irst, we perform the folding of  ( ) (  )  (  ). The pair
( ) ( ) is
::
cn inverted
( ( ))

$$\mathbf{\Sigma}^{1}$$

9.

ethus unite it into a block. This block does not pass the ad acenc  restriction test (
uation

$$\begin{array}{l l l}{{c\overline{{{o s t}}}\left(}&{{\left(\begin{array}{l}{{}}\\ {{}}\end{array}\right)}}&{{\left(\begin{array}{l}{{}}\\ {{}}\end{array}\right)}}\end{array}\right)}}\\ {{c\overline{{{o s t}}}\left(}}&{{\left(\begin{array}{l}{{}}\\ {{}}\end{array}\right)}}&{{\left(\begin{array}{l}{{}}\\ {{}}\end{array}\right)}}\end{array}\right)}}\end{array}$$

Therefore, this ordering is contradicting and can be discarded.

e perform the folding of   (   )   (   )  .    (   (   ))
( ( )) d(X)
1, the pair is cn inverted, and we unite it into a block. This block does not pass the ad acenc  restriction test

$$\begin{array}{l l l}{{c\overline{{{o s t}}}\left(}&{{\left(\begin{array}{l}{{}}\\ {{}}\end{array}\right)}}&{{\left(\begin{array}{l}{{}}\\ {{}}\end{array}\right)}}\end{array}\right)}}\\ {{c\overline{{{o s t}}}\left(}}&{{\left(\begin{array}{l}{{}}\\ {{}}\end{array}\right)}}&{{\left(\begin{array}{l}{{}}\\ {{}}\end{array}\right)}}\end{array}\right)}}\end{array}$$
$\downarrow$ . 
This ordenng is re ected too, even before its folding is finished. If we continue the folding process, we shall see that the subgoal ( ) must also be added to this 4×0.5 1 block, since
( ) ( ))
( ) ) (d(x),c(x))
, and .

9.

e perform the folding of   (   )   (   )  .    (   (   ))
) 
( )

$\frac{1}{2}$ 2. 
(
1, the pair is cn inverted, and we form a block    (   )
)
), which passes the ad acenc restriction test

$$\begin{array}{l l l}{{c\overline{{{o o s}}}(}}&{{(}}&{{)}}&{{(}}&{{)}}\\ {{c\overline{{{o s}}}(}}&{{(}}&{{)}}&{{(}}&{{)}}\end{array}$$

e compute the control values of the new block

$$\begin{array}{c c c}{{c\overline{{{o s}}}s\left(}&{{\left(\begin{array}{c}{{}}\\ {{}}\end{array}\right)}\right)}}\\ {{n\overline{{{s o l}}}s\left(}&{{\left(\begin{array}{c}{{}}\\ {{}}\end{array}\right)}\right)}}\end{array}$$
$\downarrow$ . 

$$\left(\begin{array}{c c}{{}}&{{}}\\ {{}}&{{}}\end{array}\right)$$

, thus the pair . ( ), ( ) is cn ordered, no more folding
( )
) 
is needed, and we add the folded candidate   (  ). (  ) to the candidate set of e now perform merging of the candidate set of
, with the candidate set
( ). (  ) . In the resulting se uence ma  blocks must be sorted b of
( )
( )

$$\left(\begin{array}{c c}{{}}&{{}}\\ {{}}&{{}}\end{array}\right)$$

( ( )) cc(X)
The merged ordering, ( ) , ( ) , is added to the candidate set of 1.

e compare the costs of all candidates of 1, and output the cheapest one. In our case, 7.

there is onl  one candidate,     (   )
( ) . The algorithm returns this candidate unfolded, , ( ) (
(
)