Geometric primitives.
To get started, use the following geometric primitives for points and
axis-aligned rectangles in the plane.
Use the immutable data type Point2D (part of algs4.jar) for points in the plane. Here is the subset of its API that you may use:
Use the immutable data type RectHV (part of algs4.jar) for axis-aligned rectangles. Here is the subset of its API that you may use:public class Point2D implements Comparable<Point2D> { public Point2D(double x, double y) // construct the point (x, y) public double x() // x-coordinate public double y() // y-coordinate public double distanceTo(Point2D that) // Euclidean distance between two points public double distanceSquaredTo(Point2D that) // square of Euclidean distance between two points public int compareTo(Point2D that) // for use in an ordered symbol table public boolean equals(Object that) // does this point equal that object? public void draw() // draw to standard draw public String toString() // string representation }
Do not modify these data types.public class RectHV { public RectHV(double xmin, double ymin, // construct the rectangle [xmin, xmax] x [ymin, ymax] double xmax, double ymax) // throw a java.lang.IllegalArgumentException if (xmin > xmax) or (ymin > ymax) public double xmin() // minimum x-coordinate of rectangle public double ymin() // minimum y-coordinate of rectangle public double xmax() // maximum x-coordinate of rectangle public double ymax() // maximum y-coordinate of rectangle public boolean contains(Point2D p) // does this rectangle contain the point p (either inside or on boundary)? public boolean intersects(RectHV that) // does this rectangle intersect that rectangle (at one or more points)? public double distanceTo(Point2D p) // Euclidean distance from point p to closest point in rectangle public double distanceSquaredTo(Point2D p) // square of Euclidean distance from point p to closest point in rectangle public boolean equals(Object that) // does this rectangle equal that object? public void draw() // draw to standard draw public String toString() // string representation }
Brute-force implementation. Write a mutable data type PointSET.java that represents a set of points in the unit square. Implement the following API by using a red-black BST (using either SET from algs4.jar or java.util.TreeSet).
Corner cases. Throw a java.lang.NullPointerException if any argument is null. Performance requirements. Your implementation should support insert() and contains() in time proportional to the logarithm of the number of points in the set in the worst case; it should support nearest() and range() in time proportional to the number of points in the set.public class PointSET { public PointSET() // construct an empty set of points public boolean isEmpty() // is the set empty? public int size() // number of points in the set public void insert(Point2D p) // add the point to the set (if it is not already in the set) public boolean contains(Point2D p) // does the set contain point p? public void draw() // draw all points to standard draw public Iterable<Point2D> range(RectHV rect) // all points that are inside the rectangle public Point2D nearest(Point2D p) // a nearest neighbor in the set to point p; null if the set is empty public static void main(String[] args) // unit testing of the methods (optional) }
2d-tree implementation. Write a mutable data type KdTree.java that uses a 2d-tree to implement the same API (but replace PointSET with KdTree). A 2d-tree is a generalization of a BST to two-dimensional keys. The idea is to build a BST with points in the nodes, using the x- and y-coordinates of the points as keys in strictly alternating sequence.
insert (0.7, 0.2) insert (0.5, 0.4) insert (0.2, 0.3) insert (0.4, 0.7) insert (0.9, 0.6)
The prime advantage of a 2d-tree over a BST is that it supports efficient implementation of range search and nearest neighbor search. Each node corresponds to an axis-aligned rectangle in the unit square, which encloses all of the points in its subtree. The root corresponds to the unit square; the left and right children of the root corresponds to the two rectangles split by the x-coordinate of the point at the root; and so forth.
Clients. You may use the following interactive client programs to test and debug your code.
Analysis of running time and memory usage (optional and not graded).
Submission. Submit only the files PointSET.java and KdTree.java. We will supply algs4.jar. Your may not call library functions except those in those in java.lang, java.util, and algs4.jar.
This assignment was developed by Kevin Wayne.


Succulents are life, that's a pretty universal statement. Succulents come in all different shapes, sizes and colors; like blue, green, red, pink yellow, and more! They can live pretty much anywhere, they're like that super cool friend who's just down to have a good time. Their name comes from the Latin word "sucus," which means juice or sap since that's what their leaves are filled with. They originate in dry climates so they need all of that sap to keep them going during those long dry spells. All cacti are succulents, but not all succulents are cacti.
Come pick yours up today! They're super easy to maintain so you don't have to worry about killing them (just water them every couple weeks and let them soak up some sunshine).


Miguel Benitez
Full Stack DeveloperI am a full stack developer, with more than 10 years experience in software development. I also have experience as a system administrator in linux servers. The languages which i have experience (in order of recent use) are: Java, elixir (functional language based in erlang), erlang (around 9 months experience with erlang), elixir and phoenix (a web framework built on top of elixir language). I also have experience in python-django, before elixir phoenix, i have worked around 2 years with python django. In the frontend part i have experience with angular1 and also i have worked with jquery, bootstrap, i also had the opportunity to work with WebRTC in the last project that I assumed. In the database section i have experience with postgresql, mysql, db2, mongdb, riak and rethinkdb.


Nick Cave announces illustrated autobiography ‘Stranger Than Kindness’ and new exhibition
*Books tickets at once*.
A new exhibition dedicated to the music and work of Nick Cave has been announced, in conjunction with his new autobiography.
The music icon’s memoir will provide an illustrated journey through his life, according to publishers Canongate.
“Stranger Than Kindness’ is a journey in images and words into the creative world of musician, storyteller and cultural icon Nick Cave,” an official release states.
“This highly collectible book contains images selected by Cave from ‘Stranger Than Kindness: The Nick Cave Exhibition’, presented by the Royal Danish Library in partnership with Arts Centre Melbourne.
“Featuring full-colour reproductions of original artwork, handwritten lyrics, photographs and collected personal artefacts, it presents Cave’s life, work and inspiration and explores his many real and imagined universes.”
Next year, the Stranger Than Kindness: The Nick Cave Exhibition will also run at The Black Diamond in Copenhagen between March 23rd and October 3rd.
Sponsored by Gucci, the exhibition will feature extensive installations which are contrasted against atmospheric soundscapes which have been recorded by Nick Cave and Warren Ellis.
It will also feature more than 300 objects from Cave’s own collection, alongside the collections of the Royal Danish Library.
The exhibition also focuses on more than 50 years in Cave’s life across eight rooms featuring a selection of handwritten lyrics, literature, videos, set designs and personal artefacts.
Cave and The Bad Seeds will also head out on a European tour next year, in support of their acclaimed 2019 album ‘Ghosteen’.


--- layout: post authors: ["Greg Wilson"] title: "Changes to Workshop Administration Fees" date: 2015-07-20 time: "15:00:00" tags: ["Workshops"] ---
After discussion with our Advisory Council and Data Carpentry, we have agreed to make some changes to the administration fee we charge for workshops that we help organize in order to reflect the real cost of our staff's time and our overheads, and to reflect the value of the training. Please see the workshop request page for the updated pricing information.
These rules keep us in step with Data Carpentry, which recently decided to make similar changes in their administrative fees. The only significant differences is that Data Carpentry doesn't yet have a partnership or affiliation scheme, so there are no discounts for DC workshops.
The new administrative fees will apply to workshops requested after September 1, 2015. (Note that the transition is based on when the workshop request is received, not when the workshop takes place.) As per our current policy, if a not-for-profit organization such as a university or government institution has already paid $1250 for a workshop this year, we will only charge $750 per workshop for their subsequent workshops until December 31st. If you have any questions, please get in touch.


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{% import 'headers.html' as headers %} {% import 'footers.html' as footers %} {% import 'algo.html' as algo %}
This article is intended to show properties of multiplicative orders, primitive roots, and Carmichael function. Some of the problems will concentrate on proofs rather than algorithms; first, however, the definitions shall be formulated:
To formulate certain theorems concerning primitive roots, it may be wise to visualize first what are their values \(ord_m(g)\) for fixed \(m\) and what is their distribution.
Input format: value \(m\), for which all orders will be calculated
\(m\)
A number \(m\ (m \le 10^{12})\) is given. Is there any primitive root modulo \(m\)?
Those lemmas will be heavily used in all further paragraphs:
To prove by contradiction \(x^k \equiv 1 \pmod n \implies ord_n(x)|k\): Let \(k\) be miniml such number, such that \(x^k \equiv 1 \pmod n \land ord_n(x) \nmid k\). Either \(k \lt ord_n(x)\) - this cannot be, as \(ord_n(x)\) is minimal such number, that \(x^{ord_n(x)} \equiv 1 \pmod n\); or \(k \gt ord_n(x)\) - then, \(x^k \equiv x^{ord_n(x)} \equiv 1 \pmod n\) \(x^{k-ord_n(x)}x^{ord_n(x)} \equiv x^k \equiv 1 \pmod n\) \(x^{k-ord_n(x)}*1 \equiv 1 \pmod n\) \(x^{k-ord_n(x)} \equiv 1 \pmod n\) Which contradicts premise, that \(k\) is minimal number contradicting thesis, because \(k \equiv k-ord_n(x) \pmod {ord_n(x)}\), and so \(ord_n(x) \nmid k-ord_n(x)\)
Second part of the proof: \(ord_n(x)|k \implies x^k \equiv 1 \pmod n\) is here: \(x^k=x^{d*ord_n(k)}=(x^{ord_n(k)})^d \equiv 1^d \equiv 1 \pmod n\)
\(y^{ord_{dx}(y)} \equiv 1 \pmod {dx}\) \(y^{ord_{dx}(y)}=1+(mx)d\) \(y^{ord_{dx}(y)} \equiv 1 \pmod {d}\) And so, from lemma \(0\), \(ord_d(y)|ord_{dx}(y)\)
{% set post_lcm="lcm(ord_{p_1^{q_1}}(g), ord_{p_2^{q_2}}(g), \dots, ord_{p_n^{q_n}}(g))" %}Later in this lemma, \(m = p_1^{q_1} \dots p_n^{q_n}\)
Proof: Define \(c_1, \dots c_n\) in such way, that \({{post_lcm}}=ord_{p_1^{q_1}}(g)c_1 = ord_{p_2^{q_2}}(g)c_2 = \dots = ord_{p_n^{q_n}}(g)c_n \) $$ \begin{cases} {% for x in [1,2] %} g^{ord_{p_{{x}}^{q_{{x}} }}(g) c_{{x}} } \equiv (g^{ord_{p_{{x}}^{q_{{x}} }}(g) })^{c_{{x}} } \equiv 1^{c_{{x}} } \equiv 1 \pmod {p_{{x}}^{q_{{x}} }} \\ {% endfor %} ... \\ g^{ord_{p_n^{q_n }}(g) c_n } \equiv (g^{ord_{p_n^{q_n }}(g) })^{c_n } \equiv 1^{c_n } \equiv 1 \pmod {p_n^{q_n} } \\ \end{cases} $$
Thus, from chinese remainder theorem, $$g^{ {{post_lcm}} } \equiv 1 \pmod {p_1^{q_1} p_2^{q_2} \dots p_n^{q_n} }$$
By definition, some \(ord_{p_i^{q_i}}(g) \nmid h\). And from lemma 0.5, $$ord_{p_i^{q_i}}(g)| ord_{m} (g)$$ Thus, \(g^h \not \equiv 1 \pmod m\).
For what \(n\) primitive root exist? There are several cases, which will be covered separately:
It may seem worth a note that this lemma cannot work for not prime modulus - for example, \((x-3)(x-4) \pmod {12}\) has 0, 3, 4 and 7 as its zeros - in case of \(7\) despite \(12\) does not divide any bracket directly, it divides their product - which is possible, because it is not prime, and so \(12\) can be shown as product of certain numbers greater than \(1\). Prime can be shown only as one bracket
From lemma 2, there are \(v_i^{q_i}\) solutions to \(x^{v_i^{q_i}} \equiv 1 \pmod p\). For any solution \(s\), \(ord_p(s)=q^j, j \in \lt 0;q_i \gt\) - this follows from lemma 0, as power of a prime number is divisible only by its lower powers. To exclude all numbers, whose order is \(ord_p(s)=q^j, j \in \lt 0;q_i-1 \gt\), one can exclude all \(v_i^{q_i-1}\) solutions to \(x^{v_i^{q_i-1}} \equiv 1 \pmod p\) - and so, there are \(v_i^{q_i}-v_i^{q_i-1}=\phi(v_i^{q_i}) \gt 0\) such numbers, that \(ord_p(s)=v_i^{q_i}\)
Suppose \(ord_p(ab)=m\); then, \(a^{my} \equiv a^{my}(b^{y})^m \equiv (ab)^{my} \equiv ((ab)^m)^y \equiv 1\pmod p\) By lemma 0, \(x|my\). Analogously, \(y|mx\). However, \(gcd(x, y)=1\) - so, \(x|m \land y|m\). Minimal \(m\) divisible by both \(x, y\) is \(lcm(x, y)=xy\) - so \(ord_p(ab) \ge xy\). \((ab)^{xy} \equiv (a^x)^y(b^y)^x \equiv 1 \pmod p\) And so, \(ord_p(ab)=xy\)
Let \(\phi(p)=v_1^{q_1} \dots v_n^{q_n}\). Let \(s_1, \dots s_n\) be such numbers, such that \(ord_p(s_i)=v_i^{q_i}\) - by lemma 3, such numbers \(s_i\) must exist. Then, \(ord_p(s_1 \dots s_n)=v_1^{q_1} \dots v_n^{q_n}=\phi(p)\) by lemma 4, which finishes the proof.
Suppose that \(g^{\phi(p)} \equiv 1 \pmod {p^2}\) (otherwise the problem is already solved). Then, \((g+p)^{\phi(p)}={ {\phi(p)}\choose{0} }g^{\phi(p)}+{ {\phi(p)}\choose{1} }g^{\phi(p)-1}p^1+{ {\phi(p)}\choose{2} }g^{\phi(p)-2}p^2+ \dots \equiv 1+(p-1)g^{\phi(p)}p+0+\dots \pmod {p^2}\) Neither \(p-1\) nor \(g\) have any common divisors with \(p\), so \(p \nmid (p-1)g^{\phi(p)}\) - and so, \((g+p)^{\phi(p)} \neq mp^2+1\).
It's worth noting that \(g+p\) is primitive root modulo \(p\), because \(g+p \equiv g \pmod p\)
It's worth noting that \(3k \lt k+2 \implies k \lt 1\), which cannot be the case by assumption stated on start - and so, all further elements in this sum are equivalent to \(0 \pmod {p^{k+2}}\). Furthermore, this proof won't work for \(p=2\), as in this case: $$ \frac{p(p-1)}{2}b^2 p^{2k} \equiv \frac{2*1}{2} b^2 2^{2k} \equiv b^2 2^{2k} \pmod {2^{k+2}}$$ in the last equation either \(p\) or \(p-1\) is divisible by \(2\) - in case of \(p=2\), it is \(p\), and so, the third element in sum cannot be reduced to \(0 \pmod {p^{k+2}}\) for \(k = 1\).
For \(4\), \(3\) is primitive root. For \(1\), any number is a primitive root straight outta definition.
The numbers having primitive roots are:
This leads to possibility of an algorithm checking, whether a number has primitive root by checking whether it has proper factorization in \(O(fac(m))\) - time to factorize \(m\).
One can ask, and benefit from asking the question about number of primitive roots \(\pmod m\). It is enough to establish two theses to answer this problem
Proof by contradiction: \(g^x \equiv g^y \pmod m \implies g^{y-x} \equiv 1 \pmod m\) - but \(y-x \lt \phi(m)\), which contradicts the assumption that \(ord_m(g)=\phi(m)\)
One can check, for each subsequent number \(x\) in range \(\lt 1;m-1 \gt\), whether it is primitive root - by taking all numbers \(y\) from range\(\lt 1;\phi(m)-1 \gt\) and checking, whether \(x^y \equiv 1 \pmod m\), and if so, then \(ord_m(x) \lt \phi(m)\) - this can be checked with binary exponentation. This solution works in \(O(fac(m)+Ans*m*log(m))\) - \(fac(m)\) comes from need to find \(\phi(m)\).
From lemma 0 it can be seen, that if \(x^y \equiv 1 \pmod m\), then \(ord_m(a)|y\). \(y=\phi(m)\) satisfies this equation, and so, \(ord_m(a)|\phi(m)\). What follows, one needs to check only divisors \(d:d|\phi(m)\) as exponents to check, whether \(\exists_{y \lt \phi(m)} x^y \equiv 1 \pmod m\). Number of divisors of a number \(y\) can be estimated by \(\sqrt{y}\), and so, this solution works in \(O(fac(m)+Ans*\sqrt{m}*log(m))\)
Suppose \(\phi(m)=p_1^{q_1}p_2^{q_2} \dots p_n^{q^n}\) and \(ord_m(x)=p_1^{r_1}p_2^{r_2} \dots p_n^{r_n}\) (from algorithm above, \(ord_m(x)|\phi(m)\)). Either \(\forall_i r_i=q_i\) - then \(x\) is a primitive root - or \(\exists_i r_i \lt q_i\) - then \(x\) is not a primitive root. In the second case, let \(d=p_1^{q_1} \dots p_i^{q_i-1} \dots p_n^{q_n}=\frac{\phi(m)}{p_i}\). Then \(ord_m(x)|d\), because \(\forall_j r_j \le q_j\) and \(r_i \le q_i-1 \lt q_i\). And from lemma 0 \(ord_m(x)|d \implies x^d = x^{\frac{\phi(m)}{p_i}} \equiv 1 \pmod m\). What follows, one can take all primes from factorizing \(\phi(m)\) (whose number may be estimated as \(O(log(\phi(m)))\) or \(O(log(m))\) - and check only for them, whether \(x^{\frac{\phi(m)}{p_i}} \equiv 1 \pmod m\) in order to find out, whether \(x\) is a primitive root. Overall complexity will be \(O(Ans*log(m)*log(m))=O(Ans*log^2(m))\)
For numbers in form \(xp^k\), where \(x=1 \lor x=2\), one can find only primitive root \(\pmod p\), and then, using formulas from proofs find primitive root for \(xp^k\) - this will allow for very specific simplification in the last algorithm.
One can find - after factorization - primitive roots of \(p\) by subsequently choosing random numbers from interval \(\lt 1;p-1 \gt\) and checking them for being primitive roots in \(O(log^2(p))\). This method is independent of distribution of primitive roots over interval \(\lt 1;p-1 \gt\). Probability that random number from given range is primitive root is \(\frac{\phi(\phi(p))}{p-1}=\frac{\phi(p-1)}{p-1}\).
One can show, that \(2\frac{\phi(n)}{n} \ge \frac{1}{\log_2(n)}\), or \(\phi(n)\log_2(n) \ge \frac{n}{2}\): \(\phi(n)\log_2(n)= n\frac{(p_1-1)(p_2-1) \dots (p_m-1)}{p_1p_2 \dots p_m}*log_2(n) \ge n\frac{(p_1-1)(p_2-1) \dots (p_m-1)}{p_1p_2 \dots p_m}*m \ge \frac{1*2 \dots m}{2*3 \dots (m+1)} m = n\frac{m}{m+1} \ge \frac{n}{2}\)
So, probability, that drawn number is a primitive root is not lower than \(\frac{1}{2\log_2(n)}\). After how many moves the probability of drawing primitive root will be higher than \(\frac{1}{2}\)? When probability, that no primitive root was drawn in last \(k\) moves will be lower than \(\frac{1}{2}\). This probability after \(k\) moves is lower than \((\frac{2log_2(n)-1}{2log_2(n)})^k\). To do anything with this equation, it is wise to know the cult-classic definition of Euler number.
As \(\lim_{n -\gt \infty} (\frac{n-1}{n})^n=\frac{1}{e} \lt \frac{1}{2}\), this allows to estimate expected number of operations of given algorithm by \(O(fac(m)+\log^3(m))\), and the algorithm will terminate with probability \(2^{-l}\) in expected number of steps being at most \(O(l*\log^3(m))\) (note that if probability, that in \(kl\) random numbers there will be no primitive roots is \(p^{kl}\) - and if probability, that after \(k\) steps there will be no primitive root found is \(\le \frac{1}{2}\), then the probability of not finding primitive root in \(kl\) steps is \((p^k)^l \le (\frac{1}{2})^l\)).
{{ algo.neo_algorithm(2, 'Primitive root finding', 'It\'s worth noting You can change, whether the algorithm checks numbers:
Of course, number written on the input at start was not coincidentally chosen: \(21\) is the least positive integer being primitive root of 409, so deterministic approach will be almost certainly slower than probabilistic
Input format: One number \(m\), such that primitive root modulo \(m\) has to be found
\(m\)
'
, '334562', {'add_on':'radio', 'added_elements':['Probabilistic', 'Deterministic'],
'color_description':{'standard':'', 'change':'', 'crucial':'Primitive root shown along with its order, also candidate for primitive root.'} }) }}
Note, that modification of the method described above allows to find quickly \(ord_m(g)\) for given \(m, g\): one can start from \(s=\phi(m)=p_1^{q_1} \dots p_n^{q_n}\), find out, for subsequent \(p_i\), whether \(g^{s/p_i} \equiv 1 \pmod m\): if so, then set \(s=p_1^{r_1} \dots p_i^{r_i-1} \dots p_n^{r_n}\), where \(r_i\) were previous exponents for \(s\) (in the beginning, \(r_i=q_i\)) and try to mow down either exponent by \(p_i\) if currently \(r_i \ge 0\), or move to \(r_{i+1}=q_{i+1}\). In the end of this process \(ord_m(g) = s\), and this method works in \(O(\log^2(m))\) - one \(\log\) for binary exponentation, one for passing through exponents.
For given \(m\), what is the maximal value of \(ord_m(g)\) for any \(g\)? Carmichael function denoted as \(\lambda(x)\) is defined as an answer to this problem. More formally, $$ \lambda(m) = \max_{g: gcd(g,m)=1} ord_m(g) $$ There are 3 cases that will be discussed separately:
Note, that above means, that for each \(k \ge 3\) there exists such \(g\), such that: $$ g^{2^{k-3}} \not \equiv 1 \pmod {2^k}$$ As proven in 2.2.5 for \(k \ge 3\) \(ord_{2^k}(g) \le \frac{\phi(m)}{2}\), and also, as \(\lambda(m) | \phi(m)\) and \(\phi(2^k)=2^{k-1}\), then \(\lambda(2^k)\) has form \(2^l\); showing, that \(\lambda(2^k) \gt 2^{k-3}\) will show that \(\lambda(2^k) = 2^{k-2}\) finishing the proof.
Proof will consist of showing, that if there exists \(g\) for which \(g^{2^{k-3}} \equiv 2^{k-1}+1 \pmod {2^{k}}\), then \(g^{2^{k-2}} \equiv 2^k+1 \pmod {2^{k+1}}\)
For \(k=3\), let \(g=5\): then, \(g^{2^{k-3}} \equiv 5^1 \equiv 5 \equiv 2^2+1 \pmod 8\)
Thus $$ \exists_{k} 5^{2^{k-3}} \equiv 2^{k-1}+1 \pmod {2^k} $$ Induction hypothesis consists of: $$ \forall_{k \ge k_0} 5^{2^{k-2}} \equiv 2^k+1 \pmod {2^{k+1}} $$ As for the proof itself, for \(k \gt 3\): $$ 5^{2^{k-2}} \equiv (5^{2^{k-3}})^2 \equiv ((2^{k-1}+1)+c2^k)^2 \equiv $$ $$ \equiv (2^{k-1}+1)^2 + 2(2^{k-1}+1)c2^k + c^22^{2k} \equiv (2^{k-1}+1)^2 + (2^{k-1}+1)c2^{k+1} + c^22^{2k} \equiv $$ $$ \equiv (2^{k-1}+1)^2 + 0 + 0 \equiv 2^{2k-2}+2*2^{k-1}+1 \equiv 2^k+1 \pmod {2^{k+1}}$$ Which shows existence of a number \(g\), for which \(ord_m(g)=\frac{\phi(2^k)}{2}\), and thus, \(\lambda(2^k) = \frac{\phi(2^k)}{2}\) for \(k \ge 3\).
{% set lcm="lcm(\lambda(p_1^{q_1}), \dots, \lambda(p_n^{q_n}))" %}From lemma 0.75: $$ord_m(g) = lcm(ord_{p_1^{q_1}}(g), ord_{p_2^{q_2}}(g), \dots, ord_{p_n^{q_n}}(g))$$ Note, that there exists such \(g\), that for each \(i\): \(ord_{p_i^{q_i}}(g)=\lambda(p_i^{q_i})\) - this comes straight from chinese remainder theorem: namely, one can choose numbers \(c_1, c_2 \dots c_n\), such that \(ord_{p_i^{q_i}}(c_i)=\lambda(p_i^{q_i})\) - and from chinese remainder theorem, there exists exactly one such number \(g\), that $$ \begin{cases} {% for x in [1,2] %} g \equiv c_{{x}} \pmod {p_{{x}}^{q_{{x}} }} \\ {% endfor %} ... \\ g \equiv c_n \pmod {p_n^{q_n }} \\ \end{cases} $$ This implies, that there exists such number \(g\), that \(ord_m(g)={{lcm}}\), and there cannot be any number \(h\) with higher \(ord_m(h)\).
Wisdom regarding orders and primitive roots is useful for understanding theory of NTT and discrete roots, besides they can tell something about existence of discrete logarithm. Examples are:
Prefix Product Sequence,
Levko and Sets,
Tokitsukaze and Powers,


Tuesday - August 11, 2020
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Thank you for your interest in sponsoring the Flourish! 2016 open source computing conference! The conference seeks to bring together some of the brightest developers and thinkers from the open source community to Chicago. We hope that through community-driven presentations, workshops, and other activities, we can all promote a feeling of interconnectedness and inspire others.
The open source philosophy states that software source code should be free, easily shared, and available to anyone. This philosophy of openness has allowed for the creation of software such as the Linux operating system, the Firefox web browser, Wikipedia and the Apache web server. Presently, it has permeated almost every facet of the computing world, as well as attracted a variety of talented developers and progressive businesses into embracing the open source philosophy.
Started in 2006 by a group of computer hobbyists at the University of Illinois at Chicago, the Flourish! Conference was designed to fill a void in the open source community. For the past four years, the conference has attracted students, enthusiasts and the business community alike and has allowed them to connect, share ideas and discuss recent developments. This year, an even larger group of volunteers from all over the community has taken up the mantle to make the conference our best yet.
With your contribution, we can afford to create an amazing conference and, once again, bring together the vibrant Chicago open source community. As a sponsor, you'll be able to introduce yourself to the talent of the community and become a part of it.
We've included a guide of what you can expect with your donation along with this document. If you have questions or concerns, please contact the Flourish Organizers, by email at<EMAIL_ADDRESS>We're glad you asked. First of all, you will receive the gratitude and goodwill of the open source community, UIC students, and that warm fuzzy feeling that you've done something really really good. But, of course, it doesn't stop there as we offer the following perks based on the amount of your generous donation.
We will be happy to work with you in devising any "substitute" promotions if you find it in your company's best interest.
Send an email to<EMAIL_ADDRESS>with a description of which level as well as any pertinent details.


In Fate universe long ago mages of powerful bloodlines created a battle royal known as The Holy Grail War. The Holy Grail is an omnipotent existence that can grant any wish and is the subject of the war. Each of The Holy Grail Wars usually revolves around seven Masters chosen by the Holy Grail and the seven Servants that they summon. Each Servant is a Heroic Spirit based on a real historical or legendary figure, such as King Arthur, Hercules, or William Shakespeare, but are usually referred to using the name of their class.
For the third Holy Grail war we get see it through these main characters
Kiritsugu Emiya
Aurtoria pendragon
Irisviel von Einzbern
Kirei kotomine


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Annoucement
See Annoucements made on the system
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See Annoucements made on the system
Annoucement
See Annoucements made on the system
Annoucement
See Annoucements made on the system
Annoucement
See Annoucements made on the system
Annoucement
See Annoucements made on the system
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See Annoucements made on the system
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See Annoucements made on the system
Annoucement
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AIDE.JS - Video Consultations
iborn.net
Ohrid 2015
OUTLINE
Overview
Advantages
Application
Forecast
Demo
Team
Overview
Seamless high quality video consultations for telemedicine, using only a browser, without the need of a plugin based technology.
WebRTC - Web Real-Time Communications
Advantages
No plugins needed
Peer to peer
Efficiency
Encryption
Privacy
Ease of use
Disadvantages
Scaling group conversation
Still in development
Necessary codecs
Application
Help doctors offer their expertise
Tele-medicine
Online mentoring
Online education
Enables users to seamlessly connect with experts for high quality video advice without technical hurdles by just using their favourite device.
Doctors can reach patients where the physical distances are big and internet availability is still present.
Same concepts can be applied in online education, mentoring and continous learning programmes
Forecast
Year 1
Year 2
Year 3
Year 4
Year 5
# of Customers
2-4
6
16
30
50
# of Employees
Sales
6
10
20
25
30
Expenses
Profits
Demo
http://aide.iborn.net
Our Team
Igor Stamatovski, CEO
Zlate Dodevski, COO
Mihail Dimitrovski, CTO
Marija Alagjozovska, CQAO
Petar Efnushev, Team Lead
Dejan Mladenovski, Software Engineer
Martin Markovski, Software Engineer
Sandra Stanisavljevic, Software Engineer
Monika Jandrevska, Frontend Developer
Mitko Cakalov, Software Engineer
THANK YOU!


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Notice
Summaries of the Further Education Commissioner's findings and recommendations following his review of specific intervention cases.
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https://www.gov.uk/government/publications/review-of-post-16-provision-in-north-east-norfolk-and-north-suffolk
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Summary assessment reports of further education colleges and institutions from the FE Commissioner, giving findings and recommendations following an intervention visit.
These are published alongside copies of a minister’s letter to the college or institution calling for an improvement action plan.
These reports are produced in line with our commitments in ‘intervention in further education – the strengthened intervention process’ .
Published: 19 June 2014
Updated: 11 May 2016


This is a searchable catalogue of all known emoji kitchen combos available on Google's keyboard: Gboard. If you find some that aren't here, please let us know here.💡 You can click on any image to copy it to your clipboard if your device supports it.


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CrimPapers
Articles published on 5 February 2020
On 5 February 2020, 5 new articles were published in 5 journals.
Those Who Counter Match-Fixing Fraudsters: Voices From A Multistakeholder Ecosystem
A Token For Your Thoughts? Perceptions of Tokenism Among Female Corrections Executives by Kimberly Collica-Cox, Dorothy M. Schulz
Policing Domestic Violence: Strategy, Competence, Training by Jarmo Tapio Houtsonen
The Intersection of Intellectual and Developmental Disabilities With Child Exposure to Intimate Partner Violence: Implications For Research and Practice
State Variation in the Drug Felony Lifetime Ban on Temporary Assistance For Needy Families: Why the Modified Ban Matters by Brittany T Martin, Sarah Ks Shannon


tnetwork.DCD.smoothed_graph¶
-
tnetwork.DCD.
smoothed_graph
(dynNetSN, alpha=0.9, match_function=<function jaccard>, threshold=0.3, **kwargs)[source]¶ Smoothed graph approach
This approach is a naive implementation of the idea proposed in [1]. To sum up, at each snapshot, a new graph is create which is the combination of the graph at this step and a graph in which edges are present between any two nodes belonging to the same community in the previous step. Note than in the original paper, a method is proposed to greatly reduce the complexity of the solution, but this method is not implemented here.
Alpha is a parameter to tune how important is the weight of the current topology compared with previous partition.
The label attribution process is the same described in the paper XXX, see method simple_matching for details.
Internally, it calls the simple_matching method, the same parameters can be passed to it.
[1]Guo, C., Wang, J., & Zhang, Z. (2014). Evolutionary community structure discovery in dynamic weighted networks. Physica A: Statistical Mechanics and its Applications, 413, 565-576.
- Parameters
dynNetSN –
alpha – parameter setting relative importance of past VS current graph. 1: only current, 0: only previous
- Returns


Vuku Chief Driko:
I am Driko, the Chief of the Vuku Orcs. Our tribe moved here after we were deprived of our base camp in the Cruma Marshlands by the insects.
Quest


Pediatric LRTI hospitalizations are a significant burden on patients, families, and healthcare systems. This study determined the burden of pediatric LRTIs on hospital settings in British Columbia and the benefits of prevention strategies as they relate to healthcare resource demand.
LRTI inpatient episodes for patients <19 years of age during 2008–2010 were extracted from the BC Discharge Abstract Database. The annual number of acute care beds required to treat pediatric LRTIs was estimated. Sub-analyses determined the burden due to infants <1 year of age and high-risk infants. Population projections were used to forecast LRTI hospitalizations and the effectiveness of public health initiatives to reduce the incidence of LRTIs to 2020 and 2030.
During 2008–2010, LRTI as the primary diagnosis accounted for 32.0 and 75.9% hospitalizations for diseases of the respiratory system in children <19 years of age and infants <1 year of age, respectively. Infants <1 year of age accounted for 47 and 77% hospitalizations due to pediatric LRTIs and pediatric LRTI hospitalizations specifically due to respiratory syncytial virus (RSV), respectively. The average length of stay was 3.1 days for otherwise healthy infants <1 year of age and 9.1 days for high-risk infants (P <0.0001). 73.1% pediatric LRTI hospitalizations occurred between November and April. Over the study timeframe, 19.6 acute care beds were required on average to care for pediatric LRTIs which increased to 64.0 beds at the peak of LRTI hospitalizations. Increases in LRTI bed-days of 5.5 and 16.2% among <19 year olds by 2020 and 2030, respectively, were predicted. Implementation of appropriate prevention strategies could cause 307 and 338 less LRTI hospitalizations in <19 year olds in 2020 and 2030, respectively.
Pediatric LRTI hospitalizations require significant use of acute care infrastructure particularly between November and April. Population projections show the burden may increase in the next 20 years, but implementation of effective public health prevention strategies may contribute to reducing the acute care demand and to supporting efforts for overall pediatric healthcare sustainability.