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Brave Leo
Brave Leo is a large language model-based chatbot developed by Brave Software and included with the Brave browser. == History == In November 2023, the company said versions for iOS and Android would be available "in the coming months". == Features == Since January 2024, Leo has used the open-source Mixtral 8x7B from Mistral AI as its default large language model, in addition to LLaMA 2 from Meta Platforms and Claude from Anthropic, both of which have been used previously. Leo can suggest follow-up questions, and summarize webpages, PDFs, and videos. Leo has a $15 (US) per month premium version that enables more requests and uses larger LLMs. == Privacy == The answers given by Leo are not saved. Brave uses the slogan Love Privacy to emphasize its focus on user privacy and data protection. The phrase has been featured in Brave's official marketing campaigns and has been cited in media coverage of the browser's privacy-first approach. == Controversies == In 2023, PC World reported that Leo evades questions about US elections.
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Talking Angela
Talking Angela is a mobile game (formerly a chatbot), developed by Slovenian studio Outfit7 as part of the Talking Tom & Friends series. It was released on 13 November 2012 and December 2012 for iPhone, iPod and iPad, January 2013 for Android, and January 2014 for Google Play. The game's successor, the My Talking Angela game, was released in December 2014. The game takes place in a café in Paris and allows players to interact with Angela, an anthropomorphic white cat in different ways. Players can use coins to purchase makeup, accessories and items, as well as drinks that will trigger different visual effects. The fortune cookie button causes Angela to read out a fortune cookie, while the bird icon will prompt birds to fly around the screen, or have Angela feed them. Players can also pet or poke Angela, as well the café's sign. Prior to their removal, the game featured a chat system and a camera button. Users can engage in conversations with Angela, ask for quizzes or initiate a short snippet of the song "That's Falling In Love". If the player was to type in "Who is an idiot?", Angela would respond with a random swear word. Additionally, inquiring Angela about sexual topics would cause her to reply with "Do you want to talk about sex?", though she will quickly change the topic regardless of what the player writes next. A hoax claiming that Angela's eyes were hidden cameras that enabled hackers or paedophiles to watch children was spread. Despite the claims, Snopes and The Guardian found no evidence. Due to the hoax, Angela received a blue dress, as well as an altered eye asset with a different reflection, and later the chat and camera functions were removed altogether. == Hoaxes == In February 2014, Talking Angela was the subject of an Internet hoax alleging that the application was a front for child predators to exploit children. The rumor, which was widely circulated on Facebook and various websites claiming to be dedicated to parenting, claims that a sinister sexual predator or hacker, asked children for private personal information using the game's text-chat feature. Other versions of the rumour even attributed the disappearance of a child to the game; one news report claimed that a seven year old boy disappeared after downloading the app. Another variation included that it was run by a paedophile ring, citing a man that could be seen in Angela's eyes. The app's developers, Outfit7, later gave a statement refuting the hoaxes. The hoax was eventually debunked by Snopes, a fact-checking website. The site's owners, Barbara and David Mikkelson, reported that they had tried to "prompt" it to give responses asking for private information, but were unsuccessful, even when asking it explicitly sexual questions. While it is true that, in the game with child mode off, Angela does ask for the user's name, age and personal preferences to determine conversation topics, Outfit7 has said that this information is all "anonymized" and all personal information is removed from it. It is also impossible for a person to take control of what Angela says in the game, since the game is based on chatbot software. When the mode was turned on, the chat feature was disabled, meaning no personal questions could be asked. In 2015, the hoax was revived on Facebook, which prompted online security company Sophos and The Guardian to debunk it again. Sophos employee Paul Ducklin wrote that the message being posted on Facebook promoting the hoax was "close to 600 rambling, repetitious words, despite claiming at the start that it didn't have words to describe the situation. It's ill-written, and borders on being illiterate and incomprehensible." Bruce Wilcox, one of the game's programmers, attributed the hoax's popularity to the fact that the chatbot program in Talking Angela aimed to sound realistic. Concern was raised that the game's child mode may have been too easy for children to turn off. It allowed them to purchase "coins", premium currency in the game, via iTunes, and enabled the chat feature. While not "connecting your children to paedophiles", this still raised concerns according to The Guardian. === Impact === The scare significantly boosted the game's popularity, and was credited with helping the app enter the top 10 free iPhone apps soon after the hoax became widely known in February 2015,In the truth the reason there is a man in Angela’s eyes is because of pareidoila, the ability to see through diamonds and other minerals and water bodies and shiny objects,which is the reason why players notice a man in her eyes,The truth is that being Angela’s eyes simply serve as a reflective surface,Because of the low quality of this reflection the reflection was mistaken for a humanoid figure. oref>Smith, Josh (19 February 2014). "Talking Angela App Scare Skyrockets App to Top of Charts". GottaBeMobile.com. Archived from the original on 2 April 2016. Retrieved 10 May 2014. and third most popular for all iPhone apps at the start of the following month. In 2016, Outfit7 removed the chat feature along with the camera function from the app due to this controversy, though this decision was met with criticism.
Pumping lemma for regular languages
In the theory of formal languages, the pumping lemma for regular languages is a lemma that describes an essential property of all regular languages. Informally, it says that all sufficiently long strings in a regular language may be pumped—that is, have a middle section of the string repeated an arbitrary number of times—to produce a new string that is also part of the language. The pumping lemma is useful for proving that a specific language is not a regular language, by showing that the language does not have the property. Specifically, the pumping lemma says that for any regular language L {\displaystyle L} , there exists a constant p {\displaystyle p} such that any string w {\displaystyle w} in L {\displaystyle L} with length at least p {\displaystyle p} can be split into three substrings x {\displaystyle x} , y {\displaystyle y} and z {\displaystyle z} ( w = x y z {\displaystyle w=xyz} , with y {\displaystyle y} being non-empty), such that the strings x z , x y z , x y y z , x y y y z , . . . {\displaystyle xz,xyz,xyyz,xyyyz,...} are also in L {\displaystyle L} . The process of repeating y {\displaystyle y} zero or more times is known as "pumping". Moreover, the pumping lemma guarantees that the length of x y {\displaystyle xy} will be at most p {\displaystyle p} , thus giving a "small" substring x y {\displaystyle xy} that has the desired property. Languages with a finite number of strings vacuously satisfy the pumping lemma by having p {\displaystyle p} equal to the maximum string length in L {\displaystyle L} plus one. By doing so, no strings at all in L {\displaystyle L} have length at least p {\displaystyle p} . The pumping lemma was first proven by Michael Rabin and Dana Scott in 1959, and rediscovered shortly after by Yehoshua Bar-Hillel, Micha A. Perles, and Eli Shamir in 1961, as a simplification of their pumping lemma for context-free languages. == Formal statement == Let L {\displaystyle L} be a regular language. Then there exists an integer p ≥ 1 {\displaystyle p\geq 1} depending only on L {\displaystyle L} such that every string w {\displaystyle w} in L {\displaystyle L} of length at least p {\displaystyle p} ( p {\displaystyle p} is called the "pumping length") can be written as w = x y z {\displaystyle w=xyz} (i.e., w {\displaystyle w} can be divided into three substrings), satisfying the following conditions: | y | ≥ 1 {\displaystyle |y|\geq 1} | x y | ≤ p {\displaystyle |xy|\leq p} ( ∀ n ≥ 0 ) ( x y n z ∈ L ) {\displaystyle (\forall n\geq 0)(xy^{n}z\in L)} y {\displaystyle y} is the substring that can be pumped (removed or repeated any number of times, and the resulting string is always in L {\displaystyle L} ). (1) means the loop y {\displaystyle y} to be pumped must be of length at least one, that is, not an empty string; (2) means the loop must occur within the first p {\displaystyle p} characters. | x | {\displaystyle |x|} must be smaller than p {\displaystyle p} (conclusion of (1) and (2)), but apart from that, there is no restriction on x {\displaystyle x} and z {\displaystyle z} . In simple words, for any regular language L {\displaystyle L} , any sufficiently long string w {\displaystyle w} (in L {\displaystyle L} ) can be split into 3 parts, i.e. w = x y z {\displaystyle w=xyz} , such that all the strings x y n z {\displaystyle xy^{n}z} for n ≥ 0 {\displaystyle n\geq 0} are also in L {\displaystyle L} . Below is a formal expression of the pumping lemma. ∀ L ⊆ Σ ∗ , regular ( L ) ⟹ ∃ p ≥ 1 , ∀ w ∈ L , | w | ≥ p ⟹ ∃ x , y , z ∈ Σ ∗ , ( w = x y z ) ∧ ( | y | ≥ 1 ) ∧ ( | x y | ≤ p ) ∧ ( ∀ n ≥ 0 , x y n z ∈ L ) {\displaystyle {\begin{array}{l}\forall L\subseteq \Sigma ^{},{\mbox{regular}}(L)\implies \\\quad \exists p\geq 1,\forall w\in L,|w|\geq p\implies \\\qquad \exists x,y,z\in \Sigma ^{},(w=xyz)\land (|y|\geq 1)\land (|xy|\leq p)\land (\forall n\geq 0,xy^{n}z\in L)\end{array}}} == Use of the lemma to prove non-regularity == The pumping lemma is often used to prove that a particular language is non-regular: a proof by contradiction may consist of exhibiting a string (of the required length) in the language that lacks the property outlined in the pumping lemma. Example: The language L = { a n b n : n ≥ 0 } {\displaystyle L=\{a^{n}b^{n}:n\geq 0\}} over the alphabet Σ = { a , b } {\displaystyle \Sigma =\{a,b\}} can be shown to be non-regular as follows: Assume that some constant p ≥ 1 {\displaystyle p\geq 1} exists as required by the lemma. Let w {\displaystyle w} in L {\displaystyle L} be given by w = a p b p {\displaystyle w=a^{p}b^{p}} , which is a string longer than p {\displaystyle p} . By the pumping lemma, there must exist a decomposition w = x y z {\displaystyle w=xyz} with | x y | ≤ p {\displaystyle |xy|\leq p} and | y | ≥ 1 {\displaystyle |y|\geq 1} such that x y i z {\displaystyle xy^{i}z} in L {\displaystyle L} for every i ≥ 0 {\displaystyle i\geq 0} . Since | x y | ≤ p {\displaystyle |xy|\leq p} , the string y {\displaystyle y} only consists of instances of a {\displaystyle a} . Because | y | ≥ 1 {\displaystyle |y|\geq 1} , it contains at least one instance of the letter a {\displaystyle a} . Pumping y {\displaystyle y} to give x y 2 z {\displaystyle xy^{2}z} gives a word with more instances of the letter a {\displaystyle a} than the letter b {\displaystyle b} , since some instances of a {\displaystyle a} but none of b {\displaystyle b} were added. Therefore, x y 2 z {\displaystyle xy^{2}z} is not in L {\displaystyle L} which contradicts the pumping lemma. Therefore, L {\displaystyle L} cannot be regular. The proof that the language of balanced (i.e., properly nested) parentheses is not regular follows the same idea. Given p {\displaystyle p} , there is a string of balanced parentheses that begins with more than p {\displaystyle p} left parentheses, so that y {\displaystyle y} will consist entirely of left parentheses. By repeating y {\displaystyle y} , a string can be produced that does not contain the same number of left and right parentheses, and so they cannot be balanced. == Proof of the pumping lemma == For every regular language there is a finite-state automaton (FSA) that accepts the language. The number of states in such an FSA are counted and that count is used as the pumping length p {\displaystyle p} . For a string of length at least p {\displaystyle p} , let q 0 {\displaystyle q_{0}} be the start state and let q 1 , . . . , q p {\displaystyle q_{1},...,q_{p}} be the sequence of the next p {\displaystyle p} states visited as the string is emitted. Because the FSA has only p {\displaystyle p} states, within this sequence of p + 1 {\displaystyle p+1} visited states there must be at least one state that is repeated. Write q s {\displaystyle q_{s}} for such a state. The transitions that take the machine from the first encounter of state q s {\displaystyle q_{s}} to the second encounter of state q s {\displaystyle q_{s}} match some string. This string is called y {\displaystyle y} in the lemma, and since the machine will match a string without the y {\displaystyle y} portion, or with the string y {\displaystyle y} repeated any number of times, the conditions of the lemma are satisfied. For example, the following image shows an FSA. The FSA accepts the string: abcd. Since this string has a length at least as large as the number of states, which is four (so the total number of states that the machine passes through to scan abcd would be 5), the pigeonhole principle indicates that there must be at least one repeated state among the start state and the next four visited states. In this example, only q 1 {\displaystyle q_{1}} is a repeated state. Since the substring bc takes the machine through transitions that start at state q 1 {\displaystyle q_{1}} and end at state q 1 {\displaystyle q_{1}} , that portion could be repeated and the FSA would still accept, giving the string abcbcd. Alternatively, the bc portion could be removed and the FSA would still accept giving the string ad. In terms of the pumping lemma, the string abcd is broken into an x {\displaystyle x} portion a, a y {\displaystyle y} portion bc and a z {\displaystyle z} portion d. As a side remark, the problem of checking whether a given string can be accepted by a given nondeterministic finite automaton without visiting any state repeatedly, is NP hard. == General version of pumping lemma for regular languages == If a language L {\displaystyle L} is regular, then there exists a number p ≥ 1 {\displaystyle p\geq 1} (the pumping length) such that every string u w v {\displaystyle uwv} in L {\displaystyle L} with | w | ≥ p {\displaystyle |w|\geq p} can be written in the form u w v = u x y z v {\displaystyle uwv=uxyzv} with strings x {\displaystyle x} , y {\displaystyle y} and z {\displaystyle z} such that | x y | ≤ p {\displaystyle |xy|\leq p} , | y | ≥ 1 {\displaystyle |y|\geq 1} and u x y i z v {\displaystyle uxy^{i}zv} is in L {\displaystyle L} for every integer i ≥ 0 {\displaystyle i\geq 0} . From this, the above standard v