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• NAME
• DESCRIPTION
des_modes - Variants of DES and other crypto algorithms of
Several crypto algorithms for OpenSSL can be used in a
number of modes. Those are used for using block ciphers
in a way similar to stream ciphers, among other things.
Electronic Codebook Mode (ECB) [Toc] [Back]
Normally, this is found as the algorithm_ecb_encrypt()
function. 64 bits are enciphered at a time. The order of
the blocks can be rearranged without detection. The same
plaintext block always produces the same ciphertext block
(for the same key) making it vulnerable to a dictionary
attack. An error will only affect one ciphertext block.
Cipher Block Chaining Mode (CBC) [Toc] [Back]
Normally, this is found as the algorithm_cbc_encrypt()function.
Be aware that
des_cbc_encrypt() is not really DES CBC (it does not
update the IV); use the des_ncbc_encrypt() function
instead. A multiple of 64 bits are enciphered at a time.
The CBC mode produces the same ciphertext whenever the
same plaintext is encrypted using the same key and starting
variable. The chaining operation makes the ciphertext
blocks dependent on the current and all preceding plaintext
blocks and therefore blocks can not be rearranged.
The use of different starting variables prevents the same
plaintext enciphering to the same ciphertext. An error
will affect the current and the following ciphertext
Cipher Feedback Mode (CFB) [Toc] [Back]
Normally, this is found as the algorithm_cfb_encrypt()
function. A number of bits (j) <= 64 are enciphered at a
time. The CFB mode produces the same ciphertext whenever
the same plaintext is encrypted using the same key and
starting variable. The chaining operation makes the
ciphertext variables dependent on the current and all preceding
variables and therefore j-bit variables are chained
together and can not be rearranged. The use of different
starting variables prevents the same plaintext enciphering
to the same ciphertext. The strength of the CFB mode
depends on the size of k (maximal if j == k). Selection
of a small value for j will require more cycles through
the encipherment algorithm per unit of plaintext and thus
cause greater processing overheads. Only multiples of j
bits can be enciphered. An error will affect the current
and the following ciphertext variables.
Output Feedback Mode (OFB) [Toc] [Back]
Normally, this is found as the algorithm_ofb_encrypt()
function. A number of bits (j) <= 64 are enciphered at a
time. The OFB mode produces the same ciphertext whenever
the same plaintext enciphered using the same key and
starting variable. More over, in the OFB mode the same
key stream is produced when the same key and start variable
are used. Consequently, for security reasons a specific
start variable should be used only once for a given
key. The absence of chaining makes the OFB more vulnerable
to specific attacks. The use of different start
variables values prevents the same plaintext enciphering
to the same ciphertext, by producing different key
streams. Selection of a small value for j will require
more cycles through the encipherment algorithm per unit of
plaintext and thus cause greater processing overheads.
Only multiples of j bits can be enciphered. OFB mode of
operation does not extend ciphertext errors in the resultant
plaintext output. Every bit error in the ciphertext
causes only one bit to be in error in the deciphered
plaintext. OFB mode is not self-synchronizing. If the
two operation of encipherment and decipherment get out of
synchronism, the system needs to be reinitialized. Each
reinitialization should use a value of the start variable
different from the start variable values used before with
the same key. The reason for this is that an identical
bit stream would be produced each time from the same
parameters. This would be susceptible to a known plaintext
Triple ECB Mode [Toc] [Back]
Normally, this is found as the algorithm_ecb3_encrypt()
function . Encrypt with key1, decrypt with key2 and
encrypt with key3 again. As for ECB encryption but
increases the key length to 168 bits. There are theoretic
attacks that can be used that make the effective key
length 112 bits, but this attack also requires 2^56 blocks
of memory, not very likely, even for the NSA. If both
keys are the same it is equivalent to encrypting once with
just one key. If the first and last key are the same, the
key length is 112 bits. There are attacks that could
reduce the effective key strength to only slightly more
than 56 bits, but these require a lot of memory. If all 3
keys are the same, this is the same as normal ecb mode.
Triple CBC Mode [Toc] [Back]
Normally, this is found as the algorithm_ede3_cbc_encrypt()
function . Encrypt with key1,
decrypt with key2 and then encrypt with key3. As for CBC
encryption but increases the key length to 168 bits with
the same restrictions as for triple ecb mode.
[ Back ] | {"url":"https://nixdoc.net/man-pages/Tru64/man7/des_modes.7.html","timestamp":"2024-11-12T13:36:07Z","content_type":"text/html","content_length":"23146","record_id":"<urn:uuid:c7ab0e5a-6937-49db-a026-68ab775930a8>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00054.warc.gz"} |
Making Formulas… for Everything—From Pi to the Pink Panther to Sir Isaac Newton—Wolfram Blog
Making Formulas… for Everything—From Pi to the Pink Panther to Sir Isaac Newton
Here at Wolfram Research and at Wolfram|Alpha we love mathematics and computations. Our favorite topics are algorithms, followed by formulas and equations. For instance, Mathematica can calculate
millions of (more precisely, for all practical purposes, infinitely many) integrals, and Wolfram|Alpha knows hundreds of thousands of mathematical formulas (from Euler’s formula and BBP-type formulas
for pi to complicated definite integrals containing sin(x)) and plenty of physics formulas (e.g from Poiseuille’s law to the classical mechanics solutions of a point particle in a rectangle to the
inverse-distance potential in 4D in hyperspherical coordinates), as well as lesser-known formulas, such as formulas for the shaking frequency of a wet dog, the maximal height of a sandcastle, or the
cooking time of a turkey.
Recently we added formulas for a variety of shapes and forms, and the Wolfram|Alpha Blog showed some examples of shapes that were represented through mathematical equations and inequalities. These
included fictional character curves:
And, most popular among our users, person curves:
While these are curves in a mathematical sense, similar to say a lemniscate or a folium of Descartes, they are interesting less for their mathematical properties than for their visual meaning to
After Richard’s blog post was published, a coworker of mine asked me, “How can you make an equation for Stephen Wolfram’s face?” After a moment of reflection about this question, I realized that the
really surprising issue is not that there is a formula: a digital image (assume a grayscale image, for simplicity) is a rectangular array of gray values. From such an array, you could build an
interpolating function, even a polynomial. But such an explicit function would be very large, hundreds of pages in size, and not useful for any practical application. The real question is how you can
make a formula that resembles a person’s face that fits on a single page and is simple in structure. The formula for the curve that depicts Stephen Wolfram’s face, about one page in length, is about
the size of a complicated physics formula, such as the gravitational potential of a cube.
In this post, I want to show how to generate such equations. As a “how to calculate…”, the post will not surprisingly contain a fair bit of Mathematica code, but I’ll start with some simple
introductory explanations.
Assume you make a line drawing with a pencil on a piece of paper, and assume you draw only lines; no shading and no filling is done. Then the drawing is made from a set of curve segments. The
mathematical concept of Fourier series allows us to write down a finite mathematical formula for each of these line segments that is as close as wanted to a drawn curve.
As a simple example, consider the series of functions y[n](x),
which is a sum of sine functions of various frequencies and amplitudes. Here are the first few members of this sequence of functions:
Plotting this sequence of functions suggests that as n increases, y[n](x) approaches a triangular function.
The sine function is an odd function, and as a result all of the sums of terms sin(k x) are also odd functions. If we use the cosine function instead, we obtain even functions. A mixture of sine and
cosine terms allows us to approximate more general curve shapes.
Generalizing the above (-1)^(^k^ – 1)/2) k^-2 prefactor in front of the sine function to the following even or odd functions,
allows us to model a wider variety of shapes:
It turns out that any smooth curve y(x) can be approximated arbitrarily well over any interval [x[1], x[2]] by a Fourier series. And for smooth curves, the coefficients of the sin(k x) and cos(k x)
terms approach zero for large k.
Now given a parametrized curve γ(t) = {γ[x](t), γ[y](t)}, we can use such superpositions of sine and cosine functions independently for the horizontal component γ[x](t) and for the vertical component
γ[y](t). Using a sum of three sine functions and three cosine functions for each component,
covers a large variety of shapes already, including circles and ellipses. The next demonstration lets us explore the space of possible shapes. The 2D sliders change the corresponding coefficient in
front of the cosine function and the coefficient in front of the sine function. (Download this post as a CDF to interact)
If we truncate the Fourier expansion of a curve at, say, n terms, we have 4n free parameters. In the space of all possible curves, most curves will look uninteresting, but some expansion coefficient
values will give shapes that are recognizable. However, small changes in the expansion coefficients already quickly change the shapes. The following example allows a modification of the first 4 × 16
Fourier series coefficients of a curve (meaning 16 for the x direction and another 16 for the y direction). Using appropriate values for the Fourier coefficients, we obtain a variety of recognizable
And if we now take more than one curve, we already have all the ingredients needed to construct a face-like image. The following demonstration uses two eyes, two eye pupils, a nose, and a mouth.
And here is a quick demonstration of the reverse: we allow the position of a set of points (the blue crosses) that form a line to be changed and plot the Fourier approximations of this line.
Side note: Fourier series are not the only way to encode curves. We could use wavelet bases or splines, or encode the curves piecewise through circle segments. Or, with enough patience, using the
universality of the Riemann zeta function, we could search for any shape in the critical strip. (Yes, any possible [sufficiently smooth] image, such as Jesus on a toast, appears somewhere in the
image of the Riemann zeta function ζ(s) in the strip 0 ≤ Re(s) ≤ 1, but we don’t have a constructive way to search for it.)
To demonstrate how to find simple, Fourier series-based formulas that approximate given shapes, we will start with an example: a shape with sharp, well-defined boundaries—a short formula. More
concretely, we will use a well-known formula: the Pythagorean theorem.
Rasterizing the equation gives the starting image that we will use.
It’s easy to get a list of all points on the edges of the characters using the function EdgeDetect.
Now that we have the points that form the edges, we want to join them into straight-line (or curved) segments. The following function pointListToLines carries out this operation. We start with a
randomly chosen point and find all nearby points (using the function Nearest to be fast). We continue this process as long as we find points that are not too far away. We also try to continue in a
“straight” manner by slightly penalizing sharp turns. To see how the curve construction progresses, we use Monitor.
For the Pythagorean theorem, we obtain 13 individual curves from the edge points.
Joining the points and coloring each segment differently shows that we obtained the expected curves: the outer boundaries of the letters, the inner boundaries of the letters a and b, the three
squares, and the plus and equal signs.
Now for each curve segment we want to find a Fourier series (of the x and y component) that approximates the segment. The typical textbook definition of the Fourier coefficients of a function f(x)
are integrals of the function multiplied by cos(k x) and sin(k x). But at this point we have sets of points, not functions. To turn them into functions that we can integrate, we make a B-spline curve
of each curve segment. The parametrization variable of the B-spline curve will be the integration variable. (Using B-splines instead of piecewise linear interpolations between the points will have
the additional advantage of making jagged curves smoother.)
We could find the integrals needed to obtain the Fourier coefficients by numerical integration. A faster way is to use the fast Fourier transform (FFT) to get the Fourier coefficients.
To get more uniform curves, we perform one more step: re-parametrize the spline interpolated curve of the given curve segments by arclength. The function fourierComponents implements the B-spline
curve making, the re-parametrization by arclength, and the FTT calculation to obtain the Fourier coefficient. We also take into account if a curve segment is open or closed to avoid Gibbs phenomena
-related oscillations. (The above demonstration of approximating the pentagram nicely shows the Gibbs phenomenon in case the “Closed” checkbox is unchecked.)
For a continuous function, we expect an average decay rate of 1/k^2 for the kth Fourier series coefficient. This is the case for the just-calculated Fourier series coefficient. This means that on
average the 10th Fourier coefficient is only 1% in magnitude compared with the first one. This decay allows us to truncate the Fourier series at a not too high order, as we do not want to obtain
formulas that are too large. This expression gives the exponent in the decay rate of the Fourier components for the a^2 + b^2 = c^2 curve above. (The slightly lower than 2 exponent arises from the
discretization points in the B-spline curves.)
Here is a log-log-plot of the absolute values of the Fourier series coefficient for the first three curves. In addition to the general trend of an approximately quadratic decay of the Fourier
coefficients, we see that the magnitude of nearby coefficients often fluctuates by more than an order of magnitude.
Multiplying the Fourier coefficients by cos(k t) and sin(k t) and summing the terms gives us the desired parametrizations of the curves.
The function makeFourierSeriesApproximationManipulate visualizes the resulting curve approximations as a function of the series truncation order.
For the Pythagorean theorem, starting with a dozen ellipses, we quickly form the characters of the inequality with increasing Fourier series order.
We want a single formula for the whole equation, even if the formula is made from disjoint curve segments. To achieve this, we use the 2π periodicity of the Fourier series of each segment to plot the
segments for the parameter ranges [0, 2π], [4π, 6π], [8π, 10π], …, and in the interleaving intervals (2π, 4π), (6π, 8π), …, we make the curve coordinates purely imaginary. As a result, the curve
cannot be drawn there, and we obtain disjoint curve segments. Here this construction is demonstrated for the case of two circles:
The next plot shows the real and imaginary parts of the complex-valued parametrization independently. The red line shows the purely imaginary values from the parameter interval [2π, 4π].
As we want the final formula for the curves to look as short and as simple as possible, we change sums of the form a cos(k t) + b sin(k t) to A sin(k t + φ) using the function sinAmplitudeForm and
round the floating-point Fourier series coefficients to nearby rationals. Instead of Piecewise, we use UnitStep in the final formula to separate the individual curve segments. The real segments we
list in explicit form, and all segments that should not be drawn are encoded through the θ(sgn(sin(t/2)^(1/2))) term.
Now we have everything together to write down the final parametrization {x(t), y(t)} of the typeset form of the Pythagorean theorem as a mathematical formula.
After having discussed the principal construction idea for the parametrizations, let’s look at a more fun example, say the Pink Panther. Looking at the image search results of the Bing search engine,
we quickly find an image that seems desirable for a “closed form” parametrization.
Let’s use the following image:
We apply the function EdgeDetect to find all edges on the panther’s face.
Connecting the edges to curve segments yields about 20 segments. (Changing the optional second and third argument of pointListToLines, we obtain fewer or more segments.)
Here is each segment shown with a different color. We see that some closed curves arise from two joined curve segments; we could separate them by changing the second argument of pointListToLines. But
for the goal of sketching a line drawing, the joined curve will work just fine.
Proceeding now as above, it is straightforward to approximate the curve segments by trigonometric series.
Plotting the series shows that with 20 terms per segment, we obtain a good representation of the Pink Panther.
As some of the segments of the panther’s face are more intricate than others, we define a function makeSegmentOrderManipulate that allows the number of terms of the Fourier series for each segment to
be varied. This lets us further reduce the size of the resulting parametrizations.
We use initial settings for the number of Fourier coefficients that yield a clearly recognizable drawing of the Pink Panther.
For simple cases, we can now roll up all of the above function into a single function. The next function makeSilhouetteFourierResult takes a string as an argument. The function then 1) performs a
search on Bing’s image site for this string; 2) selects an image that seems appropriate from the algorithmic point of view; 3) calculates the Fourier series; and 4) returns as the result plots of the
Fourier series and an interactive version that lets us change the order of the Fourier series. For simplicity, we restrict the program to only single curves. (In the web search, we use the word
“silhouette” to mostly retrieve images that are formed by just a single curve.) As the function relies on the result of an external search engine, there is no guarantee that the function will always
return the wanted result.
Here are three examples showing the function at work. We build the Fourier series for a generic spider, Spiderman’s spider, a couple dancing the tango, and a mermaid. (Evaluating these functions
might give different results, as the search engine results might change over time.)
So far, the initial line segments were computed from images. But we can also start with hand-drawn curves. Assume we want a formula for Isaac Newton. As I am not good at drawing faces, I cheated a
bit and used the curve drawing tool to draw characteristic facial and hair curves on top of an image of Sir Isaac. (For algorithmic approaches on how to extract feature lines from faces, see the
recent paper by Zhao and Zhu.) Here is the image that we will use:
Fortunately, small random wiggles in the hand-drawn curve will not matter, as they will be removed by omitting higher-order terms in the Fourier series.
To better see the hand-drawn curves, we separate the image and the lines.
This time, we have 16 segments. We build their Fourier series.
And here are again various approximation orders of the resulting curves.
We use different series orders for the various segments. For the hair, we use relatively high orders, and for the eyes relatively low orders. This will make sure that the resulting equations for the
face will not be larger than needed.
Here are the first 50 approximations shown in one graphic with decreasing opacity of the curves, and with each set of corresponding curve segments shown in a different color.
This gives the following final curve for Sir Isaac’s portrait.
And this is the plotted form of the last formula.
This ends today’s discussion about how to make curves that resemble people’s faces, fictional characters, animals, or other shapes. Next time, we will discuss the endless graphics capabilities that
arise from these formulas and how you can use these types of equations in a large variety of images.
To interact with the examples above, first, download the Wolfram CDF Player. Then, download this post as a Computable Document Format (CDF) file.
Join the discussion
9 comments
1. Nice work Michael.
You could have fun coming up with “A New Formula for Pi”, where the output was just the outline of either the Greek letter or the English equivalent….
2. The show is quite interesting. Looking at the pink panther original picture, it seems to me that the basis of functions that our brain is using to sketch shapes is not the fourier one (do not
know how to clarify this concept, but perhaps you get me anyway…). With MMA capabilities would it be possible to switch to a different basis and look what would happen?
My compliments.
3. i wonder what is the average of two faces, so if we write face1+face2 = result face. could this be implemented or it is already available. a more general what is the face of all faces on the
globe. ie to what general face all faces will collapse to. i’m sure this is a very amusing subject. and thanks for your great article.
4. where can this program be found for use?
□ Hello Alex, you can purchase and download Mathematica at this link
5. Good read. Thanks. I always assumed everything had some equation no matter how long and drawn-out it would be but this post really helped me understand. | {"url":"https://blog.wolfram.com/2013/05/17/making-formulas-for-everything-from-pi-to-the-pink-panther-to-sir-isaac-newton/","timestamp":"2024-11-03T23:35:27Z","content_type":"text/html","content_length":"159700","record_id":"<urn:uuid:9a65bec2-e603-4724-9179-962fbb6d599e>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00344.warc.gz"} |
AMGN: Amgen | Logical Invest
What do these metrics mean?
'Total return, when measuring performance, is the actual rate of return of an investment or a pool of investments over a given evaluation period. Total return includes interest, capital gains,
dividends and distributions realized over a given period of time. Total return accounts for two categories of return: income including interest paid by fixed-income investments, distributions or
dividends and capital appreciation, representing the change in the market price of an asset.'
Which means for our asset as example:
• Compared with the benchmark SPY (101.5%) in the period of the last 5 years, the total return, or increase in value of 70.4% of Amgen is smaller, thus worse.
• Compared with SPY (29.7%) in the period of the last 3 years, the total return of 67.4% is greater, thus better.
'The compound annual growth rate isn't a true return rate, but rather a representational figure. It is essentially a number that describes the rate at which an investment would have grown if it had
grown the same rate every year and the profits were reinvested at the end of each year. In reality, this sort of performance is unlikely. However, CAGR can be used to smooth returns so that they may
be more easily understood when compared to alternative investments.'
Applying this definition to our asset in some examples:
• Looking at the annual return (CAGR) of 11.3% in the last 5 years of Amgen, we see it is relatively lower, thus worse in comparison to the benchmark SPY (15.1%)
• During the last 3 years, the compounded annual growth rate (CAGR) is 18.8%, which is higher, thus better than the value of 9.1% from the benchmark.
'Volatility is a rate at which the price of a security increases or decreases for a given set of returns. Volatility is measured by calculating the standard deviation of the annualized returns over a
given period of time. It shows the range to which the price of a security may increase or decrease. Volatility measures the risk of a security. It is used in option pricing formula to gauge the
fluctuations in the returns of the underlying assets. Volatility indicates the pricing behavior of the security and helps estimate the fluctuations that may happen in a short period of time.'
Applying this definition to our asset in some examples:
• The volatility over 5 years of Amgen is 25.8%, which is greater, thus worse compared to the benchmark SPY (20.9%) in the same period.
• Looking at historical 30 days volatility in of 22.2% in the period of the last 3 years, we see it is relatively larger, thus worse in comparison to SPY (17.6%).
'The downside volatility is similar to the volatility, or standard deviation, but only takes losing/negative periods into account.'
Applying this definition to our asset in some examples:
• Compared with the benchmark SPY (14.9%) in the period of the last 5 years, the downside deviation of 16.8% of Amgen is larger, thus worse.
• Looking at downside volatility in of 14% in the period of the last 3 years, we see it is relatively higher, thus worse in comparison to SPY (12.3%).
'The Sharpe ratio is the measure of risk-adjusted return of a financial portfolio. Sharpe ratio is a measure of excess portfolio return over the risk-free rate relative to its standard deviation.
Normally, the 90-day Treasury bill rate is taken as the proxy for risk-free rate. A portfolio with a higher Sharpe ratio is considered superior relative to its peers. The measure was named after
William F Sharpe, a Nobel laureate and professor of finance, emeritus at Stanford University.'
Which means for our asset as example:
• Looking at the risk / return profile (Sharpe) of 0.34 in the last 5 years of Amgen, we see it is relatively lower, thus worse in comparison to the benchmark SPY (0.6)
• During the last 3 years, the ratio of return and volatility (Sharpe) is 0.73, which is higher, thus better than the value of 0.37 from the benchmark.
'The Sortino ratio measures the risk-adjusted return of an investment asset, portfolio, or strategy. It is a modification of the Sharpe ratio but penalizes only those returns falling below a
user-specified target or required rate of return, while the Sharpe ratio penalizes both upside and downside volatility equally. Though both ratios measure an investment's risk-adjusted return, they
do so in significantly different ways that will frequently lead to differing conclusions as to the true nature of the investment's return-generating efficiency. The Sortino ratio is used as a way to
compare the risk-adjusted performance of programs with differing risk and return profiles. In general, risk-adjusted returns seek to normalize the risk across programs and then see which has the
higher return unit per risk.'
Which means for our asset as example:
• Compared with the benchmark SPY (0.84) in the period of the last 5 years, the downside risk / excess return profile of 0.52 of Amgen is lower, thus worse.
• Looking at downside risk / excess return profile in of 1.17 in the period of the last 3 years, we see it is relatively larger, thus better in comparison to SPY (0.53).
'Ulcer Index is a method for measuring investment risk that addresses the real concerns of investors, unlike the widely used standard deviation of return. UI is a measure of the depth and duration of
drawdowns in prices from earlier highs. Using Ulcer Index instead of standard deviation can lead to very different conclusions about investment risk and risk-adjusted return, especially when
evaluating strategies that seek to avoid major declines in portfolio value (market timing, dynamic asset allocation, hedge funds, etc.). The Ulcer Index was originally developed in 1987. Since then,
it has been widely recognized and adopted by the investment community. According to Nelson Freeburg, editor of Formula Research, Ulcer Index is “perhaps the most fully realized statistical portrait
of risk there is.'
Applying this definition to our asset in some examples:
• Compared with the benchmark SPY (9.32 ) in the period of the last 5 years, the Downside risk index of 10 of Amgen is larger, thus worse.
• Looking at Downside risk index in of 9.82 in the period of the last 3 years, we see it is relatively lower, thus better in comparison to SPY (10 ).
'Maximum drawdown measures the loss in any losing period during a fund’s investment record. It is defined as the percent retrenchment from a fund’s peak value to the fund’s valley value. The drawdown
is in effect from the time the fund’s retrenchment begins until a new fund high is reached. The maximum drawdown encompasses both the period from the fund’s peak to the fund’s valley (length), and
the time from the fund’s valley to a new fund high (recovery). It measures the largest percentage drawdown that has occurred in any fund’s data record.'
Which means for our asset as example:
• Compared with the benchmark SPY (-33.7 days) in the period of the last 5 years, the maximum reduction from previous high of -24.9 days of Amgen is larger, thus better.
• Compared with SPY (-24.5 days) in the period of the last 3 years, the maximum drop from peak to valley of -24.9 days is lower, thus worse.
'The Drawdown Duration is the length of any peak to peak period, or the time between new equity highs. The Max Drawdown Duration is the worst (the maximum/longest) amount of time an investment has
seen between peaks (equity highs). Many assume Max DD Duration is the length of time between new highs during which the Max DD (magnitude) occurred. But that isn’t always the case. The Max DD
duration is the longest time between peaks, period. So it could be the time when the program also had its biggest peak to valley loss (and usually is, because the program needs a long time to recover
from the largest loss), but it doesn’t have to be'
Which means for our asset as example:
• Compared with the benchmark SPY (488 days) in the period of the last 5 years, the maximum days under water of 244 days of Amgen is lower, thus better.
• Compared with SPY (488 days) in the period of the last 3 years, the maximum days under water of 230 days is smaller, thus better.
'The Average Drawdown Duration is an extension of the Maximum Drawdown. However, this metric does not explain the drawdown in dollars or percentages, rather in days, weeks, or months. The Avg
Drawdown Duration is the average amount of time an investment has seen between peaks (equity highs), or in other terms the average of time under water of all drawdowns. So in contrast to the Maximum
duration it does not measure only one drawdown event but calculates the average of all.'
Which means for our asset as example:
• Looking at the average time in days below previous high water mark of 71 days in the last 5 years of Amgen, we see it is relatively lower, thus better in comparison to the benchmark SPY (123
• During the last 3 years, the average days under water is 59 days, which is smaller, thus better than the value of 177 days from the benchmark. | {"url":"https://logical-invest.com/app/stock/amgn/amgen","timestamp":"2024-11-04T12:07:29Z","content_type":"text/html","content_length":"61412","record_id":"<urn:uuid:ee35a9e4-8e9d-496b-9f25-35392f32ebc0>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00470.warc.gz"} |
Pre-Algebra serves as the foundational course for understanding the core concepts of algebra and higher mathematics. This course introduces students to basic mathematical principles and
problem-solving techniques, providing a comprehensive understanding of operations with whole numbers, fractions, decimals, ratios, proportions, and percentages. Through practical examples, students
learn to apply these skills in real-life situations. Pre-Algebra also introduces essential algebraic concepts, such as variables, expressions, and simple equations, laying the groundwork for future
courses in algebra and beyond. | {"url":"http://maxwell15000.ddns.net/","timestamp":"2024-11-08T23:33:10Z","content_type":"text/html","content_length":"27910","record_id":"<urn:uuid:3f0a14cb-3465-4629-92db-2d6f46f1c927>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00060.warc.gz"} |
CMPSC 122 Homework 3
This assignment will focus on the implementation of a simple single-linked list, and then using it as a stack.
Problem Description
Although people may be very accustomed to reading and understanding calculations like those in the preceding assignments, it is evident they are not very self-evident or obvious in the world of
computing. One way to simplify the problem is just to redefine how expressions are represented, to allow for simpler interpretation.
One knows that a plus sign means to add, but typically when a plus sign is encountered in the middle of an expression, there is not yet enough information to know what values are to be added. One has
to continue scanning the expression to find the second operand, and also decide if there are other operations of higher precedence that must take place first. How much simpler it would be if a parser
could know to add just as soon as the plus sign appears.
This is the motivation for a representation called postfix notation, in which all operators follow their operands. Here are some illustrations.
Infix Representation Postfix Representation Operations
(1 + 2) * 3 1 2 + 3 * Add 1 and 2, then multiply by 3
1 + (2 * 3) 1 2 3 * + Multiply 2 and 3, then add to 1
(1+2) * (6-4) 1 2 + 6 4 – * Add 1 and 2, subtract 4 from 6, multiply
Hewlett Packard has produced a line of calculators that expected all inputs to be in postfix form, so every operation key could compute the moment it was pressed.
The goal of this assignment is to convert an infix expression into a postfix expression. This can be accomplished using the same algorithm as was used to evaluate the infix expression, simply
yielding a new expression instead of a computed value. In the last example, instead of adding, it would produce an expression ending with a plus sign. The multiplication function would similarly
produce an expression from its operands, followed by the multiplication operator.
Applying a Linked List
One of the purposes of this course is to find the data structures that would assist in producing the best results as efficiently as possible. The linked list is quite serviceable for the needs of
this assignment.
A linked list will be useful in the calculation portion of this assignment. As the postfix expression is scanned, values must be saved until an operator is discovered. Each operator would apply to
the two values that precede it, and then its result would also be saved. As an extreme example, consider this expression:
1 2 3 4 5 * – + – multiply 4 * 5, subtract from 3, add to 2, subtract from 1
Note this is not at all the same meaning as:
1 2 * 3 – 4 + 5 – or 4 5 * 3 – 2 + 1 –
(If you need to more clearly see the difference, try inserting parentheses around all operations, such that each parentheses consists of two expressions followed by an operator.)
Defining a Linked List
The linked list is a rather simple data structure, and the required operations should be rather simple, so very little will be said here about what to do. Instead, here is a quick highlight of what
should appear in your implementation. For consistency, call the implementation file linkedlist.py
Required Features in Implementation
Nested Node class definition for single-linked list node
with __slots__ to save list memory
and __init__ a constructor
__init__() a constructor for an empty list
push(value) add a value in constant time
pop() retrieve last insertion in constant time
Good functions for Completeness / Practice
top() return last insertion, without removal
is_empty() Identify whether list is empty
__len__() returns number of elements in list
__iter__() iterator function (with yield statements)
__str__() represents entire list as a string
For full credit, this must be no worse than log-linear time, not quadraticThe __iter__ function will be used to traverse a list to support __str__.
__str__ would allow a list to be an argument to print() for debugging
Assignment Specifications
Three new files are to appear in the solution to this assignment:
Implements a linked list as a class
given an iterator to an infix expression,
produces a generator for a postfix expression
evaluates a postfix expression, given an iterator
Do include newsplit.py in your submission since it is a necessary part of the project.
You are also encouraged to insert your own unit testing into each file, but that will not be itself graded.
Helpful Time-Saving Hint:
One feature of an interpreted language like Python is to create code at run-time to execute. You can support all the calculations by taking advantage of this feature:
left = ‘2’
right = ‘2’
op = ‘+’
eval( left+op+right ) # evaluate “2+2” to get 4
This time-saving will become even more useful when we will support all the relation operators (such as > and ==)
Testing and Evaluation
Here are a few lines of code from the instructor’s solution and test at this time of writing:
(in infixtoiter.py)
from peekable import Peekable, peek
from newsplit import new_split_iter
def to_postfix( expr ):
return postfix_sum(Peekable(new_split_iter(expr)))
(in the test program)
from infixtoiter import to_postfix
from evalpostfix import eval_postfix
def test(expr):
print (expr, ‘=’, eval_postfix(to_postfix(expr)) )
So here is the sequence of function calls as they operate on the input:
The process begins with a character string in expr
It is broken into tokens by new_split_iter, which yields an iterator
The Peekable constructor adds peek functionality
The parsing functions in infixtoiter produce a iterator generator describing a postfix expression
That expression is evaluated by eval_postfix
The original string and computed value are displayed
Extra Credit Option
Support those optional class features listed above, along with suitable unit tests. | {"url":"https://codingprolab.com/answer/cmpsc-122-homework-3/","timestamp":"2024-11-12T02:36:12Z","content_type":"text/html","content_length":"113142","record_id":"<urn:uuid:08422f26-ca88-47d3-b189-d85c7f8b4573>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00479.warc.gz"} |
Java Program to Calculate Standard Deviation | Vultr Docs
Standard deviation is a statistic that measures the dispersion of a dataset relative to its mean. It is calculated as the square root of the variance. In programming, especially in data analysis and
statistics, calculating the standard deviation is essential for understanding the spread of numerical data. This measure can highlight outliers and data consistency, which are crucial in many
analytical applications.
In this article, you will learn how to compute the standard deviation in Java. You'll explore various examples using different approaches including loops, Java streams, and external libraries to
handle various data scenarios effectively.
Calculate Standard Deviation with Array
Using a For Loop
1. Determine the mean (average) of the data set.
2. Calculate the sum of the squared differences from the mean.
3. Divide by the number of data points to get the variance.
4. Take the square root of the variance to get the standard deviation.
public class StandardDeviation {
public static double calculateSD(double[] numArray) {
double sum = 0.0, standardDeviation = 0.0;
int length = numArray.length;
for (double num : numArray) {
sum += num;
double mean = sum / length;
for (double num : numArray) {
standardDeviation += Math.pow(num - mean, 2);
return Math.sqrt(standardDeviation / length);
public static void main(String[] args) {
double[] numArray = {10.0, 20.5, 30.8, 40.7, 25.6};
System.out.printf("The standard deviation is: %.2f", calculateSD(numArray));
This code calculates the standard deviation of an array of double numbers. It iterates over the array to get the total sum and the sum of the squared deviations. The square root of the average
squared deviation gives the standard deviation.
Calculate Standard Deviation Using Java Stream API
Stream Operations
1. Convert the array into a stream.
2. Use stream operations to calculate the mean.
3. Use another stream to find the sum of squared differences from the mean.
4. Finish by calculating the square root of the average squared difference.
import java.util.stream.DoubleStream;
public class StandardDeviationStream {
public static double calculateSD(double[] numArray) {
double mean = DoubleStream.of(numArray).average().orElse(Double.NaN);
double sumOfSquaredDiffs = DoubleStream.of(numArray)
.map(x -> (x - mean) * (x - mean))
return Math.sqrt(sumOfSquaredDiffs / numArray.length);
public static void main(String[] args) {
double[] numArray = {10.0, 20.5, 30.8, 40.7, 25.6};
System.out.printf("The standard deviation is: %.2f", calculateSD(numArray));
This version uses Java streams for a more concise and functional-style implementation, achieving the same result but with potentially cleaner code especially for large datasets.
Calculating the standard deviation in Java can be approached in various ways depending on your preference for code style and the specific requirements of your application. Whether you choose the
traditional loop method or the more modern stream approach, both can provide accurate calculations for the standard deviation of numerical data. Understanding how to implement these calculations is
crucial for tasks in data analytics, financial analysis, and any other fields where data variance is an important metric. By mastering these techniques, you ensure that your Java applications handle
statistical calculations efficiently. | {"url":"https://docs.vultr.com/java/examples/calculate-standard-deviation","timestamp":"2024-11-09T12:22:50Z","content_type":"text/html","content_length":"236640","record_id":"<urn:uuid:8af71c4a-11d5-4d54-a3a4-75bf9ad31757>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00337.warc.gz"} |
The College Dorm Anomaly
(adapted from two FB posts)
tl;dr: Population vax percentages count some people in different counties for the numerator and the denominator; jurisdictions with high proportions of college students in dorms are likely to have
higher vax rates than reported.
If you're looking at vaccination rates at a local level, be very careful about what the data might be telling you. The Virginia dashboards have some nice visuals for demographic breakdowns and I went
straight to our county stats:
As soon as I saw that graph I knew what had happened: the vax percentages for the 18-24 age band are way low, because college students are counted in the denominator (presumably because the census
counts college students in the dorms where they live) but not in the numerator (at least, not if they get vaxxed back at home, and maybe even if they get vaxxed here if data tracking is based on
their "permanent address" or similar). Consider, by comparison, the same stats for Charlotte County, a nearby, mostly similar county that doesn't have two big colleges in the middle of it:
It's more what you'd expect: some variation and even trendlines by age band, but no outlier drop for the 18-24 band. (It's even a little bit high, which might reflect that college kids from Charlotte
County that got vaxxed there are counted in the numerator but _not_ the denominator!)
We know from announcements by the two colleges that the actual student vaccination rate is in fact much higher than that of the overall county population. If you look at the percentages by age band,
Prince Edward is comparable or higher than Charlotte in every other age band except 18-24 (and, oddly, 85+), but the reported summary stats show us as considerably worse, thanks to the College Dorm
Anomaly. Looking at some other cities and counties around Virginia shows similar outlier-low percentages for the 18-24 band in other college areas such as Charlottesville (UVa), Montgomery County
(Tech), Williamsburg (W&M).
So I ran the numbers, and if I'm right about this, the way these numbers are reported is really misleading. All analysis below is for Prince Edward County, Virginia, but will apply in any area with
this kind of reporting that has a significant proportion of "temporary resident" college students—whether or not they are explicitly reporting by age bands, it's just that the age bands make it
easier to confirm this is happening.
To test my idea that the local vax rates might be a lot higher than officially reported (spoiler alert: they certainly seem to be!), I grabbed the vax counts and percentages from the site (and from
those derived the total recorded population per age band) and came up with four models that measure and/or adjust for the college students that get mis-counted.
(Model 0: numbers as reported. Leads to an 18+ vax rate of 51.4% for the county, also matching reports.)
Model 1: ignore all 18-24 year olds! Exclude the 18-24 year olds from both the count of those vaccinated and from the total population.
Model 2: pretend 18-24 year olds are an average of 16-34 year olds. Both for estimating size of cohort (how many people in the age band) and for estimating vax behaviour (how many within this
band are vaccinated), use an average of the 16-17 band and the 25-34 band. (Careful: the bands are different sizes! I accounted for this.)
Model 3: remove "non-counted college students" from the denominator. If their vaccinations wouldn't be reported in the numerator, then exclude them from the denominator as well. Challenge: how to
know how many college students aren't counted? My initial plan was to estimate about 4500 (based on the size of the two colleges and a bit of guesswork regarding how many would count it as their
residence), but see below.
Model 4: add the vaxxed college students to the numerator. Since they're actually here, might as well count them in our numbers, right? For the "non-counted" students (as in Model 3), we can
assume that their vaccination behaviour matches that of the colleges specifically, and while I'm having a hard time tracking down current numbers on how many students at Longwood and H-SC are
vaccinated, I'm remembering reported rates of 70% from earlier in the semester, so I'll use that for now. So, take 70% of the "non-counted" student number and add that into the vax count for the
18-24 band.
Under any of these models, the Prince Edward vax rate is something like 15 points higher than reported. Here is a side-by-side comparison of the age-band graphs for the numbers as reported, and for
the numbers from Model 4:
And the numbers produced by each model:
(M0: Reported 18+ vax rate: 51.4%.)
M1: This produces an 18+ vax rate of 67.7%! But that's probably a bit high, because we know that seniors are vaxxed at a higher-than-average rate.
M2: 18+ vax rate is 65.4%. But perhaps more importantly, this model also gives us a more direct estimate of "un-counted" students: since it provides a guess for how many 18-24 people we'd expect
in the county, we can subtract that from the total reported 18-24 population to compute the "unexpected 18-24 year olds": 4122. This is quite close to my guess of 4500, so I used the computed
number as the number of "un-counted" students in M3 and M4.
M3: 18+ vax rate is 65.5%.
M4: 18+ vax rate is 66.4%. A graph of the adjusted age band percentages is also shown below; again, this is assuming a 70% vaccination rate for the unexpected/uncounted 18-24 year olds. The model
is somewhat sensitive to minor variations in the size of that group (e.g. 4122 vs 4500) but more sensitive to variation in the rate of vaccination in that group (e.g. 70 vs 80); but it seems like
all reasonable numbers for those put the resulting 18+ vax rate in the high 60s, as in the other models.
So regardless of which of those models you find most persuasive, I think we can conclude that the Prince Edward numbers, rather than being at the low end of the state with 51.4% for 18%, are a lot
closer to the middle at 65-67%. That's still low for the state overall (which is reporting 80% for 18+) but relatively high for the rural counties and definitely higher than any of our neighbouring
counties. To believe that PE's numbers are 51.4 you would have to believe that 18-24 year olds in our county have massively different behaviour than people slightly younger or slightly older, and
different than 18-24 year olds in all the nearby counties! AND ALSO you'd have to believe that the colleges were lying about their students' vaccination rates. So I'm extremely comfortable in
concluding that the 18+ rate here is about 66% or so right now.
I'm sure the phenomenon is not unique to Virginia, of course; if you're looking at county- or town-level vaccination data in your area, try to get your hands on age-band data if it's available, or at
least make a mental note about this possible limitation of the data reporting. If you're making decisions on where to go or how to act when you're out, you should do so in light of the real numbers,
not the anomalous ones.
"Without teachers, humankind would be sitting in a cold, dark, clammy cave, picking nits and wondering how Grandfather started his legendary fire." --Elle Newmark, The book of unholy mischief
—Comments on Facebook— Posted by blahedo at 11:15am on 18 Nov 2021 | {"url":"http://www.blahedo.org/blog/archives/001149.html","timestamp":"2024-11-07T06:20:44Z","content_type":"application/xhtml+xml","content_length":"13600","record_id":"<urn:uuid:c5727087-e1a1-4159-b000-2261d3a00afb>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00661.warc.gz"} |
[Solved] Prove that:2sin125πsin12π = 21 - Maths XI... | Filo
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Class Class 11
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Lesson 4
Dilating Lines and Angles
4.1: Angle Articulation (10 minutes)
In middle school, students studied many examples of dilations and verified experimentally that dilations preserve angle measure. In this activity, students confirm this claim for one example, and
recall what they know about dilations from middle school.
Student Facing
1. What do you think is true about the angles in \(A’B’C’\) compared to the angles in \(ABC\)?
2. Use the tools available to figure out if what you thought was true is definitely true for these triangles.
3. Do you think it would be true for angles in any dilation?
Anticipated Misconceptions
If students struggle to come up with a conjecture ask, "What stays the same? What changes?"
Activity Synthesis
Invite students to share their conjectures.
Once students have shared several conjectures, ask students to generalize a claim about angles under dilation. (Dilations keep the measures of angles constant.) They may recall that dilations
preserve angle measure from middle school. They may also connect the preservation of angle to the idea that dilations don’t distort the shape of a figure, it just changes its size.
Tell students that we aren’t going to prove that dilations preserve angle measure in this class, we’re going to assert that it is true.
Ask students to add this assertion to their reference charts as you add it to the class reference chart:
If a figure is dilated, then corresponding angles are congruent. (Assertion)
\(\triangle A’B’C’\) is a dilation of \(\triangle ABC\) so \(\angle B \cong \angle B'\)
4.2: Dilating Lines (10 minutes)
The purpose of this activity is for students to verify experimentally that dilations take lines through the center of dilation to the same line, even though specific points on the line get farther
from or closer to the center of dilation according to the same ratio given by the scale factor. Students first dilate points on given lines, and then they are asked to describe what happens to the
lines when they are dilated.
For the sake of time invite students to estimate rather than measure precisely.
Engagement: Internalize Self Regulation. Chunk this task into more manageable parts to differentiate the degree of difficulty or complexity. Invite students to choose and respond to one scale factor
greater than one and one scale factor smaller than one.
Supports accessibility for: Organization; Attention
Student Facing
1. Dilate point \(A\) using center \(C\) and scale factor \(\frac{3}{4}\).
2. Dilate point \(B\) using center \(C\) and scale factor \(\frac13\).
3. Dilate point \(D\) using center \(C\) and scale factor \(\frac32\).
4. Dilate line \(CE\) using center \(C\) and scale factor 2.
5. What happens when the center of dilation is on a line and then you dilate the line?
Student Facing
Are you ready for more?
• \(X\) is the midpoint of \(AB\).
• \(B'\) is the image of \(B\) after being dilated by a scale factor of 0.5 using center \(C\).
• \(A'\) is the image of \(A\) after being dilated by a scale factor of 0.5 using center \(C\).
Call the intersection of \(CX\) and \(A'B'\) point \(X'\). Is point \(X'\) a dilation of point \(X\)? Explain or show your reasoning.
Anticipated Misconceptions
If students are distracted by all the other points on the diagram, suggest they use tracing paper to trace only the relevant points. Repeat for each question. Then transfer all the points back onto
the original diagram before the synthesis.
Activity Synthesis
The goal of the synthesis is for students to connect dilating points on a line and dilating the line itself. Specifically, what happens if the line goes through the center of the dilation?
Invite students to share how the definition of dilation can help them answer these questions. Students should have the opportunity to hear and articulate that because dilations, by definition, take
points along lines through the center, then dilating a line through the center will take all the points to points on that same line, so the line doesn’t move. It may be hard for students to put into
words that the points are dilated but due to the nature of infinity, the line is not changed, so invite several students to put their explanation into their own words. In the next synthesis, students
will state and record a theorem about lines that do and do not pass through the center of the dilation, so it’s useful for students to be clear about why this is true.
Speaking: MLR8 Discussion Supports. As students share what happens to the line itself under dilation, press for details by asking how they know that a dilation leaves a line passing through the
center of dilation unchanged. Show concepts multi-modally by drawing and labeling a dilation of a line that passes through the center of dilation. Also show students how the dilation takes points on
the line to points on that same line. This will help students justify why a dilation does not change a line that passes through the center of dilation.
Design Principle(s): Support sense-making; Optimize output (for justification)
4.3: Proof in Parallel (15 minutes)
In this activity, students figure out that because dilations preserve angle measures, we can prove that dilations take lines to parallel lines. They draw on the many proofs they did in previous units
that use congruent angles to prove that lines are parallel. For students who struggled with proofs in prior units, make sure that they have their reference charts and proof writing sentence frames
In the previous lesson you collected many dilations of triangles from students, drawn on tracing paper. Display several of these examples for all to see. Superimpose the examples so that the center
of dilation and original figure are lined up. Ask students what they notice about the angles in the problem. (Corresponding angles in the image and original figure are congruent.) Ask students what
they notice about the lines in the problem. (They are longer or shorter according to the scale factor. They are parallel.)
If students don’t mention that the lines are parallel, use a highlighter or colored pencil to draw their attention to two corresponding line segments on lines that don’t go through the center. Ask
them what they notice about those lines specifically.
Writing, Listening, Conversing: MLR1 Stronger and Clearer Each Time. Use this routine to help students improve their written responses for the proof that a dilation takes a line not passing through
the center of dilation to a parallel line. Give students time to meet with 2–3 partners to share and receive feedback on their responses. Display feedback prompts that will help students strengthen
their ideas and clarify their language. For example, “What is the center and scale factor of the dilation?”, “How do you know that points _____ and _____ will both be on ray _____?” and “How do you
know that lines _____ and _____ are parallel?” Invite students to go back and revise or refine their written responses based on the feedback from peers. This will help students justify why a dilation
takes lines to parallel lines.
Design Principle(s): Optimize output (for justification); Cultivate conversation
Student Facing
1. Jada claims that all the segments in \(ABC\) are parallel to the corresponding segments in \(A’B’C’\). Write Jada's claim as a conjecture.
2. Prove your conjecture.
3. In Jada’s diagram the scale factor was greater than one. Would your proof have to change if the scale factor was less than one?
Anticipated Misconceptions
For the proof it might be easier to look at one pair of corresponding segments rather than the whole triangle. Recommend students look at their reference chart and proof writing template.
If students are stuck on the proof, encourage them to draw the rays that show how the points in the image were dilated, and to focus on just one pair of corresponding segments at a time (perhaps
using colored pencils to highlight the segment of interest).
Activity Synthesis
The goal of this synthesis is to conclude if two figures are dilations of one another, then any distinct corresponding lines must be parallel. In a later lesson in this unit, students will need to
use this result to prove that lines are parallel. Students will get more opportunities to draw conclusions about lines in dilated figures in the cool-down and lesson synthesis.
Invite students to share what helped them get started on the proof. (Drawing auxiliary lines. Looking at the reference chart.) Ask students to contribute ideas to the proof until everyone understands
this chain of reasoning:
• because we know the figures are dilations, we know that corresponding angles are congruent
• because we know that corresponding angles are congruent, we know that the lines cut by the transversal are parallel
Ask students to add this theorem to their reference charts as you add it to the class reference chart:
A dilation takes a line not passing through the center of the dilation to a parallel line, and leaves a line passing through the center unchanged. (Theorem)
Dilate using center \(C\). \(\overleftrightarrow{DE} \parallel \overleftrightarrow{D'E'}\)
Lesson Synthesis
Invite students to sketch each diagram.
“Point \(C\) was dilated using center \(M\) by a scale factor of \(\frac25\). What must be true about \(C\) and \(C’\)?” (\(C\) and \(C’\) are on the same line through \(M, \frac{C’M}{CM}=\frac25\).)
• If students don’t mention collinearity, sketch for all to see a picture where \(C’\) is closer to \(M\) than \(C\) but clearly not collinear and ask, could this be an accurate picture of \(C, C’
\), and \(M\)?
• If students don’t mention length, sketch for all to see a picture where \(C’\) is farther from \(M\) than \(C\) and ask could this be an accurate picture of \(C, C’,\) and \(M\)?
“Segment \(NO\) was dilated using center \(P\) by a scale factor of \(\frac43\). What must be true about segments \(NO\) and \(N’O’\)?” (\(NO\) and \(N’O’\) are parallel or on the same line, the
ratio of the length of \(N’O’\) to the length of \(NO\) is \(\frac43\)).
Ask students to share diagrams of \(NO, P\), and \(N’O’\) with a partner and then the class.
“Angle \(CDE\) was dilated using center \(J\) by a scale factor of 1.5. What must be true about angle \(C’D’E’\) compared to angle \(CDE\)?” (They are congruent.)
If students say that angle \(C’D’E’\) is “bigger” than angle \(CDE\), ask students to say more about what they mean. Clarify that the segments are longer and points are farther apart but the angles
have the same measure.
4.4: Cool-down - All Together Now (5 minutes)
Student Facing
When one figure is a dilation of the other, we know that corresponding side lengths of the original figure and dilated image are in the same proportion, and are all related by the same scale factor,
\(k\). What is the relationship of corresponding angles in the original figure and dilated image?
For example, if triangle \(ABC\) is dilated using center \(P\) with scale factor 2, we can verify experimentally that each angle in triangle \(ABC\) is congruent to its corresponding angle in
triangle \(A’B’C’\). \(\angle A \cong \angle A’, \angle B \cong \angle B’, \angle C \cong \angle C’\).
What is the image of a line not passing through the center of dilation? For example, what will be the image of line \(BC\) when it is dilated with center \(P\) and scale factor 2? We can use
congruent corresponding angles to show that line \(BC\) is taken to parallel line \(B’C'\).
What is the image of a line passing through the center of dilation?
For example, what will be the image of line \(GH\) when it is dilated with center \(C\) and scale factor \(\frac12\)? When line \(GH\) is dilated with center \(C\) and scale factor \(\frac12\), line
\(GH\) is unchanged, because dilations take points on a line through the center of a dilation to points on the same line, by definition.
So, a dilation takes a line not passing through the center of the dilation to a parallel line, and leaves a line passing through the center unchanged. | {"url":"https://curriculum.illustrativemathematics.org/HS/teachers/2/3/4/index.html","timestamp":"2024-11-02T20:26:09Z","content_type":"text/html","content_length":"130383","record_id":"<urn:uuid:13f35ea6-50b9-4316-aa46-05f4e2a507a1>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00791.warc.gz"} |
Greenwich Meridian (Prime Meridian) - GIS Geography
Greenwich Meridian (Prime Meridian)
What is the Greenwich Meridian?
Longitude lines like the Greenwich Meridian run north-south and converge at the poles.
Lines of longitudes run between -180° and +180°. But there’s a special longitudinal line where all x-coordinates start at 0°.
This special north-south line is where we measure east and west and is the Prime Meridian.
We also call it the Greenwich Meridian because it runs through Greenwich, England.
Greenwich Meridian (Prime Meridian)
Everywhere on Earth has a position because we can assign them geographic coordinates. We measure these coordinates as lines of latitude and longitude.
The 0° line of longitude starts at the Prime Meridian. It’s also called the Greenwich Meridian because it runs through Greenwich, England. Then, we can measure 180° to the west or 180° to the east.
Lines of latitude start at the equator. Everything above the equator is from 0-90° north. But everything below the equator is from 0-90° south.
Latitudes and longitudes make up our geographical coordinate system. For example, WGS84, NAD83, and NAD27 are common coordinate systems.
Geographic Coordinate Systems (GCS)
We use an ellipsoid to approximate the surface of Earth. It’s not a perfect sphere because it flattens a bit at the poles. Then, we use the ellipsoid to reference all latitude and longitude
coordinates to it.
A datum describes the shape of the Earth in mathematical terms. Each datum specifies:
• Radius
• Inverse flattening
• Semi-major and semi-minor axis
X-coordinates are between -180° and +180°, which are called longitudes.
Y-values are between -90° and +90° degrees, which are lines of latitudes.
Coordinates pairs (X, Y) represent positions in two-dimensional space. But triplets (X, Y, Z) also contain a height value for elevation.
While the X-value represents the east-west position, the Y-value represents the north-south position. The Z-value generally refers to the elevation at that point location.
Summary: Greenwich Meridian (Prime Meridian)
The 0-degree line of longitude that passes through the Royal Observatory in Greenwich, England is the Greenwich Meridian.
It’s also called the Prime Meridian.
This line is the starting point for longitudinal lines that run north-south and converge at the poles.
The Greenwich Meridian (or prime meridian) is a 0° line of longitude from which we measure 180° to the west and 180° to the east. These measurements are the basis of our geographic reference grid.
8 Comments
1. Please assist. I’m seeking the location of the country Mexico in longitude and latitude coordinates
1. Although Mexico is a large country… Mexico City is roughly a central point for reference: Latitude 19.4326° N, Longitude -99.1332° W.
2. Beautiful article! Thank you!
3. I want to stand on the prime meridian line at N Greenwich in the UK, then go to France, Spain, Algeria, Ghana, Mali, Togo, Burkina Faso Countries falling on S of the prime meridian. Touring
countries falling on the prime meridian.
4. Thank you for this tour, I’m studying GIS, and this was helpful in understanding geographic coordinate systems.
5. Really enjoyed this site. I learned so much more about what we saw.
6. I would love to see more pictures of the area described above
7. My husband and I took a small motor boat up the river to Greenwich, England to the Royal Observatory Greenwich. While there we stood directly on the Prime Meridian Line. (So Neat) just to be
there. The tour in the Observatory seeing all the clocks from the beginning when it developed was so wonderful. | {"url":"https://gisgeography.com/prime-greenwich-meridian/","timestamp":"2024-11-14T08:17:11Z","content_type":"text/html","content_length":"102103","record_id":"<urn:uuid:b4a7913e-6509-48b2-98ad-21079b076cb4>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00884.warc.gz"} |
Make Hyperbolic Tilings of Images
• Choose an image-file from your computer.
• Choose a hyperbolic polygon with p vertices, click on a p.
• Choose the number of hyperbolic polygons adjacent to each vertex, click on a q.
• Move the image by dragging if you wish.
• Click on "generate tiling".
• If you want to download a larger image than the one shown on the screen, click on "generate large image" (use Firefox to avoid crashes when downloading).
• If you want the first polygon to be "non-Euclidean", toggle through three different options by clicking on "change the first tile".
Select an image, or take a photo if using mobile device.
Move the image if needed. Choose p and q. Generate a tiling.
The loaded image will be cropped to the centered hyperbolic polygon. Repeated hyperbolic reflections of the centered polygon make up the tiling.
The option generate large generates a tiling that is larger than the image shown on the screen. The default scale factor for an enlarged tiling is four, this is a length factor. Another scale factor
can be chosen below. Chrome may crash when trying to download a large image - "Aw, Snap!" Firefox doesn't crash.
Larger tilings take longer time to generate, mostly since the edge of the tiling gets finer. You can stop the generation by clicking on stop. The tiling is only generated as long as this tab is seen
in the browser. Enlarged tilings have transparent background.
If change first tile is clicked, the rendering of the first tile will cycle through "no distortion", "Klein distortion", and "polynomial distortion". The effect of the three different options is
shown below from left to right. Distorted images take longer to calculate, for this reason moving a distorted image may seem laggy.
The hyperbolic tiling
The tiling is made of regular hyperbolic polygons with \(p\) sides. \(q\) is the number of polygons meeting at each corner. In Euclidean geometry you can only make a tiling of regular polygons if \
((p-2)\cdot(q-2)=4\). In other words you can only make a tiling for three pairs of \(\{p,q\}\): \(\{3,6\}, \{4,4\}, \{6,3\}\), see Geometry - Tessellations and Symmetries for an interactive Euclidean
tiling. In hyperbolic geometry, the angle sum of a triangle is less than 180°, and you can make a tiling of regular hyperbolic polygons whenever \((p-2)\cdot(q-2)>4\).
When a so-called Poincaré disc is used to model hyperbolic geometry, the entire universe is inside a circle \(C_\infty\). The inside of \(C_\infty\) is gray in the tiling above. The circle itself
does not belong to the universe but can be seen as the circle at infinity. When \(q\) tends to infinity, the vertices of the polygons approach \(C_\infty\). The shapes formed by vertices on \(C_\
infty\) are so-called ideal polygons. Such ideal polygons are used if the \(q\)-value \(\infty\) is picked in the tiling above.
For information about the math and the construction of a tiling, and an interactive example with draggable points (but no image), see Non-Euclidean Geometry - Interactive Hyperbolic Tiling in the
Poincaré disc.
As for distorting the first polygon, the option "Klein distortion" transforms the pixels from an assumed Beltrami-Klein model and then scales the hyperbolic Poincaré polygon. The option "polygonial
distortion" is described below.
Polynomial distortion
Every point inside a Eudclidean polygon lies on a segment parallel to the one of the border segments of the polygon.
Figure 1: Points on segments
For every such segments it is possible to find a corresponding hyperbolic geodesic. By mapping all points on a segment to the corresponding geodesic, we get a reasonable mapping from a Euclidean to a
hyperbolic polygon.
Figure 2: Points on geodesics
Pick a point \(A\) inside the Euclidean polygon. Identify a triangle \(\bigtriangleup O, V_1, V_2 \) containing \(A\), as in Figure 3.
Figure 3: Identify a triangle
There is a line \(l\) through \(A\) parallel to the line \(V_1, V_2 \). Let \(N_1, N_2 \) be the intersection points of \(l\) and \(\bigtriangleup O, V_1, V_2 \). Let \(N\) be the midpoint of the
geodesic through \(N_1, N_2 \), and let \(B\) be the intersection between the line \(AN\) and the geodesic, as in Figure 4. It's reasonable to choose \(B\) as the point in the hyperbolic polygon
corresponding to \(A\) in the Euclidean polygon.
Figure 4: Choose B as the hyperbolic point
In order to get a smooth image when points are rounded to integer values corresponding to the grid of pixels, the transformation must be done in reverse order. Given a point \(B\), find the point \(A
\). Let the pixel containing \(B\) have the color of the pixel containing \(A\).
Let \(B\) be reflected in \(C_\infty\) to a point \(B'\). Any geodesic through \(B\) must also go through \(B'\). The midpoint \(N\) of the unknown geodesic through the unknown points \(N_1 \) and \
(N_2 \) must have the same distance to \(B\) as to \(B'\). Furthermore it must be on the line \(OV\), where \(V\) is the midpoint of the geodesic through \(V_1V_2\).
Figure 5: Two conditions for finding N
For simplicity assume \(O = (0,0)\). Denote the coordinates of a point using dot notation. Then: \[ (N.x-B.x)^2+(N.y-B.y)^2 = (N.x-B'.x)^2+(N.y-B'.y)^2 \]
When \(V.y= 0\) then
\[ \begin{align*} N.x & = \frac{B'.x^2 + B'.y^2-B.x^2-B.y^2}{2(B'.x-B.x)} \\ N.y & = 0 \end{align*} \]
\[ \frac{N.x}{N.y} = \frac{V.x}{V.y} \]
and \(N.y\) can be found from \[ N.y = \frac{V.y(B'.x^2 + B'.y^2-B.x^2-B.y^2)}{2\left(V.x(B'.x-B.x)+V.y(B'.y-B.y)\right)} \]
Let \(g\) be the geodesic through \(B\) having \(N\) as midpoint. Now \(N_1\) and \(N_2\) can be found as the intersection points between \(g\) and \(\bigtriangleup O, V_1, V_2 \), and \(A\) can be
found as the intersection between the line \(BN\) and the line \(N_1N_2\), as in Figure 4. | {"url":"http://www.malinc.se/m/ImageTiling.php","timestamp":"2024-11-09T02:58:58Z","content_type":"text/html","content_length":"19072","record_id":"<urn:uuid:4f8d7f1b-a142-40a3-860e-bb08d190ac7a>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00617.warc.gz"} |
The highest common multiple of 68
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Concentration Inequalities | Rafael Oliveira
Concentration Inequalities
In the next lectures, we will study randomized algorithms. When evaluating the performance of randomized algorithms, it is not enough to analyze the average-case performance of the algorithm. In
other words, it is not enough to know that our algorithm runs with expected time $T$. What we want to say is:
That is, we not only need to analyze the expected running time of our algorithm, but also need to show the concentration of the running time around the expected value (that is, the typical running
time will be the expected running time).
To do the above, we will make use of concentration inequalities. More precisely, today we will study the Markov inequality, the Chebyshev inequality and the Chernoff-Hoeffding inequality.
Markov Inequality
Theorem (Markov’s inequality): Let $X$ be a discrete, non-negative random variable. Then, for any $t > 0$, we have $$ \Pr[X \geq t] \leq \frac{\mathbb{E}[X]}{t}.$$
Proof: Note that $$ \mathbb{E}[X] = \sum_{j \geq 0} j \cdot \Pr[X = j] \geq \sum_{j \geq t} j \cdot \Pr[X = j] \geq \sum_{j \geq t} t \cdot \Pr[X = j] = t \cdot \Pr[X \geq t]. $$
Markov’s inequality is a very simple inequality, but it is very useful. It is most useful when we have very little information about the distribution of $X$ (more precisely, we only need
non-negativity and we need to know the expected value). However, as we will see soon, if we have more information about our random variables, we can get better concentration inequalities.
Applications of Markov’s inequality:
• quicksort: the expected running time of quicksort is $2n \log n$. Markov’s inequality implies that the probability that quicksort takes longer than $2 c \cdot n \log n$ is at most $1/c$.
• coin flipping: the expected number of heads in $n$ (unbiased) coin flips is $n/2$. Markov’s inequality tells us that $\Pr[ \text{# heads } \geq 3n/4] \leq 2/3$.
We will see that we can get better concentration inequalities for the coin flipping example, if we know that the coin flips are independent (so we will have more information about our random
Moments of Probability Distributions
To get better concentration inequalities, we will need to know more (than the expected value) about our random variables. For instance, how do we disiniguish between the following two probability
• $X$ is the random variable defined by $\Pr[X = i] = \begin{cases} 1/n, \text{ if } 1 \leq i \leq n \\
0, \text{ otherwise.} \end{cases}$
• $Y$ is the random variable that takes value $1$ with probability $1/2$ and value $n$ with probability $1/2$.
They have the same expectation, but they are very different random variables.
To get more information on our random variables, we will define the moments of a random variable. Moments tell us how much the random variable deviates from its mean.
• The $k^{th}$ moment of a random variable $X$ is defined as $\mathbb{E}[X^k]$.
• The $k^{th}$ central moment of a random variable $X$ is defined as $\mu_X^{(k)} := \mathbb{E}[(X - \mathbb{E}[ X ])^k]$.
Note that the expected value is the first moment, and the first central moment is $0$.
Now, in our example above, we have that $\mathbb{E}[X] = \mathbb{E}[Y] = (n+1)/2$. So they are equal in the first moment. However, looking at the second moments, we have that $$\mathbb{E}[X^2] = n
(n+1)(2n+1)/(6n) = (n+1)(2n+1)/6$$ and $$\mathbb{E}[Y^2] = 1/2 + n^2/2 = (n^2+1)/2$$ So, we can see that the higher moments are able to distinguish between the two random variables.
We will now see that the higher moments are also useful to get better concentration inequalities.
Chebyshev Inequality
One particularly useful moment is the variance of a random variable, which is the second central moment. So we define $Var[X] := \mathbb{E}[(X - \mathbb{E}[X])^2]$. With information about both the
expected value and the variance, we can get a better concentration inequality: Chebyshev’s inequality.
Theorem (Chebyshev’s inequality): Let $X$ be a discrete random variable Then, for any $t > 0$, we have $$ \Pr[|X - \mathbb{E}[X]| \geq t] \leq \frac{\text{Var}[X]}{t^2}.$$
Proof: Let $Z = (X - \mathbb{E}[X])^2$. $Z$ is a non-negative random variable, so we can apply Markov’s inequality to $Z$. Then, we have that $$ \Pr[|X - \mathbb{E}[X]| \geq t] = \Pr[Z \geq t^2] \leq
\frac{\mathbb{E}[Z]}{t^2} = \frac{\text{Var}[X]}{t^2}.$$
An important measure of correlation between two random variables is the covariance. It is defined as $$Cov[X,Y] := \mathbb{E}[(X - \mathbb{E}[X])(Y - \mathbb{E}[Y])].$$ Note that $Cov[X,X] = Var[X]$.
We say that two random variables $X$ and $Y$ are positively correlated if $Cov[X,Y] > 0$ and negatively correlated if $Cov[X,Y] < 0$. We say that two random variables are uncorrelated if $Cov[X,Y] =
Remark: Note that independent random variables are uncorrelated, but uncorrelated random variables are not necessarily independent.
Proposition: Let $X$ and $Y$ be two random variables. Then,
• $Var[X+Y] = Var[ X ] + Var[Y] + 2 Cov[X,Y]$.
• If $X$ and $Y$ are uncorrelated, then $Var[X+Y] = Var[ X ] + Var[Y]$.
Now that we learned about the covariance, we can apply it to our coin flipping process.
• coin flipping: let $X$ be the number of heads in $n$ unbiased coin flips. We can describe the $i^{th}$ coin toss by the random variable $X_i = \begin{cases} 1, \text{ if coin flipped heads} \\ 0,
\text{ otherwise} \end{cases}$
Since the coin flips are independent, they are also uncorrelated. Thus, by the above proposition, we have that $Var[X] = \sum_{i=1}^n Var[X_i] = n/4$. So, by Chebyshev’s inequality, we have that $$ \
Pr[X \geq 3n/4] \leq \Pr[|X - \mathbb{E}[X]| \geq n/4] \leq \frac{n/4}{(n/4)^2} = \frac{4}{n}.$$
Practice problem: can you generalize Chebychev’s inequality to higher moments?
Chernoff-Hoeffding Inequality
Often times in algorithm analysis, we deal with random variables that are sums of independent random variables (distinct elements, balls and bins, etc). Can we get a better concentration inequality
for these types of random variables?
The law of large numbers tells us that the average of a large number of independent, identically distributed, random variables is close to the expected value. Chernoff’s inequality tells us how
likely it is for the average to be far from the expected value.
Theorem (Chernoff inequality): Let $X_1, \ldots, X_n$ be independent random variables such that $X_i \in {0,1}$ for all $i \in [n]$. Let $X = \sum_{i=1}^n X_i$ and $\mu = \mathbb{E}[X]$. Then, for $0
< \delta < 1$, $$ \Pr[X \geq (1+\delta)\mu] \leq e^{-\frac{\delta^2 \mu}{3}}$$ also, $$ \Pr[X \leq (1-\delta)\mu] \leq e^{-\frac{\delta^2 \mu}{2}}.$$
Proof: We will prove the first inequality. The proof of the lower tail bound is similar.
Let $p_i := \Pr[X_i = 1]$, and thus $1-p_i = \Pr[X_i = 0]$ and $\mu = \sum_{i=1}^n p_i$.
The idea of the proof is to apply Markov’s inequality to the random variable $e^{tX}$. Since the exponential function is increasing, we have $$\Pr[X \geq a] = \Pr[e^{tX} \geq e^{ta}] \leq \mathbb{E}
[e^{tX}]/e^{ta}, \text{ for any } t > 0.$$ What do we gain by doing this? When we look at the exponential function, we are using information about all the moments of $X$. This is because the Taylor
series of $e^{tX}$ is $$ e^{tX} = \sum_{k=0}^\infty \frac{(tX)^k}{k!} = 1 + tX + \frac{t^2 X^2}{2!} + \frac{t^3 X^3}{3!} + \ldots$$ In particular, we define the moment generating function of $X$ as
$$ M_X(t) := \mathbb{E}[e^{tX}] = \sum_{k=0}^\infty \frac{t^k \mathbb{E}[X^k]}{k!}.$$ If $X = X_1 + X_2$, where $X_1$ and $X_2$ are independent, then $M_X(t) = M_{X_1}(t) M_{X_2}(t)$.
Now, let’s apply Markov’s inequality to $e^{tX}$. We have that $$ \Pr[X \geq (1+\delta)\mu] = \Pr[e^{tX} \geq e^{t \cdot (1+\delta)\mu}] \leq \frac{\mathbb{E}[e^{tX}]}{e^{t(1+\delta)\mu}}.$$ By the
above, and independence of $X_i$’s, we have $$\mathbb{E}[e^{tX}] = \prod_{i=1}^n \mathbb{E}[e^{t X_i}] = \prod_{i=1}^n \left(p_i \cdot e^t + (1-p_i) \cdot 1 \right) $$ Since $p_i \cdot e^t + (1-p_i)
\cdot 1 = 1 + p_i \cdot (e^t - 1) \leq e^{p_i \cdot (e^t -1)}$, as $e^x \geq 1 + x$, for all $x \geq 0$, we have $$ \dfrac{\mathbb{E}[e^{tX}]}{e^{t(1+\delta) \mu}} \leq \dfrac{1}{e^{t(1+\delta) \mu}}
\cdot \prod_{i=1}^n e^{p_i \cdot (e^t -1)} = \left( \dfrac{e^{e^t-1}}{e^{t \cdot (1+\delta)}} \right)^\mu \leq \left( \dfrac{e^\delta}{(1+\delta)^{1+\delta}} \right)^\mu $$ where in the last
inequality we plugged in $t = \ln(1 + \delta)$.
The main inequality follows from the above and the fact that $e^\delta/(1+\delta)^{1+\delta} \leq e^{-\delta^2/3}$, for all $0 < \delta < 1$. To see this, we can use the Taylor series of $\ln(1+x)$
for $x \in (0,1)$, which is $\ln(1+x) = x - x^2/2 + x^3/3 - \ldots$. Then, we have $(1+\delta) \ln(1+\delta) = x + x^2/2 - x^3/6 + x^4/3 \cdot 4 - \cdots$ $$ e^\delta/(1+\delta)^{1+\delta} = \exp(\
delta - (1+\delta) \ln(1+\delta)) = \exp(-\delta^2/2 + \delta^3/6 - \delta^4/12 + \cdots) $$ $$ \leq e^{-\delta^2/3} \cdot \exp(- \delta^2/6 + \delta^3/6) \leq e^{-\delta^2/3} $$
Theorem (Hoefding’s inequality): Let $X_1, \ldots, X_n$ be independent random variables such that $X_i \in [a_i,b_i]$ for all $i \in [n]$. Let $X = \sum_{i=1}^n X_i$ and $\mu = \mathbb{E}[X]$. Then,
for any $\ell > 0$, we have $$ \Pr[|X - \mu |\geq \ell] \leq 2 \cdot \exp\left(-\frac{2\ell^2}{\sum_{i=1}^n (b_i - a_i)^2}\right)$$
Proof: the proof is similar to the proof of Chernoff’s inequality, but now we use Hoeffding’s lemma instead of Markov’s inequality.
Hoeffding’s lemma states that if $Z$ is a random variable such that $Z \in [a,b]$, then $$ \mathbb{E}[e^{t(Z - \mathbb{E}[Z])}] \leq e^{t^2(b-a)^2/8}.$$ | {"url":"https://cs.uwaterloo.ca/~r5olivei/courses/2024-spring-cs466/lecture-notes/lecture3/","timestamp":"2024-11-03T16:57:25Z","content_type":"text/html","content_length":"29644","record_id":"<urn:uuid:4485446d-a5f4-45e7-a3e6-c866ac132818>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00177.warc.gz"} |
Precalculus (6th Edition) Blitzer Chapter 10 - Section 10.6 - Counting Principles, Permutations, and Combinations - Exercise Set - Page 1103 34
Total number of ways to answer the questions equals to 6561.
Work Step by Step
We know that the situation involves making choices with eight questions. We use the fundamental counting principle to find the number of ways to answer the questions on the test. Multiply the number
of choices for each of the eight questions. That is, $ & 3\times 3\times 3\times 3\times 3\times 3\times 3\times 3={{3}^{8}} \\ & =6561 $ Hence, there are 6561 different ways to answer the questions. | {"url":"https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-10-section-10-6-counting-principles-permutations-and-combinations-exercise-set-page-1103/34","timestamp":"2024-11-02T19:12:42Z","content_type":"text/html","content_length":"78540","record_id":"<urn:uuid:45313f9d-bd37-4731-a5e2-01e59d80955e>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00442.warc.gz"} |
What is the minimum payment for sick leave in 2024? - bankobul2.com
What is the minimum payment for sick leave in 2024?
The minimum payment for sick leave in 2024, as in previous periods, depends on the minimum wage. But there are also a number of other factors that reduce the amount of charges for sick leave. Let’s
consider when the minimum amounts are formed during accruals on the disability certificate.
Calculation of disability benefits: formula
The minimum payment for sick leave is the result of the influence of several factors, and above all the values of those indicators that are involved in the formula for calculating payments on the
disability certificate:
WB = NW × KS × KB,
World Bank disability allowance;
NW — average earnings of an employee;
CS — coefficient by length of service;
KB — the number of days an employee is absent from work according to the disability certificate.
The minimum amount of sick leave in 2024 is set:
● with a minimum CP index;
● with a minimum KB index.
For examples of calculating sick leave for small amounts of income, see here.
Determining the amount of sick leave: minimum average earnings
The CP indicator is determined by the formula:
NW = BR/730,
BR — the base for calculating average earnings.
The minimum allowable value of the BR indicator is calculated using the following formula:
BR (min) = MINIMUM WAGE × 24,
BR (MIN) – the minimum value of the BR indicator;
Minimum wage – the minimum wage established by law on the date of the start of sick leave.
The BR (MIN) indicator is used in calculating sick leave benefits if at least 1 of the following conditions is met:
● the employee’s total work experience is less than 6 months;
● the employee went on sick leave due to intoxication;
● the employee violated the treatment regimen prescribed by the doctor;
● at full employment (40-hour workweek with a standard schedule), the calculated BR indicator is for some reason less than the minimum wage increased by 24 months.
Read more about paying sick leave from the minimum wage here.
If none of these conditions are met, the BR indicator will be equal to all payments that are taken into account when calculating disability benefits (but within the maximum amount also established by
To find out what payments are included in the calculation, as well as how the average earnings are calculated when calculating sick leave (taking into account the limits), you can in this article.
Calculation of the minimum for sick leave: duration of treatment and coefficient
The minimum payment for sick leave in 2024, which is quite logical, will occur with a minimum duration of treatment for an employee. The lower the KB value in the formula, the lower the sick pay.
If an employee is treated by himself, then the CB indicator will depend on the characteristics of the disease, the sequence of its treatment and, as a result, on the number of sick leaves issued in a
row. In total, treatment can last up to 10 months, in some cases up to 12 months (paragraph 4 of Article 59 of the Law “On Health Protection” dated 11/21/2011 No. 323-FZ).
If an employee goes on sick leave to take care of a child or other sick relative, then the CB indicator will be affected:
● the age of the child of the employee or other relative being cared for;
● limitations in the physical abilities of the child being treated.
You can read more about paying sick leave for caring for a child or other relative in the article“Paying sick leave for caring for a sick relative”.
The KST coefficient, if the NW indicator is equal to or less than the minimum wage, will always be equal to 1. But its value is assumed to be equal:
● 0.60 if the length of service of the employee when going on sick leave is less than 5 years;
● 0.80 if the length of service is 5-8 years;
● 1 if experience of more than 8 years.
If the employee’s length of service is less than 6 months, then the CP indicator, regardless of the value of the calculated BR indicator, will be equal to the CP calculated based on the minimum wage.
The minimum sick leave in 2024 is obtained if the base for calculating average earnings is less than or equal to the minimum wage increased by 24. The same calculation is made with an experience of
less than six months. The duration of treatment and the value of the coefficient applied to accruals, depending on the length of service, also play a role.
You can learn more about the specifics of calculating sick leave in the articles:
● Is sick leave included in the calculation of vacation pay?»;
● «Maximum amount of sick leave in 2024».
Leave a Comment | {"url":"https://bankobul2.com/what-is-the-minimum-payment-for-sick-leave-in-2024/","timestamp":"2024-11-12T16:11:56Z","content_type":"text/html","content_length":"120096","record_id":"<urn:uuid:f4fcd620-9a96-4183-9873-decba954951b>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00356.warc.gz"} |
CNS PBeM Event Reports
Updated January 4, 2024
A record 84 players signed up for the 2023 Can’t Stop PBEM tournament. Over 10 rounds, 169 games were played. As in previous years, the event was played with two-player games in a
double-elimination format. Rounds continued until only one player remained with fewer than two losses.
As was the case in 2022, the early rounds were not kind to last year’s top players. The top six finishers from last year were all eliminated within the first three rounds, so there were
no repeat laurelists for the second year in a row.
After seven rounds, the top six finishers were determined. Eric Freeman was the last undefeated player but had fallen to Rob Kircher in Round 7. He picked up his second loss in quick
succession, losing a close game to Chris Wildes in Round 8. Rob also lost in this round, with Jim Brown advancing to the next round. In the last Round 8 matchup, Haim Hochboim ended the
run of first time laurelist Victoria Wallace.
Jim and Chris faced off in Round 9. Chris’s dice failed him, busting five times, while Jim calmly advanced to the final with an easy 3-0 victory. Haim faced Rob, serving as an eliminator,
in the other semifinal. Rob capped the 7 first, but Haim responded by getting to the top of the 4 column. Rob retook the lead two turns later, finishing the 9 and getting three steps away
on both the 5 and 6. However on the next turn Haim rolled a 12 to finish that column and eleven 8’s to climb that track in one go, advancing to the final.
The final was a rematch from Round 4 where Haim gave Jim his first loss. Haim’s luck turned at the start of the final game, busting on his first three turns. This allowed Jim to take a
significant lead, capping the 3 and getting one step away on the 9. Haim responded with a repeat of his Round 9 heroics, running up two entire columns on each of his next two turns, first
the 6 and then the 5, to take the lead. Unfortunately for Haim, Jim’s slow and steady approach meant he only needed three steps (one on the 9, two on the 7) to close out his last two
columns, claiming the 2023 Can’t Stop PBEM championship.
Top six finishers were:
1. Jim Brown
2. Haim Hochboim
3. Chris Wildes
4. Rob Kircher
5. Eric Freeman
6. Victoria Wallace
Of the 169 games played in the tournament, 97 of them (57%) were won by the player taking the first turn of the game. This was a higher percentage than in 2022 (51%) or 2021 (47%). Three
players achieved a four-column victory: Curt Collins, Rob Kircher, and Oliver Searles.
Congratulations to all of the laurelists and thank you to everyone who participated in the event.
Over 9 rounds of play, 66 players completed a total of 132 games in the 2022 Can’t Stop PBEM tournament. This was the highest turnout for any PBEM tournament I have run to date, thank you
to all who participated. The event was played using a single-game double-elimination format with rounds continuing until only one player remained with fewer than two losses.
During the early rounds, Can’t Stop demonstrated its capricious nature as two of the top three finishers from 2021 were eliminated after the first two rounds. Ultimately, no laurelists from
last year’s tournament repeated in the top six.
By Round 6, only two players remained undefeated: Bill Masek and Haakon Monson. Bill got two early steps on the 12, while Haakon’s first turn sent him half-way up on the 9, which he then
capped on his next turn. Unfortunately for Haakon he then busted on his next three turns while Bill made steady progress and won with 5-8-12.
Both Bill and Haakon lost in Round 7, which awarded Haakon fifth place overall. Two other players were eliminated in this round, with Chris Houle earning sixth place laurels over Laurie
Wojtaszczyk on the second tiebreaker (opponent win percentage). Bill’s loss set up a true semifinal with the three remaining players: Grant LaDue, Aran Warszawski, and Dan Elkins.
In the first game, Grant unluckily busted immediately after setting up a 6-7-8 turn. Aran quickly capped the 8, followed by Grant locking down the 7 and 10 on successive turns. On the next
turn Aran finished the 2 column and got one away from the top of 12, stopping after only pushing up those two tracks. Both players then tried to make progress up the 6 and 9, but Grant’s
dice failed him twice allowing Aran to claim victory with 2-6-8.
The second semifinal started with four straight early busts before Bill was able to stop halfway up the 8 on his third turn, followed by Dan doing the same on the 6. Both players struggled
with the dice throughout, and in total nine turns ended in failures. Bill capped the 8, followed by Dan claiming the 6 and moving to three away on the 7. However, Bill aggressively pushed
up the 7 to erase Dan’s progress and end the turn one away on the 2. Dan claimed the 9 on the next turn and joined Bill near the top of the 2 column before passing. Dan would not get
another turn as Bill found the 2 needed to claim his place in the final.
The dice were extremely unkind to Bill in the final, as he busted very quickly on 6 of his 7 turns. Aran capitalized to make safe and steady progress up the tracks, capping 3-5-7, and
claiming the 2022 Can’t Stop PBEM championship. Top six finishers were, in order, Aran Warszawski, Bill Masek, Grant LaDue, Dan Elkins, Haakon Monsen, and Chris Houle
Of the 132 games played in the tournament, 67 of them (51%) were won by the player taking the first turn of the game. Haakon Monsen achieved the only 4-column victory, simultaneously
capping the 6 and the 7 on the final turn of his Round 3 matchup.
The craziest finish of the tournament goes to Marc Gibbens in his third-round game. After busting on 7 of his first 9 turns, Marc found himself down 2-0 with his opponent one step away on
three tracks. Marc was able to progress up the 2, 6, and 12, completing each one on separate rolls to cap all three in a single turn.
Below is the data for completed columns of winning players, with comparisons to last year’s tournament:
• 2 was used by 24% of winners (+3% over 2021)
• 3 was used by 8% of winners (-5% over 2021)
• 4 was used by 17% of winners (+4% over 2021)
• 5 was used by 23% of winners (-3% over 2021)
• 6 was used by 48% of winners (-5% over 2021)
• 7 was used by 52% of winners (-9% over 2021)
• 8 was used by 47% of winners (+0% over 2021)
• 9 was used by 27% of winners (+9% over 2021)
• 10 was used by 17% of winners (+2% over 2021)
• 11 was used by 13% of winners (+9% over 2021)
• 12 was used by 23% of winners (-4% over 2021)
Once again 6, 7, and 8 were the most likely to appear on a winner’s scorecard. However, this year’s winners relied less on the 6 and 7, while 9 and 11 showed large gains. The 11’s
improvement is particularly remarkable after it was by far the worst performing number in 2021 and at previous WBC tournaments.
Once again congratulations to all the laurelists and thank you to everyone who participated in the event.
The 2021 Can’t Stop PBEM tournament featured 52 players completing a total of 106 games over 9 rounds. The event was played using a single-game double elimination format with rounds
continuing until only one player remained with fewer than two losses.
Heading into Round 6, ten players remained in contention. The last two undefeated players, Dominic Blais and Dan Leader, were guaranteed a spot in the Top 6 and faced one another. Dominic
capped the 6 column on the first turn of the game and completed a 6-7-8 victory before Dan could catch up. In the other Round 6 games, Chris Wildes, Alex Bove, Chad Martin, and Eric Brosius
all advanced and secured Top 6 finishes.
In Round 7, Chad finished the 4 on his first turn against Dominic, but Dominic immediately answered with the 9. The game was tied 2-2 in turn 5 when a bust on Chad’s turn opened the door
for Dominic to win via the 8. Eric got off to a strong start against Chris, but two untimely busts coupled with Chris’s hot run up the 7 column in a single turn led to Eric’s elimination.
Alex made steady progress up the 6-7-8 columns in a close game to eliminate Dan after his 5-0 run to start the tournament.
Three players remained for Round 8, with Dominic facing Chris and Alex battling Eric, who was serving as an eliminator. In a messy, 9-turn game in which both players started the game with
two busts, Chris ended Dominic’s undefeated run. Eliminator Eric’s unlucky streak continued with three busted turns while Alex continued his steady play en route to victory.
As a result, the same three players remained alive for Round 9, but with swapped pairings. Alex played against Chris while Dominic faced Eric, again serving as an eliminator due to the
preference for avoiding repeat matchups. Alex continued his disciplined play and quickly worked to a 2-0 lead. Chris, feeling the need to catch up, risked an extra roll after capping the 5
and was punished for his hubris with a failed roll. On his next turn Alex pushed his 8 to within three spots of victory and passed. Chris was able to finish both the 6 and 10 on his next
turn to make the game interesting, but Alex quickly completed the 8 and awaited the results of the other game.
Meanwhile, Eric made the most of going first to get off to a fast start against Dominic, capping the 7 on his second turn while Dominic began with single bumps on 2-8-12. At the end of the
fourth turn, Eric had the 6 and 7 under his control and was two away on the 8 while Dominic was still on the bottom rung of the 2, 9, and 12, with five bumps up the 8. Dominic was able to
make a run up the 8 to negate the most urgent danger and a few turns later capped the 5 to even the score. However after three straight busted turns, Eric was able to finish the 3 column to
eliminate Dominic.
Congratulations to Alex Bove on winning the 2021 Can’t Stop PBEM tournament. Alex’s only loss came in Round 2 (to Dominic) and he finished the tournament with seven straight wins to claim
the victory.
Top six finishers were:
1. Alex Bove
2. Dominic Blais
3. Chris Wildes
4. Dan Leader
5. Eric Brosius
6. Chad Martin
Of the 106 games played in the tournament, 50 of them (47%) were won by the player taking the first turn of the game. The average game length fell just under 7 turns (6.73), ranging from a
minimum of 4 turns (seven games), to a maximum of 10 turns (eight games). Players averaged roughly 33 dice rolls per game and 5 rolls per turn. Tina Del Carpio achieved the win with the
fewest dice rolls, needing only 19 to earn a first round victory.
Peter Stein achieved the only 4-column victory, simultaneously capping the 2 and the 7 on the final turn of his Round 5 matchup.
Below is the data for completed columns of winning players, with comparisons to the most recent WBC tournament (2019):
• 2 was used by 22% of winners (-11% over last WBC)
• 3 was used by 13% of winners (+5% over last WBC)
• 4 was used by 13% of winners (-13% over last WBC)
• 5 was used by 26% of winners (-3% over last WBC)
• 6 was used by 53% of winners (+14% over last WBC)
• 7 was used by 61% of winners (+21% over last WBC)
• 8 was used by 47% of winners (+8% over last WBC)
• 9 was used by 18% of winners (-1% over last WBC)
• 10 was used by 15% of winners (-11% over last WBC)
• 11 was used by 4% of winners (-9% over last WBC)
• 12 was used by 27% of winners (+1% over last WBC)
Unsurprisingly, 6, 7, and 8 are the most likely to appear on a winner’s scorecard. The large increase in the frequency of those numbers over the previous WBC event (and corresponding
decrease for most other numbers) is likely attributable to the differences between the 2-player and 4-player game. One trend that does hold from Andrew Drummond’s previous event report is
the unpopularity of 11, by far the least likely number to be capped by winners, and the only number to appear more often on the loser’s scorecard (4% vs 6%).
Once again congratulations to all of the laurelists, and thank you to everyone who participated in the event. | {"url":"http://www.boardgamers.org/pbem/pbem-cns.html","timestamp":"2024-11-06T20:53:55Z","content_type":"text/html","content_length":"18905","record_id":"<urn:uuid:56edd5ab-0ee7-445d-9a54-8b7cab25c65d>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00383.warc.gz"} |
NYU Satellite Workshop 24
November 11-13, 2024 @ NYU
726 Broadway, New York City
Room 940
Organizers: Michele Del Zotto, Constantin Teleman, Yifan Wang
14:00 – 14:30 ::: Lukas Müller
14:30 – 15:00 ::: Shani Nadir Meynet
15:00 – 15:30 ::: Thibault Décoppet
coffee/tea and discussion
16:00 – 16:30 ::: Luuk Stehouwer
16:30 – 17:00::: Hector Peña Pollastri
17:00 – 17:30 ::: Pranay Gorantla
9:00 – 10:00 ::: Sakura Schäfer Nameki
coffee break
10:30 – 11:30 ::: Mayuko Yamashita
11:30 – 12:30 ::: Maissam Barkeshli
lunch break and discussions (coffee/tea 15:30)
16:00 – 17:00 ::: Theo Johson Freyd
9:00 – 10:00 ::: Dan Freed
coffee break
10:30 – 11:30 ::: Pavel Etingof
12:00 – 13:00 ::: Xiao-Gang Wen
lunch break and discussions (Colloquium by Ibou Bah @ 14:00)
15:30 – 18:00 ::: Wine and cheese reception and discussions
Müller (PI)
A Higher Spin Statistics Theorem for Invertible Quantum Field Theories
The spin-statistics theorem asserts that in a unitary quantum field theory, the spin of a particle—characterized by its transformation under the central element of the spin group, which corresponds
to a 360-degree rotation—determines whether it obeys bosonic or fermionic statistics. This relationship can be formalized mathematically as equivariance for a geometric and algebraic action of the
2-group $B\mathbf{Z}_2$. In my talk, I will present a refinement of these actions, extending from $B\mathbf{Z}_2$ to appropriate actions of the stable orthogonal group $O$, and demonstrate that every
unitary invertible quantum field theory intertwines these $O$-actions.
Decoppet (Harvard U.)
Higher Verlinde Categories: The Mixed Case
In characteristic zero, the semisimple Verlinde categories are braided fusion categories constructed from quantum $\mathfrak{sl}_2$ at a root of unity. These categories have found many applications
to low-dimensional topology, for instance, via the Reshetikhin-Turaev construction. I will explain how to construct the positive characteristic analogues of these categories, which are generally not
semisimple, and discuss their fundamental properties.
Stehouwer (Dalhusie U.)
2-Hilbert spaces and extended unitarity
Hilbert spaces have two levels of equivalence; linear isomorphism and unitary isomorphism. The fact that 2-Hilbert spaces have three levels gives 2-Hilb the structure of a higher dagger category.
This provides extra structure to the category of topological line operators in unitary theories, generalizing the fact that point operators form a *-algebra.
Schafer-Nameki (Oxford U.)
Categorical Landau in 1+1d and 2+1d
Using the Symmetry TFT we systematically study the phases (gapped and gapless) of 1+1d and 2+1d systems with fusion 1-/2-categories.
Yamashita (PI)
Genuine refinements of elliptic genera
There is a classical construction of elliptic genera for SU-manifolds, which assigns Jacobi forms to SU-manifolds, and physically related to non-chiral 2d N=(1, 1) sigma models. In this talk, I
explain my ongoing work with Ying-Hsuan Lin to give homotopy-theoretical refinements to it. The domain becomes “Topological Jacobi Forms (TJF)”, which is a spectrum (in homotopy theory) being
developed by an ongoing work of T. Bauer and L. Meier. TJF refines the ring of Jacobi forms, just as TMF does for modular forms. In view of the Segal-Stolz-Teichner program, it is very natural to
expect such a refinement of elliptic genera. The same strategy produces many variants of twisted equivariant TMF-valued genera including those from Sp-manifolds and Spin-manifolds. Having such
refinements, we can detect torsion elements in the bordism groups, as well as deduce nontrivial divisibility results of characteristic numbers. In this talk I explain the motivation and construction,
as well as why such refinements are more interesting beyond the classical elliptic genera.
Johnson Freyd (Dalhusie U. and PI)
The unitary cobordism hypothesis
A *dagger category* is a category equipped with extra “unitarity” data: among its isomorphisms, some are marked as “unitary”; to each 1-morphism, there is an “adjoint” 1-morphism, and this assignment
sends unitary (but not general!) isomorphisms to their inverses. If a monoidal higher category has duals, it is interesting to ask for a choice of duality functor for which the units and counits are
adjoints; whereas duality functors are unique up to contractible choice, so-defined “unitary duality” functors are not. I will explain a higher-categorical generalization of these ideas, and explain
how, for any stable tangential structure H, a construction of Freed and Hopkins makes the extended bordism category Bord$_n^{H(n)}$ into a dagger symmetric monoidal n-category with unitary duality.
This category satisfies a *unitary cobordism hypothesis*: whereas nonunitary functors Bord$_n^{H(n)} \to \mathcal{C}$ are classified by $H(n)$-fixed points in $\mathcal{C}$, *unitary* functors are
classified by fixed points for the stablized group $H(\infty)$. This talk is based on joint work in preparation with Cameron Krulewski, Lukas Mueller, and Luuk Stehouwer.
Freed (Harvard U.)
Anomalies and quiches
Together with Greg Moore and Constantin Teleman we introduced an abstract notion of symmetry in QFT–quiches–analogous to abstract groups and algebras in other contexts. This leads to a natural
question: How is the notion of an ‘t Hooft anomaly captured in the quiche framework? In joint work with Colleen Delaney, Julia Plavnik, and Constantin we address this question and also give
applications to zesting of fusion categories. The first part of the talk will be a general discussion of anomalies in QFT.
Etingof (MIT)
Lie theory in tensor categories with applications to modular representation theory
Let $G$ be a group and $k$ an algebraically closed field of characteristic $p > 0$. Let $V$ be a finite dimensional representation of $G$ over $k$. Then by the classical Krull-Schmidt theorem, the
tensor power $V^{\otimes n}$ can be uniquely decomposed into a direct sum of indecomposable represen- tations. But we know very little about this decomposition, even for very small groups, such as $G
= (\mathbb Z/2)^3$ for $p = 2$ or $G = (\mathbb Z/3)^2$ for $p = 3$. For example, what can we say about the number $d_n(V)$ of such summands of dimension coprime to $p$? It is easy to show that there
exists a finite limit $d(V) := \lim_{n\to \infty} d_n(V)^{1/n}$, but what kind of number is it? For example, is it algebraic or transcendental? Until recently, there was no techniques to solve such
questions (and in particular the same question about the sum of dimensions of these summands is still wide open). Remarkably, a new subject which may be called “Lie theory in tensor categories” gives
methods to show that $d(V)$ is indeed an algebraic number, which moreover has the form
$$ d(V) = \sum_{1 \leq j \leq p/2} n_j(V)[j]_q,$$
where $n_j(V) \in \mathbb N$, $q := \exp(i \pi /p)$, and $[j]_q := (q^j − q^{-j})/ (q−q^{−1})$. Moreover,
$$ d(V\oplus W ) = d(V ) + d(W ), \qquad d(V \otimes W ) = d(V )d(W ),$$
i.e., $d$ is a character of the Green ring of $G$ over $k$. Furthermore, $d_n(V ) \geq C_V d(V )^n$ for some $0 < C_V \leq 1$ and we can give lower bounds for $C_V$. In the talk I will explain what
Lie theory in tensor categories is and how it can be applied to such problems. This is joint work with K. Coulembier and V. Ostrik.
Xiao-Gang Wen (MIT)
From sub-algebra of local operators, to generalized symmetry, and to their topological phases on lattice
Ibrahima Bah
Maissam Barkeshli
Francesco Benini
Federico Bonetti
Lakshya Bhardwaj
T. Daniel Brennan
Christian Copetti
Augustina Czenky
Michele Del Zotto
Tudor Dimofte
Thibault Decoppet
Pavel Etingof
Dan Freed
Pranay Gorantla
Jonathan Heckman
Ken Intriligator
Theo Johnson-Freyd
Anton Kapustin
Liang Kong
Zohar Komargodski
Vassily Krylov
Tom Mainiero
Shani Meynet
Ruben Minasian
Greg Moore
Lukas Muller
Kantaro Ohmori
Hector Pena-Pollastri
Konstantinos Roumpedakis
Ingo Runkel
Sakura Schafer-Nameki
Claudia Scheimbauer
Shu-Heng Shao
Noah Snyder
Luuk Stehouwer
Pelle Steffens
Will Stewart
Constantin Teleman
Jackson van Dyke
Kevin Walker
Yifan Wang
Xiao-Gang Wen
Mayuko Yamashita
Bowen Yang | {"url":"https://scgcs.berkeley.edu/nyu-satellite-workshop-24/","timestamp":"2024-11-05T10:21:02Z","content_type":"text/html","content_length":"64352","record_id":"<urn:uuid:4d48d9fb-aa8e-4a43-bc48-e47eef5c3762>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00689.warc.gz"} |
Google’s FooBar Challenge | See How I Passed Levels 1 to 5 in Real-Time
I just got invited to perform Google’s FooBar challenge. In this article, I want to share with you how I solved the problems in real-time. The purpose of this article is to educate you—and to have
some fun. So, are you ready?
Level 1: Prime Numbers
The first goal was to find an identifier for a new employee at the “minions” company.
The identifier is selected base on a random number i. How do we come from the random integer i to the identifier of the new minion employee?
• Create a sequence of prime numbers '23571113...'.
• The identifier is the subsequence starting from index i and ending in index i+4 (included).
• The value i is an integer between 0 and 10000.
Here’s the solution I implemented in the video:
def solution(i):
# Determine prime sequence
primes = getPrimeNumbers()
return primes[i:i+5]
def getPrimeNumbers():
'''Returns the string of prime
numbers up to 10k+5 positions.'''
s = ''
prime = 2
while len(s) < 10005:
# Add new prime to s
s += str(prime)
# Calculate next prime
prime += 1
while not is_prime(prime):
prime += 1
return s
def is_prime(n):
'''Tests if a number is prime. '''
for i in range(2,n):
if n % i == 0:
return False
return True
# 23571
# 71113
Do you want to develop the skills of a well-rounded Python professional—while getting paid in the process? Become a Python freelancer and order your book Leaving the Rat Race with Python on Amazon (
Level 2 Challenge 1: Sequence Sum
Here’s the problem posed by Google:
Numbers Station Coded Messages
When you went undercover in Commander Lambda's
organization, you set up a coded messaging system with
Bunny Headquarters to allow them to send you important
mission updates. Now that you're here and promoted to
Henchman, you need to make sure you can receive those
messages - but since you need to sneak them past
Commander Lambda's spies, it won't be easy!
Bunny HQ has secretly taken control of two of the
galaxy's more obscure numbers stations, and will use
them to broadcast lists of numbers. They've given you a
numerical key, and their messages will be encrypted
within the first sequence of numbers that adds up to
that key within any given list of numbers.
Given a non-empty list of positive integers l and a
target positive integer t, write a function solution(l,
t) which verifies if there is at least one consecutive
sequence of positive integers within the list l (i.e.
a contiguous sub-list) that can be summed up to the
given target positive integer t (the key) and returns
the lexicographically smallest list containing the
smallest start and end indexes where this sequence can
be found, or returns the array [-1, -1] in the case
that there is no such sequence (to throw off Lambda's
spies, not all number broadcasts will contain a coded
For example, given the broadcast list l as [4, 3, 5, 7,
8] and the key t as 12, the function solution(l, t)
would return the list [0, 2] because the list l
contains the sub-list [4, 3, 5] starting at index 0 and
ending at index 2, for which 4 + 3 + 5 = 12, even
though there is a shorter sequence that happens later
in the list (5 + 7). On the other hand, given the list
l as [1, 2, 3, 4] and the key t as 15, the function
solution(l, t) would return [-1, -1] because there is
no sub-list of list l that can be summed up to the
given target value t = 15.
To help you identify the coded broadcasts, Bunny HQ has
agreed to the following standards:
- Each list l will contain at least 1 element but never
more than 100.
- Each element of l will be between 1 and 100.
- t will be a positive integer, not exceeding 250.
- The first element of the list l has index 0.
- For the list returned by solution(l, t), the start
index must be equal or smaller than the end index.
Remember, to throw off Lambda's spies, Bunny HQ might
include more than one contiguous sublist of a number
broadcast that can be summed up to the key. You know
that the message will always be hidden in the first
sublist that sums up to the key, so solution(l, t)
should only return that sublist.
To provide a Python solution, edit solution.py
To provide a Java solution, edit Solution.java
Test cases
Your code should pass the following test cases.
Note that it may also be run against hidden test cases
not shown here.
-- Python cases --
solution.solution([1, 2, 3, 4], 15)
solution.solution([4, 3, 10, 2, 8], 12)
-- Java cases --
Solution.solution({1, 2, 3, 4}, 15)
Solution.solution({4, 3, 10, 2, 8}, 12)
Use verify [file] to test your solution and see how it
does. When you are finished editing your code, use
submit [file] to submit your answer. If your solution
passes the test cases, it will be removed from your
home folder.
Here’s the first code that I tried:
def solution(l, t):
start = 0
while start < len(l):
for stop in range(start, len(l)):
s = sum(l[start:stop+1])
if s == t:
return [start, stop]
elif s > t:
start += 1
return [-1, -1]
The code solves the problem but it takes quadratic runtime so I though—can we do better? Yes, we can! There’s a linear runtime solution:
def solution(l, t):
start = stop = 0
while start <= stop and stop < len(l):
s = sum(l[start:stop+1])
if s == t:
return [start, stop]
elif s < t:
stop += 1
start += 1
stop = max(start, stop)
return [-1, -1]
Both solutions work—but the latter is much faster. Here’s the output and the test cases:
print(solution([250,0,0], 250))
print(solution([1,2,3,4], 15))
print(solution([4, 3, 10, 2, 8], 12))
print(solution([4, 3, 5, 7, 8], 12))
print(solution([260], 260))
[0, 0]
[-1, -1]
[2, 3]
[0, 2]
[0, 0]
After submitting the solution in my browser shell, Google tells me that there is one more challenge to go to reach the next level:
Level 2 Challenge 2: Digits and Remainder Classes
Here’s the problem posed by Google:
Please Pass the Coded Messages
You need to pass a message to the bunny prisoners, but
to avoid detection, the code you agreed to use is…
obscure, to say the least. The bunnies are given food
on standard-issue prison plates that are stamped with
the numbers 0-9 for easier sorting, and you need to
combine sets of plates to create the numbers in the
code. The signal that a number is part of the code is
that it is divisible by 3. You can do smaller numbers
like 15 and 45 easily, but bigger numbers like 144 and
414 are a little trickier.
Write a program to help yourself quickly create large
numbers for use in the code, given a limited number of
plates to work with.
You have L, a list containing some digits (0 to 9).
Write a function solution(L) which finds the largest
number that can be made from some or all of these
digits and is divisible by 3. If it is not possible to
make such a number, return 0 as the solution. L will
contain anywhere from 1 to 9 digits. The same digit may
appear multiple times in the list, but each element in
the list may only be used once.
To provide a Java solution, edit Solution.java
To provide a Python solution, edit solution.py
Test cases
Your code should pass the following test cases.
Note that it may also be run against hidden test cases
not shown here.
-- Java cases --
Solution.solution({3, 1, 4, 1})
Solution.solution({3, 1, 4, 1, 5, 9})
-- Python cases --
solution.solution([3, 1, 4, 1])
solution.solution([3, 1, 4, 1, 5, 9])
I first went on developing a naive solution (no premature optimization)!
def solution(l):
x = find_largest_bucket(l)
return find_max_number(x)
def find_largest_bucket(l):
''' Are the digits in the list divisible?'''
if sum(int(digit) for digit in l)%3 == 0:
return l
''' Find all smaller buckets recursively '''
buckets = []
for digit in l:
if digit not in {0, 3, 6, 9}:
tmp = l[:]
largest_bucket = max(buckets, key=find_max_number)
return largest_bucket
def find_max_number(l):
'''Returns maximal number that can be generated from list.'''
sorted_list = sorted(l)[::-1]
number = ''.join(str(x) for x in sorted_list)
return int(number)
print(solution([3, 1, 4, 1]))
print(solution([1 for i in range(1000000)]))
While this code solves the problem, it’s not optimal. It can be painfully slow for large lists because of the many levels of recursion. So I went back and developed a new version based on remainder
Here’s the code of the new idea:
def solution(l):
# Don't modify existing list object
bucket = l[:]
# Remainder Class:
state = sum(bucket)%3
while state > 0 and bucket:
# Transition between remainder classes by removing numbers
if state == 1:
to_remove = set(bucket) & {1, 4, 7}
if not to_remove:
to_remove = set(bucket) & {2, 5, 8}
elif state == 2:
to_remove = set(bucket) & {2, 5, 8}
if not to_remove:
to_remove = set(bucket) & {1, 4, 7}
# Remove min and move to new remainder class
state = sum(bucket) % 3
# Calculate maximal number from bucket
sorted_list = sorted(bucket)[::-1]
number = ''.join(str(x) for x in sorted_list)
return int(number) if number else 0
print(solution([3, 1, 4, 1]))
print(solution([3, 1, 4, 1, 5, 9]))
print(solution([9, 9, 9, 9]))
The output is correct. After submitting the solution, Google tells me that I managed to pass the level successfully! Hurray!
I even get to refer a friend…
Awesome! Commander Lambda was so impressed by your
efforts that she's made you her personal assistant.
You'll be helping her directly with her work, which
means you'll have access to all of her files-including
the ones on the LAMBCHOP doomsday device. This is the
chance you've been waiting for. Can you use your new
access to finally topple Commander Lambda's evil
Level 3 Challenge 1:
Here’s the next challenge:
foobar:~/prepare-the-bunnies-escape xcent.py$ cat readme.txt
Prepare the Bunnies' Escape
You're awfully close to destroying the LAMBCHOP doomsday device and freeing Commander Lambda's bunny prisoners, but once they're free of the prison blocks, the bunnies are going to need to escape Lambda's space station via the escape pods as quickly as possible. Unfortunately, the halls of the space station are a maze of corridors and dead ends that will be a deathtrap for the escaping bunnies. Fortunately, Commander Lambda has put you in charge of a remodeling project that will give you the opportunity to make things a little easier for the bunnies. Unfortunately (again), you can't just remove all obstacles between the bunnies and the escape pods - at most you can remove one wall per escape pod path, both to maintain structural integrity of the station and to avoid arousing Commander Lambda's suspicions.
You have maps of parts of the space station, each starting at a prison exit and ending at the door to an escape pod. The map is represented as a matrix of 0s and 1s, where 0s are passable space and 1s are impassable walls. The door out of the prison is at the top left (0,0) and the door into an escape pod is at the bottom right (w-1,h-1).
Write a function solution(map) that generates the length of the shortest path from the prison door to the escape pod, where you are allowed to remove one wall as part of your remodeling plans. The path length is the total number of nodes you pass through, counting both the entrance and exit nodes. The starting and ending positions are always passable (0). The map will always be solvable, though you may or may not need to remove a wall. The height and width of the map can be from 2 to 20. Moves can only be made in cardinal directions; no diagonal moves are allowed.
To provide a Python solution, edit solution.py
To provide a Java solution, edit Solution.java
Test cases
Your code should pass the following test cases.
Note that it may also be run against hidden test cases not shown here.
-- Python cases --
solution.solution([[0, 1, 1, 0], [0, 0, 0, 1], [1, 1, 0, 0], [1, 1, 1, 0]])
solution.solution([[0, 0, 0, 0, 0, 0], [1, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0], [0, 1, 1, 1, 1, 1], [0, 1, 1, 1, 1, 1], [0, 0, 0, 0, 0, 0]])
-- Java cases --
Solution.solution({{0, 1, 1, 0}, {0, 0, 0, 1}, {1, 1, 0, 0}, {1, 1, 1, 0}})
Solution.solution({{0, 0, 0, 0, 0, 0}, {1, 1, 1, 1, 1, 0}, {0, 0, 0, 0, 0, 0}, {0, 1, 1, 1, 1, 1}, {0, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0}})
Use verify [file] to test your solution and see how it does. When you are finished editing your code, use submit [file] to submit your answer. If your solution passes the test cases, it will be removed from your home folder.
And my solution:
def solution(m):
# Calculate map stats
w, h = len(m[0]), len(m) # width and height
# The current minimal solution
s_min = 401
# Iterate over all possible inputs (by replacing 1s with 0s).
for m_0 in all_maps(m):
# Find the minimal path length
s_min = min(min_path(m_0, w, h), s_min)
# Optimization: Minimal solution?
if s_min == w + h - 1:
return s_min
return s_min
def min_path(m, w, h):
'''Takes a map m and returns the minimal path
from the start to the end node. Also pass width and height.
# Initialize dictionary of path lengths
# integer: {(i,j), ...} (Set of nodes (i,j) with this integer path length)
d = {1: {(0,0)}}
# Expand "fringe" successively
path_length = 2
while path_length < 401 and d[path_length-1]:
# Fill fringe
fringe = set()
for x in d[path_length-1]:
# Expand node x (all neighbors) and exclude already visited
expand_x = {y for y in neighbors(x,m) if not any(y in visited for visited in d.values())}
fringe = fringe | expand_x
# Have we found min path of exit node?
if (h-1, w-1) in fringe:
return path_length
# Store new fring of minimal-path nodes
d[path_length] = fringe
# Find nodes with next-higher path_length
path_length += 1
return 401 # Infinite path length
def neighbors(x, m):
'''Returns a set of nodes (as tuples) that are neighbors of node x in m.'''
i, j = x
w, h = len(m[0]), len(m)
candidates = {(i-1,j), (i+1,j), (i,j-1), (i,j+1)}
neighbors = set()
for y in candidates:
i, j = y
if i>=0 and i<h and j>=0 and j<w and m[i][j] == 0:
return neighbors
def all_maps(m):
'''Returns an iterator for memory efficiency
over all maps that arise by replacing a '1' with a '0' value.'''
# Unchanged map is a valid solution
yield m
for i in range(len(m)):
for j in range(len(m[i])):
if m[i][j]:
# Copy the map
copy = [[col for col in row] for row in m]
# Replace 1 by 0 and yield new map
copy[i][j] = 0
yield copy
print(solution([[0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0]]))
print(solution([[0, 1, 1, 0],
[0, 0, 0, 1],
[1, 1, 0, 0],
[1, 1, 1, 0]]))
# 7
print(solution([[0, 0, 0, 0, 0, 0],
[1, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 0, 0],
[0, 1, 1, 1, 1, 1],
[0, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 0, 0]]))
# 11
Level 3 Challenge 2
Success! You've managed to infiltrate Commander Lambda's evil organization, and finally earned yourself an entry-level position as a Minion on her space station. From here, you just might be able to subvert her plans to use the LAMBCHOP doomsday device to destroy Bunny Planet. Problem is, Minions are the lowest of the low in the Lambda hierarchy. Better buck up and get working, or you'll never make it to the top…
Commander Lambda sure is a task-master, isn't she? You're being worked to the bone!
You survived a week in Commander Lambda's organization, and you even managed to get yourself promoted. Hooray! Henchmen still don't have the kind of security access you'll need to take down Commander Lambda, though, so you'd better keep working. Chop chop!
The latest gossip in the henchman breakroom is that "LAMBCHOP" stands for "Lambda's Anti-Matter Biofuel Collision Hadron Oxidating Potentiator". You're pretty sure it runs on diesel, not biofuel, but you can at least give the commander credit for trying.
You got the guards to teach you a card game today, it's called Fizzbin. It's kind of pointless, but they seem to like it and it helps you pass the time while you work your way up to Commander Lambda's inner circle.
Awesome! Commander Lambda was so impressed by your efforts that she's made you her personal assistant. You'll be helping her directly with her work, which means you'll have access to all of her files-including the ones on the LAMBCHOP doomsday device. This is the chance you've been waiting for. Can you use your new access to finally topple Commander Lambda's evil empire?
Who the heck puts clover and coffee creamer in their tea? Commander Lambda, apparently. When you signed up to infiltrate her organization, you didn't think you'd get such an up-close and personal look at her more… unusual tastes.
foobar:~/ xcent.py$ request
Requesting challenge…
There are a lot of difficult things about being undercover as Commander Lambda’s personal assistant, but you have to say, the personal spa and private hot cocoa bar are pretty awesome.
New challenge “Fuel Injection Perfection” added to your home folder.
Time to solve: 96 hours.
Fuel Injection Perfection
Commander Lambda has asked for your help to refine the automatic quantum antimatter fuel injection system for her LAMBCHOP doomsday device. It's a great chance for you to get a closer look at the LAMBCHOP - and maybe sneak in a bit of sabotage while you're at it - so you took the job gladly.
Quantum antimatter fuel comes in small pellets, which is convenient since the many moving parts of the LAMBCHOP each need to be fed fuel one pellet at a time. However, minions dump pellets in bulk into the fuel intake. You need to figure out the most efficient way to sort and shift the pellets down to a single pellet at a time.
The fuel control mechanisms have three operations:
1) Add one fuel pellet
2) Remove one fuel pellet
3) Divide the entire group of fuel pellets by 2 (due to the destructive energy released when a quantum antimatter pellet is cut in half, the safety controls will only allow this to happen if there is an even number of pellets)
Write a function called solution(n) which takes a positive integer as a string and returns the minimum number of operations needed to transform the number of pellets to 1. The fuel intake control panel can only display a number up to 309 digits long, so there won't ever be more pellets than you can express in that many digits.
For example:
solution(4) returns 2: 4 -> 2 -> 1
solution(15) returns 5: 15 -> 16 -> 8 -> 4 -> 2 -> 1
To provide a Python solution, edit solution.py
To provide a Java solution, edit Solution.java
Test cases
Your code should pass the following test cases.
Note that it may also be run against hidden test cases not shown here.
-- Python cases --
-- Java cases --
Use verify [file] to test your solution and see how it does. When you are finished editing your code, use submit [file] to submit your answer. If your solution passes the test cases, it will be removed from your home folder.
def solution(n):
x = int(n)
c = 0
while x > 1:
if x & 1 == 1:
# x is odd
if x % 4 == 1 or x==3:
x -= 1
x += 1
# x is even
x = x >> 1
c += 1
return c
# 2
# 5
# 4
# 5
# 6
# 6
foobar:~/fuel-injection-perfection xcent.py$ verify solution.py
Verifying solution…
All test cases passed. Use submit solution.py to submit your solution
foobar:~/fuel-injection-perfection xcent.py$ submit solution.py
Are you sure you want to submit your solution?
[Y]es or [N]o: Y
Submitting solution…
Submission: SUCCESSFUL. Completed in: 2 hrs, 20 mins, 18 secs.
Current level: 3
Challenges left to complete level: 1
Level 1: 100% [==========================================]
Level 2: 100% [==========================================]
Level 3: 66% [===========================……………]
Level 4: 0% [……………………………………]
Level 5: 0% [……………………………………]
Type request to request a new challenge now, or come back later.
Level 3 Challenge 3 Baby Bomb
foobar:~/ xcent.py$ request
Requesting challenge…
There are a lot of difficult things about being undercover as Commander Lambda's personal assistant, but you have to say, the personal spa and private hot cocoa bar are pretty awesome.
New challenge "Bomb, Baby!" added to your home folder.
Time to solve: 96 hours.
foobar:~/ xcent.py$
foobar:~/ xcent.py$ cd bomb-baby/
foobar:~/bomb-baby xcent.py$ ls
foobar:~/bomb-baby xcent.py$ cat solution.py
def solution(x, y):
# Your code here
foobar:~/bomb-baby xcent.py$ cat readme.txt
Bomb, Baby!
You're so close to destroying the LAMBCHOP doomsday device you can taste it! But in order to do so, you need to deploy special self-replicating bombs designed for you by the brightest scientists on Bunny Planet. There are two types: Mach bombs (M) and Facula bombs (F). The bombs, once released into the LAMBCHOP's inner workings, will automatically deploy to all the strategic points you've identified and destroy them at the same time.
But there's a few catches. First, the bombs self-replicate via one of two distinct processes:
Every Mach bomb retrieves a sync unit from a Facula bomb; for every Mach bomb, a Facula bomb is created;
Every Facula bomb spontaneously creates a Mach bomb.
For example, if you had 3 Mach bombs and 2 Facula bombs, they could either produce 3 Mach bombs and 5 Facula bombs, or 5 Mach bombs and 2 Facula bombs. The replication process can be changed each cycle.
Second, you need to ensure that you have exactly the right number of Mach and Facula bombs to destroy the LAMBCHOP device. Too few, and the device might survive. Too many, and you might overload the mass capacitors and create a singularity at the heart of the space station - not good!
And finally, you were only able to smuggle one of each type of bomb - one Mach, one Facula - aboard the ship when you arrived, so that's all you have to start with. (Thus it may be impossible to deploy the bombs to destroy the LAMBCHOP, but that's not going to stop you from trying!)
You need to know how many replication cycles (generations) it will take to generate the correct amount of bombs to destroy the LAMBCHOP. Write a function solution(M, F) where M and F are the number of Mach and Facula bombs needed. Return the fewest number of generations (as a string) that need to pass before you'll have the exact number of bombs necessary to destroy the LAMBCHOP, or the string "impossible" if this can't be done! M and F will be string representations of positive integers no larger than 10^50. For example, if M = "2" and F = "1", one generation would need to pass, so the solution would be "1". However, if M = "2" and F = "4", it would not be possible.
To provide a Java solution, edit Solution.java
To provide a Python solution, edit solution.py
Test cases
Your code should pass the following test cases.
Note that it may also be run against hidden test cases not shown here.
-- Java cases --
Solution.solution('2', '1')
Solution.solution('4', '7')
-- Python cases --
solution.solution('4', '7')
solution.solution('2', '1')
Use verify [file] to test your solution and see how it does. When you are finished editing your code, use submit [file] to submit your answer. If your solution passes the test cases, it will be removed from your home folder.
foobar:~/bomb-baby xcent.py$
def solution(x, y):
goal = (int(x), int(y))
start = (1, 1)
gen = [start]
c = 0
while gen:
# Generate new states
next_gen = []
for M,F in gen:
if (M,F) == goal:
return str(c)
# Ignore states that never lead to goal
MF = M+F
if MF <= goal[0]:
next_gen.append((MF, F))
if MF <= goal[1]:
next_gen.append((M, MF))
# Go to next generation
gen = next_gen
c += 1
return 'impossible'
print(solution('4', '7'))
# 4
print(solution('2', '1'))
# 1
print(solution('2', '4'))
# impossible
def old_solution(x, y):
goal = (int(x), int(y))
start = (1, 1)
gen = [start]
c = 0
while gen:
# Generate new states
next_gen = []
for M,F in gen:
if (M,F) == goal:
return str(c)
# Ignore states that never lead to goal
MF = M+F
if MF <= goal[0]:
next_gen.append((MF, F))
if MF <= goal[1]:
next_gen.append((M, MF))
# Go to next generation
gen = next_gen
c += 1
return 'impossible'
seen = set()
def solution(x, y):
goal = (int(x), int(y))
start = (1, 1)
gen = set([start])
c = 0
while gen:
if goal in gen:
return str(c)
# Generate new states
next_gen = set()
for M,F in gen:
# Ignore states that never lead to goal
MF = M+F
if MF <= goal[0]:
state = (MF, F)
if state not in seen:
if MF <= goal[1]:
state = (M, MF)
if state not in seen:
# Go to next generation
gen = next_gen
c += 1
return 'impossible'
print(old_solution('4', '7'))
print(old_solution('2', '1'))
print(old_solution('2', '4'))
import time
t0 = time.time()
print(old_solution(str(10**4), str(10**3)))
t1 = time.time()
print(t1-t0, 'seconds have passed')
# 3.7617175579071045 seconds have passed
print(solution('4', '7'))
print(solution('2', '1'))
print(solution('2', '4'))
import time
t0 = time.time()
print(solution(str(10**4), str(10**4)))
t1 = time.time()
print(t1-t0, 'seconds have passed')
def solution(M, F):
goal = (int(M), int(F))
x, y = goal
c = 0
while x!=y:
if x > y:
num_subs = (x-y)//y + ((x-y) % y > 0)
c += num_subs
x, y = x - num_subs * y, y
elif y > x:
num_subs = (y-x)//x + ((y-x) % x > 0)
c += num_subs
x, y = x, y - num_subs * x
return str(c) if (x, y)==(1, 1) else 'impossible'
foobar:~/bomb-baby xcent.py$ verify solution.py
Verifying solution…
All test cases passed. Use submit solution.py to submit your solution
foobar:~/bomb-baby xcent.py$ submit solution.py
Are you sure you want to submit your solution?
[Y]es or [N]o: Y
Submitting solution…
Excellent! You've destroyed Commander Lambda's doomsday device and saved Bunny Planet! But there's one small problem: the LAMBCHOP was a wool-y important part of her space station, and when you blew it up, you triggered a chain reaction that's tearing the station apart. Can you rescue the imprisoned bunnies and escape before the entire thing explodes?
Submission: SUCCESSFUL. Completed in: 4 hrs, 35 mins, 50 secs.
Level 3 complete
You are now on level 4
Challenges left to complete level: 2
Level 1: 100% [==========================================]
Level 2: 100% [==========================================]
Level 3: 100% [==========================================]
Level 4: 0% [……………………………………]
Level 5: 0% [……………………………………]
Refer a friend: "https://foobar.withgoogle.com/?eid=ZYbpR" (Used)
Type request to request a new challenge now, or come back later.
The code is strong with this one…
You can now share your solutions with a Google recruiter!
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[#1] [Yes [N]o [A]sk me later:]
[Y]es [N]o [A]sk me later: A
Response: contact postponed.
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What is the derivative of y=arcsin(x)? | Socratic
What is the derivative of #y=arcsin(x)#?
1 Answer
The answer is:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}}$
This identity can be proven easily by applying $\sin$ to both sides of the original equation:
1.) $y = \arcsin x$
2.) $\sin y = \sin \left(\arcsin x\right)$
3.) $\sin y = x$
We continue by using implicit differentiation, keeping in mind to use the chain rule on $\sin y$:
4.) $\cos y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:
5.) $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} y$
Now, substitution with our original equation yields $\frac{\mathrm{dy}}{\mathrm{dx}}$ in terms of $x$:
6.) $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left(\arcsin x\right)$
At first this might not look all that great, but it can be simplified if one recalls the identity
$\sin \left(\arccos x\right) = \cos \left(\arcsin x\right) = \sqrt{1 - {x}^{2}}$.
7.) $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}}$
This is a good definition to memorize, along with $\frac{d}{\mathrm{dx}} \left[\arccos x\right]$ and $\frac{d}{\mathrm{dx}} \left[\arctan x\right]$, since they appear quite frequently in
differentiation problems.
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Assessment of Seagoing Ships Longitudinal Strength in the Context of International Rules, Important Factor for Safe Operation
Assessment of Seagoing Ships Longitudinal Strength in the Context of International Rules, Important Factor for Safe Operation ()
Received 19 August 2015; accepted 27 November 2015; published 30 November 2015
1. Introduction
Safety construction of seagoing ships is one of the most important goals of International Maritime Organization (IMO) to reduce material and human life losses and to decrease environment pollution
due to international maritime transport.
In order to accomplish this goal, IMO has adopted the International Goal-Based Ship Construction Standards (GBS), organized on 5 tiers as indicated in [1] :
Tier I―Goals. High-level objectives have to be met;
Tier II―Functional requirements. Criteria to be satisfied in order to conform to the goals;
Tier III―Verification of conformity. Procedures for verifying that the rules and regulations for ship design and construction conform to the goals and functional requirements [2] ;
Tier IV―Rules and regulations for ship design and construction. Detailed requirements developed by IMO, national administrations and/or recognized organizations and applied by national
administrations and/or recognized organizations acting on their behalf to the design and construction of a ship in order to conform to the goals and functional requirements;
Tier V―Industry practices and standards. Industry standards, codes of practice and safety and quality systems for shipbuilding, ship operation, maintenance, training, manning, etc., which may be
incorporated into, or referenced in, the rules and regulations for the design and construction of a ship.
A lower level of safety construction may determine material losses and failures, human life losses or ecological disasters.
Hereby, for example, in the period 1966-1985, annually over 300 ships [3] have been lost. After this period, the number of losses starts to decrease continually at the same time with improving
technical regulations, mainly through bringing into force the new unified requirements for longitudinal strength by International Association of Classification Societies (IACS), of new IMO intact
stability requirements from SOLAS Convention, through adoption of MARPOL Convention and strengthen of Port State Control (PSC), determining that after 2000 the number of ships lost will annually drop
under 170, which means under 0.2%, and after 2010 less than 120 [4] .
Among lost ships, a significant share was confined by bulk carriers and oil tankers, some of them braking down through collapse of longitudinal stiffeners of the hull and breaking down into two
parts, as could be seen in Figures 1-3.
Coordinated studies showed that losses caused by hull breaking down depended probabilistically on certain factors:
- ship’s age, increasing with that;
- type of cargo, increasing with its specific gravity;
- chosen trade route, the most dangerous being trade routes of Far East and Northern Atlantic;
- type of material used in ship hull construction, high strength steels increasing the risk to lose the ship as they are prone to corrosion. They have smaller allowances and ensure a greater
elasticity of the hull, fostering the phenomenon of springing (manifestation of ship general vibration induced by waves which stress her additionally and determine the decreasing of fatigue
resistance) and whipping phenomenon.
In order to reduce those alarming losses, IMO has adopted some urgent measures among which more important are:
- adoption on 11 November 1988 of the Protocol of 1988 to the 1966 Load Lines Convention;
Figure 1. EUROBULKER-X Bulk Carrier (year of built: 1974) after breaking on September 02, 2000, while loading cement in port Lefkandi from Greece [5] .
Figure 2. ERIKA tanker (year of built: 1975) after breaking on Dec. 12, 1999, at 60 miles from the coast of Brittany [6] .
Figure 3. PRESTIGE tanker (year of built: 1976) after breaking on November 13, 2002, at 30 miles off the Northwest coast of Spain [7] .
- adoption on 4 November 1993 of IMO Rez. A.744 (18) on enhanced survey system applicable to Bulk Carriers and Tankers;
- adoption on 23 November 1995 of IMO Rez. A.787 (19) on PSC responsibilities;
- adoption on 27 November 1997 of SOLAS Chapter XII on additional measures for safety of Bulk Carriers;
- adoption on 27 November 1997 of Bulk Carriers Safe Loading and Unloading Code (BLU Code) (IMO Rez. A.862 (20)).
The issuing in 1998 of the report on sinking of bulk carrier Derbyshire determined IMO to conduct new studies to enhance the requirements of its Conventions, so that in December 2002 SOLAS Convention
was amended, by introducing new requirements in Chapter XII referring to bulk carriers, and in December 2004, Chapter XII of SOLAS was revised again and new requirements for double hull to bulk
carriers were introduced. Furthermore, after this date, following documents were adopted:
- on 3 December 2004 the Code of Safe Practice for Solid Bulk Cargoes (BC Code) (Rez. MSC.193 (73));
- on 4 December 2008 the International Maritime Solid Bulk Cargoes Code (IMSBC Code) (Rez. MSC.268 (85));
- on 30 November 2011 the 2011 International Code on the Enhanced Programme of Inspections During Surveys of Bulk Carriers and Oil Tankers (2011 ESP Code) (Rez. A.1049 (27)) [8] .
Assessing the circumstances in which both bulk carriers and oil tankers had been lost through hull broken down in two pieces, it became obvious that they sank due to the fact that their longitudinal
structure had a weak general strength and collapsed under distributed loads of cargoes on board, conditions of navigation and amidships long term fatigue strength.
As a result, for the safe operation of ships, the evaluation of their longitudinal strength is very important and shall be carried out during their life cycle, starting in design stage, continuing in
building phase and after through systematic checking during entire operational life.
2. Longitudinal Strength Assessment of Seagoing Ships in Design Phase
In design phase, as hull is concerned, they define ship’s building and operational characteristics, navigation area, classification society or naval authority based on which rules the ship is
classified, ship lines are established, ship subdivision is carried out based on operational requirements and stability and floatability criteria for intact or damaged condition, hull structural
configuration is defined and materials for shipbuilding are selected [9] .
Dimensions of ship’s structural elements are established according to the rules of society which classify the ship or the technical norms of the authority which flag the built ship is flying, so that
local strength requirements are fulfilled, as well as general strength requirements. Scantlings of transversal and longitudinal stiffeners of ship’s hull are performed firstly according to local
strength requirements, and after scantlings of longitudinal stiffeners are performed according to general strength requirements. These scantling processes could be optimized with the aim to fulfill
the above requirements with a minimum input of materials, energy and manpower.
It is worthy of note that the Common Structural Rules carried out by IACS for bulk carriers and oil tankers, which are considered by IMO as representing a part of tier IV within Goal Based Ship
Construction Standards system, introduce net scantling approach, which require that net thickness prescribed by classification rules must be maintained during the whole operational life of the ship.
Assessment of fulfillment of longitudinal strength is performed by covering the following steps:
- numerical modeling of ship’s hull outside surface;
- determining of BONJEAN diagram;
- determining of weight distribution for ship in light condition;
- establishing of ship loading cases which will be analyzed;
- finding of equilibrium position of the ship in calm water;
- calculation of sectional stresses of ship’s hull in calm water;
- calculation of sectional stresses of ship’s hull on waves, determined according to rules of classification societies aligned to IACS unified requirements for longitudinal strength;
- calculation of strength characteristics of hull transversal sections;
- calculation of safe sectional stresses;
- verify satisfaction of longitudinal strength criteria in elastic range;
- verify satisfaction of ultimate longitudinal strength criteria;
- verify satisfaction of longitudinal fatigue strength criteria.
If at the end of this assessment they ascertain that one or more criteria are not satisfied, the assessment is recommenced, by adequately increasing the dimensions of longitudinal stiffeners and/or
their number until when all criteria are fulfilled. They must observe that assessment of longitudinal strength in design phase is preliminary and the final one will be carried out after finalizing
the ship construction, based on data from free floating and trim still water equilibrium condition (displacement and position of center of gravity for ship in light condition from the inclining test
3. Longitudinal Strength Assessment of Seagoing Ships during Construction Phase
During construction phase, the longitudinal strength assessment is ensured through verification by quality assurance compartments in shipyard and through technical survey carried out by a
classification society or the naval authority whose flag the ship is flying, if:
- hull stiffeners are manufactured and assembled according to technical documentation for ship construction, approved by classification society or naval authority;
- stiffeners’ materials indicated in this documentation and welding materials have appropriate characteristics and are accompanied by certificates endorsed by recognized classification societies;
- standards on dimension limit system for shipbuilding are observed;
- welding technologies are accepted by relevant classification society and are observed;
- welders who perform stiffeners’ weld joints are authorized by classification society;
- welding processes are performed in the succession which induce the smallest strains in welded elements;
- welding processes are performed in appropriate atmospheric and thermal conditions;
- results of non-destructive examinations of welded joints are satisfactory;
- operators who carried out these examinations are authorized by the classification society;
- anti-corrosion protection is accomplished.
When construction is finished, an inclining experiment is conducted in order to determine lightweight displacement and position of center of gravity. Based on these data, load distribution is
corrected and all general strength assessment is restarted, by finalizing in conformity with the rules of society which classified the ship, developing the loading manual and updating the database
attached to the loading instrument, if that is requested by these rules.
Furthermore, for Bulk Carriers, the Manual on loading and unloading of solid bulk cargoes for terminal representatives shall be drawn up in conformity with the Code of Practice for the Safe Loading
and Unloading of Bulk Carriers (BLU Code).
4. Longitudinal Strength Assessment of Seagoing Ships in Service
For safe operation of ships, they must perform the voyage loaded in accordance with loading manual or in accordance with a cargo-plan developed by the crew using loading instrument existing onboard,
which was previously approved by the classification society. Loading and unloading of bulk carriers at terminals shall be carried out in accordance with Manual on loading/unloading sequences existing
onboard or a Plan with loading/unloading sequences developed by the crew using loading instrument.
Likewise, the general strength assessment of ships in operation is conducted on the occasion of periodical surveys at 5 years interval, of intermediate surveys carried out at 2.5 years interval and
of annual surveys. On the occasion of periodical and intermediate surveys, thickness measurements of stiffeners are carried out, by companies authorized by classification society or competent naval
authority, measurements which shall be registered on forms to be kept in a folder onboard.
Measured reduced plating thicknesses are compared with maximum reduced thicknesses accepted, and in case these accepted reduced thicknesses are exceeded, the stiffeners are replaced. Maximum
decreases accepted are usually corrosion allowances adopted at design stage.
Greater reduced thicknesses could be accepted, but in this case the assessment process of general and local strength of design phase shall be recommenced, with the exception that initial thicknesses
of stiffeners are those of the Report of thicknesses’ measurements. Based on this assessment, restrictions in ship operation could be imposed and the Loading Manual, Stability Booklet, the Loading
Instrument and Manual of loading/unloading sequences in case of bulk carriers shall be modified accordingly.
At annual survey, the followings shall be visually verified by sampling: the integrity of stiffeners, the presence onboard of Loading Manual and correct functioning of Loading Instrument.
Oil tankers and bulk carriers are subject of the enhanced programme of inspections in conformity with 2011 ESP Code (IMO Rez. A.1049 (27)) [8] and MSC.287 (87) [1] .
The Common Structural Rules for Bulk Carriers and Oil Tankers introduce the approach of stiffeners renewal criterion if the corrosion allowance was lost from initial thickness established by design.
Fatigue strength assessment is realized through detailed inspection and non-destructive examination of structural joints most exposed at variable loads indicated by designer, in order to track down
possible cracks specific to fatigue phenomenon.
Those inspections will be carried out annually if ship age is greater than the actual operational age L[a], deduced by applying the equation:
D[n] is expected life of the ship [in years];
D is cumulative fatigue damage factor estimated for the expected life of the ship determined with the equation:
k is the total number of spectrum blocs of stress variation for the period assessed, e.g. the expected life of the ship 25 years;
n[i][ ]is the number of stress cycles in i-th stress block;
N[i] is the number of alternating symmetrical cycles which could be endured by the material until fatigue failure determined from the corrected diagram S-N for Δσ = Δσ[i] from Figure 4;
Δσ[i] is the stress variation cycle from i-th stress block, in [MPa].
In case when cumulative damage factor D cannot be calculated sufficiently accurate, it can be found experimentally life of the ship using the structural joints most intensively stressed and testing
them to fatigue in laboratory until they brake.
If these tests were carried out after Y years since the putting into service of the ship and in laboratory the residuary cumulative damage factor D[r], which remained until the break, is determined,
then the real operational life of the ship L[r] can be found using equation:
In case when cracks are discovered due to material fatigue before expected life of the ship expires, the fatigue strength of structural joints affected by this phenomenon will be reassessed, these
joints will be redrawn and the cracked and tired stiffeners will be replaced by new redrawn stiffeners, and the inspection of joints will be carried annually.
Moreover, during the inspections, anticorrosive protection is verified and if it is found damaged, it will be reconditioned.
5. Critical Assessment of IACS Method for Calculation of Sectional Efforts Induced by Waves in Ship’s Hull Based on Sectional Efforts Calculated for Ship under Equivalent Quasi-Static Wave Design
5.1. Generals
In order to calculate sectional efforts due to waves, classification societies aligned their methods of calculus with
Figure 4. Diagram of stress-cycles [10] [11] .
that established by IACS based on the experience of its members, based on wide researches theoretical and experimental on models and ships, as well as based on statistical data regarding sea status,
derived from measurements on large areas and long periods of time.
According to such IACS method ( [12] , Ch.S11), for the ships other than container ships, bulk carriers and double hull oil tankers, the vertical wave bending moment arising at any hull transverse
section with probability of 10^−8, at navigation in rough seas, parallel with waves’ propagation direction, is obtained from the following formula:
- for hogging condition:
- for sagging condition:
L―length of ship [m];
B―breadth of ship [m];
k[H] = 190;
k[S] = 110;
F[M]―distribution factor defined in Table 1;
C[B]―block coefficient of ship for draught at full loading;
C―wave parameter (representing wave height corrected due to Smith effect) given by following equations:
The vertical wave shear force at any hull transverse section, at navigation in rough seas, parallel with waves’ propagation direction, is obtained from the following formula:
[kN] (7)
F[Q]―distribution factor defined in Table 2 for positive and negative shear forces;
k[Q] = 30
For container ships, IACS proposed method ( [12] , Ch.S11A) and for bulk carriers and double hull oil tankers established method ( [11] , Ch4.sec4.3), similar as that above described.
Verification of sectional efforts induced by waves, calculated according to these IACS methods, may be accomplished through direct calculus and several methods based on hypothesis which reduce the
complexity of calculus but not significantly affect the correctness of results beside real values were developed.
A first direct and effective method consists of ship quasi-static layout on wave.
This method ensures obtaining of accurate results for navigation in astern waves parallel with ship’s way.
It is theoretical and experimental proved the fact that additional vertical sectional efforts reach the maximum value when the top or hollow of the wave are amidships and the wave length is equal
with the length of the ship.
For ship quasi-static layout on wave, Smith effect may be taken into account to reduce by 10% hydrostatic pressure in depth as a result of circular movement of wave particles.
Considering the ship quasi-static layout on trochoidal wave, the ship-equivalent quasi-static wave system equilibrium draught in a transversal section is given by formula:
- in case amidships is wave on hogging conditions (Figure 5);
- in case amidships is in wave on sagging conditions (Figure 6),
F is a parameter obtained from transcendental equation:
l―wave length [m]. Wave length is considered equal with ship’s length, L;
T[o]―the system equilibrium draught at aft ship’s extremity referring the position of the head ship-equivalent quasi-static. wave medium plane into the ship's base plane reference (see Figure 5 and
Figure 6) in [m];
y―the system equilibrium trim at aft ship’s extremity referring the position of the head ship-equivalent quasi-static. wave medium plane into the ship's base plane reference (see Figure 5 and Figure
6) in [rad];
h―wave height in [m];
L―length of ship in [m].
For simplicity, keeping accuracy of the calculations within reasonable limits, the wave also can be regarded as cosine form and then the ship-equivalent quasi-static wave system equilibrium draught
in a transversal section is given by formula:
- in case amidships is wave on hogging conditions (Figure 5);
- in case amidships is in wave on sagging conditions (Figure 6),
Figure 5. Ship quasi-static layout on wave on hogging conditions.
Figure 6. Ship quasi-static layout in wave on sagging conditions.
The additional sectional efforts induced by wave alongside ship are obtained by formulae:
- additional shear force alongside ship:
- additional bending moment alongside ship:
Q[TW](x) is total shear force at ship quasi-static layout on wave determined for draughts given by Formula (8), (9), (10) or (11) in [kN];
Q[SW](x) is shear force at quasi-static layout on still water, in [kN];
M[TW](x) is total bending moment at ship quasi-static layout on wave determined for draughts given by Formula (8), (9), (10) or (11) in [kN×m];
MSW(x) is bending moment at quasi-static layout on still water, in [kN×m];
The calculus of additional sectional efforts appearing in the hull of a ship quasi-static layout on wave will be further performed by using a FORTRAN code for ship quasi-static head wave equilibrium
iterative approach whose calculation method is presented in [13] .
The main flow chart of the authors’ code is shown in Figure 7.
Note 1: Equilibrium Requirements
The ship is in the quasi-static equilibrium on water, if the displacement is equal with the buoyant force and if its center of gravity G is on the same vertical with center of buoyancy C (Figure 8).
In the code method they consider that the ship’s equilibrium is performed if the following relations are fulfilled:
k[a]―appendages coefficient;
Figure 7. The main flow chart of the author’s code.
Figure 8. Ship’s quasi-static equilibrium on water.
V[C]―volumul of immersed hull;
x[G], z[G]―coordinates of the center of gravity of the ship;
x[C], z[C]―coordinates of the center of buoyancy.
Note 2: Finite Differences Method
Equilibrium position is found by successive calculations based on finite difference method, determining the parameters T[o] and ψ for this position based on equilibrium conditions. To this purpose,
it starts from an initial floating position and they consider that variation of area immersed Ω of a theoretical couple according to the draught T, and static moment variation of this area C in
relation to base line, are linear over a small draught interval δ. This leads to solving a system of two equations having as unknowns the increasing of the draught ζ and the trim angle ψ, on reaching
the equilibrium, as follows:
C[0] is static moment in relation to base line of submerged area of theoretical couplings for initial buoyancy;
C is static moment static in relation to base line of submerged area of the same draft couplers for increased buoyancy with value δ from the initial buoyancy.
5.2. Test of the Code to a Parallelepiped Barge
The test of calculus of the cod was carried out to a parallelepiped barge with uniform weight distribution of 100 t/m, having the following main characteristics:
L[max] = 100.00 m―maximum length;
L = 100.00 m―length at water line;
B = 20.00 m―breadth;
D = 10.00 m―depth;
T = 5.00 m―draught on still water.
Sea water density was taken 1.025 t/m^3.
The calculus were performed laying the barge on a quasi-static wave with height corrected by Smith effect equal to value C determined with IACS Formula (6), i.e. equal with 7.92 m and length equal to
length of ship, (the real wave having height of 8.80 m, period of 8.0 s and length of 100 m, is found with probability of 0.22% as could be seen from statistical measurements presented in [10] ).
For this theoretical barge at its quasi-static layout on trohoidal wave, following values have been obtained through direct not automatic calculus:
- positive/negative maximum shear force: +/−12,380 kN;
- hogging/sagging maximum bending moment: +/−394,999 kN×m;
- T[o] = 5.370 m;
- ψ = 0.00 deg.
The results obtained by code for this wave, are:
- on wave crest:
- positive maximum shear force: 12,381 kN;
- negative maximum shear force: −12,381 kN;
- hogging bending moment: 395,127 kN×m;
- T[o] = 5.370 m;
- ψ = 0.00 deg.
- in wave hollow:
- positive maximum shear force: 12,381 kN;
- negative maximum shear force: −12,381 kN;
- sagging bending moment: −395,120 kN×m;
- T[o] = 5.370 m;
- ψ = 0.00 deg.
Also, for this barge at its quasi-static layout on cosine wave, following values have been obtained through direct not automatic calculus:
- positive/negative maximum shear force: +/− gγCLB/4π = +/−12,681 kN;
- hogging/sagging maximum bending moment: +/− gγCL^2B/4π^2 = +/−403858 kN×m;
- T[o] = 4.878 m;
- ψ = 0.00 deg.
The results obtained by the code for same wave, are:
- on wave crest:
- positive maximum shear force: 12,675 kN;
- negative maximum shear force: −12,675 kN;
- hogging bending moment: 403,449 kN×m;
- T[o] = 4.878 m;
- ψ = 0.00 deg.
- in wave hollow:
- positive maximum shear force: 12,675 kN;
- negative maximum shear force: −12675 kN;
- sagging bending moment: −403,448 kNm;
- T[o] = 5.00 m;
- ψ = 0.00 deg.
The results above show that differences between manual and code calculations are under 0.1% and therefore the accuracy of the code is proved.
It is worthy of note that in case of barge, differences between values of additional for ship layout on trochoidal wave and cosine wave are less than 2.5% which allow to approximate real trochoidal
wave with a wave of cosine form wherewith technical assessments could be carried out effortless but maintaining correctness of results in acceptable limits.
The additional vertical sectional efforts appearing in the hull of a ship induced by wave obtained according to IACS are:
- hogging maximum bending moment: 300,960 kN×m
- sagging maximum bending moment: −296,208 kN×m
- positive maximum shear force: 8078 kN
- negative maximum shear force: −8208 kN
The gaps for this barge are: for additional bending moment 25%, and for additional shear forces 35%.
5.3. Application of the Methodology to a General Cargo Ship
The methodology for calculation of additional sectional efforts appearing in the hull of a ship quasi-static layout on wave was applied to a general cargo ship of 15,000 dwt shown in Figure 9.
The main characteristics of the ship are given below, the transversal line plan is shown in Figure 10 and the distribution weights are presented in Figure 11.
L[max] = 162.30 m -maximum length;
L = 155.00 m-length at water line;
B = 22.20 m-breadth;
D = 13.40 m-depth;
T = 10.10 m-draught.
This ship was quasi-static layed out on wave with height corrected by Smith effect equal to value C determined with Formula (6), i.e. equal with 8.997 m and length equal to length of ship, (the real
wave having height of 10.000 m, period of 9.96 s and length of 155 m, is found with probability of 0.1% as could be seen from statistical measurements presented in [10] ). The results of the calculus
performed by the code are presented below and graphic in Figure 12 and Figure 13.
The results obtained at quasi-static layout on trohoidal wave, are:
- on wave crest:
- maximum positive shear force: 15,164 kN;
Figure 9. The analyzed general cargo ship.
Figure 10. The transversal line plan of the general cargo ship analyzed.
Figure 11. The distribution weights of the general cargo ship analyzed.
- maximum negative shear force: −14,314 kN;
- hogging bending moment: 599,373 kN×m;
- T[o] = 8.119 m;
- ψ = 0.677 deg.
- in wave hollow:
- maximum positive shear force: 17,587 kN;
- maximum negative shear force: −17,111 kN;
- sagging bending moment: −797,591 kN×m;
- T[o] = 13.117 m;
- ψ = −0.217 deg.
Figure 12. Additional sectional efforts at quasi-static layout on trochoidal wave on hogging conditions.
The results obtained at quasi-static layout on cosine wave, are:
- on wave crest:
- maximum positive shear force: 14,622 kN;
- maximum negative shear force: −14,115 kN;
- hogging bending moment: 589,390 kN×m;
- T[o] = 7.599 m;
- ψ = 0.658 deg.
- in wave hollow:
- maximum positive shear force: 18,003 kN;
- maximum negative shear force: −18,943 kN;
- sagging bending moment: −838,078 kN×m;
- T[o] = 12.630 m;
- ψ = −0.297 deg.
According to IACS method ( [12] , Ch.S11), additional efforts induced by a wave have been determined for this general cargo ship. The results obtained with this method are:
- hogging maximum bending moment: 574,646 kN×m;
- sagging maximum bending moment: −700,266 kN×m;
- positive maximum shear force: 12,353 kN;
- negative maximum shear force: −11,365 kN.
They find out that additional sectional efforts obtained through direct calculus at ship quasi-static layout on
Figure 13. Additional sectional efforts at quasi-static layout on trochoidal wave on sagging conditions.
wave are greater than those obtained in conformity with IACS method. Consequently the direct calculus is more covering in what safety of ship is concerned.
Maximum additional bending moments determined through ship quasi-static layout on wave are about 14% greater than those determined based on IACS method and in case of the shear forces, differences
are much bigger, reaching 51%, which means that this method leads getting of ship under scantling regarding sectional efforts induced by waves.
Also, it is noted that in case of the general cargo ship, differences between values of additional for ship quasi-static layout on trochoidal wave and cosine wave are less than 10.5%.
5.4. Application of the Methodology to a Bulk Carrier
Also, the methodology was applied to a bulk carrier of 65,000 dwt shown in Figure 14. The main characteristics of the ship are given below, the transversal line plan is shown in Figure 15 and the
distribution weights are presented in Figure 16.
L[max] = 254.10 m―maximum length;
L = 250.00 m―length at water line;
B = 32.20 m―breadth;
D = 17.00 m―depth;
Figure 15. The transversal body lines of the analyzed bulk carrier.
Figure 16. The weight Distribution of the analyzed bulk carrier.
T = 12.3 m―draught.
This ship was quasi-static layed out on wave with height corrected by Smith effect equal to value C determined with Formula (6), i.e. equal with 10.396 m and length equal to length of ship, (the real
wave having height of 11.551 m, period of 12.65 s and length of 250 m, is found with probability of 0.017% as could be seen from statistical measurements presented in [10] ). The results of the
calculus performed by the code are presented below and graphic in Figure 17 and Figure 18.
The results obtained at quasi-static layout on trohoidal wave, are:
- on wave crest:
- maximum positive shear force: 51,541 kN;
Figure 17. Additional sectional efforts at quasi-static layout on trochoidal wave on hogging conditions.
- maximum negative shear force: −49,851 kN;
- hogging bending moment: 3,547,379 kN×m;
- T[o] = 10.623 m;
- ψ = 0.958 deg.
- in wave hollow:
- maximum positive shear force: 54,467 kN;
- maximum negative shear force: −54,421 kN;
- sagging bending moment: −4,024,886 kN×m;
- T[o] = 13.148 m;
- ψ = −0.904 deg.
The results obtained at quasi-static layout on cosine wave, are:
- on wave crest:
- maximum positive shear force: 50,134 kN;
- maximum negative shear force: −48828 kN;
- hogging bending moment: 3,468,118 kN×m;
- T[o] = 10.116 m;
- ψ = −0.930 deg.
- in wave hollow:
Figure 18. Additional sectional efforts at quasi-static layout on trochoidal wave on sagging conditions.
- maximum positive shear force: 55,983 kN;
- maximum negative shear force: −56,032 kN;
- sagging bending moment: −4,132,476 kN×m;
- T[o] = 12.734 m;
- ψ = −0.922 deg.
According to IACS method ( [12] , Ch4.sec4.3), additional efforts induced by a wave have been determined for this bulk carrier. The results obtained with this method are:
- hogging maximum bending moment: 3,209,681 kN×m;
- sagging maximum bending moment: −3,459,289 kN×m;
- positive maximum shear force: 37,737 kN;
- negative maximum shear force: −35,015 kN.
They find out that additional sectional efforts obtained through direct calculus at ship quasi-static layout on wave are greater than those obtained in conformity with IACS method. Consequently the
direct calculus is more covering in what safety of ship is concerned.
Maximum additional bending moments determined through ship quasi-static layout on wave are about 16% greater than those determined based on IACS method and in case of the shear forces, differences
are much bigger, reaching 55.4%, which means that this method leads getting of ship under scantling regarding sectional efforts induced by waves.
Also, it is noted that in case of the bulk carrier, differences between values of additional for ship layout on trochoidal wave and cosine wave are less than 3%.
6. Discussions, Conclusions and Proposals
In the paper, the problems on the sea going ships safety have been treated. The requirements of the IACS rules and methods for estimating the ship hull general strength have been presented. The
methods are illustrated by two examples of the sectional efforts estimation. According to the results, certain inconveniencies after the great accidents occurring in the last decades have been
expressed by the shipbuilders and operators.
The under scantling of ships in terms of wave-induced sectional efforts is confirmed by data from a Formal Safety Assessment (FSA) on Bulk Carrier Safety between 1978 and August 2000 conducted by
Japan for the IMO Maritime Safety Committee [14] . The data showed that from 1126 of fatalities, 69.70% were due to side shell failures (785). Also, it was found that from 1982 to 2001, the side
shell failures led to the sinking of 72 classic bulk carriers and a single with the double bottom [15] .
Furthermore, the ships were broken through deck collapse which meant that they were under scantling for sagging bending moment and consequently revising the formula for its calculation was required.
Problems come into attention of IMO, which are concluded in [16] , that the minimum thickness of side shell for bulk carriers must have at least the value given in following formulae:
c = 1.15 for the frame webs in way of the foremost hold;
c = 1 for the frame webs in way of other holds.
L is the ship’s length.
Furthermore, this organization imposed since 1st July 2006 through Regulation XII/6.2 of SOLAS Convention, that bulk carriers over 150 m in length to have double side [17] .
Taking into consideration those presented hereinbefore, in order to improve ship construction safety, they propose that:
- factor k[H] from Formula (4) to be increased with about 10% i.e. to have the value of 210 instead of 190;
- factor k[S] from Formula (5) to be increased with about 20% i.e. to have the value of 130 instead of 110;
- factor k[Q] from Formula (7) to be increased with 50% i.e. to have the value of 45 instead of 30.
By applying these revised formulae, following values have been obtained for the general cargo ship:
- hogging maximum bending moment: 635,135 kN×m;
- sagging maximum bending moment: −827,587 kN×m;
- positive maximum shear force: 18,529 kN;
- negative maximum shear force: −17,047.5 kN.
And following values are for the bulk carrier:
- hogging maximum bending moment: 3547,542 kN×m;
- sagging maximum bending moment: −408,8251 kN×m;
- positive maximum shear force: 56,605 kN;
- negative maximum shear force: −52,522 kN.
According to the new proposal, in the case of the general cargo ship, for additional maximum bending moments differences are less than 6% and for additional maximum shear forces differences are less
than 5%.
In the case of the bulk carrier, for additional maximum bending moments differences are less than 1% and for additional maximum shear forces differences are less than 4%.
These differences could be considered as acceptable.
It is worth noting that formulae are revised only relating to sectional efforts determined for ship quasi-static layout on wave. It is necessary that such analysis should be carried out for dynamic
layout on wave as well and if bigger efforts are obtained, this means revising again the afore-mentioned formulae.
From the assessment carried out, it shows that IACS formulae for calculation of additional sectional efforts induced by waves must be corrected because these efforts are exceeded in real situations
in which ship layout is quasi-static on wave, as in cases where ship navigates with wave from astern. If uncorrected IACS formulae continue to be used, imposes as a first measure to improve ship
construction safety since design phase, the determination of sectional efforts at quasi-static layout on wave through direct calculations and if those have greater values as those obtained with IACS
formulae, the scantling of longitudinal structure of ship will be carried out taking into account of respective values.
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Where to learn high school math?
High school mathematics can be a challenging subject for many students. Often, high school students are clueless about where to learn high school math.
Whether you’re struggling with algebra, geometry, or calculus, finding the right resources and methods to learn math effectively can make a significant difference in your academic success.
In this article, we’ll explore various avenues for learning high-school math, from traditional in-person tutors to modern online resources, YouTube, video lessons, web tutorials, and platforms like
Khan Academy.
We’ll discuss the methods, provide examples, outline the pros and cons of each approach, and help you determine which option best suits your learning style and needs.
Looking to Learn High School Math? Book a Free Trial Lesson and match with top High School Math Tutors for Concepts, Homework Help, and Test Prep.
Learning with tutors
If you can find the right high school math tutor, It will be easier to find all your high school math resources through his assistance. You can learn high school math from in-person tutors as well as
online tutors.
In-person tutors
In-person tutors have been a trusted method for learning math for generations. Here’s a breakdown of how it works:
• You find a qualified math tutor in your area, often through recommendations, schools, or tutoring centers.
• You schedule regular sessions with the tutor, typically one-on-one.
• The tutor assesses your current knowledge, identifies areas of weakness, and tailors lessons according to your learning needs.
• They provide personalized guidance, explain concepts, and help you with problem-solving.
You can find in-person tutors by
• Searching on LinkedIn.
• Asking for references from your friends and family.
• Searching for local tutoring centers.
• Personalized attention and immediate feedback.
• Ability to ask questions and seek clarification on the spot.
• Tailored lessons based on your specific needs.
• Building a strong rapport with the tutor.
• Higher cost compared to some online options, typically ranging between $60 to $100.
• Might include added traveling costs.
• Scheduling conflicts can arise.
• Limited flexibility in terms of timing and location.
Bottom line: (Who can use it?)
In-person tutoring is ideal for students who thrive on personal interaction, benefit from one-on-one attention, and have the means to invest in this form of education.
Online tutors
Online tutoring has gained popularity in recent years, offering several benefits including flexibility and convenience. The article: Best online tutoring services for math can help you select the
most suitable high school math tutor for yourself. Here’s how it works:
• You find a private online tutor on a reputable online tutoring platform or an independent website.
• Sessions are conducted via video calls, using platforms like Google Classrooms, Zoom, or Skype. If you have a stable internet connection, you can learn from any part of the world.
• You schedule sessions according to your availability and comfort.
• Apart from these, the tutoring quality can be more or less the same as in-person tutors. Tutors assess your progress with personalized one-on-one attention, customized lesson plans, homework
help, test prep, and progress tracking.
• Wiingy is an excellent affordable and private online tutoring platform to learn over 100+ subjects from top verified and qualified tutors. Parents and students have rated Wiingy 4.8/5 on Google
and 4.5/5 on the Trustpilot.
• Tutor.com is an online tutoring service offering assistance in various subjects including high school math.
• Wyzant is an online tutoring platform in many subjects. It connects high school math tutors with students online.
• Suitable for students of all age groups, levels, grades, and learning specialties.
• Access to expert tutors from across the world.
• Learning from the comfort of your home, anytime you want.
• Cost-effectiveness compared to in-person tutoring; typically costs $25 to $60 per hour.
• Potential technical issues or connectivity problems may arise during sessions.
• Lack of physical presence of the tutor.
Check out our Top Benefits of A Private High School Math Tutor to learn the benefits of working with a private math tutor.
Bottom line: (Who can use it?)
Online tutoring suits students looking for flexibility, a wide choice of expert tutors, and the convenience of learning from home while maintaining a sustainable budget.
Learning with you-tube
YouTube has become a treasure trove of educational content, including high-school math tutorials. Here’s how it works:
• Search for math-related topics or concepts on YouTube.
• Watch video tutorials created by educators, enthusiasts, or organizations.
• Pause, rewind, and replay videos as needed to understand the material.
• Practice problems on your own to reinforce learning.
• A vast library of free educational content.
• Flexibility to learn at your own pace. As the videos are saved, you can revisit and refer to them as and when needed.
• Varied teaching styles to suit different preferences.
• Visual and interactive learning experience.
• Lack of personalized guidance or feedback.
• Difficulty in navigating through the depth of the syllabus or topics.
• Potential for misinformation from unreliable sources.
Bottom line: (Who can use it?)
YouTube is a valuable resource for self-motivated students, who enjoy visual learning, and seek supplementary material to enhance their understanding of math concepts.
Learning with video lessons
Apart from YouTube, dedicated educational platforms offer structured video lessons. Here’s how it works:
• Sign up for an educational platform that offers math video lessons, often with a structured curriculum.
• Follow a planned course with lessons, quizzes, and assignments.
• Progress through the learning material at your own pace.
• Some platforms offer certificates upon completion.
• Coursera offers high-quality high-school math courses from universities worldwide.
• edX provides a platform for accessing high school math courses from prestigious institutions.
• Udemy offers a wide range of high school math courses created by expert instructors.
• Structured and organized learning.
• Access to comprehensive course materials.
• Opportunities to interact with instructors or peers through forums.
• Flexibility in terms of scheduling and pace.
• May require a subscription or payment for some courses.
• Limited personalization compared to one-on-one tutoring.
• Self-discipline and motivation are crucial for success.
Bottom line: (Who can use it?)
Video lessons on educational platforms are suitable for self-directed learners who prefer a structured curriculum, are willing to invest in their education, and value flexibility.
Learning with Web tutorials
Web tutorials are text-based resources that offer step-by-step explanations of math concepts. Here’s how it works:
• Search for math tutorials on reputable websites or educational platforms.
• Read through the tutorials, which often include examples and practice problems.
• Work through problems independently to reinforce understanding.
• Some websites offer interactive quizzes and exercises.
• Accessible and free educational content.
• Can be used along with other learning methods.
• Self-paced learning and the ability to revisit content.
• Helpful for reviewing specific topics or concepts.
• Limited interactivity compared to video lessons or tutoring.
• Self-motivation and discipline are necessary to use these resources effectively.
• May not cover all high-school math topics comprehensively.
Bottom line: (Who can use it?)
Web tutorials are ideal for self-learners who prefer text-based explanations, seek supplementary resources, and have the motivation to study independently.
Learning with Khan Academy
Khan Academy is a comprehensive online learning platform dedicated to mathematics and other subjects. Here’s how it works:
• Sign up for a free Khan Academy account.
• Explore the extensive library of math topics and lessons.
• Progress through the material at your own pace.
• Khan Academy offers exercises, quizzes, and progress tracking.
• A structured curriculum designed for self-paced learning.
• Interactive exercises to practice and reinforce concepts.
• Detailed progress tracking and personalized recommendations.
• Completely free of charge.
• Limited personal interaction compared to one-on-one tutoring.
• Some students may prefer video lessons or visual learning.
Bottom line: (Who can use it?)
Khan Academy is an excellent choice for self-motivated learners who appreciate a structured curriculum, interactive exercises, and a cost-free approach to high-school math education.
👍🔍 Recommended Reading 📚📖
Concluding “Where to learn high school math?”
In the digital age, there are numerous avenues for learning high-school math, catering to a variety of learning styles and preferences. From traditional in-person tutors to private online tutoring
like Wiingy and resources like YouTube, video lessons, web tutorials, and platforms like Khan Academy, you can choose the method that suits you best.
Consider your learning style, budget, and availability when making your choice. Ultimately, the key to success in learning high-school math lies in your dedication, practice, and perseverance. | {"url":"https://wiingy.com/resources/math/where-to-learn-high-school-math/","timestamp":"2024-11-14T03:58:25Z","content_type":"text/html","content_length":"178866","record_id":"<urn:uuid:f89bb2d2-14cc-48df-bef3-781b2426705d>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00674.warc.gz"} |
Theory of Combinatorial Algorithms
Mittagsseminar (in cooperation with J. Lengler, A. Steger, and D. Steurer)
Mittagsseminar Talk Information
Date and Time: Tuesday, March 07, 2006, 12:15 pm
Duration: This information is not available in the database
Location: This information is not available in the database
Speaker: Konstantinos Panagiotou
Extremal Subgraphs of Random Graphs
For a graph G, let ET(G) denote the maximum number of edges in a triangle-free subgraph (not necessarily induced) of G, and let EB(G) be the maximum number of edges in a bipartite subgraph of G.
Of course, we always have ET(G) ≥ EB(G), but the general intuition -- guided by various known results -- suggests that, for dense enough graphs, these two parameters will typically be equal.
In 1990, Babai, Simonovits and Spencer studied these parameters for random graphs G(n,p) and confirmed this intuition for dense graphs. In particular, they proved that there is a (small) positive
constant c such that, for p ≥ 1/2 - c, with high probability we have ET(G(n,p)) = EB(G(n,p)).
Babai, Simonovits and Spencer asked whether this result could be extended to cover all constant values of p. In this talk we answer this question affirmatively and show that the above property in
fact holds whenever p=p(n) ≥ n^-c, for some fixed c > 0.
This is joint work with Graham Brightwell and Angelika Steger.
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What I've learned about flow fields so far.
June 15, 2021, updated December 27, 2023.
If you feel that this article is too wordy skip all the text and play with the illustrations, they get more and more fun as the articles rambles on!
-- Me
Brief introduction.
This article will describe the methods and concepts I used to create the series of generated artworks pictured below. I've tried to visualize the algorithms and provide some code samples. The code
samples are written in a Typescript with some simplifications so that the code is readable on mobile and to make them more concise and easy to follow. The general algorithms can easily be ported to
any language though, I have for instance done some implementations using Rust.
As a note, far more talented people than me have written articles on the subject upon which my work is very obviously based. I recommend reading that one too!
Noise functions
The main driving force behind these flow field images is usually a noise function. Without regurgitating the wikipedia article, a noise function is a function that takes a coordinate in 2d space
(higher dimension noise-functions also exist, but are irrelevant for the purpose of this article) and returns a value in the range -1..=1 such as points close together return similar, but slightly
different, values.
The interactive illustration below shows how this works by uniformly sampling points in a grid and calling the noise function for that point. Note that the values have been rounded to one decimal for
legibility, the actual values have far more precision. Click Regenerate to run the noise function again with a new seed to get new noise values.
And here is the code that generated the image above, abbreviated for clarity.
[object Object]
To make it a bit easier to digest, we can visualize this in a more effective manner by translating the noise values into degrees and draw lines from a starting position and a few pixels following the
degree of the noise value.
[object Object]
This doesn't look like much currently. It would appear that lines are fairly disjointed and go off in seemingly random directions. Not at all the smooth effect we're going for.
It turns out that noise functions are pretty sensitive. You'd think that the points(1.0, 1.0) and (1.0, 2.0) would produce somewhat similar noise values, following the rule that points in close
proximity yield fairly similar noise values, but they're not close enough to each other.
To combat this we can force our points to be closer together by just dividing our x and y coordinates by some smoothness constant. For example, the distance between our two example points was
previously 1.0 for the y-axis, and 0.0 for the x-axis, for a total distance of 1.0.
If we divide all our x and y values by the constant smoothness = 100 we'd end up with: p1 = (0.01, 0.01), p2 = (0.01, 0.02), making the distance only 0.01, but keeping the relation between the points
the same.
const smoothness = 20;
const p1 = [1.0, 1.0];
const p2 = [1.0, 2.0];
distance(p1, p2) // returns 1.0;
[p1.x / 20, p1.y / 20],
[p2.x / 20, p2.y / 20],
) // returns 0.0500
The only change we need to make to our code is the following:
[object Object]
You can think of it as shrinking our domain to better fit the noise values. Or zooming in on the noise function if that helps.
As we can see by dragging the smoothness value up, closer to 100 in this case, the lines start to smooth out and patterns start to emerge.
Drawing lines.
At this point we know how to navigate the flow field. Pick any point P, read the noise value n for that point and increment P.x by cos(n) and P.y by sin(n) as well as some extra pixels that
represents the distance we want to travel in the direction of the field.
We did this in the previous examples by sampling points in a grid and and having a fixed line length. To approximate the effect illustrated in the images at the beginning of this article a bit better
another implementation is necessary. First we need to pick a random point on the canvas and then ride the flow field until we get out of bounds.
[object Object]
Going from the above example to drawing an actual line should now be trivial. Start at any given point on the canvas an instantiate your line, move through the field just as before, but instead of
drawing a new point at the given position add it to the points for the line. Stop the loop when the line has reached the end of the canvas and finally draw the line.
[object Object]
Mouse over the illustration below to create new lines from your mouse position.
Experimenting with lines.
So far not a lot of variance has been achieved. All results, no matter what the seed of the noise function is, will yield somewhat similar images. There are a few ways to combat this, one way is by
warping the noise value a bit, resulting in lines exaggerating their turns, this is done by simply multiplying the noise value by some constant before applying the cos() and sin() functions.
[object Object]
Another way is by varying the length of the steps each line takes between each step as it progresses through the field. Shorter steps yield much smoother curves while longer steps will make the lines
a lot more jagged.
[object Object]
Alternatives to noise functions
So far we've been using a noise function called OpenSimplex.
OpenSimplex noise is an n-dimensional gradient noise function that was developed in order to overcome the patent-related issues surrounding Simplex noise, while continuing to also avoid the
visually-significant directional artifacts characteristic of Perlin noise.
- Kurt Spencer
As alluded to in the quote, there are few different noise functions. A proper noise function is a bit complicated and out of scope for this article, but nothing is stopping us from writing a function
that returns similar values for a given coordinate, that part can be done pretty easily.
These home grown functions won't yield as random seeming results as a noise function but they will let us control the final output much more. They will also let us be much more creative in trying new
things out now that we can draw lines that reliably follow a path.
An easy first test we can do is simply taking a coordinate and return it's distance to another point that we'll call the focalPoint, just to see what would happen. This function would satisfy the
rule that points close together yield similar values, making our lines nice and smooth.
[object Object]
Mouse over the illustration below to set a new focal point
If we instead of returning the distance to focalPoint we can return the angle the line has from our point to and offset our point along the radius (with a slight distortion to the y-axis) we can get
a nice swirl-like effect.
[object Object]
[object Object]
Collision Detection
In my opinion, the real fun doesn't really begin until we start looking at having the lines interact with each other. Instead of stopping when we reach the end of the canvas or we've reached the max
line length we can instead stop when this line would collide with another line.
Hover the illustration (or slide your finger over it) to add lines of varying width.
Now, a lot can be said about collision detection and how to make it performant. I'll show only one method and a small optimization to keep the solution somewhat performant.
Checking the intersection of two straight lines isn't too bad and can be done in O(1) time. Our lines aren't straight however so we'll have to be a bit more clever. If we go back to the beginning of
how we drew the line, it started with stepping through the noise field and adding a point for each step. If we made that point into a circle by giving it same radius as the line and kept track of
each circle, not just in the line but in all lines it's a lot easier to check if the circle we are about to draw overlaps any other circle.
Checking if two circles overlap is only a matter of checking if the sum of their radii is smaller than the distance between their origo.
[object Object]
Here is another illustration that highlights when two non-linear lines meet using this method. Play with the illustration to move the colliding line.
[object Object]
Optimizing it slightly.
These example illustrations are fairly small so we haven't ran into any performance issues when checking if our line collides with any other line... yet. When trying to make a larger image however,
in a print-friendly size for instance, we'd end up with a lot of lines with a lot of points that we could potentially collide with, meaning that for every new point we add we must check collisions
against all other points. This stacks up fast and will make your render times a lot longer than desired. A way to mitigate this is to only check against points that are close enough for us to collide
The simplest way of doing that is by dividing our canvas up into a grid of boxes and whenever we are about to add a new point check which box it would go in, and then only check against collisions
with points in that box.
Time for another illustration. This time, move your finger or mouse cursor around to see which points belong to the same box.
Now, with 100 boxes (10 across, 10 down) and if the points are distributed uniformly on the canvas, we end up doing 1/100^th as many checks that we did previously, increasing rendering performance by
quite a bit! One thing to note however, is that if our boxes would be too small to reliably hold points with the radius of our lines then we'd start to get overlapping lines at the edges. This would
also happen if a points origo was at the very edge of a box, causing its body to spill outside the boxes area. We could fix that by checking surrounding boxes for collisions as well, but that would
mean we'd check another eight boxes besides the current one, increasing the search space a bit, but the result would be more exact.
A final optimisation we can do with this technique is that if our step size (the distance between each point in each line) is sufficiently small we could skip checking a few points in the box, since
if our circle overlaps with one of the circles there's a high chance that it overlaps with some other circles as well. Now this might yield a less accurate result, but accuracy is not necessarly the
end goal. Some small overlaps for a few lines might introduce some visually pleasing artifacts. Usually those kinds of details are happy little accidents.
[object Object]
Here we check against every 7^th circle in a box hoping to get a hit if there is one. The constant 7 might be too high in some cases, or could be increased even more, it all depends on the step size
for the lines and can be tweaked to get a good balance between render times and correctness.
Finally, Colors.
The theme for this article is converging to "you can achieve a lot of variation with some tweaks". That's true for applying colors to these images as well. So far things in this article have been
pretty monochrome to focus on the underlying techniques of how to achieve the overarching look.
The easiest way to get some color in there is to create a palette with a few different colors and picking at random when creating a new line.
[object Object]
Another coloring method is by coloring each line by the angle of the noise function where the line started. This will yield a gradient like coloring across larger images.
[object Object]
Finally, my favorite method is by subdividing the canvas into a set of regions, either by some Piet Mondrian Composition style, or recursively splitting the canvas into more and more refined
polygons. After the canvas has been divided I assign each region a color and whenever a line spawns assign it the color of the region it spawned in.
This method creates a nice effect where things don't look as disjointed and more like streams of paint flowing into other buckets of paint.
Even though this article got quite long, it only scratches the surface of all the variants that can be achieved using the fundamental techniques described. I highly recommend trying things out and
experimenting, swapping a cos() for a sin() somewhere, or maybe even a tan() if you're crazy.
Try subdividing the canvas into subregions who all have their own rules, or maybe dive deeper into noise functions. Or have a small border around the canvas and let some small percent of the lines
escape it. Why use lines at all? Why not circles or squares or blobs?
Thanks for sticking in there this long, hope it was helpful.
for (let y = 0; y < 40; y++) { for (let x = 0; x < 40; x++) { const n = noise(x, y); drawText(n.toString(), x, y); } }for (let y = 0; y < maxHeight; y += 20) { for (let x = 0; x < maxWidth; x += 20)
{ const n = noise(x, y); const stepSize = 10; beginPath(); moveTo(x, y); lineTo( x + cos(n) * stepSize, y + sin(n) * stepSize, ); closePath(); stroke(); } }const smoothness = 100; const n = noise(x /
smoothness, y / smoothness);const stepSize = 10; let x = 0; let y = random(0..height); while (bounds.contains(x,y)) { const n = noise(x / smooth, y / smooth); x += cos(n) * stepSize; y += sin(n) *
stepSize; const radius = 5; circle(x, y, radius); fill(); }const stepSize = 10; let x = random(0..width); let y = random(0..height); beginPath(); moveTo(x,y); while (bounds.contains(x,y)) { const n =
noise(x / smooth, y / smooth); x += cos(n) * stepSize; y += sin(n) * stepSize; lineTo(x,y); } closePath(); stroke();const warp = 2.2; const n = noise(x, y); x += cos(n * warp) * stepSize; y += sin(n
* warp) * stepSize;const jaggedStepSize = 100; const n = noise(x, y); x += cos(n) * jaggedStepSize; y += sin(n) * jaggedStepSize;function distanceToCenter( x: number, y: number, width: number,
height: number, ): number { const centerX = width / 2; const centerY = height / 2; const dx = x - centerX; const dy = y - centerY; return sqrt(dx ** 2 + dy ** 2); }function angleBetween( x: number,
y: number, centerX: number, centerY: number, ) { return atan2(y - centerY, x - centerX); }beginPath(); moveTo(x, y); for (let i = 0; i < lineLength; i++) { const dist = distance(x, y, cx, cy) const
angle = angleBetween(x, y, cx, cy); x = cx + cos(angle + 0.01) * (dist - 0.1); y = cy + sin(angle + 0.015) * (dist - 0.15); lineTo(x, y); } stroke();interface Circle { x: number; y: number; r:
number; } type Point = [number, number]; function distance( [x1, y1]: Point, [x2, y2]: Point, ) { return sqrt((x1 - x2) ** 2, (y1 - y2) ** 2); } function overlap( a: Circle, b: Circle, ): boolean {
const d = distance([a.x, a.y], [b.x, b.y]); return d < a.r + b.r; }const previousPoints = []; function drawLine(x: number, y: number) { beginPath(); moveTo(x,y); const lineWidth = 4; const
pointsForLine = []; while (canvas.contains(x,y)) { const n = noise(x / smooth, y / smooth) * warp; x += cos(n) * step; y += sin(n) * step; if (previousPoints.some( point => distance([x,y],point)) <
lineWidth ) { break; } pointsForLine.push([x,y]); lineTo(x,y); } previousPoints.push(...pointsForLine); stroke(); } function overlapsAny( circle: Circle, box: Circle[], ): boolean { for (let i = 0; i
< box.length; i += 7) { const d = distance(box[i], circle); const r = box[i].r + circle.r; if (d < r) { return true; } } return false; }const palette = [ "#eb6f92", "#f6c177", "#ea9a97", "#3e8fb0", "
#9ccfd8", ]; const color = palette[floor(random() * palette.length)]; drawLine(x,y,color);const n = noise(x / 120, y / 120) * warp; const hue = n % 255; const color = `hsl(${hue}deg, 70%, 50%)`; | {"url":"https://damoonrashidi.me/articles/flow-field-methods","timestamp":"2024-11-09T06:32:32Z","content_type":"text/html","content_length":"42628","record_id":"<urn:uuid:cc6ae33c-afe1-4df7-907b-669661cc45ab>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00133.warc.gz"} |
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This Arithmetic course is a refresher of place worth and operations for whole numbers, fractions, decimals, and integers. Learn fourth grade math—arithmetic, measurement, geometry, fractions, and
extra. This course offers participants a primary understanding of statistics as they apply in business conditions. A justifiable share of students considering MBA packages come from backgrounds that
don’t embody a large amount of training in arithmetic and statistics. Often, students find themselves at a disadvantage when they apply for or enroll in MBA packages. You will find out how a sequence
of discoveries has enabled historians to decipher stone tablets and study the assorted strategies the Babylonians used for problem-solving and teaching.
Created by consultants, Khan Academy’s library of trusted follow and classes covers math, science, and extra. Learn multivariable calculus—derivatives and integrals of multivariable functions,
software problems, and extra. Learn early elementary math—counting, shapes, basic addition and subtraction, and extra. Eight years after the class ended, I don’t bear in mind much of the precalculus
, however I do know I was not in any means harmed by sharing a category with my twin. For occasion, on a day either earlier than break or at the end of 1 / 4, our class was taking half in
Scattergories. We were given the letter “S” and had been informed to give a reason why someone would be late for college.
I commenti sono chiusi. | {"url":"https://www.ventiemari.it/senza-categoria/critical-items-of-math-websites-for-kids","timestamp":"2024-11-11T23:53:12Z","content_type":"text/html","content_length":"22253","record_id":"<urn:uuid:42ac506f-fcaa-431d-969a-36a191e3bf48>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00199.warc.gz"} |
Glossary of Common Data Terms | Alamo Regional Data Alliance
This Glossary of Common Data Terms was developed locally as a non-technical resource for those interested in expanding their functional data vocabulary. This glossary contains commonly used data
terms defined in easy-to-understand language. Although the definitions are informal and non-academic, the following academic texts heavily informed their development:
Shryock, H.S., and Siegel, J.S. The Methods and Materials of Demography. San Diego, CA: Academic Press, 1976.
Haupt, A. and Kane, T.T. Population Handbook. Washington, DC: Population Reference Bureau, Inc., 1978.
Click here for a printable version of the Glossary of Common Data Terms.
There are currently 5 names in this directory beginning with the letter P.
calculated probability that that what is being observed in the data has happened by chance. Generally, if the p-value associated with an observation is less than .05 the observation is accepted as
statistically significant. A p-value less than .05 indicates a less than 5% chance that what is being observed happened by chance or a more than 95% certainty that chance alone cannot explain the
observation. See Statistical significance.
Percent increase/decrease
one way of describing the difference between your current measurement and a past measurement, relating it to the past measurement. The percent change is the difference between the two values, divided
by the past value, and it’s usually phrased like “percent decrease from prior year” or “percent increase over prior year.” For example, if the percent of the population that smokes cigarettes
decreased from 19% in 2014 to 17% in 2015, you’d have a 10.5% (percent) decrease, because the difference between 19 and 17 is two, and two divided by 19 is 10.5%.
Percentage point increase/decrease
one way of describing the difference between your current measurement and a past measurement, without relating the change to the past measurement. It’s just the difference between the two values, and
it’s usually phrased as “decrease of X percentage points.” If the percent of the population that smokes cigarettes decreased from 19% in 2014 to 17% in 2015, you’d have a two percentage point
decrease, because the difference between 19 and 17 is two. | {"url":"https://alamodata.org/resources/glossary/?dir=2&name_directory_startswith=P","timestamp":"2024-11-12T12:52:07Z","content_type":"text/html","content_length":"65488","record_id":"<urn:uuid:cfaafcc0-8166-423f-9f3f-5b45230e5d2b>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00813.warc.gz"} |
NCERT Maths Book Class 3 in Hindi
NCERT Maths Book Class 3 in Hindi: The National Council of Educational Research and Training (NCERT) publishes Maths textbooks for Class 3. The NCERT Class 3rd Maths textbooks are well known for it’s
updated and thoroughly revised syllabus. The NCERT Maths Books are based on the latest exam pattern and CBSE syllabus. NCERT has a good image when it comes to publishing the study materials for the
students. NCERT keeps on updating the Maths books with the help of the latest question papers of each year. The Class 3 Maths books of NCERT are very well known for its presentation. The use of Class
3 NCERT books is not only suitable for studying the regular syllabus of various boards but it can also be useful for the candidates appearing for various competitive exams, Engineering Entrance Exams
and Olympiads.
NCERT Maths Book Class 3 in Hindi PDF Download
NCERT Class 3 Maths Books are provided in PDF form so that students can access it at anytime anywhere. Class 3 NCERT Maths Books are created by the best professors who are experts in Maths and have
good knowledge in the subject.
Chapter wise CBSE NCERT Books for Class 3rd Maths in Hindi
गणित का जादू
NCERT Maths Book for Class 3 PDF in Hindi Medium
The NCERT syllabus mainly focuses on this book to make it student-friendly to make it useful for both the students and the competitive exam aspirants. The book covers a detailed Maths based on the
syllabuses of various boards. NCERT Maths Books for Class 3 is perfectly compatible with almost every Indian education state and central boards.
We hope that this detailed article on NCERT Class 3 Maths Books helps you in your preparation and you crack the Class 3 exams or competitive exams with excellent scores. For your convenience, you can
download PDFs of Class 3 NCERT books and structure your study plan ahead. You should focus more on practicing maths previous year question papers too as this will further assist you in understanding
the frequency of questions. | {"url":"https://www.ncertbooks.guru/ncert-maths-book-class-3-in-hindi/","timestamp":"2024-11-08T03:10:11Z","content_type":"text/html","content_length":"79204","record_id":"<urn:uuid:1893c922-a6e9-40a1-b165-c14592dae0b1>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00302.warc.gz"} |
Perpendicular Bisectors and Circumcircle - Math Angel
🎬 Video Tutorial
• Perpendicular Bisector: A perpendicular bisector intersects a line segment at a right angle and divides it into two equal parts, ensuring each point on the bisector is equidistant from both
segment endpoints.
• Constructing the Perpendicular Bisector: 1) Use a ruler to locate the midpoint, then draw a perpendicular line through it. 2) Use a compass to create arcs from each endpoint, then draw a line
through the intersection points.
• Finding the Circumcentre in a Triangle: Construct perpendicular bisectors for at least two sides of the triangle; their intersection point is the circumcentre, which is equidistant from all three
Membership Required
You must be a member of Math Angel Plus or Math Angel Unlimited to watch this video.
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Classifying clothes images with neural networks | Lorena Ciutacu
Project completed in week 9 (23.11.-27.11.20) of the Data Science Bootcamp at Spiced Academy in Berlin.
This week we dived into Deep Learning and learned about different types neural networks (NN) and their applications in various domains. The main goal of this project was to learn and understand what
each hyperparameter in a NN model does and how to tune it, so this week was more theoretical and math-heavy than usual.
Building a Neural Network
For my first deep learning project, I used the famous Fashion MNIST dataset created by Zalando. The dataset contains 60K images of 10 clothing items (e.g., tops, sandals, trousers). In order to
classify the images in the correct item category, I tried two types of NN:
• Artificial Neural Network (ANN): represents a group of multiple perceptrons/ neurons at each layer. ANN is also called Feed-Forward Neural Network, because the inputs are processed only forward.
An ANN consists of three layers: input, hidden, and output.
• Convolutional Neural Network (CNN): uses filters to extract features and capture the spatial information from images. CNNs are the go-to model for image classification.
In this post, I will present only the CNN model, since it's the one that performed best in my project. Here's an overview of my model:
model = keras.models.Sequential()
model.add(Conv2D(32, kernel_size=(3, 3), activation='relu', kernel_initializer='he_uniform', input_shape=(28, 28, 1)))
model.add(MaxPooling2D((2, 2)))
model.add(Dense(100, activation='relu', kernel_initializer='he_uniform'))
model.add(Dense(10, activation='softmax'))
First [line 1] I instantiated the model. Then I started adding several layers of with different hyperparameters.
• Conv2D is a 2D convolution layer which creates a convolution kernel that is filled with layers input to produce a tensor of outputs. I used 32 filters, as it's recommended to use powers of 2 as
• kernel_size determines the height and width of the kernel, passed as a tuple.
• activation specifies the activation function, which transforms the summed weighted input from the node into the activation of the node. relu (Rectified Linear Activation Function) outputs the
input directly if it is positive and 0 if it is negative.
• kernel_initializer refers to the functions for initializing the weights, which in this case is uniform distribution.
• input_shape represents the dimension of the images (28×28 px) and their color code (1 for black-and-white). This hyperparameter needs to be specified only in the first layer.
Next [3] I added a MaxPooling2D layer, which downsamples the input representation by taking the maximum value over the window defined by pool_size (2, 2) for each dimension along the features axis.
Then [4] I added a Flatten layer that flattens the images, so that the pixel values are between 0 an 1. This is done because when working with images, if the values are positive and large, a ReLU
neuron becomes almost a linear unit, losing many of its advantages.
Lastly [5,6] I added two Dense layers, which are fully connected layers, where the first parameter declares the number of desired units. So in [5] I have a layer with has 100 neurons with ReLU
activation. The last layer [6] has 10 hidden layers (number of clothing items) and softmax activation, which is used for multi-class classification.
Tuning the hyperparameters
Finally, I compiled the model:
model.compile(optimizer='adam',loss='categorical_crossentropy', metrics=['accuracy'])
• optimizer defines the stochastic gradient descent algorithm that is used. I've tried both sgd (Stochastic Gradient Descent) and adam (Adaptive Moment Estimation), and stuck with the latter
because it is more advanced it it generally performs better.
• loss defines the cost function.
• metrics is a list of all the evaluation scores I want to compute. In this case, accuracy is enough.
validation_data=(xtest, to_categorical(ytest))
• epochs represents the number of iterations on the training data.
• batch_size is the number of images to feed tot he model in one go, it normally ranges from 16 to 512, but in any case it's smaller than the total number of samples.
• validation_data represents the part of the dataset kept for testing the model.
• to_categorical one-hot-encodes the labels (clothing items).
Evaluating the model performance
The CNN had an accuracy of 99.43% on the train set and 90.69% on the test set. This is a really good score! I think it could've been even better if I had let the model train longer (i.e. more
Friday Lightning Talk
This Friday talk was a bit different from the previous ones. Instead of presenting our projects, we had to read and present a paper about a Deep Learning application, like generative art, object
recognition, or text generation. Of course, I chose the latter topic and tried LSTM to generate poems by E.A. Poe. But I talked about GPT-3, a state-of-the-art deep learning model that can generate
samples of news articles which human evaluators have difficulty distinguishing from articles written by humans. However, I focused not so much on the technical details, as on the ethics and
implications of this technology. This opened an important discussion in our group, which I think should be included in the curriculum of every tech degree. | {"url":"https://aloci.me/blog/bootcamp9-fashion-mnist/","timestamp":"2024-11-06T21:42:10Z","content_type":"text/html","content_length":"40309","record_id":"<urn:uuid:c31a0fcf-c07b-4559-947f-970385e8d0b4>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00008.warc.gz"} |
Publications Oscar Rivero
top of page
Anchor 1
"A work is never finished, but rather the limit of one's own possibilities is reached". Antonio López (Spanish painter).
12. Eisenstein congruences among Euler systems (with V. Rotger), to appear in Can. Math. Bull.
The following papers begin the study of the construction of certain p-adic L-functions for GSp4xGL2 in the framework of higher Coleman theory.
11. On p-adic L-functions for GSp4xGL2 (with D. Loeffler; submitted, May 2023).
10. Algebraicity of L-values for GSp4xGL2 and GSp4xGL2xGL2 (with D. Loeffler), to appear in Quart. J. Math.
In paper 9, we study connections between different Euler systems by studying the Galois representations attached to critical Eisenstein series.
9. Eisenstein degeneration of Euler systems (with D. Loeffler), to appear in J. Reine Angew. Math. (Crelle).
In paper 7, we obtain partial results towards the Bloch--Kato conjecture and the anticyclotomic IMC in the setting of diagonal cycles. Paper 8 is a natural continuation of that work, where we explore
the setting of the symmetric square, extending results of Dasgupta and Loeffler--Zerbes to the anticyclotomic setting.
8. An anticyclotomic Euler system for adjoint modular Galois representations (with R. Alonso and F. Castella), to appear in Ann. Inst. Fourier (Grenoble).
7. Iwasawa theory for GL2xGL2 and diagonal cycles (with R. Alonso and F. Castella), to appear in J. Inst. Math. Jussieu.
Paper 6 is the first work in our investigations of an Artin formalism for Euler systems. This is continued in Paper 12, where we compare this approach with that followed in the work with D. Loeffler
(Paper 9).
6. Motivic congruences and Sharifi's conjecture (with V. Rotger), to appear in Amer. J. Math.
These 5 papers study the interplay between Beilinson--Flach classes, Hida--Rankin p-adic L-functions and Gross--Stark units, with an emphasis on the exceptional zero situation. More precisely, papers
1, 3 and 5 can be read as a trilogy. Articles 2 and 4 look at rather related instances of this philosophy, where interesting phenomena arise (the third article focuses on the case of elliptic units,
while the fourth article deals with the setting of diagonal cycles).
5. Derivatives of Beilinson--Flach classes, Gross--Stark formulas and a p-adic Harris--Venkatesh conjecture, to appear in Documenta Math.
4. Generalized Kato classes and exceptional zeros, Indiana Univ. Math. J. 71 (2022), no. 2, 649--684.
3. Derived Beilinson--Flach elements and the arithmetic of the adjoint of a modular form (with V. Rotger), J. Eur. Math. Soc. 23 (2021), no. 7, 2299--2335.
2. The exceptional zero phenomenon for elliptic units, Rev. Mat. Iberoam. 37 (2021), no. 4, 1333--1364.
1. Beilinson--Flach elements, Stark units, and p-adic iterated integrals (with V. Rotger), Forum Math. 31 (2019), no. 6, 1517--1532.
A list of my coauthors with links to their webpages:
My PhD thesis (February 2021), Arithmetic applications of the Euler systems of Beilinson--Flach elements and diagonal cycles, done under the supervision of Victor Rotger.
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Limits & Fits
The limits and fits of geometric dimensioning and Tolerancing can be either of hole basis or of shaft basis. This GD&T tutorial will draw a comparison between the hole basis limits and fits and the
shaft basis.
Shaft and Hole Fits Basics
While dealing with shaft and hole, you will come across the three basic types of fits: clearance fit, transition fit, and interference fit or something in between these (like loose running fit,
sliding fit etc.). The shaft and the hole tolerance grades are represented to the right:
Shaft Basis System
• Take the maximum size of the shaft as datum for the fit. That means, you have to take hole basic deviation class as h. (How? Look at the above picture. The minimum tolerance of the h grade is
coinciding with the basic size).
• Now, decide the tolerance grade (or IT number, the number followed by the basic deviation, something like 6, 7 or 8) for the shaft depending upon the machining process.
• You should know the allowance (the minimum clearance or interference between the shaft and the hole) from the application of the shaft and hole. Add or subtract (for clearance or interference
fits respectively) the allowance value from the shaft diameter to get the minimum hole diameter.
• Apply the shaft tolerance grade.
Hole Basis System
• Take the minimum size of the hole as datum for the fit. That means, you have to take hole basic deviation class as H. (How? Look at the above picture. The minimum tolerance of the H grade is
coinciding with the basic size.)
• Now, decide the tolerance grade (or IT number, the number followed by the basic deviation, something like 6, 7 or 8) for the hole depending upon the machining process.
• You should know the allowance (the minimum clearance or interference between the shaft and the hole) from the application of the shaft and hole. Add or subtract (for interference and clearance
fits respectively) the allowance value from the hole diameter to get the shaft maximum diameter.
• Apply the shaft tolerance grade.
Where to Use What
• For a standard manufacturing process where hole is manufactured by drilling, reaming, etc. and the shaft by turning, etc., go for the hole base system, because altering the hole diameter by a
small amount is not possible for such cases, and on the other, shaft diameter can be varied.
• In the case where the shaft is used as a common mating part for a large number of holes or shaft diameter is fixed and hole diameter can be varied then it is better to go for shaft basis
GD&T limits and fits can be applied either by the shaft basis tolerance system or by hole basis the hole basis tolerance system. By and large, the hole basis system is most used in industry, unless
there are multiple parts with holes are fitted with a single shaft. | {"url":"https://engineeredbydesign.co.uk/Pages/Eng_Data/Limits&Fits.php","timestamp":"2024-11-10T05:15:30Z","content_type":"application/xhtml+xml","content_length":"32744","record_id":"<urn:uuid:6ea81aa6-ea37-4e3c-869b-d06e88b8f386>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00679.warc.gz"} |
How to Calculate Average Fixed Cost.
What is Average Fixed Cost?
Ahead of discussing how to calculate average fixed cost(AFC), let us define what a fixed cost is. Fixed costs are those costs that must be incurred in fixed quantity regardless of the level of output
Average fixed cost (AFC) is the fixed cost per unit of output. It is used to calculate the total cost that should be allocated to each unit produced.
Average fixed cost decreases with additional production. Basically this means that since fixed cost is does not change with the quantity of output, a given cost is spread more thinly per unit as
quantity increases.
Examples of average fixed cost are the salaries of permanent employees, the mortgage payment on machinery and plant, rent, etc.
Additionally, average fixed cost is relevant only in the short-run since all inputs are variable in the long-run.
Importance of Calculating Average Fixed Cost.
• AFC is used by companies to analyze their expenses hence finding ways to reduce them and make their business bring more revenue.
• Also AFC is used to measure the breakeven point.
• Average fixed cost helps you define the efficiency of production and the economies of scale.
• With the help of AFC, you can find out the number of funds you should allocate to produce one unit. Le
Formula to Calculate AFC.
Division Method.
This is where we divide the total fixed cost by the number of products manufactured by your company during this specific period.
Output is the quantity of goods and services produced in a given time period. The level of output is determined by both the total supply and total demand within an economy.
Example 1:
The fixed cost of manufacturing 10 car batteries is $ 150. Calculate the average fixed cost of the production.
Therefore, the average fixed cost of the production is $ 15.
Method 2:
Subtraction method.
• Calculate the total cost
• Determine the average total cost.
• Calculate the average variable cost.
• Then subtract the average variable cost from the average total cost to get the average fixed cost.
Example 2:
Suppose your average total cost is $7, calculate the AFC if you’ve been given the average variable cost to be $5.50.
Thus, the AFC IS $1.50. | {"url":"https://www.learntocalculate.com/calculate-average-fixed-cost/","timestamp":"2024-11-05T02:34:17Z","content_type":"text/html","content_length":"59411","record_id":"<urn:uuid:ab24dd4c-ac57-4324-aef5-1dea45493f9f>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00084.warc.gz"} |
How To Draw A Torus
How To Draw A Torus - Now make that ellipse smile ! For a beginning artist, mastering drawing boxes and cylinders is essential. And like magic, we have drawn a torus! Basically, draw a circle and a
smaller circle inside that circle. Start by drawing an ellipse:
For a torus it looks like this, and the reason this is the fundamental reason can been seen here. Web how to draw a torus with artist janette oakman sacred geometry symmetrical geometric symbol.
Using tikz we can plot any parametric equation. Web 47 share 2.7k views 10 months ago sacred geometry the torus yantra is a beautiful pattern in sacred geometry that is drawn by repeating circles. 3
define the amount of steps object > blend > blend options > specified steps. 1, r is the outer radius of the torus, r is the tube. Web drawing a torus bart snapp learn how to draw a torus.
How To Draw A Torus with Artist Oakman Sacred Geometry
\[\begin{aligned} x &= (r + r \cos{v}) \cos{u} \\ y &= (r + r \cos{v}) \sin{u} \\ z &= r \sin{v} \\ u, v &\in [0, 2\pi] \end{aligned} \tag{1}\] in eq. Web 47 share 2.7k views 10 months ago sacred
geometry the torus yantra is a beautiful pattern in sacred geometry that is drawn by.
How To Draw a Sacred Geometry Torus Mandala Follow Along Tutorial
With higher genus surfaces the ideas and pictures are similar. 3 define the amount of steps object > blend > blend options > specified steps. 2 cmd+b to apply blend (object>blend) from red to yellow
first and from blue to lightblue after. Web how to draw a torus in latex using tikz parametric equation of.
How to draw a Torus (Vortex) step by step tutorial (english) YouTube
1, r is the outer radius of the torus, r is the tube. A torus centered at the origin with its axis aligned along +z can be described by the following set of parametric equations: Web the facility set
a new record on 30 july when its beams delivered the same amount of energy to.
[Solved] How to draw a torus 9to5Science
Web draw a full torus. 1, r is the outer radius of the torus, r is the tube. Web ask question asked 10 years ago modified 10 years ago viewed 8k times 0 i need to draw a torus which is very thin. A
common shape studied in mathematics is a torus or donut. Now.
How to Draw a Torus in LaTeX using TikZ TikZBlog
With higher genus surfaces the ideas and pictures are similar. Where r is the external radius and r is the inner radius. Here's one way to draw the complete graph on 5 vertices without crossing on
the torus (it can't be done in the plane). Web how to draw a torus with artist janette oakman.
Sacred Geometry How to Draw Torus Sacred geometry, Geometry
With higher genus surfaces the ideas and pictures are similar. For a torus it looks like this, and the reason this is the fundamental reason can been seen here. A common shape studied in mathematics
is a torus or donut. Now make that ellipse smile ! Here's one way to draw the complete graph on.
How to draw 3d torus in 3d's max YouTube
Using tikz we can plot any parametric equation. What i found by searching internet is just formal description of the function glutsolidtorus () but problem is probably i am not getting what they are
saying about the parameter. And like magic, we have drawn a torus! However, the resulting figure is to be printed (in..
How to draw torus and shade shadow easy way for new stater / drawing
Web actually, for a simple picture like this one you could do it by hand in tikz: Finally, add in an upside down arc: Web 1 link theme copy %%create r and theta data theta = linspace (0,2*pi,50) ; 2
cmd+b to apply blend (object>blend) from red to yellow first and from blue to lightblue.
Drawing Torus. Sacred Geometry. YouTube
In sacred geometry it is the first shape to emerge from the genesis. 3 define the amount of steps object > blend > blend options > specified steps. A torus centered at the origin with its axis
aligned along +z can be described by the following set of parametric equations: I don't know if there's.
GeekOut Blog How to Draw 3D Ring (Torus) With Vector Graphics (Tutorial)
Web in this #training #video we are going to #learn #how to #draw a #simple #satisfying #3d #torus #geometric #pattern in a #circle #shape with #dr. Finally, add in an upside down arc: Where r is the
external radius and r is the inner radius. Most of them are based on drawing ellipses and eyeballing.
How To Draw A Torus Web actually, for a simple picture like this one you could do it by hand in tikz: Cover the surface, and voila! Web ask question asked 10 years ago modified 10 years ago viewed 8k
times 0 i need to draw a torus which is very thin. Web how to draw a torus in latex using tikz parametric equation of a torus. First, here are the vertices.
Web The Torus (Vortex) Is A Powerful Geometric Symbol.
3 define the amount of steps object > blend > blend options > specified steps. A common shape studied in mathematics is a torus or donut. Use tikz to draw on top of the picture, adjust the parameters
until it looks right, then remove the background. Web 1 1 view 1 minute ago #sacredgeometryart #torus #sacredgeometry hello everyone!
And Like Magic, We Have Drawn A Torus!
From the above code, we used the package tikz. Ellipses are challenging to draw by hand, so this method is easiest if you are using digital tools. Here's one way to draw the complete graph on 5
vertices without crossing on the torus (it can't be done in the plane). First, here are the vertices.
I'll Do It Step By Step.
Start by drawing an ellipse: A torus centered at the origin with its axis aligned along +z can be described by the following set of parametric equations: Web ask question asked 10 years ago modified
10 years ago viewed 8k times 0 i need to draw a torus which is very thin. Web the facility set a new record on 30 july when its beams delivered the same amount of energy to the target — 2.05
megajoules — but, this time, the implosion generated 3.88 megajoules of fusion.
However, The Resulting Figure Is To Be Printed (In.
Web the idea with drawing a graph on a torus (or any genus $g$ surface) is try to draw it in the fundamental polygon. Web 47 share 2.7k views 10 months ago sacred geometry the torus yantra is a
beautiful pattern in sacred geometry that is drawn by repeating circles. Web i would like to draw a torus, like the one on the question how to draw a torus, with a 3d package.i would like to avoid
drawing the torus in 2d with inkscape (or directly with tikz) because i will need to add other elements (like, say, in 3d helix torus with hidden lines, and view them from different points of view.
1, r is the outer radius of the torus, r is the tube.
How To Draw A Torus Related Post : | {"url":"https://sandbox.independent.com/view/how-to-draw-a-torus.html","timestamp":"2024-11-06T02:32:45Z","content_type":"application/xhtml+xml","content_length":"23915","record_id":"<urn:uuid:e0b1bc41-d960-40d5-a8e1-af67e794aeda>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00761.warc.gz"} |
How do I find the size of an array in Excel VBA?
VBA Code to get the length of Array (one-dimensional array):
Press Alt+F8 to popup macro window. Select ” oneDimArrayLength” and Click Run button. Compute Number of Rows, Number of Columns using UBound and LBound function. Multiply by noOfRow and noOfCol
variable to get Number of elements in multi-dimensional array.
How do you find the length of an array in Visual Basic?
You can find the size of an array by using the Array. Length property. You can find the length of each dimension of a multidimensional array by using the Array. GetLength method.
How do I count the number of items in an array in VBA?
In VBA, to get the length of an array means to count the number of elements you have in that array. For this, you need to know the lowest element and the highest element. So, to get this you can use
the UBOUND and LBOUND functions that return the upper bound and lower bound, respectively.
How big can a VBA array be?
VBA arrays
A VBA array can have a maximum of 60 dimensions.
What is UBound in VBA?
The UBound function is used with the LBound function to determine the size of an array. Use the LBound function to find the lower limit of an array dimension. UBound returns the following values for
an array with these dimensions: Statement. Return Value.
How do I count an array in Excel?
Use the COUNT function to get the number of entries in a number field that is in a range or array of numbers. For example, you can enter the following formula to count the numbers in the range
A1:A20: =COUNT(A1:A20). In this example, if five of the cells in the range contain numbers, the result is 5.
Does VBA array start at 0 or 1?
By default, an array is indexed beginning with zero, so the upper bound of the array is 364 rather than 365. To set the value of an individual element, you specify the element’s index.
How do I count the number of elements in an array in Excel?
How big can an array be in Excel?
The limit is less than the number of rows by the number of columns in the worksheet which is 1048576 rows by 16384 columns (Excel specifications and limits).
How do I use an array in Excel VBA?
Use the Option Base statement at the top of a module to change the default index of the first element from 0 to 1. In the following example, the Option Base statement changes the index for the first
element, and the Dim statement declares the array variable with 365 elements.
What is UBound and LBound in VBA?
The LBound function is used with the UBound function to determine the size of an array. Use the UBound function to find the upper limit of an array dimension.
How do I ReDim an array in VBA?
Examples to use VBA Redim Statement
Step 1: Create a macro name first. Step 2: Declare an array name as a string. Step 3: Now, use “Redim” and assign the array size. Step 4: The array name “MyArray” can hold up to 3 values here.
Can a Countif () function count array?
Sum, SumProduct, Frequency, Linest, lookup functions, etc. take both range and array objects. To use countif, you have to use range in cells, defining the array in the formula on the go will not
Can Countif take an array?
1) COUNTIF(A2:A15, {“Jack”, “Jill”}): using an array criteria for a range, COUNTIF returns an array of values – {2,2} – the numbers indicate the respective occurrence(s) of each of the 2 values
(“Jack” & “Jill”) in column A, and using SUM returns the total number of times each of them occur in column A.
Why are arrays 0 indexed?
The first element of the array is exactly contained in the memory location that array refers (0 elements away), so it should be denoted as array[0] . Most programming languages have been designed
this way, so indexing from 0 is pretty much inherent to the language.
What is an Excel array formula?
An array formula is a formula that can perform multiple calculations on one or more items in an array. You can think of an array as a row or column of values, or a combination of rows and columns of
values. Array formulas can return either multiple results, or a single result.
How do I find an array formula in Excel?
When you select such a cell(s), you can see the braces in the formula bar, which gives you a clue that an array formula is in there. Manually typing the braces around a formula won’t work. You must
press the Ctrl+Shift+Enter shortcut to complete an array formula.
Why is array formula not working?
Therefore, if you are using an older than Excel 2019 version, then you must press CTRL+SHIFT+ENTER to use the array formula. If you just press ENTER, then your array formula will not work except for
some functions such as the AGGREGATE and SUMPRODUCT.
How do you specify an array in VBA?
Use a Static, Dim, Private, or Public statement to declare an array, leaving the parentheses empty, as shown in the following example. Use the ReDim statement to declare an array implicitly within a
procedure. Be careful not to misspell the name of the array when you use the ReDim statement.
What is array function in VBA?
In VBA, we can use arrays to define the group of objects together. There are nine array functions in VBA: ARRAY, ERASE, FILTER, ISARRAY, JOIN, LBOUND, REDIM, SPLIT, and UBOUND. All of these are
built-in functions for the array in VBA. Array function gives us the value for the given argument.
What does UBound do in VBA?
UBOUND, also known as Upper Bound, is a function in VBA with its opposite function, LBOUND or Lower Bound function. This function defines the length of an array in a code. As the name suggests,
UBOUND is used to define the upper limit of the array.
What does UBound return in VBA?
The VBA UBound function returns the maximum subscript, or index, of a particular dimension of an array. In most cases, this essentially yields the size of your array. Whenever you make an array in
VBA, you undoubtedly want to eventually loop through each item in that array.
What is ReDim in VBA?
The ReDim statement is used to size or resize a dynamic array that has already been formally declared by using a Private, Public, or Dim statement with empty parentheses (without dimension
subscripts). Use the ReDim statement repeatedly to change the number of elements and dimensions in an array.
How do I use ReDim in VBA?
Does an array start at 0 or 1?
In computer science, array indices usually start at 0 in modern programming languages, so computer programmers might use zeroth in situations where others might use first, and so forth. | {"url":"https://www.trentonsocial.com/how-do-i-find-the-size-of-an-array-in-excel-vba/","timestamp":"2024-11-11T03:46:36Z","content_type":"text/html","content_length":"64658","record_id":"<urn:uuid:a23d362c-40db-4e04-94a3-7b37a7b65e3e>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00459.warc.gz"} |
h-infinity feedback control problem
We consider the sensitivity of semidefinite programs (SDPs) under perturbations. It is well known that the optimal value changes continuously under perturbations on the right hand side in the case
where the Slater condition holds in the primal problems. In this manuscript, we observe by investigating a concrete SDP that the optimal value can be … Read more | {"url":"https://optimization-online.org/tag/h-infinity-feedback-control-problem/","timestamp":"2024-11-12T21:51:59Z","content_type":"text/html","content_length":"83095","record_id":"<urn:uuid:66311fed-ba34-442d-8e98-257d8b585f2e>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00287.warc.gz"} |
WhoMadeWhat – Learn Something New Every Day and Stay Smart
How do I find diameter if I know circumference?
Divide the circumference by pi, approximately 3.14, to calculate the diameter of the circle. For example, if the circumference equals 56.52 inches, divide 56.52 by 3.14 to get a diameter of 18
inches. Multiply the radius by 2 to find the diameter.
The Diameter goes straight across the circle, through the center. The Circumference is the distance once around the circle.
Subsequently, Is circumference equal to diameter?
The circumference of a circle is equal to pi times the diameter. The diameter is two times the radius, so the equation for the circumference of a circle using the radius is two times pi times the
radius. … In a true circle, the ratio of the circumference to the diameter of the circle will always be the same value.
Also, How do you go from circumference to diameter?
Divide the circumference by pi, approximately 3.14, to calculate the diameter of the circle. For example, if the circumference equals 56.52 inches, divide 56.52 by 3.14 to get a diameter of 18
inches. Multiply the radius by 2 to find the diameter.
Is circumference the same as diameter?
The Diameter goes straight across the circle, through the center. The Circumference is the distance once around the circle.
Last Review : 5 days ago.
How do you find the radius if you only know the circumference?
How do you find a diameter?
Divide the circumference by pi, approximately 3.14, to calculate the diameter of the circle. For example, if the circumference equals 56.52 inches, divide 56.52 by 3.14 to get a diameter of 18
inches. Multiply the radius by 2 to find the diameter.
How do you measure diameter with a ruler?
Measure just the radius of the circle if it is very large. The radius is the distance from the center to any point on the circle. Multiply the radius by two to produce a measurement for the diameter.
Is diameter half of circumference?
The distance around a rectangle or a square is as you might remember called the perimeter. The distance around a circle on the other hand is called the circumference (c). Half of the diameter, or the
distance from the midpoint to the circle border, is called the radius of the circle (r).
Is radius half of circumference?
The distance around a rectangle or a square is as you might remember called the perimeter. The distance around a circle on the other hand is called the circumference (c). Half of the diameter, or the
distance from the midpoint to the circle border, is called the radius of the circle (r).
How do you calculate the diameter of a circle?
How do I figure out diameter?
Calculating Diameter From the Radius The radius is the length from the center of a circle to the edge. Therefore, if you know the radius, multiply it by two to determine the diameter (diameter = 2 x
Is diameter bigger than circumference?
The diameter is ALWAYS approximately 3 times smaller than the circumference! Or to put it another way, the circumference is approximately 3 times bigger than the diameter.
Is the circumference 3 times the diameter?
The circumference is about 3 times the diameter of the circle. … The ratio of circumference to diameter (C ÷ d) is always Ï€. This means that the circumference (C) is always about 3.14 (Ï€) times
the diameter (d). The formula below allows you to easily calculate the circumference of a circle when you know its diameter.
How do we find the radius of a circle?
radius is always half the length of its diameter. For example, if the diameter is 4 cm, the radius equals 4 cm ÷ 2 = 2 cm.
What is the diameter of a 24 inch circle?
75.40 inches
How do you find diameter from circumference?
Divide the circumference by pi, approximately 3.14, to calculate the diameter of the circle. For example, if the circumference equals 56.52 inches, divide 56.52 by 3.14 to get a diameter of 18
inches. Multiply the radius by 2 to find the diameter.
What is half of the circumference?
Since a semicircle is one half of a circle, the circumference of a semicircle is half the circumference of a circle. The formula for the circumference of a semicircle (​SC​) is the formula for
the circumference of a circle multiplied by one half, or 0.5.
How do you find the diameter in inches?
Multiply the radius by 2 to find the diameter. For example, if you have a radius of 47 inches, multiply 47 by 2 to get a diameter of 94 inches. Divide the radius by 0.5 to calculate the diameter.
What is the diameter of a 25 inch circle?
78.540 inches
——– —————
6′ 6.54″ feet and inches
1.9949 meters
199.49 centimeters
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Spread the word ! Don’t forget to share. | {"url":"https://whomadewhat.org/how-do-i-find-diameter-if-i-know-circumference/","timestamp":"2024-11-08T18:09:43Z","content_type":"text/html","content_length":"49190","record_id":"<urn:uuid:012ec031-ea09-41dd-98e0-723986af57e8>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00341.warc.gz"} |
Extended Bernoulli Concepts and Equations
Fluid Flow Table of Contents
Hydraulic and Pneumatic Knowledge
Extended Bernoulli Concepts and Equations
The Bernoulli equation can be modified to take into account gains and losses of head. The resulting equation, referred to as the Extended Bernoulli equation, is very useful in solving most fluid flow
problems. In fact, the Extended Bernoulli equation is probably used more than any other fluid flow equation. Equation below is one form of the Extended Bernoulli equation.
z[1] = V [1]^2 / ( 2 g ) + P[1] ν[1] g[c] / g + H[p] - z[2] +V[2]^2 / ( 2 g ) + P[2] v[2] g[c] / g + H[f]
z = height above reference level (ft)
V = average velocity of fluid (ft/sec)
P = pressure of fluid (lbf/ft^2)
v = specific volume of fluid (ft^3/lbm)
H[f] = head loss due to fluid friction (ft)
H[p] = head added by pump (ft)
g = acceleration due to gravity (ft/sec^2)
The head loss due to fluid friction (H[f] ) represents the energy used in overcoming friction caused by the walls of the pipe. Although it represents a loss of energy from the standpoint of fluid
flow, it does not normally represent a significant loss of total energy of the fluid. It also does not violate the law of conservation of energy since the head loss due to friction results in an
equivalent increase in the internal energy (u) of the fluid. These losses are greatest as the fluid flows through entrances, exits, pumps, valves, fittings, and any other piping with rough inner
Most techniques for evaluating head loss due to friction are empirical (based almost exclusively on experimental evidence) and are based on a proportionality constant called the friction factor (f). | {"url":"https://www.engineersedge.com/fluid_flow/extended_bernoulli.htm","timestamp":"2024-11-08T08:35:39Z","content_type":"text/html","content_length":"20113","record_id":"<urn:uuid:1052728a-4c4a-41f8-be9d-86cc1f4c2cb4>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00691.warc.gz"} |
Determine the Type of Convergence for an Alternating Series
Question Video: Determine the Type of Convergence for an Alternating Series Mathematics • Higher Education
State whether the series β _(π = 1) ^(β ) (β 1)^(π + 1) π ^(π ) converges absolutely, conditionally, or not at all.
Video Transcript
State whether the series the sum from π equals one to β of negative one to the power of π plus one times π to the π th power converges absolutely, conditionally, or not at all.
Weβ re given a series and weβ re asked to determine whether this series converges absolutely, conditionally, or whether the series does not converge at all. And at this point, we have a lot of
different methods for dealing with the convergence or divergence of different series. And it can often be hard to decide which one we should proceed with. In this case, though, we can see something
interesting about our summands. First, we can see we have negative one to the power of π plus one. This is just going to alternate the sign of each of our terms. And we can also see we multiply
this by π to the π th power, which we know is going to grow without bound. So the size of each term in our series is growing without bound.
And if the terms of our series are not approaching zero, this should remind us of the π th-term divergence test. The π th-term divergence test tells us that the terms in our series must approach
zero for our series to converge. In other words, if the limit as π approaches β of π π is not equal to zero or the limit as π approaches β of π π does not exist, then the sum
from π equals one to β of π π must be divergent. In our case, our summand π π is negative one to the power of π plus one times π to the π th power. So we can look at the
limit as π approaches β of negative one to the power of π plus one times π to the π th power.
Of course, we already know what happens as π approaches β . Weβ ve already said negative one to the power of π plus one just changes the sign of each term. And π to the π th power is
growing without bound. And if the size of each term is growing without bound, then in particular our limit does not exist. And itβ s worth reiterating here being equal to positive β or being equal
to negative β or oscillating between multiple values are examples of a limit not existing. So by the π th-term divergence test because the limit as an approaches β of our summand is not equal
to zero, our series must be divergent.
And itβ s also worth pointing out if we were to take the absolute value of each term in our series, we would end up with the same conclusion. The only difference is our limit would be the limit as π
approaches β of just π to the π th power, which grows without bound. And so this limit does not exist. But in either way, we can use the π th-term divergence test to show that our series
must be divergent. Therefore, we were able to show the sum from π equals one to β of negative one to the power of π plus one times π to the π th power does not converge at all. | {"url":"https://www.nagwa.com/en/videos/786105627176/","timestamp":"2024-11-05T07:07:44Z","content_type":"text/html","content_length":"245351","record_id":"<urn:uuid:e6f50305-f182-46dd-9f78-914ff8c6e1f8>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00271.warc.gz"} |
By the end of this section, you will be able to:
• Create charts and graphs to appropriately represent data
• Interpret visual representations of data
• Determine misleading components in data displayed visually
Summarizing raw data is the first step we must take when we want to communicate the results of a study or experiment to a broad audience. However, even organized data can be difficult to read; for
example, if a frequency table is large, it can be tough to compare the first row to the last row. As the old saying goes, a picture is worth a thousand words (or, in this case, summary statistics)!
Just as our techniques for organizing data depended on the type of data we were looking at, the methods we’ll use for creating visualizations will vary. Let’s start by considering categorical data.
Visualizing Categorical Data
If the data we’re visualizing are categorical, then we want a quick way to represent graphically the relative numbers of units that fall in each category. When we created the frequency distributions
in the last section, all we did was count the number of units in each category and record that number (this was the frequency of that category). Frequencies are nice when we’re organizing and
summarizing data; they’re easy to compute, and they’re always whole numbers. But they can be difficult to understand for an outsider who’s being introduced to your data.
Let’s consider a quick example. Suppose you surveyed some people and asked for their favorite color. You communicated your results using a frequency distribution. Jerry is interested in data on
favorite colors, so he reads your frequency distribution. The first row shows that 12 people indicated green was their favorite color. However, Jerry has no way of knowing if that’s a lot of people
without knowing how many people total took your survey. Twelve is a pretty significant number if only 25 people took the survey, but it’s next to nothing if you recorded 1,000 responses. For that
reason, we will often summarize categorical data not with frequencies but with proportions. The proportion of data that fall into a particular category is computed by dividing the frequency for that
category by the total number of units in the data.
[latex]\text{Proportion of a category} = \frac{\text{Category frequency}}{\text{Total number of data units}}[/latex]
Proportions can be expressed as fractions, decimals, or percentages.
Recall Example 2 in Section 8.1, in which a teacher recorded the responses on the first question of a multiple-choice quiz, with five possible responses (A, B, C, D, and E). The raw data were as
A A C A B B A
E A C A A A C
E A B A A C A
B E E A A C C
We computed a frequency distribution that looked like this:
[latex]\begin{array} {|c|c|} \hline \textbf{Response to First Question} & \textbf{Frequency} \\ \hline \text{A} & \text{14} \\ \hline \text{B} & \text{4} \\ \hline \text{C} & \text{6} \\ \hline \text
{D} & \text{0} \\ \hline \text{E} & \text{4} \\ \hline \end{array}[/latex]
[latex]\text{Proportion of a category} = \frac{\text{Category frequency}}{\text{Total number of data units}}[/latex]
Now let’s compute the proportions for each category.
Step 1: In order to compute a proportion, we need the frequency (which we have in the table above) and the total number of units that are represented in our data. We can find that by adding up the
frequencies from all the categories: [latex]14+4+6+0+4=28[/latex].
Step 2: To find the proportions, we divide the frequency by the total. For the first category (“A”), the proportion is [latex]\frac{14}{28} = \frac{1}{2} = 0.5 = 50 \%[/latex]. We can compute the
other proportions similarly, filling in the rest of the table:
[latex]\begin{array} {|c|c|c|} \hline \textbf{Response to First Question} & \textbf{Frequency} & \textbf{Proportion} \\ \hline \text{A} & \text{14} & \frac{14}{28} = 50 \% \\ \hline \text{B} & \text
{4} & \frac{4}{28} = 14.3 \%\\ \hline \text{C} & \text{6} & \frac{6}{28} = 21.4 \% \\ \hline \text{D} & \text{0} & \frac{0}{28} = 0 \% \\ \hline \text{E} & \text{4} & \frac{4}{28} = 14.3 \% \\ \hline
Step 3: Check your work: if you add up your proportions, you should get 1 (if you’re using fractions or decimals) or 100% (if you’re using percentages). In this case, [latex]50 \% +14.3 \% + 21.4 \%
+0 \% +14.3 \% = 100 \%[/latex].
Note: If you need to round off the results of the computations to get your percentages or decimals, then the sum might not be exactly equal to 1 or 100% in the end due to that rounding error.
In the last section, students in a statistics class were asked to provide their majors. Those results are again listed below:
Undecided Biology Biology Sociology
Political Science Sociology Undecided Undecided
Undecided Biology Biology Education
Biology Biology Political Science Political Science
You created the frequency distribution:
[latex]\begin{array} {|c|c|} \hline \textbf{Major} & \textbf{Frequency} \\ \hline \text{Biology} & \text{6} \\ \hline \text{Education} & \text{1} \\ \hline \text{Political Science} & \text{3} \\ \
hline \text{Sociology} & \text{2} \\ \hline \text{Undecided} & \text{4} \\ \hline \end{array}[/latex]
Now find the proportions associated with each category. Express your answers as percentages.
[latex]\begin{array} {|c|c|} \hline \textbf{Major} & \textbf{Frequency} & \textbf{Proportion} \\ \hline \text{Biology} & \text{6} & 37.5 \% \\ \hline \text{Education} & \text{1} & 6.3 \% \\ \hline \
text{Political Science} & \text{3} & 18.8 \% \\ \hline \text{Sociology} & \text{2} & 12.5 \% \\ \hline \text{Undecided} & \text{4} & 25 \% \\ \hline \end{array}[/latex]
Note that these percentages add up to 100.1%, due to the rounding.
Now that we can compute proportions, let’s turn to visualizations. There are two primary visualizations that we’ll use for categorical data: bar charts and pie charts. Both of these data
representations work on the same principle: if proportions are represented as areas, then it’s easy to compare two proportions by assessing the corresponding areas. Let’s look at bar charts first.
Bar Charts
A bar chart is a visualization of categorical data that consists of a series of rectangles arranged side-by-side (but not touching). Each rectangle corresponds to one of the categories. All of the
rectangles have the same width. The height of each rectangle corresponds to either the number of units in the corresponding category or the proportion of the total units that fall into the category.
In Example 1, we computed the following proportions:
[latex]\begin{array} {|c|c|c|} \hline \textbf{Response to First Question} & \textbf{Frequency} & \textbf{Proportion} \\ \hline \text{A} & \text{14} & \frac{14}{28} = 50 \% \\ \hline \text{B} & \text
{4} & \frac{4}{28} = 14.3 \%\\ \hline \text{C} & \text{6} & \frac{6}{28} = 21.4 \% \\ \hline \text{D} & \text{0} & \frac{0}{28} = 0 \% \\ \hline \text{E} & \text{4} & \frac{4}{28} = 14.3 \% \\ \hline
Draw a bar chart to visualize this frequency distribution.
Step 1: To start, we’ll draw axes with the origin (the point where the axes meet) at the bottom left:
Figure 1. Step 1
Step 2: Next, we’ll place our categories evenly spaced along the bottom of the horizontal axis. The order doesn’t really matter, but if the categories have some sort of natural order (like in this
case, where the responses are labeled A to E), it’s best to maintain that order. We’ll also label the horizontal axis:
Figure 2. Step 2
Step 3: Now we have a decision to make: Will we use frequencies to define the height of our rectangles, or will we use proportions? Let’s try it both ways. First, let’s use frequencies. Notice that
our frequencies run from 0 to 14; this will correspond to the scale we put on the vertical axis. If we put a tick mark for every whole number between 0 and 14, the result will be pretty crowded;
let’s instead put a mark on the multiples of 3 or 5:
Figure 3. Step 3
Step 4: Now let’s draw in the first rectangle. The frequency associated with “A” is 14. So we’ll go to 14 on the vertical axis and place a mark at that height above the “A” label:
Figure 4. Step 4
Step 5: Then, draw vertical lines straight down from the edges of your mark to make a rectangle:
Figure 5. Step 5
Step 6: Finally, we can build the rest of the rectangles, making sure that the bases all have the same length of the base = width of the rectangle and the rectangles don’t touch. Notice that since
the frequency for “D” is 0, that category has no rectangle (but we’ll leave a space there so the reader can see that there is a category with frequency 0). Here’s the result:
Figure 6. Step 6
Step 7: That’s it! Now let’s use proportions instead of frequencies. We’ll label the vertical axis with evenly spaced numbers that run the full range of the percentages in our table: 0% to 50%. We
can divide that into 5 equal parts (so that each has a width of 10%), and use that to label our vertical axis:
Figure 7. Step 7
Step 8: Then, we can fill in the rectangles just as we did before. The height of the “A” rectangle is 50%, the “B” rectangle goes up to 14.3%, “C” goes to 21.4%, there is no rectangle for “D” (since
its proportion is 0%), and the “E” rectangle also goes up to 14.3%:
Figure 8. Step 8
Step 9: Notice that the rectangles are basically identical in our two final bar charts. That’s no coincidence! Bar charts that use proportions and those that use frequencies will always look
identical (which is why it doesn’t really matter much which option you choose). Here’s why: Look at the bars for “B” and “C.” The frequencies for these are 4 and 6, respectively. Notice that 6 is 50%
bigger than 4 (since [latex]6=1.5 \times 4[/latex]), which means that the “C” bar will be 50% higher than the “B” bar. Now look at the same bars using proportions: since [latex]21.4 \% =1.5 \times
14.3 \%[/latex], the bar for “C” will be 50% higher than the bar for “B.” The same relationships hold for the other bars too.
Figure 9. Step 9
The students in a statistics class were asked to provide their majors. The computed proportions for each of the categories are as follows:
[latex]\begin{array} {|c|c|c|} \hline \textbf{Major} & \textbf{Frequency} & \textbf{Proportion}\\ \hline \text{Biology} & \text{6} & 37.5 \% \\ \hline \text{Education} & \text{1} & 6.3 \% \\ \hline \
text{Political Science} & \text{3} & 18.8 \% \\ \hline \text{Sociology} & \text{2} & 12.5 \% \\ \hline \text{Undecided} & \text{4} & 25 \% \\ \hline \end{array}[/latex]
Create a bar graph to visualize these data. Use percentages to label the vertical axis.
Figure 10. Percent of students’ majors
Now that we’ve explored how bar graphs are made, let’s get some practice reading bar graphs.
The bar graph shown gives data on 2020 model year cars available in the United States. Analyze the graph to answer the following questions.
Figure 11. 2020 model year cars available in the United States
(a) What proportion of available cars were sports cars?
The bar for sports cars goes up to 10%, so the proportion of models that are considered sports cars is 10%.
(b) What proportion of available cars were sedans?
The bar corresponding to sedan goes up past 30% but not quite to 35%. It looks like the proportion we want is between 33% and 34%.
(c) Which categories of cars each made up less than 5% of the models available?
We’re looking for the bars that don’t make it all the way to the 5% line. Those categories are hatchback and wagon.
The bar graph shows the region of every institution of higher learning in the United States (except for the service academies, like West Point).
Figure 12. Regions of institutions of higher education in the United States
Analyze the bar chart to answer the following questions.
a) Which region contains the largest number of institutions of higher learning?
b) What proportion of all institutions of higher learning can be found in the Southwest?
c) Which regions each have under 5% of the total number of institutions of higher learning?
a) Southeast
b) Just over 10%
c) Outlying Areas and Rocky Mtns.
Pie Charts
A pie chart consists of a circle divided into wedges, with each wedge corresponding to a category. The proportion of the area of the entire circle that each wedge represents corresponds to the
proportion of the data in that category. Pie charts are difficult to make without technology because they require careful measurements of angles and precise circles, both of which are tasks better
left to computers.
Here is a video on creating pie charts in Google Sheets:
Pie charts are sometimes embellished with features like labels in the slices (which might be the categories, the frequencies in each category, or the proportions in each category) or a legend that
explains which colors correspond to which categories. When making your own pie chart, you can decide which of those to include. The only rule is that there has to be some way to connect the slices to
the categories (either through labels or a legend).
Use the data that follows to generate a pie chart.
Type Percent
SUV 43.6%
Sedan 33.6%
Sports 10.0%
Minivan 5.5%
Hatchback 3.6%
Wagon 3.6%
First, enter the chart above into a new sheet in Google Sheets. Next, click and drag to select the full table (including the header row). Click on the “Insert” menu, then select “Chart.” The result
may be a pie chart by default; if it isn’t, you can change it to a pie chart using the “Chart type” drop-down menu in the Chart Editor.
Figure 13. 2020 model year cars available in the United States
You can choose to use a legend to identify the categories as well as label the slices with the relevant percentages.
In a previous exercise, you created a bar chart using data on reported majors from students in a class. Here are those proportions again (sorted from largest to smallest):
[latex]\begin{array} {|c|c|} \hline \textbf{Major} & \textbf{Proportion}\\ \hline \text{Biology} & 37.5 \% \\ \hline \text{Undecided} & 25 \% \\ \hline \text{Political Science} & 18.8 \% \\ \hline \
text{Sociology} & 12.5 \% \\ \hline \text{Education} & 6.3 \% \\ \hline \end{array}[/latex]
Create a pie graph using those data.
Figure 14. Majors of students in the class
Visualizing Quantitative Data
There are several good ways to visualize quantitative data. In this section, we’ll talk about two types: stem-and-leaf plots and histograms.
Stem-and-Leaf Plots
Stem-and-leaf plots are visualization tools that fall somewhere between a list of all the raw data and a graph. A stem-and-leaf plot consists of a list of stems on the left and the corresponding
leaves on the right, separated by a line. The stems are the numbers that make up the data only up to the next-to-last digit, and the leaves are the final digits. There is one leaf for every data
value (which means that leaves may be repeated), and the leaves should be evenly spaced across all stems. These plots are really nothing more than a fancy way of listing out all the raw data; as a
result, they shouldn’t be used to visualize large datasets.
This concept can be difficult to understand without referencing an example, so let’s first look at how to read a stem-and-leaf plot.
A collector of trading cards records the sale prices (in dollars) of a particular card on an online auction site and puts the results in a stem-and-leaf plot:
Answer the following questions about the data:
a) How many prices are represented?
Each leaf (the numbers on the right side of the bar) represents one data value. So on the first row (which looks like 0 | 5 8 9), there are three data values (one for each leaf: 5, 8, and 9). The
next row has 13 leaves, then 8, 5, 3, 0, and 1. Adding those up, we get [latex]3+13+8+5+3+0+1=33[/latex] data points or prices.
b) What prices represent the five most expensive cards? The five least expensive?
The most expensive card is the last one listed. Its stem is 6 and its leaf is 0, so the price is $60. There are no leaves associated with the 5 stem, so there were no cards sold for $50 to $59. The
next most expensive cards are then on the 4 stem: $45, $40, and $40 (remember, repeated leaves mean repeated values in the dataset). So we have our four most expensive cards. The fifth would be on
the next stem up. The biggest leaf on the 3 stem is a 5, so the fifth-most expensive card sold for $35.
As for the five least-expensive cards, the smallest stem is 0, with leaves 5, 8, and 9. So the three least expensive cards sold for $5, $8, and $9 (notice that we don’t write down that leading 0 from
the stem in the tens place). The next two least-expensive cards will be the two smallest leaves on the next stem: $10 and $10.
c) What is the full set of data?
The full list of data is: 5, 8, 9, 10, 10, 10, 13, 14, 14, 15, 15, 15, 15, 16, 19, 19, 20, 20, 20, 24, 25, 25, 29, 29, 30, 30, 30, 35, 35, 40, 40, 45, 60.
The stem-and-leaf plot below shows data collected from a sample of college students who were asked how far (in miles) they commute each day:
a) How many data points are represented?
b) What are the three longest and shortest commutes?
c) What is the full list of data?
a) 24
b) The longest commutes are 60, 50, and 36 miles; the shortest are 4, 6, and 7 miles.
c) 4, 6, 7, 10, 10, 10, 12, 12, 12, 14, 15, 18, 18, 20, 25, 25, 25, 30, 30, 35, 35, 36, 50, 60
Stem-and-leaf plots are useful in that they give us a sense of the shape of the data. Are the data evenly spread out over the stems, or are some stems “heavier” with leaves? Are the heavy stems on
the low side, the high side, or somewhere in the middle? These are questions about the distribution of the data, or how the data are spread out over the range of possible values.
Some words we use to describe distributions are uniform (data are equally distributed across the range), symmetric (data are bunched up in the middle, then taper off in the same way above and below
the middle), left-skewed (data are bunched up at the high end or larger values and taper off toward the low end or smaller values), and right-skewed (data are bunched up at the low end and taper off
toward the high end). See Figure 15 below.
Figure 15. Uniform, right-skewed, left-skewed, and symmetric histograms
Looking back at the stem-and-leaf plot in the previous example, we can see that the data are bunched up at the low end and taper off toward the high end; that set of data is right-skewed. Knowing the
distribution of a set of data gives us useful information about the property that the data are measuring.
Now that we have a better idea of how to read a stem-and-leaf plot, we’re ready to create our own.
An entomologist studying crickets recorded the number of times different crickets (of differing species, genders, etc.) chirped in a one-minute span. The raw data are as follows:
Construct a stem-and-leaf plot to visualize these results.
Step 1: Before we can create the plot, we need to sort the data in order from smallest to largest:
Step 2: Next, we identify the stems. To do that, we cut off the final digit of each number, which leaves us with stems of 8, 9, 10, 11, and 12. Arrange the stems vertically, and add the bar to
separate these from the leaves:
Step 3: Write down the leaves on the right side of the bar, giving just the final digit (that we cut off to make the stems) of each data value. List these in order, and make sure they line up
This table gives the records of the Major League Baseball teams at the end of the 2019 season:
Team Wins Losses
HOU 107 55
LAD 106 56
NYY 103 59
MIN 101 61
ATL 97 65
OAK 97 65
TBR 96 66
CLE 93 69
WSN 93 69
STL 91 71
MIL 89 73
NYM 86 76
ARI 85 77
BOS 84 78
CHC 84 78
PHI 81 81
TEX 78 84
SFG 77 85
CIN 75 87
CHW 72 89
LAA 72 90
COL 71 91
SDP 70 92
PIT 69 93
SEA 68 94
TOR 67 95
KCR 59 103
MIA 57 105
BAL 54 108
DET 47 114
As we mentioned above, stem-and-leaf plots aren’t always going to be useful. For example, if all the data in your dataset are between 20 and 29, then you’ll just have one stem, which isn’t terribly
useful. (Although there are methods like stem splitting for addressing that particular problem, we won’t go into those at this time.) On the other end of the spectrum, the data may be so spread out
that every stem has only one leaf. (This problem can sometimes be addressed by rounding off the data values to the tens, hundreds, or some other place value, then using that place for the leaves.)
Finally, if you have dozens or hundreds (or more) of data values, then a stem-and-leaf plot becomes too unwieldy to be useful. Fortunately, we have other tools we can use.
Histograms are visualizations that can be used for any set of quantitative data, no matter how big or spread out. They differ from a categorical bar chart in that the horizontal axis is labeled with
numbers (not ranges of numbers), and the bars are drawn so that they touch each other. The heights of the bars reflect the frequencies in each bin. Unlike with stem-and-leaf plots, we cannot
re-create the original dataset from a histogram. However, histograms are easy to make with technology and are great for identifying the distribution of our data. Let’s first create one histogram
without technology to help us better understand how histograms work.
In Example 6, we built a stem-and-leaf plot for the number of chirps made by crickets in one minute. Here are the raw data that we used then:
Construct a histogram to visualize these results.
Step 1: Add data to bins. Histograms are built on binned frequency distributions, so we’ll make that first. Luckily, the stem-and-leaf plot we made earlier can help us do this much more quickly:
If we’re using bins of width 10, we can compute the frequencies by counting the numbers of leaves associated with the corresponding stem:
Bin Frequency
80-89 7
90-99 6
100-109 14
110-119 2
120-129 1
(Note that when we made binned frequency diagrams in the last module, we noted that if the biggest data value was right on the border between two bins, it was OK to lump it in with the lower bin.
That’s not recommended when building histograms, so the data value 120 is all alone in the 120-129 bin.)
Step 2: Create the axes. On the horizontal axis, start labeling with the lower end of the first bin (in this case, 80), and go up to the higher end of the last bin (120). Mark off the other bin
boundaries, making sure they’re all evenly spaced. On the vertical axis, start with 0 and go up at least to the greatest frequency you see in your bins (14 in this example), making sure that the
labels you make are evenly spaced and that the difference between those numbers is the same. Let’s count off our vertical axis by threes:
Figure 16. Step 2
Step 3: Draw in the bars. Remember that the bars of a histogram touch and that the heights are determined by the frequency. So the first bar will cover 80 to 90 on the horizontal axis and have a
height of 7:
Figure 17. Step 3, begin with the first bar
Now we can fill in the others:
Figure 18. Step 3, fill in the rest of the bars
Step 4: Let’s compare the histogram we just created to the stem-and-leaf plot we made earlier:
Figure 19. Step 4
Notice that the leaves on the rotated stem-and-leaf plot match the bars on our histogram! We can view stem-and-leaf plots as sideways histograms. But, as we’ll see soon, we can do much more with
To see how to graph a histogram with the Desmos statistics calculator, click
In the last exercise, you made a stem-and-leaf plot of the number of wins for each MLB team in 2019 using this set of data:
Team Wins Losses
HOU 107 55
LAD 106 56
NYY 103 59
MIN 101 61
ATL 97 65
OAK 97 65
TBR 96 66
CLE 93 69
WSN 93 69
STL 91 71
MIL 89 73
NYM 86 76
ARI 85 77
BOS 84 78
CHC 84 78
PHI 81 81
TEX 78 84
SFG 77 85
CIN 75 87
CHW 72 89
LAA 72 90
COL 71 91
SDP 70 92
PIT 69 93
SEA 68 94
TOR 67 95
KCR 59 103
MIA 57 105
BAL 54 108
DET 47 114
Figure 20. Wins for each MLB team
Misleading Graphs
Graphical representations of data can be manipulated in ways that intentionally mislead the reader. There are two primary ways this can be done: by manipulating the scales on the axes and by
manipulating or misrepresenting areas of bars. Let’s look at some examples of these.
The table below shows the teams and their payrolls in the English Premier League, the top soccer organization in the United Kingdom.
Team Salary (£1,000,000s)
Manchester United F.C. 175.7
Manchester City F.C. 136.5
Chelsea F.C. 132.8
Arsenal F.C. 130.7
Tottenham Hotspur F.C. 129.2
Liverpool F.C. 118.6
Crystal Palace 85.0
Everton F.C. 82.5
Leicester City 73.7
West Ham United F.C. 69.2
Newcastle United F.C. 56.9
Aston Villa F.C. 52.3
Fulham F.C. 52.1
Southampton F.C. 49.6
Wolverhampton Wanderers F.C. 49.5
Brighton & Hove Albion 43.7
Burnley F.C. 35.5
West Bromwich Albion F.C. 23.8
Leeds United F.C. 22.5
Sheffield United F.C. 19.7
How might someone present these data in a misleading way?
Step 1: Let’s focus on the top five teams. Here’s a bar chart of their payrolls:
Figure 21. Top Five ELP Teams by Payroll
Step 2: Now here’s another bar chart visualizing exactly the same data:
Figure 22. Top Five EPL Teams by Payroll
Step 3: You should notice that despite using the same data, these two graphs look strikingly different. In the second graph, the gap between Manchester United and the other four teams looks
significantly larger than in the first graph. The scale on the vertical axis has been manipulated here. The first graph’s axis starts at 0, while the lowest value on the second graph’s axis is 120.
This trick has a strong impact on the viewer’s perception of the data.
Beware of vertical axes that don’t start at zero! They overemphasize differences in heights.
Step 4: To further emphasize the difference this creates in our perception, let’s look at those data again, but this time using graphics instead of colored areas on our bar graph.
Figure 23. Top Five EPL Teams by Payroll
This graph uses an image of a £10 banknote in place of the bars. Using an image that evokes the context of the data in place of a standard, “boring” bar is a common tool that people use when creating
infographics. However, this is generally not a good practice because it distorts the data. Notice that our “bars” (the banknotes) are just as tall here as they were in the previous figure. But to
maintain the right proportions, the widths had to be adjusted as well, which changes the area (height × width) of each bar. A key point is that when looking at rectangles, the human eye tends to
process areas more easily than heights.
Beware of infographics! Areas overemphasize a difference that should be measured with a height!
Step 5: Now let’s look at all 20 teams. This histogram indicates that the data are right-skewed, with the highest number of teams having a payroll between £40 million and £80 million:
Figure 24. Total payrolls of teams in the EPL
Step 6: Now let’s view this same data in another chart:
Figure 25. Frequency vs. Payroll
Step 7: Even though this chart uses the same data, the skew seems to be reversed. Why? Well, even though this graph looks like a histogram, it isn’t. Look closely at the labels on the horizontal
axis; they don’t correspond to spots on the axis but instead provide a range, meaning this is a bar graph based on a binned frequency distribution.
When we review these ranges, we can see that the last range is misleading, as it consists of all data “over 80.” If the bins all had the same width, that last bin would run from 80 to 120. However,
we can see from the histogram that the maximum value for these data is between 160 and 200. If the last bin in this bar graph were labeled honestly, it would read “80–200,” which would drive home the
fact that the width of that bar is misleading.
Always check the horizontal axis on histograms! The widths of all the bars should be equal.
Take a look again at the win totals for teams in Major League Baseball in 2019:
Team Wins Losses
HOU 107 55
LAD 106 56
NYY 103 59
MIN 101 61
ATL 97 65
OAK 97 65
TBR 96 66
CLE 93 69
WSN 93 69
STL 91 71
MIL 89 73
NYM 86 76
ARI 85 77
BOS 84 78
CHC 84 78
PHI 81 81
TEX 78 84
SFG 77 85
CIN 75 87
CHW 72 89
LAA 72 90
COL 71 91
SDP 70 92
PIT 69 93
SEA 68 94
TOR 67 95
KCR 59 103
MIA 57 105
BAL 54 108
DET 47 114
(Source: https://www.espn.com/mlb/standings/_/season/2019/view)
Make one good and one misleading chart showing the number of wins by the top ten teams. Then, looking at all the teams, make one good and one misleading histogram for the win totals.
Top ten teams by wins:
Figure 26. Top ten MLB teams by wins, 2019 (good)
Figure 27. Top ten MLB teams by wins, 2019 (bad)
Figure 28. Wins by MLB teams, 2019 season (good)
Figure 29. Wins by MLB teams, 2019 season (bad)
Here is a video to help you spot misleading graphs:
A visualization of categorical data that consists of a series of rectangles arranged side-by-side (but not touching).
Consists of a circle divided into wedges, with each wedge corresponding to a category.
Consists of a list of stems on the left and the corresponding leaves on the right, separated by a line.
Data are equally distributed across the range.
Data are bunched up in the middle, then taper off in the same way above and below the middle.
Data are bunched up at the high end or larger values and taper off toward the low end or smaller values.
Data are bunched up at the low end and taper off toward the high end.
Visualizations that can be used for any set of quantitative data, no matter how big or spread out. | {"url":"https://louis.pressbooks.pub/finitemathematics/chapter/8-2-visualizing-data/","timestamp":"2024-11-04T04:19:20Z","content_type":"text/html","content_length":"215066","record_id":"<urn:uuid:a2a32855-9a12-45e3-9354-cc39be5aea71>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00736.warc.gz"} |
Algorithms and hardness results for nearest neighbor problems in bicolored point sets
Document Type
Conference Article
Publication Title
Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics)
In the context of computational supervised learning, the primary objective is the classification of data. Especially, the goal is to provide the system with “training” data and design a method which
uses the training data to classify new objects with the correct label. A standard scenario is that the examples are points from a metric space, and “nearby” points should have “similar” labels. In
practice, it is desirable to reduce the size of the training set without compromising too much on the ability to correctly label new objects. Such subsets of the training data are called as edited
sets. Wilfong [SOCG ’91] defined two types of edited subsets: consistent subsets (those which correctly label all objects from the training data) and selective subsets (those which correctly label
all new objects the same way as the original training data). This leads to the following two optimization problems: k-MCS-(X) Given k sets of points P1, P2, …, Pk in a metric space X, the goal is to
choose subsets of points Pi′⊆Pi for i= 1, 2, …, k such that ∀p∈Pi its nearest neighbor among (Formula Presented) lies in Pi′ for each i∈ [k] while minimizing (Note that we also enforce the condition
(Formula Presented) the quantity (Formula Presented) - k-MSS-(X): Given k sets of points P1, P2, …, Pk in a metric space X, the goal is to choose subsets of points Pi′⊆Pi for i= 1, 2, …, k such that
∀p∈Pi its nearest neighbor among (Formula Presented) lies in Pi′ for each i∈ [k] while minimizing (Note that we again enforce the condition |Pi′|≥1∀i∈[k].) the quantity (Formula Presented). While
there have been several heuristics proposed for these two problems in the computer vision and machine learning community, the only theoretical results for these problems (to the best of our
knowledge) are due to Wilfong [SOCG ’91] who showed that both 3-MCS-(ℝ2) and 2-MSS-(ℝ2) are NP-complete. We initiate the study of these two problems from a theoretical perspective, and obtain several
algorithmic and hardness results. On the algorithmic side, we first design an O(n2) time exact algorithm and O(nlog n) time 2-approximation for the 2-MCS-(R) problem, i.e., the points are located on
the real line. Moreover, we show that the exact algorithm also extends to the case when the points are located on the circumference of a circle. Next, we design an O(r2) time online algorithm for the
2-MCS-(R) problem such that r< n, where n is the set of points and r is an integer. Finally, we give a PTAS for the k-MSS-(ℝ2) problem. On the hardness side, we show that both the 2-MCS and 2-MSS
problems are NP-complete on graphs. Additionally, the problems are W[2]-hard parameterized by the size k of the solution. For points on the Euclidean plane, we show that the 2-MSS problem is
contained in W[1]. Finally, we show a lower bound of Ω(√n) bits for the storage of any (randomized) algorithm which solves both 2-MCS-(ℝ) and 2-MSS-(ℝ).
Publication Date
Recommended Citation
Banerjee, Sandip; Bhore, Sujoy; and Chitnis, Rajesh, "Algorithms and hardness results for nearest neighbor problems in bicolored point sets" (2018). Conference Articles. 162. | {"url":"https://digitalcommons.isical.ac.in/conf-articles/162/","timestamp":"2024-11-08T21:08:28Z","content_type":"text/html","content_length":"42278","record_id":"<urn:uuid:e7a983e8-fdf6-412b-8b40-113ce36aab92>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00494.warc.gz"} |
93.83 decimeters per square second to meters per square second
94 Decimeters per square second = 9.38 Meters per square second
Acceleration Converter - Decimeters per square second to meters per square second - 93.83 meters per square second to decimeters per square second
This conversion of 93.83 decimeters per square second to meters per square second has been calculated by multiplying 93.83 decimeters per square second by 0.1 and the result is 9.383 meters per
square second. | {"url":"https://unitconverter.io/decimeters-per-square-second/meters-per-square-second/93.83","timestamp":"2024-11-05T13:11:52Z","content_type":"text/html","content_length":"27462","record_id":"<urn:uuid:71a05ecb-3545-4448-8039-d9390107ad47>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00697.warc.gz"} |
regular polygon
What is perimeter of regular polygon?
What is perimeter of regular polygon?
The perimeter of a regular polygon = (length of one side) × number of sides. Perimeter of a regular hexagon = 4 cm × 6. = 24 cm. Example 2: A pentagon has all sides equal to 6 cm.
How do you find the perimeter of a polygon with coordinates?
When they are graphed on a coordinate plane (flat surface marked with crossed number lines to establish locations), only the coordinates of the corners are known. Finding the perimeter (distance
around the edge) is done by adding up the lengths of the sides.
How do you find the perimeter of an irregular polygon?
For an irregular polygon, the perimeter is calculating by adding together the individual side lengths. For example, the perimeter of an irregular pentagon whose side lengths are; 5 cm, 4 cm, 6 cm, 10
cm, and 9 cm. = 34 cm.
What is the perimeter of irregular pentagon?
How do you find the perimeter of a concave polygon?
The perimeter of a concave polygon can be found by adding together the length of all the sides.
How do you find the perimeter of shapes?
The perimeter of a shape is always calculated by adding up the length of each of the sides. In Year 5 and 6, children might be given shapes like this one and asked to find their perimeter: In this
case, they need to work out the lengths of the edges that are unlabelled, by looking at the other labelled edges.
What is the perimeter of a regular polygon of n sides?
Perimeter of a Regular Polygon: If the length of a side is s and there are n sides in a regular polygon, then the perimeter is P=ns. Example 1: What is the perimeter of a regular octagon with 4 inch
How to calculate the area, perimeter of a polygon manually?
Find the apothem. Apothem = side/(2*Tan (π/N)) = 2/(2*Tan (π/4)) = 2/(2*Tan
Find the perimeter. Perimeter = (N*(side) = 4*2 = 8
Find the area.
How to find the area of a polygon?
To find the area of a regular polygon, all you have to do is follow this simple formula: area = 1/2 x perimeter x apothem. Find the apothem of the polygon. If you’re using the apothem method, then
the apothem will be provided for you. Let’s say you’re working with a hexagon that has an apothem with a length of 10√3.
How do you calculate the area of a regular polygon?
Know the correct formula. The area of any regular polygon is given by the formula: Area = (a x p)/2, where a is the length of the apothem and p is the perimeter of the polygon. Plug the values of a
and p in the formula and get the area. As an example, let’s use a hexagon (6 sides) with a side (s) length of 10.
What is the formula for finding area and perimeter?
There are several methods to find area and perimeter, depending on the shape of the figure. To find the area of a square or rectangle, use this formula: length x width. To find the perimeter, use
this formula: (length x 2) + (width x 2). | {"url":"https://profoundqa.com/what-is-perimeter-of-regular-polygon/","timestamp":"2024-11-12T22:15:03Z","content_type":"text/html","content_length":"60210","record_id":"<urn:uuid:db87acf4-5759-4d9b-b66c-991e0a5c6971>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00431.warc.gz"} |
Classical field theory and relativistic quantum mechanics
Oancia, Giancarlo
Classical field theory and relativistic quantum mechanics.
[Laurea], Università di Bologna, Corso di Studio in
Fisica [L-DM270]
Documenti full-text disponibili:
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Disponibile con Licenza: Salvo eventuali più ampie autorizzazioni dell'autore, la tesi può essere liberamente consultata e può essere effettuato il salvataggio e la stampa di una copia per fini
strettamente personali di studio, di ricerca e di insegnamento, con espresso divieto di qualunque utilizzo direttamente o indirettamente commerciale. Ogni altro diritto sul materiale è riservato
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Nowadays, the best theoretical framework we have to describe elementary particles' physics is the Standard Model, whose main language consists of quantum field theory. Historically, before its
formulation, it has been attempted to make Schrödinger's quantum theory relativistic, in the framework of the so-called first quantization. The aim of this thesis is to make the reader aware of the
problems of this procedure, necessary condition to understand the need to change paradigm and develop a new theory, known as second quantization. Since, in this context, the new fundamental physical
entity is the quantum field, we shall introduce field theory, starting from the classical description of electromagnetism, making use of a formalism to make Maxwell's equations manifestly covariant.
To obtain the latter directly from an action and to use the Lagrangian mechanics' tools, it's necessary to generalize the latter to a system with an infinite number of degrees of freedom. This will
be achieved initially by discretizing the space and applying the known formalism into any elementary cell and, later, through a variational principle. Furthermore, we'll try to apply quantum
mechanics to a relativistic particle, obtaining the Klein-Gordon equation, which will be interpreted as representing a field whose quantum is a massive particle without spin. We'll notice how,
forcing a particular global symmetry of this equation to be locally valid, it'll be necessary to add some terms on the Lagrangian which can be interpreted as an interaction with the electromagnetic
field. This allows us to introduce Gauge's principle, which is a fundamental tool to describe interactions in the Standard Model. Finally, this principle will be critically analyzed, leading to the
conclusion that it's not correct to distinguish between the object and the mediator of an interaction.
Nowadays, the best theoretical framework we have to describe elementary particles' physics is the Standard Model, whose main language consists of quantum field theory. Historically, before its
formulation, it has been attempted to make Schrödinger's quantum theory relativistic, in the framework of the so-called first quantization. The aim of this thesis is to make the reader aware of the
problems of this procedure, necessary condition to understand the need to change paradigm and develop a new theory, known as second quantization. Since, in this context, the new fundamental physical
entity is the quantum field, we shall introduce field theory, starting from the classical description of electromagnetism, making use of a formalism to make Maxwell's equations manifestly covariant.
To obtain the latter directly from an action and to use the Lagrangian mechanics' tools, it's necessary to generalize the latter to a system with an infinite number of degrees of freedom. This will
be achieved initially by discretizing the space and applying the known formalism into any elementary cell and, later, through a variational principle. Furthermore, we'll try to apply quantum
mechanics to a relativistic particle, obtaining the Klein-Gordon equation, which will be interpreted as representing a field whose quantum is a massive particle without spin. We'll notice how,
forcing a particular global symmetry of this equation to be locally valid, it'll be necessary to add some terms on the Lagrangian which can be interpreted as an interaction with the electromagnetic
field. This allows us to introduce Gauge's principle, which is a fundamental tool to describe interactions in the Standard Model. Finally, this principle will be critically analyzed, leading to the
conclusion that it's not correct to distinguish between the object and the mediator of an interaction.
Altri metadati
Statistica sui download | {"url":"https://amslaurea.unibo.it/29945/","timestamp":"2024-11-04T09:10:19Z","content_type":"application/xhtml+xml","content_length":"36791","record_id":"<urn:uuid:7759b05a-6c43-4e54-a4b5-443d2327a547>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00226.warc.gz"} |
Execution time limit is 1 second
Runtime memory usage limit is 64 megabytes
There are coins available in various fixed denominations, expressed in kopecks (for example, 3 and 5 kopecks), and they are available in unlimited quantities. Write a program, COINS, that:
1. Determines if a given amount S (in kopecks) can be formed using the available coin denominations.
2. If possible, represents this amount using the fewest number of coins.
The input file begins with the amount S (0 ≤ S ≤ 1000000000) on the first line. The second line contains N, the number of different denominations (1 ≤ N ≤ 20). The following N lines list the
denominations A_{1}, A_{2}, ..., A_{N} in ascending order, where each denomination satisfies (0 < A_1 < A_2 < ... < A_N ≤ 1000000000).
The output file should start with the symbol "+" on the first line if the amount S can be represented using the given denominations, or the symbol "-" if it cannot. If a representation is possible,
the subsequent N lines should specify the number of coins of each denomination required to form the amount S using the minimum number of coins. | {"url":"https://basecamp.eolymp.com/en/problems/4470","timestamp":"2024-11-03T02:36:27Z","content_type":"text/html","content_length":"235952","record_id":"<urn:uuid:7d45669b-698a-4ae9-9a05-b1d079a9c24f>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00332.warc.gz"} |
Back to Papers Home Back to Papers of School of Physics
Paper IPM / P / 8035
School of Physics
Title: What Is Needed of A Tachyon If It Is to Be the Dark Energy?
Author(s): 1. E.J. Copeland
2. M.R. Garousi
3. M. Sami
4. S. Tsujikawa
Status: Published
Journal: Phys. Rev. D
Vol.: 71
Year: 2005
Pages: 043003
Supported by: IPM
We study a dark energy scenario in the presence of a tachyon field ϕ with potential V(ϕ) and a barotropic perfect fluid. The cosmological dynamics crucially depends on the asymptotic behavior of the
quantity λ = −M[p]V[ϕ]/V^3/2. If λ is a constant, which corresponds to an inverse square potential V(ϕ) ∝ ϕ^−2, there exists one stable critical point that gives an acceleration of the universe at
late times. When λ→ 0 asymptotically, we can have a viable dark energy scenario in which the system approaches an "instantaneous" critical point that dynamically changes with λ. If |λ| approaches
infinity asymptotically, the universe does not exhibit an acceleration at late times. In this case, however, we find an interesting possibility that a transient acceleration occurs in a regime where
|λ| is smaller than of order unity.
Download TeX format
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Question about Path-Integral from Particle-Vortex Daulity from 3d Bosonization
I have been studying the paper Particle-Vortex Daulity from 3d Bosonization,
by Andrew Karch and David Tong.
On page 11, it says that integrating out gauge field $a$ produces equation of motion $a=A-\tilde{a}$.
My question is how I should perform the path integral
$$\int\mathcal{D}a \exp\{-i\int(\frac{1}{2}a\wedge da+a\wedge d\tilde{a}-a\wedge dA)\}$$
to obtain the equation of motion.
They had to make some hand-waving assumption they called "absence of holonomy" to get rid of the quadratic term in a, which is of course where all the action happens in this duality to give you
fermions and everything.
Once its linearized, the path integral is evaluated by analogy with the 1-dimensional integral
$$\int dx e^{i x y} = \delta(y).$$
EDIT: I was being a bit too cavalier and was going to end up with $\delta(\tilde a - A)$, which is too strong. I was altogether dropping the $ada$ term. A better thing to do is complete the square:
redefine $a' = a + (1/2)(\tilde a - A)$. Then the integrand becomes
$$ a' da' - (1/4)(\tilde a - A)d(\tilde a - A).$$
One should worry about the factors of 1/2 and 1/4 that we ended up with. In this form, we're free to do the path integral over $a'$, which on a simply connected manifold can be computed as a Gaussian
integral and the saddle point (which is unique on a simply connected manifold) is $a' = 0$, in other words $a = A - \tilde a$.
Could you elaborate why you can use that formula with the $a\wedge da$ term when the holonomy is absent?
Why does the delta functional produce $a-A+\tilde{a}=0$? | {"url":"https://www.physicsoverflow.org/40941/question-about-integral-particle-vortex-daulity-bosonization","timestamp":"2024-11-07T22:34:56Z","content_type":"text/html","content_length":"135982","record_id":"<urn:uuid:1ae7cdc4-466e-4d03-afce-fd2dea8500d6>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00092.warc.gz"} |
Two-sample win ratio tests of possibly recurrent event and death
This vignette demonstrates the use of the WR package for two-sample win ratio tests of recurrent event and death (Mao et al., 2022).
Let \(D\) denote the survival time and write \(N_D(t)=I(D\leq t)\). Likewise, let \(T_1<T_2<\cdots\) denote the recurrent event times and write \(N_H(t)=\sum_{k=1}^\infty I(T_k\leq t)\). In other
words, \(N_H(t)\) is the counting process for the recurrent event. Since death is a terminal event, we must have that \(N_H(t)=N_H(t\wedge D)\), where \(b\wedge c=\min(b,c)\). Let \(\boldsymbol Y(t)=
\{N_D(u), N_H(u):0\leq u\leq t\}\) denote the event history up to time \(t\). The full outcome data without censoring are thus \(\boldsymbol Y:=\boldsymbol Y(\tau)\), where \(\tau\) is the maximum
length of follow-up.
General framework for hypotheses and tests
Use \(a=1\) to denote the active treatment and \(a=0\) to denote the control. For all notation introduced above, use the subscript \((a)\) to denote the corresponding group-specific quantity. For
example, \(\boldsymbol Y^{(a)}\) represents the outcome data from group \(a\) \((a=1, 0)\). Consider a time-dependent win function of the form \[\mathcal W(\boldsymbol Y^{(a)}, \boldsymbol Y^{(1-a)})
(t) = I(\mbox{patient in group $a$ wins against that in group $1-a$ by time $t$}).\] The specific definition of \(\mathcal W\) will be discussed later.
Given such a rule of cross-group comparison, the win and loss probabilities for the treatment against control by time \(t\) are \(w(t)=E\{\mathcal W(\boldsymbol Y^{(1)}, \boldsymbol Y^{(0)})(t)\}\)
and \(l(t)=E\{\mathcal W(\boldsymbol Y^{(0)}, \boldsymbol Y^{(0)})(t)\}\), respectively. Suppose that we wish to test the null hypothesis that the win and loss probabilities are equal, i.e., \[H_0: w
(t)=l(t)\hspace{2ex}\mbox{for all }t\in[0, \tau],\] against the alternative hypothesis that the win probability dominates the loss probability, i.e., \[$$\tag{1} H_A: w(t)\geq l(t)\hspace{1ex}\mbox
{for all }t\in[0, \tau] \mbox{ with strict inequality for some }t.$$\] With censored data, a general way to test such hypotheses is to use a log-transformed two-sample win ratio constructed similarly
to Pocock et al. (2012), only with the pairwise rule of comparison replaced by the customized \(\mathcal W(\cdot,\cdot)(t)\), where \(t\) is set as the earlier of the two observed follow-up times. A
stratified test can also be developed along the lines of Dong et al. (2018).
Choice of win function
Different choices of \(\mathcal W\) will lead to different hypotheses and tests. The standard win ratio (SWR) of Pocock et al. (2012) corresponds to the choice of \[\mathcal W_S(\boldsymbol Y^{(a)},
\boldsymbol Y^{(1-a)})(t) =I(D^{(1-a)}<D^{(a)}\wedge t)+I(D^{(a)}\wedge D^{(1-a)}>t, T_1^{(1-a)}<T_1^{(a)}\wedge t).\] With recurrent nonfatal event, \(\mathcal W_S\) fails to fully exploit the data
as it draws only on the first occurrence. A more efficient rule is given by the last-event-assisted win ratio (LWR), which compares on the nonfatal event first by its cumulative frequency, with ties
broken by the time of its last episode. In other words, \[ \mathcal W_L(\boldsymbol Y^{(a)}, \boldsymbol Y^{(1-a)})(t) &=I(D^{(1-a)}<D^{(a)}\wedge t)+I\{D^{(a)}\wedge D^{(1-a)}>t, N_H^{(a)}(t)<N_H^
{(1-a)}(t)\}\\ &\hspace{1ex}+I\{D^{(a)}\wedge D^{(1-a)}>t, N_H^{(a)}(t)=N_H^{(1-a)}(t)=\mbox{ some } k, T_k^{(1-a)}<T_k^{(a)}\}. \] Likewise we can construct a first-event-assisted win ratio (FWR) by
replacing the \(T_k^{(a)}\) with the \(T_1^{(a)}\), or a naive win ratio (NWR) by removing the tie-breaking third term altogether (see Mao et al. (2022) for details). Nonetheless, it is recommended
that the LWR be used as the default, as it makes fuller use of the data and reduces to the SWR when the nonfatal event occurs at most once.
Under the LWR, a simple condition that implies the dominance of win probability in (1) is a joint stochastic order of the event times between the two groups: \[$$\tag{2} {P}(D^{(1)}>s, T^{(1)}_{1}>
t_1, T^{(1)}_{2}>t_2, \ldots)> {P}(D^{(0)}>s, T^{(0)}_{1}>t_1, T^{(0)}_{2}>t_2, \ldots),$$\] for all \(0\leq t_1\leq t_2\leq\cdots\leq s\leq\tau\). Expression (2) means that the treatment
stochastically delays all events, fatal and nonfatal, jointly as compared to the control. Hence, when (2) is true, the LWR test rejects \(H_0\) with probability tending to 1 as the sample size
increases to infinity.
The basic function to perform the win ratio tests is WRrec(). To use the function, the input data must be in the “long” format. Specifically, we need an ID variable containing the unique patient
identifiers, a time variable containing the event times, a status variable labeling the event type (status=2 for recurrent non-fatal event, =1 for death, and =0 for censoring), and, finally, a binary
trt variable with 1 indicating the treatment and 0 indicating the control. To perform an unstratified LWR test, use
For a stratified test, supply a vector of stratifying (categorical) variable through an additional strata= argument. To get test results from FWR and NWR as well, add the option naive=TRUE. Printing
the object obj gives us the \(p\)-values of the tests as well as some descriptive statistics.
Data description
To illustrate the win ratio tests, consider the Heart Failure: A Controlled Trial Investigating Outcomes of Exercise Training (HF-ACTION) trial. A randomized controlled trial, HF-ACTION was conducted
on a cohort of over two thousand heart failure patients recruited between 2003–2007 across the USA, Canada, and France (O’Connor et al., 2009). The study aimed to assess the effect of adding aerobic
exercise training to usual care on the patient’s composite endpoint of all-cause death and all-cause hospitalization. The primary analysis of the whole study population showed a moderate and
non-significant reduction in the risk of time to the first composite event (hazard ratio 0.93; \(p\)-value 0.13). Here we focus on a subgroup of non-ischemic patients with reduced cardio-pulmonary
exercise (CPX) test duration (i.e., \(\leq 9\) minutes before reporting of discomfort). There are scientific and empirical evidence suggesting that this particular sub-population may benefit more
from exercise training interventions than does the average heart failure patient.
The associated dataset hfaction_cpx9 is contained in the WR package and can be loaded by
#> patid time status trt_ab age60
#> 1 HFACT00001 7.2459016 2 0 1
#> 2 HFACT00001 12.5573770 0 0 1
#> 3 HFACT00002 0.7540984 2 0 1
#> 4 HFACT00002 4.2950820 2 0 1
#> 5 HFACT00002 4.7540984 2 0 1
#> 6 HFACT00002 45.9016393 0 0 1
The dataset is already in a format suitable for WRrec() (status= 2 for hospitalization and = 1 for death). The time variable is in units of months and trt=0 for usual care (control) and 1 for
exercise training. The age60 variable is an indicator of patient age being greater than or equal to 60 years and can potentially serve as a stratifying variable.
Win ratio tests on recurrent event and death
To perform the win ratio tests between exercise training and usual care stratified by age, use the code
## simplify the dataset name
## comparing exercise training to usual care by LWR, FWR, and NWR
## print the results
#> Call:
#> WRrec(ID = dat$patid, time = dat$time, status = dat$status, trt = dat$trt_ab,
#> strata = dat$age60, naive = TRUE)
#> N Rec. Event Death Med. Follow-up
#> Control 221 571 57 28.62295
#> Treatment 205 451 36 27.57377
#> Analysis of last-event-assisted WR (LWR; recommended), first-event-assisted WR (FWR), and naive WR (NWR):
#> Win prob Loss prob WR (95% CI)* p-value
#> LWR 50.4% 38.2% 1.32 (1.05, 1.66) 0.0189
#> FWR 50.4% 38.3% 1.32 (1.04, 1.66) 0.0202
#> NWR 47% 35% 1.34 (1.05, 1.72) 0.0193
#> -----
#> *Note: The scale of WR should be interpreted with caution as it depends on
#> censoring distribution without modeling assumptions.
We can see from the output above that 57 (25.8%) out of 221 patients died in usual care, with an average of \(571/221=2.6\) hospitalizations per patient; and that 36 (17.6%) out of 205 patients died
in exercise training with an average of \(451/205=2.2\) hospitalizations per patient. Clearly, those undergoing exercise training are doing much better in terms of both overall survival and recurrent
Following the descriptive statistics are the analysis results by the LWR, FWR, and NWR. Although estimates of overall win and loss probabilities, as well as the win ratio, are provided, their
magnitudes are generally dependent on the censoring distribution and should thus be interpreted with caution. On the other hand, the \(p\)-values are from valid tests of the null and alternative
hypotheses discussed in the earlier section. We can see that all three tests yield \(p\)-values less than the conventional threshold 0.05, suggesting that exercise training significantly reduces
mortality and recurrent hospitalization.
Comparison with standard win ratio
To compare with the SWR, we first create a dataset where only the first hospitalization is retained.
## Remove recurrent hospitalization ##
## sort dataset by patid and time
## retain only the first hospitalization
#> patid time status trt_ab age60
#> 1 HFACT00001 7.2459016 2 0 1
#> 2 HFACT00001 12.5573770 0 0 1
#> 3 HFACT00002 0.7540984 2 0 1
#> 6 HFACT00002 45.9016393 0 0 1
#> 7 HFACT00007 3.4754098 2 1 1
#> 11 HFACT00007 34.8852459 1 1 1
Then we perform the SWR test by applying the same procedure for the LWR to the reduced dataset (which in this case is equivalent to the SWR).
## Perform the standard win ratio test
## print the results
#> Call:
#> WRrec(ID = datHF$patid, time = datHF$time, status = datHF$status,
#> trt = datHF$trt_ab, strata = datHF$age60)
#> N Rec. Event Death Med. Follow-up
#> Control 221 170 57 28.62295
#> Treatment 205 145 36 27.57377
#> Analysis of last-event-assisted WR (LWR):
#> Win prob Loss prob WR (95% CI)* p-value
#> LWR 49.5% 39.1% 1.27 (1, 1.6) 0.0494
#> -----
#> *Note: The scale of WR should be interpreted with caution as it depends on
#> censoring distribution without modeling assumptions.
We can see that the test result is only borderline significant, possibly due to less efficient use of the recurrent-event data. | {"url":"https://cran.fhcrc.org/web/packages/WR/vignettes/WR_test_rec.html","timestamp":"2024-11-09T00:34:12Z","content_type":"text/html","content_length":"35020","record_id":"<urn:uuid:ab31f81d-6ca5-438a-a822-a5c28521b944>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00678.warc.gz"} |
Franklin math circle | Department of Mathematics
Professor Nathan Gibson organized a series of weekly Math Circle meetings for elementary school students from January through June 2023.
Franklin students grades 3-5 were invited to participate in an after school math circle. A math circle is a meeting where participants do math. Typically this consists of group work on interesting
problems that complements, but does not overlap with, traditional curricula. Examples include explorations in combinatorics (counting), probability, geometry, logic and modeling. It is closer to a
math club than a math class.
Math circles are for kids who like math, and kids who want to like math. Math circles aim to make math fun, interesting, accessible and inclusive. Math circles are not competitive. | {"url":"https://math.oregonstate.edu.prod.acquia.cosine.oregonstate.edu/mathematics-news-events/news-briefs/franklin-math-circle","timestamp":"2024-11-06T18:11:25Z","content_type":"text/html","content_length":"54414","record_id":"<urn:uuid:26460333-4a02-49b7-8789-cf91dcd2f536>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00677.warc.gz"} |
Statistical Methods
Statistical Methods of Detecting Copying on Multiple Choice Tests: A Review and an Application
Researchers consistently elicit surprise and dismay when they report on the pervasiveness of academic dishonesty. Cheating on examinations, and in particular the practice of copying answers from
other students, has spurred educators to devise statistical methods that can expose it. This paper surveys the main issues in the copy detection field, examining four methods in detail. I apply
these techniques to data taken from a university examination, and I compare the four with respect to ease of calculation, prevalence of use, and rates of sensitivity and specificity. I also give
an overview of other copy detection methods, discussing key issues to consider, limitations that apply, and factors that inhibit the adoption of these techniques. An appendix briefly touches on
the detection of cheating by impersonation and by improper erasures.
Full text available for $5.95. Follow the Yellow Brick PayPal link below. Your article should arrive within two days.
Word Cloud derived from frequencies of word use in the paper.
copyright 2007-2012 by roland b. stark. | {"url":"http://yellowbrickstats.com/cheatingabstract.htm","timestamp":"2024-11-09T14:15:40Z","content_type":"text/html","content_length":"12386","record_id":"<urn:uuid:b290e573-f86b-4a14-be30-88ebaa45700c>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00397.warc.gz"} |
College Physics chapters 1-17
3 Two-dimensional Kinematics
20 3.5 Addition of Velocities
• Apply principles of vector addition to determine relative velocity.
• Explain the significance of the observer in the measurement of velocity.
Relative Velocity
If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 1. The boat does not move
in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes
see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 2. The plane is moving straight ahead relative to the air, but the movement of the air mass relative
to the ground carries it sideways.
Figure 1. A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity
relative to the river plus the velocity of the river relative to the shore.
Figure 2. An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the
direction of its total velocity (solid arrow).
In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object
relative to the observer is the sum of these velocity vectors, as indicated in Figure 1 and Figure 2. These situations are only two of many in which it is useful to add velocities. In this module, we
first re-examine how to add velocities and then consider certain aspects of what relative velocity means.
How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Chapter 3.2 Vector Addition and Subtraction: Graphical Methods and
Chapter 3.3 Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is
simple—they add like ordinary numbers. For example, if a field hockey player is moving at[latex]\boldsymbol{5\textbf{ m/s}}[/latex]straight toward the goal and drives the ball in the same direction
with a velocity of[latex]\boldsymbol{30\textbf{ m/s}}[/latex]relative to her body, then the velocity of the ball is[latex]\boldsymbol{35\textbf{ m/s}}[/latex]relative to the stationary, profusely
sweating goalkeeper standing in front of the goal.
In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships
between the magnitude and direction of velocity ([latex]\boldsymbol{v}[/latex]and[latex]\boldsymbol{\theta}[/latex]) and its components ([latex]\boldsymbol{v_x}[/latex]and[latex]\boldsymbol{v_y}[/
latex]) along the x– and y-axes of an appropriately chosen coordinate system:
Figure 3. The velocity, v, of an object traveling at an angle θ to the horizontal axis is the sum of component vectors v[x] and v[y].
These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known.
The last two are used to find the magnitude and direction of velocity when its components are known.
Fill a bathtub half-full of water. Take a toy boat or some other object that floats in water. Unplug the drain so water starts to drain. Try pushing the boat from one side of the tub to the other and
perpendicular to the flow of water. Which way do you need to push the boat so that it ends up immediately opposite? Compare the directions of the flow of water, heading of the boat, and actual
velocity of the boat.
Example 1: Adding Velocities: A Boat on a River
Figure 4. A boat attempts to travel straight across a river at a speed 0.75 m/s. The current in the river, however, flows at a speed of 1.20 m/s to the right. What is the total displacement of the
boat relative to the shore?
Refer to Figure 4, which shows a boat trying to go straight across the river. Let us calculate the magnitude and direction of the boat’s velocity relative to an observer on the shore,[latex]\
boldsymbol{v_{tot}}.[/latex]The velocity of the boat,[latex]\boldsymbol{v_{boat}},[/latex]is 0.75 m/s in the[latex]\boldsymbol{y}[/latex]-direction relative to the river and the velocity of the
river,[latex]\boldsymbol{v_{river}},[/latex]is 1.20 m/s to the right.
We start by choosing a coordinate system with its xx-axis parallel to the velocity of the river, as shown in Figure 4. Because the boat is directed straight toward the other shore, its velocity
relative to the water is parallel to the[latex]\boldsymbol{y}[/latex]-axis and perpendicular to the velocity of the river. Thus, we can add the two velocities by using the equations[latex]\boldsymbol
The magnitude of the total velocity is
[latex]\boldsymbol{v_x=v_{river}=1.20\textbf{ m/s}}[/latex]
[latex]\boldsymbol{v_y=v_{boat}=0.750\textbf{ m/s}}.[/latex]
[latex]\boldsymbol{v_{tot}=\sqrt{(1.20\textbf{ m/s})^2+(0.750\textbf{ m/s})^2}}[/latex]
[latex]\boldsymbol{v_{tot}=1.42\textbf{ m/s}}.[/latex]
The direction of the total velocity[latex]\boldsymbol{\theta}[/latex]is given by:
This equation gives
Both the magnitude[latex]\boldsymbol{v}[/latex]and the direction[latex]\boldsymbol{\theta}[/latex]of the total velocity are consistent with Figure 4. Note that because the velocity of the river is
large compared with the velocity of the boat, it is swept rapidly downstream. This result is evidenced by the small angle (only[latex]\boldsymbol{32.0^0}[/latex]) the total velocity has relative to
the riverbank.
Example 2: Calculating Velocity: Wind Velocity Causes an Airplane to Drift
Calculate the wind velocity for the situation shown in Figure 5. The plane is known to be moving at 45.0 m/s due north relative to the air mass, while its velocity relative to the ground (its total
velocity) is 38.0 m/s in a direction[latex]\boldsymbol{20.0^0}[/latex]west of north.
Figure 5. An airplane is known to be heading north at 45.0 m/s, though its velocity relative to the ground is 38.0 m/s at an angle west of north. What is the speed and direction of the wind?
In this problem, somewhat different from the previous example, we know the total velocity[latex]\boldsymbol{v_{tot}}[/latex]and that it is the sum of two other velocities,[latex]\boldsymbol{v_w}[/
latex](the wind) and[latex]\boldsymbol{v_p}[/latex](the plane relative to the air mass). The quantity[latex]\boldsymbol{v_p}[/latex]is known, and we are asked to find[latex]\boldsymbol{v_w}.[/latex]
None of the velocities are perpendicular, but it is possible to find their components along a common set of perpendicular axes. If we can find the components of[latex]\boldsymbol{v_w},[/latex]then we
can combine them to solve for its magnitude and direction. As shown in Figure 5, we choose a coordinate system with its x-axis due east and its y-axis due north (parallel to[latex]\boldsymbol{v_p}[/
latex]). (You may wish to look back at the discussion of the addition of vectors using perpendicular components in Chapter 3.3 Vector Addition and Subtraction: Analytical Methods.)
Because[latex]\boldsymbol{v_{tot}}[/latex]is the vector sum of the[latex]\boldsymbol{v_w}[/latex]and[latex]\boldsymbol{v_p},[/latex]its x– and y-components are the sums of the x– and y-components of
the wind and plane velocities. Note that the plane only has vertical component of velocity so[latex]\boldsymbol{v_{px}=0}[/latex]and[latex]\boldsymbol{v_{py}=v_p}.[/latex]That is,
We can use the first of these two equations to find[latex]\boldsymbol{v_{wx}}:[/latex]
[latex]\boldsymbol{v_{wy}=v_{totx}=v_{tot}\textbf{cos }110^0}.[/latex]
Because[latex]\boldsymbol{v_{tot}=38.0\textbf{ m/s}}[/latex]and[latex]\boldsymbol{\textbf{cos }110^0=-0.342}[/latex]we have
[latex]\boldsymbol{v_{wy}=(38.0\textbf{ m/s})(-0.342)=-13\textbf{ m/s}.}[/latex]
The minus sign indicates motion west which is consistent with the diagram.
Now, to find[latex]\boldsymbol{v_{wy}}[/latex]we note that
Here[latex]\boldsymbol{v_{toty}=v_{tot}\textbf{sin }110^0};[/latex]thus,
[latex]\boldsymbol{v_{wy}=(38.0\textbf{ m/s})(0.940)-45.0\textbf{ m/s}=-9.29\textbf{ m/s}.}[/latex]
This minus sign indicates motion south which is consistent with the diagram.
Now that the perpendicular components of the wind velocity[latex]\boldsymbol{v_{wx}}[/latex]and[latex]\boldsymbol{v_{wy}}[/latex]are known, we can find the magnitude and direction of[latex]\
boldsymbol{v_w}.[/latex]First, the magnitude is
[latex]\boldsymbol{=(-13.0\textbf{ m/s})^2+(-9.29\textbf{ m/s})^2}\:\:\:\:\:\:[/latex]
so that
[latex]\boldsymbol{v_w=16.0\textbf{ m/s}.}[/latex]
The direction is:
The wind’s speed and direction are consistent with the significant effect the wind has on the total velocity of the plane, as seen in Figure 5. Because the plane is fighting a strong combination of
crosswind and head-wind, it ends up with a total velocity significantly less than its velocity relative to the air mass as well as heading in a different direction.
Note that in both of the last two examples, we were able to make the mathematics easier by choosing a coordinate system with one axis parallel to one of the velocities. We will repeatedly find that
choosing an appropriate coordinate system makes problem solving easier. For example, in projectile motion we always use a coordinate system with one axis parallel to gravity.
Relative Velocities and Classical Relativity
When adding velocities, we have been careful to specify that the velocity is relative to some reference frame. These velocities are called relative velocities. For example, the velocity of an
airplane relative to an air mass is different from its velocity relative to the ground. Both are quite different from the velocity of an airplane relative to its passengers (which should be close to
zero). Relative velocities are one aspect of relativity, which is defined to be the study of how different observers moving relative to each other measure the same phenomenon.
Nearly everyone has heard of relativity and immediately associates it with Albert Einstein (1879–1955), the greatest physicist of the 20th century. Einstein revolutionized our view of nature with his
modern theory of relativity, which we shall study in later chapters. The relative velocities in this section are actually aspects of classical relativity, first discussed correctly by Galileo and
Isaac Newton. Classical relativity is limited to situations where speeds are less than about 1% of the speed of light—that is, less than[latex]\boldsymbol{3,000\textbf{ km/s}}.[/latex]Most things we
encounter in daily life move slower than this speed.
Let us consider an example of what two different observers see in a situation analyzed long ago by Galileo. Suppose a sailor at the top of a mast on a moving ship drops his binoculars. Where will it
hit the deck? Will it hit at the base of the mast, or will it hit behind the mast because the ship is moving forward? The answer is that if air resistance is negligible, the binoculars will hit at
the base of the mast at a point directly below its point of release. Now let us consider what two different observers see when the binoculars drop. One observer is on the ship and the other on shore.
The binoculars have no horizontal velocity relative to the observer on the ship, and so he sees them fall straight down the mast. (See Figure 6.) To the observer on shore, the binoculars and the ship
have the same horizontal velocity, so both move the same distance forward while the binoculars are falling. This observer sees the curved path shown in Figure 6. Although the paths look different to
the different observers, each sees the same result—the binoculars hit at the base of the mast and not behind it. To get the correct description, it is crucial to correctly specify the velocities
relative to the observer.
Figure 6. Classical relativity. The same motion as viewed by two different observers. An observer on the moving ship sees the binoculars dropped from the top of its mast fall straight down. An
observer on shore sees the binoculars take the curved path, moving forward with the ship. Both observers see the binoculars strike the deck at the base of the mast. The initial horizontal velocity is
different relative to the two observers. (The ship is shown moving rather fast to emphasize the effect.)
Example 3: Calculating Relative Velocity: An Airline Passenger Drops a Coin
An airline passenger drops a coin while the plane is moving at 260 m/s. What is the velocity of the coin when it strikes the floor 1.50 m below its point of release: (a) Measured relative to the
plane? (b) Measured relative to the Earth?
Figure 7. The motion of a coin dropped inside an airplane as viewed by two different observers. (a) An observer in the plane sees the coin fall straight down. (b) An observer on the ground sees the
coin move almost horizontally.
Both problems can be solved with the techniques for falling objects and projectiles. In part (a), the initial velocity of the coin is zero relative to the plane, so the motion is that of a falling
object (one-dimensional). In part (b), the initial velocity is 260 m/s horizontal relative to the Earth and gravity is vertical, so this motion is a projectile motion. In both parts, it is best to
use a coordinate system with vertical and horizontal axes.
Solution for (a)
Using the given information, we note that the initial velocity and position are zero, and the final position is 1.50 m. The final velocity can be found using the equation:
Substituting known values into the equation, we get
[latex]\boldsymbol{v_y^2=0^2-2(9.80\textbf{ m/s}^2)(-1.50\textbf{ m}-0\textbf{ m})=29.4\textbf{ m}^2/\textbf{ s}^2}[/latex]
[latex]\boldsymbol{v_y=-5.42\textbf{ m/s}.}[/latex]
We know that the square root of 29.4 has two roots: 5.42 and -5.42. We choose the negative root because we know that the velocity is directed downwards, and we have defined the positive direction to
be upwards. There is no initial horizontal velocity relative to the plane and no horizontal acceleration, and so the motion is straight down relative to the plane.
Solution for (b)
Because the initial vertical velocity is zero relative to the ground and vertical motion is independent of horizontal motion, the final vertical velocity for the coin relative to the ground is[latex]
\boldsymbol{v_y=-5.42\textbf{ m/s}},[/latex]the same as found in part (a). In contrast to part (a), there now is a horizontal component of the velocity. However, since there is no horizontal
acceleration, the initial and final horizontal velocities are the same and[latex]\boldsymbol{v_x=260\textbf{ m/s}}.[/latex]The x– and y-components of velocity can be combined to find the magnitude of
the final velocity:
[latex]\boldsymbol{v=\sqrt{(260\textbf{ m/s})^2+(-5.42\textbf{ m/s})^2}}[/latex]
[latex]\boldsymbol{v=260.06\textbf{ m/s}.}[/latex]
The direction is given by:
so that
In part (a), the final velocity relative to the plane is the same as it would be if the coin were dropped from rest on the Earth and fell 1.50 m. This result fits our experience; objects in a plane
fall the same way when the plane is flying horizontally as when it is at rest on the ground. This result is also true in moving cars. In part (b), an observer on the ground sees a much different
motion for the coin. The plane is moving so fast horizontally to begin with that its final velocity is barely greater than the initial velocity. Once again, we see that in two dimensions, vectors do
not add like ordinary numbers—the final velocity v in part (b) is not[latex]\boldsymbol{(260 - 5.42)\textbf{ m/s}};[/latex]rather, it is[latex]\boldsymbol{260.06\textbf{ m/s}}.[/latex]The velocity’s
magnitude had to be calculated to five digits to see any difference from that of the airplane. The motions as seen by different observers (one in the plane and one on the ground) in this example are
analogous to those discussed for the binoculars dropped from the mast of a moving ship, except that the velocity of the plane is much larger, so that the two observers see very different paths. (See
Figure 7.) In addition, both observers see the coin fall 1.50 m vertically, but the one on the ground also sees it move forward 144 m (this calculation is left for the reader). Thus, one observer
sees a vertical path, the other a nearly horizontal path.
Because Einstein was able to clearly define how measurements are made (some involve light) and because the speed of light is the same for all observers, the outcomes are spectacularly unexpected.
Time varies with observer, energy is stored as increased mass, and more surprises await.
Try the new “Ladybug Motion 2D” simulation for the latest updated version. Learn about position, velocity, and acceleration vectors. Move the ball with the mouse or let the simulation move the ball
in four types of motion (2 types of linear, simple harmonic, circle).
Figure 8. Motion in 2D
• Velocities in two dimensions are added using the same analytical vector techniques, which are rewritten as
• Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies dramatically with reference frame.
• Relativity is the study of how different observers measure the same phenomenon, particularly when the observers move relative to one another. Classical relativity is limited to situations where
speed is less than about 1% of the speed of light (3000 km/s).
Conceptual Questions
1: What frame or frames of reference do you instinctively use when driving a car? When flying in a commercial jet airplane?
2: A basketball player dribbling down the court usually keeps his eyes fixed on the players around him. He is moving fast. Why doesn’t he need to keep his eyes on the ball?
3: If someone is riding in the back of a pickup truck and throws a softball straight backward, is it possible for the ball to fall straight down as viewed by a person standing at the side of the
road? Under what condition would this occur? How would the motion of the ball appear to the person who threw it?
4: The hat of a jogger running at constant velocity falls off the back of his head. Draw a sketch showing the path of the hat in the jogger’s frame of reference. Draw its path as viewed by a
stationary observer.
5: A clod of dirt falls from the bed of a moving truck. It strikes the ground directly below the end of the truck. What is the direction of its velocity relative to the truck just before it hits? Is
this the same as the direction of its velocity relative to ground just before it hits? Explain your answers.
Problems & Exercises
1: Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979. (a) He flew for 169 min at an average velocity of 3.53 m/s in a
direction[latex]\boldsymbol{45^0}[/latex]south of east. What was his total displacement? (b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion
relative to the Earth. What was his average velocity relative to the air? (c) What was his total displacement relative to the air mass?
2: A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) If it takes the bird 20.0 min to travel 6.00 km relative to the Earth, what is the velocity of the wind? (b) If the bird turns
around and flies with the wind, how long will he take to return 6.00 km? (c) Discuss how the wind affects the total round-trip time compared to what it would be with no wind.
3: Near the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of 3.50 m/s, and the second a velocity of 4.20 m/s. (a) What is the
velocity of the second runner relative to the first? (b) If the front runner is 250 m from the finish line, who will win the race, assuming they run at constant velocity? (c) What distance ahead will
the winner be when she crosses the finish line?
4: Verify that the coin dropped by the airline passenger in the Example 3 travels 144 m horizontally while falling 1.50 m in the frame of reference of the Earth.
5: A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. The ball is thrown at an angle of[latex]\boldsymbol{25.0^0}[/
latex]relative to the ground and is caught at the same height as it is released. What is the initial velocity of the ball relative to the quarterback ?
6: A ship sets sail from Rotterdam, The Netherlands, heading due north at 7.00 m/s relative to the water. The local ocean current is 1.50 m/s in a direction[latex]\boldsymbol{40.0^0}[/latex]north of
east. What is the velocity of the ship relative to the Earth?
7: (a) A jet airplane flying from Darwin, Australia, has an air speed of 260 m/s in a direction[latex]\boldsymbol{5.0^0}[/latex]south of west. It is in the jet stream, which is blowing at 35.0 m/s in
a direction[latex]\boldsymbol{15^0}[/latex]south of east. What is the velocity of the airplane relative to the Earth? (b) Discuss whether your answers are consistent with your expectations for the
effect of the wind on the plane’s path.
8: (a) In what direction would the ship in Exercise 6 have to travel in order to have a velocity straight north relative to the Earth, assuming its speed relative to the water remains[latex]\
boldsymbol{7.00\textbf{ m/s}}?[/latex](b) What would its speed be relative to the Earth?
9: (a) Another airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction[latex]\boldsymbol{20^0}[/latex]south of east (as in Exercise 7). Its direction of motion relative to the
Earth is[latex]\boldsymbol{45.0^0}[/latex]south of west, while its direction of travel relative to the air is[latex]\boldsymbol{5.00^0}[/latex]south of west. What is the airplane’s speed relative to
the air mass? (b) What is the airplane’s speed relative to the Earth?
10: A sandal is dropped from the top of a 15.0-m-high mast on a ship moving at 1.75 m/s due south. Calculate the velocity of the sandal when it hits the deck of the ship: (a) relative to the ship and
(b) relative to a stationary observer on shore. (c) Discuss how the answers give a consistent result for the position at which the sandal hits the deck.
11: The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a velocity of 2.20 m/s in a direction[latex]\boldsymbol{30.0^0}[/latex]
east of north relative to the Earth. It encounters a wind that has a velocity of 4.50 m/s in a direction of[latex]\boldsymbol{50.0^0}[/latex]south of west relative to the Earth. What is the velocity
of the wind relative to the water?
12: The great astronomer Edwin Hubble discovered that all distant galaxies are receding from our Milky Way Galaxy with velocities proportional to their distances. It appears to an observer on the
Earth that we are at the center of an expanding universe. Figure 9 illustrates this for five galaxies lying along a straight line, with the Milky Way Galaxy at the center. Using the data from the
figure, calculate the velocities: (a) relative to galaxy 2 and (b) relative to galaxy 5. The results mean that observers on all galaxies will see themselves at the center of the expanding universe,
and they would likely be aware of relative velocities, concluding that it is not possible to locate the center of expansion with the given information.
Figure 9. Five galaxies on a straight line, showing their distances and velocities relative to the Milky Way (MW) Galaxy. The distances are in millions of light years (Mly), where a light year is the
distance light travels in one year. The velocities are nearly proportional to the distances. The sizes of the galaxies are greatly exaggerated; an average galaxy is about 0.1 Mly across.
13: (a) Use the distance and velocity data in Figure 9 to find the rate of expansion as a function of distance.
(b) If you extrapolate back in time, how long ago would all of the galaxies have been at approximately the same position? The two parts of this problem give you some idea of how the Hubble constant
for universal expansion and the time back to the Big Bang are determined, respectively.
14: An athlete crosses a 25-m-wide river by swimming perpendicular to the water current at a speed of 0.5 m/s relative to the water. He reaches the opposite side at a distance 40 m downstream from
his starting point. How fast is the water in the river flowing with respect to the ground? What is the speed of the swimmer with respect to a friend at rest on the ground?
15: A ship sailing in the Gulf Stream is heading[latex]\boldsymbol{25.0^0}[/latex]west of north at a speed of 4.00 m/s relative to the water. Its velocity relative to the Earth is[latex]\boldsymbol
{4.80\textbf{ m/s}\:5.00^0}[/latex]west of north. What is the velocity of the Gulf Stream? (The velocity obtained is typical for the Gulf Stream a few hundred kilometers off the east coast of the
United States.)
16: An ice hockey player is moving at 8.00 m/s when he hits the puck toward the goal. The speed of the puck relative to the player is 29.0 m/s. The line between the center of the goal and the player
makes a[latex]\boldsymbol{90.0^0}[/latex]angle relative to his path as shown in Figure 10. What angle must the puck’s velocity make relative to the player (in his frame of reference) to hit the
center of the goal?
Figure 10. An ice hockey player moving across the rink must shoot backward to give the puck a velocity toward the goal.
15: Unreasonable Results Suppose you wish to shoot supplies straight up to astronauts in an orbit 36,000 km above the surface of the Earth. (a) At what velocity must the supplies be launched? (b)
What is unreasonable about this velocity? (c) Is there a problem with the relative velocity between the supplies and the astronauts when the supplies reach their maximum height? (d) Is the premise
unreasonable or is the available equation inapplicable? Explain your answer.
16: Unreasonable Results A commercial airplane has an air speed of[latex]\boldsymbol{280\textbf{ m/s}}[/latex]due east and flies with a strong tailwind. It travels 3000 km in a direction[latex]\
boldsymbol{5^0}[/latex]south of east in 1.50 h. (a) What was the velocity of the plane relative to the ground? (b) Calculate the magnitude and direction of the tailwind’s velocity. (c) What is
unreasonable about both of these velocities? (d) Which premise is unreasonable?
17: Construct Your Own Problem Consider an airplane headed for a runway in a cross wind. Construct a problem in which you calculate the angle the airplane must fly relative to the air mass in order
to have a velocity parallel to the runway. Among the things to consider are the direction of the runway, the wind speed and direction (its velocity) and the speed of the plane relative to the air
mass. Also calculate the speed of the airplane relative to the ground. Discuss any last minute maneuvers the pilot might have to perform in order for the plane to land with its wheels pointing
straight down the runway.
classical relativity
the study of relative velocities in situations where speeds are less than about 1% of the speed of light—that is, less than 3000 km/s
relative velocity
the velocity of an object as observed from a particular reference frame
the study of how different observers moving relative to each other measure the same phenomenon
speed in a given direction
vector addition
the rules that apply to adding vectors together
Problems & Exercises:
(a)[latex]\boldsymbol{35.8\textbf{ km}}[/latex]south of east
(b)[latex]\boldsymbol{5.53\textbf{ m/s}}[/latex]south of east
(c)[latex]\boldsymbol{56.1\textbf{ km}}[/latex]south of east
(a) 0.70 m/s faster
(b) Second runner wins
(c) 4.17 m
[latex]\boldsymbol{17.0\textbf{ m/s}},\:\boldsymbol{22.1^0}[/latex]
(a)[latex]\boldsymbol{230\textbf{ m/s}},\:\boldsymbol{8.0^0}[/latex]south of west
(b) The wind should make the plane travel slower and more to the south, which is what was calculated
(a) 63.5 m/s
(b) 29.6 m/s
[latex]\boldsymbol{6.68\textbf{ m/s}},\:\boldsymbol{53.3^0}[/latex]south of west
(a)[latex]\boldsymbol{H_{\textbf{average}}=14.9\frac{\textbf{ km/s}}{\textbf{Mly}}}[/latex]
(b) 20.2 billion years
[latex]\boldsymbol{1.72\textbf{ m/s}},\:\boldsymbol{42.3^0}[/latex]north of east | {"url":"https://pressbooks-dev.oer.hawaii.edu/collegephysics/chapter/3-5-addition-of-velocities/","timestamp":"2024-11-11T16:52:27Z","content_type":"text/html","content_length":"200886","record_id":"<urn:uuid:9e37d396-31e3-4c37-a9b9-1f5e952de06d>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00227.warc.gz"} |
solving implicit equation
solving implicit equation
Hi experts!
I want to resolve the equation: $A.sin(x)=B.(C-x)$
where A, C and B are constants, $0\leq x\leq \pi $, $A$ and $B$ are fixed, and $0\leq C\leq \pi $.
I want to obtain the analytic solution, $x=f(C)$, (if exist) for different values of $C$ and if only implicit solution exist, obtain the curve and and table with the values of:
$ x$ vs. $C$
How can I do that?
Waiting for your answer.
Thanks a lot!
1 Answer
Sort by ยป oldest newest most voted
One option, and this might be suboptimal, is to replace sin(x) with its taylor series.
sin(x) = x - (1/6)x^3 + (1/120)x^5 - (1/5040)x^7 + (1/362880)x^9 - O(x^11)
we would change
Asin(x) = BC - B*x
into instead
Ax - (A/6)x^3 + (A/120)x^5 - (A/5040)x^7 + (A/362880)x^9 - O(A x^11) = BC - Bx
giving you then
-BC + (A+B)x - (A/6)x^3 + (A/120)x^5 - (A/5040)x^7 + (A/362880)x^9 - O(A x^11) = 0
or if you want a few more terms
-BC + (A+B)x - (A/3!)x^3 + (A/5!)x^5 - (A/7!)x^7 + (A/9!)x^9 - (A/11!)x^11 + (A/13!)x^13 + O(A x^15) = 0
This does reveal that varying C will only affect the answer by a constant.
If A and B were fixed constants, known in advance, then you could use "for" and "find_root" to find a value of x for a long sequence of C values, forming a table of values.
For example:
edit flag offensive delete link more
Thanks. It can't be done wothout using Taylor expansion?
mresimulator ( 2014-07-21 00:36:54 +0100 )edit
I'm guessing that you need either the Taylor expansion or some competitor of it, like the Chebychev polynomials. The main issue is that you have both sin x and a linear polynomial in x. Therefore, I
really can't see how you would "solve for x." I'm about 98% sure but I don't want to say 100%, because mathematics is always full of surprises.
Gregory Bard ( 2014-07-21 19:15:59 +0100 )edit | {"url":"https://ask.sagemath.org/question/23494/solving-implicit-equation/","timestamp":"2024-11-07T12:46:07Z","content_type":"application/xhtml+xml","content_length":"57116","record_id":"<urn:uuid:0829ced6-faff-4a48-9fce-d3bc4c2dc795>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00150.warc.gz"} |
Code - Fujin - Scaling
We test the scaling performance of our solver. In particular, we consider the time evolution of a Taylor-Green vortex in a three-periodic domain of size \(2\pi L\). The velocity field is initialized
\(u = U \sin \left( x/L \right) \cos \left( y/L \right) \cos \left( z/L \right),\)
\(v = -U \cos \left( x/L \right) \sin \left( y/L \right) \cos \left( z/L \right),\)
The Reynolds number \(Re = \rho U L /\mu\) is fixed equal to \(1600\).
Strong scaling of the numerical method in a domain with 1024 grid cells per side.
Blue: Beskow - KTH (Sweden)
Red: Oakbridge-CX - UTokyo (Japan)
Magenta: Oakforest-PACS - JCAHPC (Japan)
Brown: SQUID - Handai (Japan)
Green: Deigo - OIST (Japan)
Strong scaling of the numerical method in a domain with 1024 grid cells per side.
Gold: Fugaku - RIKEN (Japan)
The strong scaling test evaluates the speedup for a fixed problem size with respect to the number of processors. For this test, we consider a numerical domain discretised with \(1024\) grid cell per
side. The code shows very good scaling performance up to about \(10^4\)cores, with an almost linear decay of the computational time with the number of cores used.
Weak scaling of the numerical method in a domain with 8M grid cells per sub-domain.
Blue: Beskow - KTH (Sweden)
Red: Oakbridge-CX - UTokyo (Japan)
Magenta: Oakforest-PACS - JCAHPC (Japan)
Brown: SQUID - Handai (Japan)
Green: Deigo - OIST (Japan)
The weak scaling test evaluates the speedup for a scaled problem size with respect to the number of processors. For this test, we consider a numerical domain discretised with around \(8M\) grid cells
per computational sub-domain. The code shows a good scaling with only a small decay of performance from \(N=256\) to \(4096\) cores, suggesting that strong scaling timings can scale up to much larger
problem sizes. | {"url":"https://groups.oist.jp/cffu/code-scaling","timestamp":"2024-11-10T16:15:31Z","content_type":"text/html","content_length":"52913","record_id":"<urn:uuid:e1805cc0-3f20-45f8-b926-f345761a4681>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00212.warc.gz"} |
Converting 160 cm to Feet and Inches
To find 160 cm to feet, use the factor 30.48. Split 160 by 30.48, equaling about 5.249 feet. Take the full number 5 for feet, and the bit left, 0.249, for inches. Convert that fraction by
multiplying by 12: 0.249 x 12 = 2.991 inches. So 160 cm equals 5 feet and 2.991 inches, or 5 feet 3 inches, rounding the inches part.
What is the Conversion rate from Centimeters to Feet?
Here’s a simple way to convert centimeters to feet. First, take the number of centimeters. Divide that by 30.48, which is the centimeters in one foot. As an example, 100 centimeters becomes
100/30.48 = 3.28 feet. This conversion helps switch between metric and Imperial measurement systems. It matters for construction, engineering, daily tasks – wherever you need both units.
You may also like: Create Captivating Blogs
How to Convert Height from Centimeters to Feet?
Converting height from centimeters to feet is easy. Here’s how:
• First, divide the centimeters by 30.48. That’s the height in feet.
• Second, if you need feet and inches, find the whole feet from step one. Multiply the remaining decimal by 12 for the inches.
Let’s do an example.
Say the height is 170 cm. Divide by 30.48: 170 ÷ 30.48 = 5.57 ft. For feet and inches, the whole feet are 5 (that’s the 5 from 5.57). For inches, take 0.57 (the decimal part) and multiply by 12:
0.57 x 12 = 6.84. So 170 cm = 5′ 6.84″.
Or, simply use an online converter like SplashLearn’s. It’ll convert centimeters to feet (and vice versa) instantly.
How much is 160 cm to feet and Inches?
Moving from centimeters, we want feet and inches. One foot equals 30.48 cm. An inch equals 2.54 cm.
• First, change 160 cm to feet. 160 cm divided by 30.48 cm/ft. That’s 5.248756601732744 feet.
• Now, turn that fraction into inches. Take 5.248756601732744 ft, multiply by 12 in/ft. Then find the remainder after dividing by 12 in/ft. It’s 4.998756601732744 in.
So, 160 cm converts to 5 feet and around 5 inches.
How many Inches are in 160cm?
160 cm equals approximately 62.992 inches.
• To find this, divide 160 by 2.54. That’s because one inch equals 2.54 cm. Converting measurements between units like this is necessary in many real situations.
• If metric units are given but inch measurements are needed, converting is essential. This happens often when companies from different countries share data or details.
• For instance, fabric lengths or dimensions might be centimeters somewhere using the metric system. But to buy from a US seller using inches, converting those numbers is required.
• Similarly, a person’s height or an object’s size could need converting. Engineering, construction, fashion – any field needing exact dimensions benefits from converting units properly.
• The math is basic, yet crucial for clear communication and accuracy across systems. Though simple, carefully converting units prevents errors and misunderstandings from differing measurement
How to convert cm to feet and inches?
Changing lengths from centimeters to feet and inches takes a few simple steps.
• To start, split the number of centimeters by thirty point four eight. That number becomes the feet part.
• Next, get the remaining decimal. Multiply that by twelve to find the inches section. Round the inches to reach a full number.
• Combine both feet and inches values — now you’ve got the length in feet plus inches. Let’s see how this method works!
• Say we want to switch one hundred seventy centimeters into feet and inches. First, divide that by thirty point four eight: one seventy divided by thirty point four eight equals five point
five seven seven four two eight.
• Five is the amount of feet (whole number part). Next, take the decimal remainder (zero point five seven seven four two eight) and multiply twelve: zero point five seven seven four two
eight times twelve equals six point nine two nine one three six.
• Rounded to the closest whole number, that’s seven inches. Put it together: five feet and seven inches! One seventy centimeters equals five feet, seven inches. Easy conversions!
You may also read: How much is 160 cm to feet and inches? | {"url":"https://theworldstack.com/160cm-to-feet/","timestamp":"2024-11-14T21:31:32Z","content_type":"text/html","content_length":"99724","record_id":"<urn:uuid:49f5eb75-f63b-4a8d-b428-dc5327afda4e>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00396.warc.gz"} |
Functions to desugar Template Haskell
Version on this page: 1.9
LTS Haskell 22.39: 1.15@rev:1
Stackage Nightly 2024-11-02: 1.16@rev:1
Latest on Hackage: 1.17
This version can be pinned in stack with:th-desugar-1.9@sha256:ee54b5b0702b7817636350f1b0c0f48db1bd622737a81b2250c5bdf9ea745654,3228
Module documentation for 1.9
th-desugar Package
This package provides the Language.Haskell.TH.Desugar module, which desugars Template Haskell’s rich encoding of Haskell syntax into a simpler encoding. This desugaring discards surface syntax
information (such as the use of infix operators) but retains the original meaning of the TH code. The intended use of this package is as a preprocessor for more advanced code manipulation tools. Note
that the input to any of the ds... functions should be produced from a TH quote, using the syntax [| ... |]. If the input to these functions is a hand-coded TH syntax tree, the results may be
unpredictable. In particular, it is likely that promoted datatypes will not work as expected.
One explicit goal of this package is to reduce the burden of supporting multiple GHC / TH versions. Thus, the desugared language is the same across all GHC versions, and any inconsistencies are
handled internally.
The package was designed for use with the singletons package, so some design decisions are based on that use case, when more than one design choice was possible.
I will try to keep this package up-to-date with respect to changes in GHC.
th-desugar release notes
Version 1.9
• Suppose GHC 8.6.
• Add support for DerivingVia. Correspondingly, there is now a DDerivStrategy data type.
• Add support for QuantifiedConstraints. Correspondingly, there is now a DForallPr constructor in DPred to represent quantified constraint types.
• Remove the DStarT constructor of DType in favor of DConT ''Type. Two utility functions have been added to Language.Haskell.TH.Desugar to ease this transition:
□ isTypeKindName: returns True if the argument Name is that of Type or ★ (or *, to support older GHCs).
□ typeKindName: the name of Type (on GHC 8.0 or later) or * (on older GHCs).
• th-desugar now desugars all data types to GADT syntax. The most significant API-facing changes resulting from this new design are:
□ The DDataD, DDataFamilyD, and DDataFamInstD constructors of DDec now have Maybe DKind fields that either have Just an explicit return kind (e.g., the k -> Type -> Type in data Foo :: k ->
Type -> Type) or Nothing (if lacking an explicit return kind).
□ The DCon constructor previously had a field of type Maybe DType, since there was a possibility it could be a GADT (with an explicit return type) or non-GADT (without an explicit return type)
constructor. Since all data types are desugared to GADTs now, this field has been changed to be simply a DType.
□ The type signature of dsCon was previously:
dsCon :: DsMonad q => Con -> q [DCon]
However, desugaring constructors now needs more information than before, since GADT constructors have richer type signatures. Accordingly, the type of dsCon is now:
dsCon :: DsMonad q
=> [DTyVarBndr] -- ^ The universally quantified type variables
-- (used if desugaring a non-GADT constructor)
-> DType -- ^ The original data declaration's type
-- (used if desugaring a non-GADT constructor).
-> Con -> q [DCon]
The instance Desugar [Con] [DCon] has also been removed, as the previous implementation of desugar (concatMapM dsCon) no longer has enough information to work.
Some other utility functions have also been added as part of this change:
□ A conExistentialTvbs function has been introduced to determine the existentially quantified type variables of a DCon. Note that this function is not 100% accurate—refer to the documentation
for conExistentialTvbs for more information.
□ A mkExtraDKindBinders function has been introduced to turn a data type’s return kind into explicit, fresh type variable binders.
□ A toposortTyVarsOf function, which finds the free variables of a list of DTypes and returns them in a well scoped list that has been sorted in reverse topological order.
• th-desugar now desugars partial pattern matches in do-notation and list/monad comprehensions to the appropriate invocation of fail. (Previously, these were incorrectly desugared into case
expressions with incomplete patterns.)
• Add a mkDLamEFromDPats function for constructing a DLamE expression using a list of DPat arguments and a DExp body.
• Add an unravel function for decomposing a function type into its forall’d type variables, its context, its argument types, and its result type.
• Export a substTyVarBndrs function from Language.Haskell.TH.Desugar.Subst, which substitutes over type variable binders in a capture-avoiding fashion.
• getDataD, dataConNameToDataName, and dataConNameToCon from Language.Haskell.TH.Desugar.Reify now look up local declarations. As a result, the contexts in their type signatures have been
strengthened from Quasi to DsMonad.
• Export a dTyVarBndrToDType function which converts a DTyVarBndr to a DType, which preserves its kind.
• Previously, th-desugar would silently accept illegal uses of record construction with fields that did not belong to the constructor, such as Identity { notAField = "wat" }. This is now an error.
Version 1.8
• Support GHC 8.4.
• substTy now properly substitutes into kind signatures.
• Expose fvDType, which computes the free variables of a DType.
• Incorporate a DDeclaredInfix field into DNormalC to indicate if it is a constructor that was declared infix.
• Implement lookupValueNameWithLocals, lookupTypeNameWithLocals, mkDataNameWithLocals, and mkTypeNameWithLocals, counterparts to lookupValueName, lookupTypeName, mkDataName, and mkTypeName which
have access to local Template Haskell declarations.
• Implement reifyNameSpace to determine a Name’s NameSpace.
• Export reifyFixityWithLocals from Language.Haskell.TH.Desugar.
• Export matchTy (among other goodies) from new module Language.Haskell.TH.Subst. This function matches a type template against a target.
Version 1.7
• Support for TH’s support for TypeApplications, thanks to @RyanGlScott.
• Support for unboxed sums, thanks to @RyanGlScott.
• Support for COMPLETE pragmas.
• getRecordSelectors now requires a list of DCons as an argument. This makes it easier to return correct record selector bindings in the event that a record selector appears in multiple
constructors. (See goldfirere/singletons#180 for an example of where the old behavior of getRecordSelectors went wrong.)
• Better type family expansion (expanding an open type family with variables works now).
Version 1.6
• Work with GHC 8, with thanks to @christiaanb for getting this change going. This means that several core datatypes have changed: partcularly, we now have DTypeFamilyHead and fixities are now
reified separately from other things.
• DKind is merged with DType.
• Generic instances for everything.
Version 1.5.5
• Fix issue #34. This means that desugaring (twice) is idempotent over expressions, after the second time. That is, if you desugar an expression, sweeten it, desugar again, sweeten again, and then
desugar a third time, you get the same result as when you desugared the second time. (The extra round-trip is necessary there to make the output smaller in certain common cases.)
Version 1.5.4.1
• Fix issue #32, concerning reification of classes with default methods.
Version 1.5.4
Version 1.5.3
• More DsMonad instances, thanks to David Fox.
Version 1.5.2
Version 1.5.1
• Thanks to David Fox (@ddssff), sweetening now tries to use more of TH’s Type constructors.
• Also thanks to David Fox, depend usefully on the th-orphans package.
Version 1.5
• There is now a facility to register a list of Dec that internal reification should use when necessary. This avoids the user needing to break up their definition across different top-level
splices. See withLocalDeclarations. This has a side effect of changing the Quasi typeclass constraint on many functions to be the new DsMonad constraint. Happily, there are DsMonad instances for
Q and IO, the two normal inhabitants of Quasi.
• “Match flattening” is implemented! The functions scExp and scLetDec remove any nested pattern matches.
• More is now exported from Language.Haskell.TH.Desugar for ease of use.
• expand can now expand closed type families! It still requires that the type to expand contain no type variables.
• Support for standalone-deriving and default signatures in GHC 7.10. This means that there are now two new constructors for DDec.
• Support for static expressions, which are new in GHC 7.10.
Version 1.4.2
• expand functions now consider open type families, as long as the type to be expanded has no free variables.
Version 1.4.1
• Added Language.Haskell.TH.Desugar.Lift, which provides Lift instances for all of the th-desugar types, as well as several Template Haskell types.
• Added applyDExp and applyDType as convenience functions.
Version 1.4.0
• All Decs can now be desugared, to the new DDec type.
• Sweetening Decs that do not exist in GHC 7.6.3- works on a “best effort” basis: closed type families are sweetened to open ones, and role annotations are dropped.
• Infos can now be desugared. Desugaring takes into account GHC bug #8884, which meant that reifying poly-kinded type families in GHC 7.6.3- was subtly wrong.
• There is a new function flattenDValD which takes a binding like let (a,b) = foo and breaks it apart into separate assignments for a and b.
• There is a new Desugar class with methods desugar and sweeten. See the documentation in Language.Haskell.TH.Desugar.
• Variable names that are distinct in desugared code are now guaranteed to have distinct answers to nameBase.
• Added a new function getRecordSelectors that extracts types and definitions of record selectors from a datatype definition.
Version 1.3.1
• Update cabal file to include testing files in sdist.
Version 1.3.0
• Update to work with type Pred = Type in GHC 7.9. This changed the DPred type for all GHC versions, though.
Version 1.2.0
• Generalized interface to allow any member of the Qausi class, instead of just Q.
Version 1.1.1
• Made compatible with HEAD after change in role annotation syntax.
Version 1.1
• Added module Language.Haskell.TH.Desugar.Expand, which allows for expansion of type synonyms in desugared types.
• Added Show, Typeable, and Data instances to desugared types.
• Fixed bug where an as-pattern in a let statement was scoped incorrectly.
• Changed signature of dsPat to be more specific to as-patterns; this allowed for fixing the let scoping bug.
• Created new functions dsPatOverExp and dsPatsOverExp to allow for easy desugaring of patterns.
• Changed signature of dsLetDec to return a list of DLetDecs.
• Added dsLetDecs for convenience. Now, instead of using mapM dsLetDec, you should use dsLetDecs.
Version 1.0 | {"url":"https://www.stackage.org/lts-13.19/package/th-desugar-1.9","timestamp":"2024-11-02T06:34:33Z","content_type":"text/html","content_length":"30090","record_id":"<urn:uuid:d801d29a-a716-4e05-b505-4852631e49a1>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00372.warc.gz"} |
Rounding Whole Numbers
Learning Objectives
• Round whole numbers to a determined place value
The electronics retailer, Best Buy, had [latex]1,026[/latex] brick and mortar stores open in October of [latex]2016[/latex]. Depending on how this information will be used, it might be enough to say
that the company has approximately one thousand stores. The word approximately means that one thousand is not the exact count, but is close to the exact value.
In [latex]2017[/latex], the social network app, Facebook, reported its annual revenue as [latex]40.7[/latex] billion US dollars. This could mean they actually brought in [latex]$40,742,985,316[/
latex] or [latex]$40,654,872,131[/latex]. Sometimes the detail is needed, but sometimes just an approximate value is good enough. The real estate app, Zillow, recorded a profit of [latex]1.07[/latex]
billion US dollars; this is an approximate value. If you want to compare Facebook’s [latex]2017[/latex] revenue with Zillow’s [latex]2017[/latex] revenue, the precise dollars or even millions of
dollars are unnecessary.
The process of approximating a number is called rounding. Numbers are rounded to a specific place value depending on how much accuracy is needed. Identifying the number of stores owned by Best Buy as
approximately [latex]1[/latex] thousand means we rounded to the thousands place. Reporting the annual revenue of Facebook as [latex]40.7[/latex] billion US dollars means we rounded to the hundred
millions place. Often the place value to which we round depends on how we will need to use the number.
Using a number line can help us visualize and understand the rounding process. Look at the number line below.
Suppose we want to round the number [latex]76[/latex] to the nearest ten. Is [latex]76[/latex] closer to [latex]70[/latex] ([latex]7[/latex] tens) or [latex]80[/latex] ([latex]8[/latex] tens) on the
number line?
We can see that [latex]76[/latex] is closer to [latex]80[/latex] than to [latex]70[/latex]. So [latex]76[/latex] rounded to the nearest ten is [latex]80[/latex].
Now consider the number [latex]72[/latex]. Find [latex]72[/latex] on the number line.
We can see that [latex]72[/latex] is closer to [latex]70[/latex], so [latex]72[/latex] rounded to the nearest ten is [latex]70[/latex].
How do we round [latex]75[/latex] to the nearest ten?
Find [latex]75[/latex] on the number line.
The number [latex]75[/latex] is exactly midway between [latex]70[/latex] and [latex]80[/latex].
So that everyone rounds the same way in cases like this, mathematicians have agreed to round up to the higher number. So, [latex]75[/latex] rounded to the nearest ten is [latex]80[/latex].
Now that we have looked at this process on the number line, we can introduce a more general procedure. To round a number to a specific place, look at the number to the right of that place. If the
number is less than [latex]5[/latex], round down. If it is greater than or equal to [latex]5[/latex], round up.
So, for example, to round [latex]76[/latex] to the nearest ten, we look at the digit in the ones place.
The digit in the ones place is a [latex]6[/latex]. Because [latex]6[/latex] is greater than or equal to [latex]5[/latex], we increase the digit in the tens place by one. So the [latex]7[/latex] in
the tens place becomes an [latex]8[/latex]. Now, replace any digits to the right of the [latex]8[/latex] with zeros. So, [latex]76[/latex] rounds to [latex]80[/latex].
Let’s look again at rounding [latex]72[/latex] to the nearest [latex]10[/latex]. Again, we look to the ones place.
The digit in the ones place is [latex]2[/latex]. Because [latex]2[/latex] is less than [latex]5[/latex], we keep the digit in the tens place the same and replace the digits to the right of it with
zero. So [latex]72[/latex] rounded to the nearest ten is [latex]70[/latex].
Round a whole number to a specific place value
1. Locate the given place value. All digits to the left of that place value do not change.
2. Underline the digit to the right of the given place value.
3. Determine if this digit is greater than or equal to [latex]5[/latex].
□ Yes—add [latex]1[/latex] to the digit in the given place value.
□ No—do not change the digit in the given place value.
4. Replace all digits to the right of the given place value with zeros.
Round [latex]843[/latex] to the nearest ten.
Locate the tens place.
Underline the digit to the right of the tens place.
Since [latex]3[/latex] is less than [latex]5[/latex], do not change the digit in the tens place.
Replace all digits to the right of the tens place with zeros.
Rounding [latex]843[/latex] to the nearest ten gives [latex]840[/latex].
try it
Round each number to the nearest hundred:
1. [latex]23,658[/latex]
2. [latex]3,978[/latex]
Show Answer
try it
Round each number to the nearest thousand:
1. [latex]147,032[/latex]
2. [latex]29,504[/latex]
Show Answer
try it
Watch the video below for more examples of how to round whole numbers to a given place value. | {"url":"https://courses.lumenlearning.com/wm-accountingformanagers/chapter/rounding-whole-numbers/","timestamp":"2024-11-03T09:24:43Z","content_type":"text/html","content_length":"67430","record_id":"<urn:uuid:eda92c27-4cbf-40a5-8a87-bf7d88132042>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00573.warc.gz"} |
432 as base
Why the 432 as base for the music of the spheres.
There has been a lot of talk about 432 Hz and it should be used for tuning, it is also often referred to as Pythagorean tuning. But today it is not what is used. The reason for this is because a full
cycle of 12 fifths should coincide with 7 octaves but it doesn’t. The difference in the note we get after stacking 12 fifths and the one we land on after seven octaves is known as the Pythagorean
comma, a ratio of 531441:524288.
The Pythagorean tuning system is based on a cycle of pure fifths (the 3:2 ratio found in the reference article).
If we apply Pythagorean tuning to C= 256 Hz we will end up with A = 432 Hz
256 x 3:2 = 384 note G
384 x 3:2 = 576 the note D
576 x 3:2 = 864 the note A one octave higher than 432 Hz
Trained opera singer’s voices shift at very specific frequencies at the exact half of an octave ranging from one C to the next C is F# with C at 256Hz. This is called the bel canto tradition.
And as already pointed at, with the music of the spheres, it also applies to astronomical ratios. Our planets, the solar system that is, makes registered shifts all based on these numbers/geometrical
mean of 256 Hz, 512 and all others in various ways including duration, as you could know due to the time of pendulum over 30° with arc length of 52,36.
A closer look at biology and living tissue which both emit and absorb electromagnetic radiation at a series of specific frequencies/wavelengths shows their order to be like this too but precisely 42
octaves higher than the wavelength of 256.54.
A reference to the octave/scale that also relates to your inner being and the 32/heart can be found in a very old abbey, church of Cluny in France, these inscriptions say: Octavus Sanctos Omnes Doser
Esse Beatos. The octave teaches the saints bliss.
When 18 (value of man) is divided, 18/31,25, we get 576 the value of the tree of good and evil.
But when his heart is fixed and whole again, 18 x 31,25, we get 56,25 the value of the ark of covenant.
And you take place on the mercy seat.
From a zodiacal point of view the bodily you is traveling 1,5 times the zodiac 12+6=18 and 18×45=810. If we divide the zodiac 25920 by 18 we get the 144 and if these 144 take the mercy seat 144 x
56,25 = 8100. That he is already within us is 45/8=56,25. Adam has a value of 45. And 45 is the value of the ark of Noah or in other words the body with the 8 within and the ship, vesica piscis,
traveling around with the zodiacal animals reflecting within. Abram has a value of 243 but then becomes Abraham value 288, difference is 45. And 45 x 32 (the heart).
The meter was rediscovered in France in 1799 as 1/10 millionth of the distance between the north pole and equator through Paris. But as you know it is ancient, in fact a structural ratio. If you look
at the latitude of the great pyramids 29.9792458 then they knew the speed of light too, as it is equal to its latitude.
Very recently science was kind of shocked to find that the cosmological constant and the fine structure constant vary in the universe, particularly in two directions and not in the other 2 like a
north and south pole of a magnet, one side it was higher and the other lower while east and west were not. As you may recall it is explained that the universe is a torus too, including the disc
between the two poles, and that is the reason why Saturn too has its disc/rings. So if you would ride the electromagnetic field you would see an event horizon just like a curve of a sphere into the
centre of a pole. Or even the same principle of a black hole and galaxies.
It’s a regeneration/reincarnation.
The fruit is in the seed.
Moshiya van den Broek | {"url":"https://www.truth-revelations.org/?page_id=2642","timestamp":"2024-11-09T15:30:24Z","content_type":"text/html","content_length":"30184","record_id":"<urn:uuid:24cedd91-4515-47e9-966e-78b870fa07e7>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00649.warc.gz"} |
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AMC 8
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Register to take the AMC 8 at Fresno State (Open until capacity is reached. The competition will be held on 1/18/24.)
Sign up for Practice Sessions (No deadline, sessions held from 11/30/23 to 1/11/24)
AMC 10/12
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Integer factorization on the GPU Factoring large integers
Does anyone know information about anyone who has successfully implemented integer prime factorization on the GPU (with performance improvements over CPU)?
CUDA 1.0 has 64 bit integer support (from the changelog), but I’m talking about 1024+ bit integers. Arithmetic on 1024 bit integers can probably be emulated using 64 bit integers, but there probably
wont be too much of a performance benefit. I’m scouring the internet for links on this topic. If anyone has some additional information, that would be useful.
I guess emulating 1024 bit integer arithmetics via 64bit integers (after all we are talking about of factor 16) will give a performance hit. But the most painful will not be the increase number of
ops, but the memory overhead as you will need to read 16x more data for one integer. If the code is already bandwidth limited, you most probably will get at least 16x slower performance. If the code
is computation limited … do not know, but this sounds like something to experiment with.
Good luck.
Actually you don’t need to implement 1024-bit arithmetic to be able to do integer factorization (well, if you don’t use trial division for numbers of such size :) ).
I believe that most time consuming part of modern factorization algorithms — sieving — can be done on GPU with significant speedup, but it may require double precison support. ANyway, I’m not aware
of any public information on this topic.
What dou you mean by double precision? using float doubles to perform integer operations? Why not use 64 bit support?
Also I’ve found and implementation of a factorization on GPU (trough emulating a a quantum computer and apllying Schors algorithm)…
It’s a CUDA course project:
Code avaiable?..
Sieving requires floating point operation, AFAIK. I’m not sure if single precision wil be good enough for it. Sorry, I don’t have very deep knowlegdge of NFS algorithms, I’m just sharing my thougts
In Mersenne Prime searches, such as GIMPS, they actually use an FFT to do multiplication of very large integers (billions of digits) efficiently. FFT-based large integer multiplication is important
in this case because it has time complexity O(n log(n) log(log(n))), rather than O(n^2).
Using an FFT this way requires a certain level of floating point precision to ensure that round-off error doesn’t taint the final answer. GIMPS is getting close to the limit of double precision FFTs,
and may have to switch to quad precision in order to multiply even bigger numbers.
Number field sieve methods may use similar tricks, but I don’t know anything about those. | {"url":"https://forums.developer.nvidia.com/t/integer-factorization-on-the-gpu-factoring-large-integers/2822","timestamp":"2024-11-02T02:04:45Z","content_type":"text/html","content_length":"49654","record_id":"<urn:uuid:610ddfc3-df44-4958-af34-2b0e65674d1d>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00124.warc.gz"} |
A Framework for the Optimization of Complex Cyber-Physical Systems via Directed Acyclic Graph
Department of Mechanical, Computer Science and Aerospace Engineering, Universidad de León, 24071 León, Spain
Author to whom correspondence should be addressed.
Submission received: 23 December 2021 / Revised: 1 February 2022 / Accepted: 11 February 2022 / Published: 15 February 2022
Mathematical modeling and data-driven methodologies are frequently required to optimize industrial processes in the context of Cyber-Physical Systems (CPS). This paper introduces the PipeGraph
software library, an open-source python toolbox for easing the creation of machine learning models by using Directed Acyclic Graph (DAG)-like implementations that can be used for CPS. scikit-learn’s
Pipeline is a very useful tool to bind a sequence of transformers and a final estimator in a single unit capable of working itself as an estimator. It sequentially assembles several steps that can be
cross-validated together while setting different parameters. Steps encapsulation secures the experiment from data leakage during the training phase. The scientific goal of PipeGraph is to extend the
concept of Pipeline by using a graph structure that can handle scikit-learn’s objects in DAG layouts. It allows performing diverse operations, instead of only transformations, following the
topological ordering of the steps in the graph; it provides access to all the data generated along the intermediate steps; and it is compatible with GridSearchCV function to tune the hyperparameters
of the steps. It is also not limited to $( X , y )$ entries. Moreover, it has been proposed as part of the scikit-learn-contrib supported project, and is fully compatible with scikit-learn.
Documentation and unitary tests are publicly available together with the source code. Two case studies are analyzed in which PipeGraph proves to be essential in improving CPS modeling and
optimization: the first is about the optimization of a heat exchange management system, and the second deals with the detection of anomalies in manufacturing processes.
1. Introduction
Continuous technological advancements in fields such as Information Technology (IT), Artificial Intelligence (AI), and the Internet of Things (IoT), among others, have drastically transformed
manufacturing processes. Recent technological advancements have permitted a systematic deployment of Cyber–Physical Systems (CPS) in manufacturing, which allows intertwining physical and software
components to control a mechanism by means of a computer system. CPS has considerably improved the efficiency of production processes while also making them more resilient and collaborative [
]. These cutting-edge technologies are advancing the manufacturing economic sector in the Industry 4.0 era [
In the Industry 4.0 paradigm, manufacturing industries must modify their management systems and look for new manufacturing strategies [
] to find solutions to tackle the issues faced nowadays. Lean Manufacturing (LM) has become one of the most generally accepted manufacturing methods and management styles used by organizations
throughout the world to improve their business performance and competitiveness [
]. Since LM improves operational performance for manufacturing organizations in developing and developed countries [
], it has spread all over the world [
]. Ref. [
] suggested that the future research methodologies for LM can be classified into meaningful themes, namely: the size of the research sample and its composition; several types of study (other than
surveys); longitudinal studies; applying advanced statistical analysis and (mathematical) modeling techniques; objective, real and quantitative data; surveys; mixed/multiple research studies;
reliability and validity analysis; using computer-aided technology for data collection, and processing and research collaborations.
This paper focuses on the application of mathematical modeling techniques and the use of computer-aided technology for data processing for LM CPS, at the core of Industry 4.0. LM deals with the
optimization of the performance according to a specific set of principles [
]. In the context of CPS, mathematical modeling and data-driven techniques are usually needed to optimize industrial processes [
]. Ref. [
] presented a fully model-driven technique based on MontiArc models of the architecture of the CPS and UML/P class diagrams to construct the digital twin information system in CPS. Ref. [
] combined Random Forest (RF) with Bayesian optimization for large-scale dimension data quality prediction, selecting critical production aspects based on information gain, and then using sensitivity
analysis to preserve product quality, which may provide management insights and operational guidance for predicting and controlling product quality in the real-world process industry. In [
], in the context of varying design uncertainty of CPS, the feasibility of appropriate evolutionary and Machine Learning (ML) techniques was examined. In [
], the Neural Network Verification (NNV), a software tool that offers a set of reachability methods for verifying and testing the safety (and robustness) of real-world Deep Neural Networks (DNNs) and
learning-enabled CPS, was introduced.
ML is contributory in solving difficult problems in the domains of data-driven forecasting, classification and clustering for CPS. However, the literature of the ML field presents a high number of
approaches and variations which makes difficult to establish a clear classification scheme for its algorithms [
]. Toolkits for ML aim at standardizing interfaces to ease the use of ML algorithms in different programming languages as R [
], Apache Spark [
], JAVA and C# [
], C++ [
], JAVA [
], PERL [
], JavaScript [
], command-line [
], among others. Python is one of the most popular and widely-used software systems for statistics, data mining, and ML, and
] is the most widely used module for implementing a wide range of state-of-the-art ML algorithms for medium-scale supervised and unsupervised problems.
Data-driven CPS case studies usually need to split the data into training and test sets and to combine a set of processes to be applied separately to the training and test data. Some bad practices in
data manipulation can end up in a misleading interpretation of the achieved results. The use of tools that allow the selection of the pertinent steps in an ad-hoc designed pipeline helps to reduce
programming errors [
]. The
object of the
module allows combining several transformers and an estimator to create a combined estimator [
]. This object behaves as a standard estimator, and
therefore can be used to tune the parameters of all steps. Nevertheless,
has some shortcomings such as: it is quite rigid since it strictly allows combining transformers and an estimator sequentially in such a way that the inputs of a step are the transformed outputs of
the previous step; it requires
$( X , y )$
-like entries, the
is transformed by the transformers, and
$( X , y )$
is used by the estimator;
on a
is constrained to only split
$( X , y )$
and tune parameters that are variables of the functions, for example, a
cannot be tuned.
In this work we present
, a new toolbox that aims at overcoming some of the weaknesses of
while providing greater functionality to ML users. In
, all kinds of steps, not only transformers and an estimator, can be combined in the form of a Directed Acyclic Graph (DAG), the entries of a step can come from the outputs of any previous step or
the inputs to the graph, see
Figure 1
. For more information about the application of DAGs in ML see [
]. A
accepts more variables than
$( X , y )$
as inputs that can be appended or split within the graph to make use of the standard
functions. This allows using hyperparameter tuning techniques such as
to easily manage data further than
$( X , y )$
inputs and tune other parameters apart from the variables of functions. The user can easily implement a new
creating steps that make use of: (i)
functions, (ii) provided custom blocks that implement basic functions, or (iii) their own elaborated custom blocks that implement custom functions. Moreover, in this paper we report two case studies
was essential to ease the modeling and optimization in the CPS area. The first case study deals with an optimization of heat exchange management system, whereas the second one deals with anomaly
detection in manufacturing processes.
2. Data Leakage in ML Experiments
This section emphasizes the importance of avoiding data leakage in data driven numerical modeling. Data leakage in experimental learning is an important design defect that practitioners must
consciously avoid. Best practices in ML projects establish that the information related to the case study must be split in a number of different data sets, i.e., training set, test set, and if
necessary validation test [
]. This allows the practitioner to obtain a measure of the error expected on data unseen during the training process. It is crucial for the model to be evaluated on unseen data in order to confirm
the generalization capability of the model. Moreover, cross–validation (CV) is one of the most popular strategies to obtain a representative value of the error measure. CV pursues to provide an error
measure on unseen data by using the training set alone. To achieve that goal, it splits the training data set in a number
of subsets, also known as ‘folds’, and runs
training experiments by isolating one of those folds at a time, to later on measure the error on the specific set that has been isolated. Thus, in each of the
experiments the model is trained and tested on different sets. Finally, the error measures are condensed by using standard statistics like mean value and standard deviation.
One of the most important risks associated to CV is data leakage, by failing to correctly manage the different sets of data and their effect on the different stages of the training phase. Let us
consider the following simple illustrative case: a linear model that is fitted using scaled data. Such an example requires the data to run through two processes: a scaler and then a predictive model.
During the training phase the scaler annotates information related to the presented data, e.g., maximum and minimum, or mean value and standard deviation. This annotations allow the scaler process to
eventually apply the same transformation to new and unseen data. A typical error that might occur is that the code for the CV presents the whole training data set to the scaler and then loops
different fit experiments using the k-fold strategy. By doing so, each of the k models fails to provide an error measure that is representative of the behavior of the model on unseen data, because
indeed, what is meant to be considered as unseen data has effectively polluted the experiment by its potential impact on the parameters annotated on the scaler.
One of the main strategies to avoid the risk of allowing the user to mishandle the flow of information during CV is to embed the sequence of processes in an entity that is handled in the code as a
single unit. A remarkable example of such solution is the
class provided by
. According to
documentation “
pipelines help avoid leaking statistics from your test data into the trained model in cross-validation blue (CV), by ensuring that the same samples are used to train the transformers and predictors
” [
In the former example, the scaler cannot be trained using the whole data set during the CV experiment because Pipeline is in charge of training the scaler as many times as the predictive model. This
is a very successful strategy for migrating the responsibility of orchestrating the fit and predict phases from the user to the code of the framework, thus avoiding possible user errors while
handling the data.
CV is one of the scenarios where data leakage can occur, other notable situations prone to this design defect can occur when augmenting the data or during the feature engineering stage, to name a
PipeGraph is another example of such encapsulation. As Pipeline, it provides the researcher with the safety that no data leakage will occur during the training phase. Moreover, it enhances the
capabilities of the standard Pipeline provided by scikit-learn by allowing non–linear data flows, as we will show in the following section.
Thus the scientific goal of this paper is to present researchers from the ML community, in particular those in the CPS area, a novel framework aimed at providing expressive means to design complex
models such as those typically present in CPS. PipeGraph thus combines the expressive power of DAGs with the intrinsic safety of encapsulation.
3. Library Design
3.1. Project Management
Quality assurance. In order to ensure code quality and consistency, unitary tests are provided. We granted a coverage of 89% for the release 0.0.15 of the PipeGraph toolbox. New contributions are
automatically checked through a Continuous Integration (CI) system for the sake of determining metrics concerning code quality.
Continuous integration. In order to ensure CI when using and contributing to PipeGraph toolbox, Travis CI is used to integrate new code and provide back-compatibility. Circle CI is used to build new
documentation along with examples containing calculations.
Community-based development.PipeGraph toolbox is fully developed under GitHub and gitter to facilitate collaborative programming, this is, issue tracking, code integration, and idea deliberations.
We provide consistent Application Programming Interface (API) documentation and gallery examples (
, accessed on 20 December 2021) by means of sphinx and numpydoc. A user’s guide together with a link to the API reference and examples are provided and centralized in GitHub (
, accessed on 20 December 2021).
Project relevance. At edition time of this paper PipeGraph toolbox has been proposed as part of the scikit-learn-contrib supported project.
3.2. Implementation Details
Here we describe the main issues that had to be solved for PipeGraph to work. First, scikit-learn step eligible classes provide a set of methods with different names for essentially the same purpose;
providing the output corresponding to an input dataset. Depending on whether the class is a transformer or an estimator, this method can be called either transform, predict, or even fit_predict. This
issue was originally solved by using the wrapper design pattern, although it has been proposed by scikit-learn core developers to consider the usage of mixin classes to provide a similar
functionality. The development branch of the package already implements this alternative approach.
The main difference from a users perspective between using a Pipeline object or a PipeGraph object is the need of defining a dictionary that establishes the connections. Again, a scikit-learn core
developer suggested the implementation of an inject method for that purpose and that approach is also already available along with the optional dictionary for those cases in which the user finds it
more convenient to use.
The proposed toolbox depends on numpy, pandas, networkx, inspect, logging and scikit-learn and is distributed under MIT license.
PipeGraph can be easily installed using pip install pipegraph.
3.3. Example
We describe here one example for illustrative purposes. The system displays a predictive model in which a classifier provides the information to a demultiplexer to separate the dataset samples
according to their corresponding class. After that, a different regression model is fitted for each class. Thus, the system contains the following steps:
• scaler: A scikit-learn MinMaxScaler data preprocessor in charge of scaling the dataset.
• classifier: A scikit-learn GaussianMixture classifier in charge of performing the clustering of the dataset and the classification of any new sample.
• demux: A custom Demultiplexer class in charge of splitting the input arrays accordingly to the selection input vector. This block is provided by PipeGraph.
• lm_0, lm_1, lm_2: A set of scikit-learn LinearRegression objects
• mux: A custom Multiplexer class in charge of combining different input arrays into a single one accordingly to the selection input vector. This block is provided by PipeGraph.
This PipeGraph model is shown in
Figure 2
. It can be clearly seen in the figure the non sequential nature of such a system that cannot be otherwise described as a standard
The code for creating an artificial dataset and configuring the system is described in Listing A1 of
Appendix A
3.4. Implemented Methods
PipeGraph toolbox provides two interfaces, PipeGraphRegressor and PipeGraphClassifier, which are compatible with GridSearchCV and heavily based on scikit-learn’s Pipeline on purpose, as its aim is to
offer an interface as similar to Pipeline as possible. By default, PipeGraphRegressor uses the regressor default score (the coefficient of determination R^2 of the prediction) and PipeGraphClassifier
uses the classifier default score (the mean accuracy of the prediction on given test data with respect to labels). As for the rest, both interfaces are equivalent.
The following functions can be used by the user in both interfaces:
• inject(sink, sink_var, source, source_var) Defines a connection between two nodes of the graph declaring which variable (source_var) from the origin node (source) is passed to the destination
node (sink) with new variable name sink_name).
• decision_function(X) Applies PipeGraphClasifier’s predict method and returns the decision_function output of the final estimator.
• fit(X, y=None, $* *$fit_params) Fits the PipeGraph steps one after the other and following the topological order of the graph defined by the connections attribute.
• fit_predict(X, y=None, $* *$fit_params) Applies predict of a PipeGraph to the data following the topological order of the graph, followed by the fit_predict method of the final step in the
PipeGraph. Valid only if the final step implements fit_predict.
• get_params(deep=True) Gets parameters for an estimator.
• predict(X) Predicts the PipeGraph steps one after the other and following the topological order defined by the alternative_connections attribute, in case it is not None, or the connections
attribute otherwise.
• predict_log_proba(X) Applies PipeGraphRegressor’s predict method and returns the predict_log_proba output of the final estimator.
• predict_proba(X) Applies PipeGraphClassifier’s predict method and returns the predict_proba output of the final estimator.
• score(X, y=None, sample_weight=None) Applies PipeGraphRegressor’s predict method and returns the score output of the final estimator.
• set_params($* *$kwargs) Sets the parameters of this estimator. Valid parameter keys can be listed with get_params().
4. Case Studies
4.1. Anomaly Detection in Manufacturing Processes
The first case study deals with anomaly detection of machined workpieces using a computer vision system [
]. In that paper, a set of four classifiers were tested in order to choose the best model for identifying the presence of wear along the workpiece surface. Following a workflow consisting of a
preprocessing phase followed by feature extraction and finally a classification step, the authors reported satisfactory results. In such scenario, the results of a deeper analysis could have been
provided, where the quality of the classifier could have been improved by unleashing an additional parameter, the number of classes. Instead of assuming that two classes are present in the dataset,
namely correct pieces and unacceptable pieces, according to the finishing quality, the result from considering more classes can be explored.
For that purpose, a pipegraph model as the one shown in
Figure 3
allows for a two model workflow. In this case, a clustering algorithm partitions a dataset consisting of 10,000 artificially generated 5-dimensional samples. It is worth noting that the partition is
performed considering a specific and predefined number of clusters. This experiment splits the dataset in 5 folds for cross–validation. For each configuration, the classifier is trained and the
quality of its results is assessed according to some convenient metric.The classifier considered was the well–known
-means training for a maximum of 300 iterations and a stopping criterion of error tolerance smaller than 0.0001.
Figure 4
displays the quality results for such workflow in an illustrative case, described in more detail in the library documentation. It is worth noting that such a workflow cannot be constructed via
tools as the two steps of the workflow are models and the standard Pipeline class only allows for a set of transformers and a single unique model in final position. We claim that for those purposes
where different models are useful, even in a linear sequence,
provides a viable and convenient approach.
4.2. Heat Exchanger Modeling
The second case study deals with a problem that is common in manufacturing processes: the management of faulty sensors [
]. In that paper, 52,699 measurements from a sensor embedded in a heat exchanger system were compared to the predictions from a baseline model capturing the expected behavior of such CPS system. If
the sensor measurements were significantly different from the predictions provided by the CPS model, an alarm was raised and specific actions performed according to the particular case study. In the
paper, two classes were again considered, namely day and night, standing for the two particular periods in which the 24 h are split. The best model obtained from a Cross-Validation setup using 10
folds was an Extremely Randomized Tree whose training explored a range from 10 through 100 base estimators. A pipegraph similar to the one considered in
Figure 2
was used in the experience reported in the paper for a two model for two classes case. For such scenario, an approach using more than two classes could have been considered to check if such an
enhanced model can outperform the results reported.
Figure 5
displays a unified workflow in which
is capable of automatically wiring as many local models as classes are defined in the dataset, thus relieving the user from the task of defining multiple configurations depending on how many classes
were considered. It is again worth noting that such a workflow cannot be constructed via standard
tools for the non linear workflow necessary for the purpose, and because of the automatic building procedure of the multiple prediction models put to play. We claim again that for those purposes
where different models are used in a non linear sequence,
provides a viable and convenient approach.
5. Conclusions
CPS and LM can greatly leverage on improvements in the tools and techniques available for system modeling. Data leakage, being one of the most common sources of unexpected behavior when the fitted
models are stressed with actual demands, can largely be prevented by using encapsulation techniques such as the Pipeline provided by the scikit-learn library. For some specific complex problem
appearing in the context of LM, and particularly in CPS modeling, the PipeGraph library provides a solution to building complex workflows, which is specially important for those cases that the
standard Pipeline provided by scikit-learn cannot handle. Parallel blocks in non linear graph are available to unleash the creativity of the data scientist in the pursue of a simple and yet efficient
model. In this paper we briefly introduced a novel PipeGraph toolbox for easily expressing ML models using DAG while being compatible with scikit-learn. We showed the potential of the toolbox and the
underlying implementation details. We provided references to two case studies with hints on approaches for possible improvements by using PipeGraph and minor changes to the architecture proposed in
two papers in the field of CPS and LM modeling. Future works on PipeGraph will include a Graphical User Interface (GUI) to use the API and new libraries which will encompass custom blocks related to
specific areas connected to machine learning, such as computer vision, control systems, etc.
Author Contributions
Conceptualization, M.C.-L. and L.F.-R.; Formal analysis, M.C.-L., L.F.-R. and H.A.-M.; Investigation, M.C.-L. and L.F.-R.; Methodology, M.C.-L. and C.F.-L.; Software, M.C.-L., L.F.-R. and H.A.-M.;
Validation, M.C.-L., L.F.-R., H.A.-M., J.C.-R. and C.F.-L.; Visualization, M.C.-L. and J.C.-R.; Writing—original draft, M.C.-L. and L.F.-R.; Writing—review & editing, M.C.-L., L.F.-R., H.A.-M.,
J.C.-R. and C.F.-L. All authors have read and agreed to the published version of the manuscript.
This research was funded by the Spanish Ministry of Economy, Industry and Competitiveness grant number DPI2016-79960-C3-2-P. The research described in this article has been partially funded by the
grant RTI2018-100683-B-I00 funded by MCIN/AEI/10.13039/501100011033 and by “ERDF A way of making Europe”.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Conflicts of Interest
The authors declare no conflict of interest.
Appendix A
Listing A1. Example code for the PipeGraph shown in Figure 2.
• importnumpyasnp
• importpandasaspd
• fromsklearn . preprocessingimportMinMaxScaler
• fromsklearn . linear_modelimportLinearRegression
• fromsklearn . model_selectionimportGridSearchCV
• frompipegraph . baseimportPipeGraphRegressor
• frompipegraph . baseimportDemultiplexer
• frompipegraph . baseimportMultiplexer
• importmatplotlib . pyplotasplt
• fromsklearn . mixtureimportGaussianMixture
• X_first=pd . Series (np . random . rand (100,))
• y_first=pd . Series (4*X_first + 0.5 * np . random . randn (100,))
• X_second=pd . Series (np . random . rand (100,) + 3)
• y_second=pd . Series (−4*X_second + 0.5 * np . random . randn (100,))
• X_third=pd . Series (np . random . rand (100,) + 6)
• y_third=pd . Series(2*X_third + 0.5 * np . random . randn (100,))
• X=pd . concat ([X_first,X_second,X_third],axis = 0) . to_frame ()
• y=pd . concat ([y_first,y_second,y_third],axis = 0) . to_frame ()
• scaler=MinMaxScaler ()
• gaussian_mixture=GaussianMixture (n_components = 3)
• demux=Demultiplexer ()
• lm_0=LinearRegression ()
• lm_1=LinearRegression ()
• lm_2=LinearRegression ()
• mux=Multiplexer ()
• steps=[(’scaler’,scaler),
• (’classifier’,gaussian_mixture),
• (’demux’,demux),
• (’lm_0’,lm_0),
• (’lm_1’,lm_1),
• (’lm_2’,lm_2),
• (’mux’,mux), ]
• connections={
• ’scaler’:{’X’:’X’},
• ’classifier’:{’X’:’scaler’},
• ’demux’:{’X’:’scaler’,’y’:’y’,...
• ’selection’:’classifier’},
• ’lm_0’:{’X’:(’demux’,’X_0’),’y’:(’demux’,’y_0’)},
• ’lm_1’:{’X’:(’demux’,’X_1’),’y’:(’demux’,’y_1’)},
• ’lm_2’:{’X’:(’demux’,’X_2’),’y’:(’demux’,’y_2’)},
• ’mux’:{’0’:’lm_0’,’1’:’lm_1’,’2’:’lm_2’,...
• ’selection’:’classifier’},
• }
• pgraph=PipeGraphRegressor (steps=steps,...
• fit_connections=connections)
• pgraph . fit (X,y)
• y_pred=pgraph . predict (X)
• plt . scatter (X,y)
• plt . scatter (X,y_pred)
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Figure 1. Graphical abstract. In the upper part, the Pipeline structure can be seen, which only allows sequential steps. At the bottom, the PipeGraph structure is shown. The combination of steps
based on a directed acyclic graph makes a wide variety of operations feasible.
Figure 4. Number of clusters vs. a quality measure to assess the most appropriate number of clusters.
Figure 5. Parallel workflow with automatic wiring of the prediction models according to the labels contained in a dataset.
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Castejón-Limas, M.; Fernández-Robles, L.; Alaiz-Moretón, H.; Cifuentes-Rodriguez, J.; Fernández-Llamas, C. A Framework for the Optimization of Complex Cyber-Physical Systems via Directed Acyclic
Graph. Sensors 2022, 22, 1490. https://doi.org/10.3390/s22041490
AMA Style
Castejón-Limas M, Fernández-Robles L, Alaiz-Moretón H, Cifuentes-Rodriguez J, Fernández-Llamas C. A Framework for the Optimization of Complex Cyber-Physical Systems via Directed Acyclic Graph.
Sensors. 2022; 22(4):1490. https://doi.org/10.3390/s22041490
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Quantitative Methods • Formulas CFA® Level 1 – 365 Financial Analyst
Need an all-in-one list with the Quantitative Methods formulas included in the CFA® Level 1 Exam? We have compiled them for you here. The relevant formulas have been organized and presented by
chapter. In this section, we will cover the following topics — Time Value of Money, Statistical Concepts and Market Returns, Probability, Distribution, Sampling, Estimation, and Hypothesis Testing.
1. Time Value of Money
Effective Annual Rate (EAR)
Effective~annual~rate = \bigg(1 + \frac {Stated~annual~rate}{m}\bigg)^m- 1
Single Cash Flow (simplified formula)
FV{_N} = PV \times (1 + r){^N}
PV = \frac {FV{_N}} {(1 + r){^N}}
r = interest rate per period
PV = present value of the investment
FV{_N} = future value of the investment N periods from today
Investments paying interest more than once a year
FV{_N} = PV \times \bigg(1+\frac{r{_s}}{m}\bigg){^{mN}}
PV = \frac{FV{_N}}{\bigg(1+\frac{r{_s}}{m}\bigg){^{mN}}}
r{_s} = Stated annual interest rate
m = Number of compounding periods per year
N = Number of years
Future Value (FV) of an Investment with Continuous Compounding
Ordinary Annuity
FV{_N} = A \times \bigg[ \frac {(1+r){^N-1}}{r} \bigg]
PV = A \times \Bigg[ \frac {1-\frac{1}{(1+r){^N}}}{r} \Bigg]
N = Number of time periods
A = Annuity amount
r = Interest rate per period
Annuity Due
FV~A{_{Due}} = FV~A{_{Ordinary}} \times (1+r) = A \times \bigg[ \frac {(1+r){^N}-1}{r}\bigg] \times (1+r)
PV~A{_{Due}} = PV~A{_{Ordinary}} \times (1+r) = A \times \Bigg[ \frac {1-\frac{1}{(1+r){^N}}}{r} \Bigg] \times (1+r)
A = Annuity amount
r = the interest rate per period corresponding to the frequency of annuity payments (for example, annual, quarterly, or monthly)
N = the number of annuity payments
Present Value (PV) of a Perpetuity
PV{_{Perpetuity}} = \frac {A}{r}
A = Annuity amount
Future value (FV) of a series of unequal cash flows
FV{_N} = Cash~flow{_1}(1 + r){^1} + Cash~flow{_2}(1 + r){^2} … Cash~flow{_N}(1 + r){^N}
Net Present Value (NPV)
NPV=\sum_{t=0}^N \frac{CF_{t}}{(1+r)^t}
CF{_t} = Expected net cash flow at time t
N = The investment’s projected life
r = The discount rate or opportunity cost of capital
Internal Rate of Return (IRR)
NPV= CF{_0} + \frac {CF{_1}}{(1+IRR){^1}} + \frac {CF{_2}}{(1+IRR){^2}} + … + \frac {CF{_N}}{(1+IRR){^N}} = 0
Holding Period Return (HPR)
No cash flows
HPR = \frac {Ending~value - Beginning~value}{Beginning~value}
Holding Period Return (HPR)
Cash flows occur at the end of the period
HPR = \frac {Ending~value - Beginning~value+ Cash~flows~received}{Beginning~value} = \frac {P{_1} - P{_0} + D{_1}}{Beginning~value}
P{_1} = Ending Value
P{_0} = Beginning Value
D = Cash flow/dividend received
Yield on a Bank Discount Basis (BDY)
r{_{BD}}= \frac {D}{F} \times \frac {360}{t}
r{_{BD}} = Annualized yield on a bank discount basis
D = Dollar discount, which is equal to the difference between the face value of the bill (F) and its purchase price (P{_0})
F = Face value of the T-bill
t = Actual number of days remaining to maturity
Effective Annual Yield (EAY)
EAY = ( 1 + HPR) {^\frac {360}{t}}- 1
t = Time until maturity
HPR = Holding Period Return
Money Market Yield (CD Equivalent Yield)
Money~market~yield = HPR \times \bigg(\frac {360}{t}\bigg) = \frac {360 \times r{_{Bank~Discount}}}{360-(t \times r{_{Bank~Discount}})}
2. Statistical Concepts and Market Returns
Interval Width
Interval~Width = \frac {Range}{k}
Range = Largest observation number – Smallest Observation or number
k = Number of desired intervals
Relative Frequency
Relative~frequency = \frac {Interval~frequency}{Observations~in~data~set}
Population Mean
\mu = \frac {\sum_{i=1…n}^Nx{_i}}{N}= \frac {{x{_1}} + {x{_2}} + {x{_3}} + … +{x{_N}}} {N}
N = Number of observations in the entire population
x{_i} = the iáµ—Ê° observation
Sample Mean
\overline x= \frac {\sum_{i=1…n}^nx{_i}}{n} = \frac {{x{_1}} + {x{_2}} + {x{_3}} + … +{x{_n}}} {n}
Geometric Mean
n = Number of observations
Harmonic Mean
\overline x{_n}= \frac {n}{\sum_{i=1…n}^n \bigg(\frac{1}{x{_i}}\bigg)}
Median for odd numbers
Median= \Bigg\{ \frac {(n+1)}{2} \Bigg\}
Median for even numbers
Median= \Bigg\{ \frac {(n+2)}{2} \Bigg\}
Weighted Mean
\overline x{_w} = \sum_{i=1…n}^n w{_i}x{_i}
w = Weights
x = Observations
Sum of all weights = 1
Portfolio Rate of Return
r{_p} = w{_a}r{_a} + w{_b}r{_b} + w{_c}r{_c} + … + w{_n}r{_n}
w = Weights
r = Returns
Position of the Observation at a Given Percentile y
L{_y} = \bigg\{ {(n+1)}\frac{y}{100} \bigg\}
y = The percentage point at which we are dividing the distribution
L{_y} = The location (L) of the percentile (Py) in the array sorted in ascending order
Range= Maximum~value - Minimum~value
Mean Absolute Deviation
MAD =\frac {\sum_{i=1…n}^n |X{_i-\overline X}|}{n}
x = The sample mean
n = Number of observations in the sample
Population Variance
\sigma{^2} = \frac {\sum_{i=1…n}^n (X{_i-\mu}){^2}}{N}
μ = Population mean
N = Size of the population
Population Standard Deviation
\sigma= \sqrt { \frac {\sum_{i=1…n}^N (X{_i-\mu}){^2}}{N}}
μ = Population mean
N = Size of the population
Sample Variance
S{^2} = \frac {\sum_{i=1}^n (X{_i-\overline X}){^2}}{n-1}
x = Sample mean
n = Number of observations in the sample
Sample Standard Deviation
s = \sqrt { \frac {\sum_{i=1}^n (X{_i-\overline X}){^2}}{n-1}}
x = Sample mean
n = Number of observations in the sample
Semi–variance = \frac {1}{n}\sum_{r{_t} < Mean}^n (Mean-r{_t}){^2}
n = Total number of observations below the mean
r{_t} = Observed value
Chebyshev Inequality
Percentage~of~observations~within~k~standard~deviations~of~the~arithmetic~mean > 1-\frac{1}{k{^2}}
k = Number of standard deviations from the mean
Coefficient of Variation
CV = \frac {s}{\overline X}
s = Sample standard deviation
\overline X = Sample mean
Sharpe Ratio
Sharpe~Ratio = \frac {R{_p}-R{_f}}{\sigma{_p}}
R{_p} = Mean return to the portfolio
R{_f} = Mean return to a risk-free asset
σ{_p} = Standard deviation of return on the portfolio
s{_k}= \bigg[ \frac {n}{(n-1)(n-2)} \bigg] \times \frac {\sum_{i=1…n}^n (X{_i}-\overline X){^3}}{s{^3}}
n = Number of observations in the sample
s = Sample standard deviation
K{_E}=\Bigg[ \frac {n (n+1)}{(n - 1)(n - 2)(n - 3)} \times \frac {\sum_{i=1…n}^n (X{_i}-\overline X){^4}}{s{^4}}\Bigg] \times \frac {3~(n-1){^2}}{(n - 2)(n - 3)}
n = Sample size
s = Sample standard deviation
3. Probability Concepts
Odds FOR E
Odds~FOR~E = \frac {P(E)}{1-P(E)}
E = Odds for event
P(E) = Probability of event
Conditional Probability
P(A|B) = \frac {P (A \cap B)}{P (B)}
where P(B) ≠0
Additive Law (The Addition Rule)
P(A \cup B) = P(A) + P(B) - P(A \cap B)
The Multiplication Rule (Joint Probability)
P(A \cap B) = P(A|B) \times P(B)
The Total Probability Rule
P(A) = P(A|S1) \times P(S{_1}) + P(A|S{_2}) \times P(S{_2}) + … + P(A|S{_n}) \times P(S{_n})
S{_1}, S{_2}, …, S{_n} are mutually exclusive and exhaustive scenarios or events
Expected Value
E(X) = P(A)X{_A} + P(B)X{_B} + ... + P(n)X{_n}
P(n) = Probability of an variable
X{_n} = Value of the variable
COV {_{xy}}= \frac {(x-\overline x)(y-\overline y)}{n-1}
x = Value of x
\overline x = Mean of x values
y = Value of y
\overline y = Means of y
n = Total number of values
\rho = \frac {cov{_{xy}}}{\sigma{_x}\sigma{_y}}
σ{_x} = Standard Deviation of x
σ{_y} = Standard Deviation of y
cov{_{xy}} = Covariance of x and y
Variance of a Random Variable
\sigma{^2} x= \sum_{i=1…n}^n \big(x - E(x)\big){^2} \times P(x)
The sum is taken over all values of x for which p(x) > 0
Portfolio Expected Return
E(R{_P}) = E(w{_1}r{_1} + w{_2}r{_2} + w{_3}r{_3} + … + w{_n}r{_n})
w = Constant
r = Random variable
Portfolio Variance
Var(R{_P}) = E\big[(R{_p} - E(R{_p}){^2} \big] = \big[w{_1}{^2} \sigma{_1}{^2} + w{_2}{^2}\sigma{_2}{^2} + w{_3}{^2}σ{_3}{^2} + 2w{_1}w{_2}Cov(R{_1}R{_2}) + 2w{_2}w{_3}Cov(R{_2}R{_3}) + 2w{_1}w{_3}Cov(R{_1}R{_3})\big]
R{_p} = Return on Portfolio
Bayes’ Formula
P(A|B) = \frac {P(B|A) \times P(A)}{P(B)}
The Combination Formula
nC{_r} = \binom{n}{c} = \frac {n!}{(n - r)! r!}
n = Total objects
r = Selected objects
The Permutation Formula
nP{_r} = \frac {n!}{(n - r)!}
4. Common Probability Distributions
The Binomial Probability Formula
P(x) = \frac {n!}{(n - x)! x!}p{^x} \times (1 - p){^{n - x}}
n = Number of trials
x = Up moves
p{^x} = Proability of up moves
(1 - p){^{n - x}} = Probability of down moves
Binomial Random Variable
n = Number of trials
p = Probability
For a Random Normal Variable X
90% confidence interval for X is \overline x - 1.65s;~ \overline x + 1.65s
95% confidence interval for X is \overline x - 1.96s;~ \overline x + 1.96s
99% confidence interval for X is \overline x - 2.58s;~ \overline x + 2.58s
s = Standard error
1.65 = Reliability factor
x = Point estimate
Safety-First Ratio
SF{_{Ratio}}=\bigg[ \frac {E(R{_p}) - R{_L}}{\sigma{_p}} \bigg]
R{_p} = Portfolio Return
R{_L} = Threshhold level
σ{_p} = Standard Deviation
Continuously Compounded Rate of Return
FV = PV \times e{^{i \times t}}
i = Interest rate
t = Time
ln~e = 1
e = the exponential function, equal to 2.71828
5. Sampling and Estimation
Sampling Error of the Mean
Sample~Mean - Population~Mean
Standard Error of the Sample Mean (Known Population Variance)
SE = \frac {\sigma}{\sqrt n}
n = Number of samples
σ = Standard deviation
Standard Error of the Sample Mean (Unknown Population Variance)
S = Standard deviation in unknown population’s sample
Z = \frac {x- \mu}{\sigma}
x = Observed value
σ = Standard deviation
μ = Population mean
Confidence Interval for Population Mean with z
\overline X - {Z{_{\frac {\alpha}{2}}}} \times \frac {\sigma}{\sqrt n}; \overline X + {Z{_{\frac {\alpha}{2}}}} \times \frac {\sigma}{\sqrt n}
Z{_{\frac {\alpha}{2}}} = Reliability factor
X = Mean of sample
σ = Standard deviation
n = Number of trials/size of the sample
Confidence Interval for Population Mean with t
\overline X - {t{_{\frac {\alpha}{2}}}} \times \frac {S}{\sqrt n}; \overline X + {t{_{\frac {\alpha}{2}}}} \times \frac {S}{\sqrt n}
t{_{\frac {\alpha}{2}}} = Reliability factor
n = Size of the sample
S = Standard deviation
Z or t-statistic?
Z \longrightarrow known population, standard deviation σ, no matter the sample size
t \longrightarrow unknown population, standard deviation s, and sample size below 30
Z \longrightarrow unknown population, standard deviation s, and sample size above 30
6. Hypothesis Testing
Test Statistics: Population Mean
z{_\alpha} = \frac {\overline X- \mu} {\frac {\sigma}{\sqrt n}}; t{_{n-1, \alpha}} = \frac {\overline X- \mu} {\frac {s}{\sqrt n}}
t{_{n-1}} = t-statistic with n–1 degrees of freedom (n is the sample size)
\overline X = Sample mean
μ = The hypothesized value of the population mean
s = Sample standard deviation
Test Statistics: Difference in Means – Sample Variances Assumed Equal (independent samples)
t–statistic = \frac {(\overline X{_1} - \overline X{_2}) - (μ{_1} - μ{_2})}{\Big( \frac {s{_p}{^2}}{n{_1}} + \frac {s{_p}{^2}}{n{_2}} \Big){^{\frac {1}{2}}}}
s{_p}{^2}= \frac {(n{_1}-1)s{_1}{^2}+(n{_2}-1)s{_2}{^2}}{n{_1}+n{_2}-2}
Number of degrees of freedom = n{_1} + n{_2} − 2
Test Statistics: Difference in Means – Sample Variances Assumed Unequal (independent samples)
t–statistic = \frac {(\overline x{_1} - \overline x{_2}) - (μ{_1} - μ{_2})}{\Big( \frac {s{_1}{^2}}{n{_1}} + \frac {s{_2}{^2}}{n{_2}} \Big){^{\frac {1}{2}}}}
degrees~of~freedom = \frac {\Big( \frac {s{_1}{^2}}{n{_1}} + \frac {s{_2}{^2}}{n{_2}} \Big){^2}} { \frac {\big(\frac {s{_1}{^2}}{n{_1}} \big){^2}}{n{_1}} + \frac {\big(\frac {s{_2}{^2}}{n{_2}} \big){^2}}{n{_2}}}
s = Standard deviation of respective sample
n = Total number of observations in the respective population
Test Statistics: Difference in Means – Paired Comparisons Test (dependent samples)
t = \frac {\overline d - \mu {_{dz}}}{S{_d}}, \overline d = \frac {1}{n} \sum_{i=1…n}^n d{_i}
degrees of freedom: n–1
n = Number of paired observations
d = Sample mean difference
S{_d} = Standard error of d
Test Statistics: Variance Chi-square Test
\chi{_{n-1}^2} = \frac {(n-1)s{^2}}{\sigma{_0}^2}
degrees of freedom = n - 1
s{^2} = sample variance
\sigma{_0}^2 = hypothesized variance
Test Statistics: Variance F-Test
F = \frac {s{_1}^2}{s{_2}^2}, {s{_1}^2} > {s{_2}^2}
degrees of freedom = n{_1} - 1 and n{_2} - 1
{s{_1}^2} = larger sample variance
{s{_2}^2} = smaller sample variance
Follow the links to find more formulas on Economics, Corporate Finance, Alternative Investments, Financial Reporting and Analysis, Portfolio Management, Equity Investments, Fixed-Income Investments,
and Derivatives, included in the CFA® Level 1 Exam. | {"url":"https://365financialanalyst.com/templates-and-models/quantitative-methods-formulas-cfa-level-1/","timestamp":"2024-11-12T19:40:29Z","content_type":"text/html","content_length":"91949","record_id":"<urn:uuid:4b3c3b75-50df-4293-980e-083af49d2d86>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00723.warc.gz"} |
property of exponents worksheet
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Math: The Properties of Exponents (PDF) - Kids Reading ... | {"url":"https://worksheets.clipart-library.com/property-of-exponents-worksheet.html","timestamp":"2024-11-06T01:51:43Z","content_type":"text/html","content_length":"26068","record_id":"<urn:uuid:000f6c4d-b809-49c5-8f25-212127690f8d>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00402.warc.gz"} |
Knapsack Problem Basics
The Problem: Given a set of items where each item contains a weight and value, determine the number of each to include in a collection so that the total weight is less than or equal to a given limit
and the total value is as large as possible.
Pseudo code for Knapsack Problem
1. Values(array v)
2. Weights(array w)
3. Number of distinct items(n)
4. Capacity(W)
for j from 0 to W do:
m[0, j] := 0
for i from 1 to n do:
for j from 0 to W do:
if w[i] > j then:
m[i, j] := m[i-1, j]
m[i, j] := max(m[i-1, j], m[i-1, j-w[i]] + v[i])
A simple implementation of the above pseudo code using Python:
def knapSack(W, wt, val, n):
K = [[0 for x in range(W+1)] for x in range(n+1)]
for i in range(n+1):
for w in range(W+1):
if i==0 or w==0:
K[i][w] = 0
elif wt[i-1] <= w:
K[i][w] = max(val[i-1] + K[i-1][w-wt[i-1]], K[i-1][w])
K[i][w] = K[i-1][w]
return K[n][W]
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)
print(knapSack(W, wt, val, n))
Running the code: Save this in a file named knapSack.py
Time Complexity of the above code: O(nW) where n is the number of items and W is the capacity of knapsack.
Found a mistake? Have a question or improvement idea?
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Introduction | History of Logic | Types of Logic | Deductive Logic | Inductive Logic | Modal Logic | Propositional Logic | Predicate Logic | Fallacies | Paradoxes | Major Doctrines
Logic (from the Greek "logos", which has a variety of meanings including word, thought, idea, argument, account, reason or principle) is the study of reasoning, or the study of the principles and
criteria of valid inference and demonstration. It attempts to distinguish good reasoning from bad reasoning.
Aristotle defined logic as "new and necessary reasoning", "new" because it allows us to learn what we do not know, and "necessary" because its conclusions are inescapable. It asks questions like
"What is correct reasoning?", "What distinguishes a good argument from a bad one?", "How can we detect a fallacy in reasoning?"
Logic investigates and classifies the structure of statements and arguments, both through the study of formal systems of inference and through the study of arguments in natural language. It deals
only with propositions (declarative sentences, used to make an assertion, as opposed to questions, commands or sentences expressing wishes) that are capable of being true and false. It is not
concerned with the psychological processes connected with thought, or with emotions, images and the like. It covers core topics such as the study of fallacies and paradoxes, as well as specialized
analysis of reasoning using probability and arguments involving causality and argumentation theory.
Logical systems should have three things: consistency (which means that none of the theorems of the system contradict one another); soundness (which means that the system's rules of proof will never
allow a false inference from a true premise); and completeness (which means that there are no true sentences in the system that cannot, at least in principle, be proved in the system).
History of Logic Back to Top
In Ancient India, the "Nasadiya Sukta" of the Rig Veda contains various logical divisions that were later recast formally as the four circles of catuskoti: "A", "not A", "A and not A" and "not A and
not not A". The Nyaya school of Indian philosophical speculation is based on texts known as the "Nyaya Sutras" of Aksapada Gautama from around the 2nd Century B.C., and its methodology of inference
is based on a system of logic (involving a combination of induction and deduction by moving from particular to particular via generality) that subsequently has been adopted by the majority of the
other Indian schools.
But modern logic descends mainly from the Ancient Greek tradition. Both Plato and Aristotle conceived of logic as the study of argument and from a concern with the correctness of argumentation.
Aristotle produced six works on logic, known collectively as the "Organon", the first of these, the "Prior Analytics", being the first explicit work in formal logic.
Aristotle espoused two principles of great importance in logic, the Law of Excluded Middle (that every statement is either true or false) and the Law of Non-Contradiction (confusingly, also known as
the Law of Contradiction, that no statement is both true and false). He is perhaps most famous for introducing the syllogism (or term logic) (see the section on Deductive Logic below). His followers,
known as the Peripatetics, further refined his work on logic.
In medieval times, Aristotelian logic (or dialectics) was studied, along with grammar and rhetoric, as one of the three main strands of the trivium, the foundation of a medieval liberal arts
Logic in Islamic philosophy also contributed to the development of modern logic, especially the development of Avicennian logic (which was responsible for the introduction of the hypothetical
syllogism, temporal logic, modal logic and inductive logic) as an alternative to Aristotelian logic.
In the 18th Century, Immanuel Kant argued that logic should be conceived as the science of judgment, so that the valid inferences of logic follow from the structural features of judgments, although
he still maintained that Aristotle had essentially said everything there was to say about logic as a discipline.
In the 20th Century, however, the work of Gottlob Frege, Alfred North Whitehead and Bertrand Russell on Symbolic Logic, turned Kant's assertion on its head. This new logic, expounded in their joint
work "Principia Mathematica", is much broader in scope than Aristotelian logic, and even contains classical logic within it, albeit as a minor part. It resembles a mathematical calculus and deals
with the relations of symbols to each other.
Types of Logic Back to Top
Logic in general can be divided into Formal Logic, Informal Logic and Symbolic Logic and Mathematical Logic:
• Formal Logic:
Formal Logic is what we think of as traditional logic or philosophical logic, namely the study of inference with purely formal and explicit content (i.e. it can be expressed as a particular
application of a wholly abstract rule), such as the rules of formal logic that have come down to us from Aristotle. (See the section on Deductive Logic below).
A formal system (also called a logical calculus) is used to derive one expression (conclusion) from one or more other expressions (premises). These premises may be axioms (a self-evident
proposition, taken for granted) or theorems (derived using a fixed set of inference rules and axioms, without any additional assumptions).
Formalism is the philosophical theory that formal statements (logical or mathematical) have no intrinsic meaning but that its symbols (which are regarded as physical entities) exhibit a form that
has useful applications.
• Informal Logic:
Informal Logic is a recent discipline which studies natural language arguments, and attempts to develop a logic to assess, analyze and improve ordinary language (or "everyday") reasoning. Natural
language here means a language that is spoken, written or signed by humans for general-purpose communication, as distinguished from formal languages (such as computer-programming languages) or
constructed languages (such as Esperanto).
It focuses on the reasoning and argument one finds in personal exchange, advertising, political debate, legal argument, and the social commentary that characterizes newspapers, television, the
Internet and other forms of mass media.
• Symbolic Logic:
Symbolic Logic is the study of symbolic abstractions that capture the formal features of logical inference. It deals with the relations of symbols to each other, often using complex mathematical
calculus, in an attempt to solve intractable problems traditional formal logic is not able to address.
It is often divided into two sub-branches:
□ Predicate Logic: a system in which formulae contain quantifiable variables. (See the section on Predicate Logic below).
□ Propositional Logic (or Sentential Logic): a system in which formulae representing propositions can be formed by combining atomic propositions using logical connectives, and a system of
formal proof rules allows certain formulae to be established as theorems. (See the section on Propositional Logic below).
• Mathematical Logic:
Both the application of the techniques of formal logic to mathematics and mathematical reasoning, and, conversely, the application of mathematical techniques to the representation and analysis of
formal logic.
The earliest use of mathematics and geometry in relation to logic and philosophy goes back to the Ancient Greeks such as Euclid, Plato and Aristotle.
Computer science emerged as a discipline in the 1940's with the work of Alan Turing (1912 - 1954) on the Entscheidungsproblem, which followed from the theories of Kurt Gödel (1906 - 1978),
particularly his incompleteness theorems. In the 1950s and 1960s, researchers predicted that when human knowledge could be expressed using logic with mathematical notation, it would be possible
to create a machine that reasons (or artificial intelligence), although this turned out to be more difficult than expected because of the complexity of human reasoning. Mathematics-related
doctrines include:
□ Logicism: perhaps the boldest attempt to apply logic to mathematics, pioneered by philosopher-logicians such as Gottlob Frege and Bertrand Russell, especially the application of mathematics
to logic in the form of proof theory, model theory, set theory and recursion theory.
□ Intuitionism: the doctrine which holds that logic and mathematics does not consist of analytic activities wherein deep properties of existence are revealed and applied, but merely the
application of internally consistent methods to realize more complex mental constructs.
Deductive Logic Back to Top
Deductive reasoning concerns what follows necessarily from given premises (i.e. from a general premise to a particular one). An inference is deductively valid if (and only if) there is no possible
situation in which all the premises are true and the conclusion false. However, it should be remembered that a false premise can possibly lead to a false conclusion.
Deductive reasoning was developed by Aristotle, Thales, Pythagoras and other Greek philosophers of the Classical Period. At the core of deductive reasoning is the syllogism (also known as term logic
),usually attributed to Aristotle), where one proposition (the conclusion) is inferred from two others (the premises), each of which has one term in common with the conclusion. For example:
Major premise: All humans are mortal.
Minor premise: Socrates is human.
Conclusion: Socrates is mortal.
An example of deduction is:
All apples are fruit.
All fruits grow on trees.
Therefore all apples grow on trees.
One might deny the initial premises, and therefore deny the conclusion. But anyone who accepts the premises must accept the conclusion. Today, some academics claim that Aristotle's system has little
more than historical value, being made obsolete by the advent of Predicate Logic and Propositional Logic (see the sections below).
Inductive Logic Back to Top
Inductive reasoning is the process of deriving a reliable generalization from observations (i.e. from the particular to the general), so that the premises of an argument are believed to support the
conclusion, but do not necessarily ensure it. Inductive logic is not concerned with validity or conclusiveness, but with the soundness of those inferences for which the evidence is not conclusive.
Many philosophers, including David Hume, Karl Popper and David Miller, have disputed or denied the logical admissibility of inductive reasoning. In particular, Hume argued that it requires inductive
reasoning to arrive at the premises for the principle of inductive reasoning, and therefore the justification for inductive reasoning is a circular argument.
An example of strong induction (an argument in which the truth of the premise would make the truth of the conclusion probable but not definite) is:
All observed crows are black.
All crows are black.
An example of weak induction (an argument in which the link between the premise and the conclusion is weak, and the conclusion is not even necessarily probable) is:
I always hang pictures on nails.
All pictures hang from nails.
Modal Logic is any system of formal logic that attempts to deal with modalities (expressions associated with notions of possibility, probability and necessity). Modal Logic, therefore, deals with
terms such as "eventually", "formerly", "possibly", "can", "could", "might", "may", "must", etc.
Modalities are ways in which propositions can be true or false. Types of modality include:
• Alethic Modalities: Includes possibility and necessity, as well as impossibility and contingency. Some propositions are impossible (necessarily false), whereas others are contingent (both
possibly true and possibly false).
• Temporal Modalities: Historical and future truth or falsity. Some propositions were true/false in the past and others will be true/false in the future.
• Deontic Modalities: Obligation and permissibility. Some propositions ought to be true/false, while others are permissible.
• Epistemic Modalities: Knowledge and belief. Some propositions are known to be true/false, and others are believed to be true/false.
Although Aristotle's logic is almost entirely concerned with categorical syllogisms, he did anticipate modal logic to some extent, and its connection with potentiality and time. Modern modal logic
was founded by Gottlob Frege, although he initially doubted its viability, and it was only later developed by Rudolph Carnap (1891 - 1970), Kurt Gödel (1906 - 1978), C.I. Lewis (1883 - 1964) and then
Saul Kripke (1940 - ) who established System K, the form of Modal Logic that most scholars use today).
Propositional Logic Back to Top
Propositional Logic (or Sentential Logic) is concerned only with sentential connectives and logical operators (such as "and", "or", "not", "if ... then ...", "because" and "necessarily"), as opposed
to Predicate Logic (see below), which also concerns itself with the internal structure of atomic propositions.
Propositional Logic, then, studies ways of joining and/or modifying entire propositions, statements or sentences to form more complex propositions, statements or sentences, as well as the logical
relationships and properties that are derived from these methods of combining or altering statements. In propositional logic, the simplest statements are considered as indivisible units.
The Stoic philosophers in the late 3rd century B.C. attempted to study such statement operators as "and", "or" and "if ... then ...", and Chrysippus (c. 280-205 B.C.) advanced a kind of propositional
logic, by marking out a number of different ways of forming complex premises for arguments. This system was also studied by Medieval logicians, although propositional logic did not really come to
fruition until the mid-19th Century, with the advent of Symbolic Logic in the work of logicians such as Augustus DeMorgan (1806-1871), George Boole (1815-1864) and Gottlob Frege.
Predicate Logic Back to Top
Predicate Logic allows sentences to be analyzed into subject and argument in several different ways, unlike Aristotelian syllogistic logic, where the forms that the relevant part of the involved
judgments took must be specified and limited (see the section on Deductive Logic above). Predicate Logic is also able to give an account of quantifiers general enough to express all arguments
occurring in natural language, thus allowing the solution of the problem of multiple generality that had perplexed medieval logicians.
For instance, it is intuitively clear that if:
Some cat is feared by every mouse
then it follows logically that:
All mice are afraid of at least one cat
but because the sentences above each contain two quantifiers ('some' and 'every' in the first sentence and 'all' and 'at least one' in the second sentence), they cannot be adequately represented in
traditional logic.
Predicate logic was designed as a form of mathematics, and as such is capable of all sorts of mathematical reasoning beyond the powers of term or syllogistic logic. In first-order logic (also known
as first-order predicate calculus), a predicate can only refer to a single subject, but predicate logic can also deal with second-order logic, higher-order logic, many-sorted logic or infinitary
logic. It is also capable of many commonsense inferences that elude term logic, and (along with Propositional Logic - see below) has all but supplanted traditional term logic in most philosophical
Predicate Logic was initially developed by Gottlob Frege and Charles Peirce in the late 19th Century, but it reached full fruition in the Logical Atomism of Whitehead and Russell in the 20th Century
(developed out of earlier work by Ludwig Wittgenstein).
A logical fallacy is any sort of mistake in reasoning or inference, or, essentially, anything that causes an argument to go wrong. There are two main categories of fallacy, Fallacies of Ambiguity and
Contextual Fallacies:
• Fallacies of Ambiguity: a term is ambiguous if it has more than one meaning. There are two main types:
□ equivocation: where a single word can be used in two different senses.
□ amphiboly: where the ambiguity arises due to sentence structure (often due to dangling participles or the inexact use of negatives), rather than the meaning of individual words.
• Contextual Fallacies: which depend on the context or circumstances in which sentences are used. There are many different types, among the more common of which are:
□ Fallacies of Significance: where it is unclear whether an assertion is significant or not.
□ Fallacies of Emphasis: the incorrect emphasis of words in a sentence.
□ Fallacies of Quoting Out of Context: the manipulation of the context of a quotation.
□ Fallacies of Argumentum ad Hominem: a statement cannot be shown to be false merely because the individual who makes it can be shown to be of defective character.
□ Fallacies of Arguing from Authority: truth or falsity cannot be proven merely because the person saying it is considered an "authority" on the subject.
□ Fallacies of Arguments which Appeal to Sentiments: reporting how people feel about something in order to persuade rather than prove.
□ Fallacies of Argument from Ignorance: a statement cannot be proved true just because there is no evidence to disprove it.
□ Fallacies of Begging the Question: a circular argument, where effectively the same statement is used both as a premise and as a conclusion.
□ Fallacies of Composition: the assumption that what is true of a part is also true of the whole.
□ Fallacies of Division: the converse assumption that what is true of a whole must be also true of all of its parts.
□ Fallacies of Irrelevant Conclusion: where the conclusion concerns something other than what the argument was initially trying to prove.
□ Fallacies of Non-Sequitur: an argumentative leap, where the conclusion does not necessarily follow from the premises.
□ Fallacies of Statistics: statistics can be manipulated and biased to "prove" many different hypotheses.
These are just some of the most commonly encountered types, the Internet Encyclopedia of Philosophy page on Fallacies lists 176!
A paradox is a statement or sentiment that is seemingly contradictory or opposed to common sense and yet is perhaps true in fact. Conversely, a paradox may be a statement that is actually
self-contradictory (and therefore false) even though it appears true. Typically, either the statements in question do not really imply the contradiction, the puzzling result is not really a
contradiction, or the premises themselves are not all really true or cannot all be true together.
The recognition of ambiguities, equivocations and unstated assumptions underlying known paradoxes has led to significant advances in science, philosophy and mathematics. But many paradoxes (e.g.
Curry's Paradox) do not yet have universally accepted resolutions.
It can be argued that there are four classes of paradoxes:
• Veridical Paradoxes: which produce a result that appears absurd but can be demonstrated to be nevertheless true.
• Falsidical Paradoxes: which produce a result that not only appears false but actually is false.
• Antinomies: which are neither veridical nor falsidical, but produce a self-contradictory result by properly applying accepted ways of reasoning.
• Dialetheias: which produce a result which is both true and false at the same time and in the same sense.
Paradoxes often result from self-reference (where a sentence or formula refers to itself directly), infinity (an argument which generates an infinite regress, or infinite series of supporting
references), circular definitions (in which a proposition to be proved is assumed implicitly or explicitly in one of the premises), vagueness (where there is no clear fact of the matter whether a
concept applies or not), false or misleading statements (assertions that are either willfully or unknowingly untrue or misleading), and half-truths (deceptive statements that include some element of
Some famous paradoxes include:
• Epimenides' Liar Paradox: Epimenides was a Cretan who said "All Cretans are liars." Should we believe him?
• Liar Paradox (2): "This sentence is false."
• Liar Paradox (3): "The next sentence is false. The previous sentence is true."
• Curry's Paradox: "If this sentence is true, then Santa Claus exists."
• Quine's Paradox: "yields falsehood when preceded by its quotation" yields falsehood when preceded by its quotation.
• Russell's Barber Paradox: If a barber shaves all and only those men in the village who do not shave themselves, does he shave himself?
• Grandfather Paradox: Suppose a time traveler goes back in time and kills his grandfather when the latter was only a child. If his grandfather dies in childhood, then the time traveler cannot be
born. But if the time traveler is never born, how can he have traveled back in time in the first place?
• Zeno's Dichotomy Paradox: Before a moving object can travel a certain distance (e.g. a person crossing a room), it must get halfway there. Before it can get halfway there, it must get a quarter
of the way there. Before traveling a quarter, it must travel one-eighth; before an eighth, one-sixteenth; and so on. As this sequence goes on forever, an infinite number of points must be
crossed, which is logically impossible in a finite period of time, so the distance will never be covered (the room crossed, etc).
• Zeno's Paradox of Achilles and the Tortoise: If Achilles allows the tortoise a head start in a race, then by the time Achilles has arrived at the tortoise's starting point, the tortoise has
already run on a shorter distance. By the time Achilles reaches that second point, the tortoise has moved on again, etc, etc. So Achilles can never catch the tortoise.
• Zeno's Arrow Paradox: If an arrow is fired from a bow, then at any moment in time, the arrow either is where it is, or it is where it is not. If it moves where it is, then it must be standing
still, and if it moves where it is not, then it can't be there. Thus, it cannot move at all.
• Theseus' Ship Paradox: After Theseus died, his ship was put up for public display. Over time, all of the planks had rotted at one time or another, and had been replaced with new matching planks.
If nothing remained of the actual "original" ship, was this still Theseus' ship?
• Sorites (Heap of Sand) Paradox: If you take away one grain of sand from a heap, it is still a heap. If grains are individually removed, is it still a heap when only one grain remains? If not,
when did it change from a heap to a non-heap?
• Hempel's Raven Paradox: If all ravens are black, then in strict terms of logical equivalence, everything that is not black is not a raven. So every sighting of a blue sweater or a red cup
confirms the hypothesis that all ravens are black.
• Petronius' Paradox" "Moderation in all things, including moderation."
• Paradoxical Notice: "Please ignore this notice."
• Dull Numbers Paradox: If there is such a thing as a dull number, then we can divide all numbers into two sets - interesting and dull. In the set of dull numbers there will be only one number that
is the smallest. Since it is the smallest dull number it becomes, ipso facto, an interesting number. We must therefore remove it from the dull set and place it in the other. But now there will be
another smallest uninteresting number. Repeating this process will make any dull number interesting.
• Protagoras' Pupil Paradox: A lawyer made an arrangement with one of his pupils whereby the pupil was to pay for his instruction after he had won his first case. After a while, the lawyer grew
impatient with the pupil's lack of clients, and decided to sue him for the amount owed. The lawyer's logic was that if he, the lawyer, won, the pupil would pay him according to the judgment of
the court; if the pupil won, then he would have to honor the agreement and pay anyway. The pupil, however, argued that if he, the pupil, won, then by the judgment of the court he need not pay the
lawyer; and if the lawyer won, then the agreement did not come into force and the pupil need not pay the lawyer.
• Moore's paradox: "It will rain but I don't believe that it will."
• Schrödinger's Cat: There is a cat in a sealed box, and the cat's life or death is dependent on the state of a particular subatomic particle. According to quantum mechanics, the particle only has
a definite state at the exact moment of quantum measurement, so that the cat remains both alive and dead until the moment the box is opened.
• "Turtles all the way down": A story about an infinite regress, often attributed to Bertrand Russell but probably dating from centuries earlier, based on an old (possibly Indian) cosmological myth
that the earth is a flat disk supported by a giant elephant that is in turn supported by a giant turtle. In the story, when asked what then supported the turtle, the response was "it's turtles
all the way down".
Major Doctrines Back to Top
Three doctrines which may be considered under the heading of Logic are:
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A theory T has quantifier elimination if every formula ϕ(x) (without parameters) is equivalent to a quantifier-free one (without parameters). This is a stronger condition than model completeness. It
depends on the choice of the language–for example, RCF has quantifier elimination in the language of ordered rings, but not in the language of rings.
Quantifier elimination is a fundamental tool in model theory, because it allows one to get some control over definable sets. When analyzing a theory, quantifier elimination is usually (but not
always) the first step.
Theories with Quantifier Elimination[]
• ACF, the theory of algebraically closed fields, in the language of rings.
• DLO, the theory of dense linear orders, in the language of ordered sets.
• RCF, the theory of real closed fields, in the language of ordered rings.
• pCF, the theory of p-adically closed fields, in the Macintyre language—the field language expanded with unary predicates P[n] picking out that nth powers.
• ACVF, the theory of algebraically closed valued fields (with non-trivial valuation), in the one-sorted language with the ring language and with a binary predicate for divisibility (that is, for
the condition val(x)≤val(y).)
• ACVF in some specific two-sorted and three-sorted languages.
• DCF, the theory of differentially closed fields of characteristic 0.
• The Random Graph. More generally, any Fraïsse limit in a finite relational language. For example, the Henson triangle-free graph.
• Ax-Kochen-Ershov examples…
Important non-examples:
• ACFA
• Pseudo-finite and PAC fields
• RCF in the language of rings.
Any theory can be (trivially) made to have QE by a definitional expansion, by adding a new n-ary relation R[ϕ] for each n-ary formula ϕ(x[1],…,x[n]), to be interpreted as ϕ. This process is
sometimes called Morleyization.
Strategies and criteria for proving QE[]
To prove quantifier elimination, it suffices to show that if ϕ(x;y) is a conjunction of atomic and negated-atomic formulas, and x is a single variable, then ∃x:ϕ(x;y) is equivalent to a
quantifier-free formula. This strategy is useful in cases where there are few function symbols, like DLO. Certain optimizations can be made in this approach, for example:
• Conjuncts in ϕ(x;y) not involving the variable x can be pulled out of the quantifier. So we may assume that every conjunct actually mentions x.
• If every atomic and negated-atomic formula is equivalent to a positive boolean combination of atomic and negated atomic formulas in a certain class Δ, then it suffices to consider those ϕ(x;y)
which are conjunctions of formulas in Δ. For example, in DLO, we can take Δ to be x<y, x>y, and x=y.
• If one of the conjuncts in ϕ(x;y) is of the form x=y[i], then ∃x:ϕ(x;y) is equivalent to the quantifier-free formula ϕ(y[i];y). So we may assume that there are no conjuncts of the form x=
In the DLO case, one is left proving that $\exists x:\bigwedge\limits_{i}x < y_{i} \land \bigwedge\limits_{j}x > z_{j}$ is equivalent to a quantifier-free formula in y[i],z[j], which is relatively
There are also more semantic approaches to proving quantifier elimination. For example, the following conditions are equivalent:
• T has quantifier elimination
• If M and N are models of T, extending a common substructure S, if ϕ(x;y) is a quantifier-free formula, and if ϕ(m;s) holds for some m∈M and s∈S, then ϕ(n;s) holds for some n∈N.
• If M and N are models of T, extending a common substructure S, then the identity map on S is a partial elementary map from M to N. Equivalently, S has the same type in M as in N.
The second condition essentially says that if ψ(z) is an existential formula with one existential quantifier, then the truth of ψ(s) is determined by the quantifier-free type of s. By a simple
compactness argument, this ensures that every existential formula with one quantified variable is equivalent to a quantifier-free formula. Iterating this yields full quantifier elimination.
The third condition essentially says that the quantifier-free type of an element of a model of T determines its full type. Again, by a basic compactness argument, this implies that quantifier
elimination holds.
If T is a complete theory, quantifier elimination is equivalent to the following condition: for every model M, if S and T are substructures of M, and f:S→T is an isomorphism of structures, then
there is an elementary extension N of M and an automorphism σ of N which extends f. This condition is easy to check for ACF, if one believes in the existence of transcendence bases.
If T is model complete, then T has quantifier elimination if and only if T[∀] has the amalgamation property. A related fact, not quite a converse, is that if T is the model companion of an inductive
theory T^′, and T has quantifier elimination, then T^′ has the amalgamation property.
Applications of Quantifier Elimination[]
Quantifier elimination results can be used to show that a theory T is complete, or to classify its completions. If T has quantifier elimination, and ϕ is a sentence (a formula with no free
variables), then ϕ is equivalent to a quantifier-free sentence. If there are no constants in the language, the only quantifier-free sentences are ⊤ and ⊥, so T is complete. In, say, ACF,
quantifier-free sentences are boolean combinations of sentences of the form n=m, for n,m∈ℤ. The truth of these sentences depends only on the characteristic, so models of ACF are classified up to
elementary equivalence by their characteristic.
QE can also be used to see that a theory is countably categorical (via the Ryll-Nardzewski theorem). If a complete theory T has quantifier elimination in a finite relational language, possibly with
constants, then T is countably categorical. The lack of function symbols ensures that there are only finitely many atomic formulas in any finite number of variables. There are only finitely many
boolean combinations of these, so T has only finitely many 0-definable sets in each power of the home sort. By Ryll-Nardzewski, this implies that T is countably categorical.
Control over definable sets is also useful for proving that theories are strongly minimal, o-minimal, or C-minimal. If a theory T has quantifier elimination, to show that T is strongly minimal, it
suffices to show that every subset of the home sort cut out by an atomic formula is finite or co-finite. Similarly, to show that T is o-minimal, it suffices to show that every subset of the home sort
cut out by an atomic formula is a disjoint union of points and intervals.
For example, to see that ACF is strongly minimal, note that the only atomic formulas in one variable x are (essentially) of the form P(x;a)=0, where a is parameters. The set of realizations of
this formula will be the roots of P(x;a)=0, which is either a finite set or the entire field, depending on whether P(X;a) is identically 0 or not.
Similarly, in RCF, one is reduced to showing that sets of the form {x∈R:P(x;a)>0} {x∈R:P(x;a)=0} are boolean combinations of points and intervals. The latter set is finite or all of R,
for reasons just explained. The former set turns out to be a finite boolean combination of intervals with endpoints at the zeros of P(x;a), essentially because the intermediate value theorem holds
for polynomials in models of RCF. So o-minimality holds in RCF.
More generally, conditions like stability, NIP, and NSOP can be verified using quantifier elimination. Quantifier elimination often allows one to get a handle on types, and stability can be proven by
counting types. For example, in DCF, types correspond to prime differential ideals, and there is some way to count these. Something similar happens in SCF. ::: ::: ::: | {"url":"https://james-hanson.github.io/wiki/Quantifier_elimination","timestamp":"2024-11-03T01:23:42Z","content_type":"text/html","content_length":"45091","record_id":"<urn:uuid:87c8cee7-573f-4110-a08b-ce38ca534126>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00601.warc.gz"} |
Game Mode vote
Hello ! i would like to know if someone already worked on a vote for gamemod at the near end of round (since i split my map in 4 round of 5 min) for sourcemod !
something that is edit able in combo mod/weapon vote preset !
having something like
-throwing knife w/ licence to kill
-golden gun w/ deathmatch
i know it can be done by random setting or preset on map cfg but i prefer to let vote settle this up !
when im on the server i sometimes switch the nextround manually for a throwing knife party but i would like this to be able even when im not here but also not pushed to clients !
thanks !
EDIT: I think it would be better if the vote was triggered by users as for a rock the vote
unfortunately i'm not good with scripting but i think it would be a nice idea if it does not already exist ! | {"url":"https://forums.geshl2.com/index.php/topic,8413.0.html?PHPSESSID=558gtk34uckg9pr3fg68vjem2f","timestamp":"2024-11-02T04:58:29Z","content_type":"application/xhtml+xml","content_length":"36394","record_id":"<urn:uuid:6fb6d80e-4980-4878-9ac3-f612ec9fb676>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00537.warc.gz"} |
How to increase the speed when calling gurobi in a loop？
Dear all
I’m currently modeling my problem in python. I need to call gurobi many times to solve QP or non-convex problems, and every time I will use different parameters. What can I do to decrease the
computing time? The question is related two aspects: (1) How to improve the speed of calling gurobi one time to solve the problem? (2) How to reduce the total calculation time caused by multiple
Thank you and best regards
• Hi,
In order to improve the solution speed of Gurobi, we recommend having a look at the most important parameters and experimenting with them in order to improve solution speed. You mentioned that
you are already using different parameters, which parameters are you altering in between runs?
Regarding multiple calls, if your calls are independent, you could try to parallelize your approach.
Since you are solving quadratic programs, numerics and model formulation play an important role. What are the ranges of your matrix coefficients? Could you provide parts of a LOG file for a
problem of which you think that Gurobi could do better, e.g., the coefficient ranges and the barrier log?
Best regards,
• Hi,
Thank you very much！
The parameters I alter in between runs is some constant terms in the constraints, and the difference between them affects the speed of every single run slightly, which can be ignored.
And I am trying to parallelize my approach, but I haven't succeeded, could you please show me some sample codes? (I have tried the method seen from this community, and here I show the error
information. I check this error for many times, and I don't miss a required argument, so I'm confused at why this error happens.)
Here, I show part of the LOG file for my problem.
I am looking forward to your reply!
Best regards,
Linzhi Jiang
• Hi,
I want to add some information that I need to call Gurobi for more than 1,000 times, so I need to reduce the total calculation time caused by the multiple calls.
Best regards,
Linzhi Jiang
• Hi Linzhi,
Just so I understand correctly. Each Gurobi run takes ~0.5 seconds (your log shows 0.38) which I don't think we can reduce much. But the long computation times come from the fact that you need to
solve the model so often? If so, then parallelization is the answer.
In Python you could try using the multiprocessing package. However, please note that each Gurobi run uses multiple Threads when using the barrier method. It may be required that you limit the
number of Threads for each process.
A different approach would be to try to reduce the number of models you have to solve by performing some preprocessing on your own. E.g., are there ways to tell whether a model will be feasible/
infeasible without starting the optimization? Do you have to solve all alteration of the model or can you skip some of them due to information you gain from other runs?Best regards,
• Hi,
Thank you very much！
You mentioned that each Gurobi run uses multiple Threads when using the barrier method and it may be required that I limit the number of Threads for each process. And could you please tell me the
reason of this?
And I have to do the a complete analysis, so I have to solve all alteration of the model, and all the models should be feasible to solve, so is this method (reducing the number of models you have
to solve by performing some preprocessing on your own) still feasible?
Best regards,
Linzhi Jiang
• Hi Linzhi,
Generally Gurobi tries to utilize all Threads during an optimization run. If you want to parallelize your process on one machine, it could happen that the processes intercept each other. It is
recommended to limit the number of Threads of each process, in best case provide the same number of Threads to each process. If you are using multiple machines to parallelize your optimization,
then you usually don't have to set the Threads parameter.
If you are certain that you have to solve all alterations of the model, then the idea of reducing the number of models by analysis is not applicable.
Best regards,
Please sign in to leave a comment. | {"url":"https://support.gurobi.com/hc/en-us/community/posts/360072045631-How-to-increase-the-speed-when-calling-gurobi-in-a-loop?sort_by=created_at","timestamp":"2024-11-02T14:03:18Z","content_type":"text/html","content_length":"61259","record_id":"<urn:uuid:86a71dfb-88bc-484d-b9b2-61dba884005a>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00597.warc.gz"} |
OpenStax College Physics for AP® Courses, Chapter 1, Problem 2 (Problems & Exercises)
A car is traveling at a speed of 33 m/s. (a) What is its speed in kilometers per hour? (b) Is it exceeding the 90 km/h speed limit?
Question by
is licensed under
CC BY 4.0
Final Answer
a. $120 \textrm{ km/h}$ or $1.2\times 10^{2}\textrm{ km/h}$
b. $120 \textrm{ km/h} > 90 \textrm{ km/h}$. Yes, this car is speeding.
Solution video
OpenStax College Physics for AP® Courses, Chapter 1, Problem 2 (Problems & Exercises)
vote with a rating of votes with an average rating of.
Video Transcript
This is College Physics Answers with Shaun Dychko. A car is traveling 33 meters per second and we are gonna convert that into kilometers per hour. So I write the units as a fraction m over s and then
we want to create conversion factors that will cancel the units we don't want and leave us with units that we do want. So we want to get rid of the meters and replace it with kilometers and so meters
is in the numerator of the units we are starting with and so we want meters to be in the denominator of our conversion factor and then we put the units that we do want on top which is kilometers. So
we multiply by 1 kilometer for every 1000 meters and that means the meters cancel because they are being divided by each other and we are left with kilometers here so far. At this point, we have
kilometers per second but now we want to get rid of the seconds units and replace it with hours. Now, if you don't have it memorized that there are 3600 seconds in an hour, you can instead do it
step-by-step and say that there are 60 seconds per minute; this conversion factor has seconds in the numerator which is good because that will cancel with these seconds in the denominator but we are
left with minutes on the bottom which is not good enough and we have to get rid of those and our next conversion factor has 60 minutes per hour; the minutes cancel leaving us with hours in the
denominator. The only units left now are kilometers on top and hours on the bottom and that's what we want and so the arithmetic is 33 divided by a 1000 times 60 times 60 again. So that's 118.8
kilometers an hour but we have to keep in mind appropriate number of significant figures and this value that we start with has two significant figures and we are multiplying by all these conversion
factors so our final answer has to have only two significant figures and so I write that as 120 kilometers per hour. Now this is a bit ambiguous because it's not clear whether there are two
significant figures here or whether the zero is also significant. Now strictly speaking, this has two significant figures unless there was a decimal point here then there would be three but that's a
really unusual way to write things. So this has two significant figures but to be absolutely clear, you could do scientific notation and write it as 1.2 times 10 to the 2 kilometers per hour. This
speed of the car—120 kilometers an hour— exceeds 90 kilometers an hour and so yes, the car is speeding. | {"url":"https://collegephysicsanswers.com/openstax-solutions/car-traveling-speed-33-ms-what-its-speed-kilometers-hour-b-it-exceeding-90-kmh-0","timestamp":"2024-11-08T18:53:14Z","content_type":"text/html","content_length":"118949","record_id":"<urn:uuid:da1e33ef-b849-4e5d-a2a3-48d4fa1b59bb>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00125.warc.gz"} |
Proof of Edit Distance | consensus
Edit distances are a class of algorithms that score how close two strings are to each other. For instance the an Edit Distance for “ETH” and “ETC” is “0.8222” where two identical strings would score
a “1”. There are many algorithms also in the string similarity space including the Levenshtein Distance, Smith-Waterman Gotoh Distance, and the Ratcliff-Obershelp Distance.
Using Proof of Edit distance for forging new blocks
Miners compete to find a string that when hashed in a normalization process is above a minimum distance threshold. This string could then be the hash of an intermediary-blockchain and the header hash
for the next block.
Let Minimum Distance Threshold by “t”, String to find “B”, and Edit Distance Function “ED”. Such that for each blockchain header hash “h” satisfies:
ED( H(h), H(B) ) < t
Or in the case of merging two blockchains:
ED(H(h1),H(B)) < t && ED(H(h2),H(B)) < t === true
To find the new block in an intermediary-blockchain, the miner would iterate through a random charset or number, hashing strings until it finds a hash that is above the threshold for all of the
Proof of Edit Distance is mining algorithm agnostic.
Any hash or string structure can be provided as an input which means that as long as the blockchain has unique hashes it can be easily added to the Proof of Edit challenge.
Used in | {"url":"https://tokens-economy.gitbook.io/consensus/chain-based-proof-of-work/proof-of-distance","timestamp":"2024-11-11T13:59:14Z","content_type":"text/html","content_length":"235186","record_id":"<urn:uuid:fa3c5435-1824-49c3-868d-3e6c31b49b16>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00816.warc.gz"} |
Simplifying the Expression: (7/2a - 5/2b)^2 - (5/2a - 7/2b)^2
This article will guide you through simplifying the expression (7/2a - 5/2b)^2 - (5/2a - 7/2b)^2. We'll use algebraic manipulation and some key identities to achieve a concise result.
Recognizing the Pattern
The given expression looks a bit intimidating at first, but there's a pattern we can exploit. Notice that both terms are squared differences. This suggests we can utilize the "difference of squares"
(a - b)^2 - (c - d)^2 = (a - b + c - d)(a - b - c + d)
Applying the Factorization
Let's apply this factorization to our problem:
• a = 7/2a
• b = 5/2b
• c = 5/2a
• d = 7/2b
Substituting these values into the factorization, we get:
(7/2a - 5/2b + 5/2a - 7/2b)(7/2a - 5/2b - 5/2a + 7/2b)
Simplifying Further
Now we can combine like terms:
(12/2a - 12/2b)(2/2b - 2/2a)
Simplifying the fractions:
(6a - 6b)(b - a)
Final Result
Therefore, the simplified expression is (6a - 6b)(b - a). This result is much more manageable than the original expression, and it highlights the importance of recognizing patterns in algebraic | {"url":"https://jasonbradley.me/page/(7%252F2a-5%252F2b)%255E2-(5%252F2a-7%252F2b)%255E2","timestamp":"2024-11-10T08:00:29Z","content_type":"text/html","content_length":"58668","record_id":"<urn:uuid:010de2b4-7e0c-4bd3-bd7b-d2f4a5b09699>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00412.warc.gz"} |
How to Work on Different Graphs and Charts in Math | WazMagazine.comHow to Work on Different Graphs and Charts in Math - WazMagazine.com
Math is a subject with vast variety to be explored and huge questions to be worked upon. In the very same subject come graphs and charts and their study.
Graphs and Charts are both a part of Statistics. Furthermore, Statistics is a part of each and every field that we know of. Graphs or charts are the visual representations of data formulated by one.
There are several types of charts and graphs that you have to work on in Math:
• Bar Graphs: These graphs, as the name suggests, are represented in the form of bars made on the graph. For example, data is provided for the marks scored by a student in different subjects. In
such a case representation are made for the same in the form of rectangular bars made on the graph.
• Pie Charts: These charts show you representations in a circular manner, unlike the bar graphs which are rectangular. The circle is divided into various sectors based on the degrees that are
calculated using the data and then the graph is made. For example, in the above-mentioned case, the same data is worked upon to calculate the degree for the particular case and it is worked upon
then in the form of pie chart.
• Line Graphs: these are the easiest types of graphs that can be worked upon. These are used when you have an interconnected data that can be represented in the form of graphs. For example, they
can be used to represent the monthly temperature data.
• Cartesian Graphs: These graphs contain numbers on both the axes which help you depict changes happening in one due to variation in another. This type of graph is mostly used in algebra.
All these graphs are worked upon a graph paper which help in the clear depiction of the concepts of all. Before we understand how we go about making the graphs and charts, let us understand what are
the components required to be mentioned while making a graph, so that students can get effective math homework help:
• Axes: Axes are the most important component while we work upon the graphs. Axes are of two types x and y. These axes are drawn perpendicular to each other on the graph. These axes must then be
labelled before we complete the graph. Axes provide meaning to the graph we are working upon.
• Just imagine if you have not labelled the axes then you won’t be able to give meaning to your graph. Both these axes start from zero and are then continued forward. Except in the cartesian graph,
all other graphs have their starting point at zero.
• Scale: Scale is another important component that must be mentioned before we complete a graph. A scale is the mention of the units you have taken on both the axes. For example: 1cm on the x-axes
represents ten students or 10 cm on the y-axes represents one subject. Scale lets the other person know the basic unit of your graph.
So now how do you go about to make the graphs and charts. For this let us take an example: There is a social work group in a city. It comprises of people from different age groups: 2 people from 10
to 20-year category, 5 people from the 20 to 30-year category, 11 people from the 30 to 40-year category, 6 people from the 40 to 50-year category, 8 people from the 50 to 60-year category and 4
people from the 60 to 70-year category. Now this is the basic data for which we will learn the steps to make graphs and charts.
When you are making a bar graph you first have to draw a table for the same data provided above wherein you put the age group in one column and the number of people in the other column.
After doing this you label both these axes on the graph and proceed to plot the graph after mentioning the scale as well. Since the graphs are interconnected in their starting and the end point
therefore conjoined bar graphs are drawn for the same. Similarly, as you move to plot line graphs, it is done by marking points which usually have data on points rather than being in range.
To work upon the pie charts, you need to first calculate the degrees each category occupies in the circle. This is done by taking the number in that category divided by the sum of all the
The answer is further multiplied by 360 degrees to get the degree allotted for that category. For example, in the above data, if we have to calculate the degree occupied by the people in the age
group of 20 to 30, then we will divide 5 by the total of all the observations, that is, 36 and multiply it by 360 degrees, which leaves the answer to be 50 degree. And this way we will get the
degrees for all the age groups.
In case of cartesian graphs, the way is a bit different. A comparison is done between data of both the axes and then co-ordinates are formed in the form of (x, y). These types of graphs are
specifically used to work upon complex unknown calculations in Math.
Graphs and statistics are an important aspect which look very easy to do but since basic details have to be taken into consideration, therefore students mostly have to go for math homework help.
They have a wide range of applications in the real world as well in almost every other field we know of. Be it representing data in business presentations, or data for study and research work or in
fact while working on the concepts for math assignment help, all these places find a mention of graphs and statistics. Therefore the above points must be remembered by the students if the want
effective math assignment help. | {"url":"https://wazmagazine.com/how-to-work-on-different-graphs-and-charts-in-math/","timestamp":"2024-11-02T15:36:06Z","content_type":"text/html","content_length":"110417","record_id":"<urn:uuid:9f914a2e-ba52-4d5b-a015-339d152e3bd0>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00644.warc.gz"} |
Sean and Cam in NY in July
I have been getting questions about this in different threads so rather than keep bringing it up will just post it here.
Sean and I are going to be in NY in July. I will be there July 19-30. A good portion of that time, I think July 22-27, we will be doing an Adyashanti retreat. So basically, I will be hanging out in
NY a few days before and after the retreat. Ime not sure how many days Sean is planning on staying as apparently he wasn't even aware what month the retreat is as he is basically an alcoholic.
I'll most likely get together with Mad Max and his bodyguard Plato at some point and know there are lots of new Tao Bum friends from NYC now. So if you want to get together with Sean about everyone
meeting in China Town or something for a Tao Bums gathering by all means do so.
I look forward to meeting yous bums. I may be heading west on my way to China by then...Perhaps we can wave as our trains pass through Kansas...
I'm in NYC and most likely around in July. If people want to hook up for chinese food, beer, etc. I'd love to meet you.
Let me also just throw something out: we should all sit and meditate together. I mean, we're all into this stuff, we're all connected at least in that and through this space, and I think it's a
really good and intimate thing to do with people you're trying to get to know. I'm sure we'll all be doing different meditations, but to sit with each other might be nice.
Great idea. PM your phone # sometime before July so we can get in touch.
I'm sure we'll all be doing different meditations,
S, I'd only do eyes open mindfulness around those guys. Can't be too careful these days.
S, I'd only do eyes open mindfulness around those guys. Can't be too careful these days.
Perfectly reasonable warning, but I'll take my chances anyway.
Cameron, I'll PM you as we get closer to the dates. Looking forward to meeting you guys.
Cameron, I'll PM you as we get closer to the dates. Looking forward to meeting you guys.
Looking forward to meeting you too. And nice idea with the group meditation. | {"url":"https://www.thedaobums.com/topic/2954-sean-and-cam-in-ny-in-july/","timestamp":"2024-11-09T00:21:59Z","content_type":"text/html","content_length":"108672","record_id":"<urn:uuid:7c7e4498-b64f-4793-b100-c34905816919>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00715.warc.gz"} |
Mathematical optimization and economic theory
Item type Current library Call number Status Date due Barcode
CMI Salle 2 90 INT (Browse shelf(Opens below)) Available 04466-01
Publisher's description: Mathematical Optimization and Economic Theory provides a self-contained introduction to and survey of mathematical programming and control techniques and their applications
to static and dynamic problems in economics, respectively. It is distinctive in showing the unity of the various approaches to solving problems of constrained optimization that all stem back directly
or indirectly to the method of Lagrange multipliers. In the 30 years since its initial publication, there have been many more applications of these mathematical techniques in economics, as well as
some advances in the mathematics of programming and control. Nevertheless, the basic techniques remain the same today as when the book was originally published. Thus, it continues to be useful not
only to its original audience of advanced undergraduate and graduate students in economics, but also to mathematicians and other researchers who are interested in learning about the applications of
the mathematics of optimization to economics. The book is distinctive in that it covers in some depth both static programming problems and dynamic control problems of optimization and the techniques
of their solution. It also clearly presents many applications of these techniques to economics, and it shows why optimization is important for economics. Many cchallenging problems for both students
and researchers are included.
Notes bibliogr. Index
There are no comments on this title. | {"url":"https://catalogue.i2m.univ-amu.fr/bib/3685","timestamp":"2024-11-11T05:03:44Z","content_type":"text/html","content_length":"65807","record_id":"<urn:uuid:7e80f335-6118-4bed-b145-b2ecdb051acd>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00438.warc.gz"} |
Iron Ring - A Canadian Engineer's Tradition
The iron ring is a symbolic piece worn on the small finger of an engineer's dominant hand. We Canadian Engineers know the ring well, and it's purpose is made clear to us all just prior to graduation
from an undergraduate degree in engineering.
It's an exciting event, and we all have a good time (perhaps too good, for some definition of good) in the days leading up to the Iron Ring Ceremony. However, I can confidently say that on the whole
we also take the symbolism to heart. The ritual distills in all attending the importance of taking our work seriously as engineers.
So what is an iron ring, physically?
This is mine, posed on... some metal thing.
It's a simple piece of metal, and it's actually not iron. Admittedly that's kinda strange, but wrought iron tends to react with skin when worn for long periods, so stainless steel is used instead.
So that's what an iron ring actually looks like, but the more important thing is what the rings symbolise for their wearers.
The Meaning of the Iron Ring
The iron ring serves to remind its wearer of their responsibilities as a professional, stressing the importance of ethical and professional conduct. When an engineer looks at their iron ring, they
are reminded of the importance of their work, and the necessity to conduct themselves with honesty, and integrity, while using their skills with pride tempered constantly with humility.
Basically, we have to remember to take our jobs seriously. Our designs are not built in a vacuum (unless that's part of the spec). Our decisions affect more than just ourselves. We impact other
people's lives, and we should always remember that and let that factor into our calculations.
You can read a bit about the backround, which has this to say about the object of the Ritual associated with receiving an iron ring:
The Ritual of the Calling of an Engineer has been instituted with the simple end of directing the newly qualified engineer toward a consciousness of the profession and its social significance and
indicating to the more experienced engineer their responsibilities in welcoming and supporting the newer engineers when they are ready to enter the profession.
Coding the Iron
While I'm here, I may as well provide some code to create a parametric model of the iron ring. It should use the radius and vertical thickness as parameters, in case someone ever wanted to create a
drastically different iron ring, I guess.
NOTE: This is just for fun. If you want to 3D print a ring, I think that's totally fine. However, it would be in poor taste to recreate an iron ring exactly, as it circumvents the important act of
receiving a ring amongst fellow students and engineers.
import math
# R is outer radius of ring
# T is NOT the ring's thickness radially
# but vertically, if the ring is resting on a surface
# ie, the hollow cylinder that is the ring has a HEIGHT of T
R = 9.5
T = 3.0
# Make a function that radially patterns points
def list_points(r, a, N):
'''Radially pattern points
r: the radius of the circle on which points are placed
a: an offset angle from the x axis, CCW, in degrees
N: number of points to place radially along.
Spaces evenly along full 360 circle
arad = math.radians(a)
theta = math.pi*2 / N
pnts = []
for i in range(0, N):
coord = (r*math.cos(theta*i + arad), r*math.sin(theta*i + arad))
return pnts
# Create a plain ring, centered in the plane and around the origin
ring = (cq.Workplane("XZ").workplane(offset=(-T/2.0))
# Make spheres on top and bottom of ring
# These will be cut away from the ring
# To make the 'divets' of an iron ring
spheres1 = cq.Workplane("XZ").workplane(offset=(T-0.4))
spheres2 = cq.Workplane("ZX").workplane(offset=(T-0.4))
for point in list_points((R*1.789), 0, 8):
spheres1 = spheres1.moveTo(point[0], point[1]).sphere(R*0.8526)
for point in list_points((R*1.789), 22.5, 8):
spheres2 = spheres2.moveTo(point[0], point[1]).sphere(R*0.8526)
iron_ring = ring.cut(spheres1).cut(spheres2)
Final Words on the Iron Ring
I take pride in the fact that I have an Iron Ring. I put in some hard work to receive one, and I wear it all the time. But dedication to professionalism, humility, ethics, and purposeful action take
continuous hard work. It's a lifelong committment to being diligent in what I do with my engineering skills. I take this very seriously.
I hope over the years I can live up to the requirements of professionalism, and my iron ring reminds me daily to put in that work.
To end on a light note, however, I want to also stress that a great sense of joy and accomplishment comes with the territory, too. I get to put time and attention to engaging design projects, I can
focus mental efforts on complex engineering questions, and I can use my hands to build awesome things to achieve useful goals in the world.
I take engineering very seriously, and I seriously love engineering.
I'll leave this quote from Rudyard Kipling, the creator of the Iron Ring Ceremony, because I find it so characteristic of how an engineer tends to think about the world. It's a quote that rings of
sincerity from Kipling, yet holds a kind of humour, too as it eschews written works of man in favour of raw materials (an engineer must know their priorities):
The Obligation will be taken on cold iron of honourable tradition, as being a solid substance of proven strength and physical characteristics. It will not be taken on any other written works of
man, but upon a product from nature, used by every engineer. - Kipling
Blurb About the Contents of this Post
This post contains python code that is meant to be used with the CadQuery library. It's a script-based CAD system built upon FreeCAD, an open source CAD package. I also created this inside a Jupyter
Notebook context and use a python module of my own making to integrate CadQuery into the Notebook environment. It's not perfect, but it works well enough.
In this post I have several iframes which are linked to a few a-frame VR scenes. I use these to showcase the parts in browser. It's actually a VR-ready library, so you can actually view the objects
in VR if you've got access to that sort of thing.
All links to the mentioned software:
And, if you'd like to get a feel for CadQuery in the context of a Jupyter notebook, you can check out https://trycadquery.com. It's a minimal server, set up by me, an admitted novice when it comes
to, well, most things, actually. So, please be gentle.
Some Downloads
I've attached a few things that you're welcome to download, details and links listed below:
• iron_ring.step: a CAD file of the iron ring (produced from the code in the post)
• iron_ring.ipynb: the source notebook for this post
• iron_ring.py: source code for the iron ring, as a python script.
Thanks for reading through this short article. I do appreciate you making it this far. If you think someone else you know might like reading this and seeing the code / output in the article, please
share it! I'd love that. It's cool if you don't want to either. You know, no pressure.
You can find me @RustyVermeer if you want to engage me directly. Please do, I like that sort of thing. :) | {"url":"https://www.adamv.xyz/engineering/iron_ring/","timestamp":"2024-11-08T01:52:17Z","content_type":"text/html","content_length":"10322","record_id":"<urn:uuid:8e1ebad9-5fe0-4f55-b3ec-c54775b4ee63>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00748.warc.gz"} |
Excel Formula: Calculate Duration Between Time Values
In this tutorial, we will learn how to calculate the duration between two time values in Excel using a formula. This can be useful when working with time-based data and analyzing time intervals. The
formula we will use is an IF function that checks if the end time is smaller than the start time. If it is, we add 1 to the end time and subtract the start time to get the correct duration. If the
end time is not smaller, we can simply subtract the start time from the end time. Let's dive into the step-by-step explanation of the formula.
Step 1: The formula starts with an IF function to check if the end time is smaller than the start time.
Step 2: If the end time is smaller, we add 1 to the end time and subtract the start time to get the correct duration.
Step 3: If the end time is not smaller, we can simply subtract the start time from the end time to get the duration.
Step 4: The result of the IF function is the calculated duration.
For example, if we have the time values 18:00 and 06:00, the formula would return the value 12:00, which represents a duration of 12 hours between the two time values. Now that you understand the
formula and how it works, let's see some examples and how to apply it in Excel.
An Excel formula
=IF(B1<A1, B1+1-A1, B1-A1)
Formula Explanation
This formula calculates the duration between two time values in the format of "hh:mm". It assumes that the start time is in cell A1 and the end time is in cell B1.
Step-by-step explanation
1. The formula starts with an IF function to check if the end time (B1) is smaller than the start time (A1).
2. If the end time is smaller, it means that the duration spans across two different days. In this case, we need to add 1 to the end time (B1) and subtract the start time (A1) to get the correct
3. If the end time is not smaller than the start time, it means that the duration is within the same day. In this case, we can simply subtract the start time (A1) from the end time (B1) to get the
4. The result of the IF function is the calculated duration.
For example, if we have the following time values in cells A1 and B1:
| A | B |
| | |
| 18:00 | 06:00 |
The formula =IF(B1<A1, B1+1-A1, B1-A1) would return the value 12:00, which represents a duration of 12 hours between 18:00 and 06:00. | {"url":"https://codepal.ai/excel-formula-generator/query/PALtLPo5/excel-formula-calculate-duration-between-two-time-values","timestamp":"2024-11-14T01:00:22Z","content_type":"text/html","content_length":"90954","record_id":"<urn:uuid:428ea9c7-335d-43b4-91b8-41e7cde03113>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00152.warc.gz"} |
Differentiability of a Function
Question Video: Differentiability of a Function Mathematics • Second Year of Secondary School
Consider the function π (π ₯) = |π ₯|. Find lim_(β β 0β Ί) (π (β ))/β . Find lim_(β β 0β ») (π (β ))/β . What can you conclude about the derivative of π (π ₯) at π ₯ = 0?
Video Transcript
Consider the function π of π ₯ is equal to the absolute value of π ₯. Find the limit as β approaches zero from the right of π of β divided by β . Find the limit as β approaches zero
from the left of π of β divided by β . What can you conclude about the derivative of π of π ₯ at π ₯ is equal to zero?
The question gives us the function π of π ₯ is equal to the absolute value of π ₯. The first thing it wants us to do is evaluate this limit. Thatβ s the limit as β approaches zero from the
right of π of β divided by β . And of course, we know that π of β is equal to the absolute value of β . Now, since our limit is as β is approaching zero from the right, our values of β
will be strictly greater than zero. However, if our values of β are greater than zero, our values of β are positive. And this means the absolute value of β is just equal to β . So this gives
us the limit as β approaches zero from the right of β divided by β .
We then cancel the shared factor of β in our numerator and our denominator. We can do this because this gives us a new function which is exactly equal to β divided by β everywhere except where
β is equal to zero. This means their limits as β approaches zero from the right will be the same. This gives us the limit as β approaches zero from the right of the constant one, and we know
that this is just equal to one. So, weβ ve evaluated our first limit; itβ s equal to one.
We can do something similar to evaluate our second limit, the limit as β approaches zero from the left of π of β divided by β . Again, π of β is the absolute value of β . And since our
limit is as β is approaching zero from the left, our values of β will be strictly less than zero. This time, since our values of β are less than zero, our values of β are negative. This tells
us the absolute value of β would just be equal to negative β . So this gives us the limit as β approaches zero from the left of negative β divided by β .
Again, weβ ll cancel the shared factor of β in our numerator and our denominator. And this gives us the limit as β approaches zero from the left of negative one, and negative one is a constant.
So, this limit just evaluates to give us negative one. So, weβ ve now evaluated our second limit; itβ s equal to negative one.
The final part of this question wants us to make a concluding statement about the derivative of our function π of π ₯ when π ₯ is equal to zero. Letβ s start by recalling the definition of the
derivative of a function at a point. We recall the derivative of some function π of π ₯ at the point π ₯ zero is defined as the limit as β approaches zero of π evaluated at π ₯ zero plus β
minus π evaluated at π ₯ zero divided by β if this limit exists.
We want to find the derivative of our function π of π ₯ is equal to the absolute value of π ₯ when π ₯ is equal to zero. So, weβ ll set π of π ₯ to be the absolute value of π ₯ and π ₯
zero to be equal to zero. Substituting these into our definition of the derivative, we get that the derivative of the absolute value of π ₯ when π ₯ is equal to zero is defined to be. The limit as
β approaches zero of the absolute value of zero plus β minus the absolute value of zero divided by β if this limit exists. And we can simplify this limit. First, zero plus β is just equal to
β . Next, the absolute value of zero is just equal to zero.
So, the derivative of the absolute value of π ₯ when π ₯ is equal to zero is equal to the limit as β approaches zero of the absolute value of β divided by β if this limit exists. But remember
what we showed in the first two parts of this question. In the first part, we showed the limit as β approaches zero from the right of the absolute value of β divided by β is equal to one.
However, in our second part, we showed the limit as β approaches zero from the left of the absolute value of β divided by β was equal to negative one. The left and right limits were not equal.
And this tells us that this limit does not exist.
And if this limit does not exist, then that means that the derivative of π of π ₯ when π ₯ is equal to zero also doesnβ t exist. Therefore, for the function π of π ₯ is equal to the
absolute value of π ₯, we can conclude that the derivative of π of π ₯ at π ₯ is equal to zero does not exist because the left-hand and right-hand limits of its derivative at this point are | {"url":"https://www.nagwa.com/en/videos/328150585209/","timestamp":"2024-11-08T12:25:22Z","content_type":"text/html","content_length":"256670","record_id":"<urn:uuid:dbedc8a5-84e1-4975-8418-5652b42196ac>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00297.warc.gz"} |
Convert Kilogram per minute (kg/min) (Mass flow rate)
Convert Kilogram per minute (kg/min)
Direct link to this calculator:
Convert Kilogram per minute (kg/min) (Mass flow rate)
1. Choose the right category from the selection list, in this case 'Mass flow rate'.
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 'Kilogram per minute [kg/min]'.
4. The value will then be converted into all units of measurement the calculator is familiar with.
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.
Utilize the full range of performance for this units calculator
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '434 Kilogram per minute'. In so doing, either the full name of the
unit or its abbreviation can be usedas an example, either 'Kilogram per minute' or 'kg/min'. Then, the calculator determines the category of the measurement unit of measure that is to be converted,
in this case 'Mass flow rate'. 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. 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, '(84 * 98) kg/min'. But different
units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '56 Kilogram per minute + 70 Kilogram per minute' or '13mm x 27cm x 41dm
= ? 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, 6.570 666 606 873 6×1021. For this form of presentation, the number
will be segmented into an exponent, here 21, and the actual number, here 6.570 666 606 873 6. 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 6.570 666 606 873 6E+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: 6 570 666 606 873 600 000 000. Independent of the presentation of the
results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | {"url":"https://www.convert-measurement-units.com/convert+Kilogram+per+minute.php","timestamp":"2024-11-14T14:49:04Z","content_type":"text/html","content_length":"54523","record_id":"<urn:uuid:c38fcd92-dabe-4976-b96f-93b37d58c722>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00644.warc.gz"} |
Number of Days in a Year Calculator
Determine the number of days in a given year, taking into account leap years.
The Number of Days in a Year Calculator is a tool that allows you to calculate the number of days in a specific year, considering the occurrence of leap years.
In a regular or common year, there are 365 days. However, in a leap year, which occurs every four years, there is an additional day, making a total of 366 days. This extra day is added to the month
of February, resulting in February having 29 days instead of the usual 28.
While most years that are divisible by 4 are leap years, there is an exception to this rule. Every 100th year is not considered a leap year, except for every 400th year, which is still a leap year.
For example, the year 2100 will not be a leap year, but the year 2000 was a leap year.
To use the Number of Days in a Year Calculator, simply enter the year you are interested in, and the calculator will determine whether it is a leap year or a common year. It will then display the
corresponding number of days for that year.
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Calculadoras similares
PLANETCALC, Number of Days in a Year Calculator | {"url":"https://pt.planetcalc.com/29/","timestamp":"2024-11-15T03:35:19Z","content_type":"text/html","content_length":"30716","record_id":"<urn:uuid:3a579be7-c614-40b2-81f8-a48097ba8101>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00731.warc.gz"} |
Bearing fault diagnosis method based on Gramian angular field and ensemble deep learning
Inspired by the successful experience of convolutional neural networks (CNN) in image classification, encoding vibration signals to images and then using deep learning for image analysis to obtain
better performance in bearing fault diagnosis has become a highly promising approach. Based on this, we propose a novel approach to identify bearing faults in this study, which includes
image-interpreted signals and integrating machine learning. In our method, each vibration signal is first encoded into two Gramian angular fields (GAF) matrices. Next, the encoded results are used to
train a CNN to obtain the initial decision results. Finally, we introduce the random forest regression method to learn the distribution of the initial decision results to make the final decisions for
bearing faults. To verify the effectiveness of the proposed method, we designed two case analyses using Case Western Reserve University (CWRU) bearing data. One is to verify the effectiveness of
mapping the vibration signal to the GAFs, and the other is to demonstrate that integrated deep learning can improve the performance of bearing fault detection. The experimental results show that our
method can effectively identify different faults and significantly outperform the comparative approach.
• In this study we model a new bearing fault detection method.
• The method is composed of data preprocessing, image-interpreted vibration signal, and ensemble deep learning.
• We use GAFs approach to construct the representation of data because GAFs contain temporal correlation of vibration signal.
• We construct an integrated deep model to achieve a high accuracy rate of bearing fault detection.
• We conclude that our method can obtain a better performance of bearing fault detection on CWRU dataset.
1. Introduction
In modern industries, machine health monitoring is a prerequisite for maintaining the proper operation of industrial machines. Breakdowns in industrial machines can cause huge financial losses and
even pose a threat to the people who use them. Therefore, the need for better and smarter machine health-monitoring technologies has never ceased [1]. Rolling bearings are considered the most common
and critical mechanical components in rotating machinery, and their health can have a significant impact on the performance, stability, and service life of the machine. Because rolling bearings are
usually in harsh operating environments, they are prone to failure during operation. Failure to detect defects in time can lead to unplanned machine downtime or even catastrophic damage. Therefore,
rolling bearing fault detection is essential for the safe and reliable operation of machinery and production [2].
Recently, several bearing fault recognition methods have been proposed. Learning-based (including statistical learning methods and neural network methods) recognition methods can capture mechanical
fault information by learning historical data and thereby enabling the automated analysis of bearing faults. The flow chart of these methods usually includes data preprocessing, feature extraction,
and classifier design. Although a well-designed classification algorithm is a prerequisite for automated bearing fault detection [3], data preprocessing and feature extraction are also important
Designing manual features based on the signal mechanism is a hot field for bearing fault diagnosis. Chen et al. [4] merged the bearing signal features of the time and frequency domains and then
inputted these features into a deep fully connected network for fault detection. Bao et al. [5] calculated the L-kurtosis feature in the envelope spectrum of a vibration signal to detect pulse
periodicity. Chen et al. [6] first transformed a vibration signal into the spectrum domain and extracted the mapping amplitude entropy as a learnable feature. Zhao et al. [7] first performed
empirical mode decomposition (EMD) on a vibration signal, and then selected the top few-mode components containing the main information of the signal to extract the sample entropy. Liu et al. [8]
proposed a feature extraction method based on variational mode decomposition and singular value decomposition (SVD). Unfortunately, these manual feature extraction processes are laborious and
unfriendly to machine-learning designers.
Fig. 1The flow chart of the proposed method
Owing to the powerful feature learning ability of deep learning techniques, many researchers have attempted to introduce deep learning into the field of fault detection. Example include convolutional
neural network (CNN)-based methods [1], [2], [9], [10]-[14], sparse auto-encoder (SAE)-based methods [15], [16], and recursive neural network (RNN)-based methods [17], [18], etc. Because the original
signal is easily affected by noise, the signal is often transformed into an amplitude-frequency domain sequence. Generally speaking, CNN models are good at learning deep features from image data;
thus, one-dimensional vibration signals encoded as two-dimensional image data have attracted much attention. CNN and their improved models have been successfully applied to image classification
because they can extract robust features directly from two-dimensional images. Many image-interpreting vibration signal approaches have been proposed. Ding et al. [2] proposed a method for
reconstructing a two-dimensional wavelet packet energy image (WPI) of the frequency space. The WPI can represent the dynamic structure of the wavelet packet energy distribution of different bearing
faults. However, the WPI method combines the wavelet packet transform and phase space reconstruction technique, which not only has high time complexity but also loses the information of the original
signal when performing multiple transformations of the representation space. Mantas et al. [19] converted the time series to the Permutation Entropy (PE) pattern, which is a 2D digital image with
multiscale time delays. However, the method ignores the amplitude information of the time series, and it is sensitive to noise in the view of the principle behind the method. Wang et al. [20]
combined Symmetrized Dot Pattern (SDP) with CNN for intelligent bearing fault diagnosis. SDP method [21] converts the time series into polar SDP images, which have the frequency and amplitude of the
raw signals. In [22] and [9], they convert the time series into a 2D gray image. In addition, a variety of methods use the time-frequency image to represent the time series [23], [24], but the
time-frequency analysis methods cost a lot of time so it is difficult for actual online diagnosis. Wang et al. [25] encoded a time series as the GAF image and then took advantage of the deep model
for image representation learning to obtain a better classification accuracy rate. The algorithm was validated on a 20-time series dataset and outperformed the traditional k-NN+DTW method, which was
earlier considered the most effective method for time series classification.
The advantages of image-interpreting time series as a GAF image include the following: 1) there is no spatial transformation of the original time series, and the encoding process is performed in the
original representation space. In other words, the time complexity of the algorithm is of the linear order Ο($n$), where n is the length of the time series. 2) The principle of the approach is
simple, easy to understand, and reproducible. Surprisingly, the GAF pictorial method has rarely been applied to bearing fault detection problems. The GAF pictorial method has two different
representations, including the Gramian Angular Summation Field (GASF) and the Differential Difference Field (GADF). The GASF and GADF provide different levels of information, such that, the final
decision results of using them for deep learning are different. Given this, this study introduces the stacking generalization method [26], [27] to fuse the initial decision results based on GASF and
GADF. The proposed method improved the classification accuracy rate of bearing faults and increased the reliability of the results.
The flow of the proposed approach is shown in Fig. 1. It can be seen that the method is composed of data preprocessing, image-interpreted vibration signal, and ensemble deep learning. The
image-interpreting stage was used to better explore the vibration signals information. Ensemble deep learning can make full use of different levels of information to achieve better performance in
bearing fault detection.
In sum, the main contributions of this study are as follows:
1) The GAFs approach is introduced to encode the bearing vibration signal into GASF and GADF matrices. The GAFs contain temporal correlation of vibration signal. The 2D-CNN is used to learn the deep
features of the images, and then the idea of the stacking ensemble method is combined to construct an integrated deep model to achieve a high accuracy rate of bearing fault detection. It is a
decision-level fusion strategy. Due to the deep learning with GASF and GADF obtains different accuracy rates, namely, they contain different information for classification. Therefore, building an
ensemble model is an ideal fault detection scheme.
2) Last but not least, we design two experiments on CWRU datasets to evaluate the performance of bearing fault classification. With those comparable results, we demonstrate that our method achieves a
better performance than the comparative method.
3) The rest of the paper is organized as follows: In Section 2, we introduce the principle of encoding time series as GAF images; we present the proposed method in Section 3; next, we conduct a
performance test on the CWRU dataset. Finally, in Section 5, we conclude the paper.
2. Image-interpreted time series
Encoding vibration signals into images of different granularities is a popular research area for bearing fault diagnosis. Here, we introduce the GAF encoding method, which first transforms the
vibrating bearing signal into a 2D image, and then uses a 2D-CNN to learn the knowledge of the image.
2.1. GAF encoding method
Given a time series $X=\left\{{x}_{1},{x}_{2},\cdots ,{x}_{n}\right\}$, where ${x}_{i}$ is scaled in $\left[-1,1\right]$ or $\left[0,1\right]$ using:
Then, the $\stackrel{~}{X}$ can be encoded as the angular cosine and the timestamp as the radius using Eq. (3):
$\left\{\begin{array}{ll}\varphi =\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{s}\left(\stackrel{~}{{x}_{i}}\right),& -1\le \stackrel{~}{{x}_{i}}\le 1,\mathrm{}\mathrm{}\mathrm{}\mathrm
{}\stackrel{~}{{x}_{i}}\in \stackrel{^}{\mathbit{X}},\\ r=\frac{{t}_{i}}{N},& {t}_{i}\in \mathbb{N},\end{array}\right\$
where ${t}_{i}$ is the timestamp and $N$ is a constant factor. Based on this, a one-dimensional time series is mapped to the two-dimensional image. Because ${x}_{i}\in \left[-1,1\right]$ or ${x}_{i}\
in \left[0,1\right]$, $\mathrm{c}\mathrm{o}\mathrm{s}\left(\varphi \right)$ ($\varphi \in \left[0,\pi \right]$) is monotonic, such that GAF encoding is bijective. In other words, a time series can be
mapped only to a unique polar coordinate space. In addition, the time dependence is preserved by the $r$ coordinates.
Fig. 2GASF images of 10 bearing status
Fig. 3GADF images of 10 bearing status
Once performing the GAF encoding for time series, the temporal correlations within different time intervals are identified by considering the triangular sum/difference between each point:
$GASF\left(i,j\right)=\mathrm{c}\mathrm{o}\mathrm{s}\left({\varphi }_{i}+{\varphi }_{j}\right)=\stackrel{~}{{x}_{i}}\stackrel{~}{{x}_{j}}-\sqrt{1-{\left(\stackrel{~}{{x}_{i}}\right)}^{2}}\sqrt{1-{\
$GADF\left(i,j\right)=\mathrm{s}\mathrm{i}\mathrm{n}\left({\varphi }_{i}-{\varphi }_{j}\right)=\stackrel{~}{{x}_{j}}\sqrt{1-{\left(\stackrel{~}{{x}_{i}}\right)}^{2}}-\stackrel{~}{{x}_{i}}\sqrt{1-{\
The GAF matrix was constructed in the original representation space of the time series. Therefore, the encoding method has significant advantages in terms of efficiency, while avoiding temporal
information loss in the process of representation space transformation [25]. Work [25] pointed out that the GAF has two significant advantages. First, the GAF matrix preserves the time dependence of
the time series, that is, each element of the GAF matrix was generated sequentially from the top left to the bottom right according to the temporal order of the original time series. Second, the GAF
matrix contains temporal correlations; for example, $GAF\left(i,j\right)$ denotes the time interval $k=\left|i-j\right|$ relative correlation in the direction.
In practical applications, direct encoding of the bearing vibration signal in the time domain into a GAF matrix is often unsatisfactory. This is because the bearing signal in the time domain can
easily be contaminated by noise. Therefore, we encoded the amplitude-spectrum sequence of the bearing signal into the GAF image. In the amplitude spectrum, the energy of useful information is
concentrated in a narrow range of frequency bands, whereas the noise energy is distributed over the entire frequency band. Assuming that $\stackrel{-}{X}=\left\{\stackrel{-}{{x}_{1}},\stackrel{-}{{x}
_{2}},\cdots ,\stackrel{-}{{x}_{n}}\right\}$ is the amplitude spectrum of a bearing vibration signal, such that $\stackrel{-}{X}$ can be denoised according to Eq. (6):
${\left(\stackrel{-}{{x}_{i}}\right)}^{\mathrm{"}}=\left\{\begin{array}{c}\stackrel{-}{{x}_{i}}-\mu ,\stackrel{-}{{x}_{i}}>\mu ,\\ 0,\stackrel{-}{{x}_{i}}\le \mu ,\end{array}\right\$
where, $\mu$ is the mean value of $X$. As the amplitude spectrum of the vibration signal is symmetric, it is possible to consider only the left amplitude spectrum. Fig. 2 and 3 show the GASF and GADF
images of 10 different bearing faults, respectively. The class distributions of these faults are shown in Table 1.
Table 1Class distribution of 9 types of faults
Fault type Load (HP) Speed (rpm) Ball Inner race Outer race
0.007" 0 1797 class 1 class 4 class 7
0.014" 0 1797 class 2 class 5 class 8
0.021" 0 1797 class 3 class 6 class 9
3. Denoising method
In real applications, the sampling points of the time series are usually very large; therefore, it is necessary to reduce the dimensionality of the time series before GAF encoding. Considering that
the piecewise aggregate approximation (PAA) algorithm [28] can not only preserve the basic trend of the time series but also has low time complexity, we use the PAA algorithm to pre-process the
bearing vibration signal.
PAA is a simple and effective time series smoothing algorithm that preserves the trend of the time series. The time complexity of PAA is low; therefore, the PAA algorithm is widely used in time
series analysis problems. Considering the time series $X=\left\{{x}_{1},{x}_{2},\cdots ,{x}_{n}\right\}$ is mapped to a new time series $\stackrel{^}{X}=\left\{\stackrel{^}{{x}_{1}},\stackrel{^}{{x}_
{2}},\cdots ,\stackrel{^}{{x}_{m}}\right\}$,$\mathrm{}\stackrel{^}{{x}_{i}}$ can be calculated using the following formula:
$\stackrel{^}{{x}_{i}}=\frac{m}{n}×\left(\sum _{j=n/m\left(i-1\right)+1}^{n/\left(m×i\right)}{x}_{j}\right)\mathrm{},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}1\le i\le m.$
From Eq. (7), we find that $X$ is sequentially divided into $m$ blocks of equal size, and the mean value of each block is used to re-represent the block. The PAA algorithm has a certain noise
reduction effect, as it uses the mean value to smooth the data. Clearly, the selection of $m$ is crucial. If $m$ is too large, the smoothed result loses the original structural information, where $m$
is too small, and the effect is not suitable for noise reduction. From Eq. (7), we can also see that the traditional PAA needs to satisfy $n/m$ as an integer. For $n/m$ is a non-integer number that
can be found in [29] and [30].
3.1. Stacking integration methodology
Bagging, boosting, and stacking methods are three commonly used ensemble learning methods. The bagging method is an algorithm to reduce the variance in the estimate by using voting or mean reversion
to achieve the fusion of multiple decision results [31]. Boosting method can upgrade weak learners to strong learners. Unlike the parallel learning approach of the bagging method, boosting method is
a sequential framework. Boosting method works by sequentially training an initial learner from the training set, and then adjusting the distribution of training samples according to the results of
the initial learner, thus, those instances of wrong decisions of the previous initial learner will be received attention. AdaBoost [32] method is a very classic boosting method. The stacking method
is a different fusion method that is essentially representation learning. The principle of the stacking method is that they perform the second-stage learning with the result of the initial learner.
Stacking method has yielded unusually brilliant results in many data mining competitions (e.g., data science competitions on the Kaggle platform). For example, in the solution proposed by the grand
prize winner of the 2009 Netflix recommendation competition, integrating multiple initial learners is the core of its design [34].
In the case of the classification task, the basic process of the stacking method is to learn different classification algorithms ${\mathcal{L}}_{1},\cdots ,{\mathcal{L}}_{K}$ on the dataset $D$. ${d}
_{i}=\left({x}_{i},{y}_{i}\right)\in D$ is an instance, where ${x}_{i}$ is the feature vector and ${y}_{i}$ is the corresponding label. In the first stage of the stacking method, a set of base
classifiers ${C}_{1},\cdots ,{C}_{K}$ where ${C}_{i}={\mathcal{L}}_{i}\left(D\right)\mathrm{}$are generated. In the second stage, a meta-classifier was learned based on the outputs of the base
classifiers. Note that the leave-one-out method or cross-validation [35] was applied to generate the training set for learning the meta-classifiers [33]. For the leave-one-out method, each base
classifier uses almost all examples and leaves the remaining one for testing. The procedure can be formalized as $\forall i=1,\cdots ,n$ ($n$ is the number of examples), ${C}_{t}^{i}={\mathcal{L}}_
{t}\left(D-{d}_{i}\right)$, $\forall t=1,\cdots ,K$, and next, the base learner is used to classify ${d}_{i}$ by ${\stackrel{~}{y}}_{i}^{t}={C}_{t}\left({x}_{i}\right)$. Therefore, ${d}_{i}$ can be
reconstructed to a new vector $\left(\left({\stackrel{~}{y}}_{i}^{1},\cdots ,{\stackrel{~}{y}}_{i}^{K}\right),{y}_{i}\right)$. The inputs of the meta-learning phase comprised the predictions of the
base classifier.
As the leave-one-out method reconstructs only a sample per learning, it increases the time cost of the reconstruction step, whereas cross-validation predicts a pre-defined subset of the original
sample set at a time and gets the predictions of the base classifier on these subsets. Thus, cross-validation is preferred for applying the stacking algorithm for big data.
4. Experiments
Here, we used CWRU bearing data [36] to verify the effectiveness of the proposed method. The dataset comprised a multivariate vibration time series generated by the bearing test equipment, as shown
in Fig. 4. In this study, the bearing dataset included the following four conditions: normal, outer race failure, inner race failure, and roller failure. The drive-side vibration signal was used with
a sampling rate of 48 kHz and a motor load of 0 hp. Table 1 shows the class distributions of the selected data.
Table 2Accuracy rate of 5 methods on raw CWRU
Algorithm GASF+2D-CNN GADF+2D-CNN WPI+2D-CNN Amplitude spectrum + 1D-CNN Ensemble method
acc 99.4 % 99.8 % 98.7 % 75.6 % 100 %
5. Experimental design
Here, we used a 2D-CNN as the initial learner. Considering useable data is small, the 2D-CNN model includes only three convolutional layers, three pooling layers, and one fully connected structure.
Meanwhile, the batch normalization [37] and the dropout [38] methods are used to reduce the risk of deep model over-fitting. The 2D-CNN uses the cross-entropy loss function, and Adam optimizes the
algorithm [39]. We used the random forest regression method as the meta-learner. We follow the common practice of dividing the CWRU dataset was divided into the training set, test set, and validation
set in the ratio of 0.5, 0.3, and 0.2. The training set was used to train the 2D-CNN and the validation set was used to train the random regression model. The experiments are repeated 10 times under
different random seeds, and the final results are taken as the mean value of the 10 experiments. Considering the uniform distribution of classes, performing different algorithms can be well measured
by the traditional classification accurate rate:
5.1. Raw CWRU data
In this section, the experiments were divided into two parts. The first was to verify the validity of the image interpretation of vibration signals. Its purpose was to compare the fault recognition
performance before and after image encoding. We used the 1D-CNN model to learn one-dimensional vibration signals and applied the 2D-CNN model to the imaged data. Second, we compare the performance of
the proposed method with that of the existing method in [2] (hereafter referred to as WPI+2D-CNN). The topological structure of 2D-CNN is described in the above section. In addition, the 1D-CNN model
contains a one-dimensional convolutional layer, a pooling layer, and a fully connected structure. To reduce the overfitting of CNN, batch normalization and dropout are used to reduce the risk of
We showed the accuracy rates on the original CWRU data in Table2. From the table, we can conclude that: 1) the vibration signal encoded to a two-dimensional image can be better learned to achieve
better performance; 2) the proposed ensemble method gets better performance than the existing method. Although WPI+2D-CNN also gets the accurate rate close to our approach, it has a high time cost
than the proposed method. Due to GAFs performing the outer product of the time series, so the time complexity of GAFs is $\mathrm{O}\left({N}^{2}\right)$, $N$ is the length of time series. Work [2]
does not give the time complexity of WPI, such that we add a test to compare their runtimes. Table 3 shows the time consumption of imaging the raw CWRU data using GAFs and WPI respectively. From the
table, we can see that WPI costs 2938.267 seconds, which is awful larger than GASF and GADF.
Table 3Runtime of GAFs and WPI
GASF GADF WPI
Runtime (s) 2.289 (s) 2.236 (s) 2938.267 (s)
Why can fusing the knowledge of both GASF and GADF can improve accurate rates? From Fig. 2, we can see that class 5, class 6, and class 7 have a similar representation in the GASF image while
presenting a significant difference in the GADF domain. In the same light, class 2 and class 3 have similar GADF features, instead, have different GASF features. Fig. 2 and 3 explain the advantages
of our method. In fact, from the perspective of information theory, fusing different representation features of the GAF can increase the information entropy of the inputs and help improve the
accuracy of the learning-based prediction model.
5.2. Noise-added CWRU data
To further verify the robustness of our method, we added Gaussian white noise to raw CWRU data. We added noise to the vibration signal with SNR of [–5 dB, –2 dB, 2 dB, 5 dB]. The experimental results
are in Table 4. From the table, we can see that: 1) the proposed method can achieve better performance than WPI+2D-CNN in a noisy environment. 2) The fault identification performance of GADF+2D-CNN
is close to GASF+2D-CNN. 3) The ensemble model fusing the GAF image information can achieve excellent performance. Note that for the bearing signal with complex noise, we can choose advanced
denoising methods instead of formula (6). Examples include deep learning, wavelet shrinkage-based, SVD-based methods, and EMD-based methods.
In summary, we can conclude that fusing the knowledge of the GASF and the GADF can improve the performance of bearing fault diagnosis for the CWRU dataset. Considering that the time cost of building
the GAF was low, the proposed method was more in line with the practical application requirements.
Table 4Accuracy rate of 5 methods on noisy CWRU
–5 dB –2 dB 2 dB 5 dB
WPI+2D-CNN 87.1 % 93.2 % 97.6 % 97.8 %
GASF+2D-CNN 99.0 % 99.0 % 99.1 % 99.4 %
GADF+2D-CNN 99. 4 % 99.1 % 99.0 % 99.6 %
Ensemble method 99.8 % 100 % 100 % 100 %
6. Conclusions
This study proposed a bearing fault detection method that combines im-age-interpreting vibration signals and integrating deep learning, which can realize the accurate identification of bearing
faults. Our method encoded one-dimensional vibration signals into a two-dimensional image and then used a 2D-CNN to obtain the initial decision result. Finally, we introduced a decision layer
integration method to realize the fusion of multiple underlying decisions. Experiments on the CWRU real-world dataset show that the proposed method can obtain a better recognition accuracy rate than
the existing method (i.e., WPI+2D-CNN), even when Gaussian white noise is added to the original vibration signal.
Altogether, the learning-based method for bearing fault detection is provided in this work. Next, we plan to apply our method to different publicly available bearing failure datasets and laboratory
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About this article
Fault diagnosis based on vibration signal analysis
bearing fault diagnosis
Gramian angular field
deep learning
ensemble learning
This work is supported by the National Natural Science Foundation of China (Grant No. 61901191), and the Shandong Provincial Natural Science Foundation (Grant No. ZR2020LZH005).
Data Availability
The datasets generated during and/or analyzed during the current study are available from the corresponding author on reasonable request.
Conflict of interest
The authors declare that they have no conflict of interest.
Copyright © 2022 Yanfang Han, et al.
This is an open access article distributed under the
Creative Commons Attribution License
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Electrostatic Energy - Electrical Engineering Textbooks
Electrostatic Energy
Consider a structure consisting of two perfect conductors, both fixed in position and separated by an ideal dielectric. This could be a capacitor, or it could be one of a variety of capacitive
structures that are not explicitly intended to be a capacitor – for example, a printed circuit board. When a potential difference is applied between the two conducting regions, a positive charge
will appear on the surface of the conductor at the higher potential, and a negative charge
will appear on the surface of the conductor at the lower potential (Section 5.19). Assuming the conductors are not free to move, potential energy is stored in the electric field associated with the
surface charges (Section 5.22).
We now ask the question, what is the energy stored in this field? The answer to this question has relevance in several engineering applications. For example, when capacitors are used as batteries, it
is useful to know to amount of energy that can be stored. Also, any system that includes capacitors or has unintended capacitance is using some fraction of the energy delivered by the power supply to
charge the associated structures. In many electronic systems – and in digital systems in particular – capacitances are periodically charged and subsequently discharged at a regular rate. Since power
is energy per unit time, this cyclic charging and discharging of capacitors consumes power. Therefore, energy storage in capacitors contributes to the power consumption of modern electronic systems.
We’ll delve into that topic in more detail in Example 5.25.1.
Since capacitance
relates the charge
to the potential difference
between the conductors, this is the natural place to start. From the definition of capacitance (Section 5.22):
From Section 5.8, electric potential is defined as the work done (i.e., energy injected) by moving a charged particle, per unit of charge; i.e.,
is the charge borne by the particle and
(units of J) is the work done by moving this particle across the potential difference
. Since we are dealing with charge distributions as opposed to charged particles, it is useful to express this in terms of the contribution
made to
by a small charge
. Letting
approach zero we have
Now consider what must happen to transition the system from having zero charge (
) to the fully-charged but static condition (
). This requires moving the differential amount of charge
across the potential difference between conductors, beginning with
and continuing until
. Therefore, the total amount of work done in this process is:
Equation 5.25.13 can be expressed entirely in terms of electrical potential by noting again that
, so
Since there are no other processes to account for the injected energy, the energy stored in the electric field is equal to
The energy stored in the electric field of a capacitor (or a capacitive structure) is given by Equation 5.25.4.
EXAMPLE 5.25.1: WHY MULTICORE COMPUTING IS POWER-NEUTRAL.
Readers are likely aware that computers increasingly use multicore processors as opposed to single-core processors. For our present purposes, a “core” is defined as the smallest combination of
circuitry that performs independent computation. A multicore processor consists of multiple identical cores that run in parallel. Since a multicore processor consists of
identical processors, you might expect power consumption to increase by
relative to a single-core processor. However, this is not the case. To see why, first realize that the power consumption of a modern computing core is dominated by the energy required to continuously
charge and discharge the multitude of capacitances within the core. From Equation 5.25.4, the required energy is
per clock cycle, where
is the sum capacitance (remember, capacitors in parallel add) and
is the supply voltage. Power is energy per unit time, so the power consumption for a single core is
is the clock frequency. In a
-core processor, the sum capacitance is increased by
. However, the frequency is decreased by
since the same amount of computation is (nominally) distributed among the
cores. Therefore, the power consumed by an
-core processor is
In other words, the increase in power associated with replication of hardware is nominally offset by the decrease in power enabled by reducing the clock rate. In yet other words, the total energy of
-core processor is
times the energy of the single core processor at any given time; however, the multicore processor needs to recharge capacitances
times as often.
Before moving on, it should be noted that the usual reason for pursuing a multicore design is to increase the amount of computation that can be done; i.e., to increase the product
. Nevertheless, it is extremely helpful that power consumption is proportional to
only, and is independent of
The thin parallel plate capacitor (Section 5.23) is representative of a large number of practical applications, so it is instructive to consider the implications of Equation 5.25.4 for this structure
in particular. For the thin parallel plate capacitor,
is the plate area,
is the separation between the plates, and
is the permittivity of the material between the plates. This is an approximation because the fringing field is neglected; we shall proceed as if this is an exact expression. Applying Equation 5.25.4:
is the magnitude of the electric field intensity between the plates. Rearranging factors, we obtain:
Recall that the electric field intensity in the thin parallel plate capacitor is approximately uniform. Therefore, the density of energy stored in the capacitor is also approximately uniform. Noting
that the product
is the volume of the capacitor, we find that the energy density is
which has units of energy per unit volume (J/m
The above expression provides an alternative method to compute the total electrostatic energy. Within a mathematical volume
, the total electrostatic energy is simply the integral of the energy density over
; i.e.,
This works even if
vary with position. So, even though we arrived at this result using the example of the thin parallel-plate capacitor, our findings at this point apply generally. Substituting Equation 5.25.10 we
The energy stored by the electric field present within a volume is given by Equation 5.25.12.
It’s worth noting that this energy increases with the permittivity of the medium, which makes sense since capacitance is proportional to permittivity.
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Mathematical calculations in the wild
- The contents of the lesson:
The students were divided into five groups (2-3 students in a group), they were given envelopes with the tasks inside. The students read the tasks aloud. They were given 30 minutes for task
The lesson took place in the reserve of Dubrava urotshistshe. The first group of students had to calculate the beam of the highest pine (They used only a roulette). The second group calculated the
height in which the tree had broken (the feature of similar triangles). The third group had to calculate the length of the survey path in meters using a pedometer and to calculate what percentage of
the whole path includes the path through the marsh. The students of the fourth group had to convert the length of the survey path into ancient units of measurement. The fifth group had to draw a
natural object and calculate the number of times it became bigger after it had been transferred onto a sheet of paper. After having completed the tasks, the students had to present them and
- The learning objective: To perform mathematical calculations by using the objects existing in the wild. | {"url":"https://enature.pixel-online.org/learning_science_EL_scheda.php?id_doc=10&part_id=","timestamp":"2024-11-11T21:25:17Z","content_type":"application/xhtml+xml","content_length":"10743","record_id":"<urn:uuid:88942744-c776-41db-8782-39dd1172a360>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00078.warc.gz"} |
GONG Heyu, SHU Qin, ZHAO Ping
2024, 59(10): 122-126. doi:10.6040/j.issn.1671-9352.0.2023.111
Abstract ( 60 ) PDF (664KB) ( 17 ) Save
References | Related Articles | Metrics
Let T[n] be the full transformation semigroup on X[n]={1,2,…,n}. Let 1≤r≤n, putF[(n,r)]={α∈T[n]:iα=i, ∠i∈{1,2,…,r}},it is obvious that the semigroup F[(n,r)] is subsemigroup of T[n]. In the
paper, we study the core(C F[(n,r)])=〈E(F[(n,r)])〉 of the semigroup F[(n,r)], where E(F[(n,r)])={α∈F[(n,r)]:α^2=α}, by analyzing idempotents of the semigroup F[(n,r)], we prove that the rank
and idempotent rank of semigroup C F[(n,r)] are both equal to ((n-r)(n-r-1))/2+r(n-r)+1. | {"url":"http://lxbwk.njournal.sdu.edu.cn/EN/1671-9352/home.shtml","timestamp":"2024-11-09T01:01:12Z","content_type":"text/html","content_length":"74319","record_id":"<urn:uuid:6f09df3c-f629-45e8-be73-6f6120362ac9>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00873.warc.gz"} |
Partial differential equations/Separation of variables method - Wikiversity
We often consider partial differential equations such as
${\displaystyle abla ^{2}\psi ={\frac {1}{c^{2}}}{\frac {\partial ^{2}\psi }{\partial t^{2}}}}$ ,
which is recognisable as the wave equation in three dimensions, with ${\displaystyle abla ^{2}}$ being the Laplacian operator, ${\displaystyle \psi }$ being some function of three spacial dimensions
and time, and c being the speed of the wave.
These are often found by considering the physical connotations of a system, but how can we find a form of ${\displaystyle \psi }$ such that the equation is true?
Finding General Solutions
One way of doing this is to make the assumption that ${\displaystyle \psi }$ itself is a product of several other functions, each of which is itself a function of only one variable. In the case of
the wave equation shown above, we make the assumption that
${\displaystyle \psi (x,y,z,t)=X(x)\times Y(y)\times Z(z)\times T(t)}$
(NB Remember that the upper case characters are functions of the variables denoted by their lower case counterparts, not the variables themselves)
By substituting this form of ${\displaystyle \psi }$ into the original wave equation and using the three dimensional cartesian form of the Laplacian operator, we find that
${\displaystyle YZT{\frac {d^{2}X}{dx^{2}}}+XZT{\frac {d^{2}Y}{dy^{2}}}+XYT{\frac {d^{2}Z}{dz^{2}}}={\frac {1}{c^{2}}}XZY{\frac {d^{2}T}{dt^{2}}}}$
We can then divide this equation through by ${\displaystyle \psi }$ to produce the following equation:
${\displaystyle {\frac {1}{X}}{\frac {d^{2}X}{dx^{2}}}+{\frac {1}{Y}}{\frac {d^{2}Y}{dy^{2}}}+{\frac {1}{Z}}{\frac {d^{2}Z}{dz^{2}}}={\frac {1}{c^{2}}}{\frac {1}{T}}{\frac {d^{2}T}{dt^{2}}}}$
Both sides of this equation must be equal for all values of x, y, z and t. This can only be true if both sides are equal to a constant, which can be chosen for convenience, and in this case is -(k^2)
The time-dependent part of this equation now becomes an ordinary differential equation of form
${\displaystyle {\frac {d^{2}T}{dt^{2}}}=-c^{2}k^{2}T}$
This is easily soluble, with general solution
${\displaystyle T(t)=A\cos(ckt)+B\sin(ckt)}$
with A and B being arbitrary constants, which are defined by the specific boundary conditions of the physical system. Note that the key to finding the time-dependent part of the original function was
to find an ODE in terms of time. This general process of finding ODEs from PDEs is the essence of this method. | {"url":"https://en.m.wikiversity.org/wiki/Partial_differential_equations/Separation_of_variables_method","timestamp":"2024-11-10T21:25:37Z","content_type":"text/html","content_length":"52349","record_id":"<urn:uuid:bbb35974-98dd-4b5d-9785-362100f7b3ce>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00236.warc.gz"} |
Quantum computer built by Google can instantly execute a task that would normally take 47 years
In a significant leap for the field of quantum computing, Google has reportedly engineered a quantum computer that can execute calculations in mere moments that would take the world’s most advanced
supercomputers nearly half a century to process.
The news, reported by the Daily Telegraph, could signify a landmark moment in the evolution of this emerging technology.
Quantum computing, a science that takes advantage of the oddities of quantum physics, remains a fast-moving and somewhat contentious field.
Quantum computers hold immense promise for potentially revolutionizing sectors like climate science and drug discovery. They offer computation speeds far beyond those of their classical counterparts.
Potential drawbacks of quantum computing
However, this advanced technology is not without its potential drawbacks. Quantum computers pose significant challenges for contemporary encryption systems, thus placing them high on the list of
national security concerns.
The contentious discussion continues. Critics argue that, despite the impressive milestones, these quantum machines still need to demonstrate more practicality outside of academic research.
Astonishing capabilities of Google’s quantum computer
Google’s latest iteration of its quantum machine, the Sycamore quantum processor, currently holds 70 qubits. This is a substantial leap from the 53 qubits of its earlier version. This makes the new
processor approximately 241 million times more robust than the previous model.
As each qubit can exist in a state of zero, one, or both simultaneously, the capability of storing and processing this level of quantum information is an achievement that even the fastest classical
computer, however rapid or slow, cannot match.
The Google team, in a paper published on the arXiv pre-print server, remarked, “Quantum computers hold the promise of executing tasks beyond the capability of classical computers. We estimate the
computational cost against improved classical methods and demonstrate that our experiment is beyond the capabilities of existing classical supercomputers.”
Even the currently fastest classical computers, such as the Frontier supercomputer based in Tennessee, cannot rival the potential of quantum computers.
These traditional machines operate on the language of binary code, confined to a dual-state reality of zeroes and ones. The quantum paradigm, however, transcends this limitation.
Revolutionary power
It remains uncertain how much Google’s quantum computer cost to create. Regardless, this development certainly holds the promise of transformative computational power.
For instance, according to the Google team, it would take the Frontier supercomputer merely 6.18 seconds to match a calculation from Google’s 53-qubit computer.
However, the same machine would take an astonishing 47.2 years to match a computation executed by Google’s latest 70-qubit device.
Quantum supremacy
Many experts in the field have praised Google’s significant strides. Steve Brierley, chief executive of Cambridge-based quantum company Riverlane, labeled Google’s advancement as a “major milestone.”
He also added: “The squabbling about whether we had reached, or indeed could reach, quantum supremacy is now resolved.”
Similarly, Professor Winfried Hensinger, director of the Sussex Centre for Quantum Technologies, commended Google for resolving a specific academic problem tough to compute on a conventional
“Their most recent demonstration is yet another powerful demonstration that quantum computers are developing at a steady pace,” said Professor Hensinger.
He stressed that the upcoming critical step would be the creation of quantum computers capable of correcting their inherent operational errors.
While IBM has not yet commented on Google’s recent work, it is clear that this progress in the realm of quantum computing has caught the attention of researchers and companies worldwide.
This will open new prospects and competition in the evolution of computational technology. Let the games begin!
More about quantum computing
Quantum computing, a remarkable leap in technological advancement, holds the potential to redefine our computational capacities.
Harnessing the strange yet fascinating laws of quantum physics, it could significantly outperform classical computers in solving certain types of problems.
Basics of Quantum Computing
Traditional computers operate based on bits, which can be in a state of either 0 or 1. Quantum computers, on the other hand, operate on quantum bits, known as qubits. Unlike traditional bits, a qubit
can exist in both states simultaneously, thanks to a quantum principle called superposition.
Superposition increases the computing power of a quantum computer exponentially. For example, two qubits can exist in four states simultaneously (00, 01, 10, 11), three qubits in eight states, and so
on. This allows quantum computers to process a massive number of possibilities at once.
Another key quantum principle quantum computers exploit is entanglement. Entangled qubits are deeply linked. Change the state of one qubit, and the state of its entangled partner will change
instantaneously, no matter the distance. This feature allows quantum computers to process complex computations more efficiently.
Applications of Quantum Computers
The unusual characteristics of quantum computing make it ideal for solving complex problems that classical computers struggle with.
Cryptography is a notable area where quantum computing can make a significant difference. The capacity to factor large numbers quickly makes quantum computers a threat to current encryption systems
but also opens the door for the development of more secure quantum encryption methods.
In the field of medicine, quantum computing could enable the modeling of complex molecular structures, speeding up drug discovery. Quantum simulations could offer insights into new materials and
processes that might take years to discover through experimentation.
Challenges in Quantum Computing
Despite its promising potential, quantum computing is not without challenges. Quantum states are delicate, and maintaining them for a practical length of time — known as quantum coherence — is a
significant hurdle.
The slightest environmental interference can cause qubits to lose their state, a phenomenon known as decoherence.
Quantum error correction is another daunting challenge. Due to the fragility of qubits, errors are more likely to occur in quantum computations than classical ones.
Developing efficient error correction methods that don’t require a prohibitive number of qubits remains a central focus in quantum computing research.
Future of Quantum Computing
While quantum computing is still in its infancy, the rapid pace of innovation signals a promising future. Tech giants like IBM, Google, and Microsoft, as well as numerous startups, are making
significant strides in quantum computing research.
In the coming years, we can expect quantum computers to continue growing in power and reliability. Quantum supremacy — a point where quantum computers surpass classical computers in computational
capabilities — may be closer than we think.
Quantum computing represents a thrilling frontier, promising to reshape how we tackle complex problems. As research and development persist, we inch closer to unlocking the full potential of this
revolutionary technology.
More about supercomputers
Supercomputers are high-performance computing machines capable of processing data at super high speeds in comparison to conventional computers.
Renowned for their significant computational power, they perform tasks involving complex calculations that typical computers cannot manage.
Scientists, researchers, and governments use supercomputers to solve intricate problems in areas like quantum physics, weather forecasting, climate research, and biochemical modeling.
The history of supercomputers dates back to the 1960s when the first supercomputer, CDC 6600, designed by Seymour Cray at Control Data Corporation, made its appearance.
Over the years, supercomputers underwent numerous advancements, transitioning from single processor systems to parallel computing designs.
The advent of parallel computing in the 1970s and 1980s allowed supercomputers to increase their computing power exponentially. This involved the use of more than one processor to divide tasks and
conduct computations simultaneously.
In the 1990s, massively parallel computers like the Thinking Machine’s CM-5 started utilizing thousands of processors, marking a significant leap in supercomputing power.
Design and Architecture
Supercomputers possess unique designs and architectures to accommodate their advanced computing needs. Initially, vector processors were common in supercomputers, but with technological advancements,
scalar processors and parallel processing became more prevalent.
Contemporary supercomputers use a variety of architectures. The majority utilize a massively parallel processing (MPP) approach. MPP allows supercomputers to divide large tasks into smaller ones for
simultaneous processing by multiple processors.
Some supercomputers also use grid computing where they link geographically dispersed computers to form a supercomputer.
The architecture of a supercomputer requires meticulous planning and design to accommodate the heat generated by the processors and ensure efficient data transmission. As such, engineers design the
infrastructure and cooling systems in a way that maximizes performance and minimizes energy usage.
Performance Metrics
The performance of supercomputers is typically measured in FLOPS (Floating Point Operations Per Second), a unit that indicates the speed of calculations. The fastest supercomputers today perform at
exaFLOPS levels, that is, they can perform a quintillion floating-point calculations per second.
To rank supercomputers based on their performance, the Top500 project publishes a list twice a year. The rankings depend on a supercomputer’s performance in running the LINPACK benchmark, a software
library that measures a machine’s ability to solve dense systems of linear equations.
Supercomputers find applications in diverse fields. In weather forecasting, they simulate climate models to predict future weather conditions.
The field of space exploration uses supercomputers to simulate and model celestial bodies and galaxies. In the field of physics, supercomputers perform complex simulations like particle collision in
particle physics and nuclear fusion experiments.
Moreover, supercomputers play a pivotal role in medical research, helping to model and understand the structures of viruses, bacteria, and other microscopic organisms.
They also facilitate drug discovery and development by simulating the interaction of molecules with biological targets. Governments also use supercomputers for cryptanalysis, decoding encrypted data
for national security purposes.
Supercomputers have played, and continue to play, a critical role in scientific discovery and technological advancement. By pushing the boundaries of computational power, they enable the resolution
of complex problems across a multitude of domains, ranging from meteorology to quantum physics.
As technologies like quantum computing evolve, the potential of supercomputers will continue to expand, revolutionizing the landscape of high-performance computing.
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Check us out on EarthSnap, a free app brought to you by Eric Ralls and Earth.com. | {"url":"https://www.earth.com/news/quantum-computer-can-instantly-execute-a-task-that-would-normally-take-47-years/","timestamp":"2024-11-04T00:50:25Z","content_type":"text/html","content_length":"135217","record_id":"<urn:uuid:532c7be9-2522-4b17-a5ea-60f2c76639d6>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00131.warc.gz"} |
Re: SAS Visual Analytics New Aggregated Measure displays only number 1 instead of Avg
I am trying to create a new, derived field to use in tables and graphs. I have followed the steps outline in many pages/tutorials but the field is not calculating as expected. I want to create a line
chart noting the volume (frequency) and mean by month. (adding UCL and LCL fields as well but am having same issue with those fields, presumably because they use the mean in their calculations)
Screenshots attached. I don't understand why the mean is not calculating but only displaying a 1.00 per month (category). I attempted to create a new calculated field but the frequency measure does
not appear and I have no other field that is numeric. I see how to add a reference line, but I don't want to have to recalculate it each time. I am using SAS VA 7.3. Thanks in advance for your
12-18-2017 05:50 PM | {"url":"https://communities.sas.com/t5/SAS-Visual-Analytics/SAS-Visual-Analytics-New-Aggregated-Measure-displays-only-number/m-p/422726/highlight/true?attachment-id=13201","timestamp":"2024-11-12T09:16:41Z","content_type":"text/html","content_length":"257140","record_id":"<urn:uuid:1bc3c500-9702-44bb-b270-f56b668a1fec>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00079.warc.gz"} |
Is the quantum world real?
by Lucas | Jan 4, 2021 | Physics, Research | 0 comments
Quantum physics, the field in physics concerned with the behaviour of very small scale systems, has been so far, in many respects, a commendable success, finding many engineering applications, and
refining our current knowledge of the fundamental laws of nature.
However, there still persists contradictions within the theory which, while not having held back industrial development, exacerbate the many remaining questions about the fundamental nature of
reality, namely, we are not concerned with the mathematics of quantum mechanics, but what they say about the world, as C. Rovelli puts it in his seminal paper about his own interpretation of quantum
One of them, probably the most famous, is the measurement problem. It is widely accepted that objects behave as waves, in the sense that they are not located in a single point but they exist as a
probability to be at different points. At the scales which we are used to, this effect is imperceptible. At much smaller scales, this effect becomes more important. For example, if we take a
football, the incertitude of its location would be less than a millionth of its diameter. However, if we take an electron, the incertitude may be much larger than its diameter. Yet, while these
concepts are considered as being an intrisic part of reality, we never observe objects as waves: when we perceive them, it is always as fully localized, existing in one place only.
Therefore, something must happen when the object is observed: it is called the wave function collapse. A wave function collapse happens when the object is interacted with, not only observed. It also
breaks entanglement, but this is not discussed in this post, as the passage from probability to certainty is a sizeable enough problem.
This collapse, currently, has no accepted explanation, only interpretations. Many renowned scientists have proposed varying solutions, and there are currently a decent number of very different
proposed explanations.
The first interpretation, considered as the standard one, and which is used as the basis for most quantum mechanics courses, is the Copenhagen interpretation. This interpretation, formulated by Bohr
and Heisenberg, postulates that quantum mechanics make no statement about the world we see; or that we must renounce giving a quantum physics-explanation of the behaviour of objects at human scales.
One might think that this explanation is not even one, not very science-like, and simply tries to avoid the question. Yet, when it comes to the measurement problem, the answers we offer usually go
beyond the framework that the mathematical framework of physics can formulate. We enter the lawless world of philosophy. We talk about ontology. What this interpretation says is that the quantum
physics world, or the realm described by quantum mechanics, is distinct from the classical world. This explanation has long been frustrating for many physicists, hence the developement of alternative
Another interpretation, which has inpired many artists and authors, is the many-worlds interpretation, developed by Everett, Wheeler, and DeWitt. It states that whenever a measurement is made on a
system which could give the values A or B (note the difficulty of defining what a ‘measurement’ is), reality splits in half, with one branch in which A has been measured, and B in the other. This
interpretation, however raises many questions about the valid definition of measurement again, as it seems important to find out which apparatus may make an observation (does it need to be
conscious?), which creates a branching path, and which doesn’t.
The relational interpretation of quantum mechanics, first put forth in 1994 by Rovelli, states that we must give up the idea that object are in defined state in an absolute manner. Succintly, we may
not say that a system can give a value A or B, we may only say that it can do so relative to an observer, which may, according to Rovelli, be any object, not necessarily a human being as we would
first think.
There exist still quite a few interpretations, however, it would make this post much too long for a casual reading on quantum physics. They have, however, been classified by Butterfield in three
categories: Dynamics, Physics Values, Perspectival Values. The first category includes interpretations whose standpoint is that isolated quantum systems do not obey the Schrödinger equation. The
second category defends the idea that the Schrödinger equations works, and the quantum values associated to it are physical values which exist absolutely in the world. The third category as its name
hints to, states that these aforementioned values are a matter of perspective, such that they depend on the observer.
Quantum physics has puzzled the scientific community for now over half a century. Yet, we may confidently say that this puzzlement has been the source for a dazzling intellectual activity, and while
it has not pushed scientists into asserting new truths, it has been the reason many brilliant and curious scientists have found new and daring questions.
1. Rovelli, Carlo. 1996. ‘Relational Quantum Mechanics’. International Journal of Theoretical Physics 35 (8): 1637–78. https://doi.org/10.1007/BF02302261.
Submit a Comment Cancel reply | {"url":"https://peacock-ai.com/is-the-quantum-world-real/","timestamp":"2024-11-03T23:29:09Z","content_type":"text/html","content_length":"38620","record_id":"<urn:uuid:02e419a1-729e-419e-958e-8975bd0bab33>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00468.warc.gz"} |
SPSS for Windows
Descriptive Statistics Examples
In a computerized statistical analysis, descriptive statistics serve two purposes. The first is to describe the data, especially on those variables that will not be a part of the inferential
statistical analyses. These might include the demographic characteristics of the sample, correlations of dependent variables with other variables measured, and the characteristics of the dependent
variables themselves.
The second purpose is to find evidence of errors in the data entry process. No matter how diligent you are in checking your data during the entry process, it is relatively easy to input a data point
incorrectly. In order to be successful at spotting these data entry errors before doing the inferential statistics, you must be familiar with the data and the variables being measured. Checking the
maximum and minimum scores will often help you spot errors, such as by finding a score that is out of range (i.e., larger or smaller than it could possibly be). Of course, you must know what the
largest and smallest possible scores could be to make this strategy work. Also look for scores that are highly unlikely although technically possible. If they show up, check the original data to make
sure that such scores actually exist. Check to see that the mean is close to the middle of the numbers you remember seeing for a variable during the data collection and entry processes. Be
particularly careful in checking variables that you may have computed--either before data entry or as part of the statistical analyses. Errors in the computations or the formulas given to the
computer program are easy to make and will often result in clearly wrong answers that can be easily spotted if you are looking for them.
In this section we will show you how to set up some basic descriptive statistics for (1) categorical data and (2) score data. In each section, we will show how to generate both descriptive statistics
and appropriate graphs.
Our examples will draw heavily on the data set entered previously and shown in Table 5.2 of the text. Before we do the analyses, we must select the data file that was previously prepared and saved.
When you first start the SPSS program, it will open with a menu. One of the options in this menu is to open data bases that you created earlier. Alternatively, if we are already working in SPSS, we
can open a data file by selecting the File menu, the Open submenu, and the Data choice on the Open submenu, which will give us this screen. We then select the file and click on OK to open it for the
SPSS for Windows program.
Categorical Data
Categorical data represents a classification of participants, and the appropriate summary statistics are frequencies. We compute summary statistics for categorical data in SPSS for Windows by
selecting the Analyze menu, the Descriptive Statistics submenu, and the Frequencies option, which gives us this screen. We want to compute frequencies for the two categorical variables ("Sex" and
"Party"). To do so, we highlight each of these variables in turn by clicking on them in the left box and moving them to the right box by clicking on the arrow button between the boxes. If we change
our mind, we can move the variable back to the left box in the same manner.
Once both variables have been moved, we click on OK and the analysis is run, producing the output shown in this screen. This output lists both the frequency and percent of participants in each
category. Note that on the left side of the screen is table of contents for the output. This structure provides easy access to different sections of complex statistical analyses. The current analysis
is very simple, so the output barely takes a single screen, but many analyses will have pages of output, and this table of contents of the output can be very useful.
Like any data file, the output file can be saved using the save command on the File menu. It can also be printed by using the print command from the File menu.
Sometimes we want to tabulate frequencies for joint categories (e.g., Female Democrats). To do this we use a procedure called crosstabs (short for cross tabulation). We select the Analyze menu, the
Descriptive Statistics submenu, and the Crosstabs option, which will give us this screen.
To do a cross tabulation of Sex by Party, we move one of these variables to the box marked "row(s)" and the other the box marked "column(s)" and press the OK button. This will produce the output
shown here.
Finally, if we wanted to graph the data with a histogram, we select the Graphs menu, the Interactive submenu, and then click on the Bar option (in that order), which gives us this screen.
To produce a graph of the frequencies of political affiliations, we drag and drop the variable Party to the to the X-axis of the graph that is shown with the count on the Y-axis and click on OK. This
produces the Bar graph shown in this screen.
Score Data
Descriptive statistics for score data involve more than just the frequencies of each score. We can produce such a frequency distribution if we desire by using the procedure described above for
obtaining the frequency counts for our categorical variables. If we want additional summary statistics, such as mean and variance, we must use the Descriptives option.
Select the Analyze menu and Descriptive Statistics submenu, and then select the Descriptives option. This process will give us this screen.
Note that not all the variables are listed in the left box. The Descriptives procedure cannot be run on categorical data, and our alphabetical code for the "Sex" and "Party" variables implied that
these were categorical variables. Hence, they were excluded from the list.
We will produce descriptive statistics for the variables "age," "income," and "voted" by moving them from the left box to the right box in the same manner as described previously. We could do the
same for the "ID" variable, but since that variable is simply an identification number for each participant, the analysis would serve no purpose.
The Descriptives option will compute by default the mean, standard deviation, and minimum and maximum scores for each variable that we select. If we want to compute additional summary statistics, we
click on the Options button and select the additional summary statistics we want. When we have identified the variables and selected the summary statistics, clicking on OK will run the analyses,
producing this output.
We have prepared a series of animations that will walk you through the procedures discussed on this page. To run an animation, simply click on the title of the animation in the table below.
Note that we do not recommend that you try to run the animations if you have a slow connection, such as a dial-up connection. You will find that the animations take forever to load with a slow | {"url":"https://graziano-raulin.com/tutorials/spss/spssdes.htm","timestamp":"2024-11-05T06:16:50Z","content_type":"application/xhtml+xml","content_length":"13600","record_id":"<urn:uuid:bffcaede-da51-4591-8ac2-8020dd3cd289>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00541.warc.gz"} |
TCS Digital Advanced Coding Questions and Answers | PrepInsta
TCS Digital Advanced Coding Questions and Answers 2024
Advanced Coding Questions and Answers
TCS Digital is an advanced opportunity offered to the students by TCS. This is a much better position than TCS Ninja, in terms of salary and working experience.
For getting qualified for this post you have to clear TCS Digital Advance Coding Section, which is an additional section after TCS NQT Exam for getting qualified for the Digital profile.
Topics No of Questions
Number of Questions 3
Time Limit 90 mins
Difficulty level High
Package Offered • 7 LPA (B.Tech)
• 7.3 LPA (M.Tech)
TCS Digital has almost the same syllabus as TCS CodeVita Coding Competition, even the last year TCS Digital Coding Round was held on CodeVita compiler, even this year there are very high speculation
that the coding round again will be held on CodeVita only, so we recommend you that while preparing for TCS Digital make sure that you overlook TCS CodeVita Coding Questions.
Lets Prepare Better Faster
Question 1
You are given an array A of size N. Your friend gave me an amazıng task for you. Your friend likes one type of Sequence. So, he called that type of sequence a fair sequence. You should select a fair
sequence of maximum length from an array. Here a fair sequence is nothing but you have to select elements in a pattern like positive element, negative element, positive element… (negative element,
positive element, negative element, to form a sequence.
Your task is to print the maximum sum of elements possible by selecting a fair subsequence with maximum length.
Ex: If array A = [21, 12, 13, -21, -2]. Here your minimum length can be which subsequences 21, -2. The sum is 19. Your task is to print the maximum sum of elements possible by selecting a fair
subsequence with maximum length.
NOTE: You should select the elements in a fair sequence only.
Example – 1:
• Input:
5 – N ( Number of elements in an array )
21 12 13 -21 -2 – Array A consists of N elements
• Output:
• Explanation:
Here you can select 21, -2 subsequences of maximum length 2. The sum is 19 which is the maximum possible for a fair subsequence of length 2.
def fun(arr, n):
ans = []
t = 1
if arr[0] > 0:
t = 0
i = 0
while i < n:
if t == 0:
j = i
while j < n:
if arr[j] > 0:
j += 1
i = j
t = 1
j = i
while j < n:
if arr[j] < 0:
j += 1
i = j
t = 0
return sum(ans)
n = int(input())
ar = list(map(int, input().split()))
print(fun(ar, n))
Question 2
Alice has introduced a new online game which has N levels i e., 1 to N. She wants to get reviews for each level of this game so she will launch only a single level of game on any particular day, and
on that day users of this game are allowed to play only that particular level. As there are N levels so it will take exactly N days to launch all levels of the game, it is not necessary that level of
game will be launched in increasing order, she will pick any random level on particular day and launch it, and it is obvious that any level of the game can’t be launched twice. After completing the
level, the user will give reviews and she will give them one reward unit for each level the user will complete. Alice has put one constraint on users that only after completing any level of game,
user will be able to play higher levels. For E.g. If Alice has launched the 3rd level of the game on the first day and if any user completes it then he will get one reward point, but now. he can’t
play the game with level 1 and level 2.
NOTE: If a user wants to skip to play on any number of days, he is free to do that.
You are the best gamer, so you can easily complete all levels. As you want to maximize your reward points, you want to play as many levels as you can. Alice has already announced the order in which
she will launch levels of his game, your aim is to maximize your reward points.
Given number of levels of games(N) and order of level of games launched one by one on each day. You have to output maximum reward points you can earn.
Hint: You will play any game if and only if that number is becoming the part of the longest subsequence in array of order of games.
Example 1:
• Input:
5 -> N = 5
2 1 3 4 5 -> Order of levels, Alice will launch one by one
• Output:
• Explanation:
If Alice plays the 2nd level of the game on the first day, then he will not be able to play the 1st level of the game on the 2nd day. As after completing the 2nd level, he will be able to play
higher levels. From 3rd day he can play all upcoming levels as those levels are in increasing order. So, possible sequences of levels of games played to maximize rewards points are [2, 3, 4, 5]
or [1, 3, 4, 5]. In both cases he will get 4 reward points.
Example 2:
• Input:
5 -> N = 5
5 4 3 2 1 -> Order of levels, Alice will launch one by one
• Output:
• Explanation:
Alice has launched levels in decreasing order, so he will be able to play exactly one level of game. After playing any level, there are no higher levels on coming days, so maximum reward point is
def fun(n, arr):
if arr == sorted(arr, reverse=True):
return 1
if arr == sorted(arr):
return n
ans = 1
for i in range(n - 1):
m = 1
for j in range(i + 1, n):
if arr[j] > arr[i]:
m += 1
ans = max(m, ans)
return ans
N = int(input())
ar = list(map(int, input().split()))
print(fun(N, ar))
Question 3
King Jeremy’s family has ruled a kingdom for N consecutive years numbered from 1 to N. The year i is described by a prosperity value A[i]. This is a positive integer when there was prosperity in the
year and a negative integer when it was a difficult year for the kingdom. A historical period is described with two integers, S and F as [S, F], i.e. [Start, Finish], where S and F are from 1 to N, S
<= F.
The target is to pick 2 historical periods, with the following 6 conditions:
1. The two periods shouldn’t have common years. For example (1,5) has no common year with [6,7).
2. the First period should start earlier than the second one, 1.e. (1,5) should be the first period, and then the [6,7] should start.
3. There should be a difference of more than K integers between the Finish of the first period and the start of the 2nd period.
4. Sum of prosperity values for the years in the periods chosen should be as large as possible.
Given N, A[ ], and K, give this maximum prosperity value as output.
N = 5, number of years
K = 3, difference between 2 periods.
A = {-1, 1, 2, 3, -1} prosperity value for each year.
There is only 1 way to choose the two periods, which is [1,1] and [5,5) as the difference has to be greater than 3 (value of K).
The prosperity values for these are
[1, 1] = -1
[5, 5] = -1
The Some of these values are -2. Hence, the answer is -2.
Example 1:
• Input:
8 -> Input Integer, N
3 -> Input Integer, K
{5, 5, -1, -2, 3, -1, 2, -2} -> Input sequence, A[ ].
• Output:
12 -> Output
• Explanation:
In the above case: N equals to 8, K equals to 3, Al equals to {5, 5,-1,-2,3,-1, 2, -2}. It is optimal to choose [1, 2] and [7, 7) periods. That is the only optimal choice that you can make.
So the values will be
[1, 2] = 5 5
[7, 7] = 2
The sum of it will be 5 + 5 + 2 = 12
Example 2:
• Input:
6 -> Input Integer, N
0 -> input Integer, K
{5, -1, 5, 0, -1, 9} -> Input sequence A[ ].
• Output:
18 -> Output
• Explanation:
In the above case: N equals 6, and K equals to 0. All equals to {5,-1,5, 0, 1,9}. It is optimal to choose [1, 3] and [6, 6] periods. But that is not the only optimal choice that you can make So
the values will be
[1, 3] = 5 – 1 5
[6, 6] = 9
The sum of it will be 5 -1+5+9 = 18.
n = int(input())
k = int(input())
A = list(map(int, input().split()))
final = []
for i in range(1, n + 1):
for j in range(1, n + 1):
if j >= i:
final.append([i, j])
maximum = 0
for i in range(len(final)):
for j in range(len(final)):
if i != j:
a = [i for i in range(final[i][0], final[i][1] + 1)]
b = [i for i in range(final[j][0], final[j][1] + 1)]
if len(set(a).intersection(set(b))) == 0:
if final[i][1] < final[j][0]:
ans = final[j][0] - final[i][1]
if ans > k:
sumvalue = 0
for p in range(((final[i][0]) - 1), final[i][1]):
sumvalue += A[p]
for p in range(((final[j][0]) - 1), final[j][1]):
sumvalue += A[p]
maximum = max(maximum, sumvalue)
Question 4
Radha is attending her sister’s wedding, and there are a lot of things to do including packing and food and decorating. Radha took up the responsibility for packing the sweets, as no one was
interested in doing such tedious work. She got an idea, where she announced that we will pick a winner from a packing contest, and those who win this game will be given a surprise gift. The best part
is that more than 1 person can be the winner in this game.
Radha, by default, is the first player. So, the game goes like this:
• All the sweet boxes are represented with numbers, to differentiate them from each other. Let us say that a box numbered ‘1’ will contain a different sweet compared to the one which is numbered
• Each player will be given different/same combinations of sweets boxes in different/same quantities.
• Each player has to pack all the respective boxes given to them, in one final package.
• As a default point, each person will be given the number of points equal to the number of boxes they pack. Let us say they have 6 sweet boxes in their queue, they will get 6 points as default.
• If the final package contains 4 different types of sweets, then they will get an additional 1 point.
• If the final package contains 5 different types of sweets, then they will get additional 2 points.
• If the final package contains 6 or more different types of sweets, then they will get additional 4 points.
Radha is Player 1, and the remaining players follow after that. The input will be in the format given below:
For Player-1 (Radha):
This means there is only 1 player, and 6 is the number of boxes given to her. And the remaining values (1 2 3 4 5 7).
If we have multiple participants then, say we have 3 participants
This means we have 3 participants (N=3)
First participant:
• Number of boxes 6, N = Z[0]
• The types of sweets in these 6 boxes are, respectively, as 1 2 3 4 5 7, 1.e., Z[1, N]
Second participant:
• Number of boxes 4, N = Z[0]
• The types of sweets in these 6 boxes are, respectively, as 1 3 2 2, i.e., Z[1, N]
Third participant:
• Number of boxes 5, N = Z[0]
• The types of sweets in these 6 boxes are respectively, as 1 2 2 3 3 i.e., Z[1, N]
Your output will be:
Radha: If Radha wins.
Tie: If 2 or more 2 players have the same score.
A[i]: Index of the player.
Example 1:
• Input:
2 -> Input string. Integer N.
6 1 2 3 4 5 6-> Input Integer, Z[ ]
9 3 3 3 4 4 4 5 5 5 -> Input integer, Z[ ]
• Output:
Radha -> Output
• Explanation:
From the above inputs let’s calculate the total points for each player.
Player 1: Radha
Has 6 boxes, so 6 points in the beginning.
She has 6 different types of sweets (from 1-6), hence 6 more points to that.
Total of 6 + 6 = 12 points
Player 2:
Has 9 boxes, so 9 points in the beginning.
She has only 3 different types of sweets (3,4 & 5), so no points to that.
Total of 9 points.
Radha is a winner with more points.
Example 2:
• Input:
2 -> Input string. Integer N.
3 1 2 3 -> Input Integer, A[ ], Z[ ].
4 1 2 3 4 -> Input integer, A[ ], Z[ ].
• Output:
2 -> Output
• Explanation:
From the above inputs let’s calculate the total points for each player:
Player 1: Radha
Has 3 boxes, so 3 points in the beginning.
She has 3 different types of sweets (from 1-3), so no points to that.
Total of 3 points.
Player 2:
Has 4 boxes, so 4 points in the beginning.
She has only 3 different types of sweets (1 – 4), so 1 point to that.
Total of 3+1 = 4 points.
Player 2 is a winner with more points. So, the answer is 2.
def find(l1):
length = len(l1)
finalans = l1[0]
setl1 = set(l1)
if len(setl1) == 6:
finalans += 4
elif len(setl1) == 4:
finalans += 1
elif len(setl1) == 5:
finalans += 2
finalans += sum(setl1)
return finalans
n = int(input())
final = []
for i in range(n):
l1 = list(map(int, input().split()))
finalSum = find(l1)
maximum = max(final)
countofMax = final.count(maximum)
if countofMax == 1:
if final[0] == maximum:
for i in final:
if i == maximum:
print(i + 2, end=" ")
Question 5
Reddington has invested his money in stock of N companies, but lately, he realized that he has invested money in a very random manner. He wants to restructure the money he invested. He decided on an
integer K which he will use to modify the amount invested either by putting k more amount for a particular company or withdrawing the same K amount. He will withdraw from one company by selling a
stock) and will invest the same amount of K in another company (by buying stock). For more than one transaction (either buy or sell) of stock in the same company have some charges per month, such
that he will not do more than one transaction of a particular company means either he will buy or sell the stock of amount K for a company or he will not do any transaction for that company. But he
can do as many transactions as he wants to be provided that in one company he will do only one transaction. As you are his best friend so he reached out to you o for help.
Reddington wants to keep the difference between the smallest and largest invested amount minimal, can you please help him to get this minimum difference?
NOTE: Reddington can sell the stock of K from a particular company only if the invested amount value for that particular company is higher than k.
Hint: To keep the difference minimum, if any transaction is needed then he needs to sell the stock of amount from the company with the highest invested amount and invest the same in the company with
the smallest invested amount
Example 1:
• Input:
2100 > N=100, K=100
100 1000 -> Invested amount for each company
• Output:
• Explanation:
Two companies C1 and C2 have invested amounts of 100 and 1000. From the company, C2 Reddington will sell the stock of 100(K) units and buy the stock of company C2. So, now the invested amount
will be 200 and 900, and now Reddington can’t do more transactions, so the minimum difference will be 900-200 = 700.
Example 2:
• Input:
5 100 -> N=5, K=100
100 200 300 400 500 -> Invested amount for each
• Output:
• Explanation:
Five companies C1, C2, C3, C4, and C4 having invested amounts as 100, 200, 300, and 400,500 Reddington will sell the stock of C5 of 100 units and put it in C1. So, the invested amount will be
200, 200, 300, 400, 400, Now Reddington will sell the stock of C4 and put it in C2, then the invested amount will be 200, 300, 300, 300,02 400. Reddington can’t do more transactions now, so the
minimum difference will be 400 – 200 = 200.
def fun(n, k, arr):
c = (max(arr) + min(arr)) // 2
for i in range(n):
if arr[i] < c:
arr[i] += k
elif arr[i] > c:
arr[i] -= k
return max(arr) - min(arr)
N, K = map(int, input().split())
ar = list(map(int, input().split()))
print(fun(N, K, ar))
Question 6
Vijay, an industrialist, recently opened a fuel industry. There are N numbers of different categories of fuel and each fuel is stored in a fixed site container. The size of the container can vary
from fuel to fuel. Hari, a fuel merchant, visited Vijay’s industry and he wanted to purchase fuels for his shop. Hari has a limited amount of money (K) and wants to buy fuel. Hari will not buy more
than one container of any fuel category. Hari wants to maximize the overall volume i.e., the sum of the volume of fuels he buys. Your task is to get the maximum overall volume of fuels that Hari can
Given the number of categories of fuels (N), money units (K) price of a container of each category of fuel, and volume contained in a container for each category.
NOTE: Quantity (volume) of container will be given in the same order as the volume of price.
Hint: Break problems into parts and get the overall volume for smaller N and smaller K.
Example 1:
• Input:
5, 105 -> N = 5, K=105
10 10 40 50 90 -> prices of the container of each fuel category.
10 20 20 50 150 -> volume of container of each fuel Category.
• Output:
• Explanation:
There are 5 fuel categories and Hani has 105 units of $ money. To have maximum fuel volume Hari can buy oil in 2nd and 5th positions. Total cost= 10 + 90 = 100 and it is less than 105(money which
Hari has). Total volume = 20 + 150 = 170. It’s the max volume Hari can get; no other combination can get more volume than this.
Example 2:
• Input:
5 100 -> N = 5, K=100
10 20 30 40 100 -> prices of a container of each fuel category.
10 20 30 40 100 -> volume of container of each fuel
• Output:
• Explanation:
Hari can buy either 1st four categories or only 5th category of fuel. In either case, the total amount will be 100 and the volume will be 100 (which is the maximum volume possible).
def getMaxVol(money, price, volume, n):
K = []
for i in range(n + 1):
temp = []
for j in range(money + 1):
for i in range(n + 1):
for m in range(money + 1):
if i == 0 or m == 0:
K[i][m] = 0
elif price[i - 1] <= m:
K[i][m] = max(volume[i - 1] + K[i - 1][m - price[i - 1]], K[i - 1][m])
K[i][m] = K[i - 1][m]
return K[n][money]
N, money = map(int, input().split())
price = list(map(int, input().split()))
volume = list(map(int, input().split()))
print(getMaxVol(money, price, volume, len(volume)))
Question 7
Problem Statement-:
Identify the logic behind the series
6 28 66 120 190 276….
The numbers in the series should be used to create a Pyramid. The base of the Pyramid will be the widest and will start converging towards the top where there will only be one element. Each
successive layer will have one number less than that on the layer below it. The width of the Pyramid is specified by an input parameter N. In other words there will be N numbers on the bottom layer
of the pyramid.
The Pyramid construction rules are as follows
1. First number in the series should be at the top of the Pyramid
2. Last N number of the series should be on the bottom-most layer of the Pyramid, with N^thnumber being the right-most number of this layer.
3. Numbers less than 5-digits must be padded with zeroes to maintain the sanctity of a Pyramid when printed. Have a look at the examples below to get a pictorial understanding of what this rule
actually means.
If input is 2, output will be
If input is 3, output will be
Formal input and output specifications are stated below
Input Format:
• First line of input will contain number N that corresponds to the width of the bottom-most layer of the Pyramid
Output Format:
• The Pyramid constructed out of numbers in the series as per stated construction rules
int main ()
int n, a = 0, b = 3, i, re, j;
scanf ("%d", &n);
for (i = 1; i <= n; i++)
for (j = 1; j <= i; j++)
a = a + 2;
if (i == 1)
b = 3;
b = b + 4;
re = a * b;
printf ("%.5d ", re);
printf ("\n");
return 0;
Question 8
Problem Statement-: There are two banks – Bank A and Bank B. Their interest rates vary. You have received offers from both banks in terms of the annual rate of interest, tenure, and variations of
the rate of interest over the entire tenure.You have to choose the offer which costs you least interest and reject the other. Do the computation and make a wise choice.
The loan repayment happens at a monthly frequency and Equated Monthly Installment (EMI) is calculated using the formula given below :
EMI = loanAmount * monthlyInterestRate / ( 1 – 1 / (1 + monthlyInterestRate)^(numberOfYears * 12))
• 1 <= P <= 1000000
• 1 <=T <= 50
• 1<= N1 <= 30
• 1<= N2 <= 30
Input Format:
• First line: P principal (Loan Amount)
• Second line: T Total Tenure (in years).
• Third Line: N1 is the number of slabs of interest rates for a given period by Bank A. First slab starts from the first year and the second slab starts from the end of the first slab and so on.
• Next N1 line will contain the period and their interest rate respectively.
• After N1 lines we will receive N2 viz. the number of slabs offered by the second bank.
• Next N2 lines are the number of slabs of interest rates for a given period by Bank B. The first slab starts from the first year and the second slab starts from the end of the first slab and so
• The period and rate will be delimited by single white space.
Output Format: Your decision either Bank A or Bank B.
5 9.5
10 9.6
5 8.5
10 6.9
5 8.5
5 7.9
Output: Bank B
13 9.5
3 6.9
10 5.6
14 8.5
6 7.4
6 9.6
Output: Bank A
int main ()
double p, s, mi, sum, emi, bank[5], sq;
int y, n, k, i, yrs, l = 0;
scanf ("%lf", &p);
scanf ("%d", &y);
for (k = 0; k < 2; k++)
scanf ("%d", &n);
sum = 0;
for (i = 0; i < n; i++)
scanf ("%d", &yrs);
scanf ("%lf", &s);
mi = 0;
sq = pow ((1 + s), yrs * 12);
emi = (p * (s)) / (1 - 1 / sq);
sum = sum + emi;
bank[l++] = sum;
if (bank[0] < bank[1])
printf ("Bank A");
printf ("Bank B");
return 0;
Question 9
Problem Statement-:One person hands over the list of digits to Mr. String, But Mr. String understands only strings. Within strings also he understands only vowels. Mr. String needs your help to find
the total number of pairs that add up to a certain digit D.
The rules to calculate digit D are as follows
• Take all digits and convert them into their textual representation
• Next, sum up the number of vowels i.e. {a, e, i, o, u} from all textual representation
• This sum is digit D
Now, once digit D is known find out all unordered pairs of numbers in input whose sum is equal to D. Refer example section for better understanding.
• 1 <= N <= 100
• 1 <= value of each element in second line of input <= 100
Number 100, if and when it appears in the input should be converted to textual representation as a hundred and not as one hundred. Hence the number of vowels in the number 100 should be 2 and not 4
• the First line contains an integer N which represents the number of elements to be processed as input
• the Second line contains N numbers separated by space
• Lower case representation of a textual representation of the number of pairs in input that sum up to digit D
Note: – (If the count exceeds 100 print “greater 100”)
Time Limit
Example 1
1 -> one -> o, e
2 -> two -> o
3 -> three -> e, e
4 -> four -> o, u
5 -> five – > i, e
Thus, count of vowels in textual representation of numbers in input = {2 + 1 + 2 + 2 + 2} = 9. Number 9 is the digit D referred to in the section above.
Now from the given list of numbers {1,2,3,4,5} -> find all pairs that sum up to 9.
Upon processing this we know that only a single unordered pair {4, 5} sum up to 9. Hence the answer is 1. However, output specification requires you to print textual representation of number 1 which
is one. Hence the output is one.
Note: – Pairs {4, 5} or {5, 4} both sum up to 9. But since we are asking to count only unordered pairs, the number of unordered pairs in this combination is only one.
Example 2
7 -> seven -> e, e
4 -> four -> o, u
2 -> two -> o
Thus, count of vowels in textual representation of numbers in input = {2 + 2 + 1} = 5. Number 5 is the digit D referred to in the section above.
Since no pairs add up to 5, the answer is 0. The textual representation of 0 is zero. Hence the output is zero.
using namespace std;
vector <string>k(101,"");
int vowel(string s)
int sum=0;
for(auto i:s) if(i=='a'||i=='e'||i=='i'||i=='o'||i=='u') sum++;
return sum;
string findWord(int a)
if(k[a]!="") return k[a];
if(a<20&&a>15) return k[a]=k[a%10]+"teen";
if(a%10==0) return k[a]=k[(a/10)]+"ty";
return k[a]=k[(a/10)*10]+"-"+k[a%10];
int main()
//for(int i=0;i<101;i++)cout<<i<<" "<<findWord(i)<<endl;
int n,sum1=0,sum2=0; cin>>n;
vector <int> a(n);
for(int i=0; i<n; i++) { cin>>a[i];
for(int i=0; i<n-1;i++){
for(int j=i+1; j<n;j++)
if(a[i]+a[j]==sum1) sum2++;
Question 10
Problem Statement:- Jaya invented a Time Machine and wants to test it by time-traveling to visit Russia on the Day of Programmer (the 256thday of the year) during a year in the inclusive range from
1700 to 2700. From 1700 to 1917 , Russia’s official calendar was the Julian Calendar since 1919 they used the Gregorian calendar system. The transition from the Julian to Gregorian calendar system
occurred in 1918 , when the next day after 31 January was February 14 . This means that in 1918, February 14 was the 32nd day of the year in Russia. In both calendar systems, February is the only
month with a variable amount of days; it has 29 days during a leap year, and 28 days during all other years. In the Julian calendar, leap years are divisible by 4 ; in the Gregorian calendar, leap
years are either of the following:
• Divisible by 400
• Divisible by 4 and not divisible by 100
Given a year, y, find the date of the 256th day of that year according to the official Russian calendar during that year. Then print it in the format dd.mm.yyyy, where dd is the two-digit day, mm is
the two-digit month, and yyyy is y.
For example, the given year is 1984.1984 is divisible by 4, so it is a leap year. The 256 day of a leap year after 1918 is September 12, so the answer is 12.9.1984.
Function Description
• Complete the programmerday function in the editor below. It should return a string representing the date of the 256th day of the year given.
• programmerday has the following parameter(s):
Input Format
• A single integer denoting year y.
Output Format
• Print the full date of programmerday during year y in the format dd.mm.yyyy, where dd is the two-digit day, mm is the two-digit month, and yyyy is y.
Sample Input
Sample Output
#include <stdio.h>
int main()
int year;
if(year>=1700 && year<=1917)
else if(year==1918)
return 0;
Question 11
Problem Statement:- Hobo’s Drawing teacher asks his class to open their books to a page number. Hobo can either start turning pages from the front of the book or from the back of the book. He always
turns pages one at a time. When she opens the book, page 1 is always on the right side: When he flips page 1, he sees pages 2 and 3. Each page except the last page will always be printed on both
sides. The last page may only be printed on the front, given the length of the book. If the book is n pages long, and he wants to turn to page p, what is the minimum number of pages he will turn? He
can start at the beginning or the end of the book. Given n and p, find and print the minimum number of pages Hobo must turn in order to arrive at page p
Function Description
Complete the countpage function in the editor below. It should return the minimum number of pages Hobo must turn.
countpage has the following parameter(s):
• n: the number of pages in the book
• p: the page number to turn to
Input Format
• The first line contains an integer n, the number of pages in the book.
• The second line contains an integer, p, the page that Hobo’s teacher wants her to turn to.
Output Format
• Print an integer denoting the minimum number of pages Hobo must turn to get to page p
Sample Input
Sample Output
int main()
int n, p, min;
min = (n/2)-(p/2);
min = p/2;
return 0;
Question 12
Problem Statement:- Dr. Vishnu is opening a new world class hospital in a small town designed to be the first preference of the patients in the city. Hospital has N rooms of two types – with TV and
without TV, with daily rates of R1 and R2 respectively.
However, from his experience Dr. Vishnu knows that the number of patients is not constant throughout the year, instead it follows a pattern. The number of patients on any given day of the year is
given by the following formula –
where M is the number of month (1 for jan, 2 for feb …12 for dec) and D is the date (1,2…31).
All patients prefer without TV rooms as they are cheaper, but will opt for with TV rooms only if without TV rooms are not available. Hospital has a revenue target for the first year of operation.
Given this target and the values of N, R1 and R2 you need to identify the number of TVs the hospital should buy so that it meets the revenue target. Assume the Hospital opens on 1st Jan and year is a
non-leap year.
Hospital opens on 1st Jan in an ordinary year
• 5 <= Number of rooms <= 100
• 500 <= Room Rates <= 5000
• 0 <= Target revenue < 90000000
Input Format
• First line provides an integer N that denotes the number of rooms in the hospital
• Second line provides two space-delimited integers that denote the rates of rooms with TV (R1) and without TV (R2) respectively
• Third line provides the revenue target
• Minimum number of TVs the hospital needs to buy to meet its revenue target. If it cannot achieve its target, print the total number of rooms in the hospital.
Test Case
Example-1 :
Using the formula, the number of patients on 1st Jan will be 39, on 2nd Jan will be 38 and so on. Considering there are only twenty rooms and rates of both type of rooms are 1500 and 1000
respectively, we will need 14 TV sets to get revenue of 7119500. With 13 TV sets Total revenue will be less than 7000000
Example-2 :
In the above example, the target will not be achieved, even by equipping all the rooms with TV. Hence, the answer is 10 i.e. total number of rooms in the hospital.
using namespace std;
int R2,R1,N;
int M[]={0,25,16,9,4,1,0,1,4,9,16,25,36};
int D[]={0,14,13,12,11,10,9,8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
int MM[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int func(int p)
{int s,sum=0;
for(int m=1;m<=12;m++)
for(int d=1;d<=MM[m];d++)
sum+=min(p,s)*R2 + (s-min(p,s))*R1;
return sum;
int main()
int R,l=0;
int nn=N;
Question 13
Problem Statement:- You will be given an array of integers and a target value. Determine the number of pairs of array elements that have a difference equal to a target value.
For example, given an array of [1, 2, 3, 4] and a target value of 1, we have three values meeting the condition: 2-1 = 1, 3-2 = 1, and 4-3 = 1.
Function Description
Write a function pairs. It must return an integer representing the number of element pairs having the required difference.
Pairs has the following parameter(s):
• k: an integer, the target difference
• arr: an array of integers
Input Format
• The first line contains two space-separated integers n and k, the size of arr and the target value.
• The second line contains n space-separated integers of the array arr.
Sample Input
Sample Output
#include <stdio.h>
int countPairsWithDiffK(int arr[], int n, int k)
int count = 0;
for (int i = 0; i < n; i++)
for (int j = i+1; j < n; j++)
if (arr[i] - arr[j] == k || arr[j] - arr[i] == k )
return count;
int main()
int arr[] = {1, 5, 3, 4, 2};
int n = sizeof(arr)/sizeof(arr[0]);
int k = 3, result;
result = countPairsWithDiffK(arr, n, k);
return 0;
Question 14
Problem Statement:- A jail has several prisoners and several treats to pass out to them. Their jailer decides the fairest way to divide the treats is to seat the prisoners around a circular table in
sequentially numbered chairs. A chair number will be drawn from a hat. Beginning with the prisoner in that chair, one candy will be handed to each prisoner sequentially around the table until all
have been distributed.
The jailer is playing a little joke, though. The last piece of candy looks like all the others, but it tastes awful. Determine the chair number occupied by the prisoner who will receive that candy.
For example, there are 4 prisoners and 6 pieces of candy. The prisoners arrange themselves in seats numbered 1 to 4. Let’s suppose two are drawn from the hat. Prisoners receive candy at positions
2,3,4,1,2,3. The prisoner to be warned sits in chair number 3
Function Description
Write a function to save the prisoner. It should return an integer representing the chair number of the prisoner to warn.
save the prisoner has the following parameter(s):
• n: an integer, the number of prisoners
• m: an integer, the number of sweets
• s: an integer, the chair number to begin passing out sweets from
Input Format
• The first line contains an integer t, denoting the number of test cases.
• The next t lines each contain 3 space-separated integers:
□ -: n the number of prisoners
□ -: m the number of sweets
□ -: s the chair number to start passing out treats at
Output Format
• For each test case, print the chair number of the prisoner who receives the awful treat on a new line.
Sample Output
#include <stdio.h>
int main()
int t,j,i,count=0;
long int ncr;
long int result,diff;
long int n[t];
long int m[t];
long int s[t];
return 0;
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RBSE Solutions for Class 5 Maths गणित
RBSE Solutions for Class 5 Maths Pdf download गणित in both Hindi Medium and English Medium are part of RBSE Solutions for Class 5. Here we have given Rajasthan Board Books RBSE Class 5th Maths
Solutions Pdf Ganit.
Board RBSE
Textbook SIERT, Rajasthan
Class Class 5
Subject Maths
Chapter All Chapters
Exercise Textbook & Additional
Category RBSE Solutions
Rajasthan Board RBSE Class 5 Maths Solutions in Hindi Medium
Rajasthan Board RBSE Class 5 Maths Solutions in English Medium
We hope the given RBSE Solutions for Class 5 Maths Pdf download गणित in both Hindi Medium and English Medium will help you. If you have any query regarding Rajasthan Board Books RBSE Class 5th Maths
Solutions Pdf Ganit, drop a comment below and we will get back to you at the earliest. | {"url":"https://rbseguide.com/class-5-maths/","timestamp":"2024-11-09T22:10:34Z","content_type":"text/html","content_length":"67966","record_id":"<urn:uuid:3088891f-8f6d-4399-b136-10f870e48570>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00227.warc.gz"} |
WhoMadeWhat – Learn Something New Every Day and Stay Smart
Britain is officially metric, in line with the rest of Europe. However, imperial measures are still in use, especially for road distances, which are measured in miles. Imperial pints and gallons are
20 per cent larger than US measures.
Moreover, Is US mile same as UK Mile?
A British mile is the same distance as an American mile. You don’t have to worry about the metric system, like how a British pint is really an imperial pint, which is larger than an American pint.
In respect to this, How long is a USA mile?
1 mi. or mi in …
SI units
imperial/US units
U.S. survey mile
Is American Mile same as UK Mile?
A British mile is the same distance as an American mile. You don’t have to worry about the metric system, like how a British pint is really an imperial pint, which is larger than an American pint.
Furthermore, Why does the UK use miles?
Historically the road network in England was established by the Romans who measured in miles. The metric system was first introduced to France by Napoleon at a time when they were at war with
England. This is why the English were reluctant to adopt metrification.
Did the British ever use miles?
Since 1995, goods sold in Europe have had to be weighed or measured in metric, but the UK was temporarily allowed to continue using the imperial system. This opt-out was due to expire in 2009, with
only pints of beer, milk and cider and miles and supposed to survive beyond the cut-off.Since 1995, goods sold in Europe have had to be weighed or measured in metric, but the UK was temporarily
allowed to continue using the imperial system. This opt-out was due to expire in 2009, with only pints of beer, milk and cider and miles and supposed to survive beyond the cut-off.
Does the UK use miles for distance?
Britain is officially metric, in line with the rest of Europe. However, imperial measures are still in use, especially for road distances, which are measured in miles. Imperial pints and gallons are
20 per cent larger than US measures.Britain is officially metric, in line with the rest of Europe. However, imperial measuresimperial measuresThe metric system is routinely used in business and
technology within the United Kingdom, with imperial units remaining in widespread use amongst the public. All UK roads use the imperial system except for weight limits, and newer height or width
restriction signs give metric alongside imperial.https://en.wikipedia.org â ș wiki â ș Imperial_unitsImperial units – Wikipedia are still in use, especially for road distances, which are measured in
miles. Imperial pints and gallons are 20 per cent larger than US measures.
Is it miles or km in USA?
The United States is the only real stronghold of the imperial system in the world to-date. Here, using miles and gallons is the norm, even though scientists do use metric, new units like megabytes
and megapixels are metric as well and runners compete for 100 meters like everywhere else in the world.The United States is the only real stronghold of the imperial system in the world to-date. Here,
using miles and gallons is the norm, even though scientists do use metric, new units like megabytes and megapixels are metric as well and runners compete for 100 meters like everywhere else in the
Does UK still use miles?
Even though everyone thinks Europe has completely converted to the metric system, the United Kingdom still uses miles per hour, too â and anywhere you go in the U.K., you’ll see signs in miles per
hour. … That’s because the U.K. uses miles per hour.
Does the UK use miles per hour?
The UK remains the only country in Europe, and the Commonwealth, that still defines speed limits in miles per hour (mph).
Why do Brits use miles?
Originally Answered: Why do the Brits use miles on road signs? Finishing the job would cost visible money and lose politicians votes. Keeping the mess costs less visible money and is popular.
Did the English use miles?
Even though everyone thinks Europe has completely converted to the metric system, the United Kingdom still uses miles per hour, too â and anywhere you go in the U.K., you’ll see signs in miles per
hour. … That’s because the U.K. uses miles per hour.
Why does the UK not use the metric system?
The UK switched to metric in 1965, and this happened only because the industry forced it. UK companies were simply having too much a hard time trading with European countries. Even 50 years later,
many Britons still refuse to move entirely to metric.
Are Us miles the same as UK Miles?
A British mile is the same distance as an American mile. You don’t have to worry about the metric system, like how a British pint is really an imperial pint, which is larger than an American pint.
Is miles per hour A mph?
Miles per hour (mph or mi/h) is a British imperial and United States customary unit of speed expressing the number of miles travelled in one hour.Miles per hour (mph or mi/h) is a British imperial
and United States customary unit of speed expressing the number of miles travelled in one hour.
Does the UK use miles?
Britain is officially metric, in line with the rest of Europe. However, imperial measures are still in use, especially for road distances, which are measured in miles. Imperial pints and gallons are
20 per cent larger than US measures.Britain is officially metric, in line with the rest of Europe. However, imperial measuresimperial measuresThe metric system is routinely used in business and
technology within the United Kingdom, with imperial units remaining in widespread use amongst the public. All UK roads use the imperial system except for weight limits, and newer height or width
restriction signs give metric alongside imperial.https://en.wikipedia.org â ș wiki â ș Imperial_unitsImperial units – Wikipedia are still in use, especially for road distances, which are measured in
miles. Imperial pints and gallons are 20 per cent larger than US measures.
What is an American mile?
The mile, sometimes the international mile or statute mile to distinguish it from other miles, is a British imperial unit and US customary unit of distance; both are based on the older English unit
of length equal to 5,280 English feet, or 1,760 yards.The mile, sometimes the international mile or statute mile to distinguish it from other miles, is a British imperial unit and US customary unit
of distance; both are based on the older English unitEnglish unitEnglish units are the units of measurement used in England up to 1826 (when they were replaced by Imperial units), which evolved as a
combination of the Anglo-Saxon and Roman systems of units.https://en.wikipedia.org â ș wiki â ș English_unitsEnglish units – Wikipedia of length equal to 5,280 English feet, or 1,760 yards.
Do they use miles in the UK?
Britain is officially metric, in line with the rest of Europe. However, imperial measures are still in use, especially for road distances, which are measured in miles. Imperial pints and gallons are
20 per cent larger than US measures.Britain is officially metric, in line with the rest of Europe. However, imperial measuresimperial measuresThe metric system is routinely used in business and
technology within the United Kingdom, with imperial units remaining in widespread use amongst the public. All UK roads use the imperial system except for weight limits, and newer height or width
restriction signs give metric alongside imperial.https://en.wikipedia.org â ș wiki â ș Imperial_unitsImperial units – Wikipedia are still in use, especially for road distances, which are measured in
miles. Imperial pints and gallons are 20 per cent larger than US measures.
Does the UK really use metric?
Britain is officially metric, in line with the rest of Europe. However, imperial measures are still in use, especially for road distances, which are measured in miles. Imperial pints and gallons are
20 per cent larger than US measures.Britain is officially metric, in line with the rest of Europe. However, imperial measuresimperial measuresThe metric system is routinely used in business and
technology within the United Kingdom, with imperial units remaining in widespread use amongst the public. All UK roads use the imperial system except for weight limits, and newer height or width
restriction signs give metric alongside imperial.https://en.wikipedia.org â ș wiki â ș Imperial_unitsImperial units – Wikipedia are still in use, especially for road distances, which are measured in
miles. Imperial pints and gallons are 20 per cent larger than US measures.
Is an American mile different?
In the United States, the term statute mile formally refers to the survey mile, but for most purposes, the difference of less than 1â 8 inch (3.2 mm) between the survey mile and the international
mile (1609.344 metres exactly) is insignificantâ one international mile is 0.999998 U.S. survey milesâ so statute mile can be …
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Investment Blog
Tag: Growth Markets
Attractiveness of emerging markets by P/E and implied return on capital
Date: 22 Jan 2012 22:33
The goal of this article is to analyze attractiveness of BRIC and Vietnam stock markets using multiples.
Country attractiveness based on P/E ratio
The multiple which is often used an indicator of investment attractiveness of individual stocks or countries is P/E. Below are its values for a key index in each country plus S&P for comparison.
• China was the most expensive county by P/E since 1996 to 2011 except 2007 when Vietnam took the leadership
• Vietnam was the most expensive country by P/E in 2007, but lost its leadership to China during 2008 crisis and have never recovered since
• Russia had always the lowest P/E among BRICs and Vietnam
Questions about the past performance:
• Can we say based on this data that Russia generated the biggest investment return and China the lowest in the observed timeframe?
Questions about the future:
• Can we say based on this data that Russia and Vietnam are the most attractive for investments for 3-5 years ahead?
• Can we say that India is the least attractive?
Implied return on equity
Thinking about the questions above, everybody understands that direct comparison of P/E multiples for different stocks or countries is often incorrect as different earnings growth rates are implied.
For example, expected growth for Russia in 2012 is about 3.6% while expected growth for China is 8.5%. The multiple which tries to fix this problem is PEG:
where g is taken as g*100 (for example, g=10% is used as 10). PEG multiple is even worse as it doesn't have any financial interpretation.
As P/E is widely calculated and published it's worth using it but with the correct inclusion of g in the formula:
This formula tells that if you want to compare attractiveness of stocks or countries, you should look at the implied return on equity r:
r= 1/(P/E)+g
Then the question which investors should ask is whether the implied return on equity is justified, too high or too low. Below is the implied return on equity for the analyzed indices:
This leads to interesting implications:
• Before 2008
□ Investors assumed huge cost of equity in Russia
□ China, India and Brazil had similar r since 2001
• Since 2008 crisis the order of countries changed:
□ Vietnam has the highest implied return on equity since 2009, Russia has the second since 2010
□ Russia stopped being attractive since 2008, and then became attractive for a short period of time at the end of 2008 - beginning of 2009
□ China and Brazil have the lowest implied return on equity
□ India is in the middle and decoupled from China and Brazil
□ Any of the BRICs + Vietnam have at least twice bigger implied return than S&P
Vietnam and Russia are the most attractive countries for investment in public stocks among BRIC + Vietnam.
In calculating implied return on equity the following assumptions were used:
• g in the formula should be long term expected growth in earnings. It's hard to measure it retrospectively. Instead, g which happened/expected in the next 365 days was used.
• Also, it's hard to get exactly earnings growth, so GDP growth was used.
• This approach is theoretically not ideal, but does the job. The main goal of this exercise to broadly incorporate growth expectations in P/E analysis. For example, in 2010 Russia with g=
4.0%+11.4%=15.4% is very different from India with g=9.7%+9.6%=19.3% and P/E ratios of these countries cannot be compared directly
• g = real GDP growth + GDP deflator
• Period for the above data: g is measured for the 365 days ahead of the time of P/E. For example, for 30 March 2011 real GDP growth =~ ¼ * real GDP growth 2011 + ¾ * real GDP growth 2012
• When the growth was not available or the period is in the future, short term forecasts were used
• For GDP deflator forecasts (2011-2013) CPI forecast was used
Below is P/E ratio and implied return on equity for S&P separately as the historical data is available for longer periods than for BRICs and Vietnam:
Russia had the fastest nominal GDP growth in USD in 2000-2009 with China only second
Date: 05 Jan 2011
Many investors focus on real GDP growth while thinking about perceptiveness of investments to a particular country. And this is really strange approach as real GDP growth is
• real
• measured in local currency while what investors care about is
• nominal returns
• measured in USD
The average expected equity returns over many years are normally expressed by the formula:
Expected equity return = Expected real GDP growth + Expected inflation + Expected dividends yield
If you want to measure it in $, you should add change of exchange rates.
The good explanation and data for this you can find in many articles in the internet. Here I will just quote Warren Buffet made to Fortune Magazine back in November 1999 right before dot com bust:
“Let's say that GDP grows at an average 5% a year--3% real growth, which is pretty darn good, plus 2% inflation. If GDP grows at 5%, and you don't have some help from interest rates, the aggregate
value of equities is not going to grow a whole lot more. Yes, you can add on a bit of return from dividends. But with stocks selling where they are today, the importance of dividends to total return
is way down from what it used to be.”
The key component in this equation like Buffet mentioned is nominal GDP growth while dividends returns nowadays are small and cannot cause big differentiation. If we take a look at Nominal GDP growth
in USD for 20 biggest countries (by GDP in 2009) for the period of 2000-2009, surprisingly we will find that Russia was the first one, with China only the second and India the forth. This analysis
also gives insight about the country which will be on radar of EM investors soon: Indonesia. I have already seen BRIIC abbreviation.
Real vs. Nominal GDP growth for 20 biggest countries
In many GDP growth discussions Russia was not even mentioned in the last five years. Some want to exclude it from BRIC. Why were investors so inconsistent using real GDP growth which was not even
close for their ultimate goal – nominal $ returns?
Let’s also see what the reasons were behind so drastic differences between real and nominal GDP growth.
Growth structure of nominal GDP in $ for top-20 countries, 2000-09
In many cases this was high inflation that secured leadership in nominal GDP growth like for Russia and Indonesia. In case of China and Euro zone countries it’s also devaluation of local currencies.
It’s worth to mention that Russia was the only country out of this top-20 with negative population growth which ate 0.3% GDP growth annually. | {"url":"http://macrolion.com/blog/tags/growth_markets/","timestamp":"2024-11-04T20:41:22Z","content_type":"text/html","content_length":"12401","record_id":"<urn:uuid:5f4aa314-c533-45ae-8f11-e076c0538c6f>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00462.warc.gz"} |
DNA organization along pachytene chromosome axes and its relationship with crossover frequencies
DNA organization along pachytene chromosome axes and its relationship with crossover frequencies
Published: 29 June 2022| Version 2 | DOI: 10.17632/w3n9xp5dnp.2
Maria Pigozzi
MLH1- focus map along SC1 The construction of crossover frequency histograms from MLH1 foci in birds and other vertebrates is fully explained in several publications (Pigozzi and Solari 1999; Borodin
et al 2008). Briefly, individual SC lengths, centromeric signals, and MLH1 foci were scored using version 3.3 of the MicroMeasure program [53] which records absolute and relative distances on
digitized images. To generate the recombination maps of the SC1, we calculated the absolute position of each MLH1 focus by multiplying the relative position of each focus by the average absolute
length of the chromosome arm. These data were pooled for each arm and graphed in histogram form. The genetic map length in cM of each interval in this histogram (bin size 0.25 µm) was calculated
multiplying the average number of MLH1 foci by 50. To produce the cumulative cM distribution, the average number of MLH1 foci per interval was converted to cM and then added along the SC starting
from the tip of the short arm. Foci on microbivalentes 9-38. The number of foci is the average scored in 138 nuclei.
Steps to reproduce
The construction of crossover frequency histograms from MLH1 foci in birds and other vertebrates is fully explained in several publications (Pigozzi and Solari 1999; Borodin et al 2008). Briefly,
individual SC lengths, centromeric signals, and MLH1 foci were scored using version 3.3 of the MicroMeasure program [53] which records absolute and relative distances on digitized images. To generate
the recombination maps of the SC1, we calculated the absolute position of each MLH1 focus by multiplying the relative position of each focus by the average absolute length of the chromosome arm.
These data were pooled for each arm and graphed in histogram form. The genetic map length in cM of each interval in this histogram (bin size 0.25 µm) was calculated multiplying the average number of
MLH1 foci by 50. To produce the cumulative cM distribution, the average number of MLH1 foci per interval was converted to cM and then added along the SC starting from the tip of the short arm. Foci
on microbivalentes 9-38. The number of foci is the average scored in 138 nuclei.
Natural Sciences | {"url":"https://data.mendeley.com/datasets/w3n9xp5dnp/2","timestamp":"2024-11-11T17:38:08Z","content_type":"text/html","content_length":"104252","record_id":"<urn:uuid:252958b0-8348-4a09-b2d6-867bff3db411>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00136.warc.gz"} |
Uncertainty of matter - The theory of everything by Marek Ożarowski
Uncertainty of matter
Uncertainty of matter refers directly to the Uncertainty Principle, which was proposed by Werner Heisenberg. Uncertainty of matter is a kind of extension of the Uncertainty Principle. Our Concept
also refers to the Uncertainty Principle and tries to interpret it consistently to the ToE-Quantum Space. What does this Uncertainty consist of? What is the Uncertainty of matter? Before we go on to
answer the questions posed, perhaps we should first look for our micro-world – the world of elementary particles.
Uncertainty of matter takes place outside of real time, it is a special circumstance of imaginary time and imaginary place (see Time Quaternion). The image of our Reality – our Here and our Now, is
built of matter with certain properties. The properties of our matter, are only revealed in real time – it is the state of matter determination that will stabilize our matter just for this particular
Here and Now.
At this point in time (Here and Now), our matter will have selected the appropriate properties to map our present Reality. This means that adequate Fundamental Particles and fundamental interactions
will be revealed. For our present moment there will be a „stabilization” of the Reality built from: Fermions, treated as the building blocks of matter, and Bosons, treated as the carriers of
interactions. For this point of the Here and Now, the Uncertainty of matter will end.
Can Heisenberg’s Uncertainty Principle be extended to the Uncertainty of Matter? Perhaps something should first be said about Heisenberg’s Uncertainty Principle itself. Perhaps then it may be
possible to understand how such a principle can be extended to matter?
The uncertainty principle, also known as Heisenberg’s indeterminacy principle, is a fundamental concept in quantum mechanics. It states that there is a limit to the precision with which certain
pairs of physical properties, such as position and momentum, can be simultaneously known. In other words, the more accurately one property is measured, the less accurately the other property can
be known.
More formally, the uncertainty principle is any of a variety of mathematical inequalities asserting a fundamental limit to the product of the accuracy of certain related pairs of measurements on
a quantum system, such as position, x, and momentum, p. Such paired-variables are known as complementary variables or canonically conjugate variables.
First introduced in 1927 by German physicist Werner Heisenberg, the formal inequality relating the standard deviation of position σ[x] and the standard deviation of momentum σ[p] was derived by
Earle Hesse Kennard later that year and by Hermann Weyl in 1928:
where ℏ=h2π reduced Planck constant.
The quintessentially quantum mechanical uncertainty principle comes in many forms other than position – momentum. The energy – time relationship is widely used to relate quantum state lifetime to
measured energy widths but its formal derivation is fraught with confusing issues about the nature of time. The basic principle has been extended in numerous directions; it must be considered in
many kinds of fundamental physical measurements.
It is vital to illustrate how the principle applies to relatively intelligible physical situations since it is indiscernible on the macroscopic scales that humans experience. Two alternative
frameworks for quantum physics offer different explanations for the uncertainty principle. The wave mechanics picture of the uncertainty principle is more visually intuitive, but the more
abstract matrix mechanics picture formulates it in a way that generalizes more easily.
Uncertainty of matter. The superposition of several plane waves to form a wave packet. This wave packet becomes increasingly localized with the addition of many waves. The Fourier transform is a
mathematical operation that separates a wave packet into its individual plane waves. The waves shown here are real for illustrative purposes only; in quantum mechanics the wave function is
generally complex.
Mathematically, in wave mechanics, the uncertainty relation between position and momentum arises because the expressions of the wavefunction in the two corresponding orthonormal bases in Hilbert
space are Fourier transforms of one another (i.e., position and momentum are conjugate variables). A nonzero function and its Fourier transform cannot both be sharply localized at the same time.
A similar tradeoff between the variances of Fourier conjugates arises in all systems underlain by Fourier analysis, for example in sound waves: A pure tone is a sharp spike at a single frequency,
while its Fourier transform gives the shape of the sound wave in the time domain, which is a completely delocalized sine wave. In quantum mechanics, the two key points are that the position of
the particle takes the form of a matter wave, and momentum is its Fourier conjugate, assured by the de Broglie relation p = ħk, where k is the wavenumber.
In matrix mechanics, the mathematical formulation of quantum mechanics, any pair of non-commuting self-adjoint operators representing observables are subject to similar uncertainty limits. An
eigenstate of an observable represents the state of the wavefunction for a certain measurement value (the eigenvalue). For example, if a measurement of an observable A is performed, then the
system is in a particular eigenstate Ψ of that observable.
However, the particular eigenstate of the observable A need not be an eigenstate of another observable B: If so, then it does not have a unique associated measurement for it, as the system is not
in an eigenstate of that observable.
This principle – the principle of indeterminacy has been extended to the uncertainty of matter. Matter, its properties beyond our Here and our Now, is unstable from the point of view of our present
moment. This means that we do not know what will happen in a moment. If we heat water for tea, in a moment the water will change its state of matter – at least, some of the water will turn into steam
We don’t know exactly when this will happen – How exactly will the transition be done (Phase transition). The picture of our Reality will be built in each successive second of our real time. For our
„past” moments and „future” moments, the state of aggregation of our water, its properties (water) are uncertain from the point of view of our Here and our Now.
Let’s start with the fact that at the emergence of time, initiated by the Big Bang, our elementary particles appeared in the form of occurring „changes”. Elementary particles were not yet properly
formed – they had no mass, spin, momentum, etc. But they existed – there were some primordial forms of our matter, which for the time being was deprived of its properties. The Universe was being
born, and with the birth of our Universe, our elementary particles gained more properties. At that time, from our point of view, it was impossible to use such matter, endowed only with the existence
of „changes.”
Matter, like everything around us, had to grow into the shape that exists in our present Reality. We don’t know what that might have looked like specifically. However, the simpler the system, the
less complex, the easier it is to imagine. It seems that an elementary particle like the Photon could not have been created in an instant. The photon had to mutate, to transform into the form we know
Rome wasn’t built in a day! Everything required „time” from our point of view. But „time” according to us does not exist. At that time, our Universe was „Indeterminate” from our point of view. If we
had been there, we would not have been able to see it all, because Photons were not yet there.
Matter, therefore, must have gained further properties over our time, which determined the interactions for individual elementary particles. After all, we have no knowledge whether the Photon as we
know it today could have come into existence at once – perhaps originally, the Photon could have had Invariant mass. Later, however, such a Photon, lost its mass in favor of receiving momentum. And
momentum appeared with the advent of the expansion of our Universe. See also article the Expansion of the Universe.
So if everything had to evolve, so did matter and its properties. Today’s image of the Universe is the result of what has taken place during the expansion of our Universe. The animation below shows
the concept of the appearance of successive physical properties of one elementary particle. These particles in subsequent stages of expansion will co-create more complex forms of matter for further
physical properties. Successive chemical elements and Chemical compounds took time to evolve. Our Concept assumes that it does not exist. The measure of time is change.
Thus, each successive chemical element had to mutate with the help of the „changes” made. Each „change” resulted in added value. This added value, could consequently be responsible for the appearance
of another property of matter. Therefore, if time does not exist, then a complex Chemical compound must be the fruit of all the mutations (changes) that took place during the expansion of the
Universe. It’s quite complicated, but the analogy can be repeated with a familiar example.
Our bodies appear at the moment of conception. Even then, we gain characteristics that will be revealed at our birth. Eye color, hair, genetic code, gender, senses, etc. Other traits we acquire over
time, knowledge, experience, sensitivity, Perception, character traits, etc. Further is even more interesting. We can change the Reality around us: create ideas, concepts, design magnificent
buildings, discover unknown phenomena. Finally, something extra – we have our share of procreation, which is our „extension” – our next generation that inherits our properties from us.
We rather did not invent it, it was inherited by us. We are made up of chemical compounds that have taken over properties from chemical elements. Chemical elements have properties of atoms, atoms …
and so on, down to the smallest fragments of matter. And matter how does it know what its form and properties should be in the next Here and the next Now? If there is no time, then everything must be
accomplished by a matter of complementary information through the entanglement that occurs between copies of the same elementary particle. (See also copying the world).
The uncertainty of matter occurs permanently for every moment except the present moment – our Here and our Now. No one knows what state of matter and when it will be revealed to us in the future.
Every property of an elementary particle is achieved in due time since the beginning of our Universe. Since, according to our Concept, „time does not exist,” every property of an elementary particle
belongs to a copy of that elementary particle.
Copies of one elementary particle fill what we call our „time.” All copies, as a result of „entanglement,” inherit the received properties. Each new property of an elementary particle appears with
the elapse of our time. This means that each elementary particle in imaginary time is in contact with its copy through entanglement. Entanglement is a form of information exchange between different
copies of the same elementary particle.
The uncertainty of matter is a condition that occurs outside of our Here and our Now. The uncertainty of matter destabilizes matter outside of our Here and our Now. This means that in the present
moment, our matter will use through entanglement with its copies, which have the right qualities and properties to describe our Present Reality. If there is a copy of elementary particles that can
overcome the force of gravity, such a Reality would be offered to us.
All the properties, of each copy, of each elementary particle, are available to us. However, through the time continuum, which is co-created by the expansion vector, we are not able to obtain in our
real moment, for example, a Photon that is endowed with Invariant mass, or a Photon that has no spin. However, it is theoretically possible. To give an example, we can obtain a Bose-Einstein
Condensate. Normally this is not possible, but now scientists have succeeded in obtaining such a condensate. This is only a small step, but in the next step it is possible to create matter that is
free of gravity.
This may mean that the properties of elementary particles can be provoked and triggered in our Here and Now. Sometime in the past, from our point of view, there was a state of the photon in which it
had no spin. There may have been a version of it that had Invariant mass. At that time, for the description of the Universe and the representation of That Reality, the existence of such elementary
particles with such properties was necessary.
Today, each elementary particle has a bond with all its copies through entanglement. If this could be the case, then releasing such a property – a photon without spin, or a photon with Invariant mass
, could be real. So could obtaining a Bose-Einstein Condensate. Therefore, matter from the „past” is devoid of the attribute of gravity – in this way, our Here and our Now is devoid of gravitational
interaction from the past.
What to compare it to in order to understand it better? The example of hypnosis comes to mind. It happens that in a state of hypnosis, a person has the ability to go back to a particular moment in
his life. It could be the same with our elementary particles. It could be that in such hypnosis for an elementary particle, one could find a copy of that elementary particle when it was not yet a
photon, or such a copy that did not yet have spin. Then such a property could be revealed to us in our Here and our Now.
If the unknown properties of our elementary particles could be revealed, we could understand how our Universe came to be and whether our concept of time is correct. Perhaps such elementary particles
as photons without spin or with non-zero rest mass exist, only they are unstable in our Here and our Now? Perhaps they form tachyons, which is why they cannot be experienced in our Here and our Now.
It is not known exactly what would happen to our time continuum if such an elementary particle with properties we know nothing about were released – would our Reality then continue in our „future”
moment. That’s another topic, for another article.
Thus, if our picture of matter is indeterminate beyond our Here and our Now, it appears to be in an unstable state from our point of view. This means that matter, in order to be in an unstable state,
must get rid of some of the properties it has in our Here and our Now. For example, an electron that is in our „past” may have lost its Invariant mass. Such an electron is therefore in an imaginary
state from our point of view.
Therefore, according to the indetermination principle, we know nothing about the position of such an elementary particle if we are concerned with momentum – and vice versa. The indeterminacy of
matter does not mean that it does not exist – perhaps it is precisely there, only it is inaccessible. Unstable matter becomes dark matter – dark matter is then in a state of indeterminacy.
We will continue …
Marek Ożarowski, 16 July 2024 | {"url":"https://theoryofeverything.info/2024/07/uncertainty-of-matter/","timestamp":"2024-11-05T23:26:58Z","content_type":"text/html","content_length":"112487","record_id":"<urn:uuid:b5345476-dd70-4885-9afc-50fe76a0e9e4>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00598.warc.gz"} |
The Ultimate Guide to Lens Selection for Capturing Stunning Bokeh Balls
Capturing the perfect bokeh balls in your images requires a deep understanding of the underlying optical principles and lens characteristics. In this comprehensive guide, we will delve into the
technical details and provide you with a playbook to optimize your lens selection and camera settings for breathtaking bokeh effects.
1. Focal Length and Bokeh Size
The focal length of a lens is a crucial factor in determining the size of the bokeh balls. Longer focal lengths, such as those found in telephoto lenses, generally produce larger bokeh circles
compared to shorter focal lengths, like those in wide-angle lenses. This is due to the relationship between the focal length and the depth of field.
Equation for Bokeh Ball Diameter (B) based on Focal Length (f):
B = PD * \frac{|m – mb|}{1 + \frac{mb}{p}}
– (B) is the bokeh ball diameter
– (PD) is the entrance pupil diameter
– (p) is the pupil magnification ((P’D / PD))
– (P’D) is the exit pupil diameter
– (mb) is the bokeh magnification
– (m) is the in-focus magnification
As the focal length increases, the in-focus magnification (m) and the bokeh magnification (mb) also increase, leading to larger bokeh ball diameters (B). This relationship can be observed in the
following table:
Focal Length (mm) Bokeh Ball Diameter (mm)
50 3.2
85 5.4
105 6.7
135 8.6
By selecting a lens with a longer focal length, you can achieve larger and more pronounced bokeh balls in your images.
2. Aperture (f-number) and Bokeh Size
The aperture, or f-number, of a lens also plays a significant role in the size of the bokeh balls. Smaller f-numbers, which correspond to larger apertures, create larger bokeh circles. Conversely,
larger f-numbers (smaller apertures) result in smaller bokeh balls.
Equation for Bokeh Ball Diameter (B) based on Aperture (f-number):
B = PD * \frac{|m – mb|}{1 + \frac{mb}{p}}
– (PD) is the entrance pupil diameter, which is inversely proportional to the f-number
– (p) is the pupil magnification, which is also affected by the f-number
As the f-number decreases (larger aperture), the entrance pupil diameter (PD) increases, leading to a larger bokeh ball diameter (B). This relationship can be observed in the following table:
Aperture (f-number) Bokeh Ball Diameter (mm)
f/2.8 7.2
f/4 5.4
f/5.6 3.8
f/8 2.7
By using a lens with a larger maximum aperture (smaller f-number), you can capture images with more prominent and visually appealing bokeh balls.
3. Bokeh Magnification and Depth of Field
The relationship between bokeh magnification and depth of field is crucial in understanding the size and quality of the bokeh balls. Bokeh magnification (mb) is a measure of how much the out-of-focus
areas are magnified compared to the in-focus areas.
Equations for Bokeh Ball Diameter (B):
1. When bokeh magnification (mb) is larger than in-focus magnification (m):
B = PD * \frac{mb – m}{1 + \frac{mb}{p}}
2. When bokeh magnification (mb) is smaller than in-focus magnification (m):
B = PD * \frac{m – mb}{1 + \frac{mb}{p}}
3. Unified equation:
B = PD * \frac{|m – mb|}{1 + \frac{mb}{p}}
– (B) is the bokeh ball diameter
– (PD) is the entrance pupil diameter
– (p) is the pupil magnification ((P’D / PD))
– (P’D) is the exit pupil diameter
– (mb) is the bokeh magnification
– (m) is the in-focus magnification
As the bokeh magnification (mb) increases, the bokeh ball diameter (B) also increases, leading to larger and more pronounced bokeh effects. However, this relationship is also influenced by the depth
of field, as the in-focus magnification (m) can affect the overall bokeh quality.
By understanding and manipulating the bokeh magnification and depth of field, you can fine-tune your lens selection and camera settings to achieve the desired bokeh characteristics in your images.
4. Lens Design and Bokeh Quality
The design of the lens can also have a significant impact on the quality and appearance of the bokeh balls. Lenses with aspherical elements, for example, can produce more rounded and natural-looking
bokeh balls, while lenses without aspherical elements may result in more angular or polygonal bokeh shapes due to the influence of the iris blades.
Additionally, the number and shape of the iris blades can also affect the bokeh quality. Lenses with a higher number of rounded iris blades tend to produce smoother and more circular bokeh balls,
while lenses with fewer or more angular iris blades may create more defined or “busy” bokeh patterns.
To assess the bokeh quality of a lens, you can perform a series of tests and analyze the resulting bokeh balls. This can include capturing images of out-of-focus highlights, such as small light
sources or specular highlights, and evaluating the shape, smoothness, and overall aesthetic of the bokeh.
5. Subject Distance and Bokeh Effect
The distance between the subject and the lens also plays a role in the intensity and appearance of the bokeh effect. Generally, the closer the subject is to the lens, the more pronounced the bokeh
will be.
As the subject distance decreases, the depth of field becomes shallower, and the out-of-focus areas become more blurred. This results in larger and more prominent bokeh balls, as the background
elements are further removed from the plane of focus.
Conversely, as the subject distance increases, the depth of field becomes deeper, and the bokeh effect becomes less pronounced. The bokeh balls may appear smaller and less visually striking.
By positioning your subject closer to the lens, you can maximize the bokeh effect and create a more dramatic and visually appealing background blur.
6. Measuring Bokeh Ball Diameter
To quantify the size and characteristics of the bokeh balls, you can measure the diameter of the bokeh balls in your images. This can be done by capturing a series of images with different focus
distances and analyzing the resulting bokeh balls.
One method is to photograph a scene with a high-contrast background, such as a string of lights or a field of small highlights. By adjusting the focus distance, you can create a range of bokeh ball
sizes and shapes. Then, you can use image analysis software or manual measurement tools to determine the diameter of the bokeh balls.
By measuring the bokeh ball diameter, you can compare the performance of different lenses and optimize your lens selection and camera settings to achieve the desired bokeh effect.
7. Manufacturer Specifications
When selecting a lens for capturing bokeh balls, it’s important to consider the manufacturer’s specifications and claims about the lens’s bokeh capabilities. Lenses with larger maximum apertures
(smaller f-numbers) are generally better suited for creating prominent bokeh effects.
Additionally, some lens manufacturers may provide specific information about the bokeh quality of their lenses, such as the number and shape of the iris blades, the use of aspherical elements, or the
overall bokeh rendering characteristics.
By reviewing the manufacturer’s specifications and comparing the technical details of different lenses, you can make an informed decision about the best lens for your bokeh photography needs.
In this comprehensive guide, we have explored the key factors that influence the size, quality, and appearance of bokeh balls in your images. By understanding the relationships between focal length,
aperture, bokeh magnification, lens design, subject distance, and measurement techniques, you can optimize your lens selection and camera settings to capture stunning bokeh effects.
Remember, the art of bokeh photography is not just about technical mastery, but also about creative expression and experimentation. Embrace the principles outlined in this guide, but don’t be afraid
to explore and discover your own unique approach to capturing the perfect bokeh balls.
Happy shooting!
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How old is the oldest person you know? | R-bloggersHow old is the oldest person you know?
How old is the oldest person you know?
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Last week, we had a discussion with some colleagues about the fact that – in order to prepare for the SOA exams – we did not have time (so far) to mention results on extreme values in our actuarial
program. I did gave an introduction in my nonlife actuarial models class, but it was only an introduction, in three hours, in order to illustrate reinsurance pricing. And I told my students that if
they wanted to know more about extreme values, they should start a master program in actuarial science and finance, since I will give a course on extremes (and copulas) next winter.
But actually, extreme values are everywhere ! For instance, there is a Prudential TV commercial where has people place large, round stickers on a number line to represent the age of the oldest person
they know. This forms some kind of histogram. The message is to have Prudential prepare you to have adequate money for all these years. And actually, anyone can add his or her own sticker at the
Prudential website.
Patrick Honner, on his blog (http://mrhonner.com/…), did mention this interesting representation. But this idea is not new, as mentioned in a post, published three years ago. In 1932, Emil Gumbel
gave a talk in France on the “âge limite“. And as he wrote it “on peut donc supposer que la distribution de l’âge limite – c’est à dire la probabilité que cet âge ait une valeur donnée – soit
Gaussienne“. In 1932 (not aware of Fisher and Tippett work, he thought that the limiting distribution for a maximum would be Gaussian). But a few years after, he read about Fisher’s work, and
observed also that “la distribution d’une valeur extrêmes peut être représentée pour un nombre suffisant d’observations par la formule doublement exponentielle, pourvu que la distribution initiale se
comporte asymptotiquement comme une exponentielle. La formule devient rigoureuse si la distribution initiale est exponentielle“, as he wrote in 1935. And in 1937, he wrote a paper on “les
centennaires” that can also be related to the work of Bortkiewicz on rare events. One should also mention one of the most important paper in extreme value theory, published in 1974 by Balkema and de
Haan, on Residual Life Time at Great Age.
Because in this experiment, the question is “How Old is the Oldest Person You Know?“, so it is the distribution of a maximum. And from Fisher-Tippett theorem, if we assume that the age is bounded
(and that there exists some finite upper limit), then the limiting distribution for the maxima (or to be more rigorous, a affine transformation of the maxima) should be Weibull distribution. And this
is what it looks like
> plot(-x,dweibull(x,2.25,4),type="l",lwd=2)
As an actuary, the only thing I know about demography, is the distribution of the age of death. For instance, consider the following French life table
> alive <- read.table(
+ "http://perso.univ-rennes1.fr/arthur.charpentier/TV8890.csv",
+ sep=";",header=TRUE)$Lx
> nb= -diff(alive)
> ages=0:110
> plot(ages,nb,type="h")
This is the distribution of the age of the death in a given population. Which is not the same as the distribution mentioned above! What we look for is the following: given that someone is alive, what
could be the distribution of his-her age ? Actually, if we assume that the yearly number of birth is constant with time (as well as death probability), then we can compute easily to number of people
of age $x$ : we take everyone born (exactly) $x$ years ago, and remove all those who died at at $x$, $x-1$, etc. So the function should be
> probadeath=nb/sum(nb)
> nbx=function(x) 1-sum(probadeath[1:(x+1)])
> surv=Vectorize(nbx)(ages)
> distrage=surv/sum(surv)
which looks like
But this assumption of constant number of birth is not that relevent. And actually, what we need is the distribution of the age within a population… This is a population pyramid, actually. The French
one can be downloaded from http://www.insee.fr/fr/ppp/bases-de-donnees/….
> population <- read.table("popinsee2007.csv",sep=";",header=TRUE)$POPTOT07
> ages=0:107
> plot(ages,population/sum(population),type="h")
(the red line being the one obtained previously, using some natality assumptions). Now, let us use this population to generate acquaintances.
> agemax=function(nsim=1000,size=20){
+ agemax=rep(NA,nsim)
+ for(i in 1:nsim){
+ X=sample(ages,prob=population/sum(population),size=size,replace=TRUE)
+ agemax[i]=max(X)}
+ return(agemax)}
Here, we assume that everyone knows 20 other people, randomly chosen in the entire population, then we return the age of the oldest. And we do that for 1,000 people. Here is the distribution, we
> XS=agemax(10000,20)
> plot(table(XS)/length(XS),type="h",xlim=c(0,108))
where the red line is a Weibull distribution (a transformed one, actually, since in extremely value theory, the distance to the upper bound of the distribution has a Weibull density),
> library(MASS)
> fit=fitdistr(108-XS,dweibull,list(shape=1,scale=1))
> lines(ages,dweibull(108-ages,fit$estimate[1],fit$estimate[2]),col="red")
Which is quite close to the distribution obtained in the commercial, don’t you think ? But still, it should be possible to be more accurate, since people should think of their parents, or
grandparents. So I guess it could be possible to build a more accurate algorithm, to get something closer to the distribution obtained on the Prudential website. But first, let us wait to have more
stickers, more observations… and then I’ll be back to play with it ! | {"url":"https://www.r-bloggers.com/2013/06/how-old-is-the-oldest-person-you-know/","timestamp":"2024-11-04T17:32:41Z","content_type":"text/html","content_length":"110789","record_id":"<urn:uuid:fcf5e178-7546-4bc9-8439-608e8bb8918e>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00827.warc.gz"} |
User-defined class qualifiers in C++23
It is generally known that type qualifiers (such as const and volatile in C++) can be regarded as a form of subtyping: for instance, const T is a supertype of T because the interface (available
operations) of T are strictly wider than those of const T. Foster et al. call a qualifier q positive if q T is a supertype of T, and negative it if is the other way around. Without real loss of
generality, in what follows we only consider negative qualifiers, where q T is a subtype of (extends the interface of) T.
C++23 explicit object parameters (coloquially known as "deducing this") allow for a particularly concise and effective realization of user-defined qualifiers for class types beyond what the language
provides natively. For instance, this is a syntactically complete implementation of qualifier mut, the dual/inverse of const (not to be confused with mutable):
template<typename T>
struct mut: T
using T::T;
template<typename T>
T& as_const(T& x) { return x;}
template<typename T>
T& as_const(mut<T>& x) { return x;}
struct X
void foo() {}
void bar(this mut<X>&) {}
int main()
mut<X> x;
auto& y = as_const(x);
y.bar(); // error: cannot convert argument 1 from 'X' to 'mut<X> &'
X& z = x;
z.bar(); // error: cannot convert argument 1 from 'X' to 'mut<X> &'
The class X has a regular (generally accessible) member function foo and then bar, which is only accessible to instances of the form mut<X>. Access checking and implicit and explicit conversion
between subtype mut<X> and mut<X> work as expected.
With some help fom Boost.Mp11, the idiom can be generalized to the case of several qualifiers:
#include <boost/mp11/algorithm.hpp>
#include <boost/mp11/list.hpp>
#include <type_traits>
template<typename T,typename... Qualifiers>
struct access: T
using qualifier_list=boost::mp11::mp_list<Qualifiers...>;
using T::T;
template<typename T, typename... Qualifiers>
concept qualified =
typename std::remove_cvref_t<T>::qualifier_list,
Qualifiers>::value && ...);
// some qualifiers
struct mut;
struct synchronized;
template<typename T>
concept is_mut = qualified<T, mut>;
template<typename T>
concept is_synchronized = qualified<T, synchronized>;
struct X
void foo() {}
template<is_mut Self>
void bar(this Self&&) {}
template<is_synchronized Self>
void baz(this Self&&) {}
template<typename Self>
void qux(this Self&&)
requires qualified<Self, mut, synchronized>
int main()
access<X, mut> x;
x.baz(); // error: associated constraints are not satisfied
x.qux(); // error: associated constraints are not satisfied
X y;
y.bar(); // error: associated constraints are not satisfied
access<X, mut, synchronized> z;
One difficulty remains, though:
int main()
access<X, mut, synchronized> z;
access<X, mut>& w=z; // error: cannot convert from
// 'access<X,mut,synchronized>'
// to 'access<X,mut> &'
converts to
, but not to
is a subset of
(for the mathematically inclined, qualifiers
, ... ,
over a type
induce a
of subtypes
Q T
⊆ {
, ... ,
}, ordered by qualifier inclusion). Incurring undefined behavior, we could do the following:
template<typename T,typename... Qualifiers>
struct access: T
using qualifier_list=boost::mp11::mp_list<Qualifiers...>;
using T::T;
template<typename... Qualifiers2>
operator access<T, Qualifiers2...>&()
requires qualified<access, Qualifiers2...>
return reinterpret_cast<access<T, Qualifiers2...>&>(*this);
A more interesting challenge is the following: As laid out, this technique implements
qualifier subtyping, but does not do anything towards enforcing the semantics associated to each qualifier: for instance,
should lock a mutex automatically, and a qualifier associated to some particular invariant should assert it after each invocation to a qualifier-constraied member function. I don't know if this
functionality can be more or less easily integrated into the presented framework: feedback on the matter is much welcome. | {"url":"https://bannalia.blogspot.com/2023/08/user-defined-class-qualifiers-in-c23.html","timestamp":"2024-11-01T20:53:39Z","content_type":"application/xhtml+xml","content_length":"69577","record_id":"<urn:uuid:9bb3e8ae-889e-4638-9bcf-d342b0c3e0c7>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00021.warc.gz"} |
Function Approximations Package
Economized Rational Approximations Using Padé Approximations
gives the economized rational approximation to f that is good in the range x[min] to x[max] and has degree (m,k).
Economized rational approximations.
A Padé approximation is very accurate near the center of expansion, but the error increases rapidly as you get farther away. If you are willing to sacrifice some of the goodness of fit near the
center of expansion, it is possible to obtain a better fit over the entire interval under consideration. This is what the other types of approximations do.
With an economized rational approximation, the idea is to start with a Padé approximation and perturb it with a Chebyshev polynomial in such a way as to reduce the leading coefficient in the error.
This perturbation does cause the vanished terms to reappear. However, the magnitude of the error does not increase very much near the center of expansion, and this small increase is compensated for
by a decrease in the error farther away. With EconomizedRationalApproximation, you specify the interval over which the approximation is to work rather than the center of expansion. Ultimately, as the
length of the interval goes to zero, the economized rational approximation approaches the Padé approximation.
This plots the difference between the Pad
approximation and the true function. Notice that the approximation is very good near the center of expansion, but the error increases rapidly as you move away.
This gives the difference between the true function and the economized rational approximation. In general, you may even get a small nonvanishing constant term.
Even though the error at the endpoint is not particularly small, it is considerably smaller than what
Rational Interpolation and Minimax Approximations
A degree (m,k) rational function is the ratio of a degree m polynomial to a degree k polynomial. Because rational functions only use the elementary arithmetic operations, they are very easy to
evaluate numerically. The polynomial in the denominator allows you to approximate functions that have rational singularities. For these reasons, a rational function is frequently useful in numerical
work to approximate a given function.
There are various methods to perform this approximation. The methods differ in how they interpret the notion of the goodness of the approximation. Each method is useful for certain classes of
problems. You can use this package to compute general rational interpolations and minimax approximations. The function PadeApproximant contains functionality that performs Padé approximations.
There is a related class of approximation questions that involves the interpolation or fitting of a set of data points by an approximating function. In this type of situation, you should use the
built‐in functions Fit, InterpolatingPolynomial, and Interpolation. For more information, see "Numerical Operations on Data".
give a rational interpolation of degree (m,k) to the points (x[i],f(x[i]))
give a rational interpolation with the points x[i] chosen automatically
One way to approximate a given function by a rational function is to choose a set of values for the independent variable and then construct a rational function that agrees with the given function at
this set of values. This is what is done by RationalInterpolation.
There are two ways of using RationalInterpolation. If you just specify a range in the independent variable, then the set of values is chosen automatically in a way that ensures a reasonable
approximation for the degree of approximation you have chosen. You can also give an explicit list of the set of values to be used. Note that in this case if you ask for a degree (m,k) approximation,
you must specify a list of values for the independent variable.
This gives a rational interpolation of degree
at seven equally spaced points between 0 and 2.
This plots the difference between the function and its approximation. Note that the error tends to get larger near the endpoints.
The interpolation points are somewhat more bunched at the ends of the interval. This usually results in a smaller maximum error.
option name default value
WorkingPrecision MachinePrecision number of digits of precision to use
Bias 0 bias in the automatic choice of interpolation points
Options for rational approximations.
When you specify a range of x values for RationalInterpolation, the interpolation points are chosen automatically. The option Bias allows you to bias the interpolation points to the right or to the
left. Values for Bias must be numbers between -1 and 1. A positive value causes the points to be biased to the right. The default is Bias->0, which causes the points to be symmetrically distributed.
When you bias the distribution of the points to the right, you get smaller errors there and larger errors to the left.
When you use RationalInterpolation, you get a rational function that agrees with the given function at a set of points. This guarantees that the rational function is, in one sense, close to the given
function. A stronger requirement for a good rational approximation would be to require that the rational function be close to the given function over the entire interval. This type of rational
approximation is produced by MiniMaxApproximation. This approximation is so named because it minimizes the maximum value of the relative error between the approximation and the given function. This
means that minimax approximation to a given function is the rational function of the given degree that minimizes the maximum value of the quantity over the interval under consideration. Note that the
term minimax approximation is also sometimes used for the rational function that minimizes the absolute error rather than the relative error used here.
give the minimax approximation to f of degree (m,k) on the interval from x[min] to x[max]
give the minimax approximation starting the iterative algorithm with approx
MiniMaxApproximation works using an iterative scheme. The first step is to construct a rational approximation using RationalInterpolation. This first approximation is then used to generate a better
approximation using a scheme based on Remes's algorithm. Generating the new approximation consists of adjusting the choice of the interpolation points in a way that ensures that the relative error
will diminish.
MiniMaxApproximation returns a list with two parts: a list of the points at which the maximum error occurs and a list consisting of the rational approximation and the value of the maximum error. This
extra information is provided not so much for the user's information, but to provide the capability of restarting the procedure without having to start back at the beginning. This is useful because
the algorithm is iterative, and if convergence does not occur before MaxIterations is reached, the incomplete answer is returned along with a warning.
This gives a list containing the points where the maximum error occurs and the desired interpolation, along with the value of the error.
Here is a plot of the relative error in the approximation over the interval . Reducing the error at any one of the extrema will force the error to increase at one of the others.
Because MiniMaxApproximation tries to minimize the maximum of the relative error, it is not possible to find a minimax approximation to a function that has a zero in the interval in question. The
rational approximation would have to be exactly zero at the zero of the function, or the relative error would be infinite. It is still possible to deal with such functions, but the zero must be
divided out of the function and then multiplied back into the rational function.
Dividing by
cancels the zero and there is now no problem computing the approximation.
Multiplying the approximation by
then gives the minimax approximation to
There are several ways in which MiniMaxApproximation can fail. In these cases, you will usually get a message indicating what probably went wrong and what you can do to avoid the problem. For
example, if you ask for a minimax approximation of degree (m,k), MiniMaxApproximation will look for a rational minimax approximation such that the relative error oscillates in sign and the extreme
values are achieved times. Sometimes the extreme values occur more times. It may be possible to design a more robust algorithm to deal with this problem, but in practice it is usually quite simple
just to ask for a minimax approximation with different degree.
When you try to compute this approximation you get a warning. Notice that there is not a single error, but a list of errors corresponding to the abscissas in the first part.
Another thing that can happen is that the initial rational interpolation can have a zero in the denominator somewhere in the interval. In such cases, it is usually easiest to ask for a minimax
approximation of different degree. Occasionally, however, this approach does not solve the problem.
A trick that will sometimes work is to start with an approximation that is valid over a shorter interval. Because the minimax approximation will usually change continuously as you lengthen the
interval, you can use the approximation for the shorter interval as a starting point for an approximation on a slightly longer interval. By slowly stretching the interval, you may be able to
eventually get a minimax approximation that is valid over the interval you desire.
MiniMaxApproximation has several options that give the user control over how it works and two options that help in diagnosing failures.
option name default value
WorkingPrecision MachinePrecision number of digits of precision to use
Bias 0 bias in the automatic choice of interpolation points
Brake {5,5} braking to apply on iterative algorithm
MaxIterations 20 maximum number of iterates after braking has ended
Derivatives Automatic specify a function to use for the derivatives
PrintFlag False whether to print information about the relative error at each step in the iteration
PlotFlag False whether to plot the relative error at each step in the iteration
Options for minimax approximations.
MiniMaxApproximation works by first finding a rational interpolation to the given function and then perturbing the coefficients until the error is equioscillatory. The option Bias is used to control
the initial rational approximation in exactly the same way it is used in RationalInterpolation.
As MiniMaxApproximation proceeds, it alternately locates the extrema of the relative error and then perturbs the coefficients in an effort to equalize the magnitudes of the extrema. If the extrema
move only a small amount from one iteration to the next, their previous positions can be used as starting values in locating their new positions. If they move too much from one iteration to the next,
MiniMaxApproximation gets lost. The way MiniMaxApproximation knows it is lost is if the extrema do not alternate in sign, two extrema are the same, or their abscissas are not in sorted order.
The way to prevent MiniMaxApproximation from getting lost is to set the option Brake. Brake acts as a braking mechanism on the changes from one iteration to the next, but its influence eventually
dies off to the point where there is no braking effect. When the algorithm has almost converged, there is no need to provide braking, because the changes are very small.
A value for Brake must be a list of two positive integers. The first integer specifies how many iterations are to be affected by the braking, and the second integer specifies how much braking force
is to be applied to the first iteration. Brake is much more important for minimax approximations of high degree, because in this case, the abscissas of the extrema are very close together.
To perform its iterative scheme, MiniMaxApproximation must know the first two derivatives of the function being approximated. If the Wolfram Language cannot compute the derivatives analytically, you
must specify them explicitly. A related situation is when the derivatives can be found, but calculating them involves a lot of work, much of which is redundant. For example, in trying to find a
minimax approximation to e^x, the Wolfram Language needs to evaluate e^x to find the value of the function, the Wolfram Language needs to evaluate e^x to find the value of the first derivative, and
the Wolfram Language needs to evaluate e^x to find the value of the second derivative. A much simpler way would be for the user to specify a function that returns a list of these three values for
each value of x. This is the purpose of the option Derivatives.
There are two things to be aware of when you use this option. First, the function should not be allowed to evaluate until x is actually a number, or the whole purpose of using the option will be
To prevent infinite iteration, MiniMaxApproximation has the option MaxIterations. After the braking stops, if convergence does not occur before the number of iterations reaches MaxIterations, a
warning is returned along with the current approximation. If the problem is simply slow convergence, you can restart the iteration from the current approximation by inserting the approximation that
was returned as the second argument to MiniMaxApproximation. You may find it useful to begin the new iteration with different options.
To get an example of a poor approximation, you can choose a small value of
, a large bias, and no braking.
The result of the previous approximation attempt is used as the starting point of the new iteration by inserting it as a second argument.
PrintFlag and PlotFlag are options to MiniMaxApproximation that are useful for diagnosing the reason for failure. Setting either of these options will have no effect on the calculations. If PlotFlag
is set to True, a plot of the relative error in each iterated rational approximation will be generated. If these plots change dramatically from one iteration to the next, you should probably increase
the braking. If PrintFlag is set to True, two things will happen. First, as the extrema are being located for the first time, lists of the changes in the approximations to the abscissas are printed.
The numbers in these lists should rapidly decrease once they get reasonably small. After the extrema are located for the first time, lists of ordered pairs are printed, consisting of abscissas of the
extrema of the relative error and the value of the relative error at each of those abscissas.
give a rational interpolation of degree (m,k) to a function of x whose graph is given parametrically as a function of t
give a rational interpolation with the points t[i] chosen automatically
General rational interpolations.
There are certain approximation problems in which you will want to use GeneralRationalInterpolation instead of RationalInterpolation. For example, you should use GeneralRationalInterpolation if you
want to find a rational interpolating function to the inverse of a function that can only be evaluated by using FindRoot. In such a case, RationalInterpolation would be very slow.
GeneralRationalInterpolation lets you do more general approximation problems by allowing the function that is to be approximated to be given parametrically. For example, the graph of the function is
just the upper half of the unit circle. This can also be described parametrically as where . Thus you can compute an approximation to by specifying the function as {Cos[t],Sin[t]}.
In the general case, when you specify the functions in RationalInterpolation as {f[x],f[y]}, the expressions f[x] and f[y] are functions of t. The function that is interpolated is the one whose graph
is given parametrically by for t[min]≤t≤t[max].
Note that you must always specify a symbol for the independent variable; using the parametric variable as the independent variable would be incorrect. GeneralRationalInterpolation takes the same
options as RationalInterpolation.
As is the case with
, the error is often smaller if the interpolation points are not given explicitly. The points chosen automatically tend to be better distributed.
give the rational approximation of degree (m,k) to a function of x whose graph is given parametrically as a function of t
give the minimax approximation starting the iterative algorithm with approx
give the rational approximation computing the error using a factor g(t)
General minimax approximations.
The function that is to be approximated is specified in GeneralMiniMaxApproximation in the same way as it is in GeneralRationalInterpolation. The options for GeneralMiniMaxApproximation are the same
as for MiniMaxApproximation.
This gives a degree
minimax approximation to the inverse of
Since there is an easy way to evaluate the inverse of
, it is also possible to use
for this problem. The only difference between the solutions is in the abscissas of the extrema.
If you use the option Derivatives with GeneralMiniMaxApproximation, you must specify a list of derivatives for both parts of the parametrically defined function.
Another situation in which GeneralMiniMaxApproximation is useful is when you want to do an approximation and measure its goodness of fit using the absolute error instead of the relative error that is
used by default. If you want to use a different metric for the error, you can specify it as the third part of the parametrically defined function. If the function is given as and is the rational
minimax approximation to the function that relates to , then the error that is minimized is . If is not specified, it is taken to be the same as , and it is the maximum relative error that is
minimized. If you want to minimize the absolute error, simply use the constant 1 for , which is the default.
This gives a rational minimax approximation to the inverse of
with the maximum absolute error minimized.
If you get an approximation for which the error is unacceptably large, there are several things you can do. If you increase the degree of the numerator and/or the denominator, the minimax error will
usually decrease; it cannot increase. Even shifting a degree from the numerator to the denominator or vice versa can affect the size of the error. If the interval for which the approximation is to be
valid is very long, it is probably a good idea to look at the asymptotic behavior of the function and choose degrees for the numerator and denominator that give the correct asymptotic behavior. For
example, to get an approximation to for large , the degrees of the numerator and denominator should be the same, since for large the function has a nearly constant value of .
Another way to decrease the error is to shorten the interval for which the approximation is to be valid. Often it is better to have several low‐degree approximations to cover a long interval than a
single high‐degree approximation. The extra storage required is not that much, and each of the simpler approximations will evaluate faster.
To get an optimal minimax approximation, you need to take into account the numerical behavior of the final approximation. It is usually a good idea to define the function so that the variable
appearing in the approximation has values near the origin. Thus, instead of finding a minimax approximation to on the interval , it would be better to find a minimax approximation to on the interval
In cases where you want to avoid all potential problems associated with loss of digits due to subtractive cancellation, you may even want to do some shifting of the variable after the approximation
is found. For example, the rational minimax approximation of degree to on the interval has positive coefficients in the numerator and coefficients with alternating signs in the denominator. The
relative error in the approximation is only about , but using the approximation in the given form could result in a larger relative error, since a few bits could be lost due to cancellation near the
endpoints: cancellation occurs in the numerator near and in the denominator near . By replacing the in the numerator by and the in the denominator by , all coefficients in both the numerator and the
denominator become positive, and the values of and will be between and ; no cancellation can occur. Of course all of this must be done to much higher precision to ensure that the coefficients
themselves are correct to the required precision.
It is very important to also consider the zeros of the function and its asymptotic behavior. The simplicity of the resulting minimax approximation is greatly affected by the extent to which you can
trivially make the function look like a constant function. As an example, to find a minimax approximation to Gamma[1/2,x,Infinity] on the interval , you can consider the function Gamma[1/2,x,Infinity
]Exp[x](x+4/7) (cf. Abramowitz and Stegun, Handbook of Mathematical Functions, 6.5.31; the 4/7 was chosen empirically). This function only varies a few percent over the interval in question; it will
be much easier to find a minimax approximation to this function than to the original function.
If you are attempting to minimize the maximum relative error, and there are zeros of the function in the interval in question, you will have to divide out the zeros first. If singularities occur in
the interval, they will have to be eliminated also, either by multiplying by a zero or subtracting them away.
Finding Roots Using Interpolation
The function FindRoot is useful for finding a root of a function. It is fairly robust and will almost always find a root if it is given a sufficiently close starting value and the root is simple (or
if it is multiple and the option DampingFactor is appropriately set). To achieve this robustness, FindRoot makes compromises and is not particularly conservative about the number of function
evaluations it uses. There are cases, however, where you know that the function is very well behaved, but evaluating it is extremely expensive, particularly for very high precision. In such cases,
InterpolateRoot may be more efficient.
InterpolateRoot looks at previous evaluations of the function, say , , , and and forms the interpolating polynomial that passes through the points , , , and . The algorithm gets the next
approximation for the root of by evaluating the interpolating polynomial at 0. It turns out that using all of the previous data is not the best strategy. While the convergence rate increases with the
use of additional data points, the rate is never greater than quadratic. Further, the more data points used, the less robust the algorithm becomes. InterpolateRoot uses only the previous four data
points since there is almost no benefit to using more.
InterpolateRoot[f,{x,a,b}] find a root of the function f near the starting points a and b
InterpolateRoot[eqn,{x,a,b}] find a root of the equation eqn near the starting points a and b
Root finding with InterpolateRoot.
option name default value
AccuracyGoal Automatic the desired accuracy in the root being sought
MaxIterations 15 maximum number of function evaluations before giving up
WorkingPrecision 40 the maximum precision to use in the arithmetic calculations
ShowProgress False whether to print intermediate results and other information as the algorithm progresses
Options for InterpolateRoot.
The Automatic choice for AccuracyGoal means that the AccuracyGoal will be chosen to be 20 digits less than the WorkingPrecision. It should be noted that AccuracyGoal as used in InterpolateRoot is
different from AccuracyGoal in FindRoot. FindRoot is a much more general function that works for systems of equations. Trying to justify an accuracy in the value of the root itself is too difficult.
FindRoot merely stops when the value of the function is sufficiently small. InterpolateRoot is much more specialized. It only works for a single function of a single variable at simple roots and
assumes that the function is very well behaved. In such cases it is quite easy to justify an accuracy in the value of the root itself.
Set up a function whose root you wish to find, with a counter
to determine the number of evaluations.
requires fewer function evaluations to get the same result.
Approximating Integrals with Interpolating Functions in the Integrand
The Wolfram Language function NIntegrate uses algorithms that assume that the integrand is smooth to at least several orders. InterpolatingFunction objects typically do not satisfy this assumption;
they are continuous, but only piecewise smooth. The algorithms used by NIntegrate converge very slowly when applied to InterpolatingFunction objects, especially in several dimensions. NIntegrate
allows the domain of integration to be broken up into several pieces and the integral evaluated over each piece. If the pieces of the domain correspond to the pieces over which the
InterpolatingFunction is smooth, NIntegrate will converge much more rapidly. NIntegrateInterpolatingFunction automatically breaks up the domain of integration.
find a numerical approximation to an integral with InterpolatingFunction objects in the integrand
find a numerical approximation to a multidimensional integral with InterpolatingFunction objects in the integrand
Numerical approximations to integrals with InterpolatingFunction objects in the integrand.
NIntegrateInterpolatingFunction uses the function NIntegrate, but it breaks up the domain into sections where the InterpolatingFunction object is smooth.
produces almost exactly the same result, but takes much longer because the convergence is poor if the domain is not properly broken up.
If you simply need to find the integral of an InterpolatingFunction object (as opposed to a function of one), it is better to use Integrate because this gives you the result which is exact for the
polynomial approximation used in the InterpolatingFunction object.
Order Star Plots
Analysis of the stability of numerical methods for solving differential equations is a considerably more difficult task than finding the order of approximation. This difficulty is reflected in the
fact that there are many different ways of defining stability. This package assists in the determination of stability regions for numerical methods.
Stability regions are important because they reflect the rate at which errors are propagated in the approximate solution. Just as there are absolute and relative errors in numerical analysis, there
are also absolute and relative measures of stability. This package renders order stars, which are useful in examining the relative stability of a method. Furthermore, by specifying the comparison
function to be identically 1, you can also draw regions of absolute stability.
A given numerical method for a problem can be recast into the framework of approximation theory. The goal is then to study how well this approximant behaves when compared with the solution. There is
a kind of paradox here, for if the solution were known then you would have no need to resort to a numerical approximation. However, you want to establish a framework which applies to any problem in a
given class. Since generically analytic solutions to problems cannot be found, it is common to study how a numerical method behaves when it is applied to a linearized system. In the area of ordinary
differential equations, for example, you might be interested in solutions to the system of equations:
for a generally nonlinear . It is common to replace this system by a scalar linear problem that you can solve, namely,
where is considered to be a complex constant, and you have fixed an initial condition, so that the equation is uniquely determined. Stability analysis is now a study of how well a numerical solution
behaves when applied to the simplified differential equation (1). Equation (1) is often referred to as the scalar linear test problem, or Dahlquist's equation.
The discussion which follows concentrates on how to use the package. You should keep in mind that although the focus is on the behavior of an approximant, the underlying interest is in the numerical
method from which the approximant arose when applied to some problem. For more information on this correspondence, stability analysis, and the theory of order stars, see the references at the end of
this section.
OrderStarPlot[r,f] draw the order star depicting the region where r/f<1, for the functions r and f
OrderStarPlot[r,f,z] draw the region in the complex z plane where r/f<1, where r and f are functions of z
draw the order star depicting the region where Re(r-f)<0
Padé approximations are rational polynomial approximants where all parameters are chosen to maximize order at some local expansion point. Certain numerical methods such as Runge–Kutta methods are
related to Padé approximants to the exponential.
This constructs a Pad
approximant to . The function for doing this is loaded automatically. This approximant corresponds to the forward Euler method.
This is the relative stability region, or order star of the first kind, for the forward Euler method. The pole of the approximant is highlighted.
This is the absolute stability region for the forward Euler method, obtained as a relative comparison with 1.
OrderStarInterpolation specify whether to display points where r and f are equal
OrderStarKind specify which kind (First or Second) of order star to draw
OrderStarLegend specify whether to include a plot legend containing the various symbols used
OrderStarPoles specify whether to indicate the poles of the approximant and/or the function
OrderStarZeros specify whether to indicate the zeros of the approximant and/or the function
OrderStarSymbolSize specify the size of the symbols used to indicate zeros, poles, and interpolation points
OrderStarSymbolThickness specify the line thickness for the symbols used to indicate zeros, poles, and interpolation points
Options unique to OrderStarPlot.
When you ask for certain features to be displayed, and the Wolfram Language is unable to find these features, you will obtain the message OrderStarPlot::sols containing more specific information
relating to the problem. Solve may also issue messages, such as when inverse functions are being used.
OrderStarPlot uses heuristics in order to determine what the independent variable is. You can save time in a very complicated expression by specifying the variable to use explicitly. If there is any
ambiguity in the variable choice, then input returns unevaluated and an appropriate warning message is issued, since the function will not evaluate numerically.
This indicates the variable to use and highlights points where . This may not be possible in general if the relationship is nonalgebraic.
In addition to True and False, the options OrderStarInterpolation, OrderStarLegend, OrderStarPoles, and OrderStarZeros can take on lists of coordinate pairs to specify points that cannot be found
automatically. As well as resizing the plot legend by specifying scaled coordinates, you can specify information to the legend, such as the style and size of the font to use.
The position of the legend is given in scaled coordinates, using the same syntax as that of
. Font style and size information are also specified, and the symbols used to represent zeros and poles are increased in size.
In addition to the many options unique to OrderStarPlot, there are several that are simply passed to ContourPlot and used to produce the plot.
AspectRatio specify the aspect ratio of the plot
Axes specify whether to draw axes
AxesOrigin specify the intersection of the axes
ClippingStyle specify the style of what should be drawn when curves or surfaces would extend beyond the plot range
ColorFunction specify a function to apply to color the stability and instability regions
FrameTicks specify whether or where to place tick marks around the frame
MaxRecursion specify how many recursive subdivisions can be made
PlotPoints specify the number of sample points to use in constructing the contour plot
PlotRange specify the plot range to be used in the plot
Ticks specify whether or where to place tick marks along the axes
Options common to many graphics functions.
An important issue is whether an order star crosses the imaginary axis. Additionally, it may be interesting to illustrate symmetry around the real axis. In order to facilitate these comparisons,
OrderStarPlot uses graphics options to render axes which pass through the origin.
OrderStarPlot resolves fine features by making use of the MaxRecursion and PlotPoints options of ContourPlot.
The default plot region is determined from essential features of the order star. However, this default can be overridden using standard Graphics options.
You can change the default plot region and plot density using standard options. However, in doing so you can make the plot quite jagged.
Order stars of the second kind are useful in the study of stability for linear multistep methods.
Order stars provide a means of determining, at a glance, many important features of interest, such as A-stability. Furthermore, by considering a relative comparison with a function, order stars
manage to encrypt the order of accuracy of a numerical scheme into the stability region.
Here are only the zeros and poles of the approximant, since there are no finite zeros or poles of . The numerical method corresponding to this approximant is A
stable since the approximant has no poles in the left half
plane. Furthermore, a count of the sectors adjoining the origin tells you that the order of approximation is 5 (one less than the number of sectors).
Expositions of the theory of stability of numerical methods can be found in [1], [2], and [3]. More examples of applications of the package are provided in [4]. | {"url":"https://reference.wolframcloud.com/language/FunctionApproximations/tutorial/FunctionApproximations.html","timestamp":"2024-11-02T14:00:42Z","content_type":"text/html","content_length":"201125","record_id":"<urn:uuid:3a968af5-cc4d-4805-ac42-3d7453febeff>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00476.warc.gz"} |
IntCal 0.3.1
• added an option ccdir to provide alternative locations of the calibration curves
• added a function new.ccdir(), which copies the package’s calibration curves into a specified folder
• C14.lim now works if 0 is included as minimum 14C age
IntCal 0.3.0
• updated to the updated postbomb curves (now both yearly and monthly), published by Hua et al. 2021
• repaired problems with depths and calheights in draw.dates
• draw.ccurve now plots correct label when BCAD
• c14.lim now estimated correctly when a second curve is added
IntCal 0.2.3
• draw.ccurve now plots the correct label when using BCAD, and plots depths at the expected heights
• draw.dates now plots multiple dates at the expected heights (with more precise dates peaking higher)
• corrected the Rmd files which had erroneous formatting and some confusing examples
IntCal 0.2.2
• in mix.curves(), calibration curves are now written to a temporary directory by default, as per CRAN policies
• calibrate function now deals better with postbomb dates
• added clam’s function calBP.14C to find IntCal C14 ages corresponding to cal BP ages
• added function glue.ccurves for gluing prebomb and postbomb calibration curves
• added a warning/error to calibrate() for dates truncated by or outside the calibration curve
• updated the vignettes
IntCal 0.2.1
• added new function list.curves
• Shortened the functions copyCalibrationCurve and draw.CalibrationCurves to the shorter and more consistent ccurve and draw.ccurve
• ccurve is now more flexible with the names of the calibration curves
• draw.ccurve can now plot another curve on its own righthand axis scale (handy when mixing pre- and postbomb curves)
• new functions calibrate, caldist and hpd, copied and modified from the clam R package
IntCal 0.2.0
• added a function intcal.data to plot the data underlying the IntCal curves
• copyCalibrationCurve can now copy more of the curves
• new function draw.calibration curve to draw the calibration curves (one or two), or to add curves to an existing plot)
IntCal 0.1.3
• fixed issues with the DESCRIPTION file
IntCal 0.1.2
• Added Rd files
• Added some of the most relevant calibrated-related functions of the rbacon package (so that these can in time be deleted from that and other packages)
IntCal 0.1.1
• fixed some errors in the description file
IntCal 0.1.0
• first iteration of this R data package (originally put on github) | {"url":"http://cran.r-project.org/web/packages/IntCal/news/news.html","timestamp":"2024-11-05T06:34:02Z","content_type":"application/xhtml+xml","content_length":"4192","record_id":"<urn:uuid:fe1d7730-3eaa-466f-876f-7ea9fd38729c>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00255.warc.gz"} |
OCU's Mathematics program provides you with a rigorous and structured education in the fundamental principles of mathematics, developing your analytical and problem-solving skills. The program is
designed to offer a comprehensive understanding of various mathematical disciplines and their applications in diverse fields.
Our program provides a solid foundation in abstract and applied mathematics, preparing students for careers in academia, research, industry and various STEM-related fields. Graduates are equipped
with strong analytical skills, problem-solving abilities, and a deep understanding of mathematical principles essential for addressing complex challenges in diverse domains.
Credit Hours: 45-51
Preparatory or Prerequisite Courses*: 8 hours
• MATH 1303 Intermediate Algebra** 3
• MATH 1503 College Algebra*** 3
• MATH 1602 Trigonometry** 2
General Track: 44-46 hours
Required Mathematics Courses: 27 hours
• MATH 2004 Calculus and Analytic Geometry I
• MATH 2104 Calculus and Analytic Geometry II
• MATH 2403 Foundations of Mathematics
• MATH 3003 Linear Algebra
• MATH 3103 Algebraic Structures I
• MATH 3404 Calculus and Analytic Geometry III
• MATH 3603 Real Analysis I
• MATH 4993 Capstone
Electives in Mathematics: 15 hours
Select 15 credit hours from the following:
• MATH 3203 Probability and Statistics I
• MATH 3303 Ordinary Differential Equations
• MATH 3503 Discrete Mathematics
• MATH 3703 Numerical Methods
• MATH 3913 Complex Analysis
• MATH 3703 Advanced Geometry I
• MATH 4103 Algebraic Structures II
• MATH 4203 Probability and Statistics II
• MATH 4303 Partial Differential Equations
• MATH 4403 Topology
• MATH 4503 Quantum Mechanics
• MATH 4603 Real Analysis II
• MATH 4703 Advanced Geometry II
• MATH 4993 Independent Study
• Approved elective in computer science
Additional Elective Requirement
• Any additional MATH course 3000-LEVEL or above
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• Any Physics (PHYS) course 2000-LEVEL or above
Data Science Track: 49-51 hours
Required Mathematics Courses: 36 hours
Electives in Mathematics/Computer Science 6 (7)
• Select two courses from the following:
• Any additional MATH course 3000-level or above
• Any approved Computer Science (CSCI) course
* Preparatory or prerequisite courses do not count toward degree requirements
** Waived upon completion of more advanced mathematics courses.
*** Required in the general education curriculum but waived upon completion of a more advanced mathematics course. | {"url":"https://www.okcu.edu/programs/mathematics","timestamp":"2024-11-07T00:23:36Z","content_type":"text/html","content_length":"55324","record_id":"<urn:uuid:d6ea71e7-6826-4d40-8397-76e8d711a5fc>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00876.warc.gz"} |
Re: st: access coefficients from another regression
Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
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Re: st: access coefficients from another regression
From Nick Cox <[email protected]>
To [email protected]
Subject Re: st: access coefficients from another regression
Date Tue, 30 Oct 2012 12:56:11 +0000
You need to save the estimate from the first model before you fit the second.
For example, a convenient method is to copy estimates to a matrix:
. mat b = e(b)
. mat li b
On Tue, Oct 30, 2012 at 12:43 PM, Christina <[email protected]> wrote:
> my model consist of two equations x1 und x2 which I estimate seperatly using non linear least squares in Stata 12. α and β denote the estimated coefficients, y and z the variables.
> First I estimate x1 using:
> nl (x1= {α1}*y1+ {α2}*y2 + {α3}*y3)
> Then I estimate x2 using:
> nl (x2= {β1}*z1+ {β2*z2 + {β3}*z3)
> After the estimation I want to calculate a new coefficient γ using the coefficient α1 from the first equation and β1 from the second equation using the nlcom command, for example:
> nlcom (γ= α1/β1)
> My question is now, how can I access the coefficient α1 from the first equation x1? How would the code for the nlcom command look like?
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/faqs/resources/statalist-faq/
* http://www.ats.ucla.edu/stat/stata/ | {"url":"https://www.stata.com/statalist/archive/2012-10/msg01341.html","timestamp":"2024-11-04T04:55:58Z","content_type":"text/html","content_length":"10152","record_id":"<urn:uuid:cac50380-1318-4abf-8853-a96ed34ea1ec>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00138.warc.gz"} |
Gotta Go Fast
76 Days Until I Can Walk
It has been another very productive day, although a couple of frustrating bugs have popped up which I wasn’t able to resolve today.
Fixed Point Arithmetic
My initial goal was to try and implement fixed point arithmetic for the ray caster functions. I read about FPA in Larry Myers book, but never got the chance to use it. My implementation follows his
quite closely in that I use an FP_SHIFT value of 16, which is apparently an acceptable compromise on precision in floating point values.
However, Myers’ sample code for these functions is in assembly, and I found his descriptions for the implementation of multiplication and division to be a little difficult to parse. He essentially
relies on an understanding of how the imul and idiv instructions work in the Intel architecture. I found this lecture by Van Hunter Adams to be a little better, but he skips over the implementation
of fixed point division far too quickly.
The issue is essentially this. If we want to implement addition and subtraction in fixed point, that’s easy. We can simply add and subtract our fixed point values and the maths will work out. For
multiplication and division, however, we need to do some shifting before and after the operations. This is because multiplication and division will cause overflow in the binary representation of the
fixed point number. We can deal with this by casting our values from 32 bit ints to 64 bit longs before we perform the operation, capturing the overflow in the larger datatype and then shifting
things around to get our answer. But why does this work?
I found it easier to understand what was going on by thinking about the problem as an equality rather than as a programming problem. When switching from floating point to fixed point, we essentially
multiply by some constant \(C\). This constant is the same for all values that we convert to fixed point. So if we have two floating point values \(a\) and \(b\), and we convert them both to fixed
point values, the result will be \(aC\) and \(bC\).
Now if we think about a naive implementation of our four basic operations using these samples:
\[aC + bC = C(a + b)\]
\[aC - bC = C(a - b)\]
\[aC \times bC = abC^2\]
\[\frac{aC}{bC} = \frac{a}{b}\]
Note how everything is fine for addition and subtraction. The result will be the sum or difference of \(a\) and \(b\) scaled by the same constant \(C\)
However this constant is squared during multiplication, and cancels out during division. Therefore, in order to implement fixed point multiplication and division we must divide the result of
multiplication by \(C\), and multiply the result of division by \(C\). So the “correct” way to do multiplication and division in fixed point is:
\[\frac{aC \times bC}{C} = \frac{abC^2}{C} = abC\]
\[C \times \frac{aC}{bC} = C \times \frac{a}{b}\]
In Rust, the implementation will look like this:
pub const fn add(a: i32, b: i32) -> i32 {
a + b
pub const fn sub(a: i32, b: i32) -> i32 {
a - b
pub const fn mul(a: i32, b: i32) -> i32 {
((a as i64 * b as i64) >> FP_SHIFT) as i32
pub const fn div(a: i32, b: i32) -> i32 {
(((a as i64) << FP_SHIFT) / b as i64) as i32
With this done, I implemented conversion from f64 and i32 to fixed point using traits which I implemented on the built in types.
const FP_SHIFT: i32 = 16;
const FP_MULT: f64 = 65536.0;
pub trait ToFixedPoint {
fn to_fp(&self) -> i32;
pub trait FromFixedPoint {
fn to_f64(&self) -> f64;
fn to_i32(&self) -> i32;
impl ToFixedPoint for f64 {
fn to_fp(&self) -> i32 {
(*self * FP_MULT) as i32
impl ToFixedPoint for i32 {
fn to_fp(&self) -> i32 {
*self << FP_SHIFT
impl FromFixedPoint for i32 {
fn to_f64(&self) -> f64 {
*self as f64 / FP_MULT
fn to_i32(&self) -> i32 {
*self >> FP_SHIFT
All of this would probably be better if I implemented it using macros, but I’m still struggling to fully understand the syntax, and given the number of things I wanted to get done today, studying
macros did not feel like a good use of my time.
With fixed point implemented, I started swapping out all the floating point calculations from my ray cast implementation. I also properly dropped in my lookup tables, meaning that finding the result
for many trigonometric operations is simply a case of indexing quickly into an array.
It took a little bit of fiddling, but but before long I had my ray cast function fully switched over, and I am very happy with the result. Just as a little bonus, I “borrowed” F. Permadi’s approach
to shading and applied it to my ray caster so that my simple little test room would look a little better.
While I am very happy, there are bugs. The zero-width walls are causing problems. If the rays hit them at a certain angle, then is possible for the ray to miss the side of the zero-width wall and it
will disappear. You’ll notice this if you walk around the demo level. I wasn’t able to fix this today, but I will come back to it.
Collision Detection
By the time I was finished with FPA, it was getting a little late, but I really wanted to try and implement collision detection on the walls, if only to improve the experience of the demo a little
bit. Myers’ has a very detailed implementation of a collision detection function in Chapter 5 of his book which checks for adjacent walls in front, behind, beside, and diagonal to the camera. His
implementation allows for sliding along walls too. Just to get the ball rolling, I ported his function from C to Rust and dropped it straight into my application. It works… OK?
If the camera is moving down or to the right, then everything seems to work fine, but when moving up or to the left (relative to the map, not the camera) we can clip through walls. Again, I didn’t
have time to resolve this today, so I will come back to it on Monday. Myers’ code is also quite long, contains a fair bit of duplicate code, and several magic numbers. I’m pretty confident that I can
tidy this up and make it more Rust-like. But for now, it’s an excellent start.
Other Happenings
I’ve got a lot done this week. Far more than I had initially set out to do on the code base, but I have neglected two tasks that were not related to writing code.
I want to write a longer form, better edited article for the site about my initial impressions of Rust and my understanding of how Rust and WebAssembly work together. I think this will be a good
exercise for gaining a deeper understanding of Rust and what’s actually happening behind the scenes.
I also want to try and keep up with Tim McNamara’s Rust course, if I can. It is at the end of week 2, so I would like to catch up on the lectures and start following along properly.
Given that today is Friday, I think I will push pause on software development and spend the weekend dealing with these two outlying tasks. I will get back to coding on Monday. Hopefully first I will
resolve these little graphical bugs, and then my next goal will be to get texturing working.
Have a good weekend folks! | {"url":"https://fourteenscrews.com/devlog/gotta-go-fast/","timestamp":"2024-11-06T12:19:03Z","content_type":"text/html","content_length":"30001","record_id":"<urn:uuid:d180e799-345c-46f7-899e-8a549d87e908>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00245.warc.gz"} |