text: People should learn about life from their friends, not from textbooks. Abstract. The name of the ion, s2-, is: sulfurous ion sulfide ion sulfur ion sulfate ion, How will the behavior of the gases in prison chamber change if more pressure is applied to the piston? This description is more consistent with the experimental structure. We will use these “thinner” representations whenever the true view is too crowded to easily visualize. Figure $$\PageIndex{10}$$: This alternate way of drawing the trigonal planar sp2 hybrid orbitals is sometimes used in more crowded figures. Sulfate (ion) Sulfate radical anion. There are usually two methods to prepare sulfates. For example, an atom surrounded by three regions of electron density is sp2 hybridized, and the three sp2 orbitals are arranged in a trigonal planar fashion. You can refuse to use cookies by setting the necessary parameters in your browser. Do not use hybridisation to describe coordination compounds; it is not helpful. Sulfate (7CI,8CI,9CI) SO4(2-) Sulfate-35S. Along the axis itself, there is a node, that is, a plane with no probability of finding an electron. The valence orbitals in an oxygen atom in a water molecule differ; they consist of four equivalent hybrid orbitals that point approximately toward the corners of a tetrahedron (Figure $$\PageIndex{6}$$). However, if you have to describe the ion, you can use the phrase the “ like a polar molecule” because I3- is soluble in water. Sulphate Ion. What is the hybridization of the sulfur atom in the sulfate ion, $${\text{SO}}_{4}{}^{\text{2−}}?$$ Solution The Lewis structure of sulfate shows there are four regions of electron density. Figure $$\PageIndex{16}$$: (a) In the ethane molecule, C2H6, each carbon has four sp3 orbitals. The carbon atom is surrounded by three regions of electron density, positioned in a trigonal planar arrangement. Video $$\PageIndex{2}$$: An overview of electron involvment in bonding. As an example, let us consider the water molecule, in which we have one oxygen atom bonding to two hydrogen atoms. b) molecular geometry. In addition, the electrons begin to repel each other, as do the nuclei. a. sp b. sp2 c. sp3 d. dsp3 e. it is not hybridised ps. because neutrons have no charge, i wonât mention them. For example, the nitrogen atom in ammonia is surrounded by three bonding pairs and a lone pair of electrons directed to the four corners of a tetrahedron. Yet, it has proved feasible to make HIV resistant to dextran sulfate due to the accumulation of a number of amino acid substitutions in the viral envelope gp120 glycoprotein (including the mutations K269E, Q278H, N293D in the V3 loop) [8]. In a σ bond, there is a greater degree of orbital overlap than in a π bond. Check Your Learning What is the hybridization of the selenium atom in SeF 4? In sp³ hybridization, one s orbital and three p orbitals hybridize to form four sp³ orbitals, each consisting of 25% s character and 75% p character. This is not consistent with experimental evidence.1. If this were the case, the bond angle would be 90°, as shown in Figure $$\PageIndex{5}$$, because p orbitals are perpendicular to each other. For a theory to be accepted, it must explain experimental data and be able to predict behavior. Experimental evidence shows that the bond angle is 104.5°, not 90°. There are two regions of valence electron density in the BeCl2 molecule that correspond to the two covalent Be–Cl bonds. Questions in other subjects: The Lewis structure of sulfate shows there are four regions of electron density. The geometrical arrangements characteristic of the various sets of hybrid orbitals are shown in Figure $$\PageIndex{16}$$. What is the hybridization of the sulfur atom in the sulfate ion, $$\ce{SO4^2-}$$? Sulfate (S-35) Sulfate Group. However, in ethane an sp3 orbital of one carbon atom overlaps end to end with an sp3 orbital of a second carbon atom to form a σ bond between the two carbon atoms. The energy of the system changes as the atoms interact. This process of combining the wave functions for atomic orbitals is called hybridization and is mathematically accomplished by the linear combination of atomic orbitals, LCAO, (a technique that we will encounter again later). hope this . The hybridization in a trigonal planar electron pair geometry is sp2 (Figure $$\PageIndex{17}$$), which is the hybridization of the carbon atom in urea. Assigning Hybridization Ammonium sulfate is important as a fertilizer. We can use quantum mechanics to predict the specific regions around an atom where electrons are likely to be located: A spherical shape for an s orbital, a dumbbell shape for a p orbital, and so forth. All orbitals in a set of hybrid orbitals are equivalent in shape and energy. Example $$\PageIndex{3}$$: Assigning Hybridization. For example, we have discussed the H–O–H bond angle in H2O, 104.5°, which is more consistent with sp3 hybrid orbitals (109.5°) on the central atom than with 2p orbitals (90°). However, for larger central atoms, the valence-shell electron pairs are farther from the nucleus, and there are fewer repulsions. Now for H2SO4: We can simplify this by just doing an SO4^2- ion since the configuration around the S will be the same. The hybridization is sp 3. Each hybrid orbital is oriented primarily in just one direction. And millions of other answers 4U without ads. Figure $$\PageIndex{15}$$: The four valence atomic orbitals from an isolated carbon atom all hybridize when the carbon bonds in a molecule like CH4 with four regions of electron density. This is the quantity of energy released when the bond is formed. Adelaide Clark, Oregon Institute of Technology, Crash Course Chemistry: Crash Course is a division of. The observed structure of the borane molecule, BH3, suggests sp2 hybridization for boron in this compound. Figure 11: The sulfur atom in sulfate ion is sp 3 hybridized. Which statement describe hormones in plants? Using Hess’s law of constant heat of summation and given: However, to understand how molecules with more than two atoms form stable bonds, we require a more detailed model. When boron is in a molecule with three regions of electron density, three of the orbitals hybridize and create a set of three sp2 orbitals and one unhybridized 2p orbital. They are formed only in covalently bonded atoms. the single electrons in each orbital combine to form an electron pair. Sulfates are salts of sulfuric acid and many are prepared from that acid. Example $$\PageIndex{1}$$: Counting σ and π Bonds. Meanwhile, organic esters of sulfate like dimethyl sulfate come under esters of sulfuric acid and covalent compounds. Dextran sulfate and its congeners have been generally regarded as non-specific in their antiviral action. How do you streaks? I feel a bit like a broken record. They will also learn about the molecular geometry and the bond angles of nitrate. We illustrate the orbitals and electron distribution in an isolated carbon atom and in the bonded atom in CH4 in Figure $$\PageIndex{15}$$. Although quantum mechanics yields the “plump” orbital lobes as depicted in Figure $$\PageIndex{5}$$, sometimes for clarity these orbitals are drawn thinner and without the minor lobes, as in Figure $$\PageIndex{10}$$, to avoid obscuring other features of a given illustration. The number of sigma bonds formed by nitrogen is 4 since it is bonded to 4 hydrogen atoms. To accommodate these two electron domains, two of the Be atom’s four valence orbitals will mix to yield two hybrid orbitals. STEP-3: Calculate the number of lone pairs. A set of hybrid orbitals is generated by combining atomic orbitals. [NOTE: Take into account both, the lone pair and the bond pair of electrons for determining the hybridization of any atom. Their compounds exhibit structures that are often not consistent with VSEPR theory, and hybridized orbitals are not necessary to explain the observed data. What is the hybridization of the sulfur atom in the sulfate ion, SO 4 2−? Assign the set of hybridized orbitals from Figure $$\PageIndex{17}$$ that corresponds to this geometry. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. a) electron-domain geometry. The energy of the system would then rise (making the system destabilized), as shown at the far left of Figure $$\PageIndex{1}$$. The new orbitals that result are called hybrid orbitals. The observed angle of 104.5° is experimental evidence for which quantum-mechanical calculations give a useful explanation: Valence bond theory must include a hybridization component to give accurate predictions. Let us skip this step. HYBRIDIZATION AND SHAPE OF AMMONIUM ION, NH 4 + STEP-1: Write the Lewis structure. In gaseous BeCl2, these half-filled hybrid orbitals will overlap with orbitals from the chlorine atoms to form two identical σ bonds. Example $$\PageIndex{2}$$: Assigning Hybridization. This is the completed Lewis dot structure for one of them. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110). Butadiene, C4H6, is used to make synthetic rubber. Determine the Lewis structure of the molecule. The energy of the system depends on how much the orbitals overlap. The dots indicate the locations of the nuclei. For the sulfate ion, {eq}SO_4^{2-} {/eq}, draw the Lewis structure (by counting valence electrons of each atom). Conversely, the same amount of energy is required to break the bond. As we know, a scientific theory is a strongly supported explanation for observed natural laws or large bodies of experimental data. As the atoms move together, their orbitals begin to overlap. Have questions or comments? Since lone pairs occupy more space than bonding pairs, structures that contain lone pairs have bond angles slightly distorted from the ideal. Urea, NH2C(O)NH2, is sometimes used as a source of nitrogen in fertilizers. The versatility of CS hybridization was evaluated in various in-vitro and in-vivo models. As seen in Table $$\PageIndex{1}$$, an average carbon-carbon single bond is 347 kJ/mol, while in a carbon-carbon double bond, the π bond increases the bond strength by 267 kJ/mol. It is highly similar in sequence to both ezrin and moesin. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Each of the two sp hybrid orbitals holds one electron and is thus half filled and available for bonding via overlap with a Cl 3p orbital. UNII-7IS9N8KPMG. In any multiple bond, there will be one σ bond, and the remaining one or two bonds will be π bonds. What is the hybridization of each nitrogen and carbon atom in urea? For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. The hybridization is sp 3. The hybridization of NO 3 – is sp 2 type. In a π bond, the regions of orbital overlap lie on opposite sides of the internuclear axis. What is the hybridization of the sulfur atom in the sulfate ion, $$\ce{SO4^2-}$$? A lone pair, an unpaired electron, a single bond, or a multiple bond would each count as one region of electron density. Which sy... El Principio de Escasez, es el principio generador de la actividad? Unhybridized orbitals overlap to form π bonds. The four valence electrons of the carbon atom are distributed equally in the hybrid orbitals, and each carbon electron pairs with a hydrogen electron when the C–H bonds form. Thinking in terms of overlapping atomic orbitals is one way for us to explain how chemical bonds form in diatomic molecules. This may seem like a small number. he two delocalized electrons make multiple resonance structures possible for the sulfate ion (SO 4-2). Question sent to expert. Experimental evidence shows that rotation around σ bonds occurs easily. Which statement best describes a theme expressed in this passage? Figure $$\PageIndex{9}$$: The hybridization of an s orbital (blue) and two p orbitals (red) produces three equivalent sp2 hybridized orbitals (yellow) oriented at 120° with respect to each other. The hybrids result from the mixing of one s orbital and all three p orbitals that produces four identical sp3 hybrid orbitals (Figure $$\PageIndex{14}$$). Start studying Chem Exam 2: Ions and Hybridization. Identify each illustration as depicting a σ or π bond: (a) is a π bond with a node along the axis connecting the nuclei while (b) and (c) are σ bonds that overlap along the axis. This type of hybridization involves the mixing of one ‘s’ orbital and one ‘p’ orbital of equal energy to give a new hybrid orbital known as a sp hybridized orbital. The sp set is two equivalent orbitals that point 180° from each other. The sulfate ion in terms of hybridization and resonance theory. The hybridization of the carbon atom in the carbonate ion is to have three orbitals on the carbon atom that will be used to form sigma bonds. A more complete understanding of electron distributions requires a model that can account for the electronic structure of molecules. The dots indicate the location of the nuclei. Each sp hybridized orbital has an equal amount of s … Greater overlap is possible when orbitals are oriented such that they overlap on a direct line between the two nuclei. While the atoms are still widely separated, the attractions are slightly stronger than the repulsions, and the energy of the system decreases. This arrangement results from sp2 hybridization, the mixing of one s orbital and two p orbitals to produce three identical hybrid orbitals oriented in a trigonal planar geometry (Figure $$\PageIndex{9}$$). The number of hybrid orbitals in a set is equal to the number of atomic orbitals that were combined to produce the set. For the $$H_2$$ molecule shown in Figure $$\PageIndex{1}$$, at the bond distance of 74 pm the system is $$7.24 \times 10^{−19}\, J$$ lower in energy than the two separated hydrogen atoms. In addition to the distance between two orbitals, the orientation of orbitals also affects their overlap (other than for two s orbitals, which are spherically symmetric). $\begingroup$ A structure with four double bonds in the sulphate ion would give the sulphur atom a formal charge of -2. Students will learn about how this hybridization occurs and all the steps involved in it. 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# Could (old) LIGO have detected GW150914?

The merging black hole binary system GW150914 was detected in only 16 days of aLIGO data at a signal level that appears to be well above the detection threshold at around 5 sigma. There are no further events between 4 and 5 sigma in the same data.

Could this event have been detected by previous LIGO/VIRGO incarnations which observed for much longer, albeit with lower sensitivity? If so, does this indicate that the aLIGO team have struck lucky and that this is a comparatively rare event that may not be repeated for many years?

EDIT: The answers I have agree that LIGO couldn't have seen it, but don't yet completely explain why. GW150914 had a strain that rose from a few $10^{-22}$ to $10^{-21}$ over about 0.2 seconds. This appears to make the characteristic strain maybe a few $10^{-22}$ Hz$^{-1/2}$ and thus would appear from the published sensitivity curves to lie above original LIGO's detection sensitivity at frequencies of $\sim 100$ Hz. Is my estimate of the characteristic strain way off?

• Did they say why they picked this particular data set? Can we speculate that there are other datasets from the run that contain more events that will be published later? – CuriousOne Feb 14 '16 at 19:36
• This just lets me think how unlucky it would have been, if the ligo would have been offline due to its improvement while this event had been occured... >.<' – Zaibis Feb 15 '16 at 14:56
• @CuriousOne Possibly just because they had to pick a stretch of data so they could start writing the paper. Still, I look forward to seeing what the next 88 days of data contained :) – Will Vousden Feb 16 '16 at 16:18
• @WillVousden: I am simply asking if there are any rumors about what else is in the rest of the data... :-) – CuriousOne Feb 16 '16 at 20:59
• @CuriousOne It had crossed my mind as well... I wish I knew! I left the collaboration just after the detection :-( – Will Vousden Feb 16 '16 at 21:11

To expand on HDE's answer, initial LIGO indeed wouldn't have detected GW150914, but it's not quite as simple as the peak strain being below the curve in the sensitivity plot: the integration time also matters.

These plots can be misleading; the curves they show don't represent a minimum detectable strain. Indeed, the units on the y-axis of these plots are $\mathrm{Hz}^{-1/2}$, while the GW strain is dimensionless, so you can't actually compare them! It's entirely possible to detect a signal that peaks well below the noise curve, as long as it's in-band for sufficiently long.

The curves that you see describing LIGO detector sensitivities conventionally show the amplitude spectral density of the detector noise. Meanwhile, the threshold for a detection is determined by the signal-to-noise ratio (SNR) from matched (Wiener) filtering. Assuming we know the form of the signal $h$ in advance (see caveats below), this is defined in terms of the noise-weighted inner product of $h$ with itself: $$\mathrm{SNR}^2 = \left<h,h\right> \equiv \int_0^\infty \frac{4|\tilde{h}(f)|^2}{S_n(f)}\,\mathrm{d}f$$ where $S_n(f)$ is the noise power spectral density (i.e., the square of what's shown in the sensitivity plots). The SNR therefore depends on the spectral composition of the signal and its overlap with the detector bandwidth.

If you imagine this in the time domain (Parseval's theorem), the (squared) SNR actually accumulates in proportion to the number of cycles the waveform spends in-band. For a monochromatic source, this is proportional to the integration time. For example, if $\tilde{h}(f) = \delta(f-f_0)h_0$ and, without loss, the noise PSD is a constant $S_n(f_0)$, then the SNR is given by: $$\mathrm{SNR}^2 = \frac{2}{S_n}\int_{-\infty}^\infty|\tilde{h}(f)|^2\,\mathrm{d}f = \frac{2}{S_n}\int_{-\infty}^\infty|h(t)|^2\,\mathrm{d}t$$ Therefore, since $|h(t)| = h_0$, for a finite observation window $T$, the SNR scales with $\sqrt{T}$: $$\mathrm{SNR} = \sqrt{\frac{2T}{S_n}}h_0$$

So, let's approximate GW150914 as a monochromatic source. Reading off the plots in the detection paper, let's say it has a average frequency of $f_0 \approx 60 \,\mathrm{Hz}$, an amplitude of $h_0\approx 5\times10^{-22}$, and a duration of $T \approx 0.2\,\mathrm{sec}$. Then, reading off a strain ASD of $\sqrt{S_n(f_0)} \approx 10^{-22}$ for initial LIGO, we'd get an SNR of around 3, which doesn't meet the standard detection threshold of 8 (also, see the caveats below).

There's a much more complete discussion of detector sensitivity curves in this paper; it's worth a read! A more useful quantity, described in this paper, is the characteristic strain, which attempts to account for the frequency evolution of an inspiral signal such as GW150914, to ease comparison between detector sensitivity and strain amplitude.

Caveats: in practice, it's more complicated than the matched filter model, since the detector noise is annoyingly non-stationary and non-Gaussian. There are more sophisticated search algorithms that use things like signal quality vetoes and $\chi^2$ discriminants that reject spurious responses of the matched filter. There are also search algorithms that don't require a priori knowledge of the signal waveform and can detect unmodelled bursts. It was actually this sort of generic search that detected GW150914; references are available in the detection paper.

Also note that the SNR defined above is the optimal SNR that you get if:

1. you filter the data stream with the exact signal that you're looking for, and
2. the noise realisation is zero.

Since the mean of the noise is zero, number 2 above is equivalent to taking the expectation of the SNR over all noise realisations.

In practice, we don't know the precise signal a priori, and some SNR is lost in the approximation. For a candidate waveform $u$, the expected SNR (over all noise realisations) is then given by $$\mathrm{SNR} = \frac{\left<u,h\right>}{\|u\|}$$

• Right, to compare with the sensitivity plots you need to multiply the strain by the square root of the time the signal was observed for (roughly)? – Rob Jeffries Feb 14 '16 at 22:45
• @RobJeffries To the extent that the signal is approximately monochromatic, yes (and of course you must divide by the noise PSD at the signal's characteristic frequency). That will give you an approximate SNR, which you can compare to with whatever threshold you've picked (conventionally 8) to determine whether the signal would be detected. – Will Vousden Feb 14 '16 at 23:02
• (For more complicated/broadband signals, you have to know the waveform $h(f)$ and do the integration above.) – Will Vousden Feb 14 '16 at 23:50
• Given that, then the strain of this event was a few $10^{-22}$ up to a peak of $10^{-21}$ seen over about 0.2 secs. So isn't that characteristic strain $>10^{-22}$ Hz$^{-1/2}$? In which case the sensitivity curves I've seen (see HDE226868's answer) suggest that original LIGO would have detected it. I'm guessing then that my estimate of the characteristic strain is too high by factors of a few - why? – Rob Jeffries Feb 15 '16 at 12:10
• @RobJeffries See my update for an example calculation of the SNR for GW150914 :) – Will Vousden Feb 15 '16 at 16:10

I have a direct quote from the website:

The event would not have registered in LIGO's first-generation detectors; the fact that it appeared with striking clarity in both L1 and H1 indicates the leap in detector performance that the Advanced LIGO program has produced.

This was a sensitivity issue: in most frequencies, Advanced LIGO is more strain-sensitive than LIGO by a factor of 10.

Hild (2012) gives an overview of first- and second- generation detectors, including this graph:

The detected wave had a peak strain of $\thicksim 10^{-21} ~\mathrm{1/\sqrt{\textrm{Hz}}}$ and was detected at frequencies between $35$ and $250~\textrm{Hz}.$ A portion of this does fall in the original LIGO sensitivity range. However, as Figure 3 of the Advanced LIGO detection paper shows, most falls below it.

Regarding the odds of detection, older estimates said that Advanced LIGO should be able to observe $\thicksim 40$ neutron star "inspiral events", and $\thicksim 30$ black hole binary events of the same type. This would be attributed in part to the noise reduction, making it easier to observe larger areas. The group says that event rate detection would increase by 3000 following the sensitivity upgrades.

• Agreed original LIGO is a factor 4 less sensitive, but there were intermediate upgrades. I'd also like to know the quantitative details. It does look from published sensitivity curves that it might have been seen... – Rob Jeffries Feb 14 '16 at 19:26
• @RobJeffries I understand; I'm looking for the particulars. – HDE 226868 Feb 14 '16 at 19:27
• @RobJeffries I don't have a reference on hand, but a LIGO person the other day said in a talk this event, which was SNR=24 by some measure, would have been SNR=3 or 4 in old LIGO iirc, below their threshold of 8 for reporting. – user10851 Feb 14 '16 at 20:30
• I think that strain number you quote is a bit tricky. The peak strain was $10^{-21}$, but the equivalent strain (which is on the y-axis of your plot) was more like $3\times 10^{-22}$. – Rob Jeffries Feb 14 '16 at 22:41
• I meant characteristic strain, not equivalent strain. – Rob Jeffries Feb 14 '16 at 23:58

Despite my hopes, it appears that old LIGO would not have detected GW150914. Sigh...

From the discovery announcement of GW150914, I created a facsimile of the measured strain signal:

From that paper and also the summary of Run 6 of original LIGO, in 2009-2010, I digitized amplitude spectral densities of the two:

From these inputs I calculated the characteristic strain and noise curves:

Integrating these up per the very helpful reference identified by Will Vousden, I calculate the following results for signal-to-noise ratio $\rho$ for a matched filter:

\begin{equation*} \begin{array}{lcccc} \text{detector} & \rho & \chi_r^2 & \hat{\rho} & \hat{\rho_c} \\ \text{aLIGO} & 18.73 & 1.44 & 16.69 & 23.6 \\ \text{LIGO run 6} & 4.88 & 1.44 & 4.35 & 6.2 \end{array} \end{equation*}

Here:

• $\rho$ is the calculated signal-to-noise ratio for the detector.
• the $\chi_r^2$ statistic measures how closely the signal matches the expected waveform template. This value was not reported, that I could see; I have chosen it to get the correct final result for aLIGO, and then used the same value for LIGO Run 6.
• $\hat{\rho}$ penalizes $\rho$ based on the value of $\chi_r^2$.
• $\hat{\rho}_c$ accounts for the signal being observed by both LIGO detectors: the two signal-to-noise ratios are added in quadrature (so here the value is multiplied by $\sqrt{2}$.)

Since the Run 6 signal-to-noise ratio $\rho < 8$, it would have been very difficult to claim a detection: in this review of Run 6's search for massive black hole inspiral/merger signals, Figure 2 shows that signals with these values of $\rho$ and $\chi_r^2$ would be buried in noise events.

Curiously, that same review includes Figure 1, which appears to show a 60 total solar mass event being detectable out to about 470 Mpc (with favorable orientation), which would include most of the the reported distance range of $410_{-180}^{+160}$ Mpc. I cannot explain the discrepancy.

• Belated thanks for this. Yes it appears to be close, but not quite. – Rob Jeffries Apr 25 '16 at 7:51
• Some awesome work here! – WetSavannaAnimal May 17 '16 at 8:48

## protected by Qmechanic♦Feb 15 '16 at 17:12

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

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Timezone: »

Poster
Robust Imitation via Mirror Descent Inverse Reinforcement Learning
Dong-Sig Han · Hyunseo Kim · Hyundo Lee · JeHwan Ryu · Byoung-Tak Zhang

Wed Nov 30 02:00 PM -- 04:00 PM (PST) @ Hall J #807
Recently, adversarial imitation learning has shown a scalable reward acquisition method for inverse reinforcement learning (IRL) problems. However, estimated reward signals often become uncertain and fail to train a reliable statistical model since the existing methods tend to solve hard optimization problems directly. Inspired by a first-order optimization method called mirror descent, this paper proposes to predict a sequence of reward functions, which are iterative solutions for a constrained convex problem. IRL solutions derived by mirror descent are tolerant to the uncertainty incurred by target density estimation since the amount of reward learning is regulated with respect to local geometric constraints. We prove that the proposed mirror descent update rule ensures robust minimization of a Bregman divergence in terms of a rigorous regret bound of $\mathcal{O}(1/T)$ for step sizes $\{\eta_t\}_{t=1}^{T}$. Our IRL method was applied on top of an adversarial framework, and it outperformed existing adversarial methods in an extensive suite of benchmarks.


text:
testrmd provides facilities to enable testing of and reporting on tested R Markdown chunks.

### Why test?

As the author of an R Markdown document, there are assumptions you make along the way: that your data is a certain shape, that certain populations are represented, that your time series doesn’t have any gaps. By writing explicit tests, you can spot any violations of these assumptions earlier, and prevent readers of your document from coming to any unwarranted conclusions.

This is especially important for R Markdown documents that may be run as part of a workflow, where data may change with every render, or R package dependencies may be upgraded over the lifetime of the document. Tests let you alert readers to subtle problems long after you are done writing the document.

### Why testrmd?

Tests are just regular R code that throw errors when their expectations are broken. So you may wonder why the testrmd package is needed–why not just write tests in regular {r} chunks?

The primary reason is that testrmd presents a better user experience for your readers. Test code/output is generally not useful when the tests are passing, but when the tests fail, they become very important. This package makes it easy for tests to stay out of your readers’ way until something actually goes wrong, at which time it calls your readers’ attention to the failures.

### Installation

devtools::install_github("ropenscilabs/testrmd")

### Basic Use

To enable testing of RMarkdown chunks in your document, you will need to add a function calls in a chunk before you want to do testing.

testrmd::init()

Thereafter, for any chunk that you want to have as part of the testing, you simply need to add the option test = TRUE in the chunk header, and then have testing code that returns an error if the test fails. As a simple example, let’s test if a variable is numeric:

{r test_chunk, test = TRUE}
y <- "5"
stopifnot(is.numeric(y))

### Example Output

The default and emoji (testrmd::init(theme = "emoji")) examples of running an analysis using cranlogs are available.

### Supported Frameworks

testrmd supports validation methods that throw errors when a test fails. The following validation frameworks have been shown to easily return errors compatible with testrmd:

#### Unconfirmed Frameworks

The following validation methods have not been confirmed to work with testrmd because their core functionality does not throw errors to indicate a failed test:


text:
# Creating work files in an alternate directory

From a Windows command line, enter:

``install_dir\dtxcmdsv k:\maps\invoices.mmc -AE -T -WD=c:\tmp``
Option
Description
k:\maps\invoices.mmc
This specifies the name of the map to be executed.
-AE
This produces an execution audit log using the AuditLocation settings for MapAudit compiled into the map file.
-T
This specifies that no trace information is to be produced.
-WD=c:\tmp
This creates the work files in the c:\tmp directory and deletes the work files after the map file is executed.


text:
# Mandelbrot Art

## The Project

I wanted to create a program that would allow me to create beautiful images using data from the Mandelbrot set, a fascinating piece of mathematics.

Although I started out using my own complex number class, I discovered that doing the complex number math explicitly with two doubles runs a bit faster.

Once I had things working, I began experimenting with different drawing, coloring, and reconstruction techniques.

## Drawing Techniques

Early on while testing my program, it became clear that I needed to significantly speed up draw time to keep my program interactive while zoomed in close to the fractal's border.

Instead of doing the complex number calculations for every pixel in the image (top), I could instead do every other pixel. This would cut the number of calculations while leaving gaps in the image.

These gaps could then be filled in two different ways:

• Nearest Neighbor
• Bilinear Interpolation

I will talk more about filling in the gaps later.

With the ability to skip $$n$$ pixels, I can pixelate the image into $$n + 2$$ by $$n + 2$$ chunks (middle). This reduces the number of calculations by a factor of up to $$(n+1)^2$$. I justify this with the following scary calculus maths, where $$x^2$$ is the total number of pixels and $${\lceil \frac{x}{n+1} \rceil}^2$$ is the total number of pixels when skipping every $$n$$ pixels: $\lim_{x \to \infty} \frac {x^2}{{\lceil {x \over {n+1}} \rceil}^2} = \lim_{x \to \infty} \frac {x^2}{{ ({x \over {n+1}} + \delta) }^2} = \lim_{x \to \infty} \frac {x^2}{{ ({{x+\delta(n+1)} \over {n+1}}) }^2}$ $= \lim_{x \to \infty} \frac {x^2(n+1)^2}{{({x+\delta(n+1)}) }^2} = (n+1)^2 \cdot \lim_{x \to \infty} \frac {x^2}{{({x+\delta(n+1)}) }^2}$ $= (n+1)^2 \cdot \lim_{x \to \infty} \frac {x^2}{{({x \cdot (1+\frac{\delta(n+1)}{x}})) }^2}$ $= (n+1)^2 \cdot \lim_{x \to \infty} \frac {1}{{ (1+\frac{\delta(n+1)}{x}) }^2} = (n+1)^2 \cdot \frac{1}{1^2}$ $= (n+1)^2$

For example, by skipping every other pixel, I reduce the number of calculations by a factor of $$(1+1)^2 = 2^2 = 4$$ as the image size approaches infinity.

In an attempt to speed things up without sacrificing quality, I took my new ability to pixelate the image into $$n + 2$$ by $$n + 2$$ chunks and added functionality to recursively subdivide chunks near regions of high complexity (bottom). This ended up making things slower, but it looks cool, especially when color is used.

## Coloring Techniques

My simplest coloring technique is monochromatic. For each pixel, or chunk of pixels, I calculate a complexity value. This value is the number of iterations it took for the point to escape the set divided by the maximum number of iterations. Thus, this complexity value ranges from 0 to 1. For points that do not escape after the maximum number of iterations, the complexity is defined to be 1. The complexity value does not increase linearly as you approach the boundary of the set, so it is useful for aesthetic reasons to use the square root of the complexity value. I multiply this value by 255 and type cast it to an unsigned char (top).

For my second coloring technique, I take the previously described square root of the complexity value and rescale it to range from 0 to 360. I pretend that this new value is the hue and value of an HSV color with a saturation of 1. Finally, I convert this constructed HSV color to an RGB color (middle).

My third coloring technique, the one shown in the nine videos at the start, is a subtle variation of my second coloring technique. After I construct the HSV color and convert it to an RGB color (middle), I then pretend that the RGB color is an LAB color and convert it to an RGB color (bottom).

This third coloring technique produces both smooth gradients between some colors, and sharp bands between others. The reason it produces sharp bands is caused by negative values produced in the final LAB to RGB being type casted to unsigned chars (I think).

This could be seen as a bug, but because I found the problem and chose to keep it for aesthetic reasons, I consider it a surprise feature.

## Reconstruction Techniques

If I skip 0 pixels, no reconstruction is applied to the image. There are no gaps to fill, so reconstruction is not necessary (top).

If I skip $$n \gt 0$$ pixels, there will be gaps in the image that need to be filled (second).

I can fill these gaps two different ways:

• Nearest Neighbor (third)
• Bilinear Interpolation (bottom)

With nearest neighbor, I fill in an $$(n + 2)$$ by $$(n + 2)$$ chunk with the color of the top left pixel in the chunk. This fills in the gaps between pixels with a uniform color, giving the image a blocky appearance (third).

Bilinear interpolation is interesting. The way I implement it is fairly straightforward. Given the following:

• the current pixel $$\vec{p}$$
• the known colors $$c_{tl}$$, $$c_{tr}$$, $$c_{bl}$$, $$c_{br}$$ at pixels $$\vec{p}_{tl}$$, $$\vec{p}_{tr}$$, $$\vec{p}_{bl}$$, $$\vec{p}_{br}$$
• the vertical distance $$v$$ between $$\vec{p}$$ and $$\vec{p}_{tl}$$
• the horizontal distance $$u$$ between $$\vec{p}$$ and $$\vec{p}_{tl}$$

I can calculate an interpolated color between the known surrounding four pixel colors this way, where $$c_{f}$$ is the interpolated color: $c_{f} = v \cdot (c_{bl} \cdot (1 - u) + c_{br} \cdot u)$ $+ (1 - v) \cdot (c_{tl} \cdot (1 - u) + c_{tr} \cdot u)$

This color $$c_{f}$$ is used to color the pixel $$\vec{p}$$ (bottom).

When the image is reconstructed using square chunks of different sizes, my images using bilinear interpolation look odd. This is because color values are changing at different rates for each chunk size.

I will showcase an image like this in The Gallery section.

## Building the Program

All of these images come from a single program I wrote in C++ called mandelbrot. There are several make targets that build the program in specific ways:

• default: build for viewing/saving monochromatic images
• color: build for viewing/saving full-color images using the third coloring technique described previously
• slow: build for viewing/saving monochromatic images using long doubles for more precision (no difference?)
• movie: build for rendering a sequence of 1800 images each with increasing iterations
• pic: build for saving a single full-color image
• vid: build movie target, convert PPMs to PNGs with magick, and create a 30 second MP4 using ffmpeg (requires magick and ffmpeg)
• clean: remove mandelbrot and all PPMs/PNGs

## Using the Program

Depending on how the program was built, it can be used in different ways:

• default/color/slow: Explore the Mandelbrot Set
• Run Using: ./mandelbrot
• W/A/S/D: Move Up, Left, Down, and Right
• Q/E: Zoom Out and Zoom In
• R: Decrease Image Resolution (Increases Number of Iterations)
• F: Increase Image Resolution (Decreases Number of Iterations)
• G: Toggle Adaptive Pixel Skip (Chunk Subdividing Drawing Technique)
• B: Toggle Bilinear Interpolation
• O: Write Current Window to out.png.
• movie/pic/vid: Write a PPM / Sequence of PPMs (Please don't use these make targets. These were only used to make the high(ish)-resolution images in this report.)

## Known Problems

Some of these are genuine problems. Others are problems that I have chosen to keep because they are entertaining.

• Floating-Point Precision
• I hoped to be able to zoom in farther. However, the numbers are so very tiny. To zoom in further, I would need some kind of float with arbitrary precision.
• I did a bit of research and fairly quickly came to the conclusion that writing my own arbitrarily-precise floating-point number class is not something I can just do.
• Negative Color Values
• During my LAB to RGB conversion, floating-point numbers are converted to unsigned chars, despite the fact that these floating-point values can be negative.
• This gives a mixture of smooth gradients, where everything works nicely, and sharp boundaries, where negative values lose their signedness. This is my best guess as to what is happening, but I have heard that converting between LAB and RGB with unsigned chars is a thing, so I might be wrong here.
• I am pleased with the fun resulting colors, so it stays!
• When adaptive pixel skip is enabled, there are many 2 by 2 chunks of the same color, despite having their own positions.
• Basically, there appear to be chunks twice as large as they actually are because there is a square of four of all the same color.
• I have no idea what is happening here. It's difficult for me to even describe the problem. Therefore, I consider this a surprise feature (fun!).
• Zooming Too Fast
• If you go crazy with the zoom, zooming in or zooming out as fast as you can, the program may not be able to adjust the number of iterations fast enough to keep things running smoothly.
• This can lock up the program, but I think it's mostly your fault. Don't be so hasty!

## The Future

I am unhappy with how slow my program is. It runs on a single thread, on a single core. This was something I knew, and something I did experiment with, but I was not yet able to get multi-threading functional.

In the beginning, I planned to implement cubic interpolation, as well, but the project was already getting really big, especially when I added dynamic pixel skipping.

Dynamic pixel skipping was a huge time sink. My first idea was to allow non-square chunks, creating a kind of voronoi pattern. This ended up being an absolute nightmare, and I scrapped the idea of non-square chunks in a couple of days. I finally got square chunks to subdivide properly, but it took a lot longer than I thought it would. This was the most difficult part of the project for me.


text:
# Simplicial localization of the cofibrant-fibrant objects

Let $$M$$ be a model category. I don't assume that $$M$$ has functorial factorizations or that $$M$$ is simplicial. Write $$M^{c}$$ (respectively, $$M^{cf}$$) for the full subcategory of $$M$$ on the cofibrant objects (respectively, the cofibrant and fibrant objects).

Does the inclusion $$M^{cf} \to M^c$$ induce a Dwyer-Kan equivalence on the simplicial localizations at the class of weak equivalences?

Assuming the answer is "yes", does anyone know of a reference? I couldn't find this in the original papers of Dwyer and Kan for general $$M$$, only when $$M$$ has functorial factorizations.

• This should follow from 4.4 in "Function complexes in homotopical algebra". That is: the map on simplicial localizations is essentially surjective up to equivalence, so you just need fully faithful, and 4.4 lets you compute the mapping space in the same way for both. – Dylan Wilson Feb 21 at 0:09
• The answer is definitely yes, but I do not know any published reference. What I can say is that any proof that I know of that $M^c$ and $M$ are Dwyer-Kan equivalent may be promoted to a proof that $M^{cf}$ and $M^f$ are Dwyer-Kan equivalent. Inspection of calculus of fractions as provided by 8.1 in “Function complexes in homotopical algebra” is very closed to being a proof. – Denis-Charles Cisinski Feb 21 at 9:32

The only published reference for the full statement I know is Hinich's paper (V. Hinich, Dwyer-Kan localization revisited. Homology Homotopy Appl. 18, No. 1), where it appears as (the dual of) Proposition 1.3.8. Strictly speaking, the proof given there is "this works just like the proof of the case $$\mathscr C^c\hookrightarrow\mathscr C$$ (of which Hinich gives a new proof in Section 1.3.7), but at least to me it looks like you can indeed directly adapt his argument by just adding the word "fibrant" at appropriate places.


text:
## Building/Installing Alembic for OSX

Alembic is a new open-source computer graphics interchange framework being developed by Sony Imageworks and ILM. The basic idea is that moving animation rigs and data and whatnot between packages can be a very tricky procedure since every package has its own way to handle animation, so why not bake out all of that animation data into a common interchange format? So, for example, instead of having to import a Maya rig into Houdini, you could rig/animate in Maya, bake out the animation to Alembic, bring that into Houdini to conduct simulations with, and then bake out the animation and bring it back into Maya. This is a trend that a number of studios including Sony, ILM, Pixar, etc. have been moving toward for some time.

I’ve been working on a project lately (more on that later) that makes use of Alembic, but I found that the only way to actually get Alembic is to build it from source. That’s not terribly difficult, but there’s not really any guides out there for folks who might not be as comfortable with building things from source. So, I wrote up a little guide!

Here’s how to build Alembic for OSX (10.6 and 10.7):

1. Alembic has a lot of dependencies that can be annoying to build/install by hand, so we’re going to cheat and use Homebrew. To install Homebrew:
/usr/bin/ruby -e "\$(curl -fsSL https://raw.github.com/gist/323731)"
1. Get/build/install cmake with Homebrew:
brew install cmake
1. Get/build/install Boost with Homebrew:
brew install Boost
1. Get/build/install HDF5 with Homebrew:
brew install HDF5 HDF5 has to make install itself, so this may take some time to run. Be patient.
1. Unfortunately, ilmbase is not a standard UNIX package, so we can’t use Homebrew. We’ll have to build ilmbase manually. Get it from:
http://download.savannah.nongnu.org/releases/openexr/ilmbase-1.0.2.tar.gz Untar/unzip to a readily accessible directory and cd into the ilmbase directory. Run: ./configure After that finishes, we get to the annoying part: ilmbase by default makes use of a deprecated GCC 3.x compiler flag called Wno-long-double, which no longer exists in GCC 4.x. We’ll have to deactivate this flag in ilmbase’s makefiles manually in order to build correctly. In each and every of the following files:
/Half/Makefile
/HalfTest/Makefile
/Iex/Makefile
/IexTest/Makefile
/Imath/Makefile
/ImathTest/Makefile

Find the following line: CXXFLAGS = -g -O2 -D_THREAD_SAFE -Wno-long-double and delete it from the makefile. Once all of that is done, you can make and then make install like normal. Now move the ilmbase folder to somewhere safe. Something like /Developer/Dependencies might work, or alternatively /usr/include/
1. Time to actually build Alembic. Get the source tarball from:

http://code.google.com/p/alembic/wiki/GettingAlembic

Untar/unzip into a readily accessible directory and then create a build root directory parallel to the source root you just created:

mkdir ALEMBIC_BUILD

The build root doesn’t necesarily have to be parallel, but here we’ll assume it is for the sake of consistency.

2. Now cd into ALEMBIC_BUILD and bootstrap the Alembic build process. The bootstrap script is a python script:

python ../[Your Alembic Source Root]/build/bootstrap/alembic_bootstrap.py

The script will ask you for a whole bunch of paths:

For “Please enter the location where you would like to build the Alembic”, enter the full path to your ALEMBIC_BUILD directory.

For “Enter the path to lexical_cast.hpp:”, enter the full path to your lexical_cast.hpp, which should be something like /usr/local/include/boost/lexical_cast.hpp

For “Enter the path to libboost_thread:”, your path should be something like /usr/local/lib/libboost_thread-mt.a  For “Enter the path to zlib.h”, your path should be something like /usr/include/zlib.h

For “Enter the path to libz.a”, we’re actually not going to link against libz.a. We’ll be using libz.dylib instead, which should be at something like /usr/lib/libz.dylib

For “Enter the path to hdf5.h”, your path should be something like /usr/local/include/hdf5.h

For “Enter the path to libhdf5.a”, your path should be something like /usr/local/Cellar/hdf5/1.x.x/lib/libhdf5.a (unless you did not use Homebrew for installing hdf5, in which case libhdf5.a will be in whatever lib directory you installed it to)

For “Enter the path to ImathMath.h”, your path should be something like /usr/local/include/OpenEXR/ImathMath.h

For “Enter the path to libImath.a”, your path should be something like /usr/local/lib/libImath.a

Now hit enter, and let the script finish running!

3. If everything is bootstrapped correctly, you can now make. This will take a while, be patient.

4. Once the make finishes successfully, run make test to check for any problems.

5. Finally, run make install, and we’re done! Alembic should install to something like /usr/bin/alembic-1.x.x/.


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# Confusion about square bracket notation in the the Kronecker delta symbol

I am not sure I understand what the short-hand anti-symmetrization means. I.e. I know that $$\delta_{cd}^{[ab]} ~=~ \frac{1}{2}(\delta_{c}^{a}\delta_{d}^{b} - \delta_{c}^{b}\delta_{d}^{a})$$ but how do we expand "bigger" analogues? I.e. $$\delta_{b_1 b_2 b_3 b_4}^{[a_1 a_2 a_3 a_4]} ~=~ ?$$

What is the order of permutation in this case?

$$\delta^{[\mu_1\mu_2\ldots \mu_n]}_{\nu_1\nu_2\ldots \nu_n}~=~ \frac{1}{n!}\sum_{\pi\in S_n}{\rm sgn(\pi)} \prod_{i=1}^n \delta^{\mu_{\pi(i)}}_{\nu_i}.$$

More generally,

$$T^{[\mu_1\mu_2\ldots \mu_n]}~=~ \frac{1}{n!}\sum_{\pi\in S_n}{\rm sgn(\pi)} T^{\mu_{\pi(1)}\mu_{\pi(2)}\ldots \mu_{\pi(n)}}.$$

Here $S_n$ is the symmetric group of permutations, and $\pi\in S_n$ is a permutation with signature ${\rm sgn(\pi)}$. The order of $S_n$ is $$|S_n|~=~n!~.$$

• @Qmechanic - I suggest you write out an explicit example maybe for $n=3$ or $n=4$. – Prahar Dec 16 '14 at 20:31
• I got it. It is rather long to write for $n=4$ or even for $n=3$. I have done it already and thanks for the answers. – Marion Dec 16 '14 at 23:21
• @Marion: Perfect, that saves me the trouble of typing it out. For $n=3$, see also Prahar's answer. – Qmechanic Dec 16 '14 at 23:23


text:
Thread: cross product question View Single Post

 Originally posted by matt grime In generality, given a finite dimenisional vector space V there is the n-fold tensor product, which can saefly be thought of as: let v_i be a basis of V, then a basis of $\bigotimes^nV$ is given by the set of all symbols $$v_{i_1}\otimes\ldots \otimes v_{i_n}$$ ( I am, slightly contradictarily, being a gentleman here by not defining that in a basis free way, if you want I can but it is more confusing for the non-algebraist, trust me)
while i am not myself an algebraist, i do have a fair amount of algebra. i am, for example, familiar with the universal construction of a tensor product. as well as the coordinate definition of a tensor product of vector spaces. so whatever.

mostly, i am just looking to establish what is the correct terminology, and why.
 now the permutation group on n elements acts on the n-fold tesnor product by swapping factors in the expresion. the exterior algebra is that sub-space where swapping the elements is the same as multiplying by -1
sure.

 eg in the two fold tensor, it is the subspace spanned by expressions like $u\otimes v - v\otimes u$ It is not hard to see that in an r dimensional vector space this is only non-zero if n<=r, and it has dimension r choose n. The element $u\otimes v - v\otimes u$ is labelled $u\wedge v$ (can't recall tex for it off hand; it's probably \wedge) The reason why its so damn easy and confusing at the same time for R^3 is because 3 choose 2 is 3 and is therefore a 3 dimensional vector space. Different groups label these different things all over the place, but the wedge is usually called the exterior product.
yes, as i understand it, wedge product and exterior product are synonymous.

 Let's stick in the 3-d case: where does the rule $\imath\times\jmath=k$ come from? Note i'm using x for ease of type setting, it is the wedge above. well, the answer is: there is an exact pairing from between the first and second exterior algebra
sure, because $\binom{3}{1}$ and $\binom{3}{2}$ are equal, there is an isomorphism between the two given by the Hodge dual (which depends on the orientation and the inner product we give to the vector space)

 basically, there is a unique element in the third exterior algebra's base: ixjxk corresponding to the invariant element $\hat{\imath}\otimes\hat{\jmath} \otimes \hat{k} -\hat{\jmath}\otimes \hat{\imath}\otimes\hat{k} + \hat{\jmath}\otimes\hat{k}\otimes\hat{\imath} -\hat{\imath}\otimes \hat{k}\otimes\hat{\jmath} + \hat{k}\otimes\hat{\imath}\otimes\hat{\jmath} - \hat{k}\otimes\hat{\jmath} \otimes \hat{\imath}$ in the tensor algebra.
but this exactly illustrates my point of why the exterior algebra is not the cross product algebra. first of all, the exterior algebra is associative, and the cross product algebra is not. it follows that in the cross product algebra,

$$(\hat{\imath}\times\hat{\jmath})\times\hat{k}=\hat{\imath}\times(\hat{\ jmath}\times\hat{k})=0$$

whereas in the exterior algebra, you get the invariant element $e_1\wedge e_2\wedge e_3\neq 0$.

if we assign a product on the exterior algebra $e_i\times e_j=\star(e_i\wedge e_j)$, then we get an explicit correspondence with the vector cross product algebra in R3

 so the paring matches ixj with k and jxk with i and so on. Is that clear in the slightest? I think that might answer the question but I'm not sure.
well, this isn't really my question. i know what the exterior algebra is and how it relates to the cross product.

i guess all i really want to know is: what does the term "outer product" mean to you? some people are saying "outer product"="exterior product", i think saski might have even said "outer product = Clifford product (geometric product)", but i guess we ruled that out.

on the other hand, i think "outer product"="tensor product"

i really just wanted you to weigh in with your opinion of what the term outer product should mean.


text:
Type control
03-13-2015, 03:18 PM (This post was last modified: 03-13-2015 06:05 PM by salvomic.)
Post: #21
 salvomic Senior Member Posts: 1,365 Joined: Jan 2015
RE: Type control
(03-13-2015 03:04 PM)Han Wrote:  Ensure there are no conflicts between the CAS version of myprog and the non-CAS version of myprog. In other words, name them differently to make sure that when you type myprog you aren't running the wrong program (since different programs can actually have the same name). As for the error (w/ respect to the second code block in your post), it is being generated much further inside your program. What you have posted there does not produce any errors.

thank you!
I'll control soon.
However, please, see my ANOVA program (here)
If I use the function ANOVA_list() (that need values in D1, D2, D3 of Statistic 1var) without parameters, in CAS I get "Error: Bad argument value", in Home I get "Error: Syntax error", but with something like ANOVA_list(0.01) the program seems to work...
I've in the code some controls like:
Code:
 local spar; spar:= SIZE(a); IF (TYPE(spar)==6 OR TYPE(a)<>0) THEN RETURN("Usage: ANOVA_list(a) a=significance level; data in Stat 1var");  END; IF ((spar==0)) THEN RETURN("Usage: ANOVA_list(a) a=significance level; data in Stat 1var"); END; IF (a <=0 OR a>= 1) THEN RETURN("Significance level must be > 0 AND <1"); END; IF ((size(D1) == 0) OR (size(D1) == 0) OR (size(D3) = 0)) THEN RETURN("Statistics 1var, D1 means, D2 stddev, D3 numerosity"); END;
I would like to give a message when the user input ANOVA_list() without parameter... and also when it gives wrong parameter (but this seems to work).

But the line
Code:
 IF ((spar==0)) THEN RETURN("Usage: ANOVA_list(a) a=significance level; data in Stat 1var"); END;

in this non CAS program perhaps does nothing and it doesn't intercept the error: with input ANOVA_list() without a number the program give the error and not the message...

∫aL√0mic (IT9CLU), HP Prime 50g 41CX 71b 42s 12C 15C - DM42 WP34s :: Prime Soft. Lib
03-13-2015, 06:14 PM
Post: #22
 Han Senior Member Posts: 1,811 Joined: Dec 2013
RE: Type control
(03-13-2015 03:18 PM)salvomic Wrote:
(03-13-2015 03:04 PM)Han Wrote:  Ensure there are no conflicts between the CAS version of myprog and the non-CAS version of myprog. In other words, name them differently to make sure that when you type myprog you aren't running the wrong program (since different programs can actually have the same name). As for the error (w/ respect to the second code block in your post), it is being generated much further inside your program. What you have posted there does not produce any errors.

thank you!
I'll control soon.
However, please, see my ANOVA program (here)
If I use the function ANOVA_list() (that need values in D1, D2, D3 of Statistic 1var) without parameters, in CAS I get "Error: Bad argument value", in Home I get "Error: Syntax error", but with something like ANOVA_list(0.01) the program seems to work...
I've in the code some controls like:
Code:
 local spar; spar:= SIZE(a); IF (TYPE(spar)==6 OR TYPE(a)<>0) THEN RETURN("Usage: ANOVA_list(a) a=significance level; data in Stat 1var");  END; IF ((spar==0)) THEN RETURN("Usage: ANOVA_list(a) a=significance level; data in Stat 1var"); END; IF (a <=0 OR a>= 1) THEN RETURN("Significance level must be > 0 AND <1"); END; IF ((size(D1) == 0) OR (size(D1) == 0) OR (size(D3) = 0)) THEN RETURN("Statistics 1var, D1 means, D2 stddev, D3 numerosity"); END;
I would like to give a message when the user input ANOVA_list() without parameter... and also when it gives wrong parameter (but this seems to work).

You need to use CASE and not separate IF statements. Suppose a is not a number object. Without using CASE, then your program will reach the third IF statement. However, the check IF (a<=0 OR a>=1) would not make sense and this is the cause of your error, since the <= and >= operators may not properly return the correct result to be used by the OR command. For example, try:

{1,2} >= {0,5} and you will see {1,0}

{1,0} OR {0,0} and the result is {1,0}

However, an IF statement does not know what to do with a list -- IF ({1,0}) makes no sense. To avoid this, you need to use CASE blocks surrounding your IF statements so that a list input in captured by only one IF statement and then program control is passed to the point beyond the end of the CASE block (i.e. skips all other IF tests).

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03-13-2015, 06:29 PM (This post was last modified: 03-13-2015 06:30 PM by salvomic.)
Post: #23
 salvomic Senior Member Posts: 1,365 Joined: Jan 2015
RE: Type control
(03-13-2015 06:14 PM)Han Wrote:  You need to use CASE and not separate IF statements. Suppose a is not a number object. Without using CASE, then your program will reach the third IF statement. However, the check IF (a<=0 OR a>=1) would not make sense and this is the cause of your error...

To avoid this, you need to use CASE blocks surrounding your IF statements so that a list input in captured by only one IF statement and then program control is passed to the point beyond the end of the CASE block (i.e. skips all other IF tests).

I should control that a value is from 0 (excluded) and 1 (excluded): (0..1)

CASE. Do you mean like so?
Code:
 CASE IF (TYPE(spar)==6 OR TYPE(a)<>0) THEN RETURN("Usage: ANOVA_list(a) a=significance level; data in Stat 1var");  END; IF ((spar==0)) THEN RETURN("Usage: ANOVA_list(a) a=significance level; data in Stat 1var"); END; IF (a <=0 OR a>= 1) THEN RETURN("Significance level must be > 0 AND <1"); END; IF ((size(D1) == 0) OR (size(D1) == 0) OR (size(D3) = 0)) THEN RETURN("Statistics 1var, D1 means, D2 stddev, D3 numerosity"); END; END; // case

A priority list should be: control if value "a" is given, then if D1, D2 and D3 are set, than if a is from 0 to 1 (non include the limits), then if a is a number and only a number...

thanks!

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03-13-2015, 06:34 PM
Post: #24
 Han Senior Member Posts: 1,811 Joined: Dec 2013
RE: Type control
(03-13-2015 06:29 PM)salvomic Wrote:  A priority list should be: control if value "a" is given, then if D1, D2 and D3 are set, than if a is from 0 to 1 (non include the limits), then if a is a number and only a number...

That doesn't make sense. How can you check if a is from 0 to 1 without first checking to see that a is in fact a number? I didn't read much into your code, except to look for how control flow was giving you errors.

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03-13-2015, 06:44 PM (This post was last modified: 03-13-2015 06:46 PM by salvomic.)
Post: #25
 salvomic Senior Member Posts: 1,365 Joined: Jan 2015
RE: Type control
(03-13-2015 06:34 PM)Han Wrote:  That doesn't make sense. How can you check if a is from 0 to 1 without first checking to see that a is in fact a number? I didn't read much into your code, except to look for how control flow was giving you errors.

in fact I'm trying doing that: control if "a" is a number, but first if it was given (and this only doesn't works):

new code:
Code:
 BEGIN local spar; spar:= SIZE(a); CASE IF ((spar==0)) THEN RETURN("Usage: ..."); END IF (TYPE(spar)==6 OR TYPE(a)<>0) THEN RETURN("Usage: ...");  END // control to avoid lists {} and []... IF (a <=0 OR a>= 1) THEN RETURN("Significance level must be > 0 AND <1"); END // control for interval IF ((size(D1) == 0) OR (size(D1) == 0) OR (size(D3) = 0)) THEN RETURN("Statistics 1var, D1 ..."); END DEFAULT END; // case

only the IF spar==0 doesn't control for no input, IMHO...

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03-13-2015, 10:39 PM
Post: #26
 salvomic Senior Member Posts: 1,365 Joined: Jan 2015
RE: Type control
Han,
so it goes better:

Code:
 EXPORT ANOVA(a) ... BEGIN CASE IF (SIZE(a)==0) THEN a:=0.05; END; IF (TYPE(a)<>0) THEN ANOVA_help(); RETURN("Usage: ANOVA(a)");  END; IF (a <=0 OR a>= 1) THEN RETURN("Significance level must be > 0 AND <1"); END; DEFAULT END; // case ...

I control that "a" be a number (or give the user a function with help), and if it's a number that its value be in the interval.
But SIZE(a)==0 still doesn't work.
I thought to make "a" parameter option, but I don't remember if it's possible in EXPORT myfunction()... or not.
i.e. EXPORT ANOVA(a:=0.05) doesn't work.

Salvo

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03-14-2015, 09:49 AM (This post was last modified: 03-14-2015 09:53 AM by Han.)
Post: #27
 Han Senior Member Posts: 1,811 Joined: Dec 2013
RE: Type control
(03-13-2015 10:39 PM)salvomic Wrote:  Han,
so it goes better:

Code:
 EXPORT ANOVA(a) ... BEGIN CASE IF (SIZE(a)==0) THEN a:=0.05; END; IF (TYPE(a)<>0) THEN ANOVA_help(); RETURN("Usage: ANOVA(a)");  END; IF (a <=0 OR a>= 1) THEN RETURN("Significance level must be > 0 AND <1"); END; DEFAULT END; // case ...

I control that "a" be a number (or give the user a function with help), and if it's a number that its value be in the interval.
But SIZE(a)==0 still doesn't work.
I thought to make "a" parameter option, but I don't remember if it's possible in EXPORT myfunction()... or not.
i.e. EXPORT ANOVA(a:=0.05) doesn't work.

Salvo

It appears I misunderstood you, then. If you want to have a sequence of control flow in which you add more conditions that must be met, then you need to nest your if statements:

Code:
IF (TYPE(a)<>0) THEN     ANOVA_help(); RETURN("Usage: ANOVA(a)");  ELSE     IF (a <=0 OR a>= 1) THEN       RETURN("Significance level must be > 0 AND <1");     ELSE     ... (rest of your code here)     END; END;

There's no point to checking the size of 'a' since multiple arguments in Home view would produce an argument count error (no list behavior like in CAS view) as your program only specifies one argument (which must be given in order for the program to even run).

Non-CAS programs cannot be "overloaded" to handle both cases in which an argument is passed, or when no arguments are passed. You have to specify one or the other. This is why the SIZE(a)==0 check is redundant. Just to be clear, this is regarding the behavior of a non-CAS program.

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03-14-2015, 10:00 AM (This post was last modified: 03-14-2015 10:02 AM by salvomic.)
Post: #28
 salvomic Senior Member Posts: 1,365 Joined: Jan 2015
RE: Type control
(03-14-2015 09:49 AM)Han Wrote:  If you want to have a sequence of control flow in which you add more conditions that must be met, then you need to nest your if statements:
I'll try your solution in the afternoon, thanks
Quote:There's no point to checking the size of 'a' since multiple arguments in Home view would produce an argument count error (no list behavior like in CAS view) as your program only specifies one argument (which must be given in order for the program to even run).

Non-CAS programs cannot be "overloaded" to handle both cases in which an argument is passed, or when no arguments are passed. You have to specify one or the other. This is why the SIZE(a)==0 check is redundant. Just to be clear, this is regarding the behavior of a non-CAS program.

In fact, my problem was here: I would to "overload" and to handle (as I do in CAS programs) is a parameter is passed, and (if yes) whether it's only one, it's a number and so on...
Ok, if there is no solution, my controls (with your advice) will be "better than nothing"

EDIT: a trick could be give the parameter a default value, but...

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# zbMATH — the first resource for mathematics

Holomorphic dynamical systems on manifolds. (Sistemas dinamicos holomorfos en superficies.) (Spanish) Zbl 0855.58049
Aiming at the description of global properties of solutions of holomorphic dynamical systems on compact surfaces, this paper follows ideas of Il’yashenko’s seminar [Yu. S. Il’yashenko, Sel. Math. Sov. 5, 141-199 (1986; Zbl 0585.34003); translation from Tr. Semin. Im. I. G. Petrovskogo 4, 83-136 (1978; Zbl 0418.34007)] devoted to the study of the ergodic and topological properties of polynomial differential equations in two complex variables. By the authors’ specification, the contribution of the present paper is to incorporate to Il’yashenko’s ideas the intrinsic point of view of modern algebraic geometry. So, the authors prove that the ergodic and topological properties of polynomial differential equations on the projective complex tangent space to a line described by Il’yashenko hold also for the holomorphic differential equation tangent to an ample curve in a projective surface.
In the first chapter, the relation between ordinary differential equations and holomorphic foliations by curves is given: the solutions of an holomorphic ordinary differential equation define a foliation by curves and, on the other hand, for a foliation by curves, a class of differential equations the solutions of which form the given foliation is defined. By using this local correspondence, an analytical global description of the foliations by curves on a connected complex manifold $$M$$ is given. Namely, such a foliation (possibly with singularities) is described as a vector bundle morphism $$\alpha: L\to TM$$ which is not identically zero on any connected component of $$M$$, $$L$$ being a linear complex bundle over $$M$$. By this description, the concepts of the family of foliation by curves and of universal family follow in a natural manner. These notions are concretized in the case of foliations by curves in the projective spaces $$\mathbb{C} P^n$$ which can be described by homogeneous polynomial fields on $$\mathbb{C}^{n+1}$$. In the last part of this chapter, the linearization of Poincaré’s theorem is proved by using a homological method.
The second chapter presents the dynamical concepts: the transverse dynamics of a foliation by curves on a complex manifold $$M$$, without singularities, and holonomy groups.
Il’yashenko’s results related to the density, ergodicity and rigidity of the action of the group of germs of conformal transformations on $$\mathbb{C}$$ are presented in chapters III and IV. The authors give here a new proof of the ergodicity theorem by using Koebe’s theorem. Il’yashenko’s rigidity theorem for foliations by curves tangent to a line in $$\mathbb{C} P^2$$ is proved in chapter V. Remarkable is an example of a foliation with non-abelian holonomy.
The extension of Il’yashenko’s results to the case of foliations tangent to ample curves in projective surfaces are given in the last chapter of the book as an application of the ideas from the previous chapters.
The authors were preoccupied by a clear and rigorous exposition of the results. We consider the work as an excellent reference to the problem of holomorphic dynamical systems on projective surfaces.

##### MSC:
 37F99 Dynamical systems over complex numbers 37-02 Research exposition (monographs, survey articles) pertaining to dynamical systems and ergodic theory 37C85 Dynamics induced by group actions other than $$\mathbb{Z}$$ and $$\mathbb{R}$$, and $$\mathbb{C}$$ 34M99 Ordinary differential equations in the complex domain 34D20 Stability of solutions to ordinary differential equations 37A99 Ergodic theory 54C70 Entropy in general topology


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# Prove a chord of a parabola passes through a fixed point

I am doing the end of chapter questions on parabolas in a pure maths book out of interest. I am struggling with this:

A fixed point $$P(ap^2,2ap)$$ is taken on the parabola $$y^2=4ax$$. Two points Q and R are chosen on the parabola such that the lines PQ and PR are perpendicular. Prove that the line QR passes through a fixed point, F, independent of Q and R, and that PF is normal to the parabola at P.

I started by working out the slopes of PQ,QR and PR:

$$(\frac{2}{p+q},\frac{2}{r+q},\frac{-(p+q)}{2})$$ and then I worked out the equation of QR, which I found to be:

$$y-2aq=\frac{2}{r+q}(x-aq^2) \rightarrow y=\frac{2x+2aqr}{r+q}$$

The question wants an answer independent of Q and R so I reasoned I need to find an expression in p only. I did find such an expression but even then I couldn't find where the fixed point should be. But even then, the question only gives the fixed point F after we are supposed to have found that such a point exists.

• That $PF$ is the normal at $P$ can be proved independently: in the limit $Q\to P$ line $PQ$ is tangent and chord $QR=PR$ is by construction perpendicular to it. Aug 20 at 19:32

Point $$P$$ on the parabola is $$(ap^2, 2ap)$$. Say, coordinates of $$Q$$ and $$R$$ are $$(aq^2, 2aq)$$ and $$(ar^2, 2ar)$$.

Equation of line passing through $$QR$$,

$$y - 2aq = \cfrac{2aq - 2ar}{aq^2-ar^2} (x - a q^2)$$

Simplifying,

$$(q + r) y = 2x + 2 a q r \tag1$$

As $$PQ$$ and $$PR$$ are perpendicular,

$$\cfrac{2ap - 2aq}{ap^2 - aq^2} \times \cfrac{2ap - 2ar}{ap^2 - ar^2} = - 1$$

$$\implies \cfrac{4}{(p+q)(p+r)} = - 1$$

Simplifying, $$p^2 + (q+r) p + 4 + q r = 0$$

Multiply by $$- 2a$$ and rearrange to bring it in the same form as $$(1)$$.

$$- 2 a p (q+r) = 2 a (p^2 + 4) + 2 a q r \tag2$$

From $$(1)$$ and $$(2)$$, we can see that $$x = a (p^2 + 4), y = - 2 a p$$ satisfies the equation of the line $$QR$$. So, regardless of the value of $$q$$ and $$r$$, point $$F \big(a (p^2 + 4), - 2 a p \big)$$ lies on $$QR$$.

Equation of normal through point $$P$$ is,

$$y - 2 ap = - p (x - ap^2)$$

and you can check that $$F$$ lies on the normal through $$P$$.

• For anyone who's interested, I worked on a Desmos graph when I tried to solve this question a while ago. Thanks to MathLover's answer and something I completely missed, the graph is now complete, and you can visualise the fixed point if you wish (playing with the graph at first I was convinced a fixed point didn't exist, but it all checks out): here Aug 20 at 18:04
• @FShrike thank you. that's a good interactive graph. On the contrary, when the question said that the fixed point lied on the normal, I had an intuition it would be circumcenter of the right angled triangle $\triangle PQR$ but it became difficult to use that information in the working. I finally had to abandon that approach :) Aug 20 at 18:20
• and it took me a while to see that my intuition was wrong. Aug 20 at 18:34
• You're welcome. I found the quadratic equation right off, and noticed that it looked very similar to the line equation, but missed the multiplication entirely. Good job Aug 20 at 18:43
• It was a very good job indeed. I too explored the fact that PQ is the diameter of a circle but got nowhere. I certainly would never had done it without the help of MathLover. Aug 20 at 19:33


text:
# [−]Struct lumol::sys::compute::StressAtTemperature

pub struct StressAtTemperature {
pub temperature: f64,
}

Compute the stress tensor of the system from the virial definition, at the given temperature.

$$\underline{\sigma} = \frac{1}{V} \left(\frac{N_f}{3} k_B T \space \underline{1} + \underline{W} \right)$$

where $N_f$ is the number of degrees of freedom in the system, $k_B$ is the Boltzman constant, $T$ the temperature, $V$ the simulation volume, $Tr$ is the matricial trace, and $\underline{W}$ the Virial.

## Fields

temperature: f64

Temperature for the stress tensor computation

## Trait Implementations

### impl Compute for StressAtTemperature

#### type Output = Matrix3

The data type of the property

## Blanket Implementations

### impl<T> Pointable for T

#### type Init = T

The type for initializers.

### impl<T, U> TryFrom<U> for T where    U: Into<T>, [src]

#### type Error = Infallible

The type returned in the event of a conversion error.

### impl<T, U> TryInto<U> for T where    U: TryFrom<T>, [src]

#### type Error = <U as TryFrom<T>>::Error

The type returned in the event of a conversion error.


text:
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# cumulative error distribution Pineola, North Carolina

Uncover the mysteries of the marks... Asymptotic expansion A useful asymptotic expansion of the complementary error function (and therefore also of the error function) for large real x is erfc  ( x ) = e − For the first system, the reported location 2.3 meters or less away from the true location 90% of the time. The denominator terms are sequence A007680 in the OEIS.

What mistaken pronunciation gave this character its name? Generated Thu, 06 Oct 2016 01:46:26 GMT by s_hv997 (squid/3.5.20) Know These 9 Commonly Confused... If x, A, B, C, and D are arrays, they must be the same size.

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The error function is a special case of the Mittag-Leffler function, and can also be expressed as a confluent hypergeometric function (Kummer's function): erf  ( x ) = 2 x It is defined as:[1][2] erf  ( x ) = 1 π ∫ − x x e − t 2 d t = 2 π ∫ 0 x e − t A and B contain the parameter values for the distribution. y = cdf('name',x,A,B,C) returns the cdf for the three-parameter distribution family specified by 'name', evaluated at the values in Help on a Putnam Problem from the 90s What instruction on the STM32 consumes the least amount of power?

If any of A, B , C, or D are scalars, then cdf expands them into constant matrices the same size as x. When the error function is evaluated for arbitrary complex arguments z, the resulting complex error function is usually discussed in scaled form as the Faddeeva function: w ( z ) = As explained in the paragraph above, just summing each individual error would not suffice because if 11 elements had 0 error but 1 had one that is just under the threshold The second phase relies on the predicted geology and drills to an expert based pre-specified standoff distance from the geological boundary of interest, to optimize blasting and minimize its damage.

Error function From Wikipedia, the free encyclopedia Jump to: navigation, search Plot of the error function In mathematics, the error function (also called the Gauss error function) is a special function ISBN 978-0-486-61272-0. ISBN0-486-61272-4. You can also select a location from the following list: Americas Canada (English) United States (English) Europe Belgium (English) Denmark (English) Deutschland (Deutsch) España (Español) Finland (English) France (Français) Ireland (English)

In practice, CDF is described by the percentile format. What does Billy Beane mean by "Yankees are paying half your salary"? In this study two different techniques, multiple regression and adaptive neuro-fuzzy inference system (ANFIS) were used to develop the models for preliminary estimation of rock mass strength. The error and complementary error functions occur, for example, in solutions of the heat equation when boundary conditions are given by the Heaviside step function.

Despite the name "imaginary error function", erfi  ( x ) {\displaystyle \operatorname  7 (x)} is real when x is real. Data Types: single | doubleB -- Second probability distribution parameterscalar value | array of scalar values Second probability distribution parameter, specified as a scalar value, or an array of scalar values. Math. All generalised error functions for n>0 look similar on the positive x side of the graph.

The ideal would be to assign a pair to each child so the shoes fit but if that's not possible, it's okay if a couple of children are off by half normal distribution error function There is no closed expression for CDF. (This you can calculate using statistical tables, calculator, etc) In the long run, we're all dead. See [2]. ^ http://hackage.haskell.org/package/erf ^ Commons Math: The Apache Commons Mathematics Library ^ a b c Cody, William J. (1969). "Rational Chebyshev Approximations for the Error Function" (PDF). By a7890 in forum Statistics Replies: 1 Last Post: 03-18-2011, 05:18 PM percentile for (1 - cumulative distribution function) By athos in forum Probability Replies: 4 Last Post: 01-04-2011, 09:09 AM

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• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeOct 19th 2010

I split off ∞-connected (∞,1)-topos from locally ∞-connected (∞,1)-topos and added a proof that a locally ∞-connected (∞,1)-topos is ∞-connected iff the left adjoint $\Pi$ preserves the terminal object, just as in the 1-categorical case. I also added a related remark to shape of an (∞,1)-topos saying that when H is locally ∞-connected, its shape is represented by $\Pi(*)$.

I hope that these are correct, but it would be helpful if someone with a little more $\infty$-categorical confidence could make sure I’m not assuming something that doesn’t carry over from the 1-categorical world.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeOct 20th 2010
• (edited Oct 21st 2010)

Thanks, Mike!

This does look correct. For the single possibly nonevident point– that every $\infty$-groupoid is the $(\infty,1)$-colimit over itself of the diagram constant on the point – I have added a link to this proposition.

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeOct 21st 2010

Thanks.

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text:
GATE CSE 2021 Set 2 | Question: 44

2.1k views

If the numerical value of a $2$-byte unsigned integer on a little endian computer is $255$ more than that on a big endian computer, which of the following choices represent(s) the unsigned integer on a little endian computer?

1. $0\text{x}6665$
2. $0\text{x} 0001$
3. $0\text{x} 4243$
4. $0\text{x} 0100$

recategorized ago
3
Option A and D

$0x6665 – 0x6566 = FF = (255)_{10}$

$0x0100 – 0001 = FF = (255)_{10}$
3
why are you assuming that given thing is as per little endian, question just talks about an unsigned representation..

if i interpret in this way, that the given is unsigned representation over a little endian machine, then i can even say for 0x66 65, the little endian storage would be like this way..

<in sequential locations we start by storing lsb first ...policy of little endian>

so this evaluates to 0x6566 on a little endian macinge, and 0x 66 65 over  a big endian machine, now little end- big end= 6566-6665 < 0 and not equal to 255 hence 0x6665 is wrong answer...similarly doing for rest gives, opt B,C as the only correct answers.

if you find anything wrong in this interpretation then tell me...bcoz i feel this way also one can interpret...and if there are multiple interpretation then question is ambiguous..

@Arjun Sir, @zxy123
0
No you’re right, but we have to wait for the answer key to be sure though.
3
A and D is correct.
0
How many marks was alloted for this question?
0
@ACCURACY by default it is little endian

This questions is poorly framed and has interpretation ambiguity. Refer the discussion on this question in below link :

https://cs.stackexchange.com/questions/135713/representation-of-unsigned-integer-on-a-little-endian-big-endian-computer

All kind of “interpretations” are available in that discussion.

The following is my interpretation from the question :

He is asking “which of the following choices represent(s) the unsigned integer on a little endian computer?

Take Option “0x6665” :

He is saying that 0x6665 is the representation of integer on a little endian computer, so, it means that the original number must have been 0x6566.

So, for the original number 0x6566 :

On little endian(LE) : 0x6665

On Big endian(BE) : 0x6566

Clearly, LE = 255 + BE

Similarly, for 0x0100.

Take 0x0100 :

He is saying that 0x0100 is the representation of integer on a little endian computer, so, it means that the original number must have been 0x0001.

So, for the number 0x0001 :

On little endian(LE) : 0x0100

On Big endian(BE) : 0x0001

Clearly, LE = 255 + BE

Similarly for 0x4243 and 0x0001, They do not satisfy “LE = 255 + BE”, So, answer is option A,D.

Refer Slide 26 in the below article :

Nice Reference : https://www.cs.utexas.edu/~byoung/cs429/slides2-bits-bytes.pdf

NOTE that different people are having different interpretation of this question. I have asked this question on cs.stackexchange, and you can read the discussion in below link :

https://cs.stackexchange.com/questions/135713/representation-of-unsigned-integer-on-a-little-endian-big-endian-computer

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4
Perfect Explanation
0
sir i have one doubt (maybe silly).

Whatever be the representation style, will the computer not know what the number really is. So there will be no difference actually. But obviously seeing the options we can interpret what they really meant to ask.

Am i right??
0
Why are we assuming “ab” is a byte stored at lower address while “cd” is stored at higher address in 0xabcd? Isn’t it possible to have “ab” at higher address and “cd” at lower address?
2

This questions is poorly framed and has interpretation ambiguity. Refer the discussion on this question in below link :

https://cs.stackexchange.com/questions/135713/representation-of-unsigned-integer-on-a-little-endian-big-endian-computer

All kind of “interpretations” are available in that discussion.

2
Sir açcording to official answer key A,D are correct.. i marked B,C :(
3
It will be challenged. We will have to see final key for this. I think it is unlikely they will change it, But let’s see.
0
Did they change?

If not then what is the solution,

If yes, then ok.
0
Final answer of GATE was option A,D only. They did not change it.
Ans: B and C.
0x0001 (binary: 0000 0000    0000 0001 ),as the value is stored in little endian would amount to 2^8*1 which is 256, however the same sequence of bits stored in big endian would amount to 2^0*1 =1 , so the liitle endian rep-BE rep=256-1=255
0x4243, same calculations as above(trick: just swap the 8 bits and look at the data bit’s that are not matching ) in this case too the difference between the little endian and big endian representation happens to be 256-1=255 hence both B and C are correct, the value given by option A and D is -255 as for those representations the little endian has a lower value than big endian system,
PS: In a big endian system the leftmost byte (sequence of 8 bits) are the most significant bytes [this is similar to how we generally solve binary numbers], however in a little endian system the sequence of bytes to the extreme left are least signigicant , it can be termed as mirror image of bytes as present in a big endian system

$0\text{x}0001-$ On big endian computer (like humans read) this will read as $1$ and on little endian computer (lower byte comes at left most) this will read as $0\text{x}0100=256$ giving a difference of $255.$

Similarly, $0\text{x}4243$ also gives a difference of $255.$

Since, there is confusion regarding the meaning of the question, lets dissect the question.

• If the numerical value of a 2-byte unsigned integer on a little endian computer is $255$ more than that on a big endian computer, which of the following choices represent(s) the unsigned integer on a little endian computer?

In the above question, the bold portion refers to the same thing. We have a $2-$ byte integer (say $X$ and this is represented on a little endian computer and this is now given in the options. So, lets consider option B for example. It is $0\text{x}0001$ which is the little endian representation. Its numerical is $00 + 0\text{x} 01 \times 256 = 256.$ That is, value of $X = 256$ on a little endian machine. Same byte sequence if read on a big endian machine will give $00 \times 256 + 01 = 1.$  That’s a difference of $256-1 = 255.$

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1 vote

The way the question is framed, I do not agree with the concept at all.

A number is after all a number, be it big endian or little endian.

The main concept of Big endian or little endian is as follows:

Big Endian: In computer system which is big endian, the lower order byte of the data in the memory maps to the higher (big) order byte of the registers in CPU. (see figure below)

Little Endian: In computer system which is little endian, the lower order byte of the data in the memory maps to the lower (little) order byte of the registers in CPU. (see figure below)

As such, the whatever be the system type (Big or little endian) the number should actually be the same in the registers in the CPU, as the ALU knows one and only one representation of a number which is binary equivalent of a number.

The difference arises in the way the number is STORED IN THE MAIN MEMORY.

This makes the things a bit ambiguous.

Assuming they are taking about the representation of the number in main memory.

i.e. $\text{Memory[1]Memory[0]}$ [A complete word] (see figure below)

This ambiguity can be resolved by the way the question is asked.

From the above concept we can see that: what ever be byte sequence we assume for LITTLE ENDIAN, that byte sequence shall be reversed for BIG ENDIAN.

Option A:  $$0x6665–0x6566=FF=(255)_{10}$$

Option D:  $$0x0100–0x0001=FF=(255)_{10}$$

In other cases, (option B or C, assuming the given option and reversing the byte sequence gives a number greater than the option, as per the question the little endian value is greater)

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