text:
groupNormalizationLayer

Group normalization layer

Description

A group normalization layer normalizes a mini-batch of data across grouped subsets of channels for each observation independently. To speed up training of the convolutional neural network and reduce the sensitivity to network initialization, use group normalization layers between convolutional layers and nonlinearities, such as ReLU layers.

After normalization, the layer scales the input with a learnable scale factor γ and shifts by a learnable offset β.

Creation

Syntax

``layer = groupNormalizationLayer(numGroups)``
``layer = groupNormalizationLayer(numGroups,Name,Value)``

Description

example

````layer = groupNormalizationLayer(numGroups)` creates a group normalization layer.```

example

````layer = groupNormalizationLayer(numGroups,Name,Value)` creates a group normalization layer and sets the optional `'Epsilon'`, Parameters and Initialization, Learn Rate and Regularization, and `Name` properties using one or more name-value pair arguments. You can specify multiple name-value pair arguments. Enclose each property name in quotes.```

Input Arguments

expand all

Number of groups into which to divide the channels of the input data, specified as one of the following:

• Positive integer – Divide the incoming channels in to the specified number of groups. The specified number of groups must divide the number of channels of the input data exactly.

• `'all-channels'` – Group all incoming channels into a single group. This is also known as layer normalization. Alternatively, use `layerNormalizationLayer`.

• `'channel-wise'` – Treat all incoming channels as separate groups. This is also known as instance normalization. Alternatively, use `instanceNormalizationLayer`.

Properties

expand all

Group Normalization

Constant to add to the mini-batch variances, specified as a numeric scalar equal to or larger than `1e-5`.

The layer adds this constant to the mini-batch variances before normalization to ensure numerical stability and avoid division by zero.

Number of input channels, specified as `'auto'` or a positive integer.

This property is always equal to the number of channels of the input to the layer. If `NumChannels` equals `'auto'`, then the software automatically determines the correct value for the number of channels at training time.

Parameters and Initialization

Function to initialize the channel scale factors, specified as one of the following:

• `'ones'` – Initialize the channel scale factors with ones.

• `'zeros'` – Initialize the channel scale factors with zeros.

• `'narrow-normal'` – Initialize the channel scale factors by independently sampling from a normal distribution with zero mean and standard deviation 0.01.

• Function handle – Initialize the channel scale factors with a custom function. If you specify a function handle, then the function must be of the form `scale = func(sz)`, where `sz` is the size of the scale. For an example, see Specify Custom Weight Initialization Function.

The layer only initializes the channel scale factors when the `Scale` property is empty.

Data Types: `char` | `string` | `function_handle`

Function to initialize the channel offsets, specified as one of the following:

• `'zeros'` – Initialize the channel offsets with zeros.

• `'ones'` – Initialize the channel offsets with ones.

• `'narrow-normal'` – Initialize the channel offsets by independently sampling from a normal distribution with zero mean and standard deviation 0.01.

• Function handle – Initialize the channel offsets with a custom function. If you specify a function handle, then the function must be of the form `offset = func(sz)`, where `sz` is the size of the scale. For an example, see Specify Custom Weight Initialization Function.

The layer only initializes the channel offsets when the `Offset` property is empty.

Data Types: `char` | `string` | `function_handle`

Channel scale factors γ, specified as a numeric array.

The channel scale factors are learnable parameters. When training a network, if `Scale` is nonempty, then `trainNetwork` uses the `Scale` property as the initial value. If `Scale` is empty, then `trainNetwork` uses the initializer specified by `ScaleInitializer`.

At training time, `Scale` is one of the following:

• For 2-D image input, a numeric array of size 1-by-1-by-`NumChannels`

• For 3-D image input, a numeric array of size 1-by-1-by-1-by-`NumChannels`

• For feature or sequence input, a numeric array of size `NumChannels`-by-1

Channel offsets β, specified as a numeric array.

The channel offsets are learnable parameters. When training a network, if `Offset` is nonempty, then `trainNetwork` uses the `Offset` property as the initial value. If `Offset` is empty, then `trainNetwork` uses the initializer specified by `OffsetInitializer`.

At training time, `Offset` is one of the following:

• For 2-D image input, a numeric array of size 1-by-1-by-`NumChannels`

• For 3-D image input, a numeric array of size 1-by-1-by-1-by-`NumChannels`

• For feature or sequence input, a numeric array of size `NumChannels`-by-1

Learn Rate and Regularization

Learning rate factor for the scale factors, specified as a nonnegative scalar.

The software multiplies this factor by the global learning rate to determine the learning rate for the scale factors in a layer. For example, if `ScaleLearnRateFactor` is `2`, then the learning rate for the scale factors in the layer is twice the current global learning rate. The software determines the global learning rate based on the settings specified with the `trainingOptions` function.

Learning rate factor for the offsets, specified as a nonnegative scalar.

The software multiplies this factor by the global learning rate to determine the learning rate for the offsets in a layer. For example, if `OffsetLearnRateFactor` equals `2`, then the learning rate for the offsets in the layer is twice the current global learning rate. The software determines the global learning rate based on the settings specified with the `trainingOptions` function.

L2 regularization factor for the scale factors, specified as a nonnegative scalar.

The software multiplies this factor by the global L2 regularization factor to determine the learning rate for the scale factors in a layer. For example, if `ScaleL2Factor` is 2, then the L2 regularization for the offsets in the layer is twice the global L2 regularization factor. You can specify the global L2 regularization factor using the `trainingOptions` function.

L2 regularization factor for the offsets, specified as a nonnegative scalar.

The software multiplies this factor by the global L2 regularization factor to determine the learning rate for the offsets in a layer. For example, if `OffsetL2Factor` is 2, then the L2 regularization for the offsets in the layer is twice the global L2 regularization factor. You can specify the global L2 regularization factor using the `trainingOptions` function.

Layer

Layer name, specified as a character vector or a string scalar. To include a layer in a layer graph, you must specify a nonempty, unique layer name. If you train a series network with the layer and `Name` is set to `''`, then the software automatically assigns a name to the layer at training time.

Data Types: `char` | `string`

Number of inputs of the layer. This layer accepts a single input only.

Data Types: `double`

Input names of the layer. This layer accepts a single input only.

Data Types: `cell`

Number of outputs of the layer. This layer has a single output only.

Data Types: `double`

Output names of the layer. This layer has a single output only.

Data Types: `cell`

Examples

collapse all

Create a group normalization layer that normalizes incoming data across three groups of channels. Name the layer `'groupnorm'`.

`layer = groupNormalizationLayer(3,'Name','groupnorm')`
```layer = GroupNormalizationLayer with properties: Name: 'groupnorm' NumChannels: 'auto' Hyperparameters NumGroups: 3 Epsilon: 1.0000e-05 Learnable Parameters Offset: [] Scale: [] Show all properties ```

Include a group normalization layer in a `Layer` array. Normalize the incoming 20 channels in four groups.

```layers = [ imageInputLayer([28 28 3]) convolution2dLayer(5,20) groupNormalizationLayer(4) reluLayer maxPooling2dLayer(2,'Stride',2) fullyConnectedLayer(10) softmaxLayer classificationLayer]```
```layers = 8x1 Layer array with layers: 1 '' Image Input 28x28x3 images with 'zerocenter' normalization 2 '' Convolution 20 5x5 convolutions with stride [1 1] and padding [0 0 0 0] 3 '' Group Normalization Group normalization 4 '' ReLU ReLU 5 '' Max Pooling 2x2 max pooling with stride [2 2] and padding [0 0 0 0] 6 '' Fully Connected 10 fully connected layer 7 '' Softmax softmax 8 '' Classification Output crossentropyex ```

expand all

Algorithms

The group normalization operation normalizes the elements xi of the input by first calculating the mean μG and variance σG2 over spatial, time, and grouped subsets of the channel dimensions for each observation independently. Then, it calculates the normalized activations as

`${\stackrel{^}{x}}_{i}=\frac{{x}_{i}-{\mu }_{G}}{\sqrt{{\sigma }_{G}^{2}+\epsilon }},$`

where ϵ is a constant that improves numerical stability when the variance is very small. To allow for the possibility that inputs with zero mean and unit variance are not optimal for the operations that follow group normalization, the group normalization operation further shifts and scales the activations using the transformation

`${y}_{i}=\gamma {\stackrel{^}{x}}_{i}+\beta ,$`

where the offset β and scale factor γ are learnable parameters that are updated during network training.

References

[1] Wu, Yuxin, and Kaiming He. “Group Normalization.” ArXiv:1803.08494 [Cs], June 11, 2018. http://arxiv.org/abs/1803.08494.

Introduced in R2020b


text:
## Bruno Mascaro ; Gaetano Siciliano - Positive Solutions For a Schrödinger-Bopp-Podolsky system in $\mathbb R^{3}$

cm:10363 - Communications in Mathematics, December 13, 2022, Volume 31 (2023), Issue 1 - https://doi.org/10.46298/cm.10363
Positive Solutions For a Schrödinger-Bopp-Podolsky system in $\mathbb R^{3}$

Authors: Bruno Mascaro ; Gaetano Siciliano

We consider the following Schrödinger-Bopp-Podolsky system in $\mathbb R^{3}$ $\left\{ \begin{array}{c} -\varepsilon^{2} \Delta u + V(x)u + \phi u = f(u)\\ -\varepsilon^{2} \Delta \phi + \varepsilon^{4} \Delta^{2}\phi = 4\pi\varepsilon u^{2}\\ \end{array} \right.$ where $\varepsilon > 0$ with $V:\mathbb{R}^{3} \rightarrow \mathbb{R}, f:\mathbb{R} \rightarrow \mathbb{R}$ satisfy suitable assumptions. By using variational methods, we prove that the number of positive solutions is estimated below by the Ljusternick-Schnirelmann category of $M$, the set of minima of the potential $V$.

Volume: Volume 31 (2023), Issue 1
Published on: December 13, 2022
Accepted on: November 24, 2022
Submitted on: November 24, 2022
Keywords: Mathematics - Analysis of PDEs

format: plain


text:
Malware analysis tools to extract patterns of interest from files and crack obfuscation such as XOR

## Balbuzard

[Balbuzard](http://www.decalage.info/python/balbuzard) is a package of malware analysis tools in python to extract patterns of interest from suspicious files (IP addresses, domain names, known file headers, interesting strings, etc).

It can also crack malware obfuscation such as XOR, ROL, etc by bruteforcing and checking for those patterns.

## Balbuzard tools:

## When to use these tools:

## 5 minutes demo

See the [Demo](https://bitbucket.org/decalage/balbuzard/wiki/Demo/) page to see examples and test the tools by yourself in a few minutes using the provided samples.

## Help wanted:

# News

• 2014-05-22: Balbuzard v0.19 included in [REMnux v5](http://blog.zeltser.com/post/86508269224/remnux-v5-release-for-malware-analysts), pre-installed and ready to use.

• 2014-03-23 v0.19: bugfix when Yara is not installed, improved [documentation](https://bitbucket.org/decalage/balbuzard/wiki).

• 2014-02-26 v0.18: Initial release of Balbuzard tools

• 2013-03-15: added harvest mode (bbharvest)

• 2011-05-06: added bruteforce functions (bbcrack)

• 2008-06-06: first public release as [reScan](http://decalage.info/rescan) for [SSTIC08](http://decalage.info/sstic08)

• 2007-07-11: first versions of reScan

# Installation

See the [installation instructions](https://bitbucket.org/decalage/balbuzard/wiki/Installation).

# How to contribute / report bugs / ask for help:

These are open-source tools developed on my spare time. Any contribution such as code improvements, ideas, bug reports, additional patterns or transforms would be highly appreciated. You may contact me using [this online form](http://www.decalage.info/contact), by e-mail (decalage at laposte.net) or use the [issue page on Bitbucket](https://bitbucket.org/decalage/balbuzard/issues?status=new&status=open) to report bugs/ideas, or clone the project then send me pull requests to suggest changes.

This license applies to the whole Balbuzard package including balbuzard, bbcrack, bbharvest and bbtrans, apart from the thirdparty and plugins folders which contain third-party files published with their own license.

Redistribution and use in source and binary forms, with or without modification, are permitted provided that the following conditions are met:

• Redistributions of source code must retain the above copyright notice, this list of conditions and the following disclaimer.

• Redistributions in binary form must reproduce the above copyright notice, this list of conditions and the following disclaimer in the documentation and/or other materials provided with the distribution.

THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS “AS IS” AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.

## Project details

Uploaded source


text:
How should I manage the AI using lua scripts?

My game is in C++ and I want to make AI being managed by lua scripts, but I have no idea how should the scripts look like and the integration in C++.

Should the script be like

   if (whatever_happening) do_something
if (....) .....
etc

And in every frame in C++ I should call for every enemy:

   foreach(enemy)
enemy.handleAi(luascript.file)

Or it is wrong and I should handle it different?

A good approach is to design the behaviors in C++, and put the desicion parts in Lua scripts. The benefit is you can modify AI decisions without compiling your game; and with some good design, you may be able to modify AI desicions at runtime (opening up a debug window in game and change the variables to affect the decisions).

A behavior can be related to the position, the animations, and other status changing values. For example: Jump. This behavior changes the position, the velocity, and the animation of the AI, and also change the "IsInAir" and "IsJumping" flag.

A decision can be a branch of several behaviors. Here is the place you can use a state machine or behavior tree to implement the decision making system.

• I will go with this approach most likely. – user1873947 Feb 15 '13 at 23:22

This is a pretty general question with many different approaches. I'll just mention two major differences in coding AI logic into Lua.

As a prerequisite you will still need to expose C/C++ low level functions so that lua can make function calls to affect game systems or query for game information.

Movement, pathfinding, inventory, etc, etc

Script as low level decision logic

Basically you can use script for the very core decisions. How an AI may select a target, or how an AI chooses which weapon to use, etc. Think of lua functions or individual scripts as very small amounts of AI behavior.

This would mean C/C++ would call a function on the lua side, passing in values where the decision logic comes up with the result and passes the result back to the caller.

This type of approach works very well when the overall structure of what makes up an AI is in code and can be constructed either through hard coding or from a data source.

Structures like Finite State Machines with states in lua, or Behavior Trees where individual tree nodes are coded in lua fit perfectly in this model.

Script as high level decision logic

A completely different approach is to allow the lua script to completely control the AI. In this sense you could have a "think" lua function that is called every tick which represents the AI logic. The "think" function would need to be ticked either per frame or whenever is appropriate for the game you're making.

For example if lua is controlling animation playback it would probably need to be ticked per frame to have control over animation handling.

This approach isn't structured except for the known function that C/C++ calls into for the specific AI. All AI structure and logic can be written in any form you want in script with the AI storing any persistent knowledge as global variables in lua.

These are just two very different high level approaches to using scripting. Of course not taking into consideration memory or performance of the AI.

It depends. If determining_happening is expensive and well-defined, it may be beneficial to keep that calculation in C++. E.g. in C++

for (auto& enemy : enemies) {
if enemy.collide(player) enemy.callLua(collisionScript)
}

• good point. Are there any other simple approaches? – user1873947 Feb 13 '13 at 11:31


text:
# Nature of conservation laws

1. Jul 26, 2011

### QuArK21343

It seems to me that there is a difference between, say, the law of conservation of linear momentum or energy and the law of conservation of angular momentum. The first two are valid in any frame of reference and their invariance is a direct consequence of the relativity principle, whereas angular momentum may be conserved for observer 1 and not for observer 2, if they compute it around two different points (both observers are inertial). How can it be so? Is there a reason? And is it true that energy and momentum must be conserved for all inertial observers?

2. Jul 26, 2011

### xts

Really??? It is not so... Try to find an example.
Angular momentum is conserved for both observers 1 and 2. The total angular momentum may have different value for each of them, but from each one's perspective it is conserved during any evolution of isolated system.

As momentum conservation is a consequence of translational symmetry and energy conservation is consequence of symmetry of time shift, the angular momentum results from rotational symmetry.

Not both but all three conservation laws are true for all inertial observers.

3. Jul 26, 2011

### Petr Mugver

QuArK21343, can you give an example of a system whose energy, linear and angular momentum are conserved in one inertial system, but not in a different inertial system?

4. Jul 26, 2011

### QuArK21343

I believe I never said that the conservation laws are valid for one inertial observer but not for another. To verify the equivalence of all inertial observer, as far as dynamics is concerned, we can limit ourselves to the principle of conservation of momentum and the definition of force. If they are invariant, all other consequences of these two still hold in a new inertial frame. Take the principle of conservation of momentum. In an isolated system of two particles:

$m_1v_1+m_2v_2=const$

For an observer O' moving with constant velocity v, $v_1'=v_1-v$ and $v_2'=v_2-v$. Substituting, we find the total momentum is still a constant. Also, differentiating $v_1=v_1'+v$, we see that $a_1=a_1'$, so acceleration is not a relative concept in newtonian mechanics. Therefore, both inertial observers measure the same force on particle 1. From here, we can verify that also energy is a constant in both frames, that is

$\frac{1}{2}m_1v_1^2+U=\frac{1}{2}m_1'v_1'^2+U'$

where $U=U'$, if the potential energy depends only on the relative distance, as it usually is.

What I am saying is that, in an inertial frame, angular momentum may be conserved around one point but not around another. (Let's speak about only one observer now, since it is clearer) To a beginner like me, it seems strange and rather peculiar. So I was asking if there is some deep reason I still don't know.

Please, feel free to correct me if I am wrong.

Last edited: Jul 26, 2011
5. Jul 26, 2011

### mikeph

angular momentum around all points is separately conserved. So I believe this is incorrect ^

6. Jul 26, 2011

### QuArK21343

I believe angular momentum conservation depends on the choice of the fixed point around which you compute it. Let the angular momentum $L$ of a system of $n$ particles be:

$L=\sum m_i r_i \times v_i$

where $r_i$ is computed relative to a point, moving with velocity $v_0$ in our inertial frame, in general different from the origin. The rate of change of angular momentum is:

$\dot L=\sum m_i \dot r_i\times v_i + m_i r_i\times \dot v_i$

Of course, $\dot r_i= v_i - v_0$ and therefore $\dot L=\sum r_i\times F_i -v_0\times M v_{CM}$

where the forces $F_i$ are only the external ones, since if we assume the internal forces are central, their contribution is exactly zero and where $v_{CM}$ is the velocity of the center of mass of the system. So angular momentum is conserved if the RHS of the previous equation is zero. In particular, the term $v_0\times M v_{CM}$ is zero if the point is stationary, if the center of mass is stationary, if the point is moving in the same direction as the center of mass or if the point is the center of mass. Let's suppose this first condition is met. Then, the moment of external forces is zero if the system is isolated (and conservation is guaranteed for any point) or if, for example, the forces are central with respect to the fixed point or, in general, if the moment of external forces is anyway zero with respect to this point. If we choose another point, the moment of forces may not be zero. Now, if the resultant of all forces is zero, then the moment is independent of the point chosen, but in general this is not true.

Returning to my original question, isn't that weird that conservation of angular momentum depends on the point chosen? Second part of my questions is: given that energy and momentum are conserved for all inertial frames in newtonian mechanics, is it true in all fields of physics? But I think xts partly answer this second part.

7. Jul 27, 2011

### mikeph

I believe you're wrong. Conservation doesn't depend on the point chosen, that's what I'm trying to say. Angular momentum depends on the point, but for any fixed point, that angular momentum is conserved.

8. Jul 27, 2011

### QuArK21343

That would explain everything, but could you show it to me? And could you then spot the error in my previous post?

9. Jul 27, 2011

### mikeph

I think you go wrong when you introduce external forces. Nobody said angular momentum is conserved if you have a general, external torque being applied.

10. Jul 27, 2011

### QuArK21343

Could you be more precise? I can't understand what is wrong with my introduction of external forces.

Also, think about this example. Let's consider a particle going around the sun in a circular orbit. If I compute the angular momentum relative to the sun S, then I get a constant vector of magnitude $mvr$. Angular momentum is conserved! Now take a point P on the circular orbit: angular momentum is changing all the time. It is zero when the particle passes through point P, since the position vector is zero. It is not zero, for sure, in any other position. Angular momentum is not conserved! This is so because when I choose to compute angular momentum around S the external torque is zero, since the force is central. But if I choose another less special and uninteresting point, it is different because now the torque is not zero.

11. Jul 27, 2011

### mikeph

Your two body system is flawed, the sun and Earth will both orbit a common central point (somewhere between the two) called the barycentre. So the computation of the angular momentum from S cannot be done, because S moves. You're computing angular momentum of a single object when there are two objects in the system moving under constant mutual interaction.

It's like me saying "the momentum of a snooker ball is not conserved when it strikes another ball".

12. Jul 28, 2011

### QuArK21343

What I was trying to say is that, in general, whenever you have a central force, and a body moving under its action, the centre of force is special, because angular momentum is always conserved around it, but it is in no way clear if angular momentum is conserved around any other point.

Anyway, here is another one: let's imagine that a particle is accelerating along a line. First, we compute its angular momentum around a point P along the line. Then its position vector and velocity are parallel and angular momentum is equally zero. Angular momentum is conserved. Now, we take a point P' not along the line. The arm of the force is always the same throughout the motion of the particle, but its speed v is increasing. So the magnitude of its angular momentum is mvb, which is changing all the time. Angular momentum is not conserved.

13. Jul 28, 2011

### mikeph

Accepted. Angular momentum isn't conserved, in general, if you have forces acting on the body...

Again, the particle is accelerating so you've got an external force. I don't get where this is going. You're simply thinking of situations in which angular momentum of a particle is not conserved because of the action of external forces. Nobody is debating this.

14. Jul 28, 2011

### Philip Wood

The fundamental law of conservation of momentum is that in any closed system of particles acting on each other by central forces, the sum of angular momenta of bodies about any point is constant.

What's wrong with your last counterexample is that you're not dealing with a closed system. You have a force acting from outside, otherwise your particle wouldn't be accelerating.

Confusingly, for a non-closed system angular momentum may be conserved about a special point. Consider a single planet orbiting around the centre of mass (CM) of the star-planet system. The planet's angular momentum is conserved about the CM. Here we're considering the angular momentum of just the planet, which is clearly acted upon by an external force, but since this is always directed to the CM, the planet's angular momentum is conserved about this point (only). Note that this is is not a case of the P of C of AM, as defined earlier. If we wanted to apply the fundamental P of C of AM we'd need to consider a system including both the star and the planet. Then we'd find that $\sum$AM is conserved about any point.

Last edited: Jul 28, 2011
15. Jul 28, 2011

### QuArK21343

Thanks for your reply. Can I ask you what is a closed system in this context? Is it a system on which there are no external forces? Or is it a system on which there is no external torque? What I know is that the requirement for the conservation of angular momentum is that the external torque be zero, so I guess a closed system is one with no external torque on it. Is this right?

16. Jul 28, 2011

### Philip Wood

I think that for the fundamental P of C of AM to apply we need no external forces to act.

If we say 'no external torque', we would have to state about what point $\sum$G was zero. This would be a special point, and angular momentum of the system would be conserved about that point. We're back to planets' AM being conserved only about the CM of the stellar system.

17. Jul 28, 2011

### QuArK21343

I see: when I say P of C of AM I ask that the total torque must be zero, because whenever the total torque of external forces is zero (if we assume the strong principle of action-reaction!) AM is conserved (if all other conditions listed above are met about the term $v_0\times Mv_{CM}$); on the contrary, you only ask that there are no external forces acting on the system. This is a particular case, but it guarantees that conservation of AM is true about all points.

This is exactly my point. There is something about C of AM that doesn't arise in the P of C of momentum or energy. What I find counterintuitive is the fact that AM may be conserved about only one point or many points but not about all points, and so in problems we must carefully choose our pole to exploit this principle of conservation. It is not something we can count in only one way. My questions: is there a reason for that? Maybe does it depend on the fact the torque, as a vector, is unlike force or velocity?

18. Jul 28, 2011

### Philip Wood

But if you state the P of C of AM as applying to a closed system, meaning one whose particles experience no forces from outside, then $\sum$AM is conserved about any point. That's fine isn't it?

Whose idea was it to weaken the requirement to that of no external torque acting?

19. Jul 28, 2011

### Oudeis Eimi

When computing the angular momentum from a point moving with respect to the origin, you need to use the velocities of the particles as seen from that point, otherwise you're mixing up quantities as seen from different inertial systems.

If we have $R_i$ and $r_i$ as the positions as seen from the origin and the point O, respectively, both considered inertial, with the point O at coordinates $r_0$ moving with velocity $v_0$ (as seen from the origin), then

$r_i = R_i - r_0$, and
$v_i = V_i - v_0$.

Denoting $L_\mathrm{orig}$ the angular momentum as seen from the origin, which we assume is conserved, and $R_\mathrm{CM}$ and $V_\mathrm{CM}$ the position and velocity of the centre of mass as seen from the origin, we have:

$L = \sum_i m_i r_i \times v_i = \sum_i m_i (R_i - r_0) \times (V_i - v_0) = L_\mathrm{orig} + M r_0 \times v_0 - M R_\mathrm{CM} \times v_0 - M r_0 \times V_\mathrm{CM}$,

$\dot L = \dot L_\mathrm{orig} -M V_\mathrm{CM} \times v_0 - M v_0 \times V_\mathrm{CM} = \dot L_\mathrm{orig} -M V_\mathrm{CM} \times v_0 + M V_\mathrm{CM} \times v_0 = \dot L_\mathrm{orig} = 0$,

where in the last equation we've made use of the antisymmetry of the vector product.

20. Jul 28, 2011

### magison

thanks Oudeis Eimi


text:
# How do you simplify ab(b^2-0.4a^2b)?

To simplify this expression multiply each term within parenthesis by $\textcolor{red}{a b}$:
$\textcolor{red}{a b} \left({b}^{2} - 0.4 {a}^{2} b\right) = \left(\textcolor{red}{a b} \times {b}^{2}\right) - \left(\textcolor{red}{a b} \times 0.4 {a}^{2} b\right) =$
$a {b}^{3} - 0.4 {a}^{3} {b}^{2}$


text:
# Purification, Classification and Nonemeclature of Organic Compound Test 1

Total Questions:50 Total Time: 75 Min

Remaining:

## Questions 1 of 50

Question:An organic compound contains $$C = 36\%$$ $$H = 6\%$$ and rest oxygen. Its Empirical formula is

$$C{H_2}O$$

$${C_2}{H_3}{O_3}$$

$$C{H_2}{O_2}$$

$${C_2}{H_2}{O_2}$$

## Questions 2 of 50

Question:Empirical formula of a compound is $$C{H_2}O$$ and its vapour density is 30. Molecular formula of the compound is

$${C_3}{H_6}{O_3}$$

$${C_2}{H_4}{O_2}$$

$${C_2}{H_4}O$$

$$C{H_2}O$$

## Questions 3 of 50

Question:An organic compound on analysis gave C = 48 gm, H = 8 gm and N = 56 gm. Volume of 1.0 g of the compound was found to be 200 ml at NTP. Molecular formula of the compound is

$${C_4}{H_8}{N_4}$$

$${C_2}{H_4}{N_2}$$

$${C_{12}}{H_{24}}{N_{12}}$$

$${C_{16}}{H_{32}}{N_{16}}$$

## Questions 4 of 50

Question:Insulin contains 3.4% sulphur. The minimum molecular weight of insulin is

350

470

560

940

## Questions 5 of 50

Question:Which element is estimated by Carius method

Carbon

Hydrogen

Halogen

Nitrogen

## Questions 6 of 50

Question:In Kjeldahl's method for the estimation of nitrogen, the formula used is

$$\% N = \frac{{1.4{\rm{ }}V{\rm{ }}W}}{N}$$

$$\% N = \frac{{1.4{\rm{ }}N{\rm{ }}W}}{V}$$

$$\% N = \frac{{V{\rm{ }}N{\rm{ }}W}}{{1.8}}$$

$$\% N = \frac{{1.4{\rm{ }}V{\rm{ }}N}}{W}$$

## Questions 7 of 50

Question:An organic compound on analysis gave the following results : C = 54.5%, O = 36.4%, H = 9.1%. The Empirical formula of the compound is

$$C{H_3}O$$

$${C_2}{H_4}O$$

$${C_3}{H_4}O$$

$${C_4}{H_8}O$$

## Questions 8 of 50

Question:An organic compound gave C = 92.31% and $$H = 7.69\%$$. If molecular weight of the compound is 78, its molecular formula is

$${C_6}{H_6}$$

$${C_7}{H_7}$$

$${C_6}{H_{18}}$$

$${C_8}{H_{20}}$$

## Questions 9 of 50

Question:An organic compound gave the following results $$C = 53.3\% ,{\rm{ }}H = 15.6,{\rm{ }}N = 31.1\% ,$$ mol. wt. = 45, What is molecular formula of the compound ?

$${C_2}{H_5}{N_2}$$

$${C_2}{H_5}N$$

$${C_2}{H_7}N$$

$${C_2}{H_6}N$$

## Questions 10 of 50

Question:A compound gave 80% carbon and 20 % hydrogen on analysis. The compound is possibly

$${C_6}{H_6}$$

$${C_2}{H_5}OH$$

$${C_2}{H_6}$$

$$CHC{l_3}$$

## Questions 11 of 50

Question:If a compound on analysis was found to contain C = 18.5%, H = 1.55%, Cl = 55.04% and O = 24.81%, then its empirical formula is

$$CHClO$$

$$C{H_2}ClO$$

$${C_2}{H_2}OCl$$

$$ClC{H_2}O$$

## Questions 12 of 50

Question:An organic compound has % of C and % of H in the ratio 6 : 1 and % of C and % of O in the ratio 3 : 4. The compound is

$$HCHO$$

$$C{H_3}OH$$

$$C{H_3}C{H_2}OH$$

$${H_2} \to {H^ \bullet } + {H^ \bullet }$$

## Questions 13 of 50

Question:0.2595g of an organic substance in a quantitative analysis yielded 0.35 g of the barium sulphate. The percentage of sulphur in the substance is

18.52g

182.2 g

17. 5 g

175.2g

## Questions 14 of 50

Question:In kjeldahl's method, $$CuS{O_4}$$ acts as

Oxidising agent

Reducing agent

Hydrolysing agent

Catalytic agent

## Questions 15 of 50

Question:The Empirical formula of a compound is $$C{H_2}O$$ and its molecular weight is 120. The molecular formula of the compound is

$${C_2}{H_4}{O_2}$$

$${C_3}{H_6}{O_3}$$

$${C_4}{H_8}{O_4}$$

$$C{H_2}O$$

## Questions 16 of 50

Question:In Victor Mayer's method 0.2 gm of an organic substance displaced 56 ml of air at STP the molecular weight of the compound

56

112

80

28

## Questions 17 of 50

Question:If we want to study relative arrangement of atoms in a molecule we study

Empirical formula

Molecular formula

Structural formula

None of these

## Questions 18 of 50

Question:Which one of the following reagents is used for detection of unsaturation in alkenes

$$NaOH + CaO$$

Cold dilute alkaline $$KMn{O_4}$$

$$C{l_2}/hv$$

$$KOH/{C_2}{H_5}OH$$

## Questions 19 of 50

Question:The study of organic compounds even at present is done separate from other compounds because

The formation of organic compounds is not based on chemical combination

Organic compounds are covalent

Catenation is the main characteristics

It is the easiest method of study

## Questions 20 of 50

Question:Which of the following pair of the species has the same percentage of carbon

$$C{H_3} - C{H_2} - \mathop {\mathop {C\,\,\,\,\,\,\,\, - \,\,\,\,\,\,}\limits_{\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}^{\mathop {\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\limits^{\,C{H_3}C{H_2}\,\,\,\,} } }\limits_{\,\,C{H_2}C{H_3}\,\,\,\,\,} \mathop {CH\, - \,}\limits^{\mathop {|\,\,\,\,\,\,}\limits^{C{H_3}} \,\,\,\,\,\,} C{H_2} - {(C{H_2})_2} - C{H_3}$$ and $${C_2}{H_5}OH$$

$${C_6}{H_{12}}{O_6}$$ and $${C_{12}}{H_{22}}{O_{11}}$$

$${C_6}{H_{12}}{O_6}$$ and $${C_{12}}{O_{22}}{O_{11}}$$

$${C_{12}}{O_{22}}{O_{11}}$$ and $${C_6}{H_{12}}{O_6}$$

## Questions 21 of 50

Question:In Kjeldahl's method of estimation of N,$$CuS{O_4}$$acts as

Oxidising agent

Reducing agent

Catalytic agent

Hydrolysis agent

## Questions 22 of 50

Question:An organic compound having molecular mass 60 is found to contain C = 20%, H= 6.67% and N=46.67% while rest is oxygen. On heating it gives $$N{H_3}$$ alongwith a solid residue. The solid residue give violet colour with alkaline copper sulphate solution. the compound is

$$C{H_3}NCO$$

$$C{H_3}CON{H_2}$$

$${(N{H_2})_2}CO$$

$$C{H_3}C{H_2}CON{H_2}$$

## Questions 23 of 50

Question:IUPAC name of the compound is $$C{H_3} - \mathop {\mathop {\mathop {\mathop {CH}\limits_{|\,\,\,\,\,} }\limits_{C{H_2}} }\limits_{|\,\,\,\,\,\,} }\limits_{C{H_3}} - C{H_2} - CH(OH) - C{H_3}$$ is

4-ethyl-2-pentanol

4-methyl-2-hexanol

2-ethyl-2-pentanol

3-methyl-2-hexanol

## Questions 24 of 50

Question:IUPAC name of the compound is $$C{H_3} - CH = \mathop {\mathop {C - C{H_3}}\limits_{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,} }\limits_{C{H_2} - C{H_2}}$$

2-ethyl-2-butene

3-ethyl-2-butene

3-Methyl-3-pentene

3-methyl-2-pentene

## Questions 25 of 50

Question:The IUPAC name of $$C{H_3}C \equiv N$$is

Acetonitrile

Ethanenitrile

Methyl cyanide

Cyanoethane

## Questions 26 of 50

Question:Which compound is 2, 2, 3-trimethylhexane

$$\begin{array}{*{20}{c}} {C{H_3}\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,} \\ {|\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {C{H_3} - C - CH - C{H_2} - C{H_3}} \\ {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\end{array}$$

$$\begin{array}{*{20}{c}} {\,\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,} \\ {C{H_3} - C - C{H_2} - CH - C{H_3}} \\ {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\end{array}$$

$$\begin{array}{*{20}{c}} {C{H_3}\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {|\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {C{H_3} - C - CH - C{H_2} - C{H_2} - C{H_3}} \\ {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\end{array}$$

$$\begin{array}{*{20}{c}} {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ {C{H_3} - CH - C{H_2} - C{H_2} - C - C{H_3}} \\ {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,} \\ {\,\,\,\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\\end{array}$$

## Questions 27 of 50

Question:IUPAC name of the compound is $$C{H_3}C{H_2}C{H_2}C{H_2}C{H_2} - \mathop {\mathop {CH - }\limits_{|\,\,\,\,\,\,\,\,\,} }\limits_{C{H_3}\,\,} \mathop {\mathop {\mathop {\mathop {C - C{H_2}C{H_3}\,\,\,\,\,\,\,}\limits^{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} }\limits^{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} }\limits_{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} }\limits_{C{H_2}\,\,C{H_2}C{H_3}\,\,\,\,\,}$$

3, 4-dimethyl-3-n-propyl nonane

5, 7-dimethyl-7-n-propyl nonane

4, 5-dimethyl-4-ethyl decane

6, 7-dimethyl-7-ethyl decane

## Questions 28 of 50

Question:IUPAC name of $$C{H_3} - \mathop {\mathop {CH - C{H_2}}\limits_{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} }\limits_{C{H_3}\,\,\,\,\,\,\,\,\,\,\,} - CH = C{H_2}$$ is

2-methyl pentene

4-methyl pentene-1

1-hexene

2-methyl pentene-1

## Questions 29 of 50

Question:In the structure $$^1{H_3}C - \mathop {\mathop {\mathop {\mathop {^2C - {\,^3}C{H_2} - {\,^4}C{H_3}}\limits^{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} }\limits^{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} }\limits_{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} }\limits_{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}$$ Which one is quarternary carbon atom

C – 1

C – 2

C – 3

C – 5

## Questions 30 of 50

Question:The IUPAC name of $$C{H_3} - C{H_2} - \mathop {\mathop {C = }\limits_{|\,\,\,\,\,\,} }\limits_{C{H_3}} C{H_2}$$ is

2-methylbutene-1

3-methylbutene-1

Vinyl methylethane

Propylethene-1

## Questions 31 of 50

Question:2-methyl-2-butene will be represented as

$$C{H_3} - \mathop {\mathop {CH}\limits_{|\,\,\,\,} }\limits_{C{H_3}} - C{H_2} - C{H_3}$$

$$C{H_3} - \mathop {\mathop {C = }\limits_{|\,\,\,\,} }\limits_{C{H_3}} CH - C{H_3}$$

$$C{H_3} - C{H_2} - \mathop {\mathop {C = }\limits^{|\,\,\,\,\,} }\limits^{C{H_3}} C{H_2}$$

$$C{H_3} - \mathop {\mathop {CH - }\limits_{|\,\,\,\,} }\limits_{C{H_3}} CH = C{H_2}$$

## Questions 32 of 50

Question:$$Cl - C - Cl$$ angle in 1, 1, 2, 2 - tetrachloroethene and tetrachloromethane respectively are about

$${120^o}$$ and $${109.5^o}$$

$${90^o}$$ and $${109.5^o}$$

$${109.5^o}$$ and $${90^o}$$

$${109.5^o}$$ and $${120^o}$$

## Questions 33 of 50

Question:The IUPAC name of succinic acid is

1, 4-butanedioic acid

Dimethyl-2-acid

1, 2-dimethyldioic acid

None of these

## Questions 34 of 50

Question:IUPAC name of $${(C{H_3})_2}CH - C{H_2} - C{H_2}Br$$ is

1-bromopentane

2-methyl-4-bromobutane

1-bromo-3-methylbutane

2-methyl-3-romopropane

## Questions 35 of 50

Question:IUPAC name of $$C{H_3} - \mathop {\mathop {\mathop {\mathop {C - }\limits^{|\,\,\,\,} }\limits^{OH} }\limits_{|\,\,\,\,\,} }\limits_{C{H_3}} C{H_2} - \mathop {\mathop {CH - }\limits_{|\,\,\,\,\,\,\,} }\limits_{C{H_3}\,} C{H_3}$$ is

2, 4-dimethyl pentanol-2

2, 4-dimethyl pentanol-4

2, 2-dimethyl butanol-2

None of these

## Questions 36 of 50

Question:Which is correct IUPAC name of the following compound $$C{H_3} - \mathop {\mathop {CH - }\limits^{|\,\,\,\,} }\limits^{C{H_3}} \mathop {\mathop {CH - \mathop {\mathop {CH - C{H_3}}\limits^{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} }\limits^{C{H_3}\,\,\,\,\,\,\,\,\,\,\,} }\limits_{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} }\limits_{C{H_2} - C{H_3}\,\,\,\,\,\,\,\,}$$

3-isopropyl-2-methylpentane

3-ethyl-2,4-dimethylpentane

2,4-dimethyl-3-ethylpentane

3-isopropyl-4-methylpentane

## Questions 37 of 50

Question:IUPAC name of $$C{H_3} - CH = CH - C \equiv CH$$ is

Pent-2-en-4-yne

Pent-3-en-1-yne

Pent-3-yne-1-en

Pent-2-yne-1-en

## Questions 38 of 50

Question:The IUPAC name for the formula $$C{H_3} - \mathop {\mathop {C = \mathop {\mathop C\limits^| }\limits^H }\limits_{|\,\,\,\,\,\,\,\,\,} }\limits_{C{H_3}\,\,\,} - COOH$$

2-methyl-2-butenoic acid

3-methyl-3-butenoic acid

3-methyl-2-butenoic acid

2-methyl-3-butenoic acid

## Questions 39 of 50

Question:Which of the following alkanes contains primary, secondary, tertiary and quaternary carbon atoms together

$${(C{H_3})_4}C$$

$$HCHO$$

$$C{H_3}CHO$$

$${(C{H_3})_4}C$$

## Questions 40 of 50

Question:The number of tertiary carbon atoms in the compound $${(C{H_3})_2}CHC{H_2}C\;{(C{H_3})_3}$$ is

2

3

1

4

## Questions 41 of 50

Question:The compound which has one isopropyl group is

2, 2, 3, 3-tetramethyl pentane

3, 3-dimethyl pentane

2, 2, 3-trimethyl pentane

2-methyl pentane

## Questions 42 of 50

Question:Write the IUPAC name of $$C{H_3} - \mathop {\mathop {\,C\, - \,}\limits_{|\,\,\,\,\,\,}^{|\,\,\,\,\,} }\limits_{OH\,}^{H\,} C{H_2} - C{H_2} - C{H_2} - \mathop {\mathop {C\,}\limits_{|\,\,\,}^{|\,\,\,\,} }\limits_{Br}^{Br} - C{H_3}$$

6, 6-dibromoheptane-2-ol

2, 2-dibromoheptane-6-ol

6, 6 dibromoheptane-2-ol

None of these

## Questions 43 of 50

Question:The name of $$Cl{H_2}C - \mathop {\mathop C\limits_{|\,\,} }\limits_{Br} = \mathop {\mathop C\limits_{|\,\,} }\limits_{Br} - C{H_2}Cl$$ according to IUPAC nomenclature system is :

2,3 dibromo-1, 4-dichlorobutene-2

1, 4-dichloro-2, 3-bromobutene-2

Dichlorodibromobutene

Dichlorodibromobutane

## Questions 44 of 50

Question:The IUPAC name of acraldehyde is

Prop-2-ene–1–al

Propenyl aldehyde

But-2-ene–1–al

Propenal

## Questions 45 of 50

Question:IUPAC name of the compound $$C{H_3} - \mathop {\mathop {CH}\limits_{|\,\,\,\,\,\,} }\limits_{OH} - C{H_2} - \mathop {\mathop {CH}\limits_{|\,\,\,\,\,\,\,} }\limits_{C{H_3}} - C{H_3}$$ is.....

4-methyl pentene-2-ol

2-methyl pentanol-4

4, 4-dimethyl butan-2-ol

4-methyl pentane-2-ol

## Questions 46 of 50

Question:Cycloalkane has the formula

$${C_n}{H_{2n + 2}}$$

$${C_n}{H_{2n - 2}}$$

$${C_n}{H_{2n}}$$

$${C_{2n}}{H_2}$$

## Questions 47 of 50

Question:The IUPAC name of $$C{H_3} - \mathop {\mathop {CH}\limits^{|\,\,\,\,} }\limits^{Cl\,} - C{H_3}$$ is

2-chloropropane

Chloropropane

1-chloropropane

2-chlorobutane

## Questions 48 of 50

Question:The IUPAC name of $$C{H_3} - \mathop {\mathop {CH - }\limits_{|\,\,\,\,\,\,\,\,\,\,} }\limits_{OH\,\,\,} C{H_2} - \mathop {\mathop {\mathop {\mathop {C - }\limits_{|\,\,\,\,\,} }\limits_{OH} }\limits^{|\,\,\,\,\,} }\limits^{C{H_3}} C{H_3}$$ and $$\mathop {\mathop {C{H_2} - CH - C{H_2}}\limits_{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,} }\limits_{CN\,\,\,\,\,\,\,CN\,\,\,\,\,\,CN\,\,}$$

1, 1-dimethyl-1, 3-butanediol and propanetricarbyl amine

4-methyl-2, 4-pentanediol and 1, 2, 3 propanetrinitrile

2-methyl 2, 4-pentanediol and 3 cyano 1, 5-pentanedinitrile

1, 3, 3-trimethyl 1,3-propanediol and 1, 2, 3 tricyano propane

## Questions 49 of 50

Question:The IUPAC name of $$C{H_3}C{H_2}C(Br) = CH - Cl$$ is

2-bromo-1-chloro butane

1-chloro-2-bromo-butene

3-chloro-2-bromo butene-2

None of these

## Questions 50 of 50

Question:Which of the following compounds is not chiral


text:
Restricted Direct Products in Koch's Number Theory

On p.353 of Number Theory: Algebraic Numbers and Functions by Helmut Koch, he considers a group $G$ which is the restricted direct product of the locally compact abelian groups $G_i$ with respect to subgroups $H_i$ for $i \in I$, some index set. He considers a compact subset $T$ of the form $\prod_{i \notin S} H_i \times \prod_{i \in S} T_i$ where $T_i$ is a compact subset of $G_i$ for $i \in S$, and he consider a function $f$ supported on $T$.

He then writes: "For each $i \in I$ for which $H_i$ is defined we select that Haar measure on $H_i$ with $\int_{H_i} d \xi_i = 1$...On the compact group $G^S = \prod_{i \notin S} H_i$ we take the Haar measure $d \xi_S$ with $\int_{G^S} d\xi_S = 1$ and on $\prod_{i \in S} G_i$ the product measure $\prod_{i \in S} d\xi_i$. Then by $$\int_G f(\xi) d\xi = \int_{G^S} f(\xi_S) d\xi_S \times \prod_{i \in S} \int_{G_i} f(\xi_i) d\xi_i,$$ where $\xi_S$, respectively $\xi_i$, denotes the projection of $G$ onto $G^S$, respectively $G_i$, the Haar measure on $G$ is determined uniquely and independently of the choice of $S$. We denote it by $\prod_{i \in I} d\xi_i$.

I don't understand where the equality of the integrals is from. This works if $f$ is a product of functions on each of the $G_i$, just like in $\mathbb{R}^n$ if $f(x,y,z)=g(x)h(y)j(z)$, for example. But Fubini's theorem involves multiple steps of integration, so I don't understand where this comes from.

-
Such "restricted direct products" are colimits of the subgroups $G_S$ where all but the factors in $S$ are in the subgroups $H_i$. The topology on the colimit is uniquely determined, and for $f$ to be continuous requires that for some $S$ it be expressible as characteristic function of $G^S$ times continuous function on $\prod_{i\in S} G_i$. This reduces the issue to a more down-to-earth characterization of Haar measure on a finite product of topological groups. Probably Urysohn's lemma suffices to show that finite linear combinations of "monomial" functions are dense, in that case.


text:
### accidentallygivenfuck's blog

By accidentallygivenfuck, 4 years ago, ,

Hi Codeforces,

Does anyone know a way to download problems in a PDF file for later offline use?

I know a somewhat not-so-good way:

• Open codeforces.com/contest/<contestid>/problems
• Press ctrl+p
• Set the correct preferences, so that the result will be readable
• Print to file

But I remember that I downloaded 482B - Interesting Array in a PDF file at the beginning of the contest, when I tried to open the problem page. I don't know how, but maybe it was because the servers were too busy. That is why, I thought there might be a way to download problems in PDF format.

I still have the file. This file is more convenient than printing the problems page.

•
• +6
•

 » 4 years ago, # | ← Rev. 2 →   +3 Hello, I think you can use wkhtmltopdf, It justs requires an url as input, and it gives you the required pdf.
 » 4 years ago, # | ← Rev. 3 →   +5 I just wrote a code for this C++0x:Using this you can remove surplus elements from problem and have a clean page instead. You can also change  and style the document. It saves the document into saved.html. After saving it, Use your own method to PDFify your new HTML. // // main.cpp // CF to PDF // // Created by Hossein Yousefi on 1/11/15. // Copyright (c) 2015 IranGeek. All rights reserved. // #include using namespace std; int main(int argc, const char * argv[]) { cout << "Enter URL of problem: "; string url; cin >> url; string download_cmd = "curl " + url + " > _cfprob.html"; system(download_cmd.c_str()); ifstream fin("_cfprob.html"); string tmp, parsed_html; while (fin >> tmp) { parsed_html = parsed_html + " " + tmp; } system("rm _cfprob.html"); unsigned long begin, length; begin = parsed_html.find(""); length = parsed_html.find("") - begin + 7; string head = parsed_html.substr(begin, length); begin = parsed_html.find(""); length = parsed_html.find("") - begin; string problem = parsed_html.substr(begin, length); string html_to_be_saved = " " + head + " " + problem + " "; ofstream fout("saved.html"); fout << html_to_be_saved; system("open saved.html"); } TESTED in OS X.This that problem, converted to PDF using this C++ code: https://drive.google.com/file/d/0B6dkBXoLJA10Q1ZpUTNCejA2T0k/view?usp=sharing
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 You know that this only works in mac, right ? Linux and Windows don't have open or curl, or windows doesn't have rm.
•  » » » 4 years ago, # ^ |   +13 You can convert commands into your OS.rm is del in windows.open is xdg-open in linux. open is start in windows.curl is available here to download it for windows: http://curl.haxx.se/dlwiz/?type=bin&os=Win64curl should work on linux too as far as I know.
•  » » » » 4 years ago, # ^ |   0 You should install it (curl on linux) :)
•  » » » » 4 years ago, # ^ |   0 I converted it to Windows but when it s finished it says that the path I am using is being used by another process...
 » 4 years ago, # |   0 thanks
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 You are welcome. And thanks to you too :P But I didn't mean to be informative. I rather asked a question.So can someone please tell if it is possible to download problems in exactly the same format as in this file?


text:
# Qudits

try:
import cirq
except ImportError:
print("installing cirq...")
!pip install --quiet cirq
print("installed cirq.")

Most of the time in quantum computation, we work with qubits, which are 2-level quantum systems. However, it is possible to also define quantum computation with higher dimensional systems. A qu-d-it is a generalization of a qubit to a d-level or d-dimension system. For example, the state of a single qubit is a superposition of two basis states, $$|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$$, whereas the state of a qudit for a three dimensional system is a superposition of three basis states $$|\psi\rangle=\alpha|0\rangle+\beta|1\rangle+\gamma|2\rangle$$.

Qudits with known values for d have specific names. A qubit has dimension 2, a qutrit has dimension 3, a ququart has dimension 4, and so on. In Cirq, qudits work exactly like qubits except they have a dimension attribute different than 2, and they can only be used with gates specific to that dimension. In cirq, both qubits and qudits are subclasses of the class cirq.Qid.

To apply a gate to some qudits, the dimensions of the qudits must match the dimensions it works on. For example, consider gate represents a unitary evolution on three qudits,. Further suppose that there are a qubit, a qutrit, and another qutrit. Then the gate's "qid shape" is (2, 3, 3) and its on method will accept exactly 3 Qids with dimension 2, 3, and 3, respectively.

This is an example single qutrit gate acting on a single qutrit in a simple quantum circuit:

import cirq
import numpy as np

class QutritPlusGate(cirq.Gate):
"""A gate that adds one in the computational basis of a qutrit.

This gate acts on three-level systems. In the computational basis of
this system it enacts the transformation U|x〉 = |x + 1 mod 3〉, or
in other words U|0〉 = |1〉, U|1〉 = |2〉, and U|2> = |0〉.
"""

def _qid_shape_(self):
# By implementing this method this gate implements the
# cirq.qid_shape protocol and will return the tuple (3,)
# when cirq.qid_shape acts on an instance of this class.
# This indicates that the gate acts on a single qutrit.
return (3,)

def _unitary_(self):
# Since the gate acts on three level systems it has a unitary
# effect which is a three by three unitary matrix.
return np.array([[0, 0, 1],
[1, 0, 0],
[0, 1, 0]])

def _circuit_diagram_info_(self, args):
return '[+1]'

# Here we create a qutrit for the gate to act on.
q0 = cirq.LineQid(0, dimension=3)

# We can now enact the gate on this qutrit.
circuit = cirq.Circuit(
QutritPlusGate().on(q0)
)

# When we print this out we see that the qutrit is labeled by its dimension.
print(circuit)

0 (d=3): ───[+1]───

## cirq.Qid

cirq.Qid is the type that represents both qubits and qudits.

Cirq has the built-in qubit types, cirq.NamedQubit, cirq.GridQubit, and cirq.LineQubit, and it also provides corresponding cirq.Qid types:

By default cirq.Qid classes in cirq will default to qubits unless their dimension parameter is specified in creation. Thus a cirq.Qid like cirq.NamedQid('a') is a qubit.

### The cirq.qid_shape protocol

Quantum gates, operations, and other types that act on a sequence of qudits can specify the dimension of each qudit they act on by implementing the _qid_shape_ magic method. This method returns a tuple of integers corresponding to the required dimension of each qudit it operates on, e.g. (2, 3, 3) means an object that acts on a qubit, a qutrit, and another qutrit. When you specify _qid_shape_ we say that the object implements the qid_shape protocol.

When cirq.Qids are used with cirq.Gates, cirq.Operations, and cirq.Circuits, the dimension of each qid must match the corresponding entry in the qid shape. An error is raised otherwise.

Callers can query the qid shape of an object or a list of Qids by calling cirq.qid_shape on it. By default, cirq.qid_shape will return the equivalent qid shape for qubits if _qid_shape_ is not defined. In particular, for a qubit-only gate the qid shape is a tuple of 2s containing one 2 for each qubit e.g. (2,) * cirq.num_qubits(gate).

# Create an instance of the qutrit gate defined above.
gate = QutritPlusGate()

# Verify that it acts on a single qutrit.
print(cirq.qid_shape(gate))

(3,)

### Unitaries, mixtures, and channels on qudits

The magic methods _unitary_, _apply_unitary_, _mixture_, and _kraus_ can be used to define unitary gates, mixtures, and channels can be used with qudits (see protocols for how these work.)

Because the state space for qudits for $$d>2$$ live on larger dimensional spaces, the corresponding objects returned by the magic methods will be of corresponding higher dimension.

# Create an instance of the qutrit gate defined above. This gate implements _unitary_.
gate = QutritPlusGate()

# Because it acts on qutrits, its unitary is a 3 by 3 matrix.
print(cirq.unitary(gate))

[[0 0 1]
[1 0 0]
[0 1 0]]

For a single qubit gate, its unitary is a 2x2 matrix, whereas for a single qutrit gate its unitary is a 3x3 matrix. A two qutrit gate will have a unitary that is a 9x9 matrix (3 * 3 = 9) and a qubit-ququart gate will have a unitary that is an 8x8 matrix (2 * 4 = 8). The size of the matrices involved in defining mixtures and channels follow the same pattern.

### Simulating qudits

Cirq's simulators can be used to simulate or sample from circuits which act on qudits.

Simulators like cirq.Simulator and cirq.DensityMatrixSimulator will return simulation results with larger states than the same size qubit circuit when simulating qudit circuits. The size of the state returned is determined by the product of the dimensions of the qudits being simulated. For example, the state vector output of cirq.Simulator after simulating a circuit on a qubit, a qutrit, and a qutrit will have 2 * 3 * 3 = 18 elements. You can call cirq.qid_shape(simulation_result) to check the qudit dimensions.

# Create a circuit from the gate we defined above.
q0 = cirq.LineQid(0, dimension=3)
circuit = cirq.Circuit(QutritPlusGate()(q0))

# Run a simulation of this circuit.
sim = cirq.Simulator()
result = sim.simulate(circuit)

# Verify that the returned state is that of a qutrit.
print(cirq.qid_shape(result))

(3,)

Circuits on qudits are always assumed to start in the $$|0\rangle$$ computational basis state, and all the computational basis states of a qudit are assumed to be $$|0\rangle$$, $$|1\rangle$$, ..., $$|d-1\rangle$$. Correspondingly, measurements of qudits are assumed to be in the computational basis and for each qudit return an integer corresponding to these basis states. Thus measurement results for each qudit are assumed to run from $$0$$ to $$d-1$$.

# Create a circuit with three qutrit gates.
q0, q1 = cirq.LineQid.range(2, dimension=3)
circuit = cirq.Circuit([
QutritPlusGate()(q0),
QutritPlusGate()(q1),
QutritPlusGate()(q1),
cirq.measure(q0, q1, key="x")
])

# Sample from this circuit.
result = cirq.sample(circuit, repetitions=3)

# See that the results are all integers from 0 to 2.
print(result)

x=111, 222

[{ "type": "thumb-down", "id": "missingTheInformationINeed", "label":"Missing the information I need" },{ "type": "thumb-down", "id": "tooComplicatedTooManySteps", "label":"Too complicated / too many steps" },{ "type": "thumb-down", "id": "outOfDate", "label":"Out of date" },{ "type": "thumb-down", "id": "samplesCodeIssue", "label":"Samples / code issue" },{ "type": "thumb-down", "id": "otherDown", "label":"Other" }]
[{ "type": "thumb-up", "id": "easyToUnderstand", "label":"Easy to understand" },{ "type": "thumb-up", "id": "solvedMyProblem", "label":"Solved my problem" },{ "type": "thumb-up", "id": "otherUp", "label":"Other" }]


text:
You are currently offline. Some features of the site may not work correctly.

# Optical conductivity

The optical conductivity is a material property, which links the current density to the electric field for general frequencies. In this sense, this… Expand
Wikipedia

## Papers overview

Semantic Scholar uses AI to extract papers important to this topic.
2017
2017
• 2017
• Corpus ID: 118645855
Multi-Weyl semimetals are new types of Weyl semimetals which have anisotropic non-linear energy dispersion and a topological… Expand
Is this relevant?
2016
2016
• 2016
• Corpus ID: 119307715
The interband optical response of a three-dimensional Dirac cone is linear in photon energy ($\Omega$). Here, we study the… Expand
Is this relevant?
Highly Cited
2014
Highly Cited
2014
• 2014
• Corpus ID: 16601486
A method for analysis of spectroscopic ellipsometry data is demonstrated to extract the optical conductivity of mono- and bilayer… Expand
Is this relevant?
Highly Cited
2012
Highly Cited
2012
• 2012
• Corpus ID: 55473989
A bstractWe add a gravitational background lattice to the simplest holographic model of matter at finite density and calculate… Expand
Is this relevant?
Highly Cited
2010
Highly Cited
2010
• 2010
• Corpus ID: 118418977
Within the tight-binding approximation, we study the dependence of the electronic band structure and of the optical conductivity… Expand
Is this relevant?
2010
2010
• 2010
• Corpus ID: 73631050
Recent experimental data on the optical conductivity of niobium-doped ${\text{SrTiO}}_{3}$ are interpreted in terms of a gas of… Expand
Is this relevant?
Highly Cited
2008
Highly Cited
2008
• 2008
• Corpus ID: 55151443
We compute the optical conductivity of graphene beyond the usual Dirac-cone approximation, giving results that are valid in the… Expand
Is this relevant?
Highly Cited
2008
Highly Cited
2008
• 2008
• Corpus ID: 52086241
When a bilayer of graphene is placed in a suitably configured field effect device, an asymmetry gap can be generated and the… Expand
Is this relevant?
Highly Cited
2007
Highly Cited
2007
• 2007
• Corpus ID: 119638159
Landau level quantization in graphene reflects the Dirac nature of its quasiparticles and has been found to exhibit an unusual… Expand
Is this relevant?
Highly Cited
2003
Highly Cited
2003
In certain materials with strong electron correlations a quantum phase transition (QPT) at zero temperature can occur, in the… Expand
Is this relevant?


text:
# How to prove $\left\lceil\frac{x}y\right\rceil=\left\lfloor\frac{x-1}y\right\rfloor+1$ for positive integers $x,y$? [closed]

I have to prove that

$$\left\lceil\frac{x}y\right\rceil=\left\lfloor\frac{x-1}y\right\rfloor+1\;?$$

For any positive integers $x, y$.

Can anyone help me?

## closed as off-topic by Najib Idrissi, Joey Zou, Parcly Taxel, Watson, quid♦Sep 8 '16 at 9:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Najib Idrissi, Joey Zou, Parcly Taxel, Watson, quid
If this question can be reworded to fit the rules in the help center, please edit the question.

• Have you checked a few cases? Seen what happens? – Arthur Sep 7 '16 at 7:17
• The statement is false for $x=1,y=\frac12$ – 5xum Sep 7 '16 at 7:19
• Do you have to prove this for integer $x, y$? – 6005 Sep 7 '16 at 7:22
• You probably want to specify that $x,y\in\mathbb{Z}^{+}$. – barak manos Sep 7 '16 at 7:22
• @6005: Not true for all $x,y\in\mathbb{Z}^{-}$ (i.e., false for some $x,y\in\mathbb{Z}^{-}$). – barak manos Sep 7 '16 at 7:24

For this to be true, you need to specify that $\boldsymbol{x,y \in \mathbb{Z}}$ and $\boldsymbol{y > 0}$.

Then: if $$\left\lceil \frac{x}{y} \right\rceil = m$$ then $$m - 1 < \frac{x}{y} \le m$$ which means $$my - y < x \le my.$$ which we can rewrite $$my -y \le x - 1 < my$$ since $x$ is an integer and both bounds are integers.

Can you finish? You need to reach the conclusion that $$\left\lfloor \frac{x-1}{y} \right\rfloor = m - 1.$$

• oh i see, sorry – JonMark Perry Sep 7 '16 at 7:54

As @6005 clearly said, you need to agree that the $x,y$ are positive integers with the constraint $y \gt 0$.

This is a hint regarding the right side of the inequality:

$$\left\lceil\frac{x}y\right\rceil=\left\lfloor\frac{x-1}y\right\rfloor+1=\left\lfloor\frac{x-1}y+1\right\rfloor=\left\lfloor\frac{x-1+y}y\right\rfloor=\left\lfloor\frac{x}{y} + \frac{y-1}y\right\rfloor=\left\lfloor\frac{x}{y} + 1 - \frac{1}y\right\rfloor$$

• The statement $\lceil x/y \rceil = \lfloor x/ y + 1\rfloor$ is only true if $x/y$ is not an integer. – 6005 Sep 7 '16 at 7:47
• Your modification did not fix it :) – 6005 Sep 7 '16 at 7:49
• @6005 you were right, thank you for the feedback! I have removed that part of the explanation, I am keeping the part that can help to study what happens with the right side term. – iadvd Sep 7 '16 at 7:54
• Well you were almost there before, if you just split into cases based on whether or not $\frac{x}{y}$ was an integer. – 6005 Sep 7 '16 at 7:55
• @6005 good hint for the OP, thank you again! – iadvd Sep 7 '16 at 7:56

Let $x=py+q$, with $0\le q<y$.

Then if $q=0$, $$\left\lceil\frac{x}y\right\rceil=p,\left\lfloor\frac{x-1}y\right\rfloor+1=p+\left\lfloor\frac{y-1}y\right\rfloor+1=p.$$ and if $q>0$,

$$\left\lceil\frac{x}y\right\rceil=p+1,\left\lfloor\frac{x-1}y\right\rfloor+1=p+\left\lfloor\frac{q-1}y\right\rfloor+1=p+1.$$


text:
10

# Questions

4
0
111
views

### From universal measurability to measurability

may 13 at 7:21 Michael Greinecker 3,88411125

1
1
vote
0
83
views

### Can random elements be defined in terms of a measure algebra?

mar 10 at 15:07 Michael Greinecker 3,88411125

3
7
2
405
views

### Can one view the Independent Product in Probability categorially?

feb 6 at 9:16 Uwe Franz 1,294215

3
1
234
views

### Atoms of a sequence of Sigma-algebras

may 24 at 14:35 Nik Weaver 4,4171720

1
5
2
1k
views

### Can we put a Probability Measure on every $\sigma$-Algebra?

feb 21 12 at 15:04 Ramiro de la Vega 3,580715

1
3
0
160
views

### Proof of Lomnicki and Ulam on Infinite Product Probability Spaces

jan 9 12 at 20:29 Michael Greinecker 3,88411125

3
6
1
242
views

### Random Functions and Transition Probabilities

jul 26 at 13:17 Michael Greinecker 3,88411125

4
15
6
1k
views

### Unique Existence and the Axiom of Choice

mar 14 12 at 21:35 Ramiro de la Vega 3,580715

3
9
0
427
views

### Quotients of Measurable Spaces?

feb 8 12 at 19:06 Michael Greinecker 3,88411125

23
33


text:
• Save

# Numbers Problems with Solutions for SSC and Banking Exams

Last month 934 Views
NEW

Many students are puzzled about the numbers problems in their exams. Problems are part of our life but every problem has a solution. If you are preparing for SSC exam, here you can learn the numbers problems with solutions for SSC and banking exam.

In this blog, you will understand how to face or solve numbers problems which will be beneficial for your upcoming competitive exams. After learning the use of numbers problems with solutions you can practice problem on number aptitude question and answers.

You can read Number System Questions in Hindi if you want to prepare in Hindi.

## Numbers Problems with Solutions for Competitive Exams

Q.1. A number is as much greater than 36 as is less than 86. Find the number.

Solution. Let the number be x. Then, x – 36 = 86 – x ⟺ 2x = 86 + 36 = 122 ⟺ x = 61.

Hence, the required number is 61.

Q.2. The Sum of a rational number and its reciprocal is . Find the number.

Solution. Let the number be x.

Then,

$${x} + \frac{1}{x} = \frac{13}{6} ⟺\frac{x^2+1}x=\frac{13}{6}⟺6x^2-13x+6 = 0$$

$$⟺ 6x^2-9x-4x+6=0⟺(3x-2)(2x-3)=0$$

$$⟺x=\frac{2}{3} or\frac{3}{2}$$

Hence the required number is $$⟺x=\frac{2}{3} or\frac{3}{2}$$

Q.3. The difference of two numbers is 11 and one-fifth of their sum is 9. Find the numbers.

Solution. Let the number be x and y. Then.

x – y = 11            …(i)

and $$\frac{1}{5} (x + y)= 9 ⟺ x + y = 45$$…….(ii)

Adding (i) and (ii), We get : 2x = 56 or x = 28. Putting x = 28 in (i), we get : y = 17.

Hence, the numbers are 28 and 17.

Q.4. The average of four consecutive even numbers is 27. Find the largest of these numbers.

Solution. Let the four consecutive even number be x, x+2, x+4 and x +6

Then, the sum of these numbers = (27 x 4) = 108.

So, x + (x +2) + (x+4) + (x+6) = 108 or 4x = 96 or x = 24.

Largest number = (x + 6) = 30.

Q.5. Of two numbers, 4 times the smaller one is less than 3 times the larger one by 5. If the sum of the numbers is larger than 6 times their difference by 6, find the two numbers.

Solution. Let the number be x and y, such that x > y.

Then, 3x – 4y = 5 ……….(i)

and (x + y) – 6 ( x – y ) = 6 ⟺ - 5x + 7y = 6………(ii)

Solving (i) and (ii), we get : x = 59 and y = 43.

Hence, the required numbers are 59 and 43.

Q.6. Find a number such that when 15 is subtracted from 7 time the number, the result is 10 more than twice the number.

Solution. Let the number be x. Then, 7x – 15 = 2x + 10 ⟺ 5x = 25 ⟺ x = 5.

Hence, the required number is 5

Q.7. The sum of number is 184. If one-third of the one exceeds one-seventh of the other by 8, find the smaller number.

Solution. Let the numbers be x and (184-x). Then,

-  = 8  7x – 3(184-x) = 168 10x = 720  x = 72.

So, the numbers are 72 and 112. Hence, smaller number = 72.

Q.8. The sum of two number is 15 and the sum of their squares is 113. Find the numbers.

Solution. Let the numbers be x and (15 – x).

Then,  x2 + (15 – x)2 = 113 ⟺ x2 + 225 + x– 30x = 113

⟺ 2x2 – 30x + 112 = 0  ⟺ x2 – 15x + 56 = 0

⟺ (x – 7) (x – 8) = 0 ⟺ x = 7 or x = 8.

So, the numbers are 7 and 8.

Q.9 The sum of the squares of three consecutive odd numbers is 2531. Find the numbers.

Solution. Let the number be x, x+2 and x+4.

Then, x2 + (x+2)2+(x+4)2= 2531 ⟺ 3x2 + 12x – 2511 = 0

⟺ x2 + 4x – 837 = 0 ⟺ ( x – 27) (x +31) = 0 ⟺ x = 27.

Hence, the required numbers are 27, 29 and 31.

Q.10. The ratio between a two-digit number and the sum of the digit of that number is 4 : 1. If the digit in the unit’s place is 3 more than the digit in the ten’s place, what is the number.

Solution. Let the ten’s digit be x. Then, unit’s digit = (x+3).

Sum of the digits = x + (x +3) = 2x + 3. Number = 10x + (x+3) = 11x + 3.

= ⟺ 11x+3 = 4(2x+3)  3x = 9  x = 3.

Hence, required number = 11x + 3 = 36.

I hope these given number problems with solutions help you to solve number problems quickly and correctly in competitive exams. Practice more problems on numbers question and answers.

You can ask me in the comment section if you are facing problems related to number problems with solutions and its use.

All the best for Exams.

## About author

### Vikram Singh

Providing knowledgable questions of Reasoning and Aptitude for the competitive exams.

## Report Error : Numbers Problems with Solutions for SSC and Banking Exams

Please Enter Message
Error Reported Successfully


text:
# CAC201 Not Not Easy, CRACK-A-CODE 2.0 Editorial

Setter & Editorialist: Mayuresh Patle
Testers: Uttkarsh Bawankar, Vallabh Deshpande

EASY

# PREREQUISITES

Prefix Sum, Binary Search

# PROBLEM

You are given an array A of length N, containing values in range [0,2000], having weight W_i associated with i^{th} position of the array (for 1\leq i \leq N). You have to peform Q queries of two types:

• Type 1: 1 \ X, in this query, you have to add X to current weights of all position.
• Type 2: 2 \ V \ L \ R, in this query, you have to print the sum of weights of all positions in range [L,R] which have value V in array A.

# QUICK EXPLANATION

Maintain a variable, say extra to keep track of total increment. For type 1 queries, simply add X to extra. For each i \in [0,2001], maintain 2 arrays, one array should contain the positions of elements having value i and other array should contain the prefix sum of weights corresponding to the saved positions. For each query of type 2, use binary search to get the range of indices of favourable positions in the array maintained for value V, print the difference of prefix sums for new range of indices +\ count*extra, where count = count of indices in the new range.

# EXPLANATION

Let’s jump directly to the solution!

Let’s say that pos[i] is an array that contains all the positions in array A which have element value i, i.e. pos[i] contains all positions such that A[position] = i. Let’s say another array pre[i] contains the prefix sums corresponding to value i, such that pre[i][j]= \sum_{x=0}^{x=j}{W[pos[i][x]]}

We have to calculate and store this information for all i such that 0 \leq i \leq 2000. To avoid the overhead of handling exceptions, let’s initialize pos[i][0]=0 and pre[i][0]=0.

Assume that arr.back() gives the value of last element of array arr, and arr.push\_back(x) adds new elelment x at the end of arr. Now, to store the above information, we just have to do the following operations

for i = 1 to N
pos[A[i]].push_back(i)
pre[A[i]].push_back(pre[A[i]].back() + W[i])

Now we are ready with the structure that was discussed above. Next task is to handle the queries.
Consider the qeries of type 2, we have the values of V, L and R, to handle this query, we somehow need to find largest index rt such that pos[V][rt] \leq R and largest index lt such that pos[V][lt] \lt L. This can be done in logarithmic time using binary search since the array pos[i] is already sorted.

Now , our answer is pre[V][rt] - pre[V][lt]
Great! We are done with type 2 query!

Wait, wait, wait!!!
What if some type 1 queries were asked earlier?
We see that the result of Type 2 queries depends previous type 1 queries. Now, how to handle type 1 queries? Should we use Segment Tree or Fenwick Tree?

Don’t worry, we don’t need Segment Tree or Fenwick Tree here.
Since all weights are being updated, we can simply use a variable, say extra, to keep track of the total increment.

Initially, extra=0. If we encounter any query of type 1, we can simply do
extra = extra + X

Now, final task is to calculate the correct answer for Type 2 queries. We observe that we just need to add extra*count to the previously calculated answer, where count = number of elements to be considered.

Great we have the exact answer now, i.e.
pre[V][rt] - pre[V][lt] + extra*(rt - lt)

For each testcase:
Time Complexity = O(Q*logN)
Auxiliary Space = O(N)

# SOLUTION

Setter's Solution
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector <ll> vll;
#define pb push_back
#define all(vec) vec.begin(),vec.end()
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
ll T;
cin>>T;
while(T--)
{
vector <vll> pos(2001,vll(1,0)), pre(2001,vll(1,0));

ll extra=0,n,q,i,type;
cin>>n>>q;
ll a[n],w[n];

for(ll& i:a) cin>>i;                        //Input array A
for(ll& i:w) cin>>i;                        //Input weights

for(i=0;i<n;++i)
{
pos[a[i]].pb(i+1);                      //Add current index i in the positions of a[i]

pre[a[i]].pb(pre[a[i]].back()+w[i]);    //Store the prefix sum of weights till current index
}

while(q--)
{
cin>>type;

//Type 1 Queries
if(type==1)
{
ll x;
cin>>x;
extra+=x;
continue;
}

//Type 2 Queries
ll v,l,r;
cin>>v>>l>>r;
auto lt=lower_bound(all(pos[v]),l);
auto rt=upper_bound(all(pos[v]),r);
--lt,--rt;
l=lt-pos[v].begin();
r=rt-pos[v].begin();
cout<<pre[v][r]-pre[v][l]+(r-l)*extra<<"\n";
}
}
return 0;
}


text:
# Height and Distance MCQ Question with Answer

## Height and Distance MCQ with detailed explanation for interview, entrance and competitive exams. Explanation are given for understanding.

Question No : 1

The angle of elevation of the sun, when the length of the shadow of a tree 3 times the height of the tree, is:

30º
45º
60º
90º

Question No : 2

From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is:

149 m
156 m
173 m
200 m

Question No : 3

An observer 1.6 m tall is 203 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:

21.6 m
23.2 m
24.72 m
None of these

Question No : 4

The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

2.3 m
4.6 m
7.8 m
9.2 m

Question No : 5

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?

8 units
12 units
None of these


text:
# PageRank

(Difference between revisions)
 Revision as of 21:18, 17 February 2009 (edit)← Previous diff Revision as of 21:27, 17 February 2009 (edit) (undo)Next diff → Line 30: Line 30: y = x(i) + e; y = x(i) + e; s_hat = s_hat_old + y; s_hat = s_hat_old + y; - e = (s_hat_old - s_hat) + y; + e = s_hat_old - s_hat + y; end end

## Revision as of 21:27, 17 February 2009

CSUM() in Digital Signal Processing terms: Q is a floating-point quantizer to 64 bits, z-1 is a unit delay, qi represents error due to quantization (additive by definition).  -Jon Dattorro
```function s=csum(x)
% CSUM Sum of elements using a compensated summation algorithm
%
% For large vectors, the native sum command in Matlab does
% not appear to use a compensated summation algorithm which
% can cause significant roundoff errors.
%
% This code implements a variant of Kahan's compensated
% summation algorithm which often takes about twice as long,
% but produces more accurate sums when the number of
% elements is large.
%
%
% Example:
%   v=rand(1e7,1);
%   sum1 = sum(v);
%   sum2 = csum(v);
%   fprintf('sum1 = %18.16e\nsum2 = %18.16e\n', sum1, sum2);

% David Gleich, Stanford University, 2008

s_hat=0; y=0; e=0;
for i=1:numel(x)
s_hat_old = s_hat;
y = x(i) + e;
s_hat = s_hat_old + y;
e = s_hat_old - s_hat + y;
end
```


text:
In SF2 the sulfur in the middle has SP3 hybridization. It also has the same properties of non-polar molecules such as being non-soluble in water and being soluble in non-polar organic solvents. So H3O+ has 3 polar bonds, and the overall molecule is polar too. Sulfur hexafluoride, abbreviated as SF6, is a nonpolar molecule. The polarity of each individual bond is therefore cancelled, leaving a nonpolar molecule. 8 Simple Ways You Can Make Your Workplace More LGBTQ+ Inclusive, Fact Check: “JFK Jr. Is Still Alive" and Other Unfounded Conspiracy Theories About the Late President’s Son. Nonpolar Molecules: If there is an even number of lone pairs, you must check the VSEPR structure to decide. SF6 has an octahedral molecular geometry, which means that the sulfur molecule has six fluorine atoms surrounding it. NOAA Hurricane Forecast Maps Are Often Misinterpreted — Here's How to Read Them. jpg ZnCl2 (aq) + H2 (g) What is the theoretical yield of hydrogen gas if 5.00 mol of zinc are added to an excess of hydrochloric acid? CEO Compensation and America's Growing Economic Divide. Here is a list of molecules that are considered. Answer: SF6 is a nonpolar molecule due to the symmetrical placement of all of the fluorines around the central sulfur atom. So the Is H2S Polar Or Nonpolar? CH4 is a non-polar molecular because the dipole moment between atoms cancels out each other. These forces include ionic bond, covalent bond, metallic bond, and hydrogen bonding. SF6 has an octahedral molecular geometry, which means that the sulfur molecule has six fluorine atoms surrounding it. So, 1.2 is polar bond. CO2 is a polar molecule because the dipole moment between atoms does no cancel each other, and the magnitude is greater than 0. Polar "In chemistry, polarity is a separation of electric charge leading to a molecule or its chemical groups having an electric dipole or multipole moment. Which is a good example of a contact force? Although it has an asymmetrical molecular geometry, the entire molecule is non-polar dues to the absence of any polar bonds. Molecules polar. This video discusses how to tell if a molecule / compound is polar or nonpolar. SF6 is a nonpolar compound in nature because as per VSEPR theory six fluorine atoms are arranged symmetrically with the sulfur atom such that dipole moment of S-F bond gets canceled out making the SF6 a nonpolar compound. (Hint diprotic acid), Which sample of matter is a mixture? Polar "In chemistry, polarity is a separation of electric charge leading to a molecule or its chemical groups having an electric dipole or multipole moment. Even though each of the bonds is polar, the molecule is non polar as the dipole moments cancel each other. This arrangement at perpendicular (90˚) angles ensures that the pull for electrons by fluorines on one side is balanced by fluorines pulling on the other side. It is considered to be an extreme greenhouse gas. The EN difference between hydrogen and sulfur is 0.4, so hydrogen and sulfur form non-polar bonds. Here's a look at what polar and nonpolar mean, how to predict whether a molecule will be one or the other, and examples of representative compounds. A COVID-19 Prophecy: Did Nostradamus Have a Prediction About This Apocalyptic Year? 1-butanol. I'll tell you the polar or nonpolar list below. SF6 is a nonpolar compound in nature because as per VSEPR theory six fluorine atoms are arranged symmetrically with the sulfur atom such that dipole moment of S-F bond gets canceled out making the SF6 a nonpolar compound. The molecule has a central bromine atom surrounded by five fluorides and a pair of electrons. The easiest way to determine if a molecule is polar or nonpolar is to draw its Lewis Structure and, if necessary, check its molecular geometry. SF6 has its IUPAC name Sulfur hexafluoride and it is considered as an extreme gas responsible for the greenhouse effect. It is an odorless, non-toxic, and non-flammable gas. Question = Is SnCl4 ( Tin(IV) chloride ) polar or nonpolar ? Sulfur Hexafluoride having its chemical formula SF6 is an inorganic compound that is colorless in appearance. Always subtract the smaller number from the greater one. The electronegative difference is-3.44- 2.58= 0.86. SF6 polar or nonpolar? Answer: SF4 is polar due to the presence of a lone pair of electrons leading to electron-electron repulsion and thereby a region of partial negative charge. Nonpolar. The $\mathrm{C}-\mathrm{Cl}$ bonds and the $\mathrm{C}-\mathrm{F}$ bonds have different polarities. How many protons are in a ion with 18 electrons and a - 1 charge, Which of the following element combinations will likely form polyatomic ions? Many students may have a query about whether SF6 is polar or not. Because there are six fluorine atoms, this means that each atom is 90 degrees from its neighbors. b) SO2. In your example of "SF"_4, the Lewis Structure would look … Polar: IF, PCl3, IF5 Nonpolar: CS2, SO3, SF6 Explanation: Polar molecules form when two atoms do not share electrons equally in a covalent bond. Zn (s) + 2HCl (aq) mc014-1. If you want to quickly find the word you want to search, use Ctrl + F, then type the word you want to search. Air, water, ammonia. Question = Is SF2 polar or nonpolar ? Sulfur- 2.58, Oxygen- 3.44. Answer = SnCl4 ( Tin(IV) chloride ) is Nonpolar What is polar and non-polar? As a result, it is a non-polar molecule. cesium and chlorine oxygen and carbon sodium and bromine xenon and neon, How many moles of carbon dioxide are produced from 8.73 moles of tristearin?, Find the pH of a solution that contains 3.25 g of H2SO4 (MM = 98.08 g/mol) dissolved in 2.75 liters of solution. Answer = SF2 ( Sulfur difluoride ) is Polar What is polar and non-polar? © 2021 Education Expert, All rights reserved. There are different forms of forces that bind the molecules. c. In the Lewis structure, the central Se atom has six electron groups with no lone pairs. Polar and Nonpolar Molecules. 2009-03-18 16:06:45. According to VSEPR theory, the fluorine molecules position themselves as far away from each other as possible. If there is an odd number of lone pairs of electrons around the central atom, the molecule is polar. 5.05 g 10. b. If the difference is from 0-0.4 the bond is nonpolar, but if it's from 0.5-1.9 the bond is polar. So the molecule is polar. The equation shows the reaction between zinc metal and hydrochloric acid. The two main classes of molecules are polar molecules and nonpolar molecules.Some molecules are clearly polar or nonpolar, while others fall somewhere on the spectrum between two classes. SiF6 2- is Polar. The U.S. Supreme Court: Who Are the Nine Justices on the Bench Today? Is BrF5 Polar or Nonpolar. To determine if a molecule overall is polar or nonpolar you must look at both the electronegativity differences in each bond and the shape of the molecule. Write five observation of cotton ball and pine cone of the solid. The molecules that are covalently bonded can be polar and nonpolar. This is because the VSEPR theory says that when six fluorineatoms are arranged symmetrically around the sulfur atom, the bond dipoles are cancelled. List molecules polar and non polar. According to VSEPR theory, the fluorine molecules position themselves as far away from each other as possible. Why don't libraries smell like bookstores? A molecule is classified as a polar molecule when the arrangement of the atoms is such that one end of the molecule has a positive electrical charge and the other end has a negative charge. Polar molecules must contain polar bonds due to a difference in electronegativity between the bonded atoms. SF6 is a non-polar molecule. Hydrogen sulfide is non-polar on account of its non polar H–S bonds. This further means that each fluorine atom is directly opposite from another fluorine atom. HCN is a linear molecule with a single bond between the H and the C and a triple bond between the C and the N. Looking at the electronegativities H
Can You Put Liquid Highlighter Over Powder, Unique Sports Management International, Vu Amsterdam Vacancies, Dog Itching After Bath, How Much Caffeine Is In A Medium Dunkin' Refresher, Picsart Stickers Ideas, Vortex Crossfire Hd 10x50,


text:
# Knowledge Base

### EMRAYS Integration

With the EMRAYS integration you can instantly analyze your Bynder assets to determine which assets will perform best with the sentiment you want convey in your content. EMRAYS uses their powerful AI with built-in human intuition to help marketers and content creators to make better and faster decisions on what images to use in their content.

Are you a content creator and would you like to know how which images in your Bynder DAM will be most impactful? Check out the integration below and start choosing the right image for your content creation.

#### Tell me More About the EMRAYS Integration

The EMRAYS integration leverages a powerful AI to analyze your image assets and rank them by (social) impact or similarity. Depending on which social impact you're looking for (positivity, wow or fun) the integration will rank your selection of image assets from high to low impact. Alternatively, you can also define your own models, so that you can have custom rankings implemented.

Do you have a selection of assets that you need to chose one image from but are you not exactly sure what impact you're looking for? You may have an existing image in your DAM that visualizes and expresses the message you want to send out. EMRAYS can compare this image with a selection of assets and show which images from this selection have the most similar impact. This allows you to find images with a similar impact across the entire DAM.

#### How to Enable the EMRAYS Integration?

Contact your Bynder Customer Success Manager if you would like to have the EMRAYS integration enabled for your portal. Once the integration has been enabled, you'll receive a URL to log in to the integration.

#### How to Use the EMRAYS Integration?

1. Log in to the EMRAYS integration by using the URL we've provided you with. The Compact View will open. If your company's Bynder portal URL is already filled in in the pop-up, click Connect. If not, enter the URL in the below format. A popup will open, which allows you to log in to your Bynder environment.

https://YourPortalURL.com
2. Check out the available options below.

##### How to Analyze the Social Impact of Your Bynder Image Assets?
1. Select the images you want EMRAYS to analyze in the  Compact View and click Add assets. EMRAYS will load the assets and show the thumbnails in a comparison screen.

2. Rank the images by the preferred social impact positivity, wow factor or fun and rank them by low or high intensity by clicking these options in the top bar.

Based on your selected criteria the integration will show you the images with the most impact.

##### How to Find Similar Ranking Bynder Image Assets?

Are you not exactly sure what asset you're looking for? Follow the steps below if you want to find images with an impact similar to an existing image in your DAM.

1. Select the images in the Compact View that you want to compare and click Add assets. Make sure this selection includes the base image as well as the images you want to compare. EMRAYS will load the assets and show the thumbnails in a comparison screen.

2. Select your base image and click the Similarity Ranking button in the top bar.

The other images in your selection will now be compared with your base image and ranked by a similar impact.


text:
You have formulas that allow you to differentiate $x^2\text{,}$ $2^x\text{,}$ and $2^2\text{.}$ You don't, however, have a formula to differentiate $x^x\text{.}$ In this section you are going to use a process called logarithmic differentiation to determine the derivative formula for the function $y=x^x\text{.}$ Example 7.3.1 shows this process for a different function.

With logarithmic differentiation, when $y$ is written as a function of $x\text{,}$ especially one involving only products, quotients, and powers, begin by taking the logarithm of both sides of the equation relating $y$ to $x\text{.}$ This technique will usually only be helpful if the logarithm rules allow you to break up the right side some. Proceed to differentiate each side with respect to $x\text{.}$ On the left side, respect the rules for implicit differentiation. Once this is done, algebraically solve for $\lz{y}{x}\text{.}$ If there are any instances of $y$ left, replace them with the original expression in $x\text{.}$

##### Example7.3.1Logarithmic Differentiation

\begin{align*} y&=\frac{x^{e^x}}{4x+1}\\ \fe{\ln}{y}&=\fe{\ln}{\frac{x^{e^x}}{4x+1}}\\ \fe{\ln}{y}&=\fe{\ln}{x^{e^x}}-\fe{\ln}{4x+1}\\ \fe{\ln}{y}&=e^x\fe{\ln}{x}-\fe{\ln}{4x+1}\\ \lzoo{x}{\fe{\ln}{y}}&=\lzoo{x}{e^x\fe{\ln}{x}-\fe{\ln}{4x+1}}\\ \frac{1}{y}\lz{y}{x}&=\lzoo{x}{e^x}\cdot\fe{\ln}{x}+e^x\cdot\lzoo{x}{\fe{\ln}{x}}-\frac{1}{4x+1}\cdot\lzoo{x}{4x+1}\\ \frac{1}{y}\lz{y}{x}&=e^x\cdot\fe{\ln}{x}+e^x\cdot\frac{1}{x}-\frac{1}{4x+1}\cdot4\\ \frac{1}{y}\lz{y}{x}&=e^x\fe{\ln}{x}+\frac{e^x}{x}-\frac{4}{4x+1}\\ \lz{y}{x}&=y\left(e^x\fe{\ln}{x}+\frac{e^x}{x}-\frac{4}{4x+1}\right)\\ \lz{y}{x}&=\frac{x^{e^x}}{4x+1}\left(e^x\fe{\ln}{x}+\frac{e^x}{x}-\frac{4}{4x+1}\right) \end{align*}

In the olden days (before symbolic calculators) we would use the process of logarithmic differentiation to find derivative formulas for complicated functions. The reason this process is “simpler” than straight forward differentiation is that we can obviate the need for the product and quotient rules if we completely expand the logarithmic expression before taking the derivative.

# Subsection7.3.1Exercises

##### 1

The function $y=x^x$ is only defined for positive values of $x$ (which in turn means $y$ is also positive), so we can say that $\fe{\ln}{y}=\fe{\ln}{x^x}\text{.}$ Use implicit differentiation to find a formula for $\lz{y}{x}$ after first applying the power rule of logarithms to the logarithmic expression on the right side of the equal sign. Once you have your formula for $\lz{y}{x}\text{,}$ substitute $x^x$ for $y\text{.}$ Voila! You will have the derivative formula for $x^x\text{.}$ So go ahead and do it.