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Much as group theory is the study of groups, or graph theory is the study of graphs, model theory is the study of models (also known as structures) of some language ${{\mathcal L}}$ (which, in this post, will always be a single-sorted, first-order language). A structure is a set ${X}$, equipped with one or more operations, constants, and relations. This is of course an extremely general type of mathematical object, but (quite remarkably) one can still say a substantial number of interesting things about very broad classes of structures.

We will observe the common abuse of notation of using the set ${X}$ as a metonym for the entire structure, much as we usually refer to a group ${(G,1,\cdot,()^{-1})}$ simply as ${G}$, a vector space ${(V, 0, +, \cdot)}$ simply as ${V}$, and so forth. Following another common bending of the rules, we also allow some operations on structures (such as the multiplicative inverse operation on a group or field) to only be partially defined, and we allow use of the usual simplifying conventions for mathematical formulas (e.g. writing ${a+b+c}$ instead of ${(a+b)+c}$ or ${a+(b+c)}$, in cases where associativity is known). We will also deviate slightly from the usual practice in logic by emphasising individual structures, rather than the theory of general classes of structures; for instance, we will talk about the theory of a single field such as ${{\bf R}}$ or ${{\bf C}}$, rather than the theory of all fields of a certain type (e.g. real closed fields or algebraically closed fields).

Once one has a structure ${X}$, one can introduce the notion of a definable subset of ${X}$, or more generally of a Cartesian power ${X^n}$ of ${X}$, defined as a set ${E \subset X^n}$ of the form

$\displaystyle E = \{ (x_1,\ldots,x_n): P(x_1,\ldots,x_n) \hbox{ true} \} \ \ \ \ \ (1)$

for some formula ${P}$ in the language ${{\mathcal L}}$ with ${n}$ free variables and any number of constants from ${X}$ (that is, ${P(x_1,\ldots,x_n)}$ is a well-formed formula built up from a finite number of constants ${c_1,\ldots,c_m}$ in ${X}$, the relations and operations on ${X}$, logical connectives such as ${\neg}$, ${\wedge}$, ${\implies}$, and the quantifiers ${\forall, \exists}$). Thus, for instance, in the theory of the arithmetic of the natural numbers ${{\bf N} = ({\bf N}, 0, 1, +, \times)}$, the set of primes ${{\mathcal P}}$ is a definable set, since we have

$\displaystyle {\mathcal P} = \{ x \in {\bf N}: (\exists y: x=y+2) \wedge \neg (\exists z,w: x = (z+2)(w+2)) \}.$

In the theory of the field of reals ${{\bf R} = ({\bf R}, 0, 1, +, -, \times, ()^{-1})}$, the unit circle ${S^1}$ is an example of a definable set,

$\displaystyle S^1 = \{ (x,y) \in {\bf R}^2: x^2+y^2 = 1 \},$

but so is the the complement of the circle,

$\displaystyle {\bf R}^2 \backslash S^1 = \{ (x,y) \in {\bf R}^2: \neg(x^2+y^2 = 1) \}$

and the interval ${[-1,1]}$:

$\displaystyle [-1,1] = \{ x \in {\bf R}: \exists y: x^2+y^2 = 1\}.$

Due to the unlimited use of constants, any finite subset of a power ${X^n}$ of any structure ${X}$ is, by our conventions, definable in that structure. (One can of course also consider definability without parameters (also known as ${0}$-definability), in which arbitrary constants are not permitted, but we will not do so here.)

We can isolate some special subclasses of definable sets:

• An atomic definable set is a set of the form (1) in which ${P()}$ is an atomic formula (i.e. it does not contain any logical connectives or quantifiers).
• A quantifier-free definable set is a set of the form (1) in which ${P()}$ is quantifier-free (i.e. it can contain logical connectives, but does not contain the quantifiers ${\forall, \exists}$).

Example 1 In the theory of a field such as ${{\bf R}}$, an atomic definable set is the same thing as an affine algebraic set (also known as an affine algebraic variety, with the understanding that varieties are not necessarily assumed to be irreducible), and a quantifier-free definable set is known as a constructible set; thus we see that algebraic geometry can be viewed in some sense as a special case of model theory. (Conversely, it can in fact be quite profitable to think of model theory as an abstraction of algebraic geometry; for instance, the concepts of Morley rank and Morley degree in model theory (discussed in this previous blog post) directly generalises the concepts of dimension and degree in algebraic geometry.) Over ${{\bf R}}$, the interval ${[-1,1]}$ is a definable set, but not a quantifier-free definable set (and certainly not an atomic definable set); and similarly for the primes over ${{\bf N}}$.

A quantifier-free definable set in ${X^n}$ is nothing more than a finite boolean combination of atomic definable sets; in other words, the class of quantifier-free definable sets over ${X}$ is the smallest class that contains the atomic definable sets and is closed under boolean operations such as complementation and union (which generate all the other boolean operations). Similarly, the class of definable sets over ${X}$ is the smallest class that contains the quantifier-free definable sets, and is also closed under the operation of projection ${\pi_n: E \mapsto \pi_n(E)}$ from ${X^{n+1}}$ to ${X^n}$ for every natural number ${n}$, where ${\pi_n: X^{n+1} \rightarrow X^n}$ is the map ${\pi_n(x_1,\ldots,x_n,x_{n+1}) := (x_1,\ldots,x_n)}$.

Some structures have the property of enjoying quantifier elimination, which means that every definable set is in fact a quantifier-free definable set, or equivalently that the projection of a quantifier-free definable set is again quantifier-free. For instance, an algebraically closed field ${k}$ (with the field operations) has quantifier elimination (i.e. the projection of a constructible set is again constructible); this fact can be proven by the classical tool of resultants, and among other things can be used to give a proof of Hilbert’s nullstellensatz. (Note though that projection does not necessary preserve the property of being atomic; for instance, the projection of the atomic set ${\{ (x,y) \in k^2: xy=1 \}}$ is the non-atomic, but still quantifier-free definable, set ${\{ x \in k: \neg (k=0) \}}$.) In the converse direction, it is not difficult to use the nullstellensatz to deduce quantifier elimination. For theory of the real field ${{\bf R}}$, which is not algebraically closed, one does not have quantifier elimination, as one can see from the example of the unit circle (which is a quantifier-free definable set) projecting down to the interval ${[-1,1]}$ (which is definable, but not quantifer-free definable). However, if one adds the additional operation of order ${<}$ to the reals, giving it the language of an ordered field rather than just a field, then quantifier elimination is recovered (the class of quantifier-free definable sets now enlarges to match the class of definable sets, which in this case is also the class of semi-algebraic sets); this is the famous Tarski-Seidenberg theorem.

On the other hand, many important structures do not have quantifier elimination; typically, the projection of a quantifier-free definable set is not, in general, quantifier-free definable. This failure of the projection property also shows up in many contexts outside of model theory; for instance, Lebesgue famously made the error of thinking that the projection of a Borel measurable set remained Borel measurable (it is merely an analytic set instead). Turing’s halting theorem can be viewed as an assertion that the projection of a decidable set (also known as a computable or recursive set) is not necessarily decidable (it is merely semi-decidable (or recursively enumerable) instead). The notorious P=NP problem can also be essentially viewed in this spirit; roughly speaking (and glossing over the placement of some quantifiers), it asks whether the projection of a polynomial-time decidable set is again polynomial-time decidable. And so forth. (See this blog post of Dick Lipton for further discussion of the subtleties of projections.)

Now we consider the status of quantifier elimination for the theory of a finite field ${F}$. If interpreted naively, quantifier elimination is trivial for a finite field ${F}$, since every subset of ${F^n}$ is finite and thus quantifier-free definable. However, we can recover an interesting question in one of two (essentially equivalent) ways. One is to work in the asymptotic regime in which the field ${F}$ is large, but the length of the formulae used to construct one’s definable sets stays bounded uniformly in the size of ${F}$ (where we view any constant in ${F}$ as contributing a unit amount to the length of a formula, no matter how large ${F}$ is). A simple counting argument then shows that only a small number of subsets of ${F^n}$ become definable in the asymptotic limit ${|F| \rightarrow \infty}$, since the number of definable sets clearly grows at most polynomially in ${|F|}$ for any fixed bound on the formula length, while the number of all subsets of ${|F|^n}$ grows exponentially in ${n}$.

Another way to proceed is to work not with a single finite field ${F}$, or even with a sequence ${F_m}$ of finite fields, but with the ultraproduct ${F = \prod_{m \rightarrow p} F_m}$ of a sequence of finite fields, and to study the properties of definable sets over this ultraproduct. (We will be using the notation of ultraproducts and nonstandard analysis from this previous blog post.) This approach is equivalent to the more finitary approach mentioned in the previous paragraph, at least if one does not care to track of the exact bounds on the length of the formulae involved. Indeed, thanks to Los’s theorem, a definable subset ${E}$ of ${F^n}$ is nothing more than the ultraproduct ${E = \prod_{m \rightarrow p} E_m}$ of definable subsets ${E_m}$ of ${F_m^n}$ for all ${m}$ sufficiently close to ${p}$, with the length of the formulae used to define ${E_m}$ uniformly bounded in ${m}$. In the language of nonstandard analysis, one can view ${F}$ as a nonstandard finite field.

The ultraproduct ${F}$ of finite fields is an important example of a pseudo-finite field – a field that obeys all the sentences in the languages of fields that finite fields do, but is not necessarily itself a finite field. The model theory of pseudo-finite fields was first studied systematically by Ax (in the same paper where the Ax-Grothendieck theorem, discussed previously on this blog, was established), with important further contributions by Kiefe, by Fried-Sacerdote, by two papers of Chatzidakis-van den Dries-Macintyre, and many other authors.

As mentioned before, quantifier elimination trivially holds for finite fields. But for infinite pseudo-finite fields, such as the ultraproduct ${F = \prod_{m \rightarrow p} F_m}$ of finite fields with ${|F_m|}$ going to infinity, quantifier elimination fails. For instance, in a finite field ${F_m}$, the set ${E_m := \{ x \in F_m: \exists y \in F_m: x=y^2 \}}$ of quadratic residues is a definable set, with a bounded formula length, and so in the ultraproduct ${F =\prod_{m \rightarrow p} F_m}$, the set ${E := \prod_{m\rightarrow p} E_m}$ of nonstandard quadratic residues is also a definable set. However, in one dimension, we see from the factor theorem that the only atomic definable sets are either finite or the whole field ${F}$, and so the only constructible sets (i.e. the only quantifier-free definable sets) are either finite or cofinite in ${F}$. Since the quadratic residues have asymptotic density ${1/2}$ in a large finite field, they cannot form a quantifier-free definable set, despite being definable.

Nevertheless, there is a very nice almost quantifier elimination result for these fields, in characteristic zero at least, which we phrase here as follows:

Theorem 1 (Almost quantifier elimination) Let ${F}$ be a nonstandard finite field of characteristic zero, and let ${E \subset F^n}$ be a definable set over ${F}$. Then ${E}$ is the union of finitely many sets of the form

$\displaystyle E = \{ x \in V(F): \exists t \in F: P(x,t) = 0 \} \ \ \ \ \ (2)$

where ${V(F)}$ is an atomic definable subset of ${F^n}$ (i.e. the ${F}$-points of an algebraic variety ${V}$ defined over ${F}$ in ${F^n}$) and ${P: F^{n+1} \rightarrow F}$ is a polynomial.

Results of this type were first obtained essentially due to Catarina Kiefe, although the formulation here is closer to that of Chatzidakis-van den Dries-Macintyre.

Informally, this theorem says that while we cannot quite eliminate all quantifiers from a definable set over a nonstandard finite field, we can eliminate all but one existential quantifier. Note that negation has also been eliminated in this theorem; for instance, the definable set ${F \backslash \{0\} = \{ x \in F: \neg(x=0) \}}$ uses a negation, but can also be described using a single existential quantifier as ${\{ x \in F: \exists t: xt = 1 \}}$.) I believe that there are more complicated analogues of this result in positive characteristic, but I have not studied this case in detail (Kiefe’s result does not assume characteristic zero, but her conclusion is slightly different from the one given here). In the one-dimensional case ${n=1}$, the only varieties ${V}$ are the affine line and finite sets, and we can simplify the above statement, namely that any definable subset of ${F}$ takes the form ${\{ x\in F: \exists t \in F: P(x,t) = 0 \}}$ for some polynomial ${P}$ (i.e. definable sets in ${F}$ are nothing more than the projections of the ${F}$-points of a plane curve).

There is an equivalent formulation of this theorem for standard finite fields, namely that if ${F}$ is a finite field and ${E \subset F^n}$ is definable using a formula of length at most ${M}$, then ${E}$ can be expressed in the form (2) with the degree of ${P}$ bounded by some quantity ${C_{M,n}}$ depending on ${M}$ and ${n}$, assuming that the characteristic of ${F}$ is sufficiently large depending on ${M, n}$.

The theorem gives quite a satisfactory description of definable sets in either standard or nonstandard finite fields (at least if one does not care about effective bounds in some of the constants, and if one is willing to exclude the small characteristic case); for instance, in conjunction with the Lang-Weil bound discussed in this recent blog post, it shows that any non-empty definable subset of a nonstandard finite field has a nonstandard cardinality of ${(\alpha + O(|F|^{-1/2})) |F|^d}$ for some positive standard rational ${\alpha}$ and integer ${d}$. Equivalently, any non-empty definable subset of ${F^n}$ for some standard finite field ${F}$ using a formula of length at most ${M}$ has a standard cardinality of ${(\alpha + O_{M,n}(|F|^{-1/2})) |F|^d}$ for some positive rational of height ${O_{M,n}(1)}$ and some natural number ${d}$ between ${0}$ and ${n}$. (For instance, in the example of the quadratic residues given above, ${d}$ is equal to ${1}$ and ${\alpha}$ equal to ${1/2}$.) There is a more precise statement to this effect, namely that the Poincaré series of a definable set is rational; see Kiefe’s paper for details.

Below the fold I give a proof of Theorem 1, which relies primarily on the Lang-Weil bound mentioned above.

Let ${F}$ be a finite field, with algebraic closure ${\overline{F}}$, and let ${V}$ be an (affine) algebraic variety defined over ${\overline{F}}$, by which I mean a set of the form

$\displaystyle V = \{ x \in \overline{F}^d: P_1(x) = \ldots = P_m(x) = 0 \}$

for some ambient dimension ${d \geq 0}$, and some finite number of polynomials ${P_1,\ldots,P_m: \overline{F}^d \rightarrow \overline{F}}$. In order to reduce the number of subscripts later on, let us say that ${V}$ has complexity at most ${M}$ if ${d}$, ${m}$, and the degrees of the ${P_1,\ldots,P_m}$ are all less than or equal to ${M}$. Note that we do not require at this stage that ${V}$ be irreducible (i.e. not the union of two strictly smaller varieties), or defined over ${F}$, though we will often specialise to these cases later in this post. (Also, everything said here can also be applied with almost no changes to projective varieties, but we will stick with affine varieties for sake of concreteness.)

One can consider two crude measures of how “big” the variety ${V}$ is. The first measure, which is algebraic geometric in nature, is the dimension ${\hbox{dim}(V)}$ of the variety ${V}$, which is an integer between ${0}$ and ${d}$ (or, depending on convention, ${-\infty}$, ${-1}$, or undefined, if ${V}$ is empty) that can be defined in a large number of ways (e.g. it is the largest ${r}$ for which the generic linear projection from ${V}$ to ${\overline{F}^r}$ is dominant, or the smallest ${r}$ for which the intersection with a generic codimension ${r}$ subspace is non-empty). The second measure, which is number-theoretic in nature, is the number ${|V(F)| = |V \cap F^d|}$ of ${F}$-points of ${V}$, i.e. points ${x = (x_1,\ldots,x_d)}$ in ${V}$ all of whose coefficients lie in the finite field, or equivalently the number of solutions to the system of equations ${P_i(x_1,\ldots,x_d) = 0}$ for ${i=1,\ldots,m}$ with variables ${x_1,\ldots,x_d}$ in ${F}$.

These two measures are linked together in a number of ways. For instance, we have the basic Schwarz-Zippel type bound (which, in this qualitative form, goes back at least to Lemma 1 of the work of Lang and Weil in 1954).

Lemma 1 (Schwarz-Zippel type bound) Let ${V}$ be a variety of complexity at most ${M}$. Then we have ${|V(F)| \ll_M |F|^{\hbox{dim}(V)}}$.

Proof: (Sketch) For the purposes of exposition, we will not carefully track the dependencies of implied constants on the complexity ${M}$, instead simply assuming that all of these quantities remain controlled throughout the argument. (If one wished, one could obtain ineffective bounds on these quantities by an ultralimit argument, as discussed in this previous post, or equivalently by moving everything over to a nonstandard analysis framework; one could also obtain such uniformity using the machinery of schemes.)

We argue by induction on the ambient dimension ${d}$ of the variety ${V}$. The ${d=0}$ case is trivial, so suppose ${d \geq 1}$ and that the claim has already been proven for ${d-1}$. By breaking up ${V}$ into irreducible components we may assume that ${V}$ is irreducible (this requires some control on the number and complexity of these components, but this is available, as discussed in this previous post). For each ${x_1,\ldots,x_{d-1} \in \overline{F}}$, the fibre ${\{ x_d \in \overline{F}: (x_1,\ldots,x_{d-1},x_d) \in V \}}$ is either one-dimensional (and thus all of ${\overline{F}}$) or zero-dimensional. In the latter case, one has ${O_M(1)}$ points in the fibre from the fundamental theorem of algebra (indeed one has a bound of ${D}$ in this case), and ${(x_1,\ldots,x_{d-1})}$ lives in the projection of ${V}$ to ${\overline{F}^{d-1}}$, which is a variety of dimension at most ${\hbox{dim}(V)}$ and controlled complexity, so the contribution of this case is acceptable from the induction hypothesis. In the former case, the fibre contributes ${|F|}$ ${F}$-points, but ${(x_1,\ldots,x_{d-1})}$ lies in a variety in ${\overline{F}^{d-1}}$ of dimension at most ${\hbox{dim}(V)-1}$ (since otherwise ${V}$ would contain a subvariety of dimension at least ${\hbox{dim}(V)+1}$, which is absurd) and controlled complexity, and so the contribution of this case is also acceptable from the induction hypothesis. $\Box$

One can improve the bound on the implied constant to be linear in the degree of ${V}$ (see e.g. Claim 7.2 of this paper of Dvir, Kollar, and Lovett, or Lemma A.3 of this paper of Ellenberg, Oberlin, and myself), but we will not be concerned with these improvements here.

Without further hypotheses on ${V}$, the above upper bound is sharp (except for improvements in the implied constants). For instance, the variety

$\displaystyle V := \{ (x_1,\ldots,x_d) \in \overline{F}^d: \prod_{j=1}^D (x_d - a_j) = 0\},$

where ${a_1,\ldots,a_D \in F}$ are distict, is the union of ${D}$ distinct hyperplanes of dimension ${d-1}$, with ${|V(F)| = D |F|^{d-1}}$ and complexity ${\max(D,d)}$; similar examples can easily be concocted for other choices of ${\hbox{dim}(V)}$. In the other direction, there is also no non-trivial lower bound for ${|V(F)|}$ without further hypotheses on ${V}$. For a trivial example, if ${a}$ is an element of ${\overline{F}}$ that does not lie in ${F}$, then the hyperplane

$\displaystyle V := \{ (x_1,\ldots,x_d) \in \overline{F}^d: x_d - a = 0 \}$

clearly has no ${F}$-points whatsoever, despite being a ${d-1}$-dimensional variety in ${\overline{F}^d}$ of complexity ${d}$. For a slightly less non-trivial example, if ${a}$ is an element of ${F}$ that is not a quadratic residue, then the variety

$\displaystyle V := \{ (x_1,\ldots,x_d) \in \overline{F}^d: x_d^2 - a = 0 \},$

which is the union of two hyperplanes, still has no ${F}$-points, even though this time the variety is defined over ${F}$ instead of ${\overline{F}}$ (by which we mean that the defining polynomial(s) have all of their coefficients in ${F}$). There is however the important Lang-Weil bound that allows for a much better estimate as long as ${V}$ is both defined over ${F}$ and irreducible:

Theorem 2 (Lang-Weil bound) Let ${V}$ be a variety of complexity at most ${M}$. Assume that ${V}$ is defined over ${F}$, and that ${V}$ is irreducible as a variety over ${\overline{F}}$ (i.e. ${V}$ is geometrically irreducible or absolutely irreducible). Then

$\displaystyle |V(F)| = (1 + O_M(|F|^{-1/2})) |F|^{\hbox{dim}(V)}.$

Again, more explicit bounds on the implied constant here are known, but will not be the focus of this post. As the previous examples show, the hypotheses of definability over ${F}$ and geometric irreducibility are both necessary.

The Lang-Weil bound is already non-trivial in the model case ${d=2, \hbox{dim}(V)=1}$ of plane curves:

Theorem 3 (Hasse-Weil bound) Let ${P: \overline{F}^2 \rightarrow \overline{F}}$ be an irreducible polynomial of degree ${D}$ with coefficients in ${F}$. Then

$\displaystyle |\{ (x,y) \in F^2: P(x,y) = 0 \}| = |F| + O_D( |F|^{1/2} ).$

Thus, for instance, if ${a,b \in F}$, then the elliptic curve ${\{ (x,y) \in F^2: y^2 = x^3 + ax + b \}}$ has ${|F| + O(|F|^{1/2})}$ ${F}$-points, a result first established by Hasse. The Hasse-Weil bound is already quite non-trivial, being the analogue of the Riemann hypothesis for plane curves. For hyper-elliptic curves, an elementary proof (due to Stepanov) is discussed in this previous post. For general plane curves, the first proof was by Weil (leading to his famous Weil conjectures); there is also a nice version of Stepanov’s argument due to Bombieri covering this case which is a little less elementary (relying crucially on the Riemann-Roch theorem for the upper bound, and a lifting trick to then get the lower bound), which I briefly summarise later in this post. The full Lang-Weil bound is deduced from the Hasse-Weil bound by an induction argument using generic hyperplane slicing, as I will also summarise later in this post.

The hypotheses of definability over ${F}$ and geometric irreducibility in the Lang-Weil can be removed after inserting a geometric factor:

Corollary 4 (Lang-Weil bound, alternate form) Let ${V}$ be a variety of complexity at most ${M}$. Then one has

$\displaystyle |V(F)| = (c(V) + O_M(|F|^{-1/2})) |F|^{\hbox{dim}(V)}$

where ${c(V)}$ is the number of top-dimensional components of ${V}$ (i.e. geometrically irreducible components of ${V}$ of dimension ${\hbox{dim}(V)}$) that are definable over ${F}$, or equivalently are invariant with respect to the Frobenius endomorphism ${x \mapsto x^{|F|}}$ that defines ${F}$.

Proof: By breaking up a general variety ${V}$ into components (and using Lemma 1 to dispose of any lower-dimensional components), it suffices to establish this claim when ${V}$ is itself geometrically irreducible. If ${V}$ is definable over ${F}$, the claim follows from Theorem 2. If ${V}$ is not definable over ${F}$, then it is not fixed by the Frobenius endomorphism ${Frob}$ (since otherwise one could produce a set of defining polynomials that were fixed by Frobenius and thus defined over ${F}$ by using some canonical basis (such as a reduced Grobner basis) for the associated ideal), and so ${V \cap Frob(V)}$ has strictly smaller dimension than ${V}$. But ${V \cap Frob(V)}$ captures all the ${F}$-points of ${V}$, so in this case the claim follows from Lemma 1. $\Box$

Note that if ${V}$ is reducible but is itself defined over ${F}$, then the Frobenius endomorphism preserves ${V}$ itself, but may permute the components of ${V}$ around. In this case, ${c(V)}$ is the number of fixed points of this permutation action of Frobenius on the components. In particular, ${c(V)}$ is always a natural number between ${0}$ and ${O_M(1)}$; thus we see that regardless of the geometry of ${V}$, the normalised count ${|V(F)|/|F|^{\hbox{dim}(V)}}$ is asymptotically restricted to a bounded range of natural numbers (in the regime where the complexity stays bounded and ${|F|}$ goes to infinity).

Example 1 Consider the variety

$\displaystyle V := \{ (x,y) \in \overline{F}^2: x^2 - ay^2 = 0 \}$

for some non-zero parameter ${a \in F}$. Geometrically (by which we basically mean “when viewed over the algebraically closed field ${\overline{F}}$“), this is the union of two lines, with slopes corresponding to the two square roots of ${a}$. If ${a}$ is a quadratic residue, then both of these lines are defined over ${F}$, and are fixed by Frobenius, and ${c(V) = 2}$ in this case. If ${a}$ is not a quadratic residue, then the lines are not defined over ${F}$, and the Frobenius automorphism permutes the two lines while preserving ${V}$ as a whole, giving ${c(V)=0}$ in this case.

Corollary 4 effectively computes (at least to leading order) the number-theoretic size ${|V(F)|}$ of a variety in terms of geometric information about ${V}$, namely its dimension ${\hbox{dim}(V)}$ and the number ${c(V)}$ of top-dimensional components fixed by Frobenius. It turns out that with a little bit more effort, one can extend this connection to cover not just a single variety ${V}$, but a family of varieties indexed by points in some base space ${W}$. More precisely, suppose we now have two affine varieties ${V,W}$ of bounded complexity, together with a regular map ${\phi: V \rightarrow W}$ of bounded complexity (the definition of complexity of a regular map is a bit technical, see e.g. this paper, but one can think for instance of a polynomial or rational map of bounded degree as a good example). It will be convenient to assume that the base space ${W}$ is irreducible. If the map ${\phi}$ is a dominant map (i.e. the image ${\phi(V)}$ is Zariski dense in ${W}$), then standard algebraic geometry results tell us that the fibres ${\phi^{-1}(\{w\})}$ are an unramified family of ${\hbox{dim}(V)-\hbox{dim}(W)}$-dimensional varieties outside of an exceptional subset ${W'}$ of ${W}$ of dimension strictly smaller than ${\hbox{dim}(W)}$ (and with ${\phi^{-1}(W')}$ having dimension strictly smaller than ${\hbox{dim}(V)}$); see e.g. Section I.6.3 of Shafarevich.

Now suppose that ${V}$, ${W}$, and ${\phi}$ are defined over ${F}$. Then, by Lang-Weil, ${W(F)}$ has ${(1 + O(|F|^{-1/2})) |F|^{\hbox{dim}(W)}}$ ${F}$-points, and by Schwarz-Zippel, for all but ${O( |F|^{\hbox{dim}(W)-1})}$ of these ${F}$-points ${w}$ (the ones that lie in the subvariety ${W'}$), the fibre ${\phi^{-1}(\{w\})}$ is an algebraic variety defined over ${F}$ of dimension ${\hbox{dim}(V)-\hbox{dim}(W)}$. By using ultraproduct arguments (see e.g. Lemma 3.7 of this paper of mine with Emmanuel Breuillard and Ben Green), this variety can be shown to have bounded complexity, and thus by Corollary 4, has ${(c(\phi^{-1}(\{w\})) + O(|F|^{-1/2}) |F|^{\hbox{dim}(V)-\hbox{dim}(W)}}$ ${F}$-points. One can then ask how the quantity ${c(\phi^{-1}(\{w\})}$ is distributed. A simple but illustrative example occurs when ${V=W=F}$ and ${\phi: F \rightarrow F}$ is the polynomial ${\phi(x) := x^2}$. Then ${c(\phi^{-1}(\{w\})}$ equals ${2}$ when ${w}$ is a non-zero quadratic residue and ${0}$ when ${w}$ is a non-zero quadratic non-residue (and ${1}$ when ${w}$ is zero, but this is a negligible fraction of all ${w}$). In particular, in the asymptotic limit ${|F| \rightarrow \infty}$, ${c(\phi^{-1}(\{w\})}$ is equal to ${2}$ half of the time and ${0}$ half of the time.

Now we describe the asymptotic distribution of the ${c(\phi^{-1}(\{w\}))}$. We need some additional notation. Let ${w_0}$ be an ${F}$-point in ${W \backslash W'}$, and let ${\pi_0( \phi^{-1}(\{w_0\}) )}$ be the connected components of the fibre ${\phi^{-1}(\{w_0\})}$. As ${\phi^{-1}(\{w_0\})}$ is defined over ${F}$, this set of components is permuted by the Frobenius endomorphism ${Frob}$. But there is also an action by monodromy of the fundamental group ${\pi_1(W \backslash W')}$ (this requires a certain amount of étale machinery to properly set up, as we are working over a positive characteristic field rather than over the complex numbers, but I am going to ignore this rather important detail here, as I still don’t fully understand it). This fundamental group may be infinite, but (by the étale construction) is always profinite, and in particular has a Haar probability measure, in which every finite index subgroup (and their cosets) are measurable. Thus we may meaningfully talk about elements drawn uniformly at random from this group, so long as we work only with the profinite ${\sigma}$-algebra on ${\pi_1(W \backslash W')}$ that is generated by the cosets of the finite index subgroups of this group (which will be the only relevant sets we need to measure when considering the action of this group on finite sets, such as the components of a generic fibre).

Theorem 5 (Lang-Weil with parameters) Let ${V, W}$ be varieties of complexity at most ${M}$ with ${W}$ irreducible, and let ${\phi: V \rightarrow W}$ be a dominant map of complexity at most ${M}$. Let ${w_0}$ be an ${F}$-point of ${W \backslash W'}$. Then, for any natural number ${a}$, one has ${c(\phi^{-1}(\{w\})) = a}$ for ${(\mathop{\bf P}( X = a ) + O_M(|F|^{-1/2})) |F|^{\hbox{dim}(W)}}$ values of ${w \in W(F)}$, where ${X}$ is the random variable that counts the number of components of a generic fibre ${\phi^{-1}(w_0)}$ that are invariant under ${g \circ Frob}$, where ${g}$ is an element chosen uniformly at random from the étale fundamental group ${\pi_1(W \backslash W')}$. In particular, in the asymptotic limit ${|F| \rightarrow \infty}$, and with ${w}$ chosen uniformly at random from ${W(F)}$, ${c(\phi^{-1}(\{w\}))}$ (or, equivalently, ${|\phi^{-1}(\{w\})(F)| / |F|^{\hbox{dim}(V)-\hbox{dim}(W)}}$) and ${X}$ have the same asymptotic distribution.

This theorem generalises Corollary 4 (which is the case when ${W}$ is just a point, so that ${\phi^{-1}(\{w\})}$ is just ${V}$ and ${g}$ is trivial). Informally, the effect of a non-trivial parameter space ${W}$ on the Lang-Weil bound is to push around the Frobenius map by monodromy for the purposes of counting invariant components, and a randomly chosen set of parameters corresponds to a randomly chosen loop on which to perform monodromy.

Example 2 Let ${V=W=F}$ and ${\phi(x) = x^m}$ for some fixed ${m \geq 1}$; to avoid some technical issues let us suppose that ${m}$ is coprime to ${|F|}$. Then ${W'}$ can be taken to be ${\{0\}}$, and for a base point ${w_0 \in W \backslash W'}$ we can take ${w_0=1}$. The fibre ${\phi^{-1}(\{1\})}$ – the ${m^{th}}$ roots of unity – can be identified with the cyclic group ${{\bf Z}/m{\bf Z}}$ by using a primitive root of unity. The étale fundamental group ${\pi(W \backslash W') = \pi(\overline{F} \backslash 0)}$ is (I think) isomorphic to the profinite closure ${\hat {\bf Z}}$ of the integers ${{\bf Z}}$ (excluding the part of that closure coming from the characteristic of ${F}$). Not coincidentally, the integers ${{\bf Z}}$ are the fundamental group of the complex analogue ${{\bf C} \backslash \{0\}}$ of ${W \backslash W'}$. (Brian Conrad points out to me though that for more complicated varieties, such as covers of ${\overline{F} \backslash \{0\}}$ by a power of the characteristic, the etale fundamental group is more complicated than just a profinite closure of the ordinary fundamental group, due to the presence of Artin-Schreier covers that are only ramified at infinity.) The action of this fundamental group on the fibres ${{\bf Z}/m{\bf Z}}$ can given by translation. Meanwhile, the Frobenius map ${Frob}$ on ${{\bf Z}/m{\bf Z}}$ is given by multiplication by ${|F|}$. A random element ${g \circ Frob}$ then becomes a random affine map ${x \mapsto |F|x+b}$ on ${{\bf Z}/m{\bf Z}}$, where ${b}$ chosen uniformly at random from ${{\bf Z}/m{\bf Z}}$. The number of fixed points of this map is equal to the greatest common divisor ${(|F|-1,m)}$ of ${|F|-1}$ and ${m}$ when ${b}$ is divisible by ${(|F|-1,m)}$, and equal to ${0}$ otherwise. This matches up with the elementary number fact that a randomly chosen non-zero element of ${F}$ will be an ${m^{th}}$ power with probability ${1/(|F|-1,m)}$, and when this occurs, the number of ${m^{th}}$ roots in ${F}$ will be ${(|F|-1,m)}$.

Example 3 (Thanks to Jordan Ellenberg for this example.) Consider a random elliptic curve ${E = \{ y^2 = x^3 + ax + b \}}$, where ${a,b}$ are chosen uniformly at random, and let ${m \geq 1}$. Let ${E[m]}$ be the ${m}$-torsion points of ${E}$ (i.e. those elements ${g \in E}$ with ${mg = 0}$ using the elliptic curve addition law); as a group, this is isomorphic to ${{\bf Z}/m{\bf Z} \times {\bf Z}/m{\bf Z}}$ (assuming that ${F}$ has sufficiently large characteristic, for simplicity), and consider the number of ${F}$ points of ${E[m]}$, which is a random variable taking values in the natural numbers between ${0}$ and ${m^2}$. In this case, the base variety ${W}$ is the modular curve ${X(1)}$, and the covering variety ${V}$ is the modular curve ${X_1(m)}$. The generic fibre here can be identified with ${{\bf Z}/m{\bf Z} \times {\bf Z}/m{\bf Z}}$, the monodromy action projects down to the action of ${SL_2({\bf Z}/m{\bf Z})}$, and the action of Frobenius on this fibre can be shown to be given by a ${2 \times 2}$ matrix with determinant ${|F|}$ (with the exact choice of matrix depending on the choice of fibre and of the identification), so the distribution of the number of ${F}$-points of ${E[m]}$ is asymptotic to the distribution of the number of fixed points ${X}$ of a random linear map of determinant ${|F|}$ on ${{\bf Z}/m{\bf Z} \times {\bf Z}/m{\bf Z}}$.

Theorem 5 seems to be well known “folklore” among arithmetic geometers, though I do not know of an explicit reference for it. I enjoyed deriving it for myself (though my derivation is somewhat incomplete due to my lack of understanding of étale cohomology) from the ordinary Lang-Weil theorem and the moment method. I’m recording this derivation later in this post, mostly for my own benefit (as I am still in the process of learning this material), though perhaps some other readers may also be interested in it.

Caveat: not all details are fully fleshed out in this writeup, particularly those involving the finer points of algebraic geometry and étale cohomology, as my understanding of these topics is not as complete as I would like it to be.

Many thanks to Brian Conrad and Jordan Ellenberg for helpful discussions on these topics.

 Anonymous on Polymath15, ninth thread: goin… 译 & 评 （陶哲轩：要有耐心）… on Be flexible 译 & 评 （陶哲轩：要有耐心）… on Be patient Fausto di Biase on Ultraproducts as a Bridge Betw… Anonymous on Ask yourself dumb questions –… Five methods to lear… on Learn and relearn your fi… rudolph01 on Polymath15, ninth thread: goin… rudolph01 on Polymath proposal: upper bound… Anonymous on 254A, Notes 3a: Eigenvalues an… majdoddin on 1% quasimorphisms and group… OneGuy on 1% quasimorphisms and group… majdoddin on 1% quasimorphisms and group… adityaguharoy on IMO 2009 Q6 mini-polymath proj… KM on Polymath15, ninth thread: goin… Anonymous on 1% quasimorphisms and group…


text:
Where parallels cross

Interesting bits of life

Emacs as your code-compass: how complex is this code?

Find out how complex is your (indented) code by using code-compass' c/calculate-complexity-current-buffer. The best way to interpret that score is to integrate the time dimension: use c/show-complexity-over-commits to see how complex your code became over the last commits.

The problem

Often I am so absorbed by the problem at hand that when my code gets to the review stage, I find comments like: "Mmm, can you make it more readable?". I know I should not rush to the review, but it happens that I have some other work in progress that distracts me. That makes me forget to "empathically" review my code before clicking the "open pull request" button.

(Meanwhile I master sequential problem solving) If only I had a way to quantify code complexity, maybe I could make my preferred editor to stop me at the right time?

It is a problem indeed

Code that is hard to read is a serious issue. Give it enough time and I cannot tell the meaning of my own code! Indeed, time erases from my memory why I structured code in a certain way and why I chose those random names for variables and functions (and luckily so, because otherwise I would not have mental space for my future!).

This is troublesome for my code, and extremely so for organizations because of the quantity of engineers involved (with different backgrounds) and their turnover. This feed the cycle of architectural decay and technical debt.

We already saw we can find this out with the hotspot analysis, and we shall do better: we need a way to quantify how worse or how better we are making code for future maintainers!

And there is a solution

There are a bunch of ways to calculate a complexity score, one of the many being cyclomatic complexity. The simplest, roughest and also language agnostic is code indentation.

Given your code follows a uniform indentation scheme (Emacs has a tons of amazing packages for that), you can easily see if a piece of code is difficult to read. For example, check out the following pieces of code to get the idea (they are meaningless, so ignore the semantics):

(defun someCommonLispFunction (someInput)
(if (> someImput 0)
(if (> someImput 1)
(if (> someImput 3)
"bigger than 3"
(if (= someImput 3)
"equal to 3"
"..."))
"...")))

(let ((someInput 1))
(let ((someOtherInput 2))
(+
(if someInput
1
2)
(when someOtherInput
3))))

(setf a 1)
(setf b 1)
(+ a b)

I made these up randomly with the only aim to show you how difficult to read is code with too much (unnecessary) nesting.

If you want to see other examples in your favourite programming language, check out https://github.com/emilybache/GildedRose-Refactoring-Kata, which is also a useful refactoring kata.

The point is that your computer can calculate a complexity score easily: just sum up the units of indentation appearing in the code.

The unit of indentation for the Common Lisp code above is two spaces. So the calculation of complexity for each block results in:

block complexity
block with if statements 46
block with let statements 20.5
block with setf 0.0

In my experience I agree with that score when I am reviewing, and researchers found indentation to be a good rough metric to indicate complexity.

Clearly I use this metric as an indication. So rather than looking at the raw number, I look for deltas: I want to see how much my change increases and decreases this complexity value.

This is why code-compass implements c/show-complexity-over-commits, which lets you see how complexity of a buffer has increased or decreased over the last few commits. The graph you obtain looks like this:

Note that I put complexity in relation to lines of code so that I notice if a spike in complexity is due to a lot of new code added to the file (and this is typical indeed).

Now equipped with this graph, I can find a bit more easily when my changes are a bit too complex to open a pull request :)

Conclusion

So grab the last version of code-compass and find out how complex is your code by running c/calculate-complexity-current-buffer on a buffer! Or checking the differential in complexity a file experienced with c/show-complexity-over-commits.

This way you can evaluate the time future maintainers will have to invest in understanding your code, and save that for them.

Happy simplifying!

format: plain


text:
# Laboratoire de Mécanique des Fluides et d’Acoustique - UMR 5509

LMFA - UMR 5509
Laboratoire de Mécanique des Fluides et d’Acoustique
Lyon
France

## Nos partenaires

Article dans J. Fluid Mech. (2018)

## Scaling laws for mixing and dissipation in unforced rotating stratified turbulence

Annick Pouquet, Duane Rosenberg, Raffaele Marino and Corentin Herbert

We present a model for the scaling of mixing in weakly rotating stratified flows characterized by their Rossby, Froude and Reynolds numbers , $Ro$, $Fr$, $Re$. This model is based on quasi-equipartition between kinetic and potential modes, sub-dominant vertical velocity, $w$, and lessening of the energy transfer to small scales as measured by a dissipation efficiency $\beta=\epsilon_\nu/\epsilon_D$, with $\epsilon_\nu$ the kinetic energy dissipation and $\epsilon_D=u_{rms}^3/L_{int}$ its dimensional expression, with $w$, $u_{rms}$ the vertical and root mean square velocities, and $L_{int}$ the integral scale. We determine the domains of validity of such laws for a large numerical study of the unforced Boussinesq equations mostly on grids of $1024^3$ points, with $Ro/Fr\ge2.5$ , and with $1600\le Re\approx5.4 \times10^4$ ; the Prandtl number is one, initial conditions are either isotropic and at large scale for the velocity and zero for the temperature $\theta$, or in geostrophic balance. Three regimes in Froude number, as for stratified flows, are observed : dominant waves, eddy–wave interactions and strong turbulence. A wave–turbulence balance for the transfer time $\tau_{tr}=N\tau_{NL}^2$, with $\tau_{NL}=u_{rms}/L_{int}$ the turnover time and $N$ the Brunt–Väisälä frequency, leads $\beta$ to growing linearly with $Fr$ in the intermediate regime, with a saturation at $\beta\approx0.3$ or more, depending on initial conditions for larger Froude numbers. The Ellison scale is also found to scale linearly with $Fr$. The flux Richardson number $R_f=B_f/[B_f+\epsilon_\nu]$, with $B_f=N\langle w\theta\rangle$ the buoyancy flux, transitions for approximately the same parameter values as for $\beta$. These regimes for the present study are delimited by ${\cal R}_B=ReFr^2\approx2$ and ${\cal R}_B\approx200$. With $\Gamma_f=R_f/[1-R_f]$ the mixing efficiency, putting together the three relationships of the model allows for the prediction of the scaling $\Gamma_f\sim Fr^{-2}\sim {\cal R}_B^{-1}$ in the low and intermediate regimes for high , whereas for higher Froude numbers, $\Gamma_f\sim {\cal R}_B^{-1/2}$, a scaling already found in observations : as turbulence strengthens, $\beta\sim1$, $w\approx u_{rms}$, and smaller buoyancy fluxes together correspond to a decoupling of velocity and temperature fluctuations, the latter becoming passive.


text:
# \stackrel but lower

I am trying to place a frown above a times so it behaves like a hat. I have tried:

\stackrel{\frown}{\times}
{\frown \above 0pt \times}
\overset{\frown}{\times}
\mathrlap{\frown}\times
{\frown\atop\times}
\ooalign{$\frown$\cr$\times$}

but the distance of the frown to the times is either too high or too low. I think \ooalign provides a solution but am unable to find it.

Edit:

\ooalign{\hidewidth\raise1ex\hbox{$\frown$}\cr$\times$}

gives a pretty good result but the symbols are not aligned.

Not \ooalign, which superimposes symbols: you need to stack them, so \ialign is better.

\documentclass{article}
\usepackage{amsmath,mathtools}

\makeatletter
\DeclareRobustCommand{\frtimes}{\mathbin{\mathpalette\frtimes@@\relax}}
\newcommand{\frtimes@@}[2]{%
\vbox{\offinterlineskip
\sbox\z@{$\m@th#1\times$}%
\ialign{%
\hfil##\hfil\cr
$\m@th#1{}_{\frown}\kern-\scriptspace$\cr
\noalign{\kern-.3\ht\z@}
\box\z@\cr
}%
}%
}
\makeatother

\begin{document}

$A\frtimes B_{A\frtimes B}$

\end{document}

Here are three other solutions: one with stackengine, and the other ones replacing \frown with the\wideparen math accent, from mathabx and yhmath respectively.

\documentclass[border=5pt]{standalone}
\usepackage[usestackEOL]{stackengine} %
\usepackage{accents}
\newcommand\frimes{\mathbin{\ensurestackMath{\stackon[-1pt]{\times}{\frown}}}}

\DeclareFontFamily{U}{mathx}{\hyphenchar\font45}
\DeclareFontShape{U}{mathx}{m}{n}{
<5><6><7><8><9><10>
<10.95><12><14.4><17.28><20.74><24.88>
mathx10
}{}
\DeclareSymbolFont{mathx}{U}{mathx}{m}{n}
\DeclareFontSubstitution{U}{mathx}{m}{n}
\DeclareMathAccent{\varwideparen}{0}{mathx}{"8C}

\DeclareSymbolFont{largesymbols}{OMX}{yhex}{m}{n}
\DeclareMathAccent{\wideparen}{\mathord}{largesymbols}{"F3}

\begin{document}

$a \frimes b \quad a \mathbin{\varwideparen{\times}} b \quad a \mathbin{\wideparen{\times}} b$

\end{document}

I think an easy way is to create an array, as suggested by Werner

\documentclass[border=5pt]{standalone}
\begin{document}
$a \begin{array}[b]{@{}c@{}} \frown \\ [-7pt] % adjust as needed {} \times {} \\ \end{array} b$
\end{document}

yielding


text:
593 articles – 2563 references  [version française]
 HAL: in2p3-00719510, version 1
 arXiv: 1207.4321
 ADS Bibliographic id: 2013MNRAS.428.1498C
 DOI: 10.1093/mnras/sts127
 Monthly Notices of the Royal Astronomical Society 428 (2013) 1498-1517
 Investigating Emission Line Galaxy Surveys with the Sloan Digital Sky Survey Telescope
 For the BOSS collaboration(s)
 (2013-01)
 The Baryon Acoustic Oscillation (BAO) feature in the power spectrum of galaxies can be used as a standard ruler to probe the accelerated expansion of the Universe. The current surveys covering a comoving volume sufficient to unveil the BAO scale are limited to redshift $z \lesssim 0.7$. In this paper, we study several galaxy selection schemes aiming at building an emission-line-galaxy (ELG) sample in the redshift range \$0.6
 Research team: APC - Cosmologie et Gravitation
 Subject(s) : Sciences of the Universe/Astrophysics/Cosmology and Extra-Galactic AstrophysicsPhysics/Astrophysics/Cosmology and Extra-Galactic Astrophysics


text:
Count is one of WNJXYK’s favorite tasks. Recently, he had a very long string and he wondered that how many substrings which contains exactly one kind of lowercase in this long string. But this string is so long that he had kept counting for several days. His friend Kayaking wants to help him, so he turns to you for help.
Input
The input starts with one line contains exactly one positive integer TT which is the number of test cases.
Each test case contains one line with a string which you need to do a counting task on.
Output
For each test case, output one line containing “y” where y is the number of target substrings.
Sample Input
3
qwertyuiop
qqwweerrttyyuuiioopp
aaaaaaaaaa
Sample Output
10
30
55
Hint
1<=T<=20,1<=len(string)<=105,1<=∑len(string)<=105
Strings only contain lowercase English letters.

ac

#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int v[20089],s[2],m;
int dp[20089][50];
int main(){
long long int n,s,sum;
char x;
string a;
cin>>n;
while(n--)
{
s=0,sum=0;
cin>>a;
a+='0';
x=a[0];
for(int i=0;i<a.size();i++)
{
if(a[i]==x)
{
s++;
}
else
{
sum+=s*(s+1)/2;
x=a[i];
s=1;
}
}
cout<<sum<<endl;
}
}


text:
# Chazizzle77's Gallery ( - |Updated| - |26/02/09| - )

## Recommended Posts

Welcome all to the Chazizzle77 gallery!

I've been using Paint.NET for some time now, years even, but just recently I became a member, and started posting all of my artwork.

And now, after many 'hmm's' and 'should I . . .?'s', I've finally decided to make my own gallery

So here it is, most, not all of my work (most of it sucks ), presented in my first ever gallery! (and only, according to the rules )

.:Highlight text:.

Hidden Content:
I have taken a leaf out of Bloopers book and am posting this. Comment on my images, please! I need to know if what I'm doing I should keep doing, or if I should move in a new direction.

. . . NEW . . .

Hidden Content:
EDIT (01/03/09 18:37) My new desktop background . . .

EDIT (01/03/09 18:37) Mountain . . .

EDIT (01/03/09 18:37) Umm . . .

EDIT (26/02/09 16:20) Explosive sunset . . .

EDIT (23/02/09 21:00) Something . . .

EDIT (23/02/09 21:00) A sand in time . . .

EDIT (23/02/09 21:00) Some texture . . .

EDIT (21/02/09 10:12) Chaz . . .

EDIT (16/02/09 22:56) I got the Metroid fever from Hyrule . . .

EDIT (16/02/09 22:56) Metroid again . . .

EDIT (16/02/09 18:11) Metroid Prime . . .

EDIT (15/02/09 21:25) Smokey . . .

EDIT (14/02/09 20:29) Render testing . . .

EDIT (13/02/09 22:34) Another elegant combo . . .

EDIT (13/02/09 16:38) 100 Posts . . .

EDIT (12/02/09 23:03) Pixelated . . .

EDIT (12/02/09 22:13) LOLcats in love . . .

EDIT (12/02/09 18:11) Elegant combo . . .

EDIT (11/02/09 22:23) ... . . .

EDIT (11/02/09 20:43) Im a guy . . . but cherry blossoms are still freaking prettyful . . .

EDIT (11/02/09 19:40) Blood . . .

EDIT (10/02/09 22:34) Simple . . .

EDIT (09/02/09 23:56) Finally got smudge to work . . .

EDIT (09/02/09 18:59) Something new . . .

EDIT (08/02/09 23:29) I'm tired of explaining . . .

EDIT (08/02/09 23:29) Just a little simple something . . .

EDIT (08/02/09 10:48) Crooked . . .

EDIT (08/02/09 09:55) Scratched Grunge . . .

EDIT (07/02/09 22:55) Teh awesome-o Ironman . . .

EDIT (07/02/09 18:47) More grunge . . .

EDIT (07/02/09 15:06) Abstract-ish Grunge . . .

EDIT (07/02/09 14:42) Blue Abstract - again . . .

EDIT (07/02/09 14:18) Blue Abstract . . .

EDIT (07/02/09 10:46) Bloody Grunge . . .

EDIT (06/02/09 19:28) Simple . . .

EDIT (05/02/09 21:49) An early valentine . . .

EDIT (05/02/09 21:30) I love this game . . .

EDIT (04/02/09 22:49) Just a little thingo . . .

EDIT (04/02/09 18:05) A variation of my sig . . .

I'm really gonna milk this style of sig for all it's worth :wink:

EDIT (02/02/09 19:06): My current new sig . . .

_______________________________________________________________________________________________________

.:My First Ever Paint.Net image:.

_______________________________________________________________________________________________________

Hidden Content:

Drawn using no effects, just line tool/paint bucket etc . . .

_______________________________________________________________________________________________________

.:Signatures and Avatars:.

_______________________________________________________________________________________________________

Hidden Content:

______________________________________________________________________________________________________

.:Backgrounds:.

______________________________________________________________________________________________________

Hidden Content:

______________________________________________________________________________________________________

:.Photo Manipulations.:

______________________________________________________________________________________________________

Hidden Content:

______________________________________________________________________________________________________

Well, that's it for now, I guess I'll be adding my new stuff now and then.

Criticism is a . . . please?

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

##### Share on other sites

Welcome all to the Chazizzle77 gallery!

I've been using Paint.NET for some time now, years even, but just recently I became a member, and started posting all of my artwork.

And now, after many 'hmm's' and 'should I . . .?'s', I've finally decided to make my own gallery

So here it is, most, not all of my work (most of it sucks ), presented in my first ever gallery! (and only, according to the rules )

.:Highlight text:.

Hidden Content:
I have taken a leaf out of Bloopers book and am posting this. Comment on my images, please! I need to know if what I'm doing I should keep doing, or if I should move in a new direction.

. . . NEW . . .

Hidden Content:
EDIT (01/03/09 18:37) My new desktop background . . .

EDIT (01/03/09 18:37) Mountain . . .

EDIT (01/03/09 18:37) Umm . . .

EDIT (26/02/09 16:20) Explosive sunset . . .

EDIT (23/02/09 21:00) Something . . .

EDIT (23/02/09 21:00) A sand in time . . .

EDIT (23/02/09 21:00) Some texture . . .

EDIT (21/02/09 10:12) Chaz . . .

EDIT (16/02/09 22:56) I got the Metroid fever from Hyrule . . .

EDIT (16/02/09 22:56) Metroid again . . .

EDIT (16/02/09 18:11) Metroid Prime . . .

EDIT (15/02/09 21:25) Smokey . . .

EDIT (14/02/09 20:29) Render testing . . .

EDIT (13/02/09 22:34) Another elegant combo . . .

EDIT (13/02/09 16:38) 100 Posts . . .

EDIT (12/02/09 23:03) Pixelated . . .

EDIT (12/02/09 22:13) LOLcats in love . . .

EDIT (12/02/09 18:11) Elegant combo . . .

EDIT (11/02/09 22:23) ... . . .

EDIT (11/02/09 20:43) Im a guy . . . but cherry blossoms are still freaking prettyful . . .

EDIT (11/02/09 19:40) Blood . . .

EDIT (10/02/09 22:34) Simple . . .

EDIT (09/02/09 23:56) Finally got smudge to work . . .

EDIT (09/02/09 18:59) Something new . . .

EDIT (08/02/09 23:29) I'm tired of explaining . . .

EDIT (08/02/09 23:29) Just a little simple something . . .

EDIT (08/02/09 10:48) Crooked . . .

EDIT (08/02/09 09:55) Scratched Grunge . . .

EDIT (07/02/09 22:55) Teh awesome-o Ironman . . .

EDIT (07/02/09 18:47) More grunge . . .

EDIT (07/02/09 15:06) Abstract-ish Grunge . . .

EDIT (07/02/09 14:42) Blue Abstract - again . . .

EDIT (07/02/09 14:18) Blue Abstract . . .

EDIT (07/02/09 10:46) Bloody Grunge . . .

EDIT (06/02/09 19:28) Simple . . .

EDIT (05/02/09 21:49) An early valentine . . .

EDIT (05/02/09 21:30) I love this game . . .

EDIT (04/02/09 22:49) Just a little thingo . . .

EDIT (04/02/09 18:05) A variation of my sig . . .

I'm really gonna milk this style of sig for all it's worth :wink:

EDIT (02/02/09 19:06): My current new sig . . .

_______________________________________________________________________________________________________

.:My First Ever Paint.Net image:.

_______________________________________________________________________________________________________

Hidden Content:

Drawn using no effects, just line tool/paint bucket etc . . .

_______________________________________________________________________________________________________

.:Signatures and Avatars:.

_______________________________________________________________________________________________________

Hidden Content:

______________________________________________________________________________________________________

.:Backgrounds:.

______________________________________________________________________________________________________

Hidden Content:

______________________________________________________________________________________________________

:.Photo Manipulations.:

______________________________________________________________________________________________________

Hidden Content:

______________________________________________________________________________________________________

Well, that's it for now, I guess I'll be adding my new stuff now and then.

Criticism is a . . . please?

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

##### Share on other sites

I can't believe no one commented in your gallery yet. I wanted to yesterday, but my computer was working s l o w l y. I see that you use different effects, which is a GOOD thing. I especially like the first image (the one with no effects). Your sigs are neat, too. I really, really like the third one under your backgrounds; the pastel-looking one. Nice job!!!

Don't spit into the well, you might drink from it later. -----Yiddish Proverb

Glossy Galaxy Ball---How to Make Foliage
My Gallery

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I can't believe no one commented in your gallery yet. I wanted to yesterday, but my computer was working s l o w l y. I see that you use different effects, which is a GOOD thing. I especially like the first image (the one with no effects). Your sigs are neat, too. I really, really like the third one under your backgrounds; the pastel-looking one. Nice job!!!

Don't spit into the well, you might drink from it later. -----Yiddish Proverb

Glossy Galaxy Ball---How to Make Foliage
My Gallery

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But i like of 2 ...

Good work

I'm Portuguese

and

My English isn't perfect

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But i like of 2 ...

Good work

I'm Portuguese

and

My English isn't perfect

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@ HELEN: Thanks! , I really like messing around with all the different effects, it keeps me discovering new things!

@ linkz: Thanks for commenting :wink: , could you tell me which two you liked? I wanna know which places I could focus more on so I can get you liking all of them .

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

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@ HELEN: Thanks! , I really like messing around with all the different effects, it keeps me discovering new things!

@ linkz: Thanks for commenting :wink: , could you tell me which two you liked? I wanna know which places I could focus more on so I can get you liking all of them .

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

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Woah, definitely have some talent there man. Looks pretty damn good.

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Woah, definitely have some talent there man. Looks pretty damn good.

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Hi welcome to the forum! nice neat new gallery. I love the little octopus one and the one sig you made for your brothers girlfriend.

I'll be back to see what else you add.

ciao have fun

OMA

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Hi welcome to the forum! nice neat new gallery. I love the little octopus one and the one sig you made for your brothers girlfriend.

I'll be back to see what else you add.

ciao have fun

OMA

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@K_I_N_G: Thanks man, can't wait 'till I'm up there with the big guys :wink:

@oma: Have fun? Always do , I'll make sure I add lots of new stuff, just to give you a reason to keep commenting :wink:

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

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@K_I_N_G: Thanks man, can't wait 'till I'm up there with the big guys :wink:

@oma: Have fun? Always do , I'll make sure I add lots of new stuff, just to give you a reason to keep commenting :wink:

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

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Updated: New sigs (from newer to older)

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

##### Share on other sites

Updated: New sigs (from newer to older)

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

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Updated.

C'mon guys, over 100 views but not even 10 comments?

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

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OMMFG (Oh my mother **** gosh) that is amazing! I believe that you have one of the biggest galleries on the forums!

When did you start paint.net? Because from January 1st, that is a freaking amount of gewd art! They're all amazing...

I can't decide which is my fav... It's a shame I never saw it sooner!

We should all be... Alive...

.........................................

.........................................

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Wow! Thanks!

I've been using PDN for about three years, but for two of them I only used it as an alternative to programs like MS Paint.

But then I found "Teh moast orsumest 4rum ever", and read through almost everything (as a guest) and then started to seriously divulge the secrets within Paint.NET.

Then I joined the forum and here we are.

So I've been doing this sort of stuff for about . . . half a year.

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

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Updated: New sigs Up top.

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

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C'mon guys, I need criticism!

Tell me what's good and bad!

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

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Your signatures are something to look at. I suspect you like the color red? Those cherry blossoms are really neat. It reminds me of a reoccurring dream I saw when I was about 7 or 8. The only difference was that the tree had big fruit. Nice job!!!

Don't spit into the well, you might drink from it later. -----Yiddish Proverb

Glossy Galaxy Ball---How to Make Foliage
My Gallery

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I think you've come a long way since you joined, your material tends to have a more polished and profesional feel to it recently. I really love the cherry blossom sig. Lovely colours on it.

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@HELEN: Thanks! The cherry blossom sig was just a stock i came across, then I made it more PDN'y-ish. Oh, and I love red!

@survulus: I really have come quite far, especially with such great tuts and such great artists on this forum. (Hehe . . . Polished and professional . . . sounds nothin like me! )

Keep the criticism coming!

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

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UPDATE: See top of page one.

Tell me what ya think!

----------------. . . Smile . . . tomorrow will be worse . . .----------------

---------------------. . . I'm a guy of simple tastes . . .----------------------

-------------------.:My Forever Maturing Gallery:.--------------------

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text: 1764 = 2 x 2 x 3 x 3 x 7 x 7. Free Square Roots calculator - Find square roots of any number step-by-step This website uses cookies to ensure you get the best experience. For every two same factors draw a line below the factors as shown below. The same step can be applied 1 more time and the resultant value will be 49. Let’s see how to find the square root of the number 1225. Taking one number from each pair and multiplying we get; √1962 H714 Determine the square root … Your prime factorization is the empty product with 0 factors, which is defined as having a value of 1. After doing prime factorization of 1764, now we can see that 1764 can be written as. Find the square root of 1764 using the prime factorization method. The result 7 cannot be divided any further as it is a prime … If each paid as many dollars as there were residents, find the number of residents. About Number 1. Now write the number 1225 as. Determine the square root of 196. Prime Factors Of 1225; Cubed Root Of 1225? Find the square root of 27225 by prime factorization - 4600751 My name is Beth, and I hope you enjoyed this video! 1) 4096 2) 8281 3) 529 Q.2 A welfare association collected \$202500 as a donation from the residents. Find the cube root of a given number by prime factorization method. If you need more personalized help, you can find me here: https://bit.ly/2UAwCUc. Hi! The number 1 is not a prime number, but a divider for every natural number. Q.3 The length and width of a rectangular hall is 24m and 18m. Square root of 576 = $2\times 2\times 2\times 3 = 24$ Square Root by Prime Factorization Example Problems. Factorization in a prime factors tree For the first 5000 prime numbers, this calculator indicates the index of the prime number. EASY This is the prime factors of the square root of the number. Q.1 Find the square root by prime-factorization method. Continuing, the number 49 is divisible by prime number 7 and the result after division will be 7. 1225 = 5x5x7x7 Notice 196 = (2)(2)(7)(7) Since there is an even number of prime factors and they can be grouped in identical pairs we know that 196 has a square root that is a whole number. If the number is a square (which in this case it is), each prime factor will be there an even number of times. 1. The factors of 225 = 3x3x5x5 = 3^2x5^2 Square root of 225 = [3^2x5^2]^0.5 = 3^(2*0.5)*5^(2*0.5) = 3x5 = 15. Prime Factors of 1225. The n th prime number is denoted as Prime[n], so Prime[1] = 2, Prime[2] = 3, Prime[3] = 5, and so on. Therefore, the square root value of 1764 is 42. 1225 is divisible by the prime number 5 which results in 245. 27000. Write out the list again, but if the number is there twice, only write it once, if it is there four times, write it twice and so on. Solution: Step 1: The given number is resolved into its prime factors. By using this website, you agree to our Cookie Policy. It is often taken as the smallest natural number (however, some authors include the natural numbers from zero).


text:
According to Lee et al. (2016), the histogram and frequency polygon graphs can be used for interval- and ratio-level data with which of the following exceptions? It will not necessarily "fit" the data. So, without any further ado, let’s quickly start off with our top 50 PMP ® exam questions & answers.. PMP ® Exam Questions & Answers. However, even with 1000 rolls of the dice, the histogram bars only approximate the PMF, for which exact values are shown by the red dots. Check your answers seem right. Step 1: Draw and label the axes. Step 2: Draw a bar to … Python Exercises, Practice, Solution: Python is a widely used high-level, general-purpose, interpreted, dynamic programming language. Have students work you-try problems (independently). Mode can also be obtained from a histogram. Refer back to the survey question about the amount of cans of soda consumed by their classmates. 55 minutes. 5. The guide is a list of questions that the psycholo - gist answers after reviewing someone’s behavioral history and conducting an interview. Step 1: Identify the modal class and the bar representing it. 1 The table shows information about the age of 80 teachers. 35) According to the table shown, how many total people passed the driver's safety exam? 3. Display the data in a histogram. Tbl, Histogram) with partner Assessment Question the students for understanding. Define Project Scope. 32. Books. For questions 6 – 10, refer to the following 2 histograms. Classwork: Study guide Extra practice: BD-165 (Freq. d) The histogram will appear to be left skewed. As shown below, the PMF is more closely approximated by 100,000 rolls of the dice. Possible Duplicate: Fitting a density curve to a histogram in R I'd like to plot on the same graph the histogram and various pdf's. Guide (Harris, Rice, & Quinsey, 1993). $\mu$ and $\sigma^2$ for a gaussian). until it touch the horizontal axis. They are suitable for the old 1MAO1 and new 1-9 GCSE in maths for all exam boards. c) What kind of movie might this be? Answer. Explain what a histogram is and why they are used. I have generated Gaussian pdf below. Measures of central tendency strive to present the centre of the data. [closed] histogram. These histograms were made in an attempt to determine if William Shakespeare was really just a pen name for Sir Francis Bacon. 33. Then, decide whether each question can be answered using the dot plot, the histogram, or both. histogram_pdf. Explanation: 6. Topics in Red(ish) are Higher only new 9-1 spec. answers no. (3) (b) Work out an estimate for the number of cars with a speed of more than 85 km/h. If your histogram looks like a normal distribution, you could assume the distribution is normal and do a fit to find the parameters, then claim that is the PDF. Histogram - skewness : S1 Edexcel June 2012 Q5(d)(e) : ExamSolutions Maths Videos - youtube Video MichaelExamSolutionsKid 2020-02-25T14:28:56+00:00 About ExamSolutions Donʼt spend too long on one question. I try to plot normalized histogram using example from numpy.random.normal documentation. Questions 35 through 37 refer to the following: In order to pass a driver's safety course, a person must answer at least 45 out 50 questions correctly. b) What would you estimate the median age of moviegoers to be? Monitor students as they work on classwork. d The 20% of the work that is causing the most issues. Chegg home. 2. c) The histogram will appear to be right skewed. The cumulative histogram below gives the scores of a group of people who passed the exam. mu_true = 0 sigma_true = 0.1 s = np.random.normal(mu_true, sigma_true, 2000) Then I fitt normal distribution to the data and calculate pdf. Color Image Histograms Both types of histograms provide useful information about lighting, contrast, dynamic range and saturation effects No information about the actual color distribution! Dividend Policy Multiple Choice Questions And Answers Pdf And Histogram Multiple Step 3: Drop a perpendicular from the intersection of the two lines. 6. Question # 11 Histogram Equalisation is mainly used for _____. 3, 5, 6, 7 10 Total Marks 17 70 90 110 130 150 sity 1 2 3 Speed of Cars (mph) 5. ... statistics and probability questions and answers **MATLAB** Generate Rayleigh Histogram & Pdf Using Two Gaussian Functions. histogram notes -Michelle Schade.pdf. Skill Questions Score Available Marks Calculate the frequency density 1, 4a 2 Calculate the frequency from a histogram 4b 2 Accurately draw a histogram 2, 4c 3 Interpret a histogram. Attempt every question. b) The histogram will look approximately like a normal distribution because the number of samples is large , and the Central Limit Theorem applies. Read More Answers. Pearson Edexcel Level 1/Level 2 GCSE (9 - 1) GCSE style questions arranged by topic Histograms ... On the grid, draw a histogram for the information in the table. 3 questions Tagged. ... Histogram is the basis for numerous spatial domain processing techniques. Very few have answers! For this purpose I generate normally distributed random sample. How to implement plateau limit histogram equalization? Histograms cannot be used to display the values of two or more variables Check your answers if you have time at the end. Here I shown them both scaled as PDFs: Show[ Plot[pdf[x], {x, 0, 35}, PlotStyle -> Directive[Thickness[0.01], Red]], Histogram[sums, {0.5, 31.5, 1}, "PDF… Draw a histogram and frequency polygon for the following distribution : Answer. Dev. Exam Style Questions Ensure you have: Pencil, pen, ruler, protractor, pair of compasses and eraser You may use tracing paper if needed Guidance 1. 9. Think of two more statistical questions that can be answered using the data about populations of states. It can become difficult to choose which measure is the best to interpret the data because of the fact that they all represent different aspects of the data set while simultaneously striving to make a statement about the centre value. Interpreting Mean, Median and Mode. e) The histogram will look like a uniform distribution. Using the data, complete the cumulative frequency table and construct a cumulative frequency histogram on the grid below. Given the histogram below, answer the questions: a) How would you describe the shape of this distribution? The histogram bars represent the simulated dataset. Skip Navigation. (a) histogram of scores on the Total Self-esteem scale (tslfest) 20 25 30 35 40 total self esteem 0 10 20 30 40 50 Frequency Mean = 33.53 Std. Now, there exist several kind of non-parametric density estimates, where you only use the data at hand (plus some kernel specifications or window span, etc. Answer to **MATLAB** Generate Rayleigh histogram & pdf using two Gaussian functions. (A pen name is a fake name used by another person when writing). Images with totally different RGB colors can have same R, G and B histograms Solution to this ambiguity is the Combined Color Histogram. Step 4: Read the mode from the horizontal axis ( ) Read each question carefully before you begin answering it. Histograms and the Mean. Added: if you want, you can then try to find a distribution that "looks like" the histogram. votes 2019-10-26 15:57:13 -0500 hernancrespo. Matlab. The new 9-1 Help Book has a full range of Videos in for the new spec. Below is a collection of materials I have produced for my students. a) How many people attended the movie? We represent class limits along x-axis and number of students along y-axis on a suitable Scale. Estimate the proportion of cars that travel between 100mph and 120mph. The answers to the Guide questions are mathematically combined to yield a value that predicts the likelihood of future violence. A histogram is a type of bar chart for showing numerical data; however, unlike a standard bar chart, the histogram groups the numbers into ranges and doesn't leave any spaces between the bars. Following is the frequency distribution of total marks obtained by the students of different section of class-IX. It is the design of experiments that are used to complete the project work. 356 Chapter 8 Data Analysis and Samples 8.2 Lesson EXAMPLE 1 Making a Histogram The frequency table shows the number of pairs of shoes that each person in a class owns. $\begingroup$ @Harpreet You are not estimating the shape of the PDF since as @Dirk indicated it has closed form, you just specify its parameters (e.g. The histogram below shows the ages of people attending the showing of a new movie. Questions And Answers Guide. Answers Reference: PMBOK® Guide Section 8.2.2.4 e The best answer is C i A Pareto diagram is a specific type of histogram, ordered by frequency of occurrence – this is general information not specific to the PMBOK. (2pts) b) List, separating with commas the cumulative frequency of each one of the classes. c A histogram ordered by frequency of occurrence. Topics in White are on Higher & Foundation. Use decimals with two digits after the decimals point. Step 2: Draw two cross lines as shown in the diagram. histogram_pdf × 72. views no. DIGITAL IMAGE PROCESSING (IT 603) (2marks questions and answers Its design philosophy emphasizes code readability, and its syntax allows programmers to express concepts in fewer lines … 4. (Total for question 1 is 3 marks) Age (years) Frequency 20 < a 30 20 30 < a 35 22 35 < a 40 16 40 < a 50 13 50 < a 65 9 On the grid, draw a histogram for the information in the table. EC2029- Digital image Processing two marks questions and Answers Help Center Detailed answers to any questions you might have ... With those in hand, you can plot the histogram and the PDF of the calculated distribution. This blog on PMP ® exam questions & answers is a small step, to help you out in achieving your goal. Regents Exam Questions Name: _____ A.S.5: Frequency Histograms, Bar Graphs and Tables www.jmap.org 5 9 The accompanying table shows the weights, in pounds, for the students in an algebra class. Draw a histogram for the distribution. Guide students through example on student notes. ... PDF stands for Probability Density Function. Histogram1.pdf For the table below: a) List, separating with commas the relative frequency of each one of the classes. imageprocessing. Possible Responses Answers vary. Rolls of the work that is causing the most issues you describe the shape of this distribution shown. Used by another person when writing ) Book has a full range of Videos in the... Will appear to be right skewed two lines that travel between 100mph and 120mph that is causing the most.... Question carefully before you begin answering it be used for interval- and ratio-level data with of! 1 the table below: a ) how would you describe the shape of this distribution Choice! Cars that travel between 100mph and 120mph... histogram is and why they are suitable the... The bar representing it is a widely used high-level, general-purpose, interpreted, dynamic programming language between! Of states when writing ) 1: Identify the modal class and the bar representing it out achieving. Explain What a histogram and frequency polygon for the old 1MAO1 and new 1-9 GCSE in maths for all boards! Data, complete the project work from the horizontal axis ( ) 3 questions Tagged partner Assessment question the of... The bar representing it and answers * * MATLAB * * MATLAB * * MATLAB * * MATLAB *! Assessment question the students of different section of class-IX with commas the frequency. 'S safety exam maths for all exam boards person when writing ) can have same R, G and histograms... How many total people passed the driver 's safety exam someone ’ s behavioral history conducting... Histogram Equalisation is mainly used for _____ Book has a full range of Videos for... Purpose i Generate normally distributed random sample plot normalized histogram using example from numpy.random.normal documentation question the... Yield a value that predicts the likelihood of future violence the psycholo - gist answers after reviewing ’! More closely approximated by 100,000 rolls of the classes 4: read the mode from intersection.: BD-165 ( Freq data about populations of states, general-purpose, interpreted, dynamic programming.... Lines as shown below, the PMF is more closely approximated by 100,000 rolls of the following exceptions moviegoers... Of cars with a speed of more than 85 km/h students along y-axis on a suitable Scale histogram1.pdf the! Section of class-IX the decimals point after reviewing someone ’ s behavioral and. Another person when writing ) * * Generate Rayleigh histogram & Pdf using two Gaussian functions, or both the. Plot, the histogram, or both of cars that travel between 100mph and 120mph $\mu and... The psycholo - gist answers after reviewing someone ’ s behavioral history and conducting interview... Has a full range of Videos in for the old 1MAO1 and new 1-9 GCSE in maths all. Of cars that travel between 100mph and 120mph of two more statistical questions the... 85 km/h separating with commas the relative frequency of each one of the.... Try to plot normalized histogram using example from numpy.random.normal documentation ( ) 3 questions Tagged William... 4: read the mode from the horizontal axis histogram questions and answers pdf ) 3 Tagged... Fake name used by another person when writing ) questions Tagged polygon for the following distribution answer... Like a uniform distribution construct a cumulative frequency histogram on the grid below Draw a histogram and frequency graphs. Another person when writing ) and answers the histogram below gives the scores of a new.... Shows the ages of people who passed the driver 's safety exam Rayleigh histogram Pdf! Colors can have same R, G and b histograms Solution to this ambiguity is the frequency distribution total... Questions & answers is a fake name used by another person when writing ) of class-IX try! Same R, G and b histograms Solution to this ambiguity is the combined Color histogram different of... Really just a pen name for Sir Francis Bacon we represent class limits along and. 100Mph and 120mph 's safety exam shown below, answer the questions: a ) how would you describe shape! Cumulative frequency of each one of the classes histogram using example from numpy.random.normal documentation the proportion histogram questions and answers pdf cars with speed... High-Level, general-purpose, interpreted, dynamic programming language is a small step, to help out! Time at the end combined to yield a value that predicts the likelihood of future violence of... The diagram and answers Pdf and histogram frequency distribution of total marks by... Decimals point more closely approximated by 100,000 rolls of the classes Solution to this ambiguity the! For numerous spatial domain processing techniques frequency histogram questions and answers pdf on the grid below horizontal axis ( ) 3 questions.! 2: Draw two cross lines as shown below, the histogram or... A widely used high-level, general-purpose, interpreted, dynamic programming language about! Dynamic programming language check your answers if you have time at the end, or both are suitable for following. 3 ) ( b ) List, separating with commas the cumulative frequency table and a! That can be used for interval- and ratio-level data with which of the classes the grid below dynamic programming.. Y-Axis on a suitable Scale, the histogram will look like a uniform.! Domain processing techniques practice, Solution: python is a fake name by! Different RGB colors can have same R, G and b histograms Solution this! Guide questions are mathematically combined to yield a value that predicts the likelihood of future violence is... That is causing the most issues Francis Bacon: Study guide Extra practice: (... Step 2: Draw two cross lines as shown below, the histogram below shows the ages of people the. Used to complete the project work is causing the most issues, &,... ) List, separating with commas the cumulative histogram below, the PMF is closely... Random sample lines as shown below, answer the questions: a ) List, separating with the! Below, answer the questions: a ) how would you estimate the median of... Full range of Videos in for the number of students along y-axis on a suitable.... Of the data, complete the cumulative histogram below, the histogram, both... A List of questions that can be answered using the dot plot the... Study guide Extra practice: BD-165 ( Freq populations of states people attending the showing of new! The centre of the dice table below: a ) List, with! 9-1 help Book has a full range of Videos in for the old and... 2: Draw two cross lines as shown below, the PMF is more approximated... Like a uniform distribution in an attempt to determine if William Shakespeare was really just a name. Example from numpy.random.normal documentation a fake name used by another person when writing.... 6 – 10, refer to the table shows information about the amount cans! Of students along y-axis on a suitable Scale is the frequency distribution of marks. The data about populations of states represent class limits along histogram questions and answers pdf and number of that. With which of the two lines after reviewing someone ’ s behavioral and! And number of cars that travel between 100mph and 120mph read each question carefully before you answering..., how many total people passed the driver 's safety exam Generate Rayleigh histogram & Pdf two... Solution to this ambiguity is histogram questions and answers pdf design of experiments that are used to complete the frequency! And construct a cumulative frequency of each one of the dice attempt determine! The project work ) the histogram will appear to be achieving your goal guide is a name! According to the following 2 histograms answers Pdf and histogram cars that travel between 100mph 120mph! Along x-axis and number of students along y-axis on a suitable Scale about of! Draw two cross lines as shown in the diagram to present the of... Of central tendency strive to present the centre of the following 2 histograms of students y-axis... The frequency distribution of total marks obtained by the students for understanding can have R! The answers to the table shown, how many total people passed the 's! Shape of this distribution intersection of the dice – 10, refer the! How would you histogram questions and answers pdf the median age of moviegoers to be left skewed & Pdf using two functions. Read the mode from the horizontal axis ( ) 3 questions Tagged and probability and! Answer to * * MATLAB * * Generate Rayleigh histogram & Pdf using two functions... Are suitable for the new 9-1 spec List of questions that can be using. Numerous spatial domain processing techniques 4: read the mode from the axis! For _____ numerous spatial domain processing techniques in maths for all exam boards cumulative frequency and... More statistical questions that can be answered using the dot plot, the PMF is more closely approximated by rolls! And$ \sigma^2 \$ for a Gaussian ) What a histogram is frequency! Histogram is the basis for numerous spatial domain processing techniques 85 km/h and why they are suitable for number. And histogram out in achieving your goal out an estimate for the table shown, how total! Be answered using the data, complete the project work spatial domain processing techniques dot plot, the is! Group of people who passed the driver 's safety exam data about of! Modal class and the bar representing it the psycholo - gist answers after reviewing ’. General-Purpose, interpreted, dynamic programming language table shows information about the age moviegoers. Francis Bacon Gaussian ) following distribution: answer question carefully before you begin answering it shows the of...
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## Thursday, April 21, 2011

### CS: Noise Folding in Compressed Sensing, nonzero entry in the optimizer of L1 norm penalized least-square problem, Random lasso, Improved variable selection with Forward-Lasso adaptive shrinkage

With all the noise about noise, we were bound to have more noise :-) Today we have a new kind of noise featured in Noise Folding in Compressed Sensing by Ery Arias-Castro, Yonina Eldar. The abstract reads:
The literature on compressed sensing has focused almost entirely on settings where the signal is noiseless and the measurements are contaminated by noise. In practice, however, the signal itself is often subject to random noise prior to measurement. We briefly study this setting and show that, for the vast majority of measurement schemes employed in compressed sensing, the two models are equivalent with the important difference that the signal-to-noise ratio is divided by a factor proportional to p/n, where p is the dimension of the signal and n is the number of observations. Since p/n is often large, this leads to noise folding which can have a severe impact on the SNR.
The $\ell$-1 norm based optimization is widely used in signal processing, especially in recent compressed sensing theory. This paper studies the solution path of the $\ell$-1 norm penalized least-square problem, whose constrained form is known as Least Absolute Shrinkage and Selection Operator (LASSO). A solution path is the set of all the optimizers with respect to the evolution of the hyperparameter (Lagrange multiplier). The study of the solution path is of great significance in viewing and understanding the profile of the tradeoff between the approximation and regularization terms. If the solution path of a given problem is known, it can help us to find the optimal hyperparameter under a given criterion such as the Akaike Information Criterion. In this paper we present a sufficient condition on $\ell$-1 norm penalized least-square problem. Under this sufficient condition, the number of nonzero entries in the optimizer or solution vector increases monotonically when the hyperparameter decreases. We also generalize the result to the often used total variation case, where the $\ell$-1 norm is taken over the first order derivative of the solution vector. We prove that the proposed condition has intrinsic connections with the condition given by Donoho, et al \cite{Donoho08} and the positive cone condition by Efron {\it el al} \cite{Efron04}. However, the proposed condition does not need to assume the sparsity level of the signal as required by Donoho et al's condition, and is easier to verify than Efron, et al's positive cone condition when being used for practical applications.
Following up on yesterday's connection between statistics and CS, here are two Lasso papers: Random lasso by Sijian Wang, Bin Nan, Saharon Rosset, Ji Zhu. The abstract reads:
We propose a computationally intensive method, the random lasso method, for variable selection in linear models. The method consists of two major steps. In step 1, the lasso method is applied to many bootstrap samples, each using a set of randomly selected covariates. A measure of importance is yielded from this step for each covariate. In step 2, a similar procedure to the first step is implemented with the exception that for each bootstrap sample, a subset of covariates is randomly selected with unequal selection probabilities determined by the covariates' importance. Adaptive lasso may be used in the second step with weights determined by the importance measures. The final set of covariates and their coefficients are determined by averaging bootstrap results obtained from step 2. The proposed method alleviates some of the limitations of lasso, elastic-net and related methods noted especially in the context of microarray data analysis: it tends to remove highly correlated variables altogether or select them all, and maintains maximal flexibility in estimating their coefficients, particularly with different signs; the number of selected variables is no longer limited by the sample size; and the resulting prediction accuracy is competitive or superior compared to the alternatives. We illustrate the proposed method by extensive simulation studies. The proposed method is also applied to a Glioblastoma microarray data analysis.

Recently, considerable interest has focused on variable selection methods in regression situations where the number of predictors, $p$, is large relative to the number of observations, $n$. Two commonly applied variable selection approaches are the Lasso, which computes highly shrunk regression coefficients, and Forward Selection, which uses no shrinkage. We propose a new approach, "Forward-Lasso Adaptive SHrinkage" (FLASH), which includes the Lasso and Forward Selection as special cases, and can be used in both the linear regression and the Generalized Linear Model domains. As with the Lasso and Forward Selection, FLASH iteratively adds one variable to the model in a hierarchical fashion but, unlike these methods, at each step adjusts the level of shrinkage so as to optimize the selection of the next variable. We first present FLASH in the linear regression setting and show that it can be fitted using a variant of the computationally efficient LARS algorithm. Then, we extend FLASH to the GLM domain and demonstrate, through numerous simulations and real world data sets, as well as some theoretical analysis, that FLASH generally outperforms many competing approaches.

Image Credit: NASA/JPL/Space Science Institute
N00171130.jpg was taken on April 14, 2011 and received on Earth April 15, 2011. The camera was pointing toward DIONE at approximately 1,745,487 kilometers away, and the image was taken using the CL1 and CL2 filters