-1,1,-1,1)
A <- as.factor(A)
B <- as.factor(B)
C <- as.factor(C)
D <- as.factor(D)
E <- as.factor(E)
F <- as.factor(F)
response<-c(0.0167,0.0062,0.0041,0.0073,0.0047,0.0219,0.0121,0.0255,0.0032,0.0078,0.0043,0.0186,0.0110,0.0065,0.0155,0.0093,0.0128,0.0066,0.0043,0.0081,0.0047,0.0258,0.0090,0.0250,0.0023,0.0158,0.0027,0.0137,0.0086,0.0109,0.0158,0.0124,0.0149,0.0044,0.0042,0.0039,0.0040,0.0147,0.0092,0.0226,0.0077,0.0060,0.0028,0.0158,0.0101,0.0126,0.0145,0.0110,0.0185,0.0020,0.0050,0.0030,0.0089,0.0296,0.0086,0.0169,0.0069,0.0045,0.0028,0.0159,0.0158,0.0071,0.0145,0.0133)
Data <- data.frame(A,B,C,D,E,F,response)
Model <- aov(response~A*B*C*D*E*F,data = Data)
summary(Model)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 0.0002422 0.0002422 27.793 3.17e-06 ***
## B 1 0.0000053 0.0000053 0.614 0.43725
## C 1 0.0005023 0.0005023 57.644 9.14e-10 ***
## D 1 0.0000323 0.0000323 3.712 0.05995 .
## E 1 0.0001901 0.0001901 21.815 2.45e-05 ***
## F 1 0.0009602 0.0009602 110.192 5.05e-14 ***
## A:B 1 0.0000587 0.0000587 6.738 0.01249 *
## A:C 1 0.0000803 0.0000803 9.218 0.00387 **
## B:C 1 0.0000527 0.0000527 6.053 0.01754 *
## A:D 1 0.0000239 0.0000239 2.741 0.10431
## B:D 1 0.0000849 0.0000849 9.739 0.00305 **
## C:D 1 0.0000622 0.0000622 7.139 0.01027 *
## D:E 1 0.0000088 0.0000088 1.007 0.32062
## A:B:D 1 0.0000000 0.0000000 0.005 0.94291
## B:C:D 1 0.0000481 0.0000481 5.523 0.02293 *
## Residuals 48 0.0004183 0.0000087
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Our ANOVA analysis shows some main effects and higher interactions that are significant and that would affect average camber. Talking about process variables A,C,E & F would significantly affect average camber.
Part D
A <- c(-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1)
B <- c(-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1)
C <- c(-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1)
D <- c(-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1)
E <- c(-1,1,1,-1,1,-1,-1,1,-1,1,1,-1,1,-1,-1,1)
F <- c(-1,1,-1,1,1,-1,1,-1,1,-1,1,-1,-1,1,-1,1)
A <- as.factor(A)
B <- as.factor(B)
C <- as.factor(C)
D <- as.factor(D)
E <- as.factor(E)
F <- as.factor(F)
SD <- c(24.418,20.976,4.083,25.025,22.41,63.639,16.029,39.42,26.725,50.341,7.681,20.083,31.12,29.51,6.75,17.45)
Data2 <- data.frame(A,B,C,D,E,F,SD)
Model2 <- lm(SD~A*B*C*D*E*F,data = Data2)
DanielPlot(Model2)
Model3 <- aov(SD~A+B,data = Data2)
summary(Model3)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 1012 1012 8.505 0.01202 *
## B 1 1099 1099 9.241 0.00948 **
## Residuals 13 1546 119
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1As evident from our ANOVA analysis, process variables A & B affect the variability in camber measurements.
Part E
If it is important to reduce camber as much as possible, what recommendations would you make?
Model2 <- lm(response~A*B*C*D*E*F,data = Data)
coef(Model2)## (Intercept) A1 B1 C1
## 0.015725000 0.001009375 -0.007137500 0.001784375
## D1 E1 F1 A1:B1
## -0.002953125 -0.004187500 -0.007746875 0.003725000
## A1:C1 B1:C1 A1:D1 B1:D1
## 0.004481250 0.007100000 -0.002550000 0.007968750
## C1:D1 A1:E1 B1:E1 C1:E1
## -0.000475000 NA NA NA
## D1:E1 A1:F1 B1:F1 C1:F1
## 0.001481250 NA NA NA
## D1:F1 E1:F1 A1:B1:C1 A1:B1:D1
## NA NA NA 0.000212500
## A1:C1:D1 B1:C1:D1 A1:B1:E1 A1:C1:E1
## NA -0.006937500 NA NA
## B1:C1:E1 A1:D1:E1 B1:D1:E1 C1:D1:E1
## NA NA NA NA
## A1:B1:F1 A1:C1:F1 B1:C1:F1 A1:D1:F1
## NA NA NA NA
## B1:D1:F1 C1:D1:F1 A1:E1:F1 B1:E1:F1
## NA NA NA NA
## C1:E1:F1 D1:E1:F1 A1:B1:C1:D1 A1:B1:C1:E1
## NA NA NA NA
## A1:B1:D1:E1 A1:C1:D1:E1 B1:C1:D1:E1 A1:B1:C1:F1
## NA NA NA NA
## A1:B1:D1:F1 A1:C1:D1:F1 B1:C1:D1:F1 A1:B1:E1:F1
## NA NA NA NA
## A1:C1:E1:F1 B1:C1:E1:F1 A1:D1:E1:F1 B1:D1:E1:F1
## NA NA NA NA
## C1:D1:E1:F1 A1:B1:C1:D1:E1 A1:B1:C1:D1:F1 A1:B1:C1:E1:F1
## NA NA NA NA
## A1:B1:D1:E1:F1 A1:C1:D1:E1:F1 B1:C1:D1:E1:F1 A1:B1:C1:D1:E1:F1
## NA NA NA NAsummary(Model2)##
## Call:
## lm.default(formula = response ~ A * B * C * D * E * F, data = Data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.008300 -0.001350 -0.000350 0.001744 0.007275
##
## Coefficients: (48 not defined because of singularities)
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.0157250 0.0014760 10.654 3.06e-14 ***
## A1 0.0010094 0.0016502 0.612 0.543644
## B1 -0.0071375 0.0018077 -3.948 0.000257 ***
## C1 0.0017844 0.0016502 1.081 0.284963
## D1 -0.0029531 0.0019525 -1.512 0.136976
## E1 -0.0041875 0.0010437 -4.012 0.000210 ***
## F1 -0.0077469 0.0007380 -10.497 5.05e-14 ***
## A1:B1 0.0037250 0.0020874 1.785 0.080655 .
## A1:C1 0.0044812 0.0014760 3.036 0.003866 **
## B1:C1 0.0071000 0.0020874 3.401 0.001359 **
## A1:D1 -0.0025500 0.0020874 -1.222 0.227809
## B1:D1 0.0079688 0.0025565 3.117 0.003083 **
## C1:D1 -0.0004750 0.0020874 -0.228 0.820954
## A1:E1 NA NA NA NA
## B1:E1 NA NA NA NA
## C1:E1 NA NA NA NA
## D1:E1 0.0014813 0.0014760 1.004 0.320619
## A1:F1 NA NA NA NA
## B1:F1 NA NA NA NA
## C1:F1 NA NA NA NA
## D1:F1 NA NA NA NA
## E1:F1 NA NA NA NA
## A1:B1:C1 NA NA NA NA
## A1:B1:D1 0.0002125 0.0029520 0.072 0.942912
## A1:C1:D1 NA NA NA NA
## B1:C1:D1 -0.0069375 0.0029520 -2.350 0.022926 *
## A1:B1:E1 NA NA NA NA
## A1:C1:E1 NA NA NA NA
## B1:C1:E1 NA NA NA NA
## A1:D1:E1 NA NA NA NA
## B1:D1:E1 NA NA NA NA
## C1:D1:E1 NA NA NA NA
## A1:B1:F1 NA NA NA NA
## A1:C1:F1 NA NA NA NA
## B1:C1:F1 NA NA NA NA
## A1:D1:F1 NA NA NA NA
## B1:D1:F1 NA NA NA NA
## C1:D1:F1 NA NA NA NA
## A1:E1:F1 NA NA NA NA
## B1:E1:F1 NA NA NA NA
## C1:E1:F1 NA NA NA NA
## D1:E1:F1 NA NA NA NA
## A1:B1:C1:D1 NA NA NA NA
## A1:B1:C1:E1 NA NA NA NA
## A1:B1:D1:E1 NA NA NA NA
## A1:C1:D1:E1 NA NA NA NA
## B1:C1:D1:E1 NA NA NA NA
## A1:B1:C1:F1 NA NA NA NA
## A1:B1:D1:F1 NA NA NA NA
## A1:C1:D1:F1 NA NA NA NA
## B1:C1:D1:F1 NA NA NA NA
## A1:B1:E1:F1 NA NA NA NA
## A1:C1:E1:F1 NA NA NA NA
## B1:C1:E1:F1 NA NA NA NA
## A1:D1:E1:F1 NA NA NA NA
## B1:D1:E1:F1 NA NA NA NA
## C1:D1:E1:F1 NA NA NA NA
## A1:B1:C1:D1:E1 NA NA NA NA
## A1:B1:C1:D1:F1 NA NA NA NA
## A1:B1:C1:E1:F1 NA NA NA NA
## A1:B1:D1:E1:F1 NA NA NA NA
## A1:C1:D1:E1:F1 NA NA NA NA
## B1:C1:D1:E1:F1 NA NA NA NA
## A1:B1:C1:D1:E1:F1 NA NA NA NA
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.002952 on 48 degrees of freedom
## Multiple R-squared: 0.849, Adjusted R-squared: 0.8018
## F-statistic: 18 on 15 and 48 DF, p-value: 9.012e-15Since the equation for this experiment in terms of coded variables is as under:
\(Y_{i,j,k,l}\) = 0.0157250 + 0.0010094(A) + 0.0017844(C) - 0.0041875(E) - 0.0077469(F)
Best strategy to reduce camber would be to keep A and C at low level and E and F at high level.
Question 8.40
Consider the following experiment:
Part A
How many factors did this experiment investigate?
This experiment investigated “Four” Factors.
Part B
What is the resolution of this design?
Resolution of the design employed in this experiment is “Four”.
Part C
Effects are as under:
One <- c(8)
AD <- c(10)
BD <- c(12)
AB <- c(7)
CD <- c(13)
AC <- c(6)
BC <- c(5)
ABCD <- c(11)
EffectA <- (2*(AD+AB+AC+ABCD-One-BD-CD-BC))/(16)
EffectA## [1] -0.5EffectB <- (2*(BD+AB+BC+ABCD-One-AD-CD-AC))/(16)
EffectB## [1] -0.25EffectC <- (2*(CD+AC+BC+ABCD-One-AD-BD-AB))/(16)
EffectC## [1] -0.25EffectD <- (2*(AD+BD+CD+ABCD-One-AB-AC-BC))/(16)
EffectD## [1] 2.5Part D
Defining relation for this design is I = ABCD.
Question 8.48
Consider the following design:
Part A
What is the generator for column D?
If we tally the readings in column D we can see that the design generator for column D is -ABC.
Part B
What is the generator for column E?
If we tally the readings in column E we can see that the design generator for column E is BC.
Part C
If this design were folded over, what is the resolution of the combined design?
Resolution of folded over design is “Four”.
Question 8.60
Consider a partial fold over for the design.Suppose that the partial fold over of this design is constructed using column A ( + signs only). Determine the alias relationships in the combined design.
library(FrF2)
design <- FrF2(nfactors=7,resolution=3,randomize=FALSE)
design## A B C D E F G
## 1 -1 -1 -1 1 1 1 -1
## 2 1 -1 -1 -1 -1 1 1
## 3 -1 1 -1 -1 1 -1 1
## 4 1 1 -1 1 -1 -1 -1
## 5 -1 -1 1 1 -1 -1 1
## 6 1 -1 1 -1 1 -1 -1
## 7 -1 1 1 -1 -1 1 -1
## 8 1 1 1 1 1 1 1
## class=design, type= FrF2design2 <- fold.design(design,column=1)
design2## A B C fold D E F G
## 1 -1 -1 -1 original 1 1 1 -1
## 2 1 -1 -1 original -1 -1 1 1
## 3 -1 1 -1 original -1 1 -1 1
## 4 1 1 -1 original 1 -1 -1 -1
## 5 -1 -1 1 original 1 -1 -1 1
## 6 1 -1 1 original -1 1 -1 -1
## 7 -1 1 1 original -1 -1 1 -1
## 8 1 1 1 original 1 1 1 1
## 9 1 -1 -1 mirror 1 1 1 -1
## 10 -1 -1 -1 mirror -1 -1 1 1
## 11 1 1 -1 mirror -1 1 -1 1
## 12 -1 1 -1 mirror 1 -1 -1 -1
## 13 1 -1 1 mirror 1 -1 -1 1
## 14 -1 -1 1 mirror -1 1 -1 -1
## 15 1 1 1 mirror -1 -1 1 -1
## 16 -1 1 1 mirror 1 1 1 1
## class=design, type= FrF2.foldeddesign3 <- design2[-c(1,3,5,7,10,12,14,16),]
design3## A B C fold D E F G
## 2 1 -1 -1 original -1 -1 1 1
## 4 1 1 -1 original 1 -1 -1 -1
## 6 1 -1 1 original -1 1 -1 -1
## 8 1 1 1 original 1 1 1 1
## 9 1 -1 -1 mirror 1 1 1 -1
## 11 1 1 -1 mirror -1 1 -1 1
## 13 1 -1 1 mirror 1 -1 -1 1
## 15 1 1 1 mirror -1 -1 1 -1aliasprint(design2)## $legend
## [1] A=A B=B C=C D=fold E=D F=E G=F H=G
##
## $main
## [1] B=CG=FH C=BG=EH E=CH=FG F=BH=EG G=BC=EF H=BF=CE
##
## $fi2
## [1] AB=-DE AC=-DF AD=-BE=-CF=-GH AE=-BD AF=-CD
## [6] AG=-DH AH=-DG