Context Free Grammar for unequal number of symbols

Question

I'm aware of how it is possible to construct a context free grammar with the same number of two given elements ie. if we use {0,1}

S->SS
S->0S1
S->1S0
S->ε

I am however struggling to find a way to construct a grammar that has a given amount of one element more than another. ie. consistently two more 0s than 1s. Does anyone have any ideas of how to construct such a grammar?

Yes, I do have an idea. Hint: two unequal sets are just two equal sets separated by multiple of one of the elements. — John Dvorak, Jan 24 '16 at 13:53

Gerard Rozsavolgyi · Answer 1 · 2016-01-24T15:06:24.440

2

Edit : (corrected) Something like the following forces to have at least one 0 more than 1's:

S->T0S | T0T
T->0T1T | 1T0T | ε

so now, it shouldn't be too hard to add one more 0 by repeating same pattern ...

doing so the following grammar answers the problem:

S->T0S | T0T
T->0T1T | 1T0T 
T->U0T | U0U
U->0U1U | 1U0U | ε

edited Jan 24 '16 at 15:06

answered Jan 24 '16 at 14:14

Gerard Rozsavolgyi

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But what about the following: 0T (begin), 0TT (T->TT rule), 000 (T->0 rule), Surely in this case you would break the rule? – Harry Wright Jan 24 '16 at 14:27
you're right ! we should not change the last rule and rewrite the first rules – Gerard Rozsavolgyi Jan 24 '16 at 14:30
thanks! surely you could end up with 000 though by going from T0T, then substituting 0 in for T? – Harry Wright Jan 24 '16 at 14:52
yes, if we end by ε instead of 0, we have the language with at least one more 0 than 1's so shouldn't be too hard to force a new 0 ... i'll edit – Gerard Rozsavolgyi Jan 24 '16 at 14:55

Harry Wright · Accepted Answer · 2016-01-29T14:49:54.890

1

I worked out a nice answer to this:

S->P0P0P
P->PP
P->0P1
P->1P0
P->ε

It should come up with a string of two more 0s than 1s, and can be easily extended to greater numbers.

edited Jan 29 '16 at 14:49

answered Jan 26 '16 at 17:58

Harry Wright

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Context Free Grammar for unequal number of symbols

2 Answers2