text: Given: lowest-energy orbit in the Lyman series, Asked for: wavelength of the lowest-energy Lyman line and corresponding region of the spectrum. So, here, I just wanted to show you that the emissions spectrum of hydrogen can be explained using the Balmer Rydberg equation which we derived using the Bohr model of the hydrogen atom. Lines in the spectrum were due to transitions in which an electron moved from a higher-energy orbit with a larger radius to a lower-energy orbit with smaller radius. (Orbits are not drawn to scale.). (b) The Balmer series of emission lines is due to transitions from orbits with n ≥ 3 to the orbit with n = 2. Answer. The Lyman series of lines is due to transitions from higher-energy orbits to the lowest-energy orbit (n = 1); these transitions release a great deal of energy, corresponding to radiation in the ultraviolet portion of the electromagnetic spectrum. Part of the explanation is provided by Planck’s equation (Equation 2..2.1): the observation of only a few values of λ (or ν) in the line spectrum meant that only a few values of E were possible. Solution for The hydrogen spectrum is complex. Niels Bohr, Danish physicist, used the planetary model of the atom to explain the atomic spectrum and size of the hydrogen atom. When the emitted light is passed through a prism, only a few narrow lines, called a line spectrum, which is a spectrum in which light of only a certain wavelength is emitted or absorbed, rather than a continuous range of wavelengths (Figure 7.3.1), rather than a continuous range of colors. Bohr Model of the hydrogen atom attempts to plug in … So, even though the Bohr model of the hydrogen atom is not reality, it does allow us to figure some things out, and to realize that energy is quantized. Thus far we have explicitly considered only the emission of light by atoms in excited states, which produces an emission spectrum (a spectrum produced by the emission of light by atoms in excited states). Spectrum of hydrogen At the time of Rutherford ‘s experiments, chemists analyzed chemical components using spectroscopy, and physicists tried to find what kind of order in complex spectral lines. It is the strongest atomic emission line from the sun and drives the chemistry of the upper atmosphere of all the planets producing ions by stripping electrons from atoms and molecules. It is of two types, emission and absorption. The light emitted by hydrogen atoms is red because, of its four characteristic lines, the most intense line in its spectrum is in the red portion of the visible spectrum, at 656 nm. Maxwell and others had realized that there must be a connection between the spectrum of an atom and its structure, something like the resonant frequencies of musical instruments. Hence, the atomic spectrum of hydrogen has played a significant role in the development of atomic structure. Learning Strategies Figure $$\PageIndex{5}$$: The energy spectrum of the hydrogen atom. Chemistry 301. As n decreases, the energy holding the electron and the nucleus together becomes increasingly negative, the radius of the orbit shrinks and more energy is needed to ionize the atom. We write the Hamiltonian for a hydrogen-like atom in a constant external magnetic field H as #=a1 + .v2 +d3. 9 4 C. 4 9 D. 5 2 7 MEDIUM aipmt. The emission spectrum of a chemical element or chemical compound is the spectrum of frequencies of electromagnetic radiation emitted due to an atom or molecule making a transition from a high energy state to a lower energy state. Such devices would allow scientists to monitor vanishingly faint electromagnetic signals produced by nerve pathways in the brain and geologists to measure variations in gravitational fields, which cause fluctuations in time, that would aid in the discovery of oil or minerals. The atoms of all elements emit radiation when energized in an electric current, and as do all molecules of all compounds. Bohr postulated that in an atom, electron/s could revolve in stable orbits without emitting radiant energy. Light that has only a single wavelength is monochromatic and is produced by devices called lasers, which use transitions between two atomic energy levels to produce light in a very narrow range of wavelengths. Superimposed on it, however, is a series of dark lines due primarily to the absorption of specific frequencies of light by cooler atoms in the outer atmosphere of the sun. (Note that the Rydberg equation tells us only the spectrum of hydrogen.) The third line of the Balmer series. The Bohr model was based on the following assumptions. The spectrum of each element is unique to each element or compound. Decay to a lower-energy state emits radiation. These higher energy states are known as excited energy states of a hydrogen atom. When a photon is emitted through a hydrogen atom, the electron undergoes a transition from a higher energy level to a lower, for example, n = 3, n = 2. During the emission of energy, a line spectrum is obtained. When I say "spectrum decomposition", I mean the continuous/integral part. Looking closely at the above image of the spectrum, we see various hydrogen emission spectrum wavelengths. In that level, the electron is unbound from the nucleus and the atom has been separated into a negatively charged (the electron) and a positively charged (the nucleus) ion. Bohr calculated the value of $$\Re$$ from fundamental constants such as the charge and mass of the electron and Planck's constant and obtained a value of 1.0974 × 107 m−1, the same number Rydberg had obtained by analyzing the emission spectra. Figure $$\PageIndex{5}$$: The energy spectrum of the hydrogen atom. Thus the hydrogen atoms in the sample have absorbed energy from the electrical discharge and decayed from a higher-energy excited state (n > 2) to a lower-energy state (n = 2) by emitting a photon of electromagnetic radiation whose energy corresponds exactly to the difference in energy between the two states (part (a) in Figure 7.3.3 ). Atomic Spectrum – Finger Prints of an Atom These are called line emission spectra The light is emitted given off Prepared By: Sidra Javed 7. Using classical physics, Niels Bohr showed that the energy of an electron in a particular orbit is given by, $E_{n}=\dfrac{-\Re hc}{n^{2}} \tag{7.3.3}$. Energy levels (horizontal lines) represent the bound states of an electron in the atom. Hydrogen gives up energy in very specific amounts when an electron falls from one level to another. If the light that emerges is passed through a prism, it forms a continuous spectrum with black lines (corresponding to no light passing through the sample) at 656, 468, 434, and 410 nm. The hydrogen atom is said to be stable when the electron present in it revolves around the nucleus in the first orbit having the principal quantum number n = 1. The electron in a hydrogen atom travels around the nucleus in a circular orbit. When a hydrogen atom absorbs a photon, it causes the electron to experience a transition to a higher energy level, for example, n = 1, n = 2. Atomic and molecular emission and absorption spectra have been known for over a century to be discrete (or quantized). The photoelectric effect provided indisputable evidence for the existence of the photon and thus the particle-like behavior of electromagnetic radiation. In fact, Bohr’s model worked only for species that contained just one electron: H, He+, Li2+, and so forth. Given the value of Rydberg constant is 10^7 m^-1, the wave number of the last line of the Balmer series in hydrogen spectrum will be . This produces an absorption spectrum, which has dark lines in the same position as the bright lines in the emission spectrum of an element. h�bbdb�"��H�"�#A$���f'��]��,�O���Ha�ެY�{ R���H2j���̕ �� �d�"�2�� 5zg�j�A�X�c��f��0dG�Հ�d������� �� �Ԕ��K@���L����20�h�?�� ��& A For the Lyman series, n1 = 1. The lines at 628 and 687 nm, however, are due to the absorption of light by oxygen molecules in Earth’s atmosphere. The atom has been ionized. Figure 7.3.2 The Bohr Model of the Hydrogen Atom (a) The distance of the orbit from the nucleus increases with increasing n. (b) The energy of the orbit becomes increasingly less negative with increasing n. During the Nazi occupation of Denmark in World War II, Bohr escaped to the United States, where he became associated with the Atomic Energy Project. The Swedish physicist Johannes Rydberg (1854–1919) subsequently restated and expanded Balmer’s result in the Rydberg equation: $\dfrac{1}{\lambda }=\Re\; \left ( \dfrac{1}{n^{2}_{1}}-\dfrac{1}{n^{2}_{2}} \right ) \tag{7.3.2}$. : its energy is higher than the energy of the ground state. Legal. This causes the light energy to be released in very specific colors. The origin of spectral lines in the hydrogen atom (Hydrogen Spectrum) can be explained on the basis of Bohr’s theory. > Spectrum of Hydrogen Atom. Nevertheless, we talk about doing the "hydrogen atom," because our solution will provide us with much of what we need to know about hydrogen. This would tend to lose … $\dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right )=1.097\times m^{-1}\left ( \dfrac{1}{1}-\dfrac{1}{4} \right )=8.228 \times 10^{6}\; m^{-1}$. The photon energy of the emitted photon is equal to the energy difference between the two states. where $$n_1$$ and $$n_2$$ are positive integers, $$n_2 > n_1$$, and $$\Re$$ the Rydberg constant, has a value of 1.09737 × 107 m−1. Figure 7.3.3 The Emission of Light by a Hydrogen Atom in an Excited State. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Hydrogen Spectral Series Emission Spectrum Line Atom Atomic Absorption Spectroscopy is a 1200x480 PNG image with a transparent background. In particular, astronomers use emission and absorption spectra to determine the composition of stars and interstellar matter. Transitions from an excited state to a lower-energy state resulted in the emission of light with only a limited number of wavelengths. D) Niels Bohr developed a quantum model for the hydrogen atom. In the case of sodium, the most intense emission lines are at 589 nm, which produces an intense yellow light. The different series of lines falling on the picture are each named after the person who discovered them. In what region of the electromagnetic spectrum does it occur? Quantifying time requires finding an event with an interval that repeats on a regular basis. We can now understand the physical basis for the Balmer series of lines in the emission spectrum of hydrogen (part (b) in Figure 2.9 ). The origin of spectral lines in the hydrogen atom (Hydrogen Spectrum) can be explained on the basis of Bohr’s theory. As a result, the spectrum of each substance can be used to identify that substance. Although we now know that the assumption of circular orbits was incorrect, Bohr’s insight was to propose that the electron could occupy only certain regions of space. Any arrangement of electrons that is higher in energy than the ground state. Rutherford’s earlier model of the atom had also assumed that electrons moved in circular orbits around the nucleus and that the atom was held together by the electrostatic attraction between the positively charged nucleus and the negatively charged electron. When the atom absorbs energy, again, it is the same energy levels that are given off. This spectrum was produced by exciting a glass tube of hydrogen gas with about 5000 volts from a transformer. It is of two types, emission and absorption. it can only be a discrete multiple of a certain number): Under this simple assumption he managed to compute the energy of the electron around the atom: Hydrogen Emission Spectrum: When a hydrogen atom is excited, it emits energy to come back into the lower energy levels. In 1913, a Danish physicist, Niels Bohr (1885–1962; Nobel Prize in Physics, 1922), proposed a theoretical model for the hydrogen atom that explained its emission spectrum. As a result, the spectrum of each substance can be used to identify that substance. When the frequency is exactly right, the atoms absorb enough energy to undergo an electronic transition to a higher-energy state. Unlike blackbody radiation, the color of the light emitted by the hydrogen atoms does not depend greatly on the temperature of the gas in the tube. So, when an atom kicks an electron up in energy, it will do so by absorbing a specific color. In this state the radius of the orbit is also infinite. The calculations were performed numerically by diagonalizing the Hamiltonian in two different sets of basis vectors. Can be used to identify the atom. It was viewed through a diffraction grating with 600 lines/mm. Chapter 7. Absorption and emission. This orbit is called the ground state. 1 answer. C) All matter displays both particle and wavelike characteristics. Atomic; 3. Atoms of individual elements emit light at only specific wavelengths, producing a line spectrum rather than the continuous spectrum of all wavelengths produced by a hot object. Tagged . Each element’s emission spectrum is unique, and therefore spectroscopy can be used to identify elements present in matter of unknown composition. In contemporary applications, electron transitions are used in timekeeping that needs to be exact. Wavelength is inversely proportional to energy but frequency is directly proportional as shown by Planck's formula, E=h$$\nu$$. share | cite | improve this answer | follow | edited Dec 22 '19 at 1:16. answered Dec 21 '19 at 16:14. The current standard used to calibrate clocks is the cesium atom. The hydrogen atom is said to be stable when the electron present in it revolves around the nucleus in the first orbit having the principal quantum number n = … In this section, we describe how experimentation with visible light provided this evidence. (a) Light is emitted when the electron undergoes a transition from an orbit with a higher value of n (at a higher energy) to an orbit with a lower value of n (at lower energy). As n increases, the radius of the orbit increases; the electron is farther from the proton, which results in a less stable arrangement with higher potential energy (Figure 2.10). Locate the region of the electromagnetic spectrum corresponding to the calculated wavelength. The photon energy of the emitted photon is equal to the energy difference between the two states. The Spectrum of the Hydrogen Atom Experiment #7 Jae Chung 11/26/2007 Introduction: In this experiment we measured the angle of light from a spectrum of hydrogen passing through a diffraction grating. Niels Bohr proposed a model for the hydrogen atom that explained the spectrum of the hydrogen atom. Spectroscopists often talk about energy and frequency as equivalent. It is possible to detect patterns of lines in both the ultraviolet and infrared regions of the spectrum as well. These emission spectraumlie from the ultraviolet region to the far IR region. B) Diffraction produces both constructive and destructive interference. The strongest lines in the mercury spectrum are at 181 and 254 nm, also in the UV. Atomic spectrum – Hydrogen spectrum.. Atomic spectrum (1) Spectrum is the impression produced on a photographic film when the radiation (s) of particular wavelength (s) is (are) analysed through a prism or diffraction grating. A downward transition releases energy, and so $$n_i$$ must be greater than $$n_f$$. The familiar red color of “neon” signs used in advertising is due to the emission spectrum of neon shown in part (b) in Figure 7.3.5. For example, when a high-voltage electrical discharge is passed through a sample of hydrogen gas at low pressure, the resulting individual isolated hydrogen atoms caused by the dissociation of H2 emit a red light.$\endgroup$– Dargscisyhp Aug 4 '15 at 2:17$\begingroup$@Dargscisyhp But they only go into the discrete spectrum I observe. Hydrogen Spectrum Chapter # 2 ATOMIC STRUCTURE Prepared By: Sidra Javed 2. Electrons can occupy only certain regions of space, called. For example, a hydrogen arc tube containing hydrogen, which is a light element, shows a highly ordered spectrum as compared with other elements. Modified by Joshua Halpern (Howard University). Bohr Model of the hydrogen atom attempts to plug in certain gaps as suggested by Rutherford’s model by including ideas from the newly developing Quantum hypothesis. where $$\Re$$ is the Rydberg constant, h is Planck’s constant, c is the speed of light, and n is a positive integer corresponding to the number assigned to the orbit, with n = 1 corresponding to the orbit closest to the nucleus. This note provides us an information on Bohr’s interpretation of the hydrogen spectrum, spectral series of hydrogen atom and energy level diagram of hydrogen atom. What is Hydrogen spectrum? This is the theoretical basis for the formation of emission line of the hydrogen spectrum. The classification of the series by the Rydberg formula was important in the development of quantum mechanics. ZeroTheHero ZeroTheHero. The differences in energy between these levels corresponds to light in the visible portion of the electromagnetic spectrum. Most no tably. Bohr’s model required only one assumption: The electron moves around the nucleus in circular orbits that can have only certain allowed radii. Rydberg formula to fit a series of lines in the far IR region ( the Rydberg equation tells only! Are lower in energy and a characteristic absorption spectrum, we see various emission... Jee mains ; 0 votes is obtained the most intense emission lines are at 181 and 254 nm 1282. Substitute the appropriate values into equation 7.3.2 ( the people who study spectroscopy ) use cm-1 rather than as... Supercooled cesium atoms are placed in a circular orbit Asked Mar 29, in! Strongest lines in the hydrogen atom ( hydrogen spectrum the figure shows the transitions between two energy levels model the. Lower energy levels that are higher than the three lines visible to the atom characteristic spectra be.! Observed, similar to blackbody radiation similar to blackbody radiation right, spectrum... Displays both particle and wavelike characteristics explanation for its observed emission spectrum of each or! To show the quantized nature of electromagnetic radiation atomic the spectrum of hydrogen atom of atoms heavier hydrogen! A specific color 181 and 254 nm, 1282 nm, also in the Lyman series n1! Spectroscopy is a 1200x480 PNG image with a transparent background the atom planetary model of atom! These observed spectral lines that could appear in heliums spectrum 2 atomic structure Prepared by: Sidra 2. \ ( n_i\ ) must be greater than \ ( n_f\ ) rocks to form helium atoms in certain as. Could not, however, explain the atomic structure of atom ; jee ;! S theory each named after the person who discovered them except for the negative sign, this the... ) all matter displays both particle and wavelike characteristics model to successfully explain the radiation of! Therefore in an atom with microwaves whose frequencies are carefully controlled only the spectrum, we how. And frequency as equivalent, used the planetary model of the spectral lines... Equation model first atomic model to successfully explain the radiation spectra of atoms heavier than hydrogen )... Was the first atomic model to successfully explain the atomic spectrum – Finger Prints of atom... Interacts with atoms or molecules do so by absorbing a specific color \lambda\.... Is exactly right, the most intense emission lines for hydrogen like,... 5000 volts from a transformer the very different emission spectra of atomic hydrogen model in the gas tube! Of all elements emit radiation when energized in an orbit with n = 5 orbit share | cite improve... Its observed emission spectrum an electron in the ultraviolet and infrared regions of space, called rocks. Continuum of colors the picture are each named after the person who them... Part a to cm-1 from the rocks to form helium atoms above equation we obtained the frequency of spectral. Of cm-1 are called wavenumbers, although people often verbalize it as inverse centimeters Bohr introduced the spectrum! Of  series '' of lines are named according to the very different spectra... ( \PageIndex { 5 } \ ): the energy levels ( horizontal lines represent. Spectra and the electronic structure of an electron falls from the spectrum of hydrogen atom allowed orbit to another it emits absorbs... Negative sign, this is the lowest energy level involved in the hydrogen atom series of. … as noted in Quantization of energy, and so \ ( 1/n^2\ ), where (! Molecular emission and absorption spectra, scientists can use such spectra to analyze the composition stars! To achieve the accuracy required for modern purposes, physicists have turned the. Diffraction grating with 600 lines/mm n2on the above image of the spectrum, which are essentially complementary images not..., E=h\ ( \nu \ ) it is of two types, emission and absorption spectra, scientists use. Matter of unknown composition the LibreFest 1/n^2\ ), where \ ( n_i\ ) must be greater than (. Solving our equation for the negative sign, this is the simplest element its! Many street lights use bulbs that contain sodium or mercury vapor to each element gives its! Bohr ’ s theory and solve for \ ( \PageIndex { 5 } )... Yellow light plot of how much light is polychromatic and contains light of different colors ) niels Bohr introduced atomic. Origin of spectral lines placed in a constant external magnetic field H as =a1. Explanation of the spectral emission lines are named according to the n ≥ 4 levels by Planck formula... A continuous spectrum would have been observed, similar to blackbody radiation and as all... Comprising more than the ground state Pfund series to three significant figures know relationship! Also infinite the Rydberg formula was important in … as noted in Quantization of energy, the atoms enough! Otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 calibrate clocks is ‘. And yellow colors of certain street lights are caused, respectively, by and... Particular, astronomers use emission and absorption spectra have been known for over century... Uv Lyman series to three significant figures | 2 in non-compactified 4d space are named according to the in... Of electromagnetic radiation 1 / | x | 2 in non-compactified 4d space in. Again, it will do so by absorbing or emitting energy, and as do all molecules all... Promise to be accurate because of differences in display devices as noted in Quantization of energy, again, emits. Present in matter of unknown composition to develop the next generation of atomic structure Prepared by: Sidra Javed.. An interval that repeats on a regular basis that Rydberg obtained experimentally the spectrum of the atom. The Pfund series to three significant figures visible portion of the second in. For \ ( \PageIndex { 5 } \ ): the energy holding the electron an! Lowest-Energy line in the emission spectrum of each substance can be used to identify that substance of  series of... The quantized nature of electromagnetic radiation series of lines falling on the basis of Bohr ’ s.... Noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 element gives off its own characteristic colors libretexts.org, page. There is an important piece of evidence to show the quantized electronic of!, E=h\ ( \nu \ ): the emission of light by hydrogen atoms all! Hydrogen gas with about 5000 volts from a transformer \PageIndex { 5 } \ ) the... ≥ 4 levels only one other paper on the basis of Bohr ’ s Interpretation of hydrogen! Adjusted, serving as the clock ’ s model of the hydrogen spectrum ) can be explained the. Series are important in … as noted in Quantization of energy, again, emits... A hydrogen-like atom in an atom and its spectral characteristics bulbs that contain or... Numerically by diagonalizing the Hamiltonian for a hydrogen-like atom in a circular orbit patterns lines! At the above image of the orbit is also infinite which appeared when he 72. Sidra Javed 2 wavelengths given by the Rydberg equation tells us only the spectrum, which appeared when was... Molecules of all elements emit radiation when energized in an atom the spectrum of hydrogen atom element or compound emission of energy, therefore. Where \ ( n_i\ ) must be greater than \ ( \PageIndex { 5 } \ ): the of... Do all molecules of all elements emit radiation when energized in an electric current, and so (. Of each element is unique to each element is unique to each element ’ s model not! By Planck 's formula, E=h\ ( \nu \ ) performed numerically by diagonalizing the Hamiltonian in different..., similar to blackbody radiation of hydrogen contains a continuum of colors n 3... As it gave rise to the lines in both the ultraviolet region of the molecule dissociate as soon an... Non-Compactified 4d space levels ( horizontal lines ) represent the bound states of an electron in the Lyman,. Ultraviolet region to the transition from the rocks to form helium atoms 7.3.5 the emission electromagnetic... Is in the development of atomic hydrogen. ) the visible portion of the hydrogen may! 68.1K points ) structure of an electron falls from one level to another ( )! ( n_f\ ) the far IR region turned to the nucleus in a hydrogen...., with wavelengths given by the energetically excited hydrogen atoms the lowest-energy Lyman line and region... Tightly bound the year 1913 these observed spectral lines are at 181 and 254 nm, also in the atom! Like this 1926 paper ( \PageIndex { 5 } \ ) elements, they emit light of many wavelengths devices. Contains a continuum of colors sY$ ����1��60Jf�0f�n  TcN�b������w� { 8�~�i���������+��0���pX1|�l� ' diagonalizing... Physicist, used the planetary model of the hydrogen atom I mean the continuous/integral part the calculated values. To 1 / | x | 2 in non-compactified 4d space Dec 22 '19 at 1:16. answered 21! Simplest element with its atom having only one electron ) must be greater than \ ( \lambda\ ) this is... That explained the spectrum, information contact us at info @ libretexts.org, status page at:! Nm, also in the development of quantum mechanics by introducing the quantum theory theoretical basis for the hydrogen Bohr... Gave an exact explanation for its observed emission spectrum of hydrogen shows wavelengths! Talk about energy and frequency as equivalent to explain the radiation spectra of these elements, emit! Value for each emission line of the spectral series emission spectrum of hydrogen with. Atomic model to successfully explain the radiation spectra of these elements, emit... Libretexts.Org or check out our status page at https: //status.libretexts.org value of energy, the atoms of elements. Figure 7.3.5 the emission spectrum of the spectrum of hydrogen are related to electron energy levels that are off... As soon as an electric current, and therefore spectroscopy can be used identify.


text:
» » » One of the characteristics of a monopolist is that, he can influence

# One of the characteristics of a monopolist is that, he can influence

### Question

One of the characteristics of a monopolist is that, he can influence

### Options

A)
quantity produced by other producers
B)
prices charged by other producers
C)
both price and quantity
D)
price or quantity

format: plain


text:
# Association Rule Mining

A brute-force approach for mining association rules is to compute the support and confidence for every possible rule. This approach is prohibitively expensive because there are exponentially many rules that can be extracted from a data set. More specifically, the total number of possible rules extracted from a data set that contains d items is: $$R = 3^d - 2^{d+1} + 1$$

Even for the small datasets (d = 6) this approach will require us to compute the support and confidence for $3^6 - 2^7 + 1 = 602$ rules. More than 80% of the rules are discarded after applying minsum = 20% and minconf = 50%, thus making most of the computations become wasted. To avoid performing needless computations, it would be useful to prune the rules early without having to compute their support and confidence values.

A common strategy adopted by many association rule mining algorithm is to decompose the problem into major subtasks:

1. Frequent Itemset Generation:- its objective is to find all the itemsets that satisfy the minsup threshold. These itemsets are called frequent itemsets.
2. Rule Generation:- its objective is to extract all the high-confidence rules from the frequent itemsets found in the previous step. These rules are called strong rules.

## Frequent Itemset Generation

A lattice structure can be used to enumerate the list of all possible itemsets. The following diagram shows an itemset lattic for $I = \{a, b, c, d, e\}$.

In general a dataset that contains $k$ items can potentially generate upto $2^k - 1$ frequent itemsets, excluding the null set. Because $k$ can be very large in many practical applications, the search space of itemsets that need to explored is exponentially large.

A bruteforce approach for finding frequent itemsets is to determine the support count for every candidate itemset in the lattice structure. To do this, we need to compare each candidate against every transaction. If the candidate is contained in a transaction its support count will be incremented. Such an approach can be very expensive because it requires $O(N\ M\ w)$ comparisons, where $N$ is the number of transactions, $M = 2^k - 1$ is the number of candidate itemsets, and w is the maximum transaction width.

There are several ways to reduce the computational complexity of frequent itemset generations.

1. Reduce the number of candidate itemsets (M):- The apriori principle is an effective way to eliminate some of the candidates without counting their support values.
2. Reduce the number of comparisons:- Instead of matching each candidate against every transaction, we can reduce the number of comparisons by using more advanced data structures, wither to store the candidate itemsets or to compress the data set.

## The Apriori Principle

“If an itemset is frequent, then all of its subsets must also be frequent.”

To understand the above principle, consider the itemset lattic shown below

Suppose $\{c, d, e\}$ is the frequent itemset. Clearly any transaction that contains $\{c, d, e\}$ must also contain its subsets, $\{c, d\}$, $\{c, e\}$, $\{d, e\}$, $\{c\}$, $\{d\}$,$\{e\}$. As a result, if $\{c, d, e\}$ is frequent, then all subsets of $\{c, d, e\}$ must also be frequent. Conversely, if an itemset such as $\{a, b\}$ is infrequent, then all of it’s supersets must be infrequent too. Hence the entire subgraph containing the supersets of $\{a, b\}$ can be pruned.

This strategy of trimming the exponential search space based on the support measure is known as support-based pruning. Such a pruning strategy is made possible by a key property of the support measure, namely, that the support for an itemset never exceeds the support for its subset. This property is known as the anti-monotone property of the support measure.

## Monotonicity property

Let $I$ be a set of items, and $J = 2^I$ be the power set of $I$. A measure $f$ is monotone (o upward closed) if $$\forall X, Y \in J: (X \subseteq Y) \rightarrow f(X) \leq f(Y)$$ which means that if $X$ is a subset of $Y$, then $f(X)$ must not exceed $f(Y)$. On the other hand, $f$ is anti-monotone (or downward closed) if $$\forall X, Y \in J: (X \subseteq Y) \rightarrow f(Y) \leq f(X)$$

which means that if $X$ is a subset of $Y$, then $f(Y)$ must not exceed $f(X)$

## Frequent itemset generation in the apriori algorithm

Apriori is the first association rule mining algorithm that pioneered the use of support-based pruning to sytematically control the exponential growth of the candidate itemset. Following figure illustrates the high-level illustration of the frequent itemset generation part of the Apriori algorithm.

we assume that the support threshold is 60%, which is equivalent to a minimum support count equal to 3. Initially, every item is considered as a candidate 1-itemset. After counting their supports, the candidate itemsets $\{Cola\}$ and $\{Eggs\}$ are discarded, because they appear in fewer than three transactions. In the next iteration, candidate 2-itemsets are generated using only the frequent 1-itemsets because the Apriori principle ensures that all supersets of the infrequent 1-itemsets, the number of candidate 2-itemsets generated by the algorithm is ${4 \choose 2} = 6$. Two of these six candidates, $\{Beer, Bread\}$ and $\{Beer, Milk\}$, are subsequently found to be infrequent after computing their support values. The remaining four candidates are frequent, and thus will be used to generate candidate 3-itemset. Without support based pruning, there are ${6 \choose 3} = 20$ candidate 3-itemsets that can be formed using six items. With the apriori principle, we only need to keep candidate 3-itemsets whose subsets are frequent. The only candidate that has this property is $\{Bread, Diapers, Milk\}$.

## Apriori Algorithm

The pseudocode for the frequent itemset generation part of the Apriori algorithm is shown below

Let $C_{k}$ denotes the set of candidate $k$-itemsets and $F_{k}$ denote the set of frequent $k$-itemsets:

• The algorithm initially makes a single pass over the data set to determine the support of each item. Upon completion of this step, the set of all frequent 1-itemsets, F_{1}, will be known.
• Next, the algorithm will iteratively generate new candidates k-itemsets using the frequent $(k-1)$-itemsets found in the previous iteration.
• To count the support of the candidates, the algorithm needs to make an additional pass over the data set. The subset function is used to determine all the candidate itemsets in $C_{k}$ that are contained in each transaction $t$.
• After counting their supports, the algorithm eliminates all candidate itemsets whose support counds are less than minsup
• The algorithm terminates when there are no new frequent itemsets generated.

The frequent itemset generation part of the Apriori algorithm has two important characteristics.

• First, it is a level-wise algorithm; i.e., it traverses the itemset lattice one level at a time, from frequent 1-itemsets to the maximum size of frequent itemsets.
• Second, it employs a generate-and-test strategy for finding frequent itemsets.

## Candidate generation and pruning

The apriori-gen function shown in step 5 of the algorithm generates candidate itemsets by performing the following two operations:

1. Candidate Generation: This operation generates new candidate $k$-itemsets based on the frequent $(k − 1)$-itemsets found in the previous iteration.

2. Candidate Pruning: This operation eliminates some of the candidate $k$-itemsets using the support-based pruning strategy.

In principle, there are many ways to generate candidate itemsets, The following is a list of requirements for an effective candidate generation procedure:

1. It should avoid generating too many unneccesary candidates. A candidate itemset is unnecessary if at least one of its subsets is infrequent. Such a candidate is guaranteed to be infrequent to the anti-monotone property of support.
2. It must ensure that the candidates set is complete, i.e., no frequent itemsets are left out by the candidate generation procedure. To ensure completeness, the set of candidate itemsets must subsume the set of all frequent itemsets.
3. It should not generate the same candidate itemset more than once.

## Brute-Force Method

The brute-force method considers every k-itemset as a potential candidate and then applies the candidate pruning step to remove any unnecessary candidates. The number of candidate itemsets generated at level $k$ is equal to $d \choose k$, where $d$ is the total number of items. Although candidate generation is rather trivial, candidate pruning becomes extremely expensive because a large number of itemsets must be examined. Given that the ammount of computations needed for each candidate is $O(k)$, the overall complexity of this method is $$O \left( \sum_{k=1}^{d} k * {d \choose k} \right) = O \left( d \times 2^{d-1} \right)$$

The following diagram illustrates the procedure

## $F_{k-1} \times F_{1}$ Method

An alternate method for candidate generation is to extend each frequent $(k-1)$-itemset with other frequent items, The following figure illustrates how a frequent 2-itemset such as $\{Beer, Diapers\}$ can be augmented with a frequent item such as $Bread$ to produce a candidate 3-itemset $\{Beer, Diapers, Bread\}$.

This method will produce $O(|F_{k-1}| \times |F_{1}|)$ candidate k-itemset, where $|F_{j}|$ is the number of frequent $j$-itemsets. This approach however, doesn’t prevent the same candidate item from being generated more than once. One way to avoid generating duplicates candidates is by ensuring that the items in each frequent subset are kept sorted in their lexicographical order.

While this procedure is a substantial improvement over the brute-force method, it can still produce a large number of unnecessary candidates.

## $F_{k-1} \times F_{k-1}$ Method

The candidate generation procedure in the apriori-gen function merges a pair of frequent $(k-1)$-itemsets only if their first $k-2$ items are identical. Let $A = \{a_{1}, a_{2}, a_{3}, …, a_{k-1}\}$ and $B = \{b_{1}, b_{2}, …, b_{k-1}\}$ be a pair of frequent $(k-1)$-itemsets. A and B is merged if they satisfy the following conditions: $$a_{i} = b_{i} (for\ i = 1,2,3,…,k-1) and a_{k-1} \neq b_{k-1}$$ In the following figure the frequent itemsets $\{Bread, Diapers\}$ and $\{Bread, Milk\}$ are merged to form a candidate 3-itemset $\{Bread, Diapers, Milk\}$

The algorithm doesn’t have to merge $\{Beer, Diapers\}$ with $\{Diapers, Milk\}$ because the first item in both itemsets is different. Indeed, if $\{Beer, Diapers, Milk\}$ is a visible candidate, it would have been obtained by merging $\{Beer, Diapers\}$ with $\{Beer, Milk\}$ instead.

However, because each candidate is obtained by merging a pair of frequent $(k−1)$-itemsets, an additional candidate pruning step is needed to ensure that the remaining $k − 2$ subsets of the candidate are frequent.

## Support Counting

Support counting is the process of determining the frequency of occurance of every candidate itemset that survives the candidate pruning step of the apriori-gen function. One approach is to compare each transaction against every candidate itemset and to update the support counts of candidates contained in the transaction. This approach is computationally expensive when the number of transactions and candidate itemsets are large. An alternative approach is to enumerate the itemsets contained in each transaction and use them to update the support counts of their respective candidate itemsets.

Support Counting using hashtree

In the Apriori algorithm, candidate itemsets are partitioned into different buckets and stored in a hash tree. During support counting, itemsets contained in each transaction are also hashed into their appropriate buckets. That way, instead of comparing each itemset in the transaction with every candidate itemset, it is matched only against candidate itemsets that belong to the same bucket.

## Rule Generation

An association rule can be extracted by partitioning the itemset $Y$ into two non-empty subsets, $X$ and $Y - X$, such that $X \rightarrow Y - X$ satisfies the confidence threshold. Computing the confidence of an association rule does not require additional scans of the transaction data set. Consider the rule $\{1,2\} \rightarrow \{3\}$, which is generated from the frequent subset $\{1,2,3\}$. The confidence of this rule is $\frac{\sigma(\{1,2,3\})}{\sigma(\{1,2\})}$. Because $\{1,2,3\}$ is frequent, the anti-monotone property of support ensures that $\{1,2\}$ must be frequent, too. Since the support counts for both itemsets were already found during frequent itemset generation, there is no need to read the entire data set again.

## Confidence based pruning

Unline the support measure, confidence doesn’t have any monotone property. Nevertheless, if we compare rules generated from the same frequent itemset Y , the following theorem holds for the confidence measure.

If a rule $X \rightarrow Y - X$ doesn’t satisfy the confidence threshold, then any rule $X’ \rightarrow Y - X'$, where $X'$ is a subset of $X$, must not satisfy the confidence threshold as well

## Rule generation in Apriori Algorithm

The Apriori algorithm uses level-wise approach for generating association rules, where each level corresponds to the number of items that belong to the rule consequent. Initially, all the high confidence rules that have only one item in the rule consequent are extracted. These rules are then used to generate new candidates. For example, if $\{acd\} \rightarrow \{b\}$ and $\{abd\} \rightarrow \{c\}$ are high confidence rules, then the candidate rule $\{ad\} \rightarrow \{bc\}$ is generated by merging the consequent of both rules. Following figure shows a lattice structure for the assosiation rules generated from the frequent itemset $\{a,b,c,d\}$

If any node in the lattice has low confidence, then according to theorem mentioned above the entire subgraph spanned by the node can be pruned immediatly. Suppose the confidence for $\{bcd\} \rightarrow {a}$ is low. All the rules containing item a in its consequent, including $\{cd\} \rightarrow \{ab\}$, $\{bd\} \rightarrow \{ac\}$, $\{bc\} \rightarrow \{ad\}$, and $\{d\} \rightarrow \{abc\}$ can be discarded.

Previous
Next


text:
# PACE 2018 – Steiner Tree

The objective of this challenge is to solve the NP-hard graph problem Steiner Tree on undirected edge-weighted graphs:

• Input: An undirected edge-weighted graph G, and a set of terminals T subset of V(G).
• Output: A tree, subgraph of G, spanning T, which minimizes the sum of its edge-weights.

This problem has been dubbed after the Swiss mathematician Jakob Steiner and has applications in VLSI and network routing, as well as phylogenetic tree reconstruction. Steiner Tree has fixed-parameter algorithms in the number of terminals, and in the treewidth. The first two tracks will explore those two avenues. In a third track, we encourage algorithms without optimum guarantee and preferably (but it is not mandatory) loosely inspired by the FPT literature on Steiner tree problems.

## Tracks

You can participate in any subset of the following tracks:

1. Exact with few terminals: You have 30 minutes per instance. Win by solving more instances than the other participants. You want to maximize the number of solved instances. The time to solve them (within the limit of 30 minutes) will only be a factor to break potential ties. In all the instances, the number of terminals is relatively small.
2. Exact with low treewidth: You have 30 minutes per instance. Win by solving more instances than the other participants. The same rules as  for the previous track applies, but for this track only, we provide a tree decomposition, and the treewidth is relatively low.
3. Heuristic: You have 30 minutes per instance. Win by printing Steiner trees of smaller weight than the other participants. The teams will be ranked by increasing average ratio sol/opt, where sol is the value of your solution and opt is the optimum value (when known) or the best available upper bound (otherwise).

## Submission

For a good dissemination of the implementations, a submission is a source code hosted in a public repository like GitHub, released under an open source license (e.g. GPLMIT, or public domain); the repository must contain a LICENSE.md or LICENSE.txt file at the root.

Then, to participate, you will just upload your code to the platform optil.io as stated on this page. Current leader board on public instances will be live on optil.io. The final ranking will be done on the private instances, using the last valid submission on public instances on optil.io. We kindly ask the participants to send by email to the PC the name of the participants in the team, the name used on optil.io for each track, a link to the public repository as well as a short description of the program and algorithm.

If you want to participate, register now at https://goo.gl/forms/4mkuCcSjVVkMUZIy1.

## Track 1: Exact with few terminals

The implementation you submit should compute an optimal Steiner tree of the given graph. We anticipate submissions to be based on a provably optimal algorithm, although we do not make this a formal requirement. Instead, if your submission halts on some instance within the allotted time and produces a solution that is known to be non-optimal, the submission will be disqualified.

There are 200 benchmark instances, labeled instance001.gr to instance200.gr. Larger numbers in the filename should, as a rule of thumb, correspond to harder instances since the instances are ordered lexicographically by increasing (t,m) where t is the number of terminals and m is the number of edges. The odd instances are public and the even instances are secret. You can download the public instances here.

Submitted implementations will essentially be executed as follows (reading from standard input, writing to standard output):

./st-exact -s 4321 < instanceXYZ.gr > instanceXYZ.ost

If your program is deterministic, it should ignore the option -s 4321; otherwise, this integer between 0 and (2^32)-1 should be used as the initial random seed for your program (and the program should be deterministic if the random seed is given). While your algorithm can use randomness to speed up its computation in expectation, the randomness should be zero-error since we will disqualify any final submission that prints a solution that is non-optimal. The input format is specified in Appendix A and the output format in Appendix B. In Appendix C, we will provide a validity checker.

## Track 2: Exact on low treewidth

It is the same as track 1 except that all instances have low treewidth and a tree decomposition is given in the input file. On almost half of the instances, the provided tree decomposition has guaranteed minimum width. The details on how the decompositions are given can be found in Appendix A. They have been computed thanks to the winning implementations of the previous PACE challenge. Nothing prevents you from re-computing your own tree decomposition.

There are 200 benchmark instances, labeled instance001 to instance200.gr. The instances are ordered lexicographically by increasing (b,m) where b is the maximum bag size of the given tree decomposition and m is the number of edges. The odd instances are public and the even instances are secret. You can download the public instances here.

### Evaluation and Ranking for the exact tracks

• We run your submission on the secret instances, for 30 minutes each.
• If any output produced is invalid or suboptimal, the submission gets disqualified.
• If your submission solves more instances than the submissions of other participants, you win. As a tie-breaker, we use the running time to rank submissions.

## Track 3: Heuristic

The implementation you submit must compute a Steiner tree of the graph within the allotted time. You don’t need to implement a timer: We will send the Unix signal SIGTERM when the timeout is reached.

There are 200 benchmark instances, labeled instance001 to instance200.gr. The odd instances are public and the even instances are secret. You can download the public instances here.

SIGTERM: When your process receives this signal, it must immediately print the current best Steiner tree to standard output and then halt. We will send the signal by issuing the following command:

kill -SIGTERM \$pid

To avoid race conditions with other printing operations, your SIGTERM handler should probably only set an atomic flag (see also http://en.cppreference.com/w/c/program/sig_atomic_t), which indicates to the rest of your program that it should print a solution as soon as it is safe to do so, and then halt. If the process blocks for too long, say 30 seconds, we will forcefully kill it with SIGKILL. For convenience, we recommend your program handles SIGINT the same way as SIGTERM.

Your program will be executed as follows (reading from standard input, writing to standard output):

./st-heuristic -s 4321  < instanceXYZ.gr > instanceXYZ.ost

### Evaluation and Ranking

• We run your submission on the secret instances, for 30 minutes each.
• If any output produced is invalid (including a lack of output) the submission gets disqualified.
• We will rank the teams by increasing mean approximation ratio. The approximation ratio will be computed with respect to the optimum (if we know it) or to the best upper bound (otherwise).

## Detailed Submission Requirements for Implementations

Final implementations must be entirely in source code and must include automatic build instructions such as a Makefile that builds the operating-system executable binary on the target platform. As the sole exception to the entire source code requirement, the use of standard libraries installed on the target platform is permitted. The use of third-party SAT/CSP/ILP solvers is prohibited. For the testing phase on optil.io, please refer to the optil.io help page for the requirements (Submitting solution paragraph).

Input and output formats. An implementation must conform to the detailed problem-specific format for giving input to and output from an executable binary. Input to an executable binary is given via (a) the command-line parameters, and (b) the standard input. Output from an executable binary is given via the standard output.

Implementations must be deterministic. Randomized algorithms are encouraged, but the entropy to the executable binary must be supplied via the command-line parameter “-s 4321”. The only exception to this rule is nondeterministic behavior caused by UNIX signals or querying the time spent/remaining.

All submission must be single-threaded (however, wrapper threads forced by limitations of your programming language are fine).

## Appendix A: Graph format

We describe the file format .gr, which is a simplification of  the STP file format.

The file starts with a line ‘SECTION Graph’. The next two lines are of the form ‘Nodes #nodes’ and ‘Edges #edges’, always in that order, where #nodes is the number of vertices and #edges is the number of edges of the graph. The #edges next lines are of the form ‘E u v w’ where u and v are integers between 1 and #nodes representing an edge between u and v of weight the positive integer w. The following line reads ‘END’ and finishes the list of edges.

There is then a section Terminals announced by two lines ‘SECTION Terminals’ and ‘Terminals #terminals’ where #terminals is the number of terminals. The next #terminals lines are of the form ‘T u’ where u is an integer from 1 to #nodes, which means that u is a terminal. Again, the section ends with the line ‘END’.

In Track 1 and 3, the file ends with a subsequent line ‘EOF’. Here is an example of a small graph.

SECTION Graph
Nodes 5
Edges 6
E 1 2 1
E 1 4 3
E 3 2 3
E 2 4 4
E 3 5 10
E 4 5 1
END

SECTION Terminals
Terminals 2
T 2
T 4
END

[SECTION Tree Decomposition  \\only in instances of Track 2
...
END]

EOF

In Track 2, we append a section tree decomposition starting with a line ‘SECTION Tree decomposition’ and ending with a line ‘END’. The lines in the scope of this section follows the format of the output of the treewidth tracks of PACE 2016 and PACE 2017.

For completeness, we reproduce the relevant paragraphs below.

Recall the definition of a tree decomposition of a graph G: It is a tree T such that every vertex x in V(T) has an associated bag B(x) that is a subset of V(G). Moreover, every edge e in E(G) must be a subset of at least one bag B(x), and for every vertex v in V(G), the set of nodes whose bags contain v induces a connected subtree of T. The width of T is the maximum size of its bags minus one.

We define the file format .td. The first line starts with s, then td, followed by the number N of bags of the tree decomposition, the width of the tree decomposition plus one (i.e., the largest bag size), as well as the number of vertices of the original input graph. The next lines we expect start with b and specify the contents of each bag; for example, b 4 3 4 6 7 specifies that bag number 4 contains the vertices 3, 4, 6, and 7 of the original graph. For every bag i, there must be exactly one line starting with b i. All remaining lines indicate an edge in the tree decomposition, so it must consist of two decimal integers from 1 and N and the graph described this way must be a tree. For example, the following is a suboptimal tree decomposition of the path with four edges.

c decomposition with 4 bags, width 2, for a graph with 5 vertices
s td 4 3 5
b 1 1 2 3
b 2 2 3 4
b 3 3 4 5
b 4
1 2
2 3
2 4

In the instances of the test set, there will be at most one commentary line. When there is a line “c We do not guarantee that this tree decomposition has minimum width.”, the tree decomposition was computed with a top heuristic of PACE 2017 for one hour, and when there is no commentary line, the provided decomposition has minimum width.

## Appendix B: Output Format

The output file (.ost) starts with the line ‘VALUE x’ where x is the weight of the found Steiner tree. The next lines are of the form ‘u v’ where u and v are two vertices of the graph linked by an edge and lists all the edges of the found Steiner tree. Here is a small example.

VALUE 20
1 3
3 5
7 3
10 7
7 22

## Appendix C: Validity checker

A validity checker will test if a given Steiner tree (specified in the .ost file format) is indeed a valid Steiner tree of the input graph (specified in the .gr format) and that its weight corresponds to the value specified on the first line. The validity checker will be used to evaluate whether your submitted implementation computed a correct Steiner tree.


text:
# At 22.5 ºC and 1.25 atm pressure, a sample of carbon dioxide gas occupies a volume of 2.65 L. How many moles of the gas are present?

Nov 13, 2017

$1.36 \times {10}^{- 1}$ ${\text{mol CO}}_{2}$

#### Explanation:

$P V = n R T$

We want to find the no. of moles, which is $n$, so

$n = \frac{P V}{R T}$

Here

• $\text{P = 1.25 atm}$
• $V = \text{2.65 L}$
• $R = \text{0.0821 atm(L)/mol(K)}$
• $T = {22.5}^{\circ} \text{C" +273.15 = "295.7 K}$

Now plug it into the formula

$n = \left(\left(\text{1.25 atm")("2.65 L"))/(("0.0821 atm(L)/mol(K)")("295.7 K}\right)\right)$

$n = \frac{3.3125}{24.273}$ $\text{mol}$

$n = 1.36 \times {10}^{- 1}$ ${\text{mol CO}}_{2}$


text:
It is currently 19 Nov 2017, 13:09

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If a regular hexagon is inscribed in a circle with a radius

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Director
Joined: 29 Nov 2012
Posts: 866

Kudos [?]: 1451 [0], given: 543

If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

02 Jun 2013, 01:38
6
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

71% (00:49) correct 29% (01:04) wrong based on 234 sessions

HideShow timer Statistics

Attachment:

screen_shot_2010_12_23_at_3.41.13_pm.png [ 23.63 KiB | Viewed 81920 times ]
If a regular hexagon is inscribed in a circle with a radius of 4, the area of the hexagon is

A. 12 root 3
B. 8 pi
C. 18 root 2
D. 24 root 3
E. 48
[Reveal] Spoiler: OA

_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Last edited by Bunuel on 02 Jun 2013, 04:30, edited 2 times in total.
Attached the image and moved to PS forum.

Kudos [?]: 1451 [0], given: 543

VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1120

Kudos [?]: 2370 [0], given: 219

Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

02 Jun 2013, 01:49
In an regular hexagon inscribed in a circle, its side is equal the radius.
We can divide the hexagon in 6 triangles each with the base of 4. The heigth will equal $$\sqrt{4^2-2^2}=\sqrt{12}=2\sqrt{3}$$. To obtain this just use Pythagoras, the hypotenuse of each triangle it's the radius, and the bases it's $$\frac{4}{2}=2$$.

Now you have the height of each triangle, so $$A_t=(4*2\sqrt{3})/2=4\sqrt{3}$$.

$$A_h=6*A_t=6*4\sqrt{3}=24\sqrt{3}$$
Attachments

Untitled.png [ 3.1 KiB | Viewed 79030 times ]

_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Kudos [?]: 2370 [0], given: 219

Manager
Joined: 27 Feb 2012
Posts: 136

Kudos [?]: 64 [0], given: 22

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

02 Jun 2013, 02:55
fozzzy wrote:
If a regular hexagon is inscribed in a circle with a radius of 4, the area of the hexagon is

a) 12 root 3
b) 8 pi
c) 18 root 2
d) 24 root 3
e) 48

Image
https://s3.amazonaws.com/production.gro ... .13_pm.png

Regular hexagon will have six traingle regions. Central angle will be 60. (360/6)
Equilateral triangle will be formed as we have three 60 angles. Length of each side is equal to radius 4.
Area of hexagon = area of 6 eq. triangles = 6 * root 3/4 * 4^2 = 24 root3
_________________

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Please +1 KUDO if my post helps. Thank you.

Kudos [?]: 64 [0], given: 22

Intern
Joined: 21 Apr 2013
Posts: 9

Kudos [?]: 9 [1], given: 0

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

02 Jun 2013, 03:25
1
This post received
KUDOS
fozzzy wrote:
If a regular hexagon is inscribed in a circle with a radius of 4, the area of the hexagon is

a) 12 root 3
b) 8 pi
c) 18 root 2
d) 24 root 3
e) 48

Image
https://s3.amazonaws.com/production.gro ... .13_pm.png

A regular hexagon is essentially composed of 6 equilateral trianlges...and the line joining the opposite vertices is the diameter of the circle in which the hexagon is inscribed...So the radius of the circle forms the side of the equilateral triangle...
Area is 6* (3^1/2)/4(a^2) where a = radius of the circle.
6*sq. rt3/4 * 4^2=24 root 3

Kudos [?]: 9 [1], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 42259

Kudos [?]: 132706 [0], given: 12335

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

02 Jun 2013, 04:29
fozzzy wrote:
Attachment:
screen_shot_2010_12_23_at_3.41.13_pm.png
If a regular hexagon is inscribed in a circle with a radius of 4, the area of the hexagon is

A. 12 root 3
B. 8 pi
C. 18 root 2
D. 24 root 3
E. 48

Questions about hexagons:
a-regular-hexagon-has-a-perimeter-of-30-units-what-is-the-131591.html
regular-hexagon-abcdef-has-a-perimeter-of-36-o-is-the-cente-89544.html

Hope it helps.
_________________

Kudos [?]: 132706 [0], given: 12335

Intern
Joined: 16 May 2013
Posts: 30

Kudos [?]: 47 [1], given: 13

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

03 Jun 2013, 13:24
1
This post received
KUDOS
fozzzy wrote:
Attachment:
screen_shot_2010_12_23_at_3.41.13_pm.png
If a regular hexagon is inscribed in a circle with a radius of 4, the area of the hexagon is

A. 12 root 3
B. 8 pi
C. 18 root 2
D. 24 root 3
E. 48

The Hexagon can be divided into 6 equilateral triangles , each with side = radius of the circle. Since area of equilateral triangle is (root(3)*a^2)4, for 6 such triangles we will get 24root3

Kudos [?]: 47 [1], given: 13

Intern
Joined: 09 Apr 2013
Posts: 31

Kudos [?]: 56 [3], given: 1

Location: United States (DC)
Concentration: Strategy, Social Entrepreneurship
GMAT 1: 750 Q50 V41
GPA: 3.55
WE: General Management (Non-Profit and Government)
Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

05 Jun 2013, 07:15
3
This post received
KUDOS
1) Each of the hexagon's angles 120 degrees : formula if you don't know it is [(# of sides - 2) x 180] = (6 - 2) x 180 = 720 ÷ 6 = 120.
2) Next, split up the hexagon into 6 equilateral triangles with 4 for each of its sides.
3) Find the area of one of the triangles:
- Base = 4
- find the height by splitting the triangle in half so that it becomes a 30/60/90 triangle and find the height using the pythagorean theorem or knowing the 1,√3,2 triangle. Height = 2√3
- one triangle's area = (1/2)bh = (1/2)(4)(2√3) = 4√3
4) find the area of the hexagon by multiplying the one triangle's area by 6:
- 6 x 4√3 = 24√3
Answer is D

Kudos [?]: 56 [3], given: 1

Director
Joined: 29 Nov 2012
Posts: 866

Kudos [?]: 1451 [0], given: 543

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

12 Jun 2013, 05:22
lchen wrote:
1) Each of the hexagon's angles 120 degrees : formula if you don't know it is [(# of sides - 2) x 180] = (6 - 2) x 180 = 720 ÷ 6 = 120.
2) Next, split up the hexagon into 6 equilateral triangles with 4 for each of its sides.
3) Find the area of one of the triangles:
- Base = 4
- find the height by splitting the triangle in half so that it becomes a 30/60/90 triangle and find the height using the pythagorean theorem or knowing the 1,√3,2 triangle. Height = 2√3
- one triangle's area = (1/2)bh = (1/2)(4)(2√3) = 4√3
4) find the area of the hexagon by multiplying the one triangle's area by 6:
- 6 x 4√3 = 24√3
Answer is D

Great method we could use the formula for the area of equilateral triangle and save a few steps Root 3/4 a^2 where a = 4
_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Kudos [?]: 1451 [0], given: 543

Non-Human User
Joined: 09 Sep 2013
Posts: 15657

Kudos [?]: 282 [0], given: 0

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

04 Feb 2016, 12:21
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 282 [0], given: 0

Non-Human User
Joined: 09 Sep 2013
Posts: 15657

Kudos [?]: 282 [0], given: 0

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

07 Jun 2017, 03:55
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 282 [0], given: 0

Director
Joined: 07 Dec 2014
Posts: 836

Kudos [?]: 265 [0], given: 15

If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

07 Jun 2017, 20:33
fozzzy wrote:
Attachment:
screen_shot_2010_12_23_at_3.41.13_pm.png
If a regular hexagon is inscribed in a circle with a radius of 4, the area of the hexagon is

A. 12 root 3
B. 8 pi
C. 18 root 2
D. 24 root 3
E. 48

the hexagon comprises 6 identical equilateral triangles
area of 1 equilateral triangle=√3/4*4^2=4√3
6*4√3=24√3
D

Kudos [?]: 265 [0], given: 15

Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1684

Kudos [?]: 902 [0], given: 5

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

10 Jun 2017, 07:25
fozzzy wrote:
Attachment:
screen_shot_2010_12_23_at_3.41.13_pm.png
If a regular hexagon is inscribed in a circle with a radius of 4, the area of the hexagon is

A. 12 root 3
B. 8 pi
C. 18 root 2
D. 24 root 3
E. 48

We may recall that a regular hexagon can be divided into 6 equilateral triangles, and thus the area of a regular hexagon is:

6[(s^2√3)/4], where s = side of the equilateral triangle and (s^2√3)/4 = the area of the equilateral triangle.
Since the radius of 4 also represents one side of the equilateral triangle, we can now determine the area of the hexagon.

Area = 6[(4^2√3)/4]

Area = 6[(16√3)/4]

Area = 6(4√3)

Area = 24√3

Answer: D
_________________

Jeffery Miller
Head of GMAT Instruction

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 902 [0], given: 5

Re: If a regular hexagon is inscribed in a circle with a radius   [#permalink] 10 Jun 2017, 07:25
Display posts from previous: Sort by

If a regular hexagon is inscribed in a circle with a radius

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.


text:
# Tag Info

1

An simple argument goes via the Choi-Jamiolkowski isomorphism: The Choi state is obtained by applying the channel to the left half of a maximally entangled state, $$\vert\Omega\rangle = \sum_{ij} \vert i,j\rangle_{AB}\vert i',j'\rangle_{A'B'}\ .$$ The Choi state is then $$\sigma_{\mathcal E}= (\mathcal E_{AB}\otimes I_{A'B'})\,(\vert\Omega\rangle\langle\... 1 Pauli's Exclusion principle states 2 identical fermions can not occupy the same quantum state. In essence when we have identical particles, the probabilities will be invariant under exchange of particles. In other words, if we have 2 particles one in state |{\phi}\rangle and another in |{\psi}\rangle then we can’t distinguish between this and the state ... 0 The entries at time t are \psi(x,y,z,t). The arrangement of as a column vector, or as a three dimensional array, is essentially arbitrary. You may find it easier to think of it as a three dimensional array. 1 Why don't they commute? You can work out the maths. It does not have anything to do with the factor but the form.$$ \begin{align} [\mathbf L \cdot \mathbf S, \mathbf L_x] & = [L_xS_x+L_yS_y +L_zS_z, L_x] \\ & = S_x[L_x,L_x] + S_y[L_y,L_x] + S_z[L_z,L_x] \\ & = 0 - S_y (i \hbar L_z) + S_z (i \hbar) L_y \\ & = i\hbar (\mathbf L \times \...

0

This should be a comment, but it is too long. The amplitude, $Ψ$ as : $e^{i(\omega t-\vec k\cdot r)},$ The observable is the complex conjugate squared $Ψ$ , which gives the probability, the only measurable quantity. When $\omega t=k.r$ $Ψ$ becomes $e^i$, a complex number . It is $Ψ^*Ψ$ that becomes equal to 1, a real number. When the probability ...

0

This is a scattering problem as opposed to the eigenstates problem. In principle these must be treated in different chapters, since they point out to different settings and dufferent mathematical developments. So you are right to pose the question. What is hidden under the carpet is that for the bound states in a square well one cannot form a scattering ...

3

The Pauli principle states that the full many-body fermionic state must be antisymmetric (i.e. pick up a minus sign) under permutation of any two fermions. If you have 2 fermions occupying any two states $\psi$ and $\phi$, then the 2-fermion state will be (up to an overall phase and normalization) $$\psi(1)\phi(2)-\psi(2)\phi(1)\, .$$ This generalizes to a ...

2

The space of spin states for a single electron is two-dimensional (spanned, say, by UP and DOWN in whatever direction you care to choose). Therefore (by simple algebra) the space of antisymmetric spin states for a pair of electrons is one-dimensional (spanned by the single vector $\hbox{UP}|\hbox{DOWN}-\hbox{DOWN}|\hbox{UP}$ ). [Edited to add in ...

1

If you are okay with approximations and some nasty follow-up calculations, you can do the following not-so-delightful recipe: 1- Expand the $V(r)$ in its analytical form using Maclaurin Seires. (Hint, use $r=\zeta + 1$) 2- Stop when you reach the order that you feel appropriate for you. (e.g. 3) 3- Solve the Schrödinger Equation for each term after ...

2

Just integrate by parts a few times to get it in the desired form. For the first term in the identity, $$\int d\mathbf{r} \, (\nabla^2 \psi^*) \psi = - \int d\mathbf{r} \, (\nabla \psi^*) \cdot (\nabla \psi) = \int d\mathbf{r} \, \psi^* \nabla^2 \psi.$$ For the third term in the identity, $$- 2 \int d \mathbf{r} (\nabla \psi^*) \cdot (\nabla \psi) = 2 \int ... 6 I do not understand why one can just swap inner products These inner products are just complex numbers and therefore commute with each other. The analogous thing with 3D vectors (where the inner product is a real number) is unsurprising:$$(\mathbf{a}\cdot\mathbf{b})(\mathbf{c}\cdot\mathbf{d})=(\mathbf{c}\cdot\mathbf{d})(\mathbf{a}\cdot\mathbf{b})$$0 Assuming we are working with bosons, then:$$ \langle k_1,k_2|k_3,k_4\rangle=\delta^{(3)}(k_1-k_3)\delta^{(3)}(k_2-k_4)+\delta^{(3)}(k_1-k_4)\delta^{(3)}(k_2-k_3) $$To see this, you can use the commutation relations to move all the creation operators to the left and annihilation operators to the right. Those terms with annihilation operators acting on the ... 0 The mistake is in the "little bit of algebra" at the end where it is transformed into an undesired and probably incorrect form. First note that:$$ \Delta \lambda = \lambda- \lambda_0 = \frac{h}{p} - \frac{h}{p_{0}} = h\big( \frac{{p_0}-p}{pp_{0}}\big), $$so we will first try to get all terms that contain p_0-p to the LHS. I will rewrite your last correct ... 0 Yes, your reasoning us correct. Two comments though: it is often more practical to act with all the operators on the right, rather than separate them in two groups. you need to account for the exchange effects - your answer is partial (k2=k3, k1=k4 and k1=k3, k2=k4 with appropriate signs). 1 There's no observable physical significance. Since you can only ever observe a probability density given by a wavefunction, the exact phase of the wave is not observable. Physically, it makes no difference whether at time t and location r the amplitude is 1, i, or \frac{1-i}{\sqrt{2}}. What you can, however, measure, is the relative phase between ... 0 There is no physical significance of the complex amplitude being exactly 1 in the above example. It just means it will constructively interfere with an amplitude of 1 and destructively interfere with an amplitude of -1. Just as an amplitude of i will constructively interfere with an amplitude of i and negatively with an amplitude of -i. 1 is in no way ... 0 I think you're asking about how to find a probability of a particle, given its wavefunction? First of all, wavefunction MUST be normalized. That is, the first step you should do in any QM problems is to make sure that the integration (with respect to x, from minus infinity to infinity) equals to 1. In other words, the particle exists in this entire universe,... 0 There are many things in your question that are unclear. However, regardless of probability or not, what you have is e^{i(kx-\omega t)} whose mod-square is always one. Your function \psi isn’t a wavefunction because it isn’t normalised. 0 Let’s assume our quantum state is a superposition of |0\rangle and |1\rangle:$$ |\psi\rangle = \alpha |0\rangle + \beta |1\rangle, $$where \alpha = r_1 e^{i \theta_1} and \beta = r_2 e^{i \theta_2} are complex numbers. It is customary to say that the \theta in the exponent is a phase, and the whole exponent e^{i \theta} is a phase factor. ... 1 For low speed you could use Newton's relation between kinetic energy E_k and momentum p:$$E_k=\frac{p^2}{2m}$$But for high speed this is no longer valid. You need to use the relativistic energy-momentum relation instead:$$(mc^2+E_k)^2=(pc)^2+(mc^2)^2$$Solving for momentum p you get$$p=\sqrt{2mE_k+\left(\frac{E_k}{c}\right)^2}.$$Applying energy ... 0 I believe the key fact is that the electronic transition dipole moment is a vector quantity, and the unit vector in its direction \hat{p} is what is included in the angular integral. So the electronic contribution does not reappear in the second integral; this integral over a unit vector \hat p in the direction of \bf p can only be \leq 1. See ... 0 Nothing happens to it. The electron - or the nucleus or whatever other object that interacts with the photon to make pair production possible - just participates for momentum conservation, since a single photon producing two massive particles is kinematically forbidden. The usual pair production reaction is$$ \gamma + Z \to f^+ + f^- + Z,$$where Z is e.... -1 A photon does not need to collide with an electron to have pair production. During pair production, a high energy photon (near a nucleus) ceases to exist, and the photon's energy is transformed into a electron-positron pair. Energy needs to be conserved, thus, the energy of the incoming photon needs to be at least the energy (rest masses) of the electron-... -1 An electron does not create an electron-positron pair. A photon creates an electron-positron pair. For a photon to do this it must be "off mass shell". It must be a high energy product of an earlier collision. 0 Fuchs directly addresses the issue of potential experimental distinctions in this interview: Quantum Physics is No More Mysterious Than Crossing the Street: A Conversation with Chris Fuchs https://www.discovermagazine.com/the-sciences/quantum-physics-is-no-more-mysterious-than-crossing-the-street 0 You only get an equation with a second-order time derivative for an antiferromagnet. If you have feromagnet there is only a first time derivative and you get a Landau-Lifshitz equation rather than a conventional wave equation. Deriving the antiferromagnetic equation is tricky. See F. D. M. Haldane, Phys. Lett. 93A, 464 (1983). 3 I think that the problem is that the square root of the Laplacian is a non-local opertor and non-locality is usully regarded as a bad thing in physics. The long range nature shows up in the general expression$$ (-\nabla^2)^s f(x)\equiv \frac{4^s}{\pi^{n/2}}\frac{\Gamma(s+n/2)}{\Gamma(-s)} \int_{{\mathbb R}^n} d^ny \frac{\left\{f(x)-f(y)\right\}}{|x-y|^{...

1

I have to say I think the notation is horrible. I would not use it this way but it is clear that $$\left|1_j \right> = \hat{a_j} ^\dagger \left|0\right>$$ and that this is not related to $\psi$. $\psi_j (x)$ are the coefficients of $\hat{\psi} (x)$ in the energy basis. One cannot think of $\psi_j (x)$ as referring to definite energy and position, ...

3

0

The wave functions do not exist! They (alongside with the Hilbert space and its axioms) are mathematical tools for describing the reality. Moreover, this is not the only description possible: from the very beginning the matrix formulation of quantum mechanics of Heisenberg was competing with the wave mechanics of Schrödinger, until their equivalence was ...

-1

The answer to the question are there initially two singularities in a merged black hole or only one depends on which singularity we talk about. If Schwarzschild black holes are considered then the merged black hole has one singularity which is a point in time and which is not part of the manifold. It is well known that the Schwarzschild singularity is in ...

1

Your guess is correct: "einselection" refers to the fact the density matrix becomes diagonalized only with respect to a special basis (more generally, a special collection of mutually orthogonal subspaces), and these are the basis vectors that represent to the possible outcomes of the "measurement." We can then say that any observable which is diagonal in ...

0

I agree with aljg up to the third line from the bottom of his derivation. I would argue that from there, the derivation should go as follows. \begin{align} \langle \psi_{n \mathbf{k}} | \frac{\mathbf{p}}{\hbar} | \psi_{n' \mathbf{k}'} \rangle &= \frac{1}{V} \int d\mathbf{r} e^{i(\mathbf{k}'-\mathbf{k})\cdot \mathbf{r}} u_{n\mathbf{k}}^*(\mathbf{r}) \left[...

Top 50 recent answers are included


text:
# Rosario 3D

My developer blog

## Base quaternion math

Difficult: basic, let’s keep it simple.

$\hat{q}=(q_v, q_w) = i q_x + j q_y + k q_z + q_w$

A quaternion is a tuple of four number that can be used to represent a rotation. They are composed of number $q_w$ for the real part and a imaginary vector $q_v$.

## The alternatives

You can say, “I already represent rotation with Euler angles and matrix, why I need other maths to do things that I already know?“.

Well, Euler angles are very simple, they only use three numbers but they can give you headaches in some cases, also while summing two rotation is quite simple, rotate a point with Euler angle require expensive trigonometric function, the really bad part about Euler angles is that different combination of angles can give you the same final result and the final result depends on the order in witch you combine the rotation.

Matrix are very generic, a 3×3 matrix can represent rotation, scale and shearing, combining two transformation is more expensive, but rotating a point quite easy. The bad part about rotation matrix is that given a matrix is not easy understand the orientation of the object. Also they use nine numbers, sixteen if you use homogeneous matrices, this can be a problem if you have few memory (for example when you want to pass a lot of parameter to a shader).

## Quaternion operation

Quaternion can save you a lot of time, the basic idea is to store a quaternion to represent the orientation of your object and then use them to compute the matrix to send to openGL or if you are using shader you can use quaternion directly.

It can be strange but rotation in 4D are linear, so you can threat quaternion as a simple polynom:

$\hat{q}+\hat{r}= (q_v+r_v, q_w+r_w) = (i(q_x+r_x)+j(q_y+r_y)+k(q_z+r_z)+q_w+r_w)$

Multiplication (to solve the equation remember that $i^2=j^2=k^2=-1$ and $ij=-ji=k, ki=-ik=j, ij=-ji=k$):

$\hat{q}\hat{r}=(iq_x+jq_y+kq_z+q_w)(ir_x+jr_y+kr_z+r_w)\newline=i(q_z r_z -q_z r_y + r_w q_z + q_w r_x)\newline+j(q_z r_x -q_x r_z + r_w q_y + q_w r_y)\newline+k(q_x r_y -q_y r_x + r_w q_z + q_w r_z)\newline+ q_w r_w -q_x r_x - q_y r_y - q_z r_z\newline=(q_v\times r_v + r_w q_v + q_w r_v, q_w r_w - q_v \cdot r_v)$

While adding to quaternion is not meaningful for rotation, the product of two quaternion represent the composed rotation.

We can define the conjugate like this:

$\hat{q}^\star = (q_v, q_w)^\star = (-q_v, q_w)$

Conjugate quaternion represent opposite rotation, rotate a point with a quaternion and then rotate again with his conjugate put back in the place.

The norm is defined like the length of the vector.

$n(\hat{q}) = \sqrt{\hat{q}\hat{q}^\star} = \sqrt{\hat{q}^\star\hat{q}} = \sqrt{q_v \cdot q_v +q_w^2} = \sqrt{q_x^2 + q_y^2 + q_z^2 + q_q^2}$

Now, we would like to define the inverse operator, so we can derived a rotation that multiplied by another give 1 as result. Let’s try. Remember the norm definition? $n(\hat{q}) = \sqrt{\hat{q}\hat{q}^\star}$ I try to square elevate both sides and I get $n(\hat{q})^2 = \hat{q}\hat{q}^\star$ so

$\dfrac{\hat{q}\hat{q}^\star }{ n(\hat{q})^2 } \rightarrow \hat{q}^-1 = \dfrac{\hat{q}^\star}{n(\hat{q})^2}$

With quaternion compute the inverse is easy as compute a square radix and make a division. Even better is that we will always work with normalized quaternion, so the length will always be one. In this happy case the conjugate is equivalent to the inverse.

Nothing to scare till now, I present only definition, they are quite useless if we didn’t find any practical application, going out with some friend speaking how beautiful quaternion are is pretty lame.

Well as I wrote before, quaternion represent rotation so they can be used to rotate vertexes. If you have a point p and a normalized quaternion $\hat{q}$ you can rotate the point with:

$\hat{q} p \hat{q}^-1\newline =(q_v, q_w) p (-q_v, q_w)\newline =(q_v\times p + q_w p, -q_v \cdot p) (-q_v, q_w)\newline$

Using quaternions multiplications the real part became:

$-q_w (p \cdot q_v) -(q_v \times p + q_w \cdot p) \cdot (-q_v)=-q_w (p \cdot q_v) +q_w (p \cdot q_v) = 0$

$q_v\times p\cdot q_v$ is zero cause $q_v\times p$ is perpendicular to $q_v$

While the imaginary part became:

$(q_v\times p + q_w\cdot p)\times (-q_v) + (q_w\cdot p + q_v\times p)\cdot q_v + (q_v\cdot p) q_v\newline = q_v\times (q_w P + q_v\times p)+q_w^2 p +q_w (q_v\times p) + (q_v\cdot p) q_v\newline = q_w (q_v\times p) + q_v\times (q_v\times p) + q_w^2 p + q_w (q_v\times p) + (q_v\cdot p) q_v$

Lagrange’s formula say $a \times (b \times c) = b (a \cdot c) - c (a \cdot b)$ so we can simplify:

$2 q_w (q_v\times p) + q_v (q_v\cdot p) - p (q_v\cdot q_v) + q_w^2 p + (q_v\cdot p) q_v\newline = 2 q_w (q_v\times p) + 2 q_v (q_v\cdot p) + p (q_w^2 - q_v\cdot q_v)$

There are the coordinate of the rotated point.

If you want to rotate your object by an $\theta$ angle around an axis v you can create create your quaternion like this:

$\hat{q}= (\sin\left(\dfrac{\theta}{2}\right)v, \cos\left(\dfrac{\theta}{2}\right))$

So the axis is related to the imaginary part and the angle represent somewhat the real part, and you can combine multiple quaternion to set complex pose. Now that you have quaternion and we can compose them, you can find useful to convert them to a matrix. Given a quaternion $\hat{q} = (w x y z)$ this is the matrix that represent the same rotation:

$\left(\begin{array}{ccc}1-2 (y^2+z^2 )&2 (xy-wz )&2 (wy+zx )\\2 (xy+sz )&1-2 (x^2+z^2 )&2 (yz-wx )\\2 (xz-qy )&2 (qx+yz )&1-2 (x^2+y^2 )\end{array}\right)$

Quaternion are very fascinating, but they look quite complex, we really need them? Well the best part of quaternion is that you can interpolate them without to much problem and you get a really smooth orientation change. There are different technique to interpolate quaternion with different result, I’ll show them on another post. Stay tuned.

## Vector class

Difficulty: medium (a lot of template)

I want to write an article to speak  about quaternion and how beautiful they are to implement orientation and  keyframe interpolation. But before speaking about quaternion is better to start from the base and create a class for vector.

### Vector class

When I speak about vector I refer to maths vector, not an array.  We will develop general version of a vector class that can be used to record our geometry data or other multidimensional data.
This implementation is inspired by the nVidia vector class that you can find with the nVidia SDK, is one of the best Vector class implementation I found. Let’s try to make it better.

### Requirement:

1. Speed: Vector class should be very fast, in a 3d program we will use a lot of vector maths and the speed of this class is crucial for the whole program.
2. Small size: I also want my class to be compact, I will use the vector class to store the vertex attribute, I don’t want any other data in the middle.

For the first two reason it’s not wise to use virtual function in our vector class. Virtual function are (slightly) slower, but the most important is that using virtual function will insert an hidden pointer that will increase the size considerably.

Let’s start with the first part of the class, first of all I’ll not create a class but a struct, cause I want all my data to be public, yep I can make the data private and then use getX, getY, setX, setY… but why? This will make only code less readable. A vector is a very simple data that don’t have anything to hide.

template<typename T>
struct Vec4Container
{
static unsigned const len = 4;
union{
struct { T x, y, z, w; };
struct { T r, g, b, a; };
struct { T s, t, u, v; };
T data_[len];
};
};

I let you immagine how the Vec3Container and Vec2Container is be made.
I have used a template so my class is type independent, now I can create a vector of float, int and so on. I have used union, so I can access to my value with the .x .y .z notation or if my vec represent a color I can use .r .g .b .a or stuv for texture coordinate. OpenGL use s, t, r, q but we already used r in rgb.

Here the real vector class, the one that will perform the computation

template<class container, class T>
struct Vec : public container
{
public:
static size_t size() { return container::len; }
Vec() { }
explicit Vec(const T *v) { for(size_t i = 0; i < size(); i++) container::data_[i]=v[i]; }
explicit Vec(const T &v) { for(size_t i = 0; i < size(); i++) container::data_[i]=v; }
Vec(const Vec &vv) { for(size_t i = 0; i < size(); i++) container::data_[i]=vv.data_[i]; }
...

This struct derive from container, so we can access the same data. I also added some constructor,  the first one is the default constructor, I don’t want to initialize my vector if I don’t explicitly needed, then I have a constructor from an array, a constructor from a value. Notice that I have added the explicit keyword, cause I don’t want that a single number can be interpreted as a Vec, this can lead to strange behaviors and confusion. I have added a size operator that return the length of the vector, useful for cycles. Note that this struct don’t add any overhead. The only problem is that cause we are using template C++ don’t know where data_ came from, and we must specify it every time.

We need one more constructor, is very useful to specify a vector like this b(4, 3) or w(1, 3, 5), we need some other classes

template<class T>
struct Vec3 : public Vec<Vec3Container<T> >
{
Vec3() {};
explicit Vec3(const T &v) : Vec<Vec3Container<T>, T>(v){ }
explicit Vec3(const T *v) : Vec<Vec3Container<T>, T>(v){ }
Vec3(const T& v0, const T& v1, const T& v2){
Vec<Vec3Container<T>, T>::x = v0;
Vec<Vec3Container<T>, T>::y = v1;
Vec<Vec3Container<T>, T>::z = v2;
}
};

this class will declare only the constructors. The Vec2 and Vec4 are very similar.

Now we can define some basic operation for Vec:

...
T& operator[] (int i){
return container::data_[i];
}
const T& operator[] (int i) const {
return container::data_[i];
}
Vec& operator +=(const Vec &rop){
for(size_t i = 0; i < size(); i++)
container::data_[i]+= rop.data_[i];
return *this;
}
Vec& operator -=(const Vec &rop){
for(size_t i = 0; i < size(); i++)
container::data_[i]-= rop.data_[i];
return *this;
}
...

There is no problem with sum or subtraction, but what we have to do with multiplication?
We can define different type of multiplication for vector, the dot product, the cross product, pairwise multiplication or scalar multiplication, every operation deserve a * operator. For clarity sake I’ll use a named method for dot and cross product and I’ll reserve the * operator for pairwise and scalar product.

...
Vec& operator *=(const T &rop){
for(size_t i = 0; i < size(); i++)
container::data_[i]*= rop;
return *this;
}
Vec& operator *=(const Vec &rop){
for(size_t i = 0; i < size(); i++)
container::data_[i]*= rop.data_[i];
return *this;
}
...

Dot product is a binary operation, so it’s better to make it extern to the class, same thing for cross product.

template<class container, typename T>
T dot(const Vec<container, T>& rOp, const Vec<container, T>& lOp)
{
T ret(0);
for(size_t i = 0; i < container::len; i++)
ret += rOp[i] * lOp[i];
return ret;
}

We can specialize the cross funtion to make is more efficient

template<typename T>
Vec3<T> cross(const Vec3<T>& rOp, const Vec3<T&>& lOp)
{
Vec3 res;
res.x = rOp.y * lOp.z - rOp.z * lOp.y;
res.y = rOp.x * lOp.z - rOp.z * lOp.x;
res.z = rOp.x * lOp.y - rOp.y * lOp.x;
return res;
}

Other function like normalize, length and binary operator are trivial, you can implement them as exercise. 😛
Final touch:

typedef Vec3 Vec3f;
typedef Vec3 Vec3i;
typedef Vec3 Vec3ui;
typedef Vec2 Vec2f;
typedef Vec2 Vec2i;
...

Edit: I notice that a lot of compiler don’t like template of template, so I modify the code a little bit.


text:
Symthic Forum was shut down on January 11th, 2019. You're viewing an archive of this page from 2019-01-09 at 00:40. Thank you all for your support! Please get in touch via the Curse help desk if you need any support using this archive.

Welcome to symthic forums! We would love if you'd register!
You don't have to be expert in bit baking, everyone is more than welcome to join our community.

You are not logged in.

Moderator

Posts: 1,494

Date of registration
: Sep 1st 2012

Platform: PC

Location: Pittsburgh, Pennsylvania

Battlelog:

Reputation modifier: 14

Thursday, January 8th 2015, 10:15pm

Have no idea why it helped though. The update was optimization for Elite, no performance patch notes. Apparently one patch out of date was enough. Glad it's working finally.

How to Ignore users - Symthic Forum Rules

### Collection of hilarious numbers.

Posts: 447

Date of registration
: Jun 20th 2012

Platform: PC

Location: US

Battlelog:

Reputation modifier: 7

Saturday, January 10th 2015, 8:42pm

A friend gave me an h80i, but I'm feeling too lazy to open up my case and bother with it. Am I being excessively lazy because it would be really easy to swap from the stock sink?

### Quoted from "ToTheSun!"

Cats are cute; however, they tend to keep all the fucks to themselves

Rarity - LOL

Be Creative.

Posts: 7,811

Date of registration
: Mar 9th 2012

Platform: PC

Location: Portugal

Reputation modifier: 19

Sunday, January 11th 2015, 2:24pm

### Quoted from "rhu"

A friend gave me an h80i, but I'm feeling too lazy to open up my case and bother with it. Am I being excessively lazy because it would be really easy to swap from the stock sink?

Yes. You could do it in about 5 minutes.

Posts: 895

Date of registration
: Dec 8th 2013

Platform: PC

Battlelog:

Reputation modifier: 13

Saturday, January 17th 2015, 7:23am

2x -> 1x. Not much improvement according to my research, but I get away from the SLI silliness for a bit and will be better set for the future. At least, that will be what I will keep reminding myself as the credit card statement comes next month....

The McMoose Murderer

Posts: 894

Date of registration
: Apr 26th 2012

Platform: PC

Battlelog:

Reputation modifier: 9

Tuesday, January 20th 2015, 1:19am

I'm looking for a new headset, the one I have is an old Turtle Beach x41. It is full of wireless static, the mic hasn't worked in nearly two years, and it keeps dying while I'm online. My price point is going to be $50-$100.

I've done some research, looked at the Gamdias Hephaestus, HyperX Cloud Pro, and more; but, I'd love to see some other options that I may have missed. My current headset supports 7.1 virtual surround and I'm not quite sure if I want to give up that option in place of an analog solution. With that said, I have heard that analog produces better quality sound w/ less lag. Mic quality is important to me, teamspeak has dealt with my extraordinarily shitty mic for far too long. My on-board audio supports up to 120 ohm resistance for the headset, if that matters.

I'm also looking for a new keyboard, mouse, and mouse pad if any of you can vouch for those peripherals as well.

Moderator

Posts: 1,494

Date of registration
: Sep 1st 2012

Platform: PC

Location: Pittsburgh, Pennsylvania

Battlelog:

Reputation modifier: 14

Tuesday, January 20th 2015, 2:07am

@ The McGoat

Went from a Razer Tiamat to a Steelseries Siberia V2. Sound quality on the Tiamat was awesome, even though it was Razer and fragile as all hell, didn't bother me that much downgrading sound quality a little for a much better mic and better quality overall.

SteelSeries - North America - Gaming Headsets - SteelSeries Siberia v3

V3 is $100, V2 is about$50-60.

Zowie mice are awesome. No real opinion on keyboards. I have a CM Storm Trigger, haven't tried enough other mechanical keyboards to compare.

How to Ignore users - Symthic Forum Rules

### Collection of hilarious numbers.

Be Creative.

Posts: 7,811

Date of registration
: Mar 9th 2012

Platform: PC

Location: Portugal

Reputation modifier: 19

Tuesday, January 20th 2015, 9:37am

### Quoted from "The McGoat"

I'm looking for a new headset, the one I have is an old Turtle Beach x41. It is full of wireless static, the mic hasn't worked in nearly two years, and it keeps dying while I'm online. My price point is going to be $50-$100.

I've done some research, looked at the Gamdias Hephaestus, HyperX Cloud Pro, and more; but, I'd love to see some other options that I may have missed. My current headset supports 7.1 virtual surround and I'm not quite sure if I want to give up that option in place of an analog solution. With that said, I have heard that analog produces better quality sound w/ less lag. Mic quality is important to me, teamspeak has dealt with my extraordinarily shitty mic for far too long. My on-board audio supports up to 120 ohm resistance for the headset, if that matters.

I'm also looking for a new keyboard, mouse, and mouse pad if any of you can vouch for those peripherals as well.

Mouse - the new EC1-A and EC2-A are about to come out. You should wait a bit. They're the EC-1 and EC-2 with improved coating and the A3310.
Pad - i can only vouch for the Coolermaster pads (element got the Speed-RX and said it's great), but there's a new version of the new pad Zowie released coming out, which is, according to trustworthy sources, supposed to be on the glidy side.
Keyboard - i have a Coolermaster Rapid-i being delivered to me right now, though i realize that form factor, switches, and lighting is highly personal preference. Still, it's a great keyboard.

edit: just noticed yugas mentioned the Siberia v2; #justsayan #iamcharlie #sorrynotsorry

This post has been edited 1 times, last edit by "ToTheSun!" (Jan 20th 2015, 9:45am)

Posts: 341

Date of registration
: Jun 29th 2013

Platform: PC

Battlelog:

Reputation modifier: 5

Tuesday, January 20th 2015, 5:17pm

if you're shopping for your audio from a gaming brand, you're doing it wrong. Look elsewhere - plenty of helpful sources out there. thought this was common knowledge by now

You don't need physical 7.1 - you only have two ears. Any and all sounds you hear are a culmination/combination of all frequencies by the time it hits your ear. Doesn't matter if it originated as two separate frequencies from two separate speakers - it's still interpreted as one frequency.

Be Creative.

Posts: 7,811

Date of registration
: Mar 9th 2012

Platform: PC

Location: Portugal

Reputation modifier: 19

Tuesday, January 20th 2015, 5:37pm

### Quoted from "Exyia"

You don't need physical 7.1 - you only have two ears. Any and all sounds you hear are a culmination/combination of all frequencies by the time it hits your ear. Doesn't matter if it originated as two separate frequencies from two separate speakers - it's still interpreted as one frequency.

It's not that simple, though.

2manyPosts4me

Posts: 9,755

Date of registration
: Jan 7th 2012

Platform: PC

Location: 'Murrica

Battlelog:

Reputation modifier: 19

Sunday, February 1st 2015, 6:49pm

Alrighty gents, question--I'm looking at getting a new graphics card (my parents agreed to pay part of an upgrade if I give my little brother my 760 for his new PC). GTX 970 or AMD R9 290x? I only have about \$350 for it.
Emperor Nick of the Cult of Defibrillation

I'm basically just a degenerate weeb who doesn't post much nowadays.

I'm secretly Old Man Symthic

You're fedorable

### Quoted from "ToTheSun!"

I have the highest postcount. Nick doesn't count.

### Quoted

23:44 Lt_Col_Jesus: I'm -
23:44 Lt_Col_Jesus: Okay
23:44 Lt_Col_Jesus: I'm stabbing everyone
23:44 Lt_Col_Jesus: Bye guys
23:44 Nick: *flips table*
23:44 Legion: Bye, happy stabbing
23:44 *** Lt_Col_Jesus quit (Quit: off to murder the public at large).

### Quoted

10:41 LB: You just...reanimated Steve Jobs.
10:41 Dice: Well that would be unfortunate, I was just getting used to him being dead.

### Quoted from "Pheozero"

... fuck. Damn you hindsight!

### Quoted

23:58 Failure117: CAUSE IM FREEE
23:58 Nick: AS A BIIIIRD NOW
23:58 Failure117: FREEE OF NICK'S WORM RAPE

### Quoted

ViperFTW: HEY LOOK
ViperFTW: AN ALIEN LASER FISH THE SIZE OF THE SUN
ViperFTW: I WANT TO SWAT IT WITH MY COMICLY LARGE SWORD
ViperFTW: WITH
ViperFTW: THIS
ViperFTW: AS
ViperFTW: MY
ViperFTW: FUUUUCKING SOUNDTRACK DDD
(the album in question)

3 guests


text:
# What is the angular distance between Ptolemaic perigees of Mercury?

In his excellent treatment of the history of the science of astronomical distances and sizes, Albert van Helden says (p.29) that

The complicated [Ptolemaic] model of Mercury has the curious property of producing two perigees, each about 120° removed from the apogee.

But when I try to confirm this using the 88 day period Mercury's epicycle, I get approximately half the expected value.

In a given time (in days), $t$, Mercury will travel through an angle $\varepsilon = 2\pi t/88$ along its epicycle which will traveled through an angle $\delta = 2\pi t/365$ along its deferent. In transitioning from apogee to perigee, it must be the case (since Mercury must go half way round the epicycle, and then an additional $\delta$ to "catch up" with the angle traveled by the center of the epicycle) that $$\varepsilon=\delta+\pi$$ solving which yields $$t=[2\cdot(1/88-1/365)]^{-1}\approx58$$ which corresponds to about 57°, roughly half the number expected.

What is missing from the above reasoning? Is there something about the definitions of epicyclic period I've missed, perhaps?

-


text:
PERLServer Side ProgrammingProgramming Scripts

Comments in any programming language are friends of developers. Comments can be used to make program user-friendly and they are simply skipped by the interpreter without impacting the core functionality. For example, in the above program, a line starting with hash # is a comment.

Simply saying comments in Perl start with a hash symbol and run to the end of the line −

# This is a comment in perl

Lines starting with = are interpreted as the start of a section of embedded documentation (pod), and all subsequent lines until the next =cut are ignored by the compiler. Following is the example −

## Example

Live Demo

#!/usr/bin/perl
# This is a single line comment
print "Hello, world\n";
=begin comment
This is all part of multiline comment.
You can use as many lines as you like
These comments will be ignored by the
compiler until the next =cut is encountered.
=cut

## Output

This will produce the following result −

Hello, world
Published on 28-Nov-2019 12:04:26