text:
# Variational GPs w/ Multiple Outputs¶

## Introduction¶

In this example, we will demonstrate how to construct approximate/variational GPs that can model vector-valued functions (e.g. multitask/multi-output GPs).

[1]:

import math
import torch
import gpytorch
import tqdm
from matplotlib import pyplot as plt

%matplotlib inline

### Set up training data¶

In the next cell, we set up the training data for this example. We’ll be using 100 regularly spaced points on [0,1] which we evaluate the function on and add Gaussian noise to get the training labels.

We’ll have four functions - all of which are some sort of sinusoid. Our train_targets will actually have two dimensions: with the second dimension corresponding to the different tasks.

[2]:

train_x = torch.linspace(0, 1, 100)

train_y = torch.stack([
torch.sin(train_x * (2 * math.pi)) + torch.randn(train_x.size()) * 0.2,
torch.cos(train_x * (2 * math.pi)) + torch.randn(train_x.size()) * 0.2,
torch.sin(train_x * (2 * math.pi)) + 2 * torch.cos(train_x * (2 * math.pi)) + torch.randn(train_x.size()) * 0.2,
-torch.cos(train_x * (2 * math.pi)) + torch.randn(train_x.size()) * 0.2,
], -1)

print(train_x.shape, train_y.shape)

torch.Size([100]) torch.Size([100, 4])

We are going to construct a batch variational GP - using a CholeskyVariationalDistribution and a VariationalStrategy. Each of the batch dimensions is going to correspond to one of the outputs. In addition, we will wrap the variational strategy to make the output appear as a MultitaskMultivariateNormal distribution. Here are the changes that we’ll need to make:

1. Our inducing points will need to have shape 2 x m x 1 (where m is the number of inducing points). This ensures that we learn a different set of inducing points for each output dimension.
2. The CholeskyVariationalDistribution, mean module, and covariance modules will all need to include a batch_shape=torch.Size([2]) argument. This ensures that we learn a different set of variational parameters and hyperparameters for each output dimension.
3. The VariationalStrategy object should be wrapped by a variational strategy that handles multitask models. We describe them below:

### Types of Variational Multitask Models¶

The most general purpose multitask model is the Linear Model of Coregionalization (LMC), which assumes that each output dimension (task) is the linear combination of some latent functions $$\mathbf g(\cdot) = [g^{(1)}(\cdot), \ldots, g^{(Q)}(\cdot)]$$:

$f_\text{task}(\mathbf x) = \sum_{i=1}^Q a^{(i)} g^{(i)}(\mathbf x),$

where $$a^{(i)}$$ are learnable parameters.

[3]:

num_latents = 3

def __init__(self):
# Let's use a different set of inducing points for each latent function
inducing_points = torch.rand(num_latents, 16, 1)

# We have to mark the CholeskyVariationalDistribution as batch
# so that we learn a variational distribution for each task
variational_distribution = gpytorch.variational.CholeskyVariationalDistribution(
inducing_points.size(-2), batch_shape=torch.Size([num_latents])
)

# We have to wrap the VariationalStrategy in a LMCVariationalStrategy
# so that the output will be a MultitaskMultivariateNormal rather than a batch output
variational_strategy = gpytorch.variational.LMCVariationalStrategy(
gpytorch.variational.VariationalStrategy(
self, inducing_points, variational_distribution, learn_inducing_locations=True
),
num_latents=3,
latent_dim=-1
)

super().__init__(variational_strategy)

# The mean and covariance modules should be marked as batch
# so we learn a different set of hyperparameters
self.mean_module = gpytorch.means.ConstantMean(batch_shape=torch.Size([num_latents]))
self.covar_module = gpytorch.kernels.ScaleKernel(
gpytorch.kernels.RBFKernel(batch_shape=torch.Size([num_latents])),
batch_shape=torch.Size([num_latents])
)

def forward(self, x):
# The forward function should be written as if we were dealing with each output
# dimension in batch
mean_x = self.mean_module(x)
covar_x = self.covar_module(x)
return gpytorch.distributions.MultivariateNormal(mean_x, covar_x)

With all of the batch_shape arguments - it may look like we’re learning a batch of GPs. However, LMCVariationalStrategy objects convert this batch_dimension into a (non-batch) MultitaskMultivariateNormal.

[4]:

likelihood(model(train_x)).rsample().shape

[4]:

torch.Size([100, 4])

The LMC model allows there to be linear dependencies between outputs/tasks. Alternatively, if we want independent output dimensions, we can replace LMCVariationalStrategy with IndependentMultitaskVariationalStrategy:

[7]:

class IndependentMultitaskGPModel(gpytorch.models.ApproximateGP):
def __init__(self):
# Let's use a different set of inducing points for each task

# We have to mark the CholeskyVariationalDistribution as batch
# so that we learn a variational distribution for each task
variational_distribution = gpytorch.variational.CholeskyVariationalDistribution(
)

gpytorch.variational.VariationalStrategy(
self, inducing_points, variational_distribution, learn_inducing_locations=True
),
)

super().__init__(variational_strategy)

# The mean and covariance modules should be marked as batch
# so we learn a different set of hyperparameters
self.covar_module = gpytorch.kernels.ScaleKernel(
)

Note that all the batch sizes for IndependentMultitaskVariationalStrategy are now num_tasks rather than num_latents.

### Train the model¶

This code should look similar to the SVGP training code

[5]:

# this is for running the notebook in our testing framework
import os
smoke_test = ('CI' in os.environ)
num_epochs = 1 if smoke_test else 500

model.train()
likelihood.train()

# We use SGD here, rather than Adam. Emperically, we find that SGD is better for variational regression
{'params': model.parameters()},
{'params': likelihood.parameters()},
], lr=0.1)

# Our loss object. We're using the VariationalELBO, which essentially just computes the ELBO
mll = gpytorch.mlls.VariationalELBO(likelihood, model, num_data=train_y.size(0))

# We use more CG iterations here because the preconditioner introduced in the NeurIPS paper seems to be less
# effective for VI.
epochs_iter = tqdm.tqdm_notebook(range(num_epochs), desc="Epoch")
for i in epochs_iter:
# Within each iteration, we will go over each minibatch of data
output = model(train_x)
loss = -mll(output, train_y)
epochs_iter.set_postfix(loss=loss.item())
loss.backward()
optimizer.step()

### Make predictions with the model¶

[6]:

# Set into eval mode
model.eval()
likelihood.eval()

# Initialize plots

# Make predictions
test_x = torch.linspace(0, 1, 51)
predictions = likelihood(model(test_x))
mean = predictions.mean
lower, upper = predictions.confidence_region()


text:
Î

Jump to: navigation, search

Î, î (i-circumflex) is a letter in the Friulian, Kurdish, and Romanian alphabets. This letter also appears in French, Turkish, Welsh and Walloon languages as a variant of letter “i”.

Usage in various languages

Afrikaans

In Afrikaans, î is a punctuated form of i and a usage example includes "wîe", plural of "wig" (= wedge).

Friulian

Î is used to represent the /iː/ sound.

Kurdish

Î is the 12th letter of the Kurdish Kurmanji alphabet and represents /iː/.

Romanian

Î is the 12th letter of the Romanian alphabet and represents /ɨ/. This sound is also represented in Romanian as letter â.

Italian

Î is a variant of I in Italian. It should be used when forming the plural of male nouns ending in "-io", to avoid placing the stress on the wrong syllable. Example: "principio" /prinˈtʃipjo/ (principle) has the plural "principî" /prinˈtʃipi/, while "principe" /ˈprintʃipe/ (prince) has "principi" /ˈprintʃipi/ as plural. However, the usage of Î in Italian is rapidly decreasing; most Italians write "principi" for the plurals of both "principio" and "principe".

French

Î is a letter which appears in several French words, like naître(to be born), abîme(abyss), maître(master), (crème) fraîche and more. Unlike Â, Ê, and Ô, the circumflex does not alter the pronunciation of î, nor does it alter û.

Turkish

In Turkish, î is used to indicate the /iː/ sound in Arabic loanwords where it is used as an adjectival suffix that makes an adjective from a noun.

Welsh

In Welsh, î is used to represent long stressed i [iː] when, without the circumflex, the vowel would be pronounced as short [ɪ], e.g.. tîm [tiːm] "team".

Other usage

In mathematics

• The letter $\mathbf{\hat{\boldsymbol{\imath}}}$ is sometimes used to denote a unit vector in mathematics.

Character mappings

Charset Unicode ISO 8859-1, 2, 3, 4, 9, 10, 14, 15, 16
Majuscule Î U+00CE CE
Minuscule î U+00EE EE

format: plain


text:
## Refining Deep Generative Models via Discriminator Gradient Flow

### Abdul Fatir Ansari · Ming Liang Ang · Harold Soh

Keywords: [ generative models ] [ gan ] [ vae ] [ gradient flows ] [ Normalizing Flow ]

[ Abstract ]
[ Paper ]
Tue 4 May 1 a.m. PDT — 3 a.m. PDT

Abstract: Deep generative modeling has seen impressive advances in recent years, to the point where it is now commonplace to see simulated samples (e.g., images) that closely resemble real-world data. However, generation quality is generally inconsistent for any given model and can vary dramatically between samples. We introduce Discriminator Gradient $f$low (DG$f$low), a new technique that improves generated samples via the gradient flow of entropy-regularized $f$-divergences between the real and the generated data distributions. The gradient flow takes the form of a non-linear Fokker-Plank equation, which can be easily simulated by sampling from the equivalent McKean-Vlasov process. By refining inferior samples, our technique avoids wasteful sample rejection used by previous methods (DRS & MH-GAN). Compared to existing works that focus on specific GAN variants, we show our refinement approach can be applied to GANs with vector-valued critics and even other deep generative models such as VAEs and Normalizing Flows. Empirical results on multiple synthetic, image, and text datasets demonstrate that DG$f$low leads to significant improvement in the quality of generated samples for a variety of generative models, outperforming the state-of-the-art Discriminator Optimal Transport (DOT) and Discriminator Driven Latent Sampling (DDLS) methods.


text:
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# The entropy power inequality for quantum systems

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When two independent analog signals, $X$ and $Y$ are added together giving $Z=X+Y$, the entropy of $Z$, $H(Z)$, is not a simple function of the entropies $H(X)$ and $H(Y)$, but rather depends on the details of $X$ and $Y$'s distributions.  Nevertheless, the entropy power inequality (EPI), which states that $e^{2H(Z)} \geq e^{2H(X)} + e^{2H(Y)}$, gives a very tight restriction on the entropy of $Z$. This inequality has found many applications in information theory and statistics.  The quantum analogue of adding two random variables is the combination of two independent bosonic modes at a beam splitter. The purpose of this talk is to give an outline of the proof of two separate generalizations of the entropy power inequality to the quantum regime.  These inequalities provide strong new upper bounds for the classical capacity of quantum additive noise channels, including quantum analogues of the additive white Gaussian noise channels.   Our proofs are similar in spirit to standard classical proofs of the EPI, but some new quantities and ideas are needed in the quantum setting. Specifically, we find a new quantum de Bruijin identity relating entropy production under diffusion to a divergence-based quantum Fisher information. Furthermore, this Fisher information exhibits certain convexity properties in the context of beam splitters.   This is joint work with Graeme Smith.


text:
# Differential equation involving inverse function with no simple numerical solution

I am wondering if there is a solution for (and a method to solve) the following functional equation

$$\frac{d}{dx}f(x) = 1 - \beta f^{-1}(x)$$

in $[0,1]$, with $0<\beta<1$, and initial condition $f(0)=0$. Here $f^{-1}(x)$ refers to the inverse function of $f(x)$. To avoid problems with the definition of $f^{-1}$, we can think of the problem as $f'(x) = 1 - \beta H\left(f^{-1}(x)\right)$ with $H(\cdot)$ being equal to one if $f^{-1}(x)$ is not defined in $[0,1]$.

I haven't been able to find an analytical or numerical solution. The problem with a numerical solution is that if you start from $x=0$ and build to the right, after two steps you can't compute the derivative.

• How is this a differential equation? – MathematicsStudent1122 Sep 28 '16 at 0:16
• $$f(0)=0\implies f^{-1}(0)=0\\\implies f(0)\ne1-\beta f^{-1}(0)$$ – Simply Beautiful Art Sep 28 '16 at 0:24
• I solved it! $\color{white}{;)}$ – Simply Beautiful Art Sep 28 '16 at 1:13

We have the following known:

$$\require{cancel}f(0)=0$$

Take the $f^{-1}$ of both sides:

$$\xcancel{f^{-1}(f}(0))=0=f^{-1}(0)$$

$$f^{-1}(0)=0$$

Thus, look at the original functional equation and consider $x=0$.

$$f(0)=1-\beta f^{-1}(0)$$

$$0=1$$

Contradiction. Therefore, there cannot exist an analytic solution to this problem, and the solution cannot exist at $x=0$ with the given values.

If you neglect your initial condition, though, see that we have:

$$f(x)=mx+b$$

Putting in the values, we get

$$f(x)=i\sqrt\beta x+\frac{\beta-i\beta\sqrt\beta}{\beta^2+\beta}$$

where $i=\sqrt{-1}$

• If you want to see what happens when you assume the solution is real valued, check the revisions. – Simply Beautiful Art Sep 28 '16 at 1:12


text:
3 Tutor System
Starting just at 265/hour

# Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

x1 = $$\frac{(1×(-2)+2×4)}{1+2} = \frac{(-2+8)}{3} =\frac{6}{3} = 2$$

y1 = $$\frac{(1×(-3)+2×-1)}{1+2} = \frac{(-3-2)}{3} =\frac{-5}{3}$$

$$\therefore$$P = (x1, y1) = P(2, $$\frac{-5}{3}$$)

Point Q divides AB internally in the ratio 2:1.

x2 = $$\frac{(2×(-2)+1×4)}{2+1} = \frac{(-4+4)}{3} = 0$$

y2 = $$\frac{(2×(-3)+1×-1)}{2+1} = \frac{(-6-1)}{3} =\frac{-7}{3}$$

The coordinates of the point Q(x2 ,y2 ) = (0, $$\frac{-7}{3}$$ )


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History/origin: The origin of the yard as a unit is unclear. Each side of this square is equivalent to 0.91 meters. To convert yards and feet to meters, convert yard and feet values into meters separately and add them. This is a very easy to use meter to yard converter.First of all just type the meter (m) value in the text field of the conversion form to start converting m to yd, then select the decimals value and finally hit convert button if auto calculation didn't work.Yard value will be converted automatically as you type.. Some abbreviations include sq. Ounce(Avoirdupois), Long Fabric price per meter is Rs. to square meters: 1.550 (price/MSI) x 1.550 = price/m2. Price / m 3 Regular Hours - Monday to Friday 7:00 am to 5:00 pm Ottawa Area: Monday to Friday 6:00 am to 6:00 pm After regular hours and before 9:00 pm $30.00. From this post we will learn how to calculate Kgs from Yards. 0.111 (price/yd 2) x 0.111 = price/ft 2. to MSI. How do you fined Lineal Foot? The International spelling for this unit is metre. Order) or Troy), Bali(Burma), Tickal(Burma), Viss(Burma), Chevvu(Sri It is also equal to 3 feet, or 36 inches. Fabric price for 1 yard (i.e. Note that rounding errors may occur, so always check the results. (3 ft / linear yard - 9 Square Ft in a Sqr Yard)13 feet = 4.25 yards22 feet = 7.33 yards4.25 yards x 7.33 yards = 31.2 square yards.The price is (31.2 square yards) x ($19.50 per square yard) = $607.50NOTE THATIf you don't convert the feet to yards first, then divide the 'cost per yard' by 9 to get the 'cost per square foot' to avoid errors. History/origin: The origin of the yard as a unit is unclear. 07 per metre and 0.99 per yard) then it is necessary to know just what units are being used for the purchase. The converter allows the user to convert between different currencies in easy price converter, and in complex price calculator, it can also convert price between different currencies in various measurements. 1 linear meter [=meter] = 1.09361329833771 linear yard [=yard] Reverse Conversion: From: To: 1 linear yard [=yard] = 0.9144 linear meter [=meter] Quantity & Converter(s) Unit Name: Quantity: linear meter [=meter] Length/Distance: linear yard [=yard] Length/Distance to MSI 0.77 1.296 (price/yd2) x 0.77 = price/MSI or (price/yd2) / 1.296 = price/MSI. Fabric length= (1 X 1000) /(240 *1) in meter = 4.17 meter Therefore fabric price Rs. 1 Yard is equal to 0.9144 meter. or Troy), Scruple(Apothecaries or Troy), Grain(Apothecaries), Momme(Japan Tael(Chinese Medicine), Tsin(Chinese Medicine), Centi(Chinese Medicine), The symbol for meter is m. There are 0.9144 meters in a yard. Pearl), Ratti(India,Burma), Mangelin(India), baht(raw gold or bullion,Thailand), Chi(Vietnam), mesghal(Iran), Don(South Korea), Tola(India), Lathi(Burma), In order to convert the price of a square meter to a foot, then, you need to divide the cost per square meter by 10.76. It is an English unit (predecessor of imperial units) and the term was derived from "gerd" in Old English, the earliest historical form of … Let's take a closer look at the conversion formula so that you can do these conversions yourself with a calculator or with an old-fashioned pencil and paper. Use this page to learn how to convert between yards and square meters. Just tell us your specs and we’ll respond promptly. Inch, Are, Square Feet, Square Kilometre, Acre, Square A meter , or metre, is the fundamental unit of length in the metric system, from which all other length units are based. Mile, Hectare, Square Rod, Section(U.S. Government Land), Township(U.S. How many cubic feet are there in a yard? Since 1959, a yard has been defined as exactly 0.9144 meters. There are 27 cubic feet in a cubic yard. » Square meter Conversions: m2↔km2 1 km2 = 1000000 m2 m2↔cm2 1 m2 = 10000 cm2 m2↔mm2 1 m2 = 1000000 mm2 m2↔um2 1 m2 = 1000000000000 um2 m2↔nm2 1 m2 = 1.0E+18 nm2 m2↔ha 1 ha = 10000 m2 m2↔Plaza 1 Plaza = 6400 m2 m2↔in2 1 m2 = 1550.0031 in2 m2↔ft2 1 m2 = 10.76391 ft2 m2↔yd2 1 m2 = 1.19599 yd2 m2↔mi2 1 mi2 = 2589988.1103464 m2 Determine the number of meters in a yard. (adsbygoogle = window.adsbygoogle || []).push({}). In order to calculate how many cubic yards of materials you require for your project, try the following: Measure the width, length and depth of the space that you wish to calculate materials for. Meters. Mid 20th century answer - Find a simplified conversion factor and do it in your head. Dry), Quarts(US Dry), Peck(US Dry), Bushel(US Dry Level), Bushel (U.S. Dry Meters. Lambda, Breakfast Cup, Beer Gallon, Bucket (Imperial), Since 1959, a yard has been defined as exactly 0.9144 meters. Puerto Rico: Square metre prices, premier city centre, US$. Manage tenants utility usage and eliminate disputesNB: Meter operates as a sub-meter and does not replace the municipal meter. The following is a list of definitions relating to conversions between meters and yards. Fluid Dram(Imperial), Fluid Scruple(Imperial), Minim(Imperial), Pint(US miles or meters) and the currency units and enter the following: (L) The length or distance to be bought, sold or otherwise produced or consumed. I need help please a refresher on my algebra. Gram, Milligram, Microgramme, Nanogram, Metric Ton, Quintal, Pound(Avoirdupois), As the base unit of length in the SI and other m.k.s. Milli(Chinese Medicine), Picul(HONGKONG), Catty(HONGKONG), Tael(HONGKONG), Type in your own numbers in the form to convert the units! Gallon(US Chains Square m.) prices in US$/€ of 120-sq. In Ohio we charge 5.50 a yard, so technically that costs$6.58US a square meter. Fluid Dram(US However, it depends on the area; skip's keeping period and skip license. Or leave the math to us and contact one of our expert sales representatives. The metre is a unit of length in the metric system, and is the base unit of length in the International System of Units (SI). So, if an apartment for rent in Europe was $50 per square meter, the total cost per square foot would be about$4.65 ($50 divided by 10.76). Gemlite Single Phase 80AMP 220VAC-240VAC 50 Hz / 60Hz 5-80A 20-25m/A Brand: Recharger The average cost of concrete is$119 to $147 per cubic yard, which includes delivery up to 20 miles.Pouring plain concrete costs$5 to $10 per square foot depending on the quality, while colored, stamped, or stained concrete costs$8 to $18 per square foot to install. This post may contain affiliate links. (US), Pinch (Imperial), Pony ,Shot, Jigger, Fifth, 143.88 Note: In this example, Indian currency (Rs.) Drop (Imperial), Drop (US), Drop (Medical)(alt), Drop INSTRUCTIONS: Choose length or distance units (e.g. is used. 1 m is equivalent to 1.0936 yards, or 39.370 inches. Pearl Grain, ounce(troy or apothecaries), Pennyweight(Troy), Drachm(Apothecaries The formula to convert yards to meters is: m=yd X 0.9144. Kilometre, Centimetre, Millimetre, Decimetre, Micron, Nanometer, Angstrom, systems (based around metres, kilograms and seconds) the metre is used to help derive other units of measurement such as the newton, for force. In 1958 the United States and countries of the Commonwealth (Canada, New Zealand, Australia) of Nations defined the length of the international yard (yd) is to be exactly 0.9144 meters (914.4 mm ). Note that rounding errors may occur, so always check the results. This Online Calculator converts meter to yard (m to yd) and yd to m (yard to meters). It is equal to 100 centimeters, 1/1000th of a kilometer, or about 39.37 inches. Price converter and calculator. PRICE PER POUND TO PRICE PER LINEAR YARD: x 36 ÷ x =${{f09pply | number:4}} per Linear Yard: Width (in) Yield Price / lb. to square feet 0.111 (price/yd2) x 0.111 = price/ft2. Customary System and the British Imperial System. Materials measured in cubic yards. A square yard is 0.84 square meters, so 0.84 of $17.93 is So it would be$15.06 per square yard. Yard, Foot, Inch, Mile, Furlong, Fathom, Gunter's Chain, Rod(Pole; Perch), 143.88 1.20 (price/yd 2) x 1.20 = price/m 2. to square feet. Heaped), Trike (US), Sack (US), Barrel (US Dry), Seam, Wey (US), Pint(US Liquid), Ask us for a custom solution, film recommendations and more. Find local scrap metal pricing at your closest scrap yard. Government Land), Quarter Section(U.S. Government Land), Eighth Section(U.S. 1 metre is equal to 1000 mm, or 1.0936132983377 yard. This information is given under 'Comparative Costs per Unit'. 0.9144 meter) length and width 60 inch is $1.28 We can say at$ 1.28 you will get 0.195 kgs fabric. Liquid;Wine), Quarts(US Liquid), Barrel (Petroleum), Lanka), Manchadi(Sri Lanka), Catty(Thailand), Checky(Turkey), Oitava, Octavo(Brazil), Find here Plain Cotton Fabric, Plain Cotton cloth manufacturers, suppliers & exporters in India. The Price per Length or Distance calculator computes the total price for something based on the length or distance (L) at a unit price (UP) per unit length or distance.. This plugin allows you to add quantity and user-defined calculators to enable the listing of products that are sold per unit with a known measurement (for instance, tiles with an advertised square footage per box), or products which are priced by the foot, square meter, cubic yard, kilogram, etc and are purchased using a custom measurement (such as fabric). For example, to convert 2 yards and 5 feet into meters, multiply 2 by 0.9144 and 5 by 0.3048 and add the results, that makes 2 yards and 5 feet equal to 3.352 meters. Unit Definition (yard) The yard is a fundamental unit of length in both the U.S. Meters. The converter allows the user to convert between different currencies in easy price converter, and in complex price calculator, it can also convert price between different currencies in various measurements. Water ton, Last, Quarter(Imperial), Fluid Ounce(Imperial), Kenning(Imperial), Use our calculator for conversion formulas, metric equivalents and more. Conversion Formula. Let us help you explore the possibilities. /Yards Thanks for any help for example, Indian currency ( Rs. and gravel are... Room size 15 yard price to meter price 4 x 12 ' material across India corresponding.. Compared to other skips in the lighter fields on the area ; skip 's keeping period and license... 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# Suggestion for a better way to organize data to generate frequent item-sets?

I have a data of a bag of words in a document. The data has 3 columns: {document number, word number, count of the word in the number}. I am supposed to generate frequent item-sets of a particular size.

I thought that I would make list of all words that appear in a document, create a table of this list, and then generate frequent item-sets using Mlxtend or Orange . However, this approach does not seem to be efficient.

If the size is reasonable (i.e. not too many documents and not too many words in a document), you could try to build a map for each possible itemset, for instance like this:

// Assuming data is an array of size N containing all the documents
// clusters is a map associating each itemset with a set of documents
for i=0 to N-1
for j=i+1 to N-1
group = overlap(data[i], data[j])
add data[i] to the set clusters[group]
add data[j] to the set clusters[group]

An alternative version if the number of different values and size of the sets allow it and/or if it's possible to precompute the itemsets of interest:

for i=0 to N-1
for every subset S of data[i]
add data[i] to the set clusters[S]


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# Algorithm Help

## Homework Statement

In the given picture below

## The Attempt at a Solution

Im stuck after the begin
How can I carry out the printSpaces(j+1) if the value of j wasn't given???

Is there an error in the question or is there something that im not seeing :/

#### Attachments

• trace.png
157.9 KB · Views: 426

gneill
Mentor
Looks like an error in the given algorithm unless the language automatically initializes all new variables to zero (or some other specified value).

If not defining j was an oversight on the question composer's part then you might state some assumed value for it and proceed to solve the problem using that value.

Thanks man

So if I assume that j starts at 0 the output would be:

-*--------*
***********

gneill
Mentor
Thanks man

So if I assume that j starts at 0 the output would be:

-*--------*
***********

Looks right to me!

Thanks alot man


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###### Activity3.2.3

Let's begin with the basis $$\bcal = \{\vvec_1,\vvec_2\}$$ of $$\real^2$$ where

\begin{equation*} \vvec_1 = \twovec{3}{-2}, \vvec_2 = \twovec{2}{1}\text{.} \end{equation*}
1. If the coordinates of $$\xvec$$ in the basis $$\bcal$$ are $$\coords{\xvec}{\bcal} = \twovec{-2}{4}\text{,}$$ what is the vector $$\xvec\text{?}$$

2. If $$\xvec = \twovec{3}{5}\text{,}$$ find the coordinates of $$\xvec$$ in the basis $$\bcal\text{;}$$ that is, find $$\coords{\xvec}{\bcal}\text{.}$$

3. Find a matrix $$A$$ such that, for any vector $$\xvec\text{,}$$ we have $$\xvec = A\coords{\xvec}{\bcal}\text{.}$$ Explain why this matrix is invertible.

4. Using what you found in the previous part, find a matrix $$B$$ such that, for any vector $$\xvec\text{,}$$ we have $$\coords{\xvec}{\bcal} = B\xvec\text{.}$$ What is the relationship between the two matrices you have found in this and the previous part? Explain why this relationship holds.

5. Suppose we also consider the basis

\begin{equation*} \ccal = \left\{\twovec{1}{2}, \twovec{-2}{1}\right\}\text{.} \end{equation*}

Find a matrix $$C$$ that converts coordinates in the basis $$\ccal$$ into coordinates in the basis $$\bcal\text{;}$$ that is,

\begin{equation*} \coords{\xvec}{\bcal} = C \coords{\xvec}{\ccal}\text{.} \end{equation*}

You may wish to think about converting coordinates from the basis $$\ccal$$ into the standard coordinate system and then into the basis $$\bcal\text{.}$$

6. Suppose we consider the standard basis

\begin{equation*} \ecal = \{\evec_1,\evec_2\}\text{.} \end{equation*}

What is the relationship between $$\xvec$$ and $$\coords{\xvec}{\ecal}\text{?}$$

in-context


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# IIFT 2018 Question Paper | IIFT Quants

###### IIFT Quantitative Aptitude | IIFT 2018 Question Paper

The best way to boost your IIFT prep is to practice the actual IIFT Question Papers. 2IIM offers you exactly that, in a student friendly format to take value from this. In the 2018 IIFT, quants were a mixed bag of questions of varying difficulty, with some routine questions and the others were very demanding. Some beautiful questions that laid emphasis on Learning ideas from basics and being able to comprehend more than remembering gazillion formulae and shortcuts.

1. #### IIFT 2018 Question Paper Quants

A Business Group has 3 Companies X, Y, Z and a Trust P which is engaged in charitable activities. Each group company has to donate 5% of its own funds to the Trust, excluding the loan which the company has taken from other companies of the group. X has given a loan to Y which is equivalent to 10% of the funds of Y. After receiving the loan, Y has funds which are 2 times the funds of Z. If Z gave Rs. 10,000 as donation to the Trust P, how much is the approximate contribution of 'Y' to the Trust P?

1. Rs. 17,000
2. Rs. 18,000
3. Rs. 19,000
4. Rs. 20,000

2. #### IIFT 2018 Question Paper Quants

A bucket contains 200cc of liquid. A solid ball is dropped in the bucket resulting in the rise of liquid level to 1.3 times of its original level. If the radius of the base of the bucket is 3 cm and the radius of the surface of the liquid level is 1 cm more than the radius of the base of the bucket before the ball is dropped. Find the volume of the solid metal ball.

1. 68cc
2. 80cc
3. 92cc
4. Can't be determined

3. #### IIFT 2018 Question Paper Quants

P...........Q
R...........S
T...........U
V..........W
Using 5 dots in each of the lines PQ, RS, TU, and VW as the vertices, how many triangles can be drawn such that the base is on any
one of the above lines?

1. 120
2. 150
3. 200
4. 600

4. #### IIFT 2018 Question Paper Quants

In the triangle PQR, S is the midpoint of QR. X is any point on PR. T is the point on QR such that PT || SX. If the area of triangle PQR is 5.8 sq. cm. then the area of triangle RTX is

1. 2.9 sq. cm
2. 3.2 sq. cm
3. 5.8 sq. cm
4. 2.45 sq. cm

5. #### IIFT 2018 Question Paper Quants

Given P(x,y) = x2 + xy +y2; Q(x,y) = x2 - xy +y2.
Then find the value of P(7, Q(9 , 4))

1. 4169
2. 4197
3. 4089
4. 4127

In the given figure, PA = QB and PRQ is the arc of the circle, the centre of which is O such that angle POQ = 90°. If AB= $25 $sqrt{2}$ cm and the perpendicular distance of AB from centre O is 30cm. Find the area of the shaded region? 1. 625$\sqrt{2}$sq. cm 2. 625$ $$frac{1}{2}+\frac{\pi}{4}$ )sq. cm 3. 750$\sqrt{2}$ - 625$$$frac{1}{2}+\frac{\pi}{4}$)sq. cm 4. None 7. #### IIFT 2018 Question Paper Quants The roots of quadratic equation y2-8y+14 = 0 are $\alpha$ and$\beta$. Find the value of$1 + $$alpha+\beta^{2}$)$ 1 + $$beta+\alpha^{2}$) 1. 419 2. 431 3. 485 + 3$\sqrt{22}$ 4. 453 + $\sqrt{22}$ 8. #### IIFT 2018 Question Paper Quants $\frac{1}{\log _{x} y z+1}+\frac{1}{\log _{y} x z+1}+\frac{1}{\log _{x} x y+1}$ = ? 1. 0 2. 1 3. xyz 4. $\frac{\log x y z}{\log x y z+1}$ 9. #### IIFT 2018 Question Paper Quants Ram, Ravi and Ratan can alone finish an assignment in 9 days, 12 days and 15 days respectively. They decide to complete a work by working in turns. Ram works alone on Monday, Ravi does the work alone on Tuesday, followed by Ratan working alone on Wednesday & so on. What proportion of the complete work is done by Ravi? 1. 2/9 2. 12/47 3. 1/3 4. 4/9 10. #### IIFT 2018 Question Paper Quants Let $S_{1}$ be a square of side 4 cm. Circle $C_{1}$ circumscribes the square $S_{1}$ such that all its corners are on $C_{1}$. Another square $S_{2}$ circumscribes the circle $C_{1}$. Circle $C_{2}$ circumscribes the square $S_{2}$, and square $S_{3}$ circumscribes circle $C_{2}$, & so on. If $A_{N}$ is the area between the square $S_{N}$ and the circle $C_{N}$, where N is the natural number. then the ratio of sum of all $A_{N}$ to $A_{1}$ is 1. 1 2. $\frac{\pi}{2}$ - 1 3. Can't be determined 4. None of the above 11. #### IIFT 2018 Question Paper Quants Joseph diametrically crosses a semi-circular playground and takes 48 seconds less than if he crosses the playground along the semi-circular path. If he walks 50 metres in one minute, the diameter of playground is 1. 54 metres 2. 70 metres 3. 85 metres 4. 35 metres 12. #### IIFT 2018 Question Paper Quants Garima had only Rs. 200. Rs. 500 and Rs. 2000 notes in her wallet. She goes to Shoppers Stop. purchases some dresses and gives half of her Rs. 2000 notes & in turn receives the same number of Rs. 200 notes. She then goes to a restaurant and gives all her Rs. 500 notes and receives thirty Rs. 2000 notes, which increases the number of Rs. 2000 notes she had by 75%. Now if she has fifty Rs. 200 notes. What was the original number of Rs. 2000 and Rs. 200 notes she had at the start? 1. 60, 10 2. 60, 15 3. 80, 10 4. 80, 15 13. #### IIFT 2018 Question Paper Quants A metallic solid is made up of a solid cylindrical base with a solid cone on its top. The radius of the base of the cone is 5 cm. and the ratio of the height of the cylinder and the cone is 3:2. A cylindrical hole is drilled through the solid with height equal to 2/3rd of the height of solid. What should be the radius$in cm) of the hole so that the volume of the hole is 1/3rd of the volume of the metallic solid after drilling?

1. $$sqrt{\frac{45}{8}}$ 2. $\sqrt{\frac{35}{8}}$ 3. $\sqrt{\frac{65}{8}}$ 4. $\sqrt{\frac{55}{8}}$ 14. #### IIFT 2018 Question Paper Quants Nitin installed an overhead tank on the roof of his newly constructed house. Three taps are connected to the tank: 2 taps A and B to fill the tank and one tap C to empty it. Tap A alone can fill the tank in 12 hours, while tap B alone takes one and a half times more time than tap A to fill the tank completely. Tap C alone can empty a completely filled tank in 36 hours. Yesterday, to fill the tank, Nitin first opened tap A, and then after 2 hours opened tap B also. However after 6 hours he realised that tap C was open from the very beginning. He quickly closes tap C. What will be the total time required to fill the tank? 1. 8 hours 48 minutes 2. 8 hours 30 minutes 3. 9 hours 12 minutes 4. 9 hours 36 minutes 15. #### IIFT 2018 Question Paper Quants At the foot of the mountain, the angle of elevation of the summit at the top of the mountain is 45°. After ascending 100 metres, at a slope of 30° up the mountain towards the summit. the angle of elevation of the summit is 60°. Find the height of the summit. 1. 50$$$sqrt{3}+1$)metres 2. 50$$$sqrt{5}+1$)metres 3. 50$$$sqrt{3}+2$)metres 4. 50 $\sqrt{3}$metres 16. #### IIFT 2018 Question Paper Quants Land Cruiser Prado, the latest SUV from Toyota Motors, consumes diesel at the rate of $\frac{1}{400}\left\{\frac{1000}{x}+x\right\}$ litres per Km. when travelling at the speed of x km/hr. The diesel costs Rs. 65 per litre and the driver is paid Rs. 50 per hour. Find the steady speed that will minimize the total cost of a 1000 km trip? 1. 33 km/hr 2. 36 km/hr 3. 39 km/hr 4. 52 km/hr 17. #### IIFT 2018 Question Paper Quants In a survey on the viewership of the TV channels, 73% of those surveyed viewed at least one of the three Channels: Star Plus, Sab TV, and Sony. 38% of those surveyed viewed Star Plus, 39% viewed Sony, and 23% viewed Sab TV. 11% of all those surveyed viewed all the three channels. What percentage of those surveyed, viewed more than one of the three TV channels? 1. 16 2. 38 3. 27 4. Data Inadequate 18. #### IIFT 2018 Question Paper Quants A physical therapist of the Russian football team knows that the team will play 40% of its matches on artificial turf this season. Because of his vast experience, he knows that a football player's chances of incurring a knee injury is 50% higher if he is playing on artificial turf instead of grass. If the player's chances of a knee injury on artificial turf is 0.42, what is the probability that a football player with knee injury, incurred the injury while playing on grass? 1. 0.28 2. 0.336 3. 0.5 4. None of the above 19. #### IIFT 2018 Question Paper Quants The square root of $1+x^{2}+\sqrt{1+x^{2}+x^{4}}$ is 1. $\frac{1}{\sqrt{2}}\left[\sqrt{1+x+x^{2}}+\sqrt{1-x+x^{2}}\right]$ 2. $\frac{1}{\sqrt{2}}\left[\sqrt{1+x+x^{2}}-\sqrt{1-x+x^{2}}\right]$ 3. $\frac{1}{\sqrt{2}}\left[\sqrt{1+x^{2}+x^{4}+x^{8}}\right]$ 4. None of the above 20. #### IIFT 2018 Question Paper Quants $\log _{2} x \cdot \log _{\frac{x}{64}} 2=\log _{\frac{x}{16}} 2$; then x= ? 1. 2 2. 4 3. 12 4. 16 ###### Prepare for CAT 2022 with 2IIM's Daily Preparation Schedule ###### Know all about CAT Exam Syllabus and what to expect in CAT ###### Best CAT Online Coaching Try upto 40 hours for free Learn from the best! ###### Already have an Account? ###### CAT Coaching in ChennaiCAT 2021Enroll at 49,000/- 44,000/- Online Classroom Batches Starting Now! ###### Best CAT Coaching in Chennai Introductory offer of 5000/- Attend a Demo Class ###### Best Indore IPM & Rohtak IPM CoachingSignup and sample 9 full classes for free. Register now! ##### Where is 2IIM located? 2IIM Online CAT Coaching A Fermat Education Initiative, 58/16, Indira Gandhi Street, Kaveri Rangan Nagar, Saligramam, Chennai 600 093 ##### How to reach 2IIM? Phone:$91) 44 4505 8484
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text:
# Poisson parameter 𝜇 and expected occurrence of an event

In a CSV file there is data that shows only months and number of traffic accidents in each month.

What is the expected number of accidents per month if they are distributed with a Poisson distribution parameter μ?

They also ask: according to the data in CSV file, the average traffic accident per month is 7. How can we use this to estimate the μ?

• I think more information is needed. Is the $\mu$ referencing one month or some other time period? – lulu Feb 4 at 22:03
• Well, in that case you can guess that it refers to one month in which case $\mu$ is the expected number of accidents over the month, just as you said. – lulu Feb 4 at 22:30


text:
# How do you calculate (9.4 times 10^6) / (4.97 times 10^2)?

Apr 4, 2017

See the entire solution process below:

#### Explanation:

First, rewrite this expression as:

$\frac{9.4}{4.97} \times {10}^{6} / {10}^{2} \to 1.89 \times {10}^{6} / {10}^{2}$ Rounded to the nearest hundredth.

Next, use this rule of exponents to divide the 10s terms:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$

$1.89 \times {10}^{\textcolor{red}{6}} / {10}^{\textcolor{b l u e}{2}} \to 1.89 \times {10}^{\textcolor{red}{6} - \textcolor{b l u e}{2}} \to 1.89 \times {10}^{4}$


text:
Date: Jul 14, 2013 5:46 AM
Author: daly@axiom-developer.org
Subject: Re: An independent integration test suite

On Sunday, July 14, 2013 3:19:20 AM UTC-4, Nasser M. Abbasi wrote:> On 7/13/2013 11:22 PM, daly@axiom-developer.org wrote:> > > > >> > > Axiom has published a Computer Algebra Test Suite at> > > http://axiom-developer.org/axiom-website/CATS/index.html> > >> > > It includes Schaums integrals and Kamke's Ordinary Differential Equations.> > > > I am working on Kamke, and have all the ode's and have the book also which> > I check with.> > > > The problem is writing the document itself to show the result of Maple> > and Mathematica next to each other is what is taking long time,> > since it involves lots of manual work.> > > > The problem is with Maple it is not possible to export each result> > on its own to a .png file so I can include that in my Latex report.> > > > Exporting each ODE's result to Latex does not work for long results> > since sometimes the latex needs manual breaking of the generated> > equation. (if there is long expression between \left(... and \right)> > > > So I have to do each one by one and use .png files to capture> > the result.> > > > Currently there are about 150 or so done> > > > http://12000.org/my_notes/kamek/kamke_differential_equations.htm> > > > again, to make a document, in Latex, which includes many CAS's> > results, the CAS itself must help in terms of making it easy> > to export things. Mathematica is very good in this area, so one can> > automate all this in code and run over the hundreds of the ODE's and do> > everything in code. But to integrate results of other CAS'es into> > one document, this process breaks down.> > > > --Nasser> > > > > It also includes Albert Rich's integration set.> > > In all there are several thousand examples.> > >> > > The source file format is latex, the output file format is pdf.> > > The axiom.sty package is at> > > http://axiom-developer.org/axiom-website/CATS/axiom.sty> > >> > > Each problem includes the source input.> > > Axiom's output is prefixed with --R which is an Axiom comment.> > >I will be interested in seeing your results.The whole point of this exercise, at least for Axiom,is to have a validated test suite. I found a few errorsin the reference books while testing. I will compare yourresults to the posted results for Axiom.I agree that creating these test suites by hand is ahuge task. It took me a couple months to create theCATS test suite so far. That's why I'd like to see theDLMF completely machine-readable.Tim


text:
# Volume of solid $y = x−4x^2$ revolved about y-axis using shell approach.

I am attempting to solve the following problem and would like some validation in my approach/need some help on finding zeros if this is indeed the correct approach.

Problem:

Find vol of solid of revolution - The region bounded by $y = x−4x^2$ and the $x$-axis revolved about the $y$-axis.

My approach:

1. Shells
2. $V= 2\pi$ * [integral of $\int_a^b x(x-4x^2)dx$
3. evaluate from b to a, and I'm assuming answer would be in pi cubic units because we're solving for volume.

How would I determine the bounds, and is my approach the correct one? (apologies for the poor formatting, I am new to the site)

Thanks, J

• Typically to find bounds, I am given two equations, which i can set to zero and solve for, which is why this problem is throwing me off. Jul 20 '18 at 4:23
• What is the appropriate range of values for $x$ (where does the parabola intersect the $x$-axis)? Jul 20 '18 at 4:44

First of all bounds $x-4x^2=0$ has roots $x=0, 0.25$. So these are your bounds. You have the integral setup correctly. So just evaluate the following. $$\int_0^{0.25}2\pi x(x-4x^2)dx$$
Because you have $f(x)=x(1-4x)$ as the top curve and $y=0$ i.e the x-axis as the bottom curve. You need to find the intersection of these curves to get the bounds.1


text:
Introduction to CH-acidic Compounds

The Acidity of Carbonyl Compounds

Aldehydes and ketones have remarkably low $pKa$ values that range between 15 and 20. Thus, they may act as a in an acid-base reaction with a strong base. The acidic hydrogen is the hydrogen that is bound to the carbon adjacent to the carbonyl carbon. Due to the only minor difference in electronegativity between hydrogen and carbon, C-H bonds in alkanes are hardly polarized at all. Thus, hydrogens of alkanes are in fact not acidic. The $pKa$ values of alkanes amount to only approximately 50.

The acidic hydrogen of carbonyl compounds is known as α hydrogen, as it is connected with the carbonyl compound's α carbon that is directly bound to the carbonyl carbon.
Fig.1
Acidity of alkanes and ketones.

The carbonyl compounds' relatively high acidity may be explained by the resonance stabilization of the conjugate base by the carbonyl group, or, in other words, through the stabilization of the anion formed by deprotonation. This anion is called an anion. The negative charge is mainly distributed among the α carbon and the carbonyl oxygen, by resonance, which leads to the stabilization of the otherwise highly energized carbanion.

Stability of enolate anions

Both the bonding and non-bonding π orbitals of the enolate are occupied, while the antibonding π orbital remains free. The distribution of the negative charge and the nucleophilic qualities of the enolate anion are mostly represented through the occupied non-bonding π orbital, as it is the enolate's HOMO. The illustration of the non-bonding π orbital is indicative of the location of the non-bonding π electrons, which are at the α carbon and the carbonyl oxygen. Thus, the α carbon and the carbonyl oxygen are the nucleophilic positions of enolate anions. The smaller extension of the non-bonding π orbital lobe at the carbonyl oxygen clearly supports the fact that the electron (density) is bound more strongly and closely to the oxygen atom than is the case with the α carbon. Thus, the non-bonding π orbital lobes' extensions substantiate the fact that the α carbon is a soft base, while the carbonyl oxygen is a hard base. As a result, soft Lewis acids (electrophiles), such as alkyl halides and the carbonyl carbon of carbonyl compounds, tend to be nucleophilically attacked by the enolate's α carbon.

Tab.1
π orbitals of an enolate anion.
 Bonding π orbital (occupied) Non-bonding π orbital, HOMO (occupied) Antibonding π orbital, LUMO (unoccupied) Fig.2 Fig.3 Fig.4 Fig.5 Transparency: on/off Fig.6 Transparency: on/off Fig.7 Transparency: on/off

<Page 1 of 4


text:
# Semi-Standard Young Diagrams and Families

In connection with string theory I encountered the following problem:

Given the set M_N of all semi-standard Young tableaux of size N (i.e. all fillings of Ferrers diagrams with natural numbers with weakly increasing rows and strictly increasing columns, the content of which sum to N).

The conjecture, which I cannot prove and for which I know no counterexample states:

M_N is the union of disjoint families, where each family consists of a father diagram and all his daughters, which are obtained from the father diagram by deletion of one box.

The conjecture holds for N<=8

http://www.itp.uni-hannover.de/~dragon/young.pdf

There numbers a,b are attached to the Ferrers diagrams, where a is the number of semi-standard Young Tableaux of size N and b the number of fathers, who remain after the daughters have filled up their families.

Light cone string theory proves that the conjecture holds for all M_N with less than 25 rows, i.e. for N < 325. But string theory provides no counterexample for N >= 325.

Any suggestions, e.g. how to compute the number a(N,lambda) of Young diagrams with size N and shape lambda, would be helpful.

Norbert Dragon

• I do not understand the statement of the conjecture. How do you obtain the daughters precisely? May 6, 2013 at 16:10
• The daughters are obtained from their father diagram by removing one box, e.g. the diagram 3,2,1 has daughters 2,2,1 and 3,1,1 and 3,2. The looked for families of father and daughter diagrams characterize the SO(D-2)representations in string theory which are needed for the construction of SO(D-1) representations. May 6, 2013 at 16:50

I use standard facts about symmetric functions that can be found, e.g, in Chapter 7 of Enumerative Combinatorics, vol. 2. Let $s_\lambda$ denote a Schur function and $p_1=s_1=x_1+x_2+\cdots$. Then $\frac{\partial s_\lambda}{\partial p_1}=\sum_\mu s_\mu$, where the $\mu$'s are obtained by removing a single box from $\lambda$. Moreover, the coefficient of $q^n$ in $s_\lambda(q,q^2,q^3,\dots)$ is equal to the number of SSYT with entries summing to $n$. Thus if we let $c_n$ be the coefficient of $q^n$ in the symmetric function $f$ defined by $$f+\frac{\partial}{\partial p_1}f = \sum_\lambda s_\lambda\cdot s_\lambda(q,q^2,q^3,\dots),$$ and if $c_n=\sum a_{\mu,n}s_\mu$, then we need $a_{\mu,n}$ copies of the shape $\mu$ in a set of fathers generating all SSYT with entries summing to $n$. Hence we need to show that $a_{\mu,n}\geq 0$. Now $$\sum_\lambda s_\lambda\cdot s_\lambda(q,q^2,q^3,\dots) = \exp \sum_{n\geq 1} \frac{q^n}{1-q^n}p_n.$$ This leads to a simple linear first-order differential equation with solution $$f = (1-q) \sum_\lambda s_\lambda\cdot s_\lambda(q,q^2,q^3,\dots).$$ If $h_u$ denotes the hook length of the square $u$ of $\lambda$, then $$s_\lambda(q,q^2,q^3,\dots) = \frac{q^{b(\lambda)}}{\prod_{u\in \lambda} (1-q^{h_u})},$$ where $b_\lambda=\sum i\lambda_i$. Since there is always a hook length equal to one, the power series $(1-q)s_\lambda(q,q^2,q^3,\dots)$ will be a product of factors of the form $1/(1-q^h)$, $h\geq 1$, so will have nonnegative coefficients as desired. Thus we have not just an existence proof, but a precise generating function for the number of fathers of each shape.

• For $\frac{\partial}{\partial p_1}s_{\lambda} = \sum_{\mu} s_{\mu}$ so these $\mu$ are symmetric skew schur funciton ? Is it true for differentiating with repsect to $p_n$? What about the \frac{\partial}{\partial p_1}$\frac{\partial}{\partial p_2}$ means it would be removing the $2,1$ subset from $\lambda$?
– GGT
Mar 30, 2021 at 12:18
• Applying $n\frac{\partial}{\partial p_n}$ corresponds to removing border strips $B$ of size $n$ weighted by $(-1)^{\mathrm{ht}(B)}$. This follows from EC2, Theorem 7.17.3, and the fact that multiplication by $p_n$ is adjoint to $n\frac{\partial}{\partial p_n}$. Mar 31, 2021 at 18:37
• I get it now, I have the following question, so $s_{\lambda}(p_1, p_2..)$ written in power symmetric basis and if we specialise $p_i = x^i$ then we obtain $s_{(n)}\neq 0$ and it's zero for all other $\lambda$. I guess it might follow from your theorem 7.21.2(?) so I am wondering if there is a closed formula for the linearly shifted schur function by the parameters $t_i$ that is $s_\lambda(p_1 +t_1 , p_2 +t_2 ,..)$ and then specialise $p_i= x^i$? I could derive it using the differential of schur function it's getting complicated(?) Some help.
– GGT
Apr 4, 2021 at 11:49
• Specializing $p_i=x^i$ is the same as the evaluation $p_i(x,0,0,\dots)$. There are then several ways to see that $s_\lambda(x,0,0,\dots)=0$ unless $\lambda=(n)$, e.g., from the combinatorial definition in terms of SSYT. As for $s_\lambda(p_1+t_1,\dots)$, let $t_i=p_i(z_1,z_2,\dots)$ (without loss of generality since the $t_i$'s remain algebraically independent). Let $y=(y_1,y_2,\dots)$. Then $\sum_\lambda s_\lambda(x+t_1,x^2+t_2,\dots)s_\lambda(y)$ $=(\sum x^nh_n(y)\left( \sum s_\lambda(z)s_\lambda(y)\right)$. Now use Pieri's rule and then extract the coefficient of $s_\lambda(y)$. Apr 5, 2021 at 16:21
• ∑𝜆𝑠𝜆(𝑥+𝑡1,𝑥2+𝑡2,…)𝑠𝜆(𝑦) =(∑𝑥𝑛ℎ𝑛(𝑦)(∑𝑠𝜆(𝑧)𝑠𝜆(𝑦)). So RHS of the equation I would expand the product of two Schur function in terms of Schur function, so writing it in terms of the combination of Littlewood Richardson coefficient?
– GGT
Apr 11, 2021 at 0:31


text:
## Analysis and recovery of high-dimensional data with low-dimensional structures

Series:
High Dimensional Seminar
Wednesday, November 7, 2018 - 12:52pm
1 hour (actually 50 minutes)
Location:
Skiles 006
,
Georgia Tech
,
Organizer:
High-dimensional data arise in many fields of contemporary science and introduce new challenges in statistical learning and data recovery. Many datasets in image analysis and signal processing are in a high-dimensional space but exhibit a low-dimensional structure. We are interested in building efficient representations of these data for the purpose of compression and inference, and giving performance guarantees depending on the intrinsic dimension of data. I will present two sets of problems: one is related with manifold learning; the other arises from imaging and signal processing where we want to recover a high-dimensional, sparse vector from few linear measurements. In the first problem, we model a data set in $R^D$ as samples from a probability measure concentrated on or near an unknown $d$-dimensional manifold with $d$ much smaller than $D$. We develop a multiscale adaptive scheme to build low-dimensional geometric approximations of the manifold, as well as approximating functions on the manifold. The second problem arises from source localization in signal processing where a uniform array of sensors is set to collect propagating waves from a small number of sources. I will present some theory and algorithms for the recovery of the point sources with high precision.


text:
• 营养卫生 •

### 基于主成分分析与马氏距离结合运用的婴幼儿配方奶粉营养综合评价

1. （1.西北农林科技大学食品科学与工程学院，陕西 杨凌 712100；2.陕西优利士乳业有限公司，陕西 咸阳 712000；3.陕西百跃羊乳集团有限公司，陕西 西安 710000）
• 出版日期:2020-03-15 发布日期:2020-03-23
• 基金资助:
陕西省科技创新能力建设计划项目（2017XY-02）；陕西省企业创新项目（K4030218139）

### Comprehensive Nutrition Evaluation of Infant Formula Based on Principal Component Analysis with Mahalanobis Distance

ZHANG Xue, GE Wupeng, XI Menglu, YANG Li, HE Rui, CUI Xiuxiu

1. (1. College of Food Science and Engineering, Northwest A & F University, Yangling 712100, China;2. Shaanxi Youlishi Dairy Industry Co. Ltd., Xianyang 712000, China; 3. Shaanxi Baiyue Goat Group Co. Ltd., Xi’an 710000, China)
• Online:2020-03-15 Published:2020-03-23

Abstract: This study aimed to scientifically evaluate the main nutritional ingredients of infant formula in order to provide a technical basis for the development and optimization of new infant foods. Gas chromatography, high performance liquid chromatography and ion chromatography were used to detect the three macronutrients proteins, fats and carbohydrates as well as fatty acids, amino acids, lactose and oligosaccharides in infant formula. From the multi-index system, six principal components were extracted and the Mahalanobis distance (MD) was used to measure the nutritional coincidence degree between infant formula and breast milk, and a comprehensive nutritional evaluation was carried out. The results showed that: 1) The compositions and contents of the macronutrients in nine infant formula samples were similar to those in breast milk, but there was a significant difference in some individual components (P < 0.05); 2) Among the tested infant formula samples, goat milk sample D scored highest for Mahalanobis distance (MD = 90.15), followed by goat milk sample F (MD = 86.96), and bovine milk samples H and I showed lower MD scores (82.32 and 82.11, respectively). Accordingly, MD scores for goat milk infant formula were generally higher than those for bovine milk infant formula; 3) MD scores were most affected by fatty acids, and the effect of carbohydrates was the least. The interaction between various nutrients also had an effect. This method can be considered a scientific and reasonable method for the nutritional evaluation of infant formula using multiple nutritional indicators.


text:
# PLM tests

## October 5, 2015

What happens if we look at the same two high mach number tests from the previous posts (the square wave test and the shock tube test), but use a piecewise linear reconstruction?

Figure 1a: PLMC, exact solver, no dual energy, $v \approx 800 c_s$

Figure 1b: Same as Figure 1a but with dual energy

Figure 2a: PLMP, exact solver, no dual energy, $v \approx 800 c_s$

Figure 2b: Same as Figure 2a but with dual energy

Figure 3a: PLMC, exact solver, no dual energy, $v \approx 100 c_s$

Figure 3b: Same as Figure 3a but with dual energy

Figure 4a: PLMP, exact solver, no dual energy, $v \approx 100 c_s$

Figure 4b: Same as Figure 4a but with dual energy


text:
×

Cake walk

None

# PROBLEM:

You're given an array of houses some of which have a bomb and others don't. These bombs will explode simultaneously and when bomb in ith house explodes, it destroys adjacent houses. Your task is to find out the number of houses which aren't destroyed.

# QUICK EXPLANATION:

For every house check if it is destroyed or not.

# DETAILED EXPLANATION:

There are two almost similar ways of solving this problem. Key idea is to find out for each house if it is destroyed or not. How do we check if ith house will be destroyed or not? It will be destroyed iff at least one of i-1, i and i+1th houses have a bomb. Just beware of the boundaries as first and last houses have only one adjacent house. So final solution is :

saved_count = 0
for i from 1 to N
destroyed = false
if( S[i] == '1') destroyed = true
if( i > 1 and S[i-1] == '1') destroyed = true
if( i < N and S[i+1] == '1') destroyed = true
if( not destroyed)
saved_count += 1
print saved_count

An alternative solution would be to move over houses with bombs and mark those houses that will explode.

bool will_be_destroyed [N]
fill( will_be_destroyed, false)

for i from 1 to N
if S[i] == '1'
will_be_destroyed[i] = true
if(i > 1) will_be_destroyed[i-1] = true
if(i < N) will_be_destroyed[i+1] = true

saved_count = 0
for i from 1 to N
if(not will_be_destroyed[i])
saved_count += 1

print saved_count

Both of these are O(N) solutions and very comfortably fit in time limit.

# TESTER'S SOLUTION:

Can be found here.

This question is marked "community wiki".

1243837
accept rate: 0%

 0 Tell the case where it fails? solution answered 28 Jul '14, 10:21 3★tech_boy 1.2k●4●19●31 accept rate: 7%
 0 Whats wrong in my code??? Why is it showing NZEC everytime. Some one please help. enter code here  import java.io.; import java.util.; class acm { public static void main(String[] args) { FasterScanner sc=new FasterScanner(System.in); int t=sc.nextInt(); while(t>0){ int n=sc.nextInt(); String s=sc.nextString(); int count=0; for(int i=0;i= numChars) { curChar = 0; try { numChars = mIs.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public String nextLine() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isEndOfLine(c)); return res.toString(); } public String nextString() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public long nextLong() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } long res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public int nextInt() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; }  } enter code here  answered 09 Nov '16, 02:09 1★kanha95 1 accept rate: 0% I got it...i missed the case when n=1.... (09 Nov '16, 02:42) kanha951★
 0  #include using namespace std; main() { int t = 0 ; cin>>t; while(t--) { string inp; int b = 0 ; int u = 0 ; int s = 0 ; cin>>s; cin>>inp; for(int i = 0 ; i < s;i++) { if(inp.at(i) == '1' ) { b -= 2 ; if(i == 0 || i == s-1) { b += 1 ; } } else if (inp.at(i) == '0') { u++; } } cout<
 0 I'm getting WA. I don't understand on which test case its failing. Someone please help? link to solution: https://www.codechef.com/viewsolution/16605991 answered 19 Dec '17, 20:16 2★mohmum 2 accept rate: 0% @mohmum your code fails for 1 1 1 answer should be zero but your code prints 1 (19 Dec '17, 20:38)
 0 Somebody, please tell me where does my program fail? https://www.codechef.com/viewsolution/16606169 answered 19 Dec '17, 20:42 14●4 accept rate: 0%
 0 whats wrong in my C code for the problem LEBOMBS LEBOMBS https://www.codechef.com/viewsolution/17048802 can anyone tell me on which values my code fails .. Thanks answered 11 Jan, 01:36 1●1 accept rate: 0%

# include<bits stdc++.h="">

using namespace std; int main(){ int t;cin>>t; while(t--){ int n;cin>>n;int count=0; string s;cin>>s; if(s[0]=='1'){count+=1;if(s[1]=='0'){s[1]='2';count+=1;}} for(int i=1;i<n-1;i++){ if(s[i]="='0'||s[i]=='2'){continue;}" else{++count;="" if(s[i-1]="='0'){count+=1;}" if(s[i+1]="='0'){++count;s[i+1]='2';}" }}if(s[n-1]="='1'){count+=1;" if(n="">2){if(s[n-2]=='0'&&s[n-3]=='0')count+=1;} else if(s[0]=='0')count+=1;} cout<<(n-count)<<endl;} return 0;}

What is wrong with the code?

2★shubhsy
1
accept rate: 0%

 0 https://www.codechef.com/viewsolution/17555048 Can anyone help me with this code, please? answered 27 Feb, 15:00 2★shubhsy 1 accept rate: 0%

whats wrong in my solution?? \

# include<iostream>

using namespace std; int main(){ int i,n,t,count; cin>>t; while(t--){ count=0; cin>>n; char c[n]; for(i=0;i<n;i++) cin="">>c[i]; if(c[0]=='0'&&c[1]=='0'){ count++; } if(c[n-1]=='0'&&c[n-2]=='0'){ count++; } for(i=1;i<(n-1);i++){ if(c[i]=='0'){ if(c[i-1]=='0'&&c[i+1]=='0') count++; } } cout<<count<<endl; } return 0; }

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# The Chinese remainder theorem

The Chinese remainder theorem is a theorem in number theory and modulo arithmetics. As such, it doesn’t come up in regular mathematical lessons very often. It is however well-known to all people who have dabbled in cryptography and implementation of popular cryptographic algorithms and attacks.

I happen to be one of those people, so I spent a good amount of time researching the theorem and implementations. And now you get the summary of what I discovered!

All implementations are going to be in Python. I mostly used this stuff in little scripts here and there, so I never spent the time translating them to anything else.

# The theorem

Let’s consider the following set of k equations:

This is equivalent to saying that `x mod ni = ai` (for i=1…k). The notation above is common in group theory: you define the group of integers modulo some number n and then you state equivalences (or congruences) within that group. Careful not to confuse them with actual equivalences!

Here, x is the unknown; instead of knowing x, we know the remainder of the division of x by a group of numbers. If the numbers ni are pairwise coprimes (i.e. each one is coprime with all the others) then the equations have exactly one solution. Such solution will be modulo N, with N equal to the product of all the ni.

Notes on the history: the earliest known statement of the theorem is found in a book written in the 3rd century AD by a Chinese mathematician, named Sunzi

There are certain things whose number is unknown. If we count them by threes, we have two left over; by fives, we have three left over; and by sevens, two are left over. How many things are there?

However, we would have to wait about 3 more centuries for another Indian mathematician, Aryabhata, to present the first recursive algorithm able to find a solution. Aryabhata was actually trying to solve Diophantine equations of the first order. But I digress. This is another fascinating topic for another day!

# The proof

There are many ways to prove this theorem. Most of them are directly related to the algorithms we can use to find the solution. I chose the proof that I found more immediate to understand; you will see that in action when I show Gauss’ algorithm.

Let’s define a slightly simpler problem, where we have only two equations.

As before, let’s define `N = n1 * n2`.

Let’s define `p = n1^-1 (mod n2)` and `q = n2^-1 (mod n1)`. This is the operation called modular inverse, where we find the inverse of a number in the group of all the numbers mod N. If I say that p and n1 are inverse in mod n2, this means that `p * n1 = 1 (mod n2)`. Such inverse only exists when n1 and n2 are coprimes, and here they are by definition.

Now I claim that a solution y to the set of equations can be expressed as:

this is a valid solution because `y = a1 * q * n2 = a1 (mod n1)` and `y = a2 * p * n1 = a2 (mod n2)`. This follows from the definition of the modular inverse telling me that `p * n1 = 1 (mod n2)` and `q * n2 = 1 (mod n1)`.

This is easily extendible to a generic number of equations, where the final construction of y becomes:

When building p and q before, we used only n1 or only n2; what that generalizes to is the product of all moduli excluding the “current one”: the `N / ni` above. The rest is unchanged.

So we have a solution: the next step is to prove it is the unique solution. Let’s assume a second solution z exists for the same set of equations. Then `z = a1 (mod n1)`, which implies that `z - y` is a multiple of n1, since the remainder of their division by n1 is the same number. By the same reasoning, `z - y` is also a multiple of n2. But since n1 and n2 are coprimes, then it would also be a multiple of N, or as it is often written:

z must be the same as y in the mod N group.

# Algorithm 1: Gauss algorithm

This is easy: it is a direct translation to code of the construction explained above. The n and a parameters are lists with all the related factors in order, and N is the product of the moduli.

The good thing about this algorithm is that the result is guaranteed to be positive, given bi and ni both positive. This does not apply to the next implementation.

For an implementation of `invmod` (finding the modular inverse), see next section.

# Algorithm 2: Euclid

This is the direct construction procedure described by Wikipedia.

The extended Euclidean algorithm is used to find two coefficients a and b such that `a * (N / ni) + b * ni = gcd(N / ni, ni) = 1`.

Then x is computed the following way:

Translated into code:

As you can see for my specific application, I wanted only positive results; but the si coefficients can be negative in a lot of cases, making the final sum negative. What do I do in that case? What I did there is to multiply the result by -1, then add N and take the remainder of the division by N, to wrap around the modulus of the solution.

# The extended Euclidean algorithm

As explained above, the algorithm takes two numbers, x and y, and returns two coefficients a and b such that:

The implementation returns both the coefficients and the GCD itself.

Now if I take two positive integers x and y, I know I can express them as

where q is the quotient of the division (i.e. `q = x // y` where // denotes the integer division) and r is the remainder and is always strictly smaller than y. If x is a multiple or y, of course r is going to be zero.

The GCD of two integers can be found by repeating this procedure until the remainder is 0; more specifically:

The final r before getting to 0 is the GCD. Let’s see this with an example:

so the GCD is 2. Now to find the coefficients we work backwards from the second-to-last division, expressing the new remainder in terms of the other parts:

3 and -8 are the coefficients in the Bezout identity. To compute them in practice we do not work backward, but simply store them as we go, as they can be derived from the main division equation.

# The modular inverse

Ok so let’s suppose I want to find

my unknown x is the modular inverse of 17 in mod 43. First we need to verify that gcd(17, 43) is 1, otherwise the inverse does not exist. Once we have done that, we compute the two Bezout coefficients as shown above. If we work that out manually by retracing all the divisions, we get:

So we have expressed this in terms of `a*x + b*y = gcd(x, y)`: 2 and -5 are our Bezout coefficients. Let’s rewrite the last equation in mod 43.

But 2*43 is a multiple of 43 so it is irrelevant here; this leaves us with:

By comparing this with the starting equation in the unknown x, -5 is the inverse we were looking for. However we need a positive result: in this case we can do the same procedure we have done earlier, by changing -5 to 5, then adding 43 and computing the mod 43 to remain in the group. This yields 38.

The code is quite simple:

# References

Several websites contributed to me getting to the bottom of this interesting theorem and provided the numerical examples I use above:

Random lady whose eyes sparkle for science, and who loves writing. I plan to write about a large variety of interesting science topics, one bit at a time.

## More from Astarte Kraus

Random lady whose eyes sparkle for science, and who loves writing. I plan to write about a large variety of interesting science topics, one bit at a time.

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Synopsis

# Alphabet Waves

Physics 5, s136
Researchers explain where those X- and Y-shaped waves at the shore come from.

While many beachcombers head home with nothing but a few seashells and a sunburn, two scientists have made productive use of their time in the waves. At two beaches on the Pacific coast of California and Mexico, they discovered that nonlinear wave patterns, in particular X- and Y-shaped waves, are surprisingly common. As reported in Physical Review E by Mark Ablowitz and Douglas Baldwin of the University of Colorado, Boulder, these wave phenomena may contribute to the growth of tsunami waves.

Water waves are difficult to model mathematically, since the underlying equations are strongly nonlinear. Under certain conditions, these nonlinear effects can produce solitary waves called solitons that travel long distances without losing their shape. Most solitons propagate along a single-line wave front, but if two waves merge at an angle, complicated two-dimensional patterns can form. These nonlinear interactions can create wave heights that are higher than the sum of the individual waves, which may amplify the destructive power of tsunamis.

Previously, the assumption was that these interactions are rare. However, the authors have observed thousands of X and Y waves shortly before and after low tide at two flat beaches, where water depths were less than about $20$ centimeters. The researchers showed that the shallow waves recorded in their photographs and films (see examples of the authors’ videos and photos) could be accurately described by a two-dimensional nonlinear wave equation. – Michael Schirber

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# Solve in SI units. A gear reduction unit uses the countershaft shown in the figure. Gear...

###### Question:

Solve in SI units.

A gear reduction unit uses the countershaft shown in the figure. Gear A receives power from another gear with the transmitted force FA applied at the 20ᵒ pressure angle as shown. The power is transmitted through the shaft and delivered through gear B through a transmitted force FB at the pressure angle shown. a) Determine the force FB, assuming the shaft is running at a constant speed. b) Find the magnitudes of the bearing reaction forces, assuming the bearings act as simple supports. c) Draw shear-force and bending moment diagrams for the shaft. If needed, make one set for the horizontal plane and another set for the vertical plane. d) At the point of maximum bending moment, determine the bending stress and the torsional shear stress. e) At the point of maximum bending moment, determine the principal stresses and the maximum shear stress.

400 mm 350 mm Fa 225 mm 200 30 mm dia. Gear A 500 mm dia. Gear B 200 mm dia. 20°

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# Ubuntu – Screen Randomly turning off

screensuspend

So I recently installed Ubuntu onto my Surface Pro 1, and have been running fine for about 2 days, until today I saw the screen constantly turning off, as if it were inactive for the set time and turning the screen off. I would then just have to tap the power button and it would come back on and then about 20 seconds later turn back off. This used to only happen after the system was suspended, and I would just have to restart to fix it, but now it is happening all the time and making it un usable. How can I fix this? Thanks!

I was able to solve the problem by updating the file /etc/systemd/logind.conf to include the following:
 HandleLidSwitch=ignore