cykp v0.9 starting
adding production A -> 0	B	
adding production A -> 1	D	
adding production A -> 0	
adding production B -> 0	D	
adding production B -> 1	C	
adding production C -> 0	B	
adding production C -> 1	D	
adding production C -> 0	
adding production D -> 0	D	
adding production D -> 1	D	
reading input finished.
Start symbol: A
Variables: A B C D 
Terminals: 0 1 
Sorted productions:
A -> 0	B	
A -> 0	
A -> 1	D	
B -> 0	D	
B -> 1	C	
C -> 0	B	
C -> 0	
C -> 1	D	
D -> 0	D	
D -> 1	D	
lemma 4.1 variables removed: D 
deleting useless production:
A -> 1 D 
deleting useless production:
B -> 0 D 
deleting useless production:
C -> 1 D 
deleting useless production:
D -> 0 D 
deleting useless production:
D -> 1 D 
Eliminate epslion Productions
Eliminate Unit Productions
after eliminate, sorted productions:
A -> 0 B
A -> 0
B -> 1 C
C -> 0 B
C -> 0
after eliminate, Variables:
A B C 
Chomsky 1, replace terminal with variable
Chomsky part 1, sorted productions:
A ->  0
A ->  T_0  B
B ->  T_1  C
C ->  0
C ->  T_0  B
T_0 ->  0
T_1 ->  1
Chomsky Part 2 generated productions
after Chomsky, sorted productions:
A ->  0
A ->  T_0  B
B ->  T_1  C
C ->  0
C ->  T_0  B
T_0 ->  0
T_1 ->  1
after Chomsky, Variables:
A  B  C  T_0  T_1  
about to parse 0
run CYK algorithm to build array, size=1
Finished first row of VV table
VV(1,1)= A C T_0 
finish CYK algorithm, check for S in VV[1][n]and thus Accepted 
string accepted by G, string is in L(G)
about to parse 010
run CYK algorithm to build array, size=3
Finished first row of VV table
VV(1,1)= A C T_0 
VV(2,1)= T_1 
VV(3,1)= A C T_0 
Finished 2 row of VV table
VV(1,2)= 
VV(2,2)= B 
VV(3,2)= 
Finished 3 row of VV table
VV(1,3)= A C 
VV(2,3)= 
VV(3,3)= 
finish CYK algorithm, check for S in VV[1][n]and thus Accepted 
string accepted by G, string is in L(G)
about to parse 01010
run CYK algorithm to build array, size=5
Finished first row of VV table
VV(1,1)= A C T_0 
VV(2,1)= T_1 
VV(3,1)= A C T_0 
VV(4,1)= T_1 
VV(5,1)= A C T_0 
Finished 2 row of VV table
VV(1,2)= 
VV(2,2)= B 
VV(3,2)= 
VV(4,2)= B 
VV(5,2)= 
Finished 3 row of VV table
VV(1,3)= A C 
VV(2,3)= 
VV(3,3)= A C 
VV(4,3)= 
VV(5,3)= 
Finished 4 row of VV table
VV(1,4)= 
VV(2,4)= B 
VV(3,4)= 
VV(4,4)= 
VV(5,4)= 
Finished 5 row of VV table
VV(1,5)= A C 
VV(2,5)= 
VV(3,5)= 
VV(4,5)= 
VV(5,5)= 
finish CYK algorithm, check for S in VV[1][n]and thus Accepted 
string accepted by G, string is in L(G)
about to parse 111
run CYK algorithm to build array, size=3
Finished first row of VV table
VV(1,1)= T_1 
VV(2,1)= T_1 
VV(3,1)= T_1 
Finished 2 row of VV table
VV(1,2)= 
VV(2,2)= 
VV(3,2)= 
Finished 3 row of VV table
VV(1,3)= 
VV(2,3)= 
VV(3,3)= 
finish CYK algorithm, check for S in VV[1][n]and thus Accepted 
string rejected by G, string is not in L(G)
about to parse 011
run CYK algorithm to build array, size=3
Finished first row of VV table
VV(1,1)= A C T_0 
VV(2,1)= T_1 
VV(3,1)= T_1 
Finished 2 row of VV table
VV(1,2)= 
VV(2,2)= 
VV(3,2)= 
Finished 3 row of VV table
VV(1,3)= 
VV(2,3)= 
VV(3,3)= 
finish CYK algorithm, check for S in VV[1][n]and thus Accepted 
string rejected by G, string is not in L(G)
about to parse 011
run CYK algorithm to build array, size=3
Finished first row of VV table
VV(1,1)= A C T_0 
VV(2,1)= T_1 
VV(3,1)= T_1 
Finished 2 row of VV table
VV(1,2)= 
VV(2,2)= 
VV(3,2)= 
Finished 3 row of VV table
VV(1,3)= 
VV(2,3)= 
VV(3,3)= 
finish CYK algorithm, check for S in VV[1][n]and thus Accepted 
string rejected by G, string is not in L(G)