text:
On dynamics of Out(F_{n}) on {\operatorname{PSL_{2}}({\mathbb{C}})} characters

# On dynamics of Out(Fn) on PSL2(C) characters

Yair N. Minsky Yale University
August 4, 2019
###### Abstract.

We introduce and study an open set of characters of a nonabelian free group, on which the action of the outer automorphism group is properly discontinuous, and which is strictly larger than the set of discrete, faithful convex-cocompact (i.e. Schottky) characters. This implies, in particular, that the outer automorphism group does not act ergodically on the set of characters with dense image. Hence there is a difference between the geometric (discrete vs. dense) decomposition of the characters, and a natural dynamical decomposition.

Partially supported by NSF grants DMS-0504019 and DMS-0554321

## 1. Introduction

Let be the free group on generators. Its automorphism group acts naturally, by precomposition, on for any group . The outer automorphism group acts on the quotient , where is understood to act by inner automorphisms. When is a Lie group we consider instead the character variety , the quotient in the sense of geometric invariant theory.

When is a compact Lie group and , Gelander [19] showed that this action is ergodic, settling a conjecture of Goldman [23], who had proved it for . When is non compact the situation is different because there is a natural decomposition of , up to measure 0, into (characters of) dense and discrete representations, and in the cases of interest to us the action on the discrete set is not ergodic, indeed even has a nontrivial domain of discontinuity.

See Lubotzky [32] for a comprehensive survey on the dynamics of representation spaces, from algebraic, geometric and computational points of view.

We will focus on the case of , where the interior of the discrete set is the set of Schottky representations. In this case one can ask if the action is ergodic, or even topologically transitive, in the set of dense representations, but this turns out to be the wrong question. In particular:

###### Theorem 1.1.

There is an open subset of , strictly larger than the set of Schottky characters, which is invariant, and on which acts properly discontinuously.

In other words, the natural “geometric” decomposition of , in terms of discreteness vs. density of the image group, is distinct from the “dynamical” decomposition, in terms of proper discontinuity vs. chaotic action of .

The subset promised in Theorem 1.1 will be the set of primitive-stable representations (see definitions below). It is quite easy to see that this set is open and invariant, and that the action on it is properly discontinuous (Theorems 3.2, 3.3). Thus the main content of this note is the observation, via a lemma of Whitehead on free groups and a little bit of hyperbolic geometry, that it contains non-Schottky (and in particular non-discrete) elements. This will be carried out in Section 4.

One should compare this with results of Goldman [22] on the rank 2 case for characters, and with work of Bowditch [5] on the complex rank 2 case. Bowditch, and Tan-Wong-Zhang [45], studied a condition very similar to primitive stability; we will compare the two in Section 5.

#### Acknowledgements

Bill Goldman and Alex Lubotzky got the author interested in this question, and Lubotzky pointed out the notion of “redundant representations”, whose negation leads (eventually) to the primitive-stable condition. Mark Sapir and Vladimir Shpilrain pointed the author to Whitehead’s lemma and Corollary 4.3. The author is also grateful to Dick Canary, Hossein Namazi and Juan Souto for interesting conversations and perspectives. Juan Souto in particular pointed out the connection to Otal’s work. The referee made a number of helpful and incisive suggestions.

## 2. Background and notation

In the remainder of the paper we fix and let . We also fix a free generating set of .

Note that can be identified with via once the generators are fixed. The quotient , which we denote or just , is obtained by considering characters of representations, which in our case corresponds to trace functions. This quotient naturally has the structure of an algebraic variety, and differs from the purely topological quotient only at reducible points, i.e. representations whose images fix a point on (see Kapovich [26] or Morgan-Shalen [39]). Characters of reducible representations form a subset of measure 0 outside all of the sets we shall be considering, and so we shall be able to ignore the distinction between these two quotients.

#### Geometric decomposition

There is a natural -invariant decomposition of in terms of the geometry of the action of on . Namely, let denote the (characters of) discrete and faithful representations, and let denote those of representations with dense image in .

It is fairly well-known (see [6, 7]) that

###### Lemma 2.1.

is nonempty and open, is closed, and has measure 0.

The idea of the measure 0 statement is this: If is not faithful it satisfies some relation; this is a nontrivial algebraic condition, so defines a subvariety of measure 0. There are a countable number of such relations. If is not discrete, consider the identity component of the closure of in . This is a connected Lie subgroup of and hence is either all of (so is dense), or solvable (i.e. elementary, fixing a point in the Riemann sphere), or conjugate to . The latter cases are again detected by algebraic conditions.

Openness of follows from the Kazhdan-Margulis-Zassenhaus lemma [27], which furnishes a neighborhood of the identity in in which any set of elements generates either an elementary or an indiscrete group. Generating an elementary group is a nontrivial algebraic condition on (for ), and as above for an indiscrete group not to be dense is also a nontrivial algebraic condition. Hence an open dense subset of consists of dense representations. Now given any with dense image, there exist elements such that , and since is open the same is true for sufficiently close to – hence also generates a dense subgroup.

The fact that is closed follows from Jørgensen’s inequality [25], or alternately from the Kazhdan-Margulis-Zassenhaus lemma. Lemma 2.1 in fact holds for much more general target groups – see [6, 7] for details.

Note, when is compact is empty, and in this case Gelander proved that acts ergodically on and hence on . Our main theorem will show that, in general, the action on is not ergodic.

#### Schottky groups

A Schottky group (or representation) is one which is obtained by a “ping-pong” configuration in the sphere at infinity . That is, suppose that are disjoint closed (topological) disks in and are isometries such that is the closure of the complement of . Then generate a free discrete group of rank , called a Schottky group. The representation sending is discrete and faithful, and moreover, an open neighborhood of it in consists of similar representations. We let denote the open set of all characters of Schottky representations.

Sullivan [44] proved that

###### Theorem 2.2.

is the interior of .

(This theorem is not known to hold for the higher-dimensional hyperbolic setting; see §5.)

To obtain a geometric restatement of the Schottky condition, consider the limit set for the action of any discrete group of isometries on , namely the minimal closed invariant subset of . The convex hull in of this limit set is invariant, and its quotient by the group is called the convex core of the quotient manifold. The deep usefulness of the convex core in the study of hyperbolic 3-manifolds was first exploited by Thurston. The Schottky condition on is equivalent to the condition that the convex core of is a compact handlebody of genus (see Marden [33, pp. 98-99] for a discussion and references).

It is a well-known consequence of the deformation theory of Kleinian groups (see e.g. Bers [3] for background) that acts properly discontinuously on : that is, that the set is finite for any compact . We note that this will also follow from Theorem 3.3.

## 3. Primitive-stable representations

Let be a bouquet of oriented circles labeled by our fixed generating set. We let denote the set of bi-infinite (oriented) geodesics in , as in Bestvina-Feighn-Handel [4]. Each such geodesic lifts to an -invariant set of bi-infinite geodesics in , the universal covering tree (and the Cayley graph of with respect to ).

Let be the boundary at infinity of , or the space of ends of the tree . We have a natural action of on . Each element of can be identified with an -invariant subset of (with the diagonal), i.e. the pairs of endpoints of its lifts. acts naturally on (and in general on -invariant subsets of .).

Equivalently we can identify with the set of bi-infinite reduced words in the generators, modulo shift. To every conjugacy class in is associated an element of named , namely the periodic word determined by concatenating infinitely many copies of a cyclically reduced representative of .

An element of is called primitive if it is a member of a free generating set, or equivalently if it is the image of a standard generator by an element of . Let denote the subset of consisting of for conjugacy classes of primitive elements. Note that is -invariant.

Given a representation and a basepoint , there is a unique map mapping the origin of to , -equivariant, and mapping each edge to a geodesic segment. Every element of is represented by an -invariant family of leaves in , which map to a family of broken geodesic paths in .

###### Definition 3.1.

A representation is primitive-stable if there are constants and a basepoint such that takes all leaves of to -quasi geodesics.

Note that if there is one such basepoint then any basepoint will do, at the expense of increasing . This condition is invariant under conjugacy and so makes sense for . Moreover the property is -invariant since is -invariant. Primitive-stability is a strengthening of the negation of redundancy, whose relevance was explained to me by Alex Lubotzky (see §5).

Let us establish some basic facts.

###### Lemma 3.2.

1. If is Schottky then it is primitive-stable.

2. Primitive-stability is an open condition in .

3. If is primitive-stable then, for every proper free factor of , is Schottky.

###### Proof.

To see (1), note that if is discrete and faithful then is the lift to universal covers of an embedding (and homotopy-equivalence) of into the quotient manifold . If is Schottky then the convex core of is compact and hence its homotopy-equivalence to the image of lifts to a quasi-isometry of the convex hull of the group to . It follows that all leaves in map to uniform quasi geodesics, and in particular the primitive ones.

Next we prove (2). Let be primitive-stable, and fix a basepoint and quasi-geodesic constants as in the definition. Let .

Let be a primitive leaf, with vertex sequence , and let . The condition that is quasi-geodesic is equivalent to the following statement: there exist constants and such that, if is the hyperplane perpendicularly bisecting the segment , then for all separates from , and . This is an easy exercise, and we note that determine , and vice versa.

Now consider a representation close to , and let . Up to the action of there are only finitely many sequences of tree edges of length , and hence the relative position (i.e. up to isometry) of and , over all primitive leaves and all , is determined by the image of a finite number of words of . These images each vary continuously with , and hence for sufficiently close to , we have that the separation and distance properties for the still hold, with modified constants. Hence the primitive leaves are still (uniformly) quasi-geodesically mapped by .

Finally we prove (3). Let be a proper free factor, so that with and nontrivial. Suppose is primitive-stable. If is cyclic, then being Schottky is equivalent to ’s generator having loxodromic image, and this is an immediate consequence of having a quasi-geodesic orbit. Hence we may now assume has rank at least 2. By (2), there is a neighborhood of consisting of primitive-stable elements. Suppose were not Schottky. Since is the interior of by Sullivan’s theorem 2.2, and since is dense in by Lemma 2.1, we can perturb arbitrarily slightly to get a dense representation. Leaving unchanged we obtain with dense. Now let be an infinite sequence of reduced words with . For any generator of , a sequence of elementary automorphisms multiplying by generators of (Nielsen moves) takes to , which therefore is primitive. Note that each is cyclically reduced, so primitive-stability of implies that the axes of are uniformly quasi-geodesically mapped by . But this contradicts the fact that while the length of goes to infinity. ∎

Let

 PS=PS(Fn)⊂X(Fn)

be the set of conjugacy classes of primitive-stable representations. We have shown that is an open invariant set containing the Schottky set. In fact,

###### Theorem 3.3.

The action of on is properly discontinuous.

###### Proof.

Let denote the translation length of the geodesic representative of for , and let denote the minimal combinatorial length, with respect to the fixed generators of , of any element in the conjugacy class of (equivalently it is the word length of after being cyclically reduced).

Let be a compact set in . For each we have a positive lower bound for over primitive elements of , and a continuity argument as in part (2) of Lemma 3.2 implies that a uniform lower bound

 ℓρ(w)||w||>r>0

holds over all in . Now on the other hand an upper bound on this ratio holds trivially for any by the triangle inequality applied to any . Continuity again gives us a uniform upper bound

 ℓρ(w)||w||

for (here, one should choose a compact preimage of in , which is easy to do).

Now if satisfies , we apply the inequalities to conclude, for in this intersection, that

 ||Φ(w)||≤(1/r)ℓρ(Φ(w))=(1/r)ℓρ∘Φ(w)≤(R/r)||w||.

The proof is then completed by the lemma below. ∎

###### Lemma 3.4.

For any , the set

 {f∈Aut(Fn):||f(w)||≤A||w||  ∀ primitive w}

has finite image in .

###### Proof.

In fact much less is needed; it suffices to have the inequality only for with . Let be generators of , and consider the action of on the tree (its Cayley graph). For any , let denote the distance between the axis of and the axis of . A look at the action on the tree indicates, if , that . Hence, since , we get an upper bound on .

An upper bound on the pairwise distances between the axes of the implies that there is a point on which is a bounded distance from all the axes simultaneously (minimize the sum of distances to the axes, which is proper and convex unless all the axes coincide). After conjugating we may assume that this point is the origin. Now the bound implies a finite number of choices for . ∎

## 4. Whitehead’s lemma and indiscrete primitive-stable representations

In this section we will prove that is strictly bigger than the set of characters of Schottky representations. In particular we will define the notion of a blocking curve on the boundary of a handlebody, and show

###### Theorem 4.1.

Let be discrete, faithful and geometrically finite with one cusp , which is a blocking curve. Then is primitive-stable.

We’ll see below (Lemmas 4.4 and 4.5) that blocking curves are a non-empty class. Using the deformation theory of hyperbolic 3-manifolds one can then obtain primitive-stable points in the boundary of Schottky space – see the end of the section for details.

Note that, since is open, this implies the existence of a rich class of primitive-stable representations, including dense ones (by Lemma 2.1), as well as discrete faithful ones with degenerate ends (since these are topologically generic in the boundary of Schottky space – see [11] and [12, Cor. A]).

First let us recall Whitehead’s criterion and define the notion of blocking. Whitehead studied the question of which elements in the free group are primitive. He found a necessary combinatorial condition, as part of an algorithm that decides the question of primitivity.

As before, we fix a generating set of . For a word in the generators and their inverses, the Whitehead graph is the graph with vertices labeled , and an edge from to for each string that appears in or in a cyclic permutation of . For more information see Whitehead [46, 47], Stallings [43] and Otal [41].

###### Lemma 4.2.

(Whitehead) Let be a cyclically reduced word. If is connected and has no cutpoint, then is not primitive.

Define the “reduced” Whitehead graph to be the same as except that we don’t count cyclic permutations of . In other words we don’t consider the word where is the last letter of and is the first, so may have one fewer edge than .

Let us say that a reduced word is primitive-blocking if it does not appear as a subword of any cyclically reduced primitive word. An immediate corollary of Lemma 4.2 is:

###### Corollary 4.3.

If is a reduced word with connected and without cutpoints, then is primitive-blocking.

Let us also say that is blocking if some power is primitive-blocking. A curve on the surface of the handlebody of genus is blocking if a reduced representative of its conjugacy class in the fundamental group is blocking (with respect to our given generators).

An instructive example of a blocking curve occurs for even rank, when the handlebody is homeomorphic to the product of an interval with a surface with one boundary component:

###### Lemma 4.4.

Let be a surface with one boundary component. The curve in the handlebody is blocking with respect to standard generators of ; in fact its square is primitive-blocking.

###### Proof.

Using standard generators for , the boundary is represented by . is a cycle minus one edge (corresponding to ), and contains the missing edge, and so by Whitehead’s lemma is blocking. ∎

One can construct other blocking curves on the boundary of any handlebody by explicit games with train tracks. We omit this approach, and instead study the relationship of the blocking condition to the Masur Domain in the measured lamination space.

### Laminations and the Whitehead condition

Let denote Thurston’s space of projectivized measured laminations on the boundary of the handlebody [15, 18]. Within this we have the Masur domain consisting of those laminations that have positive intersection number with every lamination that is a limit of meridians of [35]. This is an open set of full measure in [35, 28].

Using Otal’s work [41], we can extend Whitehead’s condition to laminations on the boundary as follows. Any free generating set of is dual to a system of disks on , which cut it into a 3-ball (Nielsen). Given such a system of disks and a lamination , realize both and the disk boundaries in minimal position – e.g. fix a hyperbolic metric on and make them geodesics. Otal calls tight with respect to if there are no waves on which are disjoint from . A wave on is an arc properly embedded in , which is homotopic, rel endpoints, through but not through into . In particular if is tight then no arcs of are waves. Hence, if a closed curve is tight with respect to , then its itinerary through the disks describes a cyclically reduced word in with respect to the dual generators.

We define as follows: Cutting along , becomes a planar surface with boundary components, each labeled by or . Let denote the union of the boundary components and the arcs of . Then is obtained from by making the boundary components into vertices and identifying parallel arcs of . In particular is given a planar embedding. If is a single closed curve and is dual to the original generators this is equivalent to the original definition.

Otal proved the following in [41]. We give a proof, since Otal’s thesis is hard to obtain.

###### Lemma 4.5.

(Otal) If , then there is a generating set with dual disks such that is tight with respect to , and is connected and has no cutpoints.

###### Proof.

First note that , where runs over meridians of , is positive and realized. For if is a minimizing sequence such that infinitely many of the are distinct then an accumulation point in in will have intersection number 0 with , contradicting The same holds for disk systems, so we may choose a disk system that minimizes .

Now cannot have a wave with respect to . If it did, then a surgery along such a wave would produce a new whose intersection number with is strictly smaller, contradicting the choice of .

If is disconnected then there is a loop in the planar surface which separates the boundary components, and does not intersect (here denotes a regular neighborhood). This gives a meridian that misses in , again contradicting .

If is connected but has a cutpoint, this is represented by a boundary component , equal to one copy of a component of (we are blurring the distinction between the disks in and their boundaries in ). Let denote the other copy. Then separates the planar complex , and we may let be a component of which does not contain . Now cuts into a union of disks-with-holes, one of which, , must contain . Let be the boundary component of that separates from . Any arcs of passing through must pass through (since separates the interior of from the rest of the graph), but arcs of incident to do not meet ; hence is strictly less than .

Because separates from , cutting along and regluing to yields again a connected planar surface – hence is a new disk system, with strictly smaller intersection number with . Again this is a contradiction, so we conclude that is connected and without cutpoints. ∎

Call a lamination blocking, with respect to (or the dual generators), if has no waves with respect to , and there is some such that every length subword of the infinite word determined by a leaf of passing across does not appear in a cyclically reduced primitive word. Note that, for simple closed curves, this coincides with the previous definition of blocking. An immediate corollary of the above lemma is:

###### Lemma 4.6.

A connected Masur-domain lamination (e.g. a simple closed curve or a filling lamination) on the boundary of a handlebody is blocking with respect to some generating set.

###### Proof.

Given let be as in Lemma 4.5. In a connected measured lamination every leaf is dense. Thus a sufficiently long leaf of would traverse every edge of , and so the corresponding word is blocking by Corollary 4.3. Note for a simple closed curve this argument shows that its square is primitive-blocking. ∎

### Blocking cusps are primitive-stable

We can now provide the proof of Theorem 4.1, namely that a geometrically finite representation with a single blocking cusp is primitive-stable.

###### Proof of Theorem 4.1.

Let be the quotient manifold, and its convex core. The geometrically finite hypothesis means that is a union of a compact handlebody and a subset of a parabolic cusp (namely a vertical slab in a horoball modulo ) along an annulus with core curve in , which we are further assuming is a blocking curve.

We will prove that all primitive elements of are represented by geodesics in a fixed compact set . The idea is that in order to leave a compact set, a primitive element must wind around the cusp, and this will be prohibited by the blocking property.

Let be a closed geodesic in . Then . The orthogonal projection gives a retraction . Let . Note that if a component of is long then its image winds many times around (in fact the number of times is exponential in the length of ).

Claim: is uniformly quasi-geodesic in , with constants independent of . More precisely, the lift of to the universal cover is uniformly quasi-geodesic with respect to the path metric. This follows from a basic fact about any family of disjoint horoballs in :

###### Lemma 4.7.

Let be a family of disjoint open horoballs in , and let be given by orthogonal projection from to and identity in . If is a geodesic in then is a quasigeodesic in with its path metric, with constants independent of or .

###### Proof.

This is closely related to statements in Farb [17] and Klarreich [29] and can also be proved in greater generality, e.g. for uniformly separated quasiconvex subsets of a -hyperbolic space. We will sketch a proof for completeness.

Note first, there is a constant such that, if and are horoballs in with , then any two geodesic segments connecting to in their common exterior lie within -neighborhoods of each other.

If is obtained by retracting horoballs to concentric horoballs at depth bounded by , then it suffices to prove the theorem for . This is because any arc on can be approximated in a controlled way by an arc on . Moreover, given it suffices to prove the theorem for the union of horoballs that intersect , since and . We can therefore reduce to the case that any two components of are separated by distance at least 1, and peneterates each component of to depth at least .

Let be a geodesic in with endpoints on . It is therefore a concatentation of hyperbolic geodesics in with endpoints on , alternating with geodesics on in the path (Euclidean) metric. Let be one of the hyperbolic geodesic segments, with endpoints on horoballs . Then is within of the component of connecting these horoballs, and hence can be replaced by an arc traveling along , and of uniformly comparable length. Replacing all hyperbolic segments in this way, and then straightening the arcs of intersection of the resulting path with , we obtain a segment of whose length is comparable with that of Since the endpoints of were arbitrary points of , this gives uniform quasigeodesic constants for . ∎

Now, retracts to a spine of the manifold which can be identified with the bouquet , so is naturally quasi-isometric to the Cayley graph of . In a tree uniform quasi-geodesics are uniformly close to their geodesic representatives.

By Lemma 4.7 is quasi-geodesic in , and it contains a high power of the core curve of . Hence its retraction to the tree contains a high power of some representative of the curve . This subword must be uniformly close to a segment of the form , where is a cyclically reduced representative of . It follows that the geodesic representative of in the tree is uniformly close to a segment of this form as well. Since a tree is 0-hyperbolic, it follows that the geodesic representative of in the tree actually contains a high power of .

But since is blocking, this means that cannot have been primitive. We conclude that all primitive geodesics are trapped in a fixed compact core of . The retraction of this core to the spine of therefore lifts to a quasi-isometry, and stability of quasigeodesics again implies that each primitive geodesic in the tree is uniformly quasi-geodesic. ∎

###### Proof of Theorem 1.1.

In view of Lemma 3.2 and Theorem 3.3, all that remains to prove is the existence of a non-Schottky primitive stable representation. By Theorem 4.1, any discrete, faithful and geometrically finite representation with a single blocking cusp would do.

Given a curve on the boundary of a handlebody , a sufficent condition that it be realizeable as the single cusp of a geometrically finite representation is that it be homotopically nontrivial in the handlebody, and that any homotopically nontrivial proper annulus in with a boundary component on must be parallel into . This is a consequence of Thurston’s geometrization theorem (see e.g. [26]) or of Maskit [34].

Otal [41] shows that any curve in the Masur Domain has complement satisfying these topological conditions (even more strongly, that is acylindrical, so there are no essential annuli at all in ), and hence can appear as the single cusp of a geometrically finite representation. By Lemma 4.6 such a curve is also blocking with respect to some generating set, and hence gives us a primitive-stable representation.

When is even the example of Lemma 4.4, where , also suffices, even though is not in the Masur domain. Note that we can explicitly construct Fuchsian representations for which is the unique cusp, and Lemma 4.4 provides the blocking property.

## 5. Further remarks and questions

Having established that acts properly discontinuously on , and that is strictly larger than , one is naturally led to study the dynamical decomposition of . In particular we ask if is the maximal domain of discontinuity, and what happens in its complement. We have only partial results in this direction.

### Outside PS

The polar opposite of the primitive-stable characters are the redundant characters , defined (after Lubotzky) as follows: is redundant if there is a proper free factor of such that is dense. Note that is -invariant. Clearly and are disjoint, by Lemma 3.2.

The set of representations with dense image is open (Lemma 2.1), and it follows (applying this to the factors) that is open. The action of on cannot be properly discontinuous, and in fact there is a larger set for which we can show this.

Let be the set of (conjugacy classes of) representations which are Schottky on every proper free factor. Hence by Lemma 3.2, and is still in the complement of . Let .

###### Lemma 5.1.

1. If then is dense in .

2. No point of can be in a domain of discontinuity for . Equivalently, any open invariant set in on which acts properly discontinuously must be contained in .

The proof is quite analogous to part (2) of Lemma 3.2. In fact we note that part (2) of Lemma 3.2 is an immediate corollary of Lemma 5.1 and Theorem 3.3.

###### Proof.

For (1), let and let be a proper free factor such that is not Schottky. We may assume has rank at least 2 since . Hence (as in the proof of Lemma 3.2) is approximated by representations with dense image. It follows that itself is approximated by representations dense on , so is dense in .

For (2), we will show that for every neighborhood of there is an infinite set of elements such that . Since is locally compact, this implies cannot be in any open set on which acts properly discontinuously.

Suppose first . Since is dense in , it suffices to consider the case that . Again let be a proper free factor on which is dense. We can assume that where is generated by one element . Now let be a sequence such that , and let be the automorphism that is the identity on and sends to . Note that has infinite order in . The number of powers of that take to any fixed neighborhood of itself goes to as , because . This concludes the proof for .

If , is empty. However, every has primitive element mapping to a non-loxodromic, so may be approximated by a representation sending a generator to an irrational elliptic. The same argument as above can then be applied to . ∎

We remark that is indeed strictly larger than , at least for even rank: If , represent the handlebody as an -bundle over a genus surface with one puncture, and let be a degenerate surface group with no accidental parabolics (i.e. every parabolic in is conjugate to the image of an element of , is discrete and faithful, and at least one end of the resulting manifold is geometrically infinite). The puncture cannot be in any proper free factor of , because it is represented by a curve on the boundary of the handlebody whose complement is incompressible. Hence the restriction of to any proper free factor has no parabolics. By the Thurston-Canary Covering Theorem [14] and the Tameness Theorem [1, 10], cannot be geometrically infinite, so it must be Schottky, and hence . On the other hand, every nonperipheral nonseparating simple curve on is primitive, and the ending lamination of a degenerate end can be approximated by such curves. Hence the uniform quasigeodesic condition fails on primitive elements, and cannot be in .

Note that this example can be approximated by elements of , as well as by (the latter by using Cusps are Dense [36] or the Density Theorem for surface groups [9]). This leads us to ask:

###### Question 5.2.

Is the interior of ?

In particular, in view of Lemma 5.1, a positive answer would imply

###### Conjecture 5.3.

is the domain of discontinuity of acting on .

We remark that, a priori, there may not be any such domain, i.e. there may be no maximal set on which the action is properly discontinuous.

### Rank 2

For the free group of rank 2, we have already seen that some of our statements are slightly different. In particular is empty since no one-generator subgroup of is dense. Moreover, is exactly the set for which every generator is loxodromic, and this is dense in since it is the complement of countably many proper algebraic sets.

Question 5.2 in particular asks, therefore, whether is dense. It is not clear (to the author) whether this is true, but there is some evidence against it (see below).

Another important feature of rank 2 is that the conjugacy class of the commutator of the generators, and its inverse, are permuted by automorphisms. It follows that the trace of the commutator is an -invariant function on , and one can therefore study level sets of this function.

The domain of discontinuity of was studied by Bowditch and Tan-Wong-Zhang [45]. Bowditch defines the following condition on , which Tan-Wong-Zhang call condition BQ:

1. is loxodromic for all primitive .

2. The number of conjugacy classes of primitive elements such that is finite.

They show, using Bowditch’s work, that acts properly discontinuously on the invariant open set . It is still unclear whether is the largest such set.

Note that condition (1) is equivalent to membership in . It is evident that , and it seems plausible that they are equal.

Note also that computer experiments indicate that the intersection of with a level set of the commutator trace function is not dense in the level set (see Dumas [16]). In particular the slice corresponding to trace consists of type-preserving representations of the punctured-torus group, i.e. those with parabolic commutator, and empirically it seems that in this slice coincides with the quasifuchsian representations (which are all primitive-stable too, by Theorem 4.1 combined with Lemma 4.4 ). Bowditch has conjectured that this is in fact the case, and this seems to be a difficult problem. At any rate this appears to be evidence against the density of .

### Ergodicity

The question of the ergodic decomposition of on is still open. Note, in rank 2, the decomposition must occur along level sets of the commutator trace function. In rank 3 and higher our observations indicate that the simplest possible situation is that, outside , the action is ergodic, which we pose as a variation of Lubotzky’s original question:

###### Question 5.4.

Let . Is there a decomposition of into a domain where the action is properly discontinuous, and a set where it is ergodic? More pointedly, does act ergodically on the complement of in ?

In Gelander-Minsky [20] we show that in fact the action on is ergodic and topologically minimal. So if for example and have measure 0, we would have a positive answer for the above question.

### Understanding PS(Fn)

It would also be nice to have a clearer understanding of the boundary of , and of which discrete representations contains.

From Lemma 3.2 we know that any discrete faithful representation with cusp curves that have compressible complement cannot be in . We’ve also mentioned the degenerate surface groups which are in but not .

If, however, is discrete and faithful without parabolics and is not Schottky, then it has an ending lamination which must lie in the Masur domain, and hence is blocking by Lemma 4.6. Hence it would be plausible to expect:

###### Conjecture 5.5.

Every discrete faithful representation of without parabolics is primitive-stable.

More generally, a discrete faithful representation has a possibly disconnected ending lamination, whose closed curve components are parabolics. All the examples we have considered suggest this conjecture:

###### Conjecture 5.6.

A discrete faithful representation of is primitive-stable if and only if every component of its ending lamination is blocking.

Note: since the first version of this article appeared, Jeon-Kim [24] gave a positive solution to Conjectures 5.5 and 5.6, using the work of Mj [38] on Cannon-Thurston maps.

It might also be interesting to think about which representations with discrete image (but not necessarily faithful) are primitive-stable. In [37] we construct many primitive-stable representations whose images uniformize knot complements. What properties of a marked 3-manifold correspond to primitive stability?

Another interesting question is:

###### Question 5.7.

How do we produce computer pictures of ?

For rank , the character variety has complex dimension 2, and one can try to draw slices of dimension 1. Komori-Sugawa-Wada-Yamashita developed a program for drawing Bers slices, which are parts of the discrete faithful locus [31, 30], and Dumas refined this using Bowditch’s work [16]. In particular what Dumas’ program is really doing is drawing slices of Bowditch’s domain BQ. If indeed , then this produces images of as well.

### Other target groups

The discussion can be extended to other noncompact Lie groups, with moderate success. Let us consider first the case of for all , where is the case we have been considering. The definition of is unchanged, and stability of quasigeodesics works in all dimensions in the same way. Lemmas 2.1 Theorem 3.3 still hold. However, Sullivan’s theorem (Theorem 2.2) equating Schottky representations with those in the interior of is no longer available. Schottky representations in higher dimensions can be replaced by convex-cocompact representations: discrete and faithful, with convex hull of the limit set having a compact quotient. Now the conclusions of Lemma 3.2 must be changed somewhat: A convex-cocompact representation is certainly still primitive-stable, but it is not clear that For a primitive-stable , the proof of Lemma 3.2 shows that restricted to each proper free factor is in , but not that it is convex-cocompact.

For , the natural embedding of in clearly preserves primitive-stability and non-discreteness, so it is still true that contains indiscrete representations in higher dimension (and hence dense ones, by Lemma 2.1).

The case of is slightly trickier. When is even, we have given an example of a blocking curve that is the boundary of a one-holed surface, and so Theorem 4.1 shows that a Fuchsian structure on this surface, which gives an element of is primitive-stable but not Schottky. However when is odd we have no such example, and it is unclear to me if contains indiscrete elements. For and Goldman [22] has described the domain of discontinuity for the trace slice, and proved ergodicity in its complement.

For other noncompact rank-1 semisimple Lie groups, namely isometry groups of the non-homogeneous negatively curved symmetric spaces, primitive stability can again be defined in the same way. The hyperbolic plane always embeds geodesically in such a space, with its full isometry group acting. This is not easily extracted from the literature but is well-known; see Mostow [40], Bridson-Haefliger [8, Chap. II.10] and Allcock [2] for the requisite machinery, or note that these spaces are just the real, complex, and quaternionic hyperbolic spaces, and the Cayley plane, in all of which one can restrict to a real subspace. Thus, in these cases our examples for can be used.

Other cases, such as higher rank semisimple groups, presumably require a rethinking of the definitions, but there is again no reason to think that the geometric decomposition of Lemma 2.1 should be the right dynamical decomposition for the action of .

A completely different picture holds in the setting of non locally connected groups. In the case of , with a non-Archimedean local field of characteristic , as well as for a tree , Glasner [21] has shown that acts ergodically on .

### Other domain groups

If we replace by for a closed surface , is replaced by , and the primitive elements are replaced by their natural analogue, the simple curves in the surface. The Schottky representations are replaced by the quasi-Fuchsian representations . We can define in a similar way, but now there is no good reason to think that is strictly larger than the quasi-Fuschsian locus. Indeed, every boundary point of can be shown not to lie in (nor in any domain of discontinuity for – see also Souto-Storm [42] for a related result), and this is because all parabolics are simple curves, and all ending laminations are limits of simple curves. It is still open as far as I know whether in fact is equal to ; this is closely related to (but formally weaker than) Bowditch’s conjecture in this setting.

One can also consider where is any fundamental group of a hyperbolic 3-manifold, and the dynamics are by ; see Canary-Storm [13]. In general, the more complicated the group, the weaker we should expect the correspondence between the geometric and dynamical decompositions. An extreme example is when is a non-uniform lattice in . In this case has positive dimension, while Mostow rigidity tells us that is a single point. On the other hand Mostow also tells us that is finite in this case, so that it acts properly discontinuously on the whole of .

## References

• [1] I. Agol, Tameness of hyperbolic 3-manifolds, Preprint, 2004, arXiv:math.GT/0405568.
• [2] Daniel Allcock, Reflection groups on the octave hyperbolic plane, Journal of Algebra 213 (1999), no. 2, 467 – 498.
• [3] L. Bers, Spaces of Kleinian groups, Maryland conference in Several Complex Variables I, Springer-Verlag Lecture Notes in Math, No. 155, 1970, pp. 9–34.
• [4] M. Bestvina, M. Feighn, and M. Handel, The Tits alternative for . I. Dynamics of exponentially-growing automorphisms, Ann. of Math. (2) 151 (2000), no. 2, 517–623.
• [5] B. H. Bowditch, Markoff triples and quasi-Fuchsian groups, Proc. London Math. Soc. (3) 77 (1998), no. 3, 697–736.
• [6] E. Breuillard and T. Gelander, On dense free subgroups of Lie groups, J. Algebra 261 (2003), no. 2, 448–467.
• [7] E. Breuillard, T. Gelander, J. Souto, and P. Storm, Dense embeddings of surface groups, Geom. Top. 10 (2006), 1373–1389.
• [8] M. R. Bridson and A. Haefliger, Metric spaces of non-positive curvature, Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], vol. 319, Springer-Verlag, Berlin, 1999.
• [9] J. Brock, R. Canary, and Y. Minsky, The classification of Kleinian surface groups II: the ending lamination conjecture, Preprint, arXiv:math.GT/0412006.
• [10] D. Calegari and D. Gabai, Shrinkwrapping and the taming of hyperbolic 3-manifolds, J. Amer. Math. Soc. 19 (2006), no. 2, 385–446 (electronic).
• [11] R. Canary, M. Culler, S. Hersonsky, and P. Shalen, Approximation by maximal cusps in boundaries of deformation spaces of Kleinian groups, J. Differential Geom. 64 (2003), no. 1, 57–109.
• [12] R. Canary and S. Hersonsky, Ubiquity of geometric finiteness in boundaries of deformation spaces of hyperbolic 3-manifolds, Amer. J. Math. 126 (2004), no. 6, 1193–1220.
• [13] R. Canary and P. Storm, Moduli spaces of hyperbolic 3-manifolds, preprint, 2009, arXiv:0911.1418.
• [14] R. D. Canary, A covering theorem for hyperbolic 3-manifolds and its applications, Topology 35 (1996), 751–778.
• [15] A. J. Casson and S. A. Bleiler, Automorphisms of surfaces after Nielsen and Thurston, Cambridge University Press, 1988.
• [16] D. Dumas, Bear: a tool for studying Bers slices of punctured tori
• [17] B. Farb, Relatively hyperbolic and automatic groups with applications to negatively curved manifolds, Ph.D. thesis, Princeton University, 1994.
• [18] A. Fathi, F. Laudenbach, and V. Poenaru, Travaux de Thurston sur les surfaces, vol. 66-67, Asterisque, 1979.
• [19] T. Gelander, On deformations of in compact Lie groups, Preprint, 2006, math.GR/0611830.
• [20] T. Gelander and Y. Minsky, The dynamics of on the redundant part of the character variety, In preparation.
• [21] Y. Glasner, A zero-one law for random subgroups of some totally disconnected groups, Preprint, 2009, arxiv:0902.3792v1.
• [22] W. M. Goldman, The modular group action on real -characters of a one-holed torus, Geom. Topol. 7 (2003), 443–486.
• [23] W. M. Goldman, An ergodic action of the outer automorphism group of a free group, Geom. Funct. Anal. 17 (2007), 793–805.
• [24] W. Jeon and I. Kim, Primitive stable representations of geometrically infinite handlebody hyperbolic 3-manifolds, preprint, 2010, arXiv:1003.2055.
• [25] T. Jørgensen, On discrete groups of Möbius transformations, Amer. J. of Math. 98 (1976), 739–49.
• [26] M. Kapovich, Hyperbolic manifolds and discrete groups, Progress in Mathematics, vol. 183, Birkhäuser Boston Inc., Boston, MA, 2001.
• [27] D. Kazhdan and G. Margulis, A proof of Selberg’s conjecture, Math. USSR Sb. 4 (1968), 147–152.
• [28] S. P. Kerckhoff, The measure of the limit set of the handlebody group, Topology 29 (1990), no. 1, 27–40.
• [29] E. Klarreich, Semiconjugacies between Kleinian group actions on the Riemann sphere, Amer. J. Math. 121 (1999), no. 5, 1031–1078.
• [30] Y. Komori and T. Sugawa, Bers embedding of the Teichmüller space of a once-punctured torus, Conform. Geom. Dyn. 8 (2004), 115–142 (electronic).
• [31] Y. Komori, T. Sugawa, M. Wada, and Y. Yamashita, Drawing Bers embeddings of the Teichmüller space of once-punctured tori, Experiment. Math. 15 (2006), no. 1, 51–60.
• [32] A. Lubotzky, Dynamics of actions on group presentations and representations, Preprint, 2008.
• [33] A. Marden, Outer Circles: An Introduction to Hyperbolic 3-Manifolds, Cambridge U. Press, 2007.
• [34] B. Maskit, Parabolic elements on Kleinian groups, Ann. of Math. 117 (1983), 659–668.
• [35] H. A. Masur, Measured foliations and handlebodies, Ergodic Theory Dynam. Systems 6 (1986), no. 1, 99–116.
• [36] C. McMullen, Cusps are dense, Ann. of Math. 133 (1991), 217–247.
• [37] Y. Minsky and Y. Moriah, Discrete primitive-stable representations with large rank surplus, preprint, 2010, arXiv:1009.6212.
• [38] Mahan Mj, Cannon-Thurston maps for Kleinian grups, Preprint, arXiv:1002.0996, 2010.
• [39] J. W. Morgan and P. B. Shalen, Valuations, trees, and degenerations of hyperbolic structures. I, Ann. of Math. (2) 120 (1984), no. 3, 401–476.
• [40] G. D. Mostow, Strong rigidity of locally symmetric spaces, Ann. of Math. Stud. no. 78, Princeton University Press, 1973.
• [41] J.-P. Otal, Courants géodésiques et produits libres, Thèse d’Etat, Université de Paris-Sud, Orsay, 1988.
• [42] J. Souto and P. Storm, Dynamics of the mapping class group action on the variety of characters, Geom. Topol. 10 (2006), 715–736.
• [43] J. Stallings, Whitehead graphs on handlebodies, Preprint, 1996.
• [44] D. Sullivan, On the ergodic theory at infinity of an arbitrary discrete group of hyperbolic motions, Riemann Surfaces and Related Topics: Proceedings of the 1978 Stony Brook Conference, Ann. of Math. Stud. 97, Princeton, 1981.
• [45] S. P. Tan, Y. L. Wong, and Y. Zhang, Generalized Markoff maps and McShane’s identity, preprint, 2005, math.GT:0502464.
• [46] J. H. C. Whitehead, On certain sets of elements in a free group, Proc. London Math. Soc. 41 (1936), 48–56.
• [47] by same author, On equivalent sets of elements in a free group, Annals of Math. 37 (1936), 782–800.
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Today I want to talk about how much I admire Karin. I have always wanted a big round table that would surround me with displays and that I can do all my work on. Also, those hairclips are the awesomest things in the world. Ahh... some people can be so lucky.

written by astrobunny \\ behind, cute, girl, hacker, karin, loli, screens, shangri-la

The realisticness of anime knows no bounds.

written by astrobunny \\ kuniko, shangri-la

Hajimari no asa ni hikaare.

This is by far a very very big favorite for me, particularly because it is always used to wrap up some really awesome finishing scene or cliffhanger, and also because I've gotten addicted to it, I just open youtube for the sake of listening to this song. Or maybe its just because Hojo Kuniko is so awesome.

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Area of a triangle IB Maths

Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths SL Revision Notes

Area of a triangle IB Maths

IB Mathematics SL – Trigonometry, Area of a triangle

How can we find the area of a triangle ABC with sides $a=4 \ cm, b=5 \ cm$

and included angle $\hat{C}=30^{\circ}$ ?

Thanks!
Jane

Posts: 0
Joined: Tue Mar 19, 2013 8:47 pm

Re: Area of a triangle IB Maths

IB Maths SL – Trigonometry, Area of a triangle

The area of a triangle is a half of the product of two sides and the sine of the included angle.

$A=\frac{1}{2}absinC$

Therefore in your case we have the following

$A=\frac{1}{2}(4)(5)sin30^{\circ} =10 \frac{1}{2}=5 \ cm^2$

Hope these help!!
miranda

Posts: 268
Joined: Mon Jan 28, 2013 8:03 pm

Re: Area of a triangle IB Maths

Thanks a lot!
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Joined: Tue Mar 19, 2013 8:47 pm


text:
# KeeLoq

KeeLoq is a proprietary hardware-dedicated block cipher that uses a non-linear feedback shift register (NLFSR). The uni-directional command transfer protocol was designed by Frederick Bruwer of Nanoteq (Pty) Ltd., the cryptographic algorithm was created by Gideon Kuhn, and the silicon implementation was by Willem Smit at Nanoteq Pty Ltd (South Africa) in the mid 80s. KeeLoq was sold to Microchip Technology Inc in 1995 for \$10 million.[1] It is used in "code hopping" encoders and decoders such as NTQ105/106/115/125D/129D and HCS101/2XX/3XX/4XX/5XX. KeeLoq is or was used in many remote keyless entry systems by such companies as Chrysler, Daewoo, Fiat, GM, Honda, Toyota, Volvo, Volkswagen Group, Clifford, Shurlok, Jaguar, etc.[2]

## Description

KeeLoq Encryption

KeeLoq "code hopping" encoders encrypt a 0-filled 32-bit block with KeeLoq cipher to produce a 32-bit "hopping code". A 32-bit initialization vector is linearly added (XORed) to the 32 least significant bits of the key prior to encryption and after decryption.

KeeLoq cipher accepts 64-bit keys and encrypts 32-bit blocks by executing its single-bit NLFSR for 528 rounds. The NLFSR feedback function is 0x3A5C742E or

$F(a,b,c,d,e) = d \oplus e \oplus ac \oplus ae \oplus bc \oplus be \oplus cd \oplus de \oplus ade \oplus ace \oplus abd \oplus abc$

KeeLoq uses bits 1, 9, 20, 26 and 31 of the NLFSR state as its inputs during encryption and bits 0, 8, 19, 25 and 30 during decryption. Its output is linearly combined (XORed) with two of the bits of the NLFSR state (bits 0 and 16 on encryption and bits 31 and 15 on decryption) and with a key bit (bit 0 of the key state on encryption and bit 15 of the key state on decryption) and is fed back into the NLFSR state on every round.

## Attacks

### Replay attack

For simplicity, individual "code hopping" implementations do not use cryptographic nonces and clock drift excludes the possibility of using timestamping. This makes the protocol inherently vulnerable to replay attacks: For example, by jamming the channel while intercepting the code, a thief can obtain a code that may still be usable at a later stage.[3] It made KeeLoq "code grabbers"[4] quite popular among most car thieves.[who?][citation needed][5]

### Cryptanalysis

KeeLoq Decryption

KeeLoq was first cryptanalyzed by Andrey Bogdanov using sliding techniques and efficient linear approximations. Nicolas Courtois attacked KeeLoq using sliding and algebraic methods. The attacks by Bogdanov and Courtois do not pose any threat to the actual implementations that seem to be much more vulnerable to simple brute-force of the key space that is reduced in all the code-hopping implementations of the cipher known to date. Some KeeLoq "code grabbers" use FPGA-based devices to break KeeLoq-based keys by brute force within about two weeks due to the reduced key length in the real world implementations.[citation needed]

In 2007, researchers in the COSIC group at the university at Leuven, Belgium, (K.U.Leuven) in cooperation with colleagues from Israel found a new attack against the system.[6] Using the details of the algorithm that were leaked in 2006, the researchers started to analyze the weaknesses. After determining the part of the key common to cars of a specific model, the unique bits of the key can be cracked with only sniffed communication between the key and the car.

Microchip introduced in 1996[7] a version of KeeLoq ICs which use a 60-bit seed. If a 60-bit seed is being used, an attacker would require approximately 100 days of processing on a dedicated parallel brute force attacking machine before the system is broken.[8]

### Side-channel attacks

In March 2008, researchers from the Chair for Embedded Security of Ruhr University Bochum, Germany, presented a complete break of remote keyless entry systems based on the KeeLoq RFID technology.[9][10] Their attack works on all known car and building access control systems that rely on the KeeLoq cipher.

The attack by the Bochum team allows recovering the secret cryptographic keys embedded in both the receiver and the remote control. It is based on measuring the electric power consumption of a device during an encryption. Applying what is called side-channel analysis methods to the power traces, the researchers can extract the manufacturer key from the receivers, which can be regarded as a master key for generating valid keys for the remote controls of one particular manufacturer. Unlike the cryptanalytic attack described above which requires about 65536 chosen plaintext-ciphertext pairs and days of calculation on a PC to recover the key, the side-channel attack can also be applied to the so-called KeeLoq Code Hopping mode of operation (AKA rolling code) that is widely used for keyless entry systems (cars, garages, buildings, etc.).

The most devastating practical consequence of the side-channel analysis is an attack in which an attacker, having previously learned the system's master key, can clone any legitimate encoder by intercepting only two messages from this encoder from a distance of up to 100 metres (330 ft). Another attack allows to re-set the internal counter of the receiver (garage door, car door, etc.) which makes it impossible for a legitimate user to open the door.

## References

1. ^ US patent 5517187, Bruwer, Frederick J.; Smit, Willem & Kuhn, Gideon J., "Microchips and remote control devices comprising same", issued 1996-05-14, assigned to Microchip Technology Inc
2. ^ Some evidence that Chrysler indeed uses KeeLoq can be found in (this video).
3. ^ Analysis of RF Remote Security Using Software Defined Radio
4. ^ http://www.microchip.com/stellent/idcplg?IdcService=SS_GET_PAGE&nodeId=2075&param=en001022#P108_5361 stating, "It is a simple matter to build a circuit to record such transmissions for reply at the later time. Such a system is known as a code or key grabber."
5. ^ http://www.snopes.com/autos/techno/lockcode.asp
6. ^ How To Steal Cars — A Practical Attack on KeeLoq
7. ^ (Will be in web archive backup later): a Microchip press release on Dec 11, 1996 Quote: "...HCS410 KEELOQ Code Hopping Transponder and Encoder..."
8. ^ Martin Novotny and Timo Kasper. Cryptanalysis of KeeLoq with COPACOBANA. SHARCS 2009 Conference. pp. 159–164.
9. ^ A complete break of the KeeLoq access control system
10. ^ Thomas Eisenbarth, Timo Kasper, Amir Moradi, Christof Paar, Mahmoud Salmasizadeh, and Mohammad T. Manzuri Shalmani (2008-02-29). Physical Cryptanalysis of KeeLoq Code Hopping Applications. Ruhr University of Bochum, Germany. Retrieved 2009-03-22.


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# Find the density function of a random variable

A discrete random variable $X$ has the density function $f$ given by $f(-1) = \dfrac 2 {10}, f(0) = c, f(1) = \dfrac 2 {10}, f(2) = \dfrac 1 {10}$.

a) Determine $c$
b) Find the distribution function $F$
c) Show that the random variable $Y=X^2$ has the density function $g$ given by $g(0) = \dfrac 5 {10}, g(1) = \dfrac 4 {10}, g(4) = \dfrac 1 {10}$.

I am having troubles with part C), help would be greatly appreciated! Thanks!

What seems to be the problem. Since it is discrete, for $(x = 0, y = 0), (x = -1, y =1), (x = 1, y = 1)$ and $(x = 2, y = 4)$. Then it follows $g(0) = f(0), g(1) = f(-1)+f(1)$, and $g(4) = f(2)$. That gives the distribution
$g(0) = 0.5$, $g(1) = 0.4$, $g(4) = 0.1$


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# Blog

• PWC 047 › Roman Calculator
The first challenge this week tasks us with evaluating arithmetic operations using Roman numerals. The description seems to indicate that only one operation will be given (for example, II + IV), but I have elected to support arbitrary arithmetic expressions. First, I needed a way to convert to and from Roman and Arabic numerals. I … Read more
• PWC 047 › Gapful Numbers
Task 2 this week has us print the first 20 "gapful numbers," as described by OEIS sequence A108343. Gapful numbers are numbers greater than 99 that are evenly divisible by their first and last digits combined. For example, 132 is a gapful number because 132 ÷ 12 = 11. This is certainly the easier of … Read more
• PWC 046 › Cryptic Message
Challenge #1 this week is the following: The communication system of an office is broken and message received are not completely reliable. To send message Hello, it ended up sending these following: H x l 4 ! c e – l o z e 6 l g H W l v R q 9 m # … Read more
• PWC 046 › Is the Room Open? (500 Doors)
The second of the challenges this week poses the following question (paraphrased): Suppose we have 500 doors, and 500 employees. The first employee opens all the doors. The second employee closes every 2nd door (doors 2, 4, 6, … 500). The third employee closes every third door if it is open, or opens it if … Read more
• Quine
Challenge #2 this week asks for "a script that dumps its own source code". This is almost a quine, although Mohammad did not name it as such. Specifically, we are given the constraint that perl ch-2.pl | diff – ch-2.pl must return nothing. What's a Quine? A quine, otherwise known as a self-replicating program, is … Read more
• Square Secret Code
Task #1 this week is a simple cipher, described as follows: The square secret code mechanism first removes any space from the original message. Then it lays down the message in a row of 8 columns. The coded message is then obtained by reading down the columns going left to right. Given "The quick brown … Read more
• Make it $200 For challenge #2 this week, the task is to start with$1, and by either adding $1 or doubling the amount, reach$200 in the smallest possible number of steps. "Greedy never works" In a 75-minute lecture some decades ago, my Advanced Algorithms professor said, over and over, "greedy never works," while all the while … Read more
• Only 100, please
Challenge #1 this week is as follows: You are given a string “123456789”. Write a script that would insert ”+” or ”-” in between digits so that when you evaluate, the result should be 100. I am going to add the additional constraint that we want all possible solutions, because that is a much stronger statement, and not that difficult to … Read more
• Self-descriptive Numbers
Challenge #2 this week (43) is to generate Self-descriptive Numbers in arbitrary bases. A self-descriptive number, as described by Wikipedia, is an integer m that in a given base b is b digits long in which each digit d at position n (the most significant digit being at position 0 and the least significant at position b – 1) counts how many instances of digit n are … Read more
• Olympic Rings
Challenge #1 this week (43) is a number puzzle. In short, the task is to fill in the numbers 1, 2, 3, 4, and 6 into the spaces within the intersecting Olympic rings, so that the numbers in each ring sum to 11. Some of the spaces are already filled in (given). Here is the … Read more
• Balanced Parentheses
The second challenge this week is another old chestnut: Write a script to generate a string with random number of ( and ) brackets. Then make the script validate the string if it has balanced brackets. The question has two parts: Generate strings with random brackets There are many ways to go about this task. However, I'm opting to … Read more
• Octal Representation
Challenge 1 this week is an easy one: Write a script to print decimal number 0 to 50 in [the] Octal Number System. Perl and Raku We can solve this with the following polyglot (runs in both languages at once): printf "Decimal %2d = Octal %2o\n", $_,$_ for 0..50; Still, in Raku, we can do … Read more
• Leonardo Numbers
Happy new year! We are on Week 41, and this is Challenge #2. The Leonardo Numbers (A001595) are a simple recursively defined sequence: $$L(n) = \begin{cases}1 & \text{if } n \lt 2 \\1 + L(n – 1) + L(n – 2) & \text{if } n \geq 2\end{cases}$$ The sequence starts: 1, 1, 3, 5, 9, … Read more
• Attractive Numbers
This week, Mohammad asks us to output a list of all Attractive Numbers between 1 and 50. Attractive numbers, as described by the Online Encyclopedia of Integer Sequences (OEIS) are: Numbers with a prime number of prime divisors (counted with multiplicity) OEIS Sequence A063989 The first numbers are 4, 6, 8, 9, 10, 12, 14, … Read more
• Zip6
The zip6 function has long been a staple of the List::MoreUtils CPAN module. The Week 40 challenge #1 describes it very well. But even more succinctly: zip6 takes an array of arrays (AoA) and returns another AoA of all the 1st elements, then the 2nd, and so on. It's not a difficult algorithm by any … Read more
• Reverse Polish Notation
This is my first blog post regarding the Perl Weekly Challenge tirelessly maintained by Mohammad Anwar. Although I've been doing the challenge for a while now, I haven't maintained my blog for years. Let's change that! This week's challenge #2 from the Perl Weekly Challenge is a computer science classic: create a Reverse Polish Notation … Read more


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RUS  ENG JOURNALS   PEOPLE   ORGANISATIONS   CONFERENCES   SEMINARS   VIDEO LIBRARY   PERSONAL OFFICE
 Forthcoming seminars Seminar calendar List of seminars Archive by years Register a seminar Search RSS Forthcoming seminars

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Seminar on Complex Analysis (Gonchar Seminar)
November 17, 2014 18:00, Moscow, Steklov Mathematical Institute, Room 411 (8 Gubkina)

Order problem for canonical systems and a conjecture of Valent.

R. V. Romanov

Saint Petersburg State University

Abstract: A precise upper estimate for the exponential order of the monodromy matrix of $2\times 2$ canonical systems is obtained. The estimate is shown to become an equality in the case of the Krein strings. As an application we prove a conjecture by Valent on the order for a certain class of indeterminate moment problems with polynomial Jacobi parameters related to birth–death processes.


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 mixnets× 4 routing protocols that create hard-to-trace communications by using a chain of proxy servers known as mixes which take in messages from multiple senders, shuffle them, and send them b… ocb× 4 The Offset CodeBook Mode, an authenticated encryption mode of operation for a block cipher. montgomery× 4 resources× 4 material that can be referenced for further information on cryptography. simon× 4 A family of lightweight symmetric block-ciphers designed for good performance in hardware with block sizes ranging from 32 to 128 bits and key sizes ranging from 64 to 256 bits. dual-ec-drgb× 4 short for Dual Elliptic Curve Deterministic Random Bit Generator; a pseudorandom number generator based on the elliptic curve discrete logarithm problem. information-theory× 4 concerned with sending messages via electronic signals in the most efficient and error-free way. key-check-value× 4 ipsec× 4 asn.1× 4 a language that defines the way data is sent across dissimilar communication systems. camellia× 4 a 128-bit, symmetrical block cipher jointly developed by Mitsubishi and NTT of Japan. cfb× 4 an encryption mode, that builds a self-synchronizing stream-cipher from a block-cipher. constants× 4 Fixed, more or less arbitrary parameter values that appear in the definitions of cryptographic schemes, such as fixed initialization vectors for hash functions or S-box tables for block ciphers. decryption-oracle× 3 biclique-attack× 3 bijection× 3 a function $f$ from a set $X$ to a set $Y$ with the property that, for every $y$ in $Y$, there is exactly one $x$ in $X$ such that $f(x) = y$. It follows from … alternating-step× 3 a cryptographic pseudorandom number generator intended to be used in a stream cipher. avalanche× 3 evident if, when an input is changed slightly (for example, flipping a single bit) the output changes significantly (e.g., half the output bits flip). kbkdf× 3 a key derivation function ([tag:kdf]) that uses a key in the computation. embedded× 3 a computer system with a dedicated function within a larger mechanical or electrical system, often with real-time computing constraints. encoding× 3 the act of transforming information from one form to another. In cryptography, encoding is typically used to refer to an unkeyed process that does not provide any security in itself, but j… schnorr-identification× 3 The Schnorr Identification Protocol relies upon the security of the Discrete Logarithm Problem. Schnorr's protocol was introduced after, and is comparable to, the identification protocol of Fiat and S… plausible-deniability× 3 Plausible deniability may refer $1)$ to deniable encryption schemes allowing to decrypt a ciphertext for a message $m$ to some distinct message $m'$ or $2)$ to a feature provided by deniable file syst… predicate-encryption× 3 an encryption paradigm which gives a master secret key owner fine-grained control over access to encrypted data. onion-routing× 3 A method for anonymous communications over a wide area network such as the Internet. key-generation× 3 the process of generating keys for cryptographic purposes. universal-composability× 3 The framework of universal composability (UC) is a general-purpose model for the analysis of cryptographic protocols. three-pass-protocol× 3 The three-pass protocol uses commutative encryption to let two people communicate privately without having to exchange keys in advance. tea× 3 a block cipher by David Wheeler and Roger Needham of the Cambridge Computer Laboratory; notable for its simplicity of description and implementation. u-prove× 3 a Crypto SDK was developed by Microsoft to support experimentation with the foundational features of the U-Prove technology. It is made available under the Apache 2.0 open-source license, w… xml-encryption× 2 Encryption of data represented in XML according to the W3C Specification. speck× 2 A family of lightweight symmetric block-ciphers designed for good performance in constrained software devices with block sizes ranging from 32 to 128 bits and key sizes ranging from 64 to 256 bits. s-mime× 2 Secure/Multipurpose Internet Mail Extensions, or short S-MIME, is a standard for public key encryption and signing of MIME data. voip× 2 a technology that allows telephone calls to be made over computer networks like the Internet. key-escrow× 2 key-rotation× 2 Key rotation refers to a schedule or process for changing the key material.


text:
How to configure simplex method to start from a specific point

If I have a linear programming problem e.g. $$\max 2x_1 + x_2$$ with these constraints

$$x_1-2x_2 \leq 14$$ $$2x_1-x_2\leq 10$$ $$x_1-x_2 \leq 3$$

And I want to solve the problem starting from a specific point e.g. $A=(x1=5, x2=0)$ and $B=(x1=0, x2=5)$ .

How should I configure the initial tableau to allow to iterate simplex method starting from these points?

(Note that in standard form when I add the slack variables, the points must be calculated in function of the extra variables.)

• Are these two separate starting points to set up? That is, for the first one you wish to start with $x_1=5,x_2=0$? – coffeemath Feb 4 '16 at 23:35
• A=5,0 is a single starting point – AndreaF Feb 4 '16 at 23:35
• That's what I meant, and also $B=(0,5)$ would be another starting point. What version of tableaus are you to use? Just thought, maybe you mean 5,0 as a decimal like usual 5.0 (some British use comma for decimal) – coffeemath Feb 4 '16 at 23:37
• I usually convert the problem in the standard form and use the 2-phases method. No comma is a separator x1=5 x2=0 – AndreaF Feb 4 '16 at 23:41

One way would be, for a start at (5,0), to set up a new variable $x_1'$ and put the original $x_1$ equal to $5+x_1'$ in the objective and constraints. Then the usual simplex, starting at $(0,0)$ in the $x_1',x_2$ variables, would in effect be starting at $(5,0)$ in the $x_1,x_2$ variables.
• you mean a new constraint $$x6=5+x1$$ that become $$-x1+x6=5$$ to start from $$x1=5$$ and $$x2=0$$ ??? Note: I have used $$x6$$ because $$x3$$, $$x4$$ and $$x5$$ are needed for slack Have I well understand? – AndreaF Feb 4 '16 at 23:51
• Actually I just meant to substitute say $x_1=5+x_6$ into everything in the original objective and constraints, and then work from there, i.e. (since you want other lower index vars for slacks) the new system would be all in terms of $x_6,x_2$ rather than in $x_1,x_2,$ and would after the substitutions have different looking objective and constraints. The way I'm thinking of, there would be no constraint involving $x_1$ and $x_6,$ since $x_1$ would no longer appear in the "new" problem. – coffeemath Feb 5 '16 at 0:08
• So my objective function will be $$2(5+x6)+x2$$, in the same way constraints eg. the first $$x1−2x2≤14$$ becomes $$5+x6−2x2≤14$$. Now is right? And when I have to put objective function in cost row how to deal with the constant introduced? I'm a bit confused about this approach, In addition I have just noted that $$A=(5,0)$$ in the standard form after that I put the slack becomes $$A=(5,0,9,0−2)$$ while the point $$B=(0,5)$$ becomes $$B=(0,5,4,10,8)$$ if you write the first tableau would be helpful. – AndreaF Feb 5 '16 at 2:20


text:
# Week 4¶

## Class 1¶

Monday, Oct. 1. Supervised by Ms. Healey.

We have decided to do a midterm project about attacking an image recognition neural network for better understanding of the how images, GPU and matrices work. The goal of the project is to have a neural network to recognize the picture of a dog as a tree-frog while humans still recognizing the picture as a dog.

Target image.

This image has been recognized as a German Shepherd with 53% of confidence with the neural network. We cannot modify the neural network. All we can do is to modify the image. This project, in addition to being useful to computer graphics, is very useful in real life in that it is a simple manifest of the possible attacks on the image recognition networks, such as the governmental facial recognition network or the street signs recognition network on a auto-piloted car. This is partly the reason why we picked this as our midterm project.

## Class 2¶

Wednesday, Oct. 3. Supervised by Ms. Healey.

Continuing researching common vulnerabilities of Convolutional Neural Network (CNN). The first method we looked into was one-pixel attack. This is a promising and very common method to attack neural network. However, one pixel attack is designed for non-targeted attack. A non-targeted attack differs from targeted attack in the primary goal of the attacks. In a non-targeted attack, the primary goal is to make the neural network recognize objects wrongly. There isn’t a requirement on what it should end up being. Take our project as an example, a non-targeted attack would require the distorted image to be recognized as something that’s not dog, be it an electric pole or a frog or an apple. A targeted attack would ask require the image to be recognized as a specific object, and in our case a tree-frog. It may seem that targeted attack is more powerful than a non-targeted attack and that there is no reason to do a non-targeted attack. In fact, targeted attack is a subset of non-targeted attack and is indeed more powerful. However, non-targeted attack require very minimal change in the target image and may require, as the name of one pixel attack suggests, only changing a single pixel in an image.

We have tried to fit the one-pixel attack on our model, and it worked…well, kind of. The confidence of recognizing the image as a tree-frog increased dramatically –– it went from about $$10^{-9}%$$ to 0.13%, after running the program for 2 hours. There wasn’t a significant increase in confidence after it hit 0.13%. Obviously that’s not enough. So, failed.

The second method we looked into was using clever-hans. This didn’t work because, our targeted neural network is a mobile net, which isn’t compatible with the cleverhans library. The characteristic of mobilenet is that all output images are of fixed size, in our case $$241\times 241$$ pixels. We even attempted to modify the clerver-hans library so it fit the dimension of the image. However after many modifications in the library it still doesn’t work correctly –– the program wouldn’t even run. After fixing errors after errors for at least 5 hours, we decided to give up.


text:
# Understanding policy update in PPO2

I have a question regarding the functionality of the PPO2 algorithm together with the Stable Baselines implementation:

From the original paper I know that the policy parameters $$\theta$$ are updated K-times using the steps sampled (n_env * T steps):

When updating the policy parameters for a state $$s_t$$, are only the state observations $$a_t$$ and reward $$r_{t+1}$$ of this step considered, or also the state observations and rewards of the following steps ($$t+1$$) considered? My understanding is that the policy update with stochastic gradient ascent works just like in supervised learning.

I know that PPO2 uses a truncated TD($$\lambda$$) approach (T timesteps considered). So I guess that during the policy update for each state, subsequent states are only considered through the advantage function $$A_t$$ but not through the values of subsequent state observations and rewards themselves? Is that true?

I do not quite get the Stable Baselines implementation in the method _train_step() of the PPO2 implementation so therefore the question here.


text:
# BicgStab is not able to solve while Jacobi or GaussSeidel Methods can

I am trying to solve the 2D laplace equation,

$$\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0; \qquad 0 \lt x \lt 1, \quad0 \lt y \lt 1$$

Subjected to the boundary conditions,

$${\frac{d T}{d x}}|_{x=0,y}=0;$$

$${\frac{d T}{d y}}|_{x,y=0}=0;$$

$${\frac{d T}{d y}}|_{x,y=1}=0;$$

$$T|_{x=1,y} = 1;$$

I am using Finite Difference method to discretise the differential equation and foward difference in the first two B.Cs and backward difference for the 3rd B.C. The solution which is $$T=1$$ at all grid points is perfectly coming out in Gauss Seidal as well as Jacobi iterative methods, but using BicgStab algorithm it is not converging even though the same matrices as Jacobi or Gauss Seidal. Any Ideas may help. Thank you.

• What do you mean by not converging? Is it diverging, is it slow, is it stalled? It should converge if properly implemented. – EMP Sep 26 '19 at 21:02
• Without additional info, I strongly suspect the implementation of the iterative solver (or the call to it) has a problem. Is it you custom BiCGStab implementation? – Anton Menshov Sep 26 '19 at 21:38
• @EMP @ Anton; Thanks for your replyYes the was fluctuating rapidly. Now it is converging. The discretization of the laplace equation results into a Negative Definite Matrix A, so by multiplying the equation by $-1$ made it to a Positive Definite one, (and the RHS zero). The PD matrix with negative off diag terms and positive diagonal term along with the appropriate b.c.s converged rapidly. I think I read somewhere that BiCGSTAB works only for Positive Definite Symmetric or Asymmetric matrices. – Ark101 Sep 28 '19 at 16:14


text:
# Is it possible to know where the Long March 5B will be landed approximately?

I have read some articles about the uncontrolled space rocket the Long March 5B including some space tracker websites but none of them give any useful information about the rocket landing.

Is it possible theoretically to know where and when the Long March 5B will be landed approximately?

• @uhoh "We don't know yet" is exactly the correct answer, for now, and probably until Friday (or later). May 6, 2021 at 3:19
• I’m voting to close this question because it was asked prematurely. The correct answer for now is we don't know. The current answer is essentially we don't know. That answer will no longer be correct a day or two or three from now because there will be a predicted reentry point that will be fairly accurate by then. May 6, 2021 at 5:17
• Probably in the ocean as ocean covers the majority of the planet May 6, 2021 at 7:27
• If the user were to post a duplicate of the question later, would that really change the answer to "Is it possible to know...?" The booster has landed, now we can close the question because it was asked too late.
– Greg
May 9, 2021 at 10:34
• @GAD3R, Scott Manley posted a video on 5/8 about the booster. Perhaps you would find it interesting. youtube.com/watch?v=afGFmAljL5E
– Greg
May 10, 2021 at 18:22

## 3 Answers

Is it possible theoretically to know where and when the Long March 5B will be landed approximately?

# Yes!

Take it from Jonathan McDowell @planet4589's tweet:

The EU SST prediction is narrow enough that we can start to talk about location now!

In particular, none of the orbits within the predicted window cross the northwestern US or the mid Atlantic seaboard.

More generally:

At the time of asking, No, but as we get closer the answer increasingly becomes Yes.

We can see from https://aerospace.org/reentries/cz-5b-rocket-body-id-48275 that the uncertainty in the time of reentry is currently1 +/- 8 hours. About 2.1 days ago it was +/- 21 hours.

1actually about 7 hours ago 07 May 2021 18:55:50.470 UTC the latest updated prediction posted on the site as of the time of writing.

Orbit Epoch                  07 May 2021 18:55:50.470 UTC
Predicted Reentry Time       09 May 2021 04:19        UTC ± 8 hours

Contributors to uncertainty include variations in the density of the Earth's atmosphere primarily around 150 km and below (where the spacecraft does most of its velocity loss) and the aerodynamic drag of a rocket body of unknown orientation (probably tumbling).

Past performance is not indicative of future results

CNN’s Kate Bolduan: It seems even the smartest people aren’t able to calculate where this is going to land, yet. Why can’t they?

Ret. US Astronaut2 Scott Kelley: Well it’s because the satellite is spinning, and when something is spinning like that it doesn’t have a stable trajectory; it’s hard… and so the drag is always changing.

2 cmdr STS-118, EXP-26, EXP-45, EXP-46

click for larger

left: "Doppler variations with a period of 4.5 seconds detected by radar MFDR-LR during the pass, which help infer the rotation of the object" right: "Signal-to-Noise Ratio (SNR): the ratio of the signal power to the noise power detected by radar BIRALES during one pass, with periodicity of the peaks visible about every 2.5 seconds"

The orbital period is about 88 minutes now, so an uncertainty of 8 hours means that currently one can only narrow down the location to a large swath of the Earth (roughly half of the Earth's surface) where it will happen.

As we get close, this will be come narrowed down first in terms of what part of the Earth the orbits cover (each orbit moves by about 22 degrees in longitude).

So when the uncertainty narrows to +/- 1.5 hours you'll know it will reenter +/- 22 degrees in longitude from the next-to-lat orbit and +/- 41.5 degrees in latitude, since that's the inclination of the orbit and it can't go (much) farther north or south than that.

Further: if you want to consider probabilities, those orbits spend most of their times far from the equator, so the most likely areas are concentrated between say 30 and 41.5 degrees both sides of the equator.

above: Source captured at 2021-05-08 02:00 UTC below: Source captured at 2021-05-08 04:00 UTC

### We can talk latitude probabilities any time!

From the pros, from the Spaceflight 101 article Tiangong-1 Re-Entry, click for full size:

and

Even more interesting is the time-binned histogram rescaled by $$1/ \cos(\lambda)$$ for surface area rather than latitude, as recommended by @Litho's comment. If you were looking for debris, or looking to avoid getting hit by debris personally, this would be the plot for you.

• This seems a fine answer to me. Why the downvote?
– user39728
May 8, 2021 at 2:28
• Ah, OK. Good answers often start as stubs, though. Maybe we should mark them as stubs when we write them so others know "work in progress"? Wouldn't want to discourage any answer simply because the poster happened to not have the time to finish just yet (or the references to throw in, or all thoughts firmly collected). You know how it goes---if you postpone it now waiting for the right time to write it, you might never come back around to writing it.
– user39728
May 8, 2021 at 2:53
• We shouldn't be using downvotes as retribution to others who downvoted us.
– user39728
May 8, 2021 at 3:07
• @user39728_i_said_user_39728_i_ those are four close votes (not down votes) and the effect of five is to prevent anybody in the community from posting an answer until five reopen votes are then collected, and that can take days or a week or more sometimes, and other times it may never happen.
– uhoh
May 8, 2021 at 3:15
• Yea, it was at -1 when it was just a stub. (IIRC) @user39728_i_said_user_39728_i_ It was marked as a stub when originally posted. (For the record: not a down or close voter) May 8, 2021 at 3:16

You need to rephrase your question... Give a time frame?

From right after launch, can we predict where it will re-enter? No. Too many variables. (Height and density of earth's atmosphere).

Right the second before it hits the Earth? Yes.

In between? It depends, and it gets more accurate the closer to re-entry it gets.

Side note: Not landing, that is what SpaceX does with Falcon and Starship. This is uncontrolled re-entry.

• Here's a prediction, but it's currently +/- 21 hours. It will get refined as time gets closer aerospace.org/reentries/cz-5b-rocket-body-id-48275 You know the drill: "authoritative yet unsupported answers get -1, edit adding links/supporting sources get +1
– uhoh
May 6, 2021 at 0:43
• @uhoh Upvoting just to counteract your downvote. You should have voted to close the question. May 6, 2021 at 5:19
• @DavidHammen I always vote the correct way, by definition! :-)
– uhoh
May 6, 2021 at 6:41
• But I think you can run a monte carlo simulation with a good drag model and wind model along with estimates of state vector data... and get a range of places where the debris is likely to fall... You wouldn't get a precise location with great certainty, but you could get an approximate location with a statistical degree of confidence---and that's not too bad.
– user39728
May 8, 2021 at 2:25

# Yes, it is now.

The rocket has fallen down now and the approximate location where it fell is known.

According to the China Manned Space Agency (as reported on their official website and quoted by multiple news agencies worldwide):

The debris of the last stage of the Long March-5B Y2 carrier rocket reentered the atmosphere at 10:24 a.m. on Sunday (Beijing Time) […] the landing area of the debris is around a sea area with the center at 2.65 degrees north latitude and 72.47 degrees east longitude […].

This location lies in the Indian Ocean, just west of the Maldives islands.


text:
It is a usual pattern when creating a new component, to load references to other components of the same GameObject. Here is an extreme case:

    [RequireComponent(typeof(Animator))]
[RequireComponent(typeof(SpriteRenderer))]
[RequireComponent(typeof(Stats))]
[RequireComponent(typeof(Hunger))]
[RequireComponent(typeof(Thirst))]
[RequireComponent(typeof(Navigator))]
[RequireComponent(typeof(LocationDevice))]
[RequireComponent(typeof(AttacksManager))]
[RequireComponent(typeof(ShieldEffect))]
public class EnemyController : MonoBehaviour
{
Animator animator;
SpriteRenderer spriteRenderer;
Stats stats;
Hunger hunger;
Thirst thirst;
Navigator navigator;
LocationDevice locationDevice;
AttacksManager attacksManager;
ShieldEffect shieldEffect;

void Awake()
{
animator = GetComponent<Animator>();
spriteRenderer = GetComponent<SpriteRenderer>();
stats = GetComponent<Stats>();
hunger = GetComponent<Hunger>();
thirst = GetComponent<Thirst>();
navigator = GetComponent<Navigator>();
locationDevice = GetComponent<LocationDevice>();
attacksManager = GetComponent<AttacksManager>();
shieldEffect = GetComponent<ShieldEffect>();
}
}

Is there a way to write this better?

• Greetings, we don't review stub code on CodeReview, but actual code. Please place some actual code in your question. Otherwise this question will be put on hold pretty quickly. – konijn Nov 5 '19 at 17:55
• @konijn Ok, corrected – Enrique Moreno Tent Nov 5 '19 at 17:58
• Thanks! Also, sorry! I was not clear enough, this would not compile and still does not fit the guidelines, can you put the entire EnemyController  class ? – konijn Nov 5 '19 at 18:06
• Now it does compile. Notice that I am not really asking a review on the code itself, but on a pattern that repeats itself a lot in my daily routine – Enrique Moreno Tent Nov 5 '19 at 21:55
• Got ya, thanks! It's a CodeReview community thing, I am looking forward to the review of this. – konijn Nov 5 '19 at 21:58


text:
# Area of a polygon given a set of points

## Introduction

There are many ways to calculate the area of a polygon. Area of a well-known polygon like Triangles, Rectangles, Squares, Trapezoids, etc. can be calculated using simple mathematical formula. In this post, I talk how to calculate the area of a polygon given the set of vertices. The method discussed here is applicable for most of the polygons without holes.

## Set of ordered vertices

If the vertices of a polygon are ordered in a clockwise or an anti-clockwise direction, then the area can be calculated using a shoelace algorithm. The formula can be represented by the expression
$$A = \frac{1}{2}\left|\sum_{i = 1}^{n-1}x_iy_{i+1}+x_ny_1 - \sum_{i = 1}^{n - 1}x_{i + 1}y_i - x_1y_n\right|$$
The only condition to use this formula is that the vertices must be ordered in clockwise or anti-clockwise direction otherwise the area would be incorrect. Here is the python implementation of the algorithm.

The list vertices contain a list of object of type Point. Point is a data structure that represents a point in a 2D plane. A python code to represent a point would be

The following code snippet tests the above code.

## Set of unordered vertices

If the vertices are not ordered clockwise or anti-clockwise, then calculating the area can be a little tricky. That means, before calling the polygon_area(points) , we need to sort the points clockwise or anti-clockwise. To sort the vertices, we need a reference point that lies completely inside the polygon. After we have a reference point, then we calculate the angle between the reference point and each of the vertices and sort them in increasing or decreasing order. Now the question is, how do we find the reference point? the answer is simple ‘that depends on the type of the polygon’.

1. Convex Polygon
If the polygon is convex polygon i.e. a polygon whose all interior angles are less than or equal to $180^0$, any point inside the polygon can be a reference point that gives the exact same area i.e. no matter where the reference point is inside the polygon, the area is same in all cases. This is because the choice of a reference point doesn’t alter the shape of a convex polygon. In this case, we can simply average the $x$ and $y$ coordinates to find the reference point. This can be easily calculated as shown in the code below.

2. Non-convex polygon
In the case of a non-convex polygon, the choice of reference point alters the shape of the polygon and therefore we may get different areas for different reference points. Therefore, if the polygon is not a convex polygon then finding the area if the vertices are not ordered does not make any sense.

Once we have a reference point (which make sense only for the convex polygon), we can then sort all the vertices based on the angle made by a line segment joining the reference point and each vertex with x-axis in an anti-clockwise direction as shown in the figure below

The following code calculates the angle (in radian) between the line joining the reference point and a vertex with the x-axis.

We can sort the vertices using the sorted method of the standard python library. The trick here is we pass the comparator that compares the point based on the angle.

Now we are done. We pass the sorted points in the polygon_area function that gives the correct area. The code snippet below test the codes above.

## Conclusion

In this article, I talked about how to calculate the area given a set of vertices. If the vertices are ordered, the area can be calculated using the shoelace formula. If the vertices are not ordered, then the area can be calculated after sorting the vertices which are accurate only if the polygon is convex. If the polygon is not convex, the method may or may not work. Finally, python code is provided that implements all the methods discussed.

## References

1. Shoelace formula. (n.d.). Retrieved August 20, 2018, from https://en.wikipedia.org/wiki/Shoelace_formula
2. “How to sort vertices of a polygon in counter-clockwise order?”: Computing Angle? (n.d.). Retrieved August 20, 2018, from https://math.stackexchange.com/questions/1329128/how-to-sort-vertices-of-a-polygon-in-counter-clockwise-order-computing-angle?noredirect=1&lq=1


text:
Ultrasonic additive manufacturing (UAM) is an emerging solid-state fabrication process that can be used for layered creation of solid metal structures. In UAM, ultrasonic energy is used to induce plastic deformation and nascent surface formation at the interface between layers of metal foil, thus creating bonding between the layers. UAM is an inherently stochastic process with a number of unknown facets that can affect the bond quality. In order to take advantage of the unique benefits of UAM, it is necessary to understand the relationship between manufacturing parameters (machine settings) and bond quality by quantifying the mechanical strength of UAM builds. This research identifies the optimum combination of processing parameters, including normal force, oscillation amplitude, weld speed, and number of bilayers for the manufacture of commercially pure, grade 1 $titanium+1100-O$ aluminum composites. A multifactorial experiment was designed to study the effect of the above factors on the outcome measures ultimate shear strength and ultimate transverse tensile strength. Generalized linear models were used to study the statistical significance of each factor. For a given factor, the operating levels were selected to cover the full range of machine capabilities. Transverse shear and transverse tensile experiments were conducted to quantify the bond strength of the builds. Optimum levels of each parameter were established based on statistical contrast trend analyses. The results from these analyses indicate that high mechanical strength can be achieved with a process window bounded by a 1500 N normal force, $30 μm$ oscillation amplitude, about 42 mm/s weld speed, and two bilayers. The effects of each process parameter on bond strength are discussed and explained.

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text:
2019
Том 71
№ 2

# Sokolov N. V.

Articles: 1
Brief Communications (Russian)

### Generalization of the Prokhorov multidimensional analog of the Chebyshev inequality

Ukr. Mat. Zh. - 2006. - 58, № 4. - pp. 573–576

We prove two theorems on upper and lower bounds for probabilities in the multidimensional case. We generalize and improve the Prokhorov multidimensional analog of the Chebyshev inequality and establish a multidimensional analog of the generalized Kolmogorov probability estimate.


text:
Plots the distribution of p-values derived from any statistical test as a histogram.

pval_distribution_plot(data, grouping, pval)

Arguments

data a data frame that contains at least grouping identifiers (precursor, peptide or protein) and p-values derived from any statistical test. a character column in the data data frame that contains either precursor, peptide or protein identifiers. For each entry in this column there should be one unique p-value. That means the statistical test that created the p-value should have been performed on the level of the content of this column a numeric column in the data data frame that contains p-values.

Value

A histogram that shows the p-value distribution.

Examples

set.seed(123) # Makes example reproducible

# Create example data
data <- data.frame(
peptide = paste0("peptide", 1:1000),
pval = runif(n = 1000)
)

# Plot p-values
pval_distribution_plot(
data = data,
grouping = peptide,
pval = pval
)


text:
# Mathematics 2010 | Study Mode

Question 51
A
sin x - x cosx
B
sinx + x cosx
C
sinx - cosx
D
sinx + cosx
##### Explanation
If y = x sinx, then

Let u = x and v = sinx

$$\frac{du}{dx}$$ = 1 and $$\frac{dv}{dx}$$ = cosx

Hence by the product rule,

$$\frac{dy}{dx}$$ = v $$\frac{du}{dx}$$ + u$$\frac{dv}{dx}$$

= (sin x) x 1 + x cosx

= sinx + x cosx

Question 52
A
4$$\frac{5}{6}$$
B
6$$\frac{2}{3}$$
C
1$$\frac{5}{6}$$
D
2$$\frac{5}{6}$$
##### Explanation
$$\int^{2}_{0}(x^3 + x^2)$$dx = $$\int^{2}_{0}$$($$\frac{x^4}{4} + {\frac {x^3}{3}}$$)

= ($$\frac{2^4}{4} + {\frac {2^3}{3}}$$) - ($$\frac{0^4}{4} + {\frac {0^3}{3}}$$)

= ($$\frac{16}{4} + {\frac {8}{3}}$$) - 0

= $$\frac{80}{12} = {\frac {20}{3}}$$ or 6$$\frac{2}{3}$$

Question 53
A
6
B
2
C
9
D
7
##### Explanation
From the table, the number of students who failed the test is given as:

3 + 1 + 5 = 9

Question 54
A
16
B
20
C
13
D
15
##### Explanation
from the table, the number of students who took the test is 2 + 2 + 8 + 4 + 4 = 20

Question 55
A
112o
B
102o
C
82o
D
52o
##### Explanation
(x + 15)o + (2x - 45)o + (x + 10)o = (2n - 4)90o

when n = 4

x + 15o + 2x - 45o + x - 30o + x + 10o = (2 x 4 - 4) 90o

5x - 50o = (8 - 4)90o

5x - 50o = 4 x 90o = 360o

5x = 360o + 50o

5x = 410o

x = $$\frac{410^o}{5}$$

= 82o

Hence, the value of the least interior angle is (x - 30o)

= (82 - 30)o

= 52o

Question 56
A
3.1
B
3
C
3.3
D
3.2
##### Explanation
$$\begin{array} & Marks(x) & freq.(f) & fx \\1 & 2 & 2 \\ 2 & 2 & 4 \\ 3 & 8 & 24 \\ 4 & 4 & 16\\ 5 & 4 & 20 \\ \hline & \sum f = 20 & \sum fx = 66\end{array}$$
___________________________________

Mean mark ,$$\bar{x}$$ = $$\frac{\sum fx}{\sum f}$$

= $$\frac{66}{20}$$

$$\bar{x}$$ = 3.3

Question 57
A
100
B
200
C
30
D
50
##### Explanation
A committee of 2 women and 3 men can be chosen from 6 men and 5 women, in $$^{5}C_{2}$$ x $$^{6}C_{3}$$ ways

= $$\frac{5!}{(5 - 2)!2!} \times {\frac{6!}{(6 - 3)!3!}}$$

= $$\frac{5!}{3!2!} \times {\frac{6!}{3 \times 3!}}$$

= $$\frac{5 \times 4 \times 3!}{3! \times 2!} \times {\frac{6 \times 5 \times 4 \times 3!}{3! \times 3!}}$$

= $$\frac{5 \times 4}{1 \times 2} \times {\frac{6 \times 5 \times 4}{1 \times 2 \times 3}}$$

= 10 x $$\frac{6 \times 20}{6}$$

= 200

Question 58
A
$$\frac{1}{2}$$
B
$$\frac{1}{3}$$
C
$$\frac{1}{9}$$
D
$$\frac{1}{8}$$
##### Explanation
P(H) = $$\frac{1}{2}$$ and P(T) = $$\frac{1}{2}$$

Using the binomial prob. distribution,

(H + T)3 = H3 + 3H2T1 + 3HT2 + T3

Hence the probability that three heads show in a toss of the three coins is H3

= ($$\frac{1}{2}$$)3

= $$\frac{1}{8}$$

Question 59
A
6
B
10
C
$$\frac{2}{5}$$
D
$$\frac{5}{2}$$
##### Explanation
$$\begin{array}& x & x - \bar{x} & (x - \bar{x})^2 \\2 & -2 & 4 \\ 3 & -1 & 1 \\ 5 & 1 & 1 \\ 6 & 2 & 4\\ \hline \sum x = 16 & & \sum (x - \bar{x}^2) = 0 \end{array}$$
___________________________________

$$\bar{x}$$ = $$\frac{\sum x }{N}$$

= $$\frac{16}{4}$$

= 4

S = $$\sqrt{\frac {(x - \bar{x})^2}{N}}$$

= $$\sqrt{\frac {(10)}{4}}$$

= $$\sqrt{\frac {(5)}{2}}$$

Question 60
A
N37550
B
N40400
C
N41400
D
42400
##### Explanation
Total cost = N34,000 + N2,000

= N36,000

15% = N115

$$\frac{115}{100}$$ x N36,000

= N41,400

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