Scheduled Commit

data/retrieval_battle-1b2a6102-80bb-417d-a081-cf780f49c66c.jsonl

@@ -308,3 +308,5 @@
 {"tstamp": 1739485084.047, "task_type": "retrieval", "type": "tievote", "models": ["", ""], "ip": "", "0_conv_id": "9295ba56bb0f44f4a19377505bcbb4a6", "0_model_name": "GritLM/GritLM-7B", "0_prompt": "Finding words with different meanings over time without using word embeddings.", "0_output": [["Finding words with different meanings over time without using word embeddings.", "Title: An Improved Historical Embedding without Alignment\n\nAbstract: Many words have evolved in meaning as a result of cultural and social change. Understanding such changes is crucial for modelling language and cultural evolution. Low-dimensional embedding methods have shown promise in detecting words' meaning change by encoding them into dense vectors. However, when exploring semantic change of words over time, these methods require the alignment of word embeddings across different time periods. This process is computationally expensive, prohibitively time consuming and suffering from contextual variability. In this paper, we propose a new and scalable method for encoding words from different time periods into one dense vector space. This can greatly improve performance when it comes to identifying words that have changed in meaning over time. We evaluated our method on dataset from Google Books N-gram. Our method outperformed three other popular methods in terms of the number of words correctly identified to have changed in meaning. Additionally, we provide an intuitive visualization of the semantic evolution of some words extracted by our method"]], "0_corpus": "arxiv", "1_conv_id": "2822fc419c1c44d68ec2bf54b1aa4e76", "1_model_name": "embed-english-v3.0", "1_prompt": "Finding words with different meanings over time without using word embeddings.", "1_output": [["Finding words with different meanings over time without using word embeddings.", "Title: An Improved Historical Embedding without Alignment\n\nAbstract: Many words have evolved in meaning as a result of cultural and social change. Understanding such changes is crucial for modelling language and cultural evolution. Low-dimensional embedding methods have shown promise in detecting words' meaning change by encoding them into dense vectors. However, when exploring semantic change of words over time, these methods require the alignment of word embeddings across different time periods. This process is computationally expensive, prohibitively time consuming and suffering from contextual variability. In this paper, we propose a new and scalable method for encoding words from different time periods into one dense vector space. This can greatly improve performance when it comes to identifying words that have changed in meaning over time. We evaluated our method on dataset from Google Books N-gram. Our method outperformed three other popular methods in terms of the number of words correctly identified to have changed in meaning. Additionally, we provide an intuitive visualization of the semantic evolution of some words extracted by our method"]], "1_corpus": "arxiv"}
 {"tstamp": 1739515935.6951, "task_type": "retrieval", "type": "rightvote", "models": ["", ""], "ip": "", "0_conv_id": "8f076fa44dc149d39166e661ef5624e8", "0_model_name": "Salesforce/SFR-Embedding-2_R", "0_prompt": "海明威都有哪些著作？", "0_output": [["海明威都有哪些著作？", "Gao Xingjian\n\n《高行健戲劇集》 (Collection) 《高行健喜劇六種》 (Collection, 1995, published by Taiwan Dijiao Press ()\n《行路難》 (Xinglunan)\n《喀巴拉山》 (Mountain Kebala)\n《獨白》 (Soliloquy)\nFiction\n《寒夜的星辰》 (\"Constellation in a Cold Night\", 1979)\n《有隻鴿子叫紅唇兒》 (\"Such a Pigeon called Red Lips\", 1984) – a collection of novellas\n《給我老爺買魚竿》 (Buying a Fishing Rod for My Grandfather, 1986–1990) – a short story collection\n《靈山》 (Soul Mountain, 1989)\n《一個人的聖經》 (One Man's Bible, 1999)\nPoetry\nWhile being forced to work as a peasant – a form of 'education' under the Cultural Revolution – in the 1970s, Gao Xingjian produced many plays, short stories, poems and critical pieces that he had to eventually burn to avoid the consequences of his dissident literature being discovered. Of the work he produced subsequently, he published no collections of poetry, being known more widely for his drama, fiction and essays. However, one short poem exists that represents a distinctively modern style akin to his other writings:\n天葬臺\n宰了 / 割了 / 爛搗碎了 / 燃一柱香 / 打一聲呼哨 / 來了 / 就去了 / 來去都乾乾淨淨\nSky Burial\nCut / Scalped / Pounded into pieces / Light an incense / Blow the whistle / Come / Gone / Out and out\n(April 13, 1986, Beijing)\nOther texts\n《巴金在巴黎》 (Ba Jin in Paris, 1979, essay)"]], "0_corpus": "wikipedia", "1_conv_id": "94fb9f0a1bf34a40b0cfe0752478644d", "1_model_name": "voyage-multilingual-2", "1_prompt": "海明威都有哪些著作？", "1_output": [["海明威都有哪些著作？", "American literature\n\nWriting in a lyrical, flowing style that eschews excessive use of the comma and semicolon, recalling William Faulkner and Ernest Hemingway in equal measure, Cormac McCarthy seizes on the literary traditions of several regions of the United States and includes multiple genres. He writes in the Southern Gothic aesthetic in his Faulknerian 1965 debut, The Orchard Keeper, and Suttree (1979); in the Epic Western tradition, with grotesquely drawn characters and symbolic narrative turns reminiscent of Melville, in Blood Meridian (1985), which Harold Bloom styled \"the greatest single book since Faulkner's As I Lay Dying\", calling the character of Judge Holden \"short of Moby Dick, the most monstrous apparition in all of American literature\"; in a much more pastoral tone in his celebrated Border Trilogy (1992–98) of bildungsromans, including All the Pretty Horses (1992), winner of the National Book Award; and in the post-apocalyptic genre in the Pulitzer Prize-winning The Road (2007). His novels are noted for achieving both commercial and critical success, several of his works having been adapted to film.\nDon DeLillo, who rose to literary prominence with the publication of his 1985 novel, White Noise, a work broaching the subjects of death and consumerism and doubling as a piece of comic social criticism, began his writing career in 1971 with Americana. He is listed by Harold Bloom as being among the preeminent contemporary American writers, in the company of such figures as Philip Roth, Cormac McCarthy, and Thomas Pynchon. His 1997 novel Underworld chronicles American life through and immediately after the Cold War and is usually considered his masterpiece. It was also the runner-up in a survey that asked writers to identify the most important work of fiction of the last 25 years. Among his other important novels are Libra (1988), Mao II (1991) and Falling Man (2007)."]], "1_corpus": "wikipedia"}
 {"tstamp": 1739518958.0261, "task_type": "retrieval", "type": "leftvote", "models": ["", ""], "ip": "", "0_conv_id": "d5454f73de4f4a76ad0e81178c380548", "0_model_name": "text-embedding-3-large", "0_prompt": "what is imprecise probability?", "0_output": [["what is imprecise probability?", "Prior probability\n\nStatisticians sometimes use improper priors as uninformative priors.  For example, if they need a prior distribution for the mean and variance of a random variable, they may assume p(m, v) ~ 1/v (for v > 0) which would suggest that any value for the mean is \"equally likely\" and that a value for the positive variance becomes \"less likely\" in inverse proportion to its value.  Many authors (Lindley, 1973; De Groot, 1937; Kass and Wasserman, 1996) warn against the danger of over-interpreting those priors since they are not probability densities. The only relevance they have is found in the corresponding posterior, as long as it is well-defined for all observations. (The Haldane prior is a typical counterexample.)\nBy contrast, likelihood functions do not need to be integrated, and a likelihood function that is uniformly 1 corresponds to the absence of data (all models are equally likely, given no data): Bayes' rule multiplies a prior by the likelihood, and an empty product is just the constant likelihood 1. However, without starting with a prior probability distribution, one does not end up getting a posterior probability distribution, and thus cannot integrate or compute expected values or loss. See  for details.\nExamples\nExamples of improper priors include:\nThe uniform distribution on an infinite interval (i.e., a half-line or the entire real line).\nBeta(0,0), the beta distribution for α=0, β=0 (uniform distribution on log-odds scale).\nThe logarithmic prior on the positive reals (uniform distribution on log scale).\nThese functions, interpreted as uniform distributions, can also be interpreted as the likelihood function in the absence of data, but are not proper priors.\nPrior probability in statistical mechanics"]], "0_corpus": "wikipedia", "1_conv_id": "103aaeeed66e4a309fd8b69edc0c9619", "1_model_name": "sentence-transformers/all-MiniLM-L6-v2", "1_prompt": "what is imprecise probability?", "1_output": [["what is imprecise probability?", "List of idioms of improbability\n\nAs a response to an unlikely proposition, \"when pigs fly\", \"when pigs have wings\", or simply \"pigs might fly\".\n\"When Hell freezes over\" and \"on a cold day in Hell\" are based on the understanding that Hell is eternally an extremely hot place.\nThe \"Twelfth of Never\" will never come to pass. A song of the same name was written by Johnny Mathis.\n\"On Tibb's Eve\" refers to the saint's day of a saint who never existed.\n\"When two Sundays come together\"\n\"If the sky falls, we shall catch larks\" means that it is pointless to worry about things that will never happen.\nEvents that rarely or might never happen\n\"Once in a blue moon\" refers to a rare event.\n\"Don't hold your breath\" implies that if you hold your breath while waiting for a particular thing to happen, you will die first.\nTasks that are difficult or impossible to perform\nTo have \"a snowball's chance in Hell\".\n\"Like getting blood from a stone\", and \"like squeezing water from a stone\".\n\"Like finding a needle in a haystack\"\n\"Like herding cats\"\n\"Squaring a circle\"\nThings that are impossible to find\n\"As rare as hen's teeth\".\n\"As rare as rocking-horse poo\".\nPeople or things that are of no use"]], "1_corpus": "wikipedia"}
+{"tstamp": 1739528620.4971, "task_type": "retrieval", "type": "bothbadvote", "models": ["", ""], "ip": "", "0_conv_id": "6b932b4931354c2a9dcceeef5907e62f", "0_model_name": "BAAI/bge-large-en-v1.5", "0_prompt": "Переклади наступний текст у стилі Вільний для мови українська. Старайся довго не думати, а у відповіді видай тільки перекладену фразу, нічого більше.\n\nТекст:\nLet the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "0_output": [["Переклади наступний текст у стилі Вільний для мови українська. Старайся довго не думати, а у відповіді видай тільки перекладену фразу, нічого більше.\n\nТекст:\nLet the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "Ptolemy's theorem\n\nLet ABCD be a cyclic quadrilateral.\nOn the chord BC, the inscribed angles ∠BAC = ∠BDC, and on AB, ∠ADB = ∠ACB.\nConstruct K on AC such that ∠ABK = ∠CBD; since ∠ABK + ∠CBK = ∠ABC = ∠CBD + ∠ABD, ∠CBK = ∠ABD.\nNow, by common angles △ABK is similar to △DBC, and likewise △ABD is similar to △KBC.\nThus AK/AB = CD/BD, and CK/BC = DA/BD;\nequivalently, AK⋅BD = AB⋅CD, and CK⋅BD = BC⋅DA.\nBy adding two equalities we have AK⋅BD + CK⋅BD = AB⋅CD + BC⋅DA, and factorizing this gives (AK+CK)·BD = AB⋅CD + BC⋅DA.\nBut AK+CK = AC, so AC⋅BD = AB⋅CD + BC⋅DA, Q.E.D.\nThe proof as written is only valid for simple cyclic quadrilaterals. If the quadrilateral is self-crossing then K will be located outside the line segment AC. But in this case, AK−CK = ±AC, giving the expected result.\nProof by trigonometric identities\nLet the inscribed angles subtended by ,  and  be, respectively, ,  and , and the radius of the circle be , then we have , , , ,  and , and the original equality to be proved is transformed to\nfrom which the factor  has disappeared by dividing both sides of the equation by it."]], "0_corpus": "wikipedia", "1_conv_id": "04490ea9d6994da5adfb1a127ffddebb", "1_model_name": "text-embedding-004", "1_prompt": "Переклади наступний текст у стилі Вільний для мови українська. Старайся довго не думати, а у відповіді видай тільки перекладену фразу, нічого більше.\n\nТекст:\nLet the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "1_output": [["Переклади наступний текст у стилі Вільний для мови українська. Старайся довго не думати, а у відповіді видай тільки перекладену фразу, нічого більше.\n\nТекст:\nLet the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "Square pyramid\n\nSquare pyramids have appeared throughout the history of architecture, with examples being Egyptian pyramids, and many other similar buildings. They also occur in chemistry in square pyramidal molecular structures. Square pyramids are often used in the construction of other polyhedra. Many mathematicians in ancient times discovered the formula for the volume of a square pyramid with different approaches.\nSpecial cases\nRight square pyramid\nA square pyramid has five vertices, eight edges, and five faces. One face, called the base of the pyramid, is a square; the four other faces are triangles. Four of the edges make up the square by connecting its four vertices. The other four edges are known as the lateral edges of the pyramid; they meet at the fifth vertex, called the apex. If the pyramid's apex lies on a line erected perpendicularly from the center of the square, it is called a right square pyramid, and the four triangular faces are isosceles triangles. Otherwise, the pyramid has two or more non-isosceles triangular faces and is called an oblique square pyramid.\nThe slant height  of a right square pyramid is defined as the height of one of its isosceles triangles. It can be obtained via the Pythagorean theorem:\nwhere  is the length of the triangle's base, also one of the square's edges, and  is the length of the triangle's legs, which are lateral edges of the pyramid. The height  of a right square pyramid can be similarly obtained, with a substitution of the slant height formula giving:"]], "1_corpus": "wikipedia"}
+{"tstamp": 1739528638.6911, "task_type": "retrieval", "type": "bothbadvote", "models": ["", ""], "ip": "", "0_conv_id": "75250e06c0e145dcb7bce2086f7fbb50", "0_model_name": "embed-english-v3.0", "0_prompt": "Let the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "0_output": [["Let the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "Special right triangle\n\nIn plane geometry, dividing a square along its diagonal results in two isosceles right triangles, each with one right angle (90°,  radians) and two other congruent angles each measuring half of a right angle (45°, or  radians). The sides in this triangle are in the ratio 1 : 1 : , which follows immediately from the Pythagorean theorem.\nOf all right triangles, such 45° - 45° - 90° degree triangles have the smallest ratio of the hypotenuse to the sum of the legs, namely . and the greatest ratio of the altitude from the hypotenuse to the sum of the legs, namely .\nTriangles with these angles are the only possible right triangles that are also isosceles triangles in Euclidean geometry. However, in spherical geometry and hyperbolic geometry, there are infinitely many different shapes of right isosceles triangles.\n30° - 60° - 90° triangle\nThis is a triangle whose three angles are in the ratio 1 : 2 : 3 and respectively measure 30° (), 60° (), and 90° (). The sides are in the ratio 1 :  : 2.\nThe proof of this fact is clear using trigonometry.  The geometric proof is:\nDraw an equilateral triangle ABC with side length 2 and with point D as the midpoint of segment BC.  Draw an altitude line from A to D.  Then ABD is a 30°–60°–90° triangle with hypotenuse of length 2, and base BD of length 1."]], "0_corpus": "wikipedia", "1_conv_id": "06c183a4158e44c68dbfec14fed47bc0", "1_model_name": "intfloat/e5-mistral-7b-instruct", "1_prompt": "Let the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "1_output": [["Let the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "Right triangle\n\nA right triangle or right-angled triangle, sometimes called an orthogonal triangle or rectangular triangle, is a triangle in which two sides are perpendicular\nforming a right angle ( turn or 90 degrees).\nThe side opposite to the right angle is called the hypotenuse (side  in the figure). The sides adjacent to the right angle are called legs (or catheti, singular: cathetus). Side  may be identified as the side adjacent to angle  and opposite (or opposed to) angle  while side  is the side adjacent to angle  and opposite angle"]], "1_corpus": "wikipedia"}

data/retrieval_individual-1b2a6102-80bb-417d-a081-cf780f49c66c.jsonl

@@ -1160,3 +1160,9 @@
 {"tstamp": 1739523363.6352, "task_type": "retrieval", "type": "chat", "model": "intfloat/e5-mistral-7b-instruct", "gen_params": {}, "start": 1739523363.3389, "finish": 1739523363.6352, "ip": "", "conv_id": "ef8ce822ed434f3184bb023e4949d389", "model_name": "intfloat/e5-mistral-7b-instruct", "prompt": "DCS World", "output": [["DCS World", "Digital Combat Simulator\n\nDigital Combat Simulator (DCS) is a combat flight simulation game developed primarily by Eagle Dynamics and The Fighter Collection.\nSeveral labels are used when referring to the DCS line of simulation products: DCS World, Modules, and Campaigns. DCS World is a free-to-play game that includes two free aircraft and two free maps. Modules are downloadable content that expand the game with add-on aircraft, maps, and other content. Campaigns are scripted sets of missions. Modules and campaigns are produced by Eagle Dynamics as well as third-parties."]], "corpus": "wikipedia"}
 {"tstamp": 1739523419.0448, "task_type": "retrieval", "type": "chat", "model": "intfloat/e5-mistral-7b-instruct", "gen_params": {}, "start": 1739523418.854, "finish": 1739523419.0448, "ip": "", "conv_id": "4a9f5115a9ac4c7e8df957163f800637", "model_name": "intfloat/e5-mistral-7b-instruct", "prompt": "LON_SEC_FRAC", "output": [["LON_SEC_FRAC", "Fraction\n\nA fraction (from , \"broken\") represents a part of a whole or, more generally, any number of equal parts. When spoken in everyday English, a fraction describes how many parts of a certain size there are, for example, one-half, eight-fifths, three-quarters. A common, vulgar, or simple fraction (examples:  and ) consists of an integer numerator, displayed above a line (or before a slash like ), and a non-zero integer denominator, displayed below (or after) that line. If these integers are positive, then the numerator represents a number of equal parts, and the denominator indicates how many of those parts make up a unit or a whole. For example, in the fraction , the numerator 3 indicates that the fraction represents 3 equal parts, and the denominator 4 indicates that 4 parts make up a whole. The picture to the right illustrates  of a cake.\nOther uses for fractions are to represent ratios and division. Thus the fraction  can also be used to represent the ratio 3:4 (the ratio of the part to the whole), and the division  (three divided by four)."]], "corpus": "wikipedia"}
 {"tstamp": 1739523477.855, "task_type": "retrieval", "type": "chat", "model": "text-embedding-3-large", "gen_params": {}, "start": 1739523474.9619, "finish": 1739523477.855, "ip": "", "conv_id": "a1aef99bdaac475d8533333b45cae515", "model_name": "text-embedding-3-large", "prompt": "LON_SEC_FRAC", "output": [["LON_SEC_FRAC", "Decimal time\n\nFractional days have been used by astronomers ever since. For instance, the 19th century British astronomer John Herschel gave these examples:\nFractional days are commonly used to express epochs of orbital elements.  The decimal fraction is usually added to the calendar date or Julian day for natural objects, or to the ordinal date for artificial satellites in two-line elements.\nDecimal multiples and fractions of the second\nThe second is the International System of Units (SI) unit of time duration. It is also the standard single-unit time representation in many programming languages, most notably C, and part of UNIX/POSIX standards used by Linux, Mac OS X, etc.; to convert fractional days to fractional seconds, multiply the number by 86400. Fractional seconds are represented as milliseconds (ms), microseconds (μs) or nanoseconds (ns). Absolute times are usually represented relative to 1 January 1970, at midnight UT. Other systems may use a different zero point (like Unix time).\nIn principle, time spans greater than one second may be given in units such as kiloseconds (ks), megaseconds (Ms), gigaseconds (Gs), and so on. Occasionally, these units can be found in technical literature, but traditional units like minutes, hours, days and years are much more common, and are accepted for use with SI."]], "corpus": "wikipedia"}
+{"tstamp": 1739528610.1072, "task_type": "retrieval", "type": "chat", "model": "BAAI/bge-large-en-v1.5", "gen_params": {}, "start": 1739528607.7666, "finish": 1739528610.1072, "ip": "", "conv_id": "6b932b4931354c2a9dcceeef5907e62f", "model_name": "BAAI/bge-large-en-v1.5", "prompt": "Переклади наступний текст у стилі Вільний для мови українська. Старайся довго не думати, а у відповіді видай тільки перекладену фразу, нічого більше.\n\nТекст:\nLet the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "output": [["Переклади наступний текст у стилі Вільний для мови українська. Старайся довго не думати, а у відповіді видай тільки перекладену фразу, нічого б��льше.\n\nТекст:\nLet the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "Ptolemy's theorem\n\nLet ABCD be a cyclic quadrilateral.\nOn the chord BC, the inscribed angles ∠BAC = ∠BDC, and on AB, ∠ADB = ∠ACB.\nConstruct K on AC such that ∠ABK = ∠CBD; since ∠ABK + ∠CBK = ∠ABC = ∠CBD + ∠ABD, ∠CBK = ∠ABD.\nNow, by common angles △ABK is similar to △DBC, and likewise △ABD is similar to △KBC.\nThus AK/AB = CD/BD, and CK/BC = DA/BD;\nequivalently, AK⋅BD = AB⋅CD, and CK⋅BD = BC⋅DA.\nBy adding two equalities we have AK⋅BD + CK⋅BD = AB⋅CD + BC⋅DA, and factorizing this gives (AK+CK)·BD = AB⋅CD + BC⋅DA.\nBut AK+CK = AC, so AC⋅BD = AB⋅CD + BC⋅DA, Q.E.D.\nThe proof as written is only valid for simple cyclic quadrilaterals. If the quadrilateral is self-crossing then K will be located outside the line segment AC. But in this case, AK−CK = ±AC, giving the expected result.\nProof by trigonometric identities\nLet the inscribed angles subtended by ,  and  be, respectively, ,  and , and the radius of the circle be , then we have , , , ,  and , and the original equality to be proved is transformed to\nfrom which the factor  has disappeared by dividing both sides of the equation by it."]], "corpus": "wikipedia"}
+{"tstamp": 1739528610.1072, "task_type": "retrieval", "type": "chat", "model": "text-embedding-004", "gen_params": {}, "start": 1739528607.7666, "finish": 1739528610.1072, "ip": "", "conv_id": "04490ea9d6994da5adfb1a127ffddebb", "model_name": "text-embedding-004", "prompt": "Переклади наступний текст у стилі Вільний для мови українська. Старайся довго не думати, а у відповіді видай тільки перекладену фразу, нічого більше.\n\nТекст:\nLet the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "output": [["Переклади наступний текст у стилі Вільний для мови українська. Старайся довго не думати, а у відповіді видай тільки перекладену фразу, нічого більше.\n\nТекст:\nLet the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "Square pyramid\n\nSquare pyramids have appeared throughout the history of architecture, with examples being Egyptian pyramids, and many other similar buildings. They also occur in chemistry in square pyramidal molecular structures. Square pyramids are often used in the construction of other polyhedra. Many mathematicians in ancient times discovered the formula for the volume of a square pyramid with different approaches.\nSpecial cases\nRight square pyramid\nA square pyramid has five vertices, eight edges, and five faces. One face, called the base of the pyramid, is a square; the four other faces are triangles. Four of the edges make up the square by connecting its four vertices. The other four edges are known as the lateral edges of the pyramid; they meet at the fifth vertex, called the apex. If the pyramid's apex lies on a line erected perpendicularly from the center of the square, it is called a right square pyramid, and the four triangular faces are isosceles triangles. Otherwise, the pyramid has two or more non-isosceles triangular faces and is called an oblique square pyramid.\nThe slant height  of a right square pyramid is defined as the height of one of its isosceles triangles. It can be obtained via the Pythagorean theorem:\nwhere  is the length of the triangle's base, also one of the square's edges, and  is the length of the triangle's legs, which are lateral edges of the pyramid. The height  of a right square pyramid can be similarly obtained, with a substitution of the slant height formula giving:"]], "corpus": "wikipedia"}
+{"tstamp": 1739528633.3514, "task_type": "retrieval", "type": "chat", "model": "embed-english-v3.0", "gen_params": {}, "start": 1739528631.0283, "finish": 1739528633.3514, "ip": "", "conv_id": "75250e06c0e145dcb7bce2086f7fbb50", "model_name": "embed-english-v3.0", "prompt": "Let the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "output": [["Let the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "Special right triangle\n\nIn plane geometry, dividing a square along its diagonal results in two isosceles right triangles, each with one right angle (90°,  radians) and two other congruent angles each measuring half of a right angle (45°, or  radians). The sides in this triangle are in the ratio 1 : 1 : , which follows immediately from the Pythagorean theorem.\nOf all right triangles, such 45° - 45° - 90° degree triangles have the smallest ratio of the hypotenuse to the sum of the legs, namely . and the greatest ratio of the altitude from the hypotenuse to the sum of the legs, namely .\nTriangles with these angles are the only possible right triangles that are also isosceles triangles in Euclidean geometry. However, in spherical geometry and hyperbolic geometry, there are infinitely many different shapes of right isosceles triangles.\n30° - 60° - 90° triangle\nThis is a triangle whose three angles are in the ratio 1 : 2 : 3 and respectively measure 30° (), 60° (), and 90° (). The sides are in the ratio 1 :  : 2.\nThe proof of this fact is clear using trigonometry.  The geometric proof is:\nDraw an equilateral triangle ABC with side length 2 and with point D as the midpoint of segment BC.  Draw an altitude line from A to D.  Then ABD is a 30°–60°–90° triangle with hypotenuse of length 2, and base BD of length 1."]], "corpus": "wikipedia"}
+{"tstamp": 1739528633.3514, "task_type": "retrieval", "type": "chat", "model": "intfloat/e5-mistral-7b-instruct", "gen_params": {}, "start": 1739528631.0283, "finish": 1739528633.3514, "ip": "", "conv_id": "06c183a4158e44c68dbfec14fed47bc0", "model_name": "intfloat/e5-mistral-7b-instruct", "prompt": "Let the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "output": [["Let the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "Right triangle\n\nA right triangle or right-angled triangle, sometimes called an orthogonal triangle or rectangular triangle, is a triangle in which two sides are perpendicular\nforming a right angle ( turn or 90 degrees).\nThe side opposite to the right angle is called the hypotenuse (side  in the figure). The sides adjacent to the right angle are called legs (or catheti, singular: cathetus). Side  may be identified as the side adjacent to angle  and opposite (or opposed to) angle  while side  is the side adjacent to angle  and opposite angle"]], "corpus": "wikipedia"}
+{"tstamp": 1739528646.9547, "task_type": "retrieval", "type": "chat", "model": "mixedbread-ai/mxbai-embed-large-v1", "gen_params": {}, "start": 1739528646.8052, "finish": 1739528646.9547, "ip": "", "conv_id": "6faa0f23ebce4bcaabf9dc731b5611c8", "model_name": "mixedbread-ai/mxbai-embed-large-v1", "prompt": "Let the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "output": [["Let the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "Math Stackexchange\n\nQ: Finding angle in an equilateral triangular pyramid Given an equilateral triangular pyramid (refer the below diagram) $\\Delta ABC$ & $P$ is any point inside the triangle such that ${PA}^{2}={PB}^{2}+{PC}^{2}$, then $\\angle BPC$ is -\n\nI am unable to think of how to do this question\n\nA: Here's a complete solution. (I think)\nDefine an equilateral triangle on the coordinate system as follows:\n$A=(0,{\\sqrt3\\over 2}),B=(0.5,0),C=(-0.5,0),P=(x,y)$\nBy the requirement of $P$$$y^2 +(x+0.5)^2+y^2+(x-0.5)^2=x^2+(y-{\\sqrt3\\over 2})^2$$$$\\implies \\left(y+{\\sqrt3\\over 2}\\right)^2+x^2=1 $$\n$\\therefore$ The locus of $P$ is the circle with center $\\left( 0,-{\\sqrt3\\over 2}\\right)$ and radius $1$. \nLet the center be $O$ So $\\angle BPC=1/2(\\angle BOC)=150^{\\circ}$\n( Knowing the coordinates of $ B;O;C$) Problem SOLVED!!\n"]], "corpus": "stackexchange"}
+{"tstamp": 1739528646.9547, "task_type": "retrieval", "type": "chat", "model": "jinaai/jina-embeddings-v2-base-en", "gen_params": {}, "start": 1739528646.8052, "finish": 1739528646.9547, "ip": "", "conv_id": "6aa8d61afa87483eb716e935515b3fce", "model_name": "jinaai/jina-embeddings-v2-base-en", "prompt": "Let the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "output": [["Let the base of the pyramid be a right-angled triangle ABC with angle C = 90 degrees and angle A = 15 degrees (so angle B = 75 degrees). Let AB be the hypotenuse.", "Stackoverflow Stackexchange\n\nQ: error eslint@5.6.0: The engine \"node\" is incompatible with this module. when trying to create-react-app I am trying to create a React App and I am getting the following error in my console.\nInstalling packages. This might take a couple of minutes.\nInstalling react, react-dom, and react-scripts...\n\nyarn add v1.10.1\ninfo No lockfile found.\n[1/4]   Resolving packages...\nwarning react-scripts > eslint > file-entry-cache > flat-cache > circular-   json@0.3.3: CircularJSON is in maintenance only, flatted is its successor.\n[2/4]   Fetching packages...\nerror eslint@5.6.0: The engine \"node\" is incompatible with this module.    Expected version \"^6.14.0 || ^8.10.0 || >=9.10.0\". Got \"9.5.0\"\nerror Found incompatible module\ninfo Visit https://yarnpkg.com/en/docs/cli/add for documentation about this     command.\n\nAborting installation.\nyarnpkg add --exact react react-dom react-scripts --cwd      /Volumes/server2/Desktop/JavaScript/birdcage has failed.\n\n  Deleting generated file... package.json\n  Deleting birdcage / from /Volumes/server2/Desktop/JavaScript\n  Done.\n\nI have tried brew upgrade node but that didn't seem to work.  Any ideas on how this can be fixed would be greatly appreciated.\n\nA: I had the same problem, simply doing this solved the problem:\n npm install node@latest\n\n"]], "corpus": "stackexchange"}