Hot answers tagged

11

Your answer is $$\frac{22x-7}{20x-10}$$ And the books "correct" answer is $$\frac{7-22x}{10-20x}$$ Yes? Notice what happens when you multiply both the numerator and denominator in your answer by $-1$? You're very welcome.


10

I am going to write $S_i$ for the set of strings that cause the machine to go from the start state into state $i$. From the diagram of the machine, we have: $$\begin{align} \def\a{\mathtt{a}}\def\b{\mathtt{b}} S_1 & = \epsilon + S_3\a \\ S_2 & = S_1(\a+\b) + S_2\a + S_3\b \\ S_3 & = S_2\b \end{align}$$ (This part is crucial, and if you don't ...


8

You want to show that $$\frac{\sqrt{3}+1}{2\sqrt{2}} = \frac{\sqrt{2+\sqrt{3}}} {2}.$$ Rearrange this into $$2(\sqrt{3} + 1) = 2\sqrt{2}\sqrt{2+\sqrt{3}}.$$ Since both sides are evidently positive, this is equivalent to $$(2(\sqrt{3} + 1))^2 = (2\sqrt{2}\sqrt{2+\sqrt{3}})^2,$$ which simplifies to $$4(4 + 2\sqrt{3}) = 8(2+\sqrt{3}),$$ which is of course true.


8

You got the correct answer. You just need to multiply both numerator and denominator by -1 $$\frac{22x-7}{20x-10} = \frac{-1}{-1} \times \frac{7-22x}{10-20x}$$


7

Hint: Show that any two regexps over a $1$-letter alphabet commute, i.e. $PQ = QP$ for every two regexps $P,Q$. Hint to the hint: Show that any two words over a $1$-letter alphabet commute, i.e. $ab = ba$ for every two words $a,b$.


7

In order to show that $A^*B$ is "the" solution to $X=AX+B$ you have to show that (1) $A^*B$ is "a" solution to $X=AX+B$, and (2) if $X$ is a solution to $X=AX+B$ then $X = A^*B$. Unfortunately the second part is actually false in general, since $\Sigma^*$ is always a solution when $\epsilon \in A$. For a correct statement, see the Wikipedia page on Arden's ...


5

The basic idea for an automated proof is: convert both languages to their deterministic finite state machines. Calculate the cross product of the finite state machines with the transitions $F: ([x,y],\sigma) => [F_x(x,\sigma), F_y(y,\sigma)]$ Collect the list of reachable states, and their acceptivity status from both languages. If there are any [...


5

The only strings that match the regular expression $0^*$ are strings consisting entirely of $0$s. (This includes the empty string.) Thus, a string that matches $0^*1$ must end in a $1$, everything before that $1$ has to match $0^*$, so it must be a string of $0$s (or empty). Finally, a string that matches $0^*10^*$ must start with a string that matches $0^*1$...


5

The regular expression you have given gives the language $$\{\varepsilon, a, aa, aaa, \dots\} \cup \{\varepsilon, b, bb, bbb, \dots\} \cup \{\varepsilon, c, cc, ccc, \dots\}$$ $$= \{\varepsilon, a, aa, aaa, \dots, b, bb, bbb, \dots, c, cc, ccc, \dots\},$$ which is clearly not what you are after. Using only union ($\cup$), concatenation ($\circ$), and the ...


5

So we know that $A^* = \epsilon + A + A^2 + \cdots $. Are you asking to show how $X = A^* B$ from $X = AX + B$. If so, we can show it simply \begin{align} X &= AX + B \\ &= A(AX + B) = A^2 X + AB + B \\ &= A^2 (AX + B) + AB + B = A^3 X + A^2 B + A B + B \\ &\ldots \\ &= B + A B + A^2 B + \cdots \\ &= (\epsilon + A + A^2 + \cdots) B \\...


4

One easy if tedious approach is to design a finite-state automaton that recognizes $L$ and then apply the algorithm that converts such an automaton to a corresponding regular expression. Going about it directly, I observe that if $w\in L$, every $a$ in $w$ must be immediately followed by another $a$ or a $c$ or be at the end of the word; otherwise there are ...


4

Let $R$ be the set of words described by the regular expression $r$, and let $S$ be the set of words described by the regular expression $s$. Then the regular expression $(r+s)^*$ describes the words in the set $(R\cup S)^*$, and $r^*+s^*$ describes the words in the set $R^*\cup S^*$. Thus, the question boils down to asking whether it’s necessarily true that ...


4

$(1+01)^*$ generates all strings of zeroes and ones with the property that every $0$ is immediately followed by a $1$; these are the strings that do not contain $00$ and do not end in $0$. Similarly, $(0+01)^*$ generates all strings of zeroes and ones in which every $1$ is immediately preceded by a $0$; these are the strings that do not contain $11$ and do ...


4

Calling your four patterns $A_k,B_k,C_k,D_k$ respectively, note that $01101$ is in both $A_2$ and $B_1$, so you are over counting. You can fix your solution though using inclusion-exclusion


4

For completeness, I will mention: $$ \mathbf{aaa}\cup \mathbf{aab}\cup \mathbf{aac}\cup \mathbf{aba}\cup \mathbf{abb}\cup \mathbf{abc}\cup \mathbf{aca}\cup \mathbf{acb}\cup \mathbf{acc}\cup\\ \mathbf{baa}\cup \mathbf{bab}\cup \mathbf{bac}\cup \mathbf{bba}\cup \mathbf{bbb}\cup \mathbf{bbc}\cup \mathbf{bca}\cup \mathbf{bcb}\cup \mathbf{bcc}\cup\\ \mathbf{caa}\...


4

Let me show you a systematic approach. Think of each of the state names as an abbreviation for a regular expression corresponding to the set of words that end at that state when starting from the initial state. Since $B$ and $C$ are the acceptor states, we want the expression $B+C$. (I’ll use $+$ instead of a comma for or; I use $\lambda$ for the empty word.)...


4

A DFA has no $\epsilon$-transitions, so this is not a DFA. I would rather called your automaton a nondeterministic finite automaton with $\varepsilon$-moves. Your regular expression $1(11 \cup 0^*)^*$ is correct. Your informal description is almost correct: you should just modify your sentence each $1$ after the first, will be accompanied by at least ...


4

This has been an open problem for some years until a negative answer was given indenpendtly by Redko [4] and Conway [1, pp. 105-118]): every complete system of identities for the regular expressions is necessarily infinite. Conway [1, pp. 116-119] conjectured a "good" complete system and this conjecture was ultimately proved by Krob [2, 3]. Interestingly, ...


4

One way is to take it in pieces. Clearly your automaton accepts $0^*$. What else gets it back to state $a$? Only $11^*00$, after which you can have any number of zeroes again. Thus, $0^*(11^*000^*)^*$ almost does the trick. However, like your automaton, it misses the possibility that an acceptable string can end in $1$ or $10$ if no $0$ immediately precedes ...


3

It has problems in both directions: it doesn’t accept the empty word, which is in the language, and it does accept a lot of words, like $aba$ and $abb$, that aren’t in the language. The first problem is easily fixed: just make $q_0$ an acceptor state. The second requires some more significant changes. The $b$-transition from $q_3$ should go to $q_1$, the ...


3

Try this: $$ \frac{(-1)^{n}n(n+1)}{2}+(-1)^{n+1}(n+1)^2 = (-1)^n\bigg(\frac{n(n+1)}{2}-\frac{2(n+1)^2}{2}\bigg)\\ =\frac{(-1)^n(n+1)}{2}(n-2(n+1))\\ =\frac{(-1)^{n+1}(n+1)(n+2)}{2} $$


3

Suppose that I’ve a one-state machine $E_a$ that recognizes $a^*$ and another, $E_b$, that recognizes $b^*$; combining them in the second way would result in a one-state machine that recognized $(a\lor b)^*$, not $a^*\lor b^*$. This violates both (2) and (3), but you can use similar ideas to show that violating either of them individually can produce ...


3

Another approach: identities in Kleene algebra... (a) $M \subseteq M^* \subseteq M^*N^* \text{ and } N \subseteq N^* \subseteq M^*N^* \\\text{ so } (M+N) \subseteq M^*N^* \\\text{ so } (M+N)^* \subseteq (M^*N^*)^* \\\text{QED one way} $ (b) $ M^* \subseteq (M+N)^* \text{ and } N^* \subseteq (M+N)^* \\\text { so } M^*N^* \subseteq (M+N)^*(M+N)^* \...


3

(a) The requirement boils down to saying that once you get $xx$, you can never get a $y$. Your regular expression generates only valid strings, but it doesn’t generate all valid strings. For instance, it doesn’t generate $xyxx$. You want something like $y^*(xy^+)^*x^*$. (b) The second part of your regular expression is fine: it generates precisely those ...


3

A comment before I get to your (1)-(3): $p$ and $P$ are not the same symbol, so either the pumping length is $p$, or it’s $P$, but you shouldn’t keep jumping back and forth between the two. I’ll use $p$. This is correct: the pumping lemma is strictly a tool for showing that a language is not regular. This objection is not really well-taken. It’s true that $...


3

Usually, in a monoid, we define $x^0$ to be the identity element of the monoid, that is, the element $e$ such that $ex = x = xe$ for all $x$. This is so that the standard "exponent rules" such as $x^{a+b} = x^a \cdot x^b$ hold even when $a$ or $b=0$. In your case, since the operation is concatenation, it makes sense for $x^0$ to mean the empty string, since ...


3

$r$ and $s$ are regular expressions, which represent regular sets. If $r$ and $s$ are regular expressions that represent sets $R$ and $S$, then the regular expression $r+s$ denotes the set $R\cup S$. If $r$ is a regular expression denoting the set $R$, then $$r^*$$ is a shorthand for $$\epsilon + r + rr + rrr + \cdots .$$


3

For any fixed value of $n$, the language $$ L_{n}=\{w:|w|_a=2^n+273\} $$ is certainly regular, and the regular expression you described will recognize it. But the question here is whether $$ L=\bigcup_{n\in\mathbb{N}}L_n = \{w:|w|_a=2^n+273 \text{ for some } n\in\mathbb{N}\} $$ is regular. It's not; and the easiest way to prove it is to use the so-called ...


3

Let $H$ denote the unit step function. $xH(x)$ is zero when $x$ is negative and equal to $x$ when $x$ is positive. $xH(x-1)$ is zero when $x$ is less than 1 and equal to $x$ otherwise. If I'm not mistaken, $$xH(x) - (x-1)H(x-1)$$ is the function you want.


3

$$ f(x) = \frac{|x|+x}{2} - \frac{(x-1)+|x-1|}{2}. $$ (This can be simplified a bit.)



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