text:
# Neural Transfer with PyTorch¶

Author: Alexis Jacq

## Introduction¶

Welcome! This tutorial explains how to impletment the Neural-Style algorithm developed by Leon A. Gatys, Alexander S. Ecker and Matthias Bethge.

### Neural what?¶

The Neural-Style, or Neural-Transfer, is an algorithm that takes as input a content-image (e.g. a turtle), a style-image (e.g. artistic waves) and return the content of the content-image as if it was ‘painted’ using the artistic style of the style-image:

### How does it work?¶

The principle is simple: we define two distances, one for the content ($$D_C$$) and one for the style ($$D_S$$). $$D_C$$ measures how different the content is between two images, while $$D_S$$ measures how different the style is between two images. Then, we take a third image, the input, (e.g. a with noise), and we transform it in order to both minimize its content-distance with the content-image and its style-distance with the style-image.

#### OK. How does it work?¶

Well, going further requires some mathematics. Let $$C_{nn}$$ be a pre-trained deep convolutional neural network and $$X$$ be any image. $$C_{nn}(X)$$ is the network fed by $$X$$ (containing feature maps at all layers). Let $$F_{XL} \in C_{nn}(X)$$ be the feature maps at depth layer $$L$$, all vectorized and concatenated in one single vector. We simply define the content of $$X$$ at layer $$L$$ by $$F_{XL}$$. Then, if $$Y$$ is another image of same the size than $$X$$, we define the distance of content at layer $$L$$ as follow:

$D_C^L(X,Y) = \|F_{XL} - F_{YL}\|^2 = \sum_i (F_{XL}(i) - F_{YL}(i))^2$

Where $$F_{XL}(i)$$ is the $$i^{th}$$ element of $$F_{XL}$$. The style is a bit less trivial to define. Let $$F_{XL}^k$$ with $$k \leq K$$ be the vectorized $$k^{th}$$ of the $$K$$ feature maps at layer $$L$$. The style $$G_{XL}$$ of $$X$$ at layer $$L$$ is defined by the Gram produce of all vectorized feature maps $$F_{XL}^k$$ with $$k \leq K$$. In other words, $$G_{XL}$$ is a $$K$$x$$K$$ matrix and the element $$G_{XL}(k,l)$$ at the $$k^{th}$$ line and $$l^{th}$$ column of $$G_{XL}$$ is the vectorial produce between $$F_{XL}^k$$ and $$F_{XL}^l$$ :

$G_{XL}(k,l) = \langle F_{XL}^k, F_{XL}^l\rangle = \sum_i F_{XL}^k(i) . F_{XL}^l(i)$

Where $$F_{XL}^k(i)$$ is the $$i^{th}$$ element of $$F_{XL}^k$$. We can see $$G_{XL}(k,l)$$ as a measure of the correlation between feature maps $$k$$ and $$l$$. In that way, $$G_{XL}$$ represents the correlation matrix of feature maps of $$X$$ at layer $$L$$. Note that the size of $$G_{XL}$$ only depends on the number of feature maps, not on the size of $$X$$. Then, if $$Y$$ is another image of any size, we define the distance of style at layer $$L$$ as follow:

$D_S^L(X,Y) = \|G_{XL} - G_{YL}\|^2 = \sum_{k,l} (G_{XL}(k,l) - G_{YL}(k,l))^2$

In order to minimize in one shot $$D_C(X,C)$$ between a variable image $$X$$ and target content-image $$C$$ and $$D_S(X,S)$$ between $$X$$ and target style-image $$S$$, both computed at several layers , we compute and sum the gradients (derivative with respect to $$X$$) of each distance at each wanted layer:

$\nabla_{ extit{total}}(X,S,C) = \sum_{L_C} w_{CL_C}.\nabla_{ extit{content}}^{L_C}(X,C) + \sum_{L_S} w_{SL_S}.\nabla_{ extit{style}}^{L_S}(X,S)$

Where $$L_C$$ and $$L_S$$ are respectivement the wanted layers (arbitrary stated) of content and style and $$w_{CL_C}$$ and $$w_{SL_S}$$ the weights (arbitrary stated) associated with the style or the content at each wanted layer. Then, we run a gradient descent over $$X$$:

$X \leftarrow X - \alpha \nabla_{ extit{total}}(X,S,C)$

Ok. That’s enough with maths. If you want to go deeper (how to compute the gradients) we encourage you to read the original paper by Leon A. Gatys and AL, where everything is much better and much clearer explained.

For our implementation in PyTorch, we already have everything we need: indeed, with PyTorch, all the gradients are automatically and dynamically computed for you (while you use functions from the library). This is why the implementation of this algorithm becomes very comfortable with PyTorch.

## PyTorch implementation¶

If you are not sure to understand all the mathematics above, you will probably get it by implementing it. If you are discovering PyTorch, we recommend you to first read this Introduction to PyTorch.

### Packages¶

We will have recourse to the following packages:

• torch, torch.nn, numpy (indispensables packages for neural networks with PyTorch)
• torch.optim (efficient gradient descents)
• PIL, PIL.Image, matplotlib.pyplot (load and display images)
• torchvision.transforms (treat PIL images and transform into torch tensors)
• torchvision.models (train or load pre-trained models)
• copy (to deep copy the models; system package)
from __future__ import print_function

import torch
import torch.nn as nn
import torch.nn.functional as F
import torch.optim as optim

from PIL import Image
import matplotlib.pyplot as plt

import torchvision.transforms as transforms
import torchvision.models as models

import copy

### Cuda¶

If you have a GPU on your computer, it is preferable to run the algorithm on it, especially if you want to try larger networks (like VGG). For this, we have torch.cuda.is_available() that returns True if you computer has an available GPU. Then, we can set the torch.device that will be used in this script. Then, we will use the method .to(device) that moves a tensor or a module to the desired device. When we want to move back this tensor or module to the CPU (e.g. to use numpy), we can use the .cpu() method.

device = torch.device("cuda" if torch.cuda.is_available() else "cpu")

In order to simplify the implementation, let’s start by importing a style and a content image of the same dimentions. We then scale them to the desired output image size (128 or 512 in the example, depending on gpu availablity) and transform them into torch tensors, ready to feed a neural network:

Note

Here are links to download the images required to run the tutorial: picasso.jpg and dancing.jpg. Download these two images and add them to a directory with name images

# desired size of the output image
imsize = 512 if torch.cuda.is_available() else 128  # use small size if no gpu

transforms.Resize(imsize),  # scale imported image
transforms.ToTensor()])  # transform it into a torch tensor

image = Image.open(image_name)
# fake batch dimension required to fit network's input dimensions
return image.to(device, torch.float)

assert style_img.size() == content_img.size(), \
"we need to import style and content images of the same size"

Imported PIL images have values between 0 and 255. Transformed into torch tensors, their values are between 0 and 1. This is an important detail: neural networks from torch library are trained with 0-1 tensor image. If you try to feed the networks with 0-255 tensor images the activated feature maps will have no sense. This is not the case with pre-trained networks from the Caffe library: they are trained with 0-255 tensor images.

### Display images¶

We will use plt.imshow to display images. So we need to first reconvert them into PIL images:

unloader = transforms.ToPILImage()  # reconvert into PIL image

plt.ion()

def imshow(tensor, title=None):
image = tensor.cpu().clone()  # we clone the tensor to not do changes on it
image = image.squeeze(0)      # remove the fake batch dimension
plt.imshow(image)
if title is not None:
plt.title(title)
plt.pause(0.001) # pause a bit so that plots are updated

plt.figure()
imshow(style_img, title='Style Image')

plt.figure()
imshow(content_img, title='Content Image')

### Content loss¶

The content loss is a function that takes as input the feature maps $$F_{XL}$$ at a layer $$L$$ in a network fed by $$X$$ and returns the weigthed content distance $$w_{CL}.D_C^L(X,C)$$ between this image and the content image. Hence, the weight $$w_{CL}$$ and the target content $$F_{CL}$$ are parameters of the function. We implement this function as a torch module with a constructor that takes these parameters as input. The distance $$\|F_{XL} - F_{YL}\|^2$$ is the Mean Square Error between the two sets of feature maps, that can be computed using a criterion nn.MSELoss stated as a third parameter.

We will add our content losses at each desired layer as additive modules of the neural network. That way, each time we will feed the network with an input image $$X$$, all the content losses will be computed at the desired layers and, thanks to autograd, all the gradients will be computed. For that, we just need to make the forward method of our module returning the input: the module becomes a ‘’transparent layer’’ of the neural network. The computed loss is saved as a parameter of the module.

Finally, we define a fake backward method that just calls the backward method of nn.MSELoss in order to reconstruct the gradient. This method returns the computed loss: this will be useful when running the gradient descent in order to display the evolution of style and content losses.

class ContentLoss(nn.Module):

def __init__(self, target,):
super(ContentLoss, self).__init__()
# we 'detach' the target content from the tree used
# to dynamically compute the gradient: this is a stated value,
# not a variable. Otherwise the forward method of the criterion
# will throw an error.
self.target = target.detach()

def forward(self, input):
self.loss = F.mse_loss(input, self.target)
return input

Note

Important detail: this module, although it is named ContentLoss, is not a true PyTorch Loss function. If you want to define your content loss as a PyTorch Loss, you have to create a PyTorch autograd Function and to recompute/implement the gradient by the hand in the backward method.

### Style loss¶

For the style loss, we need first to define a module that compute the gram produce $$G_{XL}$$ given the feature maps $$F_{XL}$$ of the neural network fed by $$X$$, at layer $$L$$. Let $$\hat{F}_{XL}$$ be the re-shaped version of $$F_{XL}$$ into a $$K$$x$$N$$ matrix, where $$K$$ is the number of feature maps at layer $$L$$ and $$N$$ the lenght of any vectorized feature map $$F_{XL}^k$$. The $$k^{th}$$ line of $$\hat{F}_{XL}$$ is $$F_{XL}^k$$. We let you check that $$\hat{F}_{XL} \cdot \hat{F}_{XL}^T = G_{XL}$$. Given that, it becomes easy to implement our module:

def gram_matrix(input):
a, b, c, d = input.size()  # a=batch size(=1)
# b=number of feature maps
# (c,d)=dimensions of a f. map (N=c*d)

features = input.view(a * b, c * d)  # resise F_XL into \hat F_XL

G = torch.mm(features, features.t())  # compute the gram product

# we 'normalize' the values of the gram matrix
# by dividing by the number of element in each feature maps.
return G.div(a * b * c * d)

The longer is the feature maps dimension $$N$$, the bigger are the values of the Gram matrix. Therefore, if we don’t normalize by $$N$$, the loss computed at the first layers (before pooling layers) will have much more importance during the gradient descent. We dont want that, since the most interesting style features are in the deepest layers!

Then, the style loss module is implemented exactly the same way than the content loss module, but it compares the difference in Gram matrices of target and input

class StyleLoss(nn.Module):

def __init__(self, target_feature):
super(StyleLoss, self).__init__()
self.target = gram_matrix(target_feature).detach()

def forward(self, input):
G = gram_matrix(input)
self.loss = F.mse_loss(G, self.target)
return input

Now, we have to import a pre-trained neural network. As in the paper, we are going to use a pretrained VGG network with 19 layers (VGG19).

PyTorch’s implementation of VGG is a module divided in two child Sequential modules: features (containing convolution and pooling layers) and classifier (containing fully connected layers). We are just interested by features: Some layers have different behavior in training and in evaluation. Since we are using it as a feature extractor. We will use .eval() to set the network in evaluation mode.

cnn = models.vgg19(pretrained=True).features.to(device).eval()

Additionally, VGG networks are trained on images with each channel normalized by mean=[0.485, 0.456, 0.406] and std=[0.229, 0.224, 0.225]. We will use them to normalize the image before sending into the network.

cnn_normalization_mean = torch.tensor([0.485, 0.456, 0.406]).to(device)
cnn_normalization_std = torch.tensor([0.229, 0.224, 0.225]).to(device)

# create a module to normalize input image so we can easily put it in a
# nn.Sequential
class Normalization(nn.Module):
def __init__(self, mean, std):
super(Normalization, self).__init__()
# .view the mean and std to make them [C x 1 x 1] so that they can
# directly work with image Tensor of shape [B x C x H x W].
# B is batch size. C is number of channels. H is height and W is width.
self.mean = torch.tensor(mean).view(-1, 1, 1)
self.std = torch.tensor(std).view(-1, 1, 1)

def forward(self, img):
# normalize img
return (img - self.mean) / self.std

A Sequential module contains an ordered list of child modules. For instance, vgg19.features contains a sequence (Conv2d, ReLU, MaxPool2d, Conv2d, ReLU…) aligned in the right order of depth. As we said in Content loss section, we wand to add our style and content loss modules as additive ‘transparent’ layers in our network, at desired depths. For that, we construct a new Sequential module, in which we are going to add modules from vgg19 and our loss modules in the right order:

# desired depth layers to compute style/content losses :
content_layers_default = ['conv_4']
style_layers_default = ['conv_1', 'conv_2', 'conv_3', 'conv_4', 'conv_5']

def get_style_model_and_losses(cnn, normalization_mean, normalization_std,
style_img, content_img,
content_layers=content_layers_default,
style_layers=style_layers_default):
cnn = copy.deepcopy(cnn)

# normalization module
normalization = Normalization(normalization_mean, normalization_std).to(device)

# just in order to have an iterable access to or list of content/syle
# losses
content_losses = []
style_losses = []

# assuming that cnn is a nn.Sequential, so we make a new nn.Sequential
# to put in modules that are supposed to be activated sequentially
model = nn.Sequential(normalization)

i = 0  # increment every time we see a conv
for layer in cnn.children():
if isinstance(layer, nn.Conv2d):
i += 1
name = 'conv_{}'.format(i)
elif isinstance(layer, nn.ReLU):
name = 'relu_{}'.format(i)
# The in-place version doesn't play very nicely with the ContentLoss
# and StyleLoss we insert below. So we replace with out-of-place
# ones here.
layer = nn.ReLU(inplace=False)
elif isinstance(layer, nn.MaxPool2d):
name = 'pool_{}'.format(i)
elif isinstance(layer, nn.BatchNorm2d):
name = 'bn_{}'.format(i)
else:
raise RuntimeError('Unrecognized layer: {}'.format(layer.__class__.__name__))

if name in content_layers:
target = model(content_img).detach()
content_loss = ContentLoss(target)
content_losses.append(content_loss)

if name in style_layers:
target_feature = model(style_img).detach()
style_loss = StyleLoss(target_feature)
style_losses.append(style_loss)

# now we trim off the layers after the last content and style losses
for i in range(len(model) - 1, -1, -1):
if isinstance(model[i], ContentLoss) or isinstance(model[i], StyleLoss):
break

model = model[:(i + 1)]

return model, style_losses, content_losses

Note

In the paper they recommend to change max pooling layers into average pooling. With AlexNet, that is a small network compared to VGG19 used in the paper, we are not going to see any difference of quality in the result. However, you can use these lines instead if you want to do this substitution:

# avgpool = nn.AvgPool2d(kernel_size=layer.kernel_size,

### Input image¶

Again, in order to simplify the code, we take an image of the same dimensions than content and style images. This image can be a white noise, or it can also be a copy of the content-image.

input_img = content_img.clone()
# if you want to use a white noise instead uncomment the below line:
# input_img = torch.randn(content_img.data.size(), device=device)

# add the original input image to the figure:
plt.figure()
imshow(input_img, title='Input Image')

As Leon Gatys, the author of the algorithm, suggested here, we will use L-BFGS algorithm to run our gradient descent. Unlike training a network, we want to train the input image in order to minimise the content/style losses. We would like to simply create a PyTorch L-BFGS optimizer optim.LBFGS, passing our image as the Tensor to optimize. We use .requires_grad_() to make sure that this image requires gradient.

def get_input_optimizer(input_img):
# this line to show that input is a parameter that requires a gradient
return optimizer

Last step: the loop of gradient descent. At each step, we must feed the network with the updated input in order to compute the new losses, we must run the backward methods of each loss to dynamically compute their gradients and perform the step of gradient descent. The optimizer requires as argument a “closure”: a function that reevaluates the model and returns the loss.

However, there’s a small catch. The optimized image may take its values between $$-\infty$$ and $$+\infty$$ instead of staying between 0 and 1. In other words, the image might be well optimized and have absurd values. In fact, we must perform an optimization under constraints in order to keep having right vaues into our input image. There is a simple solution: at each step, to correct the image to maintain its values into the 0-1 interval.

def run_style_transfer(cnn, normalization_mean, normalization_std,
content_img, style_img, input_img, num_steps=300,
style_weight=1000000, content_weight=1):
"""Run the style transfer."""
print('Building the style transfer model..')
model, style_losses, content_losses = get_style_model_and_losses(cnn,
normalization_mean, normalization_std, style_img, content_img)
optimizer = get_input_optimizer(input_img)

print('Optimizing..')
run = [0]
while run[0] <= num_steps:

def closure():
# correct the values of updated input image
input_img.data.clamp_(0, 1)

model(input_img)
style_score = 0
content_score = 0

for sl in style_losses:
style_score += sl.loss
for cl in content_losses:
content_score += cl.loss

style_score *= style_weight
content_score *= content_weight

loss = style_score + content_score
loss.backward()

run[0] += 1
if run[0] % 50 == 0:
print("run {}:".format(run))
print('Style Loss : {:4f} Content Loss: {:4f}'.format(
style_score.item(), content_score.item()))
print()

return style_score + content_score

optimizer.step(closure)

# a last correction...
input_img.data.clamp_(0, 1)

return input_img

Finally, run the algorithm

output = run_style_transfer(cnn, cnn_normalization_mean, cnn_normalization_std,
content_img, style_img, input_img)

plt.figure()
imshow(output, title='Output Image')

# sphinx_gallery_thumbnail_number = 4
plt.ioff()
plt.show()

Out:

Building the style transfer model..
Optimizing..
run [50]:
Style Loss : 4.125270 Content Loss: 4.188891

run [100]:
Style Loss : 1.140037 Content Loss: 3.030362

run [150]:
Style Loss : 0.709859 Content Loss: 2.655770

run [200]:
Style Loss : 0.478313 Content Loss: 2.500814

run [250]:
Style Loss : 0.350412 Content Loss: 2.411426

run [300]:
Style Loss : 0.266629 Content Loss: 2.359128

Total running time of the script: ( 0 minutes 37.587 seconds)

Gallery generated by Sphinx-Gallery


text:
# SQRTCBRT - Editorial

Author: kulyash
Testers: tabr, iceknight1093
Editorialist: iceknight1093

TBD

Binary search

# PROBLEM:

F(N) denotes the difference between the number of squares and the number of cubes from 1 to N.

Given X, find the smallest value of N such that F(N) \geq X.

# EXPLANATION:

Problems of the type "find the smallest N such that \ldots" are often solved with binary search.
Unfortunately, that requires the function to be monotonic, and F(N) here isn’t monotonic.

Analyzing how it changes, we can see that:

F(N) = \begin{cases} F(N-1), & \text{ if } N \text{ is neither a square nor a cube} \\ F(N-1) + 1, & \text{ if } N \text{ is a square but not a cube} \\ F(N-1) - 1, & \text{ if } N \text{ is a cube but not a square} \\ F(N-1), & \text{ if } N \text{ is both a square and a cube} \end{cases}

In particular, notice that F can only increase at a square number, so the final answer must be a square number.
Further, the fact that cubes are more ‘spread out’ than squares means that F is actually a monotonic function when evaluated on the squares!
That is, we consider the function g(N) = F(N^2) instead, which is increasing.

Proof

It’s enough to prove that there’s at most one cube lying between any two consecutive squares, since this is what makes F increasing on the squares.

So, suppose we have 1\leq x^2 \leq y^3 \leq (y+1)^3. We just need to prove (x+1)^2 \lt (y+1)^3.

This is not hard to do by simple algebraic manipulation.
Expand out (x+1)^2 and (y+1)^3, to obtain (x^2 + 2x + 1) and (y^3 + 3y + 3y^2 + 1).

Comparing their terms, we can see that:

• y^3 \geq x_2
• They both have a 1
• 3y^2 \geq 3x \gt 2x. This follows from the fact that x^2 \leq y^3 means that y^2 \geq x^{4/3} \geq x.

So we have the required inequality.

With this information in hand, the solution is simple: use binary search to find the smallest value of N such that g(N) \geq X; the final answer is then N^2.

This requires us to quickly evaluate g(N) = F(N^2).

Note that the number of squares that are \leq N^2 is simply N, so we just need to count the number of cubes that are \leq N^2.
There are several ways to do this: for example, use std::cbrtl or yet another binary search.

### Regarding precision

You may have encountered precision errors when computing cube roots; luckily, one of the samples was specifically designed to catch several common implementations.

In particular, cbrt in C++ or something like N ** (1 / 3) in Python have precision issues.

There are several ways to get around this, some being problem dependent:

• If possible, the absolute best solution is to simply not deal with floating-point integers at all. In this problem, the number of cubes \leq N can be computed using a binary search across the integers, so this is the safest way.
• cbrtl in C++ is precise enough to get AC in this task.
• The cube root can be manually adjusted once you compute it: if c is the value you compute as the cube root of N (after rounding) and (c+1)^3 \leq N, then increase c to c+1 instead.
• After computing the cube root, add a small value to it (like 10^{-10}) before rounding to an integer.

# TIME COMPLEXITY

\mathcal{O}(\log^2 X) per testcase.

# CODE:

Setter's code (C++)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int main() {
vector<ll>cubes;
for(ll i=1;i<=1010000;i++)cubes.push_back(i*i*i);
ll T;
cin >> T;
while(T--){
ll x;
cin >> x;
ll l=1;
ll r=2e9;
ll ans;
while(l<=r){
ll mid=(r+l)/2;
ll temp=upper_bound(cubes.begin(),cubes.end(),mid*mid)-cubes.begin();
ll curr=mid-temp;
if(curr>=x){
ans=mid*mid;
r=mid-1;
}
else{
l=mid+1;
}
}
cout << ans << endl;
}
return 0;
}

Tester's code (C++)
#include <bits/stdc++.h>
using namespace std;
#ifdef tabr
#include "library/debug.cpp"
#else
#define debug(...)
#endif

struct input_checker {
string buffer;
int pos;

const string all = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
const string number = "0123456789";
const string upper = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const string lower = "abcdefghijklmnopqrstuvwxyz";

input_checker() {
pos = 0;
while (true) {
int c = cin.get();
if (c == -1) {
break;
}
buffer.push_back((char) c);
}
}

int nextDelimiter() {
int now = pos;
while (now < (int) buffer.size() && buffer[now] != ' ' && buffer[now] != '\n') {
now++;
}
return now;
}

assert(pos < (int) buffer.size());
int nxt = nextDelimiter();
string res;
while (pos < nxt) {
res += buffer[pos];
pos++;
}
// cerr << res << endl;
return res;
}

string readString(int minl, int maxl, const string &pattern = "") {
assert(minl <= maxl);
assert(minl <= (int) res.size());
assert((int) res.size() <= maxl);
for (int i = 0; i < (int) res.size(); i++) {
assert(pattern.empty() || pattern.find(res[i]) != string::npos);
}
return res;
}

int readInt(int minv, int maxv) {
assert(minv <= maxv);
assert(minv <= res);
assert(res <= maxv);
return res;
}

long long readLong(long long minv, long long maxv) {
assert(minv <= maxv);
assert(minv <= res);
assert(res <= maxv);
return res;
}

assert((int) buffer.size() > pos);
assert(buffer[pos] == ' ');
pos++;
}

assert((int) buffer.size() > pos);
assert(buffer[pos] == '\n');
pos++;
}

assert((int) buffer.size() == pos);
}
};

int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
input_checker in;
while (tt--) {
long long low = 0, high = 2e9;
while (high - low > 1) {
long long mid = (high + low) >> 1;
long long cnt = mid;
long long t = cbrt(mid * mid);
while ((t + 1) * (t + 1) * (t + 1) <= mid * mid) {
t++;
}
while (t * t * t > mid * mid) {
t--;
}
cnt -= t;
if (cnt >= x) {
high = mid;
} else {
low = mid;
}
}
cout << high * high << '\n';
}
return 0;
}

Editorialist's code (Python)
from bisect import bisect_left
cubes = [i**3 for i in range(1, 1010000)]

for _ in range(int(input())):
x = int(input())
lo, hi = 1, 2 * 10**9
def g(n): # f(n*n)
return n - bisect_left(cubes, n*n + 1)
while lo < hi:
mid = (lo + hi)//2
if g(mid) >= x: hi = mid
else: lo = mid+1
print(lo * lo)

1 Like
#include<bits/stdc++.h>
using namespace std;
#define debug(x) ""
#define ll long long
#define int long long
#define ld long double
#define INF 998244353
#define MOD 1000000007
#define Case(test)cout<<"Case #"<<test<<": ";
#define y second
#define x first
#define mem(arr,val) memset(arr,val,sizeof(arr));
#define all(x) x.begin(),x.end()
ll max(ll a,ll b){return ((a>=b)?a:b);}
int maxself(int &var1,int var2){var1=max(var1,var2);return var1;}
ll min(ll a,ll b){return ((a<=b)?a:b);}
int minself(int &var1,int var2){var1=min(var1,var2);return var1;}

int f(int n)
{
int res=sqrt(n);
int a=(cbrt(n));
return res-a;
}

ll solve(int tt)
{
int n,res=0;
cin>>n;
int l=1,r=sqrt(1ll*2e18);
while(l<=r)
{
int mid=(l+r)>>1;
int m=1ll*mid*mid;
if(f(m)>=n)
{
res=m;
r=(mid)-1;
}
else
{
l=mid+1;
}
}
cout<<res<<endl;
return 0;
}

signed main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
ll t=1;
cin>>t;
for(int i=1;i<=t;i++)
{
solve(i);
}

return 0;
}

why my code give different results for x=3151 on my local ide than codechef ide

It is because when you are declaring the function f(m) , you are simply taking the cubrt and sqrt of m and subracting it .
Instead of that, you should also check if I increase or decrease the value of a by 1{mentioned in defination of f(m)}, whether it lie within it or not.
Like for eg . cubrt(x) = 2.9 then its roundoff will give 3 , which is an error , so in this case if you have check as i mentioned above , then code will run perfectly fine.

In the Setter’s solution,

how are these upper limits decided.? (1010000, 2e9)

They aren’t specific bounds, just large enough numbers.

For a given N, f(N^2) is quite close to N, since the number of cubes grows much slower than the number of squares.
So it’s reasonable to expect that R = 2\cdot 10^9 will definitely be enough for f(N^2) to exceed 10^9, which is the limit we care about.
You could reasonably take any number that’s somewhat larger than 10^9 as well.

1010000 is chosen similarly, it’s a bit larger than 10^6 which is approximately how many cubes you care about.
You could also take 2\cdot 10^6 and it’d be correct, anything works as long as it’s large enough.

thanks

format: plain


text:
# zbMATH — the first resource for mathematics

Summary: A generalization of the Wigner function for the case of a free particle with the “relativistic” Hamiltonian $$\sqrt{{\mathbf p}^2+ m^2}$$ is given.


text:
# Boundaries Of Free Splittings

25 Jul 2021

The other day on Twitter I asked whether it might be the case that a non-proper hyperbolic metric space might still have locally compact Gromov boundary. I was particularly interested in the case of a simplicial tree equipped with the path metric where each edge has length $$1$$ with some vertices of infinite valence. The answer in this case is no. Instead, the boundary of such a tree is homeomorphic to the Baire space $$\mathbb{N}^\mathbb{N}$$. In this post I’ll talk through why this is the case. There is also a notion of a Bowditch boundary for a tree with vertices of infinite valence. I’d like to show that the Bowditch boundary of such a tree is a Cantor set.

The trees we are interested in are Bass–Serre trees associated to free splittings of countable groups with the property that some vertex group is infinite. Let $$T$$ be such a tree. Then $$T$$ admits a cocompact group action, has countably infinitely many edges and has a vertex of countably infinite valence. We can give $$T$$ the structure of a metric space by considering the path metric where each edge is assigned length $$1$$. The “identity map” from $$T$$ as a simplicial complex to $$T$$ as a metric space is continuous, but its inverse is not. The property that $$T$$ admits a cocompact group action implies that any geodesic ray (isometric embedding of $$\mathbb{R}_{\ge 0}$$) passes through infinitely many vertices of infinite valence.

## The Gromov boundary of $$T$$

Fix a point $$p$$ in $$T$$. The Gromov boundary $$\partial_\infty T$$ of $$T$$ is the set of geodesic rays in $$T$$ beginning at $$p$$. (It turns out that the Gromov boundary is independent of the choice of basepoint.) Let $$J$$ be a (finite) geodesic segment in $$T$$ beginning at $$p$$. A basic open set $$U_J$$ for the topology on $$\partial_\infty T$$ is given by the set of geodesic rays that contain $$J$$ as a subpath. I’ll leave it to show you that the intersection of basic open sets is either empty or another basic open set.

## The Baire space

The Baire space $$\mathbb{N}^\mathbb{N}$$ is homeomorphic to the irrational numbers. A theorem of Alexandrov and Urysohn says that the Baire space is characterized up to homeomorphism by the following properties. It is a (nonempty) Polish space, meaning it is separable and completely metrizable. It is zero-dimensional, meaning it has a basis of clopen sets. Finally, every compact subset has empty interior.

To show that the Gromov boundary of $$T$$ is homeomorphic to the Baire space, we will verify these properties. It is completely metrizable by general theory. There are countably infinitely many finite geodesic segments in $$T$$ beginning at $$p$$ (one for each vertex of $$T$$), so $$\partial_\infty T$$ is second countable, and in particular separable. A basic closed set of $$\partial_\infty T$$ is given by the set of those geodesic rays that do not contain a geodesic segment $$J$$ as a subpath. It’s not hard to find a collection of geodesic segments such that any geodesic ray that does not contain any of these geodesic segments as a subpath must contain $$J$$ as a subpath. I’ll leave the details to you. Therefore $$\partial_\infty T$$ has a basis of clopen sets. To show that every compact subset has empty interior, it suffices to show that our basic clopen sets are never compact. To see this, continue $$J$$ until you reach a vertex of infinite valence, giving us a new finite geodesic segment $$J'$$. Write $$U_J \setminus U_{J'}$$ as a finite disjoint union of basic open sets, and $$U_{J'}$$ as the infinite disjoint union of the $$U_{J''}$$, where $$J''$$ is one edge longer than $$J'$$. The result is an open cover of $$U_J$$ with no finite subcover.

## The Bowditch boundary

The Bowditch boundary of $$T$$ is the set $$\partial T = \partial_\infty T \cup V_\infty(T)$$, where $$V_\infty(T)$$ denotes the set of vertices of $$T$$ with infinite valence. The Bowditch boundary carries with it the observer’s topology, which we now describe. Given two points $$q$$ and $$r$$ in $$T$$, the direction of $$r$$ at $$q$$ is the component of $$T\setminus \{q\}$$ that contains $$r$$. We continue to think of points of $$\partial_\infty T$$ as geodesic rays based at $$p$$; for convenience suppose that $$p$$ is either in the interior of an edge or a vertex with finite valence. We may additionally think of a point $$v \in V_\infty(T)$$ as being represented by the geodesic segment from $$p$$ to $$v$$. We say that a point of $$\partial T$$ belongs to a direction if the corresponding geodesic starting at $$p$$ is contained in the direction. The set of directions in $$T$$ forms a subbasis for the observer’s topology on $$\partial T$$; thus a basic open set is given by a finite intersection of directions of $$T$$. Put another way, a basic open set is given by the set of those geodesics based at $$p$$ that avoid a given finite subset of $$T$$. We can always take this finite set to be a set of vertices or midpoints of edges, so the observer’s topology is second countable.

We’d like to show that the Bowditch boundary of $$T$$ is a Cantor set. It is clear that $$\partial T$$ has no isolated points. It is compact and Hausdorff by general theory. Thus it suffices to show that our basis elements are clopen. We will show that our open basic open sets are compact. Indeed, let $$U$$ be a basic open set determined by the finite set of points $$\{z_1,\dotsc,z_n\} \subset T$$, and let $$\{U_{\alpha}\}_{\alpha\in I}$$ be an open cover of $$U$$ by basic open sets, which are determined by finite sets of points containing $$\{z_1,\dotsc, z_n\}$$ as a subset. Take $$U_\alpha \in I$$. Then there are finitely many points $$z_{n+1},\dotsc,z_{n+k}$$ determining $$U_\alpha$$ and contained in $$U$$. For each $$z_{n+i}$$ for $$i$$ satisfying $$1 \le i \le k$$, we can find $$\alpha_i \in I$$ such that $$U_{\alpha_i}$$ contains $$z_{n+i}$$. Then $$U_\alpha,U_{\alpha_1},\dotsc,U_{\alpha_k}$$ is a finite subcover of $$U$$.

## A parallel example

Let the Cantor space be $$\{0,1\}^{\mathbb{N}}$$. The subspace of sequences which are eventually constant is countable and dense. The subspace of sequences which are not eventually constant is homeomorphic to the Baire space. This is more or less what is happening here: we add countably many points to our non-locally compact Gromov boundary to get the compact Bowditch boundary.


text:
Minimum Depth of Binary Tree

LeetCode 111

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

Example

return its minimum depth = 2.


text:
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 Ask Your Question

# Running total/cumulative sum of time? [closed]

Attached is an example document to help elucidate my question:

C:\fakepath\Running total:cumulative sum of time example.ods

In this document I want to use a formula to make a running total of the time logged. However, I don't want to convert the hours into minutes; I would like the figures to remain HH/MM/SS.

Obviously when the minutes reach 60, they should be converted into an hour, then reset, and when the seconds reach 60, they should be converted into a minute, then reset.

I would prefer that this could be done with the hours, minutes and seconds divided into 3 cells as shown, but if not possible, then placing all within the same cell would be acceptable.

How can I achieve this?

Cheers.

-EDIT- I have received an answer from JohnSUN that works. I am leaving this thread open a little longer for the purpose of further discussion, but I am satisfied with this solution. Will also take a look at some of the other answers when I get more time, thanks to all who have contributed.

Here is the updated document, with the problem solved using JohnSUN's directions:

C:\fakepath\Running total; cumulative sum of time example UPDATED W: JOHNSUN'S ANSWER.ods

edit retag reopen merge delete

## 2 Answers

Sort by » oldest newest most voted

You can use a simple formula for converting the hours-minutes-seconds into one value and format this value as desired

more

## Comments

I formatted the "running total" F column cells using the time format shown in your answer.

I then tried using the formula you gave for cell D2: "=A2+(B2+C2/60)/60)/24", but in doing so was confronted with this message:

"LibreOffice Calc found an error in the formula entered. Do you want to accept the correction proposed below? =A2+(B2+C2/60)/60/24"

When "no" was selected, I got an "Err:508" message. When "yes" was selected, the result was 96:38:14

-continued below-

( 2017-05-17 12:31:32 +0200 )edit

I am not sure why I got a different result to you. So, instead of using a formula for cell D2, I tried typing the time in manually: 04:38:14.

I then used the other formulas you gave for cells D3 and D4, which worked perfectly.

You have solved my problem, thank you very, very much.

Will leave this thread open for further discussion regarding the discrepancy between my result in cell D2 and yours, however I am pretty satisfied with the outcome.

( 2017-05-17 13:10:29 +0200 )edit

Please check the @JohnSUN's answer closely: there is an opening parenthesis before A2 that you missed.

( 2017-05-17 13:29:30 +0200 )edit

I don't see in your variant of the formula the first opening bracket before A2: =(A2+(B2+C2/60)/60)/24

( 2017-05-17 13:30:20 +0200 )edit

Whoops, I must've had my eye on the formula in E3, of course there's an opening parenthesis! Sorry fellas.

Aaaand JohnSUN's answer worked perfectly. Have updated my question accordingly.

Just one more question for JohnSUN though, when you say "format this value as desired"- how exactly would I go about spreading the converted time in the "running total" column into 3 separate sections, similar to the "TIME (reading)" column?

( 2017-05-17 13:56:17 +0200 )edit

You can use functions =MINUTE(D2) and =SECOND(D2) for additional columns. With the amount of hours the formula is a little more complicated =INT(D2)*24+HOUR(D2) (or just format cells in column D with code [HH])

( 2017-05-17 14:31:48 +0200 )edit

At first a question: For what reason are the durations logged in three parts (h, min, s) instead of one standard time value?
As long as you keep time values based on the default unit of 1 day (d) there is not the least problem to get your cumulative sum. You can simply add such values as you would add values of any other kind of quantity.

Ok. You have to enter durations as if they are TOD values.
Most users wouldn't care or not even regard the difference. In addition you may display the results in a format better appropriate for durations like done by the format code H" h "MM" min "SS" s".

Actually calculating the 3-column-TimeValues cannot be done by a single "formula". You either need helper columns where the "logged" numbers are converted to default time values and then added. Based on these results you can convert back to seperate results in 3 columns using mathematical functions. Or you calculate the sums starting with the least significant part (s) and looking for the carry yourself. This cannot be done (imo) by the same formula in the three columns. as an example =F1+C2+IF(CURRENT()>=60;-60;0) in F2 (seconds) should do what you want. But as long as you do not dedicate a column to the carry, you need the rather complicated =E1+B2+ROUNDUP((F1+C2-F2)/60)+IF(CURRENT()>=60;-60;0) (or something similar) in E3. ...

(Editing:)
See also this demo. (A fluke led me back, and I noticed that I had attached a bad version. The new one should do better.)
Consider to regard the difference between "time" as TimeOfDay and "time" as duration (or as a scientific / technical kind of quantity). The difference is essential, though often ignored.
Science strongly prefers to express time values in one single unit per application case: Either only seconds (s) or only years in practise, rarely hours.

The mixed format used for TOD (and angles) goes back to an about 5000 years old tradition to subdivide and to bunch by 60 which was abandoned for most applications also some millennia ago. Why? It's simply too unhandy as soon as you need real calculations.

more

## Comments

While I agree with usefulness of the single-unit approach, I wanted to mention that the different units aren't something abandoned in most applications, as well as something conceptually alien. First, the positional number representation itself is the sample of such concept (each digit place representing its own "unit" - in decimal, it's singles, tens, hundreds etc); second, there are a lot of places like heights (feet+inch) or weights (lb+oz) etc which are common now.

( 2017-05-14 08:10:36 +0200 )edit

Another (and much more important) thing to note is that the help in LO is created by volunteers, just as LO itself, and just like this ask's answers. So, each time you add an answer that you think worth of being reflected in help, and don't actually propose a patch to help, it's a sad thing happening.

( 2017-05-14 08:21:10 +0200 )edit

@Mike Kaganski: Being a developer you surely know that first of all functions need be clearly specified. The next step must be a correct implementation. With respect to the TIME function problems start here. The specification explicitly states in its last paragraph: "Hours, minutes, and seconds may be any number (they shall not be limited to the ranges 0..24, 0..59, or 0..60 respectively).
The implementation obviously ignores this. I wouldn't ennoble the bug anchoring it in the help.

( 2017-05-14 14:19:35 +0200 )edit

@Mike Kaganski (again): With respect to some statements of yours above, it might be preferable to have a private exchange. Since this site does not offer PM, I might (mis-?)use the mailing via the bugs site if you don't object. My true name is Wolfgang Jäger, and I am known by it to the https://bugs.documentfoundation.org/ .

( 2017-05-14 14:28:59 +0200 )edit

@Lupp: my email is mikekaganski@hotmail.com; I'd like to discuss whatever you'd like to :)

( 2017-05-14 15:03:04 +0200 )edit

## Stats

Asked: 2017-05-13 18:08:12 +0200

Seen: 1,323 times

Last updated: May 17 '17


text:
Mathias Brandewinder on .NET, F#, VSTO and Excel development, and quantitative analysis / machine learning.
25. August 2013 08:54

About a month ago, FSharp.Data  released version 1.1.9, which contains some very nice improvements – you can find them listed on Gustavo Guerra’s blog. I was particularly excited by the changes made to the CSV Type Provider, because they make my life digging through datasets even simpler, but couldn’t find the time to write about it, because of my recent cross-country peregrinations.

Now that I am back, let’s talk about why this update made me so happy, with a concrete example. My latest week-end project is an F# implementation of Random Forests; as part of the process, I am trying out the algorithm on various datasets, to get a sense for potential performance problems, and dog-food my own API, the best way I know to quickly spot suckiness.

One of the problems I ran into was the representation of missing values. Most datasets don’t come clean and ready to use – usually you’ll have a few records with missing data. I opted for what seemed the most straightforward representation in F#, and decided to represent every feature value as an Option – anything can either have Some value, or None.

The original CSV Type Provider introduced a bit of friction there, because it inferred types “optimistically”: if the sample used contained only integers, it would create an integer, which is great in most cases, except when you want to be “pessimistic” (which is usually a safe world-view when setting expectations regarding data).

The new-and-improved CSV Type Provider fixes that, and introduces a few niceties. Case in point: the Kaggle Titanic dataset, which contains the Titanic’s passenger list. With the new version, extracting the data is as simple as this:

type DataSet = CsvProvider<"titanic.csv",
Schema="PassengerId=int, Pclass->Class, Parch->ParentsOrChildren, SibSp->SiblingsOrSpouse",
SafeMode=true,
PreferOptionals=true>

type Passenger = DataSet.Row

This is pretty awesome. In a couple of lines, just by passing in the path to my CSV file and some (optional) schema information, I get a Passenger type:

What’s neat here is that first, I immediately get a Passenger with properties – with the correct Optional types, thanks to SafeMode and PreferOptional. Then, notice in the Schema the Pclass->Class, Parch->ParentsOrChildren, SibSp->SiblingsOrSpouse bit? This renames “on the fly” the properties; instead of the pretty obscurely named Parch feature coming from the CSV file header, I get a nice and readable ParentsOrChildren property. The Type Provider even does a few more cool things, automagically; for instance, the feature “Survived”, which is encoded in the original dataset as 0 or 1, gets automatically converted to a boolean. Really nice.

And just like that, I can now use this CSV file, and send it to my (still very much in alpha version) Decision Tree classifier:

// We read the training set into an array,
// defining the Label we want to classify on:
let training =
use data = new DataSet()
[| for passenger in data.Data ->
passenger.Survived |> Categorical, // the label
passenger |]
// We define what features should be used:
let features = [|
"Sex", (fun (x:Passenger) -> x.Sex |> Categorical);
"Class", (fun x -> x.Class |> Categorical); |]
// We run the classifier...
let classifier, report = createID3Classifier training features { DefaultID3Config with DetailLevel = Verbose }
// ... and display the resulting tree:
report.Value.Pretty()

… which produces the following results in the F# Interactive window:

> titanicDemo();;
├ Sex = male
│   ├ Class = 3 → False
│   ├ Class = 1 → False
│   └ Class = 2 → False
└ Sex = female
├ Class = 3 → False
├ Class = 1 → True
└ Class = 2 → True
val it : unit = ()
>

The morale of the story here is triple. First, it was a much better idea to be a rich lady on the Titanic, rather than a (poor) dude. Then, Type Providers are really awesome – in a couple of lines, we extracted from a CSV file a collection of Passengers, all of them statically typed, with all the benefits attached to that; in a way, this is the best of both worlds – access the data as easily as with a dynamic language, but with all the benefits of types. Finally, the F# community is just awesome – big thanks to everyone who contributed to FSharp.Data, and specifically to @ovatsus for the recent improvements to the CSV Type Provider!

You can find the full Titanic example here on GitHub.

#### Need help with F#?

The premier team for
F# training & consulting.


text:
# Convexe and serie

1. Feb 19, 2016

### Calabi

Hello every one, let be V an open convex a a normed vectorial space E.
Let be $(a_{n}) \in \mathbb{R}^{n}$ with $\sum_{i \in \mathbb{N}} a_{i} = 1$.
Let be $(v_{n}) \in V^{\mathbb{N}}$ as $\sum_{i \in \mathbb{N}} a_{i}v_{i}$ exists.
Does necessarly $\sum_{i \in \mathbb{N}} a_{i}v_{i} \in V$ please?

Thank you in advance and have a nice afternoon.

2. Feb 19, 2016

### Calabi

I suppose a positive.
In fact for my need I can suppose $a_{n} = \frac{1}{2^{n+1}}$.

3. Feb 19, 2016

### mathman

Your original statement needs proofreading. I assume you meant V an open convex set (but I'm not sure). Also $V^N$? In any case, the sum is not necessarily in V, if it exists, but in the closure of V. The sum existence depends on whether on not E is complete.

4. Feb 20, 2016

### Samy_A

Here is a theorem about open convex sets that might help (based on theorem 3.4 in Rudin's Functional Analysis).

Suppose $X$ is a topological vector space over $\mathbb R$, $A$ and $B$ disjoint, nonempty, convex subsets of $X$ with $A$ open, then there exists a continuous linear function $f: X \to \mathbb R$ and a real number $\gamma$ such that $f(a) \lt \gamma \leq f(b)$ for all $a \in A,\ b\in B$.

This should help answer the question.

For reference, the theorem from the book:

Last edited: Feb 20, 2016
5. Feb 20, 2016

### Calabi

No I suppose my sum exist . And I clearly wroght V is an open convex.

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text: GREEK NUMERALS are a system of writing numbers using the letters of the Greek alphabet Greek alphabet. These alphabetic numerals are also known as IONIC or IONIAN NUMERALS, MILESIAN NUMERALS, or ALEXANDRIAN NUMERALS. In modern Greece Greece, they are still used for ordinal numbers and in contexts similar to those in which Roman numerals Roman numeralsare still used elsewhere in the West. For ordinary cardinal numbers , however, Greece uses Arabic numerals Arabic numerals. CONTENTS * 1 History * 2 Description * 3 Table * 4 Higher numbers * 5 Zero * 6 See also * 7 References * 8 External links HISTORYThe Minoan and Mycenaean civilizations ' Linear A Linear Aand Linear B alphabets used a different system, called Aegean numerals, which included specialized symbols for numbers: 𐄇 = 1, 𐄐 = 10, 𐄙 = 100, 𐄢 = 1000, and 𐄫 = 10000. Attic numerals, which were later adopted as the basis for Roman numerals , were the first alphabetic set. They were acrophonic , derived (after the initial one) from the first letters of the names of the numbers represented. They ran = 1, = 5, = 10, = 100, = 1000, and = 10000. 50, 500, 5000, and 50000 were represented by the letter with minuscule powers of ten written in the top right corner: , , , and . The same system was used outside of Attica Attica, but the symbols varied with the local alphabets : in Boeotia Boeotia, was 1000. The present system probably developed around Miletus Miletusin Ionia. 19th-century classicists placed its development in the 3rd century BC, the occasion of its first widespread use. More thorough modern archaeology has caused the date to be pushed back at least to the 5th century BC, a little before Athens abandoned its pre-Euclidean alphabet in favor of Miletus Miletus's in 402 BC, and it may predate that by a century or two. The present system uses the 24 letters adopted by Euclid Euclidas well as three Phoenician and Ionic ones that were not carried over: digamma , koppa , and sampi . The position of those characters within the numbering system imply that the first two were still in use (or at least remembered as letters) while the third was not. The exact dating, particularly for sampi, is problematic since its uncommon value means the first attested representative near Miletus Miletusdoes not appear until the 2nd century BC and its use is unattested in Athens until the 2nd century AD. (In general, Athens resisted the use of the new numerals for the longest of any of the Greek states but had fully adopted them by AD c. 50. ) DESCRIPTION _ Greek numerals Greek numeralsin a c. 1100 Byzantine manuscript of Hero of Alexandria Alexandria's Metrika_. The first line contains the number "͵θϡϟϛ δʹ ϛʹ", i.e. " 9996 4⁄6". It features each of the special numeral symbols sampi (ϡ), koppa (ϟ), and stigma (ϛ) in their minuscule forms. _ A 14th-century Byzantine map of the British Isles from a manuscript of Ptolemy Ptolemy's Geography_ , using Greek numerals for its graticule : 52–63°N of the equator and 6–33°E from Ptolemy's Prime Meridian Prime Meridianat the Fortunate Isles. Greek numerals Greek numeralsare decimal , based on powers of 10. The units from 1 to 9 are assigned to the first nine letters of the old Ionic alphabet from alpha to theta . Instead of reusing these numbers to form multiples of the higher powers of ten, however, each multiple of ten from 10 to 90 was assigned its own separate letter from the next nine letters of the Ionic alphabetfrom iota to koppa . Each multiple of one hundred from 100 to 900 was then assigned its own separate letter as well, from rho to sampi . (The fact that this was _not_ the traditional location of sampi or its possible predecessor san has led classicists to conclude that it was no longer in use even locally by the time the system was created.) This alphabetic system operates on the additive principle in which the numeric values of the letters are added together to obtain the total. For example, 241 was represented as (200 + 40 + 1). (It was not always the case that the numbers ran from highest to lowest: a 4th-century BC inscription at Athens placed the units to the left of the tens. This practice continued in Asia Minor Asia Minorwell into the Roman period . ) In ancient and medieval manuscripts, these numerals were eventually distinguished from letters using overbars : α, β, γ, etc. In medieval manuscripts of the Book of Revelation Book of Revelation, the number of the Beast 666 is written as χξϛ (600 + 60 + 6). (Numbers larger than 1,000 reused the same letters but included various marks to note the change.) Although the Greek alphabet Greek alphabetbegan with only majuscule forms, surviving papyrus manuscripts from Egypt show that uncial and cursive minuscule forms began early. These new letter forms sometimes replaced the former ones, especially in the case of the obscure numerals. The old Q-shaped koppa (Ϙ) began to be broken up ( and ) and simplified ( and ). The numeral for 6 changed several times. During antiquity, the original letter form of digamma ( ) came to be avoided in favor of a special numerical one ( ). By the Byzantine era , the letter was known as episemon and written as or . This eventually merged with the sigma -tau ligature stigma ( or ). In modern Greek , a number of other changes have been made. Instead of extending an overbar over an entire number, the _KERAIA_ (κεραία, lit. "hornlike projection") is marked to its upper right, a development of the short marks formerly used for single numbers and fractions. The modern _keraia_ is a symbol (ʹ) similar the acute accent (´) but has its own Unicode Unicodecharacter as U+0374. Alexander the Great Alexander the Great's father Philip II of Macedon Philip II of Macedonis thus known as Φίλιππος Βʹ in modern Greek. A lower left _keraia_ (Unicode: U+0375, "Greek Lower Numeral Sign") is now standard for distinguishing thousands: 2015 is represented as ͵ΒΙΕʹ (2000 + 10 + 5). The declining use of ligatures in the 20th century also means that stigma is frequently written as the separate letters ΣΤʹ, although a single _keraia_ is used for the group. The art of assigning Greek letters also being thought of as numerals and therefore giving words/names/phrases a numeric sum that has meaning through being connected to words/names/phrases of similar sum is called isopsephy (gematria ). TABLE ANCIENT BYZANTINE MODERN VALUE ANCIENT BYZANTINE MODERN VALUE ANCIENT BYZANTINE MODERN VALUE ANCIENT BYZANTINE MODERN VALUE α Αʹ 1 ι Ιʹ 10 ρ Ρʹ 100 ">β Βʹ 2 κ Κʹ 20 σ Σʹ 200 ͵β ͵Β 2000 γ Γʹ 3 λ Λʹ 30 τ Τʹ 300 ͵ ͵Γ 3000 δ Δʹ 4 μ Μʹ 40 υ Υʹ 400 ͵ ͵Δ 4000 ε Εʹ 5 ν Νʹ 50 φ Φʹ 500 ͵ε ͵Ε 5000 & ">ξ Ξʹ 60 χ Χʹ 600 ͵ & ͵ ͵ ">ζ Ζʹ 7 ο Οʹ 70 ψ Ψʹ 700 ͵ζ ͵Z 7000 η Ηʹ 8 π Πʹ 80 ω Ωʹ 800 ͵η ͵H 8000 θ Θʹ 9 & & Ϟʹ 90 & & & & "> M o {displaystyle {stackrel {rho oepsilon }{mathrm {M} }}} ͵εωοεʹ for 1,755,875.HIGHER NUMBERSIn his text _ The Sand Reckoner_, the natural philosopher Archimedes gives an upper bound of the number of grains of sand required to fill the entire universe, using a contemporary estimation of its size. This would defy the then-held notion that it is impossible to name a number greater than that of the sand on a beach or on the entire world. In order to do that, he had to devise a new numeral scheme with much greater range. ZERO Example of the early Greek symbol for zero (lower right corner) from a 2nd-century papyrus Hellenistic astronomers extended alphabetic Greek numerals Greek numeralsinto a sexagesimal positional numbering system by limiting each position to a maximum value of 50 + 9 and including a special symbol for zero , which was also used alone like today's modern zero, more than as a simple placeholder. However, the positions were usually limited to the fractional part of a number (called minutes , seconds, thirds, fourths, etc.) — they were not used for the integral part of a number. This system was probably adapted from Babylonian numerals Babylonian numeralsby Hipparchusc. 140 BC. It was then used by Ptolemy Ptolemy(c. 140), Theon (c. 380) and Theon's daughter Hypatia Hypatia(murdered 415). In Ptolemy\'s table of chords , the first fairly extensive trigonometric table, there were 360 rows, portions of which looked as follows: '  o {displaystyle {begin{array}{ccc}pi varepsilon varrho iota varphi varepsilon varrho varepsilon iota {tilde {omega }}nu &varepsilon {overset {text{'}}{nu }}vartheta varepsilon iota {tilde {omega }}nu &{overset {text{}}{varepsilon }}xi eta kappa mathrm {o} sigma tau {tilde {omega }}nu \{begin{array}{l}hline pi delta angle '\pi varepsilon \pi varepsilon angle '\hline pi mathrm {stigma} \pi mathrm {stigma} angle '\pi zeta \hline end{array}}&{begin{array}{rrr}hline pi &mu alpha &gamma \pi alpha &delta &iota varepsilon \pi alpha &kappa zeta &kappa beta \hline pi alpha &nu &kappa delta \pi beta &iota gamma &iota vartheta \pi beta &lambda mathrm {stigma} &vartheta \hline end{array}}&{begin{array}{rrrr}hline circ &circ &mu mathrm {stigma} &kappa varepsilon \circ &circ &mu mathrm {stigma} &iota delta \circ &circ &mu mathrm {stigma} &gamma \hline circ &circ &mu varepsilon &nu beta \circ &circ &mu varepsilon &mu \circ &circ &mu varepsilon width:51.526ex; height:26.176ex;" alt="{begin{array}{ccc}pi varepsilon varrho iota varphi varepsilon varrho varepsilon iota {tilde {omega }}nu &varepsilon {overset {text{}}{nu }}vartheta varepsilon iota {tilde {omega }}nu &{overset {text{`}}{varepsilon }}xi eta kappa mathrm {o} sigma tau {tilde {omega }}nu \{begin{array}{l}hline pi delta angle \pi varepsilon \pi varepsilon angle \hline pi mathrm {stigma} \pi mathrm {stigma} angle \pi zeta \hline end{array}}&{begin{array}{rrr}hline pi &mu alpha &gamma \pi alpha &delta &iota varepsilon \pi alpha &kappa zeta &kappa beta \hline pi alpha &nu &kappa delta \pi beta &iota gamma &iota vartheta \pi beta &lambda mathrm {stigma} &vartheta \hline end{array}}&{begin{array}{rrrr}hline circ &circ &mu mathrm {stigma} &kappa varepsilon \circ &circ &mu mathrm {stigma} &iota delta \circ &circ &mu mathrm {stigma} &gamma \hline circ &circ &mu varepsilon &nu beta \circ &circ &mu varepsilon &mu \circ &circ &mu varepsilon however, there was no ambiguity, as 70 could not appear in the fractional part of a number, and zero was usually omitted when it was the integer. Some of Ptolemy's true zeros appeared in the first line of each of his eclipse tables, where they were a measure of the angular separation between the center of the Moon Moonand either the center of the Sun Sun(for solar eclipses ) or the center of Earth Earth's shadow (for lunar eclipses ). All of these zeros took the form 0 0 0, where Ptolemy actually used three of the symbols described in the previous paragraph. The vertical bar () indicates that the integral part on the left was in a separate column labeled in the headings of his tables as _digits_ (of five arc-minutes each), whereas the fractional part was in the next column labeled _minute of immersion_, meaning sixtieths (and thirty-six-hundredths) of a digit. SEE ALSO * Attic numerals * Gematria * Greek numerals Greek numeralsin Unicode Unicode(acrophonic, not alphabetic, numerals) * Isopsephy * Number of the BeastREFERENCES * ^ _A_ _B_ Samuel Verdan (20 Mar 2007). "Systèmes numéraux en Grèce ancienne: description et mise en perspective historique" (in French). Retrieved 2 Mar 2011. * ^ _A_ _B_ _C_ Heath, Thomas L. _A Manual of Greek Mathematics_, pp. 14 ff. Oxford Univ. Press (Oxford), 1931. Reprinted Dover (Mineola ), 2003. Accessed 1 November 2013. * ^ Thompson, Edward M. _Handbook of Greek and Latin Palaeography_, p. 114. D. Appleton (New York), 1893. * ^ The Packard Humanities Institute (Cornell & Ohio State Universities). _Searchable Greek Inscriptions_: "IG I³ 1387" . Accessed 1 November 2013. * ^ Jeffery, Lilian H. _The Local Scripts of Archaic Greece_, pp. 38 ff. Clarendon (Oxford), 1961. * ^ The Packard Humanities Institute (Cornell & Ohio State Universities). _Searchable Greek Inscriptions_: "Magnesia 4" . Accessed 1 November 2013. * ^ The Packard Humanities Institute (Cornell & Ohio State Universities). _Searchable Greek Inscriptions_: "IG II² 2776". Accessed 1 November 2013. * ^ Edkins, Jo (2006). "Classical Greek Numbers". Retrieved 29 Apr 2013. * ^ Nick Nicholas (9 Apr 2005). "Numerals: Stigma, Koppa, Sampi". Retrieved 2 Mar 2011. * ^ Neugebauer, Otto (1969) . _The Exact Sciences in Antiquity_ (2 ed.). Dover Publications. pp. 13–14, plate 2. ISBN 978-0-486-22332-2 . * ^ Raymond Mercier, "Consideration of the Greek symbol \'zero\'" (PDF). (1.32 MiB ) Numerous examples * ^ _Ptolemy's Almagest Almagest_, translated by G. J. Toomer, Book VI, (Princeton, NJ: Princeton University Press, 1998), pp. 306–7.EXTERNAL LINKS _ Wikimedia Commons has media related to GREEK NUMERALS _. * The Greek Number Converter * v * t * e ORIGIN AND GENEALOGY *


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# Math Help - Set up integral

1. ## Set up integral

Hey I am trying to do this problem:
Setup an integral to find the volume for the region in the first octant that touches the
xy plane and is bounded by the plane that passes through the points (1, 2, 0), (0, 0, 1)
and (0, 0, 0) and the plane that passes through the points (1, 2, 0), (0, 0, 2) and (3, 0, 0).
You do not need to integrate.

I wasn't sure what to do so I found the equation of the two planes. For the first plane(first 3 points) I got y = 2x(Is this equation for a plane possible since there is a point at (0,0,1)? Then for the second I got z = -(2/3)x - (2/3)y.
I'm not sure what to do now though as I cant see how to find the bounds for all three variables in order to set up the triple integral.
Any help would be appreciated

2. ## Re: Set up integral

I agree that finding the equations of the planes is a good idea. Remembering that a plane of the form \displaystyle \begin{align*} ax + by + cz = d \end{align*} has the same coefficients as its normal vector, we can get the normal vector by finding two vectors that lie on the plane and taking their cross product.

Two vectors that lie on the first plane are \displaystyle \begin{align*} \mathbf{u}_1 = <1 - 0, 2 - 0, 0 - 0> = <1, 2, 0> \end{align*} and \displaystyle \begin{align*} \mathbf{v}_1 = <0 - 0, 0 - 0, 1 - 0> = <0, 0, 1> \end{align*}. So the first normal vector is found by

\displaystyle \begin{align*} \mathbf{n}_1 &= \mathbf{u}_1 \times \mathbf{v}_1 \\ &= \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 0 \\ 0 & 0 & 1 \end{matrix} \right| \\ &= 0 \left| \begin{matrix} \mathbf{j} & \mathbf{k} \\ 2 & 0 \end{matrix} \right| - 0 \left| \begin{matrix} \mathbf{i} & \mathbf{k} \\ 1 & 0 \end{matrix} \right| + 1 \left| \begin{matrix} \mathbf{i} & \mathbf{j} \\ 1 & 2 \end{matrix} \right| \\ &= 2\mathbf{i} - \mathbf{j} \\ &= <2, -1, 0> \end{align*}

So your first plane will have equation \displaystyle \begin{align*} 2x - y + 0z = d \end{align*}

Substituting in any of the points, such as (0, 0, 0), finds that \displaystyle \begin{align*} d = 0 \end{align*} and so the equation of your plane is \displaystyle \begin{align*} 2x - y + 0z = 0 \end{align*}, which as you already determined can be written as \displaystyle \begin{align*} y = 2x \end{align*}. So really this is a line in the x-y plane, that is then shifted vertically indefinitely in every direction. So as long as you have any two points which satisfy \displaystyle \begin{align*} y = 2x \end{align*}, then any value of z will be fine. So yes, this equation of your plane is possible and (0, 0, 1) lies on it.

I disagree with your calculation of the second plane. You have the points (1, 2, 0), (0, 0, 2) and (3, 0, 0) lying on it.

Two vectors which lie on the plane are \displaystyle \begin{align*} \mathbf{u}_2 = <1-0, 2-0, 0-2> = <1, 2, -2> \end{align*} and \displaystyle \begin{align*} \mathbf{v}_2 = <3-0, 0-0, 0-2> = <3, 0, -2> \end{align*}. So the normal vector is given by

\displaystyle \begin{align*} \mathbf{n}_2 &= \mathbf{u}_2 \times \mathbf{v}_2 \\ &= \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \phantom{-}\mathbf{k} \\ 1 & 2 & -2 \\ 3 & 0 & -2 \end{matrix} \right| \\ &= 3 \left| \begin{matrix} \mathbf{j} & \phantom{-}\mathbf{k} \\ 2 & -2 \end{matrix} \right| - 0 \left| \begin{matrix} \mathbf{i} & \phantom{-}\mathbf{k} \\ 1 & -2 \end{matrix} \right| + (-2) \left| \begin{matrix} \mathbf{i} & \mathbf{j} \\ 1 & 2 \end{matrix} \right| \\ &= 3 \left( -2\mathbf{j} - 2\mathbf{k} \right) - 2 \left( 2\mathbf{i} - \mathbf{j} \right) \\ &= -6\mathbf{j} - 6\mathbf{k} - 4\mathbf{i} + 2\mathbf{j} \\ &= -4\mathbf{i} - 4\mathbf{j} - 6\mathbf{k} \\ &= <-4, -4, -6> \end{align*}

So the equation of your plane will be \displaystyle \begin{align*} -4x - 4y - 6z = d \end{align*}, and substituting a point on the plane, like (0, 0, 2), gives \displaystyle \begin{align*} d = -12 \end{align*}. So the equation of your plane is

\displaystyle \begin{align*} -4x - 4y - 6z &= -12 \\ 2x + 2y + 3z &= 6 \\ 3z &= 6 - 2x - 2y \\ z &= 2 - \frac{2}{3}x - \frac{2}{3}y \end{align*}

Now it would be a good idea to draw your region, so that you can decide which variables would be easiest to integrate over first.

3. ## Re: Set up integral

Thanks for your help. I drew the region in the xy plane and came up with, in order of dzdydx: z(from 0 to 2-(2/3)x-(2/3)y) y(from 3-x to 2x) and x(from 0 to 1). Can anyone verify this?

4. ## Re: Set up integral

Originally Posted by sdsu619
Thanks for your help. I drew the region in the xy plane and came up with, in order of dzdydx: z(from 0 to 2-(2/3)x-(2/3)y) y(from 3-x to 2x) and x(from 0 to 1). Can anyone verify this?
Your z limits are correct for collecting the volume under the 'roof' which is z = 2 - (2/3)x - (2/3)y. But surely the 'floor' is a triangle going between 0 and 3 in the x direction...? If you do choose x as a first direction of travel, then you'll need to split the area of the floor in two separate integrals...

$\int_0^1\ 2x\ dx\ \equiv\ \int_0^1\ \int_{0}^{2x}\ 1\ dy\ dx$

and

$\int_1^3\ 3 - x\ dx\ \equiv\ \int_1^3\ \int_0^{3 - x}\ 1\ dy\ dx$

Which is fine, but you may prefer to travel y-wards first, between 0 and 2, during which you can integrate the area of the floor in one go, from strips of height x = [3 - y] - [(1/2)y] = 3 - (3/2)y...

$\int_0^2\ 3 - \tfrac{3}{2}y\ dy\ \equiv\int_0^2\ \int_0^{3 - \tfrac{3}{2}y}\ 1\ dx\ dy$

5. ## Re: Set up integral

Ah ok i get it I was looking at the wrong region. thanks for the input


text:
Authors: Braunstein, Matthew; Bonnet, Laurent; Roncero, Octavio

Journal: PHYSICAL CHEMISTRY CHEMICAL PHYSICS

Publication date: 2022/03/02

DOI: 10.1039/d1cp04244k

Abstract: We present quasi-classical trajectory (QCT) cross sections, rate constants, and product state distributions for the D + H-3(+) -> H2D+ + H reaction. Using the same H-4(+) potential surface, the rate constants obtained from several QCT-based methods correcting for zero-point effects by Gaussian binning the product H2D+ are compared to ring polymer molecular dynamics (RPMD) rate constants [Bulut et al., J. Phys. Chem. A, 2019, 123, 8766] which include quantum effects and to recent experimentally derived rate constants [Bowen et al., J. Chem. Phys., 2021, 154, 084307]. QCT with standard binning predicts rate constants that increase slowly as the temperature decreases from 1500 to 100 K. In contrast, the RPMD rate constants decrease rapidly with decreasing temperature. By 100 K, the QCT standard binning rate constant is more than 3 orders of magnitude larger than the RPMD rate constant. We show that QCT with Gaussian binning and proper normalization captures the zero-point effects and reproduces the RPMD rate constants over a large temperature range. Furthermore, the simple technique of counting only reactive trajectories with vibrational energy above the product zero-point energy matches the RPMD results well down to similar to 300 K. The present Gaussian binned rate constants are in fair agreement with new experimentally derived rate constants from 100 to 1500 K. However, because the Gaussian binned rate constants do not include tunneling, important at lower temperatures, and the RPMD and experimentally derived rate constants have significant differences, the roles of the competing effects of zero-point energy, internal excitation of the H-3(+), and quantum tunneling are not simple and require further study for a consistent picture of the dynamics. Since rate constants for complex forming reactions, such as the title reaction, are difficult to converge with RPMD, alternative QCT-based methods, which include quantum effects and in addition provide product state distributions as described here, are highly desirable.


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Isomerism in 1D Coordination Polymers of Cu(II) Complexes of N-(2-Hydroxybenzyl)-l-valine: Influence of Solvent and Coordination Sphere on the Conformation. Crystal Growth & Design 2004, 4 • Browse Jobs https://www.icbse.com/projects/subject/chemistry • Worksheet / Test Paper https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-apxlkwaays-5186.pdf Aldehyde-radical reactions are important in atm. and combustion chem., and the reactions studied here also serve more generally to illustrate a fundamental aspect of chem. kinetics that has been relatively unexplored from a quant. point of view, in particular the roles of multiple structures and torsional anharmonicity in detg. the rate consts. and branching ratios (product yields). We consider hydrogen abstraction from four carbon sites of butanal (carbonyl-C, α-C, β-C and γ-C) by hydroperoxyl radical. We employed multi-structural variational transition state theory for studying the first three channels; this uses a multi-faceted dividing surface and allows us to include the contributions of multiple structures of both reacting species and transition states. Multi-configurational Shepard interpolation (MCSI) was used to obtain the geometries and energies of the potential energy surface along the min.-energy paths, with gradients and Hessians calcd. by the M08-HX/maug-cc-pVTZ method. We find the nos. of structures obtained for the transition states are 46, 60, 72 and 76, resp. for the H abstraction at the carbonyl C, the α position, the β position and the γ position. Our results show that neglecting the factors arising from multiple structures and torsional anharmonicity would lead to errors at 300, 1000 and 2400 K of factors of 8, 11 and 10 for abstraction at the carbonyl-O, 2, 11 and 25 at the α-C position, 2, 23 and 47 at the β-C position, and 0.6, 8 and 18 at the γ-C position. The errors would be even larger at high temp. for the reverse of the H abstraction at the β-C. Relative yields are changed as much as a factor of 7.0 at 200 K, a factor of 5.0 at 298 K, and a factor of 3.7 in the other direction at 2400 K. The strong dependence of the product ratios on the multi-structural anharmonicity factors shows that such factors play an important role in controlling branching ratios in reaction mechanism networks. Rewriting a manuscript requires reading other writers and learning from them new ways of using the language—unthinkable ways, uncomfortable ways, unlikely ways. This is a quick list of what I took away from Joseph Heller’s novel Catch-22. , 7060-7063. https://doi.org/10.1039/C7CC03333H https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-kqjaxronqx-8581.pdf https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-ndegmokpyn-5742.pdf • Absorption spectroscopy (2000) • 2 Factors affecting reaction rate A closed system cannot contemplate itself,(intelligently). 1595 ... https://www.bookdepository.com/Darker-Shade-Magic-Collectors-Edition-V-E-Schwab/9780765399113 https://www.theguardian.com/books/2016/mar/24/fiction-nonfiction-english-literature-culture-writers-other-languages-stories , 1740-1745. https://doi.org/10.1002/zaac.201300199 Chemistry World A marine biologist developed a race of genetically engineered dolphins that could live forever if they were fed a steady diet of seagulls. One day his supply of the birds ran out, so he had to go out and trap some more. On the way back, he spied two lions asleep on the road. Afraid to wake them, he gingerly stepped over them. 2 Funk JO, Maibach HI. Propylene glycol dermatitis: re-evaluation of an old problem. Contact Dermatitis. 1994;31(4):236–41. Claude Cohen-Tannoudji 1933- • chemistry ki padhai kaise kare • Biology (Class 11, 12) • Receive Alerts/Updates , 1909-1911. https://doi.org/10.1021/ic051992i Why have I been blocked? Some new graduates may extend their education as a way to weather the recession. Joseph Meany was supposed to graduate with his bachelor’s degree from Keene State College in 2008, but because of the recession, he took additional classes and graduated the following year. “I ended up coming 1,500-odd mi [2,414 km] down to Alabama, and I started a PhD program in January of 2010. It was not my first choice to go to graduate school, but in retrospect I’m glad that I did.” In 2016, Meany graduated from the University of Alabama with his PhD. He is now a postdoc at Savannah River National Laboratory. https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-wvxyrwfckp-3187.pdf introduced Lorentz transformation equations of alkoxy- • See also the dissolution of sodium chloride is explosive https://americanenglish.state.gov/resources/adventures-tom-sawyer https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-wjucqxutdq-6130.pdf • My Jobs • Ελληνικά 337 • Related Tools • (4) • Bo Xu, Xiaoxiao Ji, Yong Cai, Ling Li, Guangning Liu, Cuncheng Li. Structure and luminescent property of a zinc(II) complex assembled from 5-methylisophthalic acid and 1,2-bis-(4-pyridyl) ethane. Journal of Molecular Structure 2014, 1056-1057 , 52-55. https://doi.org/10.1016/j.molstruc.2013.10.031 https://www.britishcouncil.org/voices-magazine/brief-history-modern-polish-literature https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-kvlpjjjbgb-4450.pdf A U. of Maryland graduate student in physics works on an experiment. Only a minority of physics undergraduates go on to study the subject in graduate school. Most majors end up in other fields, a reality the discipline is working to prepare them for. https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-hfoteqyfjh-7637.pdf • flag myMalaysia Enforce all laboratory policies and practices and that all in-use procedures are current and ... • Anti-Fatigue Mats https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-cnpazbrjor-4270.pdf https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-hugolwvean-2650.pdf https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-miquhcoutj-5123.pdf This website is using a security service to protect itself from online attacks. The action you just performed triggered the security solution. There are several actions that could trigger this block including submitting a certain word or phrase, a SQL command or malformed data. , 3652-3660. https://doi.org/10.1021/cg8002564 Clip from the film Organometallics Physics and Engineering Principles of Multi-modality Molecular Imaging of Living Subjects • Bar Accessories • biological catalysts https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-jyicacnohp-5323.pdf • Examination (91 KB) * Copyright © 2020 Insider Inc. All rights reserved. • , Shuming Chen • Carboxylic • Study Centers in Qatar • Business https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-knsvfuuxnd-5586.pdf https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-onyhizcoga-2164.pdf Density - Need to solve density, mass or volume? You'll find all necessary equations and an explanation on how to find the right answers. Maximum Height • Environmental Management • Extra Crispy this link opens in a new tab developed methods, with his colleagues, https://www.scholastic.com/teachers/articles/teaching-content/interview-anthony-horowitz/ आवर्त सारणी में अभी कुल 118 तत्व है. आवर्त सारणी अभी तक के जितने भी रासायनिक तत्वों की खोज की जा चुकी है और कुछ संभावित खोज वाले तत्वों को एक स्थान में रखी गयी व्यवस्था है , जिसमे 7 आवर्त एवं 18 वर्ग या समूह है। आवर्त सारणी में तत्वों को उसके परमाणु क्रमांक, इलेक्ट्रानिक विन्यास के बढ़ते क्रम में व्यवस्थित किया गया है तथा इसके भौतिक एवं रासायनिक गुणों को भी ध्यान में रखते हुए आवर्त, समूह तथा कई भागों ब्लॉक्स में वर्गीकृत किया गया है। 20 h 19 minutes ago in Explorejobs 10 +12 https://www.cliffsnotes.com/literature/p/the-pilgrims-progress/about-the-pilgrims-progress Rate = k((CH 3) 3CBr) $x_{cm} =\frac {m_1x_1 + m_2x_2 + m_3x_3 + .... + m_nx_n}{m_1 + m_2 + m_3 + ..... + m_n}$ • Risma Remsudeen’s Chemistry IA (23/24) • First Impressions User Polls • Catch-basin • Chemical Laws 34 , 11297-11299. https://doi.org/10.1021/ic201919q • Hayato Ohwaki, Nobuto Yoshinari, Takumi Konno. Crystal-to-crystal interconversion of open and closed dicopper( https://www.booktopia.com.au/outlander-diana-gabaldon/book/9781784751371.html • ^ Levin, Gary (January 14, 2018). "George Clooney is in a 'Catch-22' on Hulu (and now you can watch 'ER' there, too)". USA Today . Retrieved January 14, 2018. But it's something I do worry about. When I develop characters, I really put in my all. I write for myself but I also write for readers my characters may represent. https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-qorcfurakb-8736.pdf REGENTS Classical Electrodynamics Lab Analyst • Find Companies Now we.re making bad chemistry jokes because all the good ones Argon B. C6H12O6 CR) • flag euEuropean Union https://www.masterclass.com/articles/what-are-the-different-genres-of-literature-a-guide-to-14-literary-genres Cycloalkanes and aromatic compounds can be treated as the main parent chain of the compound, in which case the positions of substituents are numbered around the ring structure. For example, the three isomers of xylene CH 3C 6H 4CH 3, commonly the ortho-, meta-, and para- forms, are 1,2-dimethylbenzene, 1,3-dimethylbenzene, and 1,4-dimethylbenzene. The cyclic structures can also be treated as functional groups themselves, in which case they take the prefix "cyclo alkyl-" (e.g. "cyclohexyl-") or for benzene, "phenyl-". , 599-604. https://doi.org/10.1021/cg0342415 https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-mleewtaaer-8031.pdf https://www.sciencedirect.com/topics/social-sciences/educational-reform • "Chemical kinetics and Reaction Dynamics", Paul L. Houston, Dover Press, NY (2006). Teenage cancer isn’t a subject that is easy to write about and it’s even harder to do it without resorting to clichés. John Green however avoids any sentimentality and creates a vibrant, stimulating mediation on life, death and all the big questions with intelligent, well-drawn characters and compulsive storytelling. Witty and life-affirming, this young adult romance is heartbreaking in the very best sense of the word. https://www.bustle.com/p/which-haruki-murakami-book-should-you-read-first-heres-a-guide-to-his-most-famous-works-7606716 Q. What is the syllabus of JEE Main Paper 1, Paper 2 and Paper 3? • Randy W. Larsen, Jaroslava Miksovska, Ronald L. Musselman, and Lukasz Wojtas . Ground- and Excited-State Properties of Zn(II) Tetrakis(4-tetramethylpyridyl) Pophyrin Specifically Encapsulated within a Zn(II) HKUST Metal–Organic Framework. The Journal of Physical Chemistry A 2011, 115 , 62. https://doi.org/10.3390/ijerph13010062 http://vivo.med.cornell.edu/display/journal1946-6234 • Bo Li, Hui-Tao Fan, Wen-Chao Song, Chao-Qiang Feng, Shuang-Quan Zang, Li-Ya Wang. Two Coordination Polymers Based on 3-(Carboxymethoxy)benzene-1,2-dioic Acid and N-donor Ancillary Ligands: Synthesis, Crystal structures, and Magnetic Properties. Zeitschrift für anorganische und allgemeine Chemie 2014, 640 https://www.bookdepository.com/Brief-History-Time-Stephen-Hawking/9780857501004 • Justice • Yong Cui,, Helen L. Ngo, and, Wenbin Lin. New Rigid Angular Dicarboxylic Acid for the Construction of Nanoscopic Supramolecules: From a Molecular Rectangle to a 1-D Coordination Polymer. Inorganic Chemistry 2002, 41 • Central Singapore (13) • oil chemist Singapore https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-qnnpermiyf-4820.pdf URL: https://www.youtube.com/watch?v=dK3ReB4lsfU • AIIMS Neon Lights & Other Discharge Lamps • Answer Booklet • Archaeology Result • CrystEngComm https://salimrokhaya95.files.wordpress.com/2020/06/e-book-pdf-oisomcgflk-8127.pdf Computer Science Engineering Colleges in India • © 2020 Condé Nast. All rights reserved. Use of this site constitutes acceptance of our User Agreement (updated 1/1/20) and Privacy Policy and Cookie Statement (updated 1/1/20) and Your California Privacy Rights. Do Not Sell My Personal Information GQ may earn a portion of sales from products that are purchased through our site as part of our Affiliate Partnerships with retailers. The material on this site may not be reproduced, distributed, transmitted, cached or otherwise used, except with the prior written permission of Condé Nast. Ad Choices The first woman to win a Nobel and the first person to win two separate Nobels, Curie (1867-1934) was born in Poland and won her first Nobel in 1903 with husband, Pierre, for discovering radioactivity. However, she was not allowed to participate in the keynote lecture winners give because she was a woman. After Pierre died in a road accident in 1906, she won her second Nobel in 1911 for discovering radium, though an attempt was made to rescind it when news emerged of her affair with married colleague Paul Langevin. After collecting the prize, Curie was pilloried by the French press. Langevin was ignored. • See All Other Courses • Binomial theorem for a positive integral index,