text:
# H. Chomp

## Summary

Chomp is a two-player game played using a rectangular chocolate bar made up of smaller squares (cells). The players take turns choosing one cell and eating it together with those cells that are above it and to its right. (this is a low sugar chocolate bar). The bottom left cell is poisoned and whoever is forced to eat it loses. The following diagram shows a game during a sequence of moves:

A position in a 3-by-100 CHOMP game is described by the number p of squares in the bottom row, the number q of squares in the middle row, and the number r of squares in the top row with:
$100 ≥ p ≥ q ≥ r ≥ 0$
Write a program that determines whether a given position is a winning or losing position. If it is a winning position, give the next square to eat.

## Input

The first line of input contains a single decimal integer P,(1 ≤ P ≤ 500), which is the number of datasets that follow. Each dataset should be processed identically and independently. Each dataset consists of a single line of input containing the dataset number K followed by the counts 100 ≥ p ≥ q ≥ r ≥ 0 of squares in the bottom row (p), middle row (q), and top row (r). respectively, separated by single spaces

4
1 3 3 3
2 3 1 0
3 3 2 0
4 97 64 35

## Output

For each dataset there is a single line of output. If the input position is a losing position, the output line consists of the dataset number K, followed by a single space, followed by the capital letter ‘L’. Otherwise the output line consists of the dataset number K, followed by the capital letter ‘W’, followed by the column numbder and row number of the square to eat which results in a losing position for opponent

1 W 2 2
2 W 3 1
3 L
4 W 51 1


text:
# How do you integrate int xe^(-x^2)dx from [0,1]?

Nov 6, 2016

${\int}_{0}^{1} x {e}^{- {x}^{2}} \mathrm{dx} = \frac{1}{2} \left(1 - \frac{1}{e}\right)$

#### Explanation:

To determine the definite integral ,we compute the integral itself first then work on the boundaries.

As we know if :
$\int f \left(x\right) = F \left(x\right) + C$

Then

${\int}_{a}^{b} f \left(x\right) = F \left(b\right) - F \left(a\right)$

Let us compute the integral $\int x {e}^{- {x}^{2}} \mathrm{dx}$
Let:
color(red)(u(x)=e^(-x^2)

Then

color(blue)(du(x)=-2xe^(-x^2)dx
rArrcolor(blue)(xe^(-x^2)=-1/2du(x)

$\int x {e}^{- {x}^{2}} \mathrm{dx} = \int \textcolor{b l u e}{- \frac{1}{2} \mathrm{du} \left(x\right)}$

$\int x {e}^{- {x}^{2}} \mathrm{dx} = - \frac{1}{2} \int \mathrm{du} \left(x\right)$

$\int x {e}^{- {x}^{2}} \mathrm{dx} = - \frac{1}{2} u \left(x\right) + C$

intxe^(-x^2)dx=-1/2color(red)(e^(-x^2)+C
$\text{ }$

Let us calculate the value of the definite integral:

${\int}_{0}^{1} x {e}^{- {x}^{2}} \mathrm{dx} = - \frac{1}{2} \left({e}^{- {1}^{2}} - {e}^{- {0}^{2}}\right)$

${\int}_{0}^{1} x {e}^{- {x}^{2}} \mathrm{dx} = - \frac{1}{2} \left({e}^{- 1} - {e}^{- 0}\right)$

${\int}_{0}^{1} x {e}^{- {x}^{2}} \mathrm{dx} = - \frac{1}{2} \left({e}^{- 1} - 1\right)$

Therefore,

${\int}_{0}^{1} x {e}^{- {x}^{2}} \mathrm{dx} = \frac{1}{2} \left(1 - \frac{1}{e}\right)$

format: plain


text:
# Understanding Tensors in PyTorch 🔥

In this first post, I will try to deconstruct tensors which are the fundamental data structures behind neural networks.

July 17, 2019 - 12 minute read -

So we put a tensor in your tensor and attained a higher dimension. Wait what?

Image source

For smaller dimensions, we can think in geometric analogies as shown above, like for e.g.

• A 0-d Tensor or Scalar would be a point in space.
• A 1-d Tensor or Vector would be a line in that space.
• A 2-d Tensor or Matrix would be a square (or rectangle), and
• A 3-d Tensor would be a cube (or cuboid)

Similarly we can go further,

• A 4-d Tensor would be a vector of cubes.
• A 5-d Tensor would be a matrix of cubes.
• A 6-d Tensor would be a cube made up of cubes.
• A 7-d Tensor would be a vector of cubes made up of cubes!

You get the idea.

As you can see, tensors are nothing but a generalisation of things we know and love : scalars, vectors and matrices. Specifically it is a N-dimensional data structure (or container) where $N$ is the number of dimensions. For people familiar with arrays, its just a multi-dimensional array.

As the world around us is three dimensional, we have a hard time visualising tensors, especially as the dimensions go beyond our own. Like imagine a 10-dimensional tensor in your head. What does it look like? Use the analogy of cubes and let me know in the comments below!

But why are they important for deep learning, you ask? Well they are central to linear algebra and also the fundamental data structures of a neural network. Its all N-dimensional tensors under the hood, following some predefined rules (dot products and applying some non-linearity) to compute some output given certain inputs.

Let’s see how we can create tensors in PyTorch.

## Import Libraries

A good practice before starting any experiment is to set a fixed seed. This makes things deterministic, meaning every time we run the experiment certain non-determinisitic (i.e. probabilistic) processes won’t result in different values. And yes, 42 is a reference to The Hitchhiker’s Guide to the Galaxy :)

## Generating Tensors

There are various ways to generate tensors in PyTorch. Let’s try a few and also check the resulting tensors’ datatypes,

Note : Here torch.tensor() with a single number actually returns a scalar value with int64 as the data type, whereas the others return a tensor. In general, torch.tensor() accepts a list of numbers and the dimensions/size of the tensor is determined from it.

Lets try passing a list of lists to torch.tensor() as follows,

So its a 2-d tensor of 3x3 size. We can peek into its contents just to be sure,

So we can pass lists like,

• $[..]$ which is a list (or an array). It would be a rank 1 or a 1-d tensor.
• $[[..],[..],[..]]$ which is a list of lists. It would be a rank 2 or a 2-d tensor.
• $[[[..],[..]],[[..],[..]]]$ which is a list of lists containing lists. It would be a rank 3 or a 3-d tensor. and so on.

The notion of Rank is very important when dealing with tensors. In a rigorous mathematical setting, it would have a slightly different meaning, but here I am using the term interchangeably with dimensions.

torch.empty() can be used for randomly initialising a tensor with a given size,

Similarly, torch.ones() is for creating a tensor with all the values as ones.

Both torch.empty() and torch.ones() accept dimensions of the tensor you want to create.

Note : We can use the GPU variant of tensors by just adding .cuda() at the end. This allows for faster computations by leveraging GPU memory.

## Specifying data types and type coercion

One way to force a PyTorch tensor to be of a specific datatype is to use the dtype argument as follows,

Note : The datatypes are specified with torch.<something>.

However, we cannot do this with torch.Tensor() since it is actually an alias for torch.FloatTensor() and fixes a default datatype of float rather than inferring it from the tensor passed. On the other hand, datatypes can be explicitly stated with torch.tensor() (as we saw above).

Now let’s check torch.tensor()’s default datatype,

Another cool thing is that if we just use a dot at the end of a number, the tensor becomes of type float,

Next up, we can check the sizes or shapes of tensors with .size(),

Note : Both .size() and .shape works.

Did I say tensors can only contain number? They can contain boolean values (or strings) as well. Using dtype=torch.bool we can convert a tensors of ones into a tensor of True(s) as follows,

The tensors should only contain elements of homogeneous datatypes.

## Generating tensors containing random numbers

Now let’s generate a tensor with torch.rand(), which will be composed of numbers randomly drawn from a uniform distribution $U \backsim [0,1)$

If U is a random variable uniformly distributed on $[0, 1]$, then $(r1 - r2) * U + r2$ is uniformly distributed on $[r1, r2]$. We can also use uniform_ to perform operations inplace (anything with a underscore at the end means inplace). Let’s see them in action,

## Indexing and slicing

Both indexing and slicing of PyTorch’s tensors work similar to numpy’s ndarrays and python’s lists. For e.g.

We can get a copy of the entire tensor like so,

We can specify indices for each dimension of the tensor to get the specific element as follows,

We can also slice the tensors along a specific dimension as follows,

In case you forgot how slicing worked in python, here’s a quick refresher :

• z[start:stop] - Get all items from start index all the way upto stop index-1
• z[start:] - Get all items from start index all the way to the end
• z[:stop] - Get all items from the beginning all the way to the stop index-1
• z[:] - Get the copy of the entire list

The only difference here is that we are dealing with multiple dimensions for tensors.

## 0-d Tensor or a Scalar

We can have scalars in PyTorch as follows,

or

Note that although their sizes are slightly different, however a 2-d tensor of size 1x1 contains just one number. Similarly a 3-d tensor of size 1x1x1 will also contain a single element and so on.

And since its just a single number we can retrieve it by using .item()

## 1-d Tensor or a Vector

Similarly for vectors,

or we can manually define as,

## 2-d Tensor or a Matrix

Now let’s generate a 16x16 matrix as follows,

We can visualise the matrix by plotting it as a heatmap. This allow us to quickly get a sense of the numbers contained within visually.

The intensity of each block corresponds to the values in the matrix. Doesn’t it look like the pixels of very low resolution grayscale image?

## 3-d Tensor

Now let’s create a 3-d tensor of size 16x16x16 as follows,

Now imagine a cube with all the faces having a size of 16. We can visualise each slice of this cube along its depth. For the 1st slice,

Similarly, the second slice,

and so on. In fact, our whole 16x16x16 cube is made up of 16 of these 16x16 slices (makes sense doesn’t it?).

Just for fun, let’s examine the cross-section i.e. all the 16 slices of our cube,

Our 16x16x16 3-d tensor turned out to be composites of 16 different grayscale images in way! Looking at matrices and tensors in this way is helpful, especially in the context of image processing since images have depth in the form of channels (like RGB) and are composites of 2-d tensors.

Note : The numbers on the axis in all the heatmaps represent indices in the 1st and 2nd dimensions.

A very neat feature in PyTorch is the seamless conversion between PyTorch and numpy data structures (tensors to ndarray) by just calling .numpy().

This means we can leverage all functions/methods which rely on a numpy array.

Now let’s jump a few dimensions and revisit what a 6-d tensor looked like (remember?),

Let the sizes along each dimension for the tensor be 16x16x16x16x16x16. Think about it :

• The larger cube represents a 3-d tensor of size 16x16x16.
• The smaller cubes within also represents a 3-d tensor of size 16x16x16.
• If the larger cube contains 16x16x16 smaller cubes with each having a size of 16x16x16 we get our 6-d tensor.

Tensors ain’t so scary anymore, right? We will go deeper next time with concepts such as broadcasting and tensor operations. In case you find a typo or something didn’t quite click, please leave a comment below!


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# If Car X followed Car Y across a certain bridge that is 1/2

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If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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If Car X followed Car Y across a certain bridge that is 1/2 mile long, how many seconds did it take Car X to travel across the bridge?

(1) Car X drove onto the bridge exactly 3 seconds after Car Y drove onto the bridge and drove off the bridge exactly 2 seconds after Car Y drove off the bridge.
(2) Car Y traveled across the bridge at a constant speed of 30 miles per hour.

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Re: If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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SOLUTION

If Car X followed Car Y across a certain bridge that is 1/2 mile long, how many seconds did it take Car X to travel across the bridge?

Let the time needed for car X to travel across the bridge be $$t_x$$ seconds and the time needed for car Y be $$t_y$$ seconds.

(1) Car X drove onto the bridge exactly 3 seconds after car Y drove onto the bridge and drove off the bridge exactly 2 seconds after car Y drove off the bridge --> car X needs 1 second less to travel across the bridge than car Y --> $$t_y=t_x+1$$. Not sufficient to calculate $$t_x$$.

(2) Car Y traveled across the bridge at a constant speed of 30 miles per hour. 30 miles per hour is $$\frac{30}{3600}=\frac{1}{120}$$ miles per second --> car Y needs $$t_y=\frac{(\frac{1}{2})}{(\frac{1}{120})}=60$$ seconds to travel across the bridge. Not sufficient to calculate $$t_x$$.

(1)+(2) $$t_y=t_x+1$$ and $$t_y=60$$ --> $$t_x=60-1=59$$. Sufficient.

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Re: If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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Bunuel

Can this not be vvorked out just vvith A

--> since distances are equal
then equate

SX. Tx = Sy. Ty

Is this not possible

then in that case statement 1 becomes sufficient please explain

Bunuel wrote:
SOLUTION

If Car X followed Car Y across a certain bridge that is 1/2 mile long, how many seconds did it take Car X to travel across the bridge?

Let the time needed for car X to travel across the bridge be $$t_x$$ seconds and the time needed for car Y be $$t_y$$ seconds.

(1) Car X drove onto the bridge exactly 3 seconds after car Y drove onto the bridge and drove off the bridge exactly 2 seconds after car Y drove off the bridge --> car X needs 1 second less to travel across the bridge than car Y --> $$t_y=t_x+1$$. Not sufficient to calculate $$t_x$$.

(2) Car Y traveled across the bridge at a constant speed of 30 miles per hour. 30 miles per hour is $$\frac{30}{3600}=\frac{1}{120}$$ miles per second --> car Y needs $$t_y=\frac{(\frac{1}{2})}{(\frac{1}{120})}=60$$ seconds to travel across the bridge. Not sufficient to calculate $$t_x$$.

(1)+(2) $$t_y=t_x+1$$ and $$t_y=60$$ --> $$t_x=60-1=59$$. Sufficient.

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Re: If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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16 Sep 2012, 22:48
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The question asks the time taken by X to cross the bridge.

How come? Statement 1 helps us to get the relationship between X and Y wrt time. It does not tell us the time taken by Y to cross

Statement 2 gives us the time taken by Y to cross the bridge.

Both the statements together are sufficient.

Hope this helps!

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Re: If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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02 May 2013, 20:07
Banuel,

I'm not seeing it, is there a way you can list the three or four item being tested? and what makes the answer sufficient?

Time?
Rate of Speed?
Distance?
Any other?

Thanks.

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Re: If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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03 May 2013, 03:52
laythesmack23 wrote:
Banuel,

I'm not seeing it, is there a way you can list the three or four item being tested? and what makes the answer sufficient?

Time?
Rate of Speed?
Distance?
Any other?

Thanks.

Not sure I understand what you mean...

Anyway, we need to get $$t_x$$.

From (1) we have that $$t_y=t_x+1$$ and from (2) we have that $$t_y=60$$, thus $$t_x=60-1=59$$.
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Re: If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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03 May 2013, 16:43
In such questions, i look at the short concise option 1st. I look at B and it doesn't talk about one of the two cars. Eliminate answer choice B and D. Now look at A, it gives a relation between time.
We can form two eqn for two cars using speed distance formulae. For 1st car assume all capital letters for its calculations.
S= D/T
For 2nd car , assume small letters:
s=D/t

Now we know that t=T-1 since it took one sec less to travel the bridge.
We also know the length of the bridge. However we don't know the speeds of time in absolute numbers to determine the answer. So we can't get anywhere with the 2Eqns.

Now add info from B and we can compute our answer. hence C.

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Re: If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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07 May 2013, 06:18
Bunuel wrote:
laythesmack23 wrote:
Banuel,

I'm not seeing it, is there a way you can list the three or four item being tested? and what makes the answer sufficient?

Time?
Rate of Speed?
Distance?
Any other?

Thanks.

Not sure I understand what you mean...

Anyway, we need to get $$t_x$$.

From (1) we have that $$t_y=t_x+1$$ and from (2) we have that $$t_y=60$$, thus $$t_x=60-1=59$$.

The concept being tested here is RTD right?

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Re: If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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07 May 2013, 14:05
I dont know if i am wrong but all i did was use the formula speed = distance x time and they have given us both speed ie the rate traveled and the time so am i wrong?

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Re: If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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08 May 2013, 06:04
I dont know if i am wrong but all i did was use the formula speed = distance x time and they have given us both speed ie the rate traveled and the time so am i wrong?

The formula is (speed)*(time)=(distance).
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Re: If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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08 May 2013, 06:05
laythesmack23 wrote:
Bunuel wrote:
laythesmack23 wrote:
Banuel,

I'm not seeing it, is there a way you can list the three or four item being tested? and what makes the answer sufficient?

Time?
Rate of Speed?
Distance?
Any other?

Thanks.

Not sure I understand what you mean...

Anyway, we need to get $$t_x$$.

From (1) we have that $$t_y=t_x+1$$ and from (2) we have that $$t_y=60$$, thus $$t_x=60-1=59$$.

The concept being tested here is RTD right?

Yes, that's correct.
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Re: If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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30 Jul 2013, 16:28
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If Car X followed Car Y across a certain bridge that is 1/2 mile long, how many seconds did it take Car X to travel across the bridge?

(1) Car X drove onto the bridge exactly 3 seconds after Car Y drove onto the bridge and drove off the bridge exactly 2 seconds after Car Y drove off the bridge.
We don't know how long Car Y was on the bridge for. Therefore, we don't know how long Car X was on the bridge for.
INSUFFICIENT

(2) Car Y traveled across the bridge at a constant speed of 30 miles per hour.
This tells us nothing about Car X.
INSUFFICIENT

1+2)
If the bridge is 1/2 mile long and Car Y crossed it at 30MPH it would have taken exactly one minute to cross the bridge. If Car X followed behind Car Y by 3 seconds and left the bridge just two seconds after Y it gained one second on Y. Therefore, Car X crossed the bridge in 29 seconds.
SUFFICIENT

(C)

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Re: If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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12 Jan 2014, 13:55
WholeLottaLove wrote:
If Car X followed Car Y across a certain bridge that is 1/2 mile long, how many seconds did it take Car X to travel across the bridge?

(1) Car X drove onto the bridge exactly 3 seconds after Car Y drove onto the bridge and drove off the bridge exactly 2 seconds after Car Y drove off the bridge.
We don't know how long Car Y was on the bridge for. Therefore, we don't know how long Car X was on the bridge for.
INSUFFICIENT

(2) Car Y traveled across the bridge at a constant speed of 30 miles per hour.
This tells us nothing about Car X.
INSUFFICIENT

1+2)
If the bridge is 1/2 mile long and Car Y crossed it at 30MPH it would have taken exactly one minute to cross the bridge. If Car X followed behind Car Y by 3 seconds and left the bridge just two seconds after Y it gained one second on Y. Therefore, Car X crossed the bridge in 29 seconds.
SUFFICIENT

(C)

I think you mean 59 seconds. If it took one minute for Car Y to cross and Car X gained on Car Y by one second then the time it took would be 59 seconds.

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Re: If Car X followed Car Y across a certain bridge that is 1/2 [#permalink]

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Bunuel wrote:
If Car X followed Car Y across a certain bridge that is 1/2 mile long, how many seconds did it take Car X to travel across the bridge?

(1) Car X drove onto the bridge exactly 3 seconds after Car Y drove onto the bridge and drove off the bridge exactly 2 seconds after Car Y drove off the bridge.
(2) Car Y traveled across the bridge at a constant speed of 30 miles per hour.

We are given that Car X follows Car Y across a bridge that is ½ mile long. We need to determine the time, in seconds, it took Car X to travel the ½ mile across the bridge.

Statement One Alone:

Car X drove onto the bridge exactly 3 seconds after Car Y drove onto the bridge and drove off the bridge exactly 2 seconds after Car Y drove off the bridge.

From statement one, we know that Car X gained 1 second on Car Y while the two cars traveled over the bridge. Thus, Car X took 1 second less than Car Y took to cross the bridge. However, without knowing the actual time it took Car Y to cross the bridge, we still don't know the number of seconds it took Car X to cross it. Statement one alone is insufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

Car Y traveled across the bridge at a constant speed of 30 miles per hour.

From statement two we can determine the time it took Car Y to cross the bridge.

time = distance/rate

time = (1/2)/30

time = 1/60 hour

Since 1 hour = 3,600 seconds, 1/60 hour = 3,600 x 1/60 = 60 seconds.

We know that Car Y took 60 seconds to travel across the bridge. However, we do not know anything about Car X, so statement two is insufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Because we know that that Car Y took 60 seconds to travel across the bridge and Car X took 1 second less, we know that it took Car X 59 seconds to travel across the bridge.

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text:
# Different materials for each fluid inflow?

How to assign a different color / material to each inflow object in one simulation (bake)?

In the following example we tried to mix two inflows and experience the mixing between the two fluids. However, as bake starts there is only one domain object as fluid (object!) which does not respect color / material assigned to inflow objects!

Unfortunately, I don't know of any way to do this with the main fluid simulator, however it is possible with blender's SPH fluid simulator.

Note that there is currently no way to convert fluid particles into a mesh from blender. There are some python scripts that do this, but these are mostly pretty slow. There are also a few patches, but it seems development on these has been dropped for the most part.

Some advantages of the SPH fluid simulator:

• No domain bounds

• Supports inter-fluid interactions

• Supports force fields

To mix to different colored fluids:

1. Create two emitter planes (ShiftA> Mesh > Plane):

2. Add two separate particle systems to the planes in Properties > Particles:

3. Increase the Life time of the particles in Particles > Emission to be at least equal to the scene end frame:

4. Set the particle physics to Fluid in Particles > Physics:

5. Add an object to represent the fluid particles. It should be very low-poly, a 1 division icosphere works well (ShiftA> Mesh > Icosphere). Note that you will need one for each color:

6. Set the Shading to Smooth on the icospheres and move them to another layer (M)

7. Set the particle type to Object in Particles > Render and select an icosphere in Dupli Object:

8. In Particles > Physics > Fluid interaction press the + button to add an particle system interaction and select the other emitter plane as the particle system object:

9. With both planes setup this way, add another object (e.g. a floor plane) and make it a collision object in Properties > Physics:

10. Create separate materials for each of the icospheres.

11. Set the Size (in Particles > Physics) and Number (in Particles > Emission) for each particle system. Larger particles = lower resolution, smaller = higher resolution. Adjust these settings so as to make the particles appear as a fluid without making the simulation too heavy.

12. You may also want to set the cache step to 1 in Particles > Cache for each particle system to ensure accurate caching.

Result:

• We accepted this even though SPH simulation is not even close to the blender built-in fluid simulation result (our target). Regardless of the size of particles (even very small) it is too far from what we expected. We searched extensively and watched videos (youtube) but haven't found any promising solution considered this as a limit of technology or a work under progress. – Developer Dec 21 '13 at 3:48
• Ideally there would be a way to mesh the particles together into one solid mesh (you can kind of do this with meta balls, but it is very slow and still doesn't look the greatest). There are some patches and python addons to do this, but unfortunately it doesn't seem like this will be officially in blender for a while.. – gandalf3 Apr 11 '14 at 19:05
• The first link is 404 – MicroMachine Aug 31 '16 at 14:18
• Wiki link is a 404. – Scott Milner Jan 4 '18 at 19:24

Basically a scene similar to the one gandalf created.

But 4000 particles of size 0.1

set to Halo

and Fluid properties: Stiffness 0

Leave the impression that the fluids mix:

You can just use 2 complete fluid systems. You cannot bake with multiple domains, but you CAN play back more than one. Create the first sim and bake it to a directory. Now disable the Domain and create your second sim and bake to a new directory. Now enable the first domain and ensure it points to the correct bake directory. Both sims will play back and you can put whatever material you want on each. I have had as many as 5 sims running at once and each started at a different time. All in one scene.

• But AFAIK they can't mix, like the OP seems to want. – gandalf3 Oct 28 '14 at 6:01


text:
# [SVN] r10493 (Latex)

Wed Aug 27 01:39:57 UTC 2008

Optimized the parsing of included files.  This change should dramatically improve the time it takes to find all of the packages included using the \usepackage command.
The master file is parsed line by line rather than reading the entire file into a string.  Searching for usepackage commands stops after \begin{document} is encountered.  Only files that are included as part of the preamble are searched for additional \usepackage commands.

Changed:
U   trunk/Bundles/Latex.tmbundle/Support/bin/texMate.py


text:
# How to create a surface based on spherical triangle corner points

Axcel shared this question 1 year ago

Hi there,

Imagine I have a sphere in 3D mode and I defined three points on its surface (see file attached), how can I create a surface with these three points forming the corner points and thereby allowing me to color-in that spherical triangle on the sphere? I would love to have a tool that allows me to click the center point and the three corner points in order to create the surface but I don't know how to do it...

Thanks and kind regards

Axcel

2

Doesn't look to me an easy thing to achieve...

https://www.geogebra.org/m/ZaQWGcz5

1

```u = Vektor(M, A)
v = Vektor(M, B)
w = Vektor(M, C)
r = sqrt(Skalarprodukt(M-A, M-A))```
`Oberfläche(x((cos(k) ((cos(t) u + sin(t) v) / sqrt(Skalarprodukt(cos(t) u + sin(t) v, cos(t) u + sin(t) v)) * r ) + sin(k) w) / sqrt(Skalarprodukt(cos(k) ((cos(t) u + sin(t) v) / sqrt(Skalarprodukt(cos(t) u + sin(t) v, cos(t) u + sin(t) v)) * r) + sin(k) w, cos(k) ((cos(t) u + sin(t) v) / sqrt(Skalarprodukt(cos(t) u + sin(t) v, cos(t) u + sin(t) v)) * r ) + sin(k) w)) * r + M ), y((cos(k) ((cos(t) u + sin(t) v) / sqrt(Skalarprodukt(cos(t) u + sin(t) v, cos(t) u + sin(t) v)) * r ) + sin(k) w) / sqrt(Skalarprodukt(cos(k) ((cos(t) u + sin(t) v) / sqrt(Skalarprodukt(cos(t) u + sin(t) v, cos(t) u + sin(t) v)) * r) + sin(k) w, cos(k) ((cos(t) u + sin(t) v) / sqrt(Skalarprodukt(cos(t) u + sin(t) v, cos(t) u + sin(t) v)) * r) + sin(k) w)) * r + M ), z((cos(k) ((cos(t) u + sin(t) v) / sqrt(Skalarprodukt(cos(t) u + sin(t) v, cos(t) u + sin(t) v)) * r) + sin(k) w) / sqrt(Skalarprodukt(cos(k) ((cos(t) u + sin(t) v) / sqrt(Skalarprodukt(cos(t) u + sin(t) v, cos(t) u + sin(t) v)) * r) + sin(k) w, cos(k) ((cos(t) u + sin(t) v) / sqrt(Skalarprodukt(cos(t) u + sin(t) v, cos(t) u + sin(t) v)) * r ) + sin(k) w)) * r + M ), t, 0, pi/2, k, 0, pi/2)`

Das Werkzeug zu machen überlasse ich jemand anderen (und das Vereinfachen auch).

Falls das jemand schöner machen will. Es basiert alles auf der Kurvenfunktion

$K\left(\varphi\right) = \frac{\cos\left(\varphi\right) \vec{u} + \sin\left(\varphi\right) \vec{v}}{\left|\cos\left(\varphi\right) \vec{u} + \sin\left(\varphi\right) \vec{v}\right|}\left|\vec{u} \right| \quad \forall \; \varphi\in\left[0;\frac{\pi}{2}\right]$.
Wobei die Vektoren u und v beliebige Ursprungsvektoren sind (deshalb nach Anwendung u.U. verschieben).

1
1

I did not knew of those simplifications, thanks.

Now one additional command for 3-point-patches is:

`Oberfläche(((C - M) cos(l) + ((A - M) cos(k) + (B - M) sin(k)) / sqrt(((A - M) cos(k) + (B - M) sin(k)) ((A - M) cos(k) + (B - M) sin(k))) sin(l)) / sqrt(((C - M) cos(l) + ((A - M) cos(k) + (B - M) sin(k)) / sqrt(((A - M) cos(k) + (B - M) sin(k)) ((A - M) cos(k) + (B - M) sin(k))) sin(l)) ((C - M) cos(l) + ((A - M) cos(k) + (B - M) sin(k)) / sqrt(((A - M) cos(k) + (B - M) sin(k)) ((A - M) cos(k) + (B - M) sin(k))) sin(l))) Abstand(M, A) + M, k, 0, π / 2, l, 0, π / 2)`

1

It could be written much shorter if there was a possibility to use UnitVector( <Vector> ), points or vectors at all within Formulas or Surfaces. Then it could be written as:

`F(u, v, t) = UnitVector(cos(t) u + sin(t) v)`

`Surface(  F(A-M,F(B-M,C-A,k),l) * Distance(M, A) + M , k, 0, π / 2, l, 0, π / 2)`

But unfortunately it is not possible.

1

Thanks to all of you first of all! It is in fact much more "complicado" than I would have thought. I also hope that a future version of GeoGebra will provide easier means of achieving this, just as Loco has stated above.

In fact, the genius mathmagic helped me once on a different but similar matter. Attached please find the file from back then. I appreciate the fact that the surface has a slightly larger distance from the center point than the radius of the sphere so that there is a "clean display" of the surface. It's only that the file works with four points rather than three. How must the formula in the file be modified for three corner points?

1

That's impressive, mathmagic! (As usual, I'd say 😉)

1

@Axcel

just replace one of the corner points within that formula with the name of one of the other points (a.e. replace F with E within that formula) and simplify it (see the formula posted by mathmatic bellow).

And if you had reread those posts of him, he posted that formula that time: Spherical angles on the surface of a sphere

It is a quite easy technique similar to mine posted above. He builds a curve between two Points A and B with a simple linear combination A t + (1-t) B with t from 0 to 1 and combines this twice to get a point on an planar triangular surface between three or four points. And to get the curved triangle he divides the vector of the surface point by his the distance to the center point and multiplies it with an radius. Simple and robust.

2

or equivalently use

b = Surface(A + Distance(A, C) 1.01((1 - u) (1 - v) C + (1 - u) v D + u E - A) / sqrt(((1 - u) (1 - v) C + (1 - u) v D + u E - A)²), u, 0, 1, v, 0, 1)

1

Thank you guys very much for your help. And yes, matmagic gave me the answer earlier already but I was too blind to see it... Sorry for that.

@Loco: thanks for that brief explanation of what matmagic did with his formula. That was very helpful to get the idea behind his formula.

1

see the posts of my link, it is very easy


text:
B. String inside out
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given a string S of even length s 1..s 2n . Perform the following manipulations:

• divide it into two halves s 1..s n and s n + 1..s 2n
• reverse each of them s n..s 1 and s 2n..s n + 1
• concatenate the resulting strings into s n..s 1 s 2n..s n + 1

Output the result of these manipulations.

Input

The only line of the input contains a string of lowercase Latin letters. The length of the string is between 2 and 20, inclusive, and it is even.

Output

Output the string which is the result of the described manipulations.

Examples
Input
codeforces
Output
fedocsecro
Input
qwertyasdfgh
Output
ytrewqhgfdsa


text:
Get our free extension to see links to code for papers anywhere online!

# Randomized Transferable Machine

Nov 27, 2020
Pengfei Wei, Tze Yun Leong

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Feature-based transfer is one of the most effective methodologies for transfer learning. Existing studies usually assume that the learned new feature representation is truly \emph{domain-invariant}, and thus directly train a transfer model $\mathcal{M}$ on source domain. In this paper, we consider a more realistic scenario where the new feature representation is suboptimal and small divergence still exists across domains. We propose a new learning strategy with a transfer model called Randomized Transferable Machine (RTM). More specifically, we work on source data with the new feature representation learned from existing feature-based transfer methods. The key idea is to enlarge source training data populations by randomly corrupting source data using some noises, and then train a transfer model $\widetilde{\mathcal{M}}$ that performs well on all the corrupted source data populations. In principle, the more corruptions are made, the higher the probability of the target data can be covered by the constructed source populations, and thus better transfer performance can be achieved by $\widetilde{\mathcal{M}}$. An ideal case is with infinite corruptions, which however is infeasible in reality. We develop a marginalized solution with linear regression model and dropout noise. With a marginalization trick, we can train an RTM that is equivalently to training using infinite source noisy populations without truly conducting any corruption. More importantly, such an RTM has a closed-form solution, which enables very fast and efficient training. Extensive experiments on various real-world transfer tasks show that RTM is a promising transfer model.

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text:
# Does gravity affect magnetism, vice-versa, or do they “ignore” each other?

I am suddenly struck by the question of whether gravitation affects magnetism in some way. On the other hand, gravity is a weak force, but magnetism seems to be a strong force, so would magnetism affect gravity?

Or do they "ignore" each other, being forces which do not interact?

The answer to this is related to this question: If the earth's core were to cool so that it were no longer liquid, no longer rotated, and thus produced no magnetic field, would this do anything to earth's gravity?

The electromagnetic field tensor $F_{\mu\nu}$ which encodes all the information about the electric and magnetic field, certainly contributes to the energy-stress tensor $T_{\mu\nu}$, which appears in the Einstein Field Equations: $$G_{\mu\nu}= 8\pi G T_{\mu\nu}$$ The left hand side of this equation encodes the geometry of spacetime, while the right hand side describes the 'sources' of gravity. Therefore, we can say that magnetism does have an effect on the geometry of spacetime i.e. gravity.

• I think the questioner is interested in levels of field strength easily created on (or by) the Earth. Talking about engery stress tensors is like answering the question "Will running make me lose weight?" with the equation for relativistic mass. – Foo Bar Sep 22 '14 at 19:32
• @FooBar that made me chuckle :) If you feel a lower level answer is in order, I invite you to provide one! However, you have to keep in mind that this site aims to be of use to more people than just the asker, so answers at all levels are more than welcome, and I don't feel inclined to alter my own response. – Danu Sep 22 '14 at 19:36
• @FooBar: Actually electric and magnetic fields constitute a significant part of the rest mass of ordinary matter. The fields inside the nucleus are extremely intense. The relative contribution is highest for elements with a high atomic number. – Ben Crowell Sep 22 '14 at 19:59
• Nice! This answer suggests that if you could somehow generate a magnetic field that was strong enough, you could create artificial gravity. That would have to be an unimaginably strong field to produce anything practical, however. And where would you get the energy to do it with? Wow. – Cyberherbalist Sep 22 '14 at 22:11
• @Cyberherbalist How current loops and solenoids curve space-time, A. Füzfa "We finally propose an experimental setup, achievable with current technology of superconducting coils, that produces a phase shift of light of the same order of magnitude than astrophysical signals in ground-based gravitational wave observatories." – endolith Jan 15 '16 at 4:29

Magnetic effect on gravity: In technical terms: yes. For practical purposes: no.

As Danu states, a magnetic field is a form of energy, and Einstein showed that energy can be equivalent to mass. A sufficiently large amount of energy (of any kind) collected together will produce measurable gravity (aka bending of space-time).

This source puts the total energy of Earth's magnetic field at $10^{19}$ joules. If that's correct, it's equivalent to 111 kilograms, which is about $10^{-23}$ of Earth's mass (and gravity).

Caveat: I am not a "real" physicist; I might be calculating the wrong thing. But I'm pretty confident that the number is negligible, relative to many other things that affect Earth's gravity.

As Ben states, the electromagnetic energy of protons and electrons is worth a significant amount of the mass in every atom. But commonplace magnetic fields at the human scale expose such a tiny fraction of this energy as to be irrelevant gravitationally.

Gravitational effect on magnetism: It depends on how you look at it.

Unlike an electromagnet (such as the Earth's core), you can't "turn off" a gravitational field and compare the difference.

But in a grand sense, the magnetic fields produced by large convecting masses only exist because of the great pressure and heat generated by self-gravity. Whether that counts is more a question of philosophy than physics.

## protected by Community♦Oct 19 '15 at 16:17

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

Would you like to answer one of these unanswered questions instead?


text:
3 added 117 characters in body

Note that even if the measure is only finitely additive, the ideal of null sets will still be countably saturated. The usual arguments involving Ulam matrices, such as can be found in Jech (Chapter 10) or other texts on set theory, show that if there is a countably additive, countably saturated ideal on some set then there is a weakly inaccessible cardinal. Hence, if no such cardinal exists then, as Emil has corrected me, the Hahn-Banach extension of Lebesgue measure will , of course, be countably additive on Lebesgue null sets but not a countably additive measureon all nulls. This does not answer your question, but does show that the weaker hypothesis still requires large cardinals.

By the way, you may want to look at the paper by Joan Hart and Ken Kunen titled "Weak Measure Extension Axioms" in Topology and Applications 85 (1998) 219-246 to see what can be said about extensions of measures to measure certain sets.

2 added 115 characters in body

Note that even if the measure is only finitely additive, the ideal of null sets will still be countably saturated. The usual arguments involving Ulam matrices, such as can be found in Jech (Chapter 10) or other texts on set theory, show that if there is a countably additive, countably saturated ideal on some set then there is a weakly inaccessible cardinal. Hence, if no such cardinal exists then the Hahn-Banach extension of Lebesgue measure will, of course, be countably additive on null sets but not a countably additive measure.

By the way, you may want to look at the paper by Joan Hart and Ken Kunen titled "Weak Measure Extension Axioms" in Topology and Applications 85 (1998) 219-246 to see what can be said about extensions of measures to measure certain sets.

1

The usual arguments involving Ulam matrices, such as can be found in Jech (Chapter 10) or other texts on set theory, show that if there is a countably additive, countably saturated ideal on some set then there is a weakly inaccessible cardinal. Hence, if no such cardinal exists then the Hahn-Banach extension of Lebesgue measure will, of course, be countably additive on null sets but not a countably additive measure.

By the way, you may want to look at the paper by Joan Hart and Ken Kunen titled "Weak Measure Extension Axioms" in Topology and Applications 85 (1998) 219-246 to see what can be said about extensions of measures to measure certain sets.


text:
X

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