# The language UNIMODAL is not context free

In this article, we will prove that: the language ‘UNIMODAL’ is not cfg. Also, we will prove that cfg can not be used to compare binary number.

## proposition one:“BINARY EQUAL”：

Alphabet：{S,0,1}

language BINARY_EQUAL：string such as ${(0|1)^* S (0|1)^*}$，AND the binary number before ‘#’ is equal to the the latter one.

statement： BINARY_EQUAL is not context free grammar

Prove: Assume that BINARY_EQUAL is cfl(context free language), then use the pump lemma.

For a string s=“xxxx…xxxx#yyyy…yyyy”

$$ \exists uvwxy, \text{such that} \ s=uvwxy\ ,\text{and}\ \ \forall i\ge 0, uv^iwx^iy\in \mathbb{BINARY_EQUAL} $$

We choose a special string: $s=1^k0^kS0^k1^k$, so when k is large enough, the v and x must be different. Hence, the pumped string is not BINARY_EQUAL. So the proposition is right.

## proposition two:“BINARY BIGGER”:

Alphabet：{S,0,1}
language BINARY_EQUAL：string such as ${(0|1)^* S (0|1)^*}$，AND the binary number before ‘#’ is **BIGGER** to the the latter one.
statement： BINARY_BIGGER is not context free grammar

Prove: Assume that BINARY_BIGGER is cfl, use the pump lemma for the string $1^k0^{k-1}1S1^k0^k$.

If v and x are both in the left number, the pump down it.(pump down means that we choose the $i$ in the pump lemma as 0. a.k.a $i=0$) then the digits of left num is 2k-2, while the right is 2k, so the right number is bigger than the left.

If v and x are both in the right number, just choose i as much as you need. Then the right is obvious larger than the left.

If v and x are in different sides: (1) If v is ‘1’ and x is ‘1’,pump down it. Then two num should be equal (2) If v is ‘00…1’, and x is ‘11…1’ then pump up it and the right num should be larger.

So, the proposition is true.

## proposition three:“UNIMODAL”:

Alphabet：{S,0,1}
language UNIMODAL：string such as ${(0|1)^* S (0|1)^*(S(0|1)^*)^*}$. This string will be considered as an array of binary number. And when and only when the array is UNIMODAL. The string belongs to the language.

For example: 1 3 5 3 1 is a unimodal array. The form in binary is 1#11#101#11#1. So the string ‘1#11#101#11#1’ is unimodal. statement： UNIMODAL is not context free grammar

Prove: use pump lemma. s=uvwxy. Let s be ‘$1^K0^KS1^K0^KS1^K0^KS1^K0^KS1^K0^K$’, Actually, we need $1^K0^KS1^K0^K(S1^K0^K)^k$. Because every num in this array is equal. It’s a UNIMODAL string. And if v and x belongs to one num, then pump down it. We get a decrease then increase, hence a controdiction.

And if v and x belongs to different num. Then v is ‘000…’, x is ‘111…’,(because we let the k be large enough). Hence,we pump down, we we will get a decrease and then increase and then decrease. Controdiction!