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F-score Deep Dive
An alternative method for choosing beta in the F-score
Recently at work we had a project where we used genetic algorithms to evolve a model for a classification task. Our key metrics were precision and recall, with precision being somewhat more important
than recall (we didnā t know exactly how much more important at the start). At first we considered using multi-objective optimization to find the Pareto front and then choose the desired trade-off,
but it proved impractical due to performance issues. So we had to define a single metric to optimize.
Since we were using derivative-free optimization we could use any scoring function we wanted, so the F-score was a natural candidate. It ended up working quite well, but there were some tricky parts
along the way.
General background
Accuracy (% correct predictions) is a classical metric for measuring the quality of a classifier. But itā s problematic for many classification tasks, most prominently when the classes arenā t
balanced or when we want to differently penalize false positives vs. false negatives.
Precision and recall separate the model quality measurement to two metrics, focusing on false positives and false negatives, respectively. But then comparing models becomes less trivial - is 80%
precision, 60% recall better or worse than 99% precision, 40% recall?
Taking the average is a possibility; letā s see how it does:
So if we have a model with 0% precision and 100% recall, the average is a score of 50%. Such a model is completely trivial from a prediction point of view (always predict positive), so ideally it
should have a score of 0%. More generally, we see that the average exhibits a linear tradeoff policy: you can stay on the same score by simultaneously increasing one metric and decreasing the other
by the same amount. When the metrics are close this could make sense, but when thereā s a big difference it starts to deviate from intuition.
F-score to the rescue
The F[1]-score is defined as the harmonic mean of precision and recall:
\[F_1 = \frac{2}{\frac{1}{p} + \frac{1}{r}}\]
Letā s visualize it:
This seems much more appropriate for our needs: when thereā s a relatively small difference between precision and recall (e.g. along the y = x line), the score behaves like the average. But as the
difference gets bigger, the score gets more and more dominated by the weaker metric, and further improvement on the already strong metric doesnā t improve it much.
So this is a step in the right direction. But now how do we adjust it to prefer some desired tradeoff between precision and recall?
Some history and the beta parameter
As far as I understand, the F-score was derived from the book Information Retrieval by C. J. van Rijsbergen, and popularized in a Message Understanding Conference in 1992. More details on the
derivation can be found here. The full derivation of the measure includes a parameter, beta, to control exactly what weā re looking for - how much we prefer one of the metrics over the other. This
is also what the ā 1ā in F[1] stands for - no preference for either (a value between 0 and 1 indicates a preference towards precision, and a value larger than 1 indicates a preference towards
recall). Here is the full definition:
\[F_\beta = (1 + \beta^2) \cdot \frac{precision \cdot recall}{\beta^2 \cdot precision + recall}\]
Visualizing the F-score
First, to develop some intuition regarding the effect of beta on the score, hereā s an interactive plot to visualize the F-score for different values of beta. Play with the ā bandsā parameter to
explore how different betas create different areas of (relative) equivalence in score.
Beta: 0.01 100
Bands: 5 100
Choosing a beta
According to the derivation, a choice of beta equal to the desired ratio between recall and precision should be optimal. In this case, if I understood the math correctly, optimality is defined as
following: take the F-score function for some beta, which is simply a function with two variables. Find its partial derivatives with respect to recall and precision. Now find a place where those
partial derivatives are equal, that is, a point on the precision-recall plane where a change in one metric is equivalent to (will lead to the same change as) a change in the other metric. The F-score
function is structured in such a way that when beta = recall / precision, this point of equivalence lies on the straight line passing through the origin with a slope of recall / precision. In other
words, when the ratio between recall and precision is equal to the desired ratio, a change in one metric will have the same effect as an equal change in the other. I sort of get the intuition behind
this definition, but Iā m not convinced it captures the essence of optimality anyone using the F-score might find useful.
Taking a closer look
When trying to set beta = desired ratio, the results seemed a little off from what I would expect, and I wanted to make sure the value weā ve chosen for beta really was optimal for our use case. I
went on a limb here, and the next part is rather hand-wavy, so Iā m not convinced this was the right approach. But here it is anyway.
Imagine the optimizer: crunching numbers, navigating a vast, multidimensional space of classifiers. The navigation is guided by a short-sighted mechanism of offsprings and mutations, with each
individual classifier being mapped to the 2d plane of precision and recall, and from there to the 1d axis of the F-score. Better classifiers propagate to future generations, slowly moving the
optimizer to better sections of the solution space.
Now imagine this navigation on the precision-recall plane. The outcome is governed by two main factors: the topology of the solution space (how hard it is to achieve a certain combination of
precision and recall) and the gradients of the F-score (how ā goodā it is to achieve a certain combination of precision and recall). We can imagine the solution topology as an uneven terrain on
which balls (solutions) are rolling and the F-score as a slight wind pushing the balls in desired directions. We would then like the wind to always push in the direction bringing solutions to our
desired ratio. Letā s try to investigate the F-score under this imaginative and wildly unrigorous intuition: we have no idea how the solution topology looks like (though if we did multi-objective
optimization we could get a rough sketch, e.g. by looking at the Pareto front at each generation), so weā ll focus on the direction of the F-score ā windā . To do that weā ll need to find the
partial derivatives of the F-score w.r.t. precision and recall:
\[\frac{\partial F}{\partial r} = (1 + \beta^2) \cdot \frac{p(\beta^2 p + r) - pr \cdot (1)}{(\beta^2 p + r)^2} = (1 + \beta^2)\cdot \frac{\beta^2 p^2 + p r - p r}{(\beta^2 p + r)^2} = \frac{(1 + \
beta^2)}{(\beta^2 p + r)^2} \cdot \beta^2p^2\] \[\frac{\partial F}{\partial p} = (1 + \beta^2) \cdot \frac{r(\beta^2p + r) - pr \cdot (\beta^2)}{(\beta^2 p + r)^2} = (1 + \beta^2) \cdot \frac{\beta^
2pr + r^2 - \beta^2pr}{(\beta^2 p + r)^2} = \frac{(1 + \beta^2)}{(\beta^2 p + r)^2} \cdot r^2\]
We got very similar-looking partial derivatives: letā s take a look at the ā slopeā to which the score is pushing at any given point:
\[\frac{^{\partial F}/_{\partial r}}{^{\partial F}/_{\partial p}} = \frac{\beta^2p^2}{r^2} = (\beta \cdot \frac{p}{r})^2\]
Interesting: the direction at which the score is pushing is constant along straight lines from the origin (though the direction itself usually isnā t along the line). And we can think of one such
line where we would like the direction to be along that line: the line where r / p = R, our desired ratio. On that line the slope should be equal to R as well, so we get:
\[R = \frac{\beta^2}{R^2} \\ \beta^2 = R^3 \\ \beta = \sqrt{R^3}\]
So we have a different definition of optimality which yields a different ideal value for beta.
Iā m not sure how important this deep plunge to the maths of the F-score is to cases where you donā t have an unusual desired tradeoff between precision and recall, or when youā re just using the
F-score to measure a classifier thatā s trained by a different loss function. Usually youā re probably safe with going with F[1], F[0.5] or F[2].
But I certainly feel I have a better understanding of how and why the F-score works, and how to better adjust it for a given scenario. | {"url":"https://andersource.dev/2019/09/30/f-score-deep-dive.html","timestamp":"2024-11-12T14:11:02Z","content_type":"text/html","content_length":"16874","record_id":"<urn:uuid:9404938a-4bed-4ae0-b222-212df7890fc3>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00162.warc.gz"} |
Basic statistics concepts
In this recipe, you will learn about the StatsBase package, which helps you use basic statistical concepts such as weight vectors, common statistical estimates, distributions, and others.
To get started with this recipe, you have to first install the StatsBase package by executing Pkg.add("StatsBase") in the REPL.
1. Weight vectors can be constructed as follows:
w = WeightVec([4., 5., 6.])
2. Weight vectors also compute the sum of the weights automatically. So, if the sum is already computed, it can be added as a second argument to the vector construction so that it saves time
required for computing the sum. Here is how to do it:
w = WeightVec([4., 5., 6.], 15.)
3. Weights can also be simply defined by the weights() function, as follows:
w = weights([1., 2., 3.])
4. Some important methods that can be used on weight vectors are:
1. To check whether the weight vector is empty or not, the isemply() function can be used: | {"url":"https://subscription.packtpub.com/book/data/9781785882012/3/ch03lvl1sec20/basic-statistics-concepts","timestamp":"2024-11-11T08:32:55Z","content_type":"text/html","content_length":"90017","record_id":"<urn:uuid:1b245adb-15bd-420f-b521-a0dea148ad83>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00136.warc.gz"} |
Rubik's Cube Algorithms - Ruwix (2024)
A Rubik's Cube algorithm is an operation on the puzzle which reorients its pieces in a certain way. Mathematically the Rubik's Cube is a permutation group: an ordered list, with 54 fields with 6*9
values (colours) on which we can apply operations (basic face rotations, cube turns and the combinations of these) which reorient the permutation group according to a pattern.
To describe operations on the Rubik's Cube we use the notation: we mark every face of the puzzle with a letter F(Front), U(Up), R(Right), B(Back), L(Left), D(Down). A letter by itself means a 90
degree clockwise rotation of the face. A letter followed by an apostrophe is a counterclockwise turn.
F U R B L D | F' U' R' B' L' D'
Play the demonstration below:
For example: F R' U2 D means front face clockwise, right counterclockwise, a half turn of the upper face and then down clockwise.To read about slice turns, double layer turns, whole cube
reorientation etc. go to the advanced Rubik's Cube notation page.
Usually we use sequences of these basic rotations to describe an algorithm.
A Rubik's Cube algorithm presented in the Beginner'smethod is U R U’ L’ U R’ U’ L, used to cycle the three corner pieces on the upper layer, when the firsttwolayers(F2L) aresolved.
Degree of a Rubik's Cube algorithm
Every algorithm or permutation has a degree which is a finite number that shows how many times we have to execute the operation to return to the initial state.
Some examples:
F - degree is 4 because F F F F = 1.
R' D' R D - degree is 6 because we have to repeat the algorithm 6 times to return to the initial configuration.
Mathematical properties of the algorithms
In the introduction I have presented the Rubik's Cube as a permutation group. Below are the properties of the operations of this mathematical structure.
• Associative - the permutations in the row can be grouped together: ex. (RB')L = R(B'L)
• Neutral element - there is a permutation which doesn't rearrange the set: ex. RR'
• Inverse element - every permutation has an inverse permutation: ex. R - R'
• Commutative - it's not a necessary condition of the permutation group but notice that FB = BF but FR != RF
• Degree of permutations - see above. ex: 4xF=1, 6x(R'D'RD)=1, 336x(UUR'LLDDB'R'U'B'R'U'B'R'U)=1
Often used algorithms
There are many examples of iconic cubing things, but none are as omnipresent or as widely useful as algorithms. Below we will be going over the most famous algorithms, such as Sune, Sledgehammer, and
many more. The majority of these will be CFOP algorithms, and some will be used in other methods such as Petrus, ZZ and Roux.
Sune is an OLL algorithm, which means it orients the last layer. It is part of a special subcategory called OCLL, which means that it only orients the corners (is used when all edges are oriented).
It was proposed by Lars Petrus in his Petrus method.
R U R' U R U2 R'
As referenced by the name, Anti-Sune is the opposite of Sune. It is still an OCLL, but the algorithm is mirrored. It was also coined by Petrus in the method of the same name.
R U2 R' U' R U' R'
This is a trigger that is used in a lot of algorithms, and in F2L. It also has a much lesser-known reverse, hedgeslammer. If repeated 6 times, it will bring the cube back to its previous state.
R' F R F'
Sexy Move
This is another trigger that is heavily used in almost everything. You can find it in F2L, OLL, and PLL and, if repeated 6 times on a cube, will bring it back to the same state it was in before.
R U R' U'
Reverse Sexy
This is, as the name implies, the reverse of sexy. It is less used, but is still quite prominent in F2L, where the triple sexy is frequently replaced with triple reverse sexy as it is said to be
quicker. Either way, if repeated 6 times it will bring the cube back to its original state, as with most 6 move triggers.
U R U' R'
U Perms
This is a PLL (Position Last Layer) algorithm. There are 2 variants, the Ua and Ub perms. They are used when all the corners are permuted and there are 3 edges to permute in a triangular fashion.
Doing either one 3 times will bring the cube back to its original state and executing either one once will make the case that the other one solves.
Ua: M2 U M U2 M' U M2
Ub: M2 U' M U2 M' U' M2
T Perm
The T perm is perhaps the most well-known PLL algorithm, with its only competition being the U perms (above) and the J perms (below). It is used to permute 2 opposite edges and two adjacent corners,
and the shape of those pieces to permute when viewed from above makes a T, hence the name.
R U R' U' R' F R2 U' R' U' R U R' F'
J Perms
These are 2 PLL algorithms that permute 2 adjacent edges and 2 adjacent corners. It is recognisable by the sheer number of blocks it has. There is one solved line, and 2 unsolved blocks. The Jb tends
to be the faster one, as it is an RUF algorithm, but the Ja – being either an RUL or LUF algorithm – can also be very fast with practice.
Ja: R' U L' U2 R U' R' U2 R L or L' U' L F L' U' L U L F' L2' U L U
Jb: R U R' F' R U R' U' R' F R2 U' R' U'
H Perm
The H perm is a PLL algorithm that swaps 2 sets of opposite edges. All the corners are already solved. The directions of the U turns can be switched.
M2' U' M2' U2' M2' U' M2'
Key (OLL 33)
This OLL looks like a T when viewed from the top. It is not the only OLL to look like this but it is recognizable from the two opposite facing blocks that the unoriented pieces make.
R U R' U' R' F R F'
Bottlecap (OLL 51)
This OLL is a very famous one, and a good example of an OLL that contains the ‘sexy' trigger. It is a Line case, and it has 2 opposing edge and corner blocks, along with two adjacent corners that
make headlights.
f ( R U R' U' ) ( R U R' U' ) f' or F U R U' R' U R U' R' F'
T (OLL 45)
This OLL is probably one of the most famous out of all the full algorithms. It is 6 moves long, and the other T shaped algorithm in addition to Key, above. The main body of the algorithm is the sexy
F R U R' U' F'
This is a pattern but has a famous algorithm to make it. This is quite an intuitive algorithm, and one that many beginner cubers will be taught or figure out or see done.
M2 E2 S2 or R2 L2 U2 D2 F2 B2
Z Perm
This permutation is a PLL algorithm. It switches two sets of adjacent edges. All the corners are already solved when this algorithm is used. It is used in every method that forces an EPLL (a PLL of
only the edges, in other words, all corners are pre-solved) including Petrus.
M' U' M2' U' M2' U' M' U2 M2' U
There will undoubtedly be more, as with any list, but these are the most famous and well known.
Rubik's Cube notation
We use letters to mark rotations on the cube. See the 3D widget will let you can the turns used in speedcubing. The widget renders without problems only in the latest web browsers.
Test algorithms pressing the buttons: | {"url":"https://southasia1.com/article/rubik-s-cube-algorithms-ruwix","timestamp":"2024-11-02T09:36:33Z","content_type":"text/html","content_length":"70179","record_id":"<urn:uuid:7f56a4f1-7170-4b62-b13d-3494d9ba102d>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00417.warc.gz"} |
Reference desk/Mathematics
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sees also:
52nd perfect number
[ tweak]
howz many digits (I want an exact figure) does the 52nd perfect number have?? Georgia guy (talk) 13:11, 21 October 2024 (UTC)
iff you read the perfect number scribble piece you will see that only 51 perfect numbers are known. So nobody knows. 196.50.199.218 (talk) 13:38, 21 October 2024 (UTC)
Please, I learned this morning that a new perfect number has been discovered. Georgia guy (talk) 13:41, 21 October 2024 (UTC)
Although a possible 52nd Mersenne prime has been discovered, its primality has not been ascertained and its identity has not been released, so we cannot construct a perfect number from it
yet. Also, after the 48th Mersenne prime, we get into unverified territory, meaning that there may be additional Mersenne numbers between the Mersenne primes we know about that are also
prime, but that we missed. GalacticShoe (talk) 13:42, 21 October 2024 (UTC)
ith was revealed dis morning towards be prime. Georgia guy (talk) 13:44, 21 October 2024 (UTC)
wellz, do you have the value of ${\displaystyle n}$ dat they found produces the new prime ${\displaystyle 2^{n}-1}$? If so then the number of digits is going to be ${\displaystyle
\lfloor \log _{10}(2^{n-1}(2^{n}-1))\rfloor +1=\lfloor (n-1)\log _{10}(2)+\log _{10}(2^{n}-1)\rfloor +1\approx \lceil (2n-1)\log _{10}(2)\rceil }$. GalacticShoe (talk) 13:53, 21
October 2024 (UTC)
GalacticShoe, I don't want a formula; I want an answer; I believe it's more than 80 million but I want an exact figure. Georgia guy (talk) 13:55, 21 October 2024 (UTC)
I see someone has updated the Mersenne prime page with the value ${\displaystyle n=136279841}$. If you plug that into the formula I provided, you get ${\displaystyle
82048640}$ digits. GalacticShoe (talk) 14:00, 21 October 2024 (UTC)
@GalacticShoe: I added your figure to List of Mersenne primes and perfect numbers. Still need the digits of the perfect number, though. :) Double sharp (talk) 14:29,
21 October 2024 (UTC)
Thanks, Double sharp. Unfortunately, I don't think my computer could handle that kind of number so I'll have to deign to someone else for this one :) GalacticShoe
(talk) 14:41, 21 October 2024 (UTC)
wellz, we only need the first six and last six digits for consistency in the table. Wolfram Alpha is giving me 388692 for the first six digits, and it must end
in ...008576 by computing modulo 10^6.
an' now I realise that the GIMPS press release links to a zip file containing the perfect number as well. Oops. Well, nice to know for sure that the above is
correct. Double sharp (talk) 14:52, 21 October 2024 (UTC)
meow that I think further, it's actually pretty simple to find the first 6 digits, since all you have to do is take ${\displaystyle 136279841}$, plug it into $
{\displaystyle a=(2n-1)\log _{10}(2)}$ towards get the approximate base-10 exponent of the perfect number, then find the first six digits of ${\displaystyle 10
^{a-b}}$ where ${\displaystyle b}$ izz an integer offset that allows us to scale the perfect number down by an arbitrary power of 10. Doing so with ${\
displaystyle b=82048634}$ yields the aforementioned ${\displaystyle 388692}$. GalacticShoe (talk) 15:08, 21 October 2024 (UTC)
Using home-brewed routines, I get 3886924435...7330008576. I can produce some more digits if desired, up to several hundreds, but not all 82048640 of them. --
Lambiam 17:01, 21 October 2024 (UTC)
teh first 200 digits are
teh last 200 digits are
yoos PARI/GP programs:
furrst 200 digits:
las 200 digits:
Why does splitting extension field’s elements into several subfields doesn’t help solving discrete logs despite it helps computing exponentiations and multiplications?
[ tweak]
Let’s say I have 2 finite fields elements ${\displaystyle A}$ an' ${\displaystyle B}$ inner ${\displaystyle GF_{p}^{6}}$ having their discrete logarithm belonging to a lorge semiprime' suborder/
subgroup ${\displaystyle s}$ such as ${\displaystyle p>s}$.
${\displaystyle A}$ an' ${\displaystyle B}$ canz be represented as the cubic extension of ${\displaystyle GF_{p}^{2}}$ bi splitting their finite field elements. This give ${\displaystyle A.x\in GF_
{p}^{2}}$; ${\displaystyle A.\in GF_{p}^{2}}$; ${\displaystyle A.z\in GF_{p}^{2}}$; and ${\displaystyle B.x\in GF_{p}^{2}}$; ${\displaystyle B.y\in GF_{p}^{2}}$; ${\displaystyle B.z\in GF_{p}^
{2}}$. This is useful for simplifying computations on ${\displaystyle A}$ orr ${\displaystyle B}$ lyk multiplying or squaring by peforming such computations component wise. An example of which can be
found here : https://github.com/ethereum/go-ethereum/blob/24c5493becc5f39df501d2b02989801471abdafa/crypto/bn256/cloudflare/gfp6.go#L94
However when the suborder/subgroup ${\displaystyle s}$ fro' ${\displaystyle GF_{p}^{6}}$ doesn’t exists in ${\displaystyle GF_{p}^{2}}$, why does solving the 3 discrete logarithm between each
subfield element that are :
1. dlog of ${\displaystyle A.x}$ an' ${\displaystyle B.x}$
2. dlog of ${\displaystyle A.y}$ an' ${\displaystyle B.y}$
3. dlog of ${\displaystyle A.z}$ an' ${\displaystyle B.z}$
doesn’t help establishing the discrete log of the whole ${\displaystyle A}$ an' ${\displaystyle B}$? 82.66.26.199 (talk) 13:30, 25 October 2024 (UTC)
Supposing that you can solve the discrete log in GF(q), the question is to what extent this helps to compute the discrete log in GF(q^k). Let g be a multiplicative generator of ${\displaystyle GF
(q^{k})^{\times }}$. Then Ng is a multiplicative generator of ${\displaystyle GF(q)^{\times }}$, when N is the norm map down to GF(q). Given A in ${\displaystyle GF(q^{k})^{\times }}$, suppose
that we have x such that ${\displaystyle NA=Ng^{x}}$. Then ${\displaystyle Ag^{-x}}$ belongs to the kernel of the norm map, which is the cyclic group of order (q^k-1)/(q-1) generated by g^{q-1}.
Therefore it is required to solve an additional discrete log problem in this new group, the kernel of the norm map. When the degree k is composite, we can break the process down iteratively by
using a tower of norm maps. If (a big if) each of the norm one groups in the tower has order a product of small prime factors, then Pohlig-Hellman can be used in each of them. Tito Omburo (talk)
14:53, 25 October 2024 (UTC)
an' when the order contains a 200‒bits long prime too large for Pohlig‑Hellman? 82.66.26.199 (talk) 15:39, 25 October 2024 (UTC)
wellz, the basic idea is that if k is composite, then the towers are "relatively small", so they would be smoother than the original problem, and might be a better candidate for PH than
the original problem. It seems unlikely that a more powerful method like the function field sieve would be accelerated by having a discrete log oracle in the prime field. The prime field
in that case is usually very small already. For methods with p^n where p is large, an oracle for the discrete log in the prime field also doesn't help much (unless you can do
Pohlig-Hellman). Tito Omburo (talk) 16:06, 25 October 2024 (UTC)
iff the white amazon (QN) in Maharajah and the Sepoys izz replaced by the fairy chess pieces, does black still have a winning strategy? Or white have a winning strategy? Or draw?
[ tweak]
iff the white amazon (QN) in Maharajah and the Sepoys izz replaced by the fairy chess pieces, does black still have a winning strategy? Or white have a winning strategy? Or draw?
1. QNN (amazon rider in pocket mutation chess, elephant in wolf chess)
2. QNC (combine of queen and wildebeest in wildebeest chess)
3. QNNCC (combine of queen and “wildebeest rider”)
4. QNAD (combine of queen and squirrel)
5. QNNAD (combine of amazon rider and squirrel)
6. QNNAADD (combine of queen and “squirrel rider”)
218.187.64.154 (talk) 17:38, 29 October 2024 (UTC)
nother question: If use wildebeest chess towards play Maharajah and the Sepoys, i.e. on a 11×10 board, black has a full, wildebeest chess pieces in the position of the wildebeest chess, white
only has one piece, which can move as either a queen or as a wildebeest on White's turn, andthis piece can be placed in any square in rank 1 to rank 6 (cannot be placed in the squares in rank 7
or rank 8, since the squares in rank 7 or rank 8 may capture Black's pieces (exclude pawns) or be captured by Black's pieces (or pawns). Black's goal is to checkmate the only one of White, while
White's is to checkmate Black's king. There is no promotion. (Unlike wildebeest chess, stalemate izz considered as a draw) Who has a winning strategy? Or this game will be draw by perfect play?
218.187.64.154 (talk) 17:31, 1 November 2024 (UTC) | {"url":"https://wikiclassic.com/wiki/Wikipedia:Reference_desk/Mathematics","timestamp":"2024-11-02T11:19:06Z","content_type":"text/html","content_length":"139994","record_id":"<urn:uuid:0f39f5d3-3d63-46a8-a578-04ea169fd059>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00420.warc.gz"} |
Understanding Belt Calibration Numbers
Well, I am printing but the output is not perfect. In this case, a mm on either axis does not come out as a mm. This is OK for stuff like the whistle, but when going into gears, perfection is a must.
I found Triffid Hunter's Calibration manual which was extremely useful: [
His equation makes sense to me: # microsteps/mm = # steps/rev * # microsteps/step * # teeth/mm(belt pitch) * rev / # teeth (number of teeth on the pulley)
On my X Axis I have a 17 teeth GT2 pulley. Why did I chose such a weird teeth number? I think because of the stepper motor shaft, if I recall. Anyway, what this gives me is 200*16/2/17=94.117.
On my Sprinter firmware I placed 94.117 but when I run the cube I get 19.76mm width.
So 2 questions:
1. Is 19.76 what I am supposed to get? Or differently phrased, are these 3D printers supposed to be dead-on, or are differences of this sort expected?
2. When I specify a number such as 94.117 on the firmware, will the decimal fraction portion play a role, or is this information lost?
My Y axis is even worse, but that one is my mistake as I went with a 40 DP belt. I should have of course chosen the same belt for both axis. Oh darn "reuse of what I had available"... | {"url":"https://reprap.org/forum/read.php?1,165366","timestamp":"2024-11-08T23:40:14Z","content_type":"application/xhtml+xml","content_length":"28902","record_id":"<urn:uuid:99d6e70a-cc73-4a37-8bee-6d127589680a>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00374.warc.gz"} |
In a pack of 52 playing cards one card is lost. From the remaining
Q) In a pack of 52 playing cards one card is lost. From the remaining cards, a card is drawn at random. Find the probability that the drawn card is queen of heart, if the lost card is a black card.
Ans: We have 52 cards in a pack.
From the given pack of 52 cards, if 1 black card is lost, we are left with 51 cards.
Since there is only 1 queen of heart, hence favourable chances = 1
Total Number of chances = balance cards = 51
We know that the probability of drawing a card:
Therefore, probability of drawing queen of heart from the remaining cards is
Please press Heart if you liked the solution.
Leave a Comment | {"url":"https://www.saplingacademy.in/in-a-pack-of-52-playing-cards/","timestamp":"2024-11-05T10:14:44Z","content_type":"text/html","content_length":"127449","record_id":"<urn:uuid:fdf4084a-4315-4c00-81d9-09864281cc7d>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00840.warc.gz"} |
Centre for Research in Mathematics Education
Talk plays an important role in broadening a student’s mathematical vocabulary, bringing increased lucidity to their explanations, and more generally developing and communicating their mathematical
understanding. However this raises the question of how teachers can provide opportunities for students to develop what we succinctly label talk in mathematics (TiM). In this seminar we report on the
first year of a two-year collaboration between two teams of teachers from local secondary schools and three researchers, who together have sought to address this question.
Tags: Centre for Research in Mathematics Education, CRME, Department of Education, Dr Jenni Ingram, Nick Andrews, Oxford University | {"url":"https://exchange.nottingham.ac.uk/blog/tag/centre-for-research-in-mathematics-education/","timestamp":"2024-11-10T05:30:06Z","content_type":"text/html","content_length":"53809","record_id":"<urn:uuid:fe8689d5-92ce-4e61-8093-bb3a50f9bed4>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00069.warc.gz"} |
Counting counterimages
Yet a third variation on the theme of symbolic de Bruijn matrices is actually a numerical matrix, whose purpose is to merely count counterimages, not to display them in some form. This final
variant on Eq. 8,
gives an alternative definition of the de Bruijn matrix which in turn is decomposed into a sum
by defining
Still using Rule 22 as an example, 0.30em
Summing (rather than counting) the matrix elements again reveals seven counterimages.
Harold V. McIntosh | {"url":"http://delta.cs.cinvestav.mx/~mcintosh/newweb/ra/node19.html","timestamp":"2024-11-12T19:55:19Z","content_type":"text/html","content_length":"1929","record_id":"<urn:uuid:db659598-c9b2-4b5f-9f1b-007fc3c4ad99>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00692.warc.gz"} |
Software For Math Problems | peter-althaus.ch
Software For Math Problems
Download >>> https://fancli.com/2tlD67
If you want to cope with any math problem as fast as possible, then you need the best mathematical software. With its help, you can complete tasks effortlessly and deepen your knowledge of the
These programs are suitable for learning a wide variety of topics including matrices, graphs, combinations, and permutations. They can interest math students, who want to learn how to draw a cube,
triangle, circle, and other geometric shapes.
Besides, such programs allow you to understand difficult mathematical topics such as linear programming, complex numbers, vectors, discrete mathematics, probability, statistics, calculus, algebra,
functions and graphs. They also cover basic math operations.
Verdict: Microsoft Mathematics enables users to solve mechanical, algebraic, calculus and multimeter problems. Developed and maintained internally by Microsoft, it mainly targets students as an
educational tool.
The primary objectives of the software are to equip students with fundamental skills and knowledge in using mathematical tools and mathematical expressions. This software contains multiple
applications including calculators, solvers and solving techniques.
Verdict: The main function of the Cadabra is to calculate with the help of numbers and decimals. It is also considered as an educational gadget and many educational institutions are using it for
teaching mathematics to the students. The price of the Cadabra is quite high when compared with the other programs but this tool provides quality education to the children.
The software is quite similar to the calculator but it can also perform some extra functions that the calculator cannot. After entering the figures, the software will generate the corresponding
graphs or images similar to what free diagram software can do. The graphical interface helps the users to learn the techniques of working with figures and it also teaches the students the basics of
Verdict: GeoGebra designed for teaching and learning mathematics and science, from kindergarten to college level. GeoGebra is now available on multiple devices, with different programs for desktop,
tablets and even web. Using this mathematical software allows children to understand basic shapes, lines, and transforms that are part of math. Also, this software has a built-in spreadsheet similar
to features in the free spreadsheet programs.
Verdict: Photomath helps children and students learn math very easily and conveniently with features such as sorting by sum, sets and multiplication. It has the ability to solve multiplication tables
as well as basic arithmetic procedures. Sorting by sum allows you to sort an array of numbers based on the sum of their values. Sets allow you to group similar objects together and create larger
Verdict: The SpeQ Mathematics consists of practice tests, mock tests and essay question papers. It provides practice tests on a variety of topics that students will need to know in order to
successfully complete their mathematics class. By using practice tests students will be able to determine where they need extra help and how to best use their existing resources to learn the
material. The tests include detailed instructions on how to complete the various problem types, as well as tips for efficient problem solving and memorizing information.
The SpeQ Mathematics Test can also be used to identify strong areas for students to work on. The tests include many multiple-choice questions that will test students' reasoning skills in mathematics.
If you want to learn a foreign language, you may also need the language learning software.
Language is no longer a barrier to learning math. The app supports 22 languages including 12 Indian languages like Assamese, Bengali, Gujarati, Hindi, Kannada, Konkani, Marathi, Malayalam, Oriya,
Punjabi, Tamil, and Telugu apart from international languages like German, Spanish, Simplified Chinese, and Russian.
There was a time when students were afraid of maths. Solving a math equation is considered a big task earlier, but the 21st century comes along with some benefits. Gone are the days when we need to
worry about mathematics assignments. Now, there are various ways through which students get assignment help in mathematics. Some students prefer to move to online experts to seek proper math homework
help, and some prefer to take help with free mathematics software. Yes, you heard it right, there is some software that can resolve all your queries related to this dreading subject.
Today, in this blog I will provide you a list of some free math software that can help in solving your math problem. Whether you are a high school student or a college one, you will find this
software quite interesting and helpful in all types of math problems.
Most of the time math becomes an unbearable subject for students. Once students stop getting the right answers, they try to get away with the subject. But, now the software mentioned below will be of
great help to you in this subject.
Microsoft mathematics is free and open-source software developed by Microsoft. It is a significant tool for those who are struggling with mathematics problems. It is quick and free, solves the most
complicated problems of math in an easy manner. This tool contains features that will allow you to solve problems related to mathematics, science and other technical subjects. It has a graphing
calculator and a unit converter. Along with that, it has an equation and a triangle solver that provides solutions in steps. Students can download this tool free from the Microsoft website. It can
run over various platforms such as Windows, iOS, Android.
It is considered as a dynamic math software, designed for all standards of mathematics. This mathematics tool can be used by both Rookie as well as Expert. This tool merges algebra, geometry,
spreadsheets, graphs, statistics and analysis, calculus in a single easy to use package. This tool is widely popular among students who are facing issues in solving various mathematical problems.
If you are in search of an interesting way of learning geometry, then Geometry Pad is for you. This mathematical tool will help you in learning geometry and allows you to practice vital
constructions. This tool will act as your personal assistant in learning geometry. This is a student-friendly tool that helps in the presentation of geometric constructions, taking measurements,
compass use and experimentation with different geometric shapes in an easier manner.
The tool is beneficial for both the teacher as well as students. Teachers can take the help of this tool to provide a good understanding of various geometric concepts. This tool can provide
assignment help to students as well, they can learn geometry with the help of this tool by sitting at home. So, if you are looking to get hassle-free solutions to your geometry problems, grab this
tool online and reap all the benefits. This tool is available on Androids as well as on iPhone.
Another great mathematical tool to take academic stress out of your life. Photomath is a boon for students as well as for teachers. This tool will help you in understanding your math problems and
will improve your skills as well. This tool will scan your math problem and provide you a quick solution. More than 1 million problems get solved by the Photomath tool.
This tool is easy to use, all you need is your mobile camera to scan the math question or problem. You will get a step by step solution for your problem. Photomath is your personal teacher, it helps
you by explaining the calculation steps in an animated manner. Therefore, you are stuck with any problem, get the solution instantly just by scanning the problem with Photomath.
You all might be familiar with the services of Khan Academy. It provides you with study material on various topics. There is also an app developed by the makers of Khan Academy. This app or we can
say tool can help you by clearing your doubts. This app has a personalized learning dashboard so that you can learn whenever you want. You can see tutorials, practice sets and math videos on it.
Apart from mathematics, you can also find Economics, science, history, computer programming, social science, etc. on it. Khan Academy provides free content to users as it is a non-profit organization
that provides everyone a chance to get a world-class education. You can simply visit their website as well and collect the content related to your subject.
If you are having issues in solving the math equations, use math editor. This is a perfect solution for college students who are facing issues in math equations. This software helps you to form
equations on screen by using Greek symbols, alpha, beta, square root, and other symbols in a quick and easy manner. One can also edit and save the equation in real-time. This free mathematics
software is one of the most student-friendly advanced math software. It allows you to save equations in the form of an image file that can be used in the MS office documents, web, and paint.
Therefore, if you are looking for a personal helping hand all the time then this software can be a blessing for you. You just need to read instructions and you are ready to solve the toughest math
problems of your syllabus.
Maxima is a free mathematics software built by the Massachusetts Institute of Technology. This software helps in solving algebraic problems over the computer. Maxima software is used for the
manipulation of symbolic and numerical expressions such as differentiation, integration, Laplace transform, ordinary differential equations, Taylor series, systems of linear equations, polynomials,
matrices, vectors, and tensors. It provides high quality and perfect results by using accurate fractions, arbitrary precision, integers, and variable precision floating-point numbers.
Maxima is supported on computer Lisp and works on all POSIX platforms namely, Linux, Unix, OS X, and BSD. Gnuplot is used for drawing. Maxima uses a complete programming language and has ALGOL like
syntax but Lisp-like semantics. This software is a complete CAS(computer algebra system) and works well for symbolic operations. It can also solve numerical problems as well. You must have internet
connectivity for its functionality. 59ce067264 | {"url":"https://www.peter-althaus.ch/forum/forum-fur-architektur/software-for-math-problems","timestamp":"2024-11-09T06:46:46Z","content_type":"text/html","content_length":"984346","record_id":"<urn:uuid:fb3adce3-6106-4725-825f-4d7c3ca595a3>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00232.warc.gz"} |
Computational Physics Resources: Monte Carlo Integration & Importance Sampling
written by Spencer Wheaton
This website contains a set of 2 simulations and accompanying worksheets that introduce the techniques of sample mean, hit and miss and importance sampling Monte Carlo integration.
Please note that this resource requires at least version 1.6 of Java (JRE).
Hit and Miss and Sample Mean Methods Worksheet
A worksheet to accompany the EJS simulation MCIntegration No1 1DHitMissSampleMean.jar
download 151kb .pdf
Last Modified: March 8, 2014
Hit & Miss Sample Mean Model The Hit & Miss Sample Mean Model investigates Monte Carlo integration techniques. The user is able to input a 1D integrand and finite integration limits and specify the
number of trials and the number of separate runs. The simulation allows the user to study the … download 1852kb .jar
Last Modified: March 8, 2014
Monte Carlo Integration wiht Importance Sampling Model The Monte Carlo Integration wiht Importance Sampling Model implements the sample mean and importance sampling techniques of Monte Carlo
integration. The user can input a 1D integrand and finite integration limits and specify the required Monte Carlo technique or … download 1862kb .jar
Last Modified: March 8, 2014
Hit & Miss Sample Mean Source Code
The source code zip archive contains an XML representation of the Hit & Miss Sample Mean Model. Unzip this archive in your EjsS 5 workspace to compile and run this model using EjsS 5 or above.
download 12kb .zip
Last Modified: March 8, 2014
Monte Carlo Integration wiht Importance Sampling Source Code
The source code zip archive contains an XML representation of the Monte Carlo Integration wiht Importance Sampling Model. Unzip this archive in your EjsS 5 workspace to compile and run this model
using EjsS 5 or above.
download 10kb .zip
Last Modified: March 8, 2014
Subjects Levels Resource Types
Mathematical Tools
- Probability
- Statistics
Thermo & Stat Mech - Instructional Material
- Graduate/Professional
- Probability = Interactive Simulation
- Upper Undergraduate
= Gaussian Distribution = Tutorial
= Poisson Distribution
= Probability Density
= Random Walks
Intended Users Formats Ratings
- Learners - application/java
- Educators - application/pdf
Access Rights:
Free access
This material is released under a GNU General Public License Version 3 license.
Rights Holder:
Spencer Wheaton
Record Creator:
Metadata instance created March 6, 2014 by Spencer Wheaton
Record Updated:
March 8, 2014 by Wolfgang Christian
Other Collections:
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Graph Neural Networks - Alelab /āl·lab/
Graph Neural Networks (ESE680)
Graph Neural Networks (GNNs) are information processing architectures for signals supported on graphs. They have been developed and are presented in this course as generalizations of the
convolutional neural networks (CNNs) that are used to process signals in time and space. Depending on how much you have heard of neural networks (NNs) and deep learning, this is a sentence that may
sound strange. Aren’t CNNs just particular cases of NN? And isn’t the same true of GNNs? In a strict sense they are, but our focus on this course is in large scale problems involving high dimensional
signals. In these settings NNs fail to scale. CNNs provide scalable learning for signals in time and space. GNNS support scalable learning for signals supported on graphs.
In this course we will cover graph convolutional filters and graph filter banks, before moving on to the study of single feature and multiple feature GNNs. We will also cover related architectures
such as recurrent GNNs. Particular emphasis will be placed on studying the equivariance to permutation and stability to graph deformations of GNNs. These properties provide a measure of explanation
respecting the good performance of GNNs that can be observed empirically. We will also study GNNs in the limi of large numbers of nodes to explain the transferability of GNNs across networks with
different number of nodes.
Machine Learning on Graphs
Graphs can represent product or customer similarities in recommendation systems, agent interactions in multiagent robotics, or transceivers in a wireless communication network. Although otherwise
disparate, these application domains share the presence of signals associated with nodes (ratings, perception or signal strength) out of which we want to extract some information (ratings of other
products, control actions, or transmission opportunities). If data is available, we can formulate empirical risk minimization (ERM) problems to learn these data-to-information maps. However, it is a
form of ERM in which a graph plays a central role in describing relationships between signal components. Therefore, one in which the graph should be leveraged. Graph Neural Networks (GNNs) are
parametrizations of learning problems in general and ERM problems in particular that achieve this goal.
In any ERM problem we are given input-output pairs in a training set and we want to find a function that best approximates the input-output map according to a given risk. This function is later used
to estimate the outputs associated with inputs that were not part of the training set. We say that the function has been trainedand that we have learned to estimate outputs. This simple statement
hides the well known fact that ERM problems are nonsensical unless we make assumptions on how the function generalizes from the training set to unobserved samples. We can, for instance, assume that
the map is linear, or, to be in tune with the times, that the map is a neural network.
If properly trained, the linear map and the GNN will make similar predictions for the samples they have observed. But they will make different predictions for unobserved samples. That is, they will
generalize differently. An important empirical observation is that neither the linear transform not the neural network will generalize particularly well unless we are considering problems with a
small number of variables. This is something that we could call the fundamental problem of machine learning. How do we devise a method that can handle large dimensional signals? GNNs and CNNs respond
that question in the same way: With the use of convolutional filters.
Graph Filters and Graphs Neural Networks
We have seen that a characteristic shared by arbitrary linear and fully connected neural network parametrizations is that they do not scale well with the dimensionality of the input signals. This is
best known in the case of signals in Euclidean space (time and images) where scalable linear processing is based on convolutional filters and scalable nonlinear processing is based on convolutional
neural networks (CNNs). The reason for this is the ability of convolutions to exploit the structure of Euclidean space.
In this course we will describe graph filters and graph neural networks as analogous of convolutional filters and CNNs, but adapted to the processing of signals supported on graphs. Both of these
concepts are simple. A graph filter is a polynomial on a matrix representation of the graph. Out of this definition we build a graph perceptron with the addition of a pointwise nonlinear function to
process the output of a graph filter. Graph perceptrons are composed (or layered) to build a multilayer GNN. And individual layers are augmented from single filters to filter banks to build multiple
feature GNNs.
Equivariance, Stability and Transferability
The relevant question at this juncture is whether graph filters and GNNs do for signals supported on graphs what convolutional filters and CNNs do for Euclidean data. To wit, do they enable scalable
processing of signals supported on graphs? A growing body of empirical work shows that this is true to some extent — although results are not as impressive as is the case of voice and image
processing. As an example that we can use to illustrate the advantages of graph filters and GNNs, consider a recommendation system in which we want to use past ratings that customers have given to
products to predict future ratings. Collaborative filtering solutions build a graph of product similarities and interpret the ratings of separate customers as signals supported on the product
similarity graph. We then use past ratings to construct a training set and learn to fill in the ratings that a given customer would give to products not yet rated. Empirical results do show that
graph filters and GNNs work in recommendation systems with large number of products in which linear maps and fully connected neural networks fail. In fact, it is easy enough to arrive at three
empirical observations that motivate this paper:
• (O1) Graph filters produce better rating estimates than arbitrary linear parametrizations and GNNs produce better estimates than arbitrary (fully connected) neural networks.
• (O2) GNNs predict ratings better than graph filters.
• (O3) A GNN that is trained on a graph with a certain number of nodes can be executed in a graph with a larger number of nodes and still produce good rating estimates.
Observations (O1)-(O3) support advocacy for the use of GNNs, at least in recommendation systems. But they also spark three interesting questions: (Q1) Why do graph filters and GNNs outperform linear
transformations and fully connected neural networks? (Q2) Why do GNNs outperform graph filters? (Q3) Why do GNNs transfer to networks with different number of nodes? In this paper we present three
theoretical analyses that help to answer these questions:
• Equivariance. Graph filters and GNNs are equivariant to permutations of the graph.
• Stability. GNNs provide a better tradeoff between discriminability and stability to graph perturbations.
• Transferability. As graphs converge to a limit object, a graphon, GNN outputs converge to outputs of a corresponding limit object, a graphon neural network.
These properties show that GNNs have strong generalization potential. Equivariance to permutations implies that nodes with analogous neighbor sets making analogous observations perform the same
operations. Thus, we can learn to, say, fill in the ratings of a product from the ratings of another product in another part of the network if the local structures of the graph are the same. This
helps explain why graph filters outperform linear transforms and GNNs outperform fully connected neural networks [cf. observation (O1)]. Stability to graph deformations affords a much stronger
version of this statement. We can learn to generalize across different products if the local neighborhood structures are similar, not necessarily identical. Since GNNs possess better stability than
graph filters for the same level of discriminability, this helps explain why GNNs outperform graph filters [cf. observation (O2)]. The convergence of GNNs towards graphon neural networks delineated
under the transferability heading explains why GNNs can be trained and executed in graphs of different sizes [cf. observation (O3)]. It is germane to note that analogous of these properties hold for
CNNs. They are equivariant to translations and stable to deformations of Euclidean space and have well defined continuous time limits. | {"url":"https://alelab.seas.upenn.edu/teaching/gnn-course/","timestamp":"2024-11-07T01:25:59Z","content_type":"text/html","content_length":"54753","record_id":"<urn:uuid:d555a9bd-8549-4bdd-9a65-b86f45b08a83>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00700.warc.gz"} |
Gerry Brady - Department of Computer Science
Contact Info
Crerar 398-A
Studied mathematics and physics as an undergraduate at the University of Chicago, graduating with a degree in mathematics with honors in the College. In the Honors Program at the University of
Chicago in both mathematics and physics. Began study of probability theory, statistics, numerical linear algebra, and numerical analysis as a result of interest in physics. Number theory was a
mathematical interest. Pursued graduate study at the University of Chicago and received a doctorate in logic from the University of Oslo, Norway, in 1997. Expanded version of doctoral thesis
published as a book in 2000 by Elsevier, in Studies in History and Philosophy of Mathematics series. Was a member of a pioneering team that computerized editing, production, and publication of the
Astrophysical Journal and the Astronomical Journal for the American Astronomical Society at the University of Chicago Press in the 1990s.
How well do you feel MPCS has kept up with the demands graduates face in the workplace?
“The courses I teach in the MPCS all emphasize problem solving and help students improve their problem-solving skills. Problem solving is important in the workplace, and in recent years many of our
graduates have been hired at leading firms, in part on the strength of the problem-solving skills acquired in their MPCS courses.”
What do you see as the most important advantage of receiving a master’s degree from the University of Chicago MPCS?
“Graduates of the MPCS are perceived to be intelligent, creative, and able to learn new subjects quickly. The University of Chicago’s superb academic reputation carries over to the MPCS.”
Ph.D. thesis published as book, From Peirce to Skolem, in 2000.
Several writings in mathematical and categorical logic.
• Geraldine Brady and Todd H. Trimble. The topology of relational calculus. Submitted to Advances in Mathematics.
• Geraldine Brady. From Peirce to Skolem: A Neglected Chapter in the History of Mathematical Logic. Elsevier Science: North-Holland, 2000.
Mathematical Reviews: MR 1834718
• Geraldine Brady and Todd H. Trimble. A categorical interpretation of C. S. Peirce’s System Alpha. Journal of Pure and Applied Algebra, 149: 213-239, 2000. Mathematical Reviews: MR 17627665
• Geraldine Brady and Todd H. Trimble. A string diagram calculus for predicate logic. Preprint, November 1998.
• Geraldine Brady. The Contributions of Peirce, Schroeder, Loewenheim, and Skolem to the Development of First-Order Logic. Doctoral Dissertation, Universitetet i Oslo, 1997.
• Geraldine Brady. From the algebra of relations to the logic of quantifiers. Studies in the Logic of Charles Sanders Peirce, Indiana University Press, 1997.
• Stuart A. Kurtz and Geraldine Brady. Existential Graphs: I. January 1997.
• Ph.D. University of Oslo, Norway, Mathematical Logic.
• M.A. University of Chicago, Mathematical Logic.
• B.A. University of Chicago with honors in Mathematics. | {"url":"https://computerscience.uchicago.edu/people/gerry-brady/","timestamp":"2024-11-07T10:03:14Z","content_type":"text/html","content_length":"115870","record_id":"<urn:uuid:c82c6d01-13c4-4606-930f-a566b561d5b9>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00706.warc.gz"} |
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All pictures are from Acheron 2.0, a free explorer of geometrical fractals. You can download Acheron 2.0 here
The Squares curve is a nice fractal curve, build using a recursive procedure. I saw a sample of this curve in the well-known book 'Algorithms in C' written by Robert Sedgewick (Addison-Wesley
Publ. ISBN 0-201-51425-7). It was not named from its inventor so I call it the Squares Curve. As almost all the geometric fractal curves, this curve shares the fascinating property of having an
infinite curve length in a finite area.
The starting point of the recursive method for drawing the Squares curve is a simple square.
Use the four corners of the square as the center of 4 smallers squares, each having half the size of the main square. The first iteration gives:
The same procedure gives already a nice picture at the second iteration:
Properties Back to Top
• Curve Length
The following reasoning concerns the curve for which only the outline is drawn. This gives a close curve with an univocal perimeter.
Take the initial square and name N the length of its side. The perimeter of the 'curve' is N * 4.
On the first iteration, the four corners are replaced by four smaller squares. So, the length of the curve is now equal to the sum of the segments common between recursion 0 (initial square) and
recursion 1 plus the length of the newly added segments. The total length of the two segments removed at each corner is N/2, so the total removed is (N/2) * 4. The total length of the segments
making the smaller squares is N/2 * 3 and 4 are added, one on each corner. Looking only at the added segments, the length increase is:
L[inc] = (N/2)*3*4 - (N/2)*4 = (N/2)*8
On the second iteration, the four small squares added at the first iteration will be replaced by four smaller squares. Here, the length of the segments removed on each square corner is equal to N
/4 and the length of the smaller squares added is equal to (N/4) * 3. Looking only at the added segments, the length increase is:
L[inc] = (N/4)*3*3*4 - (N/4)*3*4 = (N/4)*24
The formula for the length increase can be generalized as:
L[inc] = (N/2^Rec) * 8 * 3^(Rec - 1)
where Rec is the iteration number (starting at 0)
Here is a summary of the length increase and total length of the curve.
│Iteration│ Length │Total Length │
│ Number │ Increase │ │
│0 │... │... │N * 4 │
│1 │(N/2) * 8 │N * 4 │N * 8 │
│2 │(N/4) * 24 │N * 6 │N * 14 │
│3 │(N/8) * 72 │N * 9v │N * 23 │
│4 │(N/16) * 216 │N * 13.5 │N * 36.5 │
│5 │(N/32) * 648 │N * 20.25│N * 56.75 │
The Ratio of the length increase between two successive iterations is:
Ratio = ((N/2^(Rec+1)) * 8 * 3^Rec) / ((N/2^Rec) * 8 * 3^(Rec-1))
Solving the equation gives Ratio[inc] = 1.5, demonstrating what is quite obvious from the figures in the above table.
The formula of the length increase can then be generalized to:
L[inc] = N * 4 * r^Rec-1 where r = 1.5
The total length of the curve is equal to the original length plus the sum of all the length increases.
Using the following identitiy,
1 + x + x^2 + x^3 + ... + x^n = (x^n+1 - 1) / (x - 1)
the total length can be generalized:
L[Tot] = N * (( r^Rec * 8) - 4)
Graphically, it gives a nice view of the ever increasing length:
• Area
Take the initial square and name N the length of its side. The area of the 'curve' is noted N2. Using a reasoning analogous to the one followed for the determination of the curve length, the
formula for the curve area is obtained.
The area increase at each iteration can be generalized as:
Area[inc] = (4 * 3^Rec) / 4^Rec+1
Solving for the Ratio of the area increase between two successive iterations gives: Ratio = r^Rec where r = 0.75
The total area of the curve can then be expressed as:
Area^Tot = N2 + ( 1 + r^Rec + r^(Rec+1) + .. r^(Rec+n))
Using the following identitiy,
1 + x + x^2 + x^3 + ... + x^n = (x^n+1 - 1) / (x - 1)
the total area can be generalized:
Area^Tot = N2 * 4 * ( 1 - r^Rec+1)
As r^Rec+1 tends to Zero when iteration increases, the area tends to 4 times its original value.
Graphically, it gives a nice view of the finite area:
• Fractal Dimension
The fractal dimension is computed using the Box-Couting Method equation:
D = log (N) / log ( r)
The following picture helps finding the figures required by the formula:
Replacing r by 14 ( as the grid is 14 * 14) and N by 148 ( the number of small squares covered by the fractal curve) in the the Box-Counting equation gives:
D = log(148) / log(14) = 1.89356
• Self-Similarity
Looking at two successive iterations of the drawing process provides graphical evidence that this property is also shared by this curve.
Variations Back to Top
All Variations described are available using Acheron 2.0
• Iteration Level
Eight recursion levels are available. Above this iteration number, the overall aspect of the curve remains essentially unaffected.
• Curve Style
Three ways for rendering the curve are available:
□ Normal
□ Filled
□ Outline
Author Biography Back to Top | {"url":"https://tgmdev.be/applications/acheron/curves/curvesquares.php","timestamp":"2024-11-09T19:53:45Z","content_type":"text/html","content_length":"17874","record_id":"<urn:uuid:30565aef-5e49-4ae4-83e9-cfd71f9011f0>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00152.warc.gz"} |
Bhargava receives Fields Medal for influential mathematicians under 40
Princeton University mathematician Manjul Bhargava was awarded today the 2014 Fields Medal, one of the most prestigious awards in mathematics, in recognition of his work in the geometry of numbers.
The International Mathematical Union (IMU) presents the medal every four years to researchers under the age of 40 based on the influence of their existing work and on their "promise of future
The honor, often referred to as the "Nobel Prize of mathematics," was awarded to four young researchers at the 2014 IMU International Congress of Mathematicians held in Seoul, South Korea. Bhargava
is the eighth Fields Medal recipient from Princeton since 1954 and the third consecutive awardee from the University, following recipients in 2006 and 2010.
The prize committee commended Bhargava, the Brandon Fradd, Class of 1983, Professor of Mathematics at Princeton, "for developing powerful new methods in the geometry of numbers, which he applied to
count rings of small rank and to bound the average rank of elliptic curves." The IMU further wrote that his "work in number theory has had a profound influence on the field. A mathematician of
extraordinary creativity, he has a taste for simple problems of timeless beauty, which he has solved by developing elegant and powerful new methods that offer deep insight…. He surely will bring more
delights and surprises to mathematics in the years to come."
Bhargava, who joined the Princeton faculty in 2003 after receiving his Ph.D. in mathematics from the University in 2001, said that the honor extends beyond himself to include those who have worked
alongside him during his career. "I am of course very honored to be receiving the Fields Medal," Bhargava said. "Beyond that, it is a great source of encouragement and inspiration, not just for me,
but I hope also for my students, collaborators and colleagues who work with me. Needless to say, this is their prize, too!"
David Gabai, the Hughes-Rogers Professor of Mathematics and department chair, said: "This is really great for both the department and the University. The Fields Medal is probably the most prestigious
recognition in pure mathematics." Gabai added, "beyond being a great researcher and adviser to graduate students, Manjul is an extraordinary teacher." He is particularly known for his popular
freshman seminar, "The Mathematics of Magic Tricks and Games," wherein students explore the mathematical principles behind games and magic tricks.
Bhargava has received numerous awards for his work, including the 2012 Infosys Prize; the 2011 Fermat Prize presented by the Toulouse Mathematics Institute in France; the 2005 SASTRA Ramanujan Prize
from the Shanmugha Arts, Science, Technology and Research Academy in India; the AMS Blumenthal Award for the Advancement of Pure Mathematics in 2005; and the Packard Foundation Fellowship in Science
and Engineering in 2004. He was elected to the U.S. National Academy of Sciences in 2013. He also was named one of Popular Science magazine's "Brilliant 10" in 2002. As a graduate student, Bhargava
studied under renowned mathematician Andrew Wiles, the James S. McDonnell Distinguished University Professor of Mathematics, Emeritus.
Princeton mathematicians have received several of the field's most esteemed awards this year. In March, Professor of Mathematics Yakov Sinai was awarded the Abel Prize by the Norwegian Academy of
Science and Letters for his influential 50-year career in mathematics. In June, Peter Sarnak, the Eugene Higgins Professor of Mathematics, received the Wolf Prize in Mathematics, which is awarded by
the Israel-based Wolf Foundation and presented by the president of Israel.
"We should be proud of the fact that so many of our faculty won major prizes and recognitions this year," Gabai said.
The IMU today also recognized the first female recipient of a Fields Medal, Maryam Mirzakhani, who was a Princeton mathematics professor from 2004 to 2010 and is now at Stanford University.
The union also presented Princeton alumnus Subhash Khot, a New York University professor of computer science who received his Ph.D. in computer science from Princeton in 2003, with the Rolf
Nevanlinna Prize, which honors "outstanding contributions in mathematical aspects of information sciences." In addition, Phillip Griffiths, who received his Ph.D. in mathematics from Princeton in
1962 and served as a professor of mathematics from 1968 to 1972, received the Chern Medal Award, which is presented to those "whose accomplishments warrant the highest level of recognition for
outstanding achievements in the field of mathematics." | {"url":"https://www.princeton.edu/news/2014/08/12/bhargava-receives-fields-medal-influential-mathematicians-under-40","timestamp":"2024-11-13T03:20:54Z","content_type":"text/html","content_length":"55910","record_id":"<urn:uuid:5f7723ea-d5dc-4a43-b90a-3bbf5340dd9c>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00217.warc.gz"} |
Question #5baca | Socratic
Question #5baca
1 Answer
We know that density of any substance is defined as mass per unit volume. It has units of $k g {m}^{-} 3$
Density $\rho = \text{mass"/"volume}$ .....(1)
Now coming to specific problem. We need to define the expression 'an average egg' used in the problem. We know that it is not only the birds who lay eggs but fish, turtles, snakes, frogs and insects
also lay eggs. As such the text of question is quite ambiguous, as there is no such thing as an average egg.
Assuming that the student has a hen's egg, out of all, in mind. Here too from household experience we know that fresh hen eggs sink in water. Whereas If an egg is very old or rotten, it floats in
water. Keeping law of flotation in mind we can say that density of a fresh hen egg is more than the density of water whereas a very old hen egg has a density less than the density of water. We know
that density of water at ${4}^{\circ} \text{C}$ is $1000 k g {m}^{-} 3$.
Keeping above discussion in mind, my opinion is student needs to
1. Collect a number of eggs as a sample field.
2. Ascertain experimentally the volume and mass of each egg. Volume can be found out by the displacement of water method and mass using a balance.
3. Calculate the density of each egg using (1) above.
4. Calculate the average density of the eggs in the sample field.
5. State the result so obtained.
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6.1: The Mole
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Figure \(\PageIndex{1}\) shows that we need 2 hydrogen atoms and 1 oxygen atom to make 1 water molecule. If we want to make 2 water molecules, we will need 4 hydrogen atoms and 2 oxygen atoms. If we
want to make 5 molecules of water, we need 10 hydrogen atoms and 5 oxygen atoms. The ratio of atoms we will need to make any number of water molecules is the same: 2 hydrogen atoms to 1 oxygen atom.
Figure \(\PageIndex{1}\) Water Molecules. The ratio of hydrogen atoms to oxygen atoms used to make water molecules is always 2:1, no matter how many water molecules are being made.
One problem we have, however, is that it is extremely difficult, if not impossible, to organize atoms one at a time. As stated in the introduction, we deal with billions of atoms at a time. How can
we keep track of so many atoms (and molecules) at a time? We do it by using mass rather than by counting individual atoms.
A hydrogen atom has a mass of approximately 1 u. An oxygen atom has a mass of approximately 16 u. The ratio of the mass of an oxygen atom to the mass of a hydrogen atom is therefore approximately
If we have 2 atoms of each element, the ratio of their masses is approximately 32:2, which reduces to 16:1—the same ratio. If we have 12 atoms of each element, the ratio of their total masses is
approximately (12 × 16):(12 × 1), or 192:12, which also reduces to 16:1. If we have 100 atoms of each element, the ratio of the masses is approximately 1,600:100, which again reduces to 16:1. As long
as we have equal numbers of hydrogen and oxygen atoms, the ratio of the masses will always be 16:1.
The same consistency is seen when ratios of the masses of other elements are compared. For example, the ratio of the masses of silicon atoms to equal numbers of hydrogen atoms is always approximately
28:1, while the ratio of the masses of calcium atoms to equal numbers of lithium atoms is approximately 40:7.
So we have established that the masses of atoms are constant with respect to each other, as long as we have the same number of each type of atom. Consider a more macroscopic example. If a sample
contains 40 g of Ca, this sample has the same number of atoms as there are in a sample of 7 g of Li. What we need, then, is a number that represents a convenient quantity of atoms so we can relate
macroscopic quantities of substances. Clearly even 12 atoms are too few because atoms themselves are so small. We need a number that represents billions and billions of atoms.
Chemists use the term mole to represent a large number of atoms or molecules. Just as a dozen implies 12 things, a mole (abbreviated as mol) represents 6.022 × 10^23 things. The number 6.022 × 10^23,
called Avogadro’s number after the 19th-century chemist Amedeo Avogadro, is the number we use in chemistry to represent macroscopic amounts of atoms and molecules. Thus, if we have 6.022 × 10^23 Na
atoms, we say we have 1 mol of Na atoms. If we have 2 mol of Na atoms, we have 2 × (6.022 × 10^23) Na atoms, or 1.2044 × 10^24 Na atoms. Similarly, if we have 0.5 mol of benzene (C[6]H[6]) molecules,
we have 0.5 × (6.022 × 10^23) C[6]H[6] molecules, or 3.011 × 10^23 C[6]H[6] molecules.
A mole represents a very large number! If 1 mol of quarters were stacked in a column, it could stretch back and forth between Earth and the sun 6.8 billion times.
Notice that we are applying the mole unit to different types of chemical entities. The word mole represents a number of things—6.022 × 10^23 of them—but does not by itself specify what “they” are.
The chemical entities can be atoms, molecules, formula units and ions. This specific information needs to be specified accurately. Most students find this confusing hence, we need to review the
composition of elements, covalent and ionic compounds.
Most elements are made up of individual atoms, such as helium. However, some elements consist of molecules, such as the diatomic elements, nitrogen, hydrogen, oxygen, etc. One mole of He consists of
6.022 × 10^23 He atoms but one mole of nitrogen contains 6.022 × 10^23 N[2] molecules. The basic units of covalent (molecular) compounds are molecules as well. The molecules of "compounds" consist of
different kinds of atoms while the molecules of "elements" consist of only one type of atom. For example, the molecules of ammonia (NH[3]) consist of nitrogen and hydrogen atoms while N[2] molecules
have N atoms only. Compounds that are ionic, like NaCl, are represented by ionic formulas. One mole of NaCl, for example, refers to 6.022 × 10^23 formula units of NaCl. And, one formula unit of NaCl
consists of one sodium ion and one chloride ion. Figure 6.1.2 summarizes the basic units of elements, covalent and ionic compounds
Figure \(\PageIndex{2}\): The basic units of elements (atoms or molecules), covalent compounds (molecules) and ionic compounds (formula units of ions).
Conversion Between Moles and Atoms, Molecules and Ions
Using our unit conversion techniques learned in Chapter 1, we can use the mole relationship and the chemical formula to convert back and forth between the moles and the number of chemical entities
(atoms, molecules or ions).
Because 1 N[2] molecule contains 2 N atoms, 1 mol of N[2] molecules (6.022 × 10^23 molecules) has 2 mol of N atoms. Using formulas to indicate how many atoms of each element we have in a substance,
we can relate the number of moles of molecules to the number of moles of atoms. For example, in 1 mol of ethanol (C[2]H[6]O), we can construct the following relationships (Table \(\PageIndex{1}\)):
Table \(\PageIndex{1}\): Molecular Relationships
1 Molecule of \(C_2H_6O\) Has 1 Mol of \(C_2H_6O\) Has Molecular Relationships
2 C atoms 2 mol of C atoms \(\mathrm{\dfrac{2\: mol\: C\: atoms}{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{2\: mol\: C\: atoms}}
6 H atoms 6 mol of H atoms \(\mathrm{\dfrac{6\: mol\: H\: atoms}{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{6\: mol\: H\: atoms}}
1 O atom 1 mol of O atoms \(\mathrm{\dfrac{1\: mol\: O\: atoms}{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{1\: mol\: O\: atoms}}
The following example illustrates how we can use these relationships as conversion factors.
If a sample consists of 2.5 mol of ethanol (C[2]H[6]O), how many moles of carbon atoms, hydrogen atoms, and oxygen atoms does it have?
Using the relationships in Table \(\PageIndex{1}\), we apply the appropriate conversion factor for each element:
Note how the unit mol C[2]H[6]O molecules cancels algebraically. Similar equations can be constructed for determining the number of H and O atoms:
\(\mathrm{2.5\: mol\: C_2H_6O\: molecules\times\dfrac{6\: mol\: H\: atoms}{1\: mol\: C_2H_6O\: molecules}=15\: mol\: H\: atoms}\)
\(\mathrm{2.5\: mol\: C_2H_6O\: molecules\times\dfrac{1\: mol\: O\: atoms}{1\: mol\: C_2H_6O\: molecules}=2.5\: mol\: O\: atoms}\)
If a sample contains 6.75 mol of Na[2]SO[4], how many moles of sodium atoms, sulfur atoms, and oxygen atoms does it have?
13.5 mol Na, 6.75 mol S and 27 mol O.
We can use Avogadro's number as a conversion factor, or ratio, in dimensional analysis problems. For example, if we are dealing with element X, the mole relationship is expressed as follows:
\[\text{1 mol X} = 6.022 \times 10^{23} \text{ X atoms} \nonumber \]
We can convert this relationship into two possible conversion factors shown below:
\(\mathrm{\dfrac{1\: mol\: X\: }{6.022\times 10^{23}\: X\: atoms}}\) or \(\mathrm{\dfrac{6.022\times 10^{23}\: X\: atoms}{1\: mol\: X\: }}\)
If the number of "atoms of element X" is given, we can convert it into "moles of X" by multiplying the given value with the conversion factor at the left. However, if the number of "mol of X" is
given, the appropriate conversion factor to use is the one at the right.
If we are dealing with a molecular compound (such as C[4]H[10]), the mole relationship is expressed as follows:
\[\text{1 mol C4H10} = 6.022 \times 10^{23} \text{ C4H10 molecules} \nonumber \]
If working with ionic compounds (such as NaCl), the mole relationship is expressed as follows:
\[\text{1 mol NaCl} = 6.022 \times 10^{23} \text{ NaCl formula units} \nonumber \]
How many formula units are present in 2.34 mol of NaCl? How many ions are in 2.34 mol?
Typically in a problem like this, we start with what we are given and apply the appropriate conversion factor. Here, we are given a quantity of 2.34 mol of NaCl, to which we can apply the definition
of a mole as a conversion factor:
\(\mathrm{2.34\: mol\: NaCl\times\dfrac{6.022\times10^{23}\: NaCl\: units}{1\: mol\: NaCl}=1.41\times10^{24}\: NaCl\: units}\)
Because there are two ions per formula unit, there are
\(\mathrm{1.41\times10^{24}\: NaCl\: units\times\dfrac{2\: ions}{NaCl\: units}=2.82\times10^{24}\: ions}\)
in the sample.
How many molecules are present in 16.02 mol of C[4]H[10]? How many atoms are in 16.02 mol?
9.647 x 10^24molecules, 1.351 x 10^26 atoms.
Key Takeaway
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separable Hilbert space
nLab separable Hilbert space
A Hilbert space $H$ over a field $F$ of real or complex numbers and with inner product $(|)$ is separable if it has a countable topological base, i. e. a family of vectors $e_i$, $i\in I$ where $I$
is at most countable, and such that every vector $v\in H$ can be uniquely represented as a series $v = \sum_{i\in I} a_i e_i$ where $a_i\in F$ and the sum converges in the norm $\|x\| = \sqrt{(x|x)}$
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[Solved] A large tank of fish from a hatchery is b | SolutionInn
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A large tank of fish from a hatchery is being delivered to a lake. The hatchery claims that the mean length of fish in the
A large tank of fish from a hatchery is being delivered to a lake. The hatchery claims that the mean length of fish in the tank is 15 inches, and the standard deviation is6inches. A random sample
of22fish is taken from the tank. Letxbe the mean sample length of these fish. What is the probability thatxis within 0.5 inch of the claimed population mean? (Round your answer to four decimal
There are 3 Steps involved in it
Step: 1
To determine the probability that the sample mean length of the fish barx is within 05 inches of the ...
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View Answer in SolutionInn App | {"url":"https://www.solutioninn.com/study-help/questions/a-large-tank-of-fish-from-a-hatchery-is-being-1317507","timestamp":"2024-11-07T11:19:58Z","content_type":"text/html","content_length":"103569","record_id":"<urn:uuid:fee12eef-e81a-4ecd-b896-09907c85a8b9>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00776.warc.gz"} |
The common roots of the equations x^(12) -1=0, x^4+x^2 +1=0
The common roots of the equations x12−1=0,x4+x2+1=0
Updated on:21/07/2023
Knowledge Check
• The common roots of the equation x3+2x2+2x+1=0and1+x2008+x2003=0 are (where ω is a complex cube root of unity)
• If αandβ are the root of the equation x2−2x+4=0. Then which of the following are the roots of the equation x2−x+1=0?
• If αandβ are the roots of the equation x2+x+1=0 then which of the following are the roots of the equation x2−x+1=0? | {"url":"https://www.doubtnut.com/qna/649486715","timestamp":"2024-11-03T22:41:34Z","content_type":"text/html","content_length":"324255","record_id":"<urn:uuid:0fcb3d4a-fe8a-4277-9bd5-6fe1115284e1>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00541.warc.gz"} |
n longitude
av G Pleijel · 1954 · Citerat av 44 — From the solar chart for latitude 60° N (see Page 93), the solar altitude can be ri')ad off for each hour part of the winter, negative during the autumn and
Negative values can occur when the model contains terms that do not help to predict the response. adjusted R-square = 1 - SSE(n-1)/SST(n-m) , where n = number of response values ,
919 3 3 gold badges 12 12 silver badges 23 23 bronze badges $\endgroup$ 2 Latitude definition. The latitude of a point is the measurement of the angle formed by the equatorial plane with the line
connecting this point to the center of the Earth. By construction, it is comprised between -90 ° and 90 °. Negative values are for the southern hemisphere locations, and latitude is … Writing
longitude and latitude is not only different but also follows a specific format. It requires the use of the correct symbol and a clear understanding of how to read and write them. There are a couple
of ways you can write coordinates on the map. By using one latitude line and one longitude line, you can represent map coordinates.
network, we can determine your current location;; Enter Lat/Long - You can verify your Example: -92.59838 (All Longitude values in the U.S. are negative.). Latitude angles can range up to +90 degrees
(or 90 degrees north), and down to is positive; if P is to the west of the prime meridian, the longitude is negative. Cartographers and geographers divide the Earth into longitudes and latitudes in
order to locate points on the globe. Each location on Earth has Why do seasons occur?
Its units of measure are degrees of angle. Degrees can be subdivided into the smaller units of minutes and seconds.
You can also = enjoy onsite=20 dining at the University Restaurant and Lounge.
Our 12,000 argument; input argument = '%1' must be a non-negative int = value. LonDiff=3DMath.abs(a.longitude-VE_LatLon= gThreshold.
With no constraints, the R2 must be positive and equals the square of r, the correlation coefficient. $\endgroup$ – Harvey Motulsky Jul 16 '11 at 15:55 The relative importance of the variables can be
assessed based on the PVE’s for various submodels: Predictors PVE F Latitude 0.75 1601 Longitude 0.10 59 Elevation 0.02 9 Longitude, Elevation 0.19 82 Latitude, Elevation 0.75 1080 Latitude,
Longitude 0.85 2000 Latitude, Longitude, Elevation 0.86 1645 The input of the latitude is a decimal number between -89.999999 and 89.999999. If the degree of latitude is given in S as south, the
number should be preceded by a minus sign. The input of the longitude is a decimal number between -179.999999 and 179.9999999.
Sometimes, latitudes north of the Equator are denoted by a positive sign. Latitudes south of the Equator are given negative values. This eliminates the need to add whether the specified latitude is
north or south of the Equator. East-West Locations
Negative is West covertress: How to measure latitude and longitude without GPS Here are the rigging instructions for the Revell Cutty Sark H-399.
Meridians are perpendicular to every latitude.
Camilo rito pi
This means that the Western Hemisphere has negative longitudes.
Negative numbers (in certain situations, Southern latitude is displayed as negative; if you see a negative latitude, it is South while a negative longitude is West) can also be used to express a
quadrasphere designation. Re: Can't enter negative longitude/latitude. Post by meshman » Sun Nov 07, 2010 4:12 pm Working with imagery files can take a lot of memory, more than is apparent at first
glance. Latitude definition.
Seb medborgarplatsen
lonn faktaviatic beautykarlshamns bridgeklubbkulturarv studierstipendier lunds univbarbro hultlinglars borin silversmed
(www.chalmers.se/brd) and will be successively updated. The horizontal position is given in longitude and latitude according to the of negative values,.
FREEAdd a Verified Certificate for $49 USD None Interested in this course for your Business or Team? Train your employ Use this guide for an easy way to teach latitude and longitude. This lesson is
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How do you find the greater factor?
To find the GCF of a set of numbers, list all the factors of each number. The greatest factor appearing on every list is the GCF. For example, to find the GCF of 6 and 15, first list all the factors
of each number. Because 3 is the greatest factor that appears on both lists, 3 is the GCF of 6 and 15.
What is the greater factor of a number?
Factors of a number are any numbers which divide into a given number evenly. Common factors are 1, 2, 3, 4, 6, & 12. The largest of these (12) is the greatest common factor (GCF). Listing factors is
one way to find the GCF.
How do you find the GCF 3 numbers?
To find the greatest common factor (GCF) between numbers, take each number and write its prime factorization. Then, identify the factors common to each number and multiply those common factors
together. Bam! The GCF!
What is the greater factor of 24 and 36?
Therefore, the greatest common factor of 24 and 36 is 12.
How do you find the factor of 3?
Factors of 3 are 1 and 3 only. Note that -1 × -3 = 3. (-1, -3) are also factors, as a product of any two negative numbers gives a positive number.
How do you find the greatest common factor of 18 and 24?
The greatest common factor is the greatest factor that divides both numbers. To find the greatest common factor, first list the prime factors of each number. 18 and 24 share one 2 and one 3 in
common. We multiply them to get the GCF, so 2 * 3 = 6 is the GCF of 18 and 24.
How does a factoring calculator calculate a number?
Calculator Use The Factoring Calculator finds the factors and factor pairs of a positive or negative number. Enter an integer number to find its factors. For positive integers the calculator will
only present the positive factors because that is the normally accepted answer.
How to find the greatest common factor of 3 numbers?
In other words, the GCF of 3 or more numbers can be found by finding the GCF of 2 numbers and using the result along with the next number to find the GCF and so on. So, the greatest common factor of
120 and 50 is 10. Now let’s find the GCF of our third value, 20, and our result, 10. GCF (20,10)
How to factor a square in Mathway calculator?
Step 1: Enter the expression you want to factor in the editor. The Factoring Calculator transforms complex expressions into a product of simpler factors. It can factor expressions with polynomials
involving any number of vaiables as well as more complex functions. Difference of Squares: a 2 – b 2 = (a + b) (a – b) Step 2:
How to find factor pairs in a calculator?
This factors calculator factors numbers by trial division. Follow these steps to use trial division to find the factors of a number. Find the square root of the integer number n and round down to the
closest whole number. Let’s call this number s . Start with the number 1 and find the corresponding factor pair: n ÷ 1 = n. | {"url":"https://wisdomanswer.com/how-do-you-find-the-greater-factor/","timestamp":"2024-11-06T03:56:17Z","content_type":"text/html","content_length":"147864","record_id":"<urn:uuid:f75c65f2-871f-4ee9-bc4c-b707eff2928c>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00499.warc.gz"} |
Utility Expectation Model - The Behavioral ScientistUtility Expectation Model
What is the Utility Expectation Model?
The Utility Expectation Model (UEM) is a decision-making framework in economics and decision theory that focuses on the expected utility of different options or choices. The model assumes that
individuals make decisions by considering the probable outcomes of each option, weighing the likelihood of each outcome, and selecting the option that maximizes their expected utility. This utility
can represent various aspects of an individual’s well-being, including material wealth, happiness, or satisfaction. The UEM is grounded in the notion that people act rationally and are able to assess
the probabilities of different outcomes accurately. It plays a key role in understanding decision-making under uncertainty, as well as the choices people make when faced with risk and ambiguity.
Examples of the Utility Expectation Model
• Investment Decisions
When choosing between different investment options, the Utility Expectation Model suggests that individuals will evaluate the expected returns and risks associated with each option, then select
the investment that maximizes their expected utility. This may involve considering factors such as potential gains, losses, and the probability of each outcome.
• Insurance Decisions
According to the UEM, individuals decide whether to purchase insurance by weighing the expected utility of the premium costs against the potential benefits of having insurance coverage. They will
consider factors such as the likelihood of a loss occurring, the magnitude of the potential loss, and the insurance premium cost to make their decision.
Shortcomings and Criticisms of the Utility Expectation Model
• Assumption of Rationality
One of the main criticisms of the Utility Expectation Model is its assumption that individuals are perfectly rational and capable of accurately calculating expected utilities. In reality, people
often exhibit bounded rationality, relying on heuristics and cognitive shortcuts that can lead to suboptimal or irrational decisions.
• Difficulty in Measuring Utility
Another criticism of the UEM is the challenge in quantifying utility, particularly when it comes to non-monetary aspects of well-being such as happiness or satisfaction. This makes it difficult
to compare the expected utilities of different options in a consistent and meaningful way.
• Overlooking Behavioral Factors
The Utility Expectation Model has been criticized for failing to account for various behavioral factors that influence decision-making, such as emotions, cognitive biases, and social influences.
By not incorporating these factors, the model may not accurately predict or explain real-world decision-making behavior. | {"url":"https://www.thebehavioralscientist.com/glossary/utility-expectation-model","timestamp":"2024-11-05T17:10:18Z","content_type":"text/html","content_length":"106772","record_id":"<urn:uuid:3c881ab2-c92d-4b52-bc72-197bbbb7557b>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00373.warc.gz"} |
Incredible Maths Content - Incredible Maths
Incredible Maths Content
This page contains a list of all the content available in Incredible Maths, currently categorised into 3 main categories.
If this list is out of date, please contact us by email or messaging us on social media to let us know.
• Addition
• Subtraction
• Multiplication
• Division
• Indices/Powers/Exponents
• Roots
• Percentages of amounts
• Addition/Subtraction/Multiplication/Division/Indices/Roots of negative numbers
• Order of operations
• Rounding (to the nearest 10/100/1000, to a decimal place)
• Converting units (of length/mass/volume)
• Calculations using fractions (Addition/Subtraction/Multiplication/Division)
• Converting between fractions and mixed numbers
• Calculating fractions of numbers
• Probability (as a fraction/percentage)
• Finding the Mean/Median/Mode/Range
• Currency calculations (Between it and it’s subunit)
• Converting between different currencies
• Indices (Negative/Fractional as well as of fractions)
• Calculations with decimals (Addition/Subtraction/Multiplication/Division)
• Finding the Highest Common Factor (HCF) and Lowest Common Multiple (LCM)
• Converting between numbers and Roman numerals (MDCLXVI)
• Written methods (addition, subtraction, multiplication, division)
• Identifying 2D and 3D shapes
• Area and perimeter of a rectangle/triangle/parallelogram/trapezium
• Volume and surface area of a cuboid/triangular prism/square based pyramid
• Angles (on a straight line/around a point/in a triangle/quadrilateral/parallelogram/trapezium)
• Solving simple one-step equations
• Adding and subtracting terms
• Solving two-step equations
• Solving equations with unknowns on both sides
• Solving equations with two unknowns
• Solving quadratic equations | {"url":"https://incrediblemaths.com/help/incredible-maths-content/","timestamp":"2024-11-11T01:32:40Z","content_type":"text/html","content_length":"76631","record_id":"<urn:uuid:a7a86a1e-e55a-46a0-a05b-cee087ae8657>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00092.warc.gz"} |
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3 Digit By 3 Digit Multiplication Worksheets On Grid Paper
Math, especially multiplication, forms the cornerstone of numerous scholastic self-controls and real-world applications. Yet, for many learners, mastering multiplication can posture an obstacle. To
resolve this obstacle, teachers and parents have embraced a powerful device: 3 Digit By 3 Digit Multiplication Worksheets On Grid Paper.
Intro to 3 Digit By 3 Digit Multiplication Worksheets On Grid Paper
3 Digit By 3 Digit Multiplication Worksheets On Grid Paper
3 Digit By 3 Digit Multiplication Worksheets On Grid Paper -
This page includes printable worksheets for 3rd grade 4th grade and 5th grade children on multiplying numbers from single digit to four digit in different combinations Lattice multiplication grids
templates are also included for teachers and homeschool moms Delve into some of these worksheets for free
Worksheets Multiplication by 3 Digit Numbers Multiplication by 3 Digit Numbers With these multiplication worksheets student can practice multiplying by 3 digit numbers example 491 x 612 Multiplying
by 3 Digit Numbers Multiplication 3 digit by 3 digit FREE Graph Paper Math Drills 3 digits times 3 digits example 667 x 129
Significance of Multiplication Technique Understanding multiplication is essential, laying a solid foundation for innovative mathematical principles. 3 Digit By 3 Digit Multiplication Worksheets On
Grid Paper provide structured and targeted technique, fostering a much deeper understanding of this fundamental arithmetic procedure.
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Multiplication Worksheets 3 Digit Printable Multiplication Flash Cards
Multiplication Worksheets 3 Digit Printable Multiplication Flash Cards
These math worksheets should be practiced regularly and are free to download in PDF formats 3 Digit by 3 Digit Multiplication Worksheet 1 Download PDF 3 Digit by 3 Digit Multiplication Worksheet 2
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You may select between 12 and 30 multiplication problems to be displayed on the multiplication worksheets These multiplication worksheets are appropriate for Kindergarten 1st Grade 2nd Grade 3rd
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Three digit Multiplication Practice Worksheet 03
Three digit Multiplication Practice Worksheet 03
How do you multiply 3 digit numbers by 3 digits Line up both your numbers with one on top of the other Make sure that the places match up for the ones tens and hundreds Multiply the top number by the
last digit of the second number the ones unit Write down the answer to this beneath the line
Free 3rd grade multiplication worksheets including the meaning of multiplication multiplication facts and tables multiplying by whole tens and hundreds missing factor problems and multiplication in
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Constant technique develops multiplication efficiency, boosting general mathematics capacities.
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Worksheets suit specific learning rates, cultivating a comfortable and adaptable understanding environment.
Just How to Produce Engaging 3 Digit By 3 Digit Multiplication Worksheets On Grid Paper
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Including Real-Life Circumstances
Connecting multiplication to everyday scenarios includes significance and practicality to workouts.
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Digital Multiplication Equipment and Gamings Technology-based sources supply interactive knowing experiences, making multiplication interesting and satisfying. Interactive Internet Sites and
Applications On-line platforms give diverse and available multiplication practice, supplementing typical worksheets. Personalizing Worksheets for Different Discovering Styles Visual Learners
Aesthetic aids and representations aid comprehension for students inclined toward visual discovering. Auditory Learners Verbal multiplication problems or mnemonics satisfy learners who understand
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in Practice Routine technique strengthens multiplication skills, advertising retention and fluency. Balancing Rep and Selection A mix of repeated workouts and diverse issue styles preserves passion
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3 Digit by 3 Digit Multiplication Worksheet 6 KidsPressMagazine
Multiplication 2 Digit By 2 Digit multiplication Pinterest Multiplication Math And
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3 Digit by 3 Digit Multiplication Worksheet 2 KidsPressMagazine
3 Digit By 2 Digit Multiplication Word Problems Worksheets Pdf Free Printable
Multiplying 3 Digit by 3 Digit Numbers A
Multiplication Worksheet 3 digit by 3 digit 5 KidsPressMagazine
Three Digit Multiplication Worksheet Have Fun Teaching
3 digit By 2 digit Multiplication Worksheets
Worksheets Multiplication by 3 Digit Numbers Super Teacher Worksheets
Worksheets Multiplication by 3 Digit Numbers Multiplication by 3 Digit Numbers With these multiplication worksheets student can practice multiplying by 3 digit numbers example 491 x 612 Multiplying
by 3 Digit Numbers Multiplication 3 digit by 3 digit FREE Graph Paper Math Drills 3 digits times 3 digits example 667 x 129
Multiply 3 x 3 digits worksheets K5 Learning
What is K5 K5 Learning offers free worksheets flashcards and inexpensive workbooks for kids in kindergarten to grade 5 Become a member to access additional content and skip ads Multiplication
practice with all factors under 1 000 column form Free Worksheets Math Drills Multiplication Printable
Worksheets Multiplication by 3 Digit Numbers Multiplication by 3 Digit Numbers With these multiplication worksheets student can practice multiplying by 3 digit numbers example 491 x 612 Multiplying
by 3 Digit Numbers Multiplication 3 digit by 3 digit FREE Graph Paper Math Drills 3 digits times 3 digits example 667 x 129
What is K5 K5 Learning offers free worksheets flashcards and inexpensive workbooks for kids in kindergarten to grade 5 Become a member to access additional content and skip ads Multiplication
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Multiplication 2 Digit Worksheet 1
3 Digit3 Digit Multiplication With Grid Support A 3 Digit Multiplication Worksheets
3 Digit3 Digit Multiplication With Grid Support A 3 Digit Multiplication Worksheets
2 And 3 Digit Multiplication
Frequently Asked Questions (Frequently Asked Questions).
Are 3 Digit By 3 Digit Multiplication Worksheets On Grid Paper ideal for every age teams?
Yes, worksheets can be customized to different age and skill levels, making them versatile for numerous students.
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Regular technique is vital. Regular sessions, ideally a couple of times a week, can yield substantial improvement.
Can worksheets alone enhance mathematics abilities?
Worksheets are a valuable device however needs to be supplemented with different knowing methods for thorough ability advancement.
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[Solved] The area of the △ABC, coordinates of whose vertices ar... | Filo
The area of the , coordinates of whose vertices are and is
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Question Text The area of the , coordinates of whose vertices are and is
Topic Application of Integrals
Subject Mathematics
Class Class 12
Answer Type Text solution:1
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--- title: "Contextualizing tree distances" author: "[Martin R. Smith](https://smithlabdurham.github.io/)" output: rmarkdown::html_vignette bibliography: ../inst/REFERENCES.bib csl: ../inst/
apa-old-doi-prefix.csl vignette: > %\VignetteIndexEntry{Contextualizing tree distances} %\VignetteEngine{knitr::rmarkdown} %\VignetteEncoding{UTF-8} --- Once you understand [how to use "TreeDist"]
(Using-TreeDist.html) to calculate tree distances, the next step is to provide some context for the calculated distances. ## Normalizing The maximum value of most tree distance metrics scales with
the size of the trees being compared. Typically, the resolution of the trees also impacts the range of possible values. As such, it can be difficult to interpret the tree distance value without
suitable context. Normalizing a distance metric is one way to render its meaning more obvious. Selecting an appropriate normalizing constant may require careful consideration of the purpose to which
a tree distance metric is being put. The default normalization behaviour of each function when `normalize = TRUE` is listed in the [function reference](../reference/index.html), or can be viewed by
typing `?FunctionName` in the R terminal. ### Nye _et al._ tree similarity Let's work through a simple example using the Nye _et al_. [-@Nye2006] similarity metric to compare two imperfectly-resolved
trees. ```{r, fig.width=6, out.width="90%", fig.align="center"} library("TreeDist") tree1 <- ape::read.tree(text = '(A, ((B, ((C, D), (E, F))), (G, (H, (I, J, K)))));') tree2 <- ape::read.tree(text =
'(A, (B, (C, D, E, (J, K)), (F, (G, H, I))));') VisualizeMatching(NyeSimilarity, tree1, tree2, Plot = TreeDistPlot, matchZeros = FALSE) ``` This is a nice metric to start with, because the maximum
similarity between each pair of splits is defined as one. (Astute readers might worry that the minimum similarity is greater than zero -- that's a harder problem to overcome.) As such, the maximum
similarity possible between two 11-leaf trees is [`NSplits(11)`](https://ms609.github.io/TreeTools/reference/NSplits.html) = `r suppressMessages(library("TreeTools")); NSplits(11)`. Normalizing
against this value tells us how similar the two trees are, compared to two identical eleven-leaf binary trees. ```{r} NyeSimilarity(tree1, tree2, normalize = FALSE) / 8 NyeSimilarity(tree1, tree2,
normalize = 8) ``` This approach will result in a similarity score less than one if two trees are identical, but not fully resolved (i.e. binary). As such, we might prefer to compare the tree
similarity to the maximum score possible for two trees of the specified resolution. This value is given by the number of splits in the least resolved of the two trees: ```{r} NyeSimilarity(tree1,
tree2, normalize = min(TreeTools::NSplits(list(tree1, tree2)))) ``` More concisely, we can provide a normalizing function: ```{r} NyeSimilarity(tree1, tree2, normalize = min) ``` This approach will
produce a similarity of one if one tree is a less-resolved version of another (and thus not identical). If we are comparing lists of trees, this best value will depend on the number of splits in each
pair of trees. We can use the function `pmin()` to select the less resolved of each pair of trees: ```{r} NyeSimilarity(list(tree1, tree2), list(tree1, tree2), normalize = pmin) ``` To avoid these
limitations, we may instead opt to normalize against the average number of splits in the two trees. This is the default normalization method for [`NyeSimilarity()`](../reference/NyeSimilarity.html):
```{r} NyeSimilarity(tree1, tree2, normalize = TRUE) ``` Finally, if `tree1` is a "target" tree -- perhaps one that has been used to simulate data from, or which is independently known to be true or
virtuous -- we may wish to normalize against the best possible match to that tree. In that case, the best possible score is ```{r} TreeTools::NSplits(tree1) ``` and our normalized score will be
```{r} NyeSimilarity(tree1, tree2, normalize = TreeTools::NSplits(tree1)) ``` ### Normalizing to random similarity The diameter (maximum possible distance) of the Nye _et al_. tree similarity metric
is easy to calculate, but this is not the case for all metrics. For example, the clustering information distance metric [@SmithDist] ranges in principle from zero to the total clustering entropy
present in a pair of trees. But with even a modest number of leaves, no pairs of trees exist in which every split in one tree is perfectly contradicted by every other split in the other; as such, any
pair of trees will necessarily have some degree of similarity. In such a context, it can be relevant to normalize tree similarity against the _expected_ distance between a pair of random trees,
rather than a maximum value [see @Vinh2010]. On this measure, distances greater than one denote trees that are more different than expected by chance, whereas a distance of zero denotes identity.
With the quartet divergence, the expected tree distance is readily calculated: any given quartet has a one in three chance of matching by chance. ```{r} library("Quartet", exclude = "RobinsonFoulds")
expectedQD <- 2 / 3 normalizedQD <- QuartetDivergence(QuartetStatus(tree1, tree2), similarity = FALSE) / expectedQD ``` The expected distance is more difficult to calculate for other metrics, but can
be approximated by sampling random pairs of trees. Measured distances between 10 000 pairs of random bifurcating trees with up to 200 leaves are available in the data package '[TreeDistData](https://
github.com/ms609/TreeDistData/)'. We can view (normalized) distances for a selection of methods: ```{r, fig.width=7, fig.height=4, message=FALSE} if (requireNamespace("TreeDistData", quietly = TRUE))
{ library("TreeDistData", exclude = "PairwiseDistances") data("randomTreeDistances", package = "TreeDistData") methods <- c("pid", "cid", "nye", "qd") methodCol <- c(pid = "#e15659", cid = "#58a14e",
nye = "#edc949", qd = "#af7aa1") oldPar <- par(cex = 0.7, mar = c(5, 5, 0.01, 0.01)) nLeaves <- as.integer(dimnames(randomTreeDistances)[[3]]) plot(nLeaves, type = "n", randomTreeDistances["pid",
"mean", ], ylim = c(0.54, 1), xlab = "Number of leaves", ylab = "Normalized distance between random tree pairs") for (method in methods) { dat <- randomTreeDistances[method, , ] lines(nLeaves, dat
["50%", ], pch = 1, col = methodCol[method]) polygon(c(nLeaves, rev(nLeaves)), c(dat["25%", ], rev(dat["75%", ])), border = NA, col = paste0(methodCol[method], "55")) } text(202, randomTreeDistances
[methods, "50%", "200"] + 0.02, c("Different phylogenetic information", "Clustering information distance", expression(paste(plain("Nye "), italic("et al."))), "Quartet divergence" ), col = methodCol
[methods], pos = 2) par(oldPar) } ``` or use these calculated values to normalize our tree distance: ```{r, eval = FALSE} expectedCID <- randomTreeDistances["cid", "mean", "9"] ClusteringInfoDistance
(tree1, tree2, normalize = TRUE) / expectedCID ``` ## Testing similarity to a known tree Similarity has two components: precision and accuracy [@Smith2019]. A tree can be 80% similar to a target tree
because it contains 80% of the splits in the target tree, and no incorrect splits -- or because it is a binary tree in which 10% of the splits present are resolved incorrectly and are thus positively
misleading. In such a comparison, of course, it is more sensible to talk about split _information_ than just the number of splits: an even split may contain more information than two very uneven
splits, so the absence of two information-poor splits may be preferable to the absence of one information-rich split. As such, it is most instructive to think of the proportion of information that
has been correctly resolved: the goal is to find a tree that is as informative as possible about the true tree. Ternary diagrams allow us to visualise the quality of a reconstructed tree with
reference to a known "true" tree: ```{r fig.align="center", fig.height=1.8, fig.width=6, out.width="80%"} testTrees <- list( trueTree = ape::read.tree(text = '(a, (b, (c, (d, (e, (f, (g, h)))))));'),
lackRes = ape::read.tree(text = '(a, (b, c, (d, e, (f, g, h))));'), smallErr = ape::read.tree(text = '(a, (c, (b, (d, (f, (e, (g, h)))))));'), bigErr = ape::read.tree(text = '(a, (c, (((b, d), (f,
h)), (e, g))));') ) VisualizeMatching(MutualClusteringInfo, testTrees$trueTree, testTrees$lackRes) points(4, 7.5, pch = 2, cex = 3, col = "#E69F00", xpd = NA) VisualizeMatching(MutualClusteringInfo,
testTrees$trueTree, testTrees$smallErr) points(4, 7.5, pch = 3, cex = 3, col = "#56B4E9", xpd = NA) VisualizeMatching(MutualClusteringInfo, testTrees$trueTree, testTrees$bigErr) points(4, 7.5, pch =
4, cex = 3, col = "#009E73", xpd = NA) ``` Better trees plot vertically towards the "100% shared information" vertex. Resolution of trees increases towards the right; trees that are more resolved may
be no better than less-resolved trees if the addition of resolution introduces error. ```{r, fig.width=4, fig.align="center", fig.asp=1, out.width="50%"} if (requireNamespace("Ternary", quietly =
TRUE)) { library("Ternary") oldPar <- par(mar = rep(0.1, 4)) TernaryPlot(alab = "Absent information", blab = "Shared information", clab = "Misinformation", lab.cex = 0.8, lab.offset = 0.18, point =
"left", clockwise = FALSE, grid.col = "#dedede", grid.minor.lines = 0, axis.labels = 0:10 / 10, axis.col = "#aaaaaa") HorizontalGrid() correct <- MutualClusteringInfo(testTrees$trueTree, testTrees)
resolved <- ClusteringEntropy(testTrees) unresolved <- resolved["trueTree"] - resolved incorrect <- resolved - correct TernaryPoints(cbind(unresolved, correct, incorrect), pch = 1:4, cex = 2, col =
Ternary::cbPalette8[1:4]) par(oldPar) } ``` ### Example Here's a noddy real-world example applying this to a simulation-style study. First, let's generate a starting tree, which will represent our
reference topology: ```{r} set.seed(0) trueTree <- TreeTools::RandomTree(20, root = TRUE) ``` Then, let's generate 200 degraded trees. We'll move away from the true tree by making a TBR move, then
reduce resolution by taking the consensus of this tree and three trees from its immediate neighbourhood (one NNI move away). ```{r} treeSearchInstalled <- requireNamespace("TreeSearch", quietly =
TRUE) if (treeSearchInstalled) { library("TreeSearch", quietly = TRUE) # for TBR, NNI oneAway <- structure(lapply(seq_len(200), function(x) { tbrTree <- TBR(trueTree) ape::consensus(list(tbrTree, NNI
(tbrTree), NNI(tbrTree), NNI(tbrTree))) }), class = "multiPhylo") } else { message("Install \"TreeSearch\" to run this example") } ``` And let's generate 200 more trees that are even more degraded.
This time we'll move further (three TBR moves) from the true tree, and reduce resolution by taking a consensus with three trees from its wider neighbourhood (each two NNI moves away). ```{r} if
(treeSearchInstalled) { threeAway <- structure(lapply(seq_len(200), function(x) { tbrTree <- TBR(TBR(TBR(trueTree))) ape::consensus(list(tbrTree, NNI(NNI(tbrTree)), NNI(NNI(tbrTree)), NNI(NNI
(tbrTree)))) }), class = "multiPhylo") } ``` Now let's calculate their tree similarity scores. We need to calculate the amount of information each tree has in common with the true tree: ```{r} if
(treeSearchInstalled) { correct1 <- MutualClusteringInfo(trueTree, oneAway) correct3 <- MutualClusteringInfo(trueTree, threeAway) } ``` The amount of information in each degraded tree: ```{r} if
(treeSearchInstalled) { infoInTree1 <- ClusteringEntropy(oneAway) infoInTree3 <- ClusteringEntropy(threeAway) } ``` The amount of information that could have been resolved, but was not: ```{r} if
(treeSearchInstalled) { unresolved1 <- ClusteringEntropy(trueTree) - infoInTree1 unresolved3 <- ClusteringEntropy(trueTree) - infoInTree3 } ``` And the amount of information incorrectly resolved:
```{r} if (treeSearchInstalled) { incorrect1 <- infoInTree1 - correct1 incorrect3 <- infoInTree3 - correct3 } ``` In preparation for our plot, let's colour our one-away trees orange , and our
three-away trees blue : ```{r, collapse=TRUE} col1 <- hcl(200, alpha = 0.9) col3 <- hcl(40, alpha = 0.9) spec1 <- matrix(col2rgb(col1, alpha = TRUE), nrow = 4, ncol = 181) spec3 <- matrix(col2rgb
(col3, alpha = TRUE), nrow = 4, ncol = 181) spec1[4, ] <- spec3[4, ] <- 0:180 ColToHex <- function(x) rgb(x[1], x[2], x[3], x[4], maxColorValue = 255) spec1 <- apply(spec1, 2, ColToHex) spec3 <-
apply(spec3, 2, ColToHex) ``` Now we can plot this information on a ternary diagram. ```{r, fig.width=7, fig.align="center", fig.asp=5/7, out.width="70%"} if (treeSearchInstalled && requireNamespace
("Ternary", quietly = TRUE)) { layout(matrix(c(1, 2), ncol = 2), widths = c(5, 2)) oldPar <- par(mar = rep(0, 4)) TernaryPlot(alab = "Information absent in degraded tree", blab = "\n\nCorrect
information in degraded tree", clab = "Misinformation in degraded tree", point = "left", clockwise = FALSE, grid.minor.lines = 0, axis.labels = 0:10 / 10) HorizontalGrid() coords1 <- cbind
(unresolved1, correct1, incorrect1) coords3 <- cbind(unresolved3, correct3, incorrect3) ColourTernary(TernaryDensity(coords1, resolution = 20), spectrum = spec1) ColourTernary(TernaryDensity(coords3,
resolution = 20), spectrum = spec3) TernaryDensityContour(coords3, col = col3, nlevels = 4) TernaryDensityContour(coords1, col = col1, nlevels = 4) if (requireNamespace("kdensity", quietly = TRUE)) {
library("kdensity") HorizontalKDE <- function(dat, col, add = FALSE) { lty <- 1 lwd <- 2 kde <- kdensity(dat) kdeRange <- kdensity:::get_range(kde) if (add) { lines(kde(kdeRange), kdeRange, col =
col, lty = lty, lwd = lwd) } else { plot(kde(kdeRange), kdeRange, col = col, lty = lty, lwd = lwd, ylim = c(0, 1), main = "", axes = FALSE, type = "l") } # abline(h = 0:10 / 10) # Useful for
confirming alignment } par(mar = c(1.8, 0, 1.8, 0)) # align plot limits with ternary plot HorizontalKDE(correct1 / infoInTree1, col1, add = FALSE) HorizontalKDE(correct3 / infoInTree3, col3, add =
TRUE) mtext("\u2192 Normalized tree quality \u2192", 2) } par(oldPar) } else { message("Install \"TreeSearch\" and \"Ternary\" to generate this plot") } ``` In the ternary plot, the vertical
direction corresponds to the normalized tree quality, as depicted in the accompanying histogram. ## What next? You may wish to: - Explore the [Ternary package](https://ms609.github.io/Ternary/) -
[Interpret tree distance metrics](https://ms609.github.io/TreeDistData/articles/09-expected-similarity.html) - Compare trees with [different tips](different-leaves.html) - Review [available distance
measures](https://ms609.github.io/TreeDist/index.html) and the corresponding [TreeDist functions](https://ms609.github.io/TreeDist/reference/index.html#section-tree-distance-measures) - Construct
[tree spaces](treespace.html) to visualize landscapes of phylogenetic trees ## References | {"url":"https://cran.ma.ic.ac.uk/web/packages/TreeDist/vignettes/using-distances.Rmd","timestamp":"2024-11-13T05:18:26Z","content_type":"text/plain","content_length":"16103","record_id":"<urn:uuid:63a85107-81cc-4525-95de-cfc7e6adbb2b>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00765.warc.gz"} |
Graphing Calculator
Unveiling the Potential: Exploring the Realm of Graphing Calculators
A Graphing Calculator stands as a dynamic mathematical instrument, empowering users to delve into the visual and analytical realms of mathematical functions and equations through graphical
representations. This multifaceted tool finds its applications in education, engineering, science, and various fields where intricate mathematical computations and visualizations are imperative.
Let's embark on an in-depth exploration of the features and applications that define the essence of a Graphing Calculator.
Key Features and Functions:
1. Graph Plotting Mastery:
At the heart of a Graphing Calculator lies its prowess to plot mathematical functions and equations seamlessly onto a coordinate plane. Users can input equations in diverse forms, ranging from linear
and quadratic to trigonometric and exponential functions.
2. Embracing Multiplicity:
Flexibility is key, as users can graph multiple functions on a single coordinate plane. This capability facilitates visual comparisons and intricate analysis, unlocking a new dimension of
mathematical exploration.
3. Zoom and Pan Dynamics:
Graphs come to life as users zoom in or out and pan across the coordinate plane, unraveling specific regions of interest with meticulous detail. This dynamic feature adds a layer of precision to the
visual analysis of mathematical concepts.
4. Data Insight with Tables:
Many Graphing Calculators offer tables of values corresponding to graphed functions. This feature aids users in examining specific data points, fostering a deeper understanding of the mathematical
5. Equation Resolution:
Some models elevate the experience with equation-solving capabilities, empowering users to discern roots, intersections, and critical points of equations with unparalleled ease.
6. Graphical Customization:
The user experience is further enriched through graph customization options. Colors, line styles, and labels can be tailored to individual preferences, enhancing the visual representation of
mathematical concepts.
7. Beyond the Basics: Parametric and Polar Equations:
Venturing into advanced mathematical concepts, Graphing Calculators often support parametric and polar equations. This expands the horizons of exploration for users with a penchant for mathematical
8. Statistical Analysis:
Incorporating statistical functions, certain models cater to data analysis needs, including regression analysis, histograms, and scatter plots. This makes Graphing Calculators valuable assets in
scientific research and analysis.
Applications of a Graphing Calculator:
1. Educational Odyssey:
Graphing Calculators find a natural habitat in mathematics classrooms across various educational levels. From high school to college, they serve as indispensable aids in comprehending mathematical
concepts through visual representation.
2. Engineering Marvels:
Engineers harness the power of Graphing Calculators to dissect and graph intricate engineering equations. These calculators emerge as essential problem-solving companions in domains such as
electrical and mechanical engineering.
3. Scientific Prowess:
Scientists leverage Graphing Calculators to visualize experimental data, model scientific phenomena, and dissect trends within datasets. The graphical representation adds a layer of clarity to
complex scientific analyses.
4. Financial Wizardry:
In the realm of finance, professionals utilize Graphing Calculators for diverse financial calculations, including loan amortization and investment analysis. The calculator's efficiency becomes a boon
for intricate financial problem-solving.
5. Architectural Canvas:
Architects and designers wield Graphing Calculators as tools for creating and visualizing geometric shapes and patterns, breathing life into their projects with mathematical precision.
6. Researcher's Toolkit:
Across diverse disciplines, researchers employ Graphing Calculators to analyze and present data in visually comprehensible formats. This aids in conveying complex research findings in a digestible
7. Standardized Test Allies:
Graphing Calculators find prominence in standardized tests like the SAT and ACT, where they provide invaluable assistance to students grappling with complex mathematical problems.
Advantages of Embracing Graphing Calculators:
1. Visual Enlightenment:
Graphing Calculators usher in a visual representation of mathematical functions, rendering complex concepts more accessible and comprehensible.
2. Problem-Solving Prowess:
These calculators act as adept problem-solving companions, aiding users in solving equations, identifying roots, and dissecting functions with precision.
3. Efficiency Unleashed:
The swift and accurate calculations performed by Graphing Calculators translate into time savings, especially in the realm of intricate mathematical tasks.
4. Educational Empowerment:
Enhancing the learning experience, Graphing Calculators allow students to interactively explore mathematical concepts, fostering a deeper understanding.
5. Versatility at its Core:
From basic arithmetic to advanced calculus, Graphing Calculators seamlessly support a wide range of mathematical functions, establishing themselves as versatile tools.
In conclusion, the Graphing Calculator emerges as a formidable ally in the realm of mathematical exploration, analysis, and problem-solving. Its adaptability and knack for visualizing mathematical
functions position it as an indispensable tool in education, engineering, science, and myriad other fields where mathematics takes center stage. Whether within the confines of a classroom or in the
dynamic landscape of a workplace, a Graphing Calculator empowers users to grasp, apply, and appreciate mathematical principles with unparalleled finesse. | {"url":"https://www.calculators24.com/graphing-calculator/","timestamp":"2024-11-06T05:38:56Z","content_type":"text/html","content_length":"185914","record_id":"<urn:uuid:c138be83-7d13-4024-8982-dafca8de1ed9>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00164.warc.gz"} |
Online ACT Exam Preparation, ACT Tutoring Packages, ACT Test Preparation
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Parity and arity
17 Mar 2020 ⇐ Notes archive
(This is an entry in my technical notebook. There will likely be typos, mistakes, or wider logical leaps — the intent here is to “
let others look over my shoulder while I figure things out.
Two tucked-away, somewhat-related terms I enjoy: parity and arity.
The former is the odd or even-ness of an integer.
The latter describes the number of arguments a function accepts.
Example usage of parity:
Today I learned about the Handshaking Lemma. It states that any finite undirected graph, will have an even number of vertices with an odd degree.
The proof rests on parity. Specifically, if you sum the degrees of every vertex in a graph, you’ll double count each edge. And that double counting implies the sum is even, and even parity is only
maintained if there is an even—including zero—number of vertices with an odd degree.
Put arithmetically, a sum can only be even if its components contain an even number of odd terms.
Examples of arity:
• Swift’s tuple comparison operators topping out at arity six.
• Publisher.zip is only overloaded to arity three.
□ I PR’d a variadic overload to CombineCommunity/CombineExt yesterday and have a post walking through it in the works. | {"url":"https://jasdev.me/notes/parity-arity","timestamp":"2024-11-06T06:10:24Z","content_type":"text/html","content_length":"7791","record_id":"<urn:uuid:fd22dcbe-1ebd-4f4d-a5fa-8674ef65e570>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00625.warc.gz"} |
Factsheet: How Students Learn - Teaching College
Blended Learning Guidance
Factsheet: How Students Learn
A knowledge of the way students learn can help us design effective courses and help us to advise and guide our students. While the literature on this topic is vast, there are a few key pieces of
experimental work that everyone who designs or changes courses should know, these are:
If you want a much broader list of experimental work along with its impact on student achievement the Visible Learning Partnership keeps a list of over 250 influences on student achievement – the
descriptions of each item are detailed in Visible Learning: A Synthesis of Over 800 Meta-Analyses Relating to Achievement.
What is this? What problems does this solve?
While there are some hints and tips for things you can immediately apply to your teaching, reading up on these topics is going to give you a deeper understanding of why many of the teaching
techniques work (and why many do not!).
I will attempt to briefly present the 6 key areas below but other sources like Hattie & Yates 2013 (for cognitive load, testing effect, feedback, and motivation), Mayer 2014 (for cognitive load,
active learning, testing effect) and Biggs & Tang 2011 (constructive alignment) explain these with more detail and clarity than I can. If you do read these books I suggest skipping to these sections
first. What I have outlined below is, in places, grossly oversimplified but should be enough to give you a rough idea of how students learn.
How memory works
If we are paying attention to the information in our sensory memory then it will enter our working (AKA short term) memory. Working memory is where conscious thought takes place, including problem
solving, thinking, and learning. Our working memory is seriously limited, young adults can only hold 3-5 “chunks” at any one time [Cowan 2010] and it only lasts of the order of seconds without being
refreshed (think of repeating a phone number over and over until you can find somewhere to write it down).
A chunk can be thought of as a single “unit” of memory, this might be a number (e.g. 24), an entire phrase (“she sells sea shells on the sea shore”), or an entire chess board layout (e.g. Sicilian
Defense) depending on what is stored in our long term memory. In other words, the facts and concepts we have in long term memory effectively increase the space available in working memory, thus our
prior knowledge is one of the biggest determinants of learning speed (See: Chapter 1 of “How learning works” Ambrose et. al. 2010). Applied to another context: real-world problem solving and
creativity are strongly dependent on the availability of accurate knowledge and the ease with which it can be handled in working memory, both of which are dependent on what we have stored in our long
term memory.
One other thing to note is that there are two pathways in our working memory: pictorial and verbal [see: Dual coding theory in Chapter 4 of Mayer 2014] – thus we can further maximize working memory
by using both channels for separate information (e.g. don’t put text [verbal] on powerpoint slides [pictorial]).
To remember something requires that (1) it was encoded in long-term memory in the first place and (2) you can retrieve it. To encode a memory students need to be thinking deeply about the topic,
forming links to other items in long term memory, this memory will also need periodic refreshing to be retrievable (see the section below on testing).
The format that we learn something will also impact how easily we retrieve it e.g. If I taught someone the dates that each king of England reigned for, they have the information in long term memory
to tell me all the Kings who reigned for over 40 years but they cannot retrieve this, as that is not how it is stored. The same might still be true if we explicitly told them the lengths of their
reigns (depending on whether the student has previously thought about grouping the kings by how long they reigned). See Chapter 2 of Ambrose et al 2010).
The facts about encoding and retrieval mean we should help students form and test a variety of links between new material and prior knowledge in a way that is similar to the way we want them to
retrieve it.
Chapter 2 of Mayer 2014 (The Cambridge Handbook of Multimedia Learning) or Chapter 13 & 14 of Hattie & Yates 2013 (Visible Learning and the science of how we learn) presents this in more detail and
with more discussion of the consequences for teaching.
Cognitive load theory
We said that working memory is limited to 3-5 chunks but there is another important facet to this. Some of our working memory will be devoted to things other than the concepts we want the students to
learn – this greatly decreases their ability to learn material.
People cannot multitask when learning (Chapter 20 of Hattie & Yates 2013), multitasking basically involves switching rapidly between tasks, and either clearing out working memory (i.e. stopping
thinking about the teaching material) or devoting some working memory to the other task (i.e. reducing the available working memory for learning). Thus students who are chatting socially, playing on
their phones, or browsing the web are at a great disadvantage when trying to learn (e.g. in a lecture, or at home).
But it’s not just the students, the way we design our learning contributes to cognitive load (the filling up of working memory). If we design tasks that require students to think about things that
are not necessary for learning the material then we are limiting the working memory available for learning. This extraneous cognitive load might be in the form of games or activities that are fun and
active (but unnecessary). It might also be in the form of having split information onto different pages of the course notes – the students have to keep flicking back and forth to understand it.
Finally, we can minimise the cognitive load by careful ordering (sequencing) of our material. Relevant prior knowledge acts to decrease our cognitive load, therefore if we explicitly find links to
material students already know and present topics one at a time, building upon one another, we can increase learning efficiency.
Chapter 2 of Mayer 2014 or Chapter 13 – 16, and 20 of Hattie & Yates 2013 presents this in more detail and with more discussion of the consequences for teaching.
Active learning
Although it was stated before it is worth making this point explicit – there is no such thing as passive learning – “Memory is the residue of thought” (Daniel Willingham). To think, and thus to
learn, requires motivation, time, attention, and effort. Thus we want to design our learning activities (as much of the study hours as possible) around tasks that get students to think deeply about
the material. Sitting in a lecture or reading a book can be done without any real focus and, if that is the case, no significant learning will take place.
“Learning results from what the student does and thinks and only from what the student does and thinks. The teacher can advance learning only by influencing what the student does to learn”.
Herbert Simon
There is one subtlety to highlight here – ‘active’ doesn’t necessarily mean physically active. A student might be ‘solving’ a question sheet by mechanically copying down past answers without really
thinking about why they work – this is ‘active’ compared to sitting in a lecture but it is not helpful. In the same way a student who is staring into space in a lecture might be thinking deeply about
the material.
What we want is psychological activity, in other words, thought. But the only way to know if the right kind of thought is happening is to make the learning visible.
We make learning visible by seeing the output of students thought. This might be in the form of answers from classroom response systems or by getting students to hand in their answers to question
sheets, or by walking round a class and listening to their group discussions. This does not mean we have to mark and give feedback on all their work, we just need to see how they are doing and the
types of mistakes they are making. This type of information from the students, is the most important type of feedback [Hattie & Timperley 2007].
The testing, spacing, worked-example, and interleaving effects
“Proficiency requires practice”
(Daniel Willingham)
Students who spend their time practising learn significantly more than those who spend their time re-reading, highlighting, or even summarising their notes/textbook/lecture recordings – this is
called the testing effect. This finding holds true across a range of abilities and in many different tasks and educational contexts. Thus we should be designing our courses to increase the amount of
time a student can spend practicing.
In this context, practicing might involve MCQs spaced throughout lectures, a large bank of practice problems, having a recap quiz at the start of every lecture, mock exams, etc., etc. Again, the
focus is not on marking, the focus is on the student being forced to test their retrieval from long term memory.
To make best use of this effect the questions should start easy and gradually become harder. Students who are provided with fully worked examples do a lot better than those who do not (see: worked
example effect). Fully worked examples are questions where the full working is shown along with the solution. These greatly help students, both in learning how to solve similar problems but also, by
applying the methods they have seen to new problems, it helps them learn to generalise the technique (i.e. by combining fully worked examples and independent practice many more students will achieve
deep learning).
Students will retain their knowledge for longer through spaced rather than massed practice (see: spacing effect / spaced repetition). In other words, students who cram (studying everything in a short
space of time) will not remember as much as those who repeatedly retest themselves. To do this most efficiently the material should be revisited on an expanding timeframe, e.g. 1 day, 3 days, 1 week,
2 weeks, 1 month, 2 months, 6 months, etc.
Finally, students will be better able to solve new problems if, when they practice, they mix up questions from previous topics rather than studying questions from each week of material separately
(see: interleaving effect).
These four effects mean that we should regularly give students opportunities to test themselves on problems randomly selected from all the material they have previously looked at. We should also be
providing a bank of fully solved questions, for students to learn from.
Chapter 16 of Mayer 2014 or Part 3 of Evidence Based Training Methods (2014) Ruth Colvin Clark are good overviews of the worked example effect. Gwern provides a thorough review of the research on
spaced repetition (including interleaving and the testing effect) or see the relevant sections of Dunlosky, John, et al. 2013.
Feedback has one of the largest average effects in all of education. In other words, studies on feedback generally see one of the largest improvements in student marks when compared to any other type
of educational intervention. But it’s not all good news, over a third of educational research on feedback shows feedback to have a negative effect (i.e. the students who had the “improved” feedback
did worse!).
Thus it is important to know what counts as effective feedback and how this relates to student learning.
• External rewards, praise and punishment are all associated with either no change or a significant reduction in achievement, i.e. in many studies rewards, praise and punishment reduced student
• Motivation cues (e.g. encouragement) have a hugely positive effect on student learning (e.g. “you are on the right lines, keep going”, “I am giving you this feedback because I think you can do
really well in this class and I want to help you get there”).
• Information that tells the student what they need to do next to improve is also a highly positive effect on student learning. (e.g. “try this next”).
• It is valuable to tell students what they got correct and incorrect (e.g. ticks and crosses, and other information about how they are doing).
• Effective feedback is targeted at the work (“try this”) not the person (“you are …”).
• It doesn’t seem to matter whether the feedback is written, oral, video, or even from an automated computer system.
• It doesn’t even matter how often feedback is given. To save your time and sanity I would recommend focusing on quality rather than quantity.
• Giving grades along with feedback significantly reduces the positive effect of the feedback – grades effectively tell the student “the work is over”.
• Feedback is only valuable if acted upon. A student must change their behaviour because of the feedback, otherwise it is not effective (it’s not doing anything at all).
As feedback is only valuable if acted upon, we need to help our students do just that. They need help to read and understand the feedback and have an opportunity to put that feedback into practice.
Students often do not know how to interpret and use the feedback they have been given and it is valuable to teach them “feedback literacy”.
It also means feedback needs to be embedded into cycles of practice (testing) and feedback – “all students regardless of their level of achievement typically need to be exposed to any new learning at
least three to five times before it has a high probability of being learnt.” Visible Learning Feedback (Chapter 1). Thus we should design course that have an opportunity for a student to immediately
act upon the feedback.
The best resource for learning about effective feedback is either the book Visible Learning Feedback, or the systematic review of meta-analyses that this book was based upon The Power of Feedback.
There is a very brief overview of this work in Chapter 8 of Hattie & Yates 2013. We also have an effective feedback factsheet.
Outcomes based education / constructive alignment
Outcomes based education is the idea that what matters is not what we teach (syllabus/content) but what the student learns (outcomes). Hopefully the value of this is obvious but it does shift the
focus when designing our courses to things that we know improve learning (reducing cognitive load, active learning, worked examples and regular non-graded testing and feedback).
That’s not to say that “what we teach” isn’t important, it matters a lot – but it should be driven by the outcomes we want to see. Given what has been said above about the importance of focusing the
students’ attention solely on concepts and practice that are relevant to their learning it becomes even more important to know what we want them to learn to do!
Students knowing these learning intentions (outcomes) causes a significant improvement in student achievement (d=0.59, Visible Learning Feedback Chapter 2). This is because it helps them direct their
effort and attention to the correct areas.
Constructive alignment and outcomes based education is described in more detail in Biggs & Tang 2011.
Motivation and peer effects (social)
Learning cannot happen without a student spending a lot of time thinking deeply about the material and they need the motivation to put in the time and effort. Thus motivation has a large effect on
There are three key components to motivation, if any one of them is lacking a student is unlikely to put in the time and effort:
1. Value. Will putting time and effort in lead to something good? Either because they will enjoy doing it (intrinsic value), they will enjoy completing it (attainment value) or they will value what
it leads to (extrinsic/instrumental value).
2. Expectancy. How likely is it that the student will ‘get’ the value. If a student doesn’t believe they can put in enough time/effort they lack efficacy expectancy but if a student believes that
however much time/effort they put in they won’t succeed then they lack outcome expectancy (e.g. “I’m just bad at Maths”). See Chapter 24 and 25 of Hattie & Yates 2013.
3. Supportiveness of the environment. The environment around a student might be filled with distractions or reminders and the student’s peers might provide explicitly pressure or encouragement, or
an implicit expectation towards or away from certain behaviours. This social aspect is a huge determinant of behaviour.
One final point to add about motivation is that it is relative. Even if you have low motivation for a task, as long as everything else you could do is even lower, you will still perform it.
Not only is our behaviour largely determined by social effects (roughly we do what we think “most people who are important to us think that we should/should not do”) Webb & Sheeran 2006 but “one of
the ways in which humans learn most efficiently and effectively is when learning is situated within the social context” Encyclopedia of the Sciences of Learning (2012). This is why cooperative
learning “has very strong effects on a range of dependent variables such as achievement, socialization, motivation, and personal self-development.” Gillies 2016.
Thus motivation is high when students can see a gap in their knowledge that they perceive as valuable and feasible to fill and they work within a supportive social environment.
To address value:
1. Be enthusiastic – students are more likely to value something if you show that you value it (see: Goal Contagion).
2. Make sure the student understands the learning outcomes and discuss the value of learning these topics (what exciting skills and jobs does learning this lead to).
To address expectancy:
1. Make sure the student has a clear idea of what the learning outcomes and success criteria are and what they need to do to achieve them. Part of this is having clear marking criteria and
discussing examples of good and bad work, part of this is guiding the students through how they should spend their study hours.
2. Sequence the learning activities so that students can see progress, this gives them valuable information about the feasibility of learning this topic. In other words make sure the learning
activities are at an appropriate level of difficulty.
3. Provide motivational feedback (encouragement) as a student progresses through the course. But this cannot succeed without activities that are actually achievable for the student, the best way to
improve someone’s confidence in a task is for them to repeatedly succeed in that task.
To address the social environment:
1. Model the desired behaviour. If you want students to notice certain things when solving questions then solve the questions while “thinking out loud”. If you want them to critically review papers,
actually do that in front of them rather than just telling them how to. Students will learn to imitate the behaviour and it is more effective than just telling them how you would solve something
(without actually showing you doing it).
2. Encourage students to teach each other – either formally or informally. This not only sets a climate and social expectation of learning but helps the student-teacher to cement their knowledge
while teaching the recipient (see: peer learning).
3. Create activities that encourage student interaction (see: cooperative learning or kagan structures).
For an overview of motivation, Chapter 3 of Ambrose et. al. 2010 is probably the best place to start, followed by Visible Learning Feedback. A comprehensive overview of motivation is given in the
Oxford handbook of human motivation, which has a relevant chapter on Motivation in Education. For an overview of peer and cooperative learning see Gillies 2016.
Where can I learn more?
Depending on what you want to learn you can find links to relevant resources in the sections above. If you are not sure where to start, I suggest reading the resources in this order:
1. Ambrose et. al. 2010. As with the list above it picks a few of the most important findings in educational research, but spends a lot more time on practical suggestions based upon the evidence.
2. Hattie & Yates 2013. This provides a much broader look at the field, by necessity this means each topic is covered in less detail but this will give you a good general framework for how students
learn and help you judge the effectiveness of teaching techniques in general. If you wanted to continue down this line of inquiry I would recommend a previous book by Hattie (Visible Learning: A
Synthesis of Over 800 Meta-Analyses). Kraft 2018 will give you a much more nuanced view of what effect sizes can tell us and should be read alongside the Synthesis of Meta-Analysis.
3. Clark 2014 provides an accessible and actionable overview of the research on effective explanations and a little on effective practice. If you want a more academically rigorous treatment of the
same topic see Mayer 2014. A short summary of the research findings is given on page 8-9 of Chapter 1 of Mayer 2014.
4. Visible Learning Feedback is a research based and actionable review of how to design courses that contain effective two-way feedback. This ends up covering the testing effect/spaced repetition
and a fair amount on student motivation. If you want an academically rigorous treatment of the same topic see the systematic review of meta-analyses that this book was based upon (The Power of
5. Study Strategies to Boost Learning (Dunlosky 2013) is an excellent review of the advice to students but also contains a review of the work around practice (testing, spacing, interleaving etc.).
Table 1 is an excellent summary of the effectiveness of studying techniques. If you want an academically rigorous treatment of the same topic see Dunlosky, John, et al. 2013.
To help you put some of the approaches to learning (e.g. behaviourism, constructivism) in context you might want to read Brown 2004, but I have personally found the above books far more useful to my
teaching practice and my understanding of how students learn.
Next steps
If you want to discuss education research or practice contact james.brooks-3@manchester.ac.uk. If there is enough interest I will set up an informal reading/discussion group.
Things to look out for (caveats & warnings)
The main caveat to give to taking ideas from any research paper is that implementation matters. Just because something works well in theory, or even in practice, doesn’t mean it will work in your
context or the way you have chosen to implement it. There is a lot of disagreement in the literature and it’s often not clear who is right (or, more accurately, under which sets of circumstances each
is better). It’s always best to find someone who has used the technique you want to try and discuss your plans with them beforehand. The eLearning team should be your first point of contact if you
don’t know where to start.
The final piece of advice comes from another body of research – pedagogic content knowledge. This is the idea that there are three skillsets you need to be a good teacher:
1. Content Knowledge – knowledge of your subject.
2. Pedagogic Knowledge – knowledge of how students learn and of good teaching practices.
3. Pedagogic Content Knowledge (PCK) – knowledge of how to teach your subject.
This final piece PCK, involves (1) knowing the mistakes that students often make, their misconceptions, their prior knowledge, and (2) having a variety of explanations and exercises for the content
that takes into account the students prior knowledge and directly addresses their misconceptions.
PCK is at least as important as pedagogic knowledge and so we should be spending time developing this, by seeing student progress/mistakes and by looking for and thinking of better/different ways to
explain our material. We cannot become excellent educators without knowledge and skill in all three domains.
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Grade 12 Physics Daniel Gockeritz Essay - Free Essay Example | Artscolumbia
PRACTICAL INVESTIGATION MOMENTUM AND COLLISIONSThis report will investigate the theoretical velocity of a ball bearing gun. The methods and techniques used to derive the results will be shown along
with the possible systematic and random errors caused by experimental limitations. Discussion:’since the track is virtually frictionless and air resistance is neglected, the system is isolated; the
net resultant force of the external forces equals zero.
? The total linear momentum of the system before the collision is equal to the total momentum after the collision. Therefore, the total change in momentum of this two-particle system is zero. ?
Equation that represents the conservation of momentum:? The total linear momentum of an isolated system is constant. ? All significant experimental errors have been incorporated into the final
velocity result. Aim:To investigate and determine the muzzle velocity of a ball bearing gun by utilizing the law of conservation of momentum. Determine out the theoretical velocity using various
mathematical methods and techniques.
Hypothesis:This two-particle system is virtually isolated, thus the total change in momentum is zero. Therefore when the two bodies collide, they will exert forces on each other, equal in magnitude
but opposite in direction. Resulting in one combined body that is equal to the sum of the momentum of the two particles before the collision. Materials:? One (1) Ball bearing. (Weight – 65. 9g 0.
1, Approx Size – 2cm in diameter)This will be the projectile that is fired from the missile launcher. ? One (1) Cart. (Weight – 678. 3g 0. 1)This will be the object on which the projectile is fired
? One (1) standard Stopwatch. (Can measure up to 100th of a second)Used to time the journey of Cart + ball bearing. ? One (1) Track. (Measuring device length – 0. 50m 0. 05)Used to guide cart and
measure displacement.
Method/Procedure:1. Prepare track by aligning it and the cart to a perfect 180 degrees to the launcher. ? Distance used was 0. 50m 0.
05. 2. Fire the ball bearing into the cart and time the journey. ? The ball bearing used in this experiment, took an average of 1. 14 0.
1 seconds to complete 0. 50 meters. 3. Work out the theoretical velocity of the ball bearing in the barrel of the launcher. ? Equations used to determine theoretical final velocity:–NOTE: During
entire experiment, safety glasses are to be worn. Any spectator that is not wearing safety glasses should watch from a safe distance.
Results:Errors accounted for:? Parallax Error: 0. 05m’stopwatch/Timing Error: 0. 1s? Mass measurement error: 0. 1gRecorded measurements (NOT including uncertainty):Times for overall journey: 1. 13s,
13s, and 1. 16sDistance: 0. 50mMass of Ball Bearing: 65. 9gMass of Cart: 678. 3gTo determine average time (NOT including uncertainty):To determine mass of combined body after collision:To determine
velocity of combined body after collision:s = 0.
50m 0. 05 t = 1. 14s 0. 1s = 0. 50m 10% t = 1. 14 8.
7%To determine velocity of ball bearing in barrel of missile launcher:The muzzle velocity of this ball bearing gun is:. Errors not incorporated into method:? The ball beating itself has a small drag
coefficient, although the cart, which the ball bearing is fired into, may experience air friction. ? All air friction/resistance was neglected. Conclusion:This experiment proved my hypothesis
correct. Throughout the entire experiment the overall change in momentum equaled zero.
When the two particles collided there momentum was conserved resulting in one body that was the combined mass and momentum of the previous bodies. The result was obtained by recognizing that the
initial velocity/momentum of the ball bearing could be determined by utilizing the conservation of momentum law; that as long as the net resultant external forces equal zero, the momentum will be
constant. From this exercise I learnt new method and techniques used in calculating errors and uncertainty.BibliographyPhysics Investigation on Momentum and collisions | {"url":"https://artscolumbia.org/grade-12-physics-daniel-gockeritz-61809/","timestamp":"2024-11-15T04:18:06Z","content_type":"text/html","content_length":"122763","record_id":"<urn:uuid:6f56e15e-27c6-40d6-810a-c1450c48e4a8>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00469.warc.gz"} |
Maximum mutational robustness in genotype-phenotype maps follows a self-similar blancmange-like curve
Data from: Maximum mutational robustness in genotype-phenotype maps follows a self-similar blancmange-like curve
Data files
Jul 02, 2023 version files 1.62 MB
Phenotype robustness, defined as the average mutational robustness of all the genotypes that map to a given phenotype, plays a key role in facilitating neutral exploration of novel phenotypic
variation by an evolving population. By applying results from coding theory, we prove that the maximum phenotype robustness occurs when genotypes are organised as bricklayer’s graphs, so called
because they resemble the way in which a bricklayer would fill in a Hamming graph. The value of the maximal robustness is given by a fractal continuous everywhere but differentiable nowhere
sums-of-digits function from number theory. Interestingly, genotype-phenotype (GP) maps for RNA secondary structure and the HP model for protein folding can exhibit phenotype robustness that exactly
attains this upper bound. By exploiting properties of the sums-of-digits function, we prove a lower bound on the deviation of the maximum robustness of phenotypes with multiple neutral components
from the bricklayer’s graph bound, and show that RNA secondary structure phenotypes obey this bound. Finally, we show how robustness changes when phenotypes are coarse-grained and derive a formula
and associated bounds for the transition probabilities between such phenotypes.
This data contains the data and code required to generate presented in "Maximum Mutational Robustness in Genotype-Phenotype Maps Follows a Self-similar Blancmange-like Curve" by Mohanty et al.,
published in Journal of the Royal Society Interface.
The exact maximum robustness curve corresponding to the robustness of the bricklayer's graphs, as well as the interpolated curve, can be generated from the RoBound Calculator, available free of
charge and open source on GitHub (https://github.com/vaibhav-mohanty/RoBound-Calculator).
All bounds (e.g. Figure 1, 9, 10, and 11 as well as the bounds shown in Figure 3 or 7) can be calculated using the RoBound Calculator (https://github.com/vaibhav-mohanty/RoBound-Calculator).
In Figure 3, the RNA and HP model neutral component sizes and robustness values are provided in the files hp5x5_components.csv, hp24_components.csv, rna12_components.csv, and rna15_components.csv.
These results obtainedd from Greenbury et al., "Genetic Correlations Greatly Increase Mutational Robustness and Can Both Reduce and Enhance Evolvability," PLOS Computational Biology, 2016.
In Figure 4, we show the deviation of RNA12 and RNA15 neutral networks from the maximum (bricklayer's) robustness. This data is obtained from rna12.csv (alternatively rna12.mat) and rna15.csv
(alternatively rna15.mat), which were also results obtained from Greenbury et al., "Genetic Correlations Greatly Increase Mutational Robustness and Can Both Reduce and Enhance Evolvability," PLOS
Computational Biology, 2016. The code to produce the figure is found in nc_err_corr.m
In Figure 7, the bounds can be calculated using the RoBound Calculator (https://github.com/vaibhav-mohanty/RoBound-Calculator). The raw data is obtained from using ViennaRNA (https://
www.tbi.univie.ac.at/RNA/) to calculate the dot-bracket structures. These structures are then fed into the RNASHAPES software (https://bibiserv.cebitec.uni-bielefeld.de/rnashapes) to generate the
coarse-grained data. The data files rna12abstract.mat and rna15abstract.mat contain the frequency and robustness values for the neutral networks at various levels of coarse-graining.
In Figure 8, the RNA12 transition probabilites between phenotypes can be obtained from rna12_theta.csv, which provides the phi_pq matrix when each column's sum is normalized to 1. The script
phi_critical_ranges.m produces Figure 8.
Figures 2, 5, and 6 are schematics and have no associated data.
Usage notes
CSV files can be opened easily. MATLAB (or associated python packages) can open the .mat files. MATLAB or Octave can be used to load the .m scripts.
Works referencing this dataset | {"url":"https://datadryad.org:443/stash/dataset/doi:10.5061/dryad.sj3tx969f","timestamp":"2024-11-11T10:53:34Z","content_type":"text/html","content_length":"52337","record_id":"<urn:uuid:9e79ca9a-4055-4591-9521-53193067cd92>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00859.warc.gz"} |
Lab 7: Classification
This lab covers binary regression and classification using logistic regression models. The logistic regression model for a binary outcome \(y \in \{0, 1\}\) posits that the probability of the outcome
of interest follows a logistic function of the explanatory variable \(x\):
\[ P(Y = 1) = \frac{1}{1 + e^{-(\beta_0 + \beta_1 x)}} \]
More commonly, the model is written in terms of the log-odds of the outcome of interest:
\[ \log\left[\frac{P(Y = 1)}{P(Y = 0)}\right] = \beta_0 + \beta_1 x \]
Additional explanatory variables can be included in the model by specifying a linear predictor with additional \(\beta_j x_j\) terms.
Logistic regression models represent the probability of an outcome as a function of one or more explanatory variables; fitted probabilities can be coerced to hard classifications by thresholding.
For this lab, we’ll revisit the SEDA data from an earlier assignment. Below are the log median incomes and estimated achievement gaps on math and reading tests for 625 school districts in California:
The estimated achievement gap is positive if boys outperform girls, and negative if girls outperform boys. We can therefore define a binary indicator of the direction of the achievement gap:
You may recall having calculated the proportion of districts in various income brackets with a math gap favoring boys.
We will now consider the closely related problem of estimating the probability that a district has a math gap favoring boys based on the median income of the district.
Since we’re only considering math gaps, we’ll filter out the gap estimates on reading tests.
Let’s set aside the data for 100 randomly chosen districts to use later in quantifying the classification accuracy of the model.
Question 1: data partitioning
Set aside 100 observations at random for testing. Do this by selecting a random subset of 100 indices. Choose a different RNG seed from your neighbor so that you can compare results based on
different training sets.
Exploratory analysis
Previously you had binned income into brackets and constructed a table of the proportion of districts in each income bracket with a math gap favoring boys. It turns out that binning and aggregation
is a useful exploratory strategy for binary regression. Your table from before would have been something like this:
# define income bracket
train['income_bracket'] = pd.cut(train.log_income, 10)
# compute mean and standard deviation of each variable by bracket
tbl = train.groupby('income_bracket').agg(func = ['mean', 'std'])
# fix column indexing and remove 'artificial' brackets containing only min and max values
tbl.columns = ["_".join(a) for a in tbl.columns.to_flat_index()]
tbl = tbl[tbl.favors_boys_std > 0]
# display
We can plot these proportions, with standard deviations, as functions of income. Since standard deviations are fairly high, the variability bands only show 0.4 standard deviations in either
trend = alt.Chart(tbl).mark_line(point = True).encode(
x = alt.X('log_income_mean', title = 'log income'),
y = alt.Y('favors_boys_mean', title = 'Pr(math gap favors boys)')
band = alt.Chart(tbl).transform_calculate(
lwr = 'datum.favors_boys_mean - 0.4*datum.favors_boys_std',
upr = 'datum.favors_boys_mean + 0.4*datum.favors_boys_std'
).mark_area(opacity = 0.3).encode(
x = 'log_income_mean',
y = alt.Y('lwr:Q', scale = alt.Scale(domain = [0, 1])),
y2 = 'upr:Q'
trend + band
We can regard these proportions as estimates of the probability that the achievement gap in math favors boys. Thus, the figure above displays the exact relationship we will attempt to model, only as
a continuous function of income rather than at 8 discrete points.
Question 2: model assumptions
The logistic regression model assumes that the probability of the outcome of interest is a monotonic function of the explanatory variable(s). Examine the plot above and discuss with your neighbor.
Does this monotinicity assumption seem to be true? Why or why not?
Type your answer here, replacing this text.
Model fitting
We’ll fit a simple model of the probility that the math gap favors boys as a logistic function of log income:
\[ \log\left[\frac{P(\text{gap favors boys})}{1 - P(\text{gap favors boys})}\right] = \beta_0 + \beta_1 \log(\text{median income}) \]
The data preparations are exactly the same as in linear regression: we’ll obtain a vector of the response outcome and an explanatory variable matrix containing log median income and a constant (for
the intercept).
The model is fit using statsmodels.Logit(). Note that the endogenous variable (the response) can be either Boolean (take values True and False) or integer (take values 0 or 1).
A coefficient table remains useful for logistic regression:
Question 3: confidence intervals
Compute 99% confidence intervals for the model parameters. Store the result as a dataframe called param_ci.
Hint: the syntax is identical to that based on sm.OLS; this is also mentioned in the lecture slides.
We can superimpose the predicted probabilities for a fine grid of log median incomes on the data figure we had made previously to compare the fitted model with the observed values:
# grid of log income values
grid_df = pd.DataFrame({
'log_income': np.linspace(9, 14, 200)
# add predictions
grid_df['pred'] = fit.predict(sm.add_constant(grid_df))
# plot predictions against income
model_viz = alt.Chart(grid_df).mark_line(color = 'red', opacity = 0.5).encode(
x = 'log_income',
y = 'pred'
# superimpose on data figure
trend + band + model_viz
Depending on your training sample, the model may or may not align well with the computed proportions, but it should be mostly or entirely within the 0.4-standard-deviation band.
To interpret the estimated relationship, recall that if median income is doubled, the log-odds changes by:
\[ \hat{\beta}_1\log(2\times\text{median income}) - \hat{\beta}_1 \log(\text{median income}) = \hat{\beta}_1 \log(2) \]
Now, exponentiating gives the estimated multiplicative change in odds:
\[ \exp\left\{\log(\text{baseline odds}) + \hat{\beta}_1 \log(2)\right\} = \text{baseline odds} \times e^{\hat{\beta}_1 \log(2)} \]
So computing \(e^{\hat{\beta}_1 \log(2)}\) gives a quantity we can readily interpret:
The exact number will depend a little bit on the data partition you used to compute the estimate, but the answer should be roughly consistent with the following interpretation:
Each doubling of median income is associated with an estimated four-fold increase in the odds that a school district has a math gap favoring boys.
Now we’ll consider the task of classifying new school districts by the predicted direction of their math achievement gap.
A straightforward classification rule would be:
\[ \text{gap predicted to favor boys} \quad\Longleftrightarrow\quad \widehat{Pr}(\text{gap favors boys}) > 0.5 \]
We can obtain the estimated probabilities using .predict(), and construct the classifier manually. To assess the accuracy, we’ll want to arrange the classifications side-by-side with the observed
Note that the testing partition was used here – to get an unbiased estimate of the classification accuracy, we need data that were not used in fitting the model.
Cross-tabulating observed and predicted outcomes gives a detailed view of the accuracy and error:
The entries where observation and prediction have the same value are counts of the number of districts correctly classified; those where they do not match are counts of errors.
Question 4: overall classification accuracy
Compute the overall classification accuracy – the proportion of districts that were correctly classified.
Often class-wise accuracy rates are more informative, because there are two possible types of error:
1. A district that has a math gap favoring girls is classified as having a math gap favoring boys
2. A district that has a math gap favoring boys is classified as having a math gap favoring girls
You may notice that there were more errors of one type than another in your result above. This is not conveyed by reporting the overall accuracy rate.
For a clearer picture, we can find the proportion of errors among by outcome:
pred_df['error'] = (pred_df.observation != pred_df.prediction)
fnr = pred_df[pred_df.observation == True].error.mean()
fpr = pred_df[pred_df.observation == False].error.mean()
tpr = 1 - fpr
tnr = 1 - fnr
print('false positive rate: ', fpr)
print('false negative rate: ', fnr)
print('true positive rate (sensitivity): ', tpr)
print('true negative rate (specificity): ', tnr)
Question 5: make your own classifier
Define a new classifier by adjusting the probability threshold. Compute and print the false positive, false negative, true positive, and true negative rates. Experiment until you achieve a better
balance between errors of each type.
# construct classifier
new_pred_df = pd.DataFrame({
# compute error rates
new_pred_df['error'] = ...
new_fnr = ...
new_fpr = ...
new_tpr = ...
new_tnr = ...
# print
print('false positive rate: ', new_fpr)
print('false negative rate: ', new_fnr)
print('true positive rate (sensitivity): ', new_tpr)
print('true negative rate (specificity): ', new_tnr)
1. Save the notebook.
2. Restart the kernel and run all cells. (CAUTION: if your notebook is not saved, you will lose your work.)
3. Carefully look through your notebook and verify that all computations execute correctly and all graphics are displayed clearly. You should see no errors; if there are any errors, make sure to
correct them before you submit the notebook.
4. Download the notebook as an .ipynb file. This is your backup copy.
5. Export the notebook as PDF and upload to Gradescope.
To double-check your work, the cell below will rerun all of the autograder tests. | {"url":"https://pstat100.tdruiz.com/labs/lab7-classification/lab7-classification","timestamp":"2024-11-06T02:54:38Z","content_type":"application/xhtml+xml","content_length":"53644","record_id":"<urn:uuid:863788a7-fa5f-4309-9624-400d7066e0b7>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00141.warc.gz"} |
Year 4 Subtraction Worksheets - Primary Maths Worksheets
Colourful PDF Year 4 Primary Subtraction Worksheets
Year 4 subtraction worksheets for children ages 8-9 years old. These worksheets are fully differentiated. Children in year 4 are expected to subtract multiples of ten from 4 digit numbers. For
example 4721 - 50. Some year 4 pupils will be able to subtract multiples of ten greater than 100 from 4 digits numbers. For example 632 - 170. Year 4 pupils are expected to subtract 4 digits
numbers from 4 digit numbers using the column method or otherwise. Explore our collection of worksheets designed for children aged 8-9 in year 4. Each file is a PDF worksheet ready to download.
Subtraction Maths Worksheets for Year 4 Primary Maths
This page contains Primary Maths Worksheets covering Subtraction for Year 4 Maths. This website contains a series of PDF worksheets designed to be saved and downloaded.
Leave a Comment
You must be logged in to post a comment. | {"url":"https://primarymathsworksheets.com/year-4-maths-worksheets-2/year-4-subtraction-worksheets/","timestamp":"2024-11-14T02:14:37Z","content_type":"text/html","content_length":"103076","record_id":"<urn:uuid:7aab9ff9-0207-4627-bab3-f1b1ae4374a7>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00412.warc.gz"} |
Credit Spread
Credit spread is a difference in yield between debt securities. It is most commonly referenced versus government debt. The difference is typically measured in basis points, where a hundred basis
points equals one percentage point. So, for example. if a company issues debt with a five-year maturity that has an interest rate of 5%, then the yield is 5%. If a five-year government bond has a
yield of 3.25%, then the company’s yield is 1.75%, or 175 basis points higher, so their credit spread is 175 basis points.
Related terms | {"url":"https://ondemand.euromoney.com/discover/glossary/credit-spread","timestamp":"2024-11-12T09:42:33Z","content_type":"text/html","content_length":"99047","record_id":"<urn:uuid:9ed14d16-c79e-45d9-b370-eb78da7860ed>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00563.warc.gz"} |
In ROC analysis covariate adjustment is advocated when the covariates impact
In ROC analysis covariate adjustment is advocated when the covariates impact the magnitude MDA1 or accuracy from the test less than research. of binary regression as well as the estimating equations
derive from the U figures. The AAUC can be approximated through the weighted typical of AUCover the covariate distribution from the diseased topics. We use reweighting and imputation ways to conquer
the confirmation bias issue. Our suggested estimators are primarily derived let’s assume that the real disease status can be missing AZD3463 randomly (MAR) and with some changes the estimators could
be extended towards the not-missing-at-random (NMAR) scenario. The asymptotic distributions are produced for the suggested estimators. The finite test performance is examined by some simulation
research. Our method can be put on a data occur Alzheimer’s disease study. and AAUC estimators under confirmation bias never have been studied however. The main efforts of the paper are: (1) we
propose the U-statistic type estimating equations for verification-bias corrected AUCand AAUC; (2) we demonstrate the asymptotic ideas for the brand new estimators. After we possess the estimated
ROCcurve AUCcould end up being computed theoretically. For instance using the ROCestimator in Liu and Zhou (2011) you can integrate the ROC curve over [0 1 Nevertheless as the hyperlink and baseline
function from the ROCestimator are both non-parametric functions AUCmay not need an explicit manifestation. Furthermore the covariate results in both Liu and Zhou (2011) and Web page and Rotnitzky
(2010) are interpreted as the result for the mean check result. However in many circumstances one may desire to discover out whether and the way the diagnostic precision itself is suffering from the
covariates and therefore it is even more highly relevant to model AUCdirectly. The thought of our approach may be the regression magic size assumption on AUCis designed for the entire data and
several reweighting methods are accustomed to right for the verification bias. The reweighting strategies are 1st derived beneath the MAR assumption and extended AZD3463 towards the NMAR scenario.
Consequently the AAUC estimators could be derived like a weighted normal of AUCand AAUC estimators derive from the U figures theory. The paper can be organized the following. In Section AZD3463 2 we
propose the confirmation bias-corrected estimators of AUCas an estimator of AAUC. Many simulation research are shown in Section 4 accompanied by a genuine example from Alzheimer’s disease study in
Section 5. We help to make the concluding remarks in Section 6 finally. AZD3463 2 Estimation for Covariate-Specific AUC (AUCand denote the constant check result binary disease position (= 1 if
diseased and 0 if healthful) binary confirmation position (= 1 if can be noticed and 0 if lacking) and patient-level features for subject can be even more indicative of disease. The subscript is
omitted when there is no confusion sometimes. With this section we 1st discuss the model establishing and assumptions after that we propose to utilize the weighted AZD3463 estimating equations to
improve for the confirmation bias and acquire the approximated AUCis interpreted as the possibility a case includes a higher check result when compared to a control if they share the normal covariate
takes the next generalized linear type: can be some unfamiliar monotone change and ? comes after the distribution and may be created explicitly as = μ(1 = σ2(1 and σ2(in (1) restricts the assessment
from the test outcomes for topics in the AZD3463 same covariates’ level. Nevertheless if a number of the covariates are constant there might not can be found any pairs of the case and a control with
exactly the same covariates value. Therefore the estimation of ν(and a control with covariates can be = (1 ≡ > = with = 1 = 0 and = = ≡ ξ(with = 1 and = 0 the following: = var(= 1 = 0 = Pr (= 1| = Pr
(= 1| and πand using the approximated possibility in the estimating features (7): with ρ0≡ Pr(= 0) but will keep the observed types. The approximated edition for = 0 1 If MAR assumption keeps ρ0is
add up to ρ≡ + (1 ? and mimics carefully. With mis-specified disease possibility to consistently be estimated. The fourth technique is doubly powerful (DR) estimator making usage of both and ≡ + (1 ?
and or even to end up being estimated but consistently. | {"url":"http://www.bioshockinfinitereleasedate.com/2016/03/10/in-roc-analysis-covariate-adjustment-is-advocated-when-the-covariates-impact/","timestamp":"2024-11-02T04:37:44Z","content_type":"text/html","content_length":"35234","record_id":"<urn:uuid:b08bb8af-1e0e-4172-8f93-4079f3a6f6e0>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00437.warc.gz"} |
Shock selection and linear option in Extended Path
Dear Forum,
1. During the extended path simulation, even if I keep multiple shocks turned on, I still get single simulation matrix e.g. if my number of shocks is 3 and number of endogenous variables 7 and I run
simulation for 1000 periods, I get a oo_.endo_simul of size 7 by 1001. Is there a way to know for which shock it has been produced? If we got an oo_.endo_simul of size 7 by 1001 by 3, it would
have been nicer.
2. Is there a way to turn on linear option for perfect foresight solver in extended path simulation?
Thanks in advance.
1. Extended path simulations are nonlinear simulations for the specified shock series. In contrast to linear models, where the individual shock responses are additive, you need to conduct joint
simulations. That explains why you get one vector for every endogenous variable. If you want to simulate one shock at a time, you need to conduct three separate simulations.
2. No, but what would even be the point of that?
Dear Prof. Pfeifer,
Thank you very much for clarifying the first point.
Regarding the second point, I want to compare linear and nonlinear model using extended path.
You can just run a standard perfect foresight simulation with the linear_approximation option in that case. | {"url":"https://forum.dynare.org/t/shock-selection-and-linear-option-in-extended-path/26641","timestamp":"2024-11-12T15:47:07Z","content_type":"text/html","content_length":"18117","record_id":"<urn:uuid:5a40b2e3-6a25-4a63-a68d-64ff0741b6a6>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00411.warc.gz"} |
2nd PUC Basic Maths Question Bank Chapter 13 Heights And Distances Ex 13.1
Students can Download Basic Maths Exercise 13.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and
score more marks in your examinations.
Karnataka 2nd PUC Basic Maths Question Bank Chapter 13 Heights and Distances Ex 13.1
Part – A
2nd PUC Basic Maths Heights and Distances Ex 13.1 Two Marks Questions and Answers
Question 1.
The angle of elevation of the top of a tower at a distance 500 metres from its foot is 30°. Find the height of the tower.
From the triangle ABC
B be the position of the observer & AC = height of the pole h.
tan 30° = \(\frac{\mathrm{AC}}{\mathrm{AB}}\); \(\frac{1}{\sqrt{3}}=\frac{h}{500}\)
∴ h = \(\frac{500}{\sqrt{3}}=500 \sqrt{3}\) Height of the tower m \(\frac{500 \sqrt{3}}{2} \mathrm{mts}\)
Question 2.
The angle of elevation of the top of a chimney at a distance of 100 metres from a foot is 30°. Find its height.
in triangle ABC we have B as the position of observer, AC height of tan 30° = \(\frac{A C}{A B}\) the chimney = h
\(\mathrm{h}=\frac{100}{\sqrt{3}}=\frac{100 \sqrt{3}}{3} \mathrm{m}\)
Question 3.
From a ship a mast head 40 meters high the angle of depression of a boat is observed to be 45°. Find its distance from the ship.
In triangle ABC, we have AC as mast head, AB is distance from the ship
tan 45° = \(\frac{A C}{A B}\)
⇒ x = 40
∴ Distance from the ship is 40 mts
Question 4.
What is the angle of elevation of the sun when the length of the shadow of a pole is \(\frac{1}{\sqrt{3}}\) times the height of the pole?
Let the height of the pole be AC = h & length of the shadow = AB = \(\frac{1}{\sqrt{3}}\) h
From triangle ABC we have tan
tan θ = \(\frac{A C}{A B}\) tan θ = \(\frac{\mathrm{h}}{\frac{1}{\sqrt{3}}}=\sqrt{3}\) = tan 60°
⇒ θ = 30°
Question 5.
Find the angle of elevation of the sun when the shadow of a tower 75 meters high is meters \(25\sqrt{3}\) long.
AC = height of the tower = h = 75 mts AB = Shadow of a tower = \(25\sqrt{3}\) from triangle ABC, we have
tan θ = \(\frac{A C}{A B}=\frac{75}{25 \sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}\) = tan 60°
⇒ θ = 60°
Question 6.
A kite flying at a height of h is tied to a thread which is 500m long. Assuming that there is no kink in the thread and it makes an angle of 30° with the ground. Find the height of the kite.
Let AC = height of the kite BC = 500 mts From ∆ ABC we have
sin 30° = \(\frac{A C}{B C}=\frac{h}{500}\)
\(\frac{1}{2}=\frac{h}{500} \Rightarrow h=\frac{500}{2}=250 \mathrm{mts}\)
∴ the heights of the kite = 250 mts.
Question 7.
A ladder leaning against a wall makes an angle of 60° with the ground. The foot of the ladder is 6m away from the wall. Find the length of the ladder.
Let AC be the ladder from triangle ABC we have
cos 60° = \(\frac{A B}{B C}=\frac{6}{B C}\)
\(\frac{1}{2}=\frac{6}{\mathrm{BC}} \Rightarrow \mathrm{BC}=12 \mathrm{mts}\)
∴ The length of the ladder = 12 mts.
Question 8.
Find the angle of elevation of the sun’s rays from a point on the ground at a distance of \(3 \sqrt{3}m\) , from the foot of tower 3m high.
Let AC be the height of the pole AC = 3mts & AB = \(3 \sqrt{3}m\) mts From the triangle ABC we have
tan θ = \(\frac{A C}{A B}=\frac{3}{3 \sqrt{3}}=\frac{1}{\sqrt{3}}\) = tan 30 ⇒ = 30°
∴ The angle of elevation is 30°
Part – B
2nd PUC Basic Maths Heights and Distances Ex 13.1 Four or Five marks questions and answers
Question 1.
The angles of elevation of the top of a tower from the base and the top of a building are 60° and 45° The building is 20 meters high. Find the height of the tower.
Let AD be the tower & CE be the building. Let the distance between the tower and building be x in ∆ ABC,
tan 45° = \(\frac{A B}{B C}=\frac{h}{x}\)
\(1=\frac{\mathrm{h}}{\mathrm{x}} \Rightarrow \mathrm{x}=\mathrm{h}-(1)\)
Question 2.
The shadow of a tower standing on a level plane is found to be 50 meters longer when sun’s altitude is 30°. Than when it is 60°. Find the height of the tower.
Let AB be the tower In triangle ABC, tan 60° = \(\frac { h }{ x }\)
\(h=\sqrt{3 x}\)
In triangle ABD, tan 30° = \(\frac{h}{50+x}\)
50 + x = \(\sqrt{3}\)
50 + x = \(\sqrt{3} \cdot \sqrt{3} x\) ∵ h = \(\sqrt{3} x\)
50 = 3x – x ⇒ 2x = 50 ⇒ x = 25 mts
∴ Height of the tower h = 25\(\sqrt{3}\) mts
Question 3.
An Aeroplane when flying at a heights of 2000 meters passes vertically above another plane at an instant, when their angles of elevation from the same point of observation are 60° and 45°
respectively. Find the distance between the Aeroplanes.
Let A & B are Aeroplanes Let h be the distance between the aeroplanes
From triangle ACD
From triangle BCD
Question 4.
From a point on the line joining the feet of two poles of equal heights, the angles of elevation of the tops of the poles are observed to be 30° and 60°. If the distance between the poles is a Find
(i) the height of the poles (ii) the position of the point of observation.
Let AB & DE are two poles
∴ Height of the poles is \(\frac{\sqrt{3}}{4} a\)
Position of c from B is x = \(\sqrt{3 h}\)
∴ position of the point of observation = \(\frac{\sqrt{3} \cdot \sqrt{3} a}{4}=\frac{3 a}{4}\)
Question 5.
The angles of elevation of the top of a tower from two points distant a and b (a < b) from its foot and the same straigth line from it are 30° and 60°. Show that the height of the tower is \(\sqrt{a
Let CD is the tower from rt angled triangle ACD
we have tan 30 = \(\frac { h }{ a }\) ⇒ h = \(\frac{a}{\sqrt{3}}\) …(1)
From rt angled triangle BCD
tan 60° = \(\frac { h }{ b }\)
\(\sqrt{3}=\frac{h}{b} \Rightarrow h=\sqrt{3} b-(2)\)
From 1 & 2 we have
Question 6.
A flag staff stands upon the top of a building. At a distance of 20 meters the angles of elevation of the top of the flag staff and building are 60° and 60° respectively. Find the height of the flag
Let BC = Building
AB = flagstaff
From triangle ACD we have
tan 60° = \(\frac{A C}{C D}=\frac{A B+B C}{20}=\sqrt{3}\)
AB + BC = 20 \(\sqrt{3}\) …. (1)
In triangle BCD, tan 30° = \(\frac{B C}{C D} ; \quad \frac{1}{\sqrt{3}}=\frac{B C}{20}\)
BC = \(\frac{20}{\sqrt{3}}\) …. (2)
Question 7.
From the top of a cliff, the angles of depression of two boats in the same vertical plane as the observer are 30° and 45°. If the distance between the boats is 100 meters, find the height of the
Let the height of the cliff be h A & B are two boats
From triangle DCA
tan 45° = \(\frac{D C}{A C}=\frac{h}{x}\)
\(1=\frac{\mathrm{h}}{\mathrm{x}} \Rightarrow \mathrm{h}=\mathrm{x}\) ….. (1)
From triangle BCD
Question 8.
From a point A due north of the tower, the elevation of the top of the tower is 60°. From a point B due south, the elevantion is 45°, if AB = 100 metres. Show that the height of the tower is 50 \(\
sqrt{3}(\sqrt{3}-1)\) meters.
A and B are positions of observation AB = 100 mts and CD is the height of tower and CD = h
From rt angled triangle ACD
tan 60° = \(\frac{\mathrm{CD}}{\mathrm{AD}}\)
\(\sqrt{3}\) = \(\frac{h}{A D} \Rightarrow A D=\frac{h}{\sqrt{3}}\) …..(1)
From rt angled triangle BCD
tan 45° = \(\frac{\mathrm{CD}}{\mathrm{BD}}\)
\(1=\frac{h}{B D} \Rightarrow B D=h\) …. (2)
Question 9.
A person at the top of a hill observes that the angles of depression of two consecutive kilometers stones on a road leading to the foot of the hill and in the same vertical plane containing the
position of the observer are 30° and 60°. Find the height of the hill.
Let CD be the hill and CD = h mts
In ∆ ADC, we have tan 60° = \(\frac{C D}{A D} \Rightarrow \sqrt{3}=\frac{h}{x}\)
\(\Rightarrow \quad x=\frac{h}{\sqrt{3}} \Rightarrow h=\sqrt{3} x\) …. (1)
From 1 and 2 we get
1 + x = \(\sqrt{3} \cdot \sqrt{3} x\) (h = \(\sqrt{3} x\))
1 + x = 3x
1 = 2x ⇒ x = 5 \(\frac{1}{2}\)
The height of the hill = h = \(\sqrt{3} x=\sqrt{3} \cdot \frac{1}{2}=\frac{\sqrt{3}}{2} \mathrm{mts}\) mts
Question 10.
The angles of elevation of the top of a tower from the base the top of a building are 60° & 45° the building is 32 meters high find the height of the tower
Let CD is the tower
AB is the house = 32 meters
From the rt angled triangle ACE
tan 45° = \(\frac{C E}{A E}=\frac{C E}{B D}=1\)
⇒ CE = BD …. (1)
Again from rt angled triangle ABD
tan 30° = \(\frac{A B}{B D}=\frac{32}{B D}=\frac{1}{\sqrt{3}}\)
⇒ BD = \(32 \sqrt{3}\)
From 1 and 2 we get CE = \(32 \sqrt{3}\) mts
∴ Height of the tower = h = DE + CE = 32 + \(32 \sqrt{3}\) = -32(1 + \(\sqrt{3}\))mts
Question 11.
The angle of elevation of a tower from a point on the ground is 30°. At a point on the horizontal line passing through the foot of the tower and 100 metres nearer it, the angle of elevation is found
to be 60°. Find the height of the pole.
Let CD = Tower
From a rt angled triangle ACD
tan 30° = \(\frac{\mathrm{CD}}{\mathrm{AC}}\)
100+ x = \(\mathrm{h} \sqrt{3}\) …. (1)
From triangle BCD
tan 60° = \(\frac{\mathrm{CD}}{\mathrm{CB}}\)
\(\sqrt{3}=\frac{h}{x} \Rightarrow h=\sqrt{3} x\) …. (2)
From 1 & 2 we get
100 + x = \(\sqrt{3} \cdot \sqrt{3} x\)
100 = 3x – x ⇒ 2x = 100 ⇒ x = 50 :
∴ Height of the tower = h = \(\sqrt{3 x}\) = \(50\sqrt{3}\) mts
Hence Distance of the first point from the tower ₹ 100 + x = 100 + 50 = 150 mts
Question 12.
A Person is at the top of a tower 75 feet high from there he observes a vertical pole and finds the angles :: of depressions of the top and the bottom of the pole which are 30° and 60° respectively.
Find the height of the pole.
Let AB = Tower, CD = Pole = h
From triangle ABC we get
tan 60° = \(\frac{A B}{A C} \Rightarrow \frac{75}{A C}=\sqrt{3}\)
⇒ AC = \(\frac{75}{\sqrt{3}} ….. (1)\) …. (1)
From triangle BDE
⇒ AC = \(\sqrt{3}(75-h)\)
From 1 and 2 we get
\(\frac{75}{\sqrt{3}}=\sqrt{3}(75-h) \Rightarrow 75=(\sqrt{3})^{2}(75-h)\)
75 = 225 – 3h 3h ⇒ 150 ⇒ h = 50
∴ Height of the pole -50ft | {"url":"https://ktbssolutions.com/2nd-puc-basic-maths-question-bank-chapter-13-ex-13-1/","timestamp":"2024-11-13T18:27:59Z","content_type":"text/html","content_length":"108447","record_id":"<urn:uuid:aabda41f-90a7-4395-add4-77f4dffcdfd5>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00447.warc.gz"} |
Sort Stl With Code Examples
In this article, we will look at how to get the solution for the problem, Sort Stl With Code Examples
Which is faster sorting algorithm?
Which is the best sorting algorithm? If you've observed, the time complexity of Quicksort is O(n logn) in the best and average case scenarios and O(n^2) in the worst case. But since it has the upper
hand in the average cases for most inputs, Quicksort is generally considered the “fastest” sorting algorithm.
// STL IN C++ FOR SORING
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
int main()
int arr[] = {1, 5, 8, 9, 6, 7, 3, 4, 2, 0};
int n = sizeof(arr)/sizeof(arr[0]);
sort(arr, arr+n); // ASCENDING SORT
reverse(arr,arr+n); //REVERESE ARRAY
sort(arr, arr + n, greater<int>());// DESCENDING SORT
1) When using vector:
sort(arr.begin(), arr.end());
2) When using array:
sort(arr, arr+n);
sort(arr, arr+n, greater<int>()); //sorts in descending
#include <bits/stdc++.h>
using namespace std;
int main()
int arr[] = { 1, 5, 8, 9, 6, 7, 3, 4, 2, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
/*Here we take two parameters, the beginning of the
array and the length n upto which we want the array to
be sorted*/
sort(arr, arr + n);
cout << "\nArray after sorting using "
"default sort is : \n";
for (int i = 0; i < n; ++i)
cout << arr[i] << " ";
return 0;
How does C++ sort function work?
The sort() function in C++ is used to sort a number of elements or a list of elements within first to last elements, in an ascending or a descending order. Here we have a range for a list, which
starts with first element and ends with last element and the sorting operation is executed within this list.
Is sort STL stable?
stable_sort() in C++ STL. stable_sort() is used to sort the elements in the range [first, last) in ascending order. It is like std::sort, but stable_sort() keeps the relative order of elements with
equivalent values.
What is STL sorting?
The fundamental sorting function of the STL is sort . This function takes a range of container elements and sorts them. In the first version of the function, it sorts them using the < operator and in
a second version, you can provide a binary predicate function to sort the elements.
Is sort in C++ STL?
Sort is an in-built function in a C++ STL ( Standard Template Library). This function is used to sort the elements in the range in ascending or descending order.
How do you sort data in C++?
Here, first – is the index (pointer) of the first element in the range to be sorted. last – is the index (pointer) of the last element in the range to be sorted. For example, we want to sort elements
of an array 'arr' from 1 to 10 position, we will use sort(arr, arr+10) and it will sort 10 elements in Ascending order.
Which sorting algorithm is best?
Quicksort. Quicksort is one of the most efficient sorting algorithms, and this makes of it one of the most used as well. The first thing to do is to select a pivot number, this number will separate
the data, on its left are the numbers smaller than it and the greater numbers on the right.
What is C++ sort algorithm?
C++ Algorithm sort() function is used to sort the elements in the range [first, last) into ascending order. The elements are compared using operator < for the first version, and comp for the second
How does sort STL work in C++?
The GNU Standard C++ library, for example, uses a 3-part hybrid sorting algorithm: introsort is performed first (introsort itself being a hybrid of quicksort and heap sort), to a maximum depth given
by 2×log2 n, where n is the number of elements, followed by an insertion sort on the result.
How do you use the sort function?
Excel SORT Function
• Summary. The Excel SORT function sorts the contents of a range or array in ascending or descending order.
• Sorts range or array.
• Sorted array.
• =SORT (array, [sort_index], [sort_order], [by_col])
• array - Range or array to sort. sort_index - [optional] Column index to use for sorting.
• Excel 2021.
Writefile In Node Js With Code Examples
In this article, we will look at how to get the solution for the problem, Writefile In Node Js With Code Examples What is the difference between writeFileSync and writeFile? The only difference
between writeFile and writeFileSync is in catching and handling the errors; otherwise, all parameters mentioned are available in both functions. const fs = require('fs') const content = &#
x27;Some content!' try { fs.writeFileSync('/Users/joe/test.txt', content) //file written
Java 8 Group A Collections By 2 Property With Code Examples
In this article, we will look at how to get the solution for the problem, Java 8 Group A Collections By 2 Property With Code Examples What are the two types of streams? One method of classifying
streams is through physical, hydrological, and biological characteristics. Using these features, streams can fall into one of three types: perennial, intermittent, and ephemeral. Map<String, Map
<BlogPostType, List>> map = posts.stream() .collect(groupingBy(BlogPost::getAuthor, groupingBy(BlogPost::g
Simple Example Of Using Inline Javascript In Html With Code Examples
In this article, we will look at how to get the solution for the problem, Simple Example Of Using Inline Javascript In Html With Code Examples What is inline CSS? An inline CSS is used to apply a
unique style to a single HTML element. An inline CSS uses the style attribute of an HTML element. <!DOCTYPE html > <html> <head> <title>Softhunt.net</title> </head> <body> <p> <a href="#" onClick=
"alert('Welcome to Softhunt.net Tutorial Website');">Click Me</a> </p> <p> in this examp
Reach Every Char Java With Code Examples
In this article, we will look at how to get the solution for the problem, Reach Every Char Java With Code Examples What is nextFloat in Java? nextFloat() method scans the next token of the input as a
float. This method will throw InputMismatchException if the next token cannot be translated into a valid float value as described below. for (char ch: "xyz".toCharArray()) { } How do you read all
characters in a string in Java? To read a character in Java, we use next() method followed by charAt(
Split String On Character Vector C++ With Code Examples
In this article, we will look at how to get the solution for the problem, Split String On Character Vector C++ With Code Examples What is delimiter in C? In computer programming, a delimiter is a
character that identifies the beginning or the end of a character string (a contiguous sequence of characters). string s, tmp; stringstream ss(s); vector<string> words; // If there is one element (so
komma) then push the whole string if(getline(ss, tmp, ',').fail()) { words.push_back(s); } | {"url":"https://www.isnt.org.in/sort-stl-with-code-examples.html","timestamp":"2024-11-10T15:23:32Z","content_type":"text/html","content_length":"150239","record_id":"<urn:uuid:6e25374f-e109-4042-82a1-37a0e1ba06af>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00279.warc.gz"} |
Dyadic analysis filter bank
The dsp.DyadicAnalysisFilterBank System object™ decomposes a broadband signal into a collection of subbands with smaller bandwidths and slower sample rates. The System object uses a series of
highpass and lowpass FIR filters to provide approximate octave band frequency decompositions of the input. Each filter output is downsampled by a factor of two. With the appropriate analysis filters
and tree structure, the dyadic analysis filter bank is a discrete wavelet transform (DWT) or discrete wavelet packet transform (DWPT).
To obtain approximate octave band frequency decompositions of the input:
1. Create the dsp.DyadicAnalysisFilterBank object and set its properties.
2. Call the object with arguments, as if it were a function.
To learn more about how System objects work, see What Are System Objects?
dydan = dsp.DyadicAnalysisFilterBank constructs a dyadic analysis filter bank object, dydan, that computes the level-two discrete wavelet transform (DWT) of a column vector input. For a 2-D matrix
input, the object transforms the columns using the Daubechies third-order extremal phase wavelet. The length of the input along the first dimension must be a multiple of 4.
dydan = dsp.DyadicAnalysisFilterBank(Name,Value) returns a dyadic analysis filter bank object, with each property set to the specified value.
Unless otherwise indicated, properties are nontunable, which means you cannot change their values after calling the object. Objects lock when you call them, and the release function unlocks them.
If a property is tunable, you can change its value at any time.
For more information on changing property values, see System Design in MATLAB Using System Objects.
Filter — Type of filter used in subband decomposition
Custom (default) | Biorthogonal | Coiflets | Daubechies | Discrete Meyer | Haar | Reverse Biorthogonal | Symlets
Specify the type of filter used to determine the high and lowpass FIR filters in the dyadic analysis filter bank as Custom , Haar, Daubechies, Symlets, Coiflets, Biorthogonal, Reverse Biorthogonal,
or Discrete Meyer. All property values except Custom require Wavelet Toolbox™ software. If the value of this property is Custom, the filter coefficients are specified by the values of the
CustomLowpassFilter and CustomHighpassFilter properties. Otherwise, the dyadic analysis filter bank object uses the Wavelet Toolbox function wfilters to construct the filters. The following table
lists supported wavelet filters and example syntax to construct the filters:
Filter Example Setting Syntax for Analysis Filters
Haar N/A [Lo_D,Hi_D]=wfilters('haar');
Daubechies extremal phase WaveletOrder=3; [Lo_D,Hi_D]=wfilters('db3');
Symlets (Daubechies least-asymmetric) WaveletOrder=4; [Lo_D,Hi_D]=wfilters('sym4');
Coiflets WaveletOrder=1; [Lo_D,Hi_D]=wfilters('coif1');
Biorthogonal FilterOrder='[3/1]'; [Lo_D,Hi_D,Lo_R,Hi_R]=... wfilters('bior3.1');
Reverse biorthogonal FilterOrder='[3/1]'; [Lo_D,Hi_D,Lo_R,Hi_R]=... wfilters('rbior3.1');
Discrete Meyer N/A [Lo_D,Hi_D]=wfilters('dmey');
CustomLowpassFilter — Lowpass FIR filter coefficients
[0.0352 -0.0854 -0.1350 0.4599 0.8069 0.3327] (default) | row vector
Specify a vector of lowpass FIR filter coefficients, in powers of z^-1. Use a half-band filter that passes the frequency band stopped by the filter specified in the CustomHighpassFilter property. The
default specifies a Daubechies third-order extremal phase scaling (lowpass) filter.
This property applies when you set the Filter property to Custom.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64
CustomHighpassFilter — Highpass FIR filter coefficients
[-0.3327 0.8069 -0.4599 -0.1350 0.0854 0.0352] (default) | row vector
Specify a vector of highpass FIR filter coefficients, in powers of z^-1. Use a half-band filter that passes the frequency band stopped by the filter specified in the CustomLowpassFilter property. The
default specifies a Daubechies 3rd-order extremal phase wavelet (highpass) filter.
This property applies when you set the Filter property to Custom.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64
WaveletOrder — Order for orthogonal wavelets
2 (default) | positive integer
Specify the order of the wavelet selected in the Filter property.
This property applies when you set the Filter property to an orthogonal wavelet: Daubechies (Daubechies extremal phase), Symlets (Daubechies least-asymmetric), or Coiflets.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | logical | fi
FilterOrder — Analysis and synthesis filter orders for biorthogonal filters
1 / 1 (default) | 1 / 3 | 1 / 5 | 2 / 2 | 2 / 4 | 2 / 6 | 2 / 8 | 3 / 1 | 3 / 3 | 3 / 5 | 3 / 7 | 3 / 9 | 4 / 4 | 5 / 5 | 6 / 8
Specify the order of the analysis and synthesis filter orders for biorthogonal filter banks as 1 / 1, 1 / 3, 1 / 5, 2 / 2, 2 / 4, 2 / 6, 2 / 8, 3 / 1, 3 / 3, 3 / 5, 3 / 7, 3 / 9, 4 / 4, 5 / 5, or 6 /
8. Unlike orthogonal wavelets, biorthogonal wavelets require different filters for the analysis (decomposition) and synthesis (reconstruction) of an input. The first number indicates the order of the
synthesis (reconstruction) filter. The second number indicates the order of the analysis (decomposition) filter.
This property applies when you set the Filter property to Biorthogonal or Reverse Biorthogonal.
Data Types: char
NumLevels — Number of filter bank levels used in analysis (decomposition)
2 (default) | integer greater than or equal to 1
Specify the number of filter bank analysis levels a positive integer greater than or equal to 1. A level-N asymmetric structure produces N+1 output subbands. A level-N symmetric structure produces 2^
N output subbands. The size of the input along the first dimension must be a multiple of 2^N, where N is the number of levels.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64
TreeStructure — Structure of filter bank
Asymmetric (default) | Symmetric
Specify the structure of the filter bank as Asymmetric or Symmetric. The asymmetric structure decomposes only the lowpass filter output from each level. The symmetric structure decomposes the
highpass and lowpass filter outputs from each level. If the analysis filters are scaling (lowpass) and wavelet (highpass) filters, the asymmetric structure is the discrete wavelet transform, while
the symmetric structure is the discrete wavelet packet transform.
When this property is Symmetric, the output has 2^N subbands each of size M/2^N. In this case, M is the length of the input along the first dimension and N is the value of the NumLevels property.
When this property is Asymmetric, the output has N+1 subbands. The following equation gives the length of the output in the kth subband in the asymmetric case:
${M}_{k}=\left\{\begin{array}{ll}\frac{M}{{2}^{k}}\hfill & 1\le k\le N\hfill \\ \frac{M}{{2}^{N}}\hfill & k=N+1\hfill \end{array}$
y = dydan(x) computes the subband decomposition of the input x and outputs the dyadic subband decomposition in y as a single concatenated column vector or matrix of coefficients.
Input Arguments
x — Data input
column vector | matrix
Data input, specified as a column vector or a matrix. Each column of x is treated as an independent input, and the number of rows of x must be a multiple of ${2}^{N},$ where N is the number of levels
specified by the NumLevels property.
Data Types: single | double
Complex Number Support: Yes
Output Arguments
y — Dyadic subband decomposition output
column vector | matrix
Dyadic subband decomposition output, returned as a column vector or a matrix. The elements of y are ordered with the highest-frequency subband first followed by subbands in decreasing frequency.
When TreeStructure is set to Symmetric, the output has 2^N subbands each of size M/2^N. In this case, M is the length of the input along the first dimension, and N is the value of the NumLevels
property. When TreeStructure is set to Asymmetric, the output has N+1 subbands. The following equation gives the length of the output in the kth subband in the asymmetric case:
${M}_{k}=\left\{\begin{array}{ll}\frac{M}{{2}^{k}}\hfill & 1\le k\le N\hfill \\ \frac{M}{{2}^{N}}\hfill & k=N+1\hfill \end{array}$
Data Types: single | double
Complex Number Support: Yes
Object Functions
To use an object function, specify the System object as the first input argument. For example, to release system resources of a System object named obj, use this syntax:
Common to All System Objects
step Run System object algorithm
release Release resources and allow changes to System object property values and input characteristics
reset Reset internal states of System object
Filter Square Wave Using Dyadic Filter Banks
Denoise square wave input using dyadic analysis and synthesis filter banks.
t = 0:.0001:.0511;
x= square(2*pi*30*t);
xn = x' + 0.08*randn(length(x),1);
dydanl = dsp.DyadicAnalysisFilterBank;
The filter coefficients correspond to a haar wavelet.
dydanl.CustomLowpassFilter = [1/sqrt(2) 1/sqrt(2)];
dydanl.CustomHighpassFilter = [-1/sqrt(2) 1/sqrt(2)];
dydsyn = dsp.DyadicSynthesisFilterBank;
dydsyn.CustomLowpassFilter = [1/sqrt(2) 1/sqrt(2)];
dydsyn.CustomHighpassFilter = [1/sqrt(2) -1/sqrt(2)];
C = dydanl(xn);
Subband outputs.
C1 = C(1:256); C2 = C(257:384); C3 = C(385:512);
Set higher frequency coefficients to zero to remove the noise.
x_den = dydsyn([zeros(length(C1),1);...
Plot the original and denoised signals.
subplot(2,1,1), plot(xn); title('Original noisy Signal');
subplot(2,1,2), plot(x_den); title('Denoised Signal');
Subband Ordering For Asymmetric Tree Structure Using Dyadic Analysis Filter Bank
Sampling frequency 1 kHz input length 1024
t = 0:.001:1.023;
x = square(2*pi*30*t);
xn = x' + 0.08*randn(length(x),1);
Default asymmetric structure with order 3 extremal phase wavelet
dydan = dsp.DyadicAnalysisFilterBank;
Y = dydan(xn);
Level 2 yields 3 subbands (two detail-one approximation) Nyquist frequency is 500 Hz
D1 = Y(1:512); % subband approx. [250, 500] Hz
D2 = Y(513:768); % subband approx. [125, 250] Hz
Approx = Y(769:1024); % subband approx. [0,125] Hz
Subband Ordering For Symmetric Tree Structure Using Dyadic Analysis Filter Bank
Sampling frequency 1 kHz input length 1024.
t = 0:.001:1.023;
x = square(2*pi*30*t);
xn = x' + 0.08*randn(length(x),1);
dydan = dsp.DyadicAnalysisFilterBank('TreeStructure',...
Y = dydan(xn);
D1 = Y(1:256); % subband approx. [375,500] Hz
D2 = Y(257:512); % subband approx. [250,375] Hz
D3 = Y(513:768); % subband approx. [125,250] Hz
Approx = Y(769:1024); % subband approx. [0, 125] Hz
This object implements the algorithm, inputs, and outputs described on the Dyadic Analysis Filter Bank block reference page. The object properties correspond to the block parameters, except:
The dyadic analysis filter bank object always concatenates the subbands into a single column vector for a column vector input, or into the columns of a matrix for a matrix input. This behavior
corresponds to the block's behavior when you set the Output parameter to Single port.
Version History
Introduced in R2012a | {"url":"https://kr.mathworks.com/help/dsp/ref/dsp.dyadicanalysisfilterbank-system-object.html","timestamp":"2024-11-05T07:34:34Z","content_type":"text/html","content_length":"119303","record_id":"<urn:uuid:fb822780-55d9-4603-9999-6d45ad6c7c8c>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00367.warc.gz"} |
Question #12fd0 | Socratic
Question #12fd0
1 Answer
Your container will weigh $\text{3547.4 g}$, or $\text{3550 g}$ - rounded to three sig figs.
So, you have all the information you need to determine the weight of the container, including how much it weighs empty. However, notice that the dimensions of the container were given to you in
inches, while the density of the alcohol was given in $\text{g/mL}$.
This means that you must perform a unit conversion to get the proper units needed for density. Since you're dealing with a rectangular prism, the volume of the container will be
#V_("container") = "8.00 in" * "6.00 in" * "5.00 in" = 240. "in"^3#
I'll convert cubic inches to mililiters in order to get the proper unit for volume
$\text{240. in"^3 * ("16.387 mL")/("1 in"^3) = "3933 mL}$
Now use the formula for density to determine how much that volume of alcohol weighs
#rho = m/V => m_("alcohol") = rho * V_("container") = "0.86 g/mL" * "3933 mL"#
#m_("alcohol") = "3382 g"#
The total mass of the container will be the sum of the two masses
#m_("TOTAL") = m_("empty") + m_("alcohol") = "3547.4 g" = "3550 g"#
SIDE NOTE I recommend solving the problem by converting inches to centimeters. This will give you the volume in ${\text{cm}}^{3}$, which you'd then convert to $\text{mL}$$\to$$\text{1 cm"^3 = "1 mL}$
. The result has to be the same.
Impact of this question
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TY - JOUR T1 - Application of the Alternating Direction Method of Multipliers to Control Constrained Parabolic Optimal Control Problems and Beyond AU - Glowinski , Roland AU - Song , Yongcun AU -
Yuan , Xiaoming AU - Yue , Hangrui JO - Annals of Applied Mathematics VL - 2 SP - 115 EP - 158 PY - 2022 DA - 2022/04 SN - 38 DO - http://doi.org/10.4208/aam.OA-2022-0004 UR - https://global-sci.org/
intro/article_detail/aam/20452.html KW - Parabolic optimal control problem, control constraint, alternating direction method of multipliers, inexactness criterion, nested iteration, convergence
analysis. AB -
Control constrained parabolic optimal control problems are generally challenging, from either theoretical analysis or algorithmic design perspectives. Conceptually, the well-known alternating
direction method of multipliers (ADMM) can be directly applied to such problems. An attractive advantage of this direct ADMM application is that the control constraints can be untied from the
parabolic optimal control problem and thus can be treated individually in the iterations. At each iteration of the ADMM, the main computation is for solving an unconstrained parabolic optimal control
subproblem. Because of its inevitably high dimensionality after space-time discretization, the parabolic optimal control subproblem at each iteration can be solved only inexactly by implementing
certain numerical scheme internally and thus a two-layer nested iterative algorithm is required. It then becomes important to find an easily implementable and efficient inexactness criterion to
perform the internal iterations, and to prove the overall convergence rigorously for the resulting two-layer nested iterative algorithm. To implement the ADMM efficiently, we propose an inexactness
criterion that is independent of the mesh size of the involved discretization, and that can be performed automatically with no need to set empirically perceived constant accuracy a priori. The
inexactness criterion turns out to allow us to solve the resulting parabolic optimal control subproblems to medium or even low accuracy and thus save computation significantly, yet convergence of the
overall two-layer nested iterative algorithm can be still guaranteed rigorously. Efficiency of this ADMM implementation is promisingly validated by some numerical results. Our methodology can also be
extended to a range of optimal control problems modeled by other linear PDEs such as elliptic equations, hyperbolic equations, convection-diffusion equations, and fractional parabolic equations. | {"url":"https://global-sci.org/intro/article_detail/getRis?article_id=20452","timestamp":"2024-11-14T15:24:22Z","content_type":"text/html","content_length":"2863","record_id":"<urn:uuid:241ebc90-1776-4c00-bee2-56e54c9f132d>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00239.warc.gz"} |
In mathematics, especially in the area of abstract algebra known as module theory, an injective module is a module Q that shares certain desirable properties with the Z-module Q of all rational
numbers. Specifically, if Q is a submodule of some other module, then it is already a direct summand of that module; also, given a submodule of a module Y, any module homomorphism from this submodule
to Q can be extended to a homomorphism from all of Y to Q. This concept is dual to that of projective modules. Injective modules were introduced in (Baer 1940) and are discussed in some detail in the
textbook (Lam 1999, §3).
Injective modules have been heavily studied, and a variety of additional notions are defined in terms of them: Injective cogenerators are injective modules that faithfully represent the entire
category of modules. Injective resolutions measure how far from injective a module is in terms of the injective dimension and represent modules in the derived category. Injective hulls are maximal
essential extensions, and turn out to be minimal injective extensions. Over a Noetherian ring, every injective module is uniquely a direct sum of indecomposable modules, and their structure is well
understood. An injective module over one ring, may not be injective over another, but there are well-understood methods of changing rings which handle special cases. Rings which are themselves
injective modules have a number of interesting properties and include rings such as group rings of finite groups over fields. Injective modules include divisible groups and are generalized by the
notion of injective objects in category theory.
A left module Q over the ring R is injective if it satisfies one (and therefore all) of the following equivalent conditions:
• If Q is a submodule of some other left R-module M, then there exists another submodule K of M such that M is the internal direct sum of Q and K, i.e. Q + K = M and Q ∩ K = {0}.
• Any short exact sequence 0 →Q → M → K → 0 of left R-modules splits.
• If X and Y are left R-modules, f : X → Y is an injective module homomorphism and g : X → Q is an arbitrary module homomorphism, then there exists a module homomorphism h : Y → Q such that hf = g,
i.e. such that the following diagram commutes:
Injective right R-modules are defined in complete analogy.
First examples
Trivially, the zero module {0} is injective.
Given a field k, every k-vector space Q is an injective k-module. Reason: if Q is a subspace of V, we can find a basis of Q and extend it to a basis of V. The new extending basis vectors span a
subspace K of V and V is the internal direct sum of Q and K. Note that the direct complement K of Q is not uniquely determined by Q, and likewise the extending map h in the above definition is
typically not unique.
The rationals Q (with addition) form an injective abelian group (i.e. an injective Z-module). The factor group Q/Z and the circle group are also injective Z-modules. The factor group Z/nZ for n > 1
is injective as a Z/nZ-module, but not injective as an abelian group.
Commutative examples
More generally, for any integral domain R with field of fractions K, the R-module K is an injective R-module, and indeed the smallest injective R-module containing R. For any Dedekind domain, the
quotient module K/R is also injective, and its indecomposable summands are the localizations ${\displaystyle R_{\mathfrak {p}}/R}$ for the nonzero prime ideals ${\displaystyle {\mathfrak {p}}}$ . The
zero ideal is also prime and corresponds to the injective K. In this way there is a 1-1 correspondence between prime ideals and indecomposable injective modules.
A particularly rich theory is available for commutative noetherian rings due to Eben Matlis, (Lam 1999, §3I). Every injective module is uniquely a direct sum of indecomposable injective modules, and
the indecomposable injective modules are uniquely identified as the injective hulls of the quotients R/P where P varies over the prime spectrum of the ring. The injective hull of R/P as an R-module
is canonically an R[P] module, and is the R[P]-injective hull of R/P. In other words, it suffices to consider local rings. The endomorphism ring of the injective hull of R/P is the completion ${\
displaystyle {\hat {R}}_{P}}$ of R at P.^[1]
Two examples are the injective hull of the Z-module Z/pZ (the Prüfer group), and the injective hull of the k[x]-module k (the ring of inverse polynomials). The latter is easily described as k[x,x^−1]
/xk[x]. This module has a basis consisting of "inverse monomials", that is x^−n for n = 0, 1, 2, …. Multiplication by scalars is as expected, and multiplication by x behaves normally except that x·1
= 0. The endomorphism ring is simply the ring of formal power series.
Artinian examples
If G is a finite group and k a field with characteristic 0, then one shows in the theory of group representations that any subrepresentation of a given one is already a direct summand of the given
one. Translated into module language, this means that all modules over the group algebra kG are injective. If the characteristic of k is not zero, the following example may help.
If A is a unital associative algebra over the field k with finite dimension over k, then Hom[k](−, k) is a duality between finitely generated left A-modules and finitely generated right A-modules.
Therefore, the finitely generated injective left A-modules are precisely the modules of the form Hom[k](P, k) where P is a finitely generated projective right A-module. For symmetric algebras, the
duality is particularly well-behaved and projective modules and injective modules coincide.
For any Artinian ring, just as for commutative rings, there is a 1-1 correspondence between prime ideals and indecomposable injective modules. The correspondence in this case is perhaps even simpler:
a prime ideal is an annihilator of a unique simple module, and the corresponding indecomposable injective module is its injective hull. For finite-dimensional algebras over fields, these injective
hulls are finitely-generated modules (Lam 1999, §3G, §3J).
Computing injective hulls
If ${\displaystyle R}$ is a Noetherian ring and ${\displaystyle {\mathfrak {p}}}$ is a prime ideal, set ${\displaystyle E=E(R/{\mathfrak {p}})}$ as the injective hull. The injective hull of ${\
displaystyle R/{\mathfrak {p}}}$ over the Artinian ring ${\displaystyle R/{\mathfrak {p}}^{k}}$ can be computed as the module ${\displaystyle (0:_{E}{\mathfrak {p}}^{k})}$ . It is a module of the
same length as ${\displaystyle R/{\mathfrak {p}}^{k}}$ .^[2] In particular, for the standard graded ring ${\displaystyle R_{\bullet }=k[x_{1},\ldots ,x_{n}]_{\bullet }}$ and ${\displaystyle {\
mathfrak {p}}=(x_{1},\ldots ,x_{n})}$ , ${\displaystyle E=\oplus _{i}{\text{Hom}}(R_{i},k)}$ is an injective module, giving the tools for computing the indecomposable injective modules for artinian
rings over ${\displaystyle k}$ .
An Artin local ring ${\displaystyle (R,{\mathfrak {m}},K)}$ is injective over itself if and only if ${\displaystyle soc(R)}$ is a 1-dimensional vector space over ${\displaystyle K}$ . This implies
every local Gorenstein ring which is also Artin is injective over itself since has a 1-dimensional socle.^[3] A simple non-example is the ring ${\displaystyle R=\mathbb {C} [x,y]/(x^{2},xy,y^{2})}$
which has maximal ideal ${\displaystyle (x,y)}$ and residue field ${\displaystyle \mathbb {C} }$ . Its socle is ${\displaystyle \mathbb {C} \cdot x\oplus \mathbb {C} \cdot y}$ , which is
2-dimensional. The residue field has the injective hull ${\displaystyle {\text{Hom}}_{\mathbb {C} }(\mathbb {C} \cdot x\oplus \mathbb {C} \cdot y,\mathbb {C} )}$ .
Modules over Lie algebras
For a Lie algebra ${\displaystyle {\mathfrak {g}}}$ over a field ${\displaystyle k}$ of characteristic 0, the category of modules ${\displaystyle {\mathcal {M}}({\mathfrak {g}})}$ has a relatively
straightforward description of its injective modules.^[4] Using the universal enveloping algebra any injective ${\displaystyle {\mathfrak {g}}}$ -module can be constructed from the ${\displaystyle {\
mathfrak {g}}}$ -module
${\displaystyle {\text{Hom}}_{k}(U({\mathfrak {g}}),V)}$
for some ${\displaystyle k}$ -vector space ${\displaystyle V}$ . Note this vector space has a ${\displaystyle {\mathfrak {g}}}$ -module structure from the injection
${\displaystyle {\mathfrak {g}}\hookrightarrow U({\mathfrak {g}})}$
In fact, every ${\displaystyle {\mathfrak {g}}}$ -module has an injection into some ${\displaystyle {\text{Hom}}_{k}(U({\mathfrak {g}}),V)}$ and every injective ${\displaystyle {\mathfrak {g}}}$
-module is a direct summand of some ${\displaystyle {\text{Hom}}_{k}(U({\mathfrak {g}}),V)}$ .
Structure theorem for commutative Noetherian rings
Over a commutative Noetherian ring ${\displaystyle R}$ , every injective module is a direct sum of indecomposable injective modules and every indecomposable injective module is the injective hull of
the residue field at a prime ${\displaystyle {\mathfrak {p}}}$ . That is, for an injective ${\displaystyle I\in {\text{Mod}}(R)}$ , there is an isomorphism
${\displaystyle I\cong \bigoplus _{i}E(R/{\mathfrak {p}}_{i})}$
where ${\displaystyle E(R/{\mathfrak {p}}_{i})}$ are the injective hulls of the modules ${\displaystyle R/{\mathfrak {p}}_{i}}$ .^[5] In addition, if ${\displaystyle I}$ is the injective hull of some
module ${\displaystyle M}$ then the ${\displaystyle {\mathfrak {p}}_{i}}$ are the associated primes of ${\displaystyle M}$ .^[2]
Submodules, quotients, products, and sums, Bass-Papp Theorem
Any product of (even infinitely many) injective modules is injective; conversely, if a direct product of modules is injective, then each module is injective (Lam 1999, p. 61). Every direct sum of
finitely many injective modules is injective. In general, submodules, factor modules, or infinite direct sums of injective modules need not be injective. Every submodule of every injective module is
injective if and only if the ring is Artinian semisimple (Golan & Head 1991, p. 152); every factor module of every injective module is injective if and only if the ring is hereditary, (Lam 1999, Th.
Bass-Papp Theorem states that every infinite direct sum of right (left) injective modules is injective if and only if the ring is right (left) Noetherian, (Lam 1999, p. 80-81, Th 3.46).^[6]
Baer's criterion
In Baer's original paper, he proved a useful result, usually known as Baer's Criterion, for checking whether a module is injective: a left R-module Q is injective if and only if any homomorphism g :
I → Q defined on a left ideal I of R can be extended to all of R.
Using this criterion, one can show that Q is an injective abelian group (i.e. an injective module over Z). More generally, an abelian group is injective if and only if it is divisible. More generally
still: a module over a principal ideal domain is injective if and only if it is divisible (the case of vector spaces is an example of this theorem, as every field is a principal ideal domain and
every vector space is divisible). Over a general integral domain, we still have one implication: every injective module over an integral domain is divisible.
Baer's criterion has been refined in many ways (Golan & Head 1991, p. 119), including a result of (Smith 1981) and (Vámos 1983) that for a commutative Noetherian ring, it suffices to consider only
prime ideals I. The dual of Baer's criterion, which would give a test for projectivity, is false in general. For instance, the Z-module Q satisfies the dual of Baer's criterion but is not projective.
Injective cogenerators
Maybe the most important injective module is the abelian group Q/Z. It is an injective cogenerator in the category of abelian groups, which means that it is injective and any other module is
contained in a suitably large product of copies of Q/Z. So in particular, every abelian group is a subgroup of an injective one. It is quite significant that this is also true over any ring: every
module is a submodule of an injective one, or "the category of left R-modules has enough injectives." To prove this, one uses the peculiar properties of the abelian group Q/Z to construct an
injective cogenerator in the category of left R-modules.
For a left R-module M, the so-called "character module" M^+ = Hom[Z](M,Q/Z) is a right R-module that exhibits an interesting duality, not between injective modules and projective modules, but between
injective modules and flat modules (Enochs & Jenda 2000, pp. 78–80). For any ring R, a left R-module is flat if and only if its character module is injective. If R is left noetherian, then a left R
-module is injective if and only if its character module is flat.
Injective hulls
The injective hull of a module is the smallest injective module containing the given one and was described in (Eckmann & Schopf 1953).
One can use injective hulls to define a minimal injective resolution (see below). If each term of the injective resolution is the injective hull of the cokernel of the previous map, then the
injective resolution has minimal length.
Injective resolutions
Every module M also has an injective resolution: an exact sequence of the form
0 → M → I^0 → I^1 → I^2 → ...
where the I^ j are injective modules. Injective resolutions can be used to define derived functors such as the Ext functor.
The length of a finite injective resolution is the first index n such that I^n is nonzero and I^i = 0 for i greater than n. If a module M admits a finite injective resolution, the minimal length
among all finite injective resolutions of M is called its injective dimension and denoted id(M). If M does not admit a finite injective resolution, then by convention the injective dimension is said
to be infinite. (Lam 1999, §5C) As an example, consider a module M such that id(M) = 0. In this situation, the exactness of the sequence 0 → M → I^0 → 0 indicates that the arrow in the center is an
isomorphism, and hence M itself is injective.^[7]
Equivalently, the injective dimension of M is the minimal integer (if there is such, otherwise ∞) n such that Ext^N
[A](–,M) = 0 for all N > n.
Every injective submodule of an injective module is a direct summand, so it is important to understand indecomposable injective modules, (Lam 1999, §3F).
Every indecomposable injective module has a local endomorphism ring. A module is called a uniform module if every two nonzero submodules have nonzero intersection. For an injective module M the
following are equivalent:
• M is indecomposable
• M is nonzero and is the injective hull of every nonzero submodule
• M is uniform
• M is the injective hull of a uniform module
• M is the injective hull of a uniform cyclic module
• M has a local endomorphism ring
Over a Noetherian ring, every injective module is the direct sum of (uniquely determined) indecomposable injective modules. Over a commutative Noetherian ring, this gives a particularly nice
understanding of all injective modules, described in (Matlis 1958). The indecomposable injective modules are the injective hulls of the modules R/p for p a prime ideal of the ring R. Moreover, the
injective hull M of R/p has an increasing filtration by modules M[n] given by the annihilators of the ideals p^n, and M[n+1]/M[n] is isomorphic as finite-dimensional vector space over the quotient
field k(p) of R/p to Hom[R/p](p^n/p^n+1, k(p)).
Change of rings
It is important to be able to consider modules over subrings or quotient rings, especially for instance polynomial rings. In general, this is difficult, but a number of results are known, (Lam 1999,
p. 62).
Let S and R be rings, and P be a left-R, right-S bimodule that is flat as a left-R module. For any injective right S-module M, the set of module homomorphisms Hom[S]( P, M ) is an injective right R
-module. The same statement holds of course after interchanging left- and right- attributes.
For instance, if R is a subring of S such that S is a flat R-module, then every injective S-module is an injective R-module. In particular, if R is an integral domain and S its field of fractions,
then every vector space over S is an injective R-module. Similarly, every injective R[x]-module is an injective R-module.
In the opposite direction, a ring homomorphism ${\displaystyle f:S\to R}$ makes R into a left-R, right-S bimodule, by left and right multiplication. Being free over itself R is also flat as a left R
-module. Specializing the above statement for P = R, it says that when M is an injective right S-module the coinduced module ${\displaystyle f_{*}M=\mathrm {Hom} _{S}(R,M)}$ is an injective right R
-module. Thus, coinduction over f produces injective R-modules from injective S-modules.
For quotient rings R/I, the change of rings is also very clear. An R-module is an R/I-module precisely when it is annihilated by I. The submodule ann[I](M) = { m in M : im = 0 for all i in I } is a
left submodule of the left R-module M, and is the largest submodule of M that is an R/I-module. If M is an injective left R-module, then ann[I](M) is an injective left R/I-module. Applying this to R=
Z, I=nZ and M=Q/Z, one gets the familiar fact that Z/nZ is injective as a module over itself. While it is easy to convert injective R-modules into injective R/I-modules, this process does not convert
injective R-resolutions into injective R/I-resolutions, and the homology of the resulting complex is one of the early and fundamental areas of study of relative homological algebra.
The textbook (Rotman 1979, p. 103) has an erroneous proof that localization preserves injectives, but a counterexample was given in (Dade 1981).
Self-injective rings
Every ring with unity is a free module and hence is a projective as a module over itself, but it is rarer for a ring to be injective as a module over itself, (Lam 1999, §3B). If a ring is injective
over itself as a right module, then it is called a right self-injective ring. Every Frobenius algebra is self-injective, but no integral domain that is not a field is self-injective. Every proper
quotient of a Dedekind domain is self-injective.
A right Noetherian, right self-injective ring is called a quasi-Frobenius ring, and is two-sided Artinian and two-sided injective, (Lam 1999, Th. 15.1). An important module theoretic property of
quasi-Frobenius rings is that the projective modules are exactly the injective modules.
Generalizations and specializations
Injective objects
One also talks about injective objects in categories more general than module categories, for instance in functor categories or in categories of sheaves of O[X]-modules over some ringed space (X,O[X
]). The following general definition is used: an object Q of the category C is injective if for any monomorphism f : X → Y in C and any morphism g : X → Q there exists a morphism h : Y → Q with hf =
Divisible groups
The notion of injective object in the category of abelian groups was studied somewhat independently of injective modules under the term divisible group. Here a Z-module M is injective if and only if
n⋅M = M for every nonzero integer n. Here the relationships between flat modules, pure submodules, and injective modules is more clear, as it simply refers to certain divisibility properties of
module elements by integers.
Pure injectives
In relative homological algebra, the extension property of homomorphisms may be required only for certain submodules, rather than for all. For instance, a pure injective module is a module in which a
homomorphism from a pure submodule can be extended to the whole module.
Primary sources | {"url":"https://www.knowpia.com/knowpedia/Injective_module","timestamp":"2024-11-09T01:30:39Z","content_type":"text/html","content_length":"198754","record_id":"<urn:uuid:13da08b7-ddba-488e-a4cd-96caeab0c82f>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00717.warc.gz"} |
Do you know how the cylinder speed is calculated?
Calculation method of cylinder speed The speed of the cylinder piston is related to the air pressure, cylinder diameter, friction, and external resistance during the entire movement process, so the
speed changes. The relationship between the operating speed of the piston and the speed is the speed control valve (exhaust throttle valve), which is often used to adjust the speed of the cylinder.
It is generally recognized that the maximum speed of the cylinder when there is no load is the theoretical reference speed, and the speed of the cylinder will gradually decrease with the increase of
the load during use.
The average speed of the cylinder is divided by the movement stroke of the cylinder by the action time of the cylinder, so the action time of the cylinder should be measured first. It is almost
impossible to calculate its average speed by borrowing the formula, but it can be approximated. The relationship between it and the speed is: the speed is 1.4 times the average speed. When the
cylinder is not loaded, it is assumed that the exhaust side exhausts at the speed of sound (Note: the exhaust air flow is almost unobstructed), the theoretical reference speed of the cylinder: u0 =
1920 * S/A (mm/s) S is the effective synthesis of the exhaust circuit The cross-sectional area, A is the effective area of the piston on the exhaust side. At this time, the speed of the cylinder is
approximately equal to u0. The standard speed range of the cylinder is 50-500mm/s. Under normal circumstances, the speed of the cylinder is adjusted to the standard speed range through the throttle
valve, which is related to the load on the cylinder.
When the speed is less than 50mm/s, which is estimated to be 20-30mm/s, due to the increased influence of the frictional resistance of the cylinder and the compressibility of the gas, the smooth
movement of the piston cannot be guaranteed, and the phenomenon of stop and go will appear. for "crawling". In order for the cylinder to work at low speed, a gas-liquid damping cylinder should be
used, or a gas-liquid combined cylinder should be used for low-speed control through a gas-liquid converter. When the speed is higher than 500mm/s, the friction and heat generation of the cylinder
sealing ring will increase, which will accelerate the wear of the seals, cause air leakage, shorten the service life, and increase the impact force at the end of the stroke, which will affect the
mechanical life. In order to work at higher speeds, it is necessary to lengthen the length of the cylinder barrel, improve the machining accuracy of the cylinder barrel, improve the material of the
sealing ring to reduce frictional resistance, and improve the cushioning performance.
Therefore, it is not good to be too large or too small, and a very suitable one should be selected, and it should be within the specified range. Taking the SMC cylinder as an example, the ordinary
CM2 cylinder runs at a speed of 50-750mm/s, the speed control valve is fully open, and can reach 750mm/s, and the speed control valve is closed smaller, and the speed can reach below 50mm/s, and then
it will crawl (shakes and shakes). The air velocity is generally about 30m/s. The working pressure of the cylinder is generally 0.6 MPa, and it can be used normally at 0.3-0.8 MPa. | {"url":"http://en.allgreat.net.cn/news/439.html","timestamp":"2024-11-03T01:07:30Z","content_type":"text/html","content_length":"41078","record_id":"<urn:uuid:c54d2584-31ae-44c9-a3d3-fcea0da95abf>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00830.warc.gz"} |
Dhiya is painting her living room the total area that nees to be covered is 64.5 square feet each can of paint coat 15.25 and covers an area of 20square ft. How many cans will dhiya need to paint all 4 walls
How much will the paint coat her
Dhiya is painting her living room the total area that nees to be covered is 64.5 square feet each can of paint coat 15.25 and covers an area of 20square ft. How many cans will dhiya need to paint all
4 walls How much will the paint coat her
785 views
Answer to a math question Dhiya is painting her living room the total area that nees to be covered is 64.5 square feet each can of paint coat 15.25 and covers an area of 20square ft. How many cans
will dhiya need to paint all 4 walls How much will the paint coat her
84 Answers
1. Determine the number of cans needed by dividing the total area to be covered by the area each can covers:
\frac{64.5}{20} = 3.225
2. Since Dhiya cannot purchase a fraction of a can, she will need to round up to the nearest whole number, which is 4 cans.
3. Calculate the total cost by multiplying the number of cans by the cost per can:
4 \times 15.25 = 61.00
Therefore, Dhiya will need 4 cans of paint, and the paint will cost her $61.00.
Frequently asked questions (FAQs)
Question: Find the value of sin(2Ο /3) + cos(5Ο /6) - tan(Ο /4) * cot(Ο /3).
Math question: Find the value of sin(60Β°) + cos(45Β°).
Math question: What is the measure of the angle at the center of a circle, subtended by an arc that is one-third the circumference of the circle? | {"url":"https://math-master.org/general/dhiya-is-painting-her-living-room-the-total-area-that-nees-to-be-covered-is-64-5-square-feet-each-can-of-paint-coat-15-25-and-covers-an-area-of-20square-ft-how-many-cans-will-dhiya-need-to-paint-all","timestamp":"2024-11-12T12:27:48Z","content_type":"text/html","content_length":"246463","record_id":"<urn:uuid:ee47851f-89c3-40e5-8d97-ed10b2fa78ae>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00706.warc.gz"} |
On additive vertex labelings
In a quite general sense, additive vertex labelings are those functions that assign nonnegative integers to the vertices of a graph and the weight of each edge is obtained by adding the labels of its
end-vertices. In this work we study one of these functions, called harmonious labeling. We calculate the number of non-isomorphic harmoniously labeled graphs with n edges and at most n vertices. We
present harmonious labelings for some families of graphs that include certain unicyclic graphs obtained via the corona product. In addition, we prove that all n-cell snake polyiamonds are harmonious;
this type of graph is obtained via edge amalgamation of n copies of the cycle C[3] in such a way that each copy of this cycle shares at most two edges with other copies. Moreover, we use the
edge-switching technique on the cycle C[4t ]to generate unicyclic graphs with another type of additive vertex labeling, called strongly felicitous, which has a solid bond with the harmonious
additive vertex labeling; harmonious; corona product; unicyclic; polyiamonds
Full Text:
G. Agnarsson and R. Greenlaw, Graph Theory: Modeling, Applications, and Algorithms, Pearson, (2006).
M. Baca and M. Miller, Super Edge-Antimagic Graphs: A Wealth of Problems and Some Solutions, BrownWalker Press, 2007, Boca Raton, FL, USA
C. Barrientos, Graceful graphs with pendant edges, Australas. J. Combin. 33 (2005) 99--107.
C. Barrientos and S. Minion, Snakes: from graceful to harmonious, Bull. Institute Combin. Appl., 79 (2017) 95--107.
C. Barrientos, https://oeis.org/A005418, 2018.
C. Barrientos, https://oeis.org/A329910, 2019.
C. Barrientos, Special graceful labelings of irregular fences and lobsters, Universal Journal of Mathematics and Applications, 2 (1) (2019) 1--10.
C. Barrientos and S. Minion, Counting and labeling grid related graphs, Electronic Journal of Graph Theory and Applications, 7 (2) (2019), 349--363.
R. Cattell, Graceful labellings of paths, Discrete Math., 307 (2007) 3161--3176.
G. J. Chang, D. F. Hsu, and D. G. Rogers, Additive variations on a graceful theme: some results on harmonious and other related graphs, Cong. Numer., 32 (1981), 181--197.
G. Chartrand and L. Lesniak, Graphs & Digraphs 4th ed. CRC Press (2005).
R. Figueroa-Centeno, R. Ichishima, and F. Muntaner-Batle, The place of super edgemagic labelings among other classes of labelings, Discrete Math., 231 (2001) 153--168.
R. Figueroa-Centeno, R. Ichishima, and F. Muntaner-Batle, Labeling the vertex amalgamation of graphs, Discuss. Math. Graph Theory, 23 (2003), 129--139.
J. A. Gallian, A dynamic survey of graph labeling, Electron. J. Combin., (2019), #DS6.
T. Grace, On sequential labelings of graphs, J. Graph Theory, 7 (1983) 195--201.
R. L. Graham and N. J. A. Sloane, On additive bases and harmonious graphs, SIAM J. Alg. Discrete Methods, 1 (1980) 382--404.
A. Kotzig, On certain vertex valuations of finite graphs, Util. Math., 4 (1973) 67--73.
B. Liu and X. Zhang, On a conjecture of harmonious graphs, Systems Science and Math. Sciences, 4 (1989) 325--328.
S. C. Lopez and F. A. Muntaner-Batle, Graceful, Harmonious and Magic Type Labelings: Relations and Techniques, Springer, Cham, 2017.
J. Renuka, P. Balaganesan, P. Selvaraju, On harmonious labeling, Internat. J. Advance in Math. Math. Sci., 1(2) (2012) 65--70.
A. Rosa, On certain valuations of the vertices of a graph, Theory of Graphs (Internat. Symposium, Rome, July 1966), Gordon and Breach, N. Y. and Dunod Paris (1967) 349-355.
M. A. Seoud, A. E. I. Abdel Maqsoud and J. Sheehan, Harmonious graphs, Util. Math., 47 (1995) 225--233.
G. Sethuraman, V. Murugan, Generating graceful unicyclic graphs from a given forest, to appear in AKCE Int. J. of Graphs and Combinatorics.
M. Truszcyński, Graceful unicyclic graphs, Demonstratio Mathematica, 17 (1984), 377-387.
W. D. Wallis, Magic Graphs, Birkhauser, Boston, 2001.
Y. Yang, W. Lu, and Q. Zeng, Harmonious graphs C_{2k} cup C_{2j+1}, Util. Math., 62 (2002) 191--198.
M. Z. Youssef, Two general results on harmonious labelings, Ars Combin., 68 (2003) 225--230.
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Motivic equivalence for algebraic groups and critical varieties
(by Charles De Clercq & Anne Quéguiner-Mathieu)
The Chow motif of a projective homogeneous variety under some algebraic group contains some information on the splitting properties of the underlying algebraic objects. Considering all projective
homogeneous varieties under the action of a given group leads to the notion of motivic equivalence for algebraic groups. The aim of this course is to state and prove a criterion of motivic
equivalence in terms of Tits’ indices of algebraic groups, and to prove the existence of critical varieties, which are test-varieties for motivic equivalence.
The course will focus on algebraic groups of classical type. The first two lectures will cover required material such as classification of classical algebraic groups, twisted flag varieties, Tits’s
indices, Chow motives, Rost’s nilpotence principle, and upper motives.
p-group actions and Chern numbers of varieties (by Olivier Haution)
The course will concern the study of actions of finite p-groups on algebraic varieties, and more precisely the use of certain numerical invariants of varieties to detect fixed points. We will thus
discuss various fixed point theorems, present methods to prove them, and illustrate them by applications and examples. Among those numerical invariants are the Chern numbers, whose consideration will
lead us to introduce the cobordism ring. We will review an elementary approach to cobordism due to Merkurjev, and illustrate how the cobordism ring can be used to interpret the fixed point theorems,
and more generally to understand better how the geometry of the fixed locus is related to that of the ambient variety (time permitting).
As prerequisites we will assume familiarity with basic algebraic geometry, the Chow group and K-theory (only K_0)
Motives without A^1-invariance (by Vova Sosnilo)
A motivic oo-category contains all the information about certain cohomology theories on schemes and forgets all the non-additive information the category of schemes has. More precisely, such an
additive oo-category should admit a functor from the category of schemes and every cohomology theory on schemes in an appropriate sense should factor through this oo-category.
One common requirement for being a cohomology theory in this context is the A^1-invariance property. The oo-category of A^1-invariant motivic spectra has been studied extensively over the past 20
years, which gave us fertile soil for studying K-theory, hermitian K-theory, algebraic cobordism and many other cohomology theories of regular schemes. However, over non-regular schemes K-theory is
not A^1-invariant and some new methods are being called for.
The goal of this short series of lectures is to construct new motivic oo-categories of non-A1-invariant motivic spectra, based on the work of Annala and Iwasa, and to show how these can be used to
prove new results about K-theory of non-regular schemes.
Isotropic motives (by Alexander Vishik)
The homological properties of algebraic varieties are encoded in their motives. These can be considered as linearizations of varieties. The category of motives, although much handier than the
category of varieties themselves, is still pretty large and complicated. One may try to read motivic information by applying some realization functor with values in a small and well-understood
category. One possibility is to consider the topological realization, thus replacing algebraic varieties by topological spaces of their complex points (if the ground field is embedded into complex
numbers). The motivic version of this functor takes values in the category of "topological motives'', which is the derived category of abelian groups. This category is small and simple, but the
functor looses a lot of information. One would want to supplement it with other similar realization functors, so that the resulting family would be reasonably conservative. Isotropic realizations
provide a large supply of such functors. These are parametrized by prime numbers and (equivalence classes of) finitely generated field extensions of the ground field.
I will start by introducing Chow motives and motivic cohomology and recalling basic facts about them. Then I will move to anisotropic varieties and flexible fields and will introduce isotropic
realizations. A particular attention will be paid to isotropic Chow motives, where Hom's are described by isotropic Chow groups. The latter groups should coincide with Chow groups (with finite
coefficents) modulo numerical equivalence, and so, are much simpler than the usual Chow groups. I will discuss various corollaries of this conjecture and will prove it for divisors. One of the
consequences is that isotropic realizations should provide points for the Balmer's tensor triangulated spectrum of the Voevodsky category. Finally, I will introduce Čech simplicial schemes and will
discuss the calculation of the isotropic motivic cohomology of a point, for p=2. | {"url":"https://www.uni-regensburg.de/mathematics/motives2022/abstracts/index.html","timestamp":"2024-11-03T20:22:27Z","content_type":"text/html","content_length":"14243","record_id":"<urn:uuid:7bb9576b-ba17-41e9-893f-8831466d14d7>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00224.warc.gz"} |
Geometry (NCTM)
Use visualization, spatial reasoning, and geometric modeling to solve problems.
Use geometric models to represent and explain numerical and algebraic relationships.
Measurement (NCTM)
Apply appropriate techniques, tools, and formulas to determine measurements.
Select and apply techniques and tools to accurately find length, area, volume, and angle measures to appropriate levels of precision.
Develop and use formulas to determine the circumference of circles and the area of triangles, parallelograms, trapezoids, and circles and develop strategies to find the area of more-complex shapes.
Connections to the Grade 7 Focal Points (NCTM)
Measurement and Geometry: Students connect their work on proportionality with their work on area and volume by investigating similar objects. They understand that if a scale factor describes how
corresponding lengths in two similar objects are related, then the square of the scale factor describes how corresponding areas are related, and the cube of the scale factor describes how
corresponding volumes are related. Students apply their work on proportionality to measurement in different contexts, including converting among different units of measurement to solve problems
involving rates such as motion at a constant speed. They also apply proportionality when they work with the circumference, radius, and diameter of a circle; when they find the area of a sector of a
circle; and when they make scale drawings. | {"url":"https://newpathworksheets.com/math/grade-7/measurement-perimeter-and-circumference?dictionary=angles&did=54","timestamp":"2024-11-13T05:09:48Z","content_type":"text/html","content_length":"46703","record_id":"<urn:uuid:02025358-6719-4b39-81ef-0372419c82a7>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00854.warc.gz"} |
Angle Properties Of Triangles Worksheet - TraingleWorksheets.com
Angle Properties Of Triangles Worksheet – Triangles are one of the fundamental shapes in geometry. Understanding the triangle is essential to understanding more advanced geometric principles. In this
blog we will explore the different kinds of triangles and triangle angles, as well as how to calculate the length and width of a triangle, and offer specific examples on each. Types of Triangles
There are three kinds of triangulars: Equilateral, isosceles, and scalene. Equilateral triangles include three … Read more | {"url":"https://www.traingleworksheets.com/tag/angle-properties-of-triangles-worksheet/","timestamp":"2024-11-13T13:15:23Z","content_type":"text/html","content_length":"47714","record_id":"<urn:uuid:0c0d16b0-7b3e-4844-bab4-3249c60e7b1f>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00514.warc.gz"} |
Lesson 106: Fundamental Trigonometric Identities
Lesson: Propositional Logic and Truth Tables
Exploring the Foundations of Logical Reasoning
In this lesson, we delve into the basics of propositional logic and the use of truth tables. Understanding these fundamental concepts is essential for developing logical reasoning skills, which are
crucial in various fields such as computer science, mathematics, and philosophy.
1. Introduction to Propositional Logic
• Definition and Importance:
□ Propositional Logic: Study of propositions and their logical relationships and connections.
□ Applications: Used in computer science for algorithm design, in mathematics for proofs, and in philosophy for logical argumentation.
• Basic Components:
□ Propositions: Statements that are either true or false.
□ Logical Connectives: Symbols used to connect propositions, such as AND (∧), OR (∨), NOT (¬), IMPLIES (→), and IF AND ONLY IF (↔).
2. Constructing Propositions
• Simple Propositions:
□ Examples: “It is raining,” “The sky is blue.”
□ Truth Value: Each proposition has a truth value, either true (T) or false (F).
• Compound Propositions:
□ Combining Simple Propositions: Use logical connectives to form compound propositions.
□ Examples: “It is raining AND the sky is blue,” “It is raining OR the sky is blue.”
3. Logical Connectives and Their Meaning
• AND (∧):
□ Definition: True if both propositions are true.
□ Example: “P ∧ Q” is true if both P and Q are true.
• OR (∨):
□ Definition: True if at least one proposition is true.
□ Example: “P ∨ Q” is true if either P or Q (or both) are true.
• NOT (¬):
□ Definition: True if the proposition is false.
□ Example: “¬P” is true if P is false.
• IMPLIES (→):
□ Definition: True if the first proposition implies the second.
□ Example: “P → Q” is true if P is false or Q is true.
• IF AND ONLY IF (↔):
□ Definition: True if both propositions are either true or false.
□ Example: “P ↔ Q” is true if both P and Q are true or both are false.
4. Constructing Truth Tables
• Purpose and Structure:
□ Purpose: Used to determine the truth value of compound propositions.
□ Structure: Table that lists all possible truth values of the component propositions and the resulting truth value of the compound proposition.
• Creating Truth Tables:
□ Step-by-Step Process:
☆ List all possible truth values for the component propositions.
☆ Apply logical connectives to determine the truth value of the compound proposition.
□ Example: Constructing a truth table for “P ∧ Q” and “P → Q.”
5. Applications and Problem-Solving
• Logic in Computer Science:
□ Algorithm Design: Use logic to design and analyze algorithms.
□ Programming: Apply logic in writing and debugging code.
• Mathematical Proofs:
□ Formal Proofs: Use logical reasoning to construct and validate mathematical proofs.
• Philosophical Arguments:
□ Critical Thinking: Apply logical reasoning to analyze and construct philosophical arguments.
• Understand Propositional Logic: Comprehend the basics of propositional logic and its components.
• Construct Truth Tables: Learn how to create and use truth tables to analyze logical propositions.
• Apply Logical Reasoning: Develop the ability to apply logical reasoning in various contexts, including computer science, mathematics, and philosophy.
• Solve Logical Problems: Enhance problem-solving skills through the application of logical concepts and techniques.
• Critical Questions: What are the key components of propositional logic? How do logical connectives affect the truth value of compound propositions? How can truth tables be used to analyze logical
statements? What are the practical applications of propositional logic in different fields?
• Thematic Focus: Emphasize the importance of logical reasoning and its applications in real-world scenarios.
• Connection to Future Learning: Highlight how mastering propositional logic and truth tables will be beneficial for more advanced topics in logic, computer science, and mathematics.
By understanding propositional logic and mastering the construction of truth tables, students will develop essential logical reasoning skills. These skills are fundamental for success in various
academic and professional fields, enabling students to analyze and solve complex problems effectively.
Lesson: Propositional Logic and Truth Tables
Understanding the Foundations of Logical Reasoning
LaTeX Code:
\section*{Lesson: Propositional Logic and Truth Tables}
\textbf{Understanding the Foundations of Logical Reasoning}
\textbf{Example of Truth Table for Conjunction (AND)}:
$P$ & $Q$ & $P \land Q$ \\
T & T & T \\
T & F & F \\
F & T & F \\
F & F & F \\
above is from the July 3 2024 rewrites: https://chatgpt.com/c/34f2f62d-8b01-4999-8e57-8f49a0ed9cab | {"url":"https://pioneeronlineacademy.org/topics/lesson-106-fundamental-trigonometric-identities/","timestamp":"2024-11-03T09:08:20Z","content_type":"text/html","content_length":"160237","record_id":"<urn:uuid:97baca6d-20c2-4a35-92d6-8bbfbde52d59>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00828.warc.gz"} |
Relationship with Other Mathematical Concepts (e.g., Vectors, Matrices) in context of fractional distance
31 Aug 2024
Title: Exploring the Relationship between Fractional Distance and Other Mathematical Concepts: A Theoretical Analysis
Fractional distance, a concept that has garnered significant attention in recent years, is a mathematical construct that generalizes traditional notions of distance and similarity measures. This
article delves into the relationship between fractional distance and other fundamental mathematical concepts, including vectors, matrices, and geometric transformations. We derive novel formulas and
establish theoretical connections between these concepts, providing a deeper understanding of the underlying mathematics.
Fractional distance, denoted as $d_f(x,y)$, is a measure that quantifies the similarity or dissimilarity between two points or objects in a metric space. It has been applied in various fields,
including computer science, physics, and engineering. In this article, we investigate the relationship between fractional distance and other mathematical concepts, with a focus on vectors, matrices,
and geometric transformations.
Relationship with Vectors:
A vector $v$ can be represented as an ordered pair $(x,y)$ of real numbers. The fractional distance between two vectors $v_1 = (x_1,y_1)$ and $v_2 = (x_2,y_2)$ is given by:
$d_f(v_1,v_2) = \frac{ x_1-x_2 + y_1-y_2 }{\sqrt{2}}$
This formula can be seen as a generalization of the Euclidean distance between two points in 2D space.
Relationship with Matrices:
A matrix $M$ is a rectangular array of numbers. The fractional distance between two matrices $M_1$ and $M_2$ can be defined as:
$d_f(M_1,M_2) = \frac{\sum_{i=1}^n\sum_{j=1}^m m_{ij}-m’_{ij} }{nm}$
where $m_{ij}$ and $m’_{ij}$ are the elements of $M_1$ and $M_2$, respectively.
Relationship with Geometric Transformations:
A geometric transformation, such as rotation or scaling, can be represented by a matrix. The fractional distance between two transformed points $x’$ and $y’$, which were originally at positions $x$
and $y$, is given by:
$d_f(x’,y’) = \frac{ x’-x + y’-y }{\sqrt{2}}$
This formula can be seen as a generalization of the Euclidean distance between two points in 2D space, taking into account the geometric transformation.
In this article, we have explored the relationship between fractional distance and other fundamental mathematical concepts, including vectors, matrices, and geometric transformations. We have derived
novel formulas and established theoretical connections between these concepts, providing a deeper understanding of the underlying mathematics. These results can be applied in various fields,
including computer science, physics, and engineering.
[1] [Author], “Fractional Distance: A Generalized Measure for Similarity and Dissimilarity”, Journal of Mathematical Analysis, 2022.
[2] [Author], “Geometric Transformations and Fractional Distance”, Journal of Geometry, 2020.
Related articles for ‘fractional distance’ :
• Reading: Relationship with Other Mathematical Concepts (e.g., Vectors, Matrices) in context of fractional distance
Calculators for ‘fractional distance’ | {"url":"https://blog.truegeometry.com/tutorials/education/8d68811c91f3d270029b8aa3cb6adc24/JSON_TO_ARTCL_Relationship_with_Other_Mathematical_Concepts_e_g_Vectors_Matr.html","timestamp":"2024-11-13T11:49:40Z","content_type":"text/html","content_length":"17085","record_id":"<urn:uuid:1d53d1d9-b97a-4c64-9d9a-33694c2bb64e>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00040.warc.gz"} |
Geometric Brownian motion (GBM) model
Creates and displays a geometric Brownian motion model (GBM), which derives from the cev (constant elasticity of variance) class.
Geometric Brownian motion (GBM) models allow you to simulate sample paths of NVars state variables driven by NBrowns Brownian motion sources of risk over NPeriods consecutive observation periods,
approximating continuous-time GBM stochastic processes. Specifically, this model allows the simulation of vector-valued GBM processes of the form
$d{X}_{t}=\mu \left(t\right){X}_{t}dt+D\left(t,{X}_{t}\right)V\left(t\right)d{W}_{t}$
• X[t] is an NVars-by-1 state vector of process variables.
• μ is an NVars-by-NVars generalized expected instantaneous rate of return matrix.
• D is an NVars-by-NVars diagonal matrix, where each element along the main diagonal is the corresponding element of the state vector X[t].
• V is an NVars-by-NBrowns instantaneous volatility rate matrix.
• dW[t] is an NBrowns-by-1 Brownian motion vector.
GBM = gbm(Return,Sigma) creates a default GBM object.
Specify the required input parameters as one of the following types:
• A MATLAB^® array. Specifying an array indicates a static (non-time-varying) parametric specification. This array fully captures all implementation details, which are clearly associated with a
parametric form.
• A MATLAB function. Specifying a function provides indirect support for virtually any static, dynamic, linear, or nonlinear model. This parameter is supported via an interface, because all
implementation details are hidden and fully encapsulated by the function.
You can specify combinations of array and function input parameters as needed.
Moreover, a parameter is identified as a deterministic function of time if the function accepts a scalar time t as its only input argument. Otherwise, a parameter is assumed to be a function of time
t and state X(t) and is invoked with both input arguments.
GBM = gbm(___,Name,Value) creates a GBM object with additional options specified by one or more Name,Value pair arguments.
Name is a property name and Value is its corresponding value. Name must appear inside single quotes (''). You can specify several name-value pair arguments in any order as Name1,Value1,…,NameN,ValueN
The GBM object has the following Properties:
• StartTime — Initial observation time
• StartState — Initial state at StartTime
• Correlation — Access function for the Correlation input, callable as a function of time
• Drift — Composite drift-rate function, callable as a function of time and state
• Diffusion — Composite diffusion-rate function, callable as a function of time and state
• Simulation — A simulation function or method
• Return — Access function for the input argument Return, callable as a function of time and state
• Sigma — Access function for the input argument Sigma, callable as a function of time and state
Input Arguments
Return — Return represents the parameter μ
array or deterministic function of time or deterministic function of time and state
Return represents the parameter μ, specified as an array or deterministic function of time.
If you specify Return as an array, it must be an NVars-by-NVars matrix representing the expected (mean) instantaneous rate of return.
As a deterministic function of time, when Return is called with a real-valued scalar time t as its only input, Return must produce an NVars-by-NVars matrix. If you specify Return as a function of
time and state, it must return an NVars-by-NVars matrix when invoked with two inputs:
• A real-valued scalar observation time t.
• An NVars-by-1 state vector X[t].
Data Types: double | function_handle
Sigma — Sigma represents the parameter V
array or deterministic function of time or deterministic function of time and state
Sigma represents the parameter V, specified as an array or a deterministic function of time.
If you specify Sigma as an array, it must be an NVars-by-NBrowns matrix of instantaneous volatility rates or as a deterministic function of time. In this case, each row of Sigma corresponds to a
particular state variable. Each column corresponds to a particular Brownian source of uncertainty, and associates the magnitude of the exposure of state variables with sources of uncertainty.
As a deterministic function of time, when Sigma is called with a real-valued scalar time t as its only input, Sigma must produce an NVars-by-NBrowns matrix. If you specify Sigma as a function of time
and state, it must return an NVars-by-NBrowns matrix of volatility rates when invoked with two inputs:
• A real-valued scalar observation time t.
• An NVars-by-1 state vector X[t].
Although the gbm object enforces no restrictions on the sign of Sigma volatilities, they are specified as positive values.
Data Types: double | function_handle
StartTime — Starting time of first observation, applied to all state variables
0 (default) | scalar
Starting time of first observation, applied to all state variables, specified as a scalar
Data Types: double
StartState — Initial values of state variables
1 (default) | scalar, column vector, or matrix
Initial values of state variables, specified as a scalar, column vector, or matrix.
If StartState is a scalar, the gbm object applies the same initial value to all state variables on all trials.
If StartState is a column vector, the gbm object applies a unique initial value to each state variable on all trials.
If StartState is a matrix, the gbm object applies a unique initial value to each state variable on each trial.
Data Types: double
Correlation — Correlation between Gaussian random variates drawn to generate the Brownian motion vector (Wiener processes)
NBrowns-by-NBrowns identity matrix representing independent Gaussian processes (default) | positive semidefinite matrix | deterministic function
Correlation between Gaussian random variates drawn to generate the Brownian motion vector (Wiener processes), specified as an NBrowns-by-NBrowns positive semidefinite matrix, or as a deterministic
function C(t) that accepts the current time t and returns an NBrowns-by-NBrowns positive semidefinite correlation matrix. If Correlation is not a symmetric positive semidefinite matrix, use nearcorr
to create a positive semidefinite matrix for a correlation matrix.
A Correlation matrix represents a static condition.
As a deterministic function of time, Correlation allows you to specify a dynamic correlation structure.
Data Types: double
Simulation — User-defined simulation function or SDE simulation method
simulation by Euler approximation (simByEuler) (default) | function | SDE simulation method
User-defined simulation function or SDE simulation method, specified as a function or SDE simulation method.
Data Types: function_handle
Drift — Drift rate component of continuous-time stochastic differential equations (SDEs)
value stored from drift-rate function (default) | drift object or function accessible by (t, X[t])
This property is read-only.
Drift rate component of continuous-time stochastic differential equations (SDEs), specified as a drift object or function accessible by (t, X[t].
The drift rate specification supports the simulation of sample paths of NVars state variables driven by NBrowns Brownian motion sources of risk over NPeriods consecutive observation periods,
approximating continuous-time stochastic processes.
The drift class allows you to create drift-rate objects (using drift) of the form:
• A is an NVars-by-1 vector-valued function accessible using the (t, X[t]) interface.
• B is an NVars-by-NVars matrix-valued function accessible using the (t, X[t]) interface.
The displayed parameters for a drift object are:
• Rate: The drift-rate function, F(t,X[t])
• A: The intercept term, A(t,X[t]), of F(t,X[t])
• B: The first order term, B(t,X[t]), of F(t,X[t])
A and B enable you to query the original inputs. The function stored in Rate fully encapsulates the combined effect of A and B.
When specified as MATLAB double arrays, the inputs A and B are clearly associated with a linear drift rate parametric form. However, specifying either A or B as a function allows you to customize
virtually any drift rate specification.
You can express drift and diffusion classes in the most general form to emphasize the functional (t, X[t]) interface. However, you can specify the components A and B as functions that adhere to the
common (t, X[t]) interface, or as MATLAB arrays of appropriate dimension.
Example: F = drift(0, 0.1) % Drift rate function F(t,X)
Data Types: struct | double
Diffusion — Diffusion rate component of continuous-time stochastic differential equations (SDEs)
value stored from diffusion-rate function (default) | diffusion object or functions accessible by (t, X[t])
Diffusion rate component of continuous-time stochastic differential equations (SDEs), specified as a drift object or function accessible by (t, X[t].
The diffusion rate specification supports the simulation of sample paths of NVars state variables driven by NBrowns Brownian motion sources of risk over NPeriods consecutive observation periods,
approximating continuous-time stochastic processes.
The diffusion class allows you to create diffusion-rate objects (using diffusion):
$G\left(t,{X}_{t}\right)=D\left(t,{X}_{t}^{\alpha \left(t\right)}\right)V\left(t\right)$
• D is an NVars-by-NVars diagonal matrix-valued function.
• Each diagonal element of D is the corresponding element of the state vector raised to the corresponding element of an exponent Alpha, which is an NVars-by-1 vector-valued function.
• V is an NVars-by-NBrowns matrix-valued volatility rate function Sigma.
• Alpha and Sigma are also accessible using the (t, X[t]) interface.
The diffusion object's displayed parameters are:
• Rate: The diffusion-rate function, G(t,X[t]).
• Alpha: The state vector exponent, which determines the format of D(t,X[t]) of G(t,X[t]).
• Sigma: The volatility rate, V(t,X[t]), of G(t,X[t]).
Alpha and Sigma enable you to query the original inputs. (The combined effect of the individual Alpha and Sigma parameters is fully encapsulated by the function stored in Rate.) The Rate functions
are the calculation engines for the drift and diffusion objects, and are the only parameters required for simulation.
You can express drift and diffusion classes in the most general form to emphasize the functional (t, X[t]) interface. However, you can specify the components A and B as functions that adhere to the
common (t, X[t]) interface, or as MATLAB arrays of appropriate dimension.
Example: G = diffusion(1, 0.3) % Diffusion rate function G(t,X)
Data Types: struct | double
Object Functions
interpolate Brownian interpolation of stochastic differential equations (SDEs) for SDE, BM, GBM, CEV, CIR, HWV, Heston, SDEDDO, SDELD, or SDEMRD models
simulate Simulate multivariate stochastic differential equations (SDEs) for SDE, BM, GBM, CEV, CIR, HWV, Heston, SDEDDO, SDELD, SDEMRD, Merton, or Bates models
simByEuler Euler simulation of stochastic differential equations (SDEs) for SDE, BM, GBM, CEV, CIR, HWV, Heston, SDEDDO, SDELD, or SDEMRD models
simBySolution Simulate approximate solution of diagonal-drift GBM processes
simByMilstein Simulate diagonal diffusion for BM, GBM, CEV, HWV, SDEDDO, SDELD, or SDEMRD sample paths by Milstein approximation
simByMilstein2 Simulate BM, GBM, CEV, HWV, SDEDDO, SDELD, SDEMRD process sample paths by second order Milstein approximation
Create a gbm Object
Create a univariate gbm object to represent the model: $d{X}_{t}=0.25{X}_{t}dt+0.3{X}_{t}d{W}_{t}$.
obj = gbm(0.25, 0.3) % (B = Return, Sigma)
obj =
Class GBM: Generalized Geometric Brownian Motion
Dimensions: State = 1, Brownian = 1
StartTime: 0
StartState: 1
Correlation: 1
Drift: drift rate function F(t,X(t))
Diffusion: diffusion rate function G(t,X(t))
Simulation: simulation method/function simByEuler
Return: 0.25
Sigma: 0.3
gbm objects display the parameter B as the more familiar Return
Compute Price of European Option Using Monte Carlo Simulation with GBM Object
This example shows the workflow to compute the price of a European option using Monte Carlo simulation with a gbm object.
Set up the parameters for the Geometric Brownian Motion (GBM) model and the European option.
% Parameters for the GBM model and option
S0 = 100; % Initial stock price
K = 110; % Strike price
T = 1; % Time to maturity in years
r = 0.05; % Risk-free interest rate
sigma = 0.20; % Volatility
nTrials = 10000; % Number of Monte Carlo trials
nPeriods = 1; % Number of periods (for one year, this can be set to 1)
Create a gbm object.
% Create GBM object
gbmobj = gbm(r,sigma,'StartState',S0);
Use simulate to simulate the end-of-year stock prices using the GBM model (gbm) over nTrials trials.
% Simulate stock prices at maturity
[Paths, ~, ~] = simulate(gbmobj,nPeriods,'nTrials',nTrials,'DeltaTime',T);
% Extract the final prices for all trials
ST = Paths(end, :, 1);
Calculate the payoff for European call and put options based on the simulated prices.
% Calculate payoffs for call and put options
callPayoff = max(ST - K, 0)
putPayoff = max(K - ST, 0)
Discount these payoffs back to the present value and then average the payoff values to estimate the option prices.
% Discount payoffs back to present value and average
callPrice = exp(-r * T) * mean(callPayoff)
putPrice = exp(-r * T) * mean(putPayoff)
Display the estimated prices for both the call and put options.
% Display results
fprintf('European Call Option Price: %.4f\n', callPrice);
European Call Option Price: 5.4727
fprintf('European Put Option Price: %.4f\n', putPrice);
European Put Option Price: 0.0000
When you specify the required input parameters as arrays, they are associated with a specific parametric form. By contrast, when you specify either required input parameter as a function, you can
customize virtually any specification.
Accessing the output parameters with no inputs simply returns the original input specification. Thus, when you invoke these parameters with no inputs, they behave like simple properties and allow you
to test the data type (double vs. function, or equivalently, static vs. dynamic) of the original input specification. This is useful for validating and designing methods.
When you invoke these parameters with inputs, they behave like functions, giving the impression of dynamic behavior. The parameters accept the observation time t and a state vector X[t], and return
an array of appropriate dimension. Even if you originally specified an input as an array, gbm treats it as a static function of time and state, by that means guaranteeing that all parameters are
accessible by the same interface.
[1] Aït-Sahalia, Yacine. “Testing Continuous-Time Models of the Spot Interest Rate.” Review of Financial Studies, vol. 9, no. 2, Apr. 1996, pp. 385–426.
[2] Aït-Sahalia, Yacine. “Transition Densities for Interest Rate and Other Nonlinear Diffusions.” The Journal of Finance, vol. 54, no. 4, Aug. 1999, pp. 1361–95.
[3] Glasserman, Paul. Monte Carlo Methods in Financial Engineering. Springer, 2004.
[4] Hull, John. Options, Futures and Other Derivatives. 7th ed, Prentice Hall, 2009.
[5] Johnson, Norman Lloyd, et al. Continuous Univariate Distributions. 2nd ed, Wiley, 1994.
[6] Shreve, Steven E. Stochastic Calculus for Finance. Springer, 2004.
Version History
Introduced in R2008a
R2023b: Added simByMilstein2 method
Use the simByMilstein2 method to approximate a numerical solution of a stochastic differential equation.
R2023a: Added simByMilstein method
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FABRIC LOCKING TAPE 100M X 6MM
FABRIC LOCKING TAPE 100M X 6MM
How to order
Please order the number of items you require:
If batch quantity is 100:
State quantity as 100 for a full box of 100
State quantity as 50 if you require a cut length of 50 units
State quantity as 120 if you require a full box of 100 units + a cut length of 20 units
If batch quantity is 25 sets:
State quantity as 25 sets for a full box of 25 sets
State quantity as 5 if you require a cut length of 5 sets
State quantity as 60 if you require 2 x full boxes of 25 sets + a cut length of 10 sets
If batch quantity is 50m supplied in 5m lengths:
State quantity as 50m for a full box of 50m
State quantity as 10 if you require a cut length of 10m
State quantity as 65 if you require a full box of 50m + a cut length of 15m
Not available as a cut length | {"url":"https://tradeportal.louvolite.com/components/fabric-locking-tape-100m-x-6mm","timestamp":"2024-11-09T20:29:26Z","content_type":"text/html","content_length":"87787","record_id":"<urn:uuid:36e8a301-d570-474c-b942-b9ea67d4034b>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00741.warc.gz"} |
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network theory ebooks index1. March16,20 onthe28thofapril2012thecontentsoftheenglishaswellasgermanwikibooksandwikipedia projectswerelicensedundercreativecommonsattributionsharealike3. Jun 08, 2019
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How to Create a Prediction Interval in R | Online Tutorials Library List | Tutoraspire.com
How to Create a Prediction Interval in R
by Tutor Aspire
A linear regression model can be useful for two things:
(1) Quantifying the relationship between one or more predictor variables and a response variable.
(2)Â Using the model to predict future values.
In regards to (2), when we use a regression model to predict future values, we are often interested in predicting both an exact value as well as an interval that contains a range of likely
values. This interval is known as a prediction interval.
For example, suppose we fit a simple linear regression model using hours studied as a predictor variable and exam score as the response variable. Using this model, we might predict that a student
who studies for 6 hours will receive an exam score of 91.
However, because there is uncertainty around this prediction, we might create a prediction interval that says there is a 95% chance that a student who studies for 6 hours will receive an exam score
between 85 and 97. This range of values is known as a 95% prediction interval and it’s often more useful to us than just knowing the exact predicted value.
How to Create a Prediction Interval in R
To illustrate how to create a prediction interval in R, we will use the built-in mtcars dataset, which contains information about characteristics of several different cars:
#view first six rows of mtcars
# mpg cyl disp hp drat wt qsec vs am gear carb
#Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
#Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
#Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
#Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
#Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
#Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
First, we’ll fit a simple linear regression model using disp as the predictor variable and mpg as the response variable.
#fit simple linear regression model
model #view summary of fitted model
#lm(formula = mpg ~ disp, data = mtcars)
# Min 1Q Median 3Q Max
#-4.8922 -2.2022 -0.9631 1.6272 7.2305
# Estimate Std. Error t value Pr(>|t|)
#(Intercept) 29.599855 1.229720 24.070
Then, we’ll use the fitted regression model to predict the value of mpg based on three new values for disp.Â
#create data frame with three new values for disp
#use the fitted model to predict the value for mpg based on the three new values
#for disp
predict(model, newdata = new_disp)
# 1 2 3
#23.41759 21.35683 19.29607
The way to interpret these values is as follows:
• For a new car with a disp of 150, we predict that it will have a mpg of 23.41759.Â
• For a new car with a disp of 200, we predict that it will have a mpg of 21.35683 .Â
• For a new car with a disp of 250, we predict that it will have a mpg of 19.29607.Â
Next, we’ll use the fitted regression model to make prediction intervals around these predicted values:
#create prediction intervals around the predicted values
predict(model, newdata = new_disp, interval = "predict")
# fit lwr upr
#1 23.41759 16.62968 30.20549
#2 21.35683 14.60704 28.10662
#3 19.29607 12.55021 26.04194
The way to interpret these values is as follows:
• The 95% prediction interval of the mpg for a car with a disp of 150 is between 16.62968 and 30.20549.
• The 95% prediction interval of the mpg for a car with a disp of 200 is between 14.60704 and 28.10662.
• The 95% prediction interval of the mpg for a car with a disp of 250 is between 12.55021 and 26.04194.
By default, R uses a 95% prediction interval. However, we can change this to whatever we’d like using the level command. For example, the following code illustrates how to create 99% prediction
#create 99% prediction intervals around the predicted values
predict(model, newdata = new_disp, interval = "predict", level = 0.99)
# fit lwr upr
#1 23.41759 14.27742 32.55775
#2 21.35683 12.26799 30.44567
#3 19.29607 10.21252 28.37963
Note that the 99% prediction intervals are wider than the 95% prediction intervals. This makes sense because the wider the interval, the higher the likelihood that it will contain the predicted
How to Visualize a Prediction Interval in R
The following code illustrates how to create a chart with the following features:
• A scatterplot of the data points for disp and mpg
• A blue line for the fitted regression line
• Gray confidence bands
• Red prediction bands
#define dataset
data #create simple linear regression model
model #use model to create prediction intervals
predictions predict")
#create dataset that contains original data along with prediction intervals
all_data #load ggplot2 library
#create plot
ggplot(all_data, aes(x = disp, y = mpg)) + #define x and y axis variables
geom_point() + #add scatterplot points
stat_smooth(method = lm) + #confidence bands
geom_line(aes(y = lwr), col = "coral2", linetype = "dashed") + #lwr pred interval
geom_line(aes(y = upr), col = "coral2", linetype = "dashed") #upr pred interval
When to Use a Confidence Interval vs. a Prediction Interval
A prediction interval captures the uncertainty around a single value. A confidence interval captures the uncertainty around the mean predicted values. Thus, a prediction interval will always be
wider than a confidence interval for the same value.
You should use a prediction interval when you are interested in specific individual predictions because a confidence interval will produce too narrow of a range of values, resulting in a greater
chance that the interval will not contain the true value.
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Scattering Problem: PDE Modeler App
This example shows how to solve a simple scattering problem, where you compute the waves reflected by a square object illuminated by incident waves that are coming from the left. This example uses
the PDE Modeler app. For the programmatic workflow, see Scattering Problem.
For this problem, assume an infinite horizontal membrane subjected to small vertical displacements U. The membrane is fixed at the object boundary. The medium is homogeneous, and the phase velocity
(propagation speed) of a wave, α, is constant. The wave equation is
$\frac{{\partial }^{2}U}{\partial {t}^{2}}-{\alpha }^{2}\Delta U=0$
The solution U is the sum of the incident wave V and the reflected wave R:
When the illumination is harmonic in time, you can compute the field by solving a single steady problem. Assume that the incident wave is a plane wave traveling in the –x direction:
$V\left(x,y,t\right)={e}^{i\left(-kx-\omega t\right)}={e}^{-ikx}{e}^{-i\omega t}$
The reflected wave can be decomposed into spatial and time components:
$R\left(x,y,t\right)=r\left(x,y\right){e}^{-i\omega t}$
Now you can rewrite the wave equation as the Helmholtz equation for the spatial component of the reflected wave with the wave number k = ω/α:
The Dirichlet boundary condition for the boundary of the object is U = 0, or in terms of the incident and reflected waves, R = -V. For the time-harmonic solution and the incident wave traveling in
the –x direction, you can write this boundary condition as follows:
The reflected wave R travels outward from the object. The condition at the outer computational boundary must allow waves to pass without reflection. Such conditions are usually called nonreflecting.
As $|\stackrel{\to }{x}|$ approaches infinity, R approximately satisfies the one-way wave equation
$\frac{\partial R}{\partial t}+\alpha \stackrel{\to }{\xi }\text{\hspace{0.17em}}·\text{\hspace{0.17em}}abla R=0$
This equation considers only the waves moving in the positive ξ-direction. Here, ξ is the radial distance from the object. With the time-harmonic solution, this equation turns into the generalized
Neumann boundary condition
$\stackrel{\to }{\xi }\text{\hspace{0.17em}}·\text{\hspace{0.17em}}abla r=ikr$
To solve the scattering problem in the PDE Modeler app, follow these steps:
1. Open the PDE Modeler app by using the pdeModeler command.
2. Set the x-axis limit to [0.1 1.5] and the y-axis limit to [0 1]. To do this, select Options > Axes Limits and set the corresponding ranges.
3. Display grid lines. To do this:
1. Select Options > Grid Spacing and clear the Auto checkboxes.
2. Enter X-axis linear spacing as 0.1:0.05:1.5 and Y-axis linear spacing as 0:0.05:1.
3. Select Options > Grid.
4. Align new shapes to the grid lines by selecting Options > Snap.
5. Draw a square with sides of length 0.1 and a center in [0.8 0.5]. To do this, first click the button. Then right-click the origin and drag to draw a square. Right-clicking constrains the shape
you draw so that it is a square rather than a rectangle. If the square is not a perfect square, double-click it. In the resulting dialog box, specify the exact location of the bottom left corner
and the side length.
6. Rotate the square by 45 degrees. To do this, select Draw > Rotate... and enter 45 in the resulting dialog box. The rotated square represents the illuminated object.
7. Draw a circle with a radius of 0.45 and a center in [0.8 0.5]. To do this, first click the button. Then right-click the origin and drag to draw a circle. Right-clicking constrains the shape you
draw so that it is a circle rather than an ellipse. If the circle is not a perfect unit circle, double-click it. In the resulting dialog box, specify the exact center location and radius of the
8. Model the geometry by entering C1-SQ1 in the Set formula field.
9. Check that the application mode is set to Generic Scalar.
10. Specify the boundary conditions. To do this, switch to the boundary mode by selecting Boundary > Boundary Mode. Use Shift+click to select several boundaries. Then select Boundary > Specify
Boundary Conditions.
□ For the perimeter of the circle, the boundary condition is the Neumann boundary condition with q = -ik, where the wave number k = 60 corresponds to a wavelength of about 0.1 units. Enter g =
0 and q = -60*i.
□ For the perimeter of the square, the boundary condition is the Dirichlet boundary condition:
$r=-v\left(x,y\right)=-{e}^{ik\stackrel{\to }{a}\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}\stackrel{\to }{x}}$
In this problem, because the reflected wave travels in the –x direction, the boundary condition is r = –e^–ikx. Enter h = 1 and r = -exp(-i*60*x).
11. Specify the coefficients by selecting PDE > PDE Specification or clicking the button on the toolbar. The Helmholtz equation is a wave equation, but in Partial Differential Equation Toolbox™ you
can treat it as an elliptic equation with a = -k^2. Specify c = 1, a = -3600, and f = 0.
12. Initialize the mesh by selecting Mesh > Initialize Mesh.
For sufficient accuracy, you need about 10 finite elements per wavelength. The outer boundary must be located a few object diameters away from the object itself. Refine the mesh by selecting Mesh
> Refine Mesh. Refine the mesh two more times to achieve the required resolution.
13. Solve the PDE by selecting Solve > Solve PDE or clicking the button on the toolbar.
The solution is complex. When plotting the solution, you get a warning message.
14. Plot the reflected waves. Change the colormap to jet by selecting Plot > Parameters and then selecting jet from the Colormap drop-down menu.
15. Animate the solution for the time-dependent wave equation. To do this:
1. Export the mesh data and the solution to the MATLAB^® workspace by selecting Mesh > Export Mesh and Solve > Export Solution, respectively.
2. Enter the following commands in the MATLAB Command Window.
maxu = max(abs(u));
m = 10;
for j = 1:m,
uu = real(exp(-j*2*pi/10*sqrt(-1))*u);
caxis([-maxu maxu]);
axis tight
ax = gca;
ax.DataAspectRatio = [1 1 1];
axis off
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How can we compare several populations with unknown distributions (the Kruskal-Wallis test)?
7. Product and Process Comparisons
7.4. Comparisons based on data from more than two processes
7.4.1. How can we compare several populations with unknown distributions (the Kruskal-Wallis test)?
The Kruskal-Wallis (KW) Test for Comparing Populations with Unknown Distributions
A nonparametric test for The KW procedure tests the null hypothesis that \(k\) samples from possibly different populations actually originate from similar populations, at least as far as
comparing population medians by their central tendencies, or medians, are concerned. The test assumes that the variables under consideration have underlying continuous distributions.
Kruskal and Wallis
In what follows assume we have \(k\) samples, and the sample size of the \(i\)-th sample is \(n_i, \,\, i=1, \, 2, \, \ldots, \, k\).
Test based on ranks of combined In the computation of the KW statistic, each observation is replaced by its rank in an ordered combination of all the \(k\) samples. By this we mean that the data
data from the \(k\) samples combined are ranked in a single series. The minimum observation is replaced by a rank of 1, the next-to-the-smallest by a rank of 2, and the
largest or maximum observation is replaced by the rank of \(N\), where \(N\) is the total number of observations in all the samples (\(N\) is the sum of the \(n_i
Compute the sum of the ranks for The next step is to compute the sum of the ranks for each of the original samples. The KW test determines whether these sums of ranks are so different by sample that
each sample they are not likely to have all come from the same population.
Test statistic follows a \(\chi^ It can be shown that if the \(k\) samples come from the same population, that is, if the null hypothesis is true, then the test statistic, \(H\), used in the KW
2\) distribution procedure is distributed approximately as a chi-square statistic with df = \(k-1\), provided that the sample sizes of the \(k\) samples are not too small (say, \(n_i
> 4\), for all \(i\)). \(H\) is defined as follows: $$ H = \frac{12}{N(N+1)} \, \sum_{i=1}^k \frac{R_i^2}{n_i} - 3(N+1) \, , $$ where
• \(k\) = number of samples (groups)
• \(n_i\) = number of observations for the \(i\)-th sample or group
• \(N\) = total number of observations (sum of all the \(n_i\))
• \(R_i\) = sum of ranks for group \(i\)
An illustrative example The following data are from a comparison of four investment firms. The observations represent percentage of growth during a three month period.for recommended funds.
A B C D
4.2 3.3 1.9 3.5
4.6 2.4 2.4 3.1
3.9 2.6 2.1 3.7
4.0 3.8 2.7 4.1
2.8 1.8 4.4
Step 1: Express the data in terms of their ranks
A B C D
19 4.5 4.5 9
SUM 65 41.5 17.5 66
Compute the test statistic The corresponding \(H\) test statistic is $$ H = \frac{12}{19(20)} \left[ \frac{65^2}{4} + \frac{41.5^2}{5} + \frac{17.5^2}{5} + \frac{66^2}{5} \right] - 3(20) =
13.678 \, . $$ From the chi-square table in Chapter 1, the critical value for 1 - \(\alpha\) = 0.95 with df = \(k\) - 1 = 3 is 7.812. Since 13.678 > 7.812, we reject
the null hypothesis.
Note that the rejection region for the KW procedure is one-sided, since we only reject the null hypothesis when the \(H\) statistic is too large. | {"url":"https://itl.nist.gov/div898/handbook/prc/section4/prc41.htm","timestamp":"2024-11-06T19:55:01Z","content_type":"text/html","content_length":"10492","record_id":"<urn:uuid:b0e8dc13-d70d-44d8-b4d2-6b63fd209597>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00459.warc.gz"} |
How do you solve x+3y=y+110 for y? | HIX Tutor
How do you solve #x+3y=y+110# for y?
Answer 1
Solve #x+3y=y+110# for #y#.
Subract #x# from both sides of the equation.
Subtract #y# from both sides.
Simplify #3y-y# to #2y#.
Divide both sides by #2#.
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Answer 2
To solve ( x + 3y = y + 110 ) for ( y ), you can start by isolating the variable ( y ) on one side of the equation. Subtract ( y ) from both sides of the equation:
( x + 3y - y = 110 )
This simplifies to:
( x + 2y = 110 )
Then, subtract ( x ) from both sides of the equation:
( x - x + 2y = 110 - x )
This simplifies to:
( 2y = 110 - x )
Finally, divide both sides of the equation by 2 to solve for ( y ):
( \frac{2y}{2} = \frac{110 - x}{2} )
This gives:
( y = \frac{110 - x}{2} )
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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Contrasts and Multiple Testing
3 Contrasts and Multiple Testing
3.1 Contrasts
3.1.1 Introduction
The \(F\)-test is rather unspecific. It basically gives us a “Yes/No” answer to the question: “Is there any treatment effect at all?”. It does not tell us what specific treatment or treatment
combination is special. Quite often, we have a more specific question than the aforementioned global null hypothesis. For example, we might want to compare a set of new treatments vs. a control
treatment or we want to do pairwise comparisons between many (or all) treatments.
Such kinds of questions can typically be formulated as a so-called contrast. Let us start with a toy example based on the PlantGrowth data set. If we only wanted to compare trt1 (\(\mu_2\)) with ctrl
(\(\mu_1\)), we could set up the null hypothesis \[ H_0: \mu_1 - \mu_2 = 0 \] vs. the alternative \[ H_A: \mu_1 - \mu_2 \neq 0. \] We can encode this with a vector \(c \in \mathbb{R}^g\) \[ H_0: \
sum_{i=1}^g c_i \mu_i = 0. \tag{3.1}\] In this example, we have \(g = 3\) and the vector \(c\) is given by \(c = (1, -1, 0)\), with respect to ctrl, trt1 and trt2. Hence, a contrast is nothing more
than an encoding of our own specific research question. A more sophisticated example would be \(c = (1, -1/2, -1/2)\) which compares ctrl vs. the average value of trt1 and trt2 which we would write
as \(H_0: \mu_1 - \frac{1}{2}(\mu_2 + \mu_3) = 0\).
Typically, we have the side constraint \[ \sum_{i=1}^g c_i = 0 \] which ensures that the contrast is about differences between treatments and not about the overall level of the response.
A contrast can also be thought of as one-dimensional “aspect” of the multi-dimensional treatment effect, if we have \(g > 2\) different treatments.
We estimate the corresponding true, but unknown, value \(\sum_{i=1}^g c_i \mu_i\) (a linear combination of model parameters!) with \[ \sum_{i=1}^g c_i \widehat{\mu}_i = \sum_{i=1}^g c_i \overline{y}_
{i\cdot} \,. \] In addition, we could derive its accuracy (standard error), construct confidence intervals and do tests. We omit the theoretical details and continue with our example.
In R, we use the function glht (general linear hypotheses) of the package multcomp (Hothorn, Bretz, and Westfall 2008). It uses the fitted one-way ANOVA model, which we refit here for the sake of
fit.plant <- aov(weight ~ group, data = PlantGrowth)
We first have to specify the contrast for the factor group with the function mcp (multiple comparisons; for the moment we only consider a single test here) and use the corresponding output as
argument linfct (linear function) in glht. All these steps together look as follows:
## ...
## Linear Hypotheses:
## Estimate Std. Error t value Pr(>|t|)
## 1 == 0 -0.0615 0.2414 -0.25 0.8
## ...
This means that we estimate the difference between ctrl and the average value of trt1 and trt2 as \(-0.0615\) and we are not rejecting the null hypothesis because the p-value is large. The annotation
1 == 0 means that this line tests whether the first (here, and only) contrast is zero or not (if needed, we could also give a custom name to each contrast). We get a confidence interval by using the
function confint.
## ...
## Linear Hypotheses:
## Estimate lwr upr
## 1 == 0 -0.0615 -0.5569 0.4339
Hence, the 95% confidence interval for \(\mu_1- \frac{1}{2}(\mu_2 + \mu_3)\) is given by \([-0.5569, 0.4339]\).
An alternative to package multcomp is package emmeans. One way of getting statistical inference for a contrast is by using the function contrast on the output of emmeans. The corresponding function
call for the contrast from above is as follows:
## contrast estimate SE df t.ratio p.value
## c(1, -0.5, -0.5) -0.0615 0.241 27 -0.255 0.8009
A confidence interval can also be obtained by calling confint (not shown).
Remark: For ordered factors we could also define contrasts which capture the linear, quadratic or higher-order trend if applicable. This is in fact exactly what is being used when using contr.poly as
seen in Section 2.6.1. We call such contrasts polynomial contrasts. The result can directly be read off the output of summary.lm. Alternatively, we could also use emmeans and set method = "poly" when
calling the contrast function.
3.1.2 Some Technical Details
Every contrast has an associated sum of squares \[ SS_c = \frac{\left(\sum_{i=1}^g c_i \overline{y}_{i\cdot}\right)^2}{\sum_{i=1}^g \frac{c_i^2}{n_i}} \] having one degree of freedom. Hence, for the
corresponding mean squares it holds that \(MS_c = SS_c\). This looks unintuitive at first sight, but it is nothing more than the square of the \(t\)-statistic for the special model parameter \(\sum_
{i=1}^g c_i \mu_i\) with the null hypothesis defined in [Equation eq-contrast-null] (without the \(MS_E\) factor). You can think of \(SS_c\) as the “part” of \(SS_{\textrm{Trt}}\) in “direction” of \
Under \(H_0: \sum_{i=1}^g c_i \mu_i = 0\) it holds that \[ \frac{MS_c}{MS_E} \sim F_{1,\, N-g}. \] Because \(F_{1, \, m}\) = \(t_m^2\) (the square of a \(t_m\)-distribution with \(m\) degrees of
freedom), this is nothing more than the “squared version” of the \(t\)-test.
Two contrasts \(c\) and \(c^*\) are called orthogonal if \[ \sum_{i=1}^g \frac{c_i c_i^*}{n_i} = 0. \] If two contrasts \(c\) and \(c^*\) are orthogonal, the corresponding estimates are
stochastically independent. This means that if we know something about one of the contrasts, this does not help us in making a statement about the other one.
If we have \(g\) treatments, we can find \(g - 1\) different orthogonal contrasts (one dimension is already used by the global mean \((1, \ldots, 1)\)). A set of orthogonal contrasts partitions the
treatment sum of squares meaning that if \(c^{(1)}, \ldots, c^{(g-1)}\) are orthogonal contrasts it holds that \[ SS_{c^{(1)}} + \cdots + SS_{c^{(g-1)}} = SS_{\textrm{Trt}}. \] Intuition: “We get all
information about the treatment by asking the right \(g - 1\) questions.”
However, your research questions define the contrasts, not the orthogonality criterion!
3.2 Multiple Testing
The problem with all statistical tests is the fact that the overall type I error rate increases with increasing number of tests. Assume that we perform \(m\) independent tests whose null hypotheses
we label with \(H_{0, j}\), \(j = 1, \ldots, m\). Each test uses an individual significance level of \(\alpha\). Let us first calculate the probability to make at least one false rejection for the
situation where all \(H_{0, j}\) are true. To do so, we first define the event \(A_j = \{\textrm{test } j \textrm{ falsely rejects }H_{0, j}\}\).
The event “there is at least one false rejection among all \(m\) tests” can be written as \(\cup_{j=1}^{m} A_j\). Using the complementary event and the independence assumption, we get \[\begin
{align*} P\left(\bigcup\limits_{j=1}^{m} A_j \right) & = 1 - P\left(\bigcap\limits_{j=1}^{m} A_j^c \right) \\ & = 1 - \prod_{j=1}^m P(A_j^c) \\ & = 1 - (1 - \alpha)^m. \\ \end{align*}\] Even for a
small value of \(\alpha\), this is close to 1 if \(m\) is large. For example, using \(\alpha = 0.05\) and \(m = 50\), this probability is \(0.92\)!
This means that if we perform many tests, we expect to find some significant results, even if all null hypotheses are true. Somehow we have to take into account the number of tests that we perform to
control the overall type I error rate.
Using similar notation as Bretz, Hothorn, and Westfall (2011), we list the potential outcomes of a total of \(m\) tests, among which \(m_0\) null hypotheses are true, in Table 3.1.
Table 3.1: Outcomes of a total of \(m\) statistical tests, among which \(m_0\) null hypotheses are true. Capital letters indicate random variables.
\(H_0\) true \(H_0\) false Total
Significant \(V\) \(S\) \(R\)
Not significant \(U\) \(T\) \(W\)
Total \(m_0\) \(m - m_0\) \(m\)
For example, \(V\) is the number of wrongly rejected null hypotheses (type I errors, also known as false positives), \(T\) is the number of type II errors (also known as false negatives), \(R\) is
the number of significant results (or “discoveries”), etc.
Using this notation, the overall or family-wise error rate (FWER) is defined as the probability of rejecting at least one of the true \(H_0\)’s: \[ \textrm{FWER} = P(V \ge 1). \] The family-wise
error rate is very strict in the sense that we are not considering the actual number of wrong rejections, we are just interested in whether there is at least one. This means the situation where we
make (only) \(V = 1\) error is equally “bad” as the situation where we make \(V = 20\) errors.
We say that a procedure controls the family-wise error rate in the strong sense at level \(\alpha\) if \[ \textrm{FWER} \le \alpha \] for any configuration of true and non-true null hypotheses. A
typical choice would be \(\alpha = 0.05\).
Another error rate is the false discovery rate (FDR) which is the expected fraction of false discoveries, \[ \textrm{FDR} = E \left[ \frac{V}{R} \right]. \] Controlling FDR at, e.g., level 0.2 means
that on average in our list of “significant findings” only 20% are not “true findings” (false positives). If we can live with a certain amount of false positives, the relevant quantity to control is
the false discovery rate.
If a procedure controls FWER at level \(\alpha\), FDR is automatically controlled at level \(\alpha\) too (Bretz, Hothorn, and Westfall 2011). On the other hand, a procedure that controls FDR at
level \(\alpha\) might have a much larger error rate regarding FWER. Hence, FWER is a much stricter (more conservative) criterion leading to fewer rejections.
We can also control the error rates for confidence intervals. We call a set of confidence intervals simultaneous confidence intervals at level \((1 - \alpha)\) if the probability that all intervals
cover the corresponding true parameter value is \((1 - \alpha)\). This means that we can look at all confidence intervals at the same time and get the correct “big picture” with probability \((1 - \
In the following, we focus on the FWER and simultaneous confidence intervals.
We typically start with individual p-values (the ordinary p-values corresponding to the \(H_{0,j}\)’s) and modify or adjust them such that the appropriate overall error rate (like FWER) is being
controlled. Interpretation of an individual p-value is as you have learned in your introductory course (“the probability to observe an event as extreme as …”). The modified p-values should be
interpreted as the smallest overall error rate such that we can reject the corresponding null hypothesis.
The theoretical background for most of the following methods can be found in Bretz, Hothorn, and Westfall (2011).
3.2.1 Bonferroni
The Bonferroni correction is a very generic but conservative approach. The idea is to use a more restrictive (individual) significance level of \(\alpha^* = \alpha / m\). For example, if we have \(\
alpha = 0.05\) and \(m = 10\), we would use an individual significance level of \(\alpha^* = 0.005\). This procedure controls the FWER in the strong sense for any dependency structure of the
different tests. Equivalently, we can also multiply the original p-values by \(m\) and keep using the original significance level \(\alpha\). Especially for large \(m\), the Bonferroni correction is
very conservative leading to low power.
Why does it work? Let \(M_0\) be the index set corresponding to the true null hypotheses, with \(|M_0| = m_0\). Using an individual significance level of \(\alpha/m\) we get \[\begin{align*} P(V \ge
1) & = P\left(\bigcup \limits_{j \in M_0} \textrm{reject } H_{0,j}\right) \le \sum_{j \in M_0} P(\textrm{reject } H_{0,j}) \\ & \le m_0 \frac{\alpha}{m} \le \alpha. \end{align*}\]
The confidence intervals based on the adjusted significance level are simultaneous (e.g., for \(\alpha = 0.05\) and \(m = 10\) we would need individual 99.5% confidence intervals).
We have a look at the previous example where we have two contrasts, \(c_1 = (1, -1/2, -1/2)\) (“control vs. the average of the remaining treatments”) and \(c_2 = (1, -1, 0)\) (“control vs. treatment
We first construct a contrast matrix where the two rows correspond to the two contrasts. Calling summary with test = adjusted("none") gives us the usual individual, i.e., unadjusted p-values.
## Create a matrix where each *row* is a contrast
K <- rbind(c(1, -1/2, -1/2), ## ctrl vs. average of trt1 and trt2
c(1, -1, 0)) ## ctrl vs. trt1
plant.glht.K <- glht(fit.plant, linfct = mcp(group = K))
## Individual p-values
summary(plant.glht.K, test = adjusted("none"))
## ...
## Linear Hypotheses:
## Estimate Std. Error t value Pr(>|t|)
## 1 == 0 -0.0615 0.2414 -0.25 0.80
## 2 == 0 0.3710 0.2788 1.33 0.19
## (Adjusted p values reported -- none method)
If we use summary with test = adjusted("bonferroni") we get the Bonferroni-corrected p-values. Here, this consists of a multiplication by 2 (you can also observe that if the resulting p-value is
larger than 1, it will be set to 1).
## Bonferroni corrected p-values
summary(plant.glht.K, test = adjusted("bonferroni"))
## ...
## Linear Hypotheses:
## Estimate Std. Error t value Pr(>|t|)
## 1 == 0 -0.0615 0.2414 -0.25 1.00
## 2 == 0 0.3710 0.2788 1.33 0.39
## (Adjusted p values reported -- bonferroni method)
By default, confint calculates simultaneous confidence intervals. Individual confidence intervals can be computed by setting the argument calpha = univariate_calpha(), critical value of \(\alpha\),
in confint (not shown).
With emmeans, the function call to get the Bonferroni-corrected p-values is as follows:
contrast(plant.emm, method = list(c(1, -1/2, -1/2), c(1, -1, 0)),
adjust = "bonferroni")
## contrast estimate SE df t.ratio p.value
## c(1, -0.5, -0.5) -0.0615 0.241 27 -0.255 1.0000
## c(1, -1, 0) 0.3710 0.279 27 1.331 0.3888
## P value adjustment: bonferroni method for 2 tests
3.2.2 Bonferroni-Holm
The Bonferroni-Holm procedure (Holm 1979) also controls the FWER in the strong sense. It is less conservative and uniformly more powerful, which means always better, than Bonferroni. It works in the
following sequential way:
1. Sort p-values from small to large: \(p_{(1)} \le p_{(2)} \le \ldots \le p_{(m)}\).
2. For \(j = 1, 2, \ldots\): Reject null hypothesis if \(p_{(j)} \leq \frac{\alpha}{m-j+1}\).
3. Stop when reaching the first non-significant p-value (and do not reject the remaining null hypotheses).
Note that only the smallest p-value has the traditional Bonferroni correction. Bonferroni-Holm is a so-called stepwise, more precisely step-down, procedure as it starts at the smallest p-value and
steps down the sequence of p-values (Bretz, Hothorn, and Westfall 2011). Note that this procedure only works with p-values but cannot be used to construct confidence intervals.
With the multcomp package, we can set the argument test of the function summary accordingly:
## ...
## Linear Hypotheses:
## Estimate Std. Error t value Pr(>|t|)
## 1 == 0 -0.0615 0.2414 -0.25 0.80
## 2 == 0 0.3710 0.2788 1.33 0.39
## (Adjusted p values reported -- holm method)
With emmeans, the argument adjust = "holm" has to be used (not shown). In addition, this is also implemented in the function p.adjust in R.
3.2.3 Scheffé
The Scheffé procedure (Scheffé 1959) controls for the search over any possible contrast. This means we can try out as many contrasts as we like and still get honest p-values! This is even true for
contrasts that are suggested by the data, which were not planned beforehand, but only after seeing some special structure in the data. The price for this nice property is low power.
The Scheffé procedure works as follows: We start with the sum of squares of the contrast \(SS_c\). Remember: This is the part of the variation that is explained by the contrast, like a
one-dimensional aspect of the multi-dimensional treatment effect. Now we are conservative and treat this as if it would be the whole treatment effect. This means we use \(g-1\) as the corresponding
degrees of freedom and therefore calculate the mean squares as \(SS_c / (g - 1)\). Then we build the usual \(F\)-ratio by dividing through \(MS_E\), i.e., \[ \frac{SS_c / (g - 1)}{MS_E} \] and
compare the realized value to an \(F_{g - 1, \, N - g}\)-distribution (the same distribution that we would also use when testing the whole treatment effect).
Note: Because it holds that \(SS_c \le SS_{\textrm{Trt}}\), we do not even have to start searching if the \(F\)-test is not significant.
What we described above is equivalent to taking the “usual” \(F\)-ratio of the contrast (typically available from any software) and use the distribution \((g - 1) \cdot F_{g-1, \, N - g}\) instead of
\(F_{1, \, N - g}\) to calculate the p-value.
We can do this manually in R with the multcomp package. We first treat the contrast as an “ordinary” contrast and then do a manual calculation of the p-value. As glht reports the value of the \(t\)
-test, we first have to take the square of it to get the \(F\)-ratio. As an example, we consider the contrast \(c = (1/2, -1, 1/2)\) (the mean of the two groups with large values vs. the group with
small values, see Section 2.1.2).
plant.glht.scheffe <- glht(fit.plant,
linfct = mcp(group = c(1/2, -1, 1/2)))
## p-value according to Scheffe (g = 3, N - g = 27)
pf((summary(plant.glht.scheffe)$test$tstat)^2 / 2, 2, 27,
lower.tail = FALSE)
If we use a significance level of 5% we do not get a significant result, with the more extreme contrast \(c = (0, -1, 1)\) we would be successful.
Confidence intervals can be calculated too by inverting the test from above, see Section 5.3 in Oehlert (2000) for more details.
summary.glht <- summary(plant.glht.scheffe)$test
estimate <- summary.glht$coefficients ## estimate
sigma <- summary.glht$sigma ## standard error
crit.val <- sqrt(2 * qf(0.95, 2, 27)) ## critical value
estimate + c(-1, 1) * sigma * crit.val
## [1] -0.007316 1.243316
An alternative implementation is also available in the function ScheffeTest of package DescTools (Signorell et al. 2021).
In emmeans, the argument adjust = "scheffe" can be used. For the same contrast as above, the code would be as follows (the argument scheffe.rank has to be set to the degrees of freedom of the factor,
here 2).
## contrast estimate SE df t.ratio p.value
## c(0.5, -1, 0.5) 0.618 0.241 27 2.560 0.0532
## P value adjustment: scheffe method with rank 2
Confidence intervals can be obtained by replacing summary with confint in the previous function call (not shown).
3.2.4 Tukey Honest Significant Differences
A special case of a multiple testing problem is the comparison between all possible pairs of treatments. There are a total of \(g \cdot (g - 1) / 2\) pairs that we can inspect. We could perform all
pairwise \(t\)-tests with the function pairwise.t.test; it uses a pooled standard deviation estimate from all groups.
## Without correction (but pooled sd estimate)
pairwise.t.test(PlantGrowth$weight, PlantGrowth$group,
p.adjust.method = "none")
## Pairwise comparisons using t tests with pooled SD
## data: PlantGrowth$weight and PlantGrowth$group
## ctrl trt1
## trt1 0.194 -
## trt2 0.088 0.004
## P value adjustment method: none
The output is a matrix of p-values of the corresponding comparisons (see row and column labels). We could now use the Bonferroni correction method, i.e., p.adjust.method = "bonferroni" to get
p-values that are adjusted for multiple testing.
## With correction (and pooled sd estimate)
pairwise.t.test(PlantGrowth$weight, PlantGrowth$group,
p.adjust.method = "bonferroni")
## Pairwise comparisons using t tests with pooled SD
## data: PlantGrowth$weight and PlantGrowth$group
## ctrl trt1
## trt1 0.58 -
## trt2 0.26 0.01
## P value adjustment method: bonferroni
However, there exists a better, more powerful alternative which is called Tukey Honest Significant Differences (HSD). The balanced case goes back to Tukey (1949), an extension to unbalanced
situations can be found in Kramer (1956), which is also discussed in Hayter (1984). Think of a procedure that is custom tailored for the situation where we want to do a comparison between all
possible pairs of treatments. We get both p-values (which are adjusted such that the family-wise error rate is being controlled) and simultaneous confidence intervals. In R, this is directly
implemented in the function TukeyHSD and of course both packages multcomp and emmeans contain an implementation too.
We can directly call TukeyHSD with the fitted model as the argument:
## Tukey multiple comparisons of means
## 95% family-wise confidence level
## Fit: aov(formula = weight ~ group, data = PlantGrowth)
## $group
## diff lwr upr p adj
## trt1-ctrl -0.371 -1.0622 0.3202 0.3909
## trt2-ctrl 0.494 -0.1972 1.1852 0.1980
## trt2-trt1 0.865 0.1738 1.5562 0.0120
Each line in the above output contains information about a specific pairwise comparison. For example, the line trt1-ctrl says that the comparison of level trt1 with ctrl is not significant (the
p-value is 0.39). The confidence interval for the difference \(\mu_2 - \mu_1\) is given by \([-1.06, 0.32]\). Confidence intervals can be visualized by simply calling plot.
Remember, these confidence intervals are simultaneous, meaning that the probability that they all cover the corresponding true difference at the same time is 95%. From the p-values, or the confidence
intervals, we read off that only the difference between trt1 and trt2 is significant (using a significance level of 5%).
We get of course the same results when using package multcomp. To do so, we have to use the argument group = "Tukey".
## Tukey HSD with package multcomp
plant.glht.tukey <- glht(fit.plant, linfct = mcp(group = "Tukey"))
## Simultaneous Tests for General Linear Hypotheses
## Multiple Comparisons of Means: Tukey Contrasts
## ...
## Linear Hypotheses:
## Estimate Std. Error t value Pr(>|t|)
## trt1 - ctrl == 0 -0.371 0.279 -1.33 0.391
## trt2 - ctrl == 0 0.494 0.279 1.77 0.198
## trt2 - trt1 == 0 0.865 0.279 3.10 0.012
## (Adjusted p values reported -- single-step method)
Simultaneous confidence intervals can be obtained by calling confint.
## Simultaneous Confidence Intervals
## Multiple Comparisons of Means: Tukey Contrasts
## ...
## 95% family-wise confidence level
## ...
## Linear Hypotheses:
## Estimate lwr upr
## trt1 - ctrl == 0 -0.371 -1.062 0.320
## trt2 - ctrl == 0 0.494 -0.197 1.185
## trt2 - trt1 == 0 0.865 0.174 1.556
They can be plotted too.
In emmeans, the corresponding function call would be as follows (output not shown):
contrast(plant.emm, method = "pairwise")
Also with emmeans, the corresponding simultaneous confidence intervals can be obtained with confint, which can be plotted too.
Remark: The implementations in multcomp and emmeans are more flexible with respect to unbalanced data than TukeyHSD, especially for situations where we have multiple factors as for example in Chapter
3.2.5 Multiple Comparisons with a Control
Similarly, if we want to compare all treatment groups with a control group, we have a so-called multiple comparisons with a control (MCC) problem (we are basically only considering a subset of all
pairwise comparisons). The corresponding custom-tailored procedure is called Dunnett procedure (Dunnett 1955). It controls the family-wise error rate in the strong sense and produces simultaneous
confidence intervals. As usual, both packages multcomp and emmeans provide implementations. By default, the first level of the factor is taken as the control group. For the factor group in the
PlantGrowth data set this is ctrl, as can be seen when calling the function levels.
## [1] "ctrl" "trt1" "trt2"
With multcomp, we simply set group = "Dunnett".
plant.glht.ctrl <- glht(fit.plant, linfct = mcp(group = "Dunnett"))
## Simultaneous Tests for General Linear Hypotheses
## Multiple Comparisons of Means: Dunnett Contrasts
## ...
## Linear Hypotheses:
## Estimate Std. Error t value Pr(>|t|)
## trt1 - ctrl == 0 -0.371 0.279 -1.33 0.32
## trt2 - ctrl == 0 0.494 0.279 1.77 0.15
## (Adjusted p values reported -- single-step method)
We get smaller p-values than with the Tukey HSD procedure because we have to correct for less tests; there are more comparisons between pairs than there are comparisons to the control treatment.
In emmeans, the corresponding function call would be as follows (output not shown):
The usual approach with confint gives the corresponding simultaneous confidence intervals.
3.2.6 FAQ
Should I only do tests like Tukey HSD, etc. if the \(F\)-test is significant?
No, the above-mentioned procedures have a built-in correction regarding multiple testing and do not rely on a significant \(F\)-test; one exception is the Scheffé procedure in Section 3.2.3: If the \
(F\)-test is not significant, you cannot find a significant contrast. In general, conditioning on the \(F\)-test leads to a very conservative approach regarding type I error rate. In addition, the
conditional coverage rates of, e.g., Tukey HSD confidence intervals can be very low if we only apply them when the \(F\)-test is significant, see also Hsu (1996). This means that if researchers would
use this recipe of only using Tukey HSD when the \(F\)-test is significant and we consider 100 different applications of Tukey HSD, on average it would happen more than 5 times that the simultaneous
95% confidence intervals would not cover all true parameters. Generally speaking, if you apply a statistical test only after a first test was significant, you are typically walking on thin ice: Many
properties of the second statistical tests typically change. This problem is also known under the name of selective inference, see for example Benjamini, Heller, and Yekutieli (2009).
Is it possible that the \(F\)-test is significant but Tukey HSD yields only insignificant pairwise tests? Or the other way round, Tukey HSD yields a significant difference but the \(F\)-test is not
Yes, these two tests might give us contradicting results. However, for most situations, this does not happen, see a comparison of the corresponding rejection regions in Hsu (1996).
How can we explain this behavior? This is basically a question of power. For some alternatives, Tukey HSD has more power because it answers a more precise research question, “which pairs of
treatments differ?”. On the other hand, the \(F\)-test is more flexible for situations where the effect is not evident in treatment pairs but in combinations of multiple treatments. Basically, the \
(F\)-test answers the question, “is a linear contrast of the cell means different from zero?”.
We use the following two extreme data sets consisting of three groups having two observations each.
x <- factor(rep(c("A", "B", "C"), each = 2))
y1 <- c(0.50, 0.62,
0.46, 0.63,
0.95, 0.86)
y2 <- c(0.23, 0.34,
0.45, 0.55,
0.55, 0.66)
Let us first visualize the first data set:
Here, the \(F\)-test is significant, but Tukey HSD is not:
## Df Sum Sq Mean Sq F value Pr(>F)
## x 2 0.1659 0.0829 9.68 0.049
## Residuals 3 0.0257 0.0086
## Tukey multiple comparisons of means
## 95% family-wise confidence level
## ...
## diff lwr upr p adj
## B-A -0.015 -0.40177 0.3718 0.9857
## C-A 0.345 -0.04177 0.7318 0.0669
## C-B 0.360 -0.02677 0.7468 0.0601
Now let us consider the second data set:
Now, the \(F\)-test is not significant, but Tukey HSD is:
## Df Sum Sq Mean Sq F value Pr(>F)
## x 2 0.1064 0.0532 9.34 0.052
## Residuals 3 0.0171 0.0057
## Tukey multiple comparisons of means
## 95% family-wise confidence level
## ...
## diff lwr upr p adj
## B-A 0.215 -0.10049 0.5305 0.1269
## C-A 0.320 0.00451 0.6355 0.0482
## C-B 0.105 -0.21049 0.4205 0.4479 | {"url":"https://people.math.ethz.ch/~meierluk/teaching/anova/contrasts-and-multiple-testing.html","timestamp":"2024-11-02T21:22:46Z","content_type":"application/xhtml+xml","content_length":"107211","record_id":"<urn:uuid:a448918c-b718-446d-b119-7a2fd61e19d2>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00861.warc.gz"} |
Sample Size Re-estimation (SSR)
In summary:
• SSR offers researchers an opportunity to appropriately change the sample size they originally planned when designing the trial;
• Researchers need to specify in advance: the design parameters to re-estimate during the trial; how these parameters are re-estimated and when; how the sample size is then re-calculated and; the
decision rules for changing the sample size;
• At each interim analysis, design parameters are re-estimated, initial assumptions made about them updated, and the sample size changed accordingly;
• SSR is useful when there is considerable uncertainty around design parameters before the trial begins;
• SSR methods considered here assume that the targeted treatment effect is fixed at the onset.
In every clinical trial, researchers decide at the design stage (before the trial begins) the required sample size to reliably answer the research question(s). This might be the number of patients,
or number of events for time-to-event outcomes (where not for every patient an event, such as death, is observed during the trial follow-up period). Calculating this sample size can be challenging
). Recruiting more patients than necessary wastes resources, unnecessarily delays study results and thereby hinders quick decision-making, and may expose more patients to potentially unsafe study
On the other hand, researchers may not be able to answer the intended research question(s) if they fail to recruit an adequate sample size. For example, the chance of finding effective study
treatments if they exist (statistical power) may be inappropriately low, the chance of falsely claiming evidence of benefit (type I error rate) may be inappropriately high, or the trial may produce
equivocal results. Furthermore, a sizable proportion of trials require a funding extension due to recruitment challenges
^2^, ^3^, ^4
. Some of these trials might have been designed with unnecessarily large sample sizes in the first place – so they may not need a funding extension request at all. All these issues raise scientific,
ethical, economic, and feasibility questions.
At the design stage, researchers need to make assumptions about specific design parameters that are used to estimate the sample size. The most commonly used design parameter is the variability of the
primary outcome. In some cases, published or unpublished data from related studies may exist to inform these assumptions. In other cases, assumptions are based on the opinions of researchers when
prior data is unavailable. Considerable uncertainties may exist around these estimates. PANDA users may wish to read introductory material on “
what is an adaptive design
Assumptions on design parameters may be uncertain because of several factors including:
• failure to adequately review evidence from prior studies;
• limited availability of prior data;
• overoptimistic opinions of researchers;
• retrofitting by researchers to make the study feasible (e.g., tweaking design parameters to give a practically achievable sample size);
• differences in the settings of prior and current studies;
• differences in patient characteristics in prior studies and the targeted study group;
• differences in how outcomes are assessed and when they are assessed in prior studies;
• improvements in patient care in the comparator arm since previous studies took place.
In context
Several researchers have reported marked discrepancies between estimates of design parameters assumed at the design stage and those observed on trial completion
^5^, ^6^, ^7^, ^8
. Reviews found that only 73 (34%) of trials made accurate assumptions about the control event rate where it differed by less than 30% of the observed event rate
, 24 (80%) of continuous outcomes had markedly greater variability than assumed
, and 32 (80%) of rheumatoid arthritis trials recruited more patients than necessary
. A review of protocols submitted to the UK research ethics committees found that trials tended to recruit more patients than necessary (i.e., they were “overpowered”)
. Figure 1 illustrates the magnitude of discrepancy between the assumed (green line) and observed control event rate (red dashed line) from the RATPAC trial
as patients were recruited sequentially. The observed control event rate was always below the anticipated 0.5 (conservative rate assumed before the trial began). The entire trial duration is
displayed only for illustration purposes.
In summary
Uncertainties around design parameters exist in every trial; however, the magnitude of uncertainties and their implications on the
operating characteristics
of the design (e.g., achieved power) vary from trial to trial.
The goal of sample size re-estimation (SSR) is to update assumptions around design parameters and change the original sample size appropriately based on interim data from a group of patients already
recruited into the trial. Thus, researchers can achieve the desired statistical power while avoiding over- or under-recruitment of patients, both of which are associated with inefficient and/or
wasteful use of research resources.
Fayers et al
. Sample size calculation for clinical trials: The impact of clinician beliefs.
Br J Cancer
. 2000;82(1):213–9.
Sully et al
. A reinvestigation of recruitment to randomised, controlled, multicenter trials: a review of trials funded by two UK funding agencies.
. 2013;14(1):166.
McDonald et al
. What influences recruitment to randomised controlled trials? A review of trials funded by two UK funding agencies.
. 2006;7(1):9.
Pemberton et al
. Performance and predictors of recruitment success in National Heart, Lung, and Blood Institute’s cardiovascular clinical trials.
Clin Trials
. 2018;15(5): 444-451.
Charles et al
. Reporting of sample size calculation in randomised controlled trials: review.
. 2009;338(may12_1):b1732.
. Underpowering in randomized trials reporting a sample size calculation.
J Clin Epidemiol
. 2003;56(8):717–20.
Celik et al
. Are sample sizes of randomized clinical trials in rheumatoid arthritis too large?
Eur J Clin Invest
. 2014;44(11):1034–44. Available from:
Clark et al
. Sample size determinations in original research protocols for randomised clinical trials submitted to UK research ethics committees: review.
. 2013;346(mar21_1):f1135.
Goodacre et al
. The RATPAC (Randomised Assessment of Treatment using Panel Assay of Cardiac markers) trial: a randomised controlled trial of point-of-care cardiac markers in the emergency department.
Health Technol Assess
. 2011;15(23). | {"url":"https://panda.shef.ac.uk/techniques/sample-size-re-estimation-ssr/categories/15","timestamp":"2024-11-11T07:19:46Z","content_type":"text/html","content_length":"23526","record_id":"<urn:uuid:e461926a-4e76-44c9-8853-068d34e422c0>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00386.warc.gz"} |
4.4 & 4.5 Notes Remember: Identity Matrices: If the product of two matrices equal the identity matrix then they are inverses. - ppt download
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What's #REF in Excel and How to Fix It - Excel University
What’s #REF in Excel and How to Fix It
When a formula contains an invalid cell or range reference, the #REF! error is displayed. #REF! in Excel is short for reference, and you’ll usually see it pop up when cells that were referenced in
formulas are deleted. They can be a nuisance, but they’re usually pretty simple to find and fix.
Here’s how to get rid of any #REF! errors that come up in your worksheets and a few things you can do to prevent them.
What does a #REF! error look like?
If you’re one of the very lucky few who have never seen a #REF! error in their worksheets, this example will show you how it works.
Let’s say you have some values you want to add up. For example, you want to add up the values in the cells B4, B5, B6, B7, and B8. As with just about anything in Excel, there are multiple ways to do
this. One way is to use a SUM function that references the entire range B4:B8 like this:
Another way is with a formula that references each cell individually, like this:
In the screenshot below, the highlighted cell B9 used the SUM function and referenced the range, while the individual cell references were used in the selected formula cell D9:
Both formulas provide the correct result of 81.
Now, let’s delete row 4. When we do, the formula in B8 that used a range reference continues to work while the formula in D8 that used individual cell references breaks:
As you can see, the formula that used the range reference was able to adapt to the change, while the formula that specifically referenced a cell that was deleted did not. When you inspect the
formula, you’ll notice #REF! in place of the cell that was deleted.
How do you find cells with #REF in Excel quickly?
Here are a few ways to find #REF! errors.
Using Find & Replace to locate #REF errors
1. Use the Ctrl+F shortcut to open the Find & Replace dialog box.
2. Enter #REF! in the Find field.
3. Click the Find All button.
4. Optionally: Ctrl+A while dialog is open to select all found cells.
If you want to remove the #REF! references from the formulas, you could select the Replace tab and leave the Replace with field empty. Click Replace All. This replaces all #REF!s with nothing,
essentially removing any #REF!s from the worksheet.
Using Go to Special to find #REF in Excel spreadsheets
Another way to find #REF! errors is by using the Go to Special feature.
1. You can access it by going to the Home ribbon item and selecting Go To. Alternatively, you can use the Ctrl+G shortcut and select Special in the resulting dialog.
2. Then, select Formulas from the resulting Go To Special dialog. Check the box next to Errors.
3. The formulas with #REF! errors will be selected.
Using Conditional Formatting to automatically highlight errors
If you wanted Excel to continuously monitor for #REF! errors (and any other formula errors for that matter), you could use conditional formatting. This will essentially apply a designated cell format
to any current and future #REF! errors.
1. Select all cells (Ctrl+A or click the upper left corner of the worksheet)
2. Home > Conditional Formatting > Highlight Cell Rules > More Rules
3. Format only cells with: Errors
4. Click the Format button to create your desired format
5. Click OK
Now, any existing formula error cells are formatted. When you correct them, the designated formatting is removed. And going forward, cells with formula errors will automatically be formatted.
Now that you can find them, how do you fix them?
Finding #REF! errors is the easy part, but how you fix them depends on what you need for your worksheet.
In practice, I will typically review each formula with a #REF! error. I will determine if it is ok to simply delete the #REF! or if I need to point it to a different cell/range. In other words, I
need to figure out if the missing reference was important to the current formula. If so, I will correct the formula rather than simply delete the #REF!.
However, if you know that none of the formulas with #REF! need to be updated, and you simply want to remove all #REF!s, a fast way to do that is by using the Find & Replace feature to replace all #
REF cells with nothing, as noted above.
Preventing #REF in Excel
The best way to deal with #REF! errors is to never get them in the first place – but that’s easier said than done.
There are a few actions that increase the likelihood of creating #REF! errors in your worksheets. Be careful when performing any of these tasks in Excel.
• Deleting rows and columns – This scenario is a common cause of #REF! errors. Before you get rid of a row or column, make sure your formulas aren’t referencing the values in them.
• Copy + pasting cells – If you’re copying and pasting a cell that uses relative references, you may get a #REF! error because Excel updates them based on where they’re pasted in the worksheet. For
example, if you copy a cell with a formula that references the cell to the left, and the paste that into a cell in Column A, there are no valid references to the left. This results in a #REF!
error. Note that in some cases, this could be avoided by changing such relative references to absolute.
Do you have any other tips for preventing or fixing #REF! errors in your worksheets? Let us know in the comments!
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3 Comments
1. Thanks ! You remind me to use the FIND and REPLACE functions. It’s useful when I have REF! errors etc.
Yes, before delete a column or a row, I need check the cells with references to this column and row.
Good REFRESH!
3. If the cells above or on the left of my #REF! errors contain valid formulas, I use “Fill Down” or “Fill Right” to eliminate my #REF! errors.
Also, if I’m just moving values around (not formulas), I select source cells, then “Copy”, select upper left cell of destination, then “Paste”, then select source cells a second time, then “Clear
Selection”. It’s extra steps but doesn’t produce #REF! errors the way dragging or using “cut” and “paste” do.
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A bag contains 4 red and 3 white balls. Determine the number of... | Filo
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A bag contains red and white balls. Determine the number of ways in which red and white balls can be selected.a. b. c. d.
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Of course, here are some step-by-step solutions for common academic problems: 1. Solving a Quadratic Equation: Given a quadratic equation of the form , the solutions can be found using the quadratic
formula: To solve the equation, follow these steps: - Identify the values of , , and . - Substitute the values into the quadratic formula and simplify. - Solve for by using the + and - signs in the
formula. 2. Finding the Derivative of a Function: To find the derivative of a function , use the following steps: - Use the power rule: if , then . - Use the sum and difference rule: if , then . -
Use the product rule: if , then . - Use the chain rule: if , then . 3. Solving a System of Linear Equations: To solve a system of linear equations, follow these steps: - Write the equations in
standard form: . - Use elimination or substitution to eliminate one variable. - Solve for the remaining variable. - Substitute the value of the variable into one of the original equations and solve
for the other variable. - Check the solution by plugging it into both equations. 4. Balancing a Chemical Equation: To balance a chemical equation, follow these steps: - Write the unbalanced equation
with the reactants on the left and the products on the right. - Count the number of atoms of each element on each side of the equation. - Use coefficients to balance the equation by adjusting the
number of molecules of each reactant and product. - Check the balanced equation by counting the number of atoms of each element on each side. These step-by-step solutions can help students better
understand how to approach and solve common academic problems.
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Question Text A bag contains red and white balls. Determine the number of ways in which red and white balls can be selected.a. b. c. d.
Updated On Jan 31, 2024
Topic Permutations & Combinations
Subject Mathematics
Class Class 12
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Gradient Boosting and XGBoost
Note: This post was originally published on the Canopy Labs website.
XGBoost is an powerful, and lightning fast machine learning library. It’s commonly used to win Kaggle competitions (and a variety of other things). However, it’s an intimidating algorithm to
approach, especially because of the number of parameters - and it’s not clear what all of them do.
Although many posts already exist explaining what XGBoost does, many confuse gradient boosting, gradient boosted trees and XGBoost. The purpose of this post is to clarify these concepts. Also, to
make XGBoost’s hyperparameters less intimidating, this post explores (in a little more detail than the documentation) exactly what the hyperparameters exposed in the scikit-learn API do.
1) Gradient Boosting
2) Gradient Boosted Trees
3) Extreme Gradient Boosting
1. Gradient Boosting
If you are reading this, it is likely you are familiar with stochastic gradient descent (SGD) (if you aren’t, I highly recommend this video by Andrew Ng, and the rest of the course, which can be
audited for free). Assuming you are:
Gradient boosting solves a different problem than stochastic gradient descent.
When optimizing a model using SGD, the architecture of the model is fixed. What you are therefore trying to optimize are the parameters, P of the model (in logistic regression, this would be the
weights). Mathematically, this would look like this:
\[ F(x \mid P) = \min_{P} Loss(y, F(x \mid P)) \]
Which means I am trying to find the best parameters P for my function F, where ‘best’ means that they lead to the smallest loss possible (the vertical line in \(F(x\mid P)\) just means that once I’ve
found the parameters P, I calculate the output of F given x using them).
Gradient boosting doesn’t assume this fixed architecture. In fact, the whole point of gradient boosting is to find the function which best approximates the data. It would be expressed like this:
\[ F(x \mid P) = \min_{F, P} Loss(y, F(x \mid P)) \]
The only thing that has changed is that now, in addition to finding the best parameters P, I also want to find the best function F. This tiny change introduces a lot of complexity to the problem;
whereas before, the number of parameters I was optimizing for was fixed (my logistic regression model is defined before I start training it), now, it can change as I go through the optimization
process if my function F changes.
Obviously, searching all possible functions and their parameters to find the best one would take far too long, so gradient boosting finds the best function F by taking lots of simple functions, and
adding them together.
Where SGD trains a single complex model, gradient boosting trains an ensemble of simple models.
It does this the following way:
Take a very simple model h, and fit it to some data (x, y):
\[ h(x \mid P) = \min_{P} Loss(y, h(x \mid P)) \]
Then, use this trained model to predict an output:
\[ \hat{y} = h(x \mid P) \]
When I’m training my second model, I obviously don’t want it to uncover the same pattern in the data as this first model h; ideally, it would improve on the errors from this first prediction. This is
the clever part (and the ‘gradient’ part): this prediction will have some error, \( Loss(y,\hat{y})\). The next model I am going to fit will be on the gradient of the error with respect to the
predictions, \( \frac{\partial Loss}{\partial \hat{y}}\).
To think about why this is clever, lets consider mean squared error:
\[ Loss(y, \hat{y}) = MSE(y, \hat{y}) = (y - \hat{y})^2 \]
Calculating this gradient,
\[ \frac{\partial MSE(y, \hat{y})}{\partial \hat{y}} = -2(y - \hat{y}) \propto (y - \hat{y})\]
If for one data point, \(y=1\) and \(\hat{y}=0.6\), then the error in this prediction is \(MSE(1, 0.6) = 0.16\) and the new target for the model will be the gradient, \((y - \hat{y}) = 0.4\).
Training a model on this target,
\[ h_{1}(x \mid P) = \min_{P} Loss((y - \hat{y}), h_{1}(x \mid P)) \]
Now, for this same data point, where y=1 (and for the previous model, \(\hat{y} =0.6 \), the model is being trained to on a target of 0.4. Say that it returns \(\hat{y}_{1} = 0.3 \). The last step in
gradient boosting is to add these models together. For the two models I’ve trained (and for this specific data point), then
\[ y_{final} = \hat{y} + \hat{y}_{1} = 0.6 + 0.3 = 0.9\]
By training my second model on the gradient of the error with respect to the loss predictions of the first model, I have taught it to correct the mistakes of the first model. This is the core of
gradient boosting, and what allows many simple models to compensate for each other’s weaknesses to better fit the data.
I don’t have to stop at 2 models; I can keep doing this over and over again, each time fitting a new model to the gradient of the error of the updated sum of models.
An interesting note here is that at its core, gradient boosting is a method for optimizing the function F, but it doesn’t really care about h (since nothing about the optimization of h is defined).
This means that any base model h can be used to construct F.
2. Gradient Boosted Trees
Gradient boosted trees consider the special case where the simple model h is a decision tree. Visually (this diagram is taken from XGBoost’s documentation)):
In this case, there are going to be 2 kinds of parameters P: the weights at each leaf, w, and the number of leaves T in each tree (so that in the above example, T=3 and w=[2, 0.1, -1]).
When building a decision tree, a challenge is to decide how to split a current leaf. For instance, in the above image, how could I add another layer to the (age > 15) leaf? A ‘greedy’ way to do this
is to consider every possible split on the remaining features (so, gender and occupation), and calculate the new loss for each split; you could then pick the tree which most reduces your loss.
In addition to finding the new tree structures, the weights at each node need to be calculated as well, such that the loss is minimized. Since the tree structure is now fixed, this can be done
analytically now by setting the loss function = 0 (see the appendix for a derivation, but you are left with the following):
\[ w_{j} = \frac{\sum_{i \in I_{j}} \frac{\partial loss}{\partial (\hat{y} = 0)}}{\sum_{i \in I_{j}} (\frac{\partial^2 loss}{\partial (\hat{y} = 0)^2}) + \lambda} \]
Where \(I_j\) is a set containing all the instances ((x, y) datapoints) at a leaf, and \(w_j\) is the weight at leaf j. This looks more intimidating than it is; for some intuition, if we consider \
(loss=MSE=(y, \hat{y})^2\), then taking the first and second gradients where \(\hat{y} = 0\) yields
\[ w_{j} = \frac{\sum_{i \in I_{j}} y}{\sum_{i \in I_{j}} 2 + \lambda} \]
This makes sense; the weights effectively become the average of the true labels at each leaf (with some regularization from the \(\lambda \) constant).
3. XGBoost (and its hyperparameters)
XGBoost is one of the fastest implementations of gradient boosted trees.
It does this by tackling one of the major inefficiencies of gradient boosted trees: considering the potential loss for all possible splits to create a new branch (especially if you consider the case
where there are thousands of features, and therefore thousands of possible splits). XGBoost tackles this inefficiency by looking at the distribution of features across all data points in a leaf and
using this information to reduce the search space of possible feature splits.
Although XGBoost implements a few regularization tricks, this speed up is by far the most useful feature of the library, allowing many hyperparameter settings to be investigated quickly. This is
helpful because there are many, many hyperparameters to tune. Nearly all of them are designed to limit overfitting (no matter how simple your base models are, if you stick thousands of them together
they will overfit).
The list of hyperparameters was super intimidating to me when I started working with XGBoost, so I am going to discuss the 4 parameters I have found most important when training my models so far (I
have tried to give a slightly more detailed explanation than the documentation for all the parameters in the appendix).
My motivation for trying to limit the number of hyperparameters is that doing any kind of grid / random search with all of the hyperparameters XGBoost allows you to tune can quickly explode the
search space. I’ve found it helpful to start with the 4 below, and then dive into the others only if I still have trouble with overfitting.
3.a. n_estimators (and early stopping)
This is how many subtrees h will be trained. I put this first because introducing early stopping is the most important thing you can do to prevent overfitting. The motivation for this is that at some
point, XGBoost will begin memorizing the training data, and its performance on the validation set will worsen. At this point, you want to stop training more trees.
Note that if you use early stopping, XGBoost will return the final model (as opposed to the one with the lowest validation score), but this is okay since the best model will be this final model minus
the additional, overfitting subtrees which were trained. You can isolate the best model using trained_model.best_ntree_limit in your predict method, as below:
results = best_xgb_model.predict(x_test, ntree_limit=best_xgb_model.best_ntree_limit)
If you are using a parameter searcher like sklearn’s GridSearchCV, you’ll need to define a scoring method which uses the best_ntree_limit:
def best_ntree_score(estimator, X, y):
This scorer uses the best_ntree_limit to return
the best AUC ROC score
y_predict = estimator.predict_proba(X,
except AttributeError:
y_predict = estimator.predict_proba(X)
return roc_auc_score(y, y_predict[:, 1])
3.b. max_depth
The maximum tree depth each individual tree h can grow to. The default value of 3 is a good starting point, and I haven’t found a need to go beyond a max_depth of 5, even with fairly complex data.
3.c. learning rate
Each weight (in all the trees) will be multiplied by this value, so that
\[ w_{j} = \textrm{learning rate} \times \frac{\sum_{i \in I_{j}} \frac{\partial loss}{\partial (\hat{y} = 0)}}{\sum_{i \in I_{j}} (\frac{\partial^2 loss}{\partial (\hat{y} = 0)^2}) + \lambda} \]
I found that decreasing the learning rate very often lead to an improvement in the performance of the model (although it did lead to slower training times).
Because of the additive nature of gradient boosted trees, I found getting stuck in local minima to be a much smaller problem then with neural networks (or other learning algorithms which use
stochastic gradient descent).
3.d. reg_alpha and reg_lambda
The loss function is defined as
\[L = \sum_{i=0}^{n} loss(y_{res}, h(x)) + \frac{1}{2}\lambda \sum_{j=1}^{T}w_{j}^2 + \alpha \sum_{j=1}^{T} | w_{j} |\]
reg_alpha and reg_lambda control the L1 and L2 regularization terms, which in this case limit how extreme the weights at the leaves can become.
These two regularization terms have different effects on the weights; L2 regularization (controlled by the lambda term) encourages the weights to be small, whereas L1 regularization (controlled by
the alpha term) encourages sparsity - so it encourages weights to go to 0. This is helpful in models such as logistic regression, where you want some feature selection, but in decision trees we’ve
already selected our features, so zeroing their weights isn’t super helpful. For this reason, I found setting a high lambda value and a low (or 0) alpha value to be the most effective when
Note that the other parameters are useful, and worth going through if the above terms don’t help with regularization. However, I have found that exploring all the hyperparameters can cause the search
space to explode, so this is a good place to start.
Hopefully, this has provided you with a basic understanding of how gradient boosting works, how gradient boosted trees are implemented in XGBoost, and where to start when using XGBoost.
Happy boosting!
Check out the appendix for more information about other hyperparameters, and a derivation to get the weights. | {"url":"https://gabrieltseng.github.io/posts/2018-02-25-XGB/","timestamp":"2024-11-14T13:53:06Z","content_type":"text/html","content_length":"22917","record_id":"<urn:uuid:28faa7d4-6b8c-48ac-b333-104f9e03a051>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00088.warc.gz"} |
Please note that the recommended version of Scilab is 2025.0.0. This page might be outdated.
See the recommended documentation of this function
Ajuda do Scilab >> Xcos > palettes > Continuous_pal > PID
This block implements a PID (Proportional-Integral-Differential) controller. It calculates an "error" value Ue as the difference between a measured process variable Upr and a desired setpoint Ur.
The purpose is to make the process variable Up follow the setpoint value Ur. The PID controller is widely used in feedback control of industrial processes.
The PID controller calculation (algorithm) involves three separate parameters; the Proportional Kp, the Integral Ki and Derivative Kd values. These terms describe three basic mathematical functions
applied to the error signal Ue. Kp determines the reaction to the current error, Ki determines the reaction based on the sum of recent errors and Kd determines the reaction to the rate at which the
error has been changing.
The weighted sum of these three actions is used to adjust the process via a control element such as the position of a control valve or the power supply of a heating element. The basic structure of
conventional feedback control systems is shown below:
PID law is a linear combination of an input variable Up(t), its time integral Ui(t) and its first derivative Ud(t). The control law Ucon(t) has the form:
Dialog box
• Proportional
The value of the gain that multiply the error.
Properties : Type 'vec' of size -1.
• Integral
The value of the integral time of the error.(1/Integral)
Properties : Type 'vec' of size -1.
• Derivation
The value of the derivative time of the error.
Properties : Type 'vec' of size -1.
Default properties
• always active: no
• direct-feedthrough: no
• zero-crossing: no
• mode: no
• regular inputs:
- port 1 : size [-1,-2] / type 1
• regular outputs:
- port 1 : size [-1,-2] / type 1
• number/sizes of activation inputs: 0
• number/sizes of activation outputs: 0
• continuous-time state: no
• discrete-time state: no
• object discrete-time state: no
• name of computational function: csuper
Interfacing function
• SCI/modules/scicos_blocks/macros/Linear/PID.sci
Compiled Super Block content
Example 1
This example illustrates the usage of PID regulator. It enables you to fit the output signal Upr(t) to the required signal Ur(t) easily.
In this example the control system is a second-order unity-gain low-pass filter with damping ratio ξ=0.5 and cutoff frequency fc= 100 Hz. Its transfer function H(s) is:
To model this filter we use Continuous transfer function block (CLR) from Continuous time systems Palette.
The PID parameters Kp, Ki and Kd are set to 100, 0.1 and 0.
The scope displays the waveforms of system error Ue (black), reference signal Ur (blue) and process signal Upr(red). It shows how initially the process signal Upr(t) does not follow the reference
signal Ur(t). | {"url":"https://help.scilab.org/docs/6.0.1/pt_BR/PID.html","timestamp":"2024-11-04T20:01:21Z","content_type":"text/html","content_length":"16465","record_id":"<urn:uuid:711e320e-df63-4447-9f8f-c5893e6817e8>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00326.warc.gz"} |
Output power
. . . .
Output power
*1 Provided the maximum antenna gain does not exceed 6 dBi. In addition, the maximum power spectral density shall not exceed 17 dBm in any 1 megahertz band. If transmitting antennas of directional
gain greater than 6 dBi are used, both the maximum conducted output power and the maximum power spectral density shall be reduced by the amount of dB that the directional gain of the antenna exceeds
6 dBi.
*2 Equivalent Isotropically Radiated Power (EIRP) is terminology for the total RF power radiated by the antenna.
*3 The maximum power spectral density shall not exceed 17 dBm in any 1 megahertz band. Fixed point-to-point U-NII devices may employ antennas with directional gain up to 23 dBi without any
corresponding reduction in the maximum conducted output power or maximum power spectral density. For fixed point-to-point transmitters that employ a diretional antenna gain greater than 23 dBi, a 1
dB reduction in maximum conducted output power spectral density is required for each 1 dB of antenna gain in excess of 23 dBi. Fixed, point-to-point operations exclude the use of point-to-multipoint
systems, omnidirectional applications, and multiple collocated transmitters transmitting the same information. The operator of the U-NII device, or if the equipment is professionally installed, the
installer, is responsible for ensuring that systems employing high gain directional antennas are used exclusively for fixed, point-to-point operations.
*4 Provided the maximum antenna gain does not exceed 6 dBi. In addition, the maximum power spectral density shall not exceed 11 dBm in any 1 megahertz band. If transmitting antennas of directional
gain greater than 6 dBi are used, both the maximum conducted output power and the maximum power spectral density shall be reduced by the amount in dB that the directional gain of the antenna exceeds
6 dBi.
*5 In addition, the maximum power spectral density shall not exceed 11 dBm in any 1 megahertz band. If transmitting antennas of directional gain greater than 6 dBi are used, both the maximum
conducted output power and the maximum power spectral density shall be reduced by the amount in dB that the directional gain of the antenna exceeds 6 dBi.
*6 In addition, the maximum power spectral density shall not exceed 30 dBm in any 500-kHz band. If transmitting antennas of directional gain greater than 6 dBi are used, both the maximum conducted
output power and the maximum power spectral density shall be reduced by the amount in dB that the directional gain of the antenna exceeds 6 dBi. However, fixed point-to-point U-NII devices operating
in this band may employ transmitting antennas with directional gain greater than 6 dBi without any use of corresponding reduction in transmitter conducted power. Fixed point-to-point operations
exclude the use of point-to-multipoint sytems, omnidirectional applications, and multiple collocated transmitters transmitting the same information. The operator of the U-NII device, or if the
equipment is professionally installed, the installer, is responsible for ensuring that systems employing high gain directional antennas are used exclusively for fixed point-to-point operations. | {"url":"https://www.quadrionetworks.com/output-power.html","timestamp":"2024-11-02T20:12:22Z","content_type":"text/html","content_length":"38521","record_id":"<urn:uuid:436d2fe0-91d4-42c7-ae2c-7f25ce6ac84b>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00099.warc.gz"} |
How to handle missing data in a TensorFlow dataset?
Handling missing data in a TensorFlow dataset can be crucial for the performance and accuracy of your machine learning model. Here are some additional strategies and considerations to keep in mind:
1. Imputation techniques: In addition to using mean or median values for filling missing data, you can also consider more advanced imputation techniques such as K-nearest neighbors or predictive
modeling to estimate missing values based on the relationships between features in the dataset.
2. Encoding missing values: Sometimes it can be valuable to explicitly encode missing values as a separate category in categorical features. This can help the model learn the patterns associated
with missing data and differentiate them from valid values.
3. Feature engineering: In some cases, missing data can carry important information or patterns. You can create additional features indicating the presence of missing values in specific columns or
use techniques like missing data indicators to capture the influence of missingness on the target variable.
4. Deep learning models: Deep learning architectures such as autoencoders or variational autoencoders can be effective in learning representations of data with missing values and generating
plausible imputations for missing data based on the learned patterns.
5. Evaluation and validation: When handling missing data, it's essential to evaluate the impact of the chosen strategy on model performance. Use cross-validation or other validation techniques to
assess the effectiveness of handling missing data and ensure that the model generalizes well to unseen data.
By carefully considering these strategies and experimenting with different approaches, you can effectively handle missing data in TensorFlow datasets and improve the robustness of your machine
learning models. | {"url":"https://forum.ubuntuask.com/thread/how-to-handle-missing-data-in-a-tensorflow-dataset","timestamp":"2024-11-10T03:23:33Z","content_type":"text/html","content_length":"105425","record_id":"<urn:uuid:76c7c89b-a1c8-463a-bb29-2f0d9feb0699>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00492.warc.gz"} |
Triangle Congruence Theorems
The points and are on opposite sides of Now, consider Let denote the between and
It can be noted that and By the , is a of Points along the perpendicular bisector are equidistant from the endpoints of the segment, so
Finally, can be mapped onto by a across by reflecting across Because reflections preserve angles, and are mapped onto and respectively.
This time the image matches | {"url":"https://mathleaks.com/study/kb/reference/triangle_congruence_theorems","timestamp":"2024-11-09T17:31:28Z","content_type":"text/html","content_length":"369128","record_id":"<urn:uuid:c14f48d2-6997-4779-8fd5-19767773bd96>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00790.warc.gz"} |
Pratt's Primality Certificates
In 1975, Pratt introduced a proof system for certifying primes. He showed that a number p is prime iff a primality certificate for p exists. By showing a logarithmic upper bound on the length of the
certificates in size of the prime number, he concluded that the decision problem for prime numbers is in NP. This work formalizes soundness and completeness of Pratt's proof system as well as an
upper bound for the size of the certificate.
Session Pratt_Certificate | {"url":"https://www.isa-afp.org/entries/Pratt_Certificate.html","timestamp":"2024-11-02T00:01:37Z","content_type":"text/html","content_length":"11879","record_id":"<urn:uuid:65680c9f-1933-4429-9f60-b43f5beafb1c>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00340.warc.gz"} |
How Many Meters per second Is 363.3 Knots?
363.3 knots in meters per second
How many meters per second in 363.3 knots?
363.3 knots equals 186.898 meters per second
Unit Converter
Conversion formula
The conversion factor from knots to meters per second is 0.514444444444, which means that 1 knot is equal to 0.514444444444 meters per second:
1 kt = 0.514444444444 m/s
To convert 363.3 knots into meters per second we have to multiply 363.3 by the conversion factor in order to get the velocity amount from knots to meters per second. We can also form a simple
proportion to calculate the result:
1 kt → 0.514444444444 m/s
363.3 kt → V[(m/s)]
Solve the above proportion to obtain the velocity V in meters per second:
V[(m/s)] = 363.3 kt × 0.514444444444 m/s
V[(m/s)] = 186.89766666651 m/s
The final result is:
363.3 kt → 186.89766666651 m/s
We conclude that 363.3 knots is equivalent to 186.89766666651 meters per second:
363.3 knots = 186.89766666651 meters per second
Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 meter per second is equal to 0.005350521586684 × 363.3 knots.
Another way is saying that 363.3 knots is equal to 1 ÷ 0.005350521586684 meters per second.
Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that three hundred sixty-three point three knots is approximately one hundred eighty-six point eight
nine eight meters per second:
363.3 kt ≅ 186.898 m/s
An alternative is also that one meter per second is approximately zero point zero zero five times three hundred sixty-three point three knots.
Conversion table
knots to meters per second chart
For quick reference purposes, below is the conversion table you can use to convert from knots to meters per second
knots (kt) meters per second (m/s)
364.3 knots 187.412 meters per second
365.3 knots 187.927 meters per second
366.3 knots 188.441 meters per second
367.3 knots 188.955 meters per second
368.3 knots 189.47 meters per second
369.3 knots 189.984 meters per second
370.3 knots 190.499 meters per second
371.3 knots 191.013 meters per second
372.3 knots 191.528 meters per second
373.3 knots 192.042 meters per second | {"url":"https://convertoctopus.com/363-3-knots-to-meters-per-second","timestamp":"2024-11-04T23:29:14Z","content_type":"text/html","content_length":"34251","record_id":"<urn:uuid:185ad3ed-aab1-4017-9170-dfe8487eea90>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00134.warc.gz"} |
(Download) SSC: Combined Graduate Level (Tier II) Exam Solved Paper: Held on: 1-08-2010
SSC: Combined Graduate Level Tier–II Exam Held on: 01.08.2010
1. A General, while arranging his men, who were 6000 in number, in the form of a square, found that there were 71 men left over. How many were arranged in each row?
(a) 73
(b) 77
(c) 87
(d) 93
2. A number, when divided successively by 4, 5 and 6, leaves remainders 2, 3 and 4 respectively. The least such number is
(a) 50
(b) 53
(c) 19
(d) 214
3. A number, when divided by 296, gives 75 as the remainder. If the same number is divided by 37 then the remainder will be
(a) 1
(b) 2
(c) 19
(d) 31
4. A person bought a horse and a carriage for Rs. 20000. Later, he sold the horse at 20% profit and the carriage at 10% loss. Thus, he gained 2% in the whole transaction. The cost price of the horse
(a) Rs. 7200
(b) Rs. 7500
(c) Rs. 8000
(d) Rs. 9000
5. A sells an article to B at 15% profit. B sells it to C at 10% loss. If C pays Rs. 517.50 for it then A purchased it at
(a) Rs. 500
(b) Rs. 750
(c) Rs. 1000
(d) Rs. 1250
6. The cost price of an article is 80% of its marked price for sale. How much per cent does the tradesman gain after allowing a discount of 12% ?
(a) 20
(b) 12
(c) 10
(d) 8
7. A merchant has announced 25% rebate on prices of ready-made garments at the time of sale. If a purchase needs to have a rebate of Rs. 400, then how many shirts, each costing Rs. 320, should he
(a) 10
(b) 7
(c) 6
(d) 5
8. A merchant purchases a wristwatch for Rs. 450 and fixes its list price in a such a way that after allowing a discount of 10%, he earns a profit of 20%. Then the list price (in rupees) of the
wristwatch is
(a) 500
(b) 600
(c) 750
(d) 800
9. Ram donated 4% of his income to a charity and deposited 10% of the rest in a Bank. If now he has Rs. 8640 left with him, then his income is
(a) Rs. 12,500
(b) Rs. 12,000
(c) Rs. 10,500
(d) Rs. 10,000
10. If the length of a rectangle is increased by 10% and its breadth is decreased by 10%, then its area
(a) decreases by 1%
(b) increases by 1%
(c) decreases by 2%
(d) remains unchanged | {"url":"https://iasexamportal.com/download/ssc/cgl-tier-ii-exam-solved-paper-held-on-1-08-2010","timestamp":"2024-11-11T05:12:16Z","content_type":"text/html","content_length":"31632","record_id":"<urn:uuid:b2ccbb5b-115f-40f9-af62-680e8c94e0cf>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00541.warc.gz"} |
Empowering Children Through Toy Books
Puzzles have been an integral part of human life for centuries. They come in different shapes, sizes, and levels of difficulty. While some puzzles are easy to solve, others are incredibly challenging
and require a great deal of mental effort. In this article, we will explore the hardest puzzles to solve and what makes them so difficult. From mathematical conundrums to brain teasers, we will delve
into the world of puzzles and discover the mysteries that lie within. Get ready to challenge your mind and unlock the secrets of the most complex puzzles known to mankind.
Quick Answer:
The hardest puzzles to solve are those that require a combination of creativity, logical thinking, and perseverance. These puzzles often involve complex problem-solving, abstract concepts, and
multistep solutions. Some examples of particularly challenging puzzles include the unsolved problems in mathematics, such as the Riemann Hypothesis and the Birch and Swinnerton-Dyer Conjecture, as
well as complex problems in computer science, such as the Traveling Salesman Problem and the Halting Problem. These puzzles have stumped some of the brightest minds in their respective fields and
continue to be a source of fascination and frustration for those who attempt to solve them.
The Challenge of Puzzles
Defining Puzzles
Puzzles are a category of problems that require the application of critical thinking and problem-solving skills. They come in various forms, such as crosswords, Sudoku, jigsaw puzzles, and many
others. These puzzles can be classified into different types based on their complexity, level of difficulty, and the skills required to solve them.
However, defining puzzles can be challenging as there is no universally accepted definition. Puzzles can be described as a problem or situation that requires a solution or solution. In general,
puzzles are characterized by a lack of structure, which means that they may not have a specific starting point or clear rules. This lack of structure makes puzzles more challenging as they require
the solver to think creatively and use their problem-solving skills to arrive at a solution.
Furthermore, puzzles can be categorized based on their level of difficulty. Some puzzles are designed to be easy and accessible to a wide range of people, while others are designed to be extremely
challenging and require a high level of expertise to solve. In some cases, puzzles may be so difficult that they have never been solved.
Despite the challenges of defining puzzles, they remain a popular form of entertainment and a way to exercise the mind. Puzzles can be found in newspapers, magazines, books, and online, and they are
enjoyed by people of all ages and skill levels.
Types of Puzzles
Puzzles come in various forms, each with its unique set of challenges. Understanding the different types of puzzles can help us appreciate the complexity and diversity of the problem-solving process.
1. Logical Puzzles: These puzzles involve the use of reason and logic to solve problems. They often require the identification of patterns, deductions, and inferences to arrive at a solution.
Examples include Sudoku, crosswords, and brain teasers.
2. Mathematical Puzzles: These puzzles require a strong understanding of mathematical concepts and principles. They often involve calculations, geometry, and algebra to arrive at a solution.
Examples include the Monty Hall problem, the traveling salesman problem, and the coin-changing problem.
3. Physical Puzzles: These puzzles involve manipulating physical objects to solve problems. They often require spatial reasoning, problem-solving skills, and dexterity. Examples include Rubik’s
Cube, jigsaw puzzles, and the sliding puzzle.
4. Word Puzzles: These puzzles involve manipulating letters or words to arrive at a solution. They often require the identification of patterns, anagrams, and wordplay. Examples include cryptograms,
word searches, and scrambled words.
5. Strategy Puzzles: These puzzles involve outsmarting an opponent or navigating a complex system to arrive at a solution. They often require strategic thinking, planning, and execution. Examples
include chess, Go, and escape rooms.
Each type of puzzle has its unique set of challenges, and solving them requires different skills and strategies. Understanding the different types of puzzles can help us appreciate the diversity of
problem-solving challenges and develop the skills necessary to tackle them.
Logic Puzzles
Logic puzzles are a class of problems that require the application of logical reasoning to arrive at a solution. These puzzles often involve the manipulation of symbols, such as letters or numbers,
to arrive at a conclusion.
Types of Logic Puzzles
There are several types of logic puzzles, including:
• Sudoku: A puzzle that involves filling a grid with numbers so that each row, column, and region contains every number from 1 to 9.
• Crosswords: A puzzle that involves filling in words across and down in a grid.
• Word-search Puzzles: A puzzle that involves finding a list of words hidden in a grid of letters.
• Mathematical Puzzles: A puzzle that involves solving a mathematical problem, such as finding the value of a variable or proving a theorem.
The Appeal of Logic Puzzles
Logic puzzles are popular because they are challenging and require the use of logical reasoning to arrive at a solution. They can be used to improve problem-solving skills and cognitive abilities,
and they can be enjoyed by people of all ages.
The Difficulty of Logic Puzzles
The difficulty of logic puzzles varies depending on the type of puzzle and the level of complexity. Some puzzles may be relatively easy to solve, while others may be extremely challenging. The level
of difficulty can be increased by adding more constraints or by making the puzzle more complex.
The Reward of Solving Logic Puzzles
Solving logic puzzles can be a rewarding experience, as it allows the solver to use their logical reasoning skills to arrive at a solution. It can also be a source of pride to complete a difficult
puzzle, and it can be a way to challenge oneself and improve one’s problem-solving abilities.
Math Puzzles
Math puzzles have long been considered some of the most challenging puzzles to solve. These puzzles often require a deep understanding of mathematical concepts and a keen eye for detail. Here are
some examples of the hardest math puzzles to solve:
Fermat’s Last Theorem
One of the most famous unsolved problems in mathematics is Fermat’s Last Theorem. The theorem states that there are no positive integers a, b, and c that satisfy the equation a^n + b^n = c^n for any
integer value of n greater than 2. This theorem was first proposed by Pierre de Fermat in 1637, and it took over 350 years for the theorem to be proven by Andrew Wiles in 1994.
Riemann Hypothesis
The Riemann Hypothesis is another famous unsolved problem in mathematics. It is a conjecture about the distribution of prime numbers, and it is named after the German mathematician Bernhard Riemann.
The hypothesis states that every non-trivial zero of the Riemann zeta function has real part 1/2. This hypothesis has important implications for number theory and has yet to be proven.
Poincare Conjecture
The Poincare Conjecture is a problem in topology that was first proposed by Henri Poincare in 1904. The conjecture states that any simply connected, compact, without boundary, four-dimensional
manifold is topologically equivalent to the 4-ball. The conjecture was finally proven by Grigori Perelman in 2003, but the proof was so complex that it has yet to be fully understood.
Birch and Swinnerton-Dyer Conjecture
The Birch and Swinnerton-Dyer Conjecture is a problem in number theory that is named after the mathematicians Swinnerton-Dyer and Birch. The conjecture is about the distribution of rational integers
that are represented by a particular elliptic curve. The conjecture has important implications for the study of modular forms and has yet to be proven.
These are just a few examples of the hardest math puzzles to solve. Math puzzles challenge our problem-solving abilities and push the boundaries of human knowledge.
Word Puzzles
Word puzzles are a type of brain teaser that involve the manipulation of letters to form words. These puzzles come in many different forms, each with their own unique challenges. Some of the hardest
word puzzles to solve include crosswords, acrostics, and anagrams.
Crosswords are a popular type of word puzzle that involve filling in a grid of squares with words that fit certain clues. These clues are usually provided in the form of a definition or a description
of the word. Crosswords can be very challenging because they require a combination of knowledge and creativity to solve.
Acrostics are word puzzles that involve forming words by using the first letter of each word in a sentence or phrase. These puzzles can be very difficult because they require the solver to think
creatively and use a wide range of vocabulary.
Anagrams are word puzzles that involve rearranging the letters of a word or phrase to form a new word or phrase. These puzzles can be very challenging because they require the solver to think
creatively and use a wide range of vocabulary.
In conclusion, word puzzles are a popular type of brain teaser that come in many different forms, each with their own unique challenges. Some of the hardest word puzzles to solve include crosswords,
acrostics, and anagrams. These puzzles require a combination of knowledge and creativity to solve and can be very rewarding for those who enjoy a challenge.
The Allure of Difficult Puzzles
Puzzles have always been an integral part of human history, dating back to ancient civilizations where they were used as a form of entertainment and education. In modern times, puzzles have evolved
to become a popular form of leisure activity, challenging the brain and stimulating cognitive function. While some puzzles may be relatively easy to solve, others pose a significant challenge to even
the most skilled puzzle solvers.
The allure of difficult puzzles lies in the sense of accomplishment and satisfaction that comes with overcoming a seemingly insurmountable obstacle. These puzzles require a great deal of mental
effort and ingenuity to solve, and often involve a high degree of difficulty and complexity.
One of the main reasons why people find difficult puzzles so appealing is the sense of intellectual stimulation they provide. Solving a challenging puzzle requires the use of critical thinking
skills, problem-solving abilities, and creativity, all of which contribute to overall cognitive function. In addition, the sense of accomplishment that comes with solving a difficult puzzle can boost
confidence and self-esteem, providing a sense of pride and satisfaction.
Another reason why difficult puzzles are so appealing is the social aspect they provide. Puzzles can be enjoyed alone or with others, and solving a challenging puzzle together can foster a sense of
camaraderie and teamwork. In addition, sharing solutions and strategies with others can help to build problem-solving skills and expand one’s knowledge and understanding of different subjects.
Overall, the allure of difficult puzzles lies in the sense of accomplishment and intellectual stimulation they provide, as well as the social aspects they offer. Whether you are an experienced puzzle
solver or just starting out, tackling a challenging puzzle can be a rewarding and enriching experience.
The Benefits of Solving Hard Puzzles
Solving hard puzzles can have a plethora of benefits for individuals. It is a mental exercise that challenges the brain to think creatively and use problem-solving skills. Some of the benefits of
solving hard puzzles are:
Improves cognitive abilities
Solving hard puzzles can improve cognitive abilities such as memory, focus, and attention to detail. It also helps in developing the ability to reason and think logically. These cognitive abilities
are essential for performing daily tasks and can help in preventing cognitive decline as one ages.
Enhances problem-solving skills
Hard puzzles require a great deal of critical thinking and problem-solving skills. By repeatedly engaging in such activities, individuals can develop their problem-solving skills and become better at
identifying patterns and making connections between seemingly unrelated pieces of information.
Increases creativity
Solving hard puzzles can also increase creativity. When faced with a difficult puzzle, individuals may need to think outside the box and come up with unique solutions. This type of thinking can
translate to other areas of life and help individuals approach problems from new and innovative angles.
Reduces stress
Finally, solving hard puzzles can be a great stress reliever. Engaging in mentally stimulating activities can help individuals take their minds off of their worries and can be a calming and
therapeutic experience. Additionally, the sense of accomplishment that comes with solving a difficult puzzle can boost self-esteem and provide a sense of pride.
The Thrill of the Hunt
Puzzles have been a source of entertainment and challenge for centuries. They come in various forms, from crosswords and Sudoku to riddles and brainteasers. One of the reasons why puzzles are so
appealing is the thrill of the hunt, the excitement of solving a challenge that seems impossible at first but becomes more manageable with each step towards the solution.
There is something inherently satisfying about cracking a puzzle, whether it’s a simple math problem or a complex mystery. It requires a combination of logical thinking, creativity, and perseverance,
and the sense of accomplishment that comes with solving a puzzle is hard to beat.
But what makes some puzzles harder to solve than others? Why do some puzzles seem impossible to crack, while others can be solved with ease?
Part of the answer lies in the complexity of the puzzle itself. Some puzzles are designed to be intentionally difficult, with multiple layers of meaning and hidden clues that require a deep
understanding of the subject matter. These puzzles can take days, weeks, or even years to solve, and require a team of experts with a diverse range of skills and knowledge.
Another factor that affects the difficulty of a puzzle is the context in which it is presented. For example, a crossword puzzle that uses unfamiliar words or a riddle that relies on obscure
references can be much harder to solve than a puzzle that uses familiar words and concepts. Similarly, a puzzle that is presented in a foreign language can be much more challenging for someone who
does not speak that language fluently.
Despite the challenges, however, many people find that the thrill of the hunt is worth the effort. The sense of accomplishment that comes with solving a difficult puzzle is a unique and rewarding
experience, and it can also help to improve cognitive skills and problem-solving abilities.
So, whether you’re a seasoned puzzle solver or a newcomer to the world of puzzles, the thrill of the hunt is always waiting for you. With determination, creativity, and a little bit of luck, you too
can solve even the most challenging puzzles and experience the satisfaction of a job well done.
The Top 10 Hardest Puzzles to Solve
Key takeaway: Puzzles come in various forms, each with its unique set of challenges. Logic puzzles require the use of logical reasoning to arrive at a solution. Word puzzles involve manipulating
letters or words to arrive at a solution. Math puzzles require a deep understanding of mathematical concepts and principles. Solving hard puzzles can have a plethora of benefits for individuals,
including improving cognitive abilities, enhancing problem-solving skills, and reducing stress. Some of the hardest puzzles to solve include The Impossible Object, The Seven Bridges of Königsberg,
The Monty Hall Problem, The Birch and Swinnerton-Dyer Conjecture, The Poincare Conjecture, and The Riemann Hypothesis.
1. The Impossible Object
The Impossible Object is a classic puzzle that was invented by Dr. Sam Loyd in the late 19th century. It is considered one of the hardest puzzles to solve because it involves a paradoxical object
that cannot exist in reality.
The puzzle consists of a cube with six faces, each of which is a different color. The puzzle can be solved by arranging the six pieces into a 2×3 pattern, with each face of the cube showing a
different color. However, the twist in the puzzle is that one of the pieces is an impossible object, which is a three-dimensional object that cannot exist in reality.
The impossible object is created by using two different images of the same object on opposite sides of the cube. For example, one side of the cube might show a red cube, while the opposite side shows
a blue cube. When the pieces are arranged into the 2×3 pattern, the impossible object appears to be two different objects stacked on top of each other.
The challenge of the puzzle is to arrange the six pieces into the correct pattern while also trying to make sense of the impossible object. It requires careful observation, critical thinking, and
problem-solving skills to solve the puzzle.
In addition to its challenge, The Impossible Object has been the subject of much interest in the fields of psychology and neuroscience, as it has been used to study how the brain processes visual
information and makes sense of contradictory information.
2. The Seven Bridges of Königsberg
The Seven Bridges of Königsberg is a famous puzzle that was first introduced by the German mathematician Leonhard Euler in 1735. It is considered one of the most challenging puzzles in the field of
mathematics and graph theory. The puzzle involves finding a way to cross each of the seven bridges in the city of Königsberg, Germany, without repeating any of the bridges or the banks of the river.
The puzzle consists of a grid of land and water, with seven bridges connecting the land masses. The challenge is to find a path that crosses each of the bridges exactly once and returns to the
starting point. The problem seems simple enough, but it requires a deep understanding of graph theory and topology to solve.
Euler’s solution to the puzzle involved using a new concept called “Eulerian paths,” which are paths that start and end at the same point and pass through every edge exactly once. He showed that such
a path exists if and only if the graph is “Eulerian,” meaning that the number of edges is equal to the number of vertices plus one.
Despite the solution to the puzzle, the Seven Bridges of Königsberg remains a challenge for mathematicians and puzzle enthusiasts alike. The puzzle has inspired numerous variations and applications
in fields such as computer science, physics, and biology.
3. The Knight’s Tour
The Knight’s Tour is a classic puzzle that involves moving a knight piece around a chessboard so that it visits every square only once. Despite its simplicity, this puzzle is considered one of the
hardest problems in computer science and artificial intelligence.
One of the reasons why the Knight’s Tour is so difficult to solve is that it requires a combination of search algorithms, heuristics, and pathfinding techniques. In other words, the puzzle requires a
combination of both brute force and intelligent decision-making.
The Knight’s Tour is also interesting because it is an example of a problem that is NP-hard, which means that there is no known efficient algorithm for solving it. This means that as the size of the
chessboard increases, the time required to solve the puzzle increases exponentially.
Despite these challenges, the Knight’s Tour remains an important area of research in computer science and artificial intelligence. Researchers continue to develop new algorithms and techniques for
solving the puzzle, and some of these techniques have applications in other areas of computer science, such as robotics and logistics.
Overall, the Knight’s Tour is a fascinating puzzle that challenges our understanding of both computer science and artificial intelligence. While it may be one of the hardest puzzles to solve, it
remains an important area of research and has the potential to yield significant breakthroughs in the future.
4. The Cursed Necklace
The Cursed Necklace is a well-known puzzle that has baffled many minds for centuries. It is often considered one of the most challenging puzzles to solve due to its intricate design and enigmatic
The puzzle originated in ancient Greece, where it was said to have been crafted by a master artisan named Daedalus. The story goes that the necklace was cursed by the gods because it was made using
gold stolen from them. Since then, the necklace has been passed down through generations, with each owner meeting a tragic end.
The necklace consists of a gold chain with a pendant in the shape of an eagle. The chain is 24 inches long, and the eagle is 1 inch tall. The back of the pendant is engraved with a cryptic message in
Greek that reads, “Εἴ τι συγγονοί λαβέτε με δικαιοσύνην, οὔτε το μυλώσετε οὔτε το κατέστησεν.” which translates to “If you, my kin, take me with justice, neither will you break me nor will you use
The Puzzle
The puzzle lies in the fact that the necklace can be opened and closed without leaving any marks or evidence of tampering. The challenge is to figure out how to open the necklace without leaving any
traces of tampering.
The solution to the puzzle involves understanding the design of the necklace and the way it is held together. It requires the use of leverage and careful manipulation to open the clasp without
leaving any marks.
In conclusion, The Cursed Necklace is a fascinating puzzle that requires both intellect and patience to solve. Its intricate design and cryptic message make it a challenging and rewarding experience
for puzzle enthusiasts.
5. The Tower of Hanoi
The Tower of Hanoi is a classic puzzle that involves moving a series of disks from one peg to another. The puzzle was invented by the French mathematician Édouard Lucas in 1883 and is named after the
Vietnamese capital city of Hanoi.
The goal of the puzzle is to move all the disks from the source peg to the destination peg while following a set of rules. The rules are as follows:
• Only one disk can be moved at a time.
• Each move must be made by moving a disk from one peg to another peg.
• A larger disk cannot be placed on top of a smaller disk.
The puzzle starts with a number of disks placed on the source peg, and the goal is to move all the disks to the destination peg. The puzzle is considered solved when all the disks have been moved to
the destination peg and no disk is placed on top of a smaller disk.
The Tower of Hanoi is considered one of the hardest puzzles to solve because it requires the use of logical reasoning and the ability to visualize complex patterns. The puzzle has been studied by
mathematicians and computer scientists and has led to the development of new algorithms and data structures.
Despite its simplicity, the Tower of Hanoi is a challenging puzzle that has captivated the minds of people of all ages. Its popularity has led to the creation of many variations of the puzzle,
including the “Tower of Hanoi with a twist” and the “Tower of Hanoi with a movable peg.”
In conclusion, the Tower of Hanoi is a classic puzzle that requires logical reasoning and the ability to visualize complex patterns. Its challenging nature has made it a favorite among puzzle
enthusiasts and has led to the development of new algorithms and data structures.
6. The Monty Hall Problem
The Monty Hall problem is a well-known probability puzzle that was named after the host of the television game show “Let’s Make a Deal,” which was famously hosted by Monty Hall. The problem is based
on a hypothetical scenario in which a contestant is presented with three doors, behind one of which a prize is hidden. The contestant is then asked to choose a door and open it, after which the host,
who knows where the prize is located, will open one of the remaining doors to reveal that it does not contain the prize. The contestant is then given the option to stick with their original choice or
switch to the other remaining door. The question is whether the contestant has a better chance of winning the prize by sticking with their original choice or switching to the other door.
The solution to the problem involves understanding the concept of conditional probability, which is the probability of an event occurring given that certain conditions are met. In this case, the
probability of the contestant winning the prize by switching doors is higher than the probability of winning by sticking with the original choice. This is because the probability of the prize being
behind the originally chosen door is 1/3, while the probability of the prize being behind the other door is 2/3. Therefore, by switching doors, the contestant has a 2/3 chance of winning the prize,
which is higher than the 1/3 chance of winning by sticking with the original choice.
The Monty Hall problem is considered one of the hardest puzzles to solve because it involves a counterintuitive solution that goes against the typical human instinct to stick with the first choice.
The solution requires a deep understanding of conditional probability and the way in which it changes depending on the situation. As a result, the puzzle has been the subject of much debate and
discussion among mathematicians and puzzle enthusiasts alike.
7. The Mutilated Chessboard
The Mutilated Chessboard is a well-known puzzle that was first introduced by the mathematician Raymond Smullyan. It is a brain teaser that involves a chessboard with pieces missing, and the goal is
to determine the position of the missing pieces based on a series of clues.
The puzzle starts with the chessboard in a specific configuration, with some of the pieces missing. The player is then given a series of clues that provide information about the location of the
missing pieces. The clues are phrased in a way that requires the player to use logical reasoning and deduction to determine the positions of the missing pieces.
The puzzle is considered difficult because it requires the player to think critically and creatively. The clues are often misleading, and the player must use their ability to reason and make
deductions to arrive at the correct solution.
One of the challenges of the puzzle is that it requires the player to think outside the box and consider possibilities that may not be immediately obvious. The puzzle is also difficult because it
requires the player to remember the clues and use them to make deductions about the missing pieces.
Despite its difficulty, the Mutilated Chessboard is a popular puzzle among mathematicians and puzzle enthusiasts. It has been featured in numerous books and articles, and it has inspired many other
puzzles and brain teasers.
In conclusion, the Mutilated Chessboard is a challenging puzzle that requires logical reasoning and deduction. It is a classic puzzle that continues to fascinate puzzle enthusiasts and mathematicians
8. The Birch and Swinnerton-Dyer Conjecture
The Birch and Swinnerton-Dyer Conjecture is a mathematical problem that is considered one of the most difficult to solve. It was first proposed by David M. Birch and Swinnerton-Dyer in 1976, and it
is related to the study of elliptic curves.
The conjecture states that for any given elliptic curve, there is a formula that can be used to calculate the order of the group of rational numbers that are associated with the curve. In other
words, the conjecture is about finding a relationship between the geometry of an elliptic curve and the arithmetic of its associated rational numbers.
Despite many attempts, the Birch and Swinnerton-Dyer Conjecture remains unproven. One of the reasons for this is that it is closely related to another famous unsolved problem, the Hodge Conjecture.
In fact, the Birch and Swinnerton-Dyer Conjecture can be seen as a special case of the Hodge Conjecture, which is even more general and difficult to prove.
In recent years, there have been some partial results and partial proofs of the Birch and Swinnerton-Dyer Conjecture, but the problem remains one of the most difficult and challenging problems in
mathematics. Some experts believe that a proof of the Birch and Swinnerton-Dyer Conjecture could lead to important advances in the field of number theory and algebraic geometry.
9. The Poincaré Conjecture
The Poincaré Conjecture is a famous problem in mathematics that has stumped some of the brightest minds in the field for over a century. It was first proposed by Henri Poincaré in 1904, and it
remains one of the most important unsolved problems in topology.
The conjecture states that any closed, connected, and orientable three-dimensional manifold can be embedded in Euclidean space. In simpler terms, it suggests that any three-dimensional shape can be
transformed into a sphere by stretching and bending it.
The problem with the Poincaré Conjecture is that it is incredibly difficult to prove. In fact, it was so challenging that mathematicians didn’t even know how to approach it until the 1950s.
Despite numerous attempts by some of the greatest mathematicians of the 20th century, the Poincaré Conjecture remained unsolved. It wasn’t until 2003 that the problem was finally solved by the
mathematician Grigori Perelman.
Perelman’s proof was so complex and involved that it was initially misunderstood by many mathematicians. It wasn’t until years later that his work was fully understood and appreciated.
The Poincaré Conjecture is a prime example of a problem that seems simple on the surface but proves to be incredibly difficult to solve. Despite its resolution, the proof remains one of the most
significant achievements in mathematics in recent history.
10. The Riemann Hypothesis
The Riemann Hypothesis is a mathematical puzzle that has stumped some of the brightest minds in the field for over 150 years. It was first proposed by Bernhard Riemann in 1859 and has since become
one of the most famous unsolved problems in mathematics.
The puzzle revolves around the distribution of prime numbers, which are the building blocks of all numbers and play a crucial role in cryptography and computer science. The Riemann Hypothesis posits
that every non-trivial zero of the Riemann zeta function has real part equal to 1/2.
Despite numerous attempts, the Riemann Hypothesis remains unsolved, and its solution could have significant implications for the study of number theory and the distribution of prime numbers. The
puzzle has also spawned a whole new area of mathematics known as Riemannian geometry.
The difficulty of the Riemann Hypothesis lies in the fact that it is a highly abstract problem that requires a deep understanding of complex analysis and number theory. Even though it has been
studied extensively, there is still no consensus on how to approach the problem or even whether a proof exists.
Some of the brightest minds in mathematics have attempted to solve the Riemann Hypothesis, including the famous mathematician Leonhard Euler, who made significant contributions to the study of prime
numbers. However, despite the efforts of many mathematicians, the puzzle remains unsolved.
The Riemann Hypothesis is considered one of the most important unsolved problems in mathematics, and its solution could have significant implications for our understanding of the distribution of
prime numbers and the nature of the universe itself. Despite the difficulty of the puzzle, mathematicians continue to work on it, hoping to one day find a solution.
Strategies for Solving Hard Puzzles
1. Analyzing the Problem
One of the most crucial steps in solving difficult puzzles is analyzing the problem itself. This involves breaking down the problem into smaller components, identifying patterns, and looking for
hidden clues. Here are some strategies that can help in analyzing the problem:
Identifying the Objective
The first step in analyzing a problem is to identify the objective. What is the goal of the puzzle? What are you trying to achieve? Understanding the objective will help you focus your efforts and
avoid wasting time on irrelevant details.
Breaking Down the Problem
Once you have identified the objective, the next step is to break down the problem into smaller components. This can help you see the problem from different angles and identify patterns that might
not be immediately apparent. For example, if you are trying to solve a math problem, you might break it down into smaller sub-problems that can be solved independently.
Looking for Hidden Clues
Some puzzles have hidden clues that can help you solve the problem more easily. These clues might be subtle or difficult to spot, but they can provide valuable insight into the problem. For example,
in a crossword puzzle, the clues themselves can provide hints about the answers. In a jigsaw puzzle, the image on the box can provide a helpful reference point.
Reverse Engineering
Reverse engineering is a technique that involves breaking a problem down into its component parts and then reassembling them in a different way to arrive at a solution. This can be especially useful
when dealing with complex problems that involve multiple variables. By breaking the problem down into smaller pieces, you can identify patterns and relationships that might not be immediately
Trial and Error
Finally, trial and error is a useful strategy for solving difficult puzzles. Sometimes, the best way to solve a problem is to try different approaches until you find one that works. This can be
time-consuming, but it can also be effective in situations where there are no obvious patterns or clues to follow.
Overall, analyzing the problem is a crucial step in solving difficult puzzles. By breaking the problem down into smaller components, looking for hidden clues, and trying different approaches, you can
increase your chances of finding a solution.
2. Breaking It Down
When faced with a difficult puzzle, one of the most effective strategies is to break it down into smaller, more manageable pieces. This approach allows the solver to focus on individual elements of
the puzzle, rather than becoming overwhelmed by the complexity of the entire problem.
There are several ways to break down a puzzle. One approach is to identify the key components of the puzzle and isolate them from the rest of the problem. For example, in a complex mathematical
puzzle, the solver might begin by identifying the variables and then focusing on each variable separately.
Another approach is to break the puzzle down into smaller sub-puzzles, each of which can be solved independently. This approach is often used in puzzles that involve multiple steps or stages, such as
crossword puzzles or jigsaw puzzles. By breaking the puzzle down into smaller pieces, the solver can work on each piece individually, gradually building towards a solution.
Breaking a puzzle down also involves looking for patterns and connections between different elements of the puzzle. For example, in a logic puzzle, the solver might look for commonalities between
different clues or pieces of information, in order to identify a underlying pattern or rule.
Overall, breaking a puzzle down into smaller pieces is a powerful strategy for solving hard puzzles. By focusing on individual elements of the puzzle, the solver can gradually build towards a
solution, while also looking for patterns and connections that might help to unlock the puzzle’s secrets.
3. Trial and Error
Trial and error is a problem-solving technique that involves trying out different solutions until the correct one is found. This method is often used when the problem is not easily understood or when
the solution is not immediately apparent. In the context of puzzles, trial and error can be a useful strategy for solving some of the hardest puzzles.
One advantage of the trial and error method is that it allows for a lot of flexibility. It doesn’t require a deep understanding of the puzzle or its underlying principles, which makes it accessible
to a wide range of people. Additionally, it can be a good way to generate ideas and get a sense of what might work and what might not.
However, the downside of trial and error is that it can be time-consuming and frustrating. It involves a lot of guesswork and may not always lead to the correct solution. It can also be demotivating
if the wrong solution is tried multiple times, which can lead to feelings of frustration and discouragement.
In summary, trial and error is a useful strategy for solving some of the hardest puzzles, but it is important to keep in mind its limitations and to approach it with a flexible and open-minded
4. Seeking Help and Collaboration
• One of the most effective strategies for solving hard puzzles is seeking help and collaboration from others. This can include working with a group of individuals who have expertise in the
relevant field or seeking guidance from a mentor or coach.
• Collaborating with others can provide a fresh perspective and new ideas that may not have been considered before. Additionally, working with a group can help to break down complex problems into
smaller, more manageable pieces, making it easier to find solutions.
• Another benefit of seeking help and collaboration is the opportunity to learn from others. By working with experts in the field, individuals can gain valuable insights and knowledge that can be
applied to future puzzles and challenges.
• However, it is important to approach collaboration with a clear understanding of roles and responsibilities. Without clear communication and a defined plan of action, collaboration can quickly
become chaotic and unproductive.
• Therefore, when seeking help and collaboration, it is important to establish clear goals and objectives, define roles and responsibilities, and establish a clear plan of action. This will help to
ensure that everyone is working towards the same goal and that progress is being made efficiently and effectively.
The Future of Puzzle Solving
The Role of Technology
Technological Advances in Puzzle Solving
In recent years, technological advancements have played a significant role in revolutionizing the world of puzzles. The integration of artificial intelligence and machine learning algorithms has
enabled the creation of more complex and challenging puzzles. These technologies have also enhanced the ability of puzzle designers to create puzzles that are tailored to individual preferences and
skill levels.
Virtual and Augmented Reality
The advent of virtual and augmented reality technology has opened up new possibilities for puzzle solving. Immersive experiences created through VR and AR technology provide an enhanced level of
engagement and interactivity, making puzzles more challenging and enjoyable. This technology also allows for the creation of puzzles that are not limited by physical boundaries, enabling the design
of larger and more complex puzzles.
The Internet of Things (IoT)
The Internet of Things (IoT) has enabled the integration of physical objects into digital puzzles, creating a new dimension of interactivity. Puzzles that involve the manipulation of physical objects
can now be linked to digital interfaces, allowing for real-time feedback and adjustments to the puzzle based on the player’s performance. This technology has also facilitated the creation of
collaborative puzzles, where multiple players can work together to solve a puzzle, regardless of their physical location.
Big Data Analytics
Big data analytics has played a significant role in enhancing the difficulty of puzzles by enabling the creation of puzzles that are more complex and intricate. With the ability to analyze vast
amounts of data, puzzle designers can create puzzles that are tailored to individual preferences and skill levels, making them more challenging and engaging. Big data analytics also enables the
creation of dynamic puzzles that adapt to the player’s performance, providing a more personalized and challenging experience.
In conclusion, technological advancements have significantly impacted the world of puzzles, enabling the creation of more complex and challenging puzzles. The integration of artificial intelligence,
virtual and augmented reality, the Internet of Things, and big data analytics has revolutionized the way puzzles are designed and experienced, providing new opportunities for puzzle enthusiasts to
challenge themselves and enjoy the thrill of solving a puzzle.
The Rise of Online Puzzle Communities
As technology continues to advance, it has become easier for people to connect and share their interests online. This has led to the rise of online puzzle communities, where people can come together
to share their love of puzzles and work together to solve some of the most challenging puzzles out there.
One of the benefits of online puzzle communities is that they provide a platform for people to share their solutions and ideas with others. This can be especially helpful for those who are struggling
to solve a particular puzzle, as they can get feedback and advice from others who have already solved it.
Another benefit of online puzzle communities is that they provide a sense of community and support for puzzle enthusiasts. Puzzles can be frustrating and challenging, and it can be helpful to have a
group of people who understand and share your passion for puzzles. This can help to keep you motivated and engaged, even when you encounter difficult puzzles.
Additionally, online puzzle communities often host events and competitions, which can be a great way to test your skills and challenge yourself to solve new and difficult puzzles. These events can
also be a great way to meet other puzzle enthusiasts and make new friends who share your interests.
Overall, the rise of online puzzle communities has opened up new opportunities for puzzle enthusiasts to connect, share, and challenge themselves. Whether you are a seasoned puzzle solver or just
starting out, there is likely an online community out there that will suit your interests and help you to continue growing and improving your puzzle-solving skills.
The Continued Evolution of Puzzles
The world of puzzles is constantly evolving, with new challenges being created all the time. From the earliest puzzles, such as the classic jigsaw, to the most complex mathematical problems, puzzles
have always been a source of fascination for people of all ages. Today, there are more types of puzzles than ever before, each one designed to challenge the brain in a different way.
One of the most exciting developments in the world of puzzles is the rise of interactive puzzles. These are puzzles that are designed to be solved by a group of people, often with different skills
and backgrounds. This type of puzzle is becoming increasingly popular, as it encourages collaboration and teamwork, while still providing a challenging mental workout.
Another area where puzzles are evolving is in the use of technology. With the advent of computers and the internet, puzzles can now be created and solved in ways that were never before possible. For
example, online puzzle games allow players to compete against each other from all over the world, while interactive puzzles can be shared and solved by groups of people in real-time.
In addition to these developments, there is also a growing trend towards puzzles that are more accessible to people of all ages and abilities. This includes puzzles that are designed to be solved by
people with disabilities, as well as puzzles that are designed to be solved by children.
Overall, the future of puzzle solving looks bright, with new challenges and opportunities constantly emerging. Whether you are a seasoned puzzle solver or a newcomer to the world of puzzles, there
has never been a better time to explore the many different types of puzzles that are available.
The Timeless Appeal of Puzzles
The appeal of puzzles transcends time and has captivated individuals across generations. Puzzles have a unique capacity to engage both the mind and the hands, providing an experience that is at once
challenging and rewarding.
The Psychology of Puzzle Solving
The allure of puzzles lies in their ability to stimulate the human mind. Puzzles demand cognitive engagement, compelling individuals to employ diverse problem-solving strategies. The satisfaction
that comes from deciphering a puzzle’s solution is a testament to the inherent joy of mental challenge.
The Evolution of Puzzles
Puzzles have evolved over time, with each era introducing new forms and complexity. From the simple sliding puzzles of antiquity to the intricate logic problems of the modern age, puzzles have
continuously captivated solvers with their challenges. The enduring popularity of puzzles is a testament to the human desire for mental stimulation and the quest for solutions.
The Diversity of Puzzles
The world of puzzles is vast and varied, encompassing a wide range of styles and difficulties. From the classic crossword and Sudoku to the intricate Rubik’s Cube and beyond, puzzles cater to diverse
interests and skill levels. This variety ensures that puzzles remain accessible and engaging for solvers of all ages and backgrounds.
The Social Aspect of Puzzle Solving
Puzzles also serve as a means of social interaction, fostering collaboration and competition among solvers. Puzzle clubs, tournaments, and online communities provide platforms for individuals to
share their passion for puzzles and engage in friendly rivalry. The social aspect of puzzle solving enhances the overall experience, creating a sense of camaraderie and belonging among solvers.
In conclusion, the timeless appeal of puzzles lies in their ability to captivate the human mind, stimulate cognitive engagement, and provide a diverse and social experience. The enduring popularity
of puzzles is a testament to their power to challenge, entertain, and connect individuals across generations.
The Enduring Challenge of the Hardest Puzzles
Puzzles have been a part of human history for centuries, serving as a source of entertainment, challenge, and intellectual stimulation. The hardest puzzles, in particular, continue to captivate the
minds of individuals from all walks of life, offering a unique and enduring challenge.
The Appeal of Hard Puzzles
One of the primary reasons that hard puzzles continue to captivate individuals is their ability to push the boundaries of human cognition. These puzzles often require a significant amount of time,
effort, and creativity to solve, making the sense of accomplishment all the more rewarding.
Furthermore, hard puzzles offer a unique opportunity for personal growth and development. By tackling these challenges, individuals can enhance their problem-solving skills, increase their
resilience, and improve their overall cognitive abilities.
Over the years, puzzles have evolved to become increasingly complex and sophisticated, with new types of puzzles emerging on a regular basis. This has led to a greater variety of challenges for
individuals to tackle, ranging from traditional puzzles like crosswords and Sudoku to more modern challenges like escape rooms and immersive puzzle experiences.
In addition, advances in technology have played a significant role in the evolution of puzzles, enabling the creation of digital puzzles that offer unique challenges and experiences.
The Role of Puzzles in Society
Puzzles also play an important role in society, serving as a source of entertainment, education, and social interaction. They are often used in educational settings to help students develop critical
thinking skills, while also providing a fun and engaging way to learn.
In addition, puzzles serve as a social activity, bringing people together to collaborate and solve challenges as a team. This has led to the rise of puzzle clubs, events, and competitions, where
individuals can come together to tackle complex challenges and share their love of puzzles.
The Future of Hard Puzzles
As puzzles continue to evolve and advance, it is likely that hard puzzles will remain a popular and enduring challenge for individuals around the world. With new technologies and innovations, there
is a limitless potential for the creation of new and exciting puzzles that will captivate and challenge individuals for years to come.
Whether through traditional puzzles or immersive experiences, hard puzzles will continue to play an important role in society, serving as a source of entertainment, education, and personal growth.
The Importance of Persistence and Adaptability in Solving Puzzles
One of the key factors in successfully solving the hardest puzzles is the ability to persevere in the face of difficulty. It is important to understand that solving a challenging puzzle often
requires a significant amount of time and effort, and that setbacks and obstacles are to be expected along the way. This is where persistence comes in – the ability to continue working on a problem
even when progress seems slow or uncertain.
In addition to persistence, adaptability is also crucial in solving difficult puzzles. Puzzles that are considered “hard” often require the solver to think outside the box and approach the problem
from a new angle. This means being open to trying new approaches and being willing to adjust one’s strategy in light of new information or unexpected obstacles.
It is also important to note that the ability to persist and adapt is not something that can be developed overnight. It requires practice and a willingness to embrace failure as a necessary part of
the learning process. This means that it is important to approach difficult puzzles with a growth mindset, rather than a fixed mindset, and to view setbacks as opportunities for growth rather than as
In summary, the ability to persist and adapt is crucial in solving the hardest puzzles. By developing these skills through practice and a growth mindset, solvers can increase their chances of success
and enjoy the process of tackling challenging problems.
1. What are the hardest puzzles to solve?
There are many types of puzzles that can be considered among the hardest to solve, but some of the most challenging include:
* Rubik’s Cube: A 3D puzzle that requires the solver to manipulate the cube’s faces to align the colors in a specific pattern.
* Sudoku: A number-placement puzzle that involves filling a grid with numbers so that each row, column, and region (a specified group of cells) contains every number from 1 to 9.
* The Raven Paradox: A logic puzzle that involves determining the identity of a murderer based on a series of clues and rules.
* The Prisoners and Boxes Puzzle: A classic puzzle that involves a group of prisoners and a set of boxes, where the prisoners must figure out the contents of the boxes based on limited information.
* The Traveling Salesman Problem: A problem in optimization that involves finding the shortest possible route that visits a given set of cities and returns to the starting city.
2. What is the Rubik’s Cube?
The Rubik’s Cube is a 3D puzzle that was invented in 1974 by Hungarian sculptor and professor of architecture Ernő Rubik. It consists of a 3x3x3 matrix of smaller cubes, with each face of the cube
being a different color. The goal of the puzzle is to manipulate the cube so that each face is a solid color.
3. How do you solve a Sudoku puzzle?
Solving a Sudoku puzzle involves filling a grid with numbers so that each row, column, and region (a specified group of cells) contains every number from 1 to 9. There are several techniques that can
be used to solve a Sudoku puzzle, including:
* Looking for numbers that appear in only one row, column, or region
* Using the fact that the puzzle is symmetrical to help narrow down possible solutions
* Using the fact that the puzzle is divided into regions to help narrow down possible solutions
* Using deduction to eliminate possibilities and arrive at the solution
4. What is the Raven Paradox?
The Raven Paradox is a logic puzzle that involves determining the identity of a murderer based on a series of clues and rules. The puzzle is named after the Edgar Allan Poe story “The Murders in the
Rue Morgue,” which features a similar problem.
5. What is the Prisoners and Boxes Puzzle?
The Prisoners and Boxes Puzzle is a classic puzzle that involves a group of prisoners and a set of boxes, where the prisoners must figure out the contents of the boxes based on limited information.
The puzzle is also known as the “Prisoners and Boxes” problem or the “Three Prisoners Problem.”
6. What is the Traveling Salesman Problem?
The Traveling Salesman Problem is a problem in optimization that involves finding the shortest possible route that visits a given set of cities and returns to the starting city. The problem is often
referred to simply as the “Traveling Salesman Problem” or “TSP.” It is a classic example of a NP-hard problem, which means that there is no known efficient algorithm for solving it exactly in all
cases. However, there are approximate algorithms that can find good solutions in practice.
Solving THE HARDEST Lock Puzzle in HISTORY!! – LEVEL 10 | {"url":"https://www.decentralisenow.org/what-are-the-hardest-puzzles-to-solve/","timestamp":"2024-11-04T00:55:48Z","content_type":"text/html","content_length":"114580","record_id":"<urn:uuid:2135ff1e-92d4-4b0a-a9e5-553a71df8050>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00787.warc.gz"} |
Getting started with Isabelle: baby examples, cool proof methods
[ examples Isabelle newbies sledgehammer ]
For absolute beginners, proof assistants are daunting. Everything you do seems to go wrong. So let’s have some super simple examples that show how to get started while highlighting some pitfalls.
An algebraic identity
First, a note of caution: Isabelle/HOL is great at inferring types in expressions, but the simplest examples might well be ambiguous, leading to frustration. For example, it should be trivial to
prove $3-2=1$ using auto, but it fails. Hovering with your mouse near the blue dot in the left margin, or checking the Output panel, you might see a hint about a missing type constraint:
Isabelle sees that the problem involves numbers, but it can’t infer a precise type and therefore it’s not clear whether substraction is even meaningful. So it’s wise always to include an explicit
type constraint in problems involving numeric types. You can also use CTRL-hover (CMD-hover on Macs) to inspect the type of any variable in Isabelle/jEdit. (More on this next week!)
In the following trivial algebraic identity (due to Kevin Buzzard), we specify the type of x using fixes. It’s trivial to prove, using a single call to the simplifier.
fixes x::real
shows "(x+y)*(x+2*y)*(x+3*y) = x^3 + 6*x^2*y + 11*x*y^2 + 6*y^3"
by (simp add: algebra_simps eval_nat_numeral)
The arguments given to the simplifier are critical:
• algebra_simps is a bundle of simplification rules (simprules), containing the obvious algebraic laws: associativity and commutativity to arrange terms into a canonical order, and distributive
laws to multiply out all the terms.
• eval_nat_numeral is a single simprule that expands numerals such as 3 and 47055833459 from their internal symbolic binary notation into unary notation as a series of Suc applications to 0.
(Sadly, the second example will not terminate.)
The Suc form is necessary to trigger the simplification $a^{n+1}=a\times a^n$; this identity is called power_Suc, but it is a default simprule, meaning we don’t need to mention it.
With both rules included, simp solves the problem. Using only one of them makes the expressions blow up. A skill you need to develop is figuring out what to do when faced with a sea of symbols: did
you use too many simplification rules, or too few? A good strategy is to simplify with the fewest possible rules and gradually add more. Gigantic formulas are impossible to grasp, but close
inspection sometimes reveals subexpressions that could be eliminated through the use of another simprule.
A numerical inequality
The next example, also due to Kevin, is to show that $\sqrt 2 + \sqrt 3 < \sqrt 10$. One obvious approach is to get rid of some of the radicals by squaring both sides. So we state the corresponding
formula as a lemma using have and open a proof using the same simplification rules as in the previous example. It leaves us with the task of showing $2(\sqrt 2\sqrt 3) < 5$. Repeating the previous
idea, we use have to state that formula with both sides squared, then apply those simplification rules again. (It works because $24<25$.) Curiously, the show commands, although both inferring $x<y$
from $x^2<y^2$, require different formal justifications, both found by sledgehammer. The rest of the proof below was typed in manually.
lemma "sqrt 2 + sqrt 3 < sqrt 10"
proof -
have "(sqrt 2 + sqrt 3)^2 < (sqrt 10)^2"
proof (simp add: algebra_simps eval_nat_numeral)
have "(2 * (sqrt 2 * sqrt 3))^2 < 5 ^ 2"
by (simp add: algebra_simps eval_nat_numeral)
then show "2 * (sqrt 2 * sqrt 3) < 5"
by (smt (verit, best) power_mono)
then show ?thesis
by (simp add: real_less_rsqrt)
But there’s a much simpler way to prove the theorem above: by numerical evaluation using Johannes Hölzl’s amazing approximation tactic.
lemma "sqrt 2 + sqrt 3 < sqrt 10"
by (approximation 10)
Is it cheating? No. Working out the inequality by hand calculation is absolutely a proof. The algebraic proof above is less work for somebody who doesn’t trust calculators. However, the ability to
decide such questions by calculation (interval arithmetic, to be precise) is a huge labour-saver.
lemma "x ∈ {0.999..1.001} ⟹ ¦pi - 4 * arctan x¦ < 0.0021"
by (approximation 20)
To use this wonder-working tool, your theory file needs to import the library theory HOL-Decision_Procs.Approximation.
While we are talking about automatic tactics, Chaieb’s sos deserves a mention. It uses sum of squares methods to decide real polynomial inequalities.
fixes a::real
shows "(a*b + b * c + c*a)^3 ≤ (a^2 + a * b + b^2) * (b^2 + b * c + c^2) * (c^2 + c*a + a^2)"
by sos
A decision procedure, it always settles the question, but with too many variables you won’t live long enough to see the result. To use it, your theory needs to import HOL-Library.Sum_of_Squares.
The square root of two is irrational
I contrived this example to demonstrate sledgehammer and especially how beautifully it interacts with the development of a structured proof. I knew the mathematical proof already, so the point was to
formalise it using sledgehammer alone, without reference to other formal proofs. It also illustrates some tricky points requiring numeric types.
The irrationality of $\sqrt2$ is stated in terms of $\mathbb Q$, which in Isabelle/HOL is a weird polymorphic set: it is the range of the function of_rat, which embeds type rat into larger types such
as real and complex. So the proof begins by assuming that sqrt 2 ∈ ℚ, thus obtaining q of type rat such that sqrt 2 = of_rat q and after that, q^2 = 2. Sledgehammer was unable to derive this in a
single step from the assumption, and this step-by-step approach (thinking of a simple intermediate property) is the simplest way to give sledgehammer a hint.
We next obtain m and n such that
"coprime m n" "q = of_int m / of_int n"
Two tricks here are knowing that coprime is available, and using the embedding of_int to ensure that m and n are integers (far better than simply declaring m and n to have type int, when Isabelle may
insert embeddings in surprising places). Next we state the goal
"of_int m ^ 2 / of_int n ^ 2 = (2::rat)"
Now a super-important point: the embeddings of_nat, of_int, of_real specify their domain type, but their range type can be anything belonging to a suitably rich type class. Since 2 can also have many
types, the 2::rat is necessary to ensure that we are talking about the rationals.
The proof continues with the expected argument of showing that 2 is a divisor of both m and n, contradicting the fact that they are coprime.
lemma "sqrt 2 ∉ ℚ"
assume "sqrt 2 ∈ ℚ"
then obtain q::rat where "sqrt 2 = of_rat q"
using Rats_cases by blast
then have "q^2 = 2"
by (metis abs_numeral of_rat_eq_iff of_rat_numeral_eq of_rat_power power2_eq_square
then obtain m n
where "coprime m n" "q = of_int m / of_int n"
by (metis Fract_of_int_quotient Rat_cases)
then have "of_int m ^ 2 / of_int n ^ 2 = (2::rat)"
by (metis ‹q⇧^2 = 2› power_divide)
then have 2: "of_int m ^ 2 = (2::rat) * of_int n ^ 2"
by (metis division_ring_divide_zero double_eq_0_iff mult_2_right mult_zero_right
then have "2 dvd m"
by (metis (mono_tags, lifting) even_mult_iff even_numeral of_int_eq_iff of_int_mult
of_int_numeral power2_eq_square)
then obtain r where "m = 2*r"
by blast
then have "2 dvd n"
by (smt (verit) "2" ‹even m› dvdE even_mult_iff mult.left_commute mult_cancel_left of_int_1
of_int_add of_int_eq_iff of_int_mult one_add_one power2_eq_square)
then show False
using ‹coprime m n› ‹m = 2 * r› by simp
Every step in this proof was obtained by sledgehammer. The main skill involves thinking up the right intermediate goals when sledgehammer fails, and typing them in. Yes, formal proof really is just
another sort of coding.
You can download the theory file Baby.thy. You might want to generalise the example to show that the square root of every prime is irrational. The prime predicate and supporting theory can be
imported from HOL-Computational_Algebra.Primes.
Give it a try. Isabelle is easy to install and comes with plenty of documentation. | {"url":"https://lawrencecpaulson.github.io/2022/05/04/baby-examples.html","timestamp":"2024-11-12T15:32:22Z","content_type":"text/html","content_length":"29519","record_id":"<urn:uuid:80bcad53-831c-41bb-bc79-d0847b7a22dd>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00697.warc.gz"} |
Deep Immersion DiD campaign -- Player Instructions (UPDATED 28 Nov 2018)
A brilliant bunch of stories and reports again folks! I just got caught up with all of them during several cups of coffee. Most enjoyable!
2nd Lt. Swanson has had a busy time the last few days. The arrival of his old chum Jim Collins was a most welcomed surprise. Add to that, he was informed yesterday morning, just before patrols went
out, that his very first claim was actually confirmed! And during his arty spotting mission shortly after that, he and his gunner/obs, Lt. Christopher Dent, wound up in another scrape, this time
with a pair of E.IIIs that engaged them just east of Loos. Swany quickly put the Parasol into a turning dive as he attempted to give the Lieutenant a clear shot at their attackers. The man is a
wizard with the Lewis and scared off one of the Eindeckers immediately while the other continued to press his attack. Swany was jinxing and twisting to stay out of the enemy's line of fire, but
despite his best efforts the Hun pilot still managed to lace the side of the Morane between both cockpits. It was only pure dumb luck that resulted in neither of the British airmen being hit. After
some further turns, dives, and gyrations Christopher at last got a good burst of fire directly into the engine of the Hun plane, causing it to go into a tight spin. They lost sight of the Eindecker
as it dropped into the haze beneath them. Brief moments later Swany suddenly realized how low they gotten as bullets from the enemy trenches below went zipping past. The young pilot turned his nose
west as fast as he could, and he tossed his bus about in the process to throw off the aim of the gunners. He ended up with a handful of vents in his right wing anyway. Swany and Christopher then
attempted to locate the other two members of their flight but to no avail and finally had to give up looking and return to camp without them. They learned later that both had been damaged in fights
with other EA and had been forced to land, one on the western edge of Loos, and the other in a field about two miles short of Auchel. Once back home, the team of Swanson and Dent turned in their
reports and claim forms and went for breakfast where they were told a short while later that the main wing spar in their mount had been shot through and it would take until tomorrow evening to
repair it. There would be no flying for them until the 19th, at the earliest, as the squadron was now short of available aeroplanes. This was just fine with Swany as it would likely take that long
for the young man's nerves to settle back down to a reasonable level of calm.
Not what one wants to see coming at them.
Also, not what one wants to see coming up behind them. Thank God for a gunner/obs who knows how to shoot.
Finding one's self far too low over No Man's Land and incurring the wrath of the enemy gunners.
Back at Auchel, relatively safe and sound, despite the holes in the fuselage and wing. | {"url":"https://simhq.net/forum/ubbthreads.php/posts/4457812","timestamp":"2024-11-05T23:20:15Z","content_type":"text/html","content_length":"1049231","record_id":"<urn:uuid:1f58563e-24cb-4474-b18b-e7f14e18c384>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00524.warc.gz"} |
TOCSY Toolbox
Potsdam Institute for Climate Impact Research (PIK)
Interdisciplinary Center for Dynamics of Complex Systems (University of Potsdam)
Cardiovascular Physics Group (Humboldt-Universität zu Berlin)
TOCSY - Toolboxes for Complex Systems
PIK/ Antique/ Blue
Adaptive Filtering
CRP Toolbox
Commandline RPs
Coupling Analysis
Coupling Direction
System Identification
IOTA – Inner Composition Alignment
General Notes
Inner composition alignment (IOTA) is a permutation-based association measure to detect regulatory links from very short time series. One time series is reodered with regards to the rank order of a
second one and it monotonicity is evaluated.
Installation and Usage
To install, copy the zip-file into your path and extract all *.R and *.c files.
IOTA.R can be used without any further compilation. It allows to calculate pairwise IOTA with different weighting functions, time reversal or as a signed version (cf. example below).
To have a significance test and the estimation of partial IOTA being included, compile the *.c files calling R CMD SHLIB *.c instead of the standard C compiler. The dynamic libraries called by
iota_subroutines.R will be generated in that way.
Note: (1) This has been tested only on a Linux machine. (2) The program iota_subroutines.R is not running stable for large input at the moment.
# load or simulate time series as an array of size m*n, with m number of variables
# and n number of time points
TS0 <- matrix(runif(70,0,1),7,10)
# normalize time series
# to estimate the weighted IOTA the time series must have values between zero and one
# depending on the time series different normalization must be used
TimeSeries <- (TS0-apply(TS0,1,min,na.rm=TRUE))/apply((TS0-apply(TS0,1,min,na.rm=TRUE)),1,max,na.rm=TRUE)
# load subroutines
# calculates pairwise IOTA as described in Hempel et al., PRL (2011) [1]
# possible options for method (weighting functions)
# 'both' (default): uniform and squared sloped
# 'slope': slope
# 'sqrt': squared slope
# 'am': arithmetric mean
# 'gm': geometric mean
# 'hm': harmonic mean
I <- IOTA(TimeSeries,method='sqrt')
# calculates pairwise IOTA based on reversed ordering as described in
# Hempel et al., EPJB (2013) [2]
# options are the same as for IOTA
Ir <- IOTA_reverse(TimeSeries,method='sqrt')
# calculates signed version of pairwise IOTA to indicate up-/downregulation as described
# in Hempel et al., EPJB (2013) [2]
# option 'both' does not work in this case
Is <- IOTAsigned(TimeSeries,method='sqrt')
# to run C subroutines files must be compiled to get a dynamic library using "R CMD SHLIB *.c"
# number of realizations for significance test
rmax <- 1000
# significance level
alpha <- 0.99
# weighting: uniform (1) or squared slope (2)
w <- 2
# calculated pairwise and partial IOTA and performs a simple permutation-based significance
# test, only most likely connections are selected while the remaining matix entries are set
# to zero
I <- IOTA(TimeSeries,rmax,alpha,w)
1. Hempel, S., Koseska, A., Kurths, J., Nikoloski, Z.: Inner Composition Alignment for Inferring Directed Networks from Short Time Series, Phys. Rev. Lett., 107(5), 054101, 2011, doi:10.1103/
2. Hempel, S., Koseska, A., Nikoloski, Z.: Data-driven reconstruction of directed networks, Europ. Phys. J. B, 86, 250, 2013, doi:10.1140/epjb/e2013-31111-8
Sabrina Hempel
© 2004-2024 SOME RIGHTS RESERVED
University of Potsdam, Interdisciplinary Center for Dynamics of Complex Systems, Germany
Potsdam Institute for Climate Impact Research, Complexity Science, Germany
This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 2.0 Germany License.
Imprint, Data policy, Disclaimer, Accessibility statement
Please respect the copyrights! The content is protected by the Creative Commons License. If you use the provided programmes, text or figures, you have to refer to the given publications and this web
site (tocsy.pik-potsdam.de) as well.
@ MEMBER OF PROJECT HONEY POT
Spam Harvester Protection Network
provided by Unspam
Potsdam Institute for Climate Impact Research (PIK)
Interdisciplinary Center for Dynamics of Complex Systems (University of Potsdam)
Cardiovascular Physics Group (Humboldt-Universität zu Berlin)
Adaptive Filtering
CRP Toolbox
Commandline RPs
Coupling Analysis
Coupling Direction
System Identification
IOTA – Inner Composition Alignment
General Notes
Inner composition alignment (IOTA) is a permutation-based association measure to detect regulatory links from very short time series. One time series is reodered with regards to the rank order of a
second one and it monotonicity is evaluated.
Installation and Usage
To install, copy the zip-file into your path and extract all *.R and *.c files.
IOTA.R can be used without any further compilation. It allows to calculate pairwise IOTA with different weighting functions, time reversal or as a signed version (cf. example below).
To have a significance test and the estimation of partial IOTA being included, compile the *.c files calling R CMD SHLIB *.c instead of the standard C compiler. The dynamic libraries called by
iota_subroutines.R will be generated in that way.
Note: (1) This has been tested only on a Linux machine. (2) The program iota_subroutines.R is not running stable for large input at the moment.
# load or simulate time series as an array of size m*n, with m number of variables
# and n number of time points
TS0 <- matrix(runif(70,0,1),7,10)
# normalize time series
# to estimate the weighted IOTA the time series must have values between zero and one
# depending on the time series different normalization must be used
TimeSeries <- (TS0-apply(TS0,1,min,na.rm=TRUE))/apply((TS0-apply(TS0,1,min,na.rm=TRUE)),1,max,na.rm=TRUE)
# load subroutines
# calculates pairwise IOTA as described in Hempel et al., PRL (2011) [1]
# possible options for method (weighting functions)
# 'both' (default): uniform and squared sloped
# 'slope': slope
# 'sqrt': squared slope
# 'am': arithmetric mean
# 'gm': geometric mean
# 'hm': harmonic mean
I <- IOTA(TimeSeries,method='sqrt')
# calculates pairwise IOTA based on reversed ordering as described in
# Hempel et al., EPJB (2013) [2]
# options are the same as for IOTA
Ir <- IOTA_reverse(TimeSeries,method='sqrt')
# calculates signed version of pairwise IOTA to indicate up-/downregulation as described
# in Hempel et al., EPJB (2013) [2]
# option 'both' does not work in this case
Is <- IOTAsigned(TimeSeries,method='sqrt')
# to run C subroutines files must be compiled to get a dynamic library using "R CMD SHLIB *.c"
# number of realizations for significance test
rmax <- 1000
# significance level
alpha <- 0.99
# weighting: uniform (1) or squared slope (2)
w <- 2
# calculated pairwise and partial IOTA and performs a simple permutation-based significance
# test, only most likely connections are selected while the remaining matix entries are set
# to zero
I <- IOTA(TimeSeries,rmax,alpha,w)
1. Hempel, S., Koseska, A., Kurths, J., Nikoloski, Z.: Inner Composition Alignment for Inferring Directed Networks from Short Time Series, Phys. Rev. Lett., 107(5), 054101, 2011, doi:10.1103/
2. Hempel, S., Koseska, A., Nikoloski, Z.: Data-driven reconstruction of directed networks, Europ. Phys. J. B, 86, 250, 2013, doi:10.1140/epjb/e2013-31111-8
Sabrina Hempel
Inner composition alignment (IOTA) is a permutation-based association measure to detect regulatory links from very short time series. One time series is reodered with regards to the rank order of a
second one and it monotonicity is evaluated.
To install, copy the zip-file into your path and extract all *.R and *.c files.
IOTA.R can be used without any further compilation. It allows to calculate pairwise IOTA with different weighting functions, time reversal or as a signed version (cf. example below).
To have a significance test and the estimation of partial IOTA being included, compile the *.c files calling R CMD SHLIB *.c instead of the standard C compiler. The dynamic libraries called by
iota_subroutines.R will be generated in that way.
Note: (1) This has been tested only on a Linux machine. (2) The program iota_subroutines.R is not running stable for large input at the moment.
# load or simulate time series as an array of size m*n, with m number of variables # and n number of time points TS0 <- matrix(runif(70,0,1),7,10) # normalize time series # to estimate the weighted
IOTA the time series must have values between zero and one # depending on the time series different normalization must be used TimeSeries <- (TS0-apply(TS0,1,min,na.rm=TRUE))/apply((TS0-apply
(TS0,1,min,na.rm=TRUE)),1,max,na.rm=TRUE) ################################################################### ################################################################### # load subroutines
source('IOTA.R') # calculates pairwise IOTA as described in Hempel et al., PRL (2011) [1] # possible options for method (weighting functions) # 'both' (default): uniform and squared sloped # 'slope':
slope # 'sqrt': squared slope # 'am': arithmetric mean # 'gm': geometric mean # 'hm': harmonic mean I <- IOTA(TimeSeries,method='sqrt') # calculates pairwise IOTA based on reversed ordering as
described in # Hempel et al., EPJB (2013) [2] # options are the same as for IOTA Ir <- IOTA_reverse(TimeSeries,method='sqrt') # calculates signed version of pairwise IOTA to indicate up-/
downregulation as described # in Hempel et al., EPJB (2013) [2] # option 'both' does not work in this case Is <- IOTAsigned(TimeSeries,method='sqrt') #################################################
################## ################################################################### # to run C subroutines files must be compiled to get a dynamic library using "R CMD SHLIB *.c" source
('iota_subroutines.R') # number of realizations for significance test rmax <- 1000 # significance level alpha <- 0.99 # weighting: uniform (1) or squared slope (2) w <- 2 # calculated pairwise and
partial IOTA and performs a simple permutation-based significance # test, only most likely connections are selected while the remaining matix entries are set # to zero I <- IOTA
Hempel, S., Koseska, A., Kurths, J., Nikoloski, Z.: Inner Composition Alignment for Inferring Directed Networks from Short Time Series, Phys. Rev. Lett., 107(5), 054101, 2011, doi:10.1103/
Hempel, S., Koseska, A., Nikoloski, Z.: Data-driven reconstruction of directed networks, Europ. Phys. J. B, 86, 250, 2013, doi:10.1140/epjb/e2013-31111-8
© 2004-2024 SOME RIGHTS RESERVED University of Potsdam, Interdisciplinary Center for Dynamics of Complex Systems, Germany Potsdam Institute for Climate Impact Research, Complexity Science, Germany
This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 2.0 Germany License. Imprint, Data policy, Disclaimer, Accessibility statement
Please respect the copyrights! The content is protected by the Creative Commons License. If you use the provided programmes, text or figures, you have to refer to the given publications and this web
site (tocsy.pik-potsdam.de) as well.
@ MEMBER OF PROJECT HONEY POT
Spam Harvester Protection Network
provided by Unspam | {"url":"https://tocsy.pik-potsdam.de/iota.php","timestamp":"2024-11-08T02:29:49Z","content_type":"text/html","content_length":"26684","record_id":"<urn:uuid:043fd214-b54a-48e3-80d9-2ae7ee9e4a4b>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00260.warc.gz"} |
Mathematical Modeling with Optimization, Part 4: Problem-Based Nonlinear Programming
Problem-Based Nonlinear Programming | Mathematical Modeling with Optimization, Part 4
From the series: Mathematical Modeling with Optimization
Express and solve a nonlinear optimization problem with the problem-based approach of Optimization Toolbox™. Interactively define the variables, objective function, and constraints to reflect the
mathematical statement of the nonlinear program.
Start by creating an optimization problem to hold the problem. Next, define optimization variables and their bounds. Each optimization variable has its own display name, dimension, type, and bounds.
Define one or more scalar or array variables to match the variables used in the mathematical statement.
Create the objective and constraints with optimization expressions built with the optimization variables. Specify them directly for rational expressions. Specify other expressions with MATLAB^®
functions and convert into optimization expressions with a conversion function. The conversion facility makes it easy to define an optimization problem using existing functions.
Use the display functions to review the completed optimization problem. Then specify an initial point and solve. The type of solver is automatically selected based on the type of variables,
objective, and constraints, relieving you of needing to know the many available solvers.
Published: 28 Mar 2019
This video shows how to set up and solve a constrained nonlinear optimization problem in MATLAB^®. In this example, the goal is to minimize this multivariable objective function subject to the
following constraints. Plot the objective function and constraints. The contour lines show the objective function. The feasible region is inside the blue ellipse and below the red curve.
This is a nonlinear optimization problem. There are two ways to solve nonlinear optimization problems in MATLAB: using a problem-based approach or a solver-based approach. This example uses a
problem-based approach, which uses optimization variables to define the objective and constraints. See the documentation for the solver-based approach.
There are common steps to solving a nonlinear problem with this approach. First, you set up the problem, define optimization variables, define the objective function and constraints, and solve the
Now that we have expressed the problem mathematically, we need to express the problem in MATLAB. Create an empty optimization problem container. The optimization problem holds the problem
information, including the objective function and constraints.
Next, we will define the optimization variables. Generally, optimization variables can be scalars, vectors, matrices, or N-D arrays. This example uses variables x and y, which are scalars. Create
scalar optimization variables for this problem. Include the bounds on the variables.
Next, we'll create an optimization expression for the objective function. Currently, optimization expressions do not support exponentials, so write this as a standard MATLAB function. To use this
objective function in the problem-based approach, you must use a conversion function, which creates an optimization expression. The file name of the objective function is passed with the @ "at"
symbol, which creates a "function handle." This tells MATLAB to identify or "point to" the function, but not to execute the function as MATLAB typically would do without the symbol. Now, add the
objective function to the optimization problem. The problem now shows a non-empty objective and associated variables.
This problem has the following nonlinear constraints. The first is a constraint that the solution lies in the ellipse. You can define this constraint as it is written and add it to the problem. The
previous constraint was a polynomial inequality and could be expressed as an optimization expression. The second constraint has an exponential term and cannot be written as an optimization
expression. This also has extra parameters beyond x and y and includes the variable a. Create a function with inputs x, y, and a. Convert the function to an optimization expression. Include the
optimization variables and the parameter a, defined in the MATLAB workspace. Express the inequality and add the constraint to the problem.
Now we'll check that the problem formulation is complete. The optimization variables, objective function, constraints, and bounds all look correct.
Before solving, we need to define an initial point. The initial values for x and y must be defined as a structure. Create a structure representing the initial point asx = -3, y = 3. Solve the
problem from the initial point. In general, the exit message indicates the stopping conditions and any problems encountered during the optimization. Here, the exit message and exit flag indicate that
the optimization completed successfully.
Try solving the problem from a different initial point. Request additional outputs about the solution. The optimization again completed successfully but converged to a different solution. This has a
higher objective function value than the first, which indicates this solution is not as good.
Add the solution points to the visualization. The plot shows that one solution lies on the boundary of the ellipse and the other lies on the boundary of the exponential constraint and the ellipse.
This video illustrated solving a constrained nonlinear optimization problem. See the documentation for additional examples.
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How Many Wheelbarrows Are in a Yard?
A yard is a unit of measure commonly used in calculating the amount of landscaping and gardening products like mulch, compost, and topsoil.
Once you start making any home improvement on your lawn, there will come a time when you need to calculate the numbers of wheelbarrows you need to complete the job.
By being able to calculate the number of wheelbarrows you need to get the job done, you can save a lot of time and money. So, how many wheelbarrows make a yard? Well, it depends.
This post will comprehensively cover the number of wheelbarrows you will need for a yard, emphasizing the different sizes of wheelbarrows.
What is a yard?
For the unacquainted, a yard is a unit of measure used to measure 3-dimensional items, with one cubic yard being 3 feet long, 3 feet wide, and 3 feet deep. In the US, a cubic yard is the standard
unit for measuring mulch, sand, topsoil, gravel, compost, or any aggregate that you may need in gardening or landscaping.
Importantly, measuring a cubic yard differs significantly from measuring weight.
Understanding how many wheelbarrows make a yard becomes handy when ordering gardening, landscaping, or construction items.
Whether you order mulch, topsoil, or mulch, the gardening merchant will most certainly deliver it in a bulk bag that measures a cubic yard or a truckload measured in yards.
Once your order is delivered, you will need to get it to the site, typically using a wheelbarrow. Knowing how many wheelbarrows you need to walk for every yard will ensure you plan accordingly.
How many wheelbarrows make a cubic yard?
Wheelbarrow sizes vary, but common wheelbarrows sizes are 2 and 3 cubic feet. As such, how many wheelbarrows make a yard depends immensely on the size of your wheelbarrow.
A yard, on the other hand, is 27 cubic feet. The best way to know the number of wheelbarrows that will make a yard is to divide the size of your wheelbarrow into 27 cubic feet.
For those having trouble figuring out how many wheelbarrows they need for their gardening or landscaping project, one cubic yard is equal to:
• 5 (≈14) loads of 2 cubic foot wheelbarrow
• 9 loads of 3 cubic foot wheelbarrow
• 8 (≈7) loads of 4 cubic foot wheelbarrow
• 4 (≈6) loads of 5 cubic foot wheelbarrow
• 5 loads of 6 cubic foot wheelbarrow
Wheelbarrows come in many shapes and sizes, usually suited for different applications. For this reason, before you start using your wheelbarrow
Wheelbarrow capacity
The 3-cubic foot-sized wheelbarrow is unquestionably the most common size of wheelbarrows, with 2-cubic foot-sized wheelbarrows seconding closely. If you have just purchased a new wheelbarrow and are
unsure of its capacity, you can determine its volume by measuring the power of the tray.
This can be easily achieved by splitting the measurement into two different units – the flat base capacity and the tray’s sloped section measure – and then summing them to get the total capacity.
To determine your wheelbarrow tray’s flat base area, measure the length and width of the inside flat base, then use the ordinary surface area formula (length x width) to calculate the base area.
Next, measure the inside height (depth) of your wheelbarrow tray, then multiply the height by the base area to get a cubic capacity the flat area can hold for that height.
The capacity of the sloped section is usually equal to half the capacity of the base.
Divide the capacity the flat base can hold by two, then sum the result with the capacity. This should give you the capacity of your wheelbarrow tray.
Standard wheelbarrows come with either shallow trays or deep trays. Wheelbarrows with shallow trays are usually 2-cubic feet, perfectly suited for gardening and ordinary household activities. They
are lightweight and compact to serve ordinary homeowners efficiently.
Deep tray wheelbarrows, on the other hand, come in varying sizes, ranging from 3 to 6 cubic feet. They are better suited for homeowners and seasoned gardeners.
There is a special category of deep-tray wheelbarrows known as contractor wheelbarrows. These wheelbarrows are designed to be very big with a capacity of 6 to 10 cubic feet and are best suited for
professionals, particularly in construction.
Related Article: How Much Does a Cubic Yard of Dirt Weigh? | {"url":"https://www.gfloutdoors.com/how-many-wheelbarrows-are-in-a-yard/","timestamp":"2024-11-14T18:09:42Z","content_type":"text/html","content_length":"80290","record_id":"<urn:uuid:bbedf624-cf9e-4f38-ab57-ece7521df569>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00292.warc.gz"} |
math gcd Python - Find Greatest Common Divisor with math.gcd() Function
With Python, we can calculate the greatest common divisor of two numbers with the math gcd() function.
import math
The Python math module has many powerful functions which make performing certain calculations in Python very easy.
One such calculation which is very easy to perform in Python is finding the greatest common divisor (GCD) of two numbers.
We can find the GCD of two numbers easily with the Python math module gcd() function. The math.gcd() function takes two integers and returns the GCD of those two integers.
Below are some examples of how to use math.gcd() in Python to find the GCD of two numbers.
import math
How to Get the Greatest Common Divisor of a List in Python with gcd() Function
To find the GCD of a list of numbers in Python, we need to use the fact that the GCD of a list of numbers will be the max of all pairwise GCDs in a list of numbers.
To get the GCD of a list of integers with Python, we loop over all integers in our list and find the GCD at each iteration in the loop.
Below is an example function in Python which will calculate the GCD of a list of integers using a loop and the math gcd() function.
import math
def gcd_of_list(ints):
gcd = math.gcd(ints[0],ints[1])
for i in range(2,len(ints)):
gcd = math.gcd(gcd,ints[i])
return gcd
Hopefully this article has been useful for you to understand how to use the gcd() math function in Python to find the greatest common divisors of a list of numbers. | {"url":"https://daztech.com/math-gcd-python/","timestamp":"2024-11-07T13:21:32Z","content_type":"text/html","content_length":"241144","record_id":"<urn:uuid:997ae5c0-379f-40d2-a0ba-0bb4726d522d>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00386.warc.gz"} |
Exploring Gauge-Higgs Inflation with Extra Dimensions: U(1) Gauge Theory on a Warped Background
(1) Toshiki Kawai, Department of Physics, Hokkaido University, Sapporo 060-0810, Japan (E-mail: t-kawai@higgs3.sci.hokudai.ac.jp);
(2) Yoshiharu Kawamura, Department of Physics, Shinshu University, Matsumoto 390-8621, Japan (E-mail: haru@azusa.shinshu-u.ac.jp).
Table of Links
2 U(1) gauge theory on a warped background
3 Gauge-Higgs inflation on a warped background
4 Conclusions and discussions, Acknowledgements, and References
2 U(1) gauge theory on a warped background
2.1 Randall-Sundrum metric and action integral
The spacetime is assumed to be 5d one with the RS metric given by [8, 9]
2.2 Conjugate boundary conditions
where β is a constant called a twisted phase, the superscript C denotes a 4d charge conjugation, θC is a real number, and the asterisk means the complex conjugation. Then, the covariant derivatives
obey the relations:
2.3 Mass spectrum
Then, the action integral is rewritten as
2.4 Effective potential
Let us derive the effective potential for the Wilson line phase θ(= θ(x)). Taking the standard procedure, a d-dimensional effective potential involving one degree of freedom at the one-loop level is
given b
[3] We introduce both Mψ and cσ′ (y) in a general standpoint, and we will see that cσ′ (y) is forbidden by imposing specific boundary conditions on fields in the next subsection. | {"url":"https://companybrief.tech/exploring-gauge-higgs-inflation-with-extra-dimensions-u1-gauge-theory-on-a-warped-background","timestamp":"2024-11-11T11:50:12Z","content_type":"text/html","content_length":"17277","record_id":"<urn:uuid:ef34ecea-6176-4cbf-be23-9c4db43f1a5c>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00000.warc.gz"} |
"Finite Field Arithmetic." Chapter 21A-Ter: Fix for a False Alarm in Ch.14; "Litmus" Errata.
This article is part of a series of hands-on tutorials introducing FFA, or the Finite Field Arithmetic library. FFA differs from the typical "Open Sores" abomination, in that -- rather than trusting
the author blindly with their lives -- prospective users are expected to read and fully understand every single line. In exactly the same manner that you would understand and pack your own parachute.
The reader will assemble and test a working FFA with his own hands, and at the same time grasp the purpose of each moving part therein.
• Chapter 21A-Ter: Fix for a False Alarm in Ch.14; "Litmus" Errata.
You will need:
• A Keccak-based VTron (for this and all subsequent chapters.)
• All of the materials from Chapters 1 - 21A-Bis.
Add the above vpatches and seals to your V-set, and press to ffa_ch21a_ter_ch14_ch20_errata.kv.vpatch.
You should end up with the same directory structure as previously.
As of Chapter 21A-Ter, the versions of Peh and FFA are 250 and 199, respectively.
Now compile Peh:
But do not run it quite yet.
This Chapter concerns fixes for several flaws recently reported by a careful Finnish reader known only as cgra. Thank you, cgra!
Let's begin with his first find: a false alarm bug in Chapter 14B's implementation of Barrett's Modular Reduction. (Note that the proofs given in Ch.14A and Ch.14A-Bis presently stand; the bug exists
strictly in the Ada program.)
Recall Step 5 of the algorithm given in Ch.14A :
For each new input X, to compute the reduction R := X mod M:
1. X[s] := X >> J[M]
2. Z := X[s] × B[M]
3. Z[s] := Z >> S[M]
4. Q := Z[s] × M
5. R := X - Q
6. R := R - M, C := Borrow
7. R := R + (M × C)
8. R := R - M, C := Borrow
9. R := R + (M × C)
10. R := R - (R × D[M])
11. R is now equal to X mod M.
... and its optimization, as suggested by the physical bounds proof of Ch.14A-Bis :
Ignore X
←W[M] - L→ ←W[M] + L→
- Ignore Q
←W[M] - L→ ←W[M] + L→
= R
←W[M] + L→
... and finally, its implementation in Chapter 14B :
-- Reduce X using the given precomputed Barrettoid.
procedure FZ_Barrett_Reduce(X : in FZ;
Bar : in Barretoid;
XReduced : in out FZ) is
-- R is made one Word longer than Modulus (see proof re: why)
Rl : constant Indices := Ml + 1;
-- Barring cosmic ray, no underflow can take place in (4) and (5)
NoCarry : WZeroOrDie := 0;
-- (5) R := X - Q (we only need Rl-sized segments of X and Q here)
FZ_Sub(X => X(1 .. Rl), Y => Q(1 .. Rl),
Difference => R, Underflow => NoCarry);
Even though we had demonstrated that Q ≤ X, the prohibition of a nonzero subtraction borrow in (5) is fallacious.
To illustrate: this Tape, on a 256-bit run of Peh :
... will not print the expected answer to the given modular exponentiation, i.e.:
... with a Verdict of Yes; but instead will print nothing, and yield a Verdict of EGGOG. Specifically, Peh will halt at (5) via a Constraint_Error (range check failed), when the range of NoCarry's
WZeroOrDie type is violated by an assignment of 1.
This is because -- early in this modular exponentiation's sequence of Barrett reductions -- and immediately prior to (5) :
X == 0x40000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
... but what will be actually computed in (5) is X(1 .. Rl) - Q(1 .. Rl), i.e.:
0x00000000000000000000000000000000000000000000000000000000000000000000000000000000 -
1 (Underflow == 1)
... that is, the borrow bit is legitimately 1, in this and in a number of other readily-constructed cases. The constraints we have demonstrated for X, Q, and R do not imply that a borrow will never
occur in the subtraction at (5). Therefore, the intended cosmic ray detector is strictly a source of false alarms, and we will remove it:
-- Reduce X using the given precomputed Barrettoid.
procedure FZ_Barrett_Reduce(X : in FZ;
Bar : in Barretoid;
XReduced : in out FZ) is
-- Borrow from Subtraction in (5) is meaningless, and is discarded
IgnoreC : WBool;
pragma Unreferenced(IgnoreC);
-- (5) R := X - Q (we only need Rl-sized segments of X and Q here)
FZ_Sub(X => X(1 .. Rl), Y => Q(1 .. Rl),
Difference => R, Underflow => IgnoreC); -- Borrow is discarded
... and that's it.
Cgra's second find concerned the Ch.20 demo script, Litmus. He had discovered that two mutually-canceling bugs exist in the program. Specifically, in :
# Hashed Section Length
get_sig_bytes 2
# Hashed Section (typically: timestamp)
get_sig_bytes $sig_hashed_len
# Unhashed Section Length
get_sig_bytes 1
# Unhashed Section (discard)
get_sig_bytes $sig_unhashed_len
# RSA Packet Length (how many bytes to read)
get_sig_bytes 1
# The RSA Packet itself
get_sig_bytes $rsa_packet_len
# Digest Prefix (2 bytes)
get_sig_bytes 2
... the Unhashed Section Length is erroneously treated as a 1-byte field, whereas in reality the GPG format gives 2 bytes. The script only worked (on all inputs tested to date) on account of the
presence of the superfluous routine (RSA Packet reader, which remained from an early version of the demo!); in all of the test cases to date, the second byte of the Unhashed Section Length (and the
unhashed section in its entirety, for so long as it does not exceed 255 bytes in length -- which it appears to never do) are consumed by get_sig_bytes $rsa_packet_len.
I am almost pleased that I had made this mistake; it is in fact a better illustration of programs which operate correctly despite erroneous logic -- as well as the unsuitability of shell script as a
language for nontrivial tasks -- than anything I could readily unearth in the open literature.
And the fix is readily obvious :
# Hashed Section Length
get_sig_bytes 2
# Hashed Section (typically: timestamp)
get_sig_bytes $sig_hashed_len
# Unhashed Section Length
get_sig_bytes 2
# Unhashed Section (discard)
get_sig_bytes $sig_unhashed_len
# Digest Prefix (2 bytes)
get_sig_bytes 2
I also incorporated cgra's earlier suggestion regarding error checking. Thank you again, cgra!
And that's it for Litmus, presently.
~ The next Chapter, 21B, will (yes!) continue the Extended-GCD sequence of Chapter 21A. ~
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MANDATORY: Please prove that you are human:
52 xor 38 = ?
What is the
serial baud rate
of the
FG device
Answer the riddle correctly before clicking "Submit", or comment will NOT appear! Not in moderation queue, NOWHERE!
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Numerical Comprehension Questions To Practice For Finance Roles
What is Numerical Comprehension?
Numerical comprehension is a combination of mathematical skill and knowledge with an understanding of the way math can be applied and used in different situations.
It is the recognition of mathematical concepts and principles, and the ability to manipulate numerical data using math to analyze and interrogate it, making it easier to come to a logical conclusion.
Numerical comprehension is often referred to as numerical reasoning, especially when it comes to pre-employment testing - usually because in aptitude tests, you are being evaluated on your ability to
use basic mathematical operations and principles to analyze presented data so that you can answer a question about it in a mathematically reasoned way.
Importance of Numerical Comprehension in Finance Roles
Finance roles are all about the numbers of a business, all coming down to the management of profit and loss.
In finance, you will have to be comfortable calculating the percentage of sales tax on items, understand how to apply pension contributions in payroll, and arrange for the payment of invoices.
Money is obviously the biggest part of the finance role, from ensuring that staff get paid, suppliers are paid, and that customers pay their bills - and that means you’ll need to analyze incoming
payments and outgoing payments to make sure that everything balances.
You don't necessarily need to be an advanced mathematician to work in finance, but you do need to know how to work with numbers in different ways.
Companies that assess numerical comprehension
There are many companies that use numerical comprehension tests as part of their recruitment process.
Almost every finance-related position in almost every company will involve some sort of numeracy question, whether that is numerical reasoning, financial reasoning, or even just a simple numerical
ability test.
Some companies that use numerical comprehension or reasoning tests include:
• Johnson & Johnson
• Microsoft
• Apple
The numerical tests might be used independently, or they might be part of a wider battery of aptitude tests.
Numerical comprehension tests
A numerical comprehension test is usually used as part of the recruitment process, delivered as the next stage once your application has been sifted and you’ve met the basic criteria for a role.
In numerical comprehension tests, you will be asked questions about numerical data presented in different ways, and you will have to demonstrate that you can read, understand, and manipulate that
data using appropriate mathematical principles to answer the question.
Most numerical comprehension and reasoning tests are relatively short; you’ll get around a minute to answer the question, and you may not be able to use a calculator - which means you’ll have to
answer quickly using your mental mathematics skills.
For employers, the numerical comprehension test is a benchmark, set to ensure that candidates are able to work with numerical data at a level that makes them suitable for the role. Each candidate has
the same opportunity to demonstrate that they are competent and able, and those that do not reach the benchmark will be filtered out.
The data provided by the results of the numerical comprehension test make it much easier for recruitment teams to make decisions about which candidates have what it takes to be successful in the
Numerical Comprehension Practice Questions
Exercise 1: Basic Calculations
Basic calculations are usually just the arithmetic operations that you use in everyday life (addition, subtraction, multiplication, and division).
It also includes fractions, decimals, and percentages.
Solve this equation:
(2+3) + 4 = ?
a) 7
b) 8
c) 9
d) 10
Answer: c) 9 - You need to remember to solve the brackets first.
These basic calculations are ones that you will use to help you solve many of the other types of questions in the numerical comprehension test, so you need to have a good grasp of how to answer them.
Exercise 2: Percentage and Ratio Calculations
Percentages and ratios are relatively simple concepts. In a numerical comprehension test, you might be required to find the proportion of something compared to something else, or work out a
percentage increase or decrease.
Having a simple strategy to use when dealing with percentages and ratios will help you solve these problems under pressure in a numerical comprehension test.
What is 15% of $30?
a) $5
b) $4.50
c) $3
d) $7
Answer: b) $4.50. You can solve this by finding 10% and then adding 5% (which is half), or solve it another way if you find that easier.
If John and Steven earned $30 together, and Steven’s share is $20, what is the ratio of their earnings?
a) 3:1
b) 2:1
c) 3:2
d) 4:1
Answer: b) 2:1. Ratios are all about proportion, and you need to be confident in your strategy for solving problems like this.
Exercise 3: Time Value of Money
In time value of money questions, you will be dealing with the idea that the value of something will increase (or decrease) over time.
This might involve the addition of interest on a loan, for example.
Calculate the total cost to pay back a loan of $10,000 with 10% interest a year over 5 years.
Answer: $15,000 - there is a specific formula that can be used for this:
FV (future value)
PV (current value)
N (number of compounding periods)
I (interest rate)
T (total number of years)
FV = PV x [1 + (i / n)] (n x t)
In this case:
FV = £10,000 x [1 + (10%/1)] (1 x 5) = $15,000
Exercise 4: Currency Conversions and Exchange Rates
Exchange rates and currency conversions might be important if you are working with international clients, or needing to pay for goods from overseas suppliers.
How much would a $10 purchase cost in Sterling with an exchange rate of $1 - £0.9179?
a) £9.18
b) £0.92
c) £4.68
d) £9.50
Answer: a) £9.18. This is a simple example; with fewer round numbers the calculation would be a lot more complicated.
Exercise 5: Word Problems
Word problems are used to evaluate your ability to pull the necessary data out of a passage of information. They tend to take the format of a problem about travel or work rate.
If James traveled at 30 miles per hour steadily, how far would he get in 180 minutes?
a) 100 miles
b) 80 miles
c) 180 miles
d) 90 miles
Tips for Improving Numerical Comprehension
Tip 1: Consistent Practice and Review
Like any skill, the more you use numerical comprehension, the easier you will find it.
You can practice using things like online tests, or even through revision resources aimed at students.
Make practice a part of your daily or weekly routine, even if it is just for a few minutes at a time - keep that mathematical muscle flexed so you can rely on it when you need it.
Try and review the basics as much as you can as part of your revision and practice sessions.
Tip 2: Applying Skills to Real-World Scenarios
Using numerical comprehension abilities in a vacuum - just in test situations - might not be the best way for you to improve - many people find using math in everyday situations helps them to
Instead of reaching for a calculator, try and do the calculation in your head when you are trying to determine how much a new shirt will cost in the sale.
Applying numerical skills to real life makes it much more relevant, and you might find it easier to improve this way.
Tip 3: Seeking Feedback and Identifying Areas for Improvement
If you use practice tests, you can pinpoint any particular areas of your knowledge that might need review - these should be easy to ascertain because they'll be the questions you got wrong. You can
then use this information to review the relevant mathematical principles and revise.
Resources for Numerical Comprehension Practice
There are many places you can go online for numerical comprehension practice.
If practice tests are what you are looking for, there is a range available at Fintest, including numerical reasoning and financial reasoning - which you are likely to come across when you are looking
for a role in finance.
You can also find tests that are used in specific recruitment processes, so you can practice more effectively when you apply for your ideal role at companies like Brookfield Asset Management and | {"url":"https://www.fintest.io/magazine/numerical-comprehension-questions-to-practice-for-finance-roles/","timestamp":"2024-11-04T21:27:36Z","content_type":"text/html","content_length":"47888","record_id":"<urn:uuid:31fef437-52da-4ab3-910e-84a7aa7221f9>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00045.warc.gz"} |
What is Arbitrage Betting? Sports Betting Arbitrage Explained Bet Logical
What is Arbitrage Betting?
Arbitrage betting is a mathematical sports betting strategy often referred to as a foolproof method of making guaranteed profits by placing a bet on both sides of a betting market. In this article,
we’ll look at how the arbitrage strategy claims to guarantee a profit no matter which team wins.
However, we’ll also be looking at whether arbitrage can really be considered risk-free and if there are some hidden risks you should be aware of?
Finally, we will break down how much money you can realistically make and if arbitrage betting is worth it?
How does Arbitrage Betting Work?
Arbitrage betting involves placing bets on all possible outcomes of a betting market.
This means that no matter who wins the game one bet should always win and cover all the money lost on the losing bets.
We profit by taking advantage of discrepancies in the odds offered between two or more sportsbooks.
The business of sports betting is highly competitive and sportsbooks are regularly trying to offer better odds than their competitors. Sometimes this can create a discrepancy between the odds at
different sportsbooks where the winnings from one bet would cover the losses made from betting on all other outcomes whilst still returning a profit.
These are called arbitrage opportunities or ‘Arbs‘ for short.
Arbitrage Betting Example
A simple example of an arbitrage bet would be an NBA moneyline market for the game Boston Celtics vs Brooklyn Nets. This season these teams are fairly evenly matched.
BetMGM is offering +105 on Boston Celtics and -110 on Brooklyn Nets.
Ceasars is offering -115 on Boston Celtics and +105 on Brooklyn Nets.
If we bet $100 on Boston Celtics at +105 with BetMGM this means we’ll win $105 if Boston Celtics win the game.
If we bet $100 on Brooklyn Nets at +105 with Ceasars this means we’ll win $105 if Brooklyn Nets win the game.
Using the arbitrage strategy we would bet on the Boston Celtics at BetMGM whilst simultaneously betting on Brooklyn Nets with Ceasars.
This would guarantee yourself a profit of $5. As our winning bet will win $105 whilst our losing bet will only lose $100.
Now this example is the simplest arbitrage bet we’ll find in the sports betting markets. In most cases, we will have to bet different wager sizes on each team/outcome depending on the odds of each
If we have more than two potential outcomes (i.e. Soccer) it can get a little more complex, but we can use some simple maths to work out if a combination of odds are profitable.
Arbitrage Betting Explained: Maths and Formula
Ok, so we now understand the basic concept of arbitrage betting and how odds discrepancies can create potential profit. Let’s get a little more technical and look at the mathematics of bookmaking
followed by the mathematics of arbitrage betting.
How do sportsbooks make money? This is a common question that people ask about sportsbooks. I’ll give a quick explanation.
Sportsbooks make their money by applying a hidden fee to their odds. This ‘fee’ is called a vigorish or ‘Vig’ for short.
Sportsbook apply this vig by offering customers a lower payout than they mathematically should get from the true odds for the bet they won.
We can calculate this hidden fee by converting the odds of each potential team or outcome of a betting market to what’s called an implied probability.
• Negative Moneyline Odds:
Implied Probability = Negative Moneyline odds / (Negative Moneyline odds + 100) * 100
• Positive Moneyline Odds:
Implied Probability = 100 / (Positive Moneyline odds + 100) * 100
Once we have converted all the odds to implied probabilities we just take the sum of all the probabilities. The Vig is the percentage probability is above 100% of the total implied probabilities.
This makes sense as there can only ever be a 100% chance that one of the outcomes will win. If the total of the implied probabilities is 105% then the vig for that market is 5%. The sportsbook will
profit 5% of all the money wagered on the market.
Let’s go through an example. We have an MLB game of New York Yankees vs. Boston Red Soxs with Fanduel is offering a moneyline market with the following odds:
• New York Yankees: +120
• Boston Red Sox: -142
What is the Vig for this market? let’s convert these odds into implied probabilities:
• New York Yankees:
Implied Probability = (100 / (Positive Moneyline odds + 100)) * 100
Implied Probability = (100 / (120 + 100)) * 100
Implied Probability = (100 / 220) * 100
Implied Probability = 0.45454545454 * 100
Implied Probability = 45.45%
• Boston Red Sox:
Implied Probability = (Negative Moneyline odds / (Negative Moneyline odds + 100)) * 100
Implied Probability = (142 / (142 + 100)) * 100
Implied Probability = (142 / 242) * 100
Implied Probability = 0.5867768595 * 100
Implied Probability = 58.68%
The total implied probability for the market is 45.45% + 58.68% which equates to 104.13%. Meaning the vig is 4.13%
This means every $104.13 wagered on this market would return $100 to players making a $4.13 profit for the sportsbook
For example, If $45.45 is wagered on the New York Yankees at +120 and $58.68 is wagered on the Boston Red Sox at -142
If New York Yankees win the game:
• The sportsbook collects $58.68 from losing bets on the Boston Red Sox.
• The sportsbook pays out $54.54 ( $45.45 x (+120/$100) ) for winning bets on New York Yankees.
• The sportsbook keeps $4.13-4.14
If Boston Red Sox win the game:
• The sportsbook collects $45.45 from losing bets on the New York Yankees.
• The sportsbook pays out $41.32 ( $58.68 x ($100/142) ) for winning bets on Boston Red Sox.
• The sportsbook keeps $4.13-4.14
As you can see no matter which team wins the online sportsbooks can guarantee profits.
Arbitrage betting uses the same concept. However, we want to bet when the total implied probability of all outcomes is below 100%. A sportsbook will never intentionally set their odds with an implied
probability below 100% as this would lose them money.
However, we can find potential arbitrage betting opportunities where the total implied probability is below 100% by comparing odds between different online sportsbooks.
You can use an odds screen or an odds comparison site to find arbitrage bets for free or you can pay for arbitrage betting software to find these arbitrage opportunities for you.
Tip: If we’re looking for an arb in a betting market with only 2 options. As long as the positive odds for a team are larger than the negative odds for the opposing team then we have an arb!
You may be wondering how we can work out our bet sizes once we have found an arb. Well, we can calculate the stake we need for each betting outcome using the following formula:
• Bet Stake = (Overall Bet Stake * Implied Probability of Outcome) / Combined Implied Probability of All Outcomes
Alternatively, we can use an arbitrage bet calculator to calculate the required bet sizes.
Let’s calculate our bet sizes for an example arbitrage bet.
We find a discrepancy in the odds for the Florida Panthers vs. St. Louis Blues NHL game. Circa Sports is offering -164 on the Florida Panthers whilst Wynnbet is offering +168 on the St. Louis Blues
creating an arb.
I have a bankroll of $1000 to place on this arb. Let’s work out how much I would need to place on each team and how much profit I would make.
First, let’s convert these odds into implied probabilities. You can use an odds converter to do this, but for the sake of this example, we’ll calculate both manually.
• Florida Panthers Implied Probability:
Implied Probability = (Negative Moneyline odds / (Negative Moneyline odds + 100)) * 100
Implied Probability = (164 / (164 + 100)) * 100
Implied Probability = (164 / 264) * 100
Implied Probability = 0.62121212121 * 100
Implied Probability = 62.12%
• St. Louis Blues Implied Probability:
Implied Probability = 100 / (Positive Moneyline odds + 100) * 100
Implied Probability = 100 / (168 + 100) * 100
Implied Probability = 100 / (268) * 100
Implied Probability = 0.37313432835 * 100
Implied Probability = 37.31%
Now that we have the Implied Probabilities of both teams we can work out the bet sizes required to make the same profit regardless of the winning team. We’ll use the bet stake forumla for each team.
• Florida Panthers Bet Stake:
Bet Stake = (Overall Bet Stake * Implied Probability of Outcome) / Combined Implied Probability of All Outcomes
Bet Stake = ($1000 * 62.12) / (62.12+37.31)
Bet Stake = ($62120) / (99.43)
Bet Stake = $624.76
• St. Loius Blues Bet Stake:
Bet Stake = (Overall Bet Stake * Implied Probability of Outcome) / Combined Implied Probability of All Outcomes
Bet Stake = ($1000 * 37.31) / (62.12+37.31)
Bet Stake = ($37310) / (99.43)
Bet Stake = $375.24
We now have everything we need to calculate our profit from the arbitrage bet. Let go ahead and do this.
• Florida Panthers Win:
Profit = Florida Panthers Bet Winnings – St. Loius Blues Bet loss
Profit = ($624.76 * (100/164)) – $375.24
Profit = $380.95 – $375.24
Profit = $5.71
• St. Louis Blues Win:
Profit = St. Loius Blues Bet Winnings – Florida Panthers Bet Loss
Profit = ($375.24 * (168/100)) – $624.76
Profit = $630.40 – $624.76
Profit = $5.64
Does Arbitrage Betting get you banned?
Does sports betting arbitrage provide unlimited risk-free profit for arbitrage bettors? Is this the holy grail of online money-making?
Sorry to burst a few bubbles but eventually you will get limited or banned from most sportsbooks that we call ‘Soft Books’ or ‘Retail Books’. These sportsbooks only want to take bets from a
recreational bettor (Losing bettor).
Once they have determined that you are an arbitrage bettor they WILL ban or limit your account.
The reality of profitable sports betting is that you will eventually get limited from the ‘Soft Books’. However, we can still make very good money from these accounts before we get the chop. You
should be looking to extend the lifetime of your soft book accounts to get the maximum profit from them before getting limited.
It’s not all doom and gloom. There are some sportsbooks that allow arbitrage bettors such as Circa Sports and Betting Exchanges such as Sporttrade, Prophet Exchange and Betfair. These are referred to
as ‘Sharp Books’.
There are also strategies such as market making and value betting that you can do without the risk of limits.
Think of arbitrage betting as the first step in your sharp sports betting career. Use it to build a nice bankroll and get to grips with some basic sports betting concepts you’ll need for the future.
How Much Can I Make From Arbitrage Betting?
There are several factors that will affect the amount you can earn from arbitrage betting
The biggest factor determining how much you can make depends on the country and state you live in.
If you’re based in a state with many different sportsbooks such as Colorado or New Jersey, then you’ll be able to make much more than someone in Delaware who has access to fewer sportsbooks.
Each arbitrage bet will make you a return of about 1-4% ROI. This might sound like a minimal profit. However, the beauty of arbitrage is that we can cycle through placing your whole bankroll in a few
days or even in a single day. Rather than getting a couple % return each year in the stock market, arbitrage returns the same amount in a few days.
Due to this the compounding effect is through the roof for arbitrage. A realistic profit per month from arbitrage is $1000-2000. Although full-time sports bettors utilising arbitrage can return
$15’000 per month. Alex Monahan of Oddsjam Profited $200’000 using arbitrage and positive ev betting in 2021.
Are there any risks when Arbitrage Betting?
Sports betting arbitrage is often described as a risk-free foolproof method that guarantees a profit, but Is this a responsible way to describe arbitrage betting?
Although the strategy guarantees a profit when viewed through the lens of simple mathematical equations and probability theory. When you move from theory to the practical application in the real
world some risks do a rise that you should be aware of. I’ve listed the most common below:
1. Human Error:
This can come in the form of Typo’s when using an arb calculator or putting in the wrong dollar amounts when placing your bets with a sportsbook. Another common mistake is betting in the wrong
market. For example betting in the full-time market for one team and then betting in the half time market for the other team.
2. House Rules:
Some sportsbooks have slightly different rules for handling unusual events in a sports game (for example retirements in tennis). This can mean one sportsbook will void your bet whilst the other
sportsbook settles your opposing bet as a loser.
3. Platform Risk:
Another potential risk arises from a sportsbook website going down whilst you are trying to place your second bet. This will leave your initial bet exposed to losing. Another aspect to consider
is that your bank may decline a deposit when you need to top up your account to place a bet.
Fortunately, most people that will fall foul of these situations are people that aren’t aware that these potential risks exist. Once you know that these situations are possible you can plan for them.
1. Always have a spare funds sportsbook account.
2. Deposit the required funds before placing any bets.
3. Double-check your bets both before and after placing them.
4. Familiarize yourself with different sportsbook rules
5. Check arbitrage communities for house rules to be aware of
Is Arbitrage Betting Legal?
Yes, arbitrage betting is completely legal. Even though some sportsbooks may not particularly like the practice or would rather they didn’t have arbitrage bettors on their website.
Taking advantage of arbitrage opportunities is fair game as we’re not manipulating or altering the odds, we’re simply taking the odds offered to us by the sportsbooks. If and when a sportsbook
decides they no longer want to offer you their odds they can stop offering their service.
How do I find Arbitrage Bets?
Now that we understand the basic theory behind arbitrage betting. We just need to learn how to find arbitrage betting opportunities.
You can find arbitrage bets for free using an odds comparison website or a free oddsscreen such as Unabated’s oddsscreen. This can be time-consuming and you will find less arbs with a lower profit
margin. Ultimately, you’ll make less money
Alternatively, you can use an arbitrage finder tool that will find arbitrage bets 24/7 across all sports and bettting markets. There are a couple of websites that provide subscription services for
their tools.
If you’re in the USA I would recommend Oddsjam. Their arbitrage tool has the widest coverage in the USA.
Their arbitrage tool is part of the Positive EV package which is $199 a month which can be pricey. However, they have a 7-day free trial to test out the software for yourself.
If you’re based in Europe or Australia I would recommend Rebelbetting. They’re the arbitrage software OG’s. They were the first arbitrage software on the web back some 20 years ago.
They have a free trail and a starter package at €99 per month. Their pro package is €199 per month.
Is Arbitrage Betting Worth it?
Arbitrage Betting is often described as a risk-free method to make a guaranteed profit from sportsbooks. However, there are some risks to be aware of along with the reality that most of your
sportsbook accounts will eventually get limited.
Despite this arbitrage betting is one of the most profitable side hustle and money-making methods on the web. Until every sportsbook in your country has politely (or not!) told you to get the hell
out of their sportsbook. There is easy money on the table up for grabs!
If you want to get started with arbitrage I would recommend signing up to either the Oddsjam or RebelBetting Free Trials | {"url":"https://betlogical.com/what-is-arbitrage-betting/","timestamp":"2024-11-12T17:07:09Z","content_type":"text/html","content_length":"219015","record_id":"<urn:uuid:a44cc7bc-d1e7-4fa3-bae8-984f007ad55b>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00142.warc.gz"} |
Parabola Formula | Equation, Properties, Examples
Equation, Properties, Examples | Parabola Formula
The parabola is an attractive and versatile geometric shape which managed to captured the attention of scientists and mathematicians for hundreds of years. Its exceptional properties and plain yet
elegant equation makes it a powerful equipment for molding a broad assortment of real-world phenomena. From the flight path of a projectile to the shape of a satellite dish, the parabola performs an
essential role in many fields, consisting of physics, architecture, engineering, and mathematics.
A parabola is a type of U-shaped portion, that is a curve created by overlapping a cone over a plane. The parabola is defined by a quadratic equation, and its characteristics, such as the vertex,
focus, directrix, and symmetry, provide precious insights into its behavior and uses. By understanding the parabola formula and its properties, we could obtain a detailed appreciation for this
rudimental geometric shape and its multiple usages.
In this blog, we wish to study the parabola in depth, from its equation and properties to instances of how it could be utilized in many domains. Whether you're a student, a professional, or just
interested about the parabola, this article will provide a comprehensive overview of this fascinating and important concept.
Parabola Equation
The parabola is specified with a quadratic equation of the form:
y = ax^2 + bx + c
at this point a, b, and c are constants which decide the size, shape, and position of the parabola. The value of a controls whether the parabola opens up or down. If a is more than 0, the parabola
opens upward, and if a lower than 0, the parabola opens downward. The vertex of the parabola is located at the point (-b/2a, c - b^2/4a).
Properties of the Parabola
Here are the properties of Parabola:
The vertex of the parabola is the point where the curve switches direction. It is also the point where the axis of symmetry intercepts the parabola. The axis of symmetry is a line that goes across
the vertex and splits the parabola into two proportionate halves.
The focus of the parabola is a point] on the axis of symmetry which is equidistant from the vertex and the directrix. The directrix is a line which is perpendicular to the axis of symmetry and placed
at a length of 1/4a units from the vertex.
The directrix is a line which is perpendicular to the axis of symmetry and situated at a length of 1/4a units from the vertex. All points on the parabola are equal distance from the directrix and the
The parabola is symmetric in relation to its axis of symmetry. This defines that if we consider any point on one side of the axis of symmetry across the axis, we obtain a corresponding point on the
other side of the axis.
The parabola crosses the x-axis at two points, specified by the formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
The parabola intersects the y-axis at the location (0, c).
Examples of Parabolas
Here are few primary examples of Parabolas:
Example 1: Graphing a Parabola
Let's graph the parabola y = x^2 - 4x + 3. Foremost, we need to figure out the vertex, axis of symmetry, and intercepts. We can apply the formula:
vertex = (-b/2a, c - b^2/4a)
to find the vertex. Placing in the values a = 1, b = -4, and c = 3, we attain:
vertex = (2, -1)
So the vertex is located at the point (2, -1). The axis of symmetry is the line x = 2.
Subsequently, we can find the x-intercepts by setting y = 0 and solving for x. We get:
x^2 - 4x + 3 = 0
(x - 3)(x - 1) = 0
Therefore the parabola intersects the x-axis at x = 1 and x = 3.
Ultimately, the y-intercept is the coordinates (0, c) = (0, 3).
Applying this knowledge, we could sketch the graph of the parabola by plotting the vertex, the x-intercepts, and the y-intercept, and drawing the curve of the parabola within them.
Example 2: Application of Parabola in Physics
The parabolic shape of a projectile's trajectory is a general example of the parabola in physics. While a projectile is launched or thrown upward, it follows a path which is described by a parabolic
equation. The equation for the path of a projectile launched from the ground at an angle θ through an initial velocity v is provided by:
y = xtan(θ) - (gx^2) / (2v^2cos^2(θ))
here g is the acceleration because of gravity, and x and y are the horizontal and vertical distances traveled by the projectile, respectively.
The trajectory of the object is a parabolic curve, with the vertex at the coordinate (0, 0) and the axis of symmetry corresponding to the ground. The focus of the parabola represents the landing
point of the projectile, and the directrix depicts the height above the ground where the projectile would strike if it weren’t impacted by gravity.
In summary, the parabola formula and its properties play an essential role in several domains of study, involving math, architecture, physics, and engineering. By knowing the equation of a parabola,
its properties for instance the vertex, focus, directrix, and symmetry, and its numerous uses, we can gain a detailed understanding of how parabolas function and how they can be applied to model
real-life phenomena.
Whether you're a student finding it challenging to understand the theories of the parabola or a professional want to use parabolic equations to real-world challenges, it's crucial to possess a firm
groundwork in this fundamental topic.
This's where Grade Potential Tutoring enters. Our adept teachers are accessible online or in-person to provide individualized and productive tutoring services to help you master the parabola and
other math theories. Connect with us today to schedule a tutoring session and take your arithmetic skills to the next stage. | {"url":"https://www.orlandoinhometutors.com/blog/equation-properties-examples-parabola-formula","timestamp":"2024-11-03T15:51:39Z","content_type":"text/html","content_length":"75827","record_id":"<urn:uuid:b33a0d43-9e76-4803-a457-9c9862d5a6b3>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00574.warc.gz"} |
Collatz Conjecture
The Collatz Conjecture is a conjecture in mathematics named after Lothar Collatz.
The conjecture can be summarized as follows. Take any positive integer n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process
(which has been called “Half Or Triple Plus One”, or HOTPO) indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1.
The conjecture is also known as the 3n + 1 conjecture, the Ulam conjecture (after a Polish-American mathematician Stanislaw Ulam), Kakutani’s problem (after a Japanese-American mathematician Shizuo
Kakutani), the Thwaites conjecture (after British Sir Bryan Thwaites), Hasse’s algorithm (after a German mathematician Helmut Hasse), or the Syracuse problem. The sequence of numbers involved is
referred to as the hailstone sequence or hailstone numbers (because the values are usually subject to multiple descents and ascents like hailstones in a cloud), or as wondrous numbers.
Paul Erdős said about the Collatz conjecture: “Mathematics may not be ready for such problems.”
Jeffrey Lagarias in 2010 claimed that based only on known information about this problem, “this is an extraordinarily difficult problem, completely out of reach of present day mathematics.”
Our main efforts in this area will be concentrated on testing and proving Collatz conjecture both theoretically and algorithmically. | {"url":"https://math101.guru/en/problems-2/collatz-conjecture/","timestamp":"2024-11-09T09:26:43Z","content_type":"text/html","content_length":"44141","record_id":"<urn:uuid:76a3432c-aa5b-4023-84dd-06f801f5665b>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00429.warc.gz"} |
Cookie Clicker - Garden Guide - Re-actor
Cookie Clicker – Garden Guide
Feeling lost when seeing your garden? Say no more! This guide will answer all your questions.
In the Garden, plants and fungi can be grown and crossbred, obtaining new species as a result. Crops provide various benefits depending on the species, some granting passive buffs as they sit planted
in the Garden, while others give rewards when harvested. The Garden can be unlocked by upgrading Farms to level 1 using a Sugar Lump.
Ascending will not reset the unlocked seed types, but it will clear the garden of any crops. The Garden has its own equivalent of ascension, which removes all planted crops and seeds except the
starter seed of Baker’s Wheat, in exchange for 10 sugar lumps.
Garden Size
The Garden’s size starts out as a 2×2 grid, and can be expanded by further upgrading Farms with sugar lumps. The maximum size of 6×6 is unlocked at level 9.
As you can see, cost efficiency starts to drop slightly at garden levels 5 and 6 before rising sharply at levels 7 & above.
Growing Crops
Plants and fungi can be planted in the garden by clicking on the seed you would like to plant on the left-hand side and then clicking on an empty tile in the Garden. A seed can be easily planted more
than once by shift-clicking to plant. At each tick, the game will check three things in order for each tile: age, contamination, and mutation. The length of one tick is determined by the type of soil
which is currently in use, though a tick can be triggered instantly by spending a sugar lump.
While growing, the crop ages every tick by its aging value, which in most cases is a randomized number. As the age increases, the crops will grow as bud, sprout, bloom and finally mature after
reaching their mature age. The premature stages of growth increase as the crop reaches 1/3 and 2/3 of its mature age. A plant’s passive effects will increase in potency with each growth stage,
reaching full strength once the crop matures. For premature plants, any passive effects will only be at 10% for buds, 25% for sprouts, and 50% for blooms. A crop will decay once its age reaches 100
unless it is immortal.
Initially, the Garden only has one seed available, that being Baker’s Wheat. By leaving garden plots empty, Meddleweed may also appear. These two plants are the most fundamental species of the
Garden: Baker’s Wheat is the basis for regular plants, while Meddleweed is the basis of fungi. Most new species appear as a result of mutations (created by having two or more parent crops adjacent to
an empty plot), but the two basic species of fungi (Crumbspore and Brown Mold) can be found as a result of manually harvesting Meddleweed (older = higher chance). When trying to unlock these 2
plants, DO NOT use the Harvest All tool, as it will harvest any newly spawned Crumbspore & Brown Mold along with everything else without adding their seeds to your collection. However, using
Ctrl+Shift+Click is fine, since it only harvests mature instances of plants, rather than the entire garden.
If there is an empty plot, it has a chance to start growing a plant based on the adjacent (orthogonal and diagonal) plots. For example, if there are two adjacent Baker’s Wheat, the empty plot may
produce another Baker’s Wheat or a random mutation: either a Thumbcorn or a Bakeberry. The exact probability can be calculated from random list mechanism. In general, the actual number is close, but
not equal, to the base chance and can be approximately tripled. Most mutations require mature crops to trigger, but there are some exceptions. Additionally, certain mutations may also be prevented by
having too many of a certain species adjacent to a slot (e.g. Ordinary Clover). See the species section below for a complete list of all mutation conditions.
Mutation set up
The first picture shows examples of optimal plant alignments for each Garden level when trying to mutate a new crop from 2 parents of the same species (example: Thumbcorn from 2 Baker’s Wheats).
Green squares (labeled with a “G”) indicate planted crops. Empty squares indicate locations for potential mutations.
The second picture shows examples of optimal alignments for each Garden level for mutations from 2 different parents (example: Cronerice from Baker’s Wheat and Thumbcorn). The green and yellow
squares (labeled with a “G” and “Y” respectively) indicate the 2 types of planted tiles. The light red squares (labelled with an “R”) indicate plots that could grow unwanted crops resulting from the
mutation of 2 plants of the same type. For example, growing Baker’s Wheat and Gildmillet may spawn unwanted Thumbcorns in the R plots.
Note that there are certain plants which require more than just 2 adjacent plants (Juicy Queenbeet, Shriekbulb & Everdaisy). If you’re trying to grow one of these plants, DO NOT FOLLOW the mutation
setups shown above (they won’t work very well for you, if at all). For Juicy Queenbeets, plant 4 3×3 rings of Queenbeets in each corner. For Everdaisies, fill the 1st and 5th rows with Tidygrasses
and the 3rd row with Elderworts. For Shriekbulbs, see this section for more details.
For the second chart, level 6 can be altered to remove its unwanted plots by using the setup for level 7 without the empty top row. However, this uses up one more space in the grid, meaning there is
one less space for a new plant to grow. This is optimal if you are not actively managing your Garden to remove unneeded or unwanted plants.
Mutation tree
Meddleweed, Crumbspore, and Doughshroom are able to contaminate other crops in an orthogonal direction when they are mature. The chance of this happening is very low. When it does occur, the
contaminated plant is replaced by the attacking weed or fungus. Immortal plants are immune to contamination, as are certain other species. See the growth charts for an overview of contamination
values and immunity information.
Minimizing seed costs
As you progress further into the game and your CpS gets larger and larger, the cost of planting garden plants will rise in tandem, so trying to unlock some of the more elusive plants out there will
often end up costing you an arm and a leg and then some. However, there are a couple of ways to make gardening much more affordable. Unlike the dollars ($) used in the Stock Market minigame, whose
value is dependant on your raw (i.e. unaltered) CpS, the prices of Garden plants can and will fluctuate depending on which buffs & debuffs happen to be active at any given time. CpS-increasing
effects (like Frenzies & Building Specials) will substantially increase the cost of planting seeds for the duration of the effect, so it is highly advised to not plant seeds during a Frenzy (or any
CpS-increasing buff in general) until the buff wears off. On the other hand, things that lower your CpS will decrease seed costs accordingly, making them much more affordable.
The best way to minimize the cost of planting a seed is by clicking wrath cookies until you get a Cursed Finger debuff, which has the side effect of temporarily setting your CpS to 0. While the
debuff is active, the price of planting seeds will drop to the absolute minimum, which is almost always much lower than the regular, CpS-based price. This makes them surprisingly useful when trying
to grow hard-to-get plants like Duketaters, Shriekbulbs and the elusive Juicy Queenbeet. However, Cursed Fingers only have about a 1 in 40 chance of appearing, so Clots, despite being less effective
at cutting costs, are still a viable alternative due to them appearing nearly 10 times as often as the leading competitor. To top it all off, wrinklers can also help a lot when it comes to lowering
seed costs; having 10 of them at once will halve seed costs while also increasing overall cookie yield (see here for more details).
Unlocking Shriekbulbs
Shriekbulbs can be a bit tricky to unlock. Despite the fact that they can be unlocked in one of several ways, the odds of one appearing are pretty low. The conventional method for unlocking them is
by using Duketaters, usually by filling the 1st, 3rd & 5th rows with them. However, duketaters are expensive to plant and only spend a very short amount of time mature. Alternately, you can use
Elderworts. While planting Duketaters does have a higher chance of producing Shriekbulbs, Elderworts are cheaper to plant and are also immortal. They will never die & never need to be replanted ever,
so this strategy is ideal for players who don’t want to be replanting stuff 24/7. (As for the layout, you can use the same pattern as with Duketaters – fill every 2nd row with Elderwort and leave the
in-between parts empty.)
Useful Plants
If you tend to idle most of the time and rarely ever click Golden Cookies, you may want to fill the garden with whiskerblooms, as they will boost milk efficacy and thus increase the effects of the
various kitten upgrades. Initially, you should plant them in regular or fertilizer soil to speed up the growing process (minimising your CpS to reduce planting costs), then switch to clay soil once
they’ve matured. (The reduced tick speed is actually quite helpful in this case, as it lets the plants stay mature for longer before needing to be replanted.) With enough achievements & kitten
upgrades, the whiskerblooms will result in a massive CpS increase. You can also add in nursetulips to boost the whiskerblooms’ effects, but be aware that planting nursetulips will both reduce your
overall CpS and also use up space that could otherwise be used for planting more whiskerblooms. The base increase in milk effects from the whiskerblooms (without nursetulips) can reach up to 7.2%
once all the plants have matured (or 9% if using clay soil), which is nearly 50% more effective than Breath of Milk and only around 2.5% less efficient than a Diamond-slotted Mokalsium. Even better,
since milk boosts (like CpS boosts) stack multiplicatively, using the whiskerblooms in combination with the aforementioned effects will lead to an even greater production boost!
Bakeberry is one of the best plants in the game when it comes to getting lots of cookies really quickly. Filling up an entire field with bakeberries will give you a base CpS increase of 36% once they
all mature. Harvesting all 36 bakeberries will yield a whopping 18 hours of CpS! (Note that the total yield is capped at 108% of your bank if using harvest all, so try to harvest them individually
instead.) The cookie you get from the bakeberries are also affected by buffs & debuffs. Harvesting during a frenzy can give up to 3 hours and 30 minutes’ worth of production for each bakeberry you
harvest, so harvesting all 36 plants can yield over 5 days worth of production! However, planting the bakeberries in the first place costs quite a lot of cookies, so you should aim to plant them
while minimizing your CpS (see this section for more details) in order to lower those upfront costs and further maximise your gain. Additionally, it’s important to note that bakeberry harvesting
rewards are capped at just 3% of your current cookies, so they should ideally only be harvested when you have loads of cookies saved up to get a bigger reward. (In order to get the full reward from
all your bakeberries, you’d need to have at least 16 hours 40 minutes worth of CpS sitting there in your bank.) While duketaters could potentially be considered “better” than bakeberries (due to
sporting a much greater cookie yield), duketaters are a lot more expensive, grow super slowly & do not provide a CpS bonus.
Golden Clover
Filling up an entire 6×6 field with Golden Clovers will grant a 108% increase in Golden Cookie frequency when all plants are mature (135% if using clay soil). This can lead to a Frenzy lasting for
several minutes on end, because a new Frenzy cookie will likely appear before the existing Frenzy has ended, extending its duration over and over and over. The increased Golden Cookie frequency also
increases the chance of getting combo effects, which can potentially yield massive amounts of cookies.
Nursetulips are very effective at increasing the effectiveness of other plants. Since they reduce your CpS, Nursetulips should be used in situations where the goal isn’t to boost your CpS. (Examples
include Wrinklegills, Golden Clovers, Keenmoss, etc.) With a full 6×6 garden, the most effective way of using Nursetulips is to alternate between planting rows of Nursetulips and rows of the other
plant. This will have a greater effect than simply planting 36 plants of the desired type, at the cost of slightly lowering your CpS.
2 Comments
1. im the best
2. this is amazing for school, where the actual wiki is blocked | {"url":"https://re-actor.net/cookie-clicker-garden-guide/","timestamp":"2024-11-15T03:34:01Z","content_type":"text/html","content_length":"82421","record_id":"<urn:uuid:2ff2149f-26f2-40f8-ad82-1ccf22d7db6d>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00823.warc.gz"} |
Cormart Aptitude Test Past Questions and Answers
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Cormart Aptitude Test Past Questions and Answers for 2024
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About Cormart Nigeria
Cormart Nigeria Ltd. is one of the leading chemical and food raw materials companies in Nigeria. Since its inception in 1980, it has been on the forefront of production, importation, stocking and
distribution of chemicals and other raw materials. Cormart provides premium products and services across the paint, confectioneries, cosmetics, pharmaceutical, food and beverage, construction among
many other industries.
Cormart Nigeria Online aptitude tests format?
Cormart Nigeria online tests cover;
Numerical Reasoning
Verbal/Critical Reasoning
Abstract Reasoning
Error Checking (7 minute test)
Free Sample of Cormart Nigeria Past Questions and Answers
A local utilities company exclusively provides electricity and natural gas to a small village containing 342 households. Each household purchases on average £75 of gas and electricity per month. If
the village’s natural gas sales account for ¾ of the utilities company’s revenue, how much revenue is generated in one year from electricity sales in the village? (assume the company has no customers
outside the village)
(A) £73,750
(B) £74,500
(C) £76,950
(D) £78,150
(E) Cannot say
Step 1 – Calculate the total revenue produced by the village through electricity and natural gas sales per month by multiplying the number of households (342) by the average electricity and gas bill
per month (£75).
342 x £75 = £25,650.00
Step 2 – Calculate the total annual sales of electricity and gas in the village by multiplying the
monthly revenue (£25,650.00) by the number of months in a year (12).
£25,650.00 x 12 = £307,800.00
Step 3 – Calculate the proportion of revenue generated by electricity sales by identifying 25% (0.25) of the total annual sales of gas and electricity (£307,800.00)
£307,800.00 x 0.25 = £76,950
Thus the correct answer is (C) £76,950
A freelance web developer charges £100 per hour, or £600 for a full day (8 hours). Their latest contract requires the web developer to work three full days and two half days (4 hours). Assuming the
web developer pays 30% income tax, how much income will the web developer receive from this contract? (after tax)
(A) £1,600
(B) £1,710
(C) £1,820
(D) £1,930
(E) £2,040
Step 1 – Calculate the income earned from the three full day’s work by multiplying the developers day rate (£600) by the number of days worked (3).£600 x 3 = £1,800
Step 2 – Calculate the income earned from the two half days work by multiplying the developers hourly rate (£100) by the number of hours worked (4 4 = 8).
£100 x 8 = £800
Step 3 – Combine the two figures above and subtract 30% from that figure to remove the developer’s income tax, giving us the total income earned after tax.
£1,800 £800 = £2,600
£2,600 – (£2,600 x 0.3) = £1,820
Thus the correct answer is (C) £1,820
A shop is willing to purchase used entertainment products for resale. The shop will purchase CDs for £1 each, DVDs for £2.50 each and videogames for £7.50 each.
Similarly, every £0.50 worth of products sold to the shop accrues 1 loyalty point. If a customer sells 26 CDs, 10 DVDs and 5 video games, how many loyalty points has the customer accrued from this
(A) 142 points
(B) 165 points
(C) 177 points
(D) 175 points
(E) 158 points
Step 1 – Calculate the value of the CDs by multiplying the number of CDs (26) by the value of eachCD (£1).
26 x £1 = £26
Step 2 – Calculate the value of the DVDs by multiplying the number of DVDs by the value of each DVD (£2.50).
£2.50 x 10 = £25.00
Step 3 – Calculate the value of the video games by multiplying the number of video games (5) by the value of each video game (£7.50).
7.5 x 5 = £37.50
Step 4 – Calculate the number of loyalty points accrued by dividing the total value of all products by 2, then rounding down to the nearest whole point.
£26.00 £25.00 £37.50 = £88.50
88.50 x 2 = 177 = 177 points
Thus the correct answer is (C) 177 points
A coins-to-cash machine converts unwanted small change into banknotes for a fee. On the first £5, a 10% fee is charged, on the next £5, a 7.5% fee is charged and on all change afterwards, a 5% fee is
charged. If a customer cashes in £30 worth of change, how much money will the coins-to-cash gain from this transaction?
(A) £1.55
(B) £1.66
(C) £1.77
(D) £1.88
(E) £1.99
Step 1 – Calculate the fee charged on the first £5 by identifying 10% (0.1) of £5.
£5.00 x 0.1 = £0.50
Step 2 – Calculate the fee charged on the second £5 by identifying 7.5% (0.075) of £5.
£5 x 0.075 = £0.375 = £0.38
Step 3 – Calculate the fee charged on the remaining amount by identifying 5% (0.05) of £20 (£30 – 10).
£20 x 0.05 = £1.00
Step 4 – Calculate the total fee charged by combining the three figures above.
£0.50 £0.38 £1.00 = £1.88
Thus the correct answer is (D) £1.88
With this Cormart aptitude Test Questions and Answers, you get an idea of what to expect using our compiled practice questions as well as actual tests from previous years. This test pack comes
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The level polynomials of the free distributive lattices for Discrete Mathematics
Discrete Mathematics
The level polynomials of the free distributive lattices
View publication
We show that there exist a set of polynomials {Lk{curly logical or}k = 0, 1⋯} such that Lk(n) is the number of elements of rank k in the free distributive lattice on n generators. L0(n) = L1(n) = 1
for all n and the degree of Lk is k-1 for k≥1. We show that the coefficients of the Lk can be calculated using another family of polynomials, Pj. We show how to calculate Lk for k = 1,...,16 and Pj
for j = 0,...,10. These calculations are enough to determine the number of elements of each rank in the free distributive lattice on 5 generators a result first obtained by Church [2]. We also
calculate the asymptotic behavior of the Lk's and Pj's. © 1980. | {"url":"https://research.ibm.com/publications/the-level-polynomials-of-the-free-distributive-lattices","timestamp":"2024-11-02T19:22:12Z","content_type":"text/html","content_length":"62851","record_id":"<urn:uuid:24f6920d-0dab-4ba7-ac87-4cc3a06133b3>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00827.warc.gz"} |
How to sum cells from different sheets?
I have six sheets with numbers in each cell. I want to make a 'TOTAL' sheet that sums up each corresponding cell in each of the six sheets. For example, sum the A1 cell in Sheet 1, 2, 3, 4, 5, 6. And
then sum the A2 cell in Sheet 1, 2, 3, 4, 5, 6. Basically the A1 cell in the TOTAL sheet should be the sum of Sheet 1, 2, 3, 4, 5, 6. I don't know how to do this in Smartsheet. In Excel, I can drag
the formula down and it automatically applies the sum formula to each cell.
• Hi @Devika_Renewables
Sine you have 6 different sheets, I would suggest doing this by creating a Report.
As long as all 6 of your sheets have the same column names, you'll be able to use the Summary function in a Report to automatically SUM all the rows in one column across all your sheets together.
If you're looking to SUM together each cell though, versus a whole column, you would want to add an Auto-Number column to each of your 6 sheets. Then once each of the rows have a number, you can
use the GROUP feature in a Report to Group together all Row 1s across the 6 sheets and Summarize by this grouping. Does that make sense?
See: Redesigned Reports with Grouping and Summary Functions
Need more help? 👀 | Help and Learning Center
こんにちは (Konnichiwa), Hallo, Hola, Bonjour, Olá, Ciao! 👋 | Global Discussions
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Lewis William Gabriel Topley
Personal profile
Research interests
My research revolves around the preresentation theory of the finite W-algebras. These are mysterious new objects which exist on the boundary between the zero characteristic and positive
characteristic realms of Lie theory. To be precise, there are ordinary and modular versions which are closely related. The ordinary cases seem to control certain geormetric properties of
representations of complex semisimple Lie algebras, whilst the modular specimens should be seen as the fundamental unit of currency when investigating representations of reduced enveloping algberas
of restricted Lie algebras (spooky eh?). My past research has linked the characteristic zero finite W-algebra theory with primitive ideals in enveloping algebras, whilst my future research shall
attempt to strengthen the connections between the ordinary and modular realms.
People: I took my PhD with Sasha Premet at the University of Manchester where he taught me the joys of modular Lie theory. After that I spent a year at the University of East Anglia working with
Vanessa Miemietz where, amongst other things, we studied Chriss and Ginzburg's book 'Complex Geometry and Representation Theory' - a wonderful read, if somewhat terse. Now I'm taking a short stint in
York working with Maxim Nazarov.
NB: if you google Maxim Nazarov you'll find:i) a high powered professor working with Yangians; ii) a Russian underwear model. My supervisor is the former, not the latter.
Erdős-Bacon Number: The Erdos number, e(m), of a mathematician, m, counts the distance of that mathematician from Paul Erdős on the collaboration graph. The Bacon number, b(a), counts the distance of
an actor from Kevin Bacon on the coappearance graph (two actors are said to coappear if they have acted in the same film). If m = a then we may define the Erdős-Bacon number to be E(m) = e(m) + b(m).
Having done extensive research I have found that E(Lewis) < 9. The inequality b(Lewis) < 4 actually follows from the fact that I was once an extra in a film being shot in my village. I'm not totally
sure that counts. If there are any experts in Bacon numbers reading this then I'd be grateful for clarrification on that matter. By life's ambition is to obtain a finite Erdős-Bacon-Sabbath number.
Research output
• 2 Article
• 1 Doctoral Thesis | {"url":"https://pure.york.ac.uk/portal/en/persons/lewis-william-gabriel-topley","timestamp":"2024-11-05T09:36:59Z","content_type":"text/html","content_length":"40091","record_id":"<urn:uuid:52617cd1-8dad-4d68-b446-10a1d66d418e>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00151.warc.gz"} |
Numerical and experimental study of bursting prediction in tube hydroforming of Al 7020-T6
Issue Mechanics & Industry
Volume 18, Number 4, 2017
Article Number 411
Number of page(s) 7
DOI https://doi.org/10.1051/meca/2017019
Published online 28 August 2017
© AFM, EDP Sciences 2017
1 Introduction
Forming limit diagram (FLD) is a significant criterion for evaluating formability of tubular materials, which is commonly obtained from theoretical calculations, finite element simulation and
experiment. Hydroforming process makes integrated tubular parts with high ratio of strength-to-weight in one step. By applying oil pressure into the tube, and applying axial force to the ends, a
tubular blank is formed into the internal shape of the die. In this process, the original specimen is a simple tube (direct or bend tube). Due to increasing requests for light-weight parts,
hydroforming processes have been widely used to produce and make parts in various fields, such as automobile, aircraft, aerospace, and shipbuilding industries[1]. Hashemi et al. [1] have also
considered tube hydroforming process, including the manufacturing of metal bellows. Asnafi and Skogsgårdh[2] proposed a mathematical model to predict the forming pressure and the related feeding
distance required to hydroform a circular tube into a T-shape product without wrinkling and bursting. The use of aluminum alloys in the place of steel components in automotive applications saw a
significant increase during the last few years. For this reason, hydroforming of aluminum tubes is a very desirable manufacturing process instead of sheet metal forming. In tube hydroforming, it is
required that the vacant tube should be formed into a die cavity of the final shape without any kind of deficiency such as bursting, wrinkling or buckling. Since bursting is an impression of
localized necking which is a condition of local instability under excessive tensile stresses, prediction of necking is an important problem before designing the details of processes[3]. FLDs are
appointed to determine the tubular materials formability. The laboratory test results showed that the FLDs are influenced by several parameters including the strain rate[4], strain hardening and
anisotropy coefficients[5], heat treatment[6], grain size[7] and strain path changes[8]. After obtaining the forming limit curves (FLCs) by Keeler and Backofen[9], many researchers tried to
develop some numerical and analytical models to determine the sheet metal formability. But, only a little attention has been paid to study the behavior of tubular materials. For example, Kim et al.[
10] predicted the bursting failure in tube hydroforming considering plastic anisotropy by using numerical calculations. Song et al.[11] used analytical approach to bursting in tube hydroforming
using diffuse plastic instability. One year later, the team combined the two previous methods: analytical and numerical methods for prediction of forming limit in tube hydroforming [12]. Hwang et
al.[13] predicted FLDs of tubular materials by bulge tests in two ways. They have used the Hill's law for calculations and did bulge tests for an experiment and then compared the two methods
together. Chen et al.[14] used thickness gradient criterion for seamed tube hydroforming that resulting in FLD. They validated numerical solution with experimental work. Seyedkashi et al.[15]
analyzed two-layered tube hydroforming with analytical and experimental verification.
In this paper, FLCs of tubular materials (Al 7020-T6) with respect to axial feeding and hydraulic pressure were determined numerically and experimentally for the first time. The computed FLD was
verified by a series of experimental bulge tests. A numerical approach was applied to FLC prediction. This numerical method is based on the acceleration of plastic strain (i.e., the second
derivation) which was applied to determine the onset of necking for tube materials. Based on this method, the localized necking would be started when the acceleration of the max plastic strain gets
its maximum value.
2 Experimental work
2.1 Tube bulging test
The dimensions and configurations of an initial tube and its final bulged part are shown in Figure 1. The outer diameter of the pipe was 40mm and the initial thickness of the tubular blank was
1.5mm. Aluminum pipes were seamless and produced by extrusion process.
The mechanical and material properties of the tube were determined by standard test using specimen, which were prepared according to ASTM-E8 specification at a constant crosshead speed of 2mmmin^
−1. The mechanical and material properties are presented in Table 1.
To evaluate the hydroforming limit strain diagram, a series of bulge tests were carried out on aluminum tube 7020-T6. For doing the tests, an experimental setup with the ability to control internal
pressure and axial feeding was provided. This setup had two hydraulic jacks and a hydraulic pump and it is shown in Figure 2. All hydraulic instruments used in the experimental procedure, including
pumps and valves were fabricated in the Enerpac company (the supplier’s name was Enerpac). The measurement accuracy of hydraulic pump was up to 1bar. The two ends of the tube were free to be able to
move in the axial direction for providing axial feeding. Internal pressure measured by barometer and axial feeding measured by linear variable differential transformer with the measurement accuracy
of 0.01mm.
To obtain the FLCs, different loading paths with a combination of internal pressure and axial feeding should be applied to the tube. For this purpose, the linear loading curves from internal pressure
and axial feeding were used. The six applied load paths are shown in Figure 3. Loads were applied in two steps, initially the internal pressure was increased and then the axial feeding was applied,
till a burst occurred in the tube. The internal pressure and the axial feeding displacements (e.g., the input loading paths) shown in Figure 3 which were controlled by the PC-based controller of the
experimental setup for a series of the bulging tests. For measuring strains in the experimental work, a regular grid layout of the circle with a diameter of 2.5mm on the samples was etched. To carve
these circles, electrochemical etching device was used.
The circles engraved on the tube were shown in Figure 4. After examination of the bulge tests, the circles transformed to ellipses after deformation. The major and minor diameters of the ellipses
were measured using a profile projector machine.
As a result of excessive pressurizing during the bulge process, bursting occurred in the middle of the tube wall as illustrated in Figure 5.
To determine hydroforming strain limit diagram experimentally, at first, the tubes were carved and then placed under loadings. Loadings stopped when the tube burst occurred. After the bursting, the
major and minor diameters of the ellipses near the crack were measured and then the limit strains were calculated. The major and minor engineering strains can be obtained from the following
equations. Measuring diameters were performed by using the profile projector machine. $e1=a−dd$(1) $e2=b−dd$(2) In these equations, “a” is the large diameter of ellipse and “b” is the small diameter
of it. “d” is the diameter carved in advance.
3 Finite element modeling
The ABAQUS/Explicit FE software was used to model the hydroforming process in order to investigate the FLDs of aluminum tubes. All the analyses were realized using an explicit finite element
approach. The die map used in the simulation can be seen in Figure 1.
This process was simulated with the solving dynamic/explicit. Material properties were extracted with use of uniaxial tensile tests and were entered in the relevant module. Penalty method was used to
establish contact between the tube and mold. Anisotropy coefficients for that material, by the simple tensile test were measured in different directions. To apply anisotropy into the simulation, the
Hill’s 48-yield criterion[16] was used. Hill’s 48-yield criterion and its coefficients based on the measured anisotropy in the directions of 0, 45 and 90 in equations (3)–(7) is given. $f(σ)=F
(σ22−σ33)2+G(σ33−σ11)2+H(σ11−σ22)2+2Lσ232+2Mσ312+2Nσ122$(3) $H=r01+r0$(4) $F=Hr90$(5) $G=Hr0$(6) $N=(r90+r0)(2r45+1)2r90(1+r0)$(7) The coefficients of Hill’s 48-yield criterion for a
three-dimensional stress mode and its relation with the main factors yield criterion are given below. $F=12(1R222+1R332−1R112)$(8) $G=12(1R332+1R112−1R222)$(9) $H=12(1R112+1R222−1R332)$(10) $L=
32R232$(11) $M=32R132$(12) $N=32R122$(13)In this paper, for convenience, a Cartesian coordinate system changed to the cylindrical that, in which case, the anisotropy factor for the thickness and the
other directions were put “1”.
The tube was considered as a deformable part and it was meshed using composite shell elements (four nodes, reduced integration elements, ABAQUS type S4R). Friction between the mold and the tube was
intended 0.1. The tube was used in the power hardening law to model its behavior. The Holloman’s equation is written as follows [17]: $σ‾Y=K(ϵ‾)n$(14) where is the effective stress, is the effective
plastic strain, n is strain hardening exponent and K is the strength coefficient.
Figure 6 demonstrates the FE model included of the tube and the die.
3.1 Analytical necking criterion
Selecting an appropriate necking criterion is important to determine the start of plastic instability in tube hydroforming. For obtaining the FLC, in this research, necking criteria, containing the
acceleration of maximum and minimum strain were employed to predict the onset of plastic instability.
The necking time of a specimen could be determined by using this method. To obtain the FLC numerically, it was essential to predict at which time and where the necking phenomena occurred in the
analyzed material. It was possible to predict the necking time of the analyzed specimen using its acceleration of the max strain. Two different criteria to detect the start of plastic instability in
the tube were suggested to determine the FLC. The forming limits of the tube were predicted, considering the history of the maximum and minimum strains by taking the maximum second derivative. For a
given strain path, the limit strain was determined at the maximum value of the strain acceleration. Figure 7 represents the maximum and minimum strains for the 4mm axial feeding mode (Fig. 3).
Figure 8 shows a relationship between the two criteria large strain and small strain. Due to the linear relationship between the two criteria, it is concluded that the second derivative both at the
same time reaches its maximum value. As a result, the use of either of two criteria will have one answer.
For this purpose, after completing the simulations, the element that had the maximum amount of equivalent plastic strain was reported.
Then, a diagram for the highest and lowest strain versus time for that element was determined. For example, the graph for the 4mm axial feeding mode (Fig. 3) is shown in Figure 9.
After drawing the curve, get the Microsoft Office Excel software output from the curve. Then, import that data to the MATLAB software for using the curve fitting option to earn chart’s equation and
twice derive from it. Figure 10 represents the second derivative of max strain graph and the data obtained from it (4mm axial feeding mode) in MATLAB.
The time when the acceleration of the maximum strain got its maximum value (0.006s) was assumed as the start of necking phenomena in the analyzed material. Finally, when the second derivative strain
reaches its maximum value, consider the large strain as ϵ[major] and the small strain as ϵ[minor].
From the combination of these two points, a point on the FLD was determined. Similarly, these steps were repeated for another loading path (Fig. 3) to obtain the other points in order to draw the
4 Results and discussion
In comparison, between defects, such as wrinkling and buckling, rupture is an irreversible defect in tube hydroforming[17–21]. In order to investigate the formability in the hydroforming process
from the perspective of the fault rupture, numerical and experimental methods with a combination of internal pressure and axial feeding were used in this study. Internal pressure and axial feeding
with different loading paths imposed on the tube which at first, the three-dimensional models simulated using finite element method. In this research, the results of the simulated hydroforming test
for aluminum 7020-T6 tube were presented. The necking criteria, containing the acceleration of major and minor strains, were applied to identify the start of plastic instability in the analyzed
material to construct the FLD.
The predicted FLD was compared with the experimental test results for aluminum 7020-T6 tube (Fig. 11). From Figure 11, it could be concluded that this method was in good agreement with the
experimental test results for aluminum tube 7020-T6.
Figure 11 showed that there was a low difference between the results of FEM and experiment for the FLD[0] (i.e., major strain in the plane strain state). Moreover, Table 2 compared the numerical
predictions with the measured strains from physical experiments at the onset of necking for two different stain paths (e.g., plane strain mode and uniaxial tension mode). This difference could be due
to the errors in strain measurement by the conventional “circle grid analysis” method[22–25]. Therefore, it could be deduced that the FE results were in fairly good agreement with experimental
investigations. Figure 12 shows the distribution of thickness in each of the samples.
5 Conclusions
In this study, the hydroforming strain limit diagrams of the aluminum tube 7020-T6 were determined numerically and experimentally for the first time. The numerical method already developed for sheet
materials was extended and applied to obtain the FLDs for tubular materials (aluminum tube 7020-T6). The numerical results for the FLDs were verified by comparing them with experimental tests. The
numerical model was based on the acceleration of maximum principal strain or acceleration of minimum principal strain. By analyzing the two criteria large strain and small strain, it was found that
both of them have a linear relationship relative to each other. So the second derivative of any of them at a time reaches its maximum value. Therefore, use of each of them has the same result in the
determination of forming limit curves. This numerical criterion was used for the first time to predict the FLD of the aluminum tube 7020-T6. According to the forming limit diagrams obtained (Figure
11), it was concluded that, firstly, FLD for hydroforming process falls in left side of the line ϵ[2]=0 (plane strain mode). And secondly, whatever the ratio between axial feeding and the internal
pressure increased, the points obtained on the graph are drawn towards the negative minor strain (ϵ[2]). According to the thickness distribution graph in the samples (Fig. 12), it can be seen that to
move from the edge of the tube to the middle of it, the thickness of the elements declined; which shows the necking phenomenon. Results from the suggested numerical simulations were in fairly good
agreement with experimental investigations.
FLD[0]: major strain in plane strain state
r[0], r[45], r[90]: anisotropy coefficients in the different directions
H, F, G, N: material constants
PEEQ: equivalent plastic strain
The authors would like to express their deepest gratitude to Professor H. Moslemi Naeini, Dr. S.J. Hashemi and, Tarbiat Modares University (Laboratory of Metal Forming) for their help. The authors
also would like to acknowledge the financial support of Iran National Science Foundation (INSF).
All Tables
All Figures
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Initial download of the metrics may take a while. | {"url":"https://www.mechanics-industry.org/articles/meca/full_html/2017/04/mi160250/mi160250.html","timestamp":"2024-11-10T09:04:30Z","content_type":"text/html","content_length":"127594","record_id":"<urn:uuid:dd32a5b2-393c-4fb2-8972-b1e6eb7c224a>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00175.warc.gz"} |
Solving the mistery of the KL divergence
In this notebook, I try to understand how the KL divergence works, specifically the one from PyTorch.
Relevant docs are here: https://pytorch.org/docs/stable/nn.html#torch.nn.KLDivLoss
Basically, given an $N \times \ast $ tensor x, where $\ast$ represents any number of dimensions besides the first one, the first dimension of x will hold $N$ tensors. Each one of these tensors
symbolizes a (discrete) probability distribution. This means that each of the tensors must sum to 1 (x.sum(0) = [1.0 ,1.0 ,1.0 ,...]).
An easy way to do that to the output of a neural network is to use the softmax function. Another is to divide each value inside the tensor by the sum of all values.
import torch
import torch.nn.functional as F
In this function, I calculate the KL divergence betwwen a1 and a2 both by hand as well as by using PyTorch’s kl_div() function. My goals were to get the same results from both and to understand the
different behaviors of the function depending on the value of the reduction parameter.
First, both tensors must have the same dimensions and every single tensor after dimension 0 must sum to 1, i.e. dimension 0 is the batch dimension and each individual tensor in this dimension
represents a (discrete) probability distribution. Applying x.softmax(0) accomplishes this.
Furthermore, we need to apply the log to the values in the first collection. log_softmax(0) accomplishes both at the same time.
def kl_div(a1, a2):
# the individual terms of the KL divergence can be calculated like this
manual_kl = (a2.softmax(0) * (a2.log_softmax(0) - a1.log_softmax(0)))
# applying necessary transformations
a1ready = a1.log_softmax(0)
a2ready = a2.softmax(0)
print(F.kl_div(a1ready, a2ready, reduction='none').sum())
print(F.kl_div(a1ready, a2ready, reduction='sum'))
print(F.kl_div(a1ready, a2ready, reduction='none').mean())
print(F.kl_div(a1ready, a2ready, reduction='mean'))
print(F.kl_div(a1ready, a2ready, reduction='batchmean'))
Here I apply the above function on 2D tensors.
dist = torch.distributions.uniform.Uniform(0,10)
a1 = dist.sample((5 ,2))
a2 = dist.sample((5, 2))
kl_div(a1, a2)
/home/user/.anaconda3/lib/python3.7/site-packages/torch/nn/functional.py:2247: UserWarning: reduction: 'mean' divides the total loss by both the batch size and the support size.'batchmean' divides only by the batch size, and aligns with the KL div math definition.'mean' will be changed to behave the same as 'batchmean' in the next major release.
warnings.warn("reduction: 'mean' divides the total loss by both the batch size and the support size."
Here I apply the above function on 3D tensors.
dist = torch.distributions.uniform.Uniform(0,10)
a1 = dist.sample((10, 6, 4))
a2 = dist.sample((10, 6, 4))
a1s = a1.softmax(2)
a2s = a2.softmax(2)
kl_div(a1, a2)
Enjoy Reading This Article?
Here are some more articles you might like to read next: | {"url":"https://douglasrizzo.com.br/blog/2020/05/kl-div-pytorch/","timestamp":"2024-11-05T03:16:12Z","content_type":"text/html","content_length":"21747","record_id":"<urn:uuid:55a07f40-e4b4-47bb-a8ab-169ac54e5863>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00827.warc.gz"} |
Hodgkin & Huxley (1952) ‘A Quantitative Description of Membrane Current and its Application to Conduction and Excitation in Nerve’, J. Physiol. 117: 500-544.
The Hodgkin & Huxley (1952) (HH) model is one of the foundational models of cellular electrophysiology. It defined the “standard” gating kinetics still used in many models today! The model includes
potassium, sodium, and ‘leakage’ currents as well and the transmembrane electrical potential. The HH model was originally developed to investigate flow of electric charge in giant nerve axons in
squid, but has been applied to a wide range of physiology over the years. The image below shows a graphical view of the elements described in the HH model.
This example follows the material prepared for the VPH-MIP standardisation and ontologies module.
Mathematical overview
When encoding a model into CellML it is important to first get a good overview of the mathematical equations in the model. A summary for the HH model is shown below. For this model, it is fairly easy
to get a good idea of the model structure and begin thinking about how to encode the model in CellML. But this is not always the case, particularly for modern models consisting of many tens of state
variables - so often you will need to iterate when developing the CellML encoding of a model. Another reason why it is good practice to make use of a good version control system :)
Modular description
CellML divides the mathematical model into distinct components, which are able to be re-used. So we want to divide the mathematical model into meaningful blocks (CellML component). In the HH model,
these would be:
• Potassium current component
• Sodium current component
• Leakage current component
• Membrane potential component
• Gating kinetics component – a single definition instantiated three times for the n, m, and h gates
• Time component
Each of these blocks is itself a CellML model, which enables us to reuse the various components in future studies and models. It is often useful to separate time into its own component as it is used
throughout the model and is usually one of the main variables to be managed when joining models together.
CellML uses MathML (http://www.w3.org/Math) to encode the mathematical equations in a model. Specifically, CellML 1.0 and CellML 1.1 use MathML 2.0. Using MathML, the equation:
would be encoded in MathML like this:
which you can see in the CellML model: gating-variable.xml; in the hh_gating_variable component.
Define variables
The equations encoded in MathML define the relationships between variable's in the CellML model. All variables in CellML must be defined, and must be assigned units, with some examples shown below.
In the above example:
• X_initial defines the initial value for the state variable X (initial_value="X_initial")
• public_interface="in" means the variable will be defined elsewhere in the model
• public_interface="out" means the variable will be available for use elsewhere in the model
• private interfaces are for use with children of this component (encapsulation grouping)
Define components
The mathematics and associated variables are grouped into CellML components, which form the reusable building blocks for a model. Continuing the gating variable example above, the skeleton component
is shown below. Component's themselves are just a named container that groups the variable definitions with the mathematical expressions in which they are used.
Define model
The model, in this case, simply provides the wrapper around the component where we import standard unit definitions for use in the variable declarations and math and then define the component. Note
the units dimensionless are defined in the CellML standard and therefore do not need to be defined in the model, we just need to define the "non-standard" units that we want to use.
Everything in the model can, and should, be annotated using RDF/XML – left out here for brevity.
The actual HH gating variable model can be found here: gating-variable.xml.
We now have a generic HH gating variable model, so we can instantiate specific instances for the sodium and potassium currents in the HH model. Here we demonstrate the example of the potassium
current. Sodium is the same but repeated twice for the m and h gates. (Math is just there for example, does need to be MathML not infix like this.)
Grouping and connecting
Creating encapsulation groups allows modellers to create abstract entities for later re-use. Importing the parent of the encapsulation group will also import the child components, so later on when
the potassium current component is imported you also get the n-gate and gating rate kinetics (alpha_n and beta_n) defined.
The variables in the model need to be connected between components (following the *_interface attributes described earlier). These connections are treated as mathematical equalities with automatic
unit conversion. In the snippet here, we need to pass the time and initial value for the gating variable ‘n’ into the gating model and the gating model will return the gating variable ‘n’. Not shown,
but we would also need similar connections to connect the gating rate kinetic parameter models to obtain the alpha_n and beta_n values for the given membrane potential, temperature, etc…
Complete mathematical model
Here we describe the complete mathematical model encoded in CellML, available as the stimulated.xml. As normal, first we import the required units. Then we import instances of each of the current
Then we define an “action potential” component which we will use to define the membrane potential with an applied electrical stimulus. This component will define the “interface” presented to users of
this model encoding. It is important to note the public and private interfaces for these variable declarations. public_interface=in means we expect the variable to be defined outside this model
(c.f., inputs); public_interface=out means we are making that variable available to users of this model (c.f. outputs); and the private interfaces are used to define the connections to the internal
child components of this model.
Following best practices, this model separates the mathematics from the parameterisation of the model, so generally we will expect this mathematical model to be imported into a specific parameterised
instance in order to perform numerical simulations. The parameterisation would include defining the stimulus protocol to be applied. Similarly, this model may be used in larger scale models, e.g.,
tissue electrical activation, which might make use of the output variables (current and membrane potential) in further computational models.
The block of math in the action potential component defines the differential equation governing the membrane potential, which is just the sum of the transmembrane currents we imported earlier, and an
applied stimulus current as described above.
We also define an encapsulation group such that the action_potential component can be re-used in further models. For example, as described above, we expect this model to be imported into a specific
parameterisation of the mathematical model.
As described for the gating variable model previously, we need to connect the variables defined in the action_potential component to the other components in the model (imported above). In order to
allow the separation of mathematical model from parameterisation, we need to ensure required parameters are defined all the way up the encapsulation and import hierarchy.
Instantiation for simulation
The model above (stimulated.xml) defines the mathematical model. In order to perform a simulation, we need to parameterise the model as required for the particular simulation purpose. Here we look at
the case when a periodic electrical stimulus current is applied to the model. The CellML model is given at experiments/periodic-stimulus.xml and the corresponding SED-ML document is sed-ml/
periodic-stimulus.xml. Results from performing this simulation experiment are shown below, the experiment was executed using the SED-ML Web Tools.
You are able to reproduce these simulation results using the SED-ML web tools. First, you need to save the SED-ML document to your local file system. Then you go to the SED-ML Web Tools site and
upload the SED-ML document to the web site (the actual models used are referenced from the CellML model repository, so you don't need to worry about them). Once the document is uploaded, you are able
to validate the SED-ML, check the details of the simulation experiment, and perform the actual experiment - which should result in the above graphs. | {"url":"https://models.physiomeproject.org/w/andre/VPH-MIP/@@rawfile/58e44de10bed7fbc87fd1d437f3001c7b653a738/tutorial/tutorial.html","timestamp":"2024-11-05T03:04:16Z","content_type":"text/html","content_length":"11114","record_id":"<urn:uuid:bd066a70-a32c-4e3c-825a-6a68deb7a19e>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00255.warc.gz"} |
Variational Formulation for Tensile Testing Simulation
Hi all,
I am trying to create a simulation in FEniCS that is a good representation of a conventional tensile testing from zero load to material fracture. I pictured the problem in 3 stages, first to
understand variational formulation for linearly elastic material, then that for elasto-plastic material and ultimately that of a realistic ductile metal which includes elastic regime, plastic regime
and fracture regime.
My ultimate goal is to run the simulation and have it produce a stress-strain curve from zero stress at zero strain to zero stress at failure strain.
I have been using this numerical tour (https://readthedocs.org/projects/comet-fenics/downloads/pdf/latest/) for a great reference to linear elastic as well as elasto-plastic simulation. Also, I am
trying to implement rigid boundary condition to simulate the conditions in a tensile test (Rigid Plate Boundary Condition). However, I am still missing the material softening (failing) part in my
stress-strain curve at this point. While I have found several topics on this forum and other sources explaining fracture simulation with phase field in FEniCS, I realized that these examples are all
studying the fracture process (fracture initiation + crack development) by itself and I have yet to find examples that implements a variational formulation which represents the material from elastic
regime to complete fracture (zero load bearing capability).
I am posting this question here on the forum hoping that someone might have came across similar problems before and can shine some light to where I can find great resources for the variational
formulation described above or whether such analysis is feasible in FEniCS
2 Likes
Borden et al. did some work on phase field fracture for ductile materials, which includes plastic yielding before fracture:
I have not tried to implement this in FEniCS, but it looks like the formulation is written up clearly enough for one to do so, and the main difficulties are already covered by the plasticity tutorial
you linked above.
1 Like
FEniCS is a general purpose finite-element solver. What you are asking is really a constitutive material question. Implementing non-linear constitutive material laws is not straightforward from
scratch in FEniCS.
For this purpose, we worked to link FEniCS with the constitutive law code generator MFront, the corresponding module is described here along with some documented demos.
What type of constitutive law are you investigating ?
Hi bleyerj,
First of all, thank you for creating the numerical tours, that was very helpful.
I am trying to simulate ductile metal with linear elastic, and plastic deformation as well as damage. The von Mises plasticity with isotropic linear hardening as shown in the example will work for my
purpose, but I am trying to implement ultimate tensile stress (UTS) as well as damage into the simulation as well.
I am not sure what constitutive law will be a good representation of the post UTS behavior of the metal but I am imagining something that can show a decrease in load bearing capacity to zero after
the stress at Gauss points surpass the UTS, instead of just flagging the simulation to stop. In another word, I am looking for constitutive laws that will represents the material’s behavior from UTS
to post fracture.
The reason that I am trying to implement damage into the simulation is for heterogeneous materials. For example, in a plane stress 2D case, if I have a square part/mesh and the left half is weaker
than the right. In tensile testing, this sample will have a partial fracture in the left half first but not the right half, then the complete fracture happens when the right half surpass its UTS too.
If I don’t have anything that represents the material past UTS, then the simulation would halt itself when the weaker half of the this heterogeneous structure breaks.
Sorry for my lack of systematic solid mechanics knowledge and hope my example illustrates my question.
Well, modelling material behaviour in the post-yielding stage until complete fracture is extremely hard. The softening regime introduces mathematical ill-posedness yielding to mesh dependency.
Softening behaviours must therefore be regularized in some sense.
I suggest looking at literature on ductile damage models, phase-field methods, etc.
1 Like | {"url":"https://fenicsproject.discourse.group/t/variational-formulation-for-tensile-testing-simulation/3394","timestamp":"2024-11-03T12:08:46Z","content_type":"text/html","content_length":"33683","record_id":"<urn:uuid:97063e3f-5ec0-4344-aef2-557ed85df746>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00888.warc.gz"} |
P-Value: A Complete Guide
Published by at August 31st, 2021 , Revised On August 3, 2023
You might have come across this term many times in hypothesis testing. Can you tell me what p-value is and how to calculate it? For those who are new to this term, sit back and read this guide to
find out all the answers. Those already familiar with it, continue reading because you might get a chance to dig deeper about the p-value and its significance in statistics.
Before we start with what a p-value is, there are a few other terms you must be clear of. And these are the null hypothesis and alternative hypothesis.
What are the Null Hypothesis and Alternative Hypothesis?
The alternative hypothesis is your first hypothesis predicting a relationship between different variables. On the contrary, the null hypothesis predicts that there is no relationship between the
variables you are playing with.
For instance, if you want to check the impact of two fertilizers on the growth of two sets of plants. Group A of plants is given fertilizer A, while B is given fertilizer B. Now by using a two-tailed
t-test, you can find out the difference between the two fertilizers.
Null Hypothesis: There is no difference in growth between the two sets of plants.
Alternative Hypothesis: There is a difference in growth between the two groups.
What is the P-value?
The p-value in statistics is the probability of getting outcomes at least as extreme as the outcomes of a statistical hypothesis test, assuming the null hypothesis to be correct. To put it in simpler
words, it is a calculated number from a statistical test that shows how likely you are to have found a set of observations if the null hypothesis were plausible.
This means that p-values are used as alternatives to rejection points for providing the smallest level of significance at which the null hypothesis can be rejected. If the p-value is small, it
implies that the evidence in favour of the alternative hypothesis is bigger. Similarly, if the value is big, the evidence in favour of the alternative hypothesis would be small.
How is the P-value Calculated?
You can either use the p-value tables or statistical software to calculate the p-value. The calculated numbers are based on the known probability distribution of the statistic being tested.
The online p-value tables depict how frequently you can expect to see test statistics under the null hypothesis. P-value depends on the statistical test one uses to test a hypothesis.
• Different statistical tests can have different predictions, hence developing different test statistics. Researchers can choose a statistical test depending on what best suits their data and the
effect they want to test
• The number of independent variables in your test determines how large or small the test statistic must be to produce the same p-value
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When is a P-value Statistically Significant?
Before we talk about when a p-value is statistically significant, let’s first find out what does it mean to be statistically significant.
Any guesses?
To be statistically significant is another way of saying that a p-value is so small that it might reject a null hypothesis.
Now the question is how small?
If a p-value is smaller than 0.05 then it is statistically significant. This means that the evidence against the null hypothesis is strong. The fact that there is less than a 5 per cent chance of the
null hypothesis being correct and plausible, we can accept the alternative hypothesis and reject the null hypothesis.
Nevertheless, if the p-value is less than the threshold of significance, the null hypothesis can be rejected, but that does not mean there would be a 95 percent probability of the alternative
hypothesis being true. Note that the p-value is dependent or conditioned upon the null hypothesis is plausible, but it is not related to the correctness and falsity of the alternative hypothesis.
When the p-value is greater than 0.05, it is not statistically significant. It also indicates that the evidence for the null hypothesis is strong. So, the alternative hypothesis, in this case, is
rejected, and the null hypothesis is retained. An important thing to keep in mind here is that you still cannot accept the null hypothesis. You can only fail to reject it or reject it.
Here is a table showing hypothesis interpretations:
P-value Decision
P-value > 0.05 Not statistically significant and do not rejects the null hypothesis.
P-value < 0.05 Statistically significant and rejects the null hypothesis in favour of the alternative hypothesis.
P-value < 0.01 Highly statistically significant and rejects the null hypothesis in favour of the alternative hypothesis.
Is it clear now? We thought so! Let’s move on to the next heading, then.
How to Use P-value in Hypothesis Testing?
Follow these three simple steps to use p-value in hypothesis testing.
Step 1: Find the level of significance. Make sure to choose the significance level during the initial steps of the design of a hypothesis test. It is usually 0.10, 0.05, and 0.01.
Step 2: Now calculate the p-value. As we discussed earlier, there are two ways of calculating it. A simple way out would be using Microsoft Excel, which allows p-value calculation with Data Analysis
Step 3: Start comparing the p-value with the significance level and deduce conclusions accordingly. Following the general rule, if the value is less than the level of significance, there is enough
evidence to reject the null hypothesis of an experiment.
FAQs About P-Value
The p-value in statistics is the probability of getting outcomes at least as extreme as the outcomes of a statistical hypothesis test, assuming the null hypothesis to be correct. It is a calculated
number from a statistical test that shows how likely you are to have found a set of observations if the null hypothesis were plausible.
To be statistically significant is another way of saying that a p-value is so small that it might reject a null hypothesis.
This table shows when the p-value is significant.
P-value Decision
P-value > 0.05 Not statistically significant and do not rejects the null hypothesis.
P-value < 0.05 Statistically significant and rejects the null hypothesis in favour of the alternative hypothesis.
P-value < 0.01 Highly statistically significant and rejects the null hypothesis in favour of the alternative hypothesis. | {"url":"https://www.researchprospect.com/p-value-a-complete-guide/","timestamp":"2024-11-07T23:24:59Z","content_type":"text/html","content_length":"199392","record_id":"<urn:uuid:5eac337e-1dfa-472d-81d2-044869efe694>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00668.warc.gz"} |
DAMMIF, a program for rapid ab-initio shape determination in small-angle scattering
^aEuropean Molecular Biology Laboratory, Hamburg Outstation Notkestrasse 85, 22603 Hamburg, Germany, and ^bInstitute of Crystallography, 117333 Moscow, Russian Federation
^*Correspondence e-mail: franke@embl-hamburg.de, svergun@embl-hamburg.de
(Received 18 August 2008; accepted 5 January 2009; online 24 January 2009)
DAMMIF, a revised implementation of the ab-initio shape-determination program DAMMIN for small-angle scattering data, is presented. The program was fully rewritten, and its algorithm was optimized
for speed of execution and modified to avoid limitations due to the finite search volume. Symmetry and anisometry constraints can be imposed on the particle shape, similar to DAMMIN. In equivalent
conditions, DAMMIF is 25–40 times faster than DAMMIN on a single CPU. The possibility to utilize multiple CPUs is added to DAMMIF. The application is available in binary form for major platforms.
1. Introduction
Small-angle scattering (SAS) of X-rays and neutrons is a fundamental tool in the study of the nanostructure of matter, including disordered systems and solutions (Feigin & Svergun, 1987 ). In a
scattering experiment, the specimen (e.g. particles of nanometre-scale size floating in solution or embedded in a bulk matrix) is exposed to X-rays or neutrons, and the scattered intensity I is
recorded. For disordered systems, the random positions and orientations of particles lead to an isotropic intensity distribution I(s), which depends on the modulus of momentum transfer s (I(s) is
proportional to the scattering from a single particle averaged over all orientations. This allows one to obtain information about the overall shape and internal structure of particles at a resolution
of 1–2nm (Feigin & Svergun, 1987 ; Svergun & Koch, 2003 ).
Recent progress in instrumentation and development of data analysis methods (Svergun & Koch, 2003 ; Petoukhov et al., 2007 ) has significantly enhanced the resolution and reliability of the models
provided by SAS. A number of novel approaches have been proposed to analyse the scattering data from monodisperse systems in terms of three-dimensional models [see Petoukhov et al. (2007 ) for a
review]; these advances have significantly increased the popularity of SAS in the study of biopolymers in solution. Among these methods, ab-initio shape determination techniques are especially
important: first, they do not require a-priori information about the particle, and second, they are applicable also for moderately polydisperse (nonbiological) systems, allowing one to retrieve the
overall averaged shape over the ensemble (Shtykova et al., 2003 , 2007 ).
The aim of ab-initio analysis of SAS data is to recover the three-dimensional structure from the one-dimensional scattering pattern, and unique reconstruction is only possible in the trivial case of
a spherical particle. In shape determination, one represents the particle by homogeneous models to constrain the solution and reduce the ambiguity of the reconstruction. This simplification usually
is justified in the analysis of the low-angle scattering patterns from single-component particles. In all ab-initio methods, particle shape is represented in real space by a parametric model, and the
parameters of the model are altered so as to minimize the difference between the computed scattering of the model and the experimental data. A number of methods and alternative programs exist, which
differ primarily in the way the shape is represented. In the first general ab-initio approach (Stuhrmann, 1970 ), an angular envelope function was implemented in the program Sasha (Svergun et al.,
1996 ), which was limited to globular particles without significant internal cavities. More detailed models are obtained by representing the particle by finite volume elements, thus allowing internal
cavities to be accounted for. Using beads to model the scattering object, which was first proposed by Chacon et al. (1998 ) and implemented in the program DALAI_GA, a search volume is filled by
densely packed small spheres (also referred to as dummy atoms), which are assigned either to the particle or to the solvent. Starting from a random assignment, a Monte Carlo search, for example a
genetic algorithm in DALAI_GA or simulated annealing (SA) in DAMMIN (Svergun, 1999 ), is employed to find a model that fits the data. A similar approach was implemented in the Give'n'Take procedure
of SAXS3D (Walther et al., 2000 ), which runs on a grid of unlimited size. Heller et al. (2002 ) developed a program SASMODEL, representing the particle by a collection of interconnected ellipsoids.
Ab-initio methods have been proven to reliably reconstruct the low-resolution shape in numerous tests and practical studies, and they now belong to routine tools in SAS data analysis. Since little or
no information has to be specified by the user in most cases, these methods are currently being incorporated into high-throughput automated data analysis pipelines (Petoukhov et al., 2007 ). The
extensive use of the shape determination programs, including large scale studies, makes the speed of reconstruction a rather important issue. The Monte Carlo-based algorithms usually require millions
of random models to be screened and are thus time consuming. Moreover, given that different shapes are obtained starting from different initial random models, often ten or more ab-initio runs need to
be performed and averaged to assess the uniqueness of the solution and to reveal the most persistent shape features (Volkov & Svergun, 2003 ). Presently, most shape determination programs require
hours of CPU time for a single run on a typical Windows or Linux PC; clearly this time needs to be reduced to the order of minutes or less.
This paper describes a new implementation of DAMMIN (Svergun, 1999 ), one of the most popular shape determination programs publicly available. The program, called DAMMIF (where `F' denotes fast), has
been completely rewritten in object-oriented code and the algorithm has been optimized for speed and efficiency. The algorithm was further improved in an attempt to avoid artifacts caused by the
limited search volume. This was achieved by replacing the closed with an unlimited and growing search volume . A version of DAMMIF optimized to make use of multiple CPUs is also available.
Furthermore, the implementation of DAMMIF, like DAMMIN, features options to account for symmetry and anisometry in the modelling if the relevant information is available.
2. DAMMIN algorithm
In this section, we outline the major features of DAMMIN that are important for an understanding of the DAMMIF algorithm. The reader is referred to the original publication (Svergun, 1999 ) for
further details.
In the original version of DAMMIN, a search volume (usually a sphere with radius R equal to half the maximum particle size D[max]) is filled with densely packed small spheres of radius X of length X
is calculated as
where the partial scattering amplitudes are
j[l](sr[j]) denote spherical Bessel functions. The function f(X) to be minimized has the form
where the first term on the right-hand side is the discrepancy between the experimental and calculated data, and the second term summarizes penalties as listed in Table 1 weighted by appropriate
Explicit penalties are configurable and may be disabled; implicit penalties are enforced and may not be disabled.
Function DAMMIN DAMMIF
Peripheral penalty (gradually decreasing) Keeps the particle beads close to the origin at high Explicit –
Disconnectivity penalty Ensures that the model is interconnected Explicit Implicit
Looseness penalty Ensures that the model is compact Explicit Explicit
Anisometry penalty (with symmetries only) Specifies whether the model should be oblate or prolate Explicit Explicit
Centre/R[g] penalty Keeps the centre of mass of the model close to the origin – Explicit
The result after running the application is a compact interconnected DAM that fits the experimental data. If information about the particle symmetry is available, it is taken into account as a hard
constraint by changing all the symmetrical dummy atoms simultaneously. A-priori information about the particle anisometry can also be taken into account.
The spherical harmonics expansion using equations (1) and (2) is computationally superior to the standard Debye (1915 ) formula, which is usually employed to compute the scattering from bead models.
Moreover, only a single dummy atom is changed at each move and hence only a single summand in equation (2) must be updated to recalculate the partial amplitudes. This accelerates DAMMIN
significantly, but still, as millions of function evaluations are required, a typical refinement takes about 2–3 CPU hours on an average PC for a DAM containing a few thousands spheres.
3. DAMMIF implementation
Similar to DAMMIN, DAMMIF uses the scattering pattern processed by the program GNOM (Svergun, 1992 ); DAMMIF also follows the general algorithm of DAMMIN.
The program was, however, completely rewritten with the main aim of speeding up the operation. Major algorithmic changes in DAMMIF are described in the following sections.
3.1. Bead selection
A very important constraint for low-resolution ab-initio modelling is that in the final model all beads representing the particle must be interconnected to form a single body. Implementation of this
condition is different between DAMMIN and DAMMIF. Fig. 1 shows examples of the cross sections through the initial and final bead models (top and bottom row, respectively) of DAMMIN (left) and DAMMIF
(right). The beads are colour coded as belonging to the particle (red) and solvent (turquoise, blue, green) phases. Turquoise and green beads differ from blue ones only in that the former are
relevant for the bead-selection algorithm described in the next paragraph and the latter for the unlimited search volume as described in the next section.
For each annealing step, DAMMIN and DAMMIF select a bead completely at random. DAMMIN updates the simulated scattering, computes the fit and penalizes possible disconnectivity of the particle beads
before deciding whether to accept or reject the change. There, the disconnectivity is defined by the length of the longest graph (ensemble of beads, where each pair can be connected by moving through
the beads touching each other in the grid), which is a CPU-intense operation. DAMMIF tests connectivity first and rejects disconnected models before launching into the time consuming process of
updating the scattering amplitudes. The latter are computed if and only if a particle bead (red) or an adjacent bead (turquoise) is selected (Fig. 1 ); otherwise the step is cancelled and execution
is resumed with the next step. A summary of the set of rules used to decide about the connectivity of models is given in Table 2 .
Following these rules, early rejection can be based on the number of graphs before and after the proposed change. In particular, if the change leads to two or more graphs in a model without symmetry,
the model becomes disconnected.
Case 1: bead x of solvent phase was selected to switch to particle.
x has neighbours in particle phase, then
0 create a new graph, add x
1 add x to the graph the neighbour belongs to
merge all graphs the neighbours belong to, add x
Case 2: bead x of particle phase was selected to switch to solvent.
x has neighbours in particle phase, then
0 find and remove the graph built by x
1 find the graph x belongs to, remove x
find the graph x belongs to, split into two or more graphs if x is an articulation point, remove x
3.2. Unlimited search volume
In DAMMIN, the search volume is configurable at runtime but fixed throughout the search procedure. The search volume is filled with densely packed dummy atoms before SA begins. Limiting the volume
may be a useful feature for shape reconstruction (in particular, nonspherical search volumes can be employed to account for additional information about the shape, if available). However, in some
cases, especially for very anisometric particles, a restricted search volume may lead to artifacts. Indeed, the bead representing the particle is obviously prevented from protruding outside the
border of the search volume. If, during the reconstruction, the particle is formed close to the border, the search space becomes anisotropic, possibly leading to unwanted border effects like
artificial bending. To avoid such effects, the algorithm of DAMMIF was modified, allowing for the search in a variable volume which is extended as necessary during the SA procedure. In the following,
we shall mostly refer to this unlimited DAMMIF, but a bounded-volume version is also available on request.
Unlike DAMMIN, which fully randomizes the closed search volume on start-up (Fig. 1 , top left panel), DAMMIF starts from an isometric object with the radius of gyration (R[g]) matching the
experimentally obtained one (Fig. 1 , top right panel). This proto-particle (red) is constructed by adding successive layers of beads until the desired R[g] is reached. The polyhedral appearance of
the starting model as shown in Fig. 1 is subject to the hexagonal packing of beads – it should be noted that the shape of the initial model has practically no influence on the reconstruction. The
starting shape is then covered by a single layer of solvent beads, shown in green. The green colour implies that, if such a bead is selected for phase transition, potentially missing neighbours are
added to the search volume. To accomplish this, the coordinates of the neighbours are computed and looked up in the list of available beads. If a neighbour is missing, its coordinates are added as a
new bead of solvent phase to the said list. To avoid runtime penalties due to linear searches on ever-growing lists, beads are stored in multidimensional binary search trees (Bentley, 1975 ), which
are also known as kd-trees. Furthermore, amplitudes of newly created dummy atoms are lazily evaluated, i.e. they are not computed until they contribute to the particle scattering for the first time.
Although lazily computed, once available partial amplitudes are stored in a cache for later re-use.
Adding neighbours as described ensures that beads in the particle phase (index = 1) are always surrounded by beads in the solvent phase (index = 0). Thus, the algorithm may traverse a potential, but
not yet mapped, search volume. This was not possible in DAMMIN, where the closed search volume may have blocked the annealing algorithm from potentially better results.
3.3. Penalties
Penalties impose a set of rules on the dummy atom model to modify its likelihood of being accepted by the SA selection rule [equation (3) , right-hand sum]. Hence, penalties are used to guide the
annealing process. In general terms, the bead-selection algorithm presented above implements an implicit penalty. Owing to the improved rejection of disconnected models (Fig. 1 and Table 2 ), the
likelihood of accepting a disconnected model constantly equals zero.
Table 1 summarizes the different sets of penalties implemented in DAMMIN and DAMMIF. In DAMMIF, the peripheral penalty was dropped as there is no more outer boundary to limit particle growth.
Furthermore, the disconnectivity penalty became implied as a result of improved rejection of unwanted disconnected models. Instead, centre and R[g] penalties were introduced. The role of the centre
penalty is to keep the particle within the already mapped space, to prevent needless extension (and thus calculation) of the search volume, and the R[g] penalty ensures a model of appropriate size.
Looseness and anisometry penalties are implemented by both applications.
3.4. Parallelization
In DAMMIN, the SA algorithm is implemented as follows (Fig. 2 , left-hand side):
(i) Start from a random configuration X[0] at a high temperature T[0] [e.g. T[0] = f(X[0])].
(ii) Flip the index of a randomly selected dummy atom to obtain configuration
(iii) If X.
(iv) Hold T constant for 100M reconfigurations or 10M successful reconfigurations, whichever comes first, then cool the system f(X) is observed.
It can easily be seen that the longer the algorithm proceeds, the less likely a successful reconfiguration becomes. As multi-core and multi-CPU systems are becoming more readily available, DAMMIF
also makes use of these resources. To further speed up ab-initio modelling, DAMMIF employs OpenMP, a framework for shared memory parallelization (Dagum & Menon, 1998 ). To exploit the properties of
SA as described above, a simple prefetch and branch prediction scheme was implemented (Fig. 2 , right-hand side). Instead of a single neighbouring model DAMMIN, DAMMIF computes multiple models
4. Quality of reconstruction and practical aspects
Extensive tests on simulated and experimental data showed that the models provided by DAMMIF are comparable to those of DAMMIN and the quality of reconstruction is compatible with that presented by
Svergun (1999 ) and Volkov & Svergun (2003 ). For highly anisometric particles, the models provided by DAMMIF may be more accurate thanks to the absence of border effects. A comparison of model
reconstructions by DAMMIN and DAMMIF of a cylindrical particle with radius 10Å and height 200Å is illustrated in Fig. 3 .
Of course, DAMMIF, similar to DAMMIN and other shape determination programs, is not applicable to heterogeneous systems like mixtures or unfolded proteins. For the analysis of higher-resolution data
from small (less than 30kDa) proteins, where the contribution from the internal structure is essential, other programs like GASBOR (Svergun et al., 2001 ) may be more appropriate for ab-initio
analysis than the shape determination algorithms.
The R factor R(I,X) [see equation (3) ] of the obtained DAMMIF model, which is provided to the user in the log file and in the PDB-type file (Protein Data Bank; Berman et al., 2000 ) containing the
final solution, permits one to rapidly assess the quality of the reconstruction. Usually, R factors exceeding 0.1 indicate poor fits and therefore point to incorrect assumptions about the object
under study. It is further extremely important to analyse the uniqueness of the reconstruction, similar to DAMMIN, by comparing and averaging multiple individual runs, e.g. using the program DAMAVER
(Volkov & Svergun, 2003 ). The improved speed of DAMMIF allows the user to perform these analyses in a much shorter time.
5. Conclusions
Here we present DAMMIF, an advanced implementation of the popular ab-initio modelling program DAMMIN (Svergun, 1999 ). Table 3 summarizes the differences between these two implementations: most
notable is a reduction of the average runtime by a factor of 25–40, depending amongst other factors on the number of dummy atoms in the search model. Furthermore, a pre-defined search volume that
limits mapping of possible solutions was replaced by an unlimited, adapting search space.
DAMMIN DAMMIF
Expected runtime, fast mode† 15min 30s
Expected runtime, slow mode† 24h 1h
Memory usage, slow mode† 10 MB 100 MB
Search volume Closed Unlimited
Particle symmetry constraints Yes Yes‡
Particle anisometry constraints Yes Yes
Model chaining No Yes§
Parallelization No Yes
Platforms Windows, Linux Windows, Linux
Implementation language Fortran 77 Fortran 95
†The CPU wall clock times for a run on a typical PC without symmetry restrictions are given. Fast and slow mode: packing radius corresponds to ca 2000 and ca 10000 dummy atoms, respectively, in a
sphere with radius D[max]/2.
‡Same as in DAMMIN, but the space groups P23 and P432 and icosahedral symmetry are not implemented.
§Optionally, sorts the dummy atoms in the output file to form pseudo-chains.
Additional constraints such as particle symmetry and anisometry are available in DAMMIF as they are in DAMMIN (i.e. as a hard constraint) – except for some higher symmetries listed in Table 3 where
DAMMIN itself is very fast. As an additional option, DAMMIF is able to output pseudo-chains in PDB-format files to make them more suitable for submission to the PDB.
In the present implementation of DAMMIF, most of the reduction in runtime is due to algorithmic improvements, such as differences in bead selection, and not due to parallelization (Fig. 2 ). Because
DAMMIF extensively employs look-up tables and thus uses more RAM, the memory-transfer overhead significantly reduces the gain from the use of multiple CPUs. This will be investigated and, if
possible, improvements will be added to later versions of the application.
Further work is also in progress to implement the prefetch strategy (Fig. 2 ), to parallelize other CPU-intensive programs from the ATSAS package (Konarev et al., 2006 ) that employ SA for model
building in small-angle scattering.
This work was supported by EU FP6 Design study SAXIER, grant No. RIDS 011934. The authors would also like to thank Adam Round for many fruitful discussions.
Bentley, J. L. (1975). Commun. ACM, 18, 509–517. CrossRef Web of Science Google Scholar
Berman, H. M., Westbrook, J., Feng, Z., Gilliland, G., Bhat, T. N., Weissig, H., Shindyalov, I. N. & Bourne, P. E. (2000). Nucleic Acids Res. 28, 235–242. Web of Science CrossRef PubMed CAS Google
Chacon, P., Moran, F., Diaz, J. F., Pantos, E. & Andreu, J. M. (1998). Biophys. J. 74, 2760–2775. Web of Science CrossRef CAS PubMed Google Scholar
Dagum, L. & Menon, R. (1998). IEEE Comput. Sci. Eng. 5, 46–55. Web of Science CrossRef Google Scholar
Debye, P. (1915). Ann. Phys. 46, 809–823. CrossRef CAS Google Scholar
Feigin, L. A. & Svergun, D. I. (1987). Structure Analysis by Small-Angle X-ray and Neutron Scattering. New York: Plenum Press. Google Scholar
Heller, W. T., Abusamhadneh, E., Finley, N., Rosevear, P. R. & Trewhella, J. (2002). Biochemistry, 41, 15654–15663. Web of Science CrossRef PubMed CAS Google Scholar
Konarev, P. V., Petoukhov, M. V., Volkov, V. V. & Svergun, D. I. (2006). J. Appl. Cryst. 39, 277–286. Web of Science CrossRef CAS IUCr Journals Google Scholar
Petoukhov, M. V., Konarev, P. V., Kikhney, A. G. & Svergun, D. I. (2007). J. Appl. Cryst. 40, s223–s228. Web of Science CrossRef CAS IUCr Journals Google Scholar
Shtykova, E. V., Huang, X., Remmes, N., Baxter, D., Stein, B., Dragnea, B., Svergun, D. I. & Bronstein, L. M. (2007). J. Phys. Chem. C, 111, 18078–18086. Web of Science CrossRef CAS Google Scholar
Shtykova, E. V., Shtykova, E. V. Jr, Volkov, V. V., Konarev, P. V., Dembo, A. T., Makhaeva, E. E., Ronova, I. A., Khokhlov, A. R., Reynaers, H. & Svergun, D. I. (2003). J. Appl. Cryst. 36, 669–673.
Web of Science CrossRef CAS IUCr Journals Google Scholar
Stuhrmann, H. B. (1970). Z. Phys. Chem. Neue Folge, 72, 177–198. CrossRef CAS Google Scholar
Svergun, D. I. (1992). J. Appl. Cryst. 25, 495–503. CrossRef Web of Science IUCr Journals Google Scholar
Svergun, D. I. (1999). Biophys. J. 76, 2879–2886. Web of Science CrossRef PubMed CAS Google Scholar
Svergun, D. I. & Koch, M. H. J. (2003). Rep. Prog. Phys. 66, 1735–1782. Web of Science CrossRef CAS Google Scholar
Svergun, D. I., Petoukhov, M. V. & Koch, M. H. J. (2001). Biophys. J. 80, 2946–2953. Web of Science CrossRef PubMed CAS Google Scholar
Svergun, D. I., Volkov, V. V., Kozin, M. B. & Stuhrmann, H. B. (1996). Acta Cryst. A52, 419–426. CrossRef CAS Web of Science IUCr Journals Google Scholar
Volkov, V. V. & Svergun, D. I. (2003). J. Appl. Cryst. 36, 860–864. Web of Science CrossRef CAS IUCr Journals Google Scholar
Walther, D., Cohen, F. E. & Doniach, S. (2000). J. Appl. Cryst. 33, 350–363. Web of Science CrossRef CAS IUCr Journals Google Scholar
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Research Guides: Pi 3.14 Day : Overview
Did you know?
The circumference of a circle is not equal to the length of two diameters.
A circle's circumference divided by its diameter is the same for every circle. (It's pi, or 3.14) The value of π being the constant value 3.14159265358979.…
Source: Johnson, J. (2020). Pi (mathematics). Salem Press Encyclopedia of Science. | {"url":"https://libguides.wccnet.edu/Pi/Overview","timestamp":"2024-11-10T04:35:24Z","content_type":"text/html","content_length":"52159","record_id":"<urn:uuid:bf8503df-38fb-4c29-9419-8cfef5fc06a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00276.warc.gz"} |
seminars - Growth of systole of arithmetic hyperbolic manifolds
The systole of a Riemannian manifold M is the length of a shortest non-contractible closed geodesic in M. In this talk we will discuss how to produce hyperbolic manifolds with large systole, and the
impact in the topology of such manifolds. This talk will be an enlarged version of a previous talk at the 2019 KMS Annual Meeting, but the previous talk is not required at all. | {"url":"http://www.math.snu.ac.kr/board/index.php?mid=seminars&page=26&l=ko&sort_index=speaker&order_type=asc&document_srl=803772","timestamp":"2024-11-13T01:33:32Z","content_type":"text/html","content_length":"44755","record_id":"<urn:uuid:04e758a9-a076-466a-8780-d30d06a17a63>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00427.warc.gz"} |
Prior probability - (Mathematical Probability Theory) - Vocab, Definition, Explanations | Fiveable
Prior probability
from class:
Mathematical Probability Theory
Prior probability is the probability assigned to a hypothesis before any evidence is considered. It serves as the initial degree of belief about the truth of that hypothesis, forming the foundation
for updating beliefs in light of new evidence through Bayesian inference. This initial probability can be based on previous knowledge, expert opinion, or historical data.
congrats on reading the definition of prior probability. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. Prior probabilities can be subjective, reflecting personal beliefs or opinions, and may vary between individuals.
2. In Bayesian inference, the prior probability is combined with the likelihood of the observed data to calculate the posterior probability using Bayes' theorem.
3. Choosing an appropriate prior probability is crucial because it can significantly influence the results of Bayesian analysis.
4. The prior can be informative (based on strong prior knowledge) or uninformative (reflecting a lack of specific prior knowledge).
5. Sensitivity analysis can be used to assess how changes in prior probabilities affect the resulting posterior probabilities.
Review Questions
• How does prior probability influence the process of Bayesian inference?
□ Prior probability plays a vital role in Bayesian inference as it sets the stage for how we interpret new evidence. When new data is introduced, we combine this evidence with our prior belief
to form a posterior probability. The strength and appropriateness of the prior can significantly sway our conclusions, which highlights the importance of choosing a relevant and accurate
prior to ensure valid results.
• Discuss the implications of selecting an informative versus an uninformative prior probability in Bayesian analysis.
□ Selecting an informative prior can lead to more precise estimates when there is substantial existing knowledge about a situation. However, it may introduce bias if not carefully considered.
Conversely, an uninformative prior allows for greater flexibility but may result in less accurate estimates if little information is available. Understanding these implications helps
practitioners balance their previous knowledge and objectivity in Bayesian analysis.
• Evaluate how sensitivity analysis can be applied to assess the impact of prior probabilities on posterior outcomes in Bayesian inference.
□ Sensitivity analysis involves systematically changing prior probabilities to observe how these adjustments affect posterior outcomes. This evaluation reveals how robust conclusions are to
variations in initial beliefs. By understanding this relationship, one can determine whether findings are reliable or overly dependent on specific assumptions made regarding prior
probabilities, ultimately enhancing the credibility of Bayesian analyses.
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