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Combinatorics on words studies finite and infinite sequences of symbols from an alphabet, especially repetition, ordering and occurrence of shorter words. A finite word is an element of the free monoid generated by its alphabet, with concatenation as its operation. For example, abab has two occurrences of ab and is the square of ab; aba is a palindrome. These questions connect algorithms on words with generating functions and symbolic descriptions of dynamical systems.

Decide whether a finite string is a palindrome by comparing corresponding symbols from opposite ends. The test needs at most half as many symbol comparisons as the string length and can use constant auxiliary space.

Let $s=k\oplus \mathrm{rev}(k)$ be the effective mask. For each position $i$,Thus the mask is always a palindrome, even though $k$ itself is uniformly random. The exclusive or of the ciphertext bits in a mirror pair isbecause the equal mask bits cancel. The observer learns the corresponding plaintext parity exactly. If $n$ is odd, the middle mask bit is also zero, so the middle plaintext bit is exposed directly.

An explicit failure of perfect secrecy works for every $n\ge 1$. Choose $X$ uniformly from $0^n$ and $10^{n-1}$. For $X=0^n$, the ciphertext is a palindrome with probability one. For $X=10^{n-1}$ and $n\ge 2$, toggling just the first bit breaks the mirror equality, so the ciphertext is a palindrome with probability zero. For $n=1$, the effective mask is zero and $C=X$. Observing $C$ therefore distinguishes these two messages perfectly.

There are only $2^{\lfloor n/2\rfloor }$ possible effective masks, each equally likely. Hence even a fresh $n$-bit key supplies at most $\lfloor n/2\rfloor$ independent masking bits in this construction. Choosing a long, uniform, one-use key is insufficient if the encryption transformation collapses its randomness.

It tests whether the input string is a palindrome: reversing the characters leaves the string unchanged. Comparisons are case-sensitive and include spaces and punctuation. For example, mystery1("level") returns True, whereas mystery1("lever") returns False.

A context-free grammar with start symbol $P$, terminal alphabet $\{0,1\}$, and the following productions generates all binary palindromes:Each production preserves the palindrome property: it either supplies a central string of length zero or one, or wraps the same symbol around a shorter palindrome. This proves soundness by structural induction on the parse tree.

Conversely, any palindrome of length at least two has equal first and last symbols. Remove them to obtain a shorter palindrome, apply mathematical induction on length, and then use the corresponding wrapping production. The three base cases are $ϵ$, $0$, and $1$, proving completeness.

These strings are the even-length palindromes $u{u}^R$. Suppose this were a regular language, and let $p$ be its length from the pumping lemma for regular languages. The stringbelongs to the formal language, with $u=0^{p}1$. In any allowed decomposition $w=abc$, where $|ab|\le p$ and $|b|>0$, the pumped block is $b=0^k$ with $1\le k\le p$. Pumping it away givesThis is not a palindrome: its initial and final runs of zeros have different lengths. Hence it cannot have the form $u{u}^R$, contradicting the pumping lemma for regular languages. Therefore

A context-free grammar is a quadruple $(N,\Sigma ,P,S)$: $N$ is a finite set of nonterminal symbols, $\Sigma$ is a disjoint finite alphabet of terminal symbols, $S\in N$ is the start symbol, and $P$ is a finite set of productions $A\to \alpha$, with $A\in N$ and $\alpha \in (N\cup \Sigma {)}^{\ast }$. Its formal language consists of the terminal strings derivable from $S$.

For the even-length palindromes in (b)(ii), take terminal alphabet $\{0,1\}$ and the context-free grammarEvery production adds identical symbols at the two ends, so every generated string is an even-length palindrome. Conversely, remove the matching outer pair of any nonempty even-length palindrome and repeat; this constructs its derivation, ending with $\epsilon$.

For the unequal-block formal language in (b)(iv), useAfter $k$ uses of $A\to 0A1$, and $j\ge 1$ zeros from $Z$, the resulting string is $0^{k+j}{1}^k$, so its zero count is larger. Every such unequal pair with $m>n$ occurs by choosing $k=n,j=m-n$. Similarly $B$ generates exactly $0^k{1}^{k+j}$ with $j\ge 1$, covering precisely $m<n$. The start alternatives cover both possibilities and never equal counts. If positive exponents are required, replace the start productions by $S\to 0A1\mid 0B1$, which adds one matched pair and ensures both counts are positive.