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Guide to Buy Solid Wood Interior Doors in Digah House Company

Guide to Buy Solid Wood Interior Doors in Digah House Company

2021-09-16
Digah Company
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What Is the Main Function of the Dining Room to Eat Desk and Chair?
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A legendary “food” location transformed into a new specialty restaurant at the three star Tulip Inn, offers fresh interpretations of classical Indian flavours. Aromas from the fresh tandoor envelope us as we approach the stretch. The open kitchen sporting shiny kebab skewers and the ultra modern décor at the almost empty restaurant looks unbelievably new! Decor The rich, red and gold brocade menu advertizes the traditional “Bawarchi khana” from the kitchens of Oudh. There’s a small problem with that promise. The beaten gold faux leather furniture, white tables with runners, dark chocolate and gold-etched glass backdrop and an over powering canopy of wrought iron filigree lighting instantly runs a flashback of the previously existing tired dining room that buzzed at the same space as Kolkata’s busiest biryani and chaanp place. Food As a first-time diner, it is a pleasure to find a simple menu prepared with great perseverance, it steered us to a slice of Lucknowi culture. Curries, tandoori and unusual vegetarian dishes dominate the options. We settle for Paya Shorba ( 150), Mawa Bhara Khumb ( 200), Galouti Kebab served with Ulta Paratha ( 300) in the first course. For second, we chose the obvious Oudh Gosht Biryani ( 300), Khusk Aloo ( 175) and Mehr-e-Aftaab ( 300), stuffed minced balls in brown sauce, very similar to the Kashmiri Rishta. Plus & Minus The flavours are great — subtle and not overpowering — and the service is efficient and fast, considering the cuisine. The delicate lamb shorba sets the mood — perfect in temperature and a mild dash of lemon makes it a fabulous blend. The char grilled button mushrooms comes with a freshness of cheese, khoya and onion stuffing, delightful for the palate. The real Oudhi jewel, melt in the mouth Galouti, comes with ample spicing and brushed in ghee. A faultless combination with the sweet touch of the Ulta Parathas. For grilled meats, you can try the Kebab Tashtari ( 475), a day’s special kebab platter, which we skipped to try the mains. The Biryani arrives in a generous portion. Though, the authenticity is a question — more “Kolkata” than “Oudh” in its manifestation, with the unique Kolkata tradition of a potato along with the well-cooked pieces of lambs! It is light and tasty in its mildness, without the strong essence. The mince balls with Mawa stuffing are dipped in a rich tomato and onion gravy, but are dry inside. The Khusk Alu fails to impress, no baby potatoes, it comes in a gravy with curry leaves and mustard tempering. We polished off a Malai Kulfi ( 125), sans the faluda and a Pista Phirnee ( 100) at the end. Daawat-E-Shiraz: Tulip Inn Kolkata , 56 Park Street, Kol: 700017 Meal for two: 1,000 Timings: Noon to 4 pm, 7 pm to 11.30 pm Alcohol: No Ratings Food: 3 Service: 3.5 Decor: 3
Australia Considers Stopping Purchasing 5g Equipment From China
Australia Considers Stopping Purchasing 5g Equipment From China
According to CNET, Australia will no longer allow Huawei to provide equipment for the country's upcoming 5g network deployment for "security reasons". The ban was first disclosed by Reuters, and neither Huawei nor the Australian government immediately responded to requests for comment. Last month, Huawei responded to concerns about an open letter that said the criticism was "not based on facts" because it claimed that the products sold to Australia were "high quality and safe".In 2012, Huawei was also unable to participate in Australia's $38 billion national broadband network project.Us congressmen have regarded Huawei as a target for many years, and tensions between Washington and Chinese companies escalated in 2018. In February this year, the directors of the Central Intelligence Agency, the FBI and the national security agency testified in the U.S. Congress to warn consumers not to buy Huawei mobile phones.In May, the Pentagon banned the sale of Huawei and ZTE mobile phones in U.S. military bases around the world. In June this year, the U.S. government wrote to Google CEO Sanda pice that the cooperative relationship between Google and Huawei in instant messaging posed a serious threat to U.S. national security and consumers.Both Huawei and ZTE have repeatedly insisted that their consumer devices will not pose a security threat to the United States.Although Huawei is struggling in the US market, it has achieved great success in the world. According to IDC's data, the company was the fourth largest smartphone seller in the world in 2017. According to a report by statista, in the first quarter of 2018, the company's global smartphone market share was 11.8%, a new high.
Modelling Lunar Motion in Python Closed
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Well, just to be clear: I haven't looked at your code, I'm not responding to that. I'm submitting in the answer section because I have a general comment.The treatment by Richard Fitzpatrick that you link to is an exercise in how to go about obtaining an analytic solution for the case of the the Sun-Earth-Moon three body problem. In the general case of three bodies with mutual interactions there is no exhaustive solution, but in some cases some well chosen approximations can make it feasible to do the computations with pen and paper. As we know: for centuries this is what astronomers did in order to do astronomy: you find ways to bring the calculations within reach of being handled with pen and paper.These approaches are not used when electronic computers are used to calculate celestial motion. On an electronic computer you use Newton's law of gravity in its direct form: $$ F fracGm_1 m_2r^2 $$(Wich in the case of the Sun, Earth, Moon system of course means that you process three interactions: Sun-Earth, Sun-Moon, Earth-Moon.)To my knowledge that is how on electronic computers all simulations of celestial motion are done.My suggestion: use cartesian coordinates, and use a non-rotating coordinate system. If you also want to plot the motion with respect to some rotating coordinate system then you can apply that coordinate transformation to the result of the numerical integration. (Again, I suppose that in past centuries in order to bring the calculations within reach of pen and paper it was in some circumstances efficient to express the equation of motion and the intended coordinate transformation in a combined formula, and proceed from there.) Let me add some general considerations.Trade-offsWith todays computing power: for the purpose of processing a three body case your computational power is effectively limitless. Only when you want to to model, say, an entire galaxy you need a supercomputer.In the treatment by Fitzpatrick there are approximations, he mentions neglecting higher order terms. Those approximations introduce error. Negligable in the short run, but over time you get accumulation of error.If the only expression you evaluate is Newton's law in bare form then there is no approximation in the formula's themselves. Then only one source of error remains, the precision level of the numerical integration. You have a trade-off there. Higher precision level may limit you to, say, computing only a couple of thousand years of Moon motion. Lower precision level may allow you to compute millions of years, but probably by that time accumulation of error has made the end state meaningless.(Also, over time the distance between the Earth and the Moon is increasing. There are tidal effects that result in the Earth slowing down and the Moon being pulled to a higher orbit. So a long term model would have include that gravitational interaction.)So it's good to be aware of the trade-offs that apply in your case, and to try and make sensible decisions.One important trade-off, I think: the simpler the expressions that the numerical integration evaluates, the more debuggable the simulation is. Newton's law of gravity is a simple, transparent expression.Displaying the resultsAgain, since your computing power is effectively limitless you can display the results of the computation in any way you want. For instance, you can choose to set up a side by side display of the motion with respect to some particular inertial coordinate system, and next to it the motion with respect to some particular rotating coordinate system. The required coordinate transformation for that is trivial.That is, given your limitless computing power the decision on how to display the results is independent from the decision of how to execute the numerical integration itself.There is no such thing as: "I want to visualize the results as motion with respect to this rotating coordinate system so I must do all the computation in that rotating coordinate system." Obviously the two are independent.• Related QuestionsQuantum version of the Galton BoardYou might like this 110-page paper by me and Alex Arkhipov, which is all about a quantum bosonic analogue of Galton's board (we even use the same graphic you did -- see Section 1.1!). In particular, we gave strong evidence that such a board (with an arbitrary configuration of "pegs," and with multiple entry points for the "balls") is exponentially hard even to simulate using a classical computer. This suggests that such a quantum Galton's board (which is now called a "BosonSampler") could be used as a rudimentary, proof-of-principle quantum computer. And indeed, within the last year the first BosonSampling experiments were done in linear optics (see here), although so far only with 3 photons.To make our argument for computational hardness work, we needed two crucial assumptions:(1) The "balls" have to be indistinguishable particles. If they're distinguishable, then the distribution for any one individual ball could still exhibit interference fringes. But once you knew the probability distribution for one ball, the distribution for n balls would just be obtained by sampling from that distribution n times independently -- thereby producing "conventional, classical" Law of Large Numbers behavior. By contrast, identical quantum particles can famously become "correlated" even if they've never explicitly interacted, as seen for example in the Hong-Ou-Mandel dip.(2) The "balls" have to be bosons. In that case, their transition amplitudes are given by nxn matrix permanents, the calculation of which is a famous hard problem in computer science. By contrast, if the balls are fermions, then their transition amplitudes are given by nxn determinants, which are easy to calculate classically.Of course, there's also a "narrower" way to interpret your question, which might be closer to what you were actually asking about! Namely, rather than an "arbitrary" Galton-like board, we could consider the specific geometry from the figure: so let's say, a network of 50/50 interferometers arranged in a diamond pattern in the plane, with a single source of particles at the top. And we could then calculate (or, if we're lazier, numerically simulate...) the particular probability distribution over n-particle outcomes that that configuration leads to, under two different assumptions:(1) That the particles are distinguishable. (In this case, of course, the problem reduces to working out the distribution for a single particle.)(2) That the particles are indistinguishable bosons.(Note that a third case, that the particles are indistinguishable fermions, never arises -- for by the Pauli exclusion principle, n identical fermions couldn't even "fit" simultaneously through the single source at the top.)If I have some time later, I might work out the answers and post them here -- but in the meantime, anyone else should feel free to do so first.Addendum: OK, so let's consider the case of a single quantum particle passing through a "diamond-shaped" network of 50/50 interferometers. In that case, the probability distribution after n steps will be determined, not by the nth row of Pascal's triangle (as in the classical case), but by the nth row of what we might call the "interferometric Pascal's triangle." The latter is defined as follows: let A(i,j) be the jth entry in row i. Then:A(0,0)1A(0,j)0 for all j0For ij positive and odd: A(i,j)A(i-1,j)A(i-1,j1)For ij positive and even: A(i,j)A(i-1,j-1)-A(i-1,j)I'm almost certain that the result will asymptotically approximate the standard behavior for the "quantum random walk on the line": see here or here for good overviews. In particular, the distribution will not look anything like the Gaussian one: instead it should be nearly uniform, except with a bunch of oscillating peaks near the two edges, with the size of the peaks getting damped as you move closer to the center. (See the linked papers for example images.)A slight caveat is that usual analyses of quantum random walks assume there's a "coin" (i.e., a spin-1/2 internal degree of freedom), whereas I've used a staggered arrangement of interferometers to remove the need for the coin. I don't think that affects the walk's qualitative behavior, but I don't have a proof
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