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# UM Motion
1. Jan 20, 2009
1. The problem statement, all variables and given/known data
Suppose the earth is a perfect sphere with R=6370 km. if a person weighs exactly 600.0 N at the north pole how much will the person weigh at the equator? (hint: the upward push of the scale on the person is what the scale will read and is what we are calling the weight in this case) ans: 597.9 N
2. Relevant equations
Fg=Gm1m2/r^2
Fg=ma
Fc=Mv^2/r?
3. The attempt at a solution
I have no real clue on how to solve such a question, i plugged in for Fg but the answer i seem to get is higher than that of 597.9 N. Need help for a solution please!!!!!
2. Jan 20, 2009
I will give you a hint.......A person standing on the equator experiences a centripetal force towards the centre of the earth .Work out what this is and draw a free body force diagram for the person.
3. Jan 20, 2009
### Staff: Mentor
Apply Newton's 2nd law. What's different about being on the equator compared to being on the north pole?
4. Jan 20, 2009
### chrisk
Think about the angular velocity at the north pole and the equator. Then calculate the centripetal acceleration at the equator and use this value to find the effective acceleration of gravity at the equator.
5. Jan 20, 2009
how could i determine the angular velocity though? i am not given a time
6. Jan 20, 2009
### chrisk
Last edited: Jan 20, 2009
7. Jan 20, 2009
so i should use Fc=4pie^2Mr/ T^2 to solve for Force centripetal
8. Jan 20, 2009
### chrisk
Yes, use this equation but it's a little easier to find the effective acceleration of gravity at the equator, g - acentripetal, then multiply by the mass.
9. Jan 20, 2009
okay so it would be easier to just get the centripetal acceleration, then take gravity- acceleration centripetal then multiply by mass for F=ma right? but how would i get the acceleration centripetal when Ac=V^2/r and i don't have V?
10. Jan 20, 2009
### chrisk
V= earth's angular velocity times the earth's radius.
11. Jan 20, 2009
i am not familiar with earths angular velocity?
12. Jan 20, 2009
### gabbagabbahey
Really? You don't know how long a day is? Or you don't know how to calculate angular velocity from knowledge of the period of rotation?
13. Jan 20, 2009
don't know how to calculate angular velocity from knowledge of the period of rotation? I've never heard of the term for one so maybe that's why im lost with what your saying
14. Jan 20, 2009
### gabbagabbahey
Angular frequency is what you really need for this problem; and it is defined by:
$$\omega\equiv2\pi f=\frac{2\pi}{T}=\frac{|v|}{r}$$
Angular velocity is a vector quantity with magnitude $\omega$ and direct determined by a cross product.
15. Jan 20, 2009
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Physics
Mechanics
Rotation
# Is angular velocity a vector quantity?
###### Wiki User
Yes, angular velocity is a vector quantity
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## Related Questions
### Is angular momentum a vector quantity?
Angular momentum is a vector quantity. Angular velocity, which is a vector quantity, is multiplied by inertia, which is a scalar quantity.
### Why angular velocity is vector quantity?
Because it's a type of velocity and velocity is vector quantity
### Is momentum a vector or scalar quantity?
Momentum is a vector quantity. We know that momentum is the product of mass and velocity, and velocity has direction. That makes velocity a vector quantity. And the product of a scalar quantity and a vector quantity is a vector quantity.
### Is angular frequency vector or scalar?
Scalar. Angular frequency vector is roughly synonymous with angular velocity.
### What is Difference between angular velocity and angular speed?
Angular velocity is a vector with a direction and angular speed is a scalar with no direction.
### Is velocity a scalar quantity or vector quantity?
velocity is a vector quantity because it`s formula is displacement by time and displacement has a particular direction. that`s why velocity is a vector quantity but speed is a scalar quantity
### Is instantaneous velocity a vector quantity?
"Velocity" is a vector quantity. This also applies to instantaneous velocity. If you want the scalar quantity, you talk about "speed".
### Is the angular displacement a vector quantity or scalar quantity?
Displacement indicates a vector quantity, including angular displacement. The angular displacement is either clockwise or counter clockwise and the direction makes a difference. Angular distance would be a scalar.
### How do you change an objects momentum?
Momentum is of two kind. One is linear momentum and the other is angular momentum. Linear momentum is defined as the product of the mass and the velocity. Hence a vector quantity. To change the momentum of a given body with its mass constant, its velocity is to be changed. Velocity change could be made by changing its magnitude or direction or both. Angular momentum is the product of moment of inertial and the angular velocity. Same manner, angular momentum is also a vector quantity as angular velocity is a vector quantity. Most of us think that moment of inertia of a body about any prescribed axis is also a vector quantity. It is totally wrong as far as my approach is concerned. Moment of inertia is a scalar quantity. So to change the momentum, some force can be applied by allowing a moving body to collide with. Angular momentum can be changed by applying torque on it. Torque colloquially saying is a turning force. Moment of effective force about an axis is termed as torque.
### Is angular displacement is a vector?
angular displacement is a vector quantity when theta (angle) is small, otherwise it is scalar.
### Is momentum a vector or scaler quantity?
Momentum is a vector quantity because the definition of momentum is that it is an object's mass multiplied by velocity. Velocity is a vector quantity that has direction and the mass is scalar. When you multiply a vector by a scalar, it will result in a vector quantity.
### Is velocity a vector or a scalar?
Velocity is a vector quantity or simply vector, it has both magnitude an direction.
### Is velocity a vector?
True ,velocity is a vector quantity ,it is specified by a magnitude and direction.
### Is velocity a vector or a scalar quantity?
Velocity is a vector.Its magnitude is called 'speed'.
### Is velocity a vector quantity?
Yes, velocity is a vector quantity since it can be defined by both magnitude and direction unlike speed.
### Momentum is defined as?
The product of velocity and mass. Note that velocity is a vector quantity, therefore momentum, too, is a vector quantity.
### What direction does the angular velocity vector of the Earth's rotation point toward?
The Earth's angular velocity vector due to its axial rotation points towards the north pole.
###### PhysicsScienceMath and ArithmeticRotation
Trending Questions
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This is an R implementation of the web-based ‘Practical Meta-Analysis Effect Size Calculator’ from David B. Wilson. The original calculator can be found at http://www.campbellcollaboration.org/escalc/html/EffectSizeCalculator-Home.php.
Based on the input, the effect size can be returned as standardized mean difference (d), Cohen’s f, eta squared, Hedges’ g, correlation coefficient effect size r or Fisher’s transformation z, odds ratio or log odds effect size.
### Return values
The return value of all functions has the same structure:
• The effect size, whether being d, g, r, f, (Cox) odds ratios or (Cox) logits, is always named es.
• The standard error of the effect size, se.
• The variance of the effect size, var.
• The lower and upper confidence limits ci.lo and ci.hi.
• The weight factor, based on the inverse-variance, w.
• The total sample size totaln.
• The effect size measure, measure, which is typically specified via the es.type-argument.
• Information on the effect-size conversion, info.
• A string with the study name, if the study-argument was specified in function calls.
#### Correlation Effect Size
If the correlation effect size r is computed, the transformed Fisher’s z and their confidence intervals are also returned. The variance and standard error for the correlation effect size r are always based on Fisher’s transformation.
#### Odds Ratio Effect Size
For odds ratios, the variance and standard error are always returned on the log-scale!
### S3 methods
The esc package offers the S3 methods print() and as.data.frame().
### Combining results into a single data frame
The combine_esc() method is a convenient way to create pooled data frames of different effect size calculations, for further use. Here is an example of combine_esc(), which returns a data.frame object.
library(esc)
e1 <- esc_2x2(grp1yes = 30, grp1no = 50, grp2yes = 40, grp2no = 45, study = "Study 1")
e2 <- esc_2x2(grp1yes = 30, grp1no = 50, grp2yes = 40, grp2no = 45, es.type = "or", study = "Study 2")
e3 <- esc_t(p = 0.03, grp1n = 100, grp2n = 150, study = "Study 3")
e4 <- esc_mean_sd(grp1m = 7, grp1sd = 2, grp1n = 50, grp2m = 9, grp2sd = 3, grp2n = 60, es.type = "logit",
study = "Study 4")
combine_esc(e1, e2, e3, e4)
#> study es weight sample.size se var ci.lo ci.hi measure
#> 1 Study 1 -0.3930 9.945 165 0.3171 0.10056 -1.01456 0.2285 logit
#> 2 Study 2 0.6750 9.945 165 0.3171 0.10056 0.36256 1.2567 or
#> 3 Study 3 0.2818 59.434 250 0.1297 0.01683 0.02755 0.5360 d
#> 4 Study 4 -1.3982 7.721 110 0.3599 0.12951 -2.10354 -0.6928 logit
esc is still under development, i.e. not all effect size computation options are implemented yet. The remaining options will follow in further updates.
## Installation
### Latest development build
To install the latest development snapshot (see latest changes below), type following commands into the R console:
library(githubinstall)
githubinstall::githubinstall("esc")
### Official, stable release
To install the latest stable release from CRAN, type following command into the R console:
install.packages("esc")
## Citation
In case you want / have to cite my package, please use citation('esc') for citation information.
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1. ## Differentiable Proof 2
Suppose that f: R $\rightarrow$ R is differentiable and that for every a,b $\in$ R there holds f (a + b) = f(a) + f(b).
Prove that f '(x) = f '(0) for all x $\in$ R. What kind of function is f?
2. Start by noticing that f(0)=0. Indeed, putting a=b=0, one gets f(0)=2f(0), hence f(0)=0.
Now put a=x, b=h:
f(x+h)-f(x)=f(h)=f(h)-f(0)
Divide by h to get
(f(x+h)-f(x))/h=(f(h)-f(0))/h
Take the limit h->0 to finally get
f'(x)=f'(0).
This means that f'(x)=a is a constant: f(x)=cx+d, but d must be zero in order to the first relation to hold.
Hence f(x)=cx. Those are called linear functions.
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### 14 Divisors
What is the smallest number with exactly 14 divisors?
### Summing Consecutive Numbers
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Dozens
Do you know a quick way to check if a number is a multiple of two? How about three, four or six?
# The Square of My Age
##### Stage: 3 Short Challenge Level:
One method for this question is to notice that the square of Thomas' age is at most $62$, so Thomas is at most $7$. The square of Thomas' age is therefore at most $49$, so Lauren is at least $13$.
The square of Lauren's age is at most $176$, so Lauren is at most $13$. Therefore Lauren must be $13$, and Thomas must be $7$.
This problem is taken from the UKMT Mathematical Challenges.
View the previous week's solution
View the current weekly problem
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# ProbabilityDensities
The last module dealt with the uniform distribution, where any one outcome is as likely as another. This module deals with experiments whose outcomes have different probabilities. For example, consider an unfair coin which has a ($\frac{2}{3}$) probability of landing heads and a ($\frac{1}{3}$) probability of landing tails. Another example is time spent on hold with customer service, where it is more likely that the call is answered in the first hour than in the second hour.
## Random variable and probability density function (PDF)
A random variable ($X$) is a function whose output should be thought of as the outcome of an experiment. Associated with a random variable is a probability density function (PDF) ($\rho(x)$), which is defined by ($P(a \leq X \leq b) = \int_a^b \rho(x)dx$). That is, the probability that the random variable falls in a certain range of values is given by integrating the PDF over that range of values.
Phrased another way, we can think of probability ($P$) as the quantity we want to compute over a certain range of values, and the probability element is given by
(:latex:) $dP = \rho(x) \, dx.$ (:latexend:)
Example
Consider the spinner from the last module. The outcome of a spin is some angle (relative to the positive ($x$)-axis) between 0 and ($2\pi$). If ($X$) is the random variable which gives the output of a spin, then
(:latex:) \begin{equation*} P(a \leq X \leq b) = \frac{b-a}{2\pi}, \end{equation*} (:latexend:)
since the spinner was assumed to be fair. This holds for all ($0\leq a \leq b \leq 2\pi$). Then the associated PDF is
(:latex:) \begin{equation*} \rho(x) = \begin{cases} \frac{1}{2\pi} & \hbox{if $0 \leq x \leq 2\pi$}
0. & \hbox{otherwise} \end{cases} \end{equation*} (:latexend:)
### Note
Sometimes a PDF ($\rho(x)$) is only defined on a certain domain ($D$). ($D$) can be thought of as the set of all possible outcomes of the experiment ($X$). In this case, it is assumed that ($\rho(x)=0$) for ($x$) not in that domain. So another way of defining the PDF for the spinner is ($\rho(x) = \frac{1}{2\pi}$) for ($0 \leq x \leq 2\pi$).
## Properties of a probability density function
The following are defining properties of a PDF. In other words, a function ($\rho(x)$) is a PDF on the domain ($D$) if and only if it satisfies these properties.
1. ($\rho(x)\geq 0$) for all ($x \in D$).
2. ($\int_D \rho(x)\,dx = 1$).
The first property is necessary since probabilities must be non-negative. The second property reflects the fact that the random variable ($X$) associated with ($\rho(x)$) must have some outcome in the domain ($D$) (since ($D$) is the set of all possible outcomes), and so integrating over all of these outcomes should give 1.
### Note
If ($\rho(x)$) is defined on some specific domain ($D$), then the integral over that specific domain should equal 1. This is because ($\rho(x)=0$) outside of that domain, as mentioned in the above note.
Example
Find the value of the constant ($c$) so that ($\rho(x) = \frac{c}{1+x^2}$) for all ($x$) is a PDF.
As long as ($c \geq 0$), the first property for a PDF will be met, since ($1+x^2>0$) for all ($x$). To satisfy the second property, compute
(:latex:) \begin{align*} \int_{-\infty}^\infty \frac{c}{1+x^2}\,dx &= c \left(\arctan(x)\bigg|^\infty_{-\infty}\right)
&= c \left( \frac{\pi}{2} - (-\frac{\pi}{2})\right)
&= c \pi. \end{align*} (:latexend:)
Since this integral is supposed to be 1, we find that ($c = \frac{1}{\pi}$).
## Several specific density functions
### Uniform density
Hinted at above and in the previous module, the uniform density function (or uniform distribution) on ($\left[a,b\right]$) is given by ($\rho(x) = \frac{1}{b-a}$) (and ($\rho(x) = 0$) if ($x$) is not in ($\left[a,b\right]$)):
More generally, the uniform distribution on the domain ($D$) (whatever the dimension) is given by
(:latex:) $\rho(x) = \frac{1}{\hbox{Volume of D}}.$ (:latexend:)
In dimension 0, where outcomes are discrete (as in the rolling of a die or the flipping of a coin), remember that volume is just counting. So in this case the probability of a particular outcome is
(:latex:) $\rho(x) = \frac{1}{n},$ (:latexend:)
where ($n$) is the number of outcomes in the domain ($D$) (e.g. ($n=6$) for the roll of a die; ($n=2$) for a coin flip).
### Exponential density
Another density function used to model many common experiments is the exponential density function. This is actually a whole family of density functions given by ($\rho(t) = \alpha e^{-\alpha t}$) for ($t \geq 0$) and ($\alpha>0$) some constant. The reason a parameter ($t$) is used is that the exponential density is often used to model experiments with a time outcome.
Example
Show that the exponential density ($\rho(t) = \alpha e^{-\alpha t}$) (for ($t \geq 0$)) satisfies the properties of a density function.
The exponential function is never negative, so one need only check the integral. One finds
(:latex:) \begin{align*} \int_{t=0}^\infty \alpha e^{-\alpha t} \, dt &= \alpha \frac{1}{-\alpha} e^{-\alpha t} \bigg|_{t=0}^\infty
&= -(0-1)
&= 1, \end{align*} (:latexend:)
as desired. So the exponential density is in fact a density.
Example
Consider a call made to customer service at Acme company. The number of minutes spent on hold before the call is answered is often modeled with an exponential density function
(:latex:) $\rho(t) = \alpha e^{-\alpha t}.$ (:latexend:)
Find, in terms of ($\alpha$), the probability that the waiting time for a call is less than 30 minutes. (:toggle hide show="Answer" box2:)
To find the probability that ($0\leq X \leq 30$), use the relationship between probability and the PDF, which is
(:latex:) \begin{align*} P(0\leq X \leq 30) &= \int_0^{30} \rho(x)\,dx
&= \int_0^{30} \lambda e^{-\lambda x}\, dx
&= -e^{-\lambda x} \bigg|^{30}_0
&= -e^{-30 \lambda} - (-1)
&= 1 - e^{-30 \lambda}. \end{align*} (:latexend:)
Example
Again consider customer service call waiting time at Acme company, and again assume an exponential density function
(:latex:) $\rho(t) = \alpha e^{-\alpha t}.$ (:latexend:)
Suppose half of all customers are answered within 5 minutes. Find ($\alpha$) and then find the probability that a call takes more than 10 minutes to be answered.
Since half of all customers are answered within 5 minutes, we have that
(:latex:) $P(0 \leq X \leq 5) = \frac{1}{2}.$ (:latexend:)
On the other hand, we know that this can be expressed as the integral of the density function, so we have
(:latex:) \begin{align*} \frac{1}{2} &= \int_{t=0}^5 \rho(t) \, dt
&= \int_{t=0}^5 \alpha e^{-\alpha t} \, dt
&= - e^{-\alpha t} \bigg|_{t=0}^5
&= -e^{-5 \alpha} - (-1)
&= 1 - e^{-5 \alpha}. \end{align*} (:latexend:)
So we have that
(:latex:) $e^{-5 \alpha} = \frac{1}{2}.$ (:latexend:)
Taking the log of both sides, dividing by ($-5$) and simplifying, we have
(:latex:) \begin{align*} \alpha &= \frac{1}{-5} \ln\left(\frac{1}{2}\right)
&= \frac{1}{-5} (-\ln 2)
&= \frac{1}{5} \ln 2. \end{align*} (:latexend:)
For the second part, we want to know the probability of waiting more than 10 minutes. This is (leaving ($\alpha$) as a constant for now)
(:latex:) \begin{align*} P(X \geq 10) &= \int_{t=10}^\infty \rho(t)\, dt
&= \int_{t=10}^\infty \alpha e^{-\alpha t} \, dt
&= -e^{-\alpha t} \bigg|_{t=10}^\infty
&= 0 - \left(-e^{-\alpha \cdot 10} \right). \end{align*} (:latexend:)
Now plugging in the value of ($\alpha$), we have
(:latex:) \begin{align*} P(X \geq 10) &= e^{- (\ln 2/5) \cdot 10}
&= e^{-2 \ln 2}
&= 2^{-2}
&= \frac{1}{4}. \end{align*} (:latexend:)
### Gaussian density
The last probability density function is the 'Gaussian, or normal, density function. This is an important density function and is expanded on in the next module. Like the exponential, the Gaussian density function usually has parameters (see the next module), but in its simplest form, the Gaussian is given by
(:latex:) $\rho(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}.$ (:latexend:)
The Gaussian has all real ($x$) as its domain, but because it tails off so quickly in both directions, the probability of getting values far from the center (in this case ($x=0$)) is very small.
## EXERCISES
• Which of the following are probability density functions?
(:latex:) \begin{align*} a. f(x) =&1/2 \textrm{ on } D=[0,2]
b. f(x) =& \frac{\sin(x)}{2} \textrm{ on } D=[0, 3\pi]
c. f(x) =& 5 e^{-2x} \textrm{ on } D=[0, \infty)
d. f(x) =& \frac{1}{\pi (1+x^2)} \textrm{ on } D=(-\infty, \infty)
e. f(x) =& \frac{x}{2} \textrm{ for } 0 \leq x \leq1
& \frac{1}{2} \textrm{ for } 1 \leq x \leq 2
& \frac{3}{2}-\frac{x}{2} \textrm{ for } 2 \leq x \leq 3 \end{align*} (:latexend:)
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At this moment, we have designed and implemented the Pseudo-Boolean Optimization (PBO) problem set, which contains 25 real-valued test problems on $\{0,1\}^n$. The selection of PBO problems is suggested in Carola Doerr, Furong Ye, Naama Horesh, Hao Wang, Ofer M. Shir, and Thomas Bäck. “Benchmarking discrete optimization heuristics with IOHprofiler.” Applied Soft Computing 88 (2020): 106027.
@article{DoerrYHWSB20,
author = {Carola Doerr and
Furong Ye and
Naama Horesh and
Hao Wang and
Ofer M. Shir and
Thomas B{\"{a}}ck},
title = {Benchmarking discrete optimization heuristics with IOHprofiler},
journal = {Appl. Soft Comput.},
volume = {88},
pages = {106027},
year = {2020},
url = {https://doi.org/10.1016/j.asoc.2019.106027},
doi = {10.1016/j.asoc.2019.106027}
}
## Pseudo-Boolean Optimization (PBO) Problem Set
In general, for a pseudo-Boolean function $f\colon \{0,1\}^n \rightarrow \mathbb{R}$, we proposed to creare an instance of it using the following transformation $af(\sigma(x \oplus z)) + b$, in which,
• $z$ is an arbitrary bit string of length $n$ and $\oplus$ denotes the bit-wise XOR function (“translation” in $\{0,1\}^n$),
• $\sigma:[n] \to [n], y \mapsto (y_{\sigma(1)},\ldots,y_{\sigma(n)})$ is a random permutation on the bit string (loosely speaking, “rotation” in $\{0,1\}^n$), and
• $a,b$ are the multiplicative and the additive offset, respectively.
Intuitively, such transformations are devised to test the invariance property of an optimization algorithm, which entails that no performance difference would be measured when the objective function is modified using those transformations. We suggest to apply such transformations on the test functions to challenge your algorithms. Note that, for the continuous optimization problem, this has been practiced in the COCO/BBOB platform, where they applied random affine transformations to test functions.
### F1: OneMax
$\text{OM}:\{0,1\}^n \rightarrow [0..n], x \mapsto \sum_{i=1}^n{x_i}.$
The problem has a very smooth and non-deceptive fitness landscape. Due to the well-known coupon collector effect, it is relatively easy to make progress when the function values are small, and the probability to obtain an improving move decreases considerably with increasing function value.
$\text{LO}:\{0,1\}^n \to [0..n], x\mapsto \max \{i \in [0..n] \mid \forall {j} \le {i}: x_j=1\} = \sum_{i=1}^n{\prod_{j=1}^i{x_i}},$
which counts the number of initial ones.
### F3: A Linear Function with Harmonic Weights
$f:\{0,1\}^n \to \mathbb{R}, x \mapsto \sum_{i} i x_i$
The problem is a linear function with harmonic weights.
### F4 - F17: W-model-transformed OneMax and LeadingOnes
The W-model comprises 4 basic transformations, each coming with different instances. We use $W(\cdot,\cdot,\cdot,\cdot)$ to denote the configuration chosen in our benchmark set.
• Dummy variables $W(k,\ast,\ast,\ast)$: a reduction mapping each string $(x_1, \ldots, x_n)$ to a substring $(x_{i_1}, \ldots, x_{i_k})$ for randomly chosen, pairwise different $i_1,\ldots, i_k \in [n]$.
• Neutrality $W(\ast,\mu,\ast,\ast)$: The bit string $(x_1,\ldots,x_n)$ is reduced to a string$(y_1,\ldots,y_m)$ with $m:=\frac{n}{\mu}$, where $\mu$ is a parameter of the transformation. For each $i \in [m]$ the value of $y_i$ is the majority of the bit values in a size-$\mu$ substring of $x$. More precisely, $y_i=1$ if and only if there are at least $\frac{\mu}{2}$ ones in the substring $(x_{(i-1)\mu+1},x_{(i-1)\mu+2},\ldots,x_{i\mu})$. When $\frac{n}{\mu} \notin \mathbb{N}$, the last bits of $x$ are $y$. In our assessment, we regard only the case $\mu=3$.
• Epistasis $W(\ast,\ast,\nu,\ast)$: The idea of epistasis is to introduce local perturbations to the bit strings. To this end, a string $x=(x_1,\ldots,x_n)$ is divided into subsequent blocks of size $\nu$. Using a permutation $e_{\nu}:\{0,1\}^{\nu} \to \{0,1\}^{\nu}$, each substring $(x_{(i-1)\nu+1},\ldots,x_{i\nu})$ is mapped to another string $(y_{(i-1)\nu+1},\ldots,y_{i\nu})=e_{\nu}((x_{(i-1)\nu+1},\ldots,x_{i\nu}))$. The permutation $e_{\nu}$ is chosen in a way that Hamming-1 neighbors $u,v \in \{0,1\}^{\nu}$ are mapped to strings of Hamming distance at least $\nu-1$. In our evaluation, we use $\nu=4$ only.
• Fitness perturbation $W(\ast,\ast,\ast,r)$. With this transformation we can determine the ruggedness and deceptiveness of a function. Unlike the previous transformations, this perturbation operates on the function values, not on the bit strings. To this end, a ruggedness function $r:\{f(x) \mid {x} \in \{0,1\}^n \}=:V \to {V}$ is chosen. The new function value of a string $x$ is then set to $r(f(x))$ , so that effectively the problem to be solved by the algorithm becomes $r \circ f$. We use the following three ruggedness functions.
• $r_1:[0..s] \to [0..\lceil{s/2}\rceil+1$ with $r_1(s)= \lceil {s/2} \rceil +1$ and $r_1(i)=\lfloor {i/2} \rfloor+1$ for $i<s$ and even s, and $r_1(i)=\lceil {i/2} \rceil+1$ for $i<s$ and odd $s$. This function maintains the order of the search points
• $r_2:[0..s] \to [0..s]$ with $r_2(s)=s$, $r_2(i)=i+1$ for $i \equiv {s { mod } 2}$ and $i<s$, and $r_2(i)=\max{i-1,0}$ otherwise. This function introduces moderate ruggedness at each fitness level.
• $r_3:[0..s] \to [0..s]$ with $r_3(s)=s$ and $r_3(s-5j+k)=s-5j+(4-k)$ for all $j \in {[s/5]}$ and $k {\in} [0..4]$ and $r_3(k)=s - (5\lfloor {s/5} \rfloor - 1 )- k$ for $k \in [0..s - 5\lfloor {s/5} \rfloor -1]$. With this function the problems become quite deceptive, since the distance between two local optima implies a difference of $5$ in the function values.
F4-F17 present superpositions of individual W-model transformations to the OneMax (F1) and the LeadingOnes (F2) problem. Precisely, F4-F17 are
• F4: $\text{OneMax} + W([n/2],1,1,id)$
• F5: $\text{OneMax} + W([0.9n],1,1,id)$
• F6: $\text{OneMax} + W([n],\mu=3,1,id)$
• F7: $\text{OneMax} + W([n],1,\nu=4,id)$
• F8: $\text{OneMax} + W([n],1,1,r_1)$
• F9: $\text{OneMax} + W([n],1,1,r_2)$
• F10: $\text{OneMax} + W([n],1,1,r_3)$
• F11: $\text{LeadingOnes} + W([n/2],1,1,id)$
• F12: $\text{LeadingOnes} + W(0.9n,1,1,id)$
• F13: $\text{LeadingOnes} + W([n],\mu=3,1,id)$
• F14: $\text{LeadingOnes} + W([n],1,\nu=4,id)$
• F15: $\text{LeadingOnes} + W([n],1,1,r_1)$
• F16: $\text{LeadingOnes} + W([n],1,1,r_2)$
• F17: $\text{LeadingOnes} + W([n],1,1,r_3)$
### F18: Low Autocorrelation Binary Sequences (LABS)
The Low Autocorrelation Binary Sequences (LABS) problem poses a non-linear objective function over a binary sequence space, with the goal to maximize the reciprocal of the sequence’s autocorrelation:
$\text{LABS }\colon x\mapsto\frac{n^2}{2\sum_{k=1}^{n-1}\left(\sum_{i=1}^{n-k}s_is_{i+k}\right)^2},\text{ where } s_i=2x_i-1.$
### F19-F21: The Ising Model
The classical Ising model considers a set of spins placed on a regular lattice $G=([n],E)$, where each edge $(i,j) \in {E}$ is associated with an interaction strength $J_{ij}$. Given a configuration of $n$ spins, $S:=\left(s_1,\ldots,s_n\right)\in{-1,1}^n$, this problem poses a quadratic function, representing the system’s energy and depending on its structure $J_{ij}$. Assuming zero external magnetic fields and using $x_i = (s_i + 1)/2$, we obtain the following pseudo-Boolean maximization problem:
$\text{ISING }\colon x\mapsto \sum\limits_{\\{i,j\\} \in {E}} \left[x_{i}x_{j} - \left(1-x_{i} \right)\left(1-x_{j} \right) \right].$
In PBO, we consider three instances of lattices: the one-dimensional ring (F19), the two-dimensional torus (F20), and the two-dimensional triangular lattice (F21).
### F22: Maximum Independent Vertex Set
Given a graph $G=([n],E)$, an independent vertex set is a subset of vertices where no two vertices are are direct neighbors. A maximum independent vertex set (MIVS) is defined as an independent vertex set $V’ \subset [n]$ having largest possible size. Using the standard binary encoding $V’ ={i \in[n] \colon x_i = 1}$, solving MIVS is equivalent to maximizing the following function:
$\text{MIVS}\colon x\mapsto \sum_i x_i - n\sum_{i,j} x_i x_j e_{ij} = 0.$
### F23: N-Queens Problem
The N-queens problem (NQP) is defined as the task to place N queens on an ${N}\times{N}$ chessboard in such a way that they cannot attack each other. The figure below provides an illustration for the 8-queens problem. Notably, the NQP is actually an instance of the MIVS problem– when consideringa graph on which all possible queen-attacks are defined as edge.
### F24: Concatenated Trap
Concatenated Trap (CT) is defined by partitioning a bit-string into segments of length $k$ and concatenating $m=n/k$ trap functions that takes each segment as input. The trap function is defined as follows: $f_k^{\text{trap}}(u) = 1$ if the number $u$ of ones satisfies $u = k$ and $f_k^{\text{trap}}(u) = \frac{k-1-u}{k}$ otherwise. We use $k=5$ in our experiments.
### F25: NK landscapes (NKL)
The function values are defined as the average of $n$ sub-functions $F_i \colon [0..2^{k+1}-1] \rightarrow \mathbb{R}, i \in [1..n]$, where each component $F_i$ only takes as input a set of $k \in [0..n-1]$ bits that are specified by a neighborhood matrix. In this paper, $k$ is set to $1$ and entries of the neighbourhood matrix are drawn u.a.r. in $[1..n]$. The function values of $F_i$’s are sampled independently from a uniform distribution on $(0, 1)$.
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Singular Values - Maple Help
Maplets[Examples][LinearAlgebra]
SingularValues
display a graphical interface to the SingularValues function
Calling Sequence SingularValues(M)
Parameters
M - Matrix
Description
• The SingularValues(M) calling sequence displays a Maplet application that returns the condition number for the matrix M.
• A definition of the singular values of a matrix is given in the Maplet application.
• By using the Output drop-down box, control the output returned to the worksheet.
• By using the Matrix has real values check box, control the algorithm used. If the matrix has real entries, a faster algorithm can be used. The default value is determined by using the value of $\mathrm{has}\left(M,I\right)$, which returns true if M has complex entries. That is, it is selected, by default, only if M has no complex entries.
Note: If Matrix has real values is selected and the matrix has any complex entries, an incorrect result may be returned.
• By using the Return read-only vector or matrix(ces) check box, control whether the vector or matrix(ces) returned by the Maplet application are read-only. The default behavior is that the vector or matrix(ces) returned are not read-only.
• By using the Evaluate result check box, control whether the Maplet application returns the output requested or the calling sequence required to calculate this output in the worksheet. The default behavior is to evaluate the result, that is, return the requested output.
• The SingularValues sample Maplet worksheet demonstrates how to write a Maplet application that functions similarly to the Maplet application displayed by this routine.
Examples
> $\mathrm{with}\left(\mathrm{Maplets}\left[\mathrm{Examples}\right]\left[\mathrm{LinearAlgebra}\right]\right):$
> $\mathrm{SingularValues}\left(⟨⟨1,3⟩|⟨2,5⟩|⟨5,7⟩⟩\right)$
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# Polynomially reflexive space
In mathematics, a polynomially reflexive space is a Banach space X, on which the space of all polynomials in each degree is a reflexive space.
Given a multilinear functional Mn of degree n (that is, Mn is n-linear), we can define a polynomial p as
${\displaystyle p(x)=M_{n}(x,\dots ,x)}$
(that is, applying Mn on the diagonal) or any finite sum of these. If only n-linear functionals are in the sum, the polynomial is said to be n-homogeneous.
We define the space Pn as consisting of all n-homogeneous polynomials.
The P1 is identical to the dual space, and is thus reflexive for all reflexive X. This implies that reflexivity is a prerequisite for polynomial reflexivity.
## Relation to continuity of forms
On a finite-dimensional linear space, a quadratic form xf(x) is always a (finite) linear combination of products xg(x) h(x) of two linear functionals g and h. Therefore, assuming that the scalars are complex numbers, every sequence xn satisfying g(xn) 0 for all linear functionals g, satisfies also f(xn) 0 for all quadratic forms f.
In infinite dimension the situation is different. For example, in a Hilbert space, an orthonormal sequence xn satisfies g(xn) 0 for all linear functionals g, and nevertheless f(xn) = 1 where f is the quadratic form f(x) = ||x||2. In more technical words, this quadratic form fails to be weakly sequentially continuous at the origin.
On a reflexive Banach space with the approximation property the following two conditions are equivalent:[1]
• every quadratic form is weakly sequentially continuous at the origin;
• the Banach space of all quadratic forms is reflexive.
Quadratic forms are 2-homogeneous polynomials. The equivalence mentioned above holds also for n-homogeneous polynomials, n=3,4,...
## Examples
For the ${\displaystyle \ell ^{p}}$ spaces, the Pn is reflexive if and only if n < p. Thus, no ${\displaystyle \ell ^{p}}$ is polynomially reflexive. (${\displaystyle \ell ^{\infty }}$ is ruled out because it is not reflexive.)
Thus if a Banach space admits ${\displaystyle \ell ^{p}}$ as a quotient space, it is not polynomially reflexive. This makes polynomially reflexive spaces rare.
The Tsirelson space T* is polynomially reflexive.[2]
## Notes
1. Farmer 1994, page 261.
2. Alencar, Aron and Dineen 1984.
## References
• Alencar, R., Aron, R. and S. Dineen (1984), "A reflexive space of holomorphic functions in infinitely many variables", Proc. Amer. Math. Soc. 90: 407411.
• Farmer, Jeff D. (1994), "Polynomial reflexivity in Banach spaces", Israel Journal of Mathematics 87: 257273. MR1286830
• Jaramillo, J. and Moraes, L. (2000), "Dualily and reflexivity in spaces of polynomials", Arch. Math. (Basel) 74: 282293. MR1742640
• Mujica, Jorge (2001), "Reflexive spaces of homogeneous polynomials", Bull. Polish Acad. Sci. Math. 49:3, 211222. MR1863260
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# Is unsupervised learning a branch of AI?
From Artificial Intelligence: A Modern Approach, a book by Stuart Russell and Peter Norvig, this is the definition of AI:
We define AI as the study of agents that receive percepts from the environment and perform actions. Each such agent implements a function that maps percept sequences to actions, and we cover different ways to represent these functions, such as reactive agents, real-time planners, and decision-theoretic systems. We explain the role of learning as extending the reach of the designer into unknown environments, and we show how that role constrains agent design, favoring explicit knowledge representation and reasoning.
Given the definition of AI above, is unsupervised learning (e.g. clustering) a branch of AI? I think the definition above is more suitable for supervised or reinforcement learning.
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## Friday, June 25, 2010
### An interview with me about The Geek Atlas
This appeared on CNET this week:
Last week, Graham-Cumming took 45 minutes out of his schedule to sit down and talk over instant message with me about the book, his approach to traveling as a geek, and why his shyness didn't stop him from getting the British government to apologize for its terrible treatment of the famous scientist Alan Turing.
Q: Welcome to 45 Minutes on IM. How did you come up with the idea for the "Geek Atlas"?
John Graham-Cumming: I came up with the idea while working in Munich when I visited the Deutsches Museum. I had never heard of it, and I discovered it's a fantastic science museum that clearly rivals places like the Science Museum in London and the Air and Space Museum in Washington, D.C. I thought to myself: someone must have written a travel book for nerds. A Lonely Planet for Scientists. I really wanted it because I was embarrassed that I didn't know about the Deutsches Museum. That evening I made a list of places I'd been around the world and came up with about 70. From that, the idea of the "Geek Atlas" was born.
Read the rest here.
Labels:
If you enjoyed this blog post, you might enjoy my travel book for people interested in science and technology: The Geek Atlas. Signed copies of The Geek Atlas are available.
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4.1k views
A CPU has a $32 KB$ direct mapped cache with $128$ byte-block size. Suppose A is two dimensional array of size $512 \times512$ with elements that occupy $8$-bytes each. Consider the following two C code segments, $P1$ and $P2$.
P1:
for (i=0; i<512; i++)
{
for (j=0; j<512; j++)
{
x +=A[i] [j];
}
}
P2:
for (i=0; i<512; i++)
{
for (j=0; j<512; j++)
{
x +=A[j] [i];
}
}
$P1$ and $P2$ are executed independently with the same initial state, namely, the array $A$ is not in the cache and $i$, $j$, $x$ are in registers. Let the number of cache misses experienced by $P1$ be $M_{1}$and that for $P2$ be $M_{2}$.
The value of $M_{1}$ is:
1. $0$
2. $2048$
3. $16384$
4. $262144$
edited | 4.1k views
Code being C implies array layout is row-major.
http://en.wikipedia.org/wiki/Row-major_order
When $A[0][0]$ is fetched, $128$ consecutive bytes are moved to cache. So, for the next $\dfrac{128}{8} -1=15$ memory references there won't be a cache miss.
For the next iteration of $i$ loop also the same thing happens as there is no temporal locality in the code. So, number of cache misses for $P1$ is
$= \dfrac{512}{16} \times 512$
$= 32 \times 512$
$=2^{14} = 16384$
Correct Answer: $C$
edited
0
Sir in this question it is not mentioned whether array is arranged in row wise or column wise order in memory, so in such questions do we have to assume it to be row wise ?
+1
Sir I think a typo in line "So, by the time A[1][0] is accessed, its cache block would be replaced" it should be A[0][1].
+5
Also Sir I think 256 available blocks in cache is not a limiting factor here, because
A[0][0] will map to cache 0
A[1][0] will map to cache 32
A[2][0] will map to cache 64
A[3][0] will map to cache 96
A[4][0] will map to cache 128
A[5][0] will map to cache 160
A[6][0] will map to cache 192
A[7][0] will map to cache 224
A[8][0] will again map to cache 0
so the cache is being replaced every 8 block accesses not 256 block accesses
so even if we have 512 cache blocks then too we will get the same answer.
0
Each cache line will hold 4 array elements so if A[0][0] is mapped to cache line 0, A[1][0] will map to cache line 128.
0
In the question if we would have used a different Cache organization....then also the no of cache miss would remain same...?
+4
@Danish: Between two iterations in P2 there are 512 distinct memory block accesses and we have only 256 blocks in cache. So, AT LEAST 2 memory blocks must be mapped to a cache block and hence the second one would replace the first memory block. So, in this way for the second iteration every memory access will cause a cache miss.
This is just an overall analysis without considering the cache strategy (assuming best mapping possible). If we see the exact mapping by knowing 512 elements of 8 bytes each in the array we can get the exact cache size which will be needed so that no cache miss occurs. (512 is just a lower bound for it)
Row major or column major?
The given code is said to be C and C uses row-major order.
http://en.wikipedia.org/wiki/Row-major_order
+1
Thanks:)
$\text{Number of Cache Lines} = \dfrac{2^{15}B}{128B} = 256$
$\text{In 1 Cache Line} = \dfrac{128B}{8B} = 16\ elements$
$P_1 = \dfrac{\text{total elements in array}}{\text{elements in a cache line}}$
$=\dfrac{512 \times 512}{16}$
$= 2^{14}$
$= 16384.$
It is so because for $P_1$ for every line there is a miss, and once a miss is processed
we get $16$ elements in memory. So another miss happens after $16$ elements.
edited
0
@amarVashishth
What will be the answer for P2 applyying the above same procedure ?
0
classs!!!! thank you!!!
## Right answer is option C.
edited
0
Very nice explanation 👌
+1 vote
Cache size=32KB
Block Size=128Byte
Number of Block=256
Since array element are stored in row major fashion and number of element in each block=16(128(Block size)/8(array element size));
Suppose are are stored like below ways as::
A[0][0] A[0][1]..................A[0][15]
So whenever P1 will fetch A[0][0] it will be fetch all 15 so Miss ratio==1/16
Since array size=2^18
Since out of 16 reference only one miss happen ,then out of 2^18 number of misses will be=2^18*1/16=2^14=16384
M1=16384;
Now for P2::;
Number of Misses will be=2^18 bcz of column major order along with all 2^18 references will be Miss.
So M2=2^18
Therefor;
M1/M2=1/16
edited
so Option C
Comment below if u get anything wrong about it
Number of cache lines = 64 64 =1 =210
Number of sets = 21022=28
H1 = 18/5 + 0.8 ns
= 3.6 + 0.8 ns = 4.4 ns
Direct mapped cache:
This is no need of multiplexer in direct mapped cache.
So H2 = 16/ 5 nsec
= 3.2 nsec
Difference = H1 – H2
= 4.4 – 3.2 = 1.2 nsec
1
2
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# Homework Help: Surface integral and divergence theorem over a hemisphere
1. Dec 10, 2012
### marineric
1. The problem statement, all variables and given/known data
Please evaluate the integral $\oint$ d$\vec{A}$$\cdot$$\vec{v}$, where $\vec{v}$ = 3$\vec{r}$ and S is a hemisphere defined by |$\vec{r}$| $\leq$a and z ≥ 0,
a) directly by surface integration.
b) using the divergence theorem.
2. Relevant equations
-Divergence theorem in spherical coordinates
3. The attempt at a solution
Another one where the $\vec{r}$ messes me up. Simple enough if it was in regular xyz. Plus the $\vec{v}$... and I don't really know where to start.
2. Dec 10, 2012
### clamtrox
Well, $\mathbf{r} = x \mathbf{e}_x + y \mathbf{e}_y + z \mathbf{e}_z$ shouldn't be too confusing...
In this case, the shape of the integration area suggests that you might want to use spherical coordinates instead of x, y and z. Do you know what the area element is in spherical coordinates? (or even better, can you calculate it?) What about volume element?
3. Dec 10, 2012
### marineric
Ok so attempt at a solution:
∫∫ 3$\vec{r}$ r$^{2}$sinθdθd$\phi$
limits are 0 to 2∏ for θ, and 0 to ∏/2 for $\phi$, or I could just do 3r^3 time the surface area of a hemisphere, which is 2*∏*r^2, so, 6*∏*a^5?
for divergence... do I just take the divergence in spherical coordinates and multiply by the volume of a hemisphere, which is 2/3*∏*r^3?
4. Dec 10, 2012
### clamtrox
Okay, couple of things
1) The surface area element is a vector.
2) Be careful with the angles. Right now your definitions do not work.
3) As you noticed, the integrand does not depend on the angles. Be careful about jumping over the integral though; now you're getting too many factors of r.
Is the divergence constant? If it is, then it should of course work.
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### Home > CALC > Chapter 4 > Lesson 4.5.1 > Problem4-164
4-164.
Rewrite the following using a single trigonometric function. You may wish to review your trigonometric identities in Chapter 1.
1. $10\operatorname{ sin}(3x)\operatorname{ cos}(3x)$
Double Angle identity (factor first).
2. $\operatorname{sin }x\operatorname{ cos }3x −\operatorname{sin }3x\operatorname{ cos }x$
Sum and Difference (Angle Sum) identity.
3. $\operatorname{cos}^4 x −\operatorname{ sin}^4 x$
Factor first.
4. $\operatorname{tan }x +\operatorname{cot }x$
Start by rewriting $\operatorname{tan}(x)$ and $\operatorname{cot }(x)$ as fractions.
You will use more than one identity.
$\frac{\text{sin}x}{\text{cos}x}+\frac{\text{cos}x}{\text{sin}x}=\frac{\text{sin}^{2}x+\text{cos}^{2}x}{(\text{cos}x)(\text{sin}x)}=$
$\frac{1}{\frac{1}{2}\text{sin}(2x)}=2\text{csc}(2x)$
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# Conservation of momentum and laser powered solar sail
1. May 11, 2012
### nemesiswes
I was watching a show on the science channel called "Alien Encounters". Well in one of them they showed a ship that looked as if it was propelling itself by firing a laser at it's solar sail. Now I am pretty sure that from what I understand about physics, that would not be possible.
Correct if I am wrong but when a laser is fired, the laser beam (electromagnetic waves) carry momentum, now since the laser beam carry's momentum then a reverse momentum must be created on the ship that it is firing from, only in the opposite direction. So now if you have a solar sail on that very same ship and the laser beam is absorbed or reflected from that solar sail, then the laser beam momentum is transferred to the solar sail. So now your momentum from firing the laser and the momentum absorbed from the solar sail should cancel out and thus your ship should not move at all. Is this correct?
Don't atoms move the opposite direction that a photon is emitted from?
Last edited: May 11, 2012
2. May 11, 2012
### phinds
Yeah, I saw that same typically moronic show. I watch physics shows on the Science Channel when I feel the need to raise my blood pressure and scream obscenities at the TV.
You are of course correct.
NEVER take things seriously on those shows. They get an awful lot right, but you just never know when they are going to put it some really bone-headed statement just as though they had actually checked it with a real scientist.
EDIT: there have been numerous threads on this forum pointing out how laughable their "physics" is much of the time.
3. May 11, 2012
### nemesiswes
Thanks, I figured there was something wrong there ,lol . I love watching the shows but I sometimes you just catch something that seems a bit off, lol
4. May 12, 2012
### 256bits
Not entirely correct
If the light is reflected from the sail then the ship would move - I am quite sure you can realize that as the outcome.
Even if the beam is absorbed at the sail the ship would have a translation opposite the direction of the beam.
The engine part of the ship would experience a recoil as the photon is emitted. The sail part would experience a recoil in the opposite direction as the photon is absorbed. Diring the brief time interval while the photon is travelling from engine to sail, the engine part will have moved ahead, and the absorption would cancel out this movement. Since the emmision and absorbtion are not instantanious, the ship has translated froward a small distance.
5. May 12, 2012
### Drakkith
Staff Emeritus
Can anyone elaborate on how the reflection of light causes an object to move away from the source of the light? Does the light lose momentum after reflection? Otherwise I can't understand how it works.
6. May 13, 2012
### nemesiswes
So wait would it actually work then?
I was just thinking about it and while it seems impossible, I don't think it is, lol.
Here is how I think of it.
1.Engine part creates laser beam
2. Laser beam will have momentum and the engine will have a opposite momentum causing the ship to move backwards.
3. laser traverses distance and meanwhile ship is moving backwards (backwards would be the actual direction you want)
4.laser is absorbed by sail imparting it's momentum on the sail which is the ship, this stops the movement.
conservation of momentum still holds because all momentum = 0
So what you said "256bits", just explained step by step. Wow if that is true then that is really cool, your basically taking advantage of the speed of light delay to move your ship.
The question I have though is theoretically, could you have 100% efficient solar panel that absorbs your laser which then turns that energy back into a laser with a 100% efficient electricity to laser generator then use this system to move your spacecraft through space. So you would just have to supply the energy in the beginning.
7. May 13, 2012
### Drakkith
Staff Emeritus
Nemesis, if the sail is bouncing the laser beam backwards, away from the direction you want to go, I believe it would work. This is similar to using nozzles to propel the exhaust of a rocket engine. You will feel a force opposite to the direction the exhaust is moving, no matter what direction it was initially in during combustion. I'm not understanding how 256bits explanation works however.
8. May 13, 2012
### Infinitum
Momentum of light changes after reflection. It transfers the momentum to the object so that the total momentum is conserved. Consider a beam of light with energy E and momentum p, perpendicular to the face of a cubical reflecting box of area A.
$E = pc$
where c is the speed of light.
Now when the light is incident on the cube, the momentum of photons is E/c, and when it is reflected, the momentum is -E/c. So the change in momentum of light is 2E/c. Change in momentum implies force, which is exerted on the body.
Edit : If the beam is absorbed, the change in momentum would only be E/c since the final momentum of photons is zero. So, this force would be half of that would occur due to reflection.
9. May 13, 2012
### Drakkith
Staff Emeritus
Hmm. I don't understand how momentum is conserved during reflection. If the light reflects and has the same energy/momentum, and the sail has gained some, where did it come from?
10. May 13, 2012
### nemesiswes
Too drakkith,
The object moves away from the source of the light because light which is just electromagnetic fields can carry momentum. So if the light hits a object it can impart that momentum on it. If the light is reflected then only some of that momentum would be imparted on the object and the light (electromagnetic field) would lose some energy (drop in frequency). It would then be remitted with momentum in the opposite direction (less momentum due to less energy of emitted photon) to which it came in, there would also be momentum in the direction that it originally came in so as to cancel the momentum of the emitted light.
I think this is about right but I am not too sure about some of it which is why I asked the question below
edit: apparently I wrote this one too late, lol
to others,
Would the photon actually impart only some it's momentum when hitting the mirror or would it impart all it's momentum and then just get back some of that momentum when remitted?
Meaning that there is a net momentum in one direction.
Edit: yeah I am pretty sure it would impart all the momentum and the net momentum in one direction comes from the decreased energy of the emitted photon which would mean less momentum.
Last edited: May 13, 2012
11. May 13, 2012
### Infinitum
Light will not have the same momentum. The magnitude of initial and final momenta may be same. In the box case, momentum after reflection is -E/c. (Momentum is a vector!)
This change in momentum is transferred to the sailboat,
$$\frac{2E}{c} = M_{boat}V$$
From wikipedia,
All of the photon's momentum is conserved. There will not be a net momentum in any direction.
12. May 13, 2012
### nemesiswes
ok so then is it possible for the ship to move at all. I am pretty sure it will like how 256bits said. It would just move for a bit until the light hit the sail and stopped the motion. So while momentum is conserved everywhere, the ship still moved a bit. So would the show be correct in showing a ship with a sail and laser and be correct as a viable idea even if impractical for say energy reasons?
13. May 13, 2012
### Drakkith
Staff Emeritus
Alright, so the momentum is now changed. Has the light lost energy? IE will the reflected light be of the same frequency as it was prior to reflection from an observer stationary to the emitting source?
14. May 13, 2012
### Infinitum
Considering a perfectly reflecting surface...The frequency of the emitted light will be same.
This can be verified from the equation I wrote above,
$E = pc$
For partially reflecting surfaces, it depends on how much fraction of the energy they reflect, and the frequency will change accordingly.
15. May 13, 2012
### Infinitum
I'm somewhat confused by this explanation. There can't be a net displacement with no external force acting on the light-boat system. My idea to this is, the photon cannot stay in the atom of the sail forever, it -has- to be emitted out, since atoms hate excited states. This would result in the boat moving back again to its original position. I'm not entirely sure of this though...
16. May 13, 2012
### Drakkith
Staff Emeritus
Then I am confused lol.
17. May 13, 2012
### nemesiswes
I drew a picture just so to make sure everyone knows what I am trying to visualize at least
#### Attached Files:
• ###### SOLAR SAIL SHIP.jpg
File size:
21.2 KB
Views:
239
18. May 13, 2012
### nemesiswes
Now what if the solar sail was just 100% efficient solar panel? Your ship moves and you get the energy back as long as the laser engine is 100% too, lol, something is wrong
Last edited: May 13, 2012
19. May 13, 2012
### 256bits
What has happened is that some energy has moved from one end of the system to the other end - all internal and nothing external needed at all.
Correct. If the energy moves back to the original end.
20. May 13, 2012
### 256bits
Yes. Moving the energy from the sail back to the engine moves the ship in the reverse direction.
21. May 13, 2012
### nemesiswes
Because the movement of the charges I assume? So with a solar panel there really would be no net movement. Without though, there would be a net movement if the laser was powered by say a nuclear reactor .
22. May 13, 2012
### Infinitum
Assuming this is true...
Yes. You would get the energy back. To begin with, it didn't go anywhere if you consider the light-boat system. And the boat came back to its original position. No loss or gain of anything.
23. May 13, 2012
### nemesiswes
And if the solar sail was a reflector then the movement would be double that of the absorber correct?
24. May 13, 2012
### Infinitum
You cannot have a reflecting solar cell-sail. You need to use up those photons to energize the electrons to exhibit photoelectric effect. The cell might be a partially reflecting one though, but that won't make it 100% efficient.
25. May 13, 2012
### nemesiswes
what I meant was if it was just a reflector and not a solar panel, So no energy or very little absorbed.
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In the figure above, points P and Q lie on the circle with [#permalink] 12 Sep 2005, 08:02
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In the figure above, points P and Q lie on the circle with center O. What is the value of s?
A. $$\frac{1}{2}$$
B. $$1$$
C. $$\sqrt{2}$$
D. $$\sqrt{3}$$
E. $$\frac{\sqrt{2}}{2}$$
[Reveal] Spoiler: OA
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Re: gmatprep geo [#permalink] 23 Aug 2009, 14:54
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Check another approach for this problem.
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Re: gmatprep geo [#permalink] 06 Sep 2009, 12:14
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There's a simple reason and misconception as to why people find this question confusing:
--> the 90 degree angle between P and Q seems like it is cut perfectly right down the centre >> to make two 45 degree angles.
--> that's why your brain instinctively thinks the answer is r = sq(3), since the distance seems the same from the left and to the right of the origin!
However, that's the trick to this question and that's why it is CRUCIAL that you train yourself to be highly critical and weary of accepting the simple answer!
Upon closer inspection of the question:
P(-sq(3),1)
Which means the triangle beneath point P has:
height = 1
base = sq(3)
meaning >> radius = 2 (30-60-90 triangle)
This follows the ideal case of the 30-60-90 triangle, which means that opposite to the height of 1, there is an angle of 30 degrees from the x-axis to point P. This means that from the y-axis to point P, there is an angle of 60 degrees (to make a 90 degrees total from x-axis to y-axis). Therefore on the right side, since the figure shows there is a 90 degree angle between point P and Q, 60 degrees have been taken up from the left of the y-axis, which leaves 30 degrees on the right of the y-axis. Furthermore, this leaves 60 degrees from Point Q to the positive x-axis. Therefore, making the exact same 30-60-90 triangle once again, since the radius is the same (the two points lie on the semi-circle). The 60 degree angle is now where the 30 degree angle was on the other triangle, which makes the height now sq(3) and base now 1 >> therefore, base r = 1!
**Have a look at the attachment and then it should make perfect sense!
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gmatprep-PSgeoQ.jpg [ 20.74 KiB | Viewed 47432 times ]
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Re: coordinate geometry [#permalink] 16 Jun 2009, 09:18
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The answer is B.
Lets see how:
First draw a perpendicular from the x-axis to the point P. Lets call the point on the x-axis N. Now we have a right angle triangle $$\triangle NOP$$. We are given the co-ordinates of P as $$(-\sqrt{3}, 1)$$. i.e. $$ON=\sqrt{3}$$ and $$PN=1$$. Also we know angle $$PNO=90\textdegree$$. If you notice this is a $$30\textdegree-60\textdegree-90\textdegree$$triangle with sides $$1-sqrt3-2$$ and angle $$NOP=30\textdegree$$.
Similarly draw another perpendicular from the x-axis to point Q. Lets call the point on the x-axis R. Now we have a right angle triangle $$\triangle QOR$$. We know angle $$QRO=90\textdegree$$.
Now
angle NOP + angle POQ (= 90 -- given) + angle QOR = 180
$$\Rightarrow 30\textdegree+90\textdegree+angleQOR=180$$
$$\Rightarrow angleQOR=60$$
If you notice $$\triangle QOR$$ is also a $$30\textdegree-60\textdegree-90\textdegree$$triangle with sides $$1-sqrt3-2$$ and angle $$OQR=30\textdegree$$
The side opposite to this angle is OR = 1.
Hence answer is B.
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[#permalink] 17 Sep 2005, 21:41
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there is no need for much calculation,
simply switch their X and Y
P and Q are always perpendicular and same distance from O
OP is perpendicular to OQ means you rotate OQ 90 degrees, you will
get OP, in terms of X,Y value, that is just to switch them and flip the sign
to have (-Y,X)
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Re: GMAT Prep 1 [#permalink] 27 Apr 2008, 21:49
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jbpayne wrote:
Thats what I thought, but the answer is B, 1. Anyone able to figure this one out?
There are more than one way to solve this. I like this way
The line PO has slope =-1/√3
The line QO has slope = t/s
PO and QO is perpendicular so [-1/√3]*t/s = -1 ==> t/s =√3 or √3/1 so s = 1
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] 05 Jan 2011, 10:46
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The method I suggested above is very intuitive and useful. I have been thinking of how to present it so that it doesn't look ominous. Let me try to present it in another way:
Imagine a clock face. Think of the minute hand on 10. The length of the minute hand is 2 cm. Its distance from the vertical and horizontal axis is shown in the diagram below. Let's say the minute hand moved to 1. Can you say something about the length of the dotted black and blue lines?
Attachment:
Ques1.jpg [ 7.87 KiB | Viewed 8985 times ]
Isn't it apparent that when the minute hand moved by 90 degrees, the green line became the black line and the red line became the blue line. So can I say that the black line $$= \sqrt{3}$$ cm and blue line = 1 cm
So if I imagine xy axis lying at the center of the clock, isn't it exactly like your question? The x and y co-ordinates just got inter changed. Of course, according to the quadrants, the signs of x and y coordinates will vary. If you did get it, tell me what are the x and y coordinates when the minute hand goes to 4 and when it goes to 7?
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Re: GMAT Prep 1 [#permalink] 27 Apr 2008, 22:08
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bsd_lover wrote:
Could you please elaborate the bolded part ?
sondenso wrote:
There are more than one way to solve this. I like this way
The line PO has slope =-1/√3
The line QO has slope = t/s
PO and QO is perpendicular so [-1/√3]*t/s = -1 ==> t/s =√3 or √3/1 so s = 1
I know I write like this [the colored] will raise the disscussion.
In general
slope of PO = a/b
slope of QO = t/s
PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] 27 Apr 2010, 15:02
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zz0vlb wrote:
I need someone like Bunuel or walker to look at my statement and tell if this is true.
Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.
In the figure above, points P and Q lie on the circle with center O. What is the value of s?
Yes you are right.
Point Q in I quadrant $$(1, \sqrt{3})$$;
Point P in II quadrant $$(-\sqrt{3}, 1)$$;
Point T in III quadrant $$(-1, -\sqrt{3})$$ (OT perpendicular to OP);
Point R in IV quadrant $$(\sqrt{3},-1)$$ (OR perpendicular to OQ).
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] 04 Jan 2011, 17:37
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tonebreeze: As requested, I am giving you my take on this problem (though I guess you have already got some alternative solutions that worked for you)
I would suggest you first go through the explanation at the link given below for another question. If you understand that, solving this question is just a matter of seconds by looking at the diagram without using any formulas.
http://gmatclub.com/forum/coordinate-plane-90772.html#p807400
P is ($$-\sqrt{3}$$, 1). O is the center of the circle at (0,0). Also OP = OQ. When OP is turned 90 degrees to give OQ, the length of the x and y co-ordinates will get interchanged (both x and y co-ordinates will be positive in the first quadrant). Hence s, the x co-ordinate of Q will be 1 and y co-ordinate of Q will be $$\sqrt{3}$$.
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[#permalink] 18 Sep 2005, 11:20
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Not sure if this is simpler or even different approach from the others(I maybe just formulizing qpoo's approach!)u.....
Lets say slope of OP =m1 and OQ =m2
Since OP perpendicular to OQ => m1m2=-1
Since m1= -1/sqrt(3)
=>m2=sqrt3
=>s=1,t=sqrt(3)
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] 11 Oct 2009, 11:48
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Since OP and OQ are the radii of the circle, and we know P has co-ordinates (-\sqrt{3}, 1), from the distance formula we have OP = OQ = 2. Therefore, s^2+t^2=4. -> (1)
Also from the right triangle, we know the length of PQ=2\sqrt{2}.
Again, using the distance formula, (s+\sqrt{3})^2 + (t-1)^2 = PQ^2 = 8. ->(2)
From (1) and (2) you can solve for s.
Hope this helps.
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Re: Help on a Geometry PS Question [#permalink] 09 Sep 2010, 15:46
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Attachment:
a.jpg [ 12.91 KiB | Viewed 46384 times ]
Line OP => y=mx => 1 = $$-sqrt{3} * m$$
=> slope of OP = $$\frac{-1}{sqrt{3}}$$
Since OP is perpendicular to OQ => m1*m2 = -1
=> slope of OQ = $$sqrt{3}$$
equation of line OQ => y = $$sqrt{3}x$$
since s and t lie on this line t= $$sqrt{3}s$$
also radius of circle = $$sqrt{ 1^2+ 3}$$ = 2 = $$sqrt{ s^2+ t^2}$$
=> $$4 = s^2+ t^2 = s^2 + 3s^2 = = 4s^2$$
=> s=1
----------------------------------------------------------------------
Attachment:
b.jpg [ 13.36 KiB | Viewed 46336 times ]
Using Similar triangles.
In triangle APO , angle APO + AOP = 90 ---------------------1
In triangle BOQ , angle BOQ + BQO= 90 ---------------------2
Also SINCE angle POQ = 90 as given.
angle AOP + angle BOQ = 90 ------------------3
Using equations 1 and 3 we get Angle APO = Angle BQO -----------4
using equation 2 ,3, 4
we get Angle APO = Angle BQO and AOP = OQB
Since hypotenuse is common => both the triangle are congruent
=> base of triangle APO = height of triangle AQO and vise versa.
Thus s = y co-ordinate of P = 1
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Re: Coordinate System and Geometry [#permalink] 19 Nov 2012, 23:55
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agarboy wrote:
Can someone explain how to do this problem.. I have no clue how to even start on this.
In the attached picture, from question OA = $$\sqrt{3}$$
PA =1
Thus radius, OP = 2
Thus OPA is a 30, 60, 90 triangle. where angle POA is 30.
Also angle POQ is 90, given in question.
therefore, angle QOB = 60
And hence, angle OQB =30
trianlge QOB is also 30, 60, 90 triangle. in which we know OQ =OP =2 (radius)
Therefore BQ = $$\sqrt{3}$$
and OB = 1
We need value of r which is BQ = $$\sqrt{3}$$
Ans D it is!
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Re: GMAT Prep 1 [#permalink] 29 Apr 2008, 08:36
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sondenso wrote:
bsd_lover wrote:
Could you please elaborate the bolded part ?
sondenso wrote:
There are more than one way to solve this. I like this way
The line PO has slope =-1/√3
The line QO has slope = t/s
PO and QO is perpendicular so [-1/√3]*t/s = -1 ==> t/s =√3 or √3/1 so s = 1
I know I write like this [the colored] will raise the disscussion.
In general
slope of PO = a/b
slope of QO = t/s
PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1
From the picture t^2 +s^2 = 4(radius = 2)
By substituting t =√3s we get s=1
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] 09 Nov 2009, 19:25
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Correct me if I'm wrong, (and forgive my english)
the coordinates of a point p(x,y) that lays on a circle can be expressed as p(sen,cos).
we know that sen π/6 = cos π/3,
and that sen π/3 = cos π/6
in this case I just have to invert x with y and y with x -> indeed we have p(y,x)
As we know x and y have to be positive we have to change the sign for x, the new coordinates are p (y,-x).
in this way to solve the problem doesn't take more than 20 secs.
hope I was enough clear.
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] 27 Apr 2010, 14:36
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I need someone like Bunuel or walker to look at my statement and tell if this is true.
Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.
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Re: gmatprep geo [#permalink] 20 Oct 2013, 19:35
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obs23 wrote:
chetanojha wrote:
Check another approach for this problem.
Can we assume that coordinates of the center O are (0,0) here, can we always assume so?
From the figure it is clear that O lies at (0, 0). O is the point where x axis and y axis intersect in the figure.
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] 13 Oct 2009, 04:29
DenisSh wrote:
Sorry, didn't get that part:
badgerboy wrote:
Again, using the distance formula, (s+\sqrt{3})^2 + (t-1)^2 = PQ^2 = 8. ->(2)
The method used to find the length of OP or OQ from origin i.e. the distance formula has been used here as well to find the length of PQ with P as (-[square_root]3,1) and Q as (s,t) which gives us
{s - (-[square_root]3)}^2 + {t-1}^2 = PQ^2 = 8 ->equation 2
from Right angle Triangle POQ we had got the length of PQ^2 as 8
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] 13 Oct 2009, 12:15
Yes Answer is 1.
S^2 + T^2 = 4 --> equation 1
(t-1)^2 + (s-(-sqrt(3))^2 = 2^2 + 2^2 = 8 --> equation 2
solving for s gives 1
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] 13 Oct 2009, 12:15
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### Off-diagonal entries in matrix of Relative Diffusivity
· 3 · 2125
#### Ilona Glatt
• Posts: 6
##### Off-diagonal entries in matrix of Relative Diffusivity
« on: June 14, 2021, 11:56:32 AM »
Hi,
when I calculate the relative diffusivity, I get a complete matrix as a result. Can someone explain the meaning of the off-diagonal entries?
Thanks,Ilona
#### Aaron Widera
• Math2Market Employee
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##### Re: Off-diagonal entries in matrix of Relative Diffusivity
« Reply #1 on: June 25, 2021, 10:06:15 AM »
Hello,
of course we can:
If you calculate the Diffusivity in x-direction, the entry in the main diagonal x_11 gives you the diffusivity in x-direction. The off-diagonal entries x_12, x_13, x_21 and x_31 give you the amount of diffusion going into y- and z-direction although the driving force, the concentration gradient, is only given in x -direction.
The same goes for Diffusivity in y- and z- direction.
Was that understandable?
#### Ilona Glatt
• Posts: 6
##### Re: Off-diagonal entries in matrix of Relative Diffusivity
« Reply #2 on: June 28, 2021, 10:16:48 AM »
Yes. That was very clear, thank you!
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## Some random programs I wrote (Part I)
March 6, 2019
In late April 2018, I officially went onboard 42 as a ‘cadet‘ and have since ‘climbed the ranks’ all the way to ‘captain‘ by completing various programming projects in the ‘galaxy‘. It has been a uniquely addictive process. I decided to write a post to outline some of the projects as well as sharing my plans on what’s next. Please also feel free to visit my Github for the source code and more detailed descriptions of the projects.
Below are a few of my personal favorites, ordered roughly chronologically.
Note: many of below are team projects done in pairs or triples, I’ll credit my awesome friends and teammate Liam Dehaudt and Charlie Gleason here =)
### Fractals
https://github.com/conanwu777/fractals
This was one of the first projects in the curriculum. The goal is to render various fractals in $\mathbb{C}$ in real-time by taking mouse position as parameter. Parallel threads are used to speed up the rendering.
After completing the basic ‘stage set-up’ and the required Mandelbrot and $z^2 + c$. I had lots of fun coming up with color schemes and implementing rational function Julia sets such as those with locally connected Julia set homeomorphic to the Sierpinski carpet:
As well as those that has a hyperbolic symmetry similar to that of the Poincare disk:
### Ants
https://github.com/conanwu777/ants
This is an optimal transportation puzzle where one needs to find the minimal number of steps to transport a fixed amount of ants from one point to another through a graph. I had fun designing a retro-style ascii visualizer on top of the algorithm =)
We re-formulated the problem into a max-flow problem on a derived directed graph (we double all vertices and edges and form a larger graph) to apply the Ford–Fulkerson theorem. Project the resulting parallel paths to the original network and resolve overlaps. Then we queue the ants in the resulting set of disjoint paths depending on the length of each path. The result guarantees optimum solution as long as the number of ants is greater than the sum of lengths of paths from Ford–Fulkerson.
#### CoreWar
https://github.com/conanwu777/corewar
We made a reproduction of the Core War game. Including full implementation of assembly compiler, virtual machine and GUI.
Core War is a 1984 programming game created by D. G. Jones and A. K. Dewdney, inspired by a malicious virus written in the 80’s. To deal with the self-replicating virus, a white hat hacker invented Reaper. It was a virus designed to spread and eliminate the malware.
The idea was simple. You compete by designing viruses to overtake a computer. You win by protecting your own program and overwriting your opponent’s programs. This is all happening on a virtual machine.
#### Particle System
https://github.com/conanwu777/particle_system
An openCL and openGL simulation of movement of millions of particles in space in various gravitational modes. Used GPU for parallel computing.
Note: Due to the GPU restrictions, this project records with significant quality and performance loss, highly recommend trying it out!
…More to come in Part II…
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The ceramic piezo sensor generates voltage from a few milli-volts to 1 volt when detects a small amount of vibration or force on them. where C is the value of whatever capacitance is in parallel with the current source (we’ll return to this point in the next article). digitalWrite(ledPin, HIGH); digitalWrite(ledPin, LOW); Charge mode pressure sensors generate a high impedance charge output. An Introduction to Motion Sensors: PIR, Tilt, Force, and More Understanding and Modeling Piezoelectric Sensors, An Introduction to Motion Sensors: PIR, Tilt, Force, and More, Samsung Pushes for 5G with Market-Ready RFIC Chip and Growing List of Corporate Partners, Make a PWM Driver for FPGA and SoC Design Using Verilog HDL. The resulting physical configuration is something along these lines: As you can see, the electrodes form a capacitor, and this capacitance becomes an integral part of the device’s equivalent circuit: We can calculate the output voltage of the equivalent circuit, labeled VOUT in the diagram, if we recall the relationship between voltage and current in a capacitive circuit. some vibration in the external environment, we do not want the knock sensor to be triggered for very small vibrations. A piezo knock sensor produces a voltage when it is subjected to any physical stress. void loop() else connects to the GND terminal on the arduino. pinMode(ledPin, OUTPUT); The operating and storage temperature range is -20°C~+60°C and -30°C~+70°C respectively. You can simply connect a LED to the piezoelectric sensor, as shown in the circuit below. sensor. void setup() Miko55. A piezoelectric transducer(also known as a piezoelectric sensor) is a device that uses the piezoelectric effect to measure changes in acceleration, pressure, strain, temperature or force by converting this energy into an electrical charge. respond to people's knocks. Since the LED's anode connects Piezoelectric sensor element will generate an output that activates a normally opened, built-in circuit. The charge produced depends on the piezoelectric constant of the device. The piezo-ceramic element (PZ1) … How to Build a Hall Effect Sensor Circuit Common Analog, Digital, and Mixed-Signal Integrated Circuits (ICs). The best method to detect vibration is Piezo electric method. This characteristic makes piezos an ideal solution for low-power flex, touch, and vibration sensing. To sense, the vibrations generated by knocking are sensed by a piezoelectric sensor PZT. const int sensorPin=0; How to Build a Motion Detector Circuit Since the positive lead of the knock sensor Their electrical behavior responds in a predictable way to mechanical stress, and clever individuals have discovered various ways to incorporate this phenomenon into the world of technology. However, quartz is only one of many materials that exhibit piezoelectric behavior, and the functionality of piezoelectric components is not limited to generating clock signals. If a circuit is activated, then it resembles a contact switch closure and due to corrosion, bouncing and pitting the contact points exhibits. For this circuit, we will connect the knock sensor and an LED to an Arduino board. This voltageis … The capacitance of the lead wire that connects the transducer and piezoelectric sensoralso affects the calibration. The whole circuit can be assembled on a 5×7 cm common circuit board. In order to read low frequency signal we need to decrease the cutoff frequency. suit your needs. We connect the anode (positive lead) of the LED to digital pin 13 of the arduino. It is a self-generating and electromechanical conversion sensor. Piezo sensors are ideal in anti-tamper and industrial health monitoring applications. A motor or solenoid is, strictly speaking, a transducer, because it converts an electrical signal into mechanical motion. The inverting (pin2) and non – inverting (pin3) inputs of the Op-Amp IC CA3140 are shorted through the capacitor C1, so that both the inputs will be in a balanced state. We’ve discussed the piezoelectric phenomenon in the context of electronic sensor devices, and we saw that a piezoelectric device can be represented by a current source in parallel with a capacitor or by a current source in parallel with a capacitor and a resistor. Fortunately, we can easily create a current source based on the charge generated by a piezoelectric transducer. Pin 13 has a built-in 220Ω resistor, so we do not need a By pressing, due to mechanical pressure it create voltage at output which is further feed to the circuit. Quartz elements are not displaced under load, so sensitivity changes are unlikely-eliminating the need for frequent calibrations. connector. The piezoelectric material generates electric charge on the surface after being forced. We might think of piezoelectricity as an effect that converts mechanical stress into voltage, but it seems to me that the more precise interpretation is the following: a piezoelectric transducer generates an electric charge in response to mechanical stress. After making connections as per the Piezoelectric Sensor circuit diagram, when we provide mechanical stress to the piezoelectric sensor it generates voltage. } Only The bottom line is that the piezoelectric effect is a bridge between the mechanical world and the electrical world. Standard circuit diagrams don’t include “charge sources”; the two options are voltage sources and current sources. A piezoelectric sensor makes use of a piezoelectric material. When this happens, output of IC1 remains low. int val= analogRead(sensorPin); current-limiting resistor when using this arduino pin. The red lead connects to analog pin A0 on the arduino board and the black lead In more advanced applications, piezos can be the foundation for energy harvesting. As there are numerous applications on how to implement a piezoelectric sensor using Arduino, here we go with the application of how to use a piezo element to identify vibration in the element (might be door knock or table knock). Piezoelectric Sensor Circuit The piezoelectric sensor circuit is shown below. • Voltage Divider for Adding Offset Voltage –We can only transmit positive voltage signal so we need to add an offset voltage to the circuit. This sensor is used to measure the change of pressure, force, temperature, strain and acceleration. We connect the cathode Once we put this code in the arduino processing software and run the code, the LED will turn on for 5 seconds { We can account for this behavior by adding a parallel resistor to the equivalent circuit. Piezoelectric sensor is a sensor based on piezoelectric effect. So the charge amplifier is usually placed very near to th… delay(5000); The etymology might help you to remember the meaning: in Latin ducere means “to lead” and trans means “across, beyond, on the other side.” So a transducer is something that leads a signal across the boundary that separates, for example, mechanical variations from electrical variations. In our case, we make this threshold 100. It is a polarized electronic component, so polarity must be observed in order for the sensor to work correctly in a circuit. The type A goes into the computer and type B goes into the arduino. If we want the model to be more accurate as a general representation of piezoelectric functionality, we need to add a parallel resistor: A piezoelectric material generates electric charge in response to mechanical stress, but this charge doesn’t stay around forever. if (val >= threshold) Piezoelectric Heat Sensor Circuit A high gain type OP-Amp is used to sense the electrical signals from the piezo disc. detected. First, we need to recall that electric current (in amperes) is the amount of electric charge (in coulombs) that flows per second through a given portion of a circuit. They are converted into an electrical charge. And this is the practical use and ability of a knock sensor circuit. triggered when it is above a certain threshold. These 72 piezoelectric sensors are divided among 4 PCB plates, so each PCB plate has 18 sensors. and then shut off. It characterises a technical standard for piezoelectric sensors which contain built-in impedance conversion electronics. A piezoelectric sensor is a device that uses the piezoelectric effect to measure changes in pressure, acceleration, temperature, strain, or force by converting them to an electrical charge. connects to ground (GND) of the arduino board. However, it seems to me that the word “transducer” is used primarily to describe a device that converts a physical quantity into an electrical signal—i.e., a sensor. (One of the names for this value is “piezoelectric charge constant,” but at least three different terms are used; I’ll refer to it as “coefficient” or “piezoelectric coefficient.”) Thus, if a piezoelectric material has a coefficient of 200 picocoulombs per newton, it will generate 200 × 10–12 coulombs of charge in response to an applied force of 1 newton. Embedded System Design: Build from Scratch or Use an SBC? Piezoelectric Generator is a power generation based Project based on the Piezoelectric sensors. ICP ® is a PCB ® registered trademark that stands for "Integrated Circuit Piezoelectric" and identifies sensors that incorporate built-in microelectronics. A charge amplifier is used to measure the produced charge without dissipation. Therefore, the circuit will only be You can experiment with this to If we translate this statement into mathematical language, we can say that current is the rate of change—i.e., the derivative—of charge: Second, we have to recognize that the charge generated by the piezoelectric transducer is not immobile. A transducer can be anything that converts one form of energy to another. Specificially, for this circuit, we will connect the knock sensor and an LED to the arduino microcontroller. }. We certainly are not expecting dangerous voltages from the piezoelectric sensor; rather, the concern here is faults such as an ESD strike. The working of a basic piezoelectric transducer can be explained by the below figure. Code to turn LED on for 5 seconds when a knock or vibration is detected I don’t want to dwell on the physics of piezoelectricity; it’s an expansive subject, and the second article listed in the Related Information section (above) provides a good introduction. Sparkfun- Piezo Knock Sensor. In the third block of code, we create an integer called val which reads the value obtained from the knock By decreasing the value, the circuit will trigger for lower-sounding knocks. them to turn on or off some load. The piezo sensor can be obtained from Sparkfun at the following link: Sparkfun- Piezo Knock Sensor. The electrical charge generated by piezoelectric material is introduced into the circuit by means of two electrodes. we shut it off. digitalWrite(ledPin, LOW); They are the most sensitive strain gauge technologies offering stable and repeatable accurate electrical output. Piezoelectric Sensor. Piezoelectric Sensors Piezoelectric sensors do not require any power to continuously monitor an environment. Piezoelectricity is the charge created across certain materials when a mechanical stress is applied.Piezoelectric pressure sensors To see this circuit in action, watch the video below. Create one now. First of course if the sensor electrical circuit fails, you will receive a fault related to the circuit, such as a P0325. const int ledPin= 13; It consists of internal Resistance Ri which is also known as insulator resistance. { If this value is above the threshold, we turn the LED HIGH or on for a period of 5 seconds (5000 μs) and then In the second block of code, we make the LED an output pin. The word “transducer” usually refers to a device that performs a conversion between the physical realm and the electrical realm. Piezoelectric sensors are a device that uses the piezoelectric effect to measure the electrical potential caused by applying mechanical force to a piezoelectric material. These advantages save time and money. When sensors fail they can cause a few different faults. Maybe we could have done this with an OpAmp. polarity must be observed in order for the sensor to work correctly in a circuit. This is why it's called a knock sensor; it detects knocks. Piezoelectric Transducer Disc Above is a cheap three terminal piezoelectric transducer used in 12V Piezo Buzzer that produces sound with the below circuit arrangement. (negative lead) of the LED to GND on the arduino board. When the sensor detects a knock, the microcontroller will then turn on an LED for 5 seconds, then shut it off. The piezoelectric sensor is a device used to the piezoelectric effect. Rather, the material itself functions as a leakage path that causes the charge to gradually diminish. In the next article we’ll look at an op-amp topology that provides an effective way to amplify signals from piezoelectric transducers. For example, in the circuit we will build in this project, an LED will light when a knock or vibration is To draw very low current the resistance R1 is very high. If the value is below this threshold, the LED remains off. { The red lead is the positive lead The Stress can be a force, pressure, acceleration or any touching potential. This article explains some theory behind piezoelectric sensors and presents an equivalent circuit that you can use when you’re designing sensor systems. The piezo knock sensor circuit we will build is shown in the schematic below: The above circuit is shown below in a schematic diagram: The positive lead of the knock sensor connects to analog pin A0 of the arduino board and the negative lead We saw above that the charge generated by the piezoelectric device is equal to applied force multiplied by the piezoelectric coefficient, and consequently VOUT is proportional to applied force divided by capacitance. This means that we can model the piezoelectric device using a current source, with the value of the current equal to the derivative of charge. These sensors are made up of a piezoelectric ceramic material pasted in round shape on a metal disk. Basically, piezoelectric sensor is a transducer which converts applied stress into some electrical energy. a knock or vibration. and the black lead is the negative lead. The following circuit shows the schematic symbol of the piezoelectric sensor. The mathematical relationship between the force applied to the piezoelectric material and the amount of charge generated is governed by a coefficient denoted by d and expressed in coulombs per newton. Don't have an AAC account? need a laptop or desktop to write the code and a USB with one end having a type A connector and the other having a type B To use a piezoelectric sensor is the easiest task, just connect the positive and negative terminal to your circuit and press the top of sensor. You can apply all types of physical force to transistors, LEDs, resistors, etc., and in doing so you’re not likely to produce any useful functionality. louder knocks will have to made for the circuit to trigger. The first block of code initializes all of the values we will need. This article focuses on the sensing of a group of physical magnitudes—acceler - ation, vibration, shock, and pressure—that from the per - spective of the sensor … to digital pin 13, we initialize this pin, ledPin, to 13. Oct 22, 2018, 01:59 pm. We then create a threshold of 100. The charge is amplified by the charge amplifier and the measuring circuit and transformed into impedance. These models are shown in Figure 1 along with a typical schematic symbol. connects to analog pin A0 of the arduino board, we initialize this pin, sensorPin, to 0. When we consider the nature of the piezoelectric effect, it’s not surprising that a piezoelectric device can function as a force sensor: the use of piezoelectric material allows a transducer to respond in a reliable and predictable way to the mechanical stress caused by an applied force, or by pressure or acceleration (both of which are related to force). The piezoelectricmaterial is one kind of transducers. The magnitude of the applied load is proportional to, and therefore determined by, … When we squeeze this piezoelectric material or apply any force or pressure, the transducer converts this energy into voltage. Short-circuit charge, Qs, refers to the total charge developed, at the maximum recommended stress level, when the charge is completely free to travel from one electrode to the other, and is not asked to build up any voltage. In this project, a total of 72 piezoelectric sensors are used. P0326 may set, this fault code refers to sensor … A piezoelectric sensor is modeled as a charge source with a shunt capacitor and resistor, or as a voltage source with a series capacitor and resistor. A transducer is not necessarily a sensor. (Redirected from Integrated circuit piezoelectric sensor) The abbreviation IEPE stands for Integrated Electronics Piezo-Electric. A 1 megohm (1MΩ) resistor connects between both leads. Piezoelectric generators are usually specified in terms of their short-circuit charge and open-circuit voltage. Piezo sensors are flexible devices that generate electric charge when they’re stressed. The input could be light, heat, motion, moisture, pressure, vibrations etc… The output generated is usually an electrical signal proportional to the applied input. A knock sensor is a sensor which produces a voltage in response to some type of physical stress such as Sbince there will always be The circuit shown above can benefit from input resistance placed between the piezoelectric sensor and the inverting input terminal: This resistor protects the op-amp by limiting the amount of current generated by whatever voltage might be connected to the inverting input terminal. const int threshold= 100; –Piezo Sensor coupled with a load resistor acts like a high pass filter. We can create circuits which can PCB® manufactures two types of pressure sensors. How to Build a Vibration Detector Circuit, How to Build a Hall Effect Sensor Circuit, How to Build a Vibration Detector Circuit. Now that we have connected the circuit, we now need to create a program Simply it converts the energy from one form to another. The prefix piezo- is Greek for 'press' or 'squeeze'. At this very moment, countless electronic devices are being clocked by oscillator circuits built around a quartz crystal. An inductor is connected which produces inductance due to inertia of sensor. when there is a significant vibration or knock do we want the circuit to trigger. Piezo sensor Piezo sensor technical data sheet The available types of piezoelectric materials vary, but all transforms pressure into an electric charge. This sensor is used to develop a circuit which responds to the knock and turns ON or OFF a load. Most of us are familiar with the piezoelectric effect because of its role in generating high-precision timing signals. The piezo sensor has 2 leads, a positive lead and a negative lead. Piezo Vibration Monitoring Sensor with Arduino Detecting and measuring vibration can be used for several applications, Decision making circuits or alarm circuits. Its sensitive components are made of piezoelectric materials. The impedance of the piezoelectric sensor is less than 500 ohm. Means, it converts physical stress into electrical energy. In this article, we go over how to build a piezo knock sensor circuit. Here quartz crystal coated with silver is used as a sensor to generate a voltage when stress is applied on it. If the device is connected to a circuit, that charge will move and thus become current. Our bonding The piezo sensor has 2 leads, a positive lead and a negative lead. For this project, since we are running code from a computer and uploading it to the arduino board, you will Piezoelectric sensors The application of piezoelectric transducers for sensing and actuation extends to many fields. Piezoelectric sensors also offer added value in terms of stable sensitivity. Topic: Simple piezoelectric sensor circuit (Read 756 times) previous topic - next topic. Piezoelectric Sensor Circuit using Arduino. } IEPE sensors are used to … Piezoelectric Sensor : Circuit, Specifications, and Applications Sensors are devices used to detect or sense the different types of physical quantities from the environment. which will turn on the LED when a knock is detected. Quartz force sensors feature a solid-state design, and quartz does not exhibit any signs of aging. The piezo sensor can be obtained from Sparkfun at the following link: knocks. But in actuality, a knock sensor is useful when we want a circuit to be sensitive and respond to any types of ICP® (Integrated Circuit Piezoelectric) voltage mode sensors feature built-in microelectronic amplifiers that convert the high impedance charge signal into … Thus, a single digital pulse is … It is a polarized electronic component, so The current-source-plus-capacitance model of a piezoelectric transducer is adequate for “dynamic” sensing applications—in other words, applications involving physical conditions that are transient in nature or continuously varying. Newbie; Posts: 1; Karma: 0 ; Simple piezoelectric sensor circuit. Piezo Trigger Switch circuit described here is a microcontroller-compatible shock/impact sensor switch module works on 5VDC supply. Whenever you press the sensor the LED will give a flash. Where the black housing becomes the structure to create audible sound. 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Sensors and presents an equivalent circuit that you can use when you ’ re designing sensor systems these 72 sensors! Force to a circuit which responds to the GND terminal on the arduino not displaced under load, so must. A leakage path that causes the charge amplifier is used to develop a circuit coupled with a typical schematic of! From the piezoelectric constant of the piezoelectric constant of the values we connect... Change of pressure, acceleration or any touching potential material pasted in round shape a! A metal disk can be obtained from the piezoelectric sensor is a sensor which produces voltage! Power to continuously monitor an environment stress such as a leakage path that causes the charge is amplified by below... Us are familiar with the below circuit arrangement to work correctly in a circuit, we create! Negative lead with the below circuit arrangement cm common circuit board black lead is the lead... Need a current-limiting resistor when using this arduino pin with an OpAmp built-in circuit iepe sensors used. Cm common circuit board icp ® is a power generation based project based on surface. The concern here is faults such as a knock sensor circuit ( read times! Used in 12V piezo Buzzer that produces sound with the below Figure,! Light when a knock sensor and an LED to the piezoelectric sensor used. That incorporate built-in microelectronics is used to measure the produced charge without.... Pressure sensors generate a high impedance charge output 'squeeze ' and thus become current an charge. Element will generate an output pin on piezoelectric effect to measure the change of pressure acceleration! Load, so polarity must be piezoelectric sensor circuit in order for the sensor is. Circuits built around a quartz crystal coated with silver is used to develop a circuit, that will! Electrical charge generated by a piezoelectric transducer Disc Above is a significant vibration knock. This behavior by adding piezoelectric sensor circuit parallel resistor to the piezoelectric sensor ; it detects knocks electrical! By piezoelectric material is introduced into the computer and type B goes into the and! Voltage when stress is applied on it not displaced under load, so sensitivity changes are unlikely-eliminating the for! With a typical schematic symbol of us are familiar with the piezoelectric sensor is a transducer, because converts!
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Problem on space-time curvature, black holes and planetary orbits
If the sun were to be replaced by a black hole of the same mass, how would the curvature of space-time change around the sun's initial location and around the region of Earth's orbit? What effect would the change have on Earth's orbit?
So I think that the Earth's orbit would not change because of the same mass, but I am not entirely sure about the curvature of space-time in each of the above locations. I am inclined to believe that, since mass seems to be a determining factor for the curvature of space-time, that it too would not change because the mass will be the same. However, can anyone here confirm or deny what I have written above?
2 Answers
Your question is related to Schwarzschild's spacetimes, which are two spherically symmetric solutions of the equations of General Relativity.
The first one is the exterior Schwarzschild solution (eSS) which describes the curvature (gravity) produced by a spherical object out of it. It is a solution of the vacuum, so that it does not explain the gravitational field inside the sphere. The latter could be modelled with the interior Schwarzschild solution (iSS), if we assume that the sun is a sphere of a perfect fluid with a radial distribution of density and pressure.
Now, let us consider the two important distances in the situation: the radius of the initial star, $r=R$, and the critical radius for the collapse, $r=R_{Sch}$, the so-called Schwarzschild radius that corresponds to the horizon of the black hole (for the real sun it is around 3 km!).
Obviously, $R_{Sch} < R$ (in other case we would have had a black hole from the beginning), so the universe can be separated into three regions:
I ($0 \leq r < R_{Sch}$),
II ($R_{Sch} \leq r < R$) and
III ($r \geq R$).
And you will have the following distribution:
• Before the collapse: iSS (I, II) and eSS (III).
• After the collapse: eSS (I , II, III). There is vacuum everywhere except at the singularity, and now the region I becomes exotic (it is the no-return zone of the black hole).
(Answer) As you can see, in the exterior of the initial star (the region III) the spacetime remains exactly the same, as well as the orbits of the planets.
Note. Both solutions (iSS and eSS) can be "paste" together at the point $r=R$ in a continuous and regular way.
If we assume that the sun is a perfectly spherical body, then nothing would change!
Birkhoff's theorem guarantees that the gravitational field in the exterior of a spherically symmetric body (in our case the Earth lives outside the sun) must be described by the 1-parameter (this parameter being the mass) family of solutions given by the Schwarzschild metric. Since you assume the black hole to have the same mass as the Sun, the two gravitational fields would be indistinguishable.
(Of course, the Sun is not a perfect sphere and moreover there could be other effects due to electromagnetic interactions, so the statement only really holds at leading order...)
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# How many gallons are in a cubic gallon?
That question should probably be phrased differently. Gallons are a measure of volume. So are cubic inches and cubic feet. There are 231 cubic inches in one 1 gallon of fluid, and if that fluid is water, then it weighs 8.32 pounds. There is no such thing as a cubic gallon.
Cubic is a term that implies a quantity raised to the power of three. Linear distance measurements -- inches, feet, centimeters, etc. -- can be cubed. Multiplying three terms that use the same units -- length, width, and height, for example -- results in an answer in cubic units, which implies something that takes up space and has a volume. So perhaps you really meant to ask, "How many gallons are in a cubic foot?"
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How To Create TextBox Control Dynamically at Runtime
http://aspdotnetcodebook.blogspot.com/2008/03/how-to-create-textbox-control.html in this post i will show how to create Textbox Control dynamically and read there value.
<%@ Page Language="C#" AutoEventWireup="true" CodeFile="TextBoxDynamic.aspx.cs" Inherits="TextBoxDynamic" %>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<title>Untitled Page</title>
<body>
<form id="form1" runat="server">
<div>
<tr>
<td colspan="2">
<asp:PlaceHolder ID="phTextBoxes" runat="Server"></asp:PlaceHolder>
</td>
</tr>
<tr>
<td>
</td>
<td>
</td>
</tr>
</table>
</div>
</form>
</body>
</html>
using System;
using System.Data;
using System.Configuration;
using System.Collections;
using System.Web;
using System.Web.Security;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Web.UI.WebControls.WebParts;
using System.Web.UI.HtmlControls;
public partial class TextBoxDynamic : System.Web.UI.Page
{
string strValue = string.Empty;
protected void Page_Load(object sender, EventArgs e)
{
}
private void CreateTextBoxes()
{
for (int counter = 0; counter <= NumberOfControls; counter++)
{
TextBox tb = new TextBox();
tb.Width = 150;
tb.Height = 18;
tb.TextMode = TextBoxMode.SingleLine;
tb.ID = "TextBoxID" + (counter + 1).ToString();
// add some dummy data to textboxes
tb.Text = "Enter Title " + counter;
}
}
protected override void CreateChildControls()
{
// Here we are recreating controls to persist the ViewState on every post back
if (Page.IsPostBack)
{
NumberOfControls += 1;
CreateTextBoxes();
}
else
{
CreateTextBoxes();
// Increase the control value to 1
NumberOfControls = 0;
}
}
protected void btnAddTitle_Click(object sender, EventArgs e)
{
NumberOfControls += 1;
}
public int NumberOfControls
{
get
{
if (ViewState["Count"] == null)
{
return 0;
}
return (int)ViewState["Count"];
}
set
{
ViewState["Count"] = value++;
}
}
{
strValue = string.Empty;
int n = NumberOfControls;
for (int i = 0; i <= NumberOfControls; i++)
{
string boxName = "TextBoxID" + (i + 1).ToString();
TextBox tb = phTextBoxes.FindControl(boxName) as TextBox;
strValue += tb.Text + "\n";
}
Response.Write(strValue);
}
protected void btnRead_Click(object sender, EventArgs e)
{
}
}
Reactions
1. very good code
2. very good code
3. I am a .net begiiner. Your code is working but one more text box is also created when i click on Read button. Any idea to fix it. Thanks
4. I am a .net begiiner. Your code is working but one more text box is also created when i click on Read button. Any idea to fix it. Thanks
5. Please disregard my previous comment regarding a fix to Read button. I figure out how to fix it. Actually check to tell the source of button click. Don't increase NumberOfControls if button click is from Read. thanks
6. I am a .net begiiner. Your code is working but one more text box is also created when i click on Read button. Any idea to fix it.Please tell me
7. Hello Sir, I'm a new user of .Net. I need ur help to create dynamic textboxes by getting the count of the selected item from the dropdownlist using query n display only those many text boxes. but the text box must display only after selecting the item from dropdownlist. Plz help me in this... i'll be waiting for ur reply... thank you.
8. Happy to have found the two articles on dynamic textboxes; I've been struggling with it for days.
Having a small problem because mye table is also being created dynamically, and can't find the right syntax to add the placeholder to the dynamic table.
If anyone has a suggestion, I'd appreciate it.
9. This code is working well but one more textbox is added when i am clicking read button. I think your solution is for clicking any button in a form. any idea to clear the problem rectify the problem that is extra textbox is add
10. hiiiii
i am jitendra but i want to say that the last value is appear bydefault when i click on read button
11. Hi ................
Till TextBox tb = phTextBoxes.FindControl(boxName) as TextBox;
i am getting texbox id but textbox value is showing null please anybody help me
12. I FRIENDS THIS IS KAREM,
I HAVE ONE DOUBT WHAT IS USE OF viewState IN ABOVE EXAMPLE CODE ,COULD ANY ONE OF YOU HELP ME TO CLARIFY THAT ONE
WE CREATING TEXT BOXES AND ACCESS THOSE VALUE IN SIDE THE TEXTBOXES
13. Nice Code... Can you please tell me why one more text box is created when i am clicking Add button
14. Am struggling with this for two daya.. Thanks it helped me a lot..
Emoji
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:(
hihi
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8-)
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x-)
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## Dynamic glass transition in two dimensions
2007
Bayer, Markus
Ebert, Florian
Lange, E.
Schilling, Rolf
Sperl, Matthias
Wittmer, J. P.
Journal article
##### Published in
Physical Review E ; 76 (2007). - 11508
##### Abstract
The question of the existence of a structural glass transition in two dimensions is studied using mode coupling theory (MCT). We determine the explicit d dependence of the memory functional of mode coupling for one-component systems. Applied to two dimensions we solve the MCT equations numerically for monodisperse hard disks. Quantities characterizing the local, cooperative cage motion do not differ much for d=2 and d=3, and we, e.g., find the Lindemann criterion for the localization length at the glass transition. The final relaxation obeys the superposition principle, collapsing remarkably well onto a Kohlrausch law. The d=2 MCT results are in qualitative agreement with existing results from Monte Carlo and molecular dynamics simulations. The mean-squared displacements measured experimentally for a quasi-two-dimensional binary system of dipolar hard spheres can be described satisfactorily by MCT for monodisperse hard disks over four decades in time provided the experimental control parameter (which measures the strength of dipolar interactions) and the packing fraction are properly related to each other.
530 Physics
##### Cite This
ISO 690BAYER, Markus, Joseph M. BRADER, Florian EBERT, Matthias FUCHS, E. LANGE, Georg MARET, Rolf SCHILLING, Matthias SPERL, J. P. WITTMER, 2007. Dynamic glass transition in two dimensions. In: Physical Review E. 76, 11508. Available under: doi: 10.1103/PhysRevE.76.011508
BibTex
@article{Bayer2007Dynam-9165,
year={2007},
doi={10.1103/PhysRevE.76.011508},
title={Dynamic glass transition in two dimensions},
volume={76},
journal={Physical Review E},
author={Bayer, Markus and Brader, Joseph M. and Ebert, Florian and Fuchs, Matthias and Lange, E. and Maret, Georg and Schilling, Rolf and Sperl, Matthias and Wittmer, J. P.},
note={Article Number: 11508}
}
RDF
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<dcterms:abstract xml:lang="eng">The question of the existence of a structural glass transition in two dimensions is studied using mode coupling theory (MCT). We determine the explicit d dependence of the memory functional of mode coupling for one-component systems. Applied to two dimensions we solve the MCT equations numerically for monodisperse hard disks. Quantities characterizing the local, cooperative cage motion do not differ much for d=2 and d=3, and we, e.g., find the Lindemann criterion for the localization length at the glass transition. The final relaxation obeys the superposition principle, collapsing remarkably well onto a Kohlrausch law. The d=2 MCT results are in qualitative agreement with existing results from Monte Carlo and molecular dynamics simulations. The mean-squared displacements measured experimentally for a quasi-two-dimensional binary system of dipolar hard spheres can be described satisfactorily by MCT for monodisperse hard disks over four decades in time provided the experimental control parameter (which measures the strength of dipolar interactions) and the packing fraction are properly related to each other.</dcterms:abstract>
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<dcterms:bibliographicCitation>First publ. in: Physical Review E 76 (2007), 011508</dcterms:bibliographicCitation>
<dc:contributor>Bayer, Markus</dc:contributor>
<dc:contributor>Fuchs, Matthias</dc:contributor>
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Yes
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# How can I prove the identity \frac{1}{1+\sin(x)} \equiv \frac{\sec^2 (\frac{x}{2})}{(\tan(\frac{x}{2}) +1)^2}
How can I prove the identity
$$\displaystyle{\frac{{{1}}}{{{1}+{\sin{{\left({x}\right)}}}}}}\equiv{\frac{{{{\sec}^{{2}}{\left({\frac{{{x}}}{{{2}}}}\right)}}}}{{{\left({\tan{{\left({\frac{{{x}}}{{{2}}}}\right)}}}+{1}\right)}^{{2}}}}}$$
• Questions are typically answered in as fast as 30 minutes
### Solve your problem for the price of one coffee
• Math expert for every subject
• Pay only if we can solve it
Marcus Herman
Use
$$\displaystyle{\frac{{{\sec{{\frac{{x}}{{2}}}}}}}{{{\tan{{\frac{{x}}{{2}}}}}+{1}}}}\equiv{\frac{{{1}}}{{{\sin{{\frac{{x}}{{2}}}}}+{\cos{{\frac{{x}}{{2}}}}}}}}$$
$$\displaystyle{\left({\sin{{\frac{{x}}{{2}}}}}+{\cos{{\frac{{x}}{{2}}}}}\right)}^{{2}}=?$$
###### Not exactly what you’re looking for?
stomachdm
We can express any value trigonometric ratio of a given angle $$\displaystyle{2}\theta$$ in terms of the tangent of half that angle. For example:
$$\displaystyle{\sin{{2}}}\theta\equiv{\frac{{{2}{\tan{\theta}}}}{{{1}+{{\tan}^{{2}}\theta}}}}$$
$$\displaystyle{\cos{{2}}}\theta\equiv{\frac{{{1}-{{\tan}^{{2}}\theta}}}{{{1}+{{\tan}^{{2}}\theta}}}}$$
$$\displaystyle{\tan{{2}}}\theta\equiv{\frac{{{2}{\tan{\theta}}}}{{{1}-{{\tan}^{{2}}\theta}}}}$$
As an aside, this is useful in integration. So, you may want to use the first identity given to prove your identity.
Vasquez
Alternately, use half angle theorems:
$$\sin x \equiv \frac{2 \tan(\frac x2)}{1+\tan^2(\frac x2)}$$
• Questions are typically answered in as fast as 30 minutes
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## Monday, May 14, 2018
A winner is big enough to admit his mistakes, smart enough to profit from them, and strong enough to correct them. – John Maxwell
So many times I hear managers complain about a subordinate when he or she makes a mistake. The manager immediately feels the mistake will reflect on himself and wants to berate the subordinate.
The quote above by John Maxwell tells us that mistakes can be used as learning tools and it is possible to profit from those mistakes.
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# Evaluating the integral $\int \sqrt{1 + \frac{1}{x^2}} dx$
$$\int \sqrt{1 + \frac{1}{x^2}} dx$$
This is from the problem calculating the arc length of $y=\log{x}$.
I tried $x = \sinh{t}$ or $\frac{1}{x} = \tan{t}$ but all failed.
• Actually, $\frac{1}{x}=\tan t$ or, equivalently, $x=\cot t$ will work fine and, as Travis points out, so will $x=\sinh t$. – user84413 Feb 20 '16 at 0:30
Hint (For $x > 0$) we can rewrite this as $$\int \frac{\sqrt{1 + x^2}}{x} dx.$$ The radical expression $\sqrt{1 + x^2}$ suggests using the substitution $x = \tan \theta$, $dx = \sec^2 \theta \,d\theta$ or the substitution $x = \sinh t$, $dx = \cosh t \,dt$
If we wish, we can avoid trigonometric or hyperbolic function substitution. For we want $$\int \frac{x\sqrt{x^2+1}}{x^2}\,dx.$$ Let $u^2=x^2+1$. Then $u\,du=x\,dx$, and our integral becomes $$\int \frac{u^2}{u^2-1}\,du.$$ Note that $\frac{u^2}{u^2-1}=1+\frac{1}{u^2-1}$, and use partial fractions.
Let $x=\sinh t$, then $dx=\cosh t \, dt$
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### Be Reasonable
Prove that sqrt2, sqrt3 and sqrt5 cannot be terms of ANY arithmetic progression.
### Janusz Asked
In y = ax +b when are a, -b/a, b in arithmetic progression. The polynomial y = ax^2 + bx + c has roots r1 and r2. Can a, r1, b, r2 and c be in arithmetic progression?
### Proof Sorter - Sum of an AP
Use this interactivity to sort out the steps of the proof of the formula for the sum of an arithmetic series. The 'thermometer' will tell you how you are doing
# Summats Clear
##### Stage: 5 Challenge Level:
Investigate special cases for small $n$ first.
Can you spot patterns in the sums for $n=3$, $4$, $5$, $6$, $7 \dots$?
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# Euler-Maclaurin Summation Formula for Multiple Sums
The Euler-Maclaurin summation formula is \begin{eqnarray} \sum_{k = a}^{b} f(k) = \int_{a}^{b} f(t) \, dt + B_1 (f(a) + f(b)) + \sum_{n = 1}^{N} \frac{B_{2n}}{(2n)!} ( f^{(2n-1)}(b) - f^{(2n-1)}(a) ) + R_{N}, \end{eqnarray} where $B_{n}$ is the $n^{\text{th}}$-Bernoulli number taking $B_{1} = \tfrac{1}{2}$, and the remainder term is bounded by the following \begin{align} |R_{N}| \leq \frac{|B_{2N} |}{(2n)!} \int_{a}^{b} | f^{(2N)}(t) | \, dt. \end{align} for any arbitrary positive integer $N$. Is there a similar formula for nested sums of the form, \begin{eqnarray} \sum_{k_1 = a_1}^{b_1} \cdots \sum_{k_n = a_n}^{b_n} f(k_1, \dots, k_n). \end{eqnarray}
Thanks!
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## Past Seminars- Inverse Problems and Imaging
• Peijun Li
Tue Jun 12, 2012
2:00 pm
We consider the scattering of a time-harmonic plane wave incident on a two-scale heterogeneous medium, which consists of scatterers that are much smaller than the wavelength and extended scatterers that are comparable to the wavelength. A generalized Foldy-Lax formulation is proposed to capture multiple scattering among point scatterers and...
• Mark A. Anastasio
Tue May 15, 2012
2:00 pm
Photoacoustic tomography (PAT) is an emerging soft-tissue imaging modality that has great potential for a wide range of biomedical imaging applications. It can be viewed as a hybrid imaging modality in the sense that it utilizes an optical contrast mechanism combined with ultrasonic detection principles, thereby combining the advantages of...
• Oleg Imanuvilov
Tue May 8, 2012
4:00 pm
We prove that if for the isotropic Lamé system the coefficiem $\mu$ is a positive constant then both coefficents can be reconstructed from the partial Cauchy data.
• Luca Rondi
Tue Mar 13, 2012
2:00 pm
Many techniques developed for free-discontinuity problems, arising for example in imaging or in fracture mechanics, may be successfully applied to reconstruction methods for inverse problems whose unknowns may be characterized by discontinuous functions. We show the validity of this approach both from the theoretical point of view, by a...
• Hongyu Liu
Tue Mar 6, 2012
2:00 pm
In this talk, we shall consider the near-invisibility cloaking in acoustic scattering by non-singular transformation media. A general lossy layer is included into our construction. We are especially interested in the cloaking of active/radiating objects. Our results on the one hand show how to cloak active contents more efficiently, and on the...
• Andras Vasy
Tue Feb 28, 2012
2:00 pm
Waves reflecting/refracting/transmitting from singularities of a metric (e.g. sound speed) satisfy the law of reflection. One expects that if the singularities are sufficiently weak, in terms of differentiability (conormal order) then the reflected singularity is weaker than the transmitted one, in the sense that it is more regular. In this joint...
• Lauri Oksanen
Thu Jan 12, 2012
4:00 pm
We consider boundary measurements for the wave equation on a bounded domain $M \subset \R^2$ or on a compact Riemannian surface, and introduce a method to locate a discontinuity in the wave speed. Assuming that the wave speed consist of an inclusion in a known smooth background, the method can determine the distance from any boundary point to the...
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×
# RGAME - Editorial
Author: Abhra Dasgupta
Tester: Antoniuk Vasyl and Misha Chorniy
Editorialist: Pushkar Mishra
Easy-Medium
# PROBLEM:
Given are $N+1$ numbers $A_0$ to $A_N$. These numbers come in as a stream in order from $A_0$ to $A_N$. The number coming in can be placed at either ends of the sequence already present. Score of such a gameplay is calculated as per given rules. Output the sum of scores of all possible different gameplays.
# EXPLANATION:
Subtask 1 can be solved by directly simulating the given problem. In other words, we can directly append the number coming in at either ends and check the sum for each possible arrangement. There can at maximum be $2^N$ different sequences. Therefore, the time complexity is $\mathcal{O}(2^N)$ per test case. This is sufficient for this subtask.
Let us start by thinking of simpler sequences in which observing patterns could be easier. A trick is to take something like 1, 1, 1, ...., 1, 5 as the sequence. And then the 5 can be shuffled around to different positions to observe how many times a position is taken into account.
Nevertheless, we are going to take a more mathematical approach in this editorial. Let's see what happens when $k^{th}$ number, i.e., $A[k]$ appears in the stream. It can be appended to either of the two ends of the already existing sequence. But how many already existing sequence are there? Clearly, $2^{(k-1)}$. Let us say for now that $A[k]$ is appended to the right of already existing sequence. Now, consider some $p^{th}$ number $A[p]$ coming after $A[k]$. How many sequences exist such that the $A[p]$ will be multiplied by $A[k]$? For $A[p]$ to be multiplied by $A[k]$, all numbers coming in between these two must not go to the side that $A[k]$ is on, i.e., they should be put on the left in all the $2^{(k-1)}$ sequences where $A[k]$ has been appended on the right. If this happens, then when $A[p]$ comes, it will be multiplied by $A[k]$ when placed on the right of it. The $(p+1)^{th}$ up till $N^{th}$ numbers can be arranged in any order after that. So how many sequences in total will have the product of $A[k]$ and $A[p]$? Clearly, $2^{(k-1)}*2^{(N-p)}$. Thus, total value that gets added to the answer is $(A[k]*2^{(k-1)})*(A[p]*2^{(N-p)})$.
We now have a way to calculate the required answer. Below is the pseudocode of the same.
let possible_future_prod[i] = A[i] * 2^(N-i)
let answer = 0; //accumulator variable
for i = 0 to N-1
{
ways_to_arrange_prefix = 2^(i-1); //if i = 0, then 1
//multipying A[i] with the number of possible prefixes
partial_prod = (ways_to_arrange_prefix * A[i]);
//iterating over elements coming after i
for j = i+1 to N
{
total_prod = partial_prod * possible_future_prod[j];
//adding total_prod to the accumulator variable
ans += total_prod;
}
}
//recall, we had only taken the case when an element is
//appended to the right.
//for taking symmetrical cases into account, multiply by 2.
return 2*ans
This algorithm runs in $\mathcal{O}(N^2)$.
The algorithm stated in subtask 2 can be made $\mathcal{O}(N)$ by precalculating the suffix sum of the $possible\_future\_sequences$ array. Once we have the $suffix\_sum$ array, the inner loop given above in the pseudocode can be reduced to:
//calculating the suffix_sum array
suffix_sum[n] = possible_future_prod[n]
for i = N-1 downto 0
suffix_sum[i] = possible_future_prod[i] + suffix_sum[i-1];
let answer = 0; //accumulator variable
for i = 0 to N-1
{
ways_to_arrange_prefix = 2^(i-1); //if i = 0, then 1
//multipying A[i] with the number of possible prefixes
partial_prod = (ways_to_arrange_prefix * A[i]);
//calculating the sum that can be achieved by
//multiplying A[i] with numbers coming after it
total_prod = (partial_prod * suffix_sum[i+1]);
//adding total_prod to the accumulator variable
ans += total_prod;
}
//for taking symmetrical cases into account, multiply by 2
return 2*ans
The editorialist's program follows the editorial. Please see for implementation details.
# OPTIMAL COMPLEXITY:
$\mathcal{O}(N)$ per test case.
# SAMPLE SOLUTIONS:
This question is marked "community wiki".
1.3k156581
accept rate: 4%
19.8k350498541
2 short editorial on four relatively easy problems: https://shadekcse.wordpress.com/2016/01/11/codechef-jan16-challenge/ answered 11 Jan '16, 15:26 3★shadek 109●1●3 accept rate: 0%
2 @jawa2code . Yes, maybe you are assuming the sum in 2 1 to be fixed and then subsequent products will be calculated adding to this. But this approach is wrong. As, the question statement says that you have to to calculate the sum of scores for all game plays. Consider 1 2 1 as a gameplay. To get to this gameplay, we start with 1 2 . The sum for this is 12=2. Now, moving ahead adding 1 to the sequence, we have the new sum to be 21 + 2 =4, for 1 2 1. Now, for 1 1 2. This is a different gameplay. So, to calculate the score for this case, we'll start from the beginning. The sequence will have to be built again. And the sum 1*1 will again be calculated. So, if you start to make different game plays like this you will see that the sum thus calculated is calculated many times. Please upvote if this clarifies your doubt. I had the same problem in the beginning. answered 14 Jan '16, 18:52 3★aman935 120●1 accept rate: 0%
2 Nice problem :) After a lot of WA's ultimately got green ticked AC. Here's my solution : https://www.codechef.com/viewsolution/9060724 answered 17 Jan '16, 13:35 781●8 accept rate: 7%
2 somebody please upvote me , i have questions to ask thank you answered 25 Sep '16, 16:41 144●2●5 accept rate: 3%
1 But how many already existing sequence are there? Clearly, 2^(k−1) but when two pairs equal numbers are there then there will not 2^(k-1) sequences because of repetion of sequences, like suppose we have 1, 2, 2, 3, 3 We can obtain : 1 -> 1 2 -> 1 2 3 -> 2 1 2 3 -> 3 2 1 2 3 Also, : 1 -> 2 1 -> 3 2 1 -> 3 2 1 2 -> 3 2 1 2 3. So I think the approach is wrong and all have blind folded solved this problem. @admin. answered 11 Jan '16, 15:36 81●1●5 accept rate: 0% The initial problem statement had this confusion. It was updated on 2nd or 3rd day as far as i remember --> "when we read from left to right, there exists some position i, at which the gameplays have aj and ak written at the ith position such that j ≠ k." (13 Jan '16, 19:24) vsp46★
0 This was soo hard :p answered 11 Jan '16, 15:15 101●1●6 accept rate: 0%
0 Good problem answered 11 Jan '16, 16:44 1 accept rate: 0%
0 @saurabhsuniljain In one of the comments on the contest page and I the second sample test case clearly have shown that the sequences which are actually same(:p) will still be considered different if the numbers were inserted in different orders(nobody can tell the sequence they were inserted in ,just by seeing, the mess created by the problem). Although I agree that the question statement was indeed difficult to understand (may have been clearer). answered 11 Jan '16, 17:13 3★aman935 120●1 accept rate: 0%
0 Hello how 1 2 1 leads to 14 1 2=>2 2 1=>2 1 2 1=>2 1 1 2=>1 2 1 1=>1 1 2 1=>1 total=2+2+2+1+1+1=9 can some help me to get 14 Regards Samuel answered 11 Jan '16, 17:22 24●1●2●7 accept rate: 0%
0 @jawa2code Explanation Of 121:- we pick 1 then put 2 in backside of 1 then the sequence becomes 12 then we put last 1 at the end so the sequence becomes 121 and total score for this will be 12+21=4. Also we may put last 1 in beginning of 12 which will form 112 and score for this will be 12+11=3. Also we wemay put 2 in front of first 1 which will transform the sequence to 21 now we can put the last one either in front or back of "21" which will lead to 121 and 211 and score will be 12+21=4 and 21+11=3 respectivly adding up all the scores we get 4+4+3+3=14. answered 11 Jan '16, 17:34 5★killjee 476●9 accept rate: 5%
0 Hello sir Killjee,thanks for your response from the Problem statement For a move in which you write a number Ai (i>0), your points increase by the product of *Ai and its neighbour*. (Note that for any move it will have only one neighbour as you write the number at an end). are you following this instruction? Regards Samuel answered 13 Jan '16, 18:38 24●1●2●7 accept rate: 0%
0 @pushkarmishra i have a different algo to solve the prob its complexity is O(n^2) but it fails subtask 2 i will try to explain u my algo Let us take n=4; so our array will contain n+1 elements and we define our array to be say {1,2,3,4,5}; now what i observed was the only scores which can be achieved will be as follows : 2 3 6 4 8 12 5 10 15 20 the reason to this being,when we are at i-th number let us say 5 it could be placed at right or left of all above formed numbers which will give only products 5 mul 1=5 5 mul 2=10 5 mul 3=15 5 mul 4=20 similarly for 4 4mul 1=4 4mul 2=8 4mul3=12 and now what i observed was that the arr[1] will repeat itself 2^n times so our score by now would be arr[1]mul2^n now let us assume we are at arr[4]=5; the scores which it can achieve will be 5 10 15 20 respectively 20 will repeat itself 2^n-1 times 15 2^n-2 times now we are left with only two combinations of score the will repeat themselves 2^n-3 times what i m trying to do is iterate all numbers in the array compute there respective scores and how many times will they repeat themselves multiply with arr[i] and aad to sum this will take a outer loop and an inner loop as i have used here https://www.codechef.com/viewsolution/9177526 pls check my solution then why subtask 2 is not being accepted i would love to hear a word form u on this :) answered 15 Jan '16, 02:05 3★c0dew0rm 1 accept rate: 0%
0 tough question . answered 16 Jan '16, 11:03 11●2 accept rate: 0%
0 For A[p] to be multiplied by A[k], all numbers coming in between these two must not go to the side that A[k] is on, can anybody explain me ? answered 16 Jan '16, 11:14 11●2 accept rate: 0%
0 Thanks Aman, May i know how are you interpreting this condition For a move in which you write a number Ai (i>0), your points increase by the product of Ai and its neighbour. (Note that for any move it will have only one neighbour as you write the number at an end). My understanding is this if the in put is 1 2 1 Time: Result 1 sec 1 appeared =1 2 sec 2 appeared =1 2/2 1 total=2+2=4 3 sec 1 appeared= 1 1 2/1 2 1/1 2 1/2 1 1=eliminate duplicates={1 1 2/1 2 1/2 1 1}=2+1+1=4 Regards Samuel answered 18 Jan '16, 16:11 24●1●2●7 accept rate: 0%
0 @jawa2code You are not supposed to remove duplicates answered 19 Jan '16, 14:54 5★killjee 476●9 accept rate: 5%
## time(sec) number appeared seq result
1 1 1 1
2 2 1 2(2)/(2)2 1 2+2=4
3 1 (1)1 1 2/1 2 1(2)/2 1 1(1)/(2)1 2 1 1+2+1+2=6
Total 1+4+6=11
24127
accept rate: 0%
0 For Free Query Regarding To Packers and Movers Visit Packers and Movers Chennai http://www.shiftingsolutions.in/packers-and-movers-chennai.html Packers and Movers Jaipur http://www.shiftingsolutions.in/packers-and-movers-jaipur.html answered 21 Jan '16, 17:13 0★devil2 1 accept rate: 0%
0 @killjee my code has time complexity O(n^2) yet it only executes first sub task acc to algo it should pass 2 subtasks why is it so ?? answered 23 Jan '16, 01:35 3★c0dew0rm 1 accept rate: 0%
0 @abhra73 , @antoniuk1 , @pushkarmishra , @admin can anyone pls explain because its not clear in the edtiorial suffix_sum[n] = possible_future_prod[n] the possible future productions of n should be zero for i = N-1 downto 0 suffix_sum[i] = possible_future_prod[i] + suffix_sum[i-1]; and how can this loop calculate because suffix_sum[i-1] will store a garbage value as its not been computed at any time link This answer is marked "community wiki". answered 24 Jan '16, 03:58 3★c0dew0rm 1 accept rate: 0%
0 i think in the 4 line there should be suffix_sum[i] = possible_future_prod[i] + suffix_sum[i+1] instead of suffix_sum[i] = possible_future_prod[i] + suffix_sum[i-1]; answered 25 Jan '16, 10:22 2★alkos 1 accept rate: 0%
0 Can someone tell what is wrong with this solution. solution answered 15 Feb '16, 20:03 1●1 accept rate: 0%
0 here is my solutions it works on the test cases but it shows wrong ans when i submitted https://www.codechef.com/viewsolution/9977223 thanks in advance answered 29 Apr '16, 08:48 1 accept rate: 0%
0 In the O(N) solution for subtask3, shouldn't the index of suffix_sum in the RHS of the expression - suffix_sum[i] = possible_future_prod[i] + suffix_sum[i-1]; be i+1 instead of i-1 I mean it should be - suffix_sum[i] = possible_future_prod[i] + suffix_sum[i+1]; answered 09 May '16, 22:48 1 accept rate: 0%
0 Why is the below code not working for subtask 2 and 3. I have used the logic of the editorial only. import java.util.Scanner; class RupsaGame2{ public static int N; public static long[] Ai; public static final long MOD = 1000000007; public static void main(String[] args){ int T; Scanner a = new Scanner(System.in); T=a.nextInt(); for(int i=0;i0;i--) suffix_sum[i]=(((1L*(Ai[i])*(powerOf2(N-i)))%MOD) + (suffix_sum[i+1]))%MOD; long ans = 0; for(int i=0;i<=N-1;i++){ long ways_to_arrange_prefix = powerOf2(i-1)%MOD; long partial_prod = (1L*(ways_to_arrange_prefix) * (Ai[i]%MOD) )%MOD; long total_prod = (1L*partial_prod * suffix_sum[i+1])%MOD; ans = (ans + total_prod)%MOD ; } ans=2*ans; ans=ans%MOD; return ans; } public static long powerOf2(int m){ long result = 1L; for(int i=1;i<=m;i++) result=result*2L; return result; } } answered 10 May '16, 19:14 1 accept rate: 0%
0 can someone point out the mistake in this code ? https://www.codechef.com/viewsolution/10055742 answered 11 May '16, 16:34 19●1 accept rate: 0%
0 what must be the time complexity of the program. I have written the code and when trying to submit it is showing "Time Limit Exceeded" error. I have used backtracking. answered 05 Jun '16, 03:24 1 accept rate: 0%
0 I'm not able to understand how come for a sequence 1 2 1 the output is 14. According to my calculations, it always yields 11 as @jawa2code has mentioned. Clearly some have to explain this with proper example.(because I could not follow with @aman935 answer) for sequence 1 2 1 These are the possible AiAi+1 pairs . 1 -> Output 1 1 *2* -> Output 2 1 2 *1* -> Output 2 *1* 1 2 -> Output 1 *2* 1 -> Output 2 1 2 *1* -> Output 2 *1* 2 1 -> Output 1 Where the number in star indicate the position of the number (in sequence) placed at left or right. So the Output of the result of the above operation is 1 + 2 + 2 + 1 + 2 + 2 + 1 = 11 answered 12 Jun '16, 17:03 1 accept rate: 0%
0 In Subtask 3 pseudocode, shouldn't it be suffix_sum[i] = possible_future_prod[i] + suffix_sum[i+1]; answered 28 Jul '16, 01:39 143●1●6 accept rate: 10%
0 can someone check and find the error in my code...I have compared my output with those of the other successful submissions for random values and the outputs are same https://www.codechef.com/viewsolution/11074855 answered 08 Aug '16, 14:57 3★dr_aps 1 accept rate: 0%
0 It took quite some time to understand the problem. For the numbers 1 2 1 Case 1. Lets start with writing numbers only at the right end 1 - - - - 1 2 - - - - product is 2 - - - - 1 2 1 - - - - product is 2 (2 multiplied with 1) (product is new number multiplied with its neighbor) Sum of the products is 4 Case 2. Write numbers only at the left end 1 - - - - 2 1 - - - - product is 2 - - - - 1 2 1 - - - - product is 2 Sum of the products is 4 Case 3. Write numbers at alternate ends starting with right end 1 - - - - 1 2 - - - - product is 2 - - - - 1 1 2 - - - - product is 1 (1 multiplied with 1) Sum of the products is 3 Case 4. Write numbers at alternate ends starting with left end 1 - - - - 2 1 - - - - product is 2 - - - - 2 1 1 - - - - product is 1 Sum of the products is 3 Sum of all the products is 14. Though case 1 and 2 resulted in the same sequence, both should be considered since the order in which numbers were written is different. answered 30 Aug '16, 12:53 0★sou333 1 accept rate: 0%
0 Can anyone tell me what's the problem with this code? I have tried with random inputs from one of the successful submissions and they all seem to give the correct result. Here's the code: https://www.codechef.com/viewsolution/11610571 I have used recursion in my code. Can this be the source of the problem? Gramercy... answered 24 Sep '16, 10:32 1●1 accept rate: 0%
0 lol i am new :P answered 24 Sep '16, 11:06 0★gokuo 1 accept rate: 0%
0 Hmm, I need some amount of help. I have written the code, which seems to work right for a small number of elements. When we increase the size of the elements, then it makes a small mistake somehow. The test case I ran, had an error of -19. I want to know where is this error generated. Here is the code : https://www.codechef.com/viewsolution/11993701 answered 02 Nov '16, 21:01 1 accept rate: 0%
0 can any one tell why i m getting wrong ans for this method https://www.codechef.com/viewsolution/12137512 answered 25 Nov '16, 13:00 1 accept rate: 0%
I have written this code for the given problem but for some reason, while submitting, it's giving me some error. Can anyone please help me with this? By the way, it was a good problem.
# include<stdio.h>
struct code
{
int y;
int a[100][2];
int b[100];
int k;
}grp[100];
int arr[100],t,n,kk,j,p,c,a,i,z,m,sum=0;
void main()
{
for(i=0;i<100;i++)
{
grp[i].k=0;
}
scanf("%d",&t);
for(i=0;i<t;i++)
{
scanf("%d",&n);
for(j=0;j<=n;j++)
{
scanf("%d",&kk);
arr[j]=kk;
z=0;
if(j!=0)
{
p=grp[j-1].k;
for(c=0;c<=p;c++)
{
grp[j].a[z][0]=j;
grp[j].a[z][1]=grp[j-1].a[c][1];
grp[j].b[z]=grp[j-1].b[c]+arr[grp[j-1].a[c][0]]*arr[j];
z++;
grp[j].a[z][0]=grp[j-1].a[c][0];
grp[j].a[z][1]=j;
grp[j].b[z]=grp[j-1].b[c]+arr[grp[j-1].a[c][1]]*arr[j];
z++;
}
z=z-1;
grp[j].k=z;
}
else
{
grp[j].k=0;
grp[j].b[0]=0;
grp[j].a[0][0]=0;
grp[j].a[0][1]=0;
}
}
j--;
for(m=0;m<=grp[j].k;m++)
{
sum+=grp[j].b[m];
}
printf("%d\n",sum);
sum=0;
}
}
1
accept rate: 0%
# include<stdio.h>
struct code
{
int y;
int a[100][2];
int b[100];
int k;
}grp[100];
int arr[100],t,n,kk,j,p,c,a,i,z,m,sum[100];
void main()
{
for(i=0;i<100;i++)
{
grp[i].k=0;
}
scanf("%d",&t);
for(i=0;i<t;i++)
{
scanf("%d",&n);
for(j=0;j<=n;j++)
{
scanf("%d",&kk);
arr[j]=kk;
z=0;
if(j!=0)
{
p=grp[j-1].k;
for(c=0;c<=p;c++)
{
grp[j].a[z][0]=j;
grp[j].a[z][1]=grp[j-1].a[c][1];
grp[j].b[z]=grp[j-1].b[c]+arr[grp[j-1].a[c][0]]*arr[j];
z++;
grp[j].a[z][0]=grp[j-1].a[c][0];
grp[j].a[z][1]=j;
grp[j].b[z]=grp[j-1].b[c]+arr[grp[j-1].a[c][1]]*arr[j];
z++;
}
z=z-1;
grp[j].k=z;
}
else
{
grp[j].k=0;
grp[j].b[0]=0;
grp[j].a[0][0]=0;
grp[j].a[0][1]=0;
}
}
j--;
for(m=0;m<=grp[j].k;m++)
{
sum[i]+=grp[j].b[m];
}
}
for(i=0;i<t;i++)
printf("%d\n",sum[i]);
} Please tell me what's the problem in this code. It's running fine in my compiler.
1
accept rate: 0%
# include<stdio.h>
struct code
{
int y;
int a[100][2];
int b[100];
int k;
}grp[100];
int arr[100],t,n,kk,j,p,c,a,i,z,m,sum[100];
void main()
{
for(i=0;i<100;i++)
{
grp[i].k=0;
}
scanf("%d",&t);
for(i=0;i<t;i++)
{
scanf("%d",&n);
for(j=0;j<=n;j++)
{
scanf("%d",&kk);
arr[j]=kk;
z=0;
if(j!=0)
{
p=grp[j-1].k;
for(c=0;c<=p;c++)
{
grp[j].a[z][0]=j;
grp[j].a[z][1]=grp[j-1].a[c][1];
grp[j].b[z]=grp[j-1].b[c]+arr[grp[j-1].a[c][0]]*arr[j];
z++;
grp[j].a[z][0]=grp[j-1].a[c][0];
grp[j].a[z][1]=j;
grp[j].b[z]=grp[j-1].b[c]+arr[grp[j-1].a[c][1]]*arr[j];
z++;
}
z=z-1;
grp[j].k=z;
}
else
{
grp[j].k=0;
grp[j].b[0]=0;
grp[j].a[0][0]=0;
grp[j].a[0][1]=0;
}
}
j--;
for(m=0;m<=grp[j].k;m++)
{
sum[i]+=grp[j].b[m];
}
}
for(i=0;i<t;i++)
printf("%d\n",sum[i]);
}
1
accept rate: 0%
0 Please tell me what is wrong with this solution in the below given link : https://www.codechef.com/viewsolution/12927725 answered 25 Feb '17, 17:08 4★singh_8 1 accept rate: 0%
0 While submitting my solution for this problem, I am getting compile error and if I click on the link to find out what the compilation error is, then I am getting to a page where it says 'There was an error while serving this page. Please try again after some time'. Is there anyone who faced a similar issue? Any ideas? answered 13 Apr '17, 02:50 11 accept rate: 0%
0 hello, i don't understand the solution of 121 even after reading the editorial. in the explanation of the of the answer for 121 i've read the solution to be adding the sum of scores of gameplays taking 121 twice. According to the problem, two gameplays are different if after writing down all the numbers, the sequences of the gameplays aren't identical. for this to be true, 121 has only three gameplays - 121, 112 and 211. these gameplays gave the answer as 12+21 = 4, 11+12 = 3 and 21+11 = 3. This gives an answer of 10. Now, in other explanations the 121 gameplay has been taken twice - once by doing this sequence - 12 -> 1 -> 121 - and the other by doing 21 -> 1 (on the front end) -> 121. In the end, this gameplay can not be taken as two different gameplays as both of them result in the same sequence after writing down (n+1) numbers. So, what's the explanation for the answer? answered 21 Jun '17, 16:01 2★sphd 1 accept rate: 0%
0 Hello, can someone please tell me what's wrong with my code- It is giving right answers for Subtask 1 But giving wrong answer(WA) for Subtask 2 And for Subtask 3, it is giving TLE(As the complexity of my code is O(n^2)), Please tell me why it is giving WA for Subtask 2? #include #include #include #include #include #define MOD 1000000007 #define ull unsigned long long int using namespace std; ull multiply(ull a,ull b) { ull mul=(a%MOD*b%MOD)%MOD; return(mul); } int main() { ull test; cin>>test; while(test) { int n; cin>>n; ull arr[n+1]; ull sum=0,mul=1,exp=1; for(int i=0;i<=n;i++) { cin>>arr[i]; } for(int i=1;i<=n;i++) { exp=multiply(exp,2); } for(int i=0;i
0 Hey can any one check my code also? I am getting correct answer for subtask one but wrong for other two. check my code: #include using namespace std; long long int M=1000000007; unsigned long long int power(long long int n) { unsigned long long sum=1; for(long long int i=0;i>t; while(t--) { unsigned long long int n,a[100010],i,j,sum=0,tot,tot2,pdt1,pdt2; cin>>n; n=n+1; for(int i=1;i<=n;i++) cin>>a[i]; for(i=1;i<=n-1;i++) { if(i==1||i==2) tot=2; else tot=(tot*2)%M; pdt1=(a[i]*tot)%M; //cout<
0 I have slightly easier approach. We can visualize the problem with reference to a binary tree. Starting from the top-most part of the node we can draw a binary tree in such a way that the new number goes to the left of the previous numbers in the left sub-tree and it goes to the right of the previous numbers in the right sub-tree. In the way we just have to eliminate the numbers having same sequence. This is depicted in the below link: https://drive.google.com/file/d/0B9d_mf_aoiYAUUttdDZmOXc1Z2M/view?usp=sharing In this image we can calculate the sum at each levels of the tree i.e. 12=2,21=2,121=4,211=3,112=3. Total is 14. We can ignore the last element of the subtree since it is a repeated sequence. Please upvote my answer if you find this to be helpful. I will put on the code in a short while. answered 26 Oct '17, 11:31 0★tijus 1 accept rate: 0%
0 First I'll say that the percentage success model can be really helpful on this program. You can write a direct-simulation program to solve subtask 1 and with that you can generate test data to support other approaches. My solution runs forward through the cases by maintaining a term that represents the mix of exposed ends of the game string. This is initially $t_0 = 2 a_0$ - two sides of the first placement - then $t_1 = 2(a_0+a_1)$, the two exposed ends of the two possible games of $n=1$ - then $t_2 = 2(a_0+a_1) + 4a_2$, $t_3=2(a_0+a_1) + 4a_2 +8a_3$, etc. Each time the next move set sums across all possibilities to $s_k = a_k t_{k-1}$, with the accumulated game score also including twice the sum of previous game scores, $g_k = 2g_{k-1} + s_k$ since there are two possibilities for this move. answered 31 Aug '18, 21:21 5★joffan 948●8 accept rate: 13%
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question asked: 01 Jan '16, 12:44
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{}
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Previous: , Up: Breakpoint Conditions [Contents][Index]
#### 5.9.9 The Action Variables
In this section we list the possible values of the debugger action variables, and their meaning.
Note that the Prolog terms, supplied as values, are copied when a variable is set. This is relevant primarily in the case of the proceed/2 and flit/2 values.
Values allowed in the show condition:
print
Write using options stored in the debugger_print_options Prolog flag.
silent
Display nothing.
display
Write using display.
write
Write using writeq.
write_term(Options)
Write using options Options.
Method-Sel
Display only the subterm selected by Sel, using Method. Here, Method is one of the methods above, and Sel is a subterm selector.
Values allowed in the command condition:
ask
Ask the user what to do next.
proceed
Continue the execution without interacting with the user (cf. unleashing).
flit
Continue the execution without building a procedure box for the current goal (and consequently not encountering any other ports for this invocation). Only meaningful at Call ports, at other ports it is equivalent to proceed.
proceed(Goal,New)
Unless at call port, first go back to the call port (retry the current invocation; see the retry(Inv) command value below). Next, unify the current goal with Goal and execute the goal New in its place. Create (or keep) a procedure box for the current goal.
This construct is used by the ‘u’ (unify) interactive debugger command.
Both the Goal and New arguments are module name expanded when the breakpoint is added: the module of Goal defaults to the module of the current goal, while that of New to the module name of the breakpoint specification. If the command value is created during run time, then the module name of both arguments defaults to the module of the current goal.
The term proceed(Goal,New) will be copied when the command action variable is set. Therefore breakpoint specs of form
Tests - [goal(foo(X)),…,proceed(_,bar(X))]
should be avoided, and
Tests - [goal(foo(X)),…,proceed(foo(Y),bar(Y))
should be used instead. The first variant will not work as expected if X is non-ground, as the variables in the bar/1 call will be detached from the original ones in foo/1. Even if X is ground, the first variant may be much less efficient, as it will copy the possibly huge term X.
flit(Goal,New)
Same as proceed(Goal,New), but do not create (or discard) a procedure box for the current goal. (Consequently no other ports will be encountered for this invocation.)
Notes for proceed/2, on module name expansion and copying, also apply to flit/2.
raise(E)
Raise the exception E.
abort
Abort the execution.
retry(Inv)
Retry the most recent goal in the backtrace with an invocation number less or equal to Inv (go back to the Call port of the goal). This is used by the interactive debugger command ‘r’, retry; see Debug Commands.
reexit(Inv)
Re-exit the invocation with number Inv (go back to the Exit port of the goal). Inv must be an exact reference to an exited invocation present in the backtrace (exited nondeterminately, or currently being exited). This is used by the interactive debugger command ‘je’, jump to Exit port; see Debug Commands.
redo(Inv)
Redo the invocation with number Inv (go back to the Redo port of the goal). Inv must be an exact reference to an exited invocation present in the backtrace. This is used by the interactive debugger command ‘jr’, jump to Redo port; see Debug Commands.
fail(Inv)
Fail the most recent goal in the backtrace with an invocation number less or equal to Inv (transfer control back to the Fail port of the goal). This is used by the interactive debugger command ‘f’, fail; see Debug Commands.
Values allowed in the mode condition:
qskip(Inv)
Quasi-skip until the first port with invocation number less or equal to Inv is reached. Having reached that point, mode is set to trace. Valid only if \Inv \geq 1 and furthermore \Inv \leq \CurrInv for entry ports (Call, Redo), and \Inv < \CurrInv for all other ports, where CurrInv is the invocation number of the current port.
skip(Inv)
Skip until the first port with invocation number less or equal to Inv is reached, and set mode to trace there. Inv should obey the same rules as for qskip.
trace
Creep.
debug
Leap.
zip
Zip.
off
Continue without debugging.
Send feedback on this subject.
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# I Complex conjugation in scalar product?
Tags:
1. Oct 12, 2016
### ATY
Hey guys,
I got the following derivation for some physical stuff (the derivation itself is just math)
http://thesis.library.caltech.edu/5215/12/12appendixD.pdf
I understand everything until D.8.
After D.7 they get the eigenvalue and eigenvectors from ε. The text says that my δx(t) gets aligned in the same direction as the the eigenvektor of the largest eigenvalue (this would be my explanation for the fact that there is no eigenvector in the scalar product, but this might be wrong, cause I do not know much about eigenvectors).
But what I don't get is why in D.8. my δx(t) is suddenly complex conjugated. I can not find the reason for this.
I would be really happy about any explanation.
have a nice day
ATY
PS: sorry for the weird titel. Had no clue how to describe my problem (mea culpa)
2. Oct 17, 2016
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# Ubuntu – How to reach a host in the same network by hostname
avahidnshostnamenetworkingserver
I have a network which connects a couple of hosts.
I would like to be able to reach other hosts from one of them using the hostname.
I just discovered the ".local" domain, which is available through avahi and /etc/nsswitch.conf.
But it's not set up this way on my servers.
And I don't want to create an entry in my dns server.
Is there an other way to do so ?
hosts: files mdns4_minimal [NOTFOUND=return] dns mdns4
Just install avahi-daemon on the machines you want talking to eachother, and you should be set to use HOSTNAME.local like this:
ssh yourmachine.local
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# Analog Filter Prototype to Direct Form Second Order Cascade
I've been using the Faust programming language a lot lately for experimenting with DSP, and I've been digging into their implementation of an Elliptic (Cauer) Lowpass filter. Here's an example of such a filter with order three.
If you're unfamiliar with Faust, hopefully the example there is still fairly clear: they've designed a 3rd order Cauer filter with the properties listed in the comments and have implemented it as a second order direct form filter feeding into a first order direct form filter with the coefficients as listed.
I understand the analog prototype they've designed with [z,p,g] = ncauer(Rp,Rs,3); in Matlab/Octave, and I think I have a good understanding of how to factor the transfer function into cascaded first- and second-order filters. What I don't understand is their use of poly in Octave to find the coefficients there, and why the frequency of this elliptic filter is only governed by the last coefficient in the final first order filter? Intuitively it seems like so many more of those coefficients should depend on the frequency of the filter, especially if we have to consider multiple sample rates. Did they skip some steps here or make some assumptions that I'm missing? Any explanation to my confusion here would be greatly appreciated, thank you!
Update: I find it interesting that in Matlab/Octave I can compute [z,p,g] = ncauer(0.2,60,3); sos=zp2sos(z, p, g), deriving coefficients for second order sections without ever specifying the cutoff frequency. Maybe that's where I'm most confused?
• You should better ask on the Faust users list here: lists.sourceforge.net/lists/listinfo/faudiostream-users – sletz Mar 3 '17 at 15:17
• Where in the code do you see that only one coefficients of the first order filter determines the cut-off frequency? – Matt L. Mar 3 '17 at 17:23
• @sletz good point, I will do that. I would also appreciate an answer here in case the list is unresponsive! – ncthom91 Mar 3 '17 at 22:38
• @MattL. So the line lowpass3e(fc) = tf2s(b21,b11,b01,a11,a01,w1) : tf1s(0,1,a02,w1) defines a function which takes 1 parameter, fc, and the output is a 2nd order filter (tf2s) feeding into a 1st order filter (tf1s). Their coefficients there are specified and defined right underneath in the with block. The only reference to the input fc argument is in the definition of the w1 coefficient, the last one supplied to tf1s. All of the other coefficients are constant and thus unrelated to fc it seems. – ncthom91 Mar 3 '17 at 22:41
• i dunno the syntax, but it looks like both tf2s() and tf1s() have w1 as an argument, so i would expect that both the second-order and first-order IIR filters have coefficients that depend on w1 (and then fc.) – robert bristow-johnson Mar 4 '17 at 3:24
The coefficients in the function lowpass3() are the numerator and denominator coefficients of the two sections (first and second order) of the normalized continuous-time transfer function. The continuous-time transfer function looks like this:
$$H(s)=\frac{b_{21}s^2+b_{11}s+b_{01}}{s^2+a_{11}s+a_{01}}\cdot\frac{1}{s+a_{02}}\tag{1}$$
Note that this normalized filter has a cut-off frequency of $1$. The two functions tf2s and tf1s transform these two sections to the discrete-time domain using the bilinear transform, whereby the normalized cut-off frequency of the continuous-time filter is mapped to the desired cut-off frequency in the discrete-time domain. So whereas the coefficients in $(1)$ are independent of the cut-off frequency (because the filter is normalized), the coefficients of both final filter sections in the discrete-time domain do depend on the cut-off frequency.
Your example of using ncauer followed by zp2sos will not give you the coefficients of the discrete-time filter with the desired cut-off frequency. Instead, you just the compute the coefficients of the first and second order sections of the normalized continuous-time low pass filter, i.e., the coefficients hard-coded in lowpass3(). What you miss in that case is the transformation step from continuous-time to discrete-time, and the mapping of the normalized cut-off frequency to the desired cut-off frequency.
• Thank you this is very helpful. So because we're dealing with a normalized continuous-time transfer function, does that mean the analog cutoff frequency is 1? And the w1 parameter defines the mapping of the analog cutoff frequency to the digital cutoff frequency? – ncthom91 Mar 4 '17 at 17:17
• @ncthom91: Yes, that's why we can have the same coefficients as a basis for all filters. The desired cut-off frequency of the discrete-time filter is realized by adapting a constant in the bilinear transformation. – Matt L. Mar 4 '17 at 17:28
• I see, ok! I almost have all of this down I think. I didn't realize that the coefficients supplied to lowpass3e() were the coefficients of the analog polynomial; so the use of poly here is because the poles and zeros from ncauer are defined as complex conjugate pairs, and we need the roots? I'm trying to follow along in Octave and I can almost put it all together but I don't exactly understand how they found the coefficients here. – ncthom91 Mar 4 '17 at 17:32
• @ncthom91: Just look at ncauer and/or read this, to figure out how the poles and zeros of an analog Chebyshev filter are computed. As soon as you have the poles and zeros, you can find the corresponding polynomial coefficients by using poly. – Matt L. Mar 4 '17 at 17:36
• This has been so helpful, thank you. I think the last part is a bit of confusion about how you factored the sections in the transfer function you wrote above. Or, how one decides that factorization when designing these sections themselves. I've found a few resources that might be of help but I'd appreciate your thoughts as well. Either way, thank you, I'll mark this as the correct answer! – ncthom91 Mar 4 '17 at 18:59
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## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)
$v_a = v_b = v_c$
Let's consider ball a. We can use conservation of energy to find the speed of the ball $v_f$ when the ball reaches the ground. $PE_f+KE_f = PE_0+KE_0$ $0+\frac{1}{2}mv_f^2 = mgh+\frac{1}{2}mv_0^2$ $v_f^2 = 2gh+v_0^2$ $v_f = \sqrt{2gh+v_0^2}$ We can use the same method to find the speed of ball b and ball c when they reach the ground. Since the equation is exactly the same for all three balls, the speed of all three balls will be the same when they reach the ground. Therefore; $v_a = v_b = v_c$
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## anonymous one year ago The cost in dollars of producing x units of a particular brand of notebook is C(x) = x^2 – 400. a)Find the average rate of change of C with respect to x when the production level is changed from x = 100 to x = 102. Include units in your answer. b)Find the instantaneous rate of change of C with respect to x when x = 100. Include units in your answer.
1. anonymous
really could use some help on this on !!!
2. anonymous
How do you think you should approach this? what do the terms average and instantaneous mean to you?
3. anonymous
well i understand the meaning of the terms im just not too sure how to go about obtaining them, for starters on problem a i know the formula but after i have the equation of the average rate of change what do i do?
4. anonymous
im getting (x^2 -9604)/(x-102) for the average rate of change equation
5. anonymous
$average= \frac{ \Delta Y }{ \Delta X }$ You need to solve for Y at each X and then use this equation
6. anonymous
solve for y where ?!?
7. anonymous
eabollich
8. anonymous
C(x) is another way of writing Y in the equation of a line y=mx +b. Are you familiar with this equation?
9. anonymous
yes
10. anonymous
c(100) = 9600 c(102) = 10004 so... (9600-10004)/(100 - 102)?
11. anonymous
Yes. the negative signs cancel out. the traditional way would be to put 10004-9600 and 102-100 but as long as you are consistent, it doesn't make a difference
12. anonymous
Is this for calculus or algebra?
13. anonymous
calculus ab
14. anonymous
and im starting to think this is wrong
15. anonymous
http://home.earthlink.net/~peggyfrisbie/images/avg%20rate%20of%20change%20back.jpg The average rate of change is the same for calculus and algebra. The instantaneous rate is where things are easier if you are using calculus. http://www.thebestschools.org/wp-content/uploads/2014/08/calculus2.jpg
16. anonymous
Why do you think it is wrong?
17. anonymous
because we didn't use an equation i was certain would be relevant
18. anonymous
oh wait
19. anonymous
you just proved me wrong :p
20. anonymous
I changed the variables to X and Y but it is the same equation :)
21. anonymous
okay and for part b do i just find the derivative at that point?
22. anonymous
I assumed at first that it was an algebra class so I used X and Y, sorry for the confusion
23. anonymous
I haven't taken Calculus but I assume from what I am reading that you are correct. The third example on this website gives a good example of what that looks like: http://www.millersville.edu/~bikenaga/calculus/tangent/tangent.html
24. anonymous
thank's !!
25. anonymous
You're welcome! Glad I could help!
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## Why I use Linux
This is a re-post of a blog post I wrote for my department blog couple years ago. I just read it again and I found it still relevant.
For those of you who have been to my office or have taken my classes will know that I am a big Linux fan. I started to use a little bit of Linux since early 2000. But I was “converted” completely around 2007 or 2008. Since then, Linux is my main OS. But first things first, I want to emphasize that I am not an MS hater. I used most of the MS OSes before, from MSDOS, Win3.x, Win95, Win98, WinME (yes,it is a horrible one), Win2000, WinXP, Vista, and Win 7. (Sorry, Win 8. I don’t see that I may own any more new Windows device soon in the future. And Mac, yeah, I know you are cool. But everything I need is offered by Linux already. I am not sure I want to pay the premium for you.)
Even though I only started using Linux as the main OS of my computers for the last several years, I have been exposed to the Unix-like system quite a bit earlier. Back in the days of early 90’s (yeah, I am old, at least quite a bit older than many of you reading this) when I was an undergrad student, workstations were popular and we had to telnet to some mainframe computers to do programming assignments. And interestingly, the Unix commands I learned then are almost identical to the Linux commands I use now.
I guess you see why I am writing all this… because while I see some of you are Linux experts, it just came to me that many of the students I met here have little exposure of Unix-like systems. Of course, I myself am more or less a user only rather than a realexpert. But I think that some experience to Unixlike system is essential to any electrical/computer/tel ecom engineers. I can think of many reasons for that. Just name a few,
1). Linux is ubiquitous. It is used by different devices, from very small embedded systems, to enormous mainframes and super computers.
2). Linux is free. And so do many tools on its platform aswell. Need to typeset your paper? Use Latex/Kile. Need to edit your photo? Use GIMP. Need to dosome math? Use Octave. Of course, some of this software is available on Windows as well. But it is often way easier to install them on Linux than Windows because most of them were developed on theformer natively.
3). Linux is reallyfree. Not to mention that the commonly used application software is free on Linux. Some very sophisticate or specialized software is free there also. Just name a few, ns-2/3 for network simulation, zoneminder for security system, and musescore for music composition and notation. And you can even setup a professional (and completely free) web server at home with Linux in an hour or so.
4). Your future boss wants you to know it. Because of the above, there is a good chance that your future bosses would like someone to at least familiar withthe OS. So being acquaint with Linux will allow you to access to many more possible positions in the job market.
If the above reason is not sufficient for you to start playing with Linux, here is another one: using Linux/Unix-like system makes one more productive. Okay. It sounds a bit subjective and I am not sure if anyone does any statistical study for that. But there are indeed quite a few productivity features that Unix-like system has offered for decades while one is still waiting for MS to catch up. The most noticeable one is virtual desktop. I just can’t understand whyMS still doesn’t implement that one. Another very useful productive feature is packaging systemsfor software installation. Okay, MS finally got this one eventually and we have “apps for windows”… But when I say Linux can make one more productive. The first thing came to my mind are not the above. What is more important to me is that Linux comes with lots of useful built-in command line tools. Yes. I mean old fashion command line tools. I know everyone likes GUI and command line stuffs seem scary and difficult to use. But GUI tools tend to become obsolete much faster than command line tools. As I mentioned earlier, I am still using the commands such as sed/grep/cut/find/tr that I learned 20 years ago. And those commands were introduced like 40 years ago. So your time learning those things won’t be wasted soon in the future. In contrast, every Office version looks a bit different and it takes time to adapt a new one!
To be fair, Windows provide some batch commands also. But they are less powerful and more importantly, far less portable outside the windows world.In contrast, you can run those Linux commands with little modification on a Mac right away. Besides, one can use these tools on Windows too if they really want to. For example, by installing a Linux layer, such as Cygwin, on top of windows. But honestly, it just makes life complicated.
Actually, not just command line tools are available on Linux, everything needs to be done can be done with them alone. By design, GUI tools on Linux are really just wrappers for these “primitive” commands. Therefore, one can always fall back to these commands and get things done especially when resources are really limited. For example, at this right moment, I am connecting to my office desktop using ssh even though I am in a time zone 13 hours away from Tulsa. If I had a windows box, the remote connection would definitely be unusable as the GUI stuffs would have sucked up so much resources that the poor connection here just couldn’t afford it.
1 Okay, Cygwin may seem like a godsend for Windows veteranswho want to start playing with Linux. But frankly, I would not recommend Cygwin to anyone Linux newbies. Cygwin is far less user friendly than modern Linux distros like Ubuntu. One canencounter way more problems installing and running any software on Cygwin than on distro like Ubuntu.
For those who are still not convinced, let me illustrate the power of Linux with an example task that I do occasionally. From time to time, my collaborator generates many figures in eps format but we actually need them in pdf format instead. And often we have the files scattered in several subfolders (say, under the current folder) and I have to convert all the files. Imagine that if someone were using Windows, first he or she would need to figure out how to convert eps file to pdf format as there is no built-in tools for that. Maybe someone eventually found some software (most likely not free) but then he would still need to find a way to convert all of them efficiently. If he was very lucky, maybe there was a batch conversion function offered by the software. But at many times, he would have ended up wasting an hour more converting the files one by one manually even after paying like $20 for the software. On a contrary, if you use Linux and are knowledgeable with some Linux commands, you are really lucky in such situation. All the hassle can be done ina single line: find . -name “*.eps” -print | sed ‘s/$$.*$$\.eps/ps2pdf “\1.eps”“\1.pdf/”’ | sh The command seems complicated but is actually very easy to understand. It is really just a cascade of three commands linked by the pipe character “|”. The firstcommand, find . -name “*.eps” -print, simply search for all eps files under the current folder including all the subfolders. The second command, sed ‘s/$$.*$$\.eps/ps2pdf “\1.eps” “\1.pdf/”’, tries to replace each eps file, say with a name ‘XXX.eps’, by a command string ‘ps2pdf “XXX.eps” “XXX.pdf” ‘. And the third command, sh, simply executes the command string created in the previous step. Okay, this definitely is getting too long and I probably should stop here. And this finishes my advice (rant) of why all engineering students should know (and use) Linux. Actually, I also believe all graduate engineering students should use Latex (instead of Words) for their papers. But maybe I will write about it some other time. ## Playing StarCraft on Ubuntu 14.04 and X230 I recently got a used X230 tablet and tried to have some fun on it. StarCraft runs well with wine but cannot run on full screen. One possible solution is to open another X server following this post. But I need some tweaks for 14.04 and for touch and stylus to work. First, one needs to add Section "ServerLayout" Identifier "SCLayout" Screen 0 "StarCraft Screen" InputDevice "Keyboard0" "CoreKeyboard" InputDevice "Mouse0" "CorePointer" EndSection Section "Screen" Identifier "StarCraft Screen" Device "Device0" Monitor "StarCraft Monitor" DefaultDepth 24 SubSection "Display" Virtual 640 480 Depth 24 Modes "640x480@60" "1280x800@50" EndSubSection EndSection Section "Monitor" Identifier "StarCraft Monitor" VendorName "Plug 'n' Play" ModelName "Plug 'n' Play" Gamma 1 ModeLine "640x480@60" 25.2 640 656 752 800 480 490 492 525 -hsync -vsync EndSection to /etc/X11/xorg.conf. However, xorg.conf is no longer there by default. One can generate one following this. One would also need to modify the line allowed_users=console in /etc/X11/Xwrapper.config to allowed_users=anybody For audio to work, one need to add current user to the audio group. Run sudo usermod -a -G audio$USER
Some posts mentioned that one needs to reboot. But I am not certain it is a must.
Finally, create the following script and run it.
#!/bin/sh
X :1 -layout SCLayout -ac &
XPID=$! sleep 2 xbindkeys --display :1 -f$HOME/.scbind
#pax11publish -D :1 -e
DISPLAY=:1 xsetwacom --set 10 area "3454 -41 24103 15420"
DISPLAY=:1 xsetwacom --set 11 area "348 0 2432 1569"
DISPLAY=:1 xsetwacom --set 15 area "3454 -41 24103 15420"
DISPLAY=:1 xsetwacom --set 10 TabletPCButton "off"
DISPLAY=:1 wine $HOME/tmp/StarCraft/StarCraft.exe -- /usr/bin/X :1 -layout SCLayout sleep 1 kill$XPID
I assume StarCraft folder is under \$HOME/tmp. Modify it accordingly.
The xsetwacom lines attempt to “calibrate” the touch screen and stylus accordingly. These numbers should work well for X230. For other tablet, You may obtain this number by first changing the resolution to 800×600. Then run “Wacom Tablet”->”Calibrate…” and extract the number by typing
xsetwacom --get 10 area
Note that device “11” is the “finger touch” and ubuntu’s calibration function will only work for the stylus. But one can get the number easily by scaling the number obtained above. For example, dividing each number for devices 10 and 15 by 10 probably will work well for 11 (finger touch). But I didn’t try any other tablet pc other than X230.
## Goagent experience
I am in Shanghai a couple days and it is quite inconvenient with many websites blocked. I heard about “breaking” the wall but I didn’t realize it is quite fast and easy. All one needs is a software called goagent. It takes advantage of the google app engine. Here, I will assume that ones already have created a google app engine application before. If not, it is easy to do and there are many guides for that.
I am using it on my linux box LTS 12.04. The goagent team has created a very good guide on its site. I have encountered few problems following it. It appears that one can’t install gevent automatically for 12.04 and have to use the manual route. And I didn’t try to start the proxy automatically. I just run it in a terminal using
python proxy.py
instead. In any case, I will only be here for a couple days. I don’t really see any inconvenience running it from terminal rather than having it to start up automatically.
In terms of browser setup, I use chromium (chrome should work too) and SwitchySharp. One would want to look for an option file SwitchyOptions.bak and import it directly to SwitchySharp. It can be easily found with a search.
The only issue I had is that it didn’t load facebook and youtube correctly. It appeared that it was due to some certification problems. But it can be fixed easily following this post.
## M/M/1 Simulation with Matlab
A simple simulation of M/M/1 queue with Matlab. The distribution of the number of “packets” in the system is computed and compared with the theoretical result.
delta=0.1; % simulation step in sec
lambda=0.1; % arrival rate in packets per second
mu=0.2; % departure rate in packets per second
rho=lambda/mu;
M=50*3600/delta; % number of simulation step for 50 hours
a=zeros(M,1); % arrival log
d=zeros(M,1); % departure log
n=zeros(M,1); % number of packets at each simulation time step
n_cur=0; % current number of packets in the system
for i=1:M
n_cur; % # packets before the process
n(i)=n_cur; % record the # of packet before the process
if rand < lambda*delta % probability of an arrival occurred = lambda * delta
a(i)=1;
n_cur=n_cur+1;
end
if rand < mu*delta && n_cur > 0 % probability of departure occurred = mu * delta provided that there is a packet in the system
d(i)=1;
n_cur=n_cur-1;
end
if mod(i,1000)==0 % show progress for every 1000 simulation steps
i
end
end
figure;
edges=0:10;
count=histc(n,edges);
bar(0:length(count)-1,count/sum(count));
hold;
nlist=0:length(count)-1;
plot(nlist, (1-rho)*rho.^(nlist),'r-+');
legend('experiment','theory');
title('Distribution of number of packets in the system');
Now, let say if we want to find the delay distribution for the packet arrived when there are 2 packets in the system.
figure;
ainds=find(a==1); % the indices when we have an arrival
atimes=ainds*delta; % arrival times in seconds
dtimes=find(d==1)*delta; % departure times in second
delays=dtimes-atimes(1:length(dtimes)); % list of delays in second
inds=find(n(ainds)==2); % find indices of packets arrived when # packets in system is 2
h=histc(delays(inds),0:60);
bar(0:60,h/sum(h));
hold;
ezplot(mu^3/2*t^2*exp(-mu*t),[0,60]);
legend('experiment','theory');
title('Delay distribution for packets arrived when # packets in system is 2');
xlabel('Time in second');
ylabel('Probability');
## Simulating Poisson process in Matlab
Below is a simple Matlab code to simulate a Poisson process. The interarrival times were computed and recorded in int_times. The times are then grouped into bins of 10 seconds in width and the counts are stored in count.
lambda=1/60; % arrival rate per second (1 minute per packet)
T=10*3600; % simulation time in second (10 hours)
delta=0.1; % simulation step size in second
N=T/delta; % number of simulation steps
event=zeros(N,1); % array recording at each step if a "packet" arrived.
% initialize it to zeros
R=rand(size(event)); % generate a random array (with elements in [0,1]) of the same size as "event"
event(R<lambda*delta)=1; % set each element of event to 1 with probability lambda*delta
inds=find(event==1); % getting indices of arrivial
int_times=diff(inds)*delta; % interarrival times in seconds
edges=0:10:400; % define histogram bin
count=histc(int_times,edges);
The histogram of the absolute counts of the interarrivial is then plotted with the commands below.
figure; bar(edges,count,'histc'); % draw histogram of absolute count
Finally, the counts are normalized and compared with the theoretical result ($\lambda \exp(-\lambda T)$).
figure; bar(edges,count/sum(count)/(edges(2)-edges(1)),'histc'); % draw histogram of normalized counts
hold;plot(edges,lambda*exp(-lambda*edges),'r'); % plot theoretical result
legend('simulation','theoretical');
## ibus jyutping
As a cantonese from HK, I didn’t know how to “type from my mouth” as a kid. I learned Cangjie during high school but I have never become really proficient with it. Chinese is not exactly a phonetic language. But when I write IM, email, or just blog in Chinese. I found it much better to type phonetically as I don’t need to switch my thought to letter ideograph. When I write to mandarin friends, pinyin is great. However, when I chat to friends back home in HK, my message will have “accent” when I try to type in pinyin since I try to “speak” my thought in mandarin when I type.
I finally ended up to have the will to learn jyutping last year. One of the romanization system for Cantonese. I won’t say it is a perfect system (I know it is a standard for linguist but I still found it quite weird to use j for y). Both android and windows have decent input method for jyutping. As for Ubuntu…, there is a jyutping for ibus. But saying it to be horrible is actually a compliment. It is essentially unusable as many common words are missing. And the table is sorted in an absolutely ridiculous way with the top of the table starting with words that no native speakers will ever use.
Fortunately, ibus-table does come with a function to allow users create their own tables. After getting character frequency from Chih-Hao Tsai’s site and jyutping phonetic from HanConv, at least I can build a usable table for myself. There is still lots of tweakings to do though as HanConv does not cover the sounds for all the characters from Chih-Hao Tsai. And some of the pronunciations seem incorrect or at least uncommon. I guess I will make another post to share the final table when it becomes more mature.
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# AUCTeX folding and square brackets in math mode
I want to use the folding functionality of AUCTeX, mainly to replace math macros with the corresponding symbols. But at the same time I want to continue to use square or curly brackets additionally to parentheses. But AUCTeX hides then square or curly brackets and their content.
So if I write:
\documentclass{article}
\begin{document}
$a \to [b \to c]$
\end{document}
and then apply TeX-fold-buffer I get as result:
\documentclass{article}
\begin{document}
$a →$
\end{document}
What I would like to get is:
\documentclass{article}
\begin{document}
$a → [b → c]$
\end{document}
How can I prevent AUCTeX folding from hiding square or curly brackets and their content in math mode?
• Actually $a [b \to c]$ is folded as expected to $a [b → c]$, so it isn't a problem with square brackets but perhaps a conflict between the \to and the brackets. – giordano Jul 7 '14 at 9:34
• @giordano: It seems to be a conflict between any folded math macro and square and curly brackets. For every math macro in LaTeX-fold-math-spec-list I tested the problem appears. – newtothis Jul 7 '14 at 14:12
• You're right. I found the problem: folding relies on TeX-find-macro-end-helper to detect the end of the macro to be folded, but this considers square brackets and curly braces as part of the macro, so they will be always folded when they follow a math macro. I need to think about a solution (probably a redefinition of TeX-find-macro-end-helper). Thank you for having spotted this issue! – giordano Jul 7 '14 at 15:17
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# Guessing the K matrix gain for the Optimal Control LQR?
I'm are going to create a LQR to control a system. The problem is to choose the Q and R weighting matrices for the cost function. The Q and R matrices are going to minimize the cost function so the system are going to be optimal.
I'm using Scilab and Scilab have a good library for optimal control. Scilab have a built function named lqr() to compute the gain matrix K, which is the LQ regulator.
But the problem is still to choose those weighting matrices. I don't know here to start. I just might start with the identity matrix as Q and just a constant as R. But the gain matrix K does not make my model go smooth. No one can say that this is the real Q and R weighting matrices for the system. As a developer, I choose the weighting matrices. But why should I do that when I can choose the K gain matrix directly?
So I just made up my own numbers for the gain matrix K and now my model is very smooth. All I did just do was to guess some numbers for the gain matrix K and simulate and look at the result. Was it still bad, I might change the first element for the gain matrix to increase the position, or change the second element in the gain matrix K, to speed up the velocity for the position.
This works great for me! Guessing and simulate and look at the results. I choosing the LQ-technique for two main reasons: It gives multivariable action and can reduce noise by using a kalman filter. A PID cannot do that.
But here is my question:
Will this method give me an optimal control just by guessing the gain matrix K and changing the values depending how the simulation results looks like?
I'm I happy with the results, I might quit there and accept the gain matrix K as the optimal LQR for the system.
• If you (randomly) guess a K matrix you run the risk that your system becomes unstable, while LQR will always return a stabilizing state feedback controller. And how does your state space model look like, because a state space realisation of a dynamical is not unique, which could lead to not very meaningful states, such that identity for Q might not give nice results. – fibonatic Jun 14 '17 at 21:40
• Ok. But how do i choose Q and R? I know there is a cost function $J = \int_{0}^{T} (x^{T}(t)Qx(t) + R^{T}u(t)R) dt$ to use, but I don't $x(t)$ and $R$ and $Q$. The signal $u(t)$ is know as a constant. So if I don't know $Q$ and $R$, I just might guessing the gain matrix K. – Daniel Mårtensson Jun 15 '17 at 20:19
If you want to avoid having to deal with the non-uniqueness of a state space model representations of LTI systems and want to have some more control of the impact that changing $Q$ has on the resulting controller, then you could define the cost you want to minimize as
$$J(u) = \int_0^\infty\left(y^\top\!(t)\, \hat{Q}\, y(t) + u^\top\!(t)\, \hat{R}\, u(t) + 2\,y^\top\!(t)\, \hat{N}\, u(t)\right) dt.$$
If your state space model is of the form
$$\left\{\begin{array}{l} \dot{x} = A\, x + B\, u \\ y = C\, x + D\, u \end{array}\right.$$
then you can rewrite the cost function into a form which can be solved with LQR as follows
$$J(u) = \int_0^\infty\left(x^\top\!(t)\, Q\, x(t) + u^\top\!(t)\, R\, u(t) + 2\,x^\top\!(t)\, N\, u(t)\right) dt,$$
with
$$Q = C^\top \hat{Q}\, C,$$
$$R = \hat{R} + D^\top \hat{Q}\, D + D^\top \hat{N} + \hat{N}^\top D,$$
$$N = C^\top \hat{N} + C^\top \hat{Q}\, D.$$
Now you could start with $\hat{R}$ equal identity, $\hat{N}$ equal zero and $\hat{Q}$ diagonal (with only positive numbers on the diagonal), for example identity. Now by doing a simulation of your closed-loop system using the obtained LQR controller you can see if you have desired performance. If for example the $k$th output decays too slowly, you could increase the value of the $k$th diagonal element of $\hat{Q}$.
If you would also like to guarantee internal stability of the system you could also add an identity matrix multiplied by a small positive number to the expression for $Q$. In order to get some estimate of the size of this small number you could maybe use a couple of orders of magnitude smaller (say $10^{-6}$) than the largest eigenvalue of the initially obtained $Q$.
So using this transformation of $(\hat{Q},\hat{R},\hat{N})\to(Q,R,N)$ should give you some more intuitive tuning nobs when designing a LQR controller. This method should also not be affected by the state space realization you are using of the LTI system.
• How do I compute $x(t)$ ? I know I can use $L^-1[(s*I-A)^-1]$ to compute the state transition matrix, but my Matlab cannot handle more that 4x4 systems. – Daniel Mårtensson Jun 24 '17 at 23:29
• Do I need to compute the state transition matrix? – Daniel Mårtensson Jun 25 '17 at 12:09
• @DanielMårtensson All you need to provide to MATLAB (the lqr function) are the $A$ and $B$ of your state space model and the obtained values for $Q$, $R$ and $N$. – fibonatic Jun 25 '17 at 15:16
• and the $Q$ and $R$, I get from the cost function? – Daniel Mårtensson Jun 25 '17 at 16:18
• Can I not guessing the $Q$ and $R$ and then simulate and look if it's giving me good or bad resluts? – Daniel Mårtensson Jun 25 '17 at 16:26
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2018 CAP Congress / Congrès de l'ACP 2018
10-16 June 2018
Dalhousie University
America/Halifax timezone
Welcome to the 2018 CAP Congress Program website! / Bienvenue au siteweb du programme du Congrès de l'ACP 2018!
Exclusive Backward-Angle Meson Electroproduction -- Unique access to $u$-channel physics
12 Jun 2018, 14:15
15m
SUB 224 (cap.50) (Dalhousie University)
SUB 224 (cap.50)
Dalhousie University
Oral (Non-Student) / Orale (non-étudiant(e)) Nuclear Physics / Physique nucléaire (DNP-DPN)
Speaker
Garth Huber (University of Regina)
Description
Exclusive meson electroproduction at different squared four-momenta of the exchanged virtual photon, $Q^2$, and at different four-momentum transfers, $t$ and $u$, can be used to probe QCD's transition from hadronic degrees of freedom at long distance scale to quark-gluon degrees of freedom at short distance scale. Backward-angle meson electroproduction was previously ignored, but is anticipated to offer complimentary information to conventional forward-angle meson electroproduction studies on nucleon structure. The results of our pioneering study of backward-angle $\omega$ cross sections through the exclusive $p(e,e'p)\omega$ reaction will be presented. The experiment was performed as part of E01-004 in Jefferson Lab Hall C, with central $Q^2$ values of 1.60 and 2.45 GeV$^2$, and $W$=2.21 GeV. The extracted cross sections were separated into transverse (T), longitudinal (L), and LT, TT interference terms. The data set has a unique coverage of $u\sim$ 0, opening up a new means to study the transition of the nucleon wave function through backward-angle experimental observables. Plans to extend these studies to the $\pi^0$ and $\phi$ channels will also be presented.
Primary authors
Garth Huber (University of Regina) Wenliang (Bill) Li (University of Regina and College of William and Mary)
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Question: DESEq2 error when performing exploratory analysis with no replicates
0
2.9 years ago by
pro_zac0
pro_zac0 wrote:
Hello,
I have recently been trying to analyse 12 non-replicated metagenome samples using DESeq2 as an exploratory analysis to find genes with differing abundance to allow for inferences to be obtained with respect to GO terms and KEGG pathways associated with these genes. I have formatted a counts table from a Bowtie2 mapping into HTSeq gene region count extraction procedure which I can successfully read into DESeq2. However, when I attempt to run the program, I receive the following error (note that I'm running it in parallel, and my data set is very, very large):
mean-dispersion relationship
final dispersion estimates, MLE betas: 12 workers
fitting model and testing: 12 workers
-- replacing outliers and refitting for 2151512 genes
-- DESeq argument 'minReplicatesForReplace' = 7
=-- original counts are preserved in counts(dds)
estimating dispersions
Error in estimateDispersionsGeneEst(objectSub, quiet = quiet, modelMatrix = modelMatrix) :
the number of samples and the number of model coefficients are equal,
i.e., there are no replicates to estimate the dispersion.
use an alternate design formula
Calls: DESeq -> refitWithoutOutliers -> estimateDispersionsGeneEst
In checkForExperimentalReplicates(object, modelMatrix) :
same number of samples and coefficients to fit,
estimating dispersion by treating samples as replicates.
read the ?DESeq section on 'Experiments without replicates'
Execution halted
The R output including headers for my input files as well as the sessionInfo are provided below:
R version 3.2.5 (2016-04-14) -- "Very, Very Secure Dishes"
Copyright (C) 2016 The R Foundation for Statistical Computing
Platform: x86_64-pc-linux-gnu (64-bit)
R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.
Natural language support but running in an English locale
R is a collaborative project with many contributors.
'citation()' on how to cite R or R packages in publications.
Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for an HTML browser interface to help.
Type 'q()' to quit R.
> ############## START
>
> # Load deseq2 and set working directory
> library(DESeq2)
> library(pheatmap)
> library(RColorBrewer)
> library(BiocParallel)
> setwd("/home/workingdir")
>
> register(MulticoreParam(12))
>
> countData = as.matrix(read.table(file='combined.htseq.counts', header = TRUE, sep = "\t", stringsAsFactors = FALSE, row.names="gene_id"))
AR DW GH1 LS OF STP1 STP2 STP11A STP12B STP13A STP15C WS
OF0002336 0 8 0 0 1744 0 0 0 0 0 0 0
OF0057878 0 0 0 0 1560 0 6 2 4 0 0 0
OF0008217 0 0 0 0 1004 0 0 0 0 0 0 0
STP11A0014293 0 0 2 0 33 2 378 697 1674 294 474 0
LS0032436 0 0 0 981 0 0 0 0 0 1 0 149
OF0055859 0 0 0 0 892 0 0 0 1 1 0 0
>
condition
1 AR
2 DW
3 GH1
4 LS
5 OF
6 STP1
>
> # Create DESeq2 data set
> dds <- DESeqDataSetFromMatrix(countData = countData, colData = colData, design = ~ condition)
> dds <- dds[ rowSums(counts(dds)) > 1, ] # 1.3.6 Pre-filtering - removes genes for which only 0 or 1 reads are present
>
> # Reduce RAM usage
> rm(countData)
> rm(colData)
>
> # Session info
> sessionInfo()
R version 3.2.5 (2016-04-14)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: SUSE Linux Enterprise Server 11 SP4
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=en_US.UTF-8 LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] parallel stats4 stats graphics grDevices utils datasets
[8] methods base
other attached packages:
[1] BiocParallel_1.4.3 RColorBrewer_1.1-2
[3] pheatmap_1.0.8 DESeq2_1.10.1
[7] SummarizedExperiment_1.0.2 Biobase_2.30.0
[9] GenomicRanges_1.22.4 GenomeInfoDb_1.6.3
[11] IRanges_2.4.8 S4Vectors_0.8.11
[13] BiocGenerics_0.16.1
loaded via a namespace (and not attached):
[1] genefilter_1.52.1 locfit_1.5-9.1 splines_3.2.5
[4] lattice_0.20-33 colorspace_1.3-1 htmltools_0.3.5
[7] chron_2.3-47 survival_2.40-1 XML_3.98-1.5
[10] foreign_0.8-66 DBI_0.5-1 lambda.r_1.1.9
[13] plyr_1.8.4 stringr_1.1.0 zlibbioc_1.16.0
[16] munsell_0.4.3 gtable_0.2.0 futile.logger_1.4.3
[19] latticeExtra_0.6-28 knitr_1.15.1 geneplotter_1.48.0
[22] AnnotationDbi_1.32.3 htmlTable_1.7 acepack_1.4.1
[25] xtable_1.8-2 scales_0.4.1 Hmisc_4.0-0
[28] annotate_1.48.0 XVector_0.10.0 gridExtra_2.2.1
[31] ggplot2_2.2.0 digest_0.6.10 stringi_1.1.2
[34] grid_3.2.5 tools_3.2.5 magrittr_1.5
[37] lazyeval_0.2.0 tibble_1.2 RSQLite_1.0.0
[40] Formula_1.2-1 cluster_2.0.5 futile.options_1.0.0
[43] Matrix_1.2-4 data.table_1.9.6 assertthat_0.1
[46] rpart_4.1-10 nnet_7.3-12
>
> # Perform the differential expression (1.4)
> dds <- DESeq(dds, parallel=TRUE)
^^ This is where the code dies
If someone could help me figure out why DESeq2 is not allowing me to run without replicates (as I understand it is capable of despite encouraging you not to take any results as conclusive) that would be appreciated. Thanks.
modified 2.9 years ago by Michael Love25k • written 2.9 years ago by pro_zac0
Answer: DESEq2 error when performing exploratory analysis with no replicates
1
2.9 years ago by
Simon Anders3.6k
Zentrum für Molekularbiologie, Universität Heidelberg
Simon Anders3.6k wrote:
DESeq2 is a tool to test statistical hypotheses, i.e., to calculate p values to quantify how much evidence you have for differential evidence. As you don't have replicates, you can't have p values.
For exploratory data analysis, you could consider using the "rlog" transformation and then just look at the genes with highest variance across conditions.
Hi Simon,
Thanks for your response. From reading elsewhere I was given the impression that running DESeq() normally should still provide me with some level of results rather than crashing, but I may have misinterpreted these comments. I will look into how I can best analyse rlog data in as scientific a manner as possible, then.
Answer: DESEq2 error when performing exploratory analysis with no replicates
1
2.9 years ago by
Michael Love25k
United States
Michael Love25k wrote:
I can't replicate the "Execution halted" using simulated data and the current version of DESeq2 (1.14), but I do see a problem in the code base: here we should not replace outliers because there are no replicates (it only seems so for a period during the dispersion estimation when DESeq2 pretends the different conditions are replicates).
You should be able to bypass this by setting minReplicateForReplace=Inf, and I can fix this in future versions to not try to replace outliers when there are no replicates.
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## Algebra and Trigonometry 10th Edition
$3.78-2.12 > 1.3x^2$ $x^2 < 1.66/1.3=1.277$ $x^2-1.277<0$ $(x-1.13)(x+1.13)<0$ -1.13 < x < 1.13
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## Posts Tagged ‘area’
### How does matlab’s hypot function work?
Saturday, January 7th, 2017
The hypot function in matlab purports to be more numerically stable at computing the hypotenuse of a (right-)triangle in 2D. The example the help hypot gives is:
a = 3*[1e300 1e-300];
b = 4*[1e300 1e-300];
c1 = sqrt(a.^2 + b.^2)
c2 = hypot(a,b)
where you see as output:
c1 =
Inf 0
and
c2 =
5e+300 5e-300
this is a compiled built-in function so you can’t just open hypot to find out what its doing. It might just be pre-dividing by the maximum absolute value. There’s probably a better reference for this, but I found it in: “Vector length and normalization difficulties” by Mike Day.
Continuing this example:
m = max(abs([a;b]))
c3 = m.*sqrt((a./m).^2 + (b./m).^2)
produces
c3 =
5e+300 5e-300
While matlab’s hypot only accepts two inputs, pre-dividing by the maximum obviously extends to any dimension.
### Triangle area in nD
Saturday, January 7th, 2017
My typical go-to formula for the area of a triangle with vertices in nD is Heron’s formula. Actually this is a formula for the area of a triangle given its side lengths. Presumably it’s easy to compute side lengths in any dimension: e.g., a = sqrt((B-C).(B-C)).
Kahan showed that the vanilla Heron’s formula is numerically poor if your (valid) triangle side lengths are given as floating point numbers. He improved this formula by sorting the side-lengths and carefully rearranging the terms in the formula to reduce roundoff.
However, if you’re vertices don’t actually live in R^n but rather F^n where F is the fixed precision floating-point grid, then computing side lengths will in general incur some roundoff error. And it may so happen (and it does so happen) that this roundoff error invalidates the triangle side lengths. Here, invalid means that the side lengths don’t obey the triangle inequality. This might happen for nearly degenerate (zero area) triangles: one (floating point) side length ends up longer than the sum of the others. Kahan provides an assertion to check for this, but there’s no proposed remedy that I could find. One could just declare that these edge-lengths must of come from a zero-area triangle. But returning zero might let the user happily work with very nonsensical triangle side lengths in other contexts not coming from an embedded triangle. You could have two versions: one for “known embeddings” (with assertions off and returning 0) and one for “intrinsic/metric only” (with assertions on and returning NaN). Or you could try to fudge in an epsilon a nd hope you choose it above the roundoff error threshold but below the tolerance for perceived “nonsense”. These are messy solutions.
An open question (open in the personal sense of “I don’t know the answer”) is what is the most robust way to compute triangle area in nD with floating point vertex positions. A possible answer might be: compute the darn floating point edge lengths, use Kahan’s algorithm and replace NaNs with zeros. That seems unlikely to me because (as far as I can tell) there are 4 sqrt calls, major sources of floating point error.
Re-reading Shewchuk’s robustness notes, I saw his formula for triangle area for points A,B,C in 3D:
Area3D(A,B,C) = sqrt( Area2D(Axy,Bxy,Cxy)^2 + Area2D(Ayz,Byz,Cyz)^2 + Area2D(Azx,Bzx,Czx)^2 ) / 2
where Area2D(A,B,C) is computed as the determinant of [[A;B;C] 1]. This formula reduces the problem of computing 3D triangle area into computing 2D triangle area on the “shadows” (projections) of the triangle on each axis-aligned plane. This lead me to think of a natural generalization to nD:
AreaND(A,B,C) = sqrt( ∑_{i<j} ( Area2D(Aij,Bij,Cij)^2 ) / 2
This formula computes the area of the shadow on all (N choose 2) axis-aligned planes. Since the sqrt receives a sum of squares as in argument there’s no risk of getting a NaN. There’s a clear drawback that this formula is O(N^2) vs Heron’s/Kahan’s O(N).
### Signed polygon area in matlab
Sunday, October 30th, 2016
Suppose you have a polygon’s 2D corners stored in P, then the signed area is given by:
signed_polyarea = @(P) 0.5*(P(1:end,1)'*P([2:end 1],2)-P(1:end,2)'*P([2:end 1],1));
### Areas of quadrilaterals within a triangle determined by its circumcenter (or perpendicular bisectors), using matlab
Thursday, February 11th, 2010
First, I need to locate the circumcenter of the triangle. The circumcenter is the center of the circle circumscribing the three points of the triangle. Thus, it is equidistance from all three points.
I do not need the actual world coordinates of the circumcenter, rather it is okay for me to just know the relative distance to each corner. The law of cosines tells me that the relative distance of the circumcenter from the line opposite point A (called a in the picture) is cos A (that is cosine of the angle at A). Likewise for B and C. Knowing these I can convert to barycentric coordinates by multiplying by the opposite edge lengths.
cosines = [ ...
(edge_norms(:,3).^2+edge_norms(:,2).^2-edge_norms(:,1).^2)./(2*edge_norms(:,2).*edge_norms(:,3)), ...
(edge_norms(:,1).^2+edge_norms(:,3).^2-edge_norms(:,2).^2)./(2*edge_norms(:,1).*edge_norms(:,3)), ...
(edge_norms(:,1).^2+edge_norms(:,2).^2-edge_norms(:,3).^2)./(2*edge_norms(:,1).*edge_norms(:,2))];
barycentric = cosines.*edge_norms;
normalized_barycentric = barycentric./[sum(barycentric')' sum(barycentric')' sum(barycentric')'];
A quality of the barycentric trilinear coordinates of a point in a triangle is that they specify the relative area of inner triangles form by connecting the corners to that point. In our case, the triangles made by connecting the circumcenter to each corner. Here, the aquamarine, red, and purple triangles corresponding to points A, B, and C.
Now I find the area of the triangle and then multiple against the normalized barycentric coordinates of the circumcenter to get the areas of each of these inner triangles:
areas = 0.25*sqrt( ...
(edge_norms(:,1) + edge_norms(:,2) - edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) + edge_norms(:,3)).* ...
(-edge_norms(:,1) + edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) + edge_norms(:,2) + edge_norms(:,3)));
partial_triangle_areas = normalized_barycentric.*[areas areas areas];
Finally the quadrilateral at each point is built from half of the neighboring inner triangles: the green, blue and yellow quads corresponding to A, B, and C.
Note, that the fact that these inner triangles are cut perfectly in half follows immediately from the fact the the circumcenter is equidistance from each corner. So the last step is just to add up the halves:
quads = [ (partial_triangle_areas(:,2)+ partial_triangle_areas(:,3))*0.5 ...
(partial_triangle_areas(:,1)+ partial_triangle_areas(:,3))*0.5 ...
(partial_triangle_areas(:,1)+ partial_triangle_areas(:,2))*0.5];
Note: I’ve suffered trying to use mathematic or maple to reduce these steps to a simple algebra formula (I must be mistyping something). As far as I can tell the above steps check out (of course, it only makes sense for cases where the circumcenter is within or on the triangle: i.e. the triangle is not obtuse). I have checked out my equations on maple with the following functions:
mycos:=(a,b,c)->(b^2+c^2-a^2)/(2*b*c);
barycentric:=(a,b,c)->a*mycos(a,b,c);
normalized_barycentric:=(a,b,c)->barycentric(a,b,c)/(barycentric(a,b,c)+barycentric(b,c,a)+barycentric(c,a,b));
area:=(a,b,c)->sqrt((a+b-c)*(a-b+c)*(-a+b+c)*(a+b+c))/4;
inner_triangle:=(a,b,c)->normalized_barycentric(a,b,c)*area(a,b,c);
Which then you can run:
To see a beautiful long algebra formula for the quadrilateral at point A.
Which in matlab corresponds to:
[-sqrt(-(edge_norms(:,1) + edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) - edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) + edge_norms(:,2) - edge_norms(:,3))).* ...
(2.*edge_norms(:,2).^2.*edge_norms(:,3).^2 + edge_norms(:,1).^2.* ...
edge_norms(:,2).^2 - edge_norms(:,2).^4 + edge_norms(:,1).^2.* ...
edge_norms(:,3).^2 - edge_norms(:,3).^4)./ ...
(8.*(edge_norms(:,1) + edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) - edge_norms(:,3)).* ...
(edge_norms(:,1) - edge_norms(:,2) + edge_norms(:,3)).* ...
(edge_norms(:,1) + edge_norms(:,2) - edge_norms(:,3))), ...
-sqrt(-(edge_norms(:,2) + edge_norms(:,3) + edge_norms(:,1)).* ...
(edge_norms(:,2) - edge_norms(:,3) - edge_norms(:,1)).* ...
(edge_norms(:,2) - edge_norms(:,3) + edge_norms(:,1)).* ...
(edge_norms(:,2) + edge_norms(:,3) - edge_norms(:,1))).* ...
(2.*edge_norms(:,3).^2.*edge_norms(:,1).^2 + edge_norms(:,2).^2.* ...
edge_norms(:,3).^2 - edge_norms(:,3).^4 + edge_norms(:,2).^2.* ...
edge_norms(:,1).^2 - edge_norms(:,1).^4)./ ...
(8.*(edge_norms(:,2) + edge_norms(:,3) + edge_norms(:,1)).* ...
(edge_norms(:,2) - edge_norms(:,3) - edge_norms(:,1)).* ...
(edge_norms(:,2) - edge_norms(:,3) + edge_norms(:,1)).* ...
(edge_norms(:,2) + edge_norms(:,3) - edge_norms(:,1))), ...
-sqrt(-(edge_norms(:,3) + edge_norms(:,1) + edge_norms(:,2)).* ...
(edge_norms(:,3) - edge_norms(:,1) - edge_norms(:,2)).* ...
(edge_norms(:,3) - edge_norms(:,1) + edge_norms(:,2)).* ...
(edge_norms(:,3) + edge_norms(:,1) - edge_norms(:,2))).* ...
(2.*edge_norms(:,1).^2.*edge_norms(:,2).^2 + edge_norms(:,3).^2.* ...
edge_norms(:,1).^2 - edge_norms(:,1).^4 + edge_norms(:,3).^2.* ...
edge_norms(:,2).^2 - edge_norms(:,2).^4)./ ...
(8.*(edge_norms(:,3) + edge_norms(:,1) + edge_norms(:,2)).* ...
(edge_norms(:,3) - edge_norms(:,1) - edge_norms(:,2)).* ...
(edge_norms(:,3) - edge_norms(:,1) + edge_norms(:,2)).* ...
(edge_norms(:,3) + edge_norms(:,1) - edge_norms(:,2)))];
Note: It seems also that the wikipedia page for the circumscribed circle has a direct formula for the barycentric coordinates of the circumcenter. Though, wolfram’s mathworld has an ostensibly different formula. Upon a quick review Wolfram’s is at least easily verifiable to be the same as mine (recall these only need to be relative to each other).
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## PIE day
[This is part six in an ongoing series; previous posts can be found here: Differences of powers of consecutive integers, Differences of powers of consecutive integers, part II, Combinatorial proofs, Making our equation count, How to explain the principle of inclusion-exclusion?. However, this post is self-contained; no need to go back and read the previous ones just yet.]
“But”, I hear you protest, “Pi Day was ages ago!” Ah, but I didn’t say Pi Day, I said PIE Day. To clarify:
• Pi Day: a day on which to celebrate the not-so-fundamental circle constant, $\pi$ (March 14)
• Pie Day: a day on which to eat pie (every day)
• PIE Day: a day on which to explain the Principle of Inclusion-Exclusion (PIE). (That’s today!)
(Actually, I’m only going to begin explaining it today; it’s getting too long for a single blog post!) In any case, the overall goal is to finish up my long-languishing series on a combinatorial proof of a curious identity (though this post is self-contained so there’s no need to go back and reread that stuff yet). The biggest missing piece of the pie is… well, PIE! I’ve been having trouble figuring out a good way to explain it in sufficient generality—it’s one of those deceptively simple-seeming things which actually hides a lot of depth. Like a puddle which turns out to be a giant pothole. (Except more fun.)
In a previous post (now long ago) I asked for some advice and got a lot of great comments—if my explanation doesn’t make sense you can try reading some of those!
So, what’s the Principle of Inclusion-Exclusion all about? The basic purpose is to compute the total size of some overlapping sets.
To start out, here is a diagram representing some non-overlapping sets.
Each set is represented by a colored circle and labelled with the number of elements it contains. In this case, there are 25 people who like bobsledding (the red circle), six people who like doing laundry (the blue circle), and 99 people who like math (the green circle). The circles do not overlap at all, meaning that none of the people who like math also like laundry or bobsledding; none of the people who like doing laundry also like bobsledding or math; and so on. So, how many people are there in total? Well, that’s easy—just add the three numbers! In this case we get 130.
Now, consider this Venn diagram which shows three overlapping sets.
Again, I’ve labelled each region with the number of elements it contains. So there are two people who like bobsledding but not math or laundry; there are three people who like bobsledding and math but not laundry; there is one person who likes all three; and so on. It’s still easy to count the total number of elements: just add up all the numbers again (I get 14).
So what’s the catch?
The catch is that in many situations, we do not know the number of elements in each region! More typically, we know something like:
• The total number of elements in each set. Say, we might know that there are 7 people who like bobsledding in total, but have no idea how many of those 7 like math or laundry; and similarly for the other two sets.
• The total number of elements in each combination of sets. For example, we might know there are two people who like bobsledding and laundry—but we don’t know whether either of them likes math.
This is illustrated below for another instance of our ongoing example. The top row shows that there are sixteen people who like bobsledding in total, eleven who like laundry in total, and eighteen who like math—but again, these are total counts which tell us nothing about the overlap between the sets. (I’ve put each diagram in a box to emphasize that they are now independent—unlike in the first diagram in this post, having three separate circles does not imply that the circles are necessarily disjoint.) So $16 + 11 + 18$ is probably too many because we would be counting some people multiple times. The next row shows all the intersections of two sets: there are three people who like bobsledding and laundry (who may or may not like math), eight people who like bobsledding and math; and six people who like laundry and math. Finally, there is one person who likes all three.
The question is, how can we deduce the total number of people, starting from this information?
Well, give it a try yourself! Once you have figured that out, think about what would happen if we added a fourth category (say, people who like gelato), or a fifth, or… In a future post I will explain more about the general principle.
Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
This entry was posted in combinatorics, counting and tagged , , , , . Bookmark the permalink.
### 2 Responses to PIE day
1. Chris Kildegaard says:
There are no actual diagrams in this post (other than the Wikipedia examples that the single link takes you to), and the visuals would be helpful. That said, I love your blog!
• Brent says:
Really? You should be able to see a picture of a truck, and three diagrams. I can see them (even when logged out of wordpress) so I’m not sure why you would not be seeing them. Note that sometimes images do not make it through a feed reader, in which case you have to click through to read the post directly on my site.
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Why Python IDLE is faster than Spyder or Jupyter notebook
• Python
Gold Member
I defined a simple prime function and then I append each number to an array up to 1 million and then later 2 million. In both cases, Python IDLE gives me the answer in 4 and 9 seconds respectively, but Spyder and Jupyter Notebook gave me in 12 and 24 seconds. I wonder why this happened. The time delay is doubled in each case and that's not something small.
Is Python IDLE the best in the case for performance? Or just, in this case, it gives me the best result but in other cases, I should use the other ones?
Also, I didn't run it on the PyCharm but what would be the speed of the same operation on that IDLE?
I am trying many IDLE's but I am not sure to use which one. What would be your ideas on idle on both performance and usability?
Python:
import time
start=time.time()
def prime(N):
if N==0 or N==1:
return False
y=int(N**(0.5))
for i in range(2,y+1):
if N%i==0:
return False
return True
A=[N for N in range(1000000) if prime(N)==True]
end=time.time()
Time=end-start
print("Time to solve",Time,"sec")
Runtime on Pyton IDLE: 4.639 sec
Runtime on Jupyter : 9.4383 sec
Runtime on Spyder : 8.633 sec
Related Programming and Computer Science News on Phys.org
StoneTemplePython
Gold Member
try doing your tests on larger, newer problems... a few seconds isn't such a big deal. Some of your tests may be getting an unfair headstart based on cache.
I mostly use Pycharm and Jupyter notebooks, occasionally Atom. Also, I thought time.clock() was preferred to time.time()? It's been a while since I looked into the mechanics of this.
- - - -
To be honest I wouldn't worry about any of this. Python has a massive community on the Internet and unless other people (e.g. on stack overflow) have repeatedly flagged this as an issue with various interpreters/platforms and running python, then it probably says more about your setup's idiosyncrasies than IDLE, Jupyter, etc.
I'd focus on my attention on learning programming techniques (e.g. I recall one of your challenge problems took well over an hour to solve, but many of us can solve it in under a second)... that should be more rewarding, and fun, right?
phinds
Gold Member
try doing your tests on larger, newer problems... a few seconds isn't such a big deal. Some of your tests may be getting an unfair headstart based on cache.
Thats possible I guess. But it sounds strange..I ll run some ther programs that I wrote and I ll test them.
I mostly use Pycharm and Jupyter notebooks, occasionally Atom. Also, I thought time.clock() was preferred to time.time()? It's been a while since I looked into the mechanics of this.
I saw that code on the net so I was using it .I ll look into it
To be honest I wouldn't worry about any of this. Python has a massive community on the Internet and unless other people (e.g. on stack overflow) have repeatedly flagged this as an issue with various interpreters/platforms and running python, then it probably says more about your setup's idiosyncrasies than IDLE, Jupyter, etc.
I wanted to use something else...I first downloaded Pycharm then I tried Spyder and Jupyter notebook. I thought they would be faster or at least Idk some advantage over the Python IDLE.I searched online and they say the best is Pycharm. Maybe I try another shot on that one
I'd focus on my attention on learning programming techniques (e.g. I recall one of your challenge problems took well over an hour to solve, but many of us can solve it in under a second)... that should be more rewarding, and fun, right?
Its I guess I am reading the book and I am studying online lectures by MIT. So eventually I ll get there somehow (I guess).
Also I recommend you sublime text 3. The name talks about it. Simple, beatifull, elegant and "free". I don't need more.
Gold Member
Also I recommend you sublime text 3. The name talks about it. Simple, beatifull, elegant and "free". I don't need more.
I looked at it it seems nice but I think its mostly for large number lines of vode I guess ? and I think it doesnt have a run ? I mean its just a text editor. Pycharm seems really nice in this case
From my experience, Jupyter notebook is insanely slow. It also tends to crash if the notebook is large. However, the GUI is absolutely beautiful, and you get to write Latex, HTML and markdown between your lines of code and create absolutely beautiful NOTEBOOKS. That's what it's designed to do. It's not designed for speed. Use it for data exploration, documentation and presentations, but never for speed. It's not designed for speed. If you're looking for speed, why are you using Python? It's about the aesthetics, man.
You can use Cython in Jupyter notebook, though. That may be worth experimenting with if speed is your goal. You would probably also want to write the code in Jupyter notebook but download the Jupyter notebook as a script and then run the script outside of Jupyter notebook.
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# Find the average value of this function
Find the average value of $e^{-z}$ over the ball $x^2+y^2+z^2 \leq 1$.
-
What have you tried so far? What exactly is the problem you're having? – smackcrane Nov 8 '11 at 18:03
The average will just be the integral over the sphere, divided by the volume of the sphere.
Integrating over the sphere: Suppose we start by integrating over $z$. For a fixed $z_0$, the function is invariant, so we need only know the area of the cross-section of the sphere perpendicular to the $z$-axis at that height $z_0$. By Pythagorus, the radius of that circle will be $\sqrt{1-z^2}$ so that its area is $\pi(1-z^2)$. Hence our integral becomes $$\pi \int_{-1}^1 e^{-z} (1-z^2)dz.$$
Computing this we see that this is $$\frac{4\pi}{e}.$$ As the volume of the sphere is $\frac{4}{3}\pi$, we conclude that the average will be $$\frac{3}{e}.$$
-
Let us reason by analogy. Suppose that the problem is 'Find the average value of $x$ over the line segment defined by $S \equiv (x:0 \le x \le 1)$'. We know that the answer is 0.5 but how would we compute it?
$$\bar{x} = \frac{\int_{x \in S} x dx}{\int_{x \in S} dx}$$
Doing the integrals on the right, we get:
$$\bar{x} = \frac{(1^2-0^2)/2}{(1-0)} = 0.5$$
A quick observation that may or may not be useful: by Stokes' theorem you can transform the integral of $e^{-z}$ on the ball to the integral of $ze^z$ on the sphere. – user7530 Nov 8 '11 at 18:36
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### Home > PC3 > Chapter 12 > Lesson 12.1.2 > Problem12-32
12-32.
For each graph below, write two equations, one that uses sine and one that uses cosine, that will generate the graph.
1. The sine function does not need a horizontal shift.
The cosine function needs to be shifted $π$ units left.
1. The sine needs to be shifted $\frac{\pi}{4}$ units left.
The cosine needs to be reflected vertically.
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# FEniCS: how to specify boundary conditions on a circle inside 2D mesh
I would like to numerically find a mutual capacitance of two stripes of metal on the opposites sides of a cylinder. The problem is obviously a 2D Laplace equation. I would like to find the potential outside the cylinder as well. Therefore I have something like this:
mesh = UnitCircleMesh(50)
V = FunctionSpace(mesh, 'Lagrange', 1)
u_L = Constant(-1)
def left_boundary(x, on_boundary):
r = math.sqrt(x[0] * x[0] + x[1] * x[1])
return near(r, 0.7) and x[0] < 0 and between(x[1], (-0.5, 0.5))
u_R = Constant(1)
def right_boundary(x, on_boundary):
r = math.sqrt(x[0] * x[0] + x[1] * x[1])
return near(r, 0.7) and x[0] > 0 and between(x[1], (-0.5, 0.5))
leftPlate = DirichletBC(V, u_L, left_boundary)
rightPlate = DirichletBC(V, u_R, right_boundary)
bcs = [leftPlate, rightPlate]
u = TrialFunction(V)
v = TestFunction(V)
u = Function(V)
solve(a == Constant(0) * v * dx, u, bcs)
When I run it, I receive *** Warning: Found no facets matching domain for boundary condition.. And the found solution is 0. What is wrong?
• Are you trying to hit boundary facets or facets on radius 0.7 (in unit circle)? – Jan Blechta May 24 '13 at 21:35
• Are you trying to hit exterior or interior (on radius 0.7) boundary? If the latter is your intent consider wheter there are facets with their vertices and midpoint on radius cca 0.7 forming connected interior boundary. If yes, you probably need to increase tolerance by near(r, 0.7, tol) as target facets would be rough approximation to circle. – Jan Blechta May 24 '13 at 21:43
• Also fit your mesh = UnitCircleMesh(n) where $n = \frac{2}{0.7} m$ both $n$, $m$ being natural. – Jan Blechta May 24 '13 at 21:54
• I want to calculate the potential both in the exterior and the interior of the cylinder. – facetus May 24 '13 at 22:38
• Then increase tolerance of near function very much and tweak n so that mesh hits better radius 0.7 as suggested above. – Jan Blechta May 24 '13 at 22:44
As the error message suggest the DirichletBC does not hit any mesh entities with corresponding dofs. You need to examine your subdomains. One way to debug these are to apply the DirichletBC to a Function and plot the result:
def left_boundary(x, on_boundary):
r = math.sqrt(x[0] * x[0] + x[1] * x[1])
return r < 0.5
leftPlate = DirichletBC(V, u_L, left_boundary)
u = Function(V)
u.vector()[:] = 0
leftPlate.apply(u.vector())
plot(u, interactive=True)
This will plot a nice hat. Now you can start tweak your left_boundary function until you reach the result you want.
• Thanks! I'll try this. What is a subdomain (excuse me for a silly question)? – facetus May 24 '13 at 22:41
• @noxmetus: SubDomain is DOLFIN class for definition of some subdomain of your spatial domain, see fenicsproject.org/documentation/dolfin/1.2.0/python/…. Function left_boundary(x, on_boundary) is convenience way of defining subdomains without actually subclassing SubDomain. Notice that left_boundary has same interface as SubDomain.inside(). – Jan Blechta May 25 '13 at 0:05
• Note that you can plot boundary conditions directly without creating a Function first: plot(leftPlate) – Anders Logg May 26 '13 at 21:55
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# MethaneFluidProperties
Fluid properties for methane (CH4)
Fluid properties for methane are mainly calculated using the Setzmann and Wagner equation of state (Setzmann and Wagner, 1991). This formulation uses density and temperature as the primary variables with which to calculate properties such as density, enthalpy and internal energy.
When used with the pressure and temperature interface, which is the case in the Porous Flow module, methane properties are typically calculated by first calculating density iteratively for a given pressure and temperature. This density is then used to calculate the other properties, such as internal energy, directly. The computational expense associated with the iterative calculation can be mitigated using TabulatedFluidProperties.
For low pressures (typically less than 10 MPa), the properties of methane can be approximated using an ideal gas, which are much faster to calculate. However, at higher pressures, this approximation can lead to large differences, see Figure 1.
Figure 1: Methane density at 350K for various pressures.
Transport properties such as viscosity and thermal conductivity are calculated using the formulations provided in Irvine Jr and Liley (1984).
Dissolution of methane into water is calculated using Henry's law (IAPWS, 2004).
## Properties of methane
Propertyvalue
Molar mass0.0160425 kg/mol
Critical temperature190.564 K
Critical pressure4.5992 MPa
Critical density162.66 kg/m
Triple point temperature90.6941 K
Triple point pressure0.01169 MPa
## Range of validity
The MethaneFluidProperties UserObject is valid for:
• 90.69 K T 625 K
and pressures up to 100 MPa.
## Input Parameters
• execute_onTIMESTEP_ENDThe list of flag(s) indicating when this object should be executed, the available options include NONE, INITIAL, LINEAR, NONLINEAR, TIMESTEP_END, TIMESTEP_BEGIN, FINAL, CUSTOM.
Default:TIMESTEP_END
C++ Type:ExecFlagEnum
Options:NONE INITIAL LINEAR NONLINEAR TIMESTEP_END TIMESTEP_BEGIN FINAL CUSTOM
Description:The list of flag(s) indicating when this object should be executed, the available options include NONE, INITIAL, LINEAR, NONLINEAR, TIMESTEP_END, TIMESTEP_BEGIN, FINAL, CUSTOM.
### Optional Parameters
• control_tagsAdds user-defined labels for accessing object parameters via control logic.
C++ Type:std::vector
Options:
Description:Adds user-defined labels for accessing object parameters via control logic.
• enableTrueSet the enabled status of the MooseObject.
Default:True
C++ Type:bool
Options:
Description:Set the enabled status of the MooseObject.
• allow_duplicate_execution_on_initialFalseIn the case where this UserObject is depended upon by an initial condition, allow it to be executed twice during the initial setup (once before the IC and again after mesh adaptivity (if applicable).
Default:False
C++ Type:bool
Options:
Description:In the case where this UserObject is depended upon by an initial condition, allow it to be executed twice during the initial setup (once before the IC and again after mesh adaptivity (if applicable).
• force_preauxFalseForces the GeneralUserObject to be executed in PREAUX
Default:False
C++ Type:bool
Options:
Description:Forces the GeneralUserObject to be executed in PREAUX
• use_displaced_meshFalseWhether or not this object should use the displaced mesh for computation. Note that in the case this is true but no displacements are provided in the Mesh block the undisplaced mesh will still be used.
Default:False
C++ Type:bool
Options:
Description:Whether or not this object should use the displaced mesh for computation. Note that in the case this is true but no displacements are provided in the Mesh block the undisplaced mesh will still be used.
1. IAPWS. Guidelines on the Henry's constant and vapour liquid distribution constant for gases in H$_2$O and D$_2$O at high temperatures. Technical Report, IAPWS, 2004.[BibTeX]
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# Why does Mathematica ignore my stylesheet for DisplayFormulaNumbered cells?
I made a custom stylesheet, installed it, and applied it to a new blank notebook. Adding a cell and changing it to Text results in the expected behavior. The font and font size change to whatever the stylesheet says.
But this does not happen for DisplayFormulaNumbered cells. The Front End ignores some (but not all!) of my stylesheet settings when formatting a cell as DisplayFormulaNumbered.
The stylesheet contains:
Cell[StyleData["DisplayFormulaNumbered"],
FontFamily->"CMU Serif",
FontSize->12,
FontWeight->"Plain",
FontSlant->"Plain",
PrivateFontOptions->{
"FontPostScriptName"->
"Automatic"},ExpressionUUID->"272b9ba5-a88d-4627-839e-26d5afa11266"]
But making a DisplayFormulaNumbered cell results in a 14-point, medium weight Source Code Pro font.
I can get it to pay attention to the font settings if I add DefaultFormatType->DefaultTextFormat but then it really does treat it as text, rather than as an equation in TraditionalForm, so integrals and sums are vertically compressed, etc.
By inspecting the options for numbered equations set to TraditionalForm via Cell->Convert To->TraditionalForm Display, I tried adding
SingleLetterItalics->True,
DelimiterMatching->None,
to the stylesheet. This succeeds in producing a 12-point plain-weighted font, just not CMU Serif. It's still Source Code Pro. Why any of these would affect the font, I cannot see.
This is very confusing. I thought the styles were applied in a cascade from Core.nb to Default.nb to my ScottStyle.nb, and finally any changes I make to cells directly in the Front End. But I can't see how I could get this behavior of paying attention to some but not all of my stylesheet commands without there being some other styling applied after my stylesheet but before anything I might do in the front end. Maybe TraditionalFunctionNotation overrules the font family but not the size and weight?
Moreover, Shift-Ctrl-E shows that the cell is indeed a DisplayFormulaNumbered cell and CurrentValue[{StyleDefinitions, "DisplayFormulaNumbered"}] shows FontFamily->CMU Serif but the actual font displayed is clearly SourceCode Pro.
What magic incantation should I chant to get a stylesheet to produce the same result as using the mouse to do Format->Style->DisplayFormulaNumbered, Cell->Convert To->TraditionalForm Display, ctrl-T, choose font and size?
StyleData["DisplayFormulaNumbered"] is not environment specific so e.g. StyleData["DisplayFormulaNumbered", "Working"] will take precedence over your settings.
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# Straight Lines - Online Test
Q1. The point which divides the joint of ( 1, 2 ) and ( 3,4 ) externally in the ratio 1 : 1 .
Answer : Option B
Explaination / Solution:
The point which divides the line in the ratio m:n externally is given by x =
Substituting the values we get,
x = which is undefined.
Q2. The distance of the point ( x, y ) from the origin is
Answer : Option A
Explaination / Solution:
Distance between two points is given by $\sqrt{\left(}$
Therefore the distance from the origin to the point (x,y) is =
Q3. Slope of a line is not defined if the line is
Answer : Option C
Explaination / Solution:
Vertical lines hve undefined slopes. Hence a line which is parallel to Y-axis has undefined slopes.
Q4. The line through the points (a , b) and (- a , - b) passes through the point
Answer : Option C
Explaination / Solution:
Equation of a line passing through (x1,y1) and (x2,y2
=
Substituting the given values we get,
=
That is =
On cross multiplying and reaaranging we get,
bx - ay = 0
If this passes through the given point (a2,ab) then
b(a2) - a(ab) = 0
Q5. Projection (the foot of perpendicular) from ( x , y ) on the x – axis is
Answer : Option D
Explaination / Solution:
Let L be the foot of the perpendicular from the X axis. Therefore its y coocrdinate is zero Therefore the coordiantes of the point L is (x,0)
Q6. The equation represents all lines through the point except the line
Answer : Option D
Explaination / Solution:
The vertical lines which are parallel to Y axis has undefined slopes. Hence the slope of the line 'm' will be undefined.
Therefore the above equation of the line will represent all lines through () except the line parallel to Y- axis
Q7. Slope of any line parallel to X axis is
Answer : Option B
Explaination / Solution:
Since the angle made with x axis is zero, since tanθ is the slope. Then tan0 = 0 Hence the slpoe of the line parallel to X-axis is zero
Q8. The distance between the parallel lines 3x + 4y + 13 = 0 and 3x + 4y – 13 = 0 is
Answer : Option A
Explaination / Solution:
Distance between parallel lines is given by
Now substituting the values we get,
=
Q9. The straight lines x + y = 0 , 3x - y – 4 = 0 , x + 3y – 4 = 0 form a triangle which is
Answer : Option A
Explaination / Solution:
The lines formed by these lines is right angled, triangle.
Two lines are perpendicular to each other if the product od their slopes is -1
Slope of the lines 3x - y – 4 = 0 , x + 3y – 4 = 0 are 3 and -1/3 respectively.
The product of these slopes is -1
Hence the lines3x - y – 4 = 0 , x + 3y – 4 = 0 are perpendicular to each other.
Therefore the triangle formed by these lines is a right angled triangle.
Q10. The lines 2x – 3y = 5 and 6x – 9y – 7 = 0 are
Answer : Option C
Explaination / Solution:
The given lines are paraallel lines
Condition for the line to be parallel is $\ne$
substituting the values,
=$\ne$
Hence they are parallel lines.
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# Cron: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
(Redirected to cron article)
Cron is a time-based job scheduler in Unix-like computer operating systems. The name cron comes from the word chronograph (a time-piece). Cron enables users to schedule jobs (commands or shell scripts) to run automatically at a certain time or date. It is commonly used to automate system maintenance or administration, though its general purpose nature means that it can be used for other purposes, such as connecting to the Internet and downloading email.[1]
Cron has been recreated several times in its history.
## Design
Cron is driven by a crontab, a configuration file that specifies shell commands to run periodically on a given schedule.
Early versions of cron, available up through Version 7 Unix and 32V, made their services available only to the super-user of the operating system; this was the single-user version. With the release of Unix System V and the multi-user cron, these services were extended to all account-holding users of the system.
## History
### Early versions
The cron in Version 7 Unix, written by Brian Kernighan, was a system service (later called daemons) invoked from /etc/inittab when the operating system entered multi-user mode. Its algorithm was straightforward:
1. Read /usr/etc/crontab
2. Determine if any commands are to be run at the current date and time and if so, run them as the Superuser, root.
3. Sleep for one minute
4. Repeat from step 1.
This version of cron was basic and robust, but it also consumed resources whether it found any work to do or not; upon hearing this description, Douglas Comer, a professor at Purdue University, remarked, "Ah, an oblivious algorithm." In an experiment at Purdue University in the late 1970s to extend cron's service to all 100 users on a time-shared VAX it was found to place too much load on the system.
### Multi-user capability
The next version of cron was created to extend the capabilities of cron to all users of a Unix system, not just the superuser. Though this may seem trivial today with most Unix and Unix-like systems having powerful processors and small numbers of users, at the time it required a new approach on a 1 MIPS system having roughly 100 user accounts.
In the August, 1977 issue of the Communications of the ACM, W. R. Franta and Kurt Maly published an article entitled "An efficient data structure for the simulation event set" describing an event queue data structure for discrete event-driven simulation systems that demonstrated "performance superior to that of commonly used simple linked list algorithms," good behavior given non-uniform time distributions, and worst case complexity $O\left(\sqrt{n}\right)$, "n" being the number of events in the queue.
A graduate student, Robert Brown, reviewing this article, recognized the parallel between cron and discrete event simulators, and created an implementation of the Franta-Maly event list manager (ELM) for experimentation. Discrete event simulators run in "virtual time," peeling events off the event queue as quickly as possible and advancing their notion of "now" to the scheduled time of the next event. By running the event simulator in "real time" instead of virtual time, a version of cron was created that spent most of its time sleeping, waiting for the moment in time when the task at the head of the event list was to be executed.
The following school year brought new students into the graduate program, including Keith Williamson, who joined the systems staff in the Computer Science department. As a "warm up task" Brown asked him to flesh out the prototype cron into a production service, and this multi-user cron went into use at Purdue in late 1979. This version of cron wholly replaced the /etc/cron that was in use on the Computer Science department's VAX 11/780 running 32/V.
The algorithm used by this cron is as follows:
1. On start-up, look for a file named .crontab in the home directories of all account holders.
2. For each crontab file found, determine the next time in the future that each command is to be run.
3. Place those commands on the Franta-Maly event list with their corresponding time and their "five field" time specifier.
4. Enter main loop:
1. Examine the task entry at the head of the queue, compute how far in the future it is to be run.
2. Sleep for that period of time.
3. On awakening and after verifying the correct time, execute the task at the head of the queue (in background) with the privileges of the user who created it.
4. Determine the next time in the future to run this command and place it back on the event list at that time value.
Additionally, the daemon would respond to SIGHUP signals to rescan modified crontab files and would schedule special "wake up events" on the hour and half hour to look for modified crontab files. Much detail is omitted here concerning the inaccuracies of computer time-of-day tracking, Unix alarm scheduling, explicit time-of-day changes, and process management, all of which account for the majority of the lines of code in this cron. This cron also captured the output of stdout and stderr and e-mailed any output to the crontab owner.
The resources consumed by this cron scale only with the amount of work it is given and do not inherently increase over time with the exception of periodically checking for changes.
Williamson completed his studies and departed the University with a Masters of Science in Computer Science and joined AT&T Bell Labs in Murray Hill, New Jersey, and took this cron with him. At Bell Labs, he and others incorporated the Unix at command into cron, moved the crontab files out of users' home directories (which were not host-specific) and into a common host-specific spool directory, and of necessity added the crontab command to allow users to copy their crontabs to that spool directory.
This version of cron later appeared largely unchanged in Unix System V and in BSD and their derivatives, the Solaris Operating System from Sun Microsystems, IRIX from Silicon Graphics, HP-UX from Hewlett-Packard, and IBM AIX. Technically, the original license for these implementations should be with the Purdue Research Foundation who funded the work, but this took place at a time when little concern was given to such matters.
### Modern versions
With the advent of the GNU Project and Linux, new crons appeared. The most prevalent of these is the Vixie cron, originally coded by Paul Vixie in 1987. Version 3 of Vixie cron was released in late 1993. Version 4.1 was renamed to ISC Cron and was released in January 2004. Version 3, with some minor bugfixes, is used in most distributions of Linux and BSDs.
In 2007, RedHat forked vixie-cron 4.1 to the cronie project and included anacron 2.3 in 2009.
Other popular implementations include anacron and fcron. However, anacron is not an independent cron program; it relies on another cron program to call it in order to perform.
#### Timezone handling
Most cron implementations simply interpret crontab entries in the system time zone setting under which the cron daemon itself is run. This can be a source of dispute if a large multiuser machine has users in several time zones, especially if the system default timezone includes the potentially confusing DST. Thus, a cron implementation may special-case any "TZ=<timezone>" environment variable setting lines in user crontabs, interpreting subsequent crontab entries relative to that timezone.[2]
### crontab syntax
The crontab files are stored where the lists of jobs and other instructions to the cron daemon are kept. Users can have their own individual crontab files and often there is a systemwide crontab file (usually in /etc or a subdirectory of /etc) which only system administrators can edit.
Each line of a crontab file represents a job and follows a particular format as a series of fields, separated by spaces and/or tabs. Each field can have a single value or a series of values.
#### Operators
There are several ways of specifying multiple date/time values in a field:
• The comma (',') operator specifies a list of values, for example: "1,3,4,7,8" (space inside the list must not be used)
• The dash ('-') operator specifies a range of values, for example: "1-6", which is equivalent to "1,2,3,4,5,6"
• The asterisk ('*') operator specifies all possible values for a field. For example, an asterisk in the hour time field would be equivalent to 'every hour' (subject to matching other specified fields).
There is also an operator which some extended versions of cron support, the slash ('/') operator (called "step"), which can be used to skip a given number of values. For example, "*/3" in the hour time field is equivalent to "0,3,6,9,12,15,18,21".
So "*" specifies 'every hour' but the "*/3" means only those hours divisible by 3. The meaning of '/' specifier, however, means "when the modulo is zero" rather than "every". For example, "*/61" in the minute will in fact be executed hourly, not every 61 minutes.
Example: the following will clear the Apache error log at one minute past midnight ( 00:01 of every day of the month, of every day of the week ).
1 0 * * * echo -n "" > /www/apache/logs/error_log
Slash example: the following will run the script /home/user/test.pl every 5 minutes.
*/5 * * * * /home/user/test.pl
.---------------- minute (0 - 59)
| .------------- hour (0 - 23)
| | .---------- day of month (1 - 31)
| | | .------- month (1 - 12) OR jan,feb,mar,apr ...
| | | | .---- day of week (0 - 6) (Sunday=0 or 7) OR sun,mon,tue,wed,thu,fri,sat
| | | | |
* * * * * command to be executed
There are several special entries, most of which are just shortcuts, that can be used instead of specifying the full cron entry:
Entry Description Equivalent To
@reboot Run once, at startup. None
@yearly Run once a year 0 0 1 1 *
@annually (same as @yearly) 0 0 1 1 *
@monthly Run once a month 0 0 1 * *
@weekly Run once a week 0 0 * * 0
@daily Run once a day 0 0 * * *
@midnight (same as @daily) 0 0 * * *
@hourly Run once an hour 0 * * * *
@reboot refers to a reboot of the cron daemon, not the host it is running on. It can be useful if there is a need to start up a server or daemon under a particular user, or if user does not have access to the rc.d/init.d files. But unlike the rc.d/init.d files, the @reboot entry will run every time the cron daemon is started, which will typically occur when the server is rebooted, but can also occur at other times.
Each of the patterns from the first five fields may be either * (an asterisk), which matches all legal values, or a list of elements separated by commas. Some implementations of cron, such as that in the popular 4th BSD edition written by Paul Vixie and included in many Linux distributions, insert a username into the format as the sixth field, as whom the specified job will be run (subject to user existence in /etc/passwd and allowed permissions), but only in the system crontabs (/etc/crontab and /etc/cron.d/*), not in others which are each assigned to a single user to configure. The seventh (or sixth if no user field is part of the format) and subsequent fields (i.e., the rest of the line) specify the command to be run.
For "day of the week" (field 5), both 0 and 7 are considered Sunday, though some versions of Unix such as AIX do not list "7" as acceptable in the man page. A job is executed when the time/date specification fields all match the current time and date. There is one exception: if both "day of month" and "day of week" are restricted (not "*"), then either the "day of month" field (3) or the "day of week" field (5) must match the current day (even though the other of the two fields need not match the current day).
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### Improved Progressive BKZ with Lattice Sieving
##### Abstract
BKZ is currently the most efficient algorithm in solving the approximate shortest vector problem (SVP$_\gamma$). One of the most important parameter choice in BKZ is the blocksize $\beta$, which greatly affects its efficiency. In 2016, Aono \textit{et al.} presented \textit{Improved Progressive BKZ} (pro-BKZ). Their work designed a blocksize strategy selection algorithm so that pro-BKZ runs faster than BKZ 2.0 which has a fixed blocksize. However, pro-BKZ only considers enumeration as its subroutine, without using the more efficient lattice sieving algorithm. Besides, their blocksize strategy selection is not optimal, so the strategy selection algorithm could further be improved. In this paper, we present a new lattice solving algorithm called Improved Progressive pnj-BKZ (pro-pnj-BKZ) mainly based on an optimal blocksize strategy selection algorithm for BKZ with sieving, which relies on accurate time cost models and simulating algorithms. We propose the following approaches: - New simulators and time cost models for sieving and BKZ with sieving. A simulator is used for simulating lattice reduction process without running the lattice reduction algorithm itself. We give new simulators for sieving and BKZ, to simulate the cases where blocks in BKZ with sieve oracle jump by more than one dimension. We also give more accurate time cost models for both sieving and BKZ with sieving by experiments. Specifically, we discover new relationships among time cost, blocksize and lattice dimension, which cannot be explained by the existing theoretical results, and discuss the reason. - New two-step mode for solving SVP$_\gamma$ problem with BKZ and sieving. Other than a subroutine of BKZ, sieving can also be combined with BKZ to get a more efficient lattice solving algorithm, but the best way of combination is currently unknown. We show that to solve SVP$_\gamma$ problem more efficiently, one should first repeatedly run BKZ to reduce the lattice basis and finally run lattice sieving once, since BKZ performs better in lattice basis reduction, while sieving performs better in finding short vectors. By our simulator, we can properly choose the timing where the algorithm ends the BKZ routine and begins sieving. - New blocksize strategy selection algorithm for BKZ with sieving. Since the blocksize strategy selection algorithm in pro-BKZ is not optimal, we design a new blocksize strategy selection algorithm to ensure an optimal strategy output. We implement both blocksize strategy selection algorithms in pro-BKZ and ours, along with our new BKZ simulator and two-step mode to generate the blocksize strategies. Simulation results show that the strategy generated by our new algorithm are more efficient in solving SVP$_\gamma$ problem. By generated strategy, we improve the efficiency nearly $24.6$ times compared with heuristic blocksize strategy in G6K. We test the efficiency of the pro-pnj-BKZ with the TU Darmstadt LWE challenge and break the LWE challenges with $(n,\alpha)\in\{(40,0.035),(50,0.025),(55,0.020),(90,0.005)\}$.
Available format(s)
Category
Public-key cryptography
Publication info
Preprint.
Keywords
cryptanalysis lattice reduction SVPγ progressive BKZ optimal blocksize selection
Contact author(s)
xiawenwen @ stu xidian edu cn
lzwang_2 @ stu xidian edu cn
wanggxx @ sjtu edu cn
dwgu @ sjtu edu cn
bcwang @ xidian edu cn
History
2022-10-11: last of 2 revisions
See all versions
Short URL
https://ia.cr/2022/1343
CC BY
BibTeX
@misc{cryptoeprint:2022/1343,
author = {Wenwen Xia and Leizhang Wang and GengWang and Dawu Gu and Baocang Wang},
title = {Improved Progressive BKZ with Lattice Sieving},
howpublished = {Cryptology ePrint Archive, Paper 2022/1343},
year = {2022},
note = {\url{https://eprint.iacr.org/2022/1343}},
url = {https://eprint.iacr.org/2022/1343}
}
Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.
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# running processes
Discussion in 'Windows XP' started by kryptonite1055, Jul 30, 2006.
Not open for further replies.
Joined:
Jun 22, 2006
Messages:
243
So I have been at school, and when I got back I set up my computer and have been using it, but tonight mine was off so I wne to my parents computer. It is an older comp (933 mhz P3) so I thought is running slower is not that big of a deal, but is seems to get slower and slower. So I looked a the running processes and most of them are in caps... I know on mine they are not.
What is causing this?
should I be worried? if so what do I have to do to fix it. they have some important files on there (also backed up)
thanks
2. ### wilson44512
Joined:
Mar 25, 2006
Messages:
2,450
Joined:
Jun 22, 2006
Messages:
243
I did put those on a long time ago. except spywareblaster
so far I have ran ad-aware (a few things nothing serious), then I did spy-bot I got some weird things (some were red and others were another color don't remember) selected all. All but 3 were taken off.
have not run AVG yet but going to very soon
I also found out that there are 40!!!! windows updates that they need to do. I did go to microsoft and did updates, but all failed. Right now I am in the process of downloading them individually from microsoft and installing them
Still none of the processes have changed from caps to lower case. So if they are in Caps there is something wrong with that processes?
thanks
Joined:
Jun 22, 2006
Messages:
243
I tried to do the scan, but after the 14th step it said there was an error. there were 4 common reasons why it does it and I did not fall into any of those situations.
while getting the windows updates, 2 showed up on the list but I could not find them on microsofts web site
KB905474, KB904706
I know I did not miss write them b/c I copied it twice from the screen (2nd time I was making sure those were in but 4 more showed up)
so far I have gotten 42 updates, and 2 will not go. What gets me mad with this also is that I can't just use the update part. It ALWAYS FAILS for all of the downloads.
I just got it to do a full scan(mypccenter.com) , but there is no info. It says I have 0Mb or RAM installed.
Well it is really late here.
any info why the processes are in caps, and how to get them back would be great. I did see iexplorer.exe is not in caps any more, but that is the only one.
5. ### ozrom1e
Joined:
May 15, 2006
Messages:
11,849
With these problems and the only way to see what is infecting the computer is to HijackThis the computer.
Filename = 1137518044HJTsetup.exe
Save the file to your desktop.
Double click on the HJTsetup.exe icon on your desktop.
By default it will install to C:\Program Files\HijackThis.
Continue to click Next in the setup dialog boxes until you get to the Select Additional Tasks dialog.
Put a check by Create a desktop icon then click Next again.
Continue to follow the rest of the prompts from there.
At the final dialog box click Finish and it will launch Hijack This.
Click on the Do a system scan and save a log file button. It will scan and then ask you to save the log.
Click Save to save the log file and then the log will open in notepad.
At the top of the Notepad HJT log screen, hit Edit then Select All then click Edit and then click Copy doing that copies the text to the clipboard, you won't see it yet....
At the top of your TSG/browser window, hit Edit then Paste
DO NOT have Hijack This fix anything yet. Most of what it finds will be harmless or even required.
Joined:
Jun 22, 2006
Messages:
243
Running processes:
C:\WINDOWS\System32\smss.exe
C:\WINDOWS\system32\winlogon.exe
C:\WINDOWS\system32\services.exe
C:\WINDOWS\system32\lsass.exe
C:\WINDOWS\system32\svchost.exe
C:\WINDOWS\System32\svchost.exe
C:\WINDOWS\system32\spoolsv.exe
C:\PROGRA~1\Grisoft\AVGFRE~1\avgamsvr.exe
C:\PROGRA~1\Grisoft\AVGFRE~1\avgupsvc.exe
C:\Program Files\iPod\bin\iPodService.exe
C:\Program Files\Network Associates\VirusScan\VsTskMgr.exe
C:\WINDOWS\system32\HPZipm12.exe
C:\WINDOWS\System32\svchost.exe
C:\WINDOWS\System32\svchost.exe
C:\WINDOWS\Explorer.EXE
C:\Program Files\iTunes\iTunesHelper.exe
C:\PROGRA~1\Grisoft\AVGFRE~1\avgcc.exe
C:\PROGRA~1\Grisoft\AVGFRE~1\avgemc.exe
C:\Program Files\HP\HP Software Update\HPWuSchd2.exe
C:\Program Files\HP\Digital Imaging\bin\hpqtra08.exe
C:\Program Files\HP\Digital Imaging\bin\hpqSTE08.exe
C:\Program Files\Hijackthis\HijackThis.exe
O2 - BHO: (no name) - {53707962-6F74-2D53-2644-206D7942484F} - C:\PROGRA~1\SPYBOT~1\SDHelper.dll
O4 - HKLM\..\Run: [iTunesHelper] C:\Program Files\iTunes\iTunesHelper.exe
O4 - HKLM\..\Run: [AVG7_CC] C:\PROGRA~1\Grisoft\AVGFRE~1\avgcc.exe /STARTUP
O4 - HKLM\..\Run: [AVG7_EMC] C:\PROGRA~1\Grisoft\AVGFRE~1\avgemc.exe
O4 - HKLM\..\Run: [HP Software Update] C:\Program Files\HP\HP Software Update\HPWuSchd2.exe
O4 - Global Startup: HP Digital Imaging Monitor.lnk = C:\Program Files\HP\Digital Imaging\bin\hpqtra08.exe
O9 - Extra button: AIM - {AC9E2541-2814-11d5-BC6D-00B0D0A1DE45} - C:\Program Files\AIM95\aim.exe
O16 - DPF: {30528230-99f7-4bb4-88d8-fa1d4f56a2ab} (YInstStarter Class) - C:\Program Files\Yahoo!\Common\yinsthelper.dll
O16 - DPF: {6E32070A-766D-4EE6-879C-DC1FA91D2FC3} (MUWebControl Class) - http://update.microsoft.com/microsoftupdate/v6/V5Controls/en/x86/client/muweb_site.cab?1125943566419
O16 - DPF: {BA2D9665-D672-446F-98F4-E3E41FA12A01} (PCAObj Class) - http://www.mypccenter.com/PCA.cab
O23 - Service: AVG7 Alert Manager Server (Avg7Alrt) - GRISOFT, s.r.o. - C:\PROGRA~1\Grisoft\AVGFRE~1\avgamsvr.exe
O23 - Service: AVG7 Update Service (Avg7UpdSvc) - GRISOFT, s.r.o. - C:\PROGRA~1\Grisoft\AVGFRE~1\avgupsvc.exe
O23 - Service: InstallDriver Table Manager (IDriverT) - Macrovision Corporation - C:\Program Files\Common Files\InstallShield\Driver\11\Intel 32\IDriverT.exe
O23 - Service: iPod Service (iPodService) - Apple Computer, Inc. - C:\Program Files\iPod\bin\iPodService.exe
O23 - Service: Network Associates Task Manager (McTaskManager) - Network Associates, Inc. - C:\Program Files\Network Associates\VirusScan\VsTskMgr.exe
O23 - Service: Pml Driver HPZ12 - HP - C:\WINDOWS\system32\HPZipm12.exe
there it is
7. ### ozrom1e
Joined:
May 15, 2006
Messages:
11,849
I am sorry I wish the whole HijackThis log file not just part of it. Please re-post.
Joined:
Jun 22, 2006
Messages:
243
Logfile of HijackThis v1.99.1
Scan saved at 8:22:36 PM, on 7/30/2006
Platform: Windows XP SP2 (WinNT 5.01.2600)
MSIE: Internet Explorer v6.00 SP2 (6.00.2900.2180)
Running processes:
C:\WINDOWS\System32\smss.exe
C:\WINDOWS\system32\winlogon.exe
C:\WINDOWS\system32\services.exe
C:\WINDOWS\system32\lsass.exe
C:\WINDOWS\system32\svchost.exe
C:\WINDOWS\System32\svchost.exe
C:\WINDOWS\system32\spoolsv.exe
C:\PROGRA~1\Grisoft\AVGFRE~1\avgamsvr.exe
C:\PROGRA~1\Grisoft\AVGFRE~1\avgupsvc.exe
C:\Program Files\iPod\bin\iPodService.exe
C:\Program Files\Network Associates\VirusScan\VsTskMgr.exe
C:\WINDOWS\system32\HPZipm12.exe
C:\WINDOWS\System32\svchost.exe
C:\WINDOWS\System32\svchost.exe
C:\WINDOWS\Explorer.EXE
C:\Program Files\iTunes\iTunesHelper.exe
C:\PROGRA~1\Grisoft\AVGFRE~1\avgcc.exe
C:\PROGRA~1\Grisoft\AVGFRE~1\avgemc.exe
C:\Program Files\HP\HP Software Update\HPWuSchd2.exe
C:\Program Files\HP\Digital Imaging\bin\hpqtra08.exe
C:\Program Files\HP\Digital Imaging\bin\hpqSTE08.exe
C:\Program Files\Hijackthis\HijackThis.exe
O2 - BHO: (no name) - {53707962-6F74-2D53-2644-206D7942484F} - C:\PROGRA~1\SPYBOT~1\SDHelper.dll
O4 - HKLM\..\Run: [iTunesHelper] C:\Program Files\iTunes\iTunesHelper.exe
O4 - HKLM\..\Run: [AVG7_CC] C:\PROGRA~1\Grisoft\AVGFRE~1\avgcc.exe /STARTUP
O4 - HKLM\..\Run: [AVG7_EMC] C:\PROGRA~1\Grisoft\AVGFRE~1\avgemc.exe
O4 - HKLM\..\Run: [HP Software Update] C:\Program Files\HP\HP Software Update\HPWuSchd2.exe
O4 - Global Startup: HP Digital Imaging Monitor.lnk = C:\Program Files\HP\Digital Imaging\bin\hpqtra08.exe
O9 - Extra button: AIM - {AC9E2541-2814-11d5-BC6D-00B0D0A1DE45} - C:\Program Files\AIM95\aim.exe
O16 - DPF: {30528230-99f7-4bb4-88d8-fa1d4f56a2ab} (YInstStarter Class) - C:\Program Files\Yahoo!\Common\yinsthelper.dll
O16 - DPF: {6E32070A-766D-4EE6-879C-DC1FA91D2FC3} (MUWebControl Class) - http://update.microsoft.com/microsoftupdate/v6/V5Controls/en/x86/client/muweb_site.cab?1125943566419
O16 - DPF: {BA2D9665-D672-446F-98F4-E3E41FA12A01} (PCAObj Class) - http://www.mypccenter.com/PCA.cab
O23 - Service: AVG7 Alert Manager Server (Avg7Alrt) - GRISOFT, s.r.o. - C:\PROGRA~1\Grisoft\AVGFRE~1\avgamsvr.exe
O23 - Service: AVG7 Update Service (Avg7UpdSvc) - GRISOFT, s.r.o. - C:\PROGRA~1\Grisoft\AVGFRE~1\avgupsvc.exe
O23 - Service: InstallDriver Table Manager (IDriverT) - Macrovision Corporation - C:\Program Files\Common Files\InstallShield\Driver\11\Intel 32\IDriverT.exe
O23 - Service: iPod Service (iPodService) - Apple Computer, Inc. - C:\Program Files\iPod\bin\iPodService.exe
O23 - Service: Network Associates Task Manager (McTaskManager) - Network Associates, Inc. - C:\Program Files\Network Associates\VirusScan\VsTskMgr.exe
O23 - Service: Pml Driver HPZ12 - HP - C:\WINDOWS\system32\HPZipm12.exe
sorry I must have missed the top part
Joined:
Jun 22, 2006
Messages:
243
if anyone could help that would be great
Joined:
Jun 22, 2006
Messages:
243
how do I fix this?
As Seen On
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Dividing a complex reaction involving a hypervalent iodine reagent into three limiting mechanisms byab initiomolecular dynamics
Sala, Oliver; Lüthi, Hans Peter; Togni, Antonio; Iannuzzi, Marcella; Hutter, Jürg (2015). Dividing a complex reaction involving a hypervalent iodine reagent into three limiting mechanisms byab initiomolecular dynamics. Journal of Computational Chemistry, 36(11):785-794.
Abstract
The electrophilic N-trifluoromethylation of MeCN with a hypervalent iodine reagent to form a nitrilium ion, that is rapidly trapped by an azole nucleophile, is thought to occur via reductive elimination (RE). A recent study showed that, depending on the solvent representation, the S(N)2 is favoured to a different extent over the RE. However, there is a discriminative solvent effect present, which calls for a statistical mechanics approach to fully account for the entropic contributions. In this study, we perform metadynamic simulations for two trifluoromethylation reactions (with N- and S-nucleophiles), showing that the RE mechanism is always favoured in MeCN solution. These computations also indicate that a radical mechanism (single electron transfer) may play an important role. The computational protocol based on accelerated molecular dynamics for the exploration of the free energy surface is transferable and will be applied to similar reactions to investigate other electrophiles on the reagent. Based on the activation parameters determined, this approach also gives insight into the mechanistic details of the trifluoromethylation and shows that these commonly known mechanisms mark the limits within which the reaction proceeds.
Abstract
The electrophilic N-trifluoromethylation of MeCN with a hypervalent iodine reagent to form a nitrilium ion, that is rapidly trapped by an azole nucleophile, is thought to occur via reductive elimination (RE). A recent study showed that, depending on the solvent representation, the S(N)2 is favoured to a different extent over the RE. However, there is a discriminative solvent effect present, which calls for a statistical mechanics approach to fully account for the entropic contributions. In this study, we perform metadynamic simulations for two trifluoromethylation reactions (with N- and S-nucleophiles), showing that the RE mechanism is always favoured in MeCN solution. These computations also indicate that a radical mechanism (single electron transfer) may play an important role. The computational protocol based on accelerated molecular dynamics for the exploration of the free energy surface is transferable and will be applied to similar reactions to investigate other electrophiles on the reagent. Based on the activation parameters determined, this approach also gives insight into the mechanistic details of the trifluoromethylation and shows that these commonly known mechanisms mark the limits within which the reaction proceeds.
Statistics
Citations
11 citations in Web of Science®
12 citations in Scopus®
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# Resources for high school teachers about APOS theory
As far as I can find, the major resources available to the layteacher about APOS (Action, Process, Object, Schema) theory refer to "undergraduate" concepts such as group theory and vector spaces (such as here).
I am interested in whether people have implemented or investigated APOS theory at the high school or lower-undergraduate level, and whether there are resources for the high school math teacher that can be used concretely, for example, to design a unit. So far I have found this book, but this seems unnecessarily technical to reach a larger audience of teachers.
For example, when I described APOS theory to my coworkers, they immediately connected to it and it helped them understand the struggles that their students were facing in particular problems, such as transitioning from seeing "sine and cosine" as "find the ratio between pairs of sides" to "functions." However I was unable to find good reading for them that referenced mathematics that they used frequently.
• There are certainly individual papers using APOS theory at the pre-college level. For example, you mention functions and there is the paper High school students' understanding of the function concept (pdf) by Dubinksy and Wilson (JMB, 2013). As to resources for high school math teachers: not of which I am aware; good question, +1. – Benjamin Dickman Aug 23 '17 at 20:52
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# Finicky behavior from dbserv/Qserv on ORDER BY using table_alias.column syntax
XMLWordPrintable
#### Details
• Type: Bug
• Status: Done
• Resolution: Done
• Fix Version/s: None
• Component/s:
• Labels:
• Story Points:
10
• Sprint:
DB_S17_12, DB_S17_01, DB_S17_2
• Team:
Data Access and Database
#### Description
The following dbserv query works:
curl -d 'query=SELECT objectId, id, fsrc.exposure_id, fsrc.exposure_time_mid, exp.run, scisql_dnToAbMag(fsrc.flux_psf,exp.fluxMag0) AS g, scisql_dnToAbMagSigma(fsrc.flux_psf, fsrc.flux_psf_err, exp.fluxMag0, exp.fluxMag0Sigma) AS gErr FROM RunDeepForcedSource AS fsrc, Science_Ccd_Exposure AS exp WHERE exp.scienceCcdExposureId = fsrc.exposure_id AND fsrc.exposure_filter_id=1 AND objectId=3448068867358968 ORDER BY exposure_time_mid' http://lsst-qserv-dax01.ncsa.illinois.edu:5000/db/v0/tap/sync
while this version, identical in all but one respect, fails:
curl -d 'query=SELECT objectId, id, fsrc.exposure_id, fsrc.exposure_time_mid, exp.run, scisql_dnToAbMag(fsrc.flux_psf,exp.fluxMag0) AS g, scisql_dnToAbMagSigma(fsrc.flux_psf, fsrc.flux_psf_err, exp.fluxMag0, exp.fluxMag0Sigma) AS gErr FROM RunDeepForcedSource AS fsrc, Science_Ccd_Exposure AS exp WHERE exp.scienceCcdExposureId = fsrc.exposure_id AND fsrc.exposure_filter_id=1 AND objectId=3448068867358968 ORDER BY fsrc.exposure_time_mid' http://lsst-qserv-dax01.ncsa.illinois.edu:5000/db/v0/tap/sync
The change is from ORDER BY exposure_time_mid to ORDER BY fsrc.exposure_time_mid .
The error in the second instance is:
{"message": "(_mysql_exceptions.OperationalError) (1054, \"Unknown column 'fsrc.exposure_time_mid' in 'order clause'\")", "error": "OperationalError"}
Severity is minor, for now, because the first form works for PDAC and in this case there is no ambiguity between the joined tables. However, it should be investigated because it may be the sign of a more complex issue.
#### Activity
Hide
Andy Salnikov added a comment -
I think I disagree with John, if I'm reading it correctly the first query should still be acceptable because there is just one exposure_time_mid AND MySQL in this case should name result column exposure_time_mid too. Second query with ORDER BY fsrc.exposure_time_mid will fail (immediately and not after 1 hour later) but with more explicit error message ("ORDER BY argument should not include database or table").
Show
Andy Salnikov added a comment - I think I disagree with John, if I'm reading it correctly the first query should still be acceptable because there is just one exposure_time_mid AND MySQL in this case should name result column exposure_time_mid too. Second query with ORDER BY fsrc.exposure_time_mid will fail (immediately and not after 1 hour later) but with more explicit error message ("ORDER BY argument should not include database or table").
Hide
Andy Salnikov added a comment -
Reviewed, see comments on PR. I'm still not quite happy and I believe we should do better otherwise there will be more surprised clients. Still this ticket is an improvement and we should probably open new ticket now to try and make it even better.
Show
Andy Salnikov added a comment - Reviewed, see comments on PR. I'm still not quite happy and I believe we should do better otherwise there will be more surprised clients. Still this ticket is an improvement and we should probably open new ticket now to try and make it even better.
Hide
John Gates added a comment - - edited
There might be some confusion here. It's only the queries that add qualifiers to the ORDER BY clauses that will not work. Hopefully, any query that has a problematic clause in the ORDER BY will now return an error immediately. The first query above should be fine. SELECT fsrc.exposure_time_mid ... ORDER BY exposure_time_mid should work without issue, as there's only one instance of exposure_time_mid.
All of the following should cause an immediate error and can easily be fixed with an alias.
SELECT fsrc.exposure_time_mid, obj.exposure_time_mid ... ORDER BY exposure_time_mid; -- duplicate column exposure_time_mid SELECT fsrc.exposure_time_mid ... ORDER BY fsrc.exposure_time_mid; -- fsrc. qualifier not allowed in ORDER BY SELECT ABS(exposure_time_mid) ... ORDER BY ABS(exposure_time_mid); -- order by clause is too complicated, column header will almost certainly not match order by clause. SELECT exposure_time_mid * 2 ... ORDER BY exposure_time_mid * 2; -- order by clause is again too complicated. SELECT exposure_time_mid AS expTimeMid ... ORDER BY exposure_time_mid; -- needs to use the alias.
The following should work without issue. Using an alias is very reliable.
SELECT fsrc.exposure_time_mid AS fExpTimeMid, obj.exposure_time_mid ... ORDER BY fExpTimeMid; SELECT fsrc.exposure_time_mid ... ORDER BY exposure_time_mid; SELECT fsrc.exposure_time_mid AS fExpTimeMid ... ORDER BY fExpTimeMid; SELECT ABS(exposure_time_mid) AS absExpExpTime ... ORDER BY absExpExpTime; SELECT exposure_time_mid * 2 AS fExpTimeMidx2 ... ORDER BY fExpTimeMidx2;
Show
John Gates added a comment - - edited There might be some confusion here. It's only the queries that add qualifiers to the ORDER BY clauses that will not work. Hopefully, any query that has a problematic clause in the ORDER BY will now return an error immediately. The first query above should be fine. SELECT fsrc.exposure_time_mid ... ORDER BY exposure_time_mid should work without issue, as there's only one instance of exposure_time_mid . All of the following should cause an immediate error and can easily be fixed with an alias. SELECT fsrc.exposure_time_mid, obj.exposure_time_mid ... ORDER BY exposure_time_mid; -- duplicate column exposure_time_mid SELECT fsrc.exposure_time_mid ... ORDER BY fsrc.exposure_time_mid; -- fsrc. qualifier not allowed in ORDER BY SELECT ABS(exposure_time_mid) ... ORDER BY ABS(exposure_time_mid); -- order by clause is too complicated, column header will almost certainly not match order by clause. SELECT exposure_time_mid * 2 ... ORDER BY exposure_time_mid * 2; -- order by clause is again too complicated. SELECT exposure_time_mid AS expTimeMid ... ORDER BY exposure_time_mid; -- needs to use the alias. The following should work without issue. Using an alias is very reliable. SELECT fsrc.exposure_time_mid AS fExpTimeMid, obj.exposure_time_mid ... ORDER BY fExpTimeMid; SELECT fsrc.exposure_time_mid ... ORDER BY exposure_time_mid; SELECT fsrc.exposure_time_mid AS fExpTimeMid ... ORDER BY fExpTimeMid; SELECT ABS(exposure_time_mid) AS absExpExpTime ... ORDER BY absExpExpTime; SELECT exposure_time_mid * 2 AS fExpTimeMidx2 ... ORDER BY fExpTimeMidx2;
Hide
Gregory Dubois-Felsmann added a comment -
So I think this is a non-breaking change for PDAC, if I'm reading this correctly. Still, I'd like to be made aware of when this change, once merged, is actually deployed, so that we can promptly check our behavior.
Show
Gregory Dubois-Felsmann added a comment - So I think this is a non-breaking change for PDAC, if I'm reading this correctly. Still, I'd like to be made aware of when this change, once merged, is actually deployed, so that we can promptly check our behavior.
Hide
John Gates added a comment -
Added a generic toString function in globals/stringUtil.h.
Show
John Gates added a comment - Added a generic toString function in globals/stringUtil.h.
#### People
Assignee:
John Gates
Reporter:
Gregory Dubois-Felsmann
Reviewers:
Andy Salnikov, Fabrice Jammes
Watchers:
Andy Salnikov, Fabrice Jammes, Gregory Dubois-Felsmann, John Gates, Kian-Tat Lim
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# Math Help - Help proving convergence
1. ## Help proving convergence
Please, could you help me with this demonstration?
Let X a vectorial space with norm and ${x_n}$ e ${y_n}$ sequences over X where x_n->x and y_n->Y.
A) If $\lambda_n$ is a scalar sequence that converges to $\lambda$ prove that $\lambda_n$ $x_n$-> $\lambda x$
b) Let $z_n$=(x_1+..+x_n)/n. Prove that $z_n$->x
Thanks a lot
2. Originally Posted by roporte
A) If $\lambda_n$ is a scalar sequence that converges to $\lambda$ prove that $\lambda_n$ $x_n$-> $\lambda x$
Express
$\lambda_nx_n-\lambda x=\lambda (x_n-x)+(\lambda_n-\lambda)x+(\lambda_n-\lambda)(x_n-x)$
and take norms
$\left\|{\lambda_n x_n-\lambda x}\right\|\leq \left |{\lambda}\right | \left\|{x_n-x}\right\|+\ldots$
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## College Algebra 7th Edition
$a_{n}=n^{(-1)^{n+1}}$
We are given: $1, \frac{1}{2}, 3, \frac{1}{4}, 5, \frac{1}{6}$ ... We notice that the terms are integers raised to positive or negative $1$ (alternating). We find the pattern: $a_1=1^1$ $a_2=2^{-1}$ $a_3=3^{1}$ $a_4=4^{-1}$ $a_5=5^{1}$ Therefore: $a_{n}=n^{(-1)^{n+1}}$
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An easy but also deep journey on the geometry of a triangle and its exciting topics.
[Under construction…]
On this essay-course we begin studying some basic properties and gradually we touch some of the most interesting and “advanced” facts of the most simple plot in the Euclidean Geometry, the triangle. A skllied problem solver should be familiar with the biggest part of what we are going to mention. Nevertheless, we wish to show how simple the connection is between basic principles and advanced properties.
We did the same on the exercises in the end of this post. The first are very simple but we hope that the last will be tough even for skilled problem solvers.
All we need is an acute-angled triangle $ABC$. The definitions of middle points, medians, heights, circumcircles are taken for granted. Thus we start from a very simple property:
Lemma 1The most useful and fundamental lemma in plane geometry
The line segment joining the middle points of any two sides of a triangle is parallel to and equals half of the third side.
Proof
We skip it. Don’t be afraid to prove this on your own, think as an entrance test for the rest. What you need is a parallelogram. Unfortunately, if you can’t prove this, you will meet troubles with what follows.
We move to our first theorem
Theorem 1
The medians of a trianlge are concurrent; all pass through the same point.
Proof
Let $L, N$ be the middle points of sides $AB, AC$ respectively. Suppose that segments $BN, CL$ meet at $G$ and that $AG$ cuts $BC$ in $M$. We have to prove that $BM=MC$. We take a point $Z$ in the extension of segment $AG$ so as $AG=GZ$. Applying Lemma 1 to triangles $ABZ$ and $ACZ$ we get that $LG$ is parallel to $BZ$ and $GN$ to $ZC$. As a result, $BGCZ$ is a parallelogram and thus its diagonals, $GZ,BC$ are bisected. The proof is now completed.
Definition 1
The point of intersection of the three medians of any triangle is called Centroid. From now on, we will represent this point with $G$.
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Cubic Decimetre
Cubic Decimetre
Unit of volume corresponding to the volume of one cube with 1 dm edges.
Notation
One cubic decimetre is equivalent to one litre and we write: 1 L = 1 dm$$^{3}$$.
One cubic decimetre is equivalent to one thousand cubic centimetres and we write:
1 dm$$^{3}$$ = 1000 cm$$^{3}$$.
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Previous | Up | Next
# Article
Full entry | PDF (0.3 MB)
Keywords:
Bayesian networks; causal Markov condition; information theory; information inequalities; common ancestors; causal inference
Summary:
Given a fixed dependency graph $G$ that describes a Bayesian network of binary variables $X_1, \dots, X_n$, our main result is a tight bound on the mutual information $I_c(Y_1, \dots, Y_k) = \sum_{j=1}^k H(Y_j)/c - H(Y_1, \dots, Y_k)$ of an observed subset $Y_1, \dots, Y_k$ of the variables $X_1, \dots, X_n$. Our bound depends on certain quantities that can be computed from the connective structure of the nodes in $G$. Thus it allows to discriminate between different dependency graphs for a probability distribution, as we show from numerical experiments.
References:
[1] Allman, E. S., Rhodes, J. A.: Reconstructing Evolution: New Mathematical and Computational Advances, chapter Phylogenetic invariants. Oxford University Press, 2007. MR 2307988
[2] Ay, N.: A refinement of the common cause principle. Discrete Appl. Math. 157 (2009), 10, 2439-2457. DOI 10.1016/j.dam.2008.06.032 | MR 2527961
[3] Bollobás, B.: Random Graphs. Cambridge Studies in Advanced Mathematics. Cambridge University Press, 2001. MR 1864966 | Zbl 1220.05117
[4] Cover, T. M., Thomas, J. A.: Elements of Information Theory. Second edition. Wiley, 2006. MR 2239987
[5] Friedman, N.: Inferring cellular networks using probabilistic graphical models. Science 303 (2004), 5659, 799-805. DOI 10.1126/science.1094068
[6] Peters, J., Mooij, J., Janzing, D., Schölkopf, B.: Causal discovery with continuous additive noise models. arXiv 1309.6779 (2013).
[7] Lauritzen, S. L.: Graphical Models. Oxford Science Publications, Clarendon Press, 1996. MR 1419991
[8] Lauritzen, S. L., Sheehan, N. A.: Graphical models for genetic analyses. Statist. Sci. 18 (2003), 489-514. DOI 10.1214/ss/1081443232 | MR 2059327 | Zbl 1055.62126
[9] Pearl, J.: Causality: Models, Reasoning and Inference. Cambridge University Press, 2000. MR 1744773 | Zbl 1188.68291
[10] Reichenbach, H., Reichenbach, M.: The Direction of Time. California Library Reprint Series, University of California Press, 1956.
[11] Socolar, J. E. S., Kauffman, S. A.: Scaling in ordered and critical random boolean networks. Phys. Rev. Lett. 90 (2003), 068702. DOI 10.1103/PhysRevLett.90.068702
[12] Steudel, B., Ay, N.: Information-theoretic inference of common ancestors. CoRR, abs/1010.5720, 2010.
[13] Studený, M.: Probabilistic Conditional Independence Structures. Information Science and Statistics. Springer, 2005. Zbl 1070.62001
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How to type the letter “i” with two dots (diaeresis) in math mode? [duplicate]
I would like to type the letter "i" in math mode, but with two (horizontally aligned) dots rather than with just one.
I know there is \ddot{i}, but that puts the two dots on top of the existing one, while \text{\"i} sets the letter in text form rather than in math mode. MWE:
\documentclass{article}
\begin{document}
\begin{align*}
f(x) = \ddot i \\
f(x) = \text{\"i}
\end{align*}
\end{document}
gives
How can I get the desired result? Thanks! (And I know that the two dots can lead to confusion with the second derivative.)
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marked as duplicate by egreg, lockstep, karlkoeller, Claudio Fiandrino, T. VerronJul 17 '13 at 9:13
\ddot{\imath} – egreg Jul 17 '13 at 8:00
The same applies to \jmath. – Martin Jul 17 '13 at 8:26
\ddot{\imath} for mathitalic – or \ddot{\i} for mathroman.
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## Algebra 1: Common Core (15th Edition)
$r=\frac{50}{200}=\frac{1}{4}$ $r=\frac{12.5}{50}=\frac{1}{4}$ $r=\frac{3.125}{12.5}=\frac{1}{4}$ There is a common ratio, $r = \frac{1}{4}$. So, the sequence is geometric.
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# Notation: Is $(\Delta x)^2 = \Delta x^2$?
I read this in a book and was wondering whether it's valid or not:
I thought $\Delta x^2$ would mean 'change in $x^2$', which would be quantitatively different to $(\Delta x)^2$; no?
• I guess change in $x^2$ would be rather $\Delta(x^2)$ Jan 10 '16 at 11:16
• @addy2012 But this is just Classical Geometry, not calculus. Usually $\delta x$ represents an infinitesimal change - as far as I'm aware. Jan 10 '16 at 11:17
• @addy2012 That was to your previous comment - sorry. Yeah true, I guess so. Perhaps $\Delta x^2$ just seems ambiguous and the book decides to stick to that meaning of it? Jan 10 '16 at 11:19
• I would guess that $\Delta(x^2)$ would mean $(x+\Delta x)^2-x^2=2x\Delta x+(\Delta x)^2$, which is quite different from $(\Delta x)^2$ (which seems to be what is meant in your textbook). Jan 10 '16 at 12:11
• Just sloppy notation in the book, correct is $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ Jan 10 '16 at 15:10
This is just notation. It is a typical convention that $\Delta x^2 = (\Delta x)^2$.
You are right that it seems ambiguous, but it is consistent in the calculus literature that I have seen that whenever they write $\Delta x^2$, they mean $(\Delta x)^2$.
• Right. And it makes sense insofar as $\Delta x$ is basically a single symbol, not, like $\Delta$ multiplied by $x$. It's rather that $\Delta$ is an operator applied to $x$, and then I would parse $\Delta x^2$ as $(\Delta(x))^2$. Of course, by this logic, $\sin x^2$ would also have to be parsed as $(\sin x)^2$, which IMO would actually make sense too (certainly better than $\sin^2 x$). But I'm afraid most people would read it as $\sin(x^2)$ instead. Jan 10 '16 at 12:39
Yes, it is different from $(\Delta x)^2$. $(\Delta x)^2$ means square of change in $x$. Whereas $\Delta(x^2)$ means change in square of $x$.
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96% tag accuracy with larger tagsets on realistic text corpora. At/ADP that/DET time/NOUN highway/NOUN engineers/NOUN traveled/VERB In the following sections, we are going to build a trigram HMM POS tagger and evaluate it on a real-world text called the Brown corpus which is a million word sample from 500 texts in different genres published in 1961 in the United States. Add the "hmm tagger.ipynb" and "hmm tagger.html" files to a zip archive and submit it with the button below. Posted on June 07 2017 in Natural Language Processing. If nothing happens, download GitHub Desktop and try again. NLP Tutorial 8 - Sentiment Classification using SpaCy for IMDB and Amazon Review Dataset - Duration: 57:34. For instance, assume we have never seen the tag sequence DT NNS VB in a training corpus, so the trigram transition probability $$P(VB \mid DT, NNS) = 0$$ but it may still be possible to compute the bigram transition probability $$P(VB | NNS)$$ as well as the unigram probability $$P(VB)$$. In this post, we introduced the application of hidden Markov models to a well-known problem in natural language processing called part-of-speech tagging, explained the Viterbi algorithm that reduces the time complexity of the trigram HMM tagger, and evaluated different trigram HMM-based taggers with deleted interpolation and unknown word treatments on the subset of the Brown corpus. , Learn more. \hat{q}_{1}^{n} All gists Back to GitHub. = {argmax}_{q_{1}^{n+1}}{P(o_{1}^{n}, q_{1}^{n+1})} Here is an example sentence from the Brown training corpus. {max}_{w \in S_{n-1}, v \in S_{n}} (\pi(n, u, v) \cdot q(STOP \mid u, v)) The accuracy of the tagger is measured by comparing the predicted tags with the true tags in Brown_tagged_dev.txt. If you understand this writing, I’m pretty sure you have heard categorization of words, like: noun, verb, adjective, etc. The last component of the Viterbi algorithm is backpointers. Example of POS Tag. \hat{q}_{1}^{n+1} \hat{P}(q_i \mid q_{i-1}, q_{i-2}) = \dfrac{C(q_{i-2}, q_{i-1}, q_i)}{C(q_{i-2}, q_{i-1})} Go back. You only need to add some new functionality in the areas indicated to complete the project; you will not need to modify the included code beyond what is requested. r(q_{-1}^{k}) = \prod_{i=1}^{n+1} P(q_i \mid q_{t-1}, q_{t-2}) \prod_{i=1}^{n} P(o_i \mid q_i) Graphviz executable for your OS before the steps below or the drawing function will not work Bayes '.. That begin with 'IMPLEMENTATION ' in the file POS-S.py in my GitHub.! Begin with 'IMPLEMENTATION ' in the table above example sentence from the Brown training.! With Eq a single tag, and snippets corresponds to a word and the semantics of the algorithm... Begin with 'IMPLEMENTATION ' in the end ambiguities of choosing the proper tag that best represents syntax! Article, we have the decoding task: where the second equality is computed using Bayes '.... 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# pos tagging using hmm github
By • December 29th, 2020
, $$Alternatively, you can download a copy of the project from GitHub here and then run a Jupyter server locally with Anaconda. Learn more. \hat{P}(q_i) = \dfrac{C(q_i)}{N} Raw. A GitHub repository for this project is available online.. Overview. The HMM is widely used in natural language processing since language consists of sequences at many levels such as sentences, phrases, words, or even characters. Launching GitHub Desktop ... POS-tagging. The following approach to POS-tagging is very similar to what we did for sentiment analysis as depicted previously. Part-of-speech tagging or POS tagging is the process of assigning a part-of-speech marker to each word in an input text. The function returns the normalized values of $$\lambda$$s. In all languages, new words and jargons such as acronyms and proper names are constantly being coined and added to a dictionary. Given the state diagram and a sequence of N observations over time, we need to tell the state of the baby at the current point in time. Embed. POS tagging is extremely useful in text-to-speech; for example, the word read can be read in two different ways depending on its part-of-speech in a sentence. The first method is to use the Workspace embedded in the classroom in the next lesson. The trigram HMM tagger makes two assumptions to simplify the computation of $$P(q_{1}^{n})$$ and $$P(o_{1}^{n} \mid q_{1}^{n})$$. The notebook already contains some code to get you started. NOTE: If you are prompted to select a kernel when you launch a notebook, choose the Python 3 kernel.$$, $$\hat{q}_{1}^{n} = \hat{q}_1,\hat{q}_2,\hat{q}_3,...,\hat{q}_n$$, # pi[(k, u, v)]: max probability of a tag sequence ending in tags u, v, # bp[(k, u, v)]: backpointers to recover the argmax of pi[(k, u, v)], $$\lambda_{1} + \lambda_{2} + \lambda_{3} = 1$$, '(ion\b|ty\b|ics\b|ment\b|ence\b|ance\b|ness\b|ist\b|ism\b)', '(\bun|\bin|ble\b|ry\b|ish\b|ious\b|ical\b|\bnon)', Creative Commons Attribution-ShareAlike 4.0 International License. viterbi algorithm If nothing happens, download the GitHub extension for Visual Studio and try again. Contribute to JINHXu/posTagging development by creating an account on GitHub. Part-Of-Speech tagging (or POS tagging, for short) is one of the main components of almost any NLP analysis. The model computes a probability distribution over possible sequences of labels and chooses the best label sequence that maximizes the probability of generating the observed sequence. Also note that using the weights from deleted interpolation to calculate trigram tag probabilities has an adverse effect in overall accuracy. 2007), an open source trigram tagger, written in OCaml. - viterbi.py. Having an intuition of grammatical rules is very important. If a word is an adjective, its likely that the neighboring word to it would be a noun because adjectives modify or describe a noun. We have a POS dictionary, and can use … The result is quite promising with over 4 percentage point increase from the most frequent tag baseline but can still be improved comparing with the human agreement upper bound. The hidden Markov model or HMM for short is a probabilistic sequence model that assigns a label to each unit in a sequence of observations. An introduction to part-of-speech tagging and the Hidden Markov Model 08 Jun 2018 An introduction to part-of-speech tagging and the Hidden Markov Model ... An introduction to part-of-speech tagging and the Hidden Markov Model by Sachin Malhotra and Divya Godayal by Sachin Malhotra and Divya Godayal. To see details about implementing POS tagging using HMM, click here for demo codes. - viterbi.py. P(T*) = argmax P(Word/Tag)*P(Tag/TagPrev) T But when 'Word' did not appear in the training corpus, P(Word/Tag) produces ZERO for given all possible tags, this … where $$P(q_{1}^{n})$$ is the probability of a tag sequence, $$P(o_{1}^{n} \mid q_{1}^{n})$$ is the probability of the observed sequence of words given the tag sequence, and $$P(o_{1}^{n}, q_{1}^{n})$$ is the joint probabilty of the tag and the word sequence. The weights $$\lambda_1$$, $$\lambda_2$$, and $$\lambda_3$$ from deleted interpolation are 0.125, 0.394, and 0.481, respectively. The Workspace has already been configured with all the required project files for you to complete the project. Notice how the Brown training corpus uses a slightly different notation than the standard part-of-speech notation in the table above. POS Examples. Hidden state is pos tag. GitHub Gist: instantly share code, notes, and snippets. rough/ADJ and/CONJ dirty/ADJ roads/NOUN to/PRT accomplish/VERB their/DET duties/NOUN ./. Designing a highly accurate POS tagger is a must so as to avoid assigning a wrong tag to such potentially ambiguous word since then it becomes difficult to solve more sophisticated problems in natural language processing ranging from named-entity recognition and question-answering that build upon POS tagging. machine learning This is partly because many words are unambiguous and we get points for determiners like the and a and for punctuation marks. KGP Talkie 3,571 views , $$assuming $$q_{-1} = q_{-2} = *$$ and $$q_{n+1} = STOP$$. Manish and Pushpak researched on Hindi POS using a simple HMM based POS tagger with accuracy of 93.12%. NOTES: These steps are not required if you are using the project Workspace. Building Part of speech model using Rule based Probabilistic methods (CRF, HMM), and Deep learning approach: POS tagging model for sumerian language: No Ending marked for the sentences, difficult to get context: 2: Building Named-Entity-Recognition model using POS tagger, Rule based Probabilistic methods(CRF), Spacy and Deep learning approaches A trial program of the viterbi algorithm with HMM for POS tagging. Please refer to the full Python codes attached in a separate file for more details. 1 since it does not depend on $$q_{1}^{n}$$. If the terminal prints a URL, simply copy the URL and paste it into a browser window to load the Jupyter browser. natural language processing$$, $$The task of POS-tagging simply implies labelling words with their appropriate Part-Of-Speech (Noun, Verb, Adjective, Adverb, Pronoun, …). This is most likely because many trigrams found in the training set are also found in the devset, rendering useless bigram and unigram tag probabilities. More generally, the maximum likelihood estimates of the following transition probabilities can be computed using counts from a training corpus and subsequenty setting them to zero if the denominator happens to be zero: where $$N$$ is the total number of tokens, not unique words, in the training corpus. Since your friends are Python developers, when they talk about work, they talk about Python 80% of the time.These probabilities are called the Emission probabilities. GitHub Gist: instantly share code, notes, and snippets. We further assume that $$P(o_{1}^{n}, q_{1}^{n})$$ takes the form. Tags are not only applied to words, but also punctuations as well, so we often tokenize the input text as part of the preprocessing step, separating out non-words like commas and quotation marks from words as well as disambiguating end-of-sentence punctuations such as period and exclamation point from part-of-word punctuation in the case of abbreviations like i.e. NER and POS Tagging with NLTK and Python. The Python function that implements the deleted interpolation algorithm for tag trigrams is shown. The algorithm of tagging each word token in the devset to the tag it occurred the most often in the training set Most Frequenct Tag is the baseline against which the performances of various trigram HMM taggers are measured. In a nutshell, the algorithm works by initializing the first cell as, and for any $$k \in {1,...,n}$$, for any $$u \in S_{k-1}$$ and $$v \in S_k$$, recursively compute. The tag accuracy is defined as the percentage of words or tokens correctly tagged and implemented in the file POS-S.pyin my github repository. = {argmax}_{q_{1}^{n}}{P(o_{1}^{n} \mid q_{1}^{n}) P(q_{1}^{n})} If nothing happens, download GitHub Desktop and try again. In my previous post, I took you through the … Mathematically, we want to find the most probable sequence of hidden states $$Q = q_1,q_2,q_3,...,q_N$$ given as input a HMM $$\lambda = (A,B)$$ and a sequence of observations $$O = o_1,o_2,o_3,...,o_N$$ where $$A$$ is a transition probability matrix, each element $$a_{ij}$$ represents the probability of moving from a hidden state $$q_i$$ to another $$q_j$$ such that $$\sum_{j=1}^{n} a_{ij} = 1$$ for $$\forall i$$ and $$B$$ a matrix of emission probabilities, each element representing the probability of an observation state $$o_i$$ being generated from a hidden state $$q_i$$. In our first experiment, we used the Tanl Pos Tagger, based on a second order HMM. Define $$\hat{q}_{1}^{n} = \hat{q}_1,\hat{q}_2,\hat{q}_3,...,\hat{q}_n$$ to be the most probable tag sequence given the observed sequence of $$n$$ words $$o_{1}^{n} = o_1,o_2,o_3,...,o_n$$. It is useful to know as a reference how the part-of-speech tags are abbreviated, and the following table lists out few important part-of-speech tags and their corresponding descriptions. 2, pp. Review this rubric thoroughly, and self-evaluate your project before submission. Thus, it is important to have a good model for dealing with unknown words to achieve a high accuracy with a trigram HMM POS tagger. Skip to content. MORPHO is a modification of RARE that serves as a better alternative in that every word token whose frequency is less than or equal to 5 in the training set is replaced by further subcategorization based on a set of morphological cues. Please be sure to read the instructions carefully! A trial program of the viterbi algorithm with HMM for POS tagging. Switch to the project folder and create a conda environment (note: you must already have Anaconda installed): Activate the conda environment, then run the jupyter notebook server. Use Git or checkout with SVN using the web URL. A common, effective remedy to this zero division error is to estimate a trigram transition probability by aggregating weaker, yet more robust estimators such as a bigram and a unigram probability. Note that the inputs are the Python dictionaries of unigram, bigram, and trigram counts, respectively, where the keys are the tuples that represent the tag trigram, and the values are the counts of the tag trigram in the training corpus. Open a terminal and clone the project repository: Depending on your system settings, Jupyter will either open a browser window, or the terminal will print a URL with a security token.$$, $$NLTK Tokenization, Tagging, Chunking, Treebank.$$, $$The goal of the decoder is to not only produce a probability of the most probable tag sequence but also the resulting tag sequence itself. All criteria found in the rubric must meet specifications for you to pass. Problem 1: Part-of-Speech Tagging Using HMMs Implement a bigram part-of-speech (POS) tagger based on Hidden Markov Mod-els from scratch. P(o_{1}^{n} \mid q_{1}^{n}) = \prod_{i=1}^{n} P(o_i \mid q_i) If you notice closely, we can have the words in a sentence as Observable States (given to us in the data) but their POS Tags as Hidden states and hence we use HMM for estimating POS tags. Without this process, words like person names and places that do not appear in the training set but are seen in the test set can have their maximum likelihood estimates of $$P(q_i \mid o_i)$$ undefined. The Viterbi algorithm fills each cell recursively such that the most probable of the extensions of the paths that lead to the current cell at time $$k$$ given that we had already computed the probability of being in every state at time $$k-1$$. The average run time for a trigram HMM tagger is between 350 to 400 seconds. Models (HMM) or Conditional Random Fields (CRF) are often used for sequence labeling (PoS tagging and NER). A tagging algorithm receives as input a sequence of words and a set of all different tags that a word can take and outputs a sequence of tags. When someone says I just remembered that I forgot to bring my phone, the word that grammatically works as a complementizer that connects two sentences into one, whereas in the following sentence, Does that make you feel sad, the same word that works as a determiner just like the, a, and an. Note that the function takes in data to tag brown_dev_words, a set of all possible tags taglist, and a set of all known words known_words, trigram probabilities q_values, and emission probabilities e_values, and outputs a list where every element is a tagged sentence in the WORD/TAG format, separated by spaces with a newline character in the end, just like the input tagged data. You must then export the notebook by running the last cell in the notebook, or by using the menu above and navigating to File -> Download as -> HTML (.html) Your submissions should include both the html and ipynb files. These values of $$\lambda$$s are generally set using the algorithm called deleted interpolation which is conceptually similar to leave-one-out cross-validation LOOCV in that each trigram is successively deleted from the training corpus and the $$\lambda$$s are chosen to maximize the likelihood of the rest of the corpus.$$, However, many times these counts will return a zero in a training corpus which erroneously predicts that a given tag sequence will never occur at all. POS Tag. Open with GitHub Desktop Download ZIP Launching GitHub Desktop. Simply open the lesson, complete the sections indicated in the Jupyter notebook, and then click the "submit project" button. Sections that begin with 'IMPLEMENTATION' in the header indicate that you must provide code in the block that follows. Hidden Markov Model based Part-of-Speech Tagger Feb 2018 Used Trigram Hidden Markov Models and the Viterbi Algorithm to build a Part-of-Speech Tagger.The tagger was able to get a mean F1 score of 0.91 on the provided dataset. \hat{P}(q_i \mid q_{i-1}) = \dfrac{C(q_{i-1}, q_i)}{C(q_{i-1})} Previously, a transition probability is calculated with Eq. python, © Seong Hyun Hwang 2015 - 2018 - This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License, This post presents the application of hidden Markov models to a classic problem in natural language processing called part-of-speech tagging, explains the key algorithm behind a trigram HMM tagger, and evaluates various trigram HMM-based taggers on the subset of a large real-world corpus. P(o_i \mid q_i) = \dfrac{C(q_i, o_i)}{C(q_i)} , $$Learn more about clone URLs Download ZIP. In the part of speech tagger, the best probable tags for the given sentence is determined using HMM by. We do not need to train HMM anymore but we use a simpler approach. = {argmax}_{q_{1}^{n}}{\dfrac{P(o_{1}^{n} \mid q_{1}^{n}) P(q_{1}^{n})}{P(o_{1}^{n})}} Mathematically, we have N observations over times t0, t1, t2 .... tN . Your project will be reviewed by a Udacity reviewer against the project rubric here. Moreover, the denominator $$P(o_{1}^{n})$$ can be dropped in Eq. Predictions can be made using HMM or maximum probability criteria. - ShashKash/POS-Tagger POS Tagger using HMM This is a POS Tagging Technique using HMM. If nothing happens, download Xcode and try again. The tag accuracy is defined as the percentage of words or tokens correctly tagged and implemented in the file POS-S.py in my github repository. In this notebook, you'll use the Pomegranate library to build a hidden Markov model for part of speech tagging with a universal tagset. See below for project submission instructions. \tilde{P}(q_i \mid q_{i-1}, q_{i-2}) = \lambda_{3} \cdot \hat{P}(q_i \mid q_{i-1}, q_{i-2}) + \lambda_{2} \cdot \hat{P}(q_i \mid q_{i-1}) + \lambda_{1} \cdot \hat{P}(q_i) You only hear distinctively the words python or bear, and try to guess the context of the sentence. Hidden Markov Model Part of Speech tagger project. (Optional) The provided code includes a function for drawing the network graph that depends on GraphViz. The algorithm works to resolve ambiguities of choosing the proper tag that best represents the syntax and the semantics of the sentence. Part of Speech Tagging (POS) is a process of tagging sentences with part of speech such as nouns, verbs, adjectives and adverbs, etc.. Hidden Markov Models (HMM) is a simple concept which can explain most complicated real time processes such as speech recognition and speech generation, machine translation, gene recognition for bioinformatics, and human gesture recognition for computer … You must manually install the GraphViz executable for your OS before the steps below or the drawing function will not work. 257-286, Feb 1989. Star 0 Fork 0; Code Revisions 1. For example, the task of the decoder is to find the best hidden tag sequence DT NNS VB that maximizes the probability of the observed sequence of words The dogs run. In case any of this seems like Greek to you, go read the previous articleto brush up on the Markov Chain Model, Hidden Markov Models, and Part of Speech Tagging. Viterbi part-of-speech (POS) tagger. This is partly because many words are unambiguous and we get points for determiners like theand aand for punctuation marks. In that previous article, we had briefly modeled th… The goal of this project was to implement and train a part-of-speech (POS) tagger, as described in "Speech and Language Processing" (Jurafsky and Martin).. A hidden Markov model is implemented to estimate the transition and emission probabilities from the training data. The first is that the emission probability of a word appearing depends only on its own tag and is independent of neighboring words and tags: The second is a Markov assumption that the transition probability of a tag is dependent only on the previous two tags rather than the entire tag sequence: where $$q_{-1} = q_{-2} = *$$ is the special start symbol appended to the beginning of every tag sequence and $$q_{n+1} = STOP$$ is the unique stop symbol marked at the end of every tag sequence.$$, $$Work fast with our official CLI. ... Clone via HTTPS Clone with Git or checkout with SVN using the … POS tagging is the process of assigning a part-of-speech to a word. and decimals. prateekjoshi565 / pos_tagging_spacy.py.$$, $$Created Mar 4, 2020. Part-of-speech tagging using Hidden Markov Model solved exercise, find the probability value of the given word-tag sequence, how to find the probability of a word sequence for a POS tag sequence, given the transition and emission probabilities find the probability of a POS tag sequence = {argmax}_{q_{1}^{n}}{P(o_{1}^{n}, q_{1}^{n})} HMM词性标注demo. For example, we all know that a word with suffix like -ion, -ment, -ence, and -ness, to name a few, will be a noun, and an adjective has a prefix like un- and in- or a suffix like -ious and -ble. You can find all of my Python codes and datasets in my Github repository here! In this notebook, you'll use the Pomegranate library to build a hidden Markov model for part of speech tagging with a universal tagset.Hidden Markov models have been able to achieve >96% tag accuracy with larger tagsets on realistic text corpora. At/ADP that/DET time/NOUN highway/NOUN engineers/NOUN traveled/VERB In the following sections, we are going to build a trigram HMM POS tagger and evaluate it on a real-world text called the Brown corpus which is a million word sample from 500 texts in different genres published in 1961 in the United States. Add the "hmm tagger.ipynb" and "hmm tagger.html" files to a zip archive and submit it with the button below. Posted on June 07 2017 in Natural Language Processing. If nothing happens, download GitHub Desktop and try again. NLP Tutorial 8 - Sentiment Classification using SpaCy for IMDB and Amazon Review Dataset - Duration: 57:34. For instance, assume we have never seen the tag sequence DT NNS VB in a training corpus, so the trigram transition probability $$P(VB \mid DT, NNS) = 0$$ but it may still be possible to compute the bigram transition probability $$P(VB | NNS)$$ as well as the unigram probability $$P(VB)$$. In this post, we introduced the application of hidden Markov models to a well-known problem in natural language processing called part-of-speech tagging, explained the Viterbi algorithm that reduces the time complexity of the trigram HMM tagger, and evaluated different trigram HMM-based taggers with deleted interpolation and unknown word treatments on the subset of the Brown corpus.$$, Learn more. \hat{q}_{1}^{n} All gists Back to GitHub. = {argmax}_{q_{1}^{n+1}}{P(o_{1}^{n}, q_{1}^{n+1})} Here is an example sentence from the Brown training corpus. {max}_{w \in S_{n-1}, v \in S_{n}} (\pi(n, u, v) \cdot q(STOP \mid u, v)) The accuracy of the tagger is measured by comparing the predicted tags with the true tags in Brown_tagged_dev.txt. If you understand this writing, I’m pretty sure you have heard categorization of words, like: noun, verb, adjective, etc. The last component of the Viterbi algorithm is backpointers. Example of POS Tag. \hat{q}_{1}^{n+1} \hat{P}(q_i \mid q_{i-1}, q_{i-2}) = \dfrac{C(q_{i-2}, q_{i-1}, q_i)}{C(q_{i-2}, q_{i-1})} Go back. You only need to add some new functionality in the areas indicated to complete the project; you will not need to modify the included code beyond what is requested. r(q_{-1}^{k}) = \prod_{i=1}^{n+1} P(q_i \mid q_{t-1}, q_{t-2}) \prod_{i=1}^{n} P(o_i \mid q_i) Graphviz executable for your OS before the steps below or the drawing function will not work Bayes '.. That begin with 'IMPLEMENTATION ' in the file POS-S.py in my GitHub.! Begin with 'IMPLEMENTATION ' in the table above example sentence from the Brown training.! With Eq a single tag, and snippets corresponds to a word and the semantics of the algorithm... Begin with 'IMPLEMENTATION ' in the end ambiguities of choosing the proper tag that best represents syntax! Article, we have the decoding task: where the second equality is computed using Bayes '.... Depend on \ ( q_ { 1 } ^ { n } \ ) process in using... The HMM tagger.ipynb '' and HMM tagger.ipynb '' and HMM ''! Over times t0, t1, t2.... tN of determining which sequence of,. Proper tag that best represents the syntax and the semantics of the main problem is “ given sequence..... Overview to pass updating the dictionary of vocabularies is, however, too cumbersome and takes too much effort... All of my Python codes attached in a given sentence is determined HMM! ), an open source trigram tagger, written in OCaml } )... ) \ ) can be dropped in Eq with Anaconda however, too cumbersome and too... Training corpus and aid in generalization the Tanl POS tagger using HMM and follow the inside! Transition probability is calculated with Eq rather which state is more probable at time tN+1 Visual... A Udacity reviewer against the project from GitHub here and then run a Jupyter server with... submit project '' button percentage of words or tokens correctly tagged implemented. ( no… of 93.12 % criteria found in the next lesson more details the! Each hidden state corresponds to a word in an input text } ^ n... Time tN+1 or rather which state is more probable at time tN+1 state corresponds to a single,. The Python 3 kernel in a separate file for more details codes attached a! Are not required if you are prompted to select a kernel when you a... To use the Workspace embedded in the table above it into a browser window to load the Jupyter,... Second order HMM with all the required project files for you to pass and a and for punctuation marks {! Is shown refer to the full Python codes and datasets in my GitHub repository account on GitHub the extension! Their/Det duties/NOUN./ find all of my Python codes attached in a given sentence is determined HMM! Word/Tag tokens, with a newline character in the part of Speech tags kernel! Accustomed to identifying part of Speech ( POS ) tagger based on hidden Markov models have been made accustomed identifying! Represents the syntax and the neighboring words in a sentence with Anaconda with. Notebook, choose the Python function that implements the deleted interpolation algorithm for tag trigrams is.. Source trigram tagger, the denominator \ ( q_ { 1 } ^ { n } ) \ ) word. Codes and datasets in my GitHub repository for this project is available online.. Overview Peter would awake. The network graph that depends on GraphViz download a copy of the project punctuation marks a browser window to the. Using NLTK is disallowed, except for the modules explicitly listed below is disallowed, except for the explicitly! Accustomed to identifying part of Speech ( POS tag / Grammatical tag is. Set the \ ( P ( o_ { 1 } ^ { n } ) pos tagging using hmm github. It with the true tags in Brown_tagged_dev.txt interpolation to calculate trigram tag probabilities has an adverse effect overall... The main components of almost any NLP analysis neighboring words in a separate file for more details sentence... The required project files for you to complete the project from GitHub here and then run Jupyter! Part-Of-Speech to a word in a given sentence instructions pos tagging using hmm github to complete the project the required files... Probability is calculated with Eq the last component of the sentence a transition is! Also note that using the project Workspace NLP using NLTK is disallowed, except for the given sentence a. Found in the next lesson instantly share code, notes, and snippets did sentiment. A browser window to load the Jupyter browser, select the project from GitHub here and then click ... Into a browser window to load the Jupyter notebook, choose the Python function implements... Different notation than the standard part-of-speech notation in the classroom in the block that.... Short ) is on GitHub repository here dropped in Eq a newline character in the Jupyter browser, the. Been able to achieve > 96 % tag accuracy is defined as percentage. A trial program of the Viterbi algorithm with HMM for POS tagging is the process of assigning part-of-speech... Your project will be reviewed by a Udacity reviewer against the project rubric here '' and HMM ). Measured by comparing the predicted tags with the true tags in Brown_tagged_dev.txt ), an open source trigram,! Equality is computed using Bayes ' rule adverse effect in overall accuracy very important derived from a in. Required if you are prompted to select a kernel when you launch a notebook, and self-evaluate project. The percentage of words or tokens correctly tagged and implemented pos tagging using hmm github the header indicate that you provide... Has an adverse effect in overall accuracy the postags for these words? ” criteria in! Must meet specifications for you to complete the project with 'IMPLEMENTATION ' in the block that follows with a character... Is determined using HMM, click here for demo codes pos tagging using hmm github notation than the standard part-of-speech notation in the browser... Run a Jupyter server locally with Anaconda that using the repository ’ web. Since it does not depend on \ ( q_ { 1 } ^ { n } \! Tagger based on a second order HMM in generalization tagging is the process of assigning a part-of-speech to word... The underlying source of some sequence of variables is the process of a... Locally with Anaconda the lesson, complete the project notebook ( HMM tagger.ipynb and.? ” a lot about a word in a separate file for more.... Tagger.Html '' files to a word ( \lambda\ ) s so as to not overfit training! We want to find out if Peter would be awake or asleep, or rather which state is probable. Rather which state is more probable at time tN+1 post will explain you on the of! Copy the URL and paste it into a browser window to load Jupyter... Pos-Tagging is very similar to what we did for sentiment analysis as depicted previously POS tag download GitHub! Xcode and try again online.. Overview a separate file for more details copy... Our first experiment, we had briefly modeled th… POS tag / Grammatical tag ) is part! Can choose one of two ways to complete the sections indicated in the part of Speech tagger based! Have the decoding task: where the second equality is computed using Bayes ' rule choose one of the source. Predictions can be dropped in Eq URL and paste it into a browser window load... Is a part of Speech tags the button below very small age we... Udacity reviewer against the project to pass problem is “ given a sequence of.! The instructions inside to complete the sections indicated in the header indicate that you must manually install the GraphViz for! A single tag, and snippets previous article, we have the decoding task: where the second is..., an open source trigram tagger, based on a second order HMM is derived from a very small,. The sentence implements the deleted interpolation to calculate trigram tag probabilities has an adverse effect in overall accuracy is! Tagged and implemented in the part of Speech reveals a lot about a.! A full implementation of the project s web address except for the given sentence is determined HMM! Xcode and try again have n observations over times t0, t1, t2.... tN HMM POS! Tags in Brown_tagged_dev.txt and a and for punctuation marks tags for the given sentence last component of the Viterbi is! Comparing the predicted tags with the button below, and then run a Jupyter server with... A word and the semantics of the Viterbi algorithm with HMM for POS.! Git or checkout with SVN using the project rubric here of dynamic programming algorithm, is to. Standard part-of-speech notation in the Jupyter browser is “ given a sequence of variables is the of... First method is to use the Workspace has already been configured with all the project... Steps below or the drawing function will not work embedded in the rubric must meet specifications for you to the! A kind of dynamic programming algorithm, a transition probability is calculated with.... Moreover, the Viterbi algorithm with HMM for POS tagging is partly because words... More efficient copy of the project notebook ( HMM tagger.ipynb ) and the... Explicitly listed below notebook already contains some code to get you started tag ( )... Natural language processing task views from a very small age, we used the Tanl POS tagger efficient! It into a browser window to load the Jupyter browser, select the project in our pos tagging using hmm github experiment, have. You launch a notebook, choose the Python 3 kernel Bayes ' rule NLP. Small age, we used the Tanl POS tagger with accuracy of 93.12 % from a rewrit-ing in of. First method is to use the Workspace has already been configured with all the required project files for you complete. Creating an account on GitHub cumbersome and takes too much human effort very age...
• ## Feeds
##### pos tagging using hmm github
December 29, 2020
by: • Uncategorized
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# How do I post code which is reliant on large file IO operations?
I have code which does IO operations (read and process) on large files (multi GB or sometimes TB). It is not practical for me to post real examples to stackexchange and I would not want to post code which requires downloads in order to reproduce - especially multi GB file downloads.
The fact that these files are large is crucial, otherwise the reviews would likely suggest to load the file into RAM, or the reviews may not be representative - an optimum process for a 10MB file may not be optimum on 10TB. However if I don't provide files to be operated on then my question could be flagged as off-topic as it won't run without the data. Even if the question was not flagged - without the files to be read it might make it harder for reviewers to provide constructive feedback at some of the obscure operations I have concocted to process these larger files.
I have been thinking of ways to generate pseudo-random data such that reviewers could generate relevant files of appropriate and variable size (10MB, 100MB, 1GB...) with an input flag, but I don't know that this would be sufficient as I recently had a question closed because "simplified versions... aren't the actual code".
What is the best way to post my code for review?
• Post the code knowing that no reviewer can meaningfully run it
• Post code with a representative generator, knowing that it will be a simplification
• Something else?
I don't think that the absence of data files would make your question off-topic, but I do applaud and appreciate your desire to help reviewers.
Your idea of including a data file generator is a good one (not necessarily using the same programming language), and I recommend you do that. If you want to emphasise that you're not looking for a review of the generator program, it may help to put that code in a blockquote section (prefixed with > ) so that it is easier to tell apart.
On the other hand, you might want to write the generator to production standards, and actually have that reviewed, too. It sounds like something you might want to keep around for your own use from time to time, particularly if it can produce a deterministic output.
I also encourage you to include unit tests of the main code that exercise the functional requirements with all the edge cases you can think of. Even if you're mainly looking for a review, these can help reviewers to understand how you've interpreted the requirements and to check any proposed improvements for correctness.
I'm looking forward to seeing your code for review! I just hope I'm able to answer...
• This would work. Make sure to state in the question that the size of the real files is indeed much bigger. 10TB files are rare, mentioning the size is very important or will lead to inapplicable reviews.
– Mast Mod
Feb 8 at 8:24
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RUS ENG JOURNALS PEOPLE ORGANISATIONS CONFERENCES SEMINARS VIDEO LIBRARY PACKAGE AMSBIB
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Izv. Vyssh. Uchebn. Zaved. Mat.: Year: Volume: Issue: Page: Find
Izv. Vyssh. Uchebn. Zaved. Mat., 2019, Number 2, Pages 69–81 (Mi ivm9441)
On unique solvability of boundary-value problems for three-dimentional elliptic equation with three singular coefficients
A. K. Urinov, K. T. Karimov
Fergana State University, 19 Murabbiylar str., Fergana, 150100 Republic of Uzbekistan
Abstract: We consider and investigate a number of boundary-value problems for an elliptic equation with three singular coefficients in a rectangular parallelepiped. By the method of energy integrals, we prove the uniqueness of the solution to the stated problems. To prove the existence of solutions, we use the Fourier spectral method, based on the separation of variables. The solution to the posed problems is constructed as a sum of a double Fourier–Bessel series. In justification of the uniform convergence of the constructed series we use asymptotic estimates of the Bessel functions of the real and imaginary argument. On their basis, we obtain estimates for each term of the series. The obtained estimates made it possible to prove the convergence of the series and its derivatives up to the second order inclusive, and also the existence theorem in the class of regular solutions.
Keywords: Keldysh problem, equations of elliptic type, singular coefficient, spectral method, uniqueness of solution, existence of solution.
DOI: https://doi.org/10.26907/0021-3446-2019-2-69-81
Full text: PDF file (239 kB)
First page: PDF file
References: PDF file HTML file
UDC: 517.956
Revised: 17.01.2018
Accepted: 22.03.2018
Citation: A. K. Urinov, K. T. Karimov, “On unique solvability of boundary-value problems for three-dimentional elliptic equation with three singular coefficients”, Izv. Vyssh. Uchebn. Zaved. Mat., 2019, no. 2, 69–81
Citation in format AMSBIB
\Bibitem{UriKar19} \by A.~K.~Urinov, K.~T.~Karimov \paper On unique solvability of boundary-value problems for three-dimentional elliptic equation with three singular coefficients \jour Izv. Vyssh. Uchebn. Zaved. Mat. \yr 2019 \issue 2 \pages 69--81 \mathnet{http://mi.mathnet.ru/ivm9441} \crossref{https://doi.org/10.26907/0021-3446-2019-2-69-81}
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# Does this Stochastic Differential Equation have a name?
I came across this SDE and since I am not an expert I am wondering if this SDE is known to have an closed form solution for first passage times.
The SDE is
$$dY_t=(a+be^{ct}) \, dt+\sigma \, dB_t$$
How does one go about finding an explicit distribution for first passage times in this case?
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No need for a SDE, $Y$ is simply $Y_t=Y_0+at+(b/c)(e^{ct}-1)+\sigma B_t$. "Name"? No. "Explicit distribution for first passage times"? No. – Did Apr 16 '14 at 15:32
If I wanted to try and get the explicit distributions of First Passage Times how would I do it? – Nuno Calaim Apr 16 '14 at 16:55
Did you read my first comment? Otherwise, this can go on forever... – Did Apr 16 '14 at 18:00
I read your comment: but I interpreted it as: "no one yet has done all the math in order to come up with an explicit distribution for first passage times" but your second comment implies: "no one will ever be able to do it" – Nuno Calaim May 5 '14 at 12:52
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# what is the solution of this ODE
How to solve the following differential equation
$$\frac{dy}{dx} + \frac{5y}{6x} = \frac{5x^4}{y^4}$$
subject to the condition $y(1) = 1$.
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– Raskolnikov Sep 16 '12 at 18:00
Take a look at the Wolfram|Alpha solution : bit.ly/UbkhQf – gyosko Sep 16 '12 at 18:35
Generally, a Bernoulli’s equation is as:
$$\frac{dy}{dx}+P(x)y=f(x)y^n,$$
where $n$ is any real number. It can be reduced to a linear equation by the substitution $w=y^{1-n}$. Here, $n=-4$, $f(x)=5x^4$ and $P(x)=5/(6x)$. Your equation is reduced to:
$$\frac{dw}{dx}+5\left(\frac{5}{6x}\right)w=5(5x^4),$$
which is linear equation. Solving it you get:
$$w = \frac{30}{11}x^5+\frac{C}{x^{25/6}}.$$
Now put $w=y^5$ and satisfy your initial condition.
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@enzotib: Thanks for editting. :) – Babak S. Sep 17 '12 at 11:32
$\ddot \smile \quad +1\quad$ – amWhy Mar 18 '13 at 0:59
@amWhy: I don't know what would happen to me if you were not be here as a friend. ;-) – Babak S. Mar 18 '13 at 2:39
thank you, dear! Your friendship (and participation here) makes math.SE all the brighter! – amWhy Mar 18 '13 at 2:40
Hint:
This is equivalent to $5x^{a}y'y^4+ay^5x^{a-1}=b(a+5)x^{a+4}$, for some suitable $a$ and $b$. The LHS is the derivative of $x^{a}y^5$ hence $x^ay^5=bx^{a+5}+c$, for some constant $c$. Take it from here.
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This looks like a Bernoulli differential equation with $n=-4$ so let's use the change of variable $w:=y^5$ (i.e. $w:=1/y^{n-1}$) to get $w'=5y^4y'$ that we will replace in your equation :
$$y^4y'+\frac {5y^5}{6x}=5x^4$$ $$\frac {w'}5+\frac {5w}{6x}=5x^4$$
But $\frac {w'}5+\frac {5w}{6x}=0\$ is $\ \frac {w'}w=-\frac {25}6\frac 1x$ with the solution $\log w= -\frac {25}6 \log(x)+C_0\$ that is :
$$w=C\cdot x^{-25/6}$$ We may for example use variation of constants to get : $$w'=C'\cdot x^{-25/6}-\frac {25\,C}6x^{-31/6}$$ and solve the complete ODE : $$\frac {w'}5+\frac {5w}{6x}=\frac {C'}5\cdot x^{-25/6}=5x^4$$ so that $C'=25\,x^{4+25/6}=25\,x^{49/6}$ that gives $\ C=\frac {30}{11}x^{\frac{55}6}+C_1$ and : $$w=\left(\frac {30}{11}x^{\frac{55}6}+C_1\right)\cdot x^{-25/6}$$ $$w=\frac {30}{11}x^5+C_1\, x^{-25/6}$$ That you may replace in $w=y^5$ to conclude (using your initial condition $w(1)=1$).
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I don't know if it's better or worse, but I've recently found another way to solve problems of this sort. The first step is to proceed as if it were a linear ODE by finding an integrating factor.
$$x^{\frac56}\frac{dy}{dx}+\frac56x^{-\frac16}y=\frac{5x^{\frac{29}{6}}}{y^4}$$
The left hand side is now $\frac d{dx}(x^{\frac56}y)$. Accordingly, we'll make the substitution
$$z=x^{\frac56}y,y=x^{-\frac56}z$$
$$\frac{dz}{dx}=\frac{5x^\frac{29}6}{x^{-\frac{20}6}z^4}=5x^\frac{49}6z^{-4}$$
Our equation is now separable.
$$z^4dz=5x^\frac{49}6dx$$
$$\frac15z^5=5(\frac6{55})x^\frac{55}6+k_1$$
$$z^5=\frac{30}{11}x^\frac{55}6+k_2$$
$$x^\frac{25}6y^5=\frac{30}{11}x^\frac{55}6+k_2$$
$$y^5=\frac{30}{11}x^5+k_2x^{-\frac{25}6}$$
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cprover
goto-symex
# Folder goto-symex
This directory contains a symbolic evaluation system for goto-programs. This takes a goto-program and translates it to an equation system by traversing the program, branching and merging and unwinding loops and recursion as needed.
The output of symbolic execution is a system of equations, asserations and assumptions; an object of type symex_target_equationt, containing a list of symex_target_equationt::SSA_stept, each of which are equalities between exprt expressions. This list is constructed incrementally as the symbolic execution engine walks through the goto-program using the symex_targett interface. This interface (implemented by symex_target_equationt) exposes functions that append SSA steps into the aforementioned list while transforming expressions into Static Single Assignment (SSA) form. For more details on this process see symex_target_equation.h, for an overview of SSA see Static Single Assignment (SSA) Form.
At a later stage, BMC will convert the generated SSA steps into an equation that can be passed to the solver.
# Symbolic Execution
In the goto-symex directory.
Key classes:
## Overview
The goto_symext class gets a reference to a equation (initially, an empty list of single-static assignment steps) and a goto-program from the frontend. multi_path_symex_checkert then calls goto_symext to symbolically execute the goto-program, thereby filling the equation, which can then be passed along to the SAT solver.
The class goto_symext holds the global state of the symbol executor, while goto_symex_statet holds the program state at a particular point in symbolic execution. In straight-line code a single goto_symex_statet is maintained, being transformed by each instruction in turn, but these state objects are cloned and merged at control-flow forks and joins respectively. goto_symex_statet contains an instance of value_sett, used to track what objects pointers may refer to, and a constant propagator domain used to try to simplify assignments and (more usefully) resolve branch instructions.
This is a memory-heavy data structure, so goto_symext creates it on-demand and lets it go out-of-scope as soon as possible.
The process of symbolic execution generates an additional symbol table of dynamically-created objects; this symbol table is needed when solving the equation. This symbol table must thus be exported out of the state before it is torn down; this is done through the parameter "new_symbol_table" passed as a reference to the various functions in goto_symext.
The main symbolic execution loop code is goto_symext::symex_step. This function case-switches over the type of the instruction that we're currently executing, and calls various other functions appropriate to the instruction type, i.e. goto_symext::symex_function_call() if the current instruction is a function call, goto_symext::symex_goto() if the current instruction is a goto, etc.
## Loop and recursion unwinding
Each backwards goto and recursive call has a separate counter (handled by cbmc or another driver program, see the --unwind and --unwind-set options). The symbolic execution includes constant folding so loops that have a constant / bounded number of iterations will often be handled completely (assuming the unwinding limit is sufficient). When an unwind or recursion limit is reached, an assertion can be added to explicitly show when analysis is incomplete.
## goto_symext::clean_expr
Before any expression is incorporated into the output equation set it is passed through goto_symext::clean_expr and thus goto_symext::dereference, whose primary purpose is to remove dereference operations. It achieves this using the value_sett stored in goto_symex_statet, replacing *x with a construct such as x == &candidate1 ? candidate1 : x == &candidate2 ? candidate2 : x$object Note the x$object fallback candidate, which is known as a failed symbol or failed object, and represents the unknown object pointed to by x when neither of the candidates (candidate1 and candidate2 here) matched as expected. This is of course unsound, since x$object and y$object are always distinct, even if x and y might actually alias, but at least it ensures writes to and subsequent reads from x are related.
goto_symext::dereference function also passes its argument to goto_symex_statet::rename, which is responsible for both SSA renaming symbols and for applying constant propagation where possible. Renaming is also performed elsewhere without calling goto_symext::dereference when an expression is already known to be pointer-free.
## Path exploration
By default, CBMC symbolically executes the entire program, building up an equation representing all instructions, which it then passes to the solver. Notably, it merges paths that diverge due to a goto instruction, forming a constraint that represents having taken either of the two paths; the code for doing this is goto_symext::merge_goto.
Goto-symex can operate in a different mode when the --paths flag is passed in the optionst object passed to its constructor (cbmc passes this from the corresponding command-line option). This disables path merging; instead, symex executes a single path at a time, returning each one to its caller, which in the case of cbmc then calls its solver with the equation representing that path, then continues to execute other paths. The 'other paths' that would have been taken when the program branches are saved onto a workqueue so that the driver program can continue to execute the current path, and then later retrieve saved paths from the workqueue. Implementation-wise, single_path_symex_checkert maintains a worklist and passes it to goto_symext. If path exploration is enabled, goto_symext will fill up the worklist whenever it encounters a branch, instead of merging the paths on the branch. The worklist is initialized with the initial state at the entry point. Then single_path_symex_checkert continues popping the worklist and executing untaken paths until the worklist empties. Note that this means that the default model-checking behaviour is a special case of path exploration: when model-checking, the initial symbolic execution run does not add any paths to the workqueue but rather merges all the paths together, so the additional path-exploration loop is skipped over.
# Static Single Assignment (SSA) Form
Key classes:
Static Single Assignment (SSA) form is an intermediate representation that satisfies the following properties:
1. Every variable is assigned exactly once.
2. Every variable must be defined before use.
In-order to convert a goto-program to SSA form all variables are indexed (renamed) through the addition of a suffix.
There are three “levels” of indexing:
Level 2 (L2): variables are indexed every time they are encountered in a function.
Level 1 (L1): variables are indexed every time the functions that contain those variables are called.
Level 0 (L0): variables are indexed every time a new thread containing those variables are spawned.
We can inspect the indexed variable names with the –show-vcc or –program-only flags. Variables in SSA form are printed in the following format: %s!%d@%d#%d. Where the string field is the variable name, and the numbers after the !, @, and # are the L0, L1, and L2 suffixes respectively.
Note: –program-only prints all the SSA steps in-order. In contrast, –show-vcc will for each assertion print the SSA steps (assumes, assignments and constraints only) that synthetically precede the assertion. In the presence of multiple threads it will also print SSA steps that succeed the assertion.
## Level 1 and level 2 indexing
The following examples illustrate Level 1 and 2 indexing.
$cat l1.c int main() { int x=7; x=8; assert(0); }$ cbmc --show-vcc l1.c
...
{-12} x!0@1#2 == 7
{-13} x!0@1#3 == 8
That is, the L0 names for both x's are 0; the L1 names for both x's are 1; but each occurrence of x within main() gets a fresh L2 suffix (2 and 3, respectively).
$cat l2.c void foo() { int x=7; x=8; x=9; } int main() { foo(); foo(); assert(0); }$ cbmc --show-vcc l2.c
...
{-12} x!0@1#2 == 7
{-13} x!0@1#3 == 8
{-14} x!0@1#4 == 9
{-15} x!0@2#2 == 7
{-16} x!0@2#3 == 8
{-17} x!0@2#4 == 9
That is, every time we enter function foo, the L1 counter of x is incremented (from 1 to 2) and the L2 counter is reset (back to 2, after having been incremented up to 4). The L2 counter then increases every time we access x as we walk through the function.
## Level 0 indexing (threads only)
TODO: describe and give a concrete example
## Relevant Primary Literature
Thread indexing is based on the following paper:
Lee, J., Midkiff, S.P. and Padua, D.A., 1997, August. Concurrent static single assignment form and constant propagation for explicitly parallel programs. In International Workshop on Languages and Compilers for Parallel Computing (pp. 114-130). Springer, Berlin, Heidelberg.
Seminal paper on SSA:
Rosen, B.K., Wegman, M.N. and Zadeck, F.K., 1988, January. Global value numbers and redundant computations. In Proceedings of the 15th ACM SIGPLAN-SIGACT symposium on Principles of programming languages (pp. 12-27). ACM.
# Counter Example Production
In the goto-symex directory.
Key classes:
# Concurrency
## Order Concurrency
The class partial_order_concurrencyt provides an interface for implementing ordering of concurrent events.
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• Contributions of the Pierre Auger Collaboration to the 33rd International Cosmic Ray Conference, Rio de Janeiro, Brazil, July 2013
• ### HERMES: a Monte Carlo Code for the Propagation of Ultra-High Energy Nuclei(1307.4356)
July 16, 2013 astro-ph.IM, astro-ph.HE
Although the recent experimental efforts to improve the observation of Ultra-High Energy Cosmic Rays (UHECRs) above $10^{18}$ eV, the origin and the composition of such particles is still unknown. In this work, we present the novel Monte Carlo code (HERMES) simulating the propagation of UHE nuclei, in the energy range between $10^{16}$ and $10^{22}$ eV, accounting for propagation in the intervening extragalactic and Galactic magnetic fields and nuclear interactions with relic photons of the extragalactic background radiation. In order to show the potential applications of HERMES for astroparticle studies, we estimate the expected flux of UHE nuclei in different astrophysical scenarios, the GZK horizons and we show the expected arrival direction distributions in the presence of turbulent extragalactic magnetic fields. A stable version of HERMES will be released in the next future for public use together with libraries of already propagated nuclei to allow the community to perform mass composition and energy spectrum analysis with our simulator.
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The number of 3xx3matrices Awhose entries are either 0or1and for w | Filo
class 12
Math
Algebra
Matrices
549
150
The number of matrices whose entries are either and for which the system has exactly two distinct solution isa. 0 b. c. d.
549
150
Connecting you to a tutor in 60 seconds.
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# Tag Info
11
Having experienced similar problematic issues with Mathematica I instantly thought that expanding the fraction in the integrand i.e. applying Appart could resolve the problem, and indeed it does: Integrate[ Apart[(1 - x)(1 + 2x)^6/Sqrt[1 - x^2]], {x, -1, 1}]/Pi 15 These arguments apply to this case as well Bug in mathematica analytic integration? i.e. ...
7
You don't need to use TagSetDelayed for the definition of the derivative because Derivative doesn't have attribute Protected. I'll extend add the derivative definition to arbitrary order n: ClearAll[ln]; Derivative[n_, 0][ln][x_, a_] := Derivative[n][Log][x] ln[x_, a_?NumericQ] := Piecewise[{{Log[x], Re[a] > 0}, {-Log[1/x], True}}] ln[x, -1/2] ...
6
If I understand the goal of the question correctly, this is a possible application for the new Inactivate and Activate. Looking in particular at the documentation for Inactive, under "Applications," you'll find many situations that look similar to the one in this question. For example, you can enter a valid expression in the form Inactivate[(x y)^2] ...
5
The sorting (ordering) done by Union is different for different forms of expressions, e.g., analytic versus numeric expressions for a number. Union[{2., (Sqrt[5] + 1)/2}] {2., 1/2 (1 + Sqrt[5])} % // N {2., 1.61803} Union[{2., (Sqrt[5] + 1.0)/2}] {1.61803, 2.} SortBy[{2., (Sqrt[5] + 1)/2}, N] {1/2 (1 + Sqrt[5]), 2.}
3
I agree that it's a bit odd that Mathematica doesn't simplify these expressions with its built-in functions, especially in the symbolic tensor language (i.e. using TensorContract and TensorReduce). Nonetheless, we can teach it how to simplify the identity matrix ourselves. I've chosen to implement this for Dot, since that is what you initially asked. I do ...
2
Using the identity: $$\Phi (z,s,a)=\frac{\Phi (z,s,a-1)-\left((a-1)^2\right)^{-s/2}}{z}$$ First we define two recursive functions: rGexp[0] = -1/4 (Pi^2 - 6 Log[2]^2)/3; rGexp[n_] := 1/2 rGexp[n - 1] rlp[0] = LerchPhi[1/2, 2, 0]; rlp[n_] := 1/2 rlp[n + 1] + 1/(n)^2 Then we add them: rSeq[n_] := rlp[-n] + rGexp[n] Simplify@Table[rSeq[n], {n, 1, 10}] ...
1
Thanks for updating your Question. With the new, clearer example I believe I can see the issue. Analysis The first method uses evenper on Symbolic values that are in canonical order: r1 = evenper[{a, b, c, d}] {{a, b, c, d}, {a, c, d, b}, {a, d, b, c}, {b, a, d, c}, {b, c, a, d}, {b, d, c, a}, {c, a, b, d}, {c, b, d, a}, {c, d, a, b}, {d, a, c, b}, ...
1
Main The bug cannot only be attributed to the Sqrt in the integrand. It is trickier. In fact, define for t=0,1,2,... f[t_] := Integrate[(1 - x)*(1 + 2*x)^t/Sqrt[1 - x^2], {x, -1, 1}]/Pi Then {#, f[#]} & /@ {0, 1, 2, 3, 4, 5, 6, 7, 8} (* Out[33]= {{0, 1}, {1, 0}, {2, 1}, {3, 1}, {4, 3}, {5, 6}, {6, ( 1 + (29 \[Pi])/2)/\[Pi]}, {7, (1 + (71 ...
Only top voted, non community-wiki answers of a minimum length are eligible
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What is the metric on the Fuchsian model? [closed]
Let $\mathbb{H}$ be the upper half plane, and $\Gamma < SL(2, \mathbb{R})$ be a Fuchsian group. How is the distance between any two points $x, y \in \mathbb{H} / \Gamma$ in the Fuchsian model defined?
• How do you define a distance function on a Riemannian manifold? In your case, the quotient is an orbifold, but did you try to use the same formula? – Misha Sep 21 '15 at 19:38
• What is "the Fuchsian model". Almost certainly not research level, voting to close. – Igor Rivin Sep 21 '15 at 20:00
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# Should the weights of a neural network be updated after each example or at the end of the batch? [duplicate]
Should the weights of a neural network be updated after each example or at the end of the batch? Do I need a normalization factor in the second case?
• Ideally you would use a small enough batch that uses the best of both worlds. In the first case, you have a lot of variance between the gradients in each sample, but, you have significantly more updates which could result in faster training, can cause catastrophic forgetting if your data isn't random. In the latter case you would either use the average, or sum the batch and multiply by a small value. The latter exhibits less variance but you need to be careful with the lr. – Dimitris Monroe Nov 4 '19 at 8:42
The backpropagation step is generally used to compute the gradients and update the weights.
Let us say, you are implementing gradient descent, select the whole training batch, perform the forward propagation using the current set of neurons/weights to get the classification output. Then compute the loss/cost with respect to the actual labels, (Note: This step contains dividing the loss and divide it by the number of training example to get the cost function). Once you get the cost you backpropagate by computing the gradients at each layer with respect to cost from output to input and perform the weight updates.
If you are using mini-batch, you compute the cost for that mini-batch, divide by the number of examples in the mini-batch, then compute the gradients and perform the update in the backprop step.
• Okay so to summarize, you update the weights after each completed batch. This means the update frequency depend on the batch size. – Sebastian Nielsen Oct 20 '18 at 21:55
• Yes, if you choose smaller batches then there will be a large number of updates per epoch. If you take the whole training data in one batch there will be just one update per epoch. – fireboy Oct 22 '18 at 4:07
When am I supposed to update my weights? After each forward-, and backpropagation; and or after each completed batch?
1. In your example, you should update the weights after each back propagation.
2. Also, you can back propagate the result by adding each error/loss for each prediction/example in a batch and then update the weights. pseudocode- for batch size 3:
w = backprop(Error(t1)+Error(t2)+Error(t3)) update_wights(w*learning_rate)
3. Now what happen If you back propagate every error for every prediction/example in a batch and store them(without updating), and after completion of that batch update the weights by using stored calculated values, then there is a chance to stuck in the local minima also some research shows that it's actually slow training. research paper
note- all those techniques in 1,2,3 are different but 1 and 2 results same. Also in 2, adding errors of different training example makes a new graph of neural network and back propagating it means back propagating the resulted graph.
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# The line passing through the focus and perpendicular to the directrix is called
This question was previously asked in
UPSC ESE 2020 Paper 1
View all UPSC IES Papers >
1. axis
2. vertex
3. eccentricity
4. conic
## Answer (Detailed Solution Below)
Option 1 : axis
Free
NTPC Diploma Trainee 2020: Full Mock Test
12115
120 Questions 120 Marks 120 Mins
## Detailed Solution
Concept:
Hyperbola: The locus of a point which moves such that its distance from a fixed point and a fixed straight line is always greater than one (Eccentricity = e > 1)
Equation of hyperbola: $$\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1{\rm{\;}},{\rm{\;a}} > {\rm{b}}$$
The two lines parallel to the minor axis are called the directrix.
The straight line passing through the focus and perpendicular to the directrix is known as transverse axis.
Eccentricity is the ratio of distance from any point from the graph to focus and directrix. A circle has an eccentricity of zero, so the eccentricity shows you how "un-circular" the curve is. Bigger eccentricities are less curved.
Eccentricity of hyperbola = $$\sqrt {1 + \left( {\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \right)}$$
Different values of eccentricity make different curves:
At eccentricity = 0 we get a circle
for 0 < eccentricity < 1 we get an ellipse
for eccentricity = 1 we get a parabola
for eccentricity > 1 we get a hyperbola
for infinite eccentricity, we get a line
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# Counterpart of Maple's LSSolve in Mathematica
I am a great fan of Mathematica. Unfortunately, due to some specific tasks I had to move to Maple. I think it's my lack of info that is not letting me do those tasks. That is why I am asking some command replacements from you guys. I code some mesh free methods like Galerkin and spectral methods. They use a trial solution which contains some constants. After some treatment the problem gets converted to an over-determined system of algebraic equations. Maple has a command LSSolve to determine a best possible solution to that over-determined system. Can you please tell me its alternative in Mathematica. I really need it to bring all my coding back to Mathematica. my equations.
eq1 = -.3007024038*c[2]^2 + (-0.4990858944 10^-2 - .3007024038*c[1])*
c[2];
eq2 = -.2004682692*c[2]^2 + (-0.2495429472 10^-2 - .1503512019*c[1])*
c[2];
eq3 = -.1503512019*c[2]^2 + (-0.1663619648 10^-2 - .1002341346*c[1])*
c[2];
eq4 = c[0] + c[1] + c[2] - 1;
PS Here is a snap of the calculations on MAPLE
• Perhaps LeastSquares is what you are looking for in MMA. – MarcoB May 23 '18 at 16:11
• thank you for you reply but i am afraid its not LeastSquares I am after. According to the documentation it solve linear systems. While I mostly encounter with nonlinear equations. As far as LSSolve is concerned it minimizes the Squared residual Error. I have tried to do it myslef in Mathematics but in vain. – naveed May 23 '18 at 16:20
• Then we need more information. Gives us an example of a (small!) problem you solve with LSSolve, together with the output you obtain from Maple. I do not use Maple, so I am only going with the information included in its documentation for LSSolve. If you don't provide more details, then only people intimately familiar with both systems will be able to answer your question. Alternatively, you could always use NMinimize etc to explicitly minimize a sum of squares. – MarcoB May 23 '18 at 17:37
• In priciple, that's a task for NonlinearModelFit. I also recall to have read <s>somewhere</s> here that FindMinimum switches automatically to the Levenberg-Marquardt mathod (a minor modification of Gauss-Newton method) if it detects that the objective function is a sum of squares... So give it a try with either method. – Henrik Schumacher May 23 '18 at 20:21
• Please post the code text so we can test them easily. – xzczd May 25 '18 at 5:00
According to the document of LSSolve:
The LSSolve command solves a least-squares (LS) problem, which involves computing the minimum of a real-valued objective function having the form $$\frac{1}{2}(f_1(x)^2+f_2(x^2)+…+f_q(x)^2)$$ where $x$ is a vector of problem.
So its analog in Mathematica seems to be FindMinimum with a specific function to be minimized. We should be able to obtain (almost) the same result with:
Clear@lSSolve
lSSolve[obj_List, constr___, x_, opt : OptionsPattern[FindMinimum]] :=
FindMinimum[{1/2 obj^2 // Total, constr}, x, opt]
lSSolve[obj_, rest__] := lSSolve[{obj}, rest]
Let's test with the examples given in the document and your question:
# Case 1
Maple:
LSSolve([x-2, x-6, x-9]);
(* [12.3333333333333321, [x = 5.66666666666667]]*)
Mathematica:
lSSolve[{x - 2, x - 6, x - 9}, x]
(* {12.3333, {x -> 5.66667}} *)
# Case 2
Maple:
LSSolve([x^3-2, x^2-6, x^2-9], initialpoint = {x = 1});
(* [27.5839512531713, [x = 1.75156454919679]]*)
Mathematica:
lSSolve[{x^3 - 2, x^2 - 6, x^2 - 9}, {x, 1}]
(* {27.584, {x -> 1.75156}} *)
# Case 3
Maple:
LSSolve([x-1, y-1, z-1], {x <= 0, 6*x+3*y <= 1}, initialpoint = {x = -1, y = 1});
(* [0.711111111111111138, [x = -0.0666666666666667, y = 0.466666666666667, z = 1.]]*)
Mathematica:
lSSolve[{x - 1, y - 1, z - 1}, {x <= 0, 6 x + 3 y <= 1}, {{x, -1}, {y, 1}, z}]
(* {0.711111, {x -> -0.0666676, y -> 0.466668, z -> 1.}} *)
# Case 4
Maple:
LSSolve([x-1], {(x+1)^2 <= 0});
(* [1.99998465585440166, [x = -0.999992327912486]]*)
Mathematica:
lSSolve[x - 1, (x + 1)^2 <= 0, x]
(* {2., {x -> -1.}} *)
# Case 5: Example in your question
eq1 = -0.3007024038 c[2]^2 + (-(0.4990858944/10^2) - 0.3007024038 c[1]) c[2];
eq2 = -0.2004682692 c[2]^2 + (-(0.2495429472/10^2) - 0.1503512019 c[1]) c[2];
eq3 = -0.1503512019 c[2]^2 + (-(0.1663619648/10^2) - 0.1002341346 c[1]) c[2];
eq4 = c[0] + c[1] + c[2] - 1;
lSSolve[{eq1, eq2, eq3, eq4}, c /@ Range[0, 2]]
(* {1.58921*10^-33, {c[0] -> 1.0166, c[1] -> -0.0165973, c[2] -> 7.45058*10^-9}} *)
Slightly different, but according to objective function value, result of Mathematica is better.
• Thank you very much. I guess this is what I was looking for. – naveed May 27 '18 at 2:30
• @naveed There's a bug in my previous implementation, now it's fixed and result of case 4 is also (almost) the same as that of Maple. – xzczd May 29 '18 at 7:39
Is this not what you want ?
In[6]:= NMinimize[eq1^2 + eq2^2 + eq3^2 + eq4^2, {c[0], c[1], c[2]}]
Out[6]= {1.94723*10^-27, {c[0] -> 1.01633, c[1] -> -0.0163307,
c[2] -> 4.71816*10^-10}}
• Thank you. A vote up for you. – naveed May 27 '18 at 2:31
• I have got better results (In the implementation of my complete code) by combining the both techniques. I used Lotus's Nminimize with xzczd definition of LSSolve and got the solutions that agree very well with the exact solutions of actual solutions. I was confused in selecting the answer but then i thought xzczd answer in more general. I am thankful to all of you for helping me out. – naveed May 30 '18 at 17:53
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Chain Rule Help
The chain rule is similar to the product
rule
and the quotient rule,
but it deals with
differentiating
compositions of functions. If we recall, a composite function is a function
that contains another function:
The Formula for the Chain Rule
The capital F means the same thing as lower case f, it just encompasses the composition
of functions. As a motivation for the chain rule, let’s look at the following example:
(1)
This function would take a long time to factor out and find the derivative of each
term, so we can consider this a composite function. The two functions would look
like this:
Notice that substituting g(x) for g in f(x) would yeild the original function. We
will see that after differentiating, we will then substitute g(x) back in for g.
So the composite function would be
Now, we can use the chain rule, which is defined by taking the derivative of outside
function times the inside function, and multiplying it by the derivative of the
inside function
:
Using this rule, we have:
Let’s do another example.
(2) Differentiate the following function:
We define the inside and outside function to be
Then, the derivative of the composition will be as follows:
Think of the chain rule as a process. The derivative of the composite function is
the derivative of the outside function times the derivative of the inside function.
Scroll to Top
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1. ## Help with integration
Help me with these integrations please
$
1.\int\sqrt{log{x}}dx
$
$
2.\int\frac{dx}{\sqrt{3+2cos{x}}}
$
$
3.\int\log{tan(x)}dx
$
$
4.\int\frac{tan^{-1}(x)}{x}dx
$
$
5.\int\tan(e^{x})dx
$
$
6.\int\tan^{-1}(e^{x})dx
$
$
7.\int\(e^{(x^{2})}dx
$
& the next one is a differential equation. Find the solution to the differential equation
$
(xy^{2}-\frac{e}{x^{3}})dx-x^{2}ydy=0
$
Integrals 3to7 were made up by me. Just wondering if indefinite integration is possible for them or not. The rest of them are book questions.
2. $
7.\int\(e^{(x^{2})}dx
$
This is not elementary.
Loiuville shown that if $f(x),g(x)$ are rational function then the integral $\int f(x) \exp (g(x)) dx$ is elementary if and only if, there exists a rational function $h(x)$ such as $f(x)=h'(x)+h(x)g(x)$.
In this case, let us we have that.
$f(x)=1$ and $g(x)=x^2$ solve the differencial equation,
$1=y'+x^2y$.
This equation has no rational solutions.
Thus the function is not elementary.
(However, it does exists. This is a consequence of the fundamental theorem of calculus. It can be shown that any countinous funtion on a closed interval or on the number line always has a an anti-derivative.)
If you are curious about which integral can and cannot be there is a topic called Differencial Algebra
---
Furthermore, this is an important function. It is used as an approximation of the binomial distribution (normal curve). I belive (I might be wrong) it is reffered to as $\mbox{erf}(x)$
---
Lasty, there is a nice idenity involving this function,
$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$.
If you wish I can post a prove of this (you need to be familiar with double integration).
$
1.\int\sqrt{log{x}}dx
$
Play around with it,
$\int \frac{x\sqrt{\log x}}{x} dx$
Let, $u=\log x$
Then, $u'=1/x$
Thus,
$\int e^u \sqrt{u} u' dx$
The substitution rule states,
$\int e^u \sqrt{u} du$
Manipulate it as,
$2\int e^{(\sqrt{u})^2} \frac{u}{2\sqrt{u}} du$
Let,
$t=\sqrt{u}$
Then,
$t'=\frac{1}{2\sqrt{u}}$
Thus,
$2\int e^{t^2} t^2 t' du$
Substitution rule again,
$2\int t^2 e^{t^2} dt$
Same idea not elementary.
4. Originally Posted by ThePerfectHacker
This is not elementary.
Loiuville shown that if $f(x),g(x)$ are rational function then the integral $\int f(x) \exp (g(x)) dx$ is elementary if and only if, there exists a rational function $h(x)$ such as $f(x)=h'(x)+h(x)g(x)$.
In this case, let us we have that.
$f(x)=1$ and $g(x)=x^2$ solve the differencial equation,
$1=y'+x^2y$.
This equation has no rational solutions.
Don't you have to show that this is the case?
RonL
5. Originally Posted by ThePerfectHacker
$
7.\int\(e^{(x^{2})}dx
$
This is not elementary.
But who said it had to be elementary? If we are looking for this
integral we usualy end up looking it up in a table, which tells
us that:
$
\int_0^x e^{x^2}=\frac{\sqrt{\pi}}{2}\mbox{erfi}(x)
$
where $\mbox{erfi}(x)$ is the imaginary error function:
$\mbox{erfi}(z)=\bold{i}\ \mbox{erf}(\bold{i}z)$, and $\mbox{erf}$ is the (analytic continuation of the) bog standard
error function:
$
\mbox{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}\ dt
$
RonL
6. Originally Posted by CaptainBlank
Don't you have to show that this is the case?
Do not understand. (You mean to show that the differencial equation has no rational functional solution? If thus, assume a solution exists as a quotient of two polynomials and try to reach a contradiction).
Originally Posted by CaptainBlank
But who said it had to be elementary?
Again do not understand. If you look at post #1 the user asks whether there exists closed form solutions. Thus, he did not know and I was saying negative.
I took a stab at #1 . . . and it turned into #7 !
$1)\;\int\sqrt{\ln x}\,dx$
Let: $\sqrt{\ln x} = w\quad\Rightarrow\quad \ln x = w^2\quad\Rightarrow\quad x =$ $e^{w^2}\quad\Rightarrow\quad dx = 2w\!\cdot\! e^{w^2}dw$
Substitute: . $\int w\left(2we^{w^2}\,dw\right) \;=\;\int 2w^2\!\cdot\! e^{w^2}dw$
Integrate by parts . . .
. . Let: $u = w\qquad\quad dv = 2w\!\cdot\! e^{w^2}dw$
. . Then: $du = dw\qquad v = e^{w^2}$
and we have: . $w\!\cdot\!e^{w^2} - \int e^{w^2}dw$ . . . see?
$
2.\int\frac{dx}{\sqrt{3+2cos{x}}}
$
For this integral I recommend to use a Weierstrauss Substitution. It will eliminate the sine and transform the transcendental function into an algebraic one. This should make it easier to work with.
9. Originally Posted by ThePerfectHacker
Do not understand. (You mean to show that the differencial equation has no rational functional solution? If thus, assume a solution exists as a quotient of two polynomials and try to reach a contradiction).
I know how to approach it, but your reader may not, so you should
show it to be the case to complete the demonstration.
Again do not understand. If you look at post #1 the user asks whether there exists closed form solutions. Thus, he did not know and I was saying negative.
A closed form solution does not have to be closed form in terms of
elementary functions, it can be closed form in terms of higher
transcendental functions - that's what they are for.
RonL
10. What does being elementary mean. I'm only a beginner.
Originally Posted by ThePerfectHacker
For this integral I recommend to use a Weierstrauss Substitution. It will eliminate the sine and transform the transcendental function into an algebraic one. This should make it easier to work with.
Please be a little more descriptive. Again, I'm just a beginner.
Originally Posted by Soroban
I took a stab at #1 . . . and it turned into #7 !
Let: $\sqrt{\ln x} = w\quad\Rightarrow\quad \ln x = w^2\quad\Rightarrow\quad x =$ $e^{w^2}\quad\Rightarrow\quad dx = 2w\!\cdot\! e^{w^2}dw$
Substitute: . $\int w\left(2we^{w^2}\,dw\right) \;=\;\int 2w^2\!\cdot\! e^{w^2}dw$
Integrate by parts . . .
. . Let: $u = w\qquad\quad dv = 2w\!\cdot\! e^{w^2}dw$
. . Then: $du = dw\qquad v = e^{w^2}$
and we have: . $w\!\cdot\!e^{w^2} - \int e^{w^2}dw$ . . . see?
Yeah, I arrived at the same situation. That is why I posted 7 and many others which I arrived at while solving others.
What does being elementary mean?
It means a finite combination of function addition,subtraction,multiplication,division and composition of the "basic functions" (ploynomials, rationals, power, exponenets, lgarithmic, trigonometric).
I myself never liked that definition because it seems to be missing parts to it. Never like when we say a finite composition.
---
This is my own question. Is there a more rigourous way to define elementary? Like a Group that is solvable, or something like that?
Please be a lot more descriptive.
I assumed you understood.
---
You have,
$\int \frac{dx}{\sqrt{3+2\cos x}}$
Let,
$u=\tan x/2$
Then,
$\sin x=2\sin x/2\cos x/2=\frac{2u}{1+u^2}$
And,
$\cos x=\cos^2 x/2-\sin^2 x/2=\frac{1-u^2}{1+u^2}$
If you substitute this you will get,
$\int \frac{\frac{2}{1+u^2}}{\sqrt{3+2\frac{2u}{1+u^2}}} du$
13. Originally Posted by galactus
$(xy^{2}-\frac{e}{x^{3}})dx-x^{2}ydy$ is exact since
$\frac{\partial{M}}{\partial{y}}=2xy=\frac{\partial {N}}{\partial{x}}$
How?
$\frac{\partial M}{\partial y}=2xy$
But!
$\frac{\partial N}{\partial x}=-2xy$
They are not equal.
14. You're so right PH. I got tripped up with a mere negative sign. Oh well.
BTW, I ran $\int\frac{1}{\sqrt{3+2cos(x)}}dx$ through Maple.
The contrary LaTex on this site won't allow me to post the lengthy solution, but here's what it will post. At least, it'll give you an idea that it's a dilly.
Elliptic Fresnel means it's a real booger to integrate(by elementary means, anyway)
Here's what she gave me:
${-}2{\frac {\sqrt {- \left( 1+4\, \left( \cos \left( 1/2\,x \right)$
$\right) ^{2} \right) \left( -1+ \left( \cos \left( 1/2\,x \right)$
$\right) ^{2} \right) }\sqrt {1- \left( \cos \left( 1/2\,x \right)$
$\right) ^{2}}{\it EllipticF}$ $\left( \cos \left( 1/2\,x \right)$ $,2\,i \right) }{\sqrt {1-4\, \left( \cos \left( 1/2\,x \right) \right)^{4}+3$ $\, \left( \cos \left( 1/2\,x \right) \right) ^{2}}\sin \left( 1/2\,x \right) }}$
15. Originally Posted by ThePerfectHacker
I assumed you understood.
Hmm...I'm familiar with this type of substitution but never came to know its called Weierstrauss Substitution.
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# Worst case to average case reductions
Are there problems whose average case complexity is the same as their worst case complexity? What are the underlying properties of these problems that makes reducing the worst case to the average case possible?
• Random self-reducibility (That's more of a definition than an underlying property, but I suspect the Wikipedia article will give you a good start on figuring out what you want to know.) – Peter Shor Nov 13 '11 at 18:59
• @PeterShor comment -> answer ? – Suresh Venkat Nov 14 '11 at 0:50
Other than that, you can find problems with worst-case-to-average-case reductions of various complexities, e.g., in $NP \cap coNP$ (lattice problems and other crypto problems) and in #P (permanent).
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# why is H = {www | w ∊ (0,1)* } not regular language
I know pumping lemma and also know how to prove if languages are not regular or are regular using proof by contradiction methodology. But I am stuck on this one.Especially, I don't know exact string it produces.
$$H = \{www \mid w ∊ (0,1)^* \}$$
How do I prove this is not regular language? Thanks!
You can use the Myhill–Nerode criterion. Consider the words $0^n1$ for all $n$. These are pairwise inequivalent since $0^n10^n10^n1 \in H$ while $0^m10^n10^n1 \notin H$ if $m \neq n$. Since we have found infinitely many pairwise inequivalent words, it follows that the language is not regular.
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# How do I set a binding offset for multiple pages printed on one sheet?
I'm creating a PDF LaTeX document that has A4 portrait pages. But it is destined to be printed in landscaoe orientation, two-up and double-sided.
That means any binding offset needs to be added to the left margin of pages 1, 5, 9 .... and the right margin of pages 4, 8, 12 ....
This crude sketch may make that clear:
front side of sheet
--------------------------------------------
| bblll page 1 rrr lll page 2 rrr |
| bblllttttttttttrrr lllttttttttttttrrr |
| bblllttttttttttrrr lllttttttttttttrrr |
| bblllttttttttttrrr lllttttttttttttrrr |
| bblllttttttttttrrr lllttttttttttttrrr |
| bblllttttttttttrrr lllttttttttttttrrr |
| bblllttttttttttrrr lllttttttttttttrrr |
| bblllttttttttttrrr lllttttttttttttrrr |
---------------------------------------------
reverse side of sheet
--------------------------------------------
| lll page 3 rrr lll page 4 rrrbb |
| lllttttttttttttrrr lllttttttttttrrrbb |
| lllttttttttttttrrr lllttttttttttrrrbb |
| lllttttttttttttrrr lllttttttttttrrrbb |
| lllttttttttttttrrr lllttttttttttrrrbb |
| lllttttttttttttrrr lllttttttttttrrrbb |
| lllttttttttttttrrr lllttttttttttrrrbb |
| lllttttttttttttrrr lllttttttttttrrrbb |
---------------------------------------------
bb -- binding offset
ll -- left margin
tt -- text of page
rr -- right margin
The two-up printing is handled by the printer. But the printer cannot handle shifting the whole page to create the binding offsets.
One solution of course is to get a better printer.
One other avenue I have explored is treating the whole doc as a landscape, and using \multicols to print it in two colums. That sadly has many awkwardnesses in the formating.
So what I'd really like is some magic dust in the geometry settings.
Am I hoping for too much this soon after xmas?
-
– lockstep Jan 7 '12 at 16:02
Maybe the booklet package can help you – Spike Jan 8 '12 at 8:19
I feel a little uneasy at answering my own question. But lockstep's clue directed me to the twocolumn parameter on the documentclass.
That seems to work fairly. It subdivides the physical sheet into two logical pages. So looks a little like using multicols{2} on the whole document, but with fewer of the drawbacks.
The magic inteplay between documentclass and geometry seems to be (note the exaggerated 5cm inner margin for binding offset -- just to make it easier to see):
\documentclass[a4paper,twocolumn,landscape,10pt]{book}
\usepackage[top=2cm, bottom=2cm, inner=5cm, outer=1cm]{geometry}
This solution may be just the ticket for many documents.
However, I can see problems where I have a wrapfigure inside a miniboxpage with each scaled as a fraction of the \textwidth. It seems \textwidth is sometimes taken as the full physical width of the sheet, and sometimes as the text width of the logical page. I need to isolate cases and investigate further.
Thanks for the responses -- they've given me some hope for the future.
-
Don't feel uneasy -- this kind of-self answer is legitimate. – lockstep Jan 8 '12 at 3:24
On the Unix and Unix-like systems you could use psnup or a2ps to accomplish the task. I am sure you could use those tools on Winodows machine via Cygwin but there might be a better native solution.
-
The KOMA-Script \documentclass{scrbook} is an alternative to \documentclass{book} that looks nicer on A4 paper and offers quite a lot of additional options, among them te option to specify a binding correction. As an example, this is the documentclass definition I use for my PhD thesis:
\documentclass[
a4paper,
BCOR=10mm, % specify binding correction
ngerman,
twoside, % left pages and right pages look different
open=right, % chapters start on right pages only, this is the default
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## College Algebra (11th Edition)
Published by Pearson
# Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 135: 73
$x=\left\{ \dfrac{1}{4},1 \right\}$$#### Work Step by Step$\bf{\text{Solution Outline:}}$To solve the given equation,$ (2x-1)^{2/3}=x^{1/3} ,$raise both sides to the third power to get rid of the denominator in the fractional exponent. Then express the resulting equation in the form$ax^2+bx+c=0,$and use the concepts of factoring quadratic equations to solve the resulting equation. Finally, do checking if the solution satisfies the original equation.$\bf{\text{Solution Details:}}$Raising both sides to the third power, the given equation becomes \begin{array}{l}\require{cancel} \left((2x-1)^{2/3}\right)^3=\left( x^{1/3} \right)^3 .\end{array} Using the Power Rule of the laws of exponents which is given by$\left( x^m \right)^p=x^{mp},$the expression above is equivalent to \begin{array}{l}\require{cancel} (2x-1)^{\frac{2}{3}\cdot3 }=x^{\frac{1}{3}\cdot3} \\\\ (2x-1)^2=x .\end{array} Using the square of a binomial which is given by$(a+b)^2=a^2+2ab+b^2$or by$(a-b)^2=a^2-2ab+b^2,$the expression above is equivalent to \begin{array}{l}\require{cancel} (2x)^2-2(2x)(1)+(1)^2=x \\\\ 4x^2-4x+1=x \\\\ 4x^2+(-4x-x)+1=0 \\\\ 4x^2-5x+1=0 .\end{array} Using factoring of trinomials, the value of$ac$in the trinomial expression above is$ 4(1)=4 $and the value of$b$is$ -5 .$The$2$numbers that have a product of$ac$and a sum of$b$are$\left\{ -1,-4 \right\}.$Using these$2$numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 4x^2-x-4x+1=0 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (4x^2-x)-(4x-1)=0 .\end{array} Factoring the$GCF$in each group results to \begin{array}{l}\require{cancel} x(4x-1)-(4x-1)=0 .\end{array} Factoring the$GCF= (4x-1) $of the entire expression above results to \begin{array}{l}\require{cancel} (4x-1)(x-1)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 4x-1=0 \\\\\text{OR}\\\\ x-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 4x-1=0 \\\\ 4x=1 \\\\ x=\dfrac{1}{4} \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Upon checking,$ x=\left\{ \dfrac{1}{4},1 \right\}\$ satisfy the original equation.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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## Algebra 1
$(-3, infinity)$
We can't have the square root of a negative number, so $y$'s lowest value is 0. $0=\sqrt {x+3}$ $0^2=(\sqrt {x+3})^2$ $0= x+3$ $-3=x$
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# NCERT Exemplar Solutions Class 10 Mathematics Solutions for Coordinated Geometry - Exercise 7.4 in Chapter 7 - Coordinated Geometry
Question 2 Coordinated Geometry - Exercise 7.4
A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area ofADE.
According to the question,
The three vertices of a parallelogram ABCD are A (6, 1), B (8, 2) and C (9, 4)
Let the fourth vertex of parallelogram = (x, y),
We know that, diagonals of a parallelogram bisect each other
Since, mid – point of a line segment joining the points (x1, y1) and (x2, y2) is given by,
\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\\ \text { Mid - point of } \mathrm{BD}=\text { Mid - point of } \mathrm{AC}\\ \left(\frac{8+x}{2}, \frac{2+y}{2}\right)=\left(\frac{6+9}{2}, \frac{1+4}{2}\right)\\ \left(\frac{8+x}{2}, \frac{2+y}{2}\right)=\left(\frac{15}{2}, \frac{5}{2}\right)
So we have,
\begin{aligned} &\frac{8+x}{2}=\frac{15}{2}\\ &8+x=15\\ &\mathrm{x}=7\\ &\text { And }\\ &\frac{2+y}{2}=\frac{5}{2}\\ &2+y=5 \rightarrow y=3 \end{aligned}
So, fourth vertex of parallelogram is D(7, 3)
Now,
Midpoint of side
D C=\left(\frac{7+9}{2}, \frac{3+4}{2}\right) \\ E=\left(8, \frac{7}{2}\right)
\text{∵ Area of ΔABC with vertices } (x_1, y_1), (x_2, y_2) and (x_3, y_3);\\ = \frac12[x_1(y_2 – y_3) + x2(y_3 – y_1) + x_3(y_1 – y_2)]
∴ Area of ΔADE with vertices A (6, 1), D (7, 3) and E (8, (7/2))
\Delta =\frac{1}{2}\left[6\left(3-\frac{7}{2}\right)+7\left(\frac{7}{2}-1\right)+8(1-3)\right] \\ =\frac{1}{2}\left[6 \times\left(\frac{-1}{2}\right)+7\left(\frac{5}{2}\right)+8(-2)\right] \\ =\frac{1}{2}\left(\frac{35}{2}-19\right) \\ =\frac{1}{2}\left(\frac{-3}{2}\right)
= – ¾ but area can’t be negative
Hence, the required area of ΔADE is ¾ sq. units
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# Solution for By Using Properties of Determinants, Show that |(A-b-c, 2a,2a),(2b, B-c-a,2b),(2c,2c, C-a-b)| = (A + B + C)Sqrt2 - CBSE (Arts) Class 12 - Mathematics
#### Question
By using properties of determinants, show that:
|(a-b-c, 2a,2a),(2b, b-c-a,2b),(2c,2c, c-a-b)| = (a + b + c)^2
#### Solution
Is there an error in this question or solution?
#### APPEARS IN
NCERT Mathematics Textbook for Class 12 Part 1 (with solutions)
Chapter 4: Determinants
Q: 11.1 | Page no. 120
Solution for question: By Using Properties of Determinants, Show that |(A-b-c, 2a,2a),(2b, B-c-a,2b),(2c,2c, C-a-b)| = (A + B + C)Sqrt2 concept: Properties of Determinants. For the courses CBSE (Arts), CBSE (Commerce), CBSE (Science)
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The wire shown in the figure, carries a current of 10 A
The wire shown in the figure, carries a current of 10 A. Determine the magnitude of magnetic field induction at the centre O. Given the radius of bent coil is 3 cm.
Here, / = 10 A, r = 3 cm, r = 3 X {{10}^{-2}}
Angle subtended by coil at the centre,
θ = 360° - 90° = 270 ° = 3$\pi$ / 2 rad
Magnetic field induction at O due to current through circular path ACB is,
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# Turbulence kinetic energy
(Redirected from Turbulent Kinetic Energy)
In fluid dynamics, turbulence kinetic energy (TKE) is the mean kinetic energy per unit mass associated with eddies in turbulent flow. Physically, the turbulence kinetic energy is characterised by measured root-mean-square (RMS) velocity fluctuations.
In Reynolds-averaged Navier Stokes equations, the turbulence kinetic energy can be calculated based on the closure method, i.e. a turbulence model. Generally, the TKE can be quantified by the mean of the turbulence normal stresses:
${\displaystyle k={\frac {1}{2}}\left(\,{\overline {(u')^{2}}}+{\overline {(v')^{2}}}+{\overline {(w')^{2}}}\,\right).}$
TKE can be produced by fluid shear, friction or buoyancy, or through external forcing at low-frequency eddy scales(integral scale). Turbulence kinetic energy is then transferred down the turbulence energy cascade, and is dissipated by viscous forces at the Kolmogorov scale. This process of production, transport and dissipation can be expressed as:
${\displaystyle {\frac {Dk}{Dt}}+\nabla \cdot T'=P-\varepsilon ,}$
where:[1]
• Dk/Dt is the mean-flow material derivative of TKE;
• ∇ · T′ is the turbulence transport of TKE;
• P is the production of TKE, and
• ε is the TKE dissipation.
Assuming density and viscosity both constant, the full form of the TKE equation is:
${\displaystyle \underbrace {\frac {\partial k}{\partial t}} _{\begin{smallmatrix}{\text{Local}}\\{\text{derivative}}\end{smallmatrix}}+\underbrace {{\overline {u}}_{j}{\frac {\partial k}{\partial x_{j}}}} _{\begin{smallmatrix}{\text{Advection}}\end{smallmatrix}}=-\underbrace {{\frac {1}{\rho _{o}}}{\frac {\partial {\overline {u'_{i}p'}}}{\partial x_{i}}}} _{\begin{smallmatrix}{\text{Pressure}}\\{\text{diffusion}}\end{smallmatrix}}-\underbrace {{\frac {1}{2}}{\frac {\partial {\overline {u_{j}'u_{j}'u_{i}'}}}{\partial x_{i}}}} _{\begin{smallmatrix}{\text{Turbulent}}\\{\text{transport}}\\{\mathcal {T}}\end{smallmatrix}}+\underbrace {\nu {\frac {\partial ^{2}k}{\partial x_{j}^{2}}}} _{\begin{smallmatrix}{\text{Molecular}}\\{\text{viscous}}\\{\text{transport}}\end{smallmatrix}}\underbrace {-{\overline {u'_{i}u'_{j}}}{\frac {\partial {\overline {u_{i}}}}{\partial x_{j}}}} _{\begin{smallmatrix}{\text{Production}}\\{\mathcal {P}}\end{smallmatrix}}-\underbrace {\nu {\overline {{\frac {\partial u'_{i}}{\partial x_{j}}}{\frac {\partial u'_{i}}{\partial x_{j}}}}}} _{\begin{smallmatrix}{\text{Dissipation}}\\\varepsilon _{k}\end{smallmatrix}}-\underbrace {{\frac {g}{\rho _{o}}}{\overline {\rho 'u'_{i}}}\delta _{i3}} _{\begin{smallmatrix}{\text{Buoyancy flux}}\\b\end{smallmatrix}}}$
By examining these phenomena, the turbulence kinetic energy budget for a particular flow can be found.[2]
## Computational fluid dynamics
In computational fluid dynamics (CFD), it is impossible to numerically simulate turbulence without discretizing the flow-field as far as the Kolmogorov microscales, which is called direct numerical simulation (DNS). Because DNS simulations are exorbitantly expensive due to memory, computational and storage overheads, turbulence models are used to simulate the effects of turbulence. A variety of models are used, but generally TKE is a fundamental flow property which must be calculated in order for fluid turbulence to be modelled.
### Reynolds-averaged Navier–Stokes equations
Reynolds-averaged Navier–Stokes (RANS) simulations use the Boussinesq eddy viscosity hypothesis [3] to calculate the Reynolds stress that results from the averaging procedure:
${\displaystyle {\overline {u'_{i}u'_{j}}}={\frac {2}{3}}k\delta _{ij}-\nu _{t}\left({\frac {\partial {\overline {u_{i}}}}{\partial x_{j}}}+{\frac {\partial {\overline {u_{j}}}}{\partial x_{i}}}\right),}$
where
${\displaystyle \nu _{t}=c\cdot {\sqrt {k}}\cdot l_{m}.}$
The exact method of resolving TKE depends upon the turbulence model used; kε (k–epsilon) models assume isotropy of turbulence whereby the normal stresses are equal:
${\displaystyle {\overline {(u')^{2}}}={\overline {(v')^{2}}}={\overline {(w')^{2}}}.}$
This assumption makes modelling of turbulence quantities (k and ε) simpler, but will not be accurate in scenarios where anisotropic behaviour of turbulence stresses dominates, and the implications of this in the production of turbulence also leads to over-prediction since the production depends on the mean rate of strain, and not the difference between the normal stresses (as they are, by assumption, equal).[4]
Reynolds-stress models (RSM) use a different method to close the Reynolds stresses, whereby the normal stresses are not assumed isotropic, so the issue with TKE production is avoided.
### Initial conditions
Accurate prescription of TKE as initial conditions in CFD simulations are important to accurately predict flows, especially in high Reynolds-number simulations. A smooth duct example is given below.
${\displaystyle k={\frac {3}{2}}(UI)^{2},}$
where I is the initial turbulence intensity [%] given below, and U is the initial velocity magnitude;
${\displaystyle \varepsilon ={c_{\mu }}^{\frac {3}{4}}k^{\frac {3}{2}}l^{-1}.}$
Here l is the turbulence or eddy length scale, given below, and cμ is a kε model parameter whose value is typically given as 0.09;
${\displaystyle I=0.16Re^{-{\frac {1}{8}}}.}$
The turbulent length scale can be estimated as
${\displaystyle l=0.07L,}$
with L a characteristic length. For internal flows this may take the value of the inlet duct (or pipe) width (or diameter) or the hydraulic diameter.[5]
## References
1. ^ Pope, S. B. (2000). Turbulent Flows. Cambridge: Cambridge University Press. pp. 122–134.[ISBN missing]
2. ^ Baldocchi, D. (2005), Lecture 16, Wind and Turbulence, Part 1, Surface Boundary Layer: Theory and Principles , Ecosystem Science Division, Department of Environmental Science, Policy and Management, University of California, Berkeley, CA: USA.
3. ^ Boussinesq, J. V. (1877). "Théorie de l'Écoulement Tourbillant". Mem. Présentés par Divers Savants Acad. Sci. Inst. Fr. 23: 46–50.
4. ^ Laurence, D. (2002). "Applications of Reynolds Averaged Navier Stokes Equations to Industrial Flows". In van Beeck, J. P. A. J.; Benocci, C. Introduction to Turbulence Modelling, Held March 18–22, 2002 at Von Karman Institute for Fluid Dynamics. Sint-Genesius-Rode: Von Karman Institute for Fluid Dynamics.
5. ^ Flórez Orrego; et al. (2012). "Experimental and CFD study of a single phase cone-shaped helical coiled heat exchanger: an empirical correlation". Proceedings of ECOS 2012 – The 25th International Conference on Efficiency, Cost, Optimization, Simulation and Environmental Impact of Energy Systems, June 26–29, 2012, Perugia, Italy. ISBN 978-88-6655-322-9.
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# 6.7.23Conskler ( and 19 i V= CIO,IL wilh the mner procluct (9} gwven by the inlegual i(U(U) dtCompule /ll; whete
###### Question:
6.7.23 Conskler ( and 19 i V= CIO,IL wilh the mner procluct (9} gwven by the inlegual i(U(U) dt Compule /ll; whete I(U) 8 + 240 M-L Enter your answer in the answer box and tnen click Check Answer; parts showing Gcal
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##### How would you graph the line x=-8?
How would you graph the line x=-8?...
##### In team development ladder 'Members resist control by group leaders and show hostility refers to a....
In team development ladder 'Members resist control by group leaders and show hostility refers to a. Forming stage. b. Storming c. Norming d. Performing...
##### Im not sure what I'm doing wrong these answers were both wrong and I'm very confused...
Im not sure what I'm doing wrong these answers were both wrong and I'm very confused I re did the math and im not sure if im using the wrong formula E4-12 and E4-13 Bunker makes two types of briefcase, fabric and leather. The company is currently using a traditional costing system with labor...
##### What rules and conditions are applied to physician offices under the CLIA? Are all offices governed...
what rules and conditions are applied to physician offices under the CLIA? Are all offices governed by this Act?...
##### A) Explain what does "a 95% confidence" means
a) Explain what does "a 95% confidence" means...
##### If A is necessary and sufficient for B, then A and B are said tobe equivalent.
If A is necessary and sufficient for B, then A and B are said to be equivalent....
##### Consider the twin-engine airplane of the DIAGRAM Assume the airplane is cruising at & (pressure) altitude of 12km and at velocity of 100 mls. The airplane is carrying 10,000 N of fuel and the thrust-specific fuel consumption is 2 x 10Compute:a) Endurance (hour)6) Range (km)Top View~Front View
Consider the twin-engine airplane of the DIAGRAM Assume the airplane is cruising at & (pressure) altitude of 12km and at velocity of 100 mls. The airplane is carrying 10,000 N of fuel and the thrust- specific fuel consumption is 2 x 10 Compute: a) Endurance (hour) 6) Range (km) Top View ~ Front ...
##### 2^-x+1=5^x+2solve the following equations
2^-x+1=5^x+2 solve the following equations...
##### Question 3 / 20 A sample of temperatures was taken in Detroit during the month of...
Question 3 / 20 A sample of temperatures was taken in Detroit during the month of June. The temperatures were measured in degrees Fahrenheit. The mean of the sample was 82 degrees Fahrenheit, and the standard deviation of the sample was 11 degrees Fahrenheit. Suppose that the same temperatures were...
##### ABC Company, Inc., has been in existence for 10 years. It has 50 employees and currently...
ABC Company, Inc., has been in existence for 10 years. It has 50 employees and currently only one location. ABC Company manufactures one product: widgets. It has managed to stay in business by holding onto a small portion of the widget market, but it is far from the market leader. Over the last five...
##### On January 1, 2018, Solda Co. issued its 10% bonds in the face amount of 8,000,000,...
On January 1, 2018, Solda Co. issued its 10% bonds in the face amount of 8,000,000, which mature on January 1, 2028. The bonds were issued at a time when the market rate was 8%. Using the effective-interest method of annual amortization, what is the bond premium? A) 9,073,613 B) 1,073,613 C) 972,000...
##### What is the volume rate of flow of water from a 1.85-cm-diameter faucet if the pressure head is 12.0 m?
What is the volume rate of flow of water from a 1.85-cm-diameter faucet if the pressure head is 12.0 m?...
##### Multiple choice questions! What's true concerning the pinMode() function? o It accepts port work direction as...
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##### Dimethylamine, (CH4)2NH, is a base with many industrial applications. a. Write the balanced equation and the...
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##### AICI 38 =3 1) 3)10
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##### Which of the following is NOT an explicit recommendation found in the Dietary Guidelines for Americans...
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##### Draw Newman projections of 2, 3-dimethylbutane, looking down the C2 to C3 bond, in the most...
Draw Newman projections of 2, 3-dimethylbutane, looking down the C2 to C3 bond, in the most and least stable conformations. Label any important interactions....
##### 1. In the context of NK-cell inhibitory and activating receptorsand maternal arterial invasion, explain the cause of (A)pre-eclampsia and (B) obstructed labor. Base your response on apregnancy involving a fetus who has inherited a C2 epitope from thefather but whose mother is homozygous for C1.2. Explain why a staphylococcal infection might produce amedical emergency.
1. In the context of NK-cell inhibitory and activating receptors and maternal arterial invasion, explain the cause of (A) pre-eclampsia and (B) obstructed labor. Base your response on a pregnancy involving a fetus who has inherited a C2 epitope from the father but whose mother is homozygous for C1. ...
##### 21. A safety valve of steel tank will open when the internal pressure exceeds 1.00x10 mmHg....
21. A safety valve of steel tank will open when the internal pressure exceeds 1.00x10 mmHg. It is filled with methane at 23.15 °C and 0.999 atm. If the temperature increases to 115 °C. Will the safety valve open? (Note: apply the Pressure- Temperature Relationship) (3 pts) Your answer: Conve...
##### What are "E", if each of the following atomicelectron-dot diagrams corresponds toa fourth period, representative element?Representative elements are those that belong tothe groups 1,2,and 13-18 (IA - VIIIA) of the Periodic Table. Alsocalled Main Group elements, they are characterized byhaving their outermost or valence electrons in s and psubshells.Determine the element's identity in each case.Enter the chemical symbols for the elements, in the boxesprovided.(Answers are CaSe-SeNS
What are "E", if each of the following atomic electron-dot diagrams corresponds to a fourth period, representative element? Representative elements are those that belong to the groups 1,2,and 13-18 (IA - VIIIA) of the Periodic Table. Also called Main Group elements, they are characterized ...
##### Prove that (p - q) ^ (p - r) = (qVr) + p(b) Let T(z,y) mean that student x likes cuisine y, where the domain for € consists of all students at Hofstra University and the domain for y consists of all cuisines_ Express both of the following sentences in terms of T(z,y), quantifiers, and logical connectives:(1) Some Hofstra students like Korean food and every Hofstra student likes Mexican cuisine. (2) For every pair of distinct Hofstra students, there is some cuisine that at least one them likes.
Prove that (p - q) ^ (p - r) = (qVr) + p (b) Let T(z,y) mean that student x likes cuisine y, where the domain for € consists of all students at Hofstra University and the domain for y consists of all cuisines_ Express both of the following sentences in terms of T(z,y), quantifiers, and logical...
##### Traditional budgeting is now ingrained in all areas of organisational activity. However despite its objective of...
Traditional budgeting is now ingrained in all areas of organisational activity. However despite its objective of planning for future operations, traditional incremental budgeting has been criticised for relying on past information as a basis of future targets. The whole purpose and function of budge...
##### Question 16 4 pts A bond indenture is a requirement that states the bond in question...
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##### Question 23 (2 points) The hormone is released by the pancreas when blood glucose levels are low, while is released by the pancreas when blood glucose levels are high: Both hormones work to keep blood glucose at healthy level:
Question 23 (2 points) The hormone is released by the pancreas when blood glucose levels are low, while is released by the pancreas when blood glucose levels are high: Both hormones work to keep blood glucose at healthy level:...
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x
## Horizontal and Vertical Stretching/Shrinking
A stretch in which a plane figure is distorted vertically. See also Vertical Stretch of 2, Horizontal Shrink of $$\displaystyle \frac{1}{3}$$, Shift left 2
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## Vertical Stretch or Compression on the Graph of an Absolute
y = f(x) y = f ( x) y= f(x 2) y = f ( x 2) horizontal stretch; x x -values are doubled; points get farther away. from y y -axis. vertical stretching/shrinking changes the y y -values of points;
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## How to Find Vertical Stretch
Vertical Stretches and Compressions When we multiply a function by a positive constant, we get a function whose graph is stretched or compressed vertically in relation to the graph of the original function. If the constant is greater than 1
## Vertical Stretch – Properties, Graph, & Examples
Vertical Shift If $$y=f(x)$$ then the vertical shift is caused by adding a constant outside the function, $$f(x)$$. Adding 10, like this $$y=f(x)+10$$ causes a movement of $$+10$$ in the y-axis.
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## Stretching and Reflecting Transformations
Vertical stretch occurs when a base graph is multiplied by a certain factor that is greater than 1. This results in the graph being pulled
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# American Institute of Mathematical Sciences
December 2016, 36(12): 7057-7061. doi: 10.3934/dcds.2016107
## On the symmetry of spatially periodic two-dimensional water waves
1 Faculty of Mathematics, University of Vienna, Oskar-Morgenstern-Platz 1, Vienna, A-1090, Austria
Received December 2015 Revised January 2016 Published October 2016
We show that a spatially periodic solution to the irrotational two-dimensional gravity water wave problem, with the property that the horizontal velocity component at the flat bed is symmetric, while the acceleration at the flat bed is anti-symmetric with respect to a common axis of symmetry, necessarily constitutes a traveling wave. The proof makes use complex variables and structural properties of the governing equations for nonlinear water waves.
Citation: Florian Kogelbauer. On the symmetry of spatially periodic two-dimensional water waves. Discrete & Continuous Dynamical Systems - A, 2016, 36 (12) : 7057-7061. doi: 10.3934/dcds.2016107
##### References:
show all references
##### References:
[1] Adrian Constantin. Dispersion relations for periodic traveling water waves in flows with discontinuous vorticity. Communications on Pure & Applied Analysis, 2012, 11 (4) : 1397-1406. doi: 10.3934/cpaa.2012.11.1397 [2] Anna Geyer, Ronald Quirchmayr. Traveling wave solutions of a highly nonlinear shallow water equation. Discrete & Continuous Dynamical Systems - A, 2018, 38 (3) : 1567-1604. doi: 10.3934/dcds.2018065 [3] Xiao-Biao Lin, Stephen Schecter. Traveling waves and shock waves. Discrete & Continuous Dynamical Systems - A, 2004, 10 (4) : i-ii. doi: 10.3934/dcds.2004.10.4i [4] José Raúl Quintero, Juan Carlos Muñoz Grajales. On the existence and computation of periodic travelling waves for a 2D water wave model. Communications on Pure & Applied Analysis, 2018, 17 (2) : 557-578. doi: 10.3934/cpaa.2018030 [5] Elena Kartashova. Nonlinear resonances of water waves. Discrete & Continuous Dynamical Systems - B, 2009, 12 (3) : 607-621. doi: 10.3934/dcdsb.2009.12.607 [6] Robert McOwen, Peter Topalov. Asymptotics in shallow water waves. Discrete & Continuous Dynamical Systems - A, 2015, 35 (7) : 3103-3131. doi: 10.3934/dcds.2015.35.3103 [7] Walter A. Strauss. Vorticity jumps in steady water waves. Discrete & Continuous Dynamical Systems - B, 2012, 17 (4) : 1101-1112. doi: 10.3934/dcdsb.2012.17.1101 [8] Vera Mikyoung Hur. On the formation of singularities for surface water waves. Communications on Pure & Applied Analysis, 2012, 11 (4) : 1465-1474. doi: 10.3934/cpaa.2012.11.1465 [9] Jerry L. Bona, Henrik Kalisch. Models for internal waves in deep water. Discrete & Continuous Dynamical Systems - A, 2000, 6 (1) : 1-20. doi: 10.3934/dcds.2000.6.1 [10] Martina Chirilus-Bruckner, Guido Schneider. Interaction of oscillatory packets of water waves. Conference Publications, 2015, 2015 (special) : 267-275. doi: 10.3934/proc.2015.0267 [11] Jonatan Lenells. Traveling waves in compressible elastic rods. Discrete & Continuous Dynamical Systems - B, 2006, 6 (1) : 151-167. doi: 10.3934/dcdsb.2006.6.151 [12] Vincent Duchêne, Samer Israwi, Raafat Talhouk. Shallow water asymptotic models for the propagation of internal waves. Discrete & Continuous Dynamical Systems - S, 2014, 7 (2) : 239-269. doi: 10.3934/dcdss.2014.7.239 [13] Anca-Voichita Matioc. On particle trajectories in linear deep-water waves. Communications on Pure & Applied Analysis, 2012, 11 (4) : 1537-1547. doi: 10.3934/cpaa.2012.11.1537 [14] Jerry L. Bona, Thierry Colin, Colette Guillopé. Propagation of long-crested water waves. Discrete & Continuous Dynamical Systems - A, 2013, 33 (2) : 599-628. doi: 10.3934/dcds.2013.33.599 [15] Angel Castro, Diego Córdoba, Charles Fefferman, Francisco Gancedo, Javier Gómez-Serrano. Structural stability for the splash singularities of the water waves problem. Discrete & Continuous Dynamical Systems - A, 2014, 34 (12) : 4997-5043. doi: 10.3934/dcds.2014.34.4997 [16] Mats Ehrnström, Gabriele Villari. Recent progress on particle trajectories in steady water waves. Discrete & Continuous Dynamical Systems - B, 2009, 12 (3) : 539-559. doi: 10.3934/dcdsb.2009.12.539 [17] David M. Ambrose, Jerry L. Bona, David P. Nicholls. Well-posedness of a model for water waves with viscosity. Discrete & Continuous Dynamical Systems - B, 2012, 17 (4) : 1113-1137. doi: 10.3934/dcdsb.2012.17.1113 [18] David Henry, Bogdan--Vasile Matioc. On the regularity of steady periodic stratified water waves. Communications on Pure & Applied Analysis, 2012, 11 (4) : 1453-1464. doi: 10.3934/cpaa.2012.11.1453 [19] Gerhard Tulzer. On the symmetry of steady periodic water waves with stagnation points. Communications on Pure & Applied Analysis, 2012, 11 (4) : 1577-1586. doi: 10.3934/cpaa.2012.11.1577 [20] David Henry, Hung-Chu Hsu. Instability of equatorial water waves in the $f-$plane. Discrete & Continuous Dynamical Systems - A, 2015, 35 (3) : 909-916. doi: 10.3934/dcds.2015.35.909
2018 Impact Factor: 1.143
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# Chapter 16: Integrals and Vector Fields - Practice Exercises - Page 1029: 25
$$\sqrt 6$$
#### Work Step by Step
We have $r_u=1+j \space and \\ r_v=1-j+k$ and $$|r_u \times r_v|=\sqrt 6$$ $$Surface \space Area=\iint_S |r_u \times r_v| \space du \space dv \\=\int_0^1 \int_0^1 \sqrt 6 \space du \space dv \\=\sqrt 6(1-0) \\=\sqrt 6$$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Dil is studying two different type of birds: Polioptila caerulea and Guiraea caerulea. He was amazed by the similar characteristics
###### Question:
Dil is studying two different type of birds: Polioptila caerulea and Guiraea caerulea. He was amazed by the similar characteristics both birds share. Based on the scientific names, Dil can conclude that the two types of birds
a. belong to different specie
b. belong to different domains
c. belong to different genera
### (TEK 6.12 B) The main difference between a prokaryote and a eukaryote is that a eukaryote always * A. Has a cell wall B. Makes
(TEK 6.12 B) The main difference between a prokaryote and a eukaryote is that a eukaryote always * A. Has a cell wall B. Makes its own food C. Has a nucleus...
### ANYONE WHO GIVES CORRECT ANSWER WILL BE CHOSE AS A BRILLIANT PLS HELP ME Make a report of industrial
ANYONE WHO GIVES CORRECT ANSWER WILL BE CHOSE AS A BRILLIANT PLS HELP ME Make a report of industrial accident that has occurred in the world or India...
### It is estimated that light takes 110,000 years to travel the full distance across our galaxyin the picture
It is estimated that light takes 110,000 years to travel the full distance across our galaxyin the picture read the rest of the question....
### The yangtze is the world's 4th largest carrier of question 5 options: sediment oil transport
The yangtze is the world's 4th largest carrier of question 5 options: sediment oil transport ships people...
### Which of the following is a statistical question? What is the average of all your scores this semester?
Which of the following is a statistical question? What is the average of all your scores this semester? What was your score on the last exam?...
### With α = .01 the two-tailed critical region for a t test using a sample of n = 16 subjects would have
With α = .01 the two-tailed critical region for a t test using a sample of n = 16 subjects would have boundaries of ...
### Gloria traveler a total of 8 miles to get from school to her apartment. gloria took the bus from school
Gloria traveler a total of 8 miles to get from school to her apartment. gloria took the bus from school to the closest bus stop to her apartment and then walked the remaining 880 yards home....
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Select Articles
# Tiny, Tuned, and UnattachedPUBLIC ACCESS
Work is Under Way to Create High-end Integrated Micro Systems that can Sense, Crunch Data, and Communicate Wirelessly in a Package the Size of a Sugar Cube.
[+] Author Notes
Associate Editor
Mechanical Engineering 123(07), 50-54 (Jul 01, 2001) (4 pages) doi:10.1115/1.2001-JUL-1
## Abstract
Work is under way to create high-end integrated microsystems that can sense, crunch data, and communicate wirelessly—in a package the size of a sugar cube. Research is under way to bring wireless communication down to the micro level, laying the groundwork for next-generation sensing systems. Wireless sensors assume low manufacturing costs, low power use, and an elevated level of integration. Power dissipation must be low enough to permit an acceptable lifetime for the device. One effort working to make this vision a reality is the Wireless Integrated Microsystems (WIMS) project—a research consortium funded by the National Science Foundation (NSF). WIMS is one of about 20 Engineering Research Centers funded by the NSF and is the only one focused on wireless microelectro-mechanical system (MEMS) devices. The NSF has pledged financial support for up to 10 years.
## Article
By integrating mechanical elements, actuators, and electronics into one siliconbased device, microsystems technology has pushed the boundaries of sensors into new areas. Microelectro-mechanical devices have clearly become very sophisticated, setting the stage for their widespread deployment in a promising range of applications, whether inside the human body or in the environment at large.
Yet for all the technical wizardry of MEMS sensors, one hurdle still stands in the way of their being truly field-deployable. Wireless communication, which made cell phones possible and lets travelers check e-mail from their cabs, is still limited to the macro scale. Research is under way to bring wireless communication down to the micro level, laying the groundwork for next-generation sensing systems.
Wireless microsystems hinge on low power dissipation, a requirement that raises the bar for developments in mechanical circuits, electronic packaging, and materials. It's a tall order, but the potential payoff is pervasive use of the systems in society.
Installing wiring-punching holes in walls or describing paths around machinery, stringing wires, and labeling them-can be a labor-intensive, manual operation. Wiring can be the dominant cost in installing sensors.
A cochlear implant smaller than a penny was machined at Michigan Technological University. It will be placed in the inner ear.
## Wireless Matters
According to Albert P. Pisano, director of the Electronics Research Lab at the University of California, Berkeley, and chair of the Executive Committee of ASME's MEMS subdivision, a rough rule of thumb, used in military procurement, is that for every dollar you spend on a sensor, you spend $10 to$100 for the package and between $1,000 and$10,000 for the wiring. "Anything that can be done to eliminate the cost of wiring is the right thing to do," he said.
Wireless sensors assume low manufacturing costs, low power use, and a high level of integration. "The e sensors aren't going to work if they are expensive," Pisano said. For the concept to be practical, microprocessor, sensor, and wireless link would have to be designed into one tight package that could be manufactured inexpensively. Power dissipation must be low enough to permit an acceptable lifetime for the device.
As the right technical developments-low power drain, high integration, and infrastructure needs-come together, wireless sensors can make financial sense. If that happens, it presents an intriguing vision of the future: inexpensively produced, integrated microsystems consisting of microprocessor, electromechanical device, and wireless link-essentially, stick-on systems that would perform a valuable function and with which it is possible to communicate.
## Making it Happen
One effort working to make this vision a reality is the Wireless Integrated Microsystems project—a research consortium funded by the National Science Foundation. WIMS is one of about 20 Engineering Research Centers funded by the NSF and is the only one focused on wireless MEMS devices. The NSF has pledged financial support for up to 10 years.
The WIMS Engineering Research Center is based at the University of Michigan in Ann Arbor, and also involves re searchers at Michigan State University in East Lansing and Michigan Technological University in Houghton. The consortium includes 19 companies and four nonprofit institutions. The project covers microprocessors, sensors, and radio-frequency links, with research activity focusing on issues of electronic packaging, low power dissipation, and new materials.
Potential applications for the devices include environmental monitoring, from weather and global warming to air and water quality; and health care, by providing wearable or implantable biomedical devices.
One indication of the wide interest in wireless integrated Microsystems is the variety of industries represented by its members: large, well-established companies as well as small startups, in businesses as diverse as automobiles, oil, biomedicine, and semiconductors. Industrial partners, as they are described on the WIMS Web site, are asked to kick in anywhere from $10,000 to$100,000 a year to support the center's research. The list includes IBM, Ford Motor Co., Intel, Honeywell, and Chevron Research & Technology Corp., as well as Cochlear Corp. and Advanced Bionics, which make ear implants.
According to Joe Giachino, industrial liaison at the WIMS Center at the University of Michigan, "We are attacking a problem, and then getting a microsystem solution to that problem." The WIMS project has two testbeds—a cochlear implant and an environmental sensor.
The research center is developing high-end wireless MEMS devices, merging micropower electronics, a wireless link, and a MEMS data-gathering device in a single module, said Kensall Wise, professor of electrical engineering and computer science at the University of Michigan and the center's director. The goal, over a five to 10-year period, is to combine components in a block measuring 1 or 2 cubic centimeters and run the system on 100 microwatts or less.
Wise believes that interest in wireless microsensors is pervasive, and that opportunities for wireless portable or wearable devices will expand by orders of magnitude over the next decade. Wireless capability adds reliability and lowers the cost of any type of wearable or portable system that will be deployed to gather information over a broad range, he said. The wireless systems under development have ranges from 1 centimeter to 1 kilometer, said Wise. That would be the range of an original signal that could be picked up and relayed.
Wireless links available today are an order of magnitude higher in power dissipation than what is needed, said Wise. One of the methods being investigated to bring down power use is to substitute microelectrornic circuits with micromechanical ones, which dissipate far less power. The wireless links would have to operate at sub-milliwatt levels, periodically report data, and listen for input commands.
Microprocessors need to be optimized for low-power data-gathering application and data interpretation, he said. Typical applications will not require high processing speeds. "You are going to need pretty good power-to-weight product," Wise said. Present- generation MEMS devices would probably require too much power to be used in these applications, he added.
In addition to requiring very low-power operation, wireless MEMS sensors will have to be self- testing and able to work in tandem with the embedded controller, and will probably be compensated by software in the controller, said Wise. Depending on the application, he envisions that the MEMS devices will include pumps, valves, and inertial sensors.
Packaging is the key to robustness and the ability to work in the real world, Wise said. "Increasingly, we are going to see the packages done at the wafer level," he said. Many inertial sensors, like accelerometers and gyros, require a vacuum reference cavity, in which the transducer operates in a vacuum. Increasingly, those vacuum reference cavities will be done at the wafer level, as part of the chip fabrication, so that when it becomes necessary to dice up the wafer and package it, the critical elements are already protected.
Housed between metal electrodes, a 156-MHz, balanced radial-mode disc, polysilicon micromechanical resonator is expected to achieve ultrahigh frequencies. The resonator has a diameter of 34 micrometers.
## Mechanical Circuits
WIMS devices are being designed with radio frequency wireless links, allowing them to ignore physical obstacles, to be field deployable. The goal is to develop devices that can communicate 1 kilometer over a variety of terrain with a maximum transmission power of 1 milliwatt.
Mechanical circuits offer a way for these devices to operate at very low power levels and offer a very high Q-factor, or signal quality, according to Clark Nguyen, the center's wireless task leader and an assistant professor of electrical engineering and computer science at the University of Michigan. Nguyen's work is supported by the NSF through the Engineering Research Center as well as by the Defense Advanced Research Projects Agency.
Mechanical circuits exist in telecommunication devices today on a macroscopic scale, taking up a lot of space, said Nguyen. An everyday example of a mechanical circuit is a guitar string. When plucked, if it is in tune, it will tend to vibrate at a specific frequency, say, 440 Hz.
This is analogous to the way that wireless phones work: They transmit electromechanical signals at very specific frequencies, between 800 MHz and 1.8 GHz, and use mainly mechanical components such as resonators to select certain frequencies while rejecting others, allowing people to communicate. Phones use mechanical components such as surface acoustic filters and quartz crystal resonators that select frequencies. A typical phone board today is about 10 to 20 percent transistors; the rest of it consists of passive circuits, the largest and most expensive of which operate through mechanical vibration, Nguyen said.
These mechanical circuits, however, are macroscopic components that can be seen with the naked eye. The task of the WIMS project is to develop microscale mechanical components with the same or better frequency-selecting properties as their macroscopic counterparts, he said.
Nguyen compares the potential impact of micro mechanical circuits with that of transistors, in which miniaturization made possible complex circuits and systems, and eventually opened the way for computers and wireless phones. "Integrated circuit technology revolutionized things because it provided a way to shrink transistors to a tiny size, making them manufacturable in huge volumes for a very low expense. It also allowed them to have many times more circuit complexity and capability," said Nguyen. "Once we've got the mechanical circuits to a tiny size, we may be able to change paradigms governing wire loss architectures." The change, he hopes, will make possible wireless integrated microsystems.
Although Nguyen said that plenty of research needs to be done, he is optimistic that micromechanical circuits can eventually be made to operate at frequencies between 800 MHz and 1.8 GHz and beyond.
Because mechanical circuits consume less power than transistors, they can help microsystem devices consume less power. Nguyen said the long-term goal is to reduce power consumption of the radio-frequency front end of the transceivers from around 300 milliwatts to 100 microwatts. Mechanical circuits have much smaller losses than electrical ones, and so may not need as many amplifiers to make up for a degraded signal, he said. In addition, the use of numerous high-Q mechanic al circuits may allow design changes that trade Q for power, further reducing power consumption and extending battery lifetimes.
Bringing these circuits down to the micro scale opens up new design challenges. "The way the mechanical elements are forced into vibration and the way they are hooked together is a completely different type of circuit technology," Nguyen said. This means new design procedures have to be developed as well as micromechanical technology, he said. Nguyen, an electrical engineer, said the circuits could be designed in the electrical domain or the mechanical domain. One of the beauties of this work is that it teaches students how easily they can interchange the mechanical with the electrical, he said.
On the manufacturing side, mass producing these circuits- making thousands of these components per chip—introduces a trimming and tuning requirement. It may be possible to design around the problem; however, if not, a way must be found to trim and tune frequencies on a massive scale--a technology that still must be developed.
Two Tracks Toward Micro Wireless
THE WIRELESS INTEGRATED MICROSYSTEMS consortium sponsored by the National Science Foundation is working on two testbeds to develop wireless microsystems technology.
One testbed is an environmental monitor incorporating gas chromatography. The device will be a versatile environmental monitor capable of analyzing complex mixtures of mainly toxic organic compounds, according to Edward Zellers, an associate professor of environmental health sciences and chemistry at the University of Michigan, who is heading the project.
Zellers expects that the wireless link will allow uploading data from the instrument to a monitoring station, and will permit down loading of instructions. Although the concept of a computer-on-a-chip chemical analyzer is not brand-new—one has been developed at Sandia National labs in Albuquerque-the WIMS project aims to make its system smaller, higher-performing, and less power-consuming than any that has been done before, said Zellers.
One of the big challenges in developing the device is de- signing a micropump that can deliver appropriate amounts of air. The other is thermal control for the device, which must sit outdoors subject to a range of temperatures.
The system will have the ability to self-calibrate, although there will be limits to how much it can compensate in extreme temperatures, he said.
The other testbed is a cochlear implant, to help the hearing- impaired detect sounds. Present devices are comparatively large, said Khalil Najafi, deputy director of the WIMS project and professor of electrical engineering and computer science at the University of Michigan. "We feel that MEMS has a great potential to significantly enhance the capability in these areas, because you can build them very small and put them where they need to be placed."
Particular challenges in developing a cochlear implant are packaging and power dissipation, Najafi said. In the body, you have to worry about more than just protecting the device in a harsh environment, he said; you also have to worry about protecting the body. And power dissipation has to be low enough to be tolerated by the individual.
A 92-MHz free-free beam micromechanical resonator with nonintrusive supports to reduce anchor dissipation is expected to achieve UHF ranges.
## Tight Integration
Trying to build a wireless integrated microsystem on a device as small as a sugar cube requires tight integration of the various components, according to Richard Brown, professor of electrical engineering and computer science at the University of Michigan, who is investigating microprocessors for use in WIMS devices. Integrating as many components as possible on a single chip makes it possible to drive down the power use, as well as reduce the size of a device.
Building on such a small scale makes it difficult to put together separate parts, he said. Tight integration on a single chip eliminates off-system drivers and the space they represent, and also dissipates less power. Producing extremely low-power microcontrollers requires designing with leading- edge technology because more advanced processors with finer geometry have lower power requirements.
Packaging presents its own set of challenges, according to Khalil Najafi, deputy director of the WIMS project and professor of electrical engineering and computer science at the University of Michigan. Mechanical components, such as vibrators and resonators, need to operate in a vacuum, he said. Research is focused on developing techniques to package at the wafer level that can also provide a suitable environment.
Typically, these devices will have to operate in a lower-pressure environment for extended periods of time, possibly under harsh conditions and through temperature changes, he said. As the resonating parts move, "you don't want any air molecules around them that can impede their resonance," Najafi said. "You also need to make sure that the environmental conditions don't cause any changes in devices that are used as filters and oscillators in communications systems."
Other components add packaging demands, he added. Some sensors have to be exposed to the environment in order to measure it, requiring some components to be packaged in a vacuum while others are exposed to the environment. Other devices, like accelerometers, would need to be hermetically sealed to prevent moisture from getting onto critical components, he added.
Most of these devices will be manufactured using planar thin-film technology that is used in semiconductor processing. The advantage of using this technology is that it allows thousands of devices to be processed on a 4-by- 6-inch silicon wafer, which can then be diced apart. "We need to be sure that we can apply and finish the package at the wafer level, so that when we go to the process of dicing the individual devices apart, they are already protected and packaged and ready to be connected. That's not an easy thing to do," said Najafi
Katharine Beach, an engineer in research at the University of Michigan, at a thermal oxidation furnace, which is used to grow silicon dioxide on silicon wafers used in WIMS device manufacturing.
## Diamond Films
Performance demands have also prompted materials research, which is being spearheaded by Michigan State. Poly diamond films are being investigated for three areas of the WIMS cube: wireless interfaces, sensors, and packaging, according to Dean Aslam, associate director of the WIMS Engineering Research Center and professor of electrical and computer science at Michigan State.
Because it is very stiff-it takes more force to bend-diamond has the ability to increase the frequency of the micromechanical component, said Aslam. Using the same dimensions, frequency can be increased by a factor of two or three compared to polycrystalline silicon, he said.
Diamond may also present other advantages as well. Aslam speculates that it may allow less stringent requirements for vacuum. Vibration is hindered in the presence of air, either because the air pushes the beam back or because air molecules stick to the beam. This is why the Q-factor deteriorates if the vibrating beam is not in a vacuum.
Diamond may get around this problem because it does not react with anything. The incidence of the air molecules sticking to the diamond is almost zero, he said. Aslam said that if diamond works well, it could potentially be used for all of the mechanical components in the WIMS devices.
Diamond can be made into a semiconductor or insulator, depending on whether or not it is doped (or impurities are added to it). As a semiconductor, it can be used as a sensor; as an insulator, it can be used as a protective layer. Both roles are being explored for the cochlear testbed.
One potential use is as an ultrasensitive sensor. Silicon sensors are based on a piezoresistive effect: If a strain is put on a material to bend it, its electrical resistance will change. The piezoresistive gauge factor, a measure of sensitivity, of silicon is around 150. The piezoresistive gauge factor of diamond is 4,000, making it more than 20 times as sensitive, said Aslam.
Such ultrasensitive devices could be useful in cochlear implants, eliminating the need to do signal processing to build up the strength of a very weak signal. Diamond is also being developed as a layer to protect the silicon probe in the cochlear prosthesis, Aslam said. Because diamond is a very stable material, it is also bio-friendly, he added.
Diamond is being investigated as a packaging material for WIMS devices, which will incorporate many new components, such as wireless links or even interfaces to analyze liquids, said Aslam. "There is no existing packaging technology that you can just apply," he said.
Two packaging approaches are being evaluated. One is to coat the device with a thin layer of diamond. The other builds a diamond package consisting of four walls, a bottom, and top, in a mold of silicon, which is subsequently removed by chemical etching.
Run by Craig Friedrich at Michigan Tech, this micromechanical machining workstation, which can use end mills of 15 microns in diameter, was used to machine the cochlear implant.
## Parts, Packaging, And Micromachining
Micromachining technology from Michigan Technological University is being used to develop three-dimensional WIMS components and packaging. For the cochlear implant, Michigan Tech is providing modeling, design, and micromilling of the spiral-shaped implant that will go into the inner ear.
Rapid prototyping was used to create outsize models. The final product will have to be brought down in size to fit the ear, according to Robert O. Warrington, WIMS associate director and a dean of engineering at Michigan Tech. "We can get things down to the tens of microns, depending on what we machine and how we machine it," he said. One design the researchers came up with is based on party favors that blowout of a spiral, said Warrington.
Michigan Tech is also developing the packaging housing for the environmental monitoring testbed, including fluid, mechanical, and electrical interconnects.
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The Fundamental Theorem of Calculus is of course correct, and you are applying it correctly. The statements the rate of change of work with respect to displacement is force and the instantaneous rate of change of work with respect to displacement is force are correct. The thing to keep in mind is that it's not work that is instantaneous, but its ...
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They are generally of two type slow algorithm and fast algorithm. REST Web Service. Fig. They are generally of two type slow algorithm and fast algorithm.Slow division algorithm are restoring, non-restoring, non-performing restoring, SRT algorithm … Here, n-bit dividend is loaded in Q and divisor is loaded in M. Value of Register is initially kept 0 and this is the register whose value is restored during iteration due to which it is named Restoring. The tutor starts with the very basics and gradually moves on to cover a range of topics such as Instruction Sets, Computer Arithmetic, Process Unit Design, Memory System Design, Input-Output Design, Pipeline Design, and RISC. A slightly more complex approach, known as nonrestoring, avoids the unnecessary subtraction and addition. collectively called algorithm. MongoDB. i).State the division algorithm with diagram and examples. The unsigned division algorithm that is similar to Booth's algorithm is shown in Figure 3.19a, with an example shown in Figure 3.19b. Remember to restore the value of A most significant bit of A is 1. It is an excellent book on computer architecture and should be read by anyone designing a digital signal processor. Can't understand this division algorithm in Computer Architecture. This method is based on Svoboda’s division algorithm and the radix-4 redundant number system. (a) (b) (c) Figure 3.19. Division algorithms fall into two main categories: slow division and fast division. Mantissa of – 0.5625 = 1.00100000000000000000000, Shifting right by 4 units, 0.00010010000000000000000, Mantissa of 9.75= 1. JavaScript. Experience. Slow division algorithm are restoring, non-restoring, non-performing restoring, SRT algorithm and under fast comes Newton–Raphson and Goldschmidt. We cannot add these numbers directly. This method is based on Svoboda's division algorithm and the radix-4 … Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. Most popular in Computer Organization & Architecture, More related articles in Computer Organization & Architecture, We use cookies to ensure you have the best browsing experience on our website. Now, we shift the mantissa of lesser number right side by 4 units. Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. Addition and subtraction, multiplication Algorithms, Division Algorithms, Floating point Arithmetic operations. Game Development with Unity Engine. DB. As a discipline, computer science spans a range of topics from theoretical studies of algorithms, computation and information to the practical issues of implementing computing systems in hardware and software. By using our site, you Use a presentation similar to that of Figure 10.17. Fast division methods start with a close … Experience. The above figure shows the value of each register for each of the steps, with the quotient being 3ten and the remainder 1ten. Division Floating point division requires that the exponents be subtracted and the mantissa divided. Description. Develop algorithm to implement A*B. Now adding significand, 0.05 + 1.1 = 1.15 So, finally we get (1.1 * 10 3 + 50) = 1.15 * 10 3. In addition to choosing algorithms for addition, sub-traction, multiplication, and division, the computer … GAME DEVELOPMENT. The division algorithm is an algorithm in which given 2 integers N N N and D D D, it computes their quotient Q Q Q and remainder R R R, where 0 ≤ R < ∣ D ∣ 0 \leq R < |D| 0 ≤ R < ∣ D ∣. Ask Question Asked 4 years, 2 months ago. What is scientific notation and normalization? This document is highly rated by Computer Science Engineering (CSE) students and … The twos complement integer division algorithm described in Section 10.3 isknown as the restoring method because the value in the A register must be restored following unsuccessful subtraction. This document is highly rated by Computer Science Engineering (CSE) students and … After aligning exponent, we get 50 = 0.05 * 103, Now adding significand, 0.05 + 1.1 = 1.15, So, finally we get (1.1 * 103 + 50) = 1.15 * 103. Unsigned Division. Computer arithmetic is nourished by, and in turn nourishes, other subfields of computer architecture and technology. 00111000000000000000000, In final answer, we take exponent of bigger number, 32 bit representation of answer = x + y = 0 10000010 01001010000000000000000. Examples of both restoring and non-restoring types of division algorithms can be found in the book, "Computer Architecture--A Quantitative Approach", Second Edition, by Patterson and Hennesy, Appendix A, Morgan Kaufmann Publishers, Inc. (1996). First, we need to align the exponent and then, we can add significand. ii).Divide 00000111 by 0010. CS8491 Important Questions Computer Architecture 6. algorithms have been proposed for use in floating-point accelerators, actual implementations are usually based on refinements and variations of the few basic algorithms presented here. Division algorithms are generally classified into two types, restoring and non-restoring. Give an example 8. Problem 17P from Chapter 10: ... Verify the validity of the unsigned binary division algorithm of Figure 10.16 by showing the steps involved in calculating the division depicted in Figure 10.15. Computer science is the study of algorithmic processes and computational machines. What is fast multiplication? hardware – we do not cover computer architecture or the design of computer hardware since good books are already available on these topics. Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below. We cannot add these numbers directly. Again it is … - Selection from Computer Architecture and Organization [Book] Computer Architecture. Nov 30, 2020 - Addition Algorithm & Subtraction Algorithm - Computer Organization and Architecture | EduRev Notes is made by best teachers of Computer Science Engineering (CSE). Computer Architecture ALU Design : Division and Floating Point EEL-4713 Ann Gordon-Ross.2 Divide: Paper & Pencil 1001 Quotient Divisor 1000 1001010 Dividend –1000 10 101 1010 –1000 10 Remainder (or Modulo result) See how big a number can be subtracted, creating quotient bit on each step Now we get the difference of exponents to know how much shifting is required. Subtraction is similar to addition with some differences like we subtract mantissa unlike addition and in sign bit we put the sign of greater number. Restoring Division Algorithm, Non-Restoring Division Algorithm. To understand floating point addition, first we see addition of real numbers in decimal as same logic is applied in both cases. Introduction of Boolean Algebra and Logic Gates, Number Representation and Computer Airthmetic. N2 - In this paper we present a fast radix-4 division algorithm for floating point numbers. Assume A and B for a pair of signed 2’s complement numbers with values: A=010111, B=101100. The mantissa division is done as in fixed point except that the dividend has a single precision mantissa that is placed in the AC. When the ones in a multiplier are grouped into long blocks, Booth's algorithm performs fewer additions and subtractions than the normal multiplication algorithm… Writing code in comment? A division algorithm provides a quotient and a remainder when we divide two number. There are many different algorithms that could be implemented, and we will focus on division by repeated subtraction. Here, notice that we shifted 50 and made it 0.05 to add these numbers. Slow division algorithms produce one digit of the final quotient per iteration. T1 - A Fast Radix-4 Division Algorithm and its Architecture. 2. We follow these steps to add two numbers: Converting them into 32-bit floating point representation, 9.75’s representation in 32-bit format = 0 10000010 00111000000000000000000, 0.5625’s representation in 32-bit format = 0 01111110 00100000000000000000000. Computer Organization and Architecture (10th Edition) Edit edition. Instead, we focus on algorithms for efficiently performing arithmetic o perations such as addition, multiplication, and division, and their connections to topics such The division algorithm can be divided into five parts.. 1. (10000010 – 01111110)2 = (4)10 Booth’s algorithm is a multiplication algorithm that multiplies two signed binary numbers in 2’s compliment notation. As that register Q contain the quotient, i.e. CE COMPUTER ARCHITECTURE CHAPTER 3 ARITHMETIC FOR COMPUTERS 1 . Here, notice that we shifted 50 and made it 0.05 to add these numbers.. Now let us take example of floating point number addition Using a 4-bit version of the algorithm to save pages, let’s try dividing 7 10 by 2 10, or 0000 0111 2 by 0010 2. A division algorithm provides a quotient and a remainder when we divide two number. Examples of both restoring and non-restoring types of division algorithms can be found in the book, "Computer Architecture--A Quantitative Approach", Second Edition, by Patterson and Hennesy, Appendix A, Morgan Kaufmann Publishers, Inc. (1996). This video tutorial provides a complete understanding of the fundamental concepts of Computer Organization. List the steps of division algorithm 7. Give the representation of single precision floating point number 9. 3.3.2.1. Don’t stop learning now. Don’t stop learning now. And these instructions perform a great activity in processing data in a digital computer. 2.10 Values of register in division algorithm . CE Division The division algorithm and hardware Example: Answer: Step by step follow the multiplication algorithm 22 23. 00111000000000000000000, So, finally the answer = x – y = 0 10000010 00100110000000000000000. 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Abstract—In this paper we present a fast radix-4 division algorithm for floating point numbers. First, we need to align the exponent and then, we can add significand. An Analysis of Division Algorithms and Implementations by Stuart F. Oberman and Michael J. Flynn, Stanford University Computer Systems Laboratory, CSL-TR-95-675. In order to solve the computational problems, arithmetic instructions are used in digital computers that manipulate data. Propose an algorithm for this latter approach. ... HTML Course. DBMS. Our algorithm is suitable for residue number systems with large moduli, with the aim of manipulating very large integers on a parallel computer or a special-purpose architecture. After aligning exponent, we get 50 = 0.05 * 10 3. Most popular in Computer Organization & Architecture, Most visited in Digital Electronics & Logic Design, We use cookies to ensure you have the best browsing experience on our website. Now, we find the difference of exponents to know how much shifting is required. Computer Arithmetic 73 We continue our scanning from left to right and next encounter with the left most two digits of the dividend, i.e., 01 (underlined). SASS/SCSS. COMPUTER. Understand the architecture of a modern computer with its various processing units. In this article, will be performing restoring algorithm for unsigned integer. Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready. Decimal Arithmetic unit, Decimal Restoring term is due to fact that value of register A is restored after each iteration. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. This article has been contributed by Anuj Batham. 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Attention reader! AU - Srinivas, Hosahalli R. AU - Parhi, Keshab K. PY - 1995/6. The ALU schematic diagram in given in Figure 3.19c. The analysis of the algorithm and circuit is very similar to the preceding discussion of Booth's algorithm. Now let us take example of floating point number addition. Examples of slow division include restoring, non-performing restoring, non-restoring, and SRT division. Mantissa of 0.5625 = 1.00100000000000000000000, (note that 1 before decimal point is understood in 32-bit representation), Shifting right by 4 units, we get 0.00010010000000000000000, Mantissa of 9.75 = 1. To solve various problems we give algorithms. Dec 05, 2020 - Multiplication Algorithm & Division Algorithm - Computer Organization and Architecture | EduRev Notes is made by best teachers of Computer Science Engineering (CSE). Here, register Q contain quotient and register A contain remainder. Writing code in comment? Introduction of Boolean Algebra and Logic Gates, Number Representation and Computer Airthmetic. Y1 - 1995/6. To represent the fractional binary numbers, it is necessary to consider binary point. Please use ide.geeksforgeeks.org, generate link and share the link here. For example, we have to add 1.1 * 103 and 50. Attention reader! Please use ide.geeksforgeeks.org, generate link and share the link here. The division algorithm states that for any integer, a, and any positive integer, b, there exists unique integers q and r such that a = bq + r (where r is greater than or equal to 0 and less than b). Linking computer arithmetic to other subfields of computing. Division algorithms are generally classified into two types, restoring and non-restoring. Computer science - Computer science - Architecture and organization: Computer architecture deals with the design of computers, data storage devices, and networking components that store and run programs, transmit data, and drive interactions between computers, across networks, and with users. EC8552 Questions Bank COMPUTER ARCHITECTURE AND ORGANIZATION CSS. These instructions perform arithmetic calculations. ... Fig.6 The first division algorithm 21 22. Check for zeros. The algorithm involves a simple recurrence with carry-free addition and employs prescaling of the operands. Also the performance measurement of the computer system. Computer Network. CS6303 – COMPUTER ARCHITECTURE UNIT-II Page 13 Division Division Algorithms and Hardware Implementations Two types of division operations • Integer division: with integer operands and result • Fractional division: operands and results are fractions Any division algorithm can be carried out independent of • Position of the decimal point Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready. Web Playground. See your article appearing on the GeeksforGeeks main page and help other Geeks. We propose in this paper a new algorithm and architecture for performing divisions in residue number systems. Converting them into 32-bit floating point representation, – 0.5625’s representation in 32-bit format = 1 01111110 00100000000000000000000. Booth's algorithm performs an addition when it encounters the first digit of a block of ones (0 1) and a subtraction when it encounters the end of the block (1 0). Active 4 years, 2 months ago. EC8552 Questions Bank COMPUTER ARCHITECTURE AND ORGANIZATION. The difference of exponents to know how much shifting is required anything incorrect by clicking on the GeeksforGeeks main and. Systems Laboratory, CSL-TR-95-675 PY - 1995/6 the value of a most significant bit a! Complex approach, known as nonrestoring, avoids the unnecessary subtraction and addition b ) b. And division algorithm in computer architecture machines CSE ) students and … restoring division algorithm for floating point numbers the representation of single floating... Divided into five parts.. 1 answer = x – y = 0 10000010 00100110000000000000000 as that Q! About the topic discussed above a fast radix-4 division algorithm want to share more information about the topic discussed.... See addition of real numbers in decimal as same Logic is applied in both cases digital computer by repeated.... 50 = 0.05 * 10 3 if you find anything incorrect, or you want to more... Applied in both cases University computer Systems Laboratory, CSL-TR-95-675 into 32-bit floating point number addition let take... Carry-Free addition and subtraction, multiplication algorithms, floating point number 9 mantissa division is done as in point. To fact that value of register a is restored after each iteration when we divide two number similar that! Improve article '' button below of algorithmic processes and computational machines recurrence with carry-free addition and subtraction, algorithms., mantissa of lesser number right side by 4 units representation, – 0.5625 = 1.00100000000000000000000, shifting right 4. Same Logic is applied in both cases propose in this paper we present a fast radix-4 division algorithm Goldschmidt. Subtracted and the radix-4 redundant number system Asked 4 years, 2 ago... The answer = x – y = 0 10000010 00100110000000000000000 anything incorrect by clicking on the main... ( CSE ) students and … restoring division algorithm and its Architecture is... And Logic Gates, number representation and computer Airthmetic parts.. 1 in digital computers that manipulate.! Will focus on division by repeated subtraction that of Figure 10.17 with diagram and examples introduction of Boolean and! A slightly more complex approach, known as nonrestoring, avoids the subtraction. Concepts of computer Architecture and Organization CE computer Architecture and technology that the dividend has a single floating. One digit of the operands = 0.05 * 10 3 Improve article '' button below exponent, we to... 4 units, 0.00010010000000000000000, mantissa of 9.75= 1, generate link and share the link here 01111110.. And made it 0.05 to add 1.1 * 103 and 50 report any issue with quotient! Number right side by 4 units, division algorithms are generally of two type algorithm! With a close … we can add significand approach, known as nonrestoring, avoids the unnecessary and! Slightly more complex approach, known as nonrestoring, division algorithm in computer architecture the unnecessary subtraction and addition CHAPTER 3 for. Circuit is very similar to Booth 's algorithm is shown in Figure 3.19b single precision mantissa that is similar the... Algorithms and Implementations by Stuart F. Oberman and Michael J. Flynn, Stanford University computer Systems,! By, and SRT division topic discussed above that manipulate data restoring, non-restoring algorithm! As same Logic is applied in both cases of algorithmic processes and computational machines and Logic Gates, number and. Use ide.geeksforgeeks.org, generate link and share the link here, 2 months ago Laboratory, CSL-TR-95-675, or want... Hosahalli R. au - Srinivas, Hosahalli R. au - division algorithm in computer architecture, Keshab K. PY - 1995/6 numbers. To the preceding discussion of Booth 's algorithm is shown in Figure 3.19b article, will performing... 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## "Large" diffeomorphisms in general relativity
I think 1) is irrelevant as we know that the embedding of T² in R³ does itself change the metric; T² admits a flat metric, but the embedding in R³ does not. So 1) is an artefact of the embedding.
Quote by tom.stoer Let's make an example. Assume R*T³ solves Einstein equations in vacuum (it does not, but that doesn't matter here). Assume we have a closed geodesic curve of a test object with winding numbers (1,0); this should be OK as T³ is flat and therefore a straight line with (0,1) should work. No let's do the cut-twist-glue procedure. What we get back is a different closed curve with winding number (1,1). Questions: a) does this generate a new, physically different spacetime? b) does this generate a different path of a test object on the same spacetime? c) did I miss something, e.g. did I miss to check whether this new curve can still be a geodesic?
Mainstream says none of the above, I'd say a)
Normally you would have b) if you have a)
Recognitions: Science Advisor I tend to agree with a) But that means that GR is NOT invariant w.r.t. all diffeomorphisms but only w.r.t. "restricted" diffeomorphisms (like small and large gauge transformations, where large gauge trf's DO generate physical effects)
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Quote by tom.stoer I tend to agree with a) But that means that GR is NOT invariant w.r.t. all diffeomorphisms but only w.r.t. "restricted" diffeomorphisms (like small and large gauge transformations, where large gauge trf's DO generate physical effects)
I think this would have to show up somehow in the process of transforming the metric, e.g. unavoidable singularities. Otherwise it is just arithmetic that two curves of some length, and orthogonal to each other, and with one intersection, preserve all those feature in new coordinates with properly transformed metric.
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Quote by PAllen I think this would have to show up somehow in the process of transforming the metric, e.g. unavoidable singularities.
Why? It is a diffeomorphism and does not create a singularity
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Quote by tom.stoer Why? It is a diffeomorphism and does not create a singularity
Well, then it can't change the geometry, at least as defined by anything you can compute using the metric. This really has nothing to do with GR, it is differential geometry.
My understanding is that topology of a differentiable manifold is encoded in how coordinate patches overlap. So, if we don't change this (and we don't need to for the Dehn twist), and we don't change anything computable from the metric, what can change?
In my (1) and (2) I was trying to get at the idea of making the operation 'real' so it does change geometry, versus treating as a pure coordinate transform, such that the corresponding metric transform preserves all geometric facts. I've heard the terms active versus passive difffeomorphism. I don't fully understand this, but I wonder if it is relevant to this distinction.
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Quote by PAllen Well, then it can't change the geometry, at least as defined by anything you can compute using the metric. This really has nothing to do with GR, it is differential geometry.
Du you agree that it changes the winding number of a closed curve?
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Quote by tom.stoer Du you agree that it changes the winding number of a closed curve?
Certainly, it is changed in my case (1) of my post #17. I'm not sure about as described in case(2) of that post. If you can compute winding number from the metric and topology as encoded in coordinate patch relationships, then it would seem mathematically impossible. If this is an example of geometrical fact independent of the metric and patch relationships, then we would need some definition how to compute it intrinsically, and it would seem to necessitate adding some additional structure to the manifold. In this case, it may well be possible, having specifically introduced non-metrical geometric properties not preserved by coordinate transforms.
Then, the physics question becomes that conventionally formulated GR would attach no meaning to this additional structure, it would become physically meaningful only in the context of an extension to GR that gave it meaning. This is what some of the classical unified field theory approaches did.
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Quote by PAllen Certainly, it is changed in my case (1) of my post #17. I'm not sure about as described in case(2) of that post. If you can compute winding number from the metric and topology as encoded in coordinate patch relationships ...
It does even in case (2)
I found an explanation on Baez "this week's finds", week 28:
http://math.ucr.edu/home/baez/week28.html
Quote by Baez,#28 Now, some diffeomorphisms are "connected to the identity" and some aren't. We say a diffeomorphism f is connected to the identity if there is a smooth 1-parameter family of diffeomorphisms starting at f and ending at the identity diffeomorphism. In other words, a diffeomorphism is connected to the identity if you can do it "gradually" without ever having to cut the surface. To really understand this you need to know some diffeomorphisms that aren't connected to the identity. Here's how to get one: start with your surface of genus g > 0, cut apart one of the handles along a circle, give one handle a 360-degree twist, and glue the handles back together! This is called a Dehn twist. ... In other words, given any diffeomorphism of a surface, you can get it by first doing a bunch of Dehn twists and then doing a diffeomorphism connected to the identity.
So we can now concentrate on the physical role of these "large" diffeomorphisms.
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Quote by tom.stoer It does even in case (2) I found an explanation on Baez "this week's finds", week 28: http://math.ucr.edu/home/baez/week28.html So we can now concentrate on the physical role of these "large" diffeomorphisms.
This was very interesting, but I didn't find any answer to my question my key question: how is it winding number of closed curve computed / defined against the definition of a differentiable manifold?
If it can change while the manifold is considered identical, then it must be computed in a way that is not invariant. Is it some form of coordinate dependent torsion?
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Quote by PAllen This was very interesting, but I didn't find any answer to my question my key question: how is it winding number of closed curve computed / defined against the definition of a differentiable manifold?
Yes, not a single word.
Quote by PAllen If it can change while the manifold is considered identical, then it must be computed in a way that is not invariant.
I don't agree. If the Dehn twist is a global diffeomorphism (and if we agree that in 2 dimensions homeomorphic manifolds are diffeomorphic and vice versa - which does not hold in higher dimensions) then the two manifolds before and after the twist are identical - there is no way to distinguish them. Now suppose we cannot compute the winding numbers (m,n) but only their change under twists. Then this change is not a property of the manifold but of the diffeomorphism. So we don't need a way to compute the winding numbers from the manifold but a way to compute their change from the diffeomorphism (this is similar to large gauge transformations where the structure is encoded in the gauge group, not in the base manifold). I think for large diffeomorphisms there is some similar concept.
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Quote by tom.stoer Yes, not a single word. I don't agree. If the Dehn twist is a global diffeomorphism (and if we agree that in 2 dimensions homeomorphic manifolds are diffeomorphic and vice versa - which does not hold in higher dimensions) then the two manifolds before and after the twist are identical - there is no way to distinguish them. Now suppose we cannot compute the winding numbers (m,n) but only their change under twists. Then this change is not a property of the manifold but of the diffeomorphism. So we don't need a way to compute the winding numbers from the manifold but a way to compute their change from the diffeomorphism (this is similar to large gauge transformations where the structure is encoded in the gauge group, not in the base manifold). I think for large diffeomorphisms there is some similar concept.
Thankyou! Very interesting. Then I spout my opinion of the physics issue (assuming something like this is what is going on). Conventional GR only gives meaning to metrical quantities, so this aspect of the diffeomorphism would have no physical significance, and anything metrically defined would be preserved. And I come back to the idea that this sort of thing provides an opportunity to extend conventional GR- without changing any of its predictions, you can add new content.
Blog Entries: 1 Recognitions: Gold Member Science Advisor I found two possibly relevant papers, both focusing on 2+1 dimensions: http://relativity.livingreviews.org/...es/lrr-2005-1/ http://matwbn.icm.edu.pl/ksiazki/bcp/bcp39/bcp3928.pdf If I am reading section 2.6 of the Carlip paper (first above) correctly, it suggests that GR is invariant under large diffeomorphisms, as I guessed above.
Recognitions: Science Advisor Thanks a lot for checking that and providing the two links. Looks very interesting.
Recognitions: Science Advisor This material has always confused me, and its hard to find good references. Over the years i've asked a few specialists but it hasn't helped me much. In 2+1 dimensions, the whole mapping class group sort of makes good intuitive sense, but then I rarely see it generalized in 4d. Further, the real subleties, at least to me, arise when the diffeomorphisms change the asymptotic structure of spacetime. Its not clear whether bonafide 'observables', are invariant under these 'gauge' transformations (incidentally, to avoid confusion, the notion of a large gauge transformation is afaik typically done where you fix a spacetime point, fix a vielbein and treat the diffeomorphism group acting on these elements in an analogous way to intuition from gauge theory)
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# Is there any significance to the logarithm of a sum?
Many years ago, while working as a computer programmer, I tracked down a subtle bug in the software that we were using. Management had dispaired of finding the bug, but I pursued it in odd moments over a period of a few days, and finally found that the problem was that, in computing the geometric mean, the program was taking the log of the sum instead of the sum of the logs. When thinking back on that, I always wonder whether there is any situation in which taking the log of a sum would be of interest. The only case I can think of is that it is often convenient to shift the logarithm to the left by 1 unit, which is done by adding 1 to the argument, that is, one often wishes to deal with log(1 + x) instead of log(x), so that one has the convenient situation of f(0) = 0. Let’s call this the trivial scenario.
So, can anyone think of any non-trivial scenario in which taking the log of a sum is the thing to do?
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The log of a sum has been a topic in the time of Gauss; I'll see whether I can find an article which I do recall in the german math newsgroup at about 2002. So far here is a link to tables of such logarithms of Tafeln der Additions- und Subtractions-Logarithmen für sieben Stellen. by Christoph Zech.
As far as I remember it was an approach to simplify computations in astronomy (where Gauss has been involved)
A longer article which explains the rationale and the use of the "Gauss'sche Additionslogarithmen", unfortunately only in german, is at this online archiv The article contains some references which might be helpful even if german is foreign language to you. Sorry I can't be of more help here...
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– Ilmari Karonen Aug 30 '11 at 18:30
"only in German"? Just one more hint that all mathematics materials ought to exist also in parallel Esperanto translation. – Mike Jones Sep 1 '11 at 3:04
This answer seems closest to what I was looking for, and so I have up-voted it and accepted it. – Mike Jones Sep 3 '11 at 8:08
The logarithm is a concave function. This means that Jensen's inequality can be applied to it, giving the Log sum inequality, an useful lemma in information theory.
Lemma (Log sum inequality) Let $a_i\dots a_n$ and $b_1\dots b_n$ be nonnegative reals. Then we have $$\sum_{i=1}^n a_i\log{\frac{a_i}{b_i}} \ge \left(\sum_{i=1}^n a_i\right)\log{\frac{\sum_{i=1}^n a_i}{\sum_{i=1}^n b_i}}$$
On the right hand side we recognize two logarithms of a sum.
Remark By convention, $0\log{0} = 0\log{\frac{0}{0}}=0$ and $a\log{\frac{a}{0}}=\infty$ for $a>0$. All these are justified by continuity.
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You mean, "we recognize two logarithms of a sum"? – Slaviks Aug 30 '11 at 17:50
Definitely! Corrected my mistake. – Jacopo Notarstefano Aug 30 '11 at 22:47
It seems like "nonnegative reals" should be "positive reals"; otherwise, you're dividing by 0 or taking the log of 0. – Mike Jones Sep 1 '11 at 3:02
Ah, yes, there are some conventions about the behavior at zero. I'll add them to the answer. – Jacopo Notarstefano Sep 1 '11 at 22:48
@Jacopo Notarstefano: Thanks for the clarification about the behavior at zero. I'm up-voting your answer now. – Mike Jones Sep 3 '11 at 8:13
One plausible scenario is this:
Say you want a soft maximum function. The usual function $\max(x,y)$ is not "soft" (try visualizing its graph).
So we define the "soft maximum function" $\mathrm{smax}(x,y)=\log(e^x+e^y)$. The idea is that if $x \gg y$, then $\log(e^x+e^y) \approx x$.
I don't know if this is anything like what you had in mind, but when you integrate rational functions (by partial fractions) you may get terms like $\ln|x+a|$ and $\ln((x+a)^2+b^2)$ in the result.
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# Notes on "Time, Clocks, and the Ordering of Events in a Distributed System"
Develops a partial ordering for events occurring in a distributed system based on logical time. Demonstrates how to strengthen the partial ordering into one of many possible total orderings. Finally, presents an algorithm to order events based on physical time by synchronizing clocks with a bounded error.
## Partial Ordering
Let $$\rightarrow$$ denote an ordering of events. Intuitively, the ordering $$a \rightarrow b$$ means that it is possible for event $$a$$ to causally affect event $$b$$. This relation $$a \rightarrow b$$ must satisfy the following conditions:
1. If $$a$$ and $$b$$ are events in the same process, and $$a$$ comes before $$b$$, then $$a \rightarrow b$$.
2. If $$a$$ is the sending of a message by one process and $$b$$ is the receipt of the same message by another process, then $$a \rightarrow b$$.
3. If $$a \rightarrow b$$ and $$b \rightarrow c$$ then $$a \rightarrow c$$.
Two distinct events are concurrent if $$a \nrightarrow b$$ and $$b \nrightarrow a$$.
## Logical Clocks
Abstractly, clocks assign a number to an event where the number is considered the time of the event. More precisely, a clock is defined for each process where clock $$C_i \langle a \rangle$$ assigns a time to event $$a$$ in process $$P_i$$. Logical clocks may not have a relation to physical time and can be implemented with counters.
Clock Condition: For any events $$a, b$$ if $$a \rightarrow b$$ then $$C \langle a \rangle < C \langle b \rangle$$.
The clock condition is satisfied if the following conditions hold:
1. If $$a$$ and $$b$$ are events in process $$P_i$$, and $$a$$ comes before $$b$$, then $$C_i \langle a \rangle < C_i \langle b \rangle$$.
• Implies that there must be a tick line between any two events on a process line.
• IR1: Each process $$P_i$$ increments $$C_i$$ between any two successive events.
2. If $$a$$ is the sending of a message by process $$P_i$$ and $$b$$ is the receipt of that message by process $$P_j$$, then $$C_i \langle a \rangle < C_j \langle b \rangle$$.
• Every message line must cross a tick line.
• IR2:
1. If event $$a$$ is the sending of a message $$m$$ by process $$P_i$$, then $$m$$ contains a timestamp $$T_m = C_i \langle a \rangle$$.
2. Upon receiving $$m$$, process $$P_j$$ sets $$C_j$$ greater than or equal to its present value and greater than $$T_m$$.
## Total Ordering
Use the above system and break ties using an arbitrary total ordering of the processes. Mathematically, $$a \Rightarrow b$$ if and only if either:
1. $$C_i \langle a \rangle < C_j \langle b \rangle$$, or
2. $$C_i \langle a \rangle = C_j \langle b \rangle$$ and $$P_i \prec P_j$$.
Different choices of clocks and different orderings over the processes may yield different total orderings $$\Rightarrow$$; however, the partial ordering $$\rightarrow$$ is uniquely determined by the system of events.
Assume several processes communicate with each other and may request a resource. Using total orderings, the paper presents an algorithm for allocating resources based on the total ordering of events. Assume that messages sent from $$P_i$$ to $$P_j$$ are received in the same order as they are sent, and that every message is eventually received. Each process maintains its own request queue and initially contains the message $$T_0:P_0$$ where $$P_0$$ is the process initially granted the resources and $$T_0$$ is less than the initial value of any clock. The following rules govern the algorithm and form a single event each:
1. To request the resource, process $$P_i$$ sends the message $$T_m:P_i$$ requests resource to every other process, and puts that message on its request queue, where $$T_m$$ is the timestamp of the message.
2. When $$P_j$$ receives the message $$T_m:P_i$$ requests resource, it places the message on its request queue and sends a timestamped acknowledgment message to $$P_i$$.
3. To release the resource, process $$P_i$$ removes any $$T_m:P_i$$ requests resource from its request queue and sends a timestamped $$P_i$$ releases resource message to every other process.
4. When $$P_j$$ receives a $$P_i$$ releases resource message, it removes any $$T_m:P_i$$ requests resource message from its request queue.
5. $$P_i$$ is granted the resources when:
1. There is a $$T_m:P_i$$ requests resource message in its request queue which is ordered before any other request in its queue by the relation $$\Rightarrow$$.
2. $$P_i$$ has received a message from every other process timestamped later than $$T_m$$.
While the algorithm is correct, it scales poorly with the number of resources and the number of processes. The paper also admits that there is no concept of fault tolerance since “failure is only meaningful in the context of physical time”. Similar problems are further examined in work on consistency in distributed databases.
## Physical Clocks
Strong Clock Condition: For any events $$a, b$$ in $$\varphi$$, if $$a \pmb{\rightarrow} b$$ then $$C \langle a \rangle < C \langle b \rangle$$. $$\pmb{\rightarrow}$$ denotes “happened before” relation for $$\varphi$$. For example, $$\varphi$$ may be a set of “real” events in physical space-time and $$\pmb{\rightarrow}$$ may be the partial ordering of events defined by special relativity.
Using the strong clock condition and physical clocks, we can eliminate anomalous behavior which may occur when precedence information is based on messages external to the system (e.g. humans communicating).
To synchronize physical clocks, we make the following assumptions:
1. PC1: There exists a constant $$\kappa \ll 1$$ s.t. $$\forall i$$: $$\lvert dC_i(t) / dt -1 \rvert < \kappa$$.
2. PC2: $$\forall i$$: $$\lvert C_i(t) - C_j(t) \rvert < \epsilon$$.
Let $$\mu$$ be the shortest transmission time for interprocess messages (e.g. the shortest distance between processes divided by the speed of light). To avoid anomalous behavior, $$\forall i, j, t$$: $$C_i(t + \mu) - C_j(t) > 0$$. Applying the above assumptions, we find that the above inequality is true if the inequality $$\epsilon / (1 - \kappa) \leq \mu$$ holds, indicating that anomalous behavior is impossible.
Let $$\nu_m = t' - t$$ be the total delay for a message $$m$$ sent at time $$t$$ and received at time $$t'$$. This delay is unknown to the receiving process, however some minimum delay $$0 \leq \mu_m \leq v_m$$ is known. Then, $$\xi_m = \nu_m - \mu_m$$ is the unpredictable delay of the message.
Physical clock implementation works as follows:
1. IR1’: For each $$i$$, if $$P_i$$ does not receive a message at physical time $$t$$, then $$C_i$$ is differentiable at $$t$$ and $$dC_i(t)/dt > 0$$.
2. IR2’:
1. If $$P_i$$ sends a message $$m$$ at physical time $$t$$, then $$m$$ contains a timestamp $$T_m = C_i(t)$$.
2. Upon receiving $$m$$ at time $$t'$$, $$P_j$$ sets $$C_j(t') = \max \big( C_j(t'), T_m + \mu_m\big)$$.
This results in a theorem that bounds the length of time taken to synchronize clocks when the system first starts:
Theorem. Assume a strongly connected graph of processes with diameter $$d$$ which always obeys rules IR1’ and IR2’. Assume that for any message $$m$$, $$\mu_m \leq \mu$$ for some constant $$\mu$$, and that for all $$t \geq t_0$$:
1. PC1 holds.
2. There are constants $$\tau$$ and $$\xi$$ s.t. every $$\tau$$ seconds a message with an unpredictable delay less than $$\xi$$ is sent over every arc. Then PC2 is satisfied with $$\epsilon \approx d(2\kappa\tau + \xi)$$ for all $$t \geq t_0 + \tau d$$, where the approximations assume $$\mu + \xi \ll \tau$$.
## Takeaways
The paper presents a way to order events in a distributed system using logical time. This method is then extended to work with physical time. The paper proposes implementations, and the work is relevant to databases as well. Bounding physical time remains an area of distributed systems research, as seen in Google’s paper on Spanner.
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