By my understanding of the question the 4th line will result in the missionary being eaten

>
>(c=cannibal, m=missionary)
>left - on boat - right
>cccmmm -
>ccmm - cm
>ccmm - m - c
>cmm - cm - c
----------^^^^^^

2 Canibals and 1 missionary - Stew!

>cmm - m - cc
>mm - cm - cc
>mm - m - ccc
>m - mm - ccc
>m - cc - cmm
>c - cm - cmm
>c - m - ccmm
> - cm - ccmm
> - cccmmm
>
>After using brute force on this, it reminds me of the Towers of Hanoi problem (Moving a stack of disks across 3 pegs where no larger disk can fit over a smaller disk)
>
>I heard a simplified version:
>A farmer has a dog, a chicken, and some grain. He must cross a stream by carrying each over his head one at a time. If left alone, the dog will eat the chicken and the chicken will eat the grain. How does he get all three across?
>
>Answer: (d=dog,c=chicken,g=grain,f=farmer)
>dcgf -
>dg - cf -
>dg - f - c
>d - gf - c
>d - cf - g
>c - df - g
>c - f - dg
> - cf - dg
> - - dcgf