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https://journal.nsps.org.ng/index.php/jnsps/article/view/589
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# Modified Gradient Flow Method for Solving One-Dimensional Optimal Control Problem Governed by Linear Equality Constraint
## Authors
• Olusegun Olotu Department of Mathematical Sciences,The Federal University of Technology Akure, Nigeria
• Charles Aladesaye Dept. of Mathematics, School of Science, College of Education, Ikere-Ekiti, Ekiti State, Nigeria
• Kazeem Adebowale Dawodu Department of Mathematical Sciences,The Federal University of Technology Akure, Nigeria
## Keywords:
Optimal Control, Gradient Flow, three-level splitting parameters, discretization scheme, linear and quadratic convergence
## Abstract
This study presents a computational technique developed for solving linearly constraint optimal control problems using the Gradient Flow Method. This proposed method, called the Modified Gradient Flow Method (MGFM), is based on the continuous gradient flow reformulation of constrained optimization problem with three-level implicit time discretization scheme. The three-level splitting parameters for the discretization of the gradient flow equations are such that the sum of the parameters equal to one (\theta1 + \theta2 +\theta3=1). The Linear and quadratic convergence of the scheme were analyzed and were shown to have first order scheme when each parameter exist in the domain [0, 1] and second order when the third parameter equal to one. Numerical experiments were carried out and the results showed that the approach is very effective for handling this class of constrained optimal control problems. It also compared favorably with the analytical solutions and performed better than the existing schemes in terms of convergence and accuracy
Dimensions
A. I. Adekunle, “Algorithm for a Class of Discretized Optimal Control Problems", M.Tech. Thesis, Federal University of Technology, Akure, Nigeria (2011) (Unpublished).
O. C. Akeremale, “Optimization of Quadratic Constrained Optimal Control Problems using Augumented Lagrangian Method", M.Tech. Thesis, Federal University of Technology, Akure, Nigeria (2012) (Unpublished)
W. Behrman, “An Effcient Gradient Flow Method for Unconstrained Optimization", PhD Thesis, Stanford University (1998) (Unpublished).
J. T. Betts, “Practical Methods for Optimal Control Problem Using Non linear programming", SIAM, Philadelphia, (2001).
Y. Evtushenko, “Generalized Lagrange Multipliers Technique for Nonlinear Programming", JOTA, 21 (1977) 121. DOI: https://doi.org/10.1007/BF00932516
Y. G. Evtushenko & V. G. Zhadan, “Stable Barrier Projection and Barrier Newton Methods", Nonlinear Programming Optimization Methods and Software, 3 (1994) 237. DOI: https://doi.org/10.1080/10556789408805567
G. T. Gilbert, “Positive Definite Matrices and Sylvester’s Criterion", The American Mathematical Monthly, Taylor & Francis, 98 (1991) 44. DOI: https://doi.org/10.2307/2324036
W. M. Haddad, S. G. Nersesov & V.S. Chellaboina, “Lyapunov Function Proof of Poincare’s Theorem", International Journal of Systems Science, 35 (2004) 287. DOI: https://doi.org/10.1080/00207720410001714824
J. B. Layton, “Efficient direct computatioion of the Pseudo-inverse and its gradient", International Journal for Numerical Methods in Engineering, 40 (1997) 4211. DOI: https://doi.org/10.1002/(SICI)1097-0207(19971130)40:22<4211::AID-NME255>3.0.CO;2-8
W. H. Morris, S. Stephen & L. D. Robert, “Di erential Equations, Dynamical Systems, and an Introduction to Chaos", Elsevier Academic Press, USA (2004) 194.
O. Olotu & K. A. Dawodu, “Quasi-Newton Embedded Augmented Lagrangian Algorithm for Discretized Optimal Proportional Control Problems", Journal of Mathematical Theory and Modeling, 3 (2013a) 67.
O. Olotu & K. A. Dawodu, “On the Discretized Algorithm for Optimal Proportional Control Problems Constrained by Delay Differential Equations", Journal of Mathematical Theory and Modeling, 3 (2013b) 157.
O. Olotu & S. A. Olorunsola, “An Algorithm for a Discretized Constrained, Continuous Quadratic Control Problem", Journal of Applied Sciences, 8 (2006) 6249. DOI: https://doi.org/10.4314/jonamp.v8i1.40017
L. S. Pontryagin, V. G. Boltyanskii, R. V. Gamkrelidze & E. F. Mishchenko, “The Mathematical Theory of Optimal Processes", Interscience Publishers, London (1962).
S. Wang, X. Q. Yang K. L. Teo “A Unified Gradient Flow Approach to Constrained Nonlinear Optimization Problems", Computational Optimization and Applications, 25 (2003) 251. DOI: https://doi.org/10.1023/A:1022973608903
2022-02-28
## How to Cite
Olotu, O., Aladesaye, C. ., & Dawodu, K. A. (2022). Modified Gradient Flow Method for Solving One-Dimensional Optimal Control Problem Governed by Linear Equality Constraint. Journal of the Nigerian Society of Physical Sciences, 4(1), 146–156. https://doi.org/10.46481/jnsps.2022.589
## Section
Original Research
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.. DO NOT EDIT. .. THIS FILE WAS AUTOMATICALLY GENERATED BY SPHINX-GALLERY. .. TO MAKE CHANGES, EDIT THE SOURCE PYTHON FILE: .. "auto_examples/inspection/plot_linear_model_coefficient_interpretation.py" .. LINE NUMBERS ARE GIVEN BELOW. .. only:: html .. note:: :class: sphx-glr-download-link-note Click :ref:here to download the full example code or to run this example in your browser via Binder .. rst-class:: sphx-glr-example-title .. _sphx_glr_auto_examples_inspection_plot_linear_model_coefficient_interpretation.py: ================================================================== Common pitfalls in interpretation of coefficients of linear models ================================================================== In linear models, the target value is modeled as a linear combination of the features (see the :ref:linear_model User Guide section for a description of a set of linear models available in scikit-learn). Coefficients in multiple linear models represent the relationship between the given feature, :math:X_i and the target, :math:y, assuming that all the other features remain constant (conditional dependence _). This is different from plotting :math:X_i versus :math:y and fitting a linear relationship: in that case all possible values of the other features are taken into account in the estimation (marginal dependence). This example will provide some hints in interpreting coefficient in linear models, pointing at problems that arise when either the linear model is not appropriate to describe the dataset, or when features are correlated. We will use data from the "Current Population Survey" _ from 1985 to predict wage as a function of various features such as experience, age, or education. .. contents:: :local: :depth: 1 .. GENERATED FROM PYTHON SOURCE LINES 30-39 .. code-block:: default print(__doc__) import numpy as np import scipy as sp import pandas as pd import matplotlib.pyplot as plt import seaborn as sns .. GENERATED FROM PYTHON SOURCE LINES 40-46 The dataset: wages ------------------ We fetch the data from OpenML _. Note that setting the parameter as_frame to True will retrieve the data as a pandas dataframe. .. GENERATED FROM PYTHON SOURCE LINES 46-51 .. code-block:: default from sklearn.datasets import fetch_openml survey = fetch_openml(data_id=534, as_frame=True) .. GENERATED FROM PYTHON SOURCE LINES 52-55 Then, we identify features X and targets y: the column WAGE is our target variable (i.e., the variable which we want to predict). .. GENERATED FROM PYTHON SOURCE LINES 55-58 .. code-block:: default X = survey.data[survey.feature_names] X.describe(include="all") .. raw:: html
EDUCATION SOUTH SEX EXPERIENCE UNION AGE RACE OCCUPATION SECTOR MARR
count 534.000000 534 534 534.000000 534 534.000000 534 534 534 534
unique NaN 2 2 NaN 2 NaN 3 6 3 2
top NaN no male NaN not_member NaN White Other Other Married
freq NaN 378 289 NaN 438 NaN 440 156 411 350
mean 13.018727 NaN NaN 17.822097 NaN 36.833333 NaN NaN NaN NaN
std 2.615373 NaN NaN 12.379710 NaN 11.726573 NaN NaN NaN NaN
min 2.000000 NaN NaN 0.000000 NaN 18.000000 NaN NaN NaN NaN
25% 12.000000 NaN NaN 8.000000 NaN 28.000000 NaN NaN NaN NaN
50% 12.000000 NaN NaN 15.000000 NaN 35.000000 NaN NaN NaN NaN
75% 15.000000 NaN NaN 26.000000 NaN 44.000000 NaN NaN NaN NaN
max 18.000000 NaN NaN 55.000000 NaN 64.000000 NaN NaN NaN NaN
.. GENERATED FROM PYTHON SOURCE LINES 59-62 Note that the dataset contains categorical and numerical variables. We will need to take this into account when preprocessing the dataset thereafter. .. GENERATED FROM PYTHON SOURCE LINES 62-65 .. code-block:: default X.head() .. raw:: html
EDUCATION SOUTH SEX EXPERIENCE UNION AGE RACE OCCUPATION SECTOR MARR
0 8.0 no female 21.0 not_member 35.0 Hispanic Other Manufacturing Married
1 9.0 no female 42.0 not_member 57.0 White Other Manufacturing Married
2 12.0 no male 1.0 not_member 19.0 White Other Manufacturing Unmarried
3 12.0 no male 4.0 not_member 22.0 White Other Other Unmarried
4 12.0 no male 17.0 not_member 35.0 White Other Other Married
.. GENERATED FROM PYTHON SOURCE LINES 66-68 Our target for prediction: the wage. Wages are described as floating-point number in dollars per hour. .. GENERATED FROM PYTHON SOURCE LINES 68-71 .. code-block:: default y = survey.target.values.ravel() survey.target.head() .. rst-class:: sphx-glr-script-out Out: .. code-block:: none 0 5.10 1 4.95 2 6.67 3 4.00 4 7.50 Name: WAGE, dtype: float64 .. GENERATED FROM PYTHON SOURCE LINES 72-77 We split the sample into a train and a test dataset. Only the train dataset will be used in the following exploratory analysis. This is a way to emulate a real situation where predictions are performed on an unknown target, and we don't want our analysis and decisions to be biased by our knowledge of the test data. .. GENERATED FROM PYTHON SOURCE LINES 77-84 .. code-block:: default from sklearn.model_selection import train_test_split X_train, X_test, y_train, y_test = train_test_split( X, y, random_state=42 ) .. GENERATED FROM PYTHON SOURCE LINES 85-90 First, let's get some insights by looking at the variable distributions and at the pairwise relationships between them. Only numerical variables will be used. In the following plot, each dot represents a sample. .. _marginal_dependencies: .. GENERATED FROM PYTHON SOURCE LINES 90-95 .. code-block:: default train_dataset = X_train.copy() train_dataset.insert(0, "WAGE", y_train) _ = sns.pairplot(train_dataset, kind='reg', diag_kind='kde') .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_001.png :alt: plot linear model coefficient interpretation :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 96-115 Looking closely at the WAGE distribution reveals that it has a long tail. For this reason, we should take its logarithm to turn it approximately into a normal distribution (linear models such as ridge or lasso work best for a normal distribution of error). The WAGE is increasing when EDUCATION is increasing. Note that the dependence between WAGE and EDUCATION represented here is a marginal dependence, i.e., it describes the behavior of a specific variable without keeping the others fixed. Also, the EXPERIENCE and AGE are strongly linearly correlated. .. _the-pipeline: The machine-learning pipeline ----------------------------- To design our machine-learning pipeline, we first manually check the type of data that we are dealing with: .. GENERATED FROM PYTHON SOURCE LINES 115-118 .. code-block:: default survey.data.info() .. rst-class:: sphx-glr-script-out Out: .. code-block:: none RangeIndex: 534 entries, 0 to 533 Data columns (total 10 columns): # Column Non-Null Count Dtype --- ------ -------------- ----- 0 EDUCATION 534 non-null float64 1 SOUTH 534 non-null category 2 SEX 534 non-null category 3 EXPERIENCE 534 non-null float64 4 UNION 534 non-null category 5 AGE 534 non-null float64 6 RACE 534 non-null category 7 OCCUPATION 534 non-null category 8 SECTOR 534 non-null category 9 MARR 534 non-null category dtypes: category(7), float64(3) memory usage: 17.2 KB .. GENERATED FROM PYTHON SOURCE LINES 119-130 As seen previously, the dataset contains columns with different data types and we need to apply a specific preprocessing for each data types. In particular categorical variables cannot be included in linear model if not coded as integers first. In addition, to avoid categorical features to be treated as ordered values, we need to one-hot-encode them. Our pre-processor will - one-hot encode (i.e., generate a column by category) the categorical columns; - as a first approach (we will see after how the normalisation of numerical values will affect our discussion), keep numerical values as they are. .. GENERATED FROM PYTHON SOURCE LINES 130-143 .. code-block:: default from sklearn.compose import make_column_transformer from sklearn.preprocessing import OneHotEncoder categorical_columns = ['RACE', 'OCCUPATION', 'SECTOR', 'MARR', 'UNION', 'SEX', 'SOUTH'] numerical_columns = ['EDUCATION', 'EXPERIENCE', 'AGE'] preprocessor = make_column_transformer( (OneHotEncoder(drop='if_binary'), categorical_columns), remainder='passthrough' ) .. GENERATED FROM PYTHON SOURCE LINES 144-146 To describe the dataset as a linear model we use a ridge regressor with a very small regularization and to model the logarithm of the WAGE. .. GENERATED FROM PYTHON SOURCE LINES 146-161 .. code-block:: default from sklearn.pipeline import make_pipeline from sklearn.linear_model import Ridge from sklearn.compose import TransformedTargetRegressor model = make_pipeline( preprocessor, TransformedTargetRegressor( regressor=Ridge(alpha=1e-10), func=np.log10, inverse_func=sp.special.exp10 ) ) .. GENERATED FROM PYTHON SOURCE LINES 162-166 Processing the dataset ---------------------- First, we fit the model. .. GENERATED FROM PYTHON SOURCE LINES 166-169 .. code-block:: default _ = model.fit(X_train, y_train) .. GENERATED FROM PYTHON SOURCE LINES 170-173 Then we check the performance of the computed model plotting its predictions on the test set and computing, for example, the median absolute error of the model. .. GENERATED FROM PYTHON SOURCE LINES 173-193 .. code-block:: default from sklearn.metrics import median_absolute_error y_pred = model.predict(X_train) mae = median_absolute_error(y_train, y_pred) string_score = f'MAE on training set: {mae:.2f} $/hour' y_pred = model.predict(X_test) mae = median_absolute_error(y_test, y_pred) string_score += f'\nMAE on testing set: {mae:.2f}$/hour' fig, ax = plt.subplots(figsize=(5, 5)) plt.scatter(y_test, y_pred) ax.plot([0, 1], [0, 1], transform=ax.transAxes, ls="--", c="red") plt.text(3, 20, string_score) plt.title('Ridge model, small regularization') plt.ylabel('Model predictions') plt.xlabel('Truths') plt.xlim([0, 27]) _ = plt.ylim([0, 27]) .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_002.png :alt: Ridge model, small regularization :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 194-208 The model learnt is far from being a good model making accurate predictions: this is obvious when looking at the plot above, where good predictions should lie on the red line. In the following section, we will interpret the coefficients of the model. While we do so, we should keep in mind that any conclusion we draw is about the model that we build, rather than about the true (real-world) generative process of the data. Interpreting coefficients: scale matters --------------------------------------------- First of all, we can take a look to the values of the coefficients of the regressor we have fitted. .. GENERATED FROM PYTHON SOURCE LINES 208-222 .. code-block:: default feature_names = (model.named_steps['columntransformer'] .named_transformers_['onehotencoder'] .get_feature_names(input_features=categorical_columns)) feature_names = np.concatenate( [feature_names, numerical_columns]) coefs = pd.DataFrame( model.named_steps['transformedtargetregressor'].regressor_.coef_, columns=['Coefficients'], index=feature_names ) coefs .. raw:: html
Coefficients
RACE_Hispanic -0.013564
RACE_Other -0.009120
RACE_White 0.022549
OCCUPATION_Clerical 0.000048
OCCUPATION_Management 0.090531
OCCUPATION_Other -0.025098
OCCUPATION_Professional 0.071967
OCCUPATION_Sales -0.046633
OCCUPATION_Service -0.091050
SECTOR_Construction -0.000180
SECTOR_Manufacturing 0.031273
SECTOR_Other -0.031008
MARR_Unmarried -0.032405
UNION_not_member -0.117154
SEX_male 0.090808
SOUTH_yes -0.033823
EDUCATION 0.054699
EXPERIENCE 0.035005
AGE -0.030867
.. GENERATED FROM PYTHON SOURCE LINES 223-235 The AGE coefficient is expressed in "dollars/hour per living years" while the EDUCATION one is expressed in "dollars/hour per years of education". This representation of the coefficients has the benefit of making clear the practical predictions of the model: an increase of :math:1 year in AGE means a decrease of :math:0.030867 dollars/hour, while an increase of :math:1 year in EDUCATION means an increase of :math:0.054699 dollars/hour. On the other hand, categorical variables (as UNION or SEX) are adimensional numbers taking either the value 0 or 1. Their coefficients are expressed in dollars/hour. Then, we cannot compare the magnitude of different coefficients since the features have different natural scales, and hence value ranges, because of their different unit of measure. This is more visible if we plot the coefficients. .. GENERATED FROM PYTHON SOURCE LINES 235-241 .. code-block:: default coefs.plot(kind='barh', figsize=(9, 7)) plt.title('Ridge model, small regularization') plt.axvline(x=0, color='.5') plt.subplots_adjust(left=.3) .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_003.png :alt: Ridge model, small regularization :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 242-253 Indeed, from the plot above the most important factor in determining WAGE appears to be the variable UNION, even if our intuition might tell us that variables like EXPERIENCE should have more impact. Looking at the coefficient plot to gauge feature importance can be misleading as some of them vary on a small scale, while others, like AGE, varies a lot more, several decades. This is visible if we compare the standard deviations of different features. .. GENERATED FROM PYTHON SOURCE LINES 253-263 .. code-block:: default X_train_preprocessed = pd.DataFrame( model.named_steps['columntransformer'].transform(X_train), columns=feature_names ) X_train_preprocessed.std(axis=0).plot(kind='barh', figsize=(9, 7)) plt.title('Features std. dev.') plt.subplots_adjust(left=.3) .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_004.png :alt: Features std. dev. :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 264-274 Multiplying the coefficients by the standard deviation of the related feature would reduce all the coefficients to the same unit of measure. As we will see :ref:after this is equivalent to normalize numerical variables to their standard deviation, as :math:y = \sum{coef_i \times X_i} = \sum{(coef_i \times std_i) \times (X_i / std_i)}. In that way, we emphasize that the greater the variance of a feature, the larger the weight of the corresponding coefficient on the output, all else being equal. .. GENERATED FROM PYTHON SOURCE LINES 274-285 .. code-block:: default coefs = pd.DataFrame( model.named_steps['transformedtargetregressor'].regressor_.coef_ * X_train_preprocessed.std(axis=0), columns=['Coefficient importance'], index=feature_names ) coefs.plot(kind='barh', figsize=(9, 7)) plt.title('Ridge model, small regularization') plt.axvline(x=0, color='.5') plt.subplots_adjust(left=.3) .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_005.png :alt: Ridge model, small regularization :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 286-313 Now that the coefficients have been scaled, we can safely compare them. .. warning:: Why does the plot above suggest that an increase in age leads to a decrease in wage? Why the :ref:initial pairplot is telling the opposite? The plot above tells us about dependencies between a specific feature and the target when all other features remain constant, i.e., **conditional dependencies**. An increase of the AGE will induce a decrease of the WAGE when all other features remain constant. On the contrary, an increase of the EXPERIENCE will induce an increase of the WAGE when all other features remain constant. Also, AGE, EXPERIENCE and EDUCATION are the three variables that most influence the model. Checking the variability of the coefficients -------------------------------------------- We can check the coefficient variability through cross-validation: it is a form of data perturbation (related to resampling _). If coefficients vary significantly when changing the input dataset their robustness is not guaranteed, and they should probably be interpreted with caution. .. GENERATED FROM PYTHON SOURCE LINES 313-335 .. code-block:: default from sklearn.model_selection import cross_validate from sklearn.model_selection import RepeatedKFold cv_model = cross_validate( model, X, y, cv=RepeatedKFold(n_splits=5, n_repeats=5), return_estimator=True, n_jobs=-1 ) coefs = pd.DataFrame( [est.named_steps['transformedtargetregressor'].regressor_.coef_ * X_train_preprocessed.std(axis=0) for est in cv_model['estimator']], columns=feature_names ) plt.figure(figsize=(9, 7)) sns.stripplot(data=coefs, orient='h', color='k', alpha=0.5) sns.boxplot(data=coefs, orient='h', color='cyan', saturation=0.5) plt.axvline(x=0, color='.5') plt.xlabel('Coefficient importance') plt.title('Coefficient importance and its variability') plt.subplots_adjust(left=.3) .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_006.png :alt: Coefficient importance and its variability :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 336-348 The problem of correlated variables ----------------------------------- The AGE and EXPERIENCE coefficients are affected by strong variability which might be due to the collinearity between the 2 features: as AGE and EXPERIENCE vary together in the data, their effect is difficult to tease apart. To verify this interpretation we plot the variability of the AGE and EXPERIENCE coefficient. .. _covariation: .. GENERATED FROM PYTHON SOURCE LINES 348-358 .. code-block:: default plt.ylabel('Age coefficient') plt.xlabel('Experience coefficient') plt.grid(True) plt.xlim(-0.4, 0.5) plt.ylim(-0.4, 0.5) plt.scatter(coefs["AGE"], coefs["EXPERIENCE"]) _ = plt.title('Co-variations of coefficients for AGE and EXPERIENCE ' 'across folds') .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_007.png :alt: Co-variations of coefficients for AGE and EXPERIENCE across folds :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 359-364 Two regions are populated: when the EXPERIENCE coefficient is positive the AGE one is negative and viceversa. To go further we remove one of the 2 features and check what is the impact on the model stability. .. GENERATED FROM PYTHON SOURCE LINES 364-386 .. code-block:: default column_to_drop = ['AGE'] cv_model = cross_validate( model, X.drop(columns=column_to_drop), y, cv=RepeatedKFold(n_splits=5, n_repeats=5), return_estimator=True, n_jobs=-1 ) coefs = pd.DataFrame( [est.named_steps['transformedtargetregressor'].regressor_.coef_ * X_train_preprocessed.drop(columns=column_to_drop).std(axis=0) for est in cv_model['estimator']], columns=feature_names[:-1] ) plt.figure(figsize=(9, 7)) sns.stripplot(data=coefs, orient='h', color='k', alpha=0.5) sns.boxplot(data=coefs, orient='h', color='cyan', saturation=0.5) plt.axvline(x=0, color='.5') plt.title('Coefficient importance and its variability') plt.xlabel('Coefficient importance') plt.subplots_adjust(left=.3) .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_008.png :alt: Coefficient importance and its variability :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 387-401 The estimation of the EXPERIENCE coefficient is now less variable and remain important for all models trained during cross-validation. .. _scaling_num: Preprocessing numerical variables --------------------------------- As said above (see ":ref:the-pipeline"), we could also choose to scale numerical values before training the model. This can be useful to apply a similar amount regularization to all of them in the Ridge. The preprocessor is redefined in order to subtract the mean and scale variables to unit variance. .. GENERATED FROM PYTHON SOURCE LINES 401-410 .. code-block:: default from sklearn.preprocessing import StandardScaler preprocessor = make_column_transformer( (OneHotEncoder(drop='if_binary'), categorical_columns), (StandardScaler(), numerical_columns), remainder='passthrough' ) .. GENERATED FROM PYTHON SOURCE LINES 411-412 The model will stay unchanged. .. GENERATED FROM PYTHON SOURCE LINES 412-424 .. code-block:: default model = make_pipeline( preprocessor, TransformedTargetRegressor( regressor=Ridge(alpha=1e-10), func=np.log10, inverse_func=sp.special.exp10 ) ) _ = model.fit(X_train, y_train) .. GENERATED FROM PYTHON SOURCE LINES 425-428 Again, we check the performance of the computed model using, for example, the median absolute error of the model and the R squared coefficient. .. GENERATED FROM PYTHON SOURCE LINES 428-447 .. code-block:: default y_pred = model.predict(X_train) mae = median_absolute_error(y_train, y_pred) string_score = f'MAE on training set: {mae:.2f} $/hour' y_pred = model.predict(X_test) mae = median_absolute_error(y_test, y_pred) string_score += f'\nMAE on testing set: {mae:.2f}$/hour' fig, ax = plt.subplots(figsize=(6, 6)) plt.scatter(y_test, y_pred) ax.plot([0, 1], [0, 1], transform=ax.transAxes, ls="--", c="red") plt.text(3, 20, string_score) plt.title('Ridge model, small regularization, normalized variables') plt.ylabel('Model predictions') plt.xlabel('Truths') plt.xlim([0, 27]) _ = plt.ylim([0, 27]) .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_009.png :alt: Ridge model, small regularization, normalized variables :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 448-449 For the coefficient analysis, scaling is not needed this time. .. GENERATED FROM PYTHON SOURCE LINES 449-459 .. code-block:: default coefs = pd.DataFrame( model.named_steps['transformedtargetregressor'].regressor_.coef_, columns=['Coefficients'], index=feature_names ) coefs.plot(kind='barh', figsize=(9, 7)) plt.title('Ridge model, small regularization, normalized variables') plt.axvline(x=0, color='.5') plt.subplots_adjust(left=.3) .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_010.png :alt: Ridge model, small regularization, normalized variables :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 460-461 We now inspect the coefficients across several cross-validation folds. .. GENERATED FROM PYTHON SOURCE LINES 461-478 .. code-block:: default cv_model = cross_validate( model, X, y, cv=RepeatedKFold(n_splits=5, n_repeats=5), return_estimator=True, n_jobs=-1 ) coefs = pd.DataFrame( [est.named_steps['transformedtargetregressor'].regressor_.coef_ for est in cv_model['estimator']], columns=feature_names ) plt.figure(figsize=(9, 7)) sns.stripplot(data=coefs, orient='h', color='k', alpha=0.5) sns.boxplot(data=coefs, orient='h', color='cyan', saturation=0.5) plt.axvline(x=0, color='.5') plt.title('Coefficient variability') plt.subplots_adjust(left=.3) .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_011.png :alt: Coefficient variability :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 479-492 The result is quite similar to the non-normalized case. Linear models with regularization --------------------------------- In machine-learning practice, Ridge Regression is more often used with non-negligible regularization. Above, we limited this regularization to a very little amount. Regularization improves the conditioning of the problem and reduces the variance of the estimates. RidgeCV applies cross validation in order to determine which value of the regularization parameter (alpha) is best suited for prediction. .. GENERATED FROM PYTHON SOURCE LINES 492-506 .. code-block:: default from sklearn.linear_model import RidgeCV model = make_pipeline( preprocessor, TransformedTargetRegressor( regressor=RidgeCV(alphas=np.logspace(-10, 10, 21)), func=np.log10, inverse_func=sp.special.exp10 ) ) _ = model.fit(X_train, y_train) .. GENERATED FROM PYTHON SOURCE LINES 507-508 First we check which value of :math:\alpha has been selected. .. GENERATED FROM PYTHON SOURCE LINES 508-511 .. code-block:: default model[-1].regressor_.alpha_ .. rst-class:: sphx-glr-script-out Out: .. code-block:: none 10.0 .. GENERATED FROM PYTHON SOURCE LINES 512-513 Then we check the quality of the predictions. .. GENERATED FROM PYTHON SOURCE LINES 513-533 .. code-block:: default y_pred = model.predict(X_train) mae = median_absolute_error(y_train, y_pred) string_score = f'MAE on training set: {mae:.2f} $/hour' y_pred = model.predict(X_test) mae = median_absolute_error(y_test, y_pred) string_score += f'\nMAE on testing set: {mae:.2f}$/hour' fig, ax = plt.subplots(figsize=(6, 6)) plt.scatter(y_test, y_pred) ax.plot([0, 1], [0, 1], transform=ax.transAxes, ls="--", c="red") plt.text(3, 20, string_score) plt.title('Ridge model, regularization, normalized variables') plt.ylabel('Model predictions') plt.xlabel('Truths') plt.xlim([0, 27]) _ = plt.ylim([0, 27]) .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_012.png :alt: Ridge model, regularization, normalized variables :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 534-536 The ability to reproduce the data of the regularized model is similar to the one of the non-regularized model. .. GENERATED FROM PYTHON SOURCE LINES 536-546 .. code-block:: default coefs = pd.DataFrame( model.named_steps['transformedtargetregressor'].regressor_.coef_, columns=['Coefficients'], index=feature_names ) coefs.plot(kind='barh', figsize=(9, 7)) plt.title('Ridge model, regularization, normalized variables') plt.axvline(x=0, color='.5') plt.subplots_adjust(left=.3) .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_013.png :alt: Ridge model, regularization, normalized variables :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 547-560 The coefficients are significantly different. AGE and EXPERIENCE coefficients are both positive but they now have less influence on the prediction. The regularization reduces the influence of correlated variables on the model because the weight is shared between the two predictive variables, so neither alone would have strong weights. On the other hand, the weights obtained with regularization are more stable (see the :ref:ridge_regression User Guide section). This increased stability is visible from the plot, obtained from data perturbations, in a cross validation. This plot can be compared with the :ref:previous one. .. GENERATED FROM PYTHON SOURCE LINES 560-581 .. code-block:: default cv_model = cross_validate( model, X, y, cv=RepeatedKFold(n_splits=5, n_repeats=5), return_estimator=True, n_jobs=-1 ) coefs = pd.DataFrame( [est.named_steps['transformedtargetregressor'].regressor_.coef_ * X_train_preprocessed.std(axis=0) for est in cv_model['estimator']], columns=feature_names ) plt.ylabel('Age coefficient') plt.xlabel('Experience coefficient') plt.grid(True) plt.xlim(-0.4, 0.5) plt.ylim(-0.4, 0.5) plt.scatter(coefs["AGE"], coefs["EXPERIENCE"]) _ = plt.title('Co-variations of coefficients for AGE and EXPERIENCE ' 'across folds') .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_014.png :alt: Co-variations of coefficients for AGE and EXPERIENCE across folds :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 582-593 Linear models with sparse coefficients -------------------------------------- Another possibility to take into account correlated variables in the dataset, is to estimate sparse coefficients. In some way we already did it manually when we dropped the AGE column in a previous Ridge estimation. Lasso models (see the :ref:lasso User Guide section) estimates sparse coefficients. LassoCV applies cross validation in order to determine which value of the regularization parameter (alpha) is best suited for the model estimation. .. GENERATED FROM PYTHON SOURCE LINES 593-607 .. code-block:: default from sklearn.linear_model import LassoCV model = make_pipeline( preprocessor, TransformedTargetRegressor( regressor=LassoCV(alphas=np.logspace(-10, 10, 21), max_iter=100000), func=np.log10, inverse_func=sp.special.exp10 ) ) _ = model.fit(X_train, y_train) .. GENERATED FROM PYTHON SOURCE LINES 608-609 First we verify which value of :math:\alpha has been selected. .. GENERATED FROM PYTHON SOURCE LINES 609-612 .. code-block:: default model[-1].regressor_.alpha_ .. rst-class:: sphx-glr-script-out Out: .. code-block:: none 0.001 .. GENERATED FROM PYTHON SOURCE LINES 613-614 Then we check the quality of the predictions. .. GENERATED FROM PYTHON SOURCE LINES 614-634 .. code-block:: default y_pred = model.predict(X_train) mae = median_absolute_error(y_train, y_pred) string_score = f'MAE on training set: {mae:.2f} $/hour' y_pred = model.predict(X_test) mae = median_absolute_error(y_test, y_pred) string_score += f'\nMAE on testing set: {mae:.2f}$/hour' fig, ax = plt.subplots(figsize=(6, 6)) plt.scatter(y_test, y_pred) ax.plot([0, 1], [0, 1], transform=ax.transAxes, ls="--", c="red") plt.text(3, 20, string_score) plt.title('Lasso model, regularization, normalized variables') plt.ylabel('Model predictions') plt.xlabel('Truths') plt.xlim([0, 27]) _ = plt.ylim([0, 27]) .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_015.png :alt: Lasso model, regularization, normalized variables :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 635-636 For our dataset, again the model is not very predictive. .. GENERATED FROM PYTHON SOURCE LINES 636-646 .. code-block:: default coefs = pd.DataFrame( model.named_steps['transformedtargetregressor'].regressor_.coef_, columns=['Coefficients'], index=feature_names ) coefs.plot(kind='barh', figsize=(9, 7)) plt.title('Lasso model, regularization, normalized variables') plt.axvline(x=0, color='.5') plt.subplots_adjust(left=.3) .. image:: /auto_examples/inspection/images/sphx_glr_plot_linear_model_coefficient_interpretation_016.png :alt: Lasso model, regularization, normalized variables :class: sphx-glr-single-img .. GENERATED FROM PYTHON SOURCE LINES 647-672 A Lasso model identifies the correlation between AGE and EXPERIENCE and suppresses one of them for the sake of the prediction. It is important to keep in mind that the coefficients that have been dropped may still be related to the outcome by themselves: the model chose to suppress them because they bring little or no additional information on top of the other features. Additionnaly, this selection is unstable for correlated features, and should be interpreted with caution. Lessons learned --------------- * Coefficients must be scaled to the same unit of measure to retrieve feature importance. Scaling them with the standard-deviation of the feature is a useful proxy. * Coefficients in multivariate linear models represent the dependency between a given feature and the target, **conditional** on the other features. * Correlated features induce instabilities in the coefficients of linear models and their effects cannot be well teased apart. * Different linear models respond differently to feature correlation and coefficients could significantly vary from one another. * Inspecting coefficients across the folds of a cross-validation loop gives an idea of their stability. .. rst-class:: sphx-glr-timing **Total running time of the script:** ( 0 minutes 11.356 seconds) .. _sphx_glr_download_auto_examples_inspection_plot_linear_model_coefficient_interpretation.py: .. only :: html .. container:: sphx-glr-footer :class: sphx-glr-footer-example .. container:: binder-badge .. image:: images/binder_badge_logo.svg :target: https://mybinder.org/v2/gh/scikit-learn/scikit-learn/main?urlpath=lab/tree/notebooks/auto_examples/inspection/plot_linear_model_coefficient_interpretation.ipynb :alt: Launch binder :width: 150 px .. container:: sphx-glr-download sphx-glr-download-python :download:Download Python source code: plot_linear_model_coefficient_interpretation.py .. container:: sphx-glr-download sphx-glr-download-jupyter :download:Download Jupyter notebook: plot_linear_model_coefficient_interpretation.ipynb .. only:: html .. rst-class:: sphx-glr-signature Gallery generated by Sphinx-Gallery _
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# Distance between Qingdao (TAO) and Yibin (YBP)
Flight distance from Qingdao to Yibin (Qingdao Liuting International Airport – Yibin Wuliangye Airport) is 1054 miles / 1697 kilometers / 916 nautical miles. Estimated flight time is 2 hours 29 minutes.
Driving distance from Qingdao (TAO) to Yibin (YBP) is 1275 miles / 2052 kilometers and travel time by car is about 21 hours 40 minutes.
## Map of flight path and driving directions from Qingdao to Yibin.
Shortest flight path between Qingdao Liuting International Airport (TAO) and Yibin Wuliangye Airport (YBP).
## How far is Yibin from Qingdao?
There are several ways to calculate distances between Qingdao and Yibin. Here are two common methods:
Vincenty's formula (applied above)
• 1054.376 miles
• 1696.854 kilometers
• 916.228 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1053.358 miles
• 1695.215 kilometers
• 915.343 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Qingdao Liuting International Airport
City: Qingdao
Country: China
IATA Code: TAO
ICAO Code: ZSQD
Coordinates: 36°15′57″N, 120°22′26″E
B Yibin Wuliangye Airport
City: Yibin
Country: China
IATA Code: YBP
ICAO Code: ZUYB
Coordinates: 28°51′28″N, 104°31′30″E
## Time difference and current local times
There is no time difference between Qingdao and Yibin.
CST
CST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 154 kg (340 pounds).
## Frequent Flyer Miles Calculator
Qingdao (TAO) → Yibin (YBP).
Distance:
1054
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1054
Round trip?
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# Analytic combinatorics by Flajolet P., Sedgewick R.
By Flajolet P., Sedgewick R.
Similar combinatorics books
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Extra resources for Analytic combinatorics
Example text
For instance, the notation (23) S EQ=k (or simply S EQk ), S EQ>k , S EQ1 . k refers to sequences whose number of components are exactly k, larger than k, or in the interval 1 . k respectively. In particular, k times S EQk (B) := B × · · · × B ≡ B k , S EQ≥k (B) = j≥k Bj ∼ = B k × S EQ(B), MS ETk (B) := S EQk (B)/R. Similarly, S EQodd , S EQeven will denote sequences with an odd or even number of components, and so on. 30 I. COMBINATORIAL STRUCTURES AND ORDINARY GENERATING FUNCTIONS Translations for such restricted constructions are available, as shown generally in Subsection I.
0, β j ∈ B , which matches our intuition as to what sequences should be. ) It is then readily checked that the construction A = S EQ(B) defines a proper class satisfying the finiteness condition for sizes if and only if B contains no object of size 0. From the definition of size for sums and products, it I. 2. ADMISSIBLE CONSTRUCTIONS AND SPECIFICATIONS 25 follows that the size of an object α ∈ A is to be taken as the sum of the sizes of its components: α = (β1 , . . , βℓ ) ⇒ |α| = |β1 | + · · · + |βℓ |.
Consider the class U of “non-empty” triangulations of the n-gon, that is, we exclude the 2-gon and the corresponding “empty” triangulation of size 0. Then U = T \ {ǫ} admits the specification U = ∇ + (∇ × U) + (U × ∇) + (U × ∇ × U) which also leads to the Catalan numbers via U = z(1 + U )2 , so that U (z) = (1 − 2z − √ 1 − 4z)/(2z) ≡ T (z) − 1. ✁ I. 4. Exploiting generating functions and counting sequences. In this book we are going to see altogether more than a hundred applications of the symbolic method.
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0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question
# (2a+3b)2=
A
4a2+9b2+12ab
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B
4a3+9b2+12ab
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C
4a2+9b2+36ab
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D
4a2+9b3+12ab
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Solution
## The correct option is A 4a2+9b2+12ab Using the identity (a+b)2=a2+b2+2ab, we get, (2a+3b)2=(2a)2+(3b)2+2×2a×3b =4a2+9b2+12ab
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# C++ program to find the Area of the circumcircle of any triangles with sides given?
To calculate the area of circumcircle of any triangles. We need to learn about basic concepts related to the problem.
Triangle − A closed figure with three sides.
Circle − A closed figure with infinite number or side or no sides.
A circle that encloses other figure inside it is a circumcircle.
A circumcircle touches the triangle from all its points. Lets say its sides are a, b, c then the radius of the circumcircle is given by the mathematical formula −
r = abc / (√((a+b+c))(a+b-c)(a+c-b)(b+c-a)))
The area of the circle with radius r is
area = 2 * (pie) * r *r.
Let’s take a few examples for this concept −
Sides of triangle : a = 4 , b = 5 , c =3
Area = 314
## Example
Live Demo
#include <iostream>
#include <math.h>
using namespace std;
int main() {
float a = 7, b = 9, c = 13;
if (a < 0 || b < 0 || c < 0)
cout<<"The figure is not a triangle";
float p = (a + b + c) / 2;
float r = (a*b*c)/ (sqrt(p * (p - a) * (p - b) * (p - c)));
float area = 3.14 * pow(r, 2);
cout<<"The area is "<<area;
return 0;
}
## Output
The area is 2347.55
Updated on: 04-Oct-2019
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## Calculate Linear Inches For Luggage? - Packing Made Easy
Before considering how to calculate linear inches for luggage, let’s actually know what are linear inches.
The term linear inches is a measurement used to determine the size of luggage. You can calculate it by adding the dimensions like length, width, and depth of your luggage in inches.
This measurement is used by airlines and other transit companies. With it, they can know if the luggage meets the size requirements for checked or carry-on baggage.
## Calculation Process
Each airline has different requirements in the case of luggage size. So, firstly make sure to check with your specific air carrier that your luggage meets their size conditions. Airlines limit the size and weight of luggage you can take on board. They have their reasons for it like:
• To ensure that the overhead compartments and storage areas are not overcrowded.
• To increase the safety and efficiency of the aircraft.
You can enjoy a hassle-free trip by measuring your luggage on your journey. Follow the steps below to calculate the linear inches for luggage.
• Take a regular measuring tape and start from the longest point of your bag or suitcase. You also have to include any wheels, handles, or other protrusions of your luggage in the measurement.
• Once you have got the measurements, start adding the height, width, and length together. For example, if your luggage measures 28 inches in length, 18 inches in width, and 12 inches in height.
Then, the linear inches will be calculated as 28 + 15 + 12 = 58 linear inches.
### Most Common Restrictions For Luggage
Every air carrier divides the luggage into three categories - personal items, carry-on bags, and checked luggage. Size restrictions vary from airline to airline. But, I have jotted down some common restrictions for luggage.
• Personal items: Most airlines allow you to carry one personal item like a purse or briefcase on the flight. Occasionally, few airlines require your personal items to be under 16 × 12 × 6 inches. But many of them accept it to just fit entirely underneath your front seat.
• Carry-ons: The carry-on bag which you can take on board your flight with most air carriers has to be under 22 × 14 × 9. US airlines usually don’t have weight limits for carry-on bags.
• Checked luggage: Almost all airlines follow the same size restriction for checked baggage. A limit of 62 linear inches is the standard.
Note: Allegiant Airlines allows you to bring a checked suitcase of 80 linear inches on your journey. Book your flight with Allegiant for cheap fares and extra size limits.
### Consequences Of Oversize Luggage
If your luggage exceeds the size limits of your airline, you may be subject to additional fees or restrictions.
You have to pay extra money to some airlines if your luggage is oversized. Check your airline’s fee structure before booking your flight. The digits can multiply into triples as a fee for oversized bags.
Some airlines would require you to check your luggage instead of carrying it on board if it is more than the size limit of carry-on baggage.
### Some Useful Insights
The fee prices for extra or oversize baggage depend on your choice of airline and your level of ticket. Your first checked bag is usually free in basic or economy fare as long as it meets the size requirements.
Know about the charges for oversize baggage of Southwest and Frontier Airlines below.
Southwest Airlines' size and weight restriction is a combination of 62 linear inches. If you are traveling with this air carrier and the limits increase at any chance. You have to pay an additional fee of a minimum of \$50 per bag.
You are flying with Frontier Airlines and the size limit of your checked bag is more than 62 linear inches. You don’t want to take anything out of your bag. Then, you have to pay an amount of \$75 for oversized luggage.
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http://www.softwareandfinance.com/CSharp/QuickSort_Iterative.html
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# C# - Sorting Algorithm - QuickSort Iterative
We often using sorting algorithm to sort numbers and strings. Also we have many sorting algorithms. I have explained here on how quicksort algorithm works in iterative mode.
For each time when partition method is called, the pivot is placed at the correct position meaning all the elements to the left are less than the pivot value and all the elements to right are greater than the pivot value.
The iteration approach requires stacking of the beg(left), end(right) positions that are used in recursion. I have used the List to store the values and made sure the loop is getting repeated until List is empty. For Partition, we need two arguments - left and right.
The complete program and test run output are given below:
## Source Code
using System;
using System.Collections.Generic;
using System.Text;
namespace CSharpSort
{
class Program
{
static public int Partition(int [] numbers, int left, int right)
{
int pivot = numbers[left];
while (true)
{
while (numbers[left] < pivot)
left++;
while (numbers[right] > pivot)
right--;
if (left < right)
{
int temp = numbers[right];
numbers[right] = numbers[left];
numbers[left] = temp;
}
else
{
return right;
}
}
}
struct QuickPosInfo
{
public int left;
public int right;
};
static public void QuickSort_Iterative(int [] numbers, int left, int right)
{
if(left >= right)
return; // Invalid index range
List<QuickPosInfo> list = new List<QuickPosInfo>();
QuickPosInfo info;
info.left = left;
info.right = right;
list.Insert(list.Count, info);
while(true)
{
if(list.Count == 0)
break;
left = list[0].left;
right = list[0].right;
list.RemoveAt(0);
int pivot = Partition(numbers, left, right);
if(pivot > 1)
{
info.left = left;
info.right = pivot - 1;
list.Insert(list.Count, info);
}
if(pivot + 1 < right)
{
info.left = pivot + 1;
info.right = right;
list.Insert(list.Count, info);
}
}
}
static void Main(string[] args)
{
int[] numbers = { 3, 8, 7, 5, 2, 1, 9, 6, 4 };
int len = 9;
Console.WriteLine("QuickSort By Iterative Method");
QuickSort_Iterative(numbers, 0, len - 1);
for (int i = 0; i < 9; i++)
Console.WriteLine(numbers[i]);
}
}
}
## Output
QuickSort By Iterative Method
1
2
3
4
5
6
7
8
9
Press any key to continue . . .
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https://brainmass.com/business/branding/where-a-competing-company-should-spend-40-billion-447656
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Share
Explore BrainMass
Where a Competing Company Should Spend \$40 Billion
Corben Inc. has a successful brand with the name Crunz. The market size in which Crunz competes is \$4 billion, and Crunz has generated sales of \$400 million. It has a contribution margin of 30% and annual fixed costs of \$20 million. Corben Inc. is thinking of introducing a new brand under the name of Zaturn. Zaturn will compete in the same market as Crunz. The annual fixed costs for this brand are expected to be \$40 million.
If it is launched, Zaturn will capture 10% of the market. It has a contribution margin of 40%. Half of the sales of Zaturn will be cannibalized from the sales of Crunz. An alternative strategy for Corben Inc. is to cancel the introduction of Zaturn and instead to spend the \$40 million (on an annual basis) to promote Crunz. This action is expected to increase the sales for Crunz by 50%. Both brands (Cruz and Zaturn) sell at the same price.
Where should the company spend the \$40 million and why? Show all calculations!
Solution Preview
Total profit from Crunz:
400 *30%- 120 million less fixed costs of 20 million or 100 million dollars.
If Zaturn is launched:
Sales (10% of 4 billion) or 400 million
Profit from ...
Solution Summary
The solution advises where the company should spend \$40 million and why in 126 words with all calculations shown.
\$2.19
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https://palass.org/publications/newsletter/palaeomath-101/palaeomath-part-9-data-blocks-and-partial-least-squares-analysis
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# PalaeoMath: Part 9 - Data Blocks and Partial Least Squares Analysis
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## 9. Data Blocks and Partial Least Squares Analysis
Written by Norm MacLeod - The Natural History Museum, London, UK (email: n.macleod@nhm.ac.uk). This article first appeared in the Nº 63 edition of Palaeontology Newsletter.
Note: This article has not been updated to the new website style, using html tables rather than embeded images, there may be presentation issues.
### Introduction
In the last four columns we've looked at problems associated with characterizing and identifying patterns in single datasets. An implicit assumption that runs across all the methods we've discussed so far (bivariate regression, multivariate regression, PCA, Factor Analysis, PCOORD, and correspondence analysis) is that the objects included in the dataset represent independent and randomly selected samples drawn from a population of interest. Using our trilobite dataset as an example, if we are asking questions about this particular assemblage of 20 trilobite genera the results we have obtained to date are perfectly valid. However, it's a big world out there and we'd often like to know how one type of data relates to another type of data. For example, in all but the last of these columns we were concerned with the analysis of simple morphological data. We first considered bivariate data (the linear regression columns), but expanded that to a (still simple) three-variable system when we came to our discussions of the various single-sample multivariate methods. Then, in the last column I wanted to show how another type of data might be handled and so introduced some ecological data in the form of hypothetical frequency counts of these 20 genera in different environments. I'd now like to ask the next most obvious question 'What can we do if we want to explore how the morphological variables relate to the ecological variables for these taxa?'.
As a matter of fact we've already discussed one approach of this situation: what to do if we want to relate one variable to a suite of others. In that case the appropriate approach is multiple regression. Using this method the pattern of linear variation in a dependent variable (e.g., a morphological variable) can be compared to linear patterns of variation in a suite of independent variables (e.g., ecological variables). The purpose of such an analysis would be to (1) assess the overall significance of the various linear relations between the dependent and independent variables and (2) obtain information about the structure of those relations (e.g., which independent variables show the strongest patterns of covariation; which the least). But this method only yields information for one dependent variable at a time. What if we want to assess the significance and structure of covariation for two different multivariate blocks of variables?
There are two approaches for addressing this data analysis situation: canonical correlation analysis (CCA) and partial least squares (PLS) analysis. The former has been around for some time while the latter is something of a new kid on the data-analysis block. I've always found it curious that neither has figured prominently in palaeontological analyses to date, though canonical correlation has been used for many years by ecologists, economists, psychometricians, and a host of others, while PLS made its impact felt first in the field of chemometrics. I think part of the problem has been that CCA requires the algebraic manipulation of complex, non-symmetric matrices that are beyond the capabilities of hand calculators and even simple spreadsheet programmes. Canonical correlation routines are also somewhat rare in various so-called 'canned' computer packages, though they are straightforward to programme in high-level computer languages or using tools such as Mathematica, Maple or MatLab. In this essay, we'll focus on PLS, in part because it's computationally simpler and illustrates many of the same principles as CCA, but mostly because it has several distinct advantages over CCA. Both methods deserve to be used much more widely in palaeontology.
First, let's review our data. You'll remember the trilobite morphological data, three variables measured on a suite of 20 trilobite specimens (Table 9.1).
Those following closely will also recall the hypothetical trilobite occurrence frequency data from a suite of seven facies arrayed along a crude onshore-offshore gradient (Table 9.2).
One of the purposes of using the frequency data in our previous discussion of correspondence analysis was to illustrate the superior data handing capabilities of that method. The scaling procedures inherent in correspondence analysis mean essentially any type of data can be submitted to this procedure. Partial least squares analysis is also a generalized descriptive technique and so makes no particular distributional assumptions about the data. Nevertheless, this seems as good a place as any to point out that all descriptive methods work better if the data exhibit some similarity to a normal distribution. Counts are always suspect from a distributional point of view because they typically follow a Poisson distribution (see Fig. 9.1A). Since we'll be making use of the correlation relation in our PLS analysis, and since correlations can be badly biased by outliers, I've transformed the ecological data using a variant of Bartlett's (1936) square-root transformation to make them more normal (Fig. 9.1B). The morphological data were also transformed by taking the log10 of their values since it is well known that this transformation makes variables more linear and removes any correlation between the variance and the mean (see the 'Data Blocks' worksheet of the PalaeoMath 101 spreadsheet for these transformed matrices).
Figure 9.1. Trilobite frequency count data prior to (A) and after (B) transformation by the equation $y=sqrt{x+0.3}$, which is variation of the Bartlett (1936) square-root transformation. Note the similarity of A to a Poisson distribution. Strictly speaking the transformation only made these data more normal (as they still do not conform to a normal distribution) but it did improve the balance of the distribution markedly and reduced the number of outlying values.
Now that we have our data in appropriate shape it's time to talk about the comparisons we want to make. PLS has many similarities to PCA, one of which is that you can base the analysis on either the covariance or correlation matrices. For these data the correlation matrix is preferred because the different data groups have different units and characteristically different magnitudes (see the Data Blocks worksheet). As with PCA, you need to consider what basis matrix to use carefully. A covariance matrix is preferred if scaling differences among the variables is something you want the data analysis to take into consideration. For example, if these were two different groups of morphometric variables and one (say the head variables) were characteristically larger than then other (say the tail variables), I might want to include this distinction in the analysis. If I chose to base my PLS analysis on the covariance matrix of raw (though transformed) values, the results would be implicitly weighted toward the larger (= more variable) head variables. On the other hand, if I didn't want these distinctions to affect the results of my analysis I'd want to standardize all my data first so the variances for all variables would be equal, in which case I'd be using a correlation matrix as the basis for my analysis. This standardized covariance, or correlation, matrix for the combined trilobite morphological and ecological variables is shown in Table 9.3.
By now you should be familiar with the general form of a correlation matrix (see the PalaeoMath 101 column in Newsletter 58 for a review). The composite matrices we use for PLS analyses are, however, a bit different. On first inspection they might look like perfectly normal correlation matrices. The diagonal is filled with 1's and the upper and lower parts are mirror images of one another. We could analyze the whole matrix and get a perfectly respectable PCA result. The difference, though lies in the fact that we know there are two different blocks of data here—the morphometric variable block and the ecological variable block. We also know that we're only interested in examining the inter-relations between these data blocks. This knowledge changes everything. Diagrammatically we can represent this block-level structure of Table 9.3 as follows.
$R_{11}$ $R_{12}$ $R_{21}$ $R_{22}$
Here $R_{11}$ refers to the $3\times3$ data block containing just the three morphological variables, $R_{22}$ refers to the $7\times7$ block containing just the seven ecological variables. Both $R_{21}$ and $R_{21}$ refer to the block containing the $3\times7$ (or $7\times3$) cross-correlation between the morphological and ecological variables with $R_{21}$ being a simple transposition of $R^{12}$ (and vice versa). Two-block PLS analysis foregoes all consideration of blocks $R_{11}$ and $R_{22}$ in favour of focusing on block $R_{12}$. In effect, our PLS analysis will be an eigenanalysis of only that part of the basis matrix both groups share. Table 9.4 shows just this section of Table 9.3.
Note this is a different type of matrix from those we've seen before. It's not square because there are many more columns than rows and it's not symmetric because the two halves of the matrix across the diagonal aren't mirror images of one another. Indeed, there isn't even a diagonal to this matrix! Although this is a common type of matrix, we can't use regular eigenanalysis to decompose it into different modes of variation. That method only works on symmetric, square matrices. Never to fear though; methods have been devised to handle this situation. As a matter of fact, you've already been introduced to the primary method for handling this matrix if you read last issue's column. Singular value decomposition (SVD) rescues us again!
Recall last time we used SVD to perform simultaneous Q-mode and R-mode analyses of the square, symmetric, $x^2$ distance matrix we used as the basis for our example correspondence analysis. That proved a convenient way to represent simultaneous ordinations of objects and variables. Recall also that SVD is an implementation of the Ekhart-Young theorem, which states that for any real matrix X, two matrices, V and U, can be found whose minor products are the identity matrix. This means matrices V and U are composed of vectors arranged at right angles to each other. These matrices are scaled to the original data (X) by matrix W, which is a matrix whose diagonal contains a set of terms called 'singular values' with all off-diagonal elements set to zero. These singular values are the square roots of the eigenvalues of both the V and the U matrices, which are identical for all non-zero singular values. Thus,
$X = VWU'$
Each eigenvalue represents an axis through the data cloud aligned with the major directions of variation. Since there are three morphological variables ($p$) and seven ecological variables ($q$) there will only be $p$ non-zero singular values (since $p<q$). Matrix V contains the R-mode loadings, which are the patterns of weights (covariance basis matrix) or angles (correlation basis matrix) that specify the directional relation between these new axes and the Q-mode variables. Matrix U' is the transpose of the Q-mode saliences (see below). Here's the bit that concerns us today, however. The Ekhart-Young theorem states the relation is true for any matrix of any shape and/or character, not just square, symmetric matrices. Table 5 shows the singular values and eigenvalues of the $R_{12}$ data block (see Table 9.3).
These were calculated using the PopTools plug-in for Excel (PC version only). As you can see, from a geometric point-of-view, this cross-variable matrix is highly elongate with very small minor axes. But remember, this is only one block of the overall matrix. Since this is a correlation matrix, we know its total variance is the sum of the number of morphological and ecological variables ($p+q=10$). Thus, this data block—or more correctly, the cross-variable substructure of the overall correlation matrix—accounts for only 17.56 percent of the total variance. Nevertheless, this is the substructure in which we are interested.
For our example analysis the directional vectors are given in Table 9.6 in their normalized (left) and scaled (right) forms. The normalized form is the most convenient for interpretation as the squares of the values always add up to 1.00. The scaled form is calculated by multiplying the normalized vector coefficients by the appropriate singular value. This operation restores the differences between the scale of the vectors.
These vectors look superficially like principal components, but there's an important difference. Whereas the coefficients or 'loadings' of principal component eigenvectors represent the angular relation between the principal component axes and the original variables, the coefficients of a PLS analysis represent the angular relations of the variables within one data block with respect to those in the other data block. In a sense they represent the variables that are most useful or salient for predicting patterns in the other data block. For this reason they are referred to as saliences.
Turning to an interpretation of these data we first need to ask ourselves how many singular values to interpret. We can approach this using the various qualitative methods discussed in the column on PCA (see the Palaeo-Math 101 column in Newsletter 58) or we can use a more sophisticated, quantitative approach that is has been developed recently for use in generalized multivariate analysis (see Morrison 2004, Zelditch et al. 2004).
$x^2=-n\displaystyle\sum_{j=1}^{r}1n\lambda_j+nr\Big(\displaystyle\sum_{j=1}^{r}\lambda_j\Big/r\Big)$
In this equation $x^2$ is the $x^2$ statistic, n is the number of objects in the sample minus 1, $r$ is the number of eigenvalues being tested and $\lambda j$ is the $j^{th}$ singular value. In its typical analytic mode singular values are tested in sequence two at a time (e.g., 1-2, 2-3, 3-4) to determine whether there is a statistically significant amount of variance being explained by the former member of the pair. For this type of test the value of the degrees of freedom is 2. For the comparison between the first and second singular values in the example analysis $x^2 = 15.196$, which means the first singular value is highly significant ($= 0.0005$) as you would expect from the high proportion of variance it explains (see Table 5). When we interpret this axis (Table 6) we see all the R-mode saliences are positive suggesting this is an allometric size axis with glabellar length exhibiting the strongest positive allometry. Environmentally, this allometric size vector is correlated most positively with the black shale facies and most negatively with the paralic shale facies, which are the deepest and shallowest environments in our ecological dataset. This is highly suggestive of a possible shallow-deep or onshore-offshore environmental gradient. Further analysis of the patterns of salience coefficients (Fig. 9.2) shows that, although the relation between size and a depth-shoreline proximity gradient is not strictly consistent, there is more than a hint this general correlation being a major source of patterning in these data.
Figure 9.2. Plot of salience coefficients for the environmental hypothetical variables used in the example analysis. While the trend in these data does not conform strictly to an onshore-offshore gradient, and is not strictly linear, there is a strong suggestion that depth-shoreline proximity is an important source of structure in the R12 block of the correlation matrix. This pattern is associated with strong and uniformly positive salience coefficients for the morphological variables (see Table 6) indicating that this depth-shoreline proximity factor is associated morphologically with an allometric size gradient. See text for discussion.
The strength of the relation between the morphological and environmental variables can also be assessed through a simple graphical device. Since we have the R-mode and Q-mode vector for the cross-variable data block we can calculate the R-mode and Q-mode scores in a manner identical to that for PCA. Table 9.7 shows these scores while Figure 9.3 plots them in a simple bivariate ordination space.
Figure 9.3. Scatterplot of PLS-1 (morphological variables) and PLS-1 (environmental variables) scores for example PLS analysis. This plot represents 97.69% of the correlation structure within the R12 data block.
Comparison of the ordination shown in Figure 3 confirms our interpretation of these results based on the V and U salience matrices. Note large-sized genera (e.g., Trimerus, Zacanthoides, Pricyclopyge, see Table 1) plot toward the upper end of PLS-1 (morphological variables) axis and small-sized genera (e.g., Acaste, Balizoma, Ormathops) toward the lower end, confirming that this axis expresses a generalized size gradient. Moreover, these two groups of genera also display strikingly different environmental occurrence patterns along the PLS-1 (ecological variables) axis with the larger-sized forms being differentially abundant in deep-water facies (see Table 2) and smaller-sized forms preferring shallow-water facies. The linear correlation between the two PLS-1 scores is 0.445, which is just significant statistically for this sample ($r_{crit., d.f. = 19, a = 0.05} = 0.433$) . Based on these results I wouldn't necessarily conclude that size-environment link represents the whole biological story for these data (e.g., the shallow water fauna is composed of mixed small and intermediated sized genera), but this is the strongest, single, linear signal in these data. More importantly for the purposes of this column, by using two-block PLS we've managed to examine the inter-relations between two datasets we've had to treat either separately or as parts of a larger analysis up to this point, and in doing this we've discovered a new patterns in these data that had been hiding there all along.
Partial least squares analysis represents a very powerful and completely generalized approach to ordination and statistical hypothesis testing. Based on a form of PCA, it extends multiple regression analysis, complements canonical correlation analysis, and allows users to test hypotheses about the inter-relations between blocks of observations made on the same objects. Unlike standard PCA which can use a variety of algorithmic approaches to obtain the eigenvalues and eigenvectors of a square, symmetric basis matrix, PLS employs singular value decomposition to obtain the singular values (square roots of eigenvalues) and eigenvectors of parts of PCA basis matrices which may or may not be square, and which will not be symmetric. Aside from the matrix of singular values, this procedure produces two sets of eigenvectors that express the orientational relations between the variables grouped by data blocks: occupying the rows and columns of the basis matrix block. The number of vectors with nonzero lengths will be equivalent to the number of basis-matrix rows ($p$) or columns ($q$), whichever is least. In the example above we employed the correlation matrix as the basis for our PLS analysis because of the nature of the variables. PLS can be performed equally well on either covariance or distance matrices.
Unlike standard multiple regression analysis in which a single dependent variable is regressed against a set of independent variables using a linear least-squares minimization criterion (see the PalaeoMath 101 column in Newsletter 55 for a review of linear least-squares minimization), PLS regresses two sets of multiple variables against one another using a major axis minimization (see the PalaeoMath 101 column in Newsletter 57 for a review of linear major axis minimization). Also, the regression coefficients (= slopes) are partial regression coefficients that represent the relation between the trend of the dependent variable and each of the independent variables when the affects of the other independent variables are held constant. Thus, if a pair of variables is highly covariant or correlated, the covariations or correlations of other pairs of variables will be correspondingly reduced since there will not be much residual covariance or correlation structure left after the effects of the first pair are held constant. In contrast, the PLS salience coefficients all represent angular relations with the complete, block-specific, covariance-correlation structure. This makes the interpretation of these coefficients less complex.
Finally, unlike CCA, which recognizes the same block structure as PLS but uses information from all blocks to create a scaled or pooled covariance-correlation basis matrix for SVD decomposition, PLS decomposes only that block which expresses the inter-relations between the variable sets. This means that PLS can focus on only the inter-block aspect of the covariance-correlation substructure irrespective of whether that substructure accounts for a large or small component of the overall covariance-correlation superstructure. Since the coefficients of a CCA, like those of PLS, are used to quantify the inter-relations between blocks of variables, both are referred to as saliences. It is important to note, however, that CCA saliences are equivalent to partial regression coefficients (see above) whereas PLS saliences are analogous to PCA loadings. In effect, CCA represents an attempt to define a set of canonical variables (= linear combinations of variables) for each data block that exhibit overall covariances-correlations that are as large as possible. Indeed, a CCA analysis in which either the set of basis matrix rows or columns contains a single variable is analogous to a major axis-based multiple regression analysis. The goal of PLS differs insofar as it tries to provide a more focused assessment of the inter-block substructure and doesn't allow within-block patterns of covariance-correlation to influence that result.
Partial least squares analysis supports a very large set of investigation types that are often encountered in palaeontological data analysis situations. The example above represents a simple situation in which a set of morphological variables are related to a set of ecological variables, allowing the morphological correlates of ecological distributions (and vice versa) to be assessed. A PLS approach could also be used to investigate inter-relations between different blocks of morphological variables, say from the anterior or posterior regions of a species (e.g., Zelditch et al. 2004) or between different regions of the same morphological structure. This type of study falls within the general 'morphological integration' research programme that tries to identify regions of correlated morphological variation within organismal Baupläne (see Olson and Miller 1958 for a classical treatment of this topic) and is related to the current interest in identifying developmental modules (see Schlosser and Wager 2004). A PLS approach could also be used to examine inter-relations between different types of ecological variables (e.g., organismal-based vs. physio-chemical), or to explore the morphological correlates of genetic variation. The possibilities are virtually endless (see Rychlik et al. 2006 for an good recent example of PLS analysis being used in a systematic context).
As for the practical matter of how to perform your own PLS analysis, unfortunately the choices here are somewhat more limited than for the other methods we've discussed to date. Of course, the PalaeoMath 101 spreadsheet contains the complete calculations for the example PLS analysis presented above. These were performed using the PopTools plug-in for the SVD calculations, but all other calculations were made using the standard MS-Excel data analysis tools. As I mentioned above, generalized mathematical packages (e.g., Mathematica, Maple, MatLab) can also be used to program your own routines. Program systems that perform PLS analysis are somewhat rare, reflecting the method's relatively recent introduction. Of these your best bets at the moment are XL-Stat (some limited PLS capability) and NT-SYS. Since PLS has a longer history of use in chemometrics some stand-alone software is available in programme packages that have been developed for that community. Of these Solo is one of the more complete and better known.
## References (cited in the text as well as recommended review articles)
Bartlett, M. S. 1936. The square root transformation in analysis of variance. Journal of the Royal Statistical Society, Supplement, 3, 68-78.
Bookstein, F. L. 1991. Morphometric tools for landmark data: geometry and biology. Cambridge University Press, Cambridge, 435 pp.
Golub, G. H. and Reinsch, C. 1971. Singular value decomposition and lest squares solutions, 134-151. In Wilkinson, J. H. and Reinsch, C., eds). Linear algebra: computer methods for mathematical computation, v. 2. Springer-Verlag, Berlin.
Jackson, J. E. 1991. A user's guide to principal components. John Wiley & Sons, New York, 592 pp.
Morrison, D. F. 2005. Multivariate statistical methods. Duxbury Press, New York, 498 pp.
Olson, E. and Miller, R. 1958. Morphological integration. University of Chicago Press, Chi-cago, 317 pp.
Rychlik, L., Ramalhino, G., and Polly, P. D. 2006. Response to environmental factors and competition: skull, mandible and tooth shapes in Polish water shrews (Neomys, Soricidae, Mammalia). Journal of the Zoological Society, 44(4), 339-351.
Rohlf, F. J. and Corti, M. 2000. Use of partial least squares to study covariation in shape. Systematic Biology, 49(4), 740-753.
Schlosser, G. and Wagner, G. 2004. Modularity in development and evolution. University of Chicago Press, Chicago, 600 pp.
Zelditch, M. L., Swiderski, D. L., Sheets, H. D., and Fink, W. L. 2004. Geometric morphomet-rics for biologists: a primer. Elsevier/Academic Press, Amsterdam, 443 pp.
### Author Information
Norm MacLeod - The Natural History Museum, London, UK (email: n.macleod@nhm.ac.uk). This article first appeared in the Nº 63 edition of Palaeontology Newsletter
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https://byjus.com/question-answer/what-is-the-value-of-x-if-the-value-of-33333-2-is-11110xxxx9-8-1/
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1
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Question
# What is the value of X, if the value of 333332 is 11110XXXX9.8
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Solution
## The correct option is A 8The square of the number 33333 can easily be found by the method as described here. Suppose the number contains n '3's. The digits of its square will be (n−1) '1's followed by '0', (n−1) '8's and '9'. In this case, the value of 333332 is 1111088889. Hence, the value of X is 8.
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https://numberworld.info/19
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# Number 19
### Properties of number 19
Cross Sum:
Factorization:
Divisors:
1, 19
Count of divisors:
Sum of divisors:
Prime number?
Yes
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 32:
j
sin(19)
0.14987720966295
cos(19)
0.98870461818667
tan(19)
0.1515894706124
ln(19)
2.9444389791664
lg(19)
1.2787536009528
sqrt(19)
4.3588989435407
Square(19)
### Number Look Up
Look Up
19 (nineteen) is a very impressive figure. The cross sum of 19 is 10. If you factorisate the number 19 you will get these result . 19 has 2 divisors ( 1, 19 ) whith a sum of 20. 19 is a prime number. The figure 19 is not a fibonacci number. The figure 19 is not a Bell Number. The figure 19 is not a Catalan Number. The convertion of 19 to base 2 (Binary) is 10011. The convertion of 19 to base 3 (Ternary) is 201. The convertion of 19 to base 4 (Quaternary) is 103. The convertion of 19 to base 5 (Quintal) is 34. The convertion of 19 to base 8 (Octal) is 23. The convertion of 19 to base 16 (Hexadecimal) is 13. The convertion of 19 to base 32 is j. The sine of the figure 19 is 0.14987720966295. The cosine of 19 is 0.98870461818667. The tangent of the number 19 is 0.1515894706124. The root of 19 is 4.3588989435407.
If you square 19 you will get the following result 361. The natural logarithm of 19 is 2.9444389791664 and the decimal logarithm is 1.2787536009528. You should now know that 19 is very unique figure!
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https://leetcode.ca/2018-08-22-996-Number-of-Squareful-Arrays/
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# 996. Number of Squareful Arrays
## Description
An array is squareful if the sum of every pair of adjacent elements is a perfect square.
Given an integer array nums, return the number of permutations of nums that are squareful.
Two permutations perm1 and perm2 are different if there is some index i such that perm1[i] != perm2[i].
Example 1:
Input: nums = [1,17,8]
Output: 2
Explanation: [1,8,17] and [17,8,1] are the valid permutations.
Example 2:
Input: nums = [2,2,2]
Output: 1
Constraints:
• 1 <= nums.length <= 12
• 0 <= nums[i] <= 109
## Solutions
• class Solution {
public int numSquarefulPerms(int[] nums) {
int n = nums.length;
int[][] f = new int[1 << n][n];
for (int j = 0; j < n; ++j) {
f[1 << j][j] = 1;
}
for (int i = 0; i < 1 << n; ++i) {
for (int j = 0; j < n; ++j) {
if ((i >> j & 1) == 1) {
for (int k = 0; k < n; ++k) {
if ((i >> k & 1) == 1 && k != j) {
int s = nums[j] + nums[k];
int t = (int) Math.sqrt(s);
if (t * t == s) {
f[i][j] += f[i ^ (1 << j)][k];
}
}
}
}
}
}
long ans = 0;
for (int j = 0; j < n; ++j) {
ans += f[(1 << n) - 1][j];
}
Map<Integer, Integer> cnt = new HashMap<>();
for (int x : nums) {
cnt.merge(x, 1, Integer::sum);
}
int[] g = new int[13];
g[0] = 1;
for (int i = 1; i < 13; ++i) {
g[i] = g[i - 1] * i;
}
for (int v : cnt.values()) {
ans /= g[v];
}
return (int) ans;
}
}
• class Solution {
public:
int numSquarefulPerms(vector<int>& nums) {
int n = nums.size();
int f[1 << n][n];
memset(f, 0, sizeof(f));
for (int j = 0; j < n; ++j) {
f[1 << j][j] = 1;
}
for (int i = 0; i < 1 << n; ++i) {
for (int j = 0; j < n; ++j) {
if ((i >> j & 1) == 1) {
for (int k = 0; k < n; ++k) {
if ((i >> k & 1) == 1 && k != j) {
int s = nums[j] + nums[k];
int t = sqrt(s);
if (t * t == s) {
f[i][j] += f[i ^ (1 << j)][k];
}
}
}
}
}
}
long long ans = 0;
for (int j = 0; j < n; ++j) {
ans += f[(1 << n) - 1][j];
}
unordered_map<int, int> cnt;
for (int x : nums) {
++cnt[x];
}
int g[13] = {1};
for (int i = 1; i < 13; ++i) {
g[i] = g[i - 1] * i;
}
for (auto& [_, v] : cnt) {
ans /= g[v];
}
return ans;
}
};
• class Solution:
def numSquarefulPerms(self, nums: List[int]) -> int:
n = len(nums)
f = [[0] * n for _ in range(1 << n)]
for j in range(n):
f[1 << j][j] = 1
for i in range(1 << n):
for j in range(n):
if i >> j & 1:
for k in range(n):
if (i >> k & 1) and k != j:
s = nums[j] + nums[k]
t = int(sqrt(s))
if t * t == s:
f[i][j] += f[i ^ (1 << j)][k]
ans = sum(f[(1 << n) - 1][j] for j in range(n))
for v in Counter(nums).values():
ans //= factorial(v)
return ans
• func numSquarefulPerms(nums []int) (ans int) {
n := len(nums)
f := make([][]int, 1<<n)
for i := range f {
f[i] = make([]int, n)
}
for j := range nums {
f[1<<j][j] = 1
}
for i := 0; i < 1<<n; i++ {
for j := 0; j < n; j++ {
if i>>j&1 == 1 {
for k := 0; k < n; k++ {
if i>>k&1 == 1 && k != j {
s := nums[j] + nums[k]
t := int(math.Sqrt(float64(s)))
if t*t == s {
f[i][j] += f[i^(1<<j)][k]
}
}
}
}
}
}
for j := 0; j < n; j++ {
ans += f[(1<<n)-1][j]
}
g := [13]int{1}
for i := 1; i < 13; i++ {
g[i] = g[i-1] * i
}
cnt := map[int]int{}
for _, x := range nums {
cnt[x]++
}
for _, v := range cnt {
ans /= g[v]
}
return
}
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OpenStudy (anonymous):
simplify √0.0016 0.04 4 0.004 0.4
2 years ago
OpenStudy (studygurl14):
What is 0.0016 in fraction form?
2 years ago
OpenStudy (anonymous):
16/10000 @Studygurl14
2 years ago
OpenStudy (studygurl14):
um, yes. But what's that simplified?
2 years ago
OpenStudy (anonymous):
i dont quite know :/
2 years ago
Or you could express it as$\sqrt{16}\sqrt{10^{-4}}$
2 years ago
OpenStudy (anonymous):
hmm i'm really confused
2 years ago
OpenStudy (studygurl14):
@radar hmm...I didn't think of that
2 years ago
OpenStudy (studygurl14):
@radar 's method is better @bayan143
2 years ago
either way will work. The method I am showing requires the student to know the rules for exponents. And scientific notation. I don't know where bayan143 is at in those subjects. How about it bayan143?
2 years ago
Then do the StudyGurl14 method expressing .0016 as a fraction.
2 years ago
OpenStudy (anonymous):
ok :/
2 years ago
.1 = 1/10 .01 = 1/100 .001=1/100 etc Now do that with .0016
2 years ago
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José Thomaz
Posted on
# Functional programming: a bit of history
## History of functional programming
Functional programming is a programming paradigm, such as object-oriented, imperative and many other programming “styles”. So, before we start to crack functional programming, let’s learn a little bit about its history.
## Computer Science background
To made functional programming possible, many mathematical and computer science theories, concepts and researches were published. Functional programming is very attached to math and functions, its roots are in mathematical logic.
Informal logic systems have been in use or over 2000 years, but the first formalization was made only in the middle of the XIX century. Hamilton, De Morgan and Boole published their works, that were the basis to the formal logic:
• Propositional Calculus;
• Predicate Calculus.
Also, in the XIX century, the number theory was introduced.
In 1936, three different approaches for computability were proposed: Turing’s Turing machines, Kleene’s recursive function theory and Church’s lambda calculus. Turing’s proposal is the foundation of the computer science and programming languages as we know today, and the recursive function theory and lambda calculus are the backbones of functional programming.
## The first functional programming language
The first programming languages were created in the late 1940s, Assembly was the first, and FORTRAN was the first high level programming language to become popular; created in 1954. In the next years new programming languages were created, but now high level, and mostly procedural.
As the computers were becoming more popular, new languages appeared, so in 1958, McCarthy created the LISP programming language, which is considered the first functional language in history.
LISP is a very simple language based on recursive functions manipulating lists of words and numbers. LISP is a non-typed language. Besides that, LISP is not considered “purely functional”, because it has some imperative elements.
The functional paradigm has become popular recently, programmers and companies started to use functional languages in their projects more frequently. Some examples of popular functional languages are:
• Elixir
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# Source code for manim.utils.space_ops
```"""Utility functions for two- and three-dimensional vectors."""
from __future__ import annotations
__all__ = [
"quaternion_mult",
"quaternion_from_angle_axis",
"angle_axis_from_quaternion",
"quaternion_conjugate",
"rotate_vector",
"thick_diagonal",
"rotation_matrix",
"z_to_vector",
"angle_of_vector",
"angle_between_vectors",
"normalize",
"get_unit_normal",
"compass_directions",
"regular_vertices",
"complex_to_R3",
"R3_to_complex",
"complex_func_to_R3_func",
"center_of_mass",
"midpoint",
"find_intersection",
"line_intersection",
"get_winding_number",
"shoelace",
"shoelace_direction",
"cross2d",
"earclip_triangulation",
"cartesian_to_spherical",
"spherical_to_cartesian",
"perpendicular_bisector",
]
import itertools as it
import math
from typing import Sequence
import numpy as np
from mapbox_earcut import triangulate_float32 as earcut
from scipy.spatial.transform import Rotation
from .. import config
from ..constants import DOWN, OUT, PI, RIGHT, TAU, UP
from ..utils.iterables import adjacent_pairs
[docs]def norm_squared(v: float) -> float:
return np.dot(v, v)
# Quaternions
# TODO, implement quaternion type
[docs]def quaternion_mult(
*quats: Sequence[float],
) -> np.ndarray | list[float | np.ndarray]:
"""Gets the Hamilton product of the quaternions provided.
<https://en.wikipedia.org/wiki/Quaternion>`__.
Returns
-------
Union[np.ndarray, List[Union[float, np.ndarray]]]
Returns a list of product of two quaternions.
"""
if config.renderer == "opengl":
if len(quats) == 0:
return [1, 0, 0, 0]
result = quats[0]
for next_quat in quats[1:]:
w1, x1, y1, z1 = result
w2, x2, y2, z2 = next_quat
result = [
w1 * w2 - x1 * x2 - y1 * y2 - z1 * z2,
w1 * x2 + x1 * w2 + y1 * z2 - z1 * y2,
w1 * y2 + y1 * w2 + z1 * x2 - x1 * z2,
w1 * z2 + z1 * w2 + x1 * y2 - y1 * x2,
]
return result
else:
q1 = quats[0]
q2 = quats[1]
w1, x1, y1, z1 = q1
w2, x2, y2, z2 = q2
return np.array(
[
w1 * w2 - x1 * x2 - y1 * y2 - z1 * z2,
w1 * x2 + x1 * w2 + y1 * z2 - z1 * y2,
w1 * y2 + y1 * w2 + z1 * x2 - x1 * z2,
w1 * z2 + z1 * w2 + x1 * y2 - y1 * x2,
],
)
[docs]def quaternion_from_angle_axis(
angle: float,
axis: np.ndarray,
axis_normalized: bool = False,
) -> list[float]:
"""Gets a quaternion from an angle and an axis.
<https://en.wikipedia.org/wiki/Conversion_between_quaternions_and_Euler_angles>`__.
Parameters
----------
angle
The angle for the quaternion.
axis
The axis for the quaternion
axis_normalized : bool, optional
Checks whether the axis is normalized, by default False
Returns
-------
List[float]
Gives back a quaternion from the angle and axis
"""
if config.renderer == "opengl":
if not axis_normalized:
axis = normalize(axis)
return [math.cos(angle / 2), *(math.sin(angle / 2) * axis)]
else:
return np.append(np.cos(angle / 2), np.sin(angle / 2) * normalize(axis))
[docs]def angle_axis_from_quaternion(quaternion: Sequence[float]) -> Sequence[float]:
"""Gets angle and axis from a quaternion.
Parameters
----------
quaternion
The quaternion from which we get the angle and axis.
Returns
-------
Sequence[float]
Gives the angle and axis
"""
axis = normalize(quaternion[1:], fall_back=np.array([1, 0, 0]))
angle = 2 * np.arccos(quaternion[0])
if angle > TAU / 2:
angle = TAU - angle
return angle, axis
[docs]def quaternion_conjugate(quaternion: Sequence[float]) -> np.ndarray:
"""Used for finding the conjugate of the quaternion
Parameters
----------
quaternion
The quaternion for which you want to find the conjugate for.
Returns
-------
np.ndarray
The conjugate of the quaternion.
"""
result = np.array(quaternion)
result[1:] *= -1
return result
[docs]def rotate_vector(
vector: np.ndarray, angle: float, axis: np.ndarray = OUT
) -> np.ndarray:
"""Function for rotating a vector.
Parameters
----------
vector
The vector to be rotated.
angle
The angle to be rotated by.
axis
The axis to be rotated, by default OUT
Returns
-------
np.ndarray
The rotated vector with provided angle and axis.
Raises
------
ValueError
If vector is not of dimension 2 or 3.
"""
if len(vector) > 3:
raise ValueError("Vector must have the correct dimensions.")
if len(vector) == 2:
vector = np.append(vector, 0)
return rotation_matrix(angle, axis) @ vector
[docs]def thick_diagonal(dim: int, thickness=2) -> np.ndarray:
row_indices = np.arange(dim).repeat(dim).reshape((dim, dim))
col_indices = np.transpose(row_indices)
return (np.abs(row_indices - col_indices) < thickness).astype("uint8")
[docs]def rotation_matrix_transpose_from_quaternion(quat: np.ndarray) -> list[np.ndarray]:
"""Converts the quaternion, quat, to an equivalent rotation matrix representation.
<https://in.mathworks.com/help/driving/ref/quaternion.rotmat.html>`_.
Parameters
----------
quat
The quaternion which is to be converted.
Returns
-------
List[np.ndarray]
Gives back the Rotation matrix representation, returned as a 3-by-3
matrix or 3-by-3-by-N multidimensional array.
"""
quat_inv = quaternion_conjugate(quat)
return [
quaternion_mult(quat, [0, *basis], quat_inv)[1:]
for basis in [
[1, 0, 0],
[0, 1, 0],
[0, 0, 1],
]
]
[docs]def rotation_matrix_from_quaternion(quat: np.ndarray) -> np.ndarray:
return np.transpose(rotation_matrix_transpose_from_quaternion(quat))
[docs]def rotation_matrix_transpose(angle: float, axis: np.ndarray) -> np.ndarray:
if all(np.array(axis)[:2] == np.zeros(2)):
return rotation_about_z(angle * np.sign(axis[2])).T
return rotation_matrix(angle, axis).T
[docs]def rotation_matrix(
angle: float,
axis: np.ndarray,
homogeneous: bool = False,
) -> np.ndarray:
"""
Rotation in R^3 about a specified axis of rotation.
"""
inhomogeneous_rotation_matrix = Rotation.from_rotvec(
angle * normalize(np.array(axis))
).as_matrix()
if not homogeneous:
return inhomogeneous_rotation_matrix
else:
rotation_matrix = np.eye(4)
rotation_matrix[:3, :3] = inhomogeneous_rotation_matrix
return rotation_matrix
[docs]def rotation_about_z(angle: float) -> np.ndarray:
"""Returns a rotation matrix for a given angle.
Parameters
----------
angle : float
Angle for the rotation matrix.
Returns
-------
np.ndarray
Gives back the rotated matrix.
"""
c, s = math.cos(angle), math.sin(angle)
return np.array(
[
[c, -s, 0],
[s, c, 0],
[0, 0, 1],
]
)
[docs]def z_to_vector(vector: np.ndarray) -> np.ndarray:
"""
Returns some matrix in SO(3) which takes the z-axis to the
(normalized) vector provided as an argument
"""
axis_z = normalize(vector)
axis_y = normalize(np.cross(axis_z, RIGHT))
axis_x = np.cross(axis_y, axis_z)
if np.linalg.norm(axis_y) == 0:
# the vector passed just so happened to be in the x direction.
axis_x = normalize(np.cross(UP, axis_z))
axis_y = -np.cross(axis_x, axis_z)
return np.array([axis_x, axis_y, axis_z]).T
[docs]def angle_of_vector(vector: Sequence[float]) -> float:
"""Returns polar coordinate theta when vector is projected on xy plane.
Parameters
----------
vector
The vector to find the angle for.
Returns
-------
float
The angle of the vector projected.
"""
return np.angle(complex(*vector[:2]))
[docs]def angle_between_vectors(v1: np.ndarray, v2: np.ndarray) -> np.ndarray:
"""Returns the angle between two vectors.
This angle will always be between 0 and pi
Parameters
----------
v1
The first vector.
v2
The second vector.
Returns
-------
np.ndarray
The angle between the vectors.
"""
return 2 * np.arctan2(
np.linalg.norm(normalize(v1) - normalize(v2)),
np.linalg.norm(normalize(v1) + normalize(v2)),
)
[docs]def normalize(vect: np.ndarray | tuple[float], fall_back=None) -> np.ndarray:
norm = np.linalg.norm(vect)
if norm > 0:
return np.array(vect) / norm
else:
return fall_back or np.zeros(len(vect))
[docs]def normalize_along_axis(array: np.ndarray, axis: np.ndarray) -> np.ndarray:
"""Normalizes an array with the provided axis.
Parameters
----------
array
The array which has to be normalized.
axis
The axis to be normalized to.
Returns
-------
np.ndarray
Array which has been normalized according to the axis.
"""
norms = np.sqrt((array * array).sum(axis))
norms[norms == 0] = 1
buffed_norms = np.repeat(norms, array.shape[axis]).reshape(array.shape)
array /= buffed_norms
return array
[docs]def get_unit_normal(v1: np.ndarray, v2: np.ndarray, tol: float = 1e-6) -> np.ndarray:
"""Gets the unit normal of the vectors.
Parameters
----------
v1
The first vector.
v2
The second vector
tol
[description], by default 1e-6
Returns
-------
np.ndarray
The normal of the two vectors.
"""
v1, v2 = (normalize(i) for i in (v1, v2))
cp = np.cross(v1, v2)
cp_norm = np.linalg.norm(cp)
if cp_norm < tol:
# Vectors align, so find a normal to them in the plane shared with the z-axis
cp = np.cross(np.cross(v1, OUT), v1)
cp_norm = np.linalg.norm(cp)
if cp_norm < tol:
return DOWN
return normalize(cp)
###
[docs]def compass_directions(n: int = 4, start_vect: np.ndarray = RIGHT) -> np.ndarray:
"""Finds the cardinal directions using tau.
Parameters
----------
n
The amount to be rotated, by default 4
start_vect
The direction for the angle to start with, by default RIGHT
Returns
-------
np.ndarray
The angle which has been rotated.
"""
angle = TAU / n
return np.array([rotate_vector(start_vect, k * angle) for k in range(n)])
[docs]def regular_vertices(
n: int, *, radius: float = 1, start_angle: float | None = None
) -> tuple[np.ndarray, float]:
"""Generates regularly spaced vertices around a circle centered at the origin.
Parameters
----------
n
The number of vertices
The radius of the circle that the vertices are placed on.
start_angle
The angle the vertices start at.
If unspecified, for even ``n`` values, ``0`` will be used.
For odd ``n`` values, 90 degrees is used.
Returns
-------
vertices : :class:`numpy.ndarray`
The regularly spaced vertices.
start_angle : :class:`float`
The angle the vertices start at.
"""
if start_angle is None:
if n % 2 == 0:
start_angle = 0
else:
start_angle = TAU / 4
start_vector = rotate_vector(RIGHT * radius, start_angle)
vertices = compass_directions(n, start_vector)
return vertices, start_angle
[docs]def complex_to_R3(complex_num: complex) -> np.ndarray:
return np.array((complex_num.real, complex_num.imag, 0))
[docs]def R3_to_complex(point: Sequence[float]) -> np.ndarray:
return complex(*point[:2])
[docs]def complex_func_to_R3_func(complex_func):
return lambda p: complex_to_R3(complex_func(R3_to_complex(p)))
[docs]def center_of_mass(points: Sequence[float]) -> np.ndarray:
"""Gets the center of mass of the points in space.
Parameters
----------
points
The points to find the center of mass from.
Returns
-------
np.ndarray
The center of mass of the points.
"""
return np.average(points, 0, np.ones(len(points)))
[docs]def midpoint(
point1: Sequence[float],
point2: Sequence[float],
) -> float | np.ndarray:
"""Gets the midpoint of two points.
Parameters
----------
point1
The first point.
point2
The second point.
Returns
-------
Union[float, np.ndarray]
The midpoint of the points
"""
return center_of_mass([point1, point2])
[docs]def line_intersection(
line1: Sequence[np.ndarray], line2: Sequence[np.ndarray]
) -> np.ndarray:
"""Returns the intersection point of two lines, each defined by
a pair of distinct points lying on the line.
Parameters
----------
line1
A list of two points that determine the first line.
line2
A list of two points that determine the second line.
Returns
-------
np.ndarray
The intersection points of the two lines which are intersecting.
Raises
------
ValueError
Error is produced if the two lines don't intersect with each other
or if the coordinates don't lie on the xy-plane.
"""
if any(np.array([line1, line2])[:, :, 2].reshape(-1)):
# checks for z coordinates != 0
raise ValueError("Coords must be in the xy-plane.")
# algorithm from https://stackoverflow.com/a/42727584
np.pad(np.array(i)[:, :2], ((0, 0), (0, 1)), constant_values=1)
for i in (line1, line2)
)
line1, line2 = (np.cross(*i) for i in padded)
x, y, z = np.cross(line1, line2)
if z == 0:
raise ValueError(
"The lines are parallel, there is no unique intersection point."
)
return np.array([x / z, y / z, 0])
[docs]def find_intersection(
p0s: Sequence[np.ndarray],
v0s: Sequence[np.ndarray],
p1s: Sequence[np.ndarray],
v1s: Sequence[np.ndarray],
threshold: float = 1e-5,
) -> Sequence[np.ndarray]:
"""
Return the intersection of a line passing through p0 in direction v0
with one passing through p1 in direction v1 (or array of intersections
from arrays of such points/directions).
For 3d values, it returns the point on the ray p0 + v0 * t closest to the
ray p1 + v1 * t
"""
# algorithm from https://en.wikipedia.org/wiki/Skew_lines#Nearest_points
result = []
for p0, v0, p1, v1 in zip(*[p0s, v0s, p1s, v1s]):
normal = np.cross(v1, np.cross(v0, v1))
denom = max(np.dot(v0, normal), threshold)
result += [p0 + np.dot(p1 - p0, normal) / denom * v0]
return result
[docs]def get_winding_number(points: Sequence[float]) -> float:
total_angle = 0
for p1, p2 in adjacent_pairs(points):
d_angle = angle_of_vector(p2) - angle_of_vector(p1)
d_angle = ((d_angle + PI) % TAU) - PI
total_angle += d_angle
[docs]def shoelace(x_y: np.ndarray) -> float:
"""2D implementation of the shoelace formula.
Returns
-------
:class:`float`
Returns signed area.
"""
x = x_y[:, 0]
y = x_y[:, 1]
return np.trapz(y, x)
[docs]def shoelace_direction(x_y: np.ndarray) -> str:
"""
Uses the area determined by the shoelace method to determine whether
the input set of points is directed clockwise or counterclockwise.
Returns
-------
:class:`str`
Either ``"CW"`` or ``"CCW"``.
"""
area = shoelace(x_y)
return "CW" if area > 0 else "CCW"
[docs]def cross2d(a, b):
if len(a.shape) == 2:
return a[:, 0] * b[:, 1] - a[:, 1] * b[:, 0]
else:
return a[0] * b[1] - b[0] * a[1]
[docs]def earclip_triangulation(verts: np.ndarray, ring_ends: list) -> list:
"""Returns a list of indices giving a triangulation
of a polygon, potentially with holes.
Parameters
----------
verts
verts is a numpy array of points.
ring_ends
ring_ends is a list of indices indicating where
the ends of new paths are.
Returns
-------
list
A list of indices giving a triangulation of a polygon.
"""
# First, connect all the rings so that the polygon
# with holes is instead treated as a (very convex)
# polygon with one edge. Do this by drawing connections
# between rings close to each other
rings = [list(range(e0, e1)) for e0, e1 in zip([0, *ring_ends], ring_ends)]
attached_rings = rings[:1]
detached_rings = rings[1:]
loop_connections = {}
while detached_rings:
i_range, j_range = (
list(
filter(
# Ignore indices that are already being
# used to draw some connection
lambda i: i not in loop_connections,
it.chain(*ring_group),
),
)
for ring_group in (attached_rings, detached_rings)
)
# Closest point on the attached rings to an estimated midpoint
# of the detached rings
tmp_j_vert = midpoint(verts[j_range[0]], verts[j_range[len(j_range) // 2]])
i = min(i_range, key=lambda i: norm_squared(verts[i] - tmp_j_vert))
# Closest point of the detached rings to the aforementioned
# point of the attached rings
j = min(j_range, key=lambda j: norm_squared(verts[i] - verts[j]))
# Recalculate i based on new j
i = min(i_range, key=lambda i: norm_squared(verts[i] - verts[j]))
# Remember to connect the polygon at these points
loop_connections[i] = j
loop_connections[j] = i
# Move the ring which j belongs to from the
# attached list to the detached list
new_ring = next(filter(lambda ring: ring[0] <= j < ring[-1], detached_rings))
detached_rings.remove(new_ring)
attached_rings.append(new_ring)
# Setup linked list
after = []
end0 = 0
for end1 in ring_ends:
after.extend(range(end0 + 1, end1))
after.append(end0)
end0 = end1
# Find an ordering of indices walking around the polygon
indices = []
i = 0
for _ in range(len(verts) + len(ring_ends) - 1):
# starting = False
if i in loop_connections:
j = loop_connections[i]
indices.extend([i, j])
i = after[j]
else:
indices.append(i)
i = after[i]
if i == 0:
break
meta_indices = earcut(verts[indices, :2], [len(indices)])
return [indices[mi] for mi in meta_indices]
[docs]def cartesian_to_spherical(vec: Sequence[float]) -> np.ndarray:
"""Returns an array of numbers corresponding to each
polar coordinate value (distance, phi, theta).
Parameters
----------
vec
A numpy array ``[x, y, z]``.
"""
norm = np.linalg.norm(vec)
if norm == 0:
return 0, 0, 0
r = norm
phi = np.arccos(vec[2] / r)
theta = np.arctan2(vec[1], vec[0])
return np.array([r, theta, phi])
[docs]def spherical_to_cartesian(spherical: Sequence[float]) -> np.ndarray:
"""Returns a numpy array ``[x, y, z]`` based on the spherical
coordinates given.
Parameters
----------
spherical
A list of three floats that correspond to the following:
r - The distance between the point and the origin.
theta - The azimuthal angle of the point to the positive x-axis.
phi - The vertical angle of the point to the positive z-axis.
"""
r, theta, phi = spherical
return np.array(
[
r * np.cos(theta) * np.sin(phi),
r * np.sin(theta) * np.sin(phi),
r * np.cos(phi),
],
)
[docs]def perpendicular_bisector(
line: Sequence[np.ndarray],
norm_vector=OUT,
) -> Sequence[np.ndarray]:
"""Returns a list of two points that correspond
to the ends of the perpendicular bisector of the
two points given.
Parameters
----------
line
a list of two numpy array points (corresponding
to the ends of a line).
norm_vector
the vector perpendicular to both the line given
and the perpendicular bisector.
Returns
-------
list
A list of two numpy array points that correspond
to the ends of the perpendicular bisector
"""
p1 = line[0]
p2 = line[1]
direction = np.cross(p1 - p2, norm_vector)
m = midpoint(p1, p2)
return [m + direction, m - direction]
```
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## The Easter Bunny Needs Your Help! April 17, 2017 Vicki McGinn Comment
There are 10 Easter baskets lined up in a row. The Easter bunny puts 1 egg in the first one, 2 in the second, 3 in the third, 4 in the fourth. The bunny continues to fill all of the baskets like this. How many eggs were used to fill
## Easter Math April 17, 2017 Vicki McGinn Comment
I was not planning on writing a post on Easter Sunday, but some real-life math occurred while preparing for dinner. I am sharing here what I posted on my Math Motivator Facebook page. My ham is 5.49 kilograms. The instructions say to cook it 12-15 minutes per pound or
## Easter Egg Tasks April 17, 2017 Vicki McGinn Comment
When engaging students in problem solving and I hope you are doing so often, it is important to think about the types of tasks you give them. Which of the following two will give you the most information about your students? Example 1: (closed) Nathan finds 6 pink eggs, 1
## Understanding Relationships Between Quantity and the Patterns Within Our Number System February 1, 2017 Vicki McGinn Comment
Working one-on-one with students who are demonstrating a fragile sense of number are opportunities I greatly value. I always approach these times from an inquiry perspective, looking for clues to help me understand their struggles. Often they are demonstrating many strengths in some areas, but something is keeping them from
## Ants! Ants! Ants! November 16, 2016 Vicki McGinn Comment
Recently in a Gr 3, 4, 5 classroom we gave the students this question: Students are learning about insects. They discover that an ant has 1 body, 2 antennae and 6 legs. They each make a model. How many bodies, antennae and legs will they need for 5, 10 and
## Natural vs Commercial Math Materials August 5, 2016 Vicki McGinn Comment
Recently I received the following question from a Kindergarten educator: Why do I get the feeling that natural materials are better than commercial ones for math? I do the exact same activity with materials we have in the classroom. Do the students actually learn more about math through the natural
## Proportional Reasoning – Division May 11, 2016 Vicki McGinn Comment
Recently I had the opportunity to work in a Grade 6 classroom as they began a unit on division and multiplication. The teacher gave the students the following problem from the Ministry support guide (division) to begin. The purpose for that day was to see what the students
## Proportional Reasoning – Measurement March 2, 2016 Vicki McGinn Comment
I love problems that provide authentic opportunities to cluster several big ideas. Recently, I was in a classroom where the teacher gave the following problem to his students. Tia is filling a bucket with water. She knows that 500 ml of water comes out of the hose every 10 seconds.
## Tasty Math Homework January 30, 2016 Vicki McGinn Comment
I just finished making brownies (from a box) with my granddaughter Charlotte who is in Junior Kindergarten. I was reminded about all of the ways both math and literacy can be injected into a fun time. I baked with my own children when they were young but I know I
## Assessment – The Key to Precision January 5, 2016 Vicki McGinn Comment
Last month I wrote a post called, “Preciseness Will Impact Student Achievement”. After posting it, I remembered something that I had written a few years ago. Due to a stroke in December 2011 that left my husband unable to speak or write we became involved in the InteRACT program at
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Page 3 of 4
### Re: Poll: How Many "Weights" Needed To Play "Known" Chess Very Well?
Posted: Tue Oct 22, 2019 4:03 am
Yes, you go and see how Human GMs play bullet chess to realize how calculation is a critical part of avoiding critical mistakes. Even at 3 0 they blunder a lot, it's just that if they're not good enough to find the opponent's tactic to avoid it's also likely their opponents don't have time to see it either.
### Re: Poll: How Many "Weights" Needed To Play "Known" Chess Very Well?
Posted: Tue Oct 22, 2019 4:06 am
Suppose it were 10,000 weights needed.
Humans would have no hope to comprehend that clearly.
We do well up to 7 distinct items, on average. After that, we start to degrade.
We can deal with bigger clumps of things by splitting into parts (especially related parts).
But past a certain point, we start to get lost even with that.
That's why menus often get split into sub-menus at a certain count on computer menu systems.
### Re: Poll: How Many "Weights" Needed To Play "Known" Chess Very Well?
Posted: Tue Oct 22, 2019 7:18 am
People can recite 30000 numbers of pi from memory so it wouldn't be out of reach for some people to just memorize the weights and apply them to positions.
If it comes to learning the weights it might be possible to applying them without needing to comprehend them.
### Re: Poll: How Many "Weights" Needed To Play "Known" Chess Very Well?
Posted: Tue Oct 22, 2019 7:51 am
fabianVDW wrote:
Mon Oct 21, 2019 9:16 pm
You are right for counting, but I would not immediately see such neurons for determining for instance passers.
More complex than counting, for sure, but still not very hard. For each square you would need a neuron that fires when there is a passer on that square. (And that for each player.) This neuron for input connects to the Pawn plane of its own square with a positive weight (say +1), and to all squares in front of it in the enemy Pawn plane on its own file and the two adjacent files, with larger negative weight (say -1). The neuron can have step-function or rectifier response. Even for a 2nd-rank square that is only 16 connections, for 7th rank it is only one. The only way to get a +1 output is if there is a passer on that square; otherwise the output will be 0.
In the next layer you can have counting cells per file for the number of passers in that file (6 inputs), or detecting the fact there is at least one passer (step-function response). In the third layer you can have a cell per pair of neighboring files with 2 inputs, to detect connected passers. All these cells can be connected to a final summing output cell with the desired evaluation weight for the feature (passer in that location, bonus or connected passers etc.).
### Re: Poll: How Many "Weights" Needed To Play "Known" Chess Very Well?
Posted: Tue Oct 22, 2019 10:10 pm
dkappe wrote:
Mon Oct 21, 2019 11:11 am
The 11258 distilled networks run all the way from 16x2, 24x3, 32x4, 48x5, etc., and will run reasonably well on CPU. You can find them here: https://github.com/dkappe/leela-chess-w ... d-Networks
Try out the various sizes on lc0 and judge for yourself.
You can also try this BOT https://github.com/dkappe/leela-chess-w ... -style-net
It’s a 32x4 looking at ~25 moves on a raspberry pi 3. Because of its source material, it plays objectively weaker moves than SF or leela, but is very effective against humans.
Thanks - just had 3 interesting games against her. By the third game, I'd learned that I have to defend my king as the top priority or it will get killed, and I thought I was holding on, but then made a losing blunder under pressure. I might be mistaken, but I think she missed a checkmate sequence in game 3 - but all that achieved was to prolong the agony!
### Re: Poll: How Many "Weights" Needed To Play "Known" Chess Very Well?
Posted: Tue Oct 22, 2019 10:36 pm
Ovyron wrote:
Tue Oct 22, 2019 7:18 am
People can recite 30000 numbers of pi from memory so it wouldn't be out of reach for some people to just memorize the weights and apply them to positions.
Someone did 100K digits. There is a difference, though, between having nearly photographic memory and knowing the meaning of the thing memorized.
https://www.foxnews.com/story/japanese- ... gits-of-pi
If it comes to learning the weights it might be possible to applying them without needing to comprehend them.
But if we still don't understand them, why not just leave them as "black box weights"?
The NN engine already knows how to use them. You could inject them into an alpha-beta searcher, but you would still need a GPU to do all the math or it would be too slow.
### Re: Poll: How Many "Weights" Needed To Play "Known" Chess Very Well?
Posted: Tue Oct 22, 2019 11:02 pm
Dann Corbit wrote:
Tue Oct 22, 2019 10:36 pm
There is a difference, though, between having nearly photographic memory and knowing the meaning of the thing memorized.
The idea would be memorizing all the weights and being able to use them to figure out what move to play in a give chess position without needing to understand their meaning. Like learning how to decode a crypto message and getting that TH is Z and being able to translate it even if you don't know what Z means by itself. The combination of pieces on a chess board could code the best move to play and you'd just need to decipher it.
Dann Corbit wrote:
Tue Oct 22, 2019 10:36 pm
But if we still don't understand them, why not just leave them as "black box weights"?
The idea would be that someone could become world chess champion by learning the weights and applying them in their chess games, even if they didn't understand their meaning.
(...in theory. In practice it may turn out the weights out of the "black box" are gibberish and unusable by humans...)
### Re: Poll: How Many "Weights" Needed To Play "Known" Chess Very Well?
Posted: Wed Oct 23, 2019 7:52 am
Dann Corbit wrote:
Tue Oct 22, 2019 10:36 pm
But if we still don't understand [the weights], why not just leave them as "black box weights"?
I believe it will be possible to use linear programming to fit numerical expressions rather than trained NNs, with the following advantages:
1. Chess without search is a complex shape that will be extremely difficult to "fit" with NN learning alone. IMO LP has a better chance
2. A set of expressions will be easier to simplify than a trained NN IMO, making for a smaller overall size, a faster run time, and more hope that humans might be able to understand it
3. IMO it is very possible that there might exist a relatively simple expression that will work in most chess positions (there probably isn't a super-simple one, or it would likely have been found already), and that it might be possible to find it. If so, it will be much easier to extract "human meaning" from a numerical expression than it would from an NN
### Re: Poll: How Many "Weights" Needed To Play "Known" Chess Very Well?
Posted: Wed Oct 23, 2019 12:58 pm
But a NN is nothing but a set of expressions...
### Re: Poll: How Many "Weights" Needed To Play "Known" Chess Very Well?
Posted: Wed Oct 23, 2019 3:05 pm
hgm wrote:
Wed Oct 23, 2019 12:58 pm
But a NN is nothing but a set of expressions...
A quick comparison of NN training with generating expressions and fitting them to a classification problem using linear prgramming. NN offers the following advantages:
1. You can download a software library like TensorFlow and, if you know what you're doing, you're good to go
2. Proven way of getting a good learning system for many problem types
3. Being used right now for chess position evaluation in many chess engines (some of them free and open source, and they mostly play very well)
4. If you want a classifier that fits the given data using Linear Programming (LP), and which would hence output a set of expressions, then right now you're building it yourself
LP generated expressions would offer the following advantages:
1. IMO LP will be more able to fit a complex shape like the solution to chess than a NN will
2. Could fit the data to the mathematical limit - the best possible fit (it may be necessary to use some LP tricks like symmetry breaking in such a large model space)
3. Having found the best fit, you could then turn the achieved fit into a model condition, and then do another optimisation to maximise the number of expressions for which the weight is zero, resulting in a smaller set of expressions
4. Having got a set of expressions and weights, it would be easy to translate this into a computer language for a program that would run on any computer - with or without a graphics card
5. If it turns out to be possible to make smallish set of expressions which can correctly classify most chess positions, it might be possible to articulate this expression in English language
Btw - if classification by LP turns out to be viable, you'll then have another optimisation problem: selection of position/evaluation pairs. The 7 piece tablebase alone has 423,836,835,667,331 positions in, which is infeasibly large for an LP model with today's technology - by many orders of magnitude. If you couldn't find a selection of around a million of them which could enable most of the others to be solved by the classifier, then the quest to solve "known chess" this way would likely be out of reach.
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# Echotastic
Username Echotastic Score 36 Stats Questions 7 Answers 3
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### An equilateral triangle shares a common side with a square as shown. What is the number of degrees in m
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### In the diagram, if \$\angle PQR = 48^\circ\$ , what is the measure of \$\angle PMN\$?
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### A standard deck of 52 cards has 13 ranks (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King) and 4 suits (, , , and ), such that
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### A standard deck of 52 cards has 13 ranks (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King) and 4 suits (, , , and ), such that
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0
# What is 15 percent of 30 percent?
Updated: 10/10/2023
Wiki User
14y ago
15% of 30%
15/100 =0.15
30/100 =0.3
0.15×0.3 = 0.045.
Prym O
Lvl 2
6mo ago
Wiki User
14y ago
15 percent of 30 percent = 0.15 x 30 percent = 4.5 percent
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### Shannon Entropy Explained
Shannon Entropy Explained
Shannon Entropy (also called Information Entropy) is a concept used in physics and information theory. Here's the scoop..
Suppose you have a system with n states i.e. whenever you make an observation of the system you find it's always in one of the n possible states.
Now make a large number of observations of the system, then use them to get the probability pi that if you make an observation the system is in state i. So for every state of the system you have a probability pi.
Now construct this crazy sum = p1*log(p1) + p2*log(p2) +... + pn*log(pn) where the sum is over all the states of the system.
If the log is base 2 then (-1)*sum is called the "information entropy" of the system.
Note that "information entropy" applies to a complete system, not individual states of a system.
Here's a simple example..
My system is a penny and a table.
I define the system to have 2 states.. penny lying stationary on the table with heads up or with tails up.
My experiment is to throw the penny and then observe which state results.
I throw the penny many times and make notes. It lands heads up 1% of the time and tails up 99% of the time (it's a biased penny).
The crazy sum is 0.01*log(0.01) + 0.99*log(0.99) = 0.01*(-6.643856) + 0.99*(-0.0145) = -0.08079356
So the information entropy of the system is (-1)*(-0.08079356) = 0.08079356
Content written and posted by Ken Abbott abbottsystems@gmail.com
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### 1 Descriptive statistics
For the exercises below download the friendsAndCars.csv file and save it on your I:-drive. Make sure that you don't have any empty lines at the end of the file.
#### 1.1 Preprocessing
The friendsAndCars.csv file contains a relation between people and the cars they own. In order to use descriptive statistics it is best to calculate frequencies:
• people and how many cars they own, or
• cars and how many people own these cars.
The following Python script counts how often each type of car is in the list:
```from networkx import *
from operator import *
from sets import Set
a = [item[1] for item in sorted(F.edges(),key=itemgetter(1))]
for item in Set(a):
print item + "," + str(a.count(item))
```
You can save this as countFreq.py and run it on the command-line using
```python countFreq.py > CarsCounted.csv
```
#### 1.2 Using Excel (or OpenOffice)
If you double click on CarsCounted.csv, it will open in Excel.
Measures of central tendency and dispersion are functions in Excel. For example, AVERAGE(B1:B6) calculates the average (arithmetic mean) of the values in cells B1 to B6.
measureformula
modeMODE()
medianMEDIAN()
mean AVERAGE()
varianceVAR()
standard deviationSTDEV()
#### 1.3 Exercises
1) Calculate the measures in the table above for the Cars data.
2) How can you interpret the data: what is the central value? Is this a normal distribution?
3) Produce a chart (diagram) of the data. In order to do this, you should highlight the data and then select the chart wizzard. You may want to create a label for each column first.
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Q:
# Evaluate P = 50x + 80y at each vertex of the feasible region. (0, 0) P = (0, 15) P = (6, 12) P =
Accepted Solution
A:
(0, 0) P = ⇒ 0(0, 15) P = ⇒ 1200(6, 12) P = ⇒ 1260
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# Binomical vs poisson distribution in Simulations
1. Sep 14, 2006
### hagen
Hey, I want to write a Computer Simulation in C++, which simulates the development of a DNA sequence with a probability to mutate x in one "generation". I do have a variable number (=n) of copies of this DNA. Now one might think, to simulate the mutation by simply:
sum(n*Poisson distributed random variable(x) )
to get the number of mutated DNA copies. But this would be too slow.
So my question is, could I also just create a
binomically distributed random variable and multiply it by n * x
to get the number of mutated DNA's? Or is this statistically incorrect?
If not, how might I set the Params for the Bin. dis.? Can I take 1 as a mean and multiply the result x or has the mean to be x? And how do I set / transform the variance of the distribution in a ratio to number of copies.
As you might probably have guessed, I'm a beginner in statistics, but i would be really grateful for any help. Thanks in advance,
Hagen
2. Sep 15, 2006
Hm, if I remember correct, there is a link between the binomial and Poisson distribution.. Poisson's probability function is given with $$f(x)=\frac{\lambda^x}{x!}e^{-\lambda}$$. Now, I think you can put $$\lambda=mp$$, where the number of repetitions of a Bernoulli scheme experiment $$m \rightarrow \infty$$ and it's probability $$p \rightarrow 0$$.
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# a wheel makes 360 revolutions in one minute. through how many radians it turns in one second.
Category :
### Mohammed
Guys, does anyone know the answer?
get a wheel makes 360 revolutions in one minute. through how many radians it turns in one second. from screen.
## A wheel make 360 revolutions in one minute .Through how many radians does it turn in one second
A wheel make 360 revolutions in one minute .Through how many radians does it turn in one second
Byju's Answer Standard XII Mathematics
Sign of Trigonometric Ratios in Different Quadrants
A wheel make ... Question
A wheel make 360 revolutions in one minute .Through how many radians does it turn in one second
Open in App Solution
Number of revolutions made by the wheel in 1 minute = 360
∴Number of revolutions made by the wheel in 1 second =360/60 = 6
In one complete revolution, the wheel turns an angle of 2π radian.
Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e.,
Thus, in one second, the wheel turns an angle of 12π radian.
Suggest Corrections 40
SIMILAR QUESTIONS
Q.
How many degrees does a minute hand of a clock turn through in one hour?
Q. A circular wheel
28
inches in diameter rotates the same number of inches per second as a circular wheel
35
inches diameter. If the smaller wheel makes x revolutions per second, how many revolutions per minute does the larger wheel make in terms of
x .
Q. A wheel makes 12 revolutions per hour The radians it turns through in 20 minutes is:Q. A fly wheel rotates about a fixed axis and slows down from 400 rpm in one minute. How many revolutions does the wheel complete in the same timeQ. A wheel makes
360
revolutions in one minute. It turns
m π
radians in one second.Find
m View More EXPLORE MORE
Sign of Trigonometric Ratios in Different Quadrants
Standard XII Mathematics
स्रोत : byjus.com
## A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Click here👆to get an answer to your question ✍️ A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
SIMILAR QUESTIONS
AssertionStatement 1: cos1ReasonStatement 2: Cosine x decreases but sine x increases for x∈(0,
2 π )
Medium View solution >
Assertion
### Statement 2: In a triangle ABC, tanA+tanB+tanC=tanAtanBtanC ,there can be at most one obtuse angle in a triangle.
Medium View solution >
Assertionsin
−1
(sin5)=5 (where 5 is in radians).
Reasonsin
−1
(sinθ)=θ for principle value.
sin
−1
(sin5)=5 (where 5 is in radians).
A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45
Medium View solution > ∘
. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30
. Then the speed (in m/s) of the bird is
If p=tan1
Hard JEE Mains View solution > 0
,q=tan1(in radians), then which of the following is true?
Medium View solution >
स्रोत : www.toppr.com
## Ex 3.1, 3
Ex3.1, 3 A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? Number of revolutions in 1 minute = 360 Number of revolutions in 60 second = 360 So,Number of revolutions in 1 second = 360/60 = 6 Angle made in 1 revolution =
Check sibling questions
## Ex 3.1, 3 - Chapter 3 Class 11 Trigonometric Functions (Term 2)
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### Transcript
Ex3.1, 3 A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? Number of revolutions in 1 minute = 360 Number of revolutions in 60 second = 360 So,Number of revolutions in 1 second = 360/60 = 6 Angle made in 1 revolution = 360 Angle made in 6 revolution = 6 360 Radians made in 6 revolution = 6 360 /180 = 6 2 =12
Next: Ex 3.1, 4 Important →
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# Two tugboats pull a disabled supertanker
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80×106N1.80×106N, one 14∘14∘ west of north and the other 14∘14∘ east of north, as they pull the tanker 0.75 km toward the north.
What is the total work they do on the supertanker?
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# Homework Help: How to calculate the tension?
1. Feb 13, 2014
### fixedglare
1. The problem statement, all variables and given/known data
Due to friction, a force of 400 N is required to pull a wooden box on the floor. The cord used to pull the box makes an angle of 56° horizontally.
How much tension should be on the cord to be able to pull the box?
2. Relevant equations
W = Fd * (cosθ)
Tension = Weight +/- Mass * Acceleration ????? (found this one online, but was never taught this) or Ft=m(a+g) (never taught this one either just found it online)
3. The attempt at a solution
I read that to find/calculate tension you should use the second formula but I'm not sure.
Should I convert the Force to mass and then multiply 9.81 m/s2?
2. Feb 13, 2014
### fixedglare
On my book the answer says it should be 715 N, I divided the Force by the angle & got that answer but everywhere I search it says to use sin & other kinds of formulas I'm confused.
Then the second question asks how much work is done if the box is moved 25 m?
In my book the answer is supposed to be 10000 but when I use W= fd* cos θ, it gives me a different answer
3. Feb 13, 2014
### fixedglare
I'm very confused because the exercises in my book are under the section of using the angle to find work but when I used the basic W= Fd formula I got the answer in the book.
How do I know when to use the angle formula to find work and the basic formula?
4. Feb 13, 2014
### jackarms
To start, what do you mean you divided the force by the angle? Do you mean by the cosine of the angle?
And for the second part, what work is it asking you to find? Work from friction for tension? Show all of your calculations, and that should help me understand your questions better.
5. Feb 13, 2014
### fixedglare
Yes by the cosine angle, that's the only way I found the tension.
It doesn't specify, that's why I'm confused as to what formula to use and when, because the equation gave me the angle so I thought to use the angle but when I did, the book said it wasn't the right answer so then I used to basic formula & that's how I got it.
The second question just asks, how much work is realized, if the box is moved in a distance of 25.0 m?
6. Feb 13, 2014
### jackarms
Okay, I'm assuming it means how much work is done by the tension, since no net work is done on the box (work-kinetic energy theorem). Please show your calculations for the work. What values are you using to arrive at the answer in the book?
7. Feb 13, 2014
### fixedglare
To get the answer from the book I used the formula W= Fd;
so 400 N * 25.0 m = 1000 J, which is the answer in my book.
8. Feb 13, 2014
### jackarms
The problem is you're using the force from friction, and the work it's asking for is tension. You can get away with it here since the two works are equal, but the reasoning is incorrect. If the works weren't equal, this wouldn't work. You have to use the magnitude of the tension force in addition to the angle the force makes with the horizontal displacement.
9. Feb 13, 2014
### Staff: Mentor
Fixedglare: Did you draw a free body diagram before you started to try to work this problem? If so, on the free body diagram, did you identify all the components of the horizontal and vertical forces acting on the box? This should have automatically cleared up many of the difficulties you have had with this problem. If you drew a FBD, please upload it so we can see it.
Chet
10. Feb 14, 2014
### PhanthomJay
what a poor statement, are you sure the problem is worded this way? It means to say apparently that when you pull on the cord directed 56 degrees above the horizontal with a certain force, then it moves at constant velocity. The friction force on the box from the floor is 400 N. What is the value of the pulling force, and how much work does it do? Or something like that.
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# Stat. Sig of a dollar amount
#### Analyst1
##### New Member
Need help measuring the statistical significance of when a customer responds to a solicitation, & the size of their purchase.
This is a direct marketing scenario where we mail 5,000 solicitations to customers & test a different package to another 5,000 customers. (We split our 10,000 customers randomly into the 2 groups)
I can easily measure the statistical significance of the # of Responses. 200 respondents in population A & 150 in population B. But I also need to factor in the size of the purchase. Avg gift of population A was $25 & the Avg gift of population B was$45.
#### Michael Schmidt
##### New Member
Dollar amounts are just numbers and can be treated as such. You could take the two samples who gave (150 and 200) and compare them directly with t-tests. Alternatively, you could take all 5000 in each sample and do the same computation. The company might be interested in an average-dollars-per-contact figure, which would 1.00 for Group A and 1.35 for B. So, in a simpleminded way, the Group B methodology has a 35% advantage.
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# Shakashaka
Translate this site.
Shakashaka (Proof of Quilt) is a logic puzzle with simple rules and challenging solutions.
The rules are simple. Shakashaka is played on a rectangular grid. The grid has both black cells and white cells in it.
The objective is to place black triangles in the white cell in such a way so that they form white rectangular (or square) areas.
- The triangles are right angled and occupy half of the white square divided diagonally.
- You can place triangles only in white cells
- The numbers in the black cells indicate how many triangles are adjacent, vertically and horizontally.
- The white rectangles can be either straight or rotated at 45°
Video Tutorial
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20x20 Shakashaka Puzzle ID: 9,128,769
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(Solved) Investment A Demand Probability Outdoor Smoker High 0.2 Moderate 0.6 Low 0.2 Outdoor Grill High 0.2 Moderate 0.6 Low 0.
You are an economist for the Vanda-Laye Corporation, which produces and distributes outdoor cooking supplies. The company has come under new ownership and management and will be undergoing changes in its product lines and operating structure. As an economist, your responsibilities include examining the market factors that affect success or failure of a product, including the supply and demand for the product, market conditions, and the behavior of competitors with similar products.
The new management has identified several possible investments for the coming year. It has asked you and your team to evaluate the possibilities and make a recommendation to the board of directors. Jorge has identified two mutually exclusive opportunities (Investment A) and two independent opportunities (Investment B) and assigned you the task of making a recommendation on the investments.
Investment A
Your company would like to increase its product lines. Two alternatives are available, a new line of outdoor smokers and a new line of outdoor grills. The two lines are mutually exclusive, meaning that only one of these investment alternatives can be selected. The projected cash flows and their respective probabilities for each alternative are given in the table. There are three possible levels of demand and their corresponding probabilities, which depend on the state of the economy.
The two alternatives carry equal risk and should be evaluated at the company's cost of capital. The cost for the new smoker line will be \$7,000,000. Also, the company has been guaranteed a buyer for the new line at the end of the fifth year. The buyer has agreed to purchase the new line for \$7,900,000. The outdoor grill alternative will cost \$3,987,000 and also has a guaranteed buyer, who has agreed to pay \$4,000,000 at the end of the fifth year.
Investment B
Investment B involves two independent investment opportunities. The decisions on these two investment alternatives are also independent of Investment A. Investment B-1 involves a new packaging machine, which will eliminate the need for a local firm for packaging Vanda-Laye's products. The cost of this machine will be \$24,000, and the expected revenues from this opportunity are given in the table and are considered to be of average risk. Investment B-2 is the purchase of a new computer system that will allow the company to sell its products on the Internet worldwide. The cost of this new system will be \$29,000, with the expected cash flows after taxes given in the table.
Jorge has asked you to provide detailed responses to the following:
• Management of Vanda-Laye has determined that the capital structure of the company will involve 30% debt and 70% common equity. This structure will be used to finance all investments by the company. Currently, the company can sell new bonds at par, with a coupon rate of 7%. Any new common stock can be sold for \$45, with a required return (or cost) of 15.57%. Using Microsoft Excel, calculate the company's cost of capital to be used in the evaluation of possible investment projects.
• For Investment A:
• Using Microsoft Excel, create a decision tree. Indicate the various levels of demand and their respective probabilities. Also, include the calculations for the expected cash flows.
• Calculate the expected NPV for each alternative. Explain the decision rules for making a selection between the two alternatives on the basis of the expected NPV.
• Assuming the two alternatives are mutually exclusive, specify which alternative you would recommend to the company. Explain why.
• If the two alternatives were independent of each other, specify which project you would select. Would you accept both projects if funding were available for both? Explain your answer.
• For Investment B:
• Using Microsoft Excel, calculate the NPV for each alternative.
• Using the decision-making criteria for the NPV, specify which alternative you would select if the two alternatives were mutually exclusive. Explain your answer.
• Given that the two alternatives are independent of each other, specify which investment you would select, if not both. Explain your answer.
• Using Microsoft Excel, calculate the IRR for each investment.
• Using the decision-making criteria for the IRR, specify which alternative you would prefer. Explain your answer.
• If funding were available, specify whether you would select both investments. Why or why not?
• Calculate the profitability index (PI) for the two investments. Which project is preferred?
• Determine whether there is a ranking conflict present in terms of the IRR and the NPV. Explain your answer. If a conflict does exist, explain how you would resolve the situation
Investment A Demand Probability
Outdoor Smoker
High
0.2
Moderate
0.6
Low
0.2
Outdoor Grill
High
0.2
Moderate
0.6
Low
0.2 Year 1 Year 2 Year 3 Year 4 Year 5 \$800,000
\$500,000
\$200,000 \$900,000
\$700,000
\$350,000 \$1,000,000
\$800,000
\$500,000 \$1,100,000
\$960,000
\$600,000 \$1,500,000
\$1,240,000
\$750,000 \$600,000
\$450,000
\$150,000 \$750,000
\$500,000
\$220,000 \$850,000
\$700,000
\$370,000 \$975,000
\$825,000
\$500,000 \$5,160,000
\$4,980,000
\$4,750,000 Page 1 of 1
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Question about Casio FX82MS Scientific Calculator
# Raise power - Casio FX82MS Scientific Calculator
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A key marked [^], [X^y] or [Y^x]
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## Related Questions:
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### How do I raise the power of 30 on calculator?
If you are using an HP 12c or have the button on the calculator "yx" you first enter the number you wish to raise (y) then enter then the power (x). For instance to find 230 (2 to the power of 30) you input the number 2 then hit the enter key on the calculator, then input 30 and hit the yx button.
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Hope it helps.
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### 99 ford expedition power window
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# Three friends, A, B and C decided to have a beer party. If each of the
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Three friends, A, B and C decided to have a beer party. If each of the [#permalink]
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17 Oct 2018, 00:46
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Three friends, A, B and C decided to have a beer party. If each of the three friends consumed equal quantities of beer, and paid equally for it, what was the price of one beer bottle?
(1) A, B and C brought along 4, 6 and 2 bottles of beer, respectively; all bottles of beer being identical.
(2) C paid a total of \$16 to A and B for his share.
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Joined: 09 Jun 2014
Posts: 246
Location: India
Concentration: General Management, Operations
Schools: Tuck '19
Re: Three friends, A, B and C decided to have a beer party. If each of the [#permalink]
### Show Tags
17 Oct 2018, 02:33
2
Bunuel wrote:
Three friends, A, B and C decided to have a beer party. If each of the three friends consumed equal quantities of beer, and paid equally for it, what was the price of one beer bottle?
(1) A, B and C brought along 4, 6 and 2 bottles of beer, respectively; all bottles of beer being identical.
(2) C paid a total of \$16 to A and B for his share.
So we need to calculate the price of each beer bottle,considering they consumed equal quantities and the price per bottle was same.
Statement 1:
The statement simply status the number of beer bottles brought.
No information about whether they consumed all bottles of beer or not.Also no information about the price of
each bottle.
Statement 2"
The statement means that A paid \$16 for his share but we dont know how many bottles he consumed.
Combing equation 1 and 2,
We still dont know how many bottles have been consumed,we just know the number of bottles that were brought.
So Insufficient.
My answer would be E for this.Waiting for OA to be posted
Press Kudos if it helps !!
Intern
Joined: 29 Apr 2017
Posts: 26
Location: India
Concentration: Operations, Other
GMAT 1: 660 Q43 V38
GPA: 4
WE: Operations (Transportation)
Re: Three friends, A, B and C decided to have a beer party. If each of the [#permalink]
### Show Tags
18 Oct 2018, 00:52
prabsahi wrote:
Bunuel wrote:
Three friends, A, B and C decided to have a beer party. If each of the three friends consumed equal quantities of beer, and paid equally for it, what was the price of one beer bottle?
(1) A, B and C brought along 4, 6 and 2 bottles of beer, respectively; all bottles of beer being identical.
(2) C paid a total of \$16 to A and B for his share.
So we need to calculate the price of each beer bottle,considering they consumed equal quantities and the price per bottle was same.
Statement 1:
The statement simply status the number of beer bottles brought.
No information about whether they consumed all bottles of beer or not.Also no information about the price of
each bottle.
Statement 2"
The statement means that A paid \$16 for his share but we dont know how many bottles he consumed.
Combing equation 1 and 2,
We still dont know how many bottles have been consumed,we just know the number of bottles that were brought.
So Insufficient.
My answer would be E for this.Waiting for OA to be posted
Press Kudos if it helps !!
self also applied same logic to get to answer E but OA says C, can anyone explain?
Manager
Joined: 09 Jun 2014
Posts: 246
Location: India
Concentration: General Management, Operations
Schools: Tuck '19
Re: Three friends, A, B and C decided to have a beer party. If each of the [#permalink]
### Show Tags
18 Oct 2018, 01:00
1
Kumar Utkarsh wrote:
prabsahi wrote:
Bunuel wrote:
Three friends, A, B and C decided to have a beer party. If each of the three friends consumed equal quantities of beer, and paid equally for it, what was the price of one beer bottle?
(1) A, B and C brought along 4, 6 and 2 bottles of beer, respectively; all bottles of beer being identical.
(2) C paid a total of \$16 to A and B for his share.
So we need to calculate the price of each beer bottle,considering they consumed equal quantities and the price per bottle was same.
Statement 1:
The statement simply status the number of beer bottles brought.
No information about whether they consumed all bottles of beer or not.Also no information about the price of
each bottle.
Statement 2"
The statement means that A paid \$16 for his share but we dont know how many bottles he consumed.
Combing equation 1 and 2,
We still dont know how many bottles have been consumed,we just know the number of bottles that were brought.
So Insufficient.
My answer would be E for this.Waiting for OA to be posted
Press Kudos if it helps !!
self also applied same logic to get to answer E but OA says C, can anyone explain?
I think there is assumption in the question then between bringing bottles and consuming bottles..
In that case since its not explicitly stated in the question I would discard this question on quality issues.
Press Kudos if it helps!!
Intern
Joined: 05 Dec 2018
Posts: 13
Three friends, A, B and C decided to have a beer party. If each of the [#permalink]
### Show Tags
14 Dec 2018, 10:33
The point is: they brought different quantities of beers, but drank them equally and thus must pay equally. it asks uf for the price
1) total beers = 12. how much is one? insufficient
2) C owes 16\$ to A and B. It means that right now A and B had paid exactly 8\$ more than what they were supposed to. but how much is one Beer ? it could be 4\$ or 2% or 1 \$. How many beers there were ? it could be 16, 8, 4 or anything. INSUFFICIENT
1+2) total beers = 12
C owes 16\$ thus 8\$ to A and 8\$ to B.
so 12 = A+B+C
total cost = X(A+B+C) = X(12)
single expenditure = X(12)/3 = 4X
but A and B spent 4X + 8 each
while C spent 16 ( the exact money he owes and thus that everyone owes) thus,
4x = 16
x=4
total expenditure = 48
single expenditure = 16
Three friends, A, B and C decided to have a beer party. If each of the [#permalink] 14 Dec 2018, 10:33
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Convert Hundredweight to Slug (Mass / Weight)
## Hundredweight into Slug
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Hundredweight+to+Slug.php
# Convert Hundredweight to Slug (Mass / Weight):
1. Choose the right category from the selection list, in this case 'Mass / Weight'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Hundredweight'.
4. Finally choose the unit you want the value to be converted to, in this case 'Slug'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '439 Hundredweight'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Mass / Weight'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '88 Hundredweight to Slug' or '58 Hundredweight into Slug' or '82 Hundredweight -> Slug' or '76 Hundredweight = Slug'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(22 * 16) Hundredweight'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '34 Hundredweight + 28 Slug' or '10mm x 4cm x 97dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 4.667 738 229 128 5×1021. For this form of presentation, the number will be segmented into an exponent, here 21, and the actual number, here 4.667 738 229 128 5. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 4.667 738 229 128 5E+21. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 4 667 738 229 128 500 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications.
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# Exact Value of cos 54°
We will learn to find the exact value of cos 36 degrees using the formula of multiple angles.
How to find exact value of cos 54°?
Solution:
Let A = 18°
Therefore, 5A = 90°
⇒ 2A + 3A = 90˚
⇒ 2θ = 90˚ - 3A
Taking sine on both sides, we get
sin 2A = sin (90˚ - 3A) = cos 3A
⇒ 2 sin A cos A = 4 cos$$^{3}$$ A - 3 cos A
⇒ 2 sin A cos A - 4 cos$$^{3}$$ A + 3 cos A = 0
⇒ cos A (2 sin A - 4 cos$$^{2}$$ A + 3) = 0
Dividing both sides by cos A = cos 18˚ ≠ 0, we get
⇒ 2 sin θ - 4 (1 - sin$$^{2}$$ A) + 3 = 0
⇒ 4 sin$$^{2}$$ A + 2 sin A - 1 = 0, which is a quadratic in sin A
Therefore, sin θ = $$\frac{-2 \pm \sqrt{- 4 (4)(-1)}}{2(4)}$$
⇒ sin θ = $$\frac{-2 \pm \sqrt{4 + 16}}{8}$$
⇒ sin θ = $$\frac{-2 \pm 2 \sqrt{5}}{8}$$
⇒ sin θ = $$\frac{-1 \pm \sqrt{5}}{4}$$
Now sin 18° is positive, as 18° lies in first quadrant.
Therefore, sin 18° = sin A = $$\frac{-1 \pm \sqrt{5}}{4}$$
Now, cos 36° = cos 2 ∙ 18°
⇒ cos 36° = 1 - 2 sin$$^{2}$$ 18°
⇒ cos 36° = 1 - 2$$(\frac{\sqrt{5} - 1}{4})^{2}$$
⇒ cos 36° = $$\frac{16 - 2(5 + 1 - 2\sqrt{5})}{16}$$
⇒ cos 36° = $$\frac{1 + 4\sqrt{5}}{16}$$
⇒ cos 36° = $$\frac{\sqrt{5} + 1}{4}$$
Therefore, sin 36° = $$\sqrt{1 - cos^{2} 36°}$$,[Taking sin 36° is positive, as 36° lies in first quadrant, sin 36° > 0]
⇒ sin 36° = $$\sqrt{1 - (\frac{\sqrt{5} + 1}{4})^{2}}$$
⇒ sin 36° = $$\sqrt{\frac{16 - (5 + 1 + 2\sqrt{5})}{16}}$$
⇒ sin 36° = $$\sqrt{\frac{10 - 2\sqrt{5}}{16}}$$
⇒ sin 36° = $$\frac{\sqrt{10 - 2\sqrt{5}}}{4}$$
Therefore, sin 36° = $$\frac{\sqrt{10 - 2\sqrt{5}}}{4}$$
Now cos 54° = cos (90° - 36°) = sin 36° = $$\frac{\sqrt{10 - 2\sqrt{5}}}{4}$$
Therefore, cos 54° = $$\frac{\sqrt{10 - 2\sqrt{5}}}{4}$$
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