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# Topology on $p$-adic period ring in an article by Fontaine

Fix a $$p$$-adic field $$K$$ with perfect residue field $$k.$$ Let $$\mathcal{C}$$ be the completion of the algebraic closure of $$K,$$ and let $$R = \varprojlim \mathcal{C}/p,$$ where the transition maps in the inverse limit are given by the Frobenius. One can then show that $$R$$ is a perfect valuation ring (of characteristic $$p.$$) We can then form $$W(R),$$ the ring of Witt vectors of $$R.$$

There is a natural topology on $$W(R),$$ making it into a topological ring, a basis of neighborhoods is given by $$p^N W(R)+ W(I)$$ for $$N\geq 0$$ and $$I$$ a non-zero ideal of $$R.$$

In this article, pg. 536, Fontaine uses the topology I just defined to define a topology on $$W(R)[1/p] = K \otimes_{W(k)} W(R),$$ by using what he calls the "tensor product topology". He then claims that the topology one get from the tensor product topology is the "same" (up to identifications under isomorphisms) as the one obtained by taking the topology coming from the inductive limit $$\cdots \rightarrow W(R) \rightarrow W(R) \rightarrow \cdots$$ where transfer maps are multiplication by $$p.$$

My questions are the following:
1) What is precisely this tensor product topology? Naively, I would say that it is the topology on $$K \otimes_{W(k)} W(R) \cong W(R)[1/p]$$ where a basis of neighborhoods are given by $$(p^N W(R)+ W(I)) \otimes_{W(k)} K + W(R) \otimes_{W(k)} p^n \mathcal{O}_K$$ (where $$\mathcal {O}_K$$ is the valuation ring of $$K).$$ What makes me think this can not be a basis of neighborhoods comes from the fact that it seems to me (maybe erroneously) that $$p^NW(R) \otimes_{W(k)} K \cong W(R)[1/p].$$ Thus, it seems to me that Fontaine must have some other sort of topology in mind for this tensor product, or I am making a silly mistake. For example, what is a basis of neighborhoods for the tensor product topology? Is it part of a more general construction?
2. Why does the topology of the tensor product and inductive limit coincide?

Inductive limit topologies are always nasty to describe: Basically you can't do better than their very definition. Namely, a subset $$U\subset W(R)[1/p]$$ is an open neighborhood of $$0$$ if and only if for all $$n$$ the intersection $$U\cap p^{-n} W(R)$$ is open. In particular, this means that it must contain $$p^N W(R)$$ for some $$N$$, but also $$p^{-n} W(\mathfrak a_n)$$ for some open ideal $$\mathfrak a_n\subset R$$. So a basis of open neighborhoods of $$0$$ is given by the subsets

$$p^N W(R) + \sum_{n\geq 0} p^{-n} W(\mathfrak a_n)$$

for varying $$N\geq 0$$ and open ideals $$\mathfrak a_n\subset R$$, $$n\geq 0$$. Note that such subsets are never contained in $$p^{-n} W(R)$$ for any $$n$$...

As Dustin notes in the other question, passing to the condensed world is helpful here, as then inductive limits are completely naive (they are just inductive limits on $$S$$-valued points for any profinite $$S$$), and in fact there is no question about which condensed structure to put on $$W(R)$$ or $$W(R)[1/p]$$, and it is clear that $$W(R)\otimes_{W(k)} K = W(R)[1/p]$$ as condensed rings. (Also, as everything lies in the essential image of the fully faithful functor from compactly generated topological spaces to condensed sets, there is no loss of information in passing to the condensed world.)

• Dear Peter, Thank you for this fascinating reply! Just three questions (maybe I should post these as separate question?) : 1. Can one construct all the period rings, such as B_{dr}, B_{st}, using the condensed perspective? – Dedalus Jun 27 at 7:22
• 2. In Fontaine’s theory you have the notion of a representation being cristalline / semi-stable / de Rham. Basically, if I recall correctly, a representation V of the Galois group G_k of a p-adic number field is semi-stable if the natural map (B_{st} \otimes_{Q_p} V)^{G_k} \to B_{st} \otimes_{Q_p} V is an isomorphism. If one just follows one’s nose, I would say that this should translate directly to the condensed world. Is this true? Namely, one embeds G_k as a condensed group, and one just does all the constructions there. – Dedalus Jun 27 at 7:23
• 3. Is there a source for this more modern perspective, or is it all forthcoming? I would love to be able to ignore topology :) – Dedalus Jun 27 at 7:23
• You're welcome! 1) Yes; just take their algebraic definition and do it internally in condensed rings. 2) Yes, everything translates directly. 3) I don't think anything has to be written, as nothing deep happens. I think you should be able to translate yourself, maybe referring to the lecture notes on my webpage for the basics. (But note that in my first p-adic Hodge theory paper, in hindsight I'm already discussing period rings from the condensed perspective, so maybe that's a little helpful.) – Peter Scholze Jun 27 at 19:17

What you are missing is that it is using the topology of K as a W(k) module (not as a topological field), so basis for K is like $$p^{-n}W(k)$$ for $$n\in\mathbb{Z}$$. So the basis of W(R)[1/p] is rather consisted of things like $$p^{-n}$$-scaled $$p^{N}W(R)+W(I)$$. Topology of inductive limit coincides with tensor product because inductive limit and tensor product commute and the inductive limit mentioned above is just W(R) tensored with the inductive limit W(k)->W(k)->... which is precisely expressing the basis of neighborhoods of 0 of K we chose, $$p^{-n}W(k)$$.

• Why is $K$ a free $W(k)$-module? If $K = \mathbb{Q}_p,$ is this really true? – Dedalus Sep 5 '19 at 11:29
• Also, maybe I am being dense, but even if say the basis is like $p^{-n}W(k),$ does this really effect the argument that says that $(p^N W(R) +W(I)) \otimes_{W(k)} K \cong W(R)[1/p]?$ This seems to be only a statement about the topology on $W(R).$ – Dedalus Sep 5 '19 at 11:54
• (The argument for $(p^NW(R) +W(I)) \otimes_{W(k)} K \cong W(R)[1/p]$ being that $K$ contains $p^{-N},$ so its imag in $W(R)[1/p]$ must be everything. ) – Dedalus Sep 5 '19 at 11:56
• (Small correction to the previous comment. Of course the argument for the isomorphism should be that $K$ contains $p^{-n}$ for all $n \geq 0.$ – Dedalus Sep 5 '19 at 12:17
• @Dedalus I put that word without much care. Thanks – GTA Sep 5 '19 at 14:34

I think a good reference is Schneider's book Nonarchimedean Funtional Analysis, but my answer is simply an expanded version of GTA's comments.

In Chapter IV, §14, A and B you find a description of two possible "tensor product topologies" with which you can endow the tensor product of two locally convex vector spaces: in your setting, $$W(R)$$ is only a $$W(k)$$-module, but this makes almost no difference if you follow the proofs. The projective limit topology has as basis of neighborhoods the products $$p^{-n}W(k)\otimes p^NW(R)\otimes W(I)$$ by definition, because $$K$$ has, as a basis of neighborhoods, the $$p^{-n}W(k)$$. That the two coincide is Proposition 17.6, which holds in your case because your spaces are complete (for $$W(R)$$ this is again on page 536 of the paper by Fontaine); you need to use the definition of "locally convex final topology" given in Schneider's book, Chapter I, § 5, E to realise that the inductive limit topology given by multiplication by $$p$$ is the inductive tensor product topology.

• Thank you for the reference. I posted a follow-up mathoverflow.net/questions/339967/… where I tried to compare two topologies which should be ”the same”, but I am not sure what to make of this. If you had the time, I would be most grateful if you looked into it since this has been bugging me for some time. – Dedalus Sep 6 '19 at 10:12
• Also, do you mean $p^{-n}W(k),$ or do you mean $W(k)[1/p^n]$ above? If the first, I do not see how the $U_{N, \mathfrak{a}}$ that Brinon and Conrad (see the link in the question I linked to) define can be contained in a product of neighborhoods as above... – Dedalus Sep 6 '19 at 10:34
• I have corrected the $[1/p^n]$ into $p^{-n}$, you are right. As for your other question, I've seen it but for the time being I don't really have an answer. – Filippo Alberto Edoardo Sep 6 '19 at 14:03
• Thanks nonetheless. It seems to me as if the two topologies are not the same, and yet, Fontaine uses the "inductive limit topology " for constructing $B_{DR},$ while Brinon-Conrad uses the topology in the question I linked. So probably I am making a very trivial mistake, but these things seem to be quite subtle. – Dedalus Sep 6 '19 at 15:16


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# Anne B. Pilkington

### Professor of the Practice

B.A. (Hons) Mathematics and Economics, 1984
M.A. University College Dublin, Ireland, 1985
Ph.D., University of Notre Dame, 1991

Email: apilking@nd.edu
Office: 242 Hayes-Healy Bldg
Phone: (574) 631-3369
Fax: (574) 631-6579

255 Hurley Bldg
Notre Dame, IN  46556-4618

Class information on Anne Pilkington's personal website.

#### Research Interests

My research uses results from Classical groups, K-theory, Number theory, Quadratic forms, Coxeter groups and representation theory. In particular, with respect to classical groups, my work has focused on the theory of the group SL_2 over number rings and the behavior of its subgroups related to ideals in the rings. I am still interested in some unresolved questions from K-theory, concerning SL_2 over rings. More recently I have done some work on convex geometries on root systems related to Coxeter groups. I have compared several natural convex closure operators on these root systems, and considered the question as to when they satisfy an anti-exchange condition, defined by Edelamn and Jamison.

I am currently working on the representation theory of an interesting algebra related to the n by n matrices over the ring R $\otimes$ R, where R is the ring of integers of a number field. The representation theory can be explored using the results of Dyer on Category C, and has interesting connections with the number theory of the rings.

#### Selected Publications

• The E_2® – Normalized Subgroups of GL_2®: Journal of Algebra 172, 584-611 (1995).
• The E_2® – Normalized Subgroups of GL_2® II: Journal of Algebra 177, 619-626 (1995)
• The highest weight representations from number ring.: In preparation.
• Convex closures on root systems: Submitted for publication.

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# Double Slit Experiment: How do scientists ensure that there's only one photon?

Many documentaries regarding the double slit experiment state that they only send a single photon through the slit. How is that achieved and can it really be ensured that it is a single photon?

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– Qmechanic Sep 3 '13 at 22:59
@Qmechanic I think this is a different question. This question is more about experimental design as far as I can tell as opposed to some conceptual difficulty with what's happening to a single photon that is fired. – joshphysics Sep 3 '13 at 23:01
Tag added in keeping with @joshphysics' take on the question. – dmckee Sep 4 '13 at 1:39
@TobiasKienzler I could have sworn it was originally a "possible duplicate" comment, but I could also be going insane. – joshphysics Sep 4 '13 at 16:48
@joshphysics That probably happened while you were observing something else - that's Quantum Mechanics for you :P – Tobias Kienzler Sep 4 '13 at 18:20

In the double slit experiment, if you decrease the amplitude of the output light gradually, you will see a transition from continuous bright and dark fringe on the screen to a single dots at a time. If you can measure the dots very accurately, you always see there is one and only one dots there. It is the proof of the existence of the smallest unit of each measurement which is called single photon: You either get a single bright dot, or not.

So, probably you may ask why it is not a single photon composite of two "sub-photon", each of them passing through the slit separately and then interference with "itself" at the screen so that we only get one dot. However, the same thing occurs for three slits, four slits, etc... but the final results is still a single dot. It means that the photon must be able to split into infinitely many "sub-photon". If you get to this point, then congratulation, you basically discover the path-integral formalism of quantum mechanics.

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I chose this answer because it explains it better to me than the highest-rated one; but thank you all for answering. I'm not a physicist, so a simple explanation like this is more clear to me than formulas. Thank you! – Felicitus Sep 8 '13 at 20:12

quantum dots. nanoscale semiconductor materials that can confine photons in 3 dimensions and release them a measurable time after. based on material used the decay time is known empirically. frequency is also known. the latter is sufficient to calculate the energy of one photon. the former is then sufficient to calculate the rate of photon re emission from the QD. if the peaks at the detector are further apart than the decay time and each peak is measurable to one photon's worth of energy then you know you have a beam of single photons.

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I'm sure this will work, but I doubt that anyone bothers. – dmckee Sep 4 '13 at 1:31
The double slit experiment was performed over a century ago, there were was no nanotechnology back then. So how do you think people made sure that just one single photon was involved in the experiment, other than deducing the existence of light as quanta by the observation of experiments? – iTunnels Sep 4 '13 at 6:46
@user1800 The double slit experiment with single photons was first performed in ... 1986, in a pioneer experiment by Aspect, Grangier and Roger, please see dx.doi.org/10.1209/0295-5075/1/4/004 especially Fig.4, which shows the appearance of the fringe for large number of photon counts. The gregsan answer is a modern version of the photon cascade effect that Aspect and Grangier used in the 80's for photons on demand. Studying quantum dots semi-conductor is still an active field of research for unique-photons-on-demand, too. – FraSchelle Sep 11 '13 at 9:11
@gregsan To comment further about the above cited Aspect/Grangier/Roger experiment alongside the question Felicitus asked: they are sure they have only one photon because they have two detectors (avalanche photo-diode) which never clic at the same time. Say differently, when they clic together at the same time, you can discard the counting controllably. Thanks you for this interesting answer. – FraSchelle Sep 11 '13 at 9:24
Indeed this has been our method of choice. We've been working particularly with CdS QDs. The great advantage is the ability to grow the QDs to any arbitrary size and therefore tune the photon frequency as desired. – John Cuna Oct 21 '13 at 6:37

The practical answer (which I also wrote in a comment on the linked question) is that you turn the intensity of the light source down until the expectation value for the number of photons on the optical path is low enough to suit you.

If $\bar{n} = 0.1$ then very few of the events that are recorded on the screen will come from events where more than one photon was present on the optical path and the data will be dominated by single photon event.

Not good enough for you? Turn it down until $\bar{n} = 0.01$. Or $0.001$ or whatever suits you.

At some point the exercise becomes silly.

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Unfortunately, this answer is incorrect. Turning the intensity off do not provide you're working with photon eigenstate of the light field. Coherent state with an average photon number is not a Fock state of photon. To go to single photon effect, you need matter-light interaction, as the experiments by Aspect/Grangier and Haroche/Raymond/Brune clearly shown. The first team using photon cascade effect, the second one using flux-number commutation relation on quasi-classical field (coherent) and matter (Rydberg). – FraSchelle Sep 11 '13 at 9:17
@Oaoa That is not material that I'm familiar with, but I found what looks like the slides from a graduate summer school presentation on the matter by Haroche, and he seem to say (lecture 1, slide 11) that for a coherent state, the distribution of fock states is Poisson, in which case very low mean occupation number does imply that the data is dominated by the single-photon fock state, which is what I said. Sure, there is some contamination, but you simply turn down the intensity until you are happy with the results. – dmckee Sep 11 '13 at 14:16
Thanks a lot for the material you shared. I'm also always puzzled with this issue. Let me try to me a little bit more clear. For me, a single photon is a wave packet. Reducing (using filter of whatever) the intensity of coherent (laser, say) field will never produce wave packet: instead you will always find a continuous in time field, at rather low intensity. Maybe this is too much naive point of view. I never heard about an experiment with low intensity which results in discrete photon state. The Haroche argument is perfectly right: low occupation means 0 or 1 photon dominant (...) – FraSchelle Sep 12 '13 at 9:24
(...) but 0 photon state is a coherent state (perhaps the easiest to do coherent state), and the 1 photon state is already in the queue of the Poisson distribution. Really puzzling issue. Of course the detection scheme you're using is of primary importance, too: you must detect one-photon energy per unit time for instance. My understanding of the Haroche/Brune/Raimond experiments is that you need an atom to make a Fock state. You start from a thermal field, and the atom-light interaction select the Fock mode. The atom clearly acts as an extremely precise clock, mandatory for the experiment. – FraSchelle Sep 12 '13 at 9:30
@Oaoa I think the difference is that I am not worrying about being able to show that any particular count came from a single photon event, just being sure that the pattern I see is dominated by them. That kind of thing won't do for quantum computing or entanglement measurements. Or maybe I just don't get the argument. As I said, I'm a bit out of my depth with that stuff. – dmckee Sep 12 '13 at 14:58

I'd like to add to gregsan's answer about using quantum dots by also raising diamond nanowire one photon sources or similar devices made on CVD-grown diamond waveguides in contemporary or future experiments - of course this this method was not used in the historical one-photon experiments!. See:

Mark P. Hiscocks, Kumaravelu Ganesan, Brant C. Gibson, Shane T. Huntington, François Ladouceur, and Steven Prawer,"Diamond waveguides fabricated by reactive ion etching", Optics Express, 16, Issue 24, pp. 19512-19519 (2008) http://dx.doi.org/10.1364/OE.16.019512

This one can be freely downloaded, and I should declare my close contact with several of the authors, although I was not part of this work.

Thomas M. Babinec, Birgit J. M. Hausmann, Mughees Khan, Yinan Zhang, Jeronimo R. Maze, Philip R. Hemmer & Marko Lončar, "A diamond nanowire single-photon source", Nature Nanotechnology 5, 195 - 199 (2010)

This one is paywalled, but independent from me.

These devices might find experimental uses in more sophisticated one photon experiments both now and in the future because one can trigger them to emit lone photons almost on demand, whereas dimmed light sources simply emit photons randomly following a Poisson process and cannot be triggered. I have heard several times the opinion of several very bright (much brighter than I am) experimentalists that "no-one would bother", but I just can't shake the feeling that triggering might be useful in yet-to-be thought of experiments - I'd call myself a lousy experimentalist but once upon a time I could drive an analogue oscilloscope pretty well and triggering is surely what makes a useful piece of kit ten times more useful!

Here's how the diamond waveguide one-photon source works. One lays down a diamond waveguide by chemical vapour deposition and shapes the surrounding environment by reactive ion etching. Then a highly controlled amount of nitrogen is included into the diamond lattice. A nitrogen atom normally only makes three covalent bonds with its neighbours, whereas carbon normally makes four, so one is left with carbon atoms with one covalent bond forming electron "dangling" in the lattice wherever there is an included nitrogen. This "dangling electron" is then a fluorophore. One controls the concentration of nitrogen carefully and dices the waveguides so that there is exactly one fluorophore centre in each waveguide device: you can do this by building many devices at once then testing each and throwing out any with none or more than one fluorophores. The waveguide device can then be coupled readily to a single mode optical fibre which links the device to the outside world. The extremely high diamond to air refractive index difference (diamond $n=2.4$) and the ability to build waveguides down to below wavelength dimensions means that we can make "nanowires", which are single moded but also their strong waveguiding properties means that there is near unity probability for the photon fluoresced from the dangling electron into the single mode waveguide. So now we have something which is very readily connected to other experimental kit and doesn't need alignment.

When the time comes to use the device, a trigger signal causes an optical pump laser in the device to pulse the device with an intense beam of light, so that the lone fluorophore in the device is almost certainly raised to its metastable state. At the same time, a very high performance optical switch gates the device's optical output so no light at all escapes to the outside world. After the pumping pulse has ebbed, the output optical gate opens and the fluorophore relaxes a short time later - the fluorescence lifetime is a few nanoseconds. Although this fluorescence is governed by an exponential time-till-emission probability distribution so that, strictly speaking, it is stochastic just like the traditional dimmed light sources, the gating process means that one can control the time of photon emission to within a few nanoseconds.

Moreover, the near unity photon coupling probability into the single mode waveguide system means that there is an extremely low coupling probability for outside photons incoherent with the waveguide mode to couple in. This system is therefore extremely immune to stray light even though one photon experiments are being done: the exposure of the waveguide system to normal room light levels means that almost no photons get in - and that's before one blackens the system or puts it in a box. There is zero contamination in this device in the hands of even the most ham fisted experimentalist.

The original and main motivation for these devices is not one-photon double slit experiments, but rather as sources for quantum cryptography protocols such as the Bennett and Brassard (BB84) quantum key distribution protocol.

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I suspect that triggered single photon sources will interest people. Perhaps interest them a lot. And if they become common, off-the shelf turn-key devices they will be used for basic interference demonstrations, but as yet such sources are a long way from that easy to use and people will only get them going for things that matter. – dmckee Sep 5 '13 at 2:47
@dmckee Alas you are right that they are a long way off, owing mainly to the difficult quest for funding! – WetSavannaAnimal aka Rod Vance Sep 5 '13 at 3:02

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2024$2,750.00$1,104.51$1,645.49$477,582.88 Oct 2024$2,750.00$1,108.31$1,641.69$476,474.57 Nov 2024$2,750.00$1,112.12$1,637.88$475,362.45 Dec 2024$2,750.00$1,115.94$1,634.06$474,246.51 Jan 2025$2,750.00$1,119.78$1,630.22$473,126.73 Feb 2025$2,750.00$1,123.63$1,626.37$472,003.10 Mar 2025$2,750.00$1,127.49$1,622.51$470,875.61 Apr 2025$2,750.00$1,131.37$1,618.63$469,744.24 May 2025$2,750.00$1,135.25$1,614.75$468,608.99 Jun 2025$2,750.00$1,139.16$1,610.84$467,469.83 Jul 2025$2,750.00$1,143.07$1,606.93$466,326.76 Aug 2025$2,750.00$1,147.00$1,603.00$465,179.76 Sep 2025$2,750.00$1,150.94$1,599.06$464,028.82 Oct 2025$2,750.00$1,154.90$1,595.10$462,873.92 Nov 2025$2,750.00$1,158.87$1,591.13$461,715.05 Dec 2025$2,750.00$1,162.85$1,587.15$460,552.20 Jan 2026$2,750.00$1,166.85$1,583.15$459,385.35 Feb 2026$2,750.00$1,170.86$1,579.14$458,214.49 Mar 2026$2,750.00$1,174.89$1,575.11$457,039.60 Apr 2026$2,750.00$1,178.93$1,571.07$455,860.67 May 2026$2,750.00$1,182.98$1,567.02$454,677.69 Jun 2026$2,750.00$1,187.05$1,562.95$453,490.64 Jul 2026$2,750.00$1,191.13$1,558.87$452,299.51 Aug 2026$2,750.00$1,195.22$1,554.78$451,104.29 Sep 2026$2,750.00$1,199.33$1,550.67$449,904.96 Oct 2026$2,750.00$1,203.45$1,546.55$448,701.51 Nov 2026$2,750.00$1,207.59$1,542.41$447,493.92 Dec 2026$2,750.00$1,211.74$1,538.26$446,282.18 Jan 2027$2,750.00$1,215.91$1,534.09$445,066.27 Feb 2027$2,750.00$1,220.08$1,529.92$443,846.19 Mar 2027$2,750.00$1,224.28$1,525.72$442,621.91 Apr 2027$2,750.00$1,228.49$1,521.51$441,393.42 May 2027$2,750.00$1,232.71$1,517.29$440,160.71 Jun 2027$2,750.00$1,236.95$1,513.05$438,923.76 Jul 2027$2,750.00$1,241.20$1,508.80$437,682.56 Aug 2027$2,750.00$1,245.47$1,504.53$436,437.09 Sep 2027$2,750.00$1,249.75$1,500.25$435,187.34 Oct 2027$2,750.00$1,254.04$1,495.96$433,933.30 Nov 2027$2,750.00$1,258.35$1,491.65$432,674.95 Dec 2027$2,750.00$1,262.68$1,487.32$431,412.27 Jan 2028$2,750.00$1,267.02$1,482.98$430,145.25 Feb 2028$2,750.00$1,271.38$1,478.62$428,873.87 Mar 2028$2,750.00$1,275.75$1,474.25$427,598.12 Apr 2028$2,750.00$1,280.13$1,469.87$426,317.99 May 2028$2,750.00$1,284.53$1,465.47$425,033.46 Jun 2028$2,750.00$1,288.95$1,461.05$423,744.51 Jul 2028$2,750.00$1,293.38$1,456.62$422,451.13 Aug 2028$2,750.00$1,297.82$1,452.18$421,153.31 Sep 2028$2,750.00$1,302.29$1,447.71$419,851.02 Oct 2028$2,750.00$1,306.76$1,443.24$418,544.26 Nov 2028$2,750.00$1,311.25$1,438.75$417,233.01 Dec 2028$2,750.00$1,315.76$1,434.24$415,917.25 Jan 2029$2,750.00$1,320.28$1,429.72$414,596.97 Feb 2029$2,750.00$1,324.82$1,425.18$413,272.15 Mar 2029$2,750.00$1,329.38$1,420.62$411,942.77 Apr 2029$2,750.00$1,333.95$1,416.05$410,608.82 May 2029$2,750.00$1,338.53$1,411.47$409,270.29 Jun 2029$2,750.00$1,343.13$1,406.87$407,927.16 Jul 2029$2,750.00$1,347.75$1,402.25$406,579.41 Aug 2029$2,750.00$1,352.38$1,397.62$405,227.03 Sep 2029$2,750.00$1,357.03$1,392.97$403,870.00 Oct 2029$2,750.00$1,361.70$1,388.30$402,508.30 Nov 2029$2,750.00$1,366.38$1,383.62$401,141.92 Dec 2029$2,750.00$1,371.07$1,378.93$399,770.85 Jan 2030$2,750.00$1,375.79$1,374.21$398,395.06 Feb 2030$2,750.00$1,380.52$1,369.48$397,014.54 Mar 2030$2,750.00$1,385.26$1,364.74$395,629.28 Apr 2030$2,750.00$1,390.02$1,359.98$394,239.26 May 2030$2,750.00$1,394.80$1,355.20$392,844.46 Jun 2030$2,750.00$1,399.60$1,350.40$391,444.86 Jul 2030$2,750.00$1,404.41$1,345.59$390,040.45 Aug 2030$2,750.00$1,409.24$1,340.76$388,631.21 Sep 2030$2,750.00$1,414.08$1,335.92$387,217.13 Oct 2030$2,750.00$1,418.94$1,331.06$385,798.19 Nov 2030$2,750.00$1,423.82$1,326.18$384,374.37 Dec 2030$2,750.00$1,428.71$1,321.29$382,945.66 Jan 2031$2,750.00$1,433.62$1,316.38$381,512.04 Feb 2031$2,750.00$1,438.55$1,311.45$380,073.49 Mar 2031$2,750.00$1,443.50$1,306.50$378,629.99 Apr 2031$2,750.00$1,448.46$1,301.54$377,181.53 May 2031$2,750.00$1,453.44$1,296.56$375,728.09 Jun 2031$2,750.00$1,458.43$1,291.57$374,269.66 Jul 2031$2,750.00$1,463.45$1,286.55$372,806.21 Aug 2031$2,750.00$1,468.48$1,281.52$371,337.73 Sep 2031$2,750.00$1,473.53$1,276.47$369,864.20 Oct 2031$2,750.00$1,478.59$1,271.41$368,385.61 Nov 2031$2,750.00$1,483.67$1,266.33$366,901.94 Dec 2031$2,750.00$1,488.77$1,261.23$365,413.17 Jan 2032$2,750.00$1,493.89$1,256.11$363,919.28 Feb 2032$2,750.00$1,499.03$1,250.97$362,420.25 Mar 2032$2,750.00$1,504.18$1,245.82$360,916.07 Apr 2032$2,750.00$1,509.35$1,240.65$359,406.72 May 2032$2,750.00$1,514.54$1,235.46$357,892.18 Jun 2032$2,750.00$1,519.75$1,230.25$356,372.43 Jul 2032$2,750.00$1,524.97$1,225.03$354,847.46 Aug 2032$2,750.00$1,530.21$1,219.79$353,317.25 Sep 2032$2,750.00$1,535.47$1,214.53$351,781.78 Oct 2032$2,750.00$1,540.75$1,209.25$350,241.03 Nov 2032$2,750.00$1,546.05$1,203.95$348,694.98 Dec 2032$2,750.00$1,551.36$1,198.64$347,143.62 Jan 2033$2,750.00$1,556.69$1,193.31$345,586.93 Feb 2033$2,750.00$1,562.04$1,187.96$344,024.89 Mar 2033$2,750.00$1,567.41$1,182.59$342,457.48 Apr 2033$2,750.00$1,572.80$1,177.20$340,884.68 May 2033$2,750.00$1,578.21$1,171.79$339,306.47 Jun 2033$2,750.00$1,583.63$1,166.37$337,722.84 Jul 2033$2,750.00$1,589.08$1,160.92$336,133.76 Aug 2033$2,750.00$1,594.54$1,155.46$334,539.22 Sep 2033$2,750.00$1,600.02$1,149.98$332,939.20 Oct 2033$2,750.00$1,605.52$1,144.48$331,333.68 Nov 2033$2,750.00$1,611.04$1,138.96$329,722.64 Dec 2033$2,750.00$1,616.58$1,133.42$328,106.06 Jan 2034$2,750.00$1,622.14$1,127.86$326,483.92 Feb 2034$2,750.00$1,627.71$1,122.29$324,856.21 Mar 2034$2,750.00$1,633.31$1,116.69$323,222.90 Apr 2034$2,750.00$1,638.92$1,111.08$321,583.98 May 2034$2,750.00$1,644.56$1,105.44$319,939.42 Jun 2034$2,750.00$1,650.21$1,099.79$318,289.21 Jul 2034$2,750.00$1,655.88$1,094.12$316,633.33 Aug 2034$2,750.00$1,661.57$1,088.43$314,971.76 Sep 2034$2,750.00$1,667.28$1,082.72$313,304.48 Oct 2034$2,750.00$1,673.02$1,076.98$311,631.46 Nov 2034$2,750.00$1,678.77$1,071.23$309,952.69 Dec 2034$2,750.00$1,684.54$1,065.46$308,268.15 Jan 2035$2,750.00$1,690.33$1,059.67$306,577.82 Feb 2035$2,750.00$1,696.14$1,053.86$304,881.68 Mar 2035$2,750.00$1,701.97$1,048.03$303,179.71 Apr 2035$2,750.00$1,707.82$1,042.18$301,471.89 May 2035$2,750.00$1,713.69$1,036.31$299,758.20 Jun 2035$2,750.00$1,719.58$1,030.42$298,038.62 Jul 2035$2,750.00$1,725.49$1,024.51$296,313.13 Aug 2035$2,750.00$1,731.42$1,018.58$294,581.71 Sep 2035$2,750.00$1,737.38$1,012.62$292,844.33 Oct 2035$2,750.00$1,743.35$1,006.65$291,100.98 Nov 2035$2,750.00$1,749.34$1,000.66$289,351.64 Dec 2035$2,750.00$1,755.35$994.65$287,596.29 Jan 2036$2,750.00$1,761.39$988.61$285,834.90 Feb 2036$2,750.00$1,767.44$982.56$284,067.46 Mar 2036$2,750.00$1,773.52$976.48$282,293.94 Apr 2036$2,750.00$1,779.61$970.39$280,514.33 May 2036$2,750.00$1,785.73$964.27$278,728.60 Jun 2036$2,750.00$1,791.87$958.13$276,936.73 Jul 2036$2,750.00$1,798.03$951.97$275,138.70 Aug 2036$2,750.00$1,804.21$945.79$273,334.49 Sep 2036$2,750.00$1,810.41$939.59$271,524.08 Oct 2036$2,750.00$1,816.64$933.36$269,707.44 Nov 2036$2,750.00$1,822.88$927.12$267,884.56 Dec 2036$2,750.00$1,829.15$920.85$266,055.41 Jan 2037$2,750.00$1,835.43$914.57$264,219.98 Feb 2037$2,750.00$1,841.74$908.26$262,378.24 Mar 2037$2,750.00$1,848.07$901.93$260,530.17 Apr 2037$2,750.00$1,854.43$895.57$258,675.74 May 2037$2,750.00$1,860.80$889.20$256,814.94 Jun 2037$2,750.00$1,867.20$882.80$254,947.74 Jul 2037$2,750.00$1,873.62$876.38$253,074.12 Aug 2037$2,750.00$1,880.06$869.94$251,194.06 Sep 2037$2,750.00$1,886.52$863.48$249,307.54 Oct 2037$2,750.00$1,893.01$856.99$247,414.53 Nov 2037$2,750.00$1,899.51$850.49$245,515.02 Dec 2037$2,750.00$1,906.04$843.96$243,608.98 Jan 2038$2,750.00$1,912.59$837.41$241,696.39 Feb 2038$2,750.00$1,919.17$830.83$239,777.22 Mar 2038$2,750.00$1,925.77$824.23$237,851.45 Apr 2038$2,750.00$1,932.39$817.61$235,919.06 May 2038$2,750.00$1,939.03$810.97$233,980.03 Jun 2038$2,750.00$1,945.69$804.31$232,034.34 Jul 2038$2,750.00$1,952.38$797.62$230,081.96 Aug 2038$2,750.00$1,959.09$790.91$228,122.87 Sep 2038$2,750.00$1,965.83$784.17$226,157.04 Oct 2038$2,750.00$1,972.59$777.41$224,184.45 Nov 2038$2,750.00$1,979.37$770.63$222,205.08 Dec 2038$2,750.00$1,986.17$763.83$220,218.91 Jan 2039$2,750.00$1,993.00$757.00$218,225.91 Feb 2039$2,750.00$1,999.85$750.15$216,226.06 Mar 2039$2,750.00$2,006.72$743.28$214,219.34 Apr 2039$2,750.00$2,013.62$736.38$212,205.72 May 2039$2,750.00$2,020.54$729.46$210,185.18 Jun 2039$2,750.00$2,027.49$722.51$208,157.69 Jul 2039$2,750.00$2,034.46$715.54$206,123.23 Aug 2039$2,750.00$2,041.45$708.55$204,081.78 Sep 2039$2,750.00$2,048.47$701.53$202,033.31 Oct 2039$2,750.00$2,055.51$694.49$199,977.80 Nov 2039$2,750.00$2,062.58$687.42$197,915.22 Dec 2039$2,750.00$2,069.67$680.33$195,845.55 Jan 2040$2,750.00$2,076.78$673.22$193,768.77 Feb 2040$2,750.00$2,083.92$666.08$191,684.85 Mar 2040$2,750.00$2,091.08$658.92$189,593.77 Apr 2040$2,750.00$2,098.27$651.73$187,495.50 May 2040$2,750.00$2,105.48$644.52$185,390.02 Jun 2040$2,750.00$2,112.72$637.28$183,277.30 Jul 2040$2,750.00$2,119.98$630.02$181,157.32 Aug 2040$2,750.00$2,127.27$622.73$179,030.05 Sep 2040$2,750.00$2,134.58$615.42$176,895.47 Oct 2040$2,750.00$2,141.92$608.08$174,753.55 Nov 2040$2,750.00$2,149.28$600.72$172,604.27 Dec 2040$2,750.00$2,156.67$593.33$170,447.60 Jan 2041$2,750.00$2,164.09$585.91$168,283.51 Feb 2041$2,750.00$2,171.53$578.47$166,111.98 Mar 2041$2,750.00$2,178.99$571.01$163,932.99 Apr 2041$2,750.00$2,186.48$563.52$161,746.51 May 2041$2,750.00$2,194.00$556.00$159,552.51 Jun 2041$2,750.00$2,201.54$548.46$157,350.97 Jul 2041$2,750.00$2,209.11$540.89$155,141.86 Aug 2041$2,750.00$2,216.70$533.30$152,925.16 Sep 2041$2,750.00$2,224.32$525.68$150,700.84 Oct 2041$2,750.00$2,231.97$518.03$148,468.87 Nov 2041$2,750.00$2,239.64$510.36$146,229.23 Dec 2041$2,750.00$2,247.34$502.66$143,981.89 Jan 2042$2,750.00$2,255.06$494.94$141,726.83 Feb 2042$2,750.00$2,262.81$487.19$139,464.02 Mar 2042$2,750.00$2,270.59$479.41$137,193.43 Apr 2042$2,750.00$2,278.40$471.60$134,915.03 May 2042$2,750.00$2,286.23$463.77$132,628.80 Jun 2042$2,750.00$2,294.09$455.91$130,334.71 Jul 2042$2,750.00$2,301.97$448.03$128,032.74 Aug 2042$2,750.00$2,309.89$440.11$125,722.85 Sep 2042$2,750.00$2,317.83$432.17$123,405.02 Oct 2042$2,750.00$2,325.80$424.20$121,079.22 Nov 2042$2,750.00$2,333.79$416.21$118,745.43 Dec 2042$2,750.00$2,341.81$408.19$116,403.62 Jan 2043$2,750.00$2,349.86$400.14$114,053.76 Feb 2043$2,750.00$2,357.94$392.06$111,695.82 Mar 2043$2,750.00$2,366.05$383.95$109,329.77 Apr 2043$2,750.00$2,374.18$375.82$106,955.59 May 2043$2,750.00$2,382.34$367.66$104,573.25 Jun 2043$2,750.00$2,390.53$359.47$102,182.72 Jul 2043$2,750.00$2,398.75$351.25$99,783.97 Aug 2043$2,750.00$2,406.99$343.01$97,376.98 Sep 2043$2,750.00$2,415.27$334.73$94,961.71 Oct 2043$2,750.00$2,423.57$326.43$92,538.14 Nov 2043$2,750.00$2,431.90$318.10$90,106.24 Dec 2043$2,750.00$2,440.26$309.74$87,665.98 Jan 2044$2,750.00$2,448.65$301.35$85,217.33 Feb 2044$2,750.00$2,457.07$292.93$82,760.26 Mar 2044$2,750.00$2,465.51$284.49$80,294.75 Apr 2044$2,750.00$2,473.99$276.01$77,820.76 May 2044$2,750.00$2,482.49$267.51$75,338.27 Jun 2044$2,750.00$2,491.02$258.98$72,847.25 Jul 2044$2,750.00$2,499.59$250.41$70,347.66 Aug 2044$2,750.00$2,508.18$241.82$67,839.48 Sep 2044$2,750.00$2,516.80$233.20$65,322.68 Oct 2044$2,750.00$2,525.45$224.55$62,797.23 Nov 2044$2,750.00$2,534.13$215.87$60,263.10 Dec 2044$2,750.00$2,542.85$207.15$57,720.25 Jan 2045$2,750.00$2,551.59$198.41$55,168.66 Feb 2045$2,750.00$2,560.36$189.64$52,608.30 Mar 2045$2,750.00$2,569.16$180.84$50,039.14 Apr 2045$2,750.00$2,577.99$172.01$47,461.15 May 2045$2,750.00$2,586.85$163.15$44,874.30 Jun 2045$2,750.00$2,595.74$154.26$42,278.56 Jul 2045$2,750.00$2,604.67$145.33$39,673.89 Aug 2045$2,750.00$2,613.62$136.38$37,060.27 Sep 2045$2,750.00$2,622.61$127.39$34,437.66 Oct 2045$2,750.00$2,631.62$118.38$31,806.04 Nov 2045$2,750.00$2,640.67$109.33$29,165.37 Dec 2045$2,750.00$2,649.74$100.26$26,515.63 Jan 2046$2,750.00$2,658.85$91.15$23,856.78 Feb 2046$2,750.00$2,667.99$82.01$21,188.79 Mar 2046$2,750.00$2,677.16$72.84$18,511.63 Apr 2046$2,750.00$2,686.37$63.63$15,825.26 May 2046$2,750.00$2,695.60$54.40$13,129.66 Jun 2046$2,750.00$2,704.87$45.13$10,424.79 Jul 2046$2,750.00$2,714.16$35.84$7,710.63 Aug 2046$2,750.00$2,723.49$26.51$4,987.14 Sep 2046$2,750.00$2,732.86$17.14$2,254.28 Oct 2046$2,262.03$2,254.28$7.75$0.00 This calculation is based on widely-accepted formulas, but it is not a recommendation for how to handle your finances. Consult with a financial professional before making financial decisions. Learn how we calculated this below ## On this page: ## How to Calculate a Mortgage Payoff Date A mortgage payoff date can be calculated when you know the remaining loan balance, interest rate, and monthly mortgage payment. The calculator above can show you how much faster you can pay off your home mortgage if you pay more than the monthly payment. Both our loan payoff calculator and the payoff calculator above can be used to calculate this. ### Formula to Calculate Interest Paid per Payment Mortgage interest can be calculated in a two-step process. First, the annual interest rate must be divided by 12 to obtain a periodic interest rate. Then the remaining principal balance will be multiplied by the periodic interest rate to calculate the monthly interest payment. The formula below summarizes the interest payment formula: interest = P \times r Where: P = remaining principal balance r = periodic interest rate For example, let’s assume you have a loan with a loan balance of$300,000, an annual interest rate of 4%, and a monthly payment of $2,000. Therefore, the periodic interest rate is 0.33% (4% divided by 12). The payoff calculator above shows that the mortgage will be paid off in 17 years and 5 months. You can use our loan payment calculator to figure out the payment. The first interest payment will be found by multiplying the principal balance of$300,000 by the periodic interest rate of 0.33%. This comes out to $1,000. The remaining principal balance fell to$299,000 after the first payment. If we multiply this new amount by 0.33%, we get an interest payment of $996.67 for the second mortgage payment. This is exactly what the amortization table shows. Try our amortization calculator to build an amortization schedule. As this shows, the interest payment decreases with each mortgage payment. This is because the principal balance falls every time a payment is made. As the remaining principal balance falls, the interest payment will fall as well. The final mortgage payment of$2,000 will have an interest payment of $8.57. ### Formula to Calculate Principal Paid per Payment The principal payment is calculated by subtracting the monthly interest payment from the monthly mortgage payment. The formula below summarizes this: principal = PMT − (P \times r) Where: PMT = payment amount P = remaining principal balance r = periodic interest rate In the example above the first interest payment was$1,000. Since the mortgage payment is $2,000, the principal payment will also be$1,000. The second interest payment fell to $996.67, so the second principal payment increased to$1,003.33.

In this case, the principal payment increased. With every payment that occurs, the principal payment will be larger than the previous principal payment.

Since the mortgage payment is fixed and because the interest payment declines, the principal payment must therefore increase. The final mortgage payment of $2,000 has a principal payment of$1,991.43.

## How to Estimate an Early Payoff Date

The estimated payoff date is slightly less than the inverse of the increase. For example, if you plan on doubling the monthly payment, the payoff date will be less than half the time. If the monthly payment is tripled, the payoff date will be slightly less than a third of the original payoff date.

In our example, we would pay off the loan in 17 years and 5 months with a $2,000 payment. If we double the payment to$4,000, the payoff time is 7 years and 3 months. If the payment becomes $6,000 per month, the payoff time is 4 years and 7 months. ## How Much Do Extra Mortgage Payments Help? There are two reasons why extra payments reduce the time it takes to pay off a loan. First, it automatically increases the amount that goes to the principal each month. Also, it reduces future interest payments because the remaining principal balance will be smaller at each time period. Let’s look at each of these in more detail. First, the principal payment automatically increases. The first payment of$2,000 in our example paid $1,000 towards the principal. If we increase the monthly payment to$2,100, that extra $100 goes straight to the principal balance, so now$1,100 goes to pay down the principal balance. This is repeated each month. The faster the principal payment goes down, the faster the loan will be paid off in full.

Also, as the remaining principal balance goes down faster, the future interest payments will also be less. This is because interest payments are based on the remaining principal balance.

Both of these work together to reduce the loan payoff time when extra payments are applied to the loan.

Let’s look back at our examples and compare the principal and interest payments of the 3rd payment when we pay $2,000 or$2,100 per month.

With a $2,000 monthly payment, the third payment calls for a principal payment of$1,006.68 and an interest payment of $993.32. With a$2,100 monthly payment, the third payment calls for a principal payment of $1,107.35 and an interest payment of$992.65.

As you can see, the amount that pays down principal is almost $101 greater with a$2,100 monthly payment, even though we only pay an extra \$100 a month.

This difference is even greater the further we are in the amortization schedule.


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Select Page
Home » Test two col

# Header for page

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[1012.2879] Tracking the precession of compact binaries from their gravitational-wave signal

Authors: Patricia Schmidt, Mark Hannam, Sascha Husa, P. Ajith

Date: 13 Dec 2010

Abstract: We present a simple method to track the precession of a black-hole-binary system, using only information from the gravitational-wave (GW) signal. Our method consists of locating the frame from which the magnitude of the $(\ell=2,|m|=2)$ modes is maximized, which we denote the "quadrupole-aligned" frame. We demonstrate the efficacy of this method when applied to waveforms from numerical simulations. In the test case of an equal-mass nonspinning binary, our method locates the direction of the orbital angular momentum to within $(\Delta \theta, \Delta \phi) = (0.05ˆ{\circ},0.2ˆ{\circ})$. We then apply the method to a $q = M_2/M_1 = 3$ binary that exhibits significant precession. In general a spinning binary's orbital angular momentum $\mathbf{L}$ is \emph{not} orthogonal to the orbital plane. Evidence that our method locates the direction of $\mathbf{L}$ rather than the normal of the orbital plane is provided by comparison with post-Newtonian (PN) results. Also, we observe that it accurately reproduces similar higher-mode amplitudes to a comparable non-spinning (and therefore non-precessing) binary, and that the frequency of the $(\ell=2,|m|=2)$ modes is consistent with the "total frequency" of the binary's motion. The simple form of the quadrupole-aligned waveform will be useful in attempts to analytically model the inspiral-merger-ringdown (IMR) signal of precessing binaries, and in standardizing the representation of waveforms for studies of accuracy and consistency of source modelling efforts, both numerical and analytical.

Dec 15, 2010

1012.2879 (/preprints)
2010-12-15, 14:27


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# Chapter 10 - Area - Mid-Chapter Quiz - Page 642: 11

28 $cm^{2}$

#### Work Step by Step

The formula for the area of a kite is A=$\frac{1}{2}$$d_{1}$$d_{2}$ A=$\frac{1}{2}$$d_{1}$$d_{2}$ A=$\frac{1}{2}$(14)(4) A=28

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# Johansen Cointegration Test

I just performed a Johansen Co-integration test on two stocks.

The results I get are:

ans =

      r0       r1
t1    true     false

I am using Matlab. Can someone interpret these for me?

If I have understood the test properly, they are a good correlated pair. With Mean reversion.

Are they mean reverting?

Also, I have read about stationary pairs but the technical definition is a bit confusing. If possible can someone help point me in the direction to a simpler explanation? Or may be a book to start off?

Did the test again with the following result:

>> [h,pValue,stat,cValue,mles] = jcitest(Y)

Results Summary (Test 1)

Data: Y
Effective sample size: 229
Model: H1
Lags: 0
Statistic: trace
Significance level: 0.05

r  h  stat      cValue   pValue   eigVal
========================================
0  0  9.6981    15.4948  0.3467   0.0411
1  0  0.0979    3.8415   0.7872   0.0004

h =

      r0       r1
t1    false    false

pValue =

      r0         r1
t1    0.34672    0.78721

stat =

      r0        r1
t1    9.6981    0.097852

cValue =

      r0        r1
t1    15.495    3.8415

mles =

      r0              r1
t1    [1x1 struct]    [1x1 struct]

I am trying to understand whether h=0 implies no cointegration? What exactly does the pValue tell us?

In short still trying to understand how exactly to interpret the results.

I will get onto generating the eigenvalues. And trying to understand them after this part gets clear.

I don't have the advantage of attending school at the moment and understanding this is very difficult.

-


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# Example: Vibrational Circular Dichroism (VCD): NHDT¶

Download VCD_COG_NHDT.run

Analytical frequencies with subsequent calculation of vibrational circular dichroism (VCD)

The VCD keyword can be used to calculate VCD spectra. It is important to note that the VCD keyword only works in combination with the keys AnalyticalFreq and symmetry NOSYM.

Recommended is use to use high accuracy for the geometry optimization which one needs to do before the frequency calculation. This simple example is an NHDT molecule, which is NH3 where one hydrogen atom is replaced with deuterium and another with tritium.

Note that the fragment t21 names, like t21.N, are made by the adf script, and here this knowledge is used.

\$ADFBIN/adf <<eor
Atoms
N         0.000000    0.000000    0.010272
H        -0.471582   -0.816803    0.407861
H.D       0.943163    0.000000    0.407861
H.T      -0.471582    0.816803    0.407861
End

AtomProps
H.D m=2.014101778
H.T m=3.01604927
End

Basis
Type TZP
Core None
End

Symmetry NOSYM

xc
GGA BP86
end

Fragments
N     t21.N
H     t21.H
H.D   t21.D
H.T   t21.T
End

BeckeGrid
quality verygood
End

AnalyticalFreq
End

VCD
end input
eor


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Competitions

# Shell Game

To pass the time, Bessie the cow and her friend Elsie like to play a version of a game they saw at the county fair.

To start, Bessie puts three inverted shells on a table and places a small round pebble under one of them (at least she hopes it is a pebble - she found it on the ground in one of the pastures). Bessie then proceeds to swap pairs of shells, while Elsie tries to guess the location of the pebble.

The standard version of the game the cows saw being played at the county fair allowed the player to see the initial location of the pebble, and then required guessing its final location after all the swaps were complete.

However, the cows like to play a version where Elsie does not know the initial location of the pebble, and where she can guess the pebble location after every swap. Bessie, knowing the right answer, gives Elsie a score at the end equal to the number of correct guesses she made.

Given the swaps and the guesses, but not the initial pebble location, please determine the highest possible score Elsie could have earned.

#### Input

The first line contains an integer n (1n100) giving the number of swaps. Each of the next n lines describes a step of the game and contains three integers a, b and g, indicating that shells a and b were swapped by Bessie, and then Elsie guessed shell g after the swap was made. All three of these integers are either 1, 2 or 3, and ab.

#### Output

Print the maximum number of points Elsie could have earned.

#### Example

In this example, Elsie could have earned at most 2 points. If the pebble started under shell 1, then she guesses right exactly once (her final guess). If the pebble started under shell 2, then she guesses right twice (the first two guesses). If the pebble started under shell 3, then she doesn't make any correct guesses.

Time limit 1 second
Memory limit 128 MiB
Input example #1
3
1 2 1
3 2 1
1 3 1

Output example #1
2

Source 2019 USACO January, Bronze


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RBSE Solutions for Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.3

Rajasthan Board RBSE Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.3

Question 1.
Given there are two bags I and II. Bag I contains 3 red and 4 black bails while another bag II contains S red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from bag II.
Solution:
Let E1 be the event that the ball is drawn from the bag I and II and E2 be the event that it is drawn from the bag II and E be the event that the drawn ball is red.
Since the drawn ball is found to be red, so we have to Find P($$\frac { { E }_{ 2 } }{ A }$$) the probability of drawing red ball from bag II. since both the bags are equally likely to be selected to draw the ball, we have

Probability of getting red ball from bag I

Probability of getting red ball from bag II

Probability of getting red ball from bag III

Hence, using Baye’s Theorem,

Question 2.
A doctor has to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively $$\frac { 3 }{ 10 }$$,$$\frac { 1 }{ 5 }$$,$$\frac { 1 }{ 10 }$$ and $$\frac { 2 }{ 5 }$$. The probabilities that he will be late are $$\frac { 1 }{ 4 }$$,$$\frac { 1 }{ 3 }$$ and $$\frac { 1 }{ 12 }$$, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, than he will not be late. When he arrives, he is late. What is the probability that he comes by train ?
Solution:
Let E be the event “Doctor comes late to visit a patient” and if Doctor comes by Train, Bus, Scooter and any-other means, then probabilities are T1, T2, T3 and T4 respectively.

Question 3.
Bag I contains 3 red and 5 black balls and by II contains 4 red and 5 black balls. One ball transferred from bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
Solution:
Bag I contain 3 red and 4 black balls, and Bag II contains 4 red and 5 black balls.
Let, E1 = Event of drawing red ball from bag I
and E2 = Event of drawing black ball from bag II
and A : Event of drawing red ball after transfering one ball from I to II

Question 4.
A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls, One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Solution:
Let E1 is the event of drawing from bag I
and E2 of drawing from bag II
and A represent the event of drawing red ball.
∴ Probability of chosing one bag = $$\frac { 1 }{ 2 }$$
Or P(E1) = P(E2) = $$\frac { 1 }{ 2 }$$
There are 4 red and 4 black balls in bag I.
∴ Probability of drawing red ball from it

In second bag, there are 2 red and 6 black balls.
∴ Probability of drawing red ball from it

Now probability of drawing red ball from first bag

Question 5.
There are three coins. One is having head on both faces, another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?
Solution:
Probability of selecting one coin out of three = $$\frac { 1 }{ 3 }$$
E1 : event that coin having head on both faces.
E2 : event that coin is biased that head come up 75% of the time.
E3 : event that third is a unbiased,
and A is the event of getting head

Hence, required probability = $$\frac { 4 }{ 9 }$$

Question 6.
A laboratory blood test is 99% effective in detecting a certain disease when it is if fact, present. However, the test also yields a false positive result for 0-5% of the healthy person tested. If 0.1 percent of the population actually has the disease. What is the probability that a person has the disease given that his test result is positive ?
Solution:
Let E1 : Event that patient is effected from desease.
E2 = Event that patient is not effected from desease and A be the event that blood is tested.
∴ Probability of the person effected from disease

Probability of the person not affected from disease

Probability of those who are patient and their blood are tested

Probability of those who were tested but not patient

∴ Probability that a person has the disease given that his test is positive

Hence, the required probability = $$\frac { 22 }{ 133 }$$

Question 7.
Of the sutdents in a college, it is known that 60% reside in hostel and 40% are day scholars. Previous year result report that 30% of all student who reside in hostel attain, A grade and 20% of day schloars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has on A grade, what is the probability that the student is a hostlier ?
Solution:
Solution : Let E1 : Event that students reside in hostel and E2 : Event that students do not reside in hostel
∴ Probability of students reside in hostel

Probability that students do not reside in hostel

Probability of students securing grade A who reside in hostel

Probability of students who do not reside in hostel but. secure grade A

∴ Probability that student securing graded is a hostelier = P ($$\frac { { E }_{ 1 } }{ A }$$)

Hence, required probability : $$\frac { 9 }{ 13 }$$

Question 8.
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident are 0-01, 0*03 and 0-15 respectively. One of the insured person meets with an accident. What is the probability that he is a scooter driver ?
Solution:
Let E1 : Event that scooter driver is insured
E2 = Event that car driver is insured
E3 = Event that truck driver is insured
and A : Even that an accident occurs.
A company insured 2000 scotter drivers, 4000 car drivers and 6000 truck drivers.
∴ Total number of drivers insured = 2000 + 4000 + 6000 = 12000
∴ Probability of scooter drivers insured

Probability of car drivers insured

Probability of truck drivers insured

Probability of scooter drivers accidents

Probability of car drivers accidents

Probability of truck drivers accidents

A person meets an accident who is insured.
∴ Probability that the persons is a scooter driver

Hence, required probability : $$\frac { 1 }{ 52 }$$

Question 9.
In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $$\frac { 3 }{ 4 }$$ be probability that he knows the answer and $$\frac { 1 }{ 4 }$$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $$\frac { 1 }{ 4 }$$. What is the probability that the student knows the answer given that he answered it correctly ?
Solution:
Let E1 : Event that student knows the answer
E2 : Event that student guesses
and A : Event that he answers correctly

∴ Probability that he answers correctly

Hence requried Probability = $$\frac { 12 }{ 13 }$$

Question 10.
Suppose 5% men and 0.25% women has white hairs. A person with white hair is chosen at random. Find the probability that a male is selected. Assume that there are equal number of men and women.
Solution:
Given that number of men and women are equal
Let events E1 = To be a man
E2 = To be a woman
A = White hair

Question 11.
Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducting a new product is 0.7 and the corresponding probability is 0-3 if the second group wins. Find the probability that the new product introduced was by the second group.
Solution:
Let events
E1 : First group wins, introduce new product.
E2 : Second group wins, introduce new product.
Given : Probability that first group wins = P(E1) = 0.6
Probability that second group wins = P(E2) = 0.4
Probability that first group wins introducting new product,

Probability that second group wins introducting new product

∴ Probability that new product introduced was by the second group

Question 12.
Suppose a girl throw a dice. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1,2,3 or 4 then tosses a coin once and notes whether a head or tall is obtained. If she obtained exactly one head, what is the probability that she threw 1,2,3 or 4 with the die ?
Solution:
Number of possible results after throwing a die = (1,2, 3, 4, 5, 6) = 6
Let E1 : Event she gets 5 or 6
E2 : Event she gets 1, 2, 3, 4
A : Event that she gets head if tosses a coin
∴ Probability that 5 or 6 acheived

Probability that 1, 2, 3,4 are acheived

If she gets 5 or 6 she tosses a coin thrice and notes the number of heads.
(HHH, HHT, HTH, THH, HIT, THT, TTH, TIT)
∴ Number of ways to get one head = (HTT, THT, TTH) = 3
∴ Probability to get one head = $$\frac { 3 }{ 8 }$$

If she gets 1, 2, 3, 4, then she tosses a coin once and notes whether a head or tail.
∴ Probability to get one head = $$\frac { 1 }{ 2 }$$

If she obtained exactly one head, than probability that she threw 1,2, 3 or 4

Question 13.
A card from a pack of 52 card is lost From remaining cards of the pack, two cards are drawn found to be both diamonds. Find probability of the lost card being a diamond.
Solution:
Let Events E1 = Lost card is diamond
E2 = Lost card is not diamond
The number of diamond card = 13 out of 52 cards.

and 39 cards are not diamonds

(i) When a card of diamond lost there remaining cards of diamond are 12 out of 52.

Here A represents the cards lost.
(ii) When diamonond card is not
Lost then probability of drawing 2 cards of diamond

∴ Probability that lost card is a diamond

Hence, required probability = $$\frac { 11 }{ 50 }$$

Question 14.
A bag contains 3 red and 7 black. Two balls are drawn one by one at a time at random without replacement. If second drawn half is red, then what is the probability that first drawn ball is also red ?
Solution:
A = Event that drawn ball is Red first time.
and B = Event that drawn ball is also red in second time.
Then P(A ∩ B) = P( 1 red and 1 red)


text:
# Centralizer product theorem for elementary abelian group

## Statement

Suppose $p,q$ are distinct primes. Suppose $P$ is an elementary abelian $p$-group, and $Q \le \operatorname{Aut}(P)$ is an abelian $q$-group that is not cyclic. Let $x_1, x_2, \dots, x_n$ be an enumeration of the non-identity elements of $Q$. Then, we have:

$P = C_P(x_1)C_P(x_2) \dots C_P(x_n)$.

where $C_P(x_i)$ denotes the set of fixed points in $P$ of the automorphism $x_i$.

## Related facts

### Generalizations

• Centralizer product theorem: The same result, except that $P$ is now an arbitrary $p$-group, rather than an elementary abelian group.

## Facts used

1. Maschke's averaging lemma

## Proof

Given: Primes $p \ne q$. An elementary abelian $p$-group $P$, a non-cyclic abelian $q$-subgroup $Q$ of $\operatorname{Aut}(P)$. $x_1, x_2, \dots, x_n$ are the non-identity elements of $Q$.

To prove: $P = C_P(x_1)C_P(x_2) \dots C_P(x_n)$.

Proof:

1. By fact (1), viewing $P$ as a vector space over the field of $p$ elements and $Q$ as a group acting on this vector space, we can decompose $P$ as a direct sum of irreducible $Q$-modules, say $P_1, P_2, \dots, P_r$.
2. Let $Q_i$ be the pointwise stabilizer of $P_i$ in $Q$. Then, $Q/Q_i$ is cyclic: $Q/Q_i$ is an abelian group of automorphisms of the vector space $P_i$, and $P_i$ is irreducible. This forces $Q/Q_i$ to be cyclic.
3. Each $Q_i$ is a nontrivial subgroup of $Q$: This follows from the previous step and the fact that $Q$ isn't cyclic.
4. For each $i$, pick a non-identity element $x$ of $Q_i$. Then, $C_P(x)$ contains $P_i$: This follows from the definition of $Q_i$.
5. $P = \sum C_P(x_i)$: Each $P_i$ is contained in $C_P(x)$ for some non-identity element $x$. Thus, $P$ is the sum of all $C_P(x)$, $x$ ranging over the non-identity elements of $P$.

Reverting to multiplicative notation yields the result.

## References

### Textbook references

• Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 69, Theorem 3.3, Chapter 3, Section 3 (Complete reducibility), More info


text:
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# A tank contains 1000 L of brine with 15 kg of dissolved salt. Pure water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank (a) after $t$ minutes and (b) after 20 minutes?

## a) $y=15 e^{-t / 100} \mathrm{kg}$b) $y \approx 12.28 \mathrm{kg}$

#### Topics

Differential Equations

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### Video Transcript

the problem. We have a time on dance from the leader off brown with 15 of the dollar. Saad. Yeah, it is like this. Yeah, 10 10,000 Later. When you 15 kids. No, it is coming on you on, But also have it their holy How you are over the right minus rate out. That is what it's coming inside, dying And what is going on time. So we have in that rate in is 10 0 equals zero on equals minus. Why don't and then people my ass? Why under now the sequel Do you? I already you asks my No what we have you do I Warney Well, miners. And why 100? Why were while he was minus DD That is Why do my am I afraid? Let's see way have why Waas devour my nasty 100 and do the bar scene. So this is equal to you. Where Posse? No, we are the after demon yet that is after Dan. We have Why 100 c r c Yeah. You know even our Why do you about minus 200 last 11 15 if one won't you know, Dynasty upon 100 in power. 15 15 You power minus? Do you go on 100? So this is the answer back. You have a problem now? This is no way. Be off the problem, man. After a friend Timmy it Why can t Even as a 15 year power minus 20 bomb and response to 12.28 that is equivalent to 12.3. Is he just answer?

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# Enzo Algorithms¶

This section provides a very short overview of the algorithms used by the Enzo code. References to texts and journal articles providing a more complete discussion of the algorithms are included at the end of this page for the interested reader, or you can go to Enzo Primary References. for a more current list. As of this writing (October 2008), a formal Enzo method paper has not been published, but is in preparation. Much of the text and images on this page have been taken from one of the Laboratory for Computational Astrophysics contributions to the 2003 University of Chicago AMR conference [1] Enzo is written in a mixture of C++ and Fortran 77. High-level functions and data structures are implemented in C++ and computationally intensive lower-level functions are written in Fortran. Enzo is parallelized using the MPI message-passing library and uses the HDF5 data format to write out data and restart files in a platform-independent format.

Enzo allows hydrodynamics in 1, 2 and 3 dimensions using the structured adaptive mesh refinement (SAMR) technique developed by Berger and Collela [2]. The code allows arbitrary integer ratios of parent and child grid resolution and mesh refinement based on a variety of criteria, including baryon and dark matter overdensity or slope, the existence of shocks, Jeans length, and cell cooling time. The code can also have fixed static nested subgrids, allowing higher initial resolution in a subvolume of the simulation. Refinement can occur anywhere within the simulation volume or in a user-specified subvolume.

The AMR grid patches are the primary data structure in Enzo. Each individual patch is treated as an individual object, and can contain both field variables and particle data. Individual patches are organized into a dynamic distributed AMR mesh hierarchy using arrays of linked lists to pointers to grid objects. The code uses a simple dynamic load-balancing scheme to distribute the workload within each level of the AMR hierarchy evenly across all processors.

Although each processor stores the entire distributed AMR hierarchy, not all processors contain all grid data. A grid is a real grid on a particular processor if its data is allocated to that processor, and a ghost grid if its data is allocated on a different processor. Each grid is a real grid on exactly one processor, and a ghost grid on all others. When communication is necessary, MPI is used to transfer the mesh or particle data between processors. The tree structure of a small illustrative 2D AMR hierachy - six total grids in a three level hierarchy distributed across two processors - is shown on the left in Figure 1.

Each data field on a real grid is an array of zones with dimensionality equal to that of the simulation (typically 3D in cosmological structure formation). Zones are partitioned into a core block of real zones and a surrounding layer of ghost zones. Real zones are used to store the data field values, and ghost zones are used to temporarily store values from surrounding areas, ie, neighboring grids, parent grids or external boundary conditions, when required for updating real zones. The ghost zone layer is three zones deep in order to accomodate the computational stencil in the hydrodynamics solver (See below), as indicated in the right panel in Figure 1. These ghost zones can lead to significant computational and storage overhead, especially for the smaller grid patches that are typically found in the deeper levels of an AMR grid hierarchy.

For more information on Enzo implementation and data structures, see references [3], [4], [5], and [6].

## Dark Matter Dynamics¶

The dynamics of large-scale structures are dominated by dark matter, which accounts for approximately 85% of the matter in the universe but can only influence baryons via gravitational interaction. There are many other astrophysical situations where gravitational physics is important as well, such as galaxy collisions, where the stars in the two galaxies tend to interact in a collisionless way.

Enzo uses the Particle-Mesh N-body method to calculate collisionless particle dynamics. This method follows trajectories of a representative sample of individual particles and is much more efficient than a direct solution of the Boltzmann equation in most astrophysical situations. The particle trajectories are controlled by a simple set of coupled equations (for simplicity, we omit cosmological terms):

and

Where xp and vp are the particle position and velocity vectors, respectively, and the term on the right-hand side of the second equation is the gravitational force term. The solution to this can be found by solving the elliptic Poisson’s equation:

where is the density of both the collisional fluid (baryon gas) and the collisionless fluid (particles).

These equations are finite-differenced and for simplicity are solved with the same timestep as the equations of hydrodynamics. The dark matter particles are sampled onto the grids using the triangular-shaped cloud (TSC) interpolation technique to form a spatially discretized density field (analogous to the baryon densities used to calculate the equations of hydrodynamics) and the elliptical equation is solved using FFTs on the triply periodic root grid and multigrid relaxation on the subgrids. Once the forces have been computed on the mesh, they are interpolated to the particle positions where they are used to update their velocities.

## Hydrodynamics¶

The primary hydrodynamic method used in Enzo is based on the piecewise parabolic method (PPM) of Woodward & Colella [7] which has been significantly modified for the study of cosmology. The modifications and several tests are described in much more detail in [8], and we recommend that the interested reader look there.

PPM is a higher-order-accurate version of Godunov’s method with third-order-accurate piecewise parabolic monotolic interpolation and a nonlinear Riemann solver for shock capturing. It does an excellent job capturing strong shocks and outflows. Multidimensional schemes are built up by directional splitting, and produce a method that is formally second-order-accurate in space and time and explicitly conserves energy, momentum and mass flux. The conservation laws for fluid mass, momentum and energy density are written in comoving coordinates for a Friedman-Robertson-Walker spacetime. Both the conservation laws and Riemann solver are modified to include gravity, which is calculated as discussed above.

There are many situations in astrophysics, such as the bulk hypersonic motion of gas, where the kinetic energy of a fluid can dominate its internal energy by many orders of magnitude. In these situations, limitations on machine precision can cause significant inaccuracy in the calculation of pressures and temperatures in the baryon gas. In order to address this issues, Enzo solves both the internal gas energy equation and the total energy equation everywhere on each grid, at all times. This dual energy formalism ensures that the method yields the correct entropy jump at strong shocks and also yields accurate pressures and temperatures in cosmological hypersonic flows. See reference [8] for more information about the dual energy formalism.

As a check on our primary hydrodynamic method, we also include an implementation of the hydro algorithm used in the Zeus astrophysical code [9], [10]. This staggered grid, finite difference method uses artificial viscosity as a shock-capturing technique and is formally first-order-accurate when using variable timesteps (as is common in structure formation simulations), and is not the preferred method in the Enzo code.

## Cooling/Heating¶

The cooling and heating of gas is extremely important in astrophysical situations. To this extent, two radiative cooling models and several uniform ultraviolet background models have been implemented in an easily extensible framework.

The simpler of the two radiative cooling models assumes that all species in the baryonic gas are in equilibrium and calculates cooling rates directly from a cooling curve assuming Z = 0.3 Zo. The second routine, developed by Abel, Zhang, Anninos & Norman [11], assumes that the gas has primordial abundances (ie, a gas which is composed of hydrogen and helium, and unpolluted by metals), and solves a reaction network of 28 equations which includes collisional and radiative processes for 9 seperate species (H, H+, He, He+, He++, H-, H2+, H2 and e-). In order to increase the speed of the calculation, this method takes the reactions with the shortest time scales (those involving H- and H2+) and decouples them from the rest of the reaction network and imposes equilibrium concentrations, which is highly accurate for cosmological processes. See [11] and [12] for more information.

The vast majority of the volume of the present-day universe is occupied by low-density gas which has been ionized by ultraviolet radiation from quasars, stars and other sources. This low density gas, collectively referred to as the Lyman-α Forest because it is primarily observed as a dense collection of absorption lines in spectra from distant quasars (highly luminous extragalactic objects), is useful because it can be used to determine several cosmological parameters and also as a tool for studying the formation and evolution of structure in the universe (see [13] for more information). The spectrum of the ultraviolet radiation background plays an important part in determining the ionization properties of the Lyman-α forest, so it is very important to model this correctly. To this end, we have implemented several models for uniform ultraviolet background radiation based upon the models of Haardt & Madau [14].

## Star Formation and Feedback¶

One of the most important processes when studying the formation and evolution of galaxies (and to a lesser extent, groups and clusters of galaxies and the gas surrounding them) is the formation and feedback of stars. We use a heuristic prescription similar to that of Cen & Ostriker [15] to convert gas which is rapidly cooling and increasing in density into star particles which represent an ensemble of stars. These particles then evolve collisionlessly while returning metals and thermal energy back into the gas in which they formed via hot, metal-enriched winds.

## Parallelization in Enzo¶

Enzo uses a grid-based parallelization scheme for load balancing. The root grid is partitioned up into N pieces (where N is the number of processors), and each processor is given a piece of the root grid, which it keeps for the duration of the simulation run. Subgrids are treated as independent objects and are distributed to the processors such that each level of grids is load-balanced across all processors. Boundary fluxes between neighboring grid patches and parent and children grids are passed back and forth using MPI commands.

The one portion of the code that is parallelized differently is the root grid gravity solver. As discussed above, the gravitational potential on the root grid is solved using a fourier transform method, which requires its own message-passing routines. The three-dimensional total density field (composed of the dark matter plus baryon density on the root grid) is decomposed into two-dimensional slabs (requiring one set of messages), which are then fourier transformed. The slabs are then transposed along another axis (requiring a second set of messages to be passed) and transformed again, and a third set of messages is required in order to obtain the original block decomposition. This is unavoidable when using a fourier transform scheme, and as a result the speed of the root grid gravity solver is very sensitive to the speed of the communication network on the platform that Enzo is being run on.

## Initial Conditions Generator¶

A somewhat detailed description of the method Enzo uses to create initial conditions can be downloaded as a postscript or PDF document. To summarize: Dark matter particles and baryon densities are laid out on a uniform Cartesian grid. Given a user-specified power spectrum P(k), the linear density fluctuation field is calculated at some initial time (typically z = 100 for high-resolution/small box simulations) by using P(k) to obtain the density fluctuations in k-space on a uniform Cartesian grid. P(k) is sampled discretely at each grid point, with the density fluctuations having a random complex phase and amplitude. The amplitude is generated such that the distribution of amplitudes is Gaussian. This cube is then fourier transformed to give physical density fluctuations. Particle positions and velocities and baryon velocities are calculated using the Zel’dovich approximate. See the document above, or read Bertschinger 1998 [16] for more information.

## References¶

Note

Some of the links to references require a subscription.

 [1] B. W. O’Shea et al. “Introducing Enzo, an AMR Cosmology Application.” To be published in Adaptive Mesh Refinement - Theory And Applications, the proceedings from the 2003 University of Chicago AMR Workshop
 [2] M. J. Berger and P. Colella. “Local adaptive mesh refinement for shock hydrodynamics,” J. Comp. Phys, 82:64-84, 1989 link
 [3] G. L. Bryan. “Fluids in the universe: Adaptive mesh in Cosmology.” Computing in Science and Engineering, 1:2 p.46, 1999 link
 [4] G. L. Bryan and M. L. Norman. “A hybrid AMR application for cosmology and astrophysics.” In Workshop on Structured Adaptive Mesh Refinement Grid Methods, p. 165. IMA Volumes in Mathematics #117, 2000 link
 [5] G. L. Bryan and M. L. Norman. In D.A. Clarke and M. Fall, editors, Computational Astrophyiscs: 12th Kingston Meeting on Theoretical Astrophysics, proceedings of a meeting held in Halifax; Nova Scotia; Canada Oct. 17-19, 1996. ASP Conference Series #123, 1997 link
 [6] M. L. Norman and G. L. Bryan. “Cosmological Adaptive Mesh Refinement.” In Kohji Tomisaka, Shoken M. Miyama and Tomoyuki Hanawa, editors, Numerical Astrophysics: Proceedings of the International Conference on Numerical Astrophysics 1998, p. 19. Kluwer Academics, 1999
 [7] P. R. Woodward and P. Colella. “A piecewise parabolic method for gas dynamical simulations,” J. Comp. Phys, 54:174, 1984 link
 [8] (1, 2) G. L. Bryan, M. L. Norman, J. M. Stone, R. Cen and J. P. Ostriker. “A piecewise parabolic method for cosmological hydrodynamics,” Comp. Phys. Comm., 89:149, 1995 link
 [9] J. M. Stone and M. L. Norman. “Zeus-2D: A radiation magnetohydrodynamics code for astrophysical flows in two space dimensions. I. The hydrodynamics algorithms and tests.” The Astrophysical Journal Supplement, 80:753, 1992 link
 [10] J. M. Stone and M. L. Norman. “Zeus-2D: A radiation magnetohydrodynamics code for astrophysical flows in two space dimensions. II. The magnetohydrodynamic algorithms and tests.” The Astrophysical Journal Supplement, 80:791, 1992 link
 [11] (1, 2) T. Abel, P. Anninos, Y. Zhang and M.L. Norman. “Modeling primordial gas in numerical cosmology.” New Astronomy, 2:181-207, August 1997 link
 [12] P. Anninos, Y. Zhang, T. Abel and M.L. Norman. “Cosmological hydrodynamics with multispecies chemistry and nonequilibrium ionization and cooling.” New Astronomy, 2:209-224, August 1997 link
 [13] M. Rauch. “The Lyman Alpha Forest in the Spectra of QSOs.” Annual Review of Astronomy and Astrophysics, 36:267-316, 1998 link
 [14] F. Haardt and P. Madau. “Radiative Transfer in a Clumpy Universe, II. The Ultraviolet Extragalactic Background.” The Astrophysical Journal, 461:20, 1996 link
 [15] R. Cen and J.P. Ostriker. “Galaxy formation and physical bias.” The Astrophysical Journal Letters, 399:L113, 1992 link
 [16] E. Bertschinger. “Computer Simulations in Cosmology.” Annual Review of Astronomy and Astrophysics, 36:599 link


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#### Volume 21, issue 5 (2021)

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 The Journal About the Journal Editorial Board Editorial Interests Subscriptions Submission Guidelines Submission Page Policies for Authors Ethics Statement ISSN (electronic): 1472-2739 ISSN (print): 1472-2747 Author Index To Appear Other MSP Journals
Essential tori in spaces of symplectic embeddings

### Julian Chaidez and Mihai Munteanu

Algebraic & Geometric Topology 21 (2021) 2489–2522
##### Abstract

Given two $2n$–dimensional symplectic ellipsoids whose symplectic sizes satisfy certain inequalities, we show that a certain map from the $n$–torus to the space of symplectic embeddings from one ellipsoid to the other induces an injective map on singular homology with mod $2$ coefficients. The proof uses parametrized moduli spaces of $J$–holomorphic cylinders in completed symplectic cobordisms.

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20 Jun 2019

# Sparse multilayer perceptrons: converting CNNs to MLPs - Part 2

This is the second part of my series about converting CNNs to MLPs, the first part can be read here. In this post i will first study the time and space complexity of the converted CNN and then i will try to verify these results using actual CNNs converted to OpenCV-MLP models.

## Complexity of CNNs and MLPs

As in part 1 i will simplify this derivation to grayscale images. I will use the Big O Notation to compare the asymptotic behaviour.

### Time complexity

For a single element of the feature map the convolution requires as many multiplications and additions (to be precise one less) as there are elements in the kernel, the number of element in the kernel is ${(2 \cdot k + 1)}^2$, this calculation needs to be done for every element of the feature map of which there are $n \cdot m$ as a result the complexity of the discrete convolution is:

For the fully connected network each element of the feature map is the result of the dot product of to vectors of size $n \cdot m$, this yields $n \cdot m$ multiplications and additions (to be precise it is one addition less as well). With an output size of $n \cdot m$ we can see that the time complexity is:

### Space complexity

The kernel of the discrete convolution is of size ${(2 \cdot k + 1)}^2$, the bias of size $1$, this means the space complexity of a single filter in a single layer is

which espacially means the space complexity is independent of the input size. For a MLP the size of the weight matrix is ${(n \cdot m)}^2$, the bias of size $n \cdot m$. This means the space complexity of a single filter in a single layer is:

For both architectures the space complexity scales linearly with the number as filters as well as the number of layers.

One thing to keep in mind, is that the conversion to a MLP requires the adaption of pooling layers to normal fully connected layers. While a pooling layer requires no learned parameter, the converted layer is another layer with parameters of complexity $\mathcal{O}(n^2 \cdot m^2)$.

## Comparison of the numbers using an actual CNN

This comparison is based on a small CNN consisting of 3 convolutional layers with kernel size $3 \times 3$, downsampling is achieved by using a stride of 3. The first layer consists of 32 filters, the following layers of 64 filters. The convolutional layers are followed by a dense layer of size 1024 and the output layer of size 29.

This network is used for a classification task on $80 \times 80$ 3-Channel images and is comparable in size to networks used for CIFAR-10 or MNIST.

### Number of calculations

For this section i will just compare the number of operations required in the convolutional layers, as the number of operations for the dense layers is the same for both representations, the calculations are based on the formulas given above.

The convolutions in the CNN require about $12\,355\,840$ Multiply-Accumulate operations, for the converted MLP the former convolution requires about $1\,340\,228\,608$ operations. We can see that the number of operations increase by a factor of $108$, this is without paying attention to the now larger bias.

### Number of weights

In this section i compare the number of weights, as in the last section i will limit this to the number of weights in the convolutional layer. In contrast to the last section these numbers are the actuall numbers from the CNN presented above, not just estimates.

The CNN consists of $56\,320$ trainable weights, the MLP of $1\,028\,130\,752$ weights. This is an increase in the number of weights by nearly factor $18\,255$

CNN MLP
Multiply-Accumulate $12\,355\,840$ $1\,340\,228\,608$
Weights $56\,320$ $1\,028\,130\,752$

To summarize one can see, that a CNN is more efficient regarding operations and space. This was expected, as most of our weight-matrices consist of zeroes or a repeating filter. It is interesting to note, that the difference for the weights is much larger than for the MAC-Operations, this is due to the reuse of the same kernel in CNNs.

## Comparison of the performance

### The conversion script

The complete script is available on my github-page at github.com/aul12/Cnn2Mlp, it reads a tensorflow .ckpt file and converts it into a OpenCV-MLP file, for more information on the usage see the my blog post From OpenCV to TensorFlow and back: fast neural networks using OpenCV and C++. Before you use the script a little disclaimer: this script is just intended as a proof of concept and comes with some limitations:

• the size of each input and the stride of the convolution is hard-coded in the script
• stride is the only supported way of downsampling
• every convolutional layer needs to have conv in it’s name to be converted
• every layer needs to consist of a layer_name/kernel and layer_name/bias tensor.
• the script orders the layers by name to get their order in the network, so keep this in mind when naming your layers (this primarily works because conv is before dense in the alphabet)
• the script produces multiple output files, named part_XY.xml, this is to reduce the memory usage of the script, these files need to be concatenated, this can be done by running cat part_*.xml > all.xml

### Performance

I tried to compare the results using the OpenCV-MLP implementation. Unluckily OpenCV is not able to load the model, probably due to the large size of the generated xml file.

This limits the evaluation to estimating the performance by extrapolating using the CNN as a base. On the test device the CNN needs about 10ms, running on the CPU (i7-6770HQ), given the number of operations the MLP is likely to need about 1s for one inference.

## Conclusion

In this series of blog post i have shown, that it is theoretically possible to convert CNNs to MLPs, but due to the huge size of the MLP the performance suffers dramatically. This can be seen as an interessting demonstrations about why CNNs got so popular for computer vision tasks.