LR(1)
-----

A->BeH

                          előreolvasási
                          szimbólum
LR(1)-elem      [A->.BeH, b]
                 ----mag

A->ε            [A->., ?]

S' -> S
S -> n S c S | ε

H0= closure([S'->.S #]) =
        [S'->.S #][S->.nScS #][S->. #]
H1= read(H0, S)=[S'->S. #]
H2= read(H0, n)=[S->n.ScS #][S->.nScS c][S->. c]
H3= read(H2, S)=[S->nS.cS #]
H4= read(H2, n)=[S->n.ScS c][S->.nScS c][S->. c]
H5= read(H3, c)=[S->nSc.S #][S->.nScS #][S->. #]
H6= read(H4, S)=[S->nS.cS c]
H4= read(H4,n)=[S->n.ScS c][S->.nScS c][S->. c]
H7= read(H5,S)=[S->nScS. #]
H2= read(H5,n)=[S->n.ScS #][S->.nScS c][S->. c]
H8= read(H6,c)=[S->nSc.S c][S->.nScS c][S->. c]
H9= read(H8,S)=[S->nScS. c]
H4= read(H8,n)=[S->n.ScS c][S->.nScS c][S->. c]

így kell kitölteni:
    H_step_ide= read(H_sor,oszlop)

    S       n       c           #
H0  step1   step2               red(S->ε)
H1                              accept
H2  step3   step4   red(S->ε)
H3                  step5
H4  step6   step4   red(S->ε)
H5  step7   step2               red(S->ε)
H6                  step8
H7                              red(S->nScS)
H8  step9   step4   red(S->ε)
H9                  red(S->nScS)

(#0 , nnccnc#) ->               0,n:step2
(#0 n2 , nccnc#) ->             2,n:step4
(#0 n2 n4 , ccnc#) ->           4,c:red(S->ε)
(#0 n2 n4 S6 , ccnc#) ->        6,c:step8
(#0 n2 n4 S6 c8 , cnc#) ->      8,c:red(S->ε)
(#0 n2 n4 S6 c8 S9 , cnc#) ->   9,c:red(S->nScS)
       -----------
(#0 n2 S3 , cnc#) ->            3,c:step5
(#0 n2 S3 c5 , nc#) ->          5,n:step2
(#0 n2 S3 c5 n2 , c#) ->        2,c:red(S->ε)
(#0 n2 S3 c5 n2 [ezittazepszilonhelye], c#) ->
(#0 n2 S3 c5 n2 S3, c#) ->      3,c:step5
(#0 n2 S3 c5 n2 S3 c5, #) ->    5,#:red(S->ε)
(#0 n2 S3 c5 n2 S3 c5 S7, #) -> 7,#:red(S->nScS)
             -----------
(#0 n2 S3 c5 S7, #) ->          7,#:red(S->nScS)
    -----------
(#0 S1, #) ->                   1,#:accept


LALR(1)
-------

H2= [S->n.ScS #][S->.nScS c][S->. c]
H4= [S->n.ScS c][S->.nScS c][S->. c]

összevonjuk őket:
J2/4=
    [S->n.ScS #][S->n.ScS c][S->.nScS c][S->. c]

most ezek az összevonások:
    2-4, 3-6 5-8 7-9

        S       n       c           #
H2      step3   step4   red(S->ε)
H4      step6   step4   red(S->ε)



J0  step1   step2               red(S->ε)
J1                              accept
J2/4    step3/6 step2/4 red(S->ε)
J3/6                step5
J5/8step7   step2               red(S->ε)
J6                  step8
J7/9                            red(S->nScS)
J8  step9   step4   red(S->ε)
J9                  red(S->nScS)