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# Revision history [back]

It seems you would prefer to use flat networking. Here are steps to create a flat network with the symbolic name flatnet1.

The ovs_quantum_plugin.ini on the node running the Quantum server needs:

network_vlan_ranges = flatnet1

The ovs_quantum_plugin.ini on the network and compute nodes need:

bridge_mappings = flatnet1:br-flatnet1

On the network and compute nodes, create a bridge named br-flatnet1 and add the interface for the physical nic to it, eg:

# ovs-vsctl add-br br-flatnet1
# ovs-vsctl add-port br-flatnet1 eth3

These physical NICs should be plugged into switch ports in the same VLAN. Now create the flat network in Quantum:

$quantum net-create fnet1 --provider:network_type flat \ --provider:physical_network flatnet1 --shared$ quantum subnet-create fnet1 10.0.1.0/24

As you are only doing flat networking, maybe it would be easier to use the Linux Bridge plugin instead of OVS.

It seems you would prefer to use flat networking. Here are steps to create a flat network with the symbolic name flatnet1.

The ovs_quantum_plugin.ini on the node running the Quantum server needs:

network_vlan_ranges = flatnet1

The ovs_quantum_plugin.ini on the network and compute nodes need:

bridge_mappings = flatnet1:br-flatnet1

On the network and compute nodes, create a bridge named br-flatnet1 and add the interface for the physical nic to it, eg:

# ovs-vsctl add-br br-flatnet1
# ovs-vsctl add-port br-flatnet1 eth3

These physical NICs should be plugged into switch ports in the same VLAN. VLAN. Configure your network startup scripts to put these interfaces UP without an IP address at boot. Now create the flat network in Quantum:

$quantum net-create fnet1 --provider:network_type flat \ --provider:physical_network flatnet1 --shared$ quantum subnet-create fnet1 10.0.1.0/24

As you are only doing flat networking, maybe it would be easier to use the Linux Bridge plugin instead of OVS.

It seems you would prefer to use flat networking. Here are steps to create a flat network with the symbolic name flatnet1.

The ovs_quantum_plugin.ini on the node running the Quantum server needs:

network_vlan_ranges = flatnet1

The ovs_quantum_plugin.ini on the network and compute nodes need:

bridge_mappings = flatnet1:br-flatnet1

On the network and compute nodes, create a bridge named br-flatnet1 and add the interface for the physical nic to it, eg:

# ovs-vsctl add-br br-flatnet1
# ovs-vsctl add-port br-flatnet1 eth3

These physical NICs should be plugged into switch ports in the same VLAN. Configure your network startup scripts to put these interfaces UP without an IP address at boot. Now create the flat network in Quantum:

$quantum net-create fnet1 --provider:network_type flat \ --provider:physical_network flatnet1 --shared$ quantum subnet-create fnet1 10.0.1.0/24

As you are only doing flat networking, maybe it would be easier to use the Linux Bridge plugin instead of OVS.

For the error see here. What OS/kernel are you on? The workaround is to add ovs_use_veth = True to l3_agent.ini, dhcp_agent.ini and lbaas_agent.ini. I don't think it will solve the error about changing the mac on br-int, but you can probably ignore that as I don't think the mac on the bridge matters. But it is important for the router and dhcp devices because it is causing exceptions in the agents.


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# Definition:Angle/Unit/Degree/Historical Note

The division of the circle into $360$ degrees originates from the Babylonians, who used a sexagesimal (base $60$) number system for the purposes of mathematics and astronomy.
Conveniently, the most commonly used angles in geometry (for example $30 \degrees$, $45 \degrees$, $60 \degrees$) are all whole numbers when measured in degrees.

format: plain


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# Tag Info

14

Based on anecdata (conversations with other quants), not much. Banks develop their own models and if they do outsource the effort, they pay someone for the code + support (there are companies like Numerix or Pricing Partners which do that). QuantLib is criticized for being poorly documented and convoluted. My own observation is that the parts of QuantLib I ...

12

This is off-topic and maybe belongs to StackOverflow, but here goes. 1. Compile QuantLib The best way is to open the Visual Studio Command Prompt using a shortcut under Programs→Microsoft Visual Studio→Visual Studio Tools. Now, you need to navigate to the QuantLib folder inside the folder where you have QuantLib (there are other folders such as ...

12

It's a very good and useful question. And it is bloody hard to answer just like many other questions relating to the proprietary nature of how banks and funds implement their core technology. A better route may be to ask on the QL lists, and/or to inquire as to who actually attented the first Quantlib forum in London last month. Another route would be to ...

12

Well, actually it's designed to use whatever discretization you throw at it---but for the time being only Euler discretization is implemented, mostly for lack of time or interest on the part of contributors. If you want to use exact numerics with a process, you can just code the corresponding discretization class (you'll have to inherit from ...

9

I've been using QuantLib for quite a while. Let me share some experience: QuantLib is a highly sophisticated quantitative framework. It can do much and much more than a simple pricing of European option. For example, in your example, you could have changed the payoff to binary payoff or giving a monte-carlo pricing engine (rather than ...

7

Assuming you're referring to the local-volatility class implemented in <QuantLib/ql/termstructures/volatility/equityfx/localvolsurface.hpp>, it's among the several classes that are not exported through SWIG. You'll have to add it to the SWIG interface files (probably in volatilities.i), regenerate the wrappers and recompile them. If you need ...

7

How at ease are you with programming, dynamic linking, Makefiles and the rest? In essence, there is no magic here. QL has a very liberal license, and you "merely" have to set up your project such that it finds the quantlib headers (and hence Boost headers) during compilation the quantlib library during linking and both of those can be automated by ...

6

I'm familiar with the library, but not with the way it is exported to R. Anyway: gearings are optional multipliers of the LIBOR fixing (some bonds might pay, for instance, 0.8 times the LIBOR) and spreads are the added spreads. In your case, the gearing is 1 and the spread is 0.0140 (that is, 140 bps; rates and spread must be expressed in decimal form). ...

6

Day-count conventions. You can't live with them, you can't live without them. The reason the prices differ is that the pricing engine can't calculate correctly the time over which the first coupon is discounted, and thus it gets slightly different discount factors to apply to the coupon amounts. Please sit down, it'll take some explaining. Ultimately, both ...

6

There's no class at this time to add two curves as you want, but it won't be much difficult to write it. The closest you'll get in the library is the ZeroSpreadedTermStructure class, that shows the general idea: it inherits from YieldTermStructure (by way of ZeroYieldStructure) takes a YieldTermStructure and a spread (constant, in this case) and override ...

5

It's because of the settlement days you passed when you initialized the flat volatility curve. You're creating the spot, forward and flat volatilities as: boost::shared_ptr<BlackVarianceSurface> volatilitySurface( new BlackVarianceSurface(todaysDate, calendar, maturityArray, strikeArray, ...

5

There are two different issues at play here. One is that, of course, you want only the future cash flows to enter the calculation. This is taken care when you set the evaluation date to 6 months from today. In C++, you would say Settings::instance().evaluationDate() = today + 6*Months; I don't remember the corresponding function in QuantLibXL, but you ...

5

Ok, I've done some digging in the code. It's an issue with the LogLinear interpolation; while trying to find the correct rate for the 1-week node, the bootstrapper wanders unchecked into a region of negative rates and the logarithms blow up. At this time, I'm afraid the workaround is just to use some other interpolation. Or recompile the library and the ...

5

Numerical methods are only approximations. So binomial trees, Monte Carlo simulations, and finite difference methods should all produce different numbers. As for whether you should install it for your customers, that can only be answered by what you think your customers want. Do retail customers really want the potential confusion? Are they going to ...

5

To add to Student T's answer, which I second: the complex setup starts making sense (and its cost gets amortized) once you start keeping the instruments around instead of throwing them away after the pricing. For instance, once the option above is built, you can change the market price of the underlying (or its volatility, or the risk free rate) by just ...

5

And don't forget that there are wrappers as eq RQuantLib which I use on the command-line here: edd@max:~$r -l RQuantLib -e 'print(EuropeanOption("call", 47, 40, 0.05, 0.0, 4/12, 0.2))' Concise summary of valuation for EuropeanOption value delta gamma vega theta rho divRho 6.4728 0.8899 0.0307 4.5139 0.7372 ... 4 I believe it's correct. However, consider that it would be easy enough, and more clear, to create a new class (at least in C++; the task is more difficult if you also want to export it to Excel). The new instrument should only inherit from Bond and implement a constructor that builds the desired cash flows via a call to FixedLeg and another to IborLeg; you ... 4 I'm guessing you're simulating rate curves etc. inside your system, and you want to reprice your instruments over the simulated curves using QuantLib. In this case, most of the logic is in your system already, and you have to plug pricing functionality in. If so, I don't think there's many steps involved besides, well, pricing the instrument on the ... 4 The formally supported way of adding extensions to QuantLib is by means of the Swig extension 'system' / library / tool. And the Swig site's page on compatibility has this to about Javascript: There is also SwigJS, a JavaScript module for SWIG so you could to familiarize yourself with a) how Swig works in the context of QuantLib and b) if/how you can ... 4 The QuantLib you installed is just a C++ library. If you were on a Windows machine, you'd need the QuantLibXL addin to use it in Excel (http://quantlib.org/quantlibxl/). But on a Mac, you've no such luck. As far as I know, Excel for Mac only allows addins written in VBA, so QuantLibXL can't be built for it. 4 At this time, there's no specific documentation for QuantLib-Python, except for a series of screencasts that I started a while ago (you can find them on YouTube at https://www.youtube.com/playlist?list=PLu_PrO8j6XAvOAlZND9WUPwTHY_GYhJVr) but which is far from exhaustive; there's just a few of them for now, and there's no definite learning path. However, the ... 3 I've never heard of it, but I've only been in the industry 2.5 years. Our C++ guys haven't even mentioned it either. They prefer using PACK/LAPACK which is mostly rooted in academia & heavily debugged. We also make heavy use of the IMSL FORTRAN libraries for hardcore statistical computation and Extreme Optimization (for .NET). One of our other ... 3 As for the first question, the schedule should start from the issue date. The bond will manage cash flows correctly based on the evaluation date. The second is a bit trickier, and I don't think I have working code handy. The general idea is: if you want to add a spread to the rate of the bond (to go from discount margin to price) you'll have to modify the ... 3 I think to have the answer: use qlBondPreviousCashFlowDate() pointing at your FloatingRateBond object to get the last date of payment; use qlInterestRateIndexFixingDate() to get the fixing date referring to the last payment date; use qlIndexAddFixings() to add a fixing rate to the fixing date you got above; repeat for each one of your bonds if they share ... 3 Tenor is just a different term for time to maturity. A schedule is generated from startDate and endDate in combination with a time to maturity and some info on calendar specifics. Here is an example from Dimitri Reiswich' Presentation from quantlib.com, I hope it makes the use of schedule clearer to you: void testingSchedule1 (){ Date begin (30 , September ... 3 There is experimental code available under https://sourceforge.net/tracker/?func=detail&atid=312740&aid=3413982&group_id=12740 Basically I tried to answer the question if you should do the riskfactor shifts on the level of the pricing engine or on the level of the market data. For me the answer is that one has to do it on the level of the ... 3 A free to use Excel Add-on providing QuantLib-backed derivatives pricing analytics directly in Excel is available at http://www.deriscope.com Disclosure: answerer is author of the package. 3 Answering my own question: use qlFloatingRateBond and pass it a range of strikes (one for each coupon) for both Caps and Floors arguments use BondEngine as pricing engine use IborCouponPricer with Type argument equal to "IborByBlack" as coupon pricer - This pricer also takes an OptionletVolatilitySurface as input the OptionletVolatilitySurface can be ... 3 It's hard to be sure without seeing the inputs, but I'm guessing that the implied curve changes shape because the original curve does (which you can see from your output: except for the 1-year and 5-years points, the actual discounts are different). The reason the original curve changes is probably the different position of weekends or holidays (so that, ... 3 The time to expiry is required, but it's included in the inputs: the two discounts$e^{-rT}$and$e^{-qT}$and the standard deviation$\sigma\sqrt{T}\$. You might argue it could be documented more clearly, and I might agree with you.

Only top voted, non community-wiki answers of a minimum length are eligible


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# What is an anonymous function in Python?

PythonServer Side ProgrammingProgramming

In Python, anonymous function is a function that is defined without a name. While normal functions are defined using the def keyword, in Python anonymous functions are defined using the lambda keyword. Hence, anonymous functions are also called lambda functions.

If we run the given code we get the following output

C:/Users/TutorialsPoint1/~.py
[(13, -3), (4, 1), (1, 2), (9, 10)]

Published on 07-Dec-2017 10:50:06


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# Thread: Basic Set Theory Question

1. ## Basic Set Theory Question

is |A| + |B| same as |A U B|?
and why if they are not, i see that |AUB| is |A| + |B| - |A n B| (A intersect B)
but why would you need to do that, cause if there are repeated elements then they can't repeat by them self cause sets are distinct elements...

2. Originally Posted by AwesomeDesiKid
is |A| + |B| same as |A U B|?
and why if they are not, i see that |AUB| is |A| + |B| - |A n B| (A intersect B)
but why would you need to do that, cause if there are repeated elements then they can't repeat by them self cause sets are distinct elements...
If $A\cap B\ne\varnothing$ then $\left|A\right|+\left|B\right|$ counts the repeated elements twice.

3. Originally Posted by Drexel28
If $A\cap B\ne\varnothing$ then $\left|A\right|+\left|B\right|$ counts the repeated elements twice.
well, i don't understand the difference between |A XOR B| and |A U B|
xor is exclusive or(you probably know) wouldn't they equal the same thing

4. Originally Posted by AwesomeDesiKid
well, i don't understand the difference between |A XOR B| and |A U B|
xor is exclusive or(you probably know) wouldn't they equal the same thing
My understanding is that $A\text{ XOR }B=A\text{ }\Delta\text{ }B=\left(A-B\right)\cup\left(B-A\right)$. Is that correct?

5. Originally Posted by Drexel28
My understanding is that $A\text{ XOR }B=A\text{ }\Delta\text{ }B=\left(A-B\right)\cup\left(B-A\right)$. Is that correct?
i have no idea what that delta means?

6. Originally Posted by AwesomeDesiKid
i have no idea what that delta means?
Th symmetric difference. I told you want it means after that. In any case, I am moderately sure that I am correct in interpreting what you mean by $A\text{ XOR }B$. Your answer is then $A\cup B=A\text{ XOR }B$ iff $A\cap B=\varnothing$.

7. Originally Posted by Drexel28
Th symmetric difference. I told you want it means after that. In any case, I am moderately sure that I am correct in interpreting what you mean by $A\text{ XOR }B$. Your answer is then $A\cup B=A\text{ XOR }B$ iff $A\cap B=\varnothing$.
i think i got the difference, if |A U B| has all the elements from a and b.
so if A = {0, 1, 2} and B = {1,2,3} then
|A U B| = |0, 1, 2, 3} but
|A XOR B| = {0. 3}
if i m right....... thanks for the help, if not...I HAVE NO HOPE ANYMORE
lol

8. Originally Posted by AwesomeDesiKid
i think i got the difference, if |A U B| has all the elements from a and b.
so if A = {0, 1, 2} and B = {1,2,3} then
|A U B| = |0, 1, 2, 3} but
|A XOR B| = {0. 3}
if i m right....... thanks for the help, if not...I HAVE NO HOPE ANYMORE
lol
! One note though, $\left|\text{some set}\right|$ usually means the cardinality.

9. so A U B DOES equal A XOR B as long as A n B equals phi?


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densregion.normal {denstrip} R Documentation

## Density regions based on normal distributions

### Description

Adds a density region to an existing plot of a normally-distributed quantity with continuously-varying mean and standard deviation, such as a time series forecast. Automatically computes a reasonable set of ordinates to evaluate the density at, which span the whole forecast space.

### Usage

## S3 method for class 'normal'
densregion(x, mean, sd, ny=20, ...)

### Arguments

 x Suppose the continuously-varying quantity varies over a space S. x is a vector of the points in S at which the posterior / predictive / fiducial distribution will be evaluated. mean Vector of normal means at each point in x. sd Vector of standard deviations at each point in x. ny Minimum number of points to calculate the density at for each x. The density is calculated for at least ny equally spaced normal quantiles for each point. The density is actually calculated at the union over x of all such points, for each x. ... Further arguments passed to densregion.

### Details

The plot is shaded by interpolating the value of the density between grid points, using the algorithm described by Cleveland (1993) as implemented in the filled.contour function.

### Author(s)

Christopher Jackson <chris.jackson@mrc-bsu.cam.ac.uk>

### References

Jackson, C. H. (2008) Displaying uncertainty with shading. The American Statistician, 62(4):340-347.

Cleveland, W. S. (1993) Visualizing Data. Hobart Press, Summit, New Jersey.

densregion, densregion.survfit, denstrip

### Examples

## Time series forecasting

(fit <- arima(USAccDeaths, order = c(0,1,1),
seasonal = list(order=c(0,1,1))))
pred <- predict(fit, n.ahead = 36)
plot(USAccDeaths, xlim=c(1973, 1982), ylim=c(5000, 15000))

## Compute normal forecast densities automatically (slow)

## Not run:
densregion.normal(time(pred$pred), pred$pred, pred$se, pointwise=TRUE, colmax="darkgreen") lines(pred$pred, lty=2)
lines(pred$pred + qnorm(0.975)*pred$se, lty=3)
lines(pred$pred - qnorm(0.975)*pred$se, lty=3)

## End(Not run)

## Compute forecast densities by hand (more efficient)

nx <- length(pred$pred) y <- seq(5000, 15000, by=100) z <- matrix(nrow=nx, ncol=length(y)) for(i in 1:nx) z[i,] <- dnorm(y, pred$pred[i], pred$se[i]) plot(USAccDeaths, xlim=c(1973, 1982), ylim=c(5000, 15000)) densregion(time(pred$pred), y, z, colmax="darkgreen", pointwise=TRUE)
lines(pred$pred, lty=2) lines(pred$pred + qnorm(0.975)*pred$se, lty=3) lines(pred$pred - qnorm(0.975)*pred$se, lty=3) densregion(time(pred$pred), y+2000, z, colmax="darkblue", pointwise=TRUE)

[Package denstrip version 1.5.4 Index]


text:
## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 11 - Counting Methods and Probability Theory - 11.3 Combinations - Exercise Set 11.3 - Page 709: 73

#### Answer

Does not make sense.

#### Work Step by Step

A password implies the importance of order of the letters. For example, "wand" and "dawn" are different orderings of the same four letters and make two different passwords. Order is important. Permutations should be used, not combinations. The formula for ${}_{n}P_{r}$ should be used.

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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## Found 26,705 Documents (Results 1–100)

100
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### Introduction to partial differential equations. For courses in engineering and natural sciences. 3rd revised edition. (Einführung in partielle Differentialgleichungen. Für die Studiengänge der Ingenieur- und Naturwissenschaften.) (German)Zbl 07589693

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text:
# Thermodynamics of a pump

1. Sep 27, 2009

### leumas614

Consider the act of evacuating a chamber with a vacuum pump. Vacuum pumping rates are usually given in volumetric pumping speeds, Sp, whose units are typically volume/unit time, e.g., 1 m3/min. In other words, the number of molecules in 1 m3 at the chamber pressure and temperature are removed in one minute. A 0.1 m3 chamber is initially at atmospheric pressure (1 bar) and 298 K. The gas may be considered to be an ideal diatomic gas. Calculate and plot the temperature and pressure in the chamber as a function of time with a pump rated at Sp= 1 m3/min if the pump down process may be considered to be

(b) isothermal.

I've done everything I can think of with energy and entropy balances. I think the problem is the temperature and pressure of the influx stream are changing as the chamber is evacuated so I don't know how to relate them to time. I need an equation for Temperature as a function of time and nothing else and the rest is easy.

I think dn/dt is the Sp but in units of mol/min with Volume held constant. Now I just need to know T without relying on P (or vice versa)

Thanks.

2. Sep 27, 2009

### Mapes

Have you applied the ideal gas law yet? It's not clear from your post.

3. Sep 27, 2009

### leumas614

Yes, I took the time derivative of $$PV=nRT$$ holding V constant.

$$V \left(\frac{dP}{dt}\right)=R\left(T \frac{dn}{dt}+n \frac{dT}{dt}\right)$$

dU=nCvdT

$$\frac{dU}{dt}=Cv\left(n\frac{dT}{dt}+T\frac{dn}{dt}\right)$$

from energy balance, dU=Hdn integrating gets U=Hn which equals U=(U+PV)n

Nothing else seems to work. Intuitively I would think the pressure would decrease non-linearly but I see no way of introducing that into the equations.

4. Sep 27, 2009

### Mapes

OK, let's apply the energy differential equation to the adiabatic case. Take the system to be the amount of gas that doesn't leave the box. What can you say about $dU$?

5. Sep 27, 2009

### leumas614

You get

$$Cv\frac{V}{R}\frac{dP}{dt}=\frac{dU}{dt}$$

So we can relate U with P but you still don't have an expression for P so only one equation with two unknowns. You can't replace P with an expression for T because T is also unknown.

Here's what I did.

$$dU=d(Hn)$$ but both H and n are not constant so

$$\frac{dU}{dt}=H\frac{dn}{dt}+n\frac{dH}{dt}$$

$$\frac{dU}{dt}=\left(U+PV\right)\frac{dn}{dt}+n\left(\frac{dU}{dt}+\frac{dP}{dt}V\right)$$

Through some algebra:

$$\frac{dU}{dt}(1-n)-U\frac{dn}{dt}=PV\frac{dn}{dt}+\frac{dP}{dt}nV$$

V and $$\frac{dn}{dt}$$ are given. I think the $$n$$ in this case is the total number of moles you start out with so that is known from $$PV=nRT$$. Now I have two first order ODE that I can't solve.

6. Sep 27, 2009

### Mapes

I was thinking more like, the rate of work that that amount of gas does is $P(dV/dt)$, which is $PF$ where $F$ is the volumetric flow rate.

It was easier for me to start with the isothermal case. What's $dn/dt$, where $n=PV/RT$? What's $dn/dt$ in terms of the volumetric flow rate? Set these equal and integrate to solve for pressure.

7. Sep 27, 2009

### leumas614

But this is what I don't understand. The volumetric flow rate is not constant because the tank is draining so presumably the amount of molecules flowing out of the tank is changing with time. So I don't know what the volumetric flow rate is. I understood the Sp in the question to mean, it takes 1 minute to evacuate a 1 m3 chamber so it would take a tenth of that time to evacuate a 0.1 m3 chamber.

8. Sep 27, 2009

### Mapes

The volumetric flow rate is constant, but the molar flow rate is not constant. A better way to think of it is that 17% of the chamber's contents are removed every second. Employing the continuum assumption, the chamber is never completely emptied.

9. Sep 28, 2009

### leumas614

I still don't quite follow what you're suggesting. The volume does not change because we took the system to be the inside of the chamber which is rigid. Using $$dV/dt$$ would only work if the chamber was not rigid. But then if the volumetric flow rate is not the molar flow rate than it has no place in any equation and that wouldn't make sense.

$$n=\frac{PV}{RT}$$

$$\frac{dn}{dt}=\frac{1}{RT}\left(V\frac{dP}{dt}+P\frac{dV}{dt}\right)$$

And the rate of work can't be $$P\dot{V}$$ because pressure is not constant.

10. Sep 29, 2009

### Mapes

OK, let's back up a second. For the isothermal case, I'm suggesting you consider the system to be the chamber, determine $dn/dt$ by the ideal gas law (which you did in your last post, $(V/RT)(dP/dt)$), and equate that to $dn/dt$ determined by the volumetric flow rate $F=dV_0/dt$. You can use the ideal gas law to get this molar flow rate because the gas removed has a constant volume and follows the ideal gas law. Set these two molar flow rates equal and solve for $P$ as a function of time.

For the adiabatic case, I'm suggesting you take the system to be the amount of gas that stays in the chamber as it pushes out an infinitesimal amount of gas during evacuation. This system is adiabatic and does work on the exiting gas, so its energy decreases, and the rate of energy decrease is $P\,dV_0/dt$. Stick in the relationship between specific heat and energy and continue from there.

11. Sep 29, 2009

### leumas614

ok, I got it.

I asked the TA and he said the key relationship that I was missing is this:

$$\frac{dn}{dt}=n\frac{Sp}{V}$$ The units work out.

So I worked it out from there. What you can do then is say $$U=H_{out}n_{out}$$ since $$H_{in}n_{in}$$ is zero because there is no mass flowing into chamber.

Then you can say dU is nCvdT so dU/dt is same as above post. Equate $$nCvdT$$ with $$H_{out}n_{out}$$n (you can replace H with U+RT) and you take it from there. I worked it out and basically you get a P=noRTo/V times an exponential e.


text:
## Lines, Line-point Incidences and Crossing Families in Dense Sets

### Author: Pavel Valtr

ABSTRACT

Let ${\cal P}$ be a set of $n$ points in the plane. We say that ${\cal P}$ is {\it dense} if the ratio between the maximum and the minimum distance in ${\cal P}$ is of order $O(\sqrt n)$. A set $C$ of line segments in the plane is called {\it a crossing family} if the relative interiors of any two line segments of $C$ intersect. Vertices of line segments of a crossing family $C$ are called {\it vertices of $C$}. It is known that for any set ${\cal P}$ of $n$ points in general position in the plane there is a crossing family of size $\Omega(\sqrt n)$ with vertices in ${\cal P}$. In this paper we show that if ${\cal P}$ is dense then there is a crossing family of almost linear size with vertices in ${\cal P}$.

The above result is related to well-known results of Beck and of Szemer\' edi and Trotter. Beck proved that any set ${\cal P}$ of $n$ points in the plane, not most of them on a line, determines at least $\Omega(n^2)$ different lines. Szemer\' edi and Trotter proved that if ${\cal P}$ is a set of $n$ points and ${\cal L}$ is a set of $m$ lines then there are at most $O(m^{2/3}n^{2/3}+m+n)$ incidences between points of ${\cal P}$ and lines of ${\cal L}$. We study whether or not the bounds shown by Beck and by Szemer\' edi and Trotter hold for any dense set ${\cal P}$ even if the notion of incidence is extended so that a point is considered to be incident to a line $l$ if it lies in a small neighborhood of $l$. In the first case we get very close to the conjectured bound $\Omega(n^2)$. In the second case we obtain a bound of order $O(\min\{m^{3/4}n^{5/8}\log^{1/4} n,n\sqrt m\})$.

Paper available at: ftp://dimacs.rutgers.edu/pub/dimacs/TechnicalReports/TechReports/1994/94-55.ps

DIMACS Home Page


text:
## Hom-Lie 2-algebras    [PDF]

Yunhe Sheng
In this paper, we introduce the notions of hom-Lie 2-algebras, which is the categorification of hom-Lie algebras, $HL_\infty$-algebras, which is the hom-analogue of $L_\infty$-algebras, and crossed modules of hom-Lie algebras. We prove that the category of hom-Lie 2-algebras and the category of 2-term $HL_\infty$-algebras are equivalent. We give a detailed study on skeletal hom-Lie 2-algebras. In particular, we construct the hom-analogues of the string Lie 2-algebras associated to any semisimple involutive hom-Lie algebras. We also proved that there is a one-to-one correspondence between strict hom-Lie 2-algebras and crossed modules of hom-Lie algebras. We give the construction of strict hom-Lie 2-algebras from hom-left-symmetric algebras and symplectic hom-Lie algebras.
View original: http://arxiv.org/abs/1110.3405


text:
# pr.chance – Constructing representations of chance revision capabilities Answer

Hello pricey customer to our community We will proffer you an answer to this query pr.chance – Constructing representations of chance revision capabilities ,and the respond will breathe typical via documented data sources, We welcome you and proffer you fresh questions and solutions, Many customer are questioning in regards to the respond to this query.

pr.chance – Constructing representations of chance revision capabilities

Let $$P$$ breathe a chance distribution over a finite Boolean algebra $$mathfrak{B}$$, and pickle a parameter $$t_{P} in (frac{2}{3}, 1)$$. Define the `revision duty of $$P$$‘, $$R_{P}: mathfrak{B}setminus{bot} rightarrow mathbb{P}(mathfrak{B})$$ as follows, (assuming that $$P$$ assigns constructive chance to all occasions aside from the null incident $$bot$$)

$$R_{P}(X) = {Y in mathfrak{B}|P(Y|X) geq t}$$

In phrases, the revision duty of $$P$$ takes an incident $$X$$ in $$mathfrak{B}$$ and returns the clique of all occasions whose conditional chance given $$X$$ (in line with $$P$$) is not less than $$t_{p}$$.

I wish to discover common system for setting up, for any given $$P$$ and any $$t_{P} in (frac{2}{3}, 1)$$, one other chance duty $$P^{*}$$ with a corresponding parameter $$t_{P^{*}} in (frac{2}{3}, 1)$$ such that $$R_{P^{*}} = R_{P}$$. In different phrases, I need a system that constructs, for any chance duty and stuck parameter within the specified meander, one other duty that (for some altenative of parameter within the specified meander) provides ascend to the identical revision duty as the unique duty.

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text:
HomeEnglishClass 12MathsChapterMatrices

show that the matrix
 A=[...

show that the matrix A=[(-5,-8,0),(3,5,0),(1,2,-1)] is involutory.

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Text Solution

Solution :
A^(2)=A.A=[(-5,-8,0),(3,5,0),(1,2,-1)]xx[(-5,-8,0),(3,5,0),(1,2,-1)]
 =[(25-24+0,40-40+0,0+0+0),(-1+15+0,-24+25+0,0+0+0),(-5+6-1,-8+10-2,0+0+1)]=[(1,0,0),(0,1,0),(0,0,1)]=I
 Hence , the given matrix A is involutory.

Transcript

TimeTranscript
00:00 - 00:59hello students in this question we have to show that the matrix A is involutory matrix so let's move to the solution so guys Matrix a matrix if a square of this matrix is equal to identity Matrix so let's first we write over here matrix A which is given in the question -531 - 85200 - 1 year we need to find out a square Matrix for that we have to multiply a matrix with itself then we can get a square Matrix ok so let's have a right over here Matrix twice so -531 - 520001
01:00 - 01:59now that's right over here is matrix again -531 - 85200 - 1 now so guys next one is to multiply two Matrix so let's multiply so here this is Rowan Row 2 and 3 we have to multiply this all rows with column one column to and column 3 ok so that's why do that so guys let's be multiplied is no one with column one column to row and column 30 - 5 x minus 5 25 - 8 x 3 - 20 40 x 10 now 2018 with column to row - 5 x
02:00 - 02:59-88 would become 40 - 8 time 5 - 40 0 x 20 again this row 1 with column 30 - 5 x 200 - 8 X 2000 x minus 10 now let's we multiply this Row 2 with column 1 ok so 3 x minus 5 minus 15 x 315 10 X 10 3 x minus 8 - 24 5 x 5 20 50 x 20 again this role to withcolumn 33 X
03:00 - 03:59005 X 000 x minus 10 2003 with column one column to row and column 3 flats first we multiply is Rotary withcolumn 11 x minus 5 minus 5 2 x 3 plus 6 - 1 X 1 -1 again 1 x minus 8 - 82 times 5 + 10 - 1 X 2 - 2 again over here One X 0 0 to 1000 - 1 x minus 1 + 1 ok now why is this is our a square Matrix so let's we add each element of this Matrix so when you add this element you will get
04:00 - 04:59one over here here are getting here I get 0 9 hair plus 40 - 4000 hair you get to one because - 24 + 25 so when you solve this you will get one ok here you are to get zero here 0 0 and 1 so guys we find that a square matrix is equal to identity Matrix so has proved that our given Matrix a given matrix A is in nutri metrics thank you guys for watching this video


text:
# Why does Equal[“string”, False] not evaluate to False?

Why does Equal["string", False] not evaluate to False? I would have thought Mathematica could prove that these two things are not equal.

-
I would think that Mathematica is unable to compare a string with a boolean; it's non-sensical. Does this usually work when comparing distinct types of variables? – Myridium Jun 1 '14 at 3:31

You have to use "===" like

 "string" === False

Or

SameQ["string",False]

-
I think OP probably knows this. The question is, why doesn't Equal give the same result? SameQ doesn't express a mathematical relation. – Oleksandr R. May 31 '14 at 15:11
@OleksandrR. Quite frankly: I don't know, all the more so since "string"==1 returns False. – eldo May 31 '14 at 15:36
Indeed, it's a puzzling inconsistency and IMO a good question. The answer seems to be something to do with strings being treated as symbols in some contexts but not others. For example, Solve can't solve with respect to strings, but Assuming["string" == False, Refine["string" == False]] works the same with a string as it would with any other symbol. – Oleksandr R. May 31 '14 at 16:37
@OleksandrR. FWIW: Assuming["string" == False, Refine["string" == False]] calls PolynomialReduce[-"string" + False, {-"string" + False}, {False}, MonomialOrder -> DegreeReverseLexicographic], – Michael E2 Jun 1 '14 at 0:19

I think it is because Equal treats symbols as having an indeterminate value, but as @M_goldberg points out this isn't the whole story.

Equal["foo", foo]
(* "foo" == foo *)

This is true when one side is numeric (but not a number per se) and the other is a string:

Equal["foo", GoldenRatio]
(* "foo" == GoldenRatio *)

Equal["foo", Sqrt[2]]  (* Sqrt can be overridden *)
(* "foo" == Sqrt[2] *)

Equal[Sqrt, Sqrt[2]]
(* Sqrt == Sqrt[2] *)

Equal also chokes on unevaluated, nonnumeric function expressions, including nonsense such as this:

Equal[2, 2[2]]
(* 2 == 2[2] *)

Equal will evaluate if both sides are numeric; also if both sides are determinate, such as comparing a string and a number. It also evaluates to True if the symbols are the same, or if one is True and the other is False. (There may be other exceptions to this rule. Thanks to @m_goldberg for point out these examples.)

Equal["foo", N@Sqrt[2]]
(* False *)

Equal[True, False]
(* False *)

Equal[x, x]
(* True *)

Override Sqrt:

Block[{Sqrt = "foo" &},
Equal["foo", Sqrt[2]]
]
(* True *)

-
Interesting. What would be the definition for "determinate"? – eldo May 31 '14 at 20:55
I suppose String or one of the number types. I don't think they have a name. It's not, for example, "atomic." – Michael E2 May 31 '14 at 21:23
It can't be as simple as "Equal treats symbols as having an indeterminate value". Consider, Equal[False, #] & /@ {True, False} which returns {False, True}. Further, Equal clearly calls on SameQ and returns True if SameQ does; i.e., Equal[x, x] is always True for any x, including symbols. – m_goldberg May 31 '14 at 21:51
Equal also considers True and False to be comparable with each other. As far as I can tell, it always remains unevaluated when the comparands are not in the same category, but what it considers a category is unclear. However it does seem that strings are in the same category as (unassigned) symbols as far as Equal is concerned, even if other functions treat them as distinct. x == x and "x" == "x" both give True. It's perhaps not as simple as falling through to SameQ, as the example in my comment under the other answer shows. – Oleksandr R. May 31 '14 at 21:53
"foo" == 1 giving False while False == 1 remains unevaluated is a puzzling case, however. – Oleksandr R. May 31 '14 at 21:58


text:
## Monday, February 20, 2017

### LeetCode OJ - Total Hamming Distance

Problem:

Analysis:

The input could have 10,000 elements, we need to figure out a way to compute the total hamming distance without going through all pairs.

Solution:

Imagine for a bit they are all 1 bit number (i.e. either 0 or 1), then it is easy. Just count the number of zeros and number of ones and multiply them together.

For a full solution, just do this for all bits.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/total-hamming-distance

#include "LEET_TOTAL_HAMMING_DISTANCE.h"
#include <map>
#include <iostream>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_TOTAL_HAMMING_DISTANCE
{
class Solution
{
public:
int totalHammingDistance(vector<int>& nums)
{
int dist = 0;
for (int i = 0; i < 32; i++)
{
int zero = 0;
int ones = 0;
for (size_t j = 0; j < nums.size(); j++)
{
bool is_zero = (nums[j] & mask) == 0;
if (is_zero) { zero++; } else { ones++; }
}
dist += zero * ones;
}
return dist;
}
};
};

using namespace _LEET_TOTAL_HAMMING_DISTANCE;

int LEET_TOTAL_HAMMING_DISTANCE()
{
Solution solution;
int input_array[] = { 4, 14, 2 };
vector<int> input(input_array, input_array + _countof(input_array));
cout << solution.totalHammingDistance(input) << endl;
return 0;
}

## Wednesday, February 15, 2017

### LeetCode OJ - Most Frequent Subtree Sum

Problem:

Analysis:

There are two subproblems, find the subtree sum and finding the ones with max frequency. Both of them are easy.

Solution:

We just need to combine them. In the unwinding phase of the tree traversal, we can get the subtree sums. We put them in a histogram, and finally we can get the ones with maximum frequency.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/most-frequent-subtree-sum/

#include "LEET_MOST_FREQUENT_SUBTREE_SUM.h"
#include <map>
#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

namespace _LEET_MOST_FREQUENT_SUBTREE_SUM
{
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution
{
public:
int findFrequentTreeSum(TreeNode* root, map<int, int>& histogram)
{
if (root == nullptr)
{
return 0;
}
int left = findFrequentTreeSum(root->left, histogram);
int right = findFrequentTreeSum(root->right, histogram);
int sum = left + right + root->val;
auto probe = histogram.find(sum);
if (probe == histogram.end())
{
histogram.insert(make_pair(sum, 1));
}
else
{
probe->second++;
}

return sum;
}
vector<int> findFrequentTreeSum(TreeNode* root)
{
map<int, int> histogram;
findFrequentTreeSum(root, histogram);
int max_freq = -1;
for (auto pair : histogram)
{
max_freq = max(max_freq, pair.second);
}
vector<int> solution;
for (auto pair : histogram)
{
if (pair.second == max_freq)
{
solution.push_back(pair.first);
}
}

return solution;
}
};
};

using namespace _LEET_MOST_FREQUENT_SUBTREE_SUM;

int LEET_MOST_FREQUENT_SUBTREE_SUM()
{
Solution solution;
TreeNode a(5);
TreeNode b(2);
TreeNode c(-3);
a.left = &b;
a.right = &c;
auto test1 = solution.findFrequentTreeSum(&a);
for (auto result : test1)
{
cout << result << " ";
}
cout << endl;

TreeNode p(5);
TreeNode q(2);
TreeNode r(-5);
p.left = &q;
p.right = &r;
auto test2 = solution.findFrequentTreeSum(&p);
for (auto result : test2)
{
cout << result << " ";
}
cout << endl;
return 0;
}

### LeetCode OJ - Find Bottom Left Tree Value

Problem:

Analysis:

This problem look very much like the previous one, with a twist. We need the leftmost entry. The easiest way to make sure we get the left most entry first is simply doing an in-order traversal.

Solution:

Just do an in-order traversal, keep track of the answer and update as necessary. As an aside, I hate pass by reference, that is so dangerous. I wouldn't use them in product code. But sometimes I do use it for competitive programming because it make the syntax so much simpler. A nice tool for a small problem could be a disaster used in scale ...

Code:

#include "stdafx.h"

// https://leetcode.com/problems/find-bottom-left-tree-value/

#include "LEET_FIND_BOTTOM_LEFT_TREE_VALUE.h"
#include <map>
#include <iostream>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_FIND_BOTTOM_LEFT_TREE_VALUE
{
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution
{
public:
void findBottomLeftValue(TreeNode* root, int depth, int& max_layer, int& answer)
{
if (root == nullptr)
{
return;
}
findBottomLeftValue(root->left, depth + 1, max_layer, answer);
if (depth > max_layer)
{
max_layer = depth;
}
findBottomLeftValue(root->right, depth + 1, max_layer, answer);
}
int findBottomLeftValue(TreeNode* root)
{
int max_layer = -1;
}
};
};

using namespace _LEET_FIND_BOTTOM_LEFT_TREE_VALUE;

int LEET_FIND_BOTTOM_LEFT_TREE_VALUE()
{
Solution solution;
TreeNode a(2);
TreeNode b(1);
TreeNode c(3);
a.left = &b;
a.right = &c;
cout << solution.findBottomLeftValue(&a) << endl;

TreeNode p(1);
TreeNode q(2);
TreeNode r(3);
TreeNode s(4);
TreeNode t(5);
TreeNode u(6);
TreeNode v(7);
p.left = &q;
p.right = &r;
q.left = &s;
r.left = &t;
r.right = &u;
t.left = &v;
cout << solution.findBottomLeftValue(&p) << endl;
return 0;
}

### LeetCode OJ - Find Largest Value in Each Tree Row

Problem:

Analysis:

Given only the root, there is no way I can tell which subtree have the largest element in a certain row, so I have to walk the whole tree anyway. To put it more technically, suppose there exist an algorithm that does not visit all nodes, in particular, it does not visit a node p. An adversary could put a huge number in the node p and therefore the algorithm output a wrong result. Therefore any algorithm must walk the whole tree.

Solution:

Now we know we have to walk the whole tree, then the solution is relatively simple. Suppose we use a tree traversal (any traversal would do) and also keep track of the current height, then we can maintain a vector of maximum values in the current row. That would be the final result.

The interesting thing about a (just about any) tree traversal is that it increase the depth only by 1. Therefore, the vector of maximum number can increase at most by 1. That make the implementation easy by a simple push_back to the vector.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/find-largest-value-in-each-tree-row/

#include "LEET_FIND_LARGEST_VALUE_IN_EACH_TREE_ROW.h"
#include <map>
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_FIND_LARGEST_VALUE_IN_EACH_TREE_ROW
{
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution
{
public:
void largestValues(TreeNode* root, int depth, vector<int>& result)
{
if (root == nullptr)
{
return;
}
if (result.size() > depth)
{
result[depth] = max(result[depth], root->val);
}
else
{
result.push_back(root->val);
}
largestValues(root->left, depth + 1, result);
largestValues(root->right, depth + 1, result);
}
vector<int> largestValues(TreeNode* root)
{
vector<int> result;
largestValues(root, 0, result);
return result;
}
};
};

using namespace _LEET_FIND_LARGEST_VALUE_IN_EACH_TREE_ROW;

int LEET_FIND_LARGEST_VALUE_IN_EACH_TREE_ROW()
{
Solution solution;
TreeNode a(1);
TreeNode b(3);
TreeNode c(2);
TreeNode d(5);
TreeNode e(3);
TreeNode f(9);
a.left = &b;
a.right = &c;
b.left = &d;
b.right = &e;
c.right = &f;
auto result = solution.largestValues(&a);
for (auto&& r : result)
{
cout << r << endl;
}
return 0;
}

### LeetCode OJ - Longest Palindromic Subsequence

Problem:

Analysis:

Recently I have been reading up about some algorithms on bioinformatics and this one does look similar. The idea is you use interval dynamic programming.

Suppose we know the answer (i.e. the length of the longest palindromic subsequence) of all substrings with length less than d. We can use that information to compute the answer for all substring with length d.

To make the description simpler, let's abbreviate the longest palindromic subsequence to just LPS

There are a few possibilities, for the LPS, one of the following must hold:

• The first character is paired with the last character.
• The first character is paired with some other character but not the last one.
• The first character is not included in the LPS.

For the first one, of course, the first and last one must be equal, and then we take the
LPS on the inner d - 2 characters. The length is therefore 2 + LPS(the inner d-2 characters)

For the second one, the last character cannot be paired with anything since the first one is paired with something else and apparently there is nothing before the first one. So this case the LPS is really just the LPS of the beginning d -1 characters.

For the third one, the first character cannot be paired with anything by design, so this case the LPS is really just the LPS of the last d -1 characters.

That make a very simple recurrence relation, define lps(i, j) be the length of the LPS of the substring of S from i to j inclusive.

lps(i, j) = max(lps(i, j - 1), lps(i + 1, j), 2 + lps(i + 1, j - 1))

Of course, the last term apply only if S[i] == S[j].

Solution:

We define lps2(i, len) = lps(i, i + len - 1). With this transformation, we know that we only need the last two lengths when computing the current length values, therefore we can save memory by keeping only the last two rows.

To avoid copying, I used a simple circular buffer scheme, that should make reading the code a little easier to understand this.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/longest-palindromic-subsequence/

#include "LEET_LONGEST_PALINDROMIC_SUBSEQUENCE.h"
#include <map>
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_LONGEST_PALINDROMIC_SUBSEQUENCE
{
class Solution
{
public:
int longestPalindromeSubseq(string s)
{
if (s.length() == 0)
{
return 0;
}

vector<int> result[3];
result[0].resize(s.length());
result[1].resize(s.length());
result[2].resize(s.length());
for (size_t i = 0; i < s.length(); i++)
{
// The maximum palindromic subsequence of a subsequence starting at i with length 0 is 0
result[0][i] = 0;
// The maximum palindromic subsequence of a subsequence starting at i with length 1 is 1
result[1][i] = 1;
}

int current_row = 2;
int last_row = 1;
int last_last_row = 0;
for (int length = 2; length <= s.length(); length++)
{
for (int i = 0; i + length <= s.length(); i++)
{
int candidate = result[last_row][i];
candidate = max(candidate, result[last_row][i + 1]);
if (s[i] == s[i + length - 1])
{
candidate = max(candidate, 2 + result[last_last_row][i + 1]);
}
result[current_row][i] = candidate;
}
current_row = (current_row + 1) % 3;
last_row = (last_row + 1) % 3;
last_last_row = (last_last_row + 1) % 3;
}
return result[last_row][0];
}
};
};

using namespace _LEET_LONGEST_PALINDROMIC_SUBSEQUENCE;

int LEET_LONGEST_PALINDROMIC_SUBSEQUENCE()
{
Solution solution;
cout << solution.longestPalindromeSubseq("") << endl;
cout << solution.longestPalindromeSubseq("bbbab") << endl;
cout << solution.longestPalindromeSubseq("cbbd") << endl;
return 0;
}

## Friday, February 10, 2017

### LeetCode OJ - Find Mode in Binary Search Tree

Problem:

Analysis:

A standard idea is to build a frequency table. Once the table is built, we go through the table to find out who are the winners. In the worst case where the tree has all distinct elements, this will lead to an extra linear space.

Solution:

The key idea is that we can do an in-order traversal of the tree to get a sorted order. We build the tree as usual with a single difference - the table only contain two entries. The current best entry, and the last number entry. Suppose an element X doesn't have enough count, and in the in-order traversal we see Y now, X is never going to win, so we can get X out of the table. Now we have a constant extra space algorithm! (Of course, you still need to save all the answers to report them, which, in the worst case, linear)

Code：

#include "stdafx.h"

// https://leetcode.com/problems/find-mode-in-binary-search-tree/

#include "LEET_FIND_MODE_IN_BINARY_SEARCH_TREE.h"
#include <map>
#include <iostream>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_FIND_MODE_IN_BINARY_SEARCH_TREE
{
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
void findMode(TreeNode* root, vector<int>& result, int& max_count, int& last_node, int& last_count)
{
if (root == nullptr)
{
return;
}

if (root->left != nullptr)
{
findMode(root->left, result, max_count, last_node, last_count);
}

if (max_count == 0)
{
// This is possible only for the very first call
result.push_back(root->val);
max_count = 1;
last_node = root->val;
last_count = 1;
}
else
{
if (root->val == last_node)
{
last_count++;
}
else
{
last_node = root->val;
last_count = 1;
}

if (last_count == max_count)
{
result.push_back(last_node);
}
else if (last_count > max_count)
{
result.clear();
result.push_back(last_node);
max_count = last_count;
}
}

if (root->right != nullptr)
{
findMode(root->right, result, max_count, last_node, last_count);
}
}

vector<int> findMode(TreeNode* root)
{
vector<int> result;
int max_count = 0;
int last_node = 0;
int last_count = 0;
findMode(root, result, max_count, last_node, last_count);
return result;
}
};
};

using namespace _LEET_FIND_MODE_IN_BINARY_SEARCH_TREE;

int LEET_FIND_MODE_IN_BINARY_SEARCH_TREE()
{
Solution solution;
TreeNode a(1);
TreeNode b(2);
TreeNode c(2);
a.left = nullptr;
a.right = &b;
b.left = &c;
b.right = nullptr;
c.left = nullptr;
c.right = nullptr;
vector<int> result = solution.findMode(&a);
for (int i = 0; i < result.size(); i++)
{
cout << result[i] << endl;
}
return 0;
}


text:
# Operators¶

Operators normally use special characters or keywords to manipulate single or multiple operands (either scalar, columns, or subqueries). They are a kind of a binary type function that requires a minimum of two (or more) operands to return results.

## Order of precedence¶

The table below shows the order of precedence for the operators.

Order of precedence for operators

Order

Operators

1

parentheses

2

3

4

comparison operators,

[NOT] LIKE, [NOT] BETWEEN, [NOT] IN,

IS [NOT] NULL, IS [NOT] TRUE, IS [NOT] FALSE

5

NOT

6

AND

7

OR

In case if the operators have the same precedence they are evaluated left-to-right.

## Arithmetic operators¶

Supported datatypes: TINYINT, SMALLINT, INT, BIGINT, DECIMAL, FLOAT, DOUBLE

Arithmetic operators provide the capability to perform arithmetic operations on columns or numbers.

The addition operator performs a sum of two numeric columns or numbers.

#### Examples¶

Perform addition of elements in two columns.

SELECT <col_1> + <col_2>
FROM <table_name>

It can also add a number to a column

SELECT <col_1> + <number>
FROM <table_name>

or add two or more numbers and columns together.

SELECT <col_1> + <col_2> + <number>
FROM <table_name>

### Subtraction¶

The subtraction operator subtracts number in one column from the values in another one or a number.

#### Examples¶

Subtract values in one column from another one.

SELECT <col_1> - <col_2>
FROM <table_name>

Subtract a number from a column.

SELECT <col_1> - <number>
FROM <table_name>

### Multiplication¶

The multiplication operator multiplies values between two or more columns or numbers.

#### Examples¶

Multiply values between two columns.

SELECT <col_1> * <col_2>
FROM <table_name>

Mutliply the values in one column and a number.

SELECT <col_1> * <number>
FROM <table_name>

### Division¶

The division operator takes values from one column and divides them by the values in another or a number.

Warning

If the divisor column contains 0 or NULL BlazingSQL will return NULL.

#### Examples¶

Divide values from one column by the values in another column.

SELECT <col_1> / <col_2>
FROM <table_name>

Divide values from one column by the a number.

Warning

Dividing explicitly by a literal 0 will throw a RunExecuteGraphError exception.

SELECT <col_1> * <number>
FROM <table_name>

### Negation¶

The negation operator returns a negative of a value in a column or of a number.

#### Examples¶

Negate values in a column.

SELECT -<col_1>
FROM <table_name>

## Comparison operators¶

Supported datatypes: TINYINT, SMALLINT, INT, BIGINT, DECIMAL, FLOAT, DOUBLE TIMESTAMP, VARCHAR, BOOLEAN

Comparison operators compare two values and return a truth value: either TRUE or FALSE. These normally require columns or literals with comparable data types.

### <¶

Operator to check if values in one column are less than the values in another, or a literal.

#### Example¶

SELECT <col_1> < <col_2>
FROM <table_name>

### <=¶

Operator to check if values in one column are less than or equal to the values in another, or a literal.

#### Example¶

SELECT <col_1> <= <col_2>
FROM <table_name>

### >¶

Operator to check if values in one column are more than the values in another, or a literal.

#### Example¶

SELECT <col_1> > <col_2>
FROM <table_name>

### >=¶

Operator to check if values in one column are more than or equal to the values in another, or a literal.

#### Example¶

SELECT <col_1> >= <col_2>
FROM <table_name>

### =¶

Operator to check if values in one column are equal to the values in another, or a literal.

#### Example¶

SELECT <col_1> = <col_2>
FROM <table_name>

### <>¶

Operator to check if values in one column are not equal to the values in another, or a literal.

#### Example¶

SELECT <col_1> <> <col_2>
FROM <table_name>

### BETWEEN¶

Operator to check if values in a column lay in or outside a specified range, specified by either literals or columns.

#### Examples¶

The range can be specified by literals.

SELECT <col_1> BETWEEN <number_1> AND <number_2>
FROM <table_name>

The range can also be specified by columns.

SELECT <col_1> BETWEEN <col_2> AND <col_3>
FROM <table_name>

To check if values lay outside of the range use NOT qualifier.

SELECT <col_1> NOT BETWEEN <number_1> AND <number_2>
FROM <table_name>

### LIKE¶

Supported datatypes: VARCHAR

Operator to check if a string matches a pattern specified in the second operand. The % sign matches any number of characters while the _ matches only a single character. To match either %, _, or the \  you need to prefix the pattern with another \  thus resulting in \% to match a percent sign, \_ to match underscore, and \\ to match a backslash.

Note

The LIKE operator is case-sensitive so to match strings with different casing all strings need to be normalized. See LOWER, UPPER or INITCAP.

#### Examples¶

Check, if the word finishes with the word WORK.

SELECT <col_1> LIKE '%WORK'
FROM <table_name>

Check, if the string matches a phone number pattern.

SELECT <col_1> LIKE '(___) ___-____'
FROM <table_name>

### IN¶

Operator to check if values in a column are found on a specified list. The list can be literals or a subquery. Check the dedicated section on IN operators.

## Logical operators¶

Supported datatypes: BOOLEAN

Logical operators allow to chain multiple boolean columns or expressions that return boolean value to obtain more complex logical expressions.

### AND¶

The logical conjunction operator returns TRUE value if and only if both inputs are TRUE.

Truth table for AND

Input 1

Input 2

Result

FALSE

FALSE

FALSE

FALSE

TRUE

FALSE

TRUE

FALSE

FALSE

TRUE

TRUE

TRUE

FALSE

NULL

FALSE

TRUE

NULL

NULL

NULL

TRUE

NULL

NULL

FALSE

FALSE

NULL

NULL

NULL

#### Examples¶

Simple conjunction of two boolean columns.

SELECT <col_1_bool> AND <col_2_bool>
FROM <table_name>

Negated column syntax.

SELECT <col_1_bool> AND NOT <col_2_bool>
FROM <table_name>

### OR¶

The logical disjunction (or alternative) operator returns TRUE value if one of the inputs is TRUE.

Truth table for OR

Input 1

Input 2

Result

FALSE

FALSE

FALSE

FALSE

TRUE

TRUE

TRUE

FALSE

TRUE

TRUE

TRUE

TRUE

FALSE

NULL

NULL

TRUE

NULL

TRUE

NULL

TRUE

TRUE

NULL

FALSE

NULL

NULL

NULL

NULL

#### Examples¶

Simple disjunction of two boolean columns.

SELECT <col_1_bool> OR <col_2_bool>
FROM <table_name>

Negated column syntax.

SELECT <col_1_bool> OR NOT <col_2_bool>
FROM <table_name>

### NOT¶

The logical negation operator returns the reverse truth value.

Truth table for NOT

Input

Result

FALSE

TRUE

TRUE

FALSE

NULL

NULL

#### Example¶

Simple negation of a boolean column.

SELECT NOT <col_1_bool>
FROM <table_name>

## IN operators¶

Supported datatypes: TINYINT, SMALLINT, INT, BIGINT, DECIMAL, FLOAT, DOUBLE TIMESTAMP, VARCHAR, BOOLEAN

The IN operator compares the values in a column with a list of allowed literals or values returned from a subquery, and returns a TRUE if the value is found.

### Finding matches¶

The syntax flags all the values that are found on the list.

#### Examples¶

Compare with a list of literals.

SELECT <col_1> IN ('literal_1', 'literal_2')
FROM <table_name>

Compare with a list generated by a subquery.

SELECT <col_1> IN (
SELECT <col_2>
FROM <table_2>
)
FROM <table_1>

### Inverting the matches¶

The syntax flags all the values that are found on the list. The comparisons can also be negated thus flagging all the values that are not found in the list

SELECT <col_1> NOT IN ('literal_1', 'literal_2')
FROM <table_name>

or query

SELECT <col_1> NOT IN (
SELECT <col_2>
FROM <table_2>
)
FROM <table_1>

## IS operators¶

The IS operators test if something is or is not. These operators never return NULL.

### IS [NOT] NULL¶

Supported datatypes: TINYINT, SMALLINT, INT, BIGINT, DECIMAL, FLOAT, DOUBLE TIMESTAMP, VARCHAR, BOOLEAN

Tests if values in the column are NULL.

#### Example¶

Check if values are NULL.

SELECT <col_1> IS NULL
FROM <table_name>

Check if values are not NULL.

SELECT <col_1> IS NULL
FROM <table_name>

### IS [NOT] TRUE and IS [NOT] FALSE¶

Supported datatypes: BOOLEAN

Tests if values in the column are TRUE or FALSE.

#### Example¶

Check if values are TRUE.

SELECT <col_1> IS TRUE
FROM <table_name>

An equivalent would be to test if the values are not FALSE

SELECT <col_1> IS NOT FALSE
FROM <table_name>

Check if values are FALSE.

SELECT <col_1> IS FALSE
FROM <table_name>

Alternatively:

SELECT <col_1> IS NOT TRUE
FROM <table_name>

## CONCAT operator¶

Supported datatypes: VARCHAR

The concatenation operator joins two or more string type values into one.

CONCAT

### || operator¶

Combines two or more string type values (columns or literals) into one.

#### Example¶

Concatenate two text columns.

SELECT <col_1> || <col_2>
FROM <table_name>

Concatenate a column with a string.

SELECT <col_1> || <string_literal>
FROM <table_name>

Add underscore between text values in two columns.

SELECT <col_1> || '_' || <col_2>
FROM <table_name>


text:
# Thread: Alternating Series Estimation Theorem

1. ## Alternating Series Estimation Theorem

Use the alternating series estimation theorem to approximate the sum of $\displaystyle (-1)^n/(n!)$ from n=0 to infinity with an error of $\displaystyle .000005$.

I know that after summing a number of terms, the remainder(error) will be less than the first omitted term, so

I've set up the problem like this:

$\displaystyle .000005 < 1/(n+1)!$

Is this right, and if so, what's next?

2. Originally Posted by JewelsofHearts
Use the alternating series estimation theorem to approximate the sum of $\displaystyle (-1)^n/(n!)$ from n=0 to infinity with an error of $\displaystyle .000005$.

I know that after summing a number of terms, the remainder(error) will be less than the first omitted term, so

I've set up the problem like this:

$\displaystyle .000005 < 1/(n+1)!$

Is this right, and if so, what's next?
So you want the least integer $\displaystyle $$n such that: \displaystyle (n+1)!>20000 To proceed you need either to use trial and error or some suitable approximation for the factorial (and then a numerical solution of the resulting equation). (look up \displaystyle$$7!$ and $\displaystyle$$8!$)

CB

3. Shouldn't it be (n+1)!<200000?

4. Originally Posted by JewelsofHearts
Shouldn't it be (n+1)!<200000?
Probably counting zeros is not something I am particularly careful about

5. Not only the zeros, but I think the direction of the inequality sign is supposed to be different. I really don't know, though. I don't understand the theorem.

6. error, $\displaystyle E < \dfrac{1}{(n+1)!} < 5 \times 10^{-6}$

$\displaystyle (n+1)! > \dfrac{1}{5 \times 10^{-6}} = 2 \times 10^5$

note the following factorial values ...

$\displaystyle 8! = 40320$

$\displaystyle 9! = 362880$

so, how many terms of the series would you need to add to get an error less than $\displaystyle 5 \times 10^{-6}$ ?

fyi, to check your work, note ...

$\displaystyle \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = \frac{1}{e}$


text:
# A Primer on Sustainable Business, v. 1.0

by Nancy E. Landrum and Sandra Edwards

## 7.1 Information Technology

According to a recent study,McKinsey & Company (2008). the carbon dioxide (CO2) emissions of the U.S. information technology (IT) industry already exceed the emissions of entire nations, such as Argentina, the Netherlands, and Malaysia. At the current pace, emissions are expected to quadruple and the IT industry is expected to exceed the airline industry in emissions by 2020. The research shows that the U.S. IT industry is increasing its energy usage at a rate of 10%–20% annually. The study estimates that at this rising rate of energy usage, the United States will need to build 30 new coal-fired or nuclear power plants by 2015 solely to support the nation’s IT usage.

The Smart 2020 reportGlobal eSustainability Initiative (2008). estimates that IT has the potential to reduce worldwide global emissions by 15% by 2020. According to this report, the greatest global opportunities for IT to help reduce emissions are in the areas of smart motor systems in China’s manufacturing industry, smart logistics in Europe’s transport and storage industries, smart building technologies in North America, and smart grid technologies in India.

In order to address growing concerns over the environmental impact of the IT industry and to take advantage of opportunities, the proactive and sustainability-focused business will develop green IT strategies. Green IT strategies are not only proactive and environmentally friendly but can also ultimately reduce the company’s energy consumption and costs.

There are a number of suggestions for green IT strategies. For example, the same McKinsey & CompanyMcKinsey & Company (2008). study suggests that most companies could double energy efficiency of data centers by 2012. The researchers propose automobile CAFE-type industry standards (corporate average fuel economy [CAFE] standards require an automaker to meet minimum average fuel efficiency across its entire fleet of manufactured vehicles). These CAFE-type industry standards would be used for measuring efficiency in conjunction with the following suggestions: creating an energy-efficiency dashboard, sealing cable cutoffs, turning off and removing excess hardware, increasing temperatures, virtualization, and upgrading equipment.

Greening the data center is often the starting point of green IT strategies. The first step in your green IT strategy is to know current energy usage, where energy is used and by what specific equipment, what usage is efficient, and what usage is wasteful. There are a number of IT-enabled energy-reduction systems (such as EnviroCube or EnerSure monitoring devices or Verdiem software tools), smart metering, and other technologies that can ultimately reduce cooling costs and electricity consumption. As if that is not incentive enough, the U.S. Environmental Protection Agency (EPA) is currently developing an ENERGY STAR rating for data center infrastructure, and the European Commission has developed a Code of Conduct for Green Data Centers. We will now look at some specific green IT strategies designed to increase efficiency and decrease energy consumption.

## Storage

Storage resource management (SRM)A method of increasing storage utilization and decreasing power needs to increase efficiency and decrease energy consumption. helps identify underutilized capacity, removes or reassigns unused storage, identifies old or noncritical data that could be moved to less expensive storage, removes inappropriate data, and helps predict future capacity requirements. SRM can increase storage utilization and decrease power needs. Companies that have used SRM have experienced utilization improvements of 30%–40%.Harrison (2008).

Storage virtualizationA method of increasing storage utilization by integrating several storage networks into one virtual storage site. allows the work of several storage networks and devices to be integrated to appear as one virtual storage site. Storage virtualization can improve storage utilization by allowing storage to be assigned where it is needed.

Another tool is continuous data protectionA method of increasing storage utilization by offering continuous or real-time byte-level backup of changes to documents., which offers continuous or real-time byte-level backup of changes to documents. This often requires less storage space than traditional file-level backups.

Yet another option for reducing storage costs is storage tieringA method of increasing storage utilization and decreasing power consumption by assigning categories of data to specific types of storage media, which can be automatically managed through software programs.. Tiered storage assigns categories of data to specific types of storage media. The categories are company-defined based on levels of security and protection, usage, performance, or other considerations. This process can be automatically managed through software programs. The benefit of tiered storage is that it allows companies to increase utilization rates and decrease power consumption and cooling costs.

## Servers

One green IT approach being used is server consolidationA means of reducing energy usage and increasing energy efficiency by reducing the number of servers used by running multiple applications on each server., which reduces the number of servers used by running multiple applications on each server. Another approach to reducing energy usage and increasing energy efficiency is server virtualizationA means of reducing energy usage and increasing energy efficiency by allowing virtual machines to run on one piece of hardware, at both the server and the PC level.. Similar to storage virtualization discussed earlier, server virtualization allows virtual machines to run on one piece of hardware, at both the server and PC level.

Cloud computingA means of accessing computer technology via the Internet without needing company investment in technology. is an option that allows access to computer technology via the Internet without your company purchasing or managing the technology. Cloud computing can be used with data centers, networks, configuration, software, hardware, infrastructure, platforms, services, and storage. Cloud computing can ultimately reduce costs while increasing utilization and efficiency. The FTC and computing professionals are beginning to address security issues in this new arena of cloud computing.Condon (2009).

## Desktops

Green PCs are designed to minimize the use of electricity and to meet the Environmental Protection Agency’s ENERGY STAR standards (new ENERGY STAR standards for computers were updated in 2007). One example is thin clients, diskless machines that consume a fraction of the power of standard desktop machines. The average desktop computer uses 4 to 8 times more energy than a thin client.Naegel (2009). Another option to consider is a laptop rather than a desktop. Laptops consume approximately 5 times less energy than desktops.Chua (n.d.). Lastly, the use of an ENERGY STAR–rated LCD monitor will reduce energy consumption.

Ideally, desktops should use 4 watts of energy or less in sleep mode and 50 watts or less when idle. For laptops, the ideal is 2 watts or less in sleep mode and 14 to 22 watts or less in idle mode.Chua (n.d.). However, the EPA estimates that fewer than 10% of computers are set to use the sleep or hibernation mode.Chua (n.d.). This power-saving feature can easily be set up on your computer through the Control Panel’s power options, although turning off your computer at the end of every workday is the best choice. Employees could also use a desktop device, such as EcoButton, to put the computer into sleep mode. Smart power strips can also conserve energy by turning off items after a period of inactivity. Smart strips are useful for printers, monitors, computers, and other items that can be powered down at the end of each day.

In addition to energy efficiency, green PCs are designed to contain fewer toxic materials (such as lead) in production and shipping and to contain more components that are made from recycled parts and that can again be recycled at the end of the machine’s usefulness. The EPA’s Electronic Product Environmental Assessment Tool allows you to compare computer models before making a purchase. See Note 7.8 "Greener Printing From Your Computer" for tips on how to be more environmentally friendly when printing from your desktop.

### Greener Printing From Your Computer

Before you print that next document, here are some ways you can achieve greener printing from your computer.

1. Make sure you are using an ENERGY STAR printer (and computer). You may think this one’s a no-brainer and you’ve got it covered, but wait . . . did you know that computer standards were revised in 2007 and new printer standards take effect this year? If your computer is older than 2007 and your printer is older than 2009, it may no longer meet ENERGY STAR standards, even though it met the standards that were in place at the time it was manufactured. If you should decide to upgrade, don’t forget to recycle the old one!
2. Change the margins. Studies at both Penn State University and Michigan State University found that changing margins can save paper. The Penn State study suggested that changing all university printer default margins to 0.75" (adding 19% more print space to the page) could save the university over $122,000 a year, and Michigan State estimated a savings of$67,512 a year.
3. Use paper with recycled content. Although both the Penn State and Michigan State studies found that switching to recycled content paper was more expensive, this has not been the case in my consulting experience. Many businesses that are not under contractual purchasing agreements do have the flexibility to comparison shop. A recent client was able to save 10% on paper costs by switching from virgin fiber to recycled content paper. Other “green” options are to look for unbleached paper or, better yet, tree-free paper!
4. Recycle and buy recycled. Recycle your paper, toner cartridges, and ink-jet cartridges. And don’t forget to buy recycled, too!
5. Install software to manage and reduce paper usage. Print management software programs (such as PaperCut, GreenPrint, and many others) can reduce printed pages and printer waste.
6. Use vegetable-based ink toner. SoyPrint is an environmentally friendly alternative to petroleum-based toner. Look for additional vegetable-based toners and ink-jet cartridges to hit the market soon.
7. Change the font. A Dutch company has created Ecofont, a new font that requires up to 20% less ink.Retrieved from http://www.ecofont.eu/english.html Ecofont is free to download and use.

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## Electronic Journal of Probability

### Aging uncoupled continuous time random walk limits

Ofer Busani

#### Abstract

Aging is a prevalent phenomenon in physics, chemistry and many other fields. In this paper we consider the aging process of uncoupled Continuous Time Random Walk Limits (CTRWLs) which are Levy processes time changed by the inverse stable subordinator of index $0 < \alpha < 1$. We apply a recent method developed by Meerscheart and Straka of finding the finite dimensional distributions of CTRWL, to obtaining the aging process’s finite dimensional distributions, self-similarity-like property, asymptotic behavior and its Fractional Fokker-Planck equation(FFPE).

#### Article information

Source
Electron. J. Probab., Volume 21 (2016), paper no. 7, 17 pp.

Dates
Accepted: 9 November 2015
First available in Project Euclid: 5 February 2016

https://projecteuclid.org/euclid.ejp/1454682888

Digital Object Identifier
doi:10.1214/16-EJP3802

Mathematical Reviews number (MathSciNet)
MR3485349

Zentralblatt MATH identifier
1338.60123

#### Citation

Busani, Ofer. Aging uncoupled continuous time random walk limits. Electron. J. Probab. 21 (2016), paper no. 7, 17 pp. doi:10.1214/16-EJP3802. https://projecteuclid.org/euclid.ejp/1454682888

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