Essay on Stage Directions in The Glass Menagerie -- Glass Menagerie e

Importance of Stage Directions in The Glass Menagerie      Ã‚  Ã‚   In Tennessee Williams' play, The Glass Menagerie, stage directions are as important to the theme of the play as the dialogue itself. Detailed stage directions intensify the unrealistic setting, foreshadow and emphasize events, and develop the characters. Dim colored lighting and symbolic melodies create the unrealistic setting for the memory play. In his opening narration Tom says, "Being a memory play, it is dimly lighted, it is sentimental, it is not realistic. In memory everything seems to happen to music. That explains the fiddle in the wings" (699). Throughout the play the stage directions call for "a turgid smokey red glow," "gloomy gray" lighting and "deep blue dusk" which create the hazy images of a memory. For a short while, as Jim enters, there is a "delicate lemony light" (688), and a soft light from the new lamp brings out Laura's "unearthly prettiness" (695). Yet, at the end of the play, and throughout its majority, the set is grim, characteristic of Tom's sad memory. Music in the play can be symbolic or simply add to the emotion of a scene. In scene four, "Ave Maria" plays softly in the background, symbolizing Amanda's duties as a mother. Throughout the play, music swells and recedes with the rising and falling of the c haracters' emotions. For example, as Tom is confronting his mother with the reality of his sister's handicap, "the music changes to a tango that has a minor and somewhat ominous tone" (687). Describing characters' appearances and presenting messages upon the screen, the stage directions foreshadow and emphasize events. The description of Tom standing on the fire escape looking "like a voyager" (692) foreshadows his escape to th... ... actions of Amanda and Laura, an audience might believe that Laura has come out of her shell for good or that Amanda is simply an overprotective mother who cannot face reality. Yet, with the elaborate stage directions, Tennessee Williams creates a distinctive memory play with each character tragically failing to reach his or her goals. Works Cited Jolemore, Nancy.   "Lecture Notes and Study Guide Questions for Tennessee William's The Glass Menagerie."   Old Dominion University.   18 January 2000.   Accessed:   29 June 2002. . Reser, Rob.   "A Touch of Glass."   29 June 2002. . Williams, Tennessee.   The Glass Menagerie.   Literature and The Writing Process. 5th ed. Elizabeth McMahan, Susan X. Day, and Robert Funk. Upper Saddle River, NJ: Prentice, 1999. 693-734.   

Looking for a hotel where kids are welcome? :: Essays Papers

Looking for a hotel where kids are welcome? GRAPH The well-known comic The Family Circus by Bil Keane, which is run in over 1500 newspapers, shows the hectic life of an American family, giving meaning to the saying, "Kids will be kids." It is based on the actual life of Bil Keane’s family and his experiences as a child. The most commonly noted features of his comic are the circular shape he often encloses the pictures in, and the dotted line that follows the children around, giving away their every move. Many other cartoonists try to capture the Family Circus mood by incorporating these features into their own works. GRAPH The Wyndham Hotels & Resorts advertisement uses this technique to show the path your child may take around the room of a hotel, where there is nothing else for them to do. It describes the creative and active characteristics of children, who require near constant entertainment that keeps them busy and uses their developing mind. At Wyndham hotels, they understand that keeping your children entertained is a challenging task for you, the parent, which is why they provide an activity package for your kids. The disapproving glare from behind the desk that is sometimes given to parents as they walk into a nice hotel with their three young children is often mirrored to pet-owners, which I can relate to. This gives you a feeling of insecurity and makes you feel unwelcome. I know I appreciate when the hotel at which I am planning to stay has a special feature for pets because then I know not only am I welcome, but that they also welcome my dog. The feeling is similar with parents wanting their children to be welcome. You know from watching Nickelodeon and Nick Jr. with your kids, how much they love Spongebob Squarepants. Moreover, you know that sleepovers are fun. Wyndham Hotels & Resorts also know this, which is why they are advertising this Spongebob Squarepants sleepover party package.

Computational Efficiency of Polar

Lecture Notes on Monte Carlo Methods Fall Semester, 2005 Courant Institute of Mathematical Sciences, NYU Jonathan Goodman, [email  protected] nyu. edu Chapter 2: Simple Sampling of Gaussians. created August 26, 2005 Generating univariate or multivariate Gaussian random variables is simple and fast. There should be no reason ever to use approximate methods based, for example, on the Central limit theorem. 1 Box Muller It would be nice to get a standard normal from a standard uniform by inverting the distribution function, but there is no closed form formula for this distribution 2 x unction N (x) = P (X < x) = v1 ? e? x /2 dx . The Box Muller method is a 2 brilliant trick to overcome this by producing two independent standard normals from two independent uniforms. It is based on the familiar trick for calculating ? 2 e? x I= /2 dx . This cannot be calculated by â€Å"integration† – the inde? nite integral does not have an algebraic expression in terms of elementary f unctions (exponentials, logs, trig functions). However, ? 2 e? x I2 = ? /2 e? y dx 2 ? /2 ? 2 e? (x dy = +y 2 )/2 dxdy . The last integral can be calculated using polar coordinates x = r cos(? ), y = r sin(? with area element dxdy = rdrd? , so that 2? I2 = r = 0? e? r 2 /2 rdrd? = 2? r = 0? e? r 2 /2 rdr . ? =0 Unlike the original x integral, this r integral is elementary. The substitution s = r2 /2 gives ds = rdr and ? e? s ds = 2? . I 2 = 2? s=0 The Box Muller algorithm is a probabilistic interpretation of this trick. If (X, Y ) is a pair of independent standard normals, then the probability density is a product: 2 2 1 1 ? (x2 +y2 )/2 1 e . f (x, y ) = v e? x /2  · v e? y /2 = 2? 2? 2? 1 Since this density is radially symmetric, it is natural to consider the polar coordinate random variables (R, ? de? ned by 0 ? ? < 2? and X = R cos(? ), and Y = R sin(? ). Clearly ? is uniformly distributed in the interval [0, 2? ] and may be sampled using ? = 2? U1 . Unlike the original dis tribution function N (x), there is a simple expression for the R distribution function: 2? r G(R) = P (R ? r) = r =0 ?=0 r 1 ? r 2 /2 e rdrd? = 2? e? r 2 /2 rdr . r =0 The same change of variable r 2 /2 = s, r dr = ds (so that r = r when s = r2 /2) allows us to calculate r 2 /2 e? s dx = 1 ? e? r G(r) = 2 /2 . s=0 Therefore, we may sample R by solving the distribution function equation1 G(R) = 1 ? e? R 2 /2 = 1 ?U2 , whose solution is R = ? 2 ln(U2 ). Altogether, the Box Muller method takes independent standard uniform random variables U1 and U2 and produces independent standard normals X and Y using the formulas ? = 2? U1 , R = ?2 ln(U2 ) , X = R cos(? ) , Y = R sin(? ) . (1) It may seem odd that X and Y in (13) are independent given that they use the same R and ?. Not only does our algebra shows that this is true, but we can test the independence computationally, and it will be con? rmed. Part of this method was generating a point â€Å"at random† on the unit circle. We sug gested doing this by choosing ? niformly in the interval [0, 2? ] then taking the point on the circle to be (cos(? ), sin(? )). This has the possible drawback that the computer must evaluate the sine and cosine functions. Another way to do this2 is to choose a point uniformly in the 2 ? 2 square ? 1 ? x ? 1, 1 ? y ? 1 then rejecting it if it falls outside the unit circle. The ? rst accepted point will be uniformly distributed in the unit disk x2 + y 2 ? 1, so its angle will be random and uniformly distributed. The ? nal step is to get a point on the unit circle x2 + y 2 = 1 by dividing by the length.The methods have equal accuracy (both are exact in exact arithmetic). What distinguishes them is computer performance (a topic discussed more in a later lecture, hopefully). The rejection method, with an acceptance probability ? ? 4 78%, seems e? cient, but rejection can break the instruction pipeline and slow a computation by a factor of ten. Also, the square root needed to compute 1 Re call that 1 ? U2 is a standard uniform if U2 is. for example, in the dubious book Numerical Recipies. 2 Suggested, 2 the length may not be faster to evaluate than sine and cosine.Moreover, the rejection method uses two uniforms while the ? method uses just one. The method can be reversed to solve another sampling problem, generating a random point on the â€Å"unit spnere† in Rn . If we generate n independent standard normals, then the vector X = (X1 , . . . , Xn ) has all angles equally n likely (because the probability density is f (x) = v1 ? exp(? (x2 + ·  ·  ·+x2 )/2), n 1 2 which is radially symmetric. Therefore X/ X is uniformly distributed on the unit sphere, as desired. 1. 1 Other methods for univariate normals The Box Muller method is elegant and reasonably fast and is ? ne for casual omputations, but it may not be the best method for hard core users. Many software packages have native standard normal random number generators, which (if they are any good) use e xpertly optimized methods. There is very fast and accurate software on the web for directly inverting the normal distribution function N (x). This is particularly important for quasi Monte Carlo, which substitutes equidistributed sequences for random sequences (see a later lecture). 2 Multivariate normals An n component multivariate normal, X , is characterized by its mean  µ = E [X ] and its covariance matrix C = E [(X ?  µ)(X ?  µ)t ].We discuss the problem of generating such an X with mean zero, since we achieve mean  µ by adding  µ to a mean zero multivariate normal. The key to generating such an X is the fact that if Y is an m component mean zero multivariate normal with covariance D and X = AY , then X is a mean zero multivariate normal with covariance t C = E X X t = E AY (AY ) = AE Y Y t At = ADAt . We know how to sample the n component multivariate normal with D = I , just take the components of Y to be independent univariate standard normals. The formula X = AY w ill produce the desired covariance matrix if we ? nd A with AAt = C .A simple way to do this in practice is to use the Choleski decomposition from numerical linear algebra. This is a simple algorithm that produces a lower triangular matrix, L, so that LLt = C . It works for any positive de? nite C . In physical applications it is common that one has not C but its inverse, H . This would happen, for example, if X had the Gibbs-Boltzmann distribution with kT = 1 (it’s easy to change this) and energy 1 X t HX , and probability 2 1 density Z exp(? 1 X t HX ). In large scale physical problems it may be impracti2 cal to calculate and store the covariance matrix C = H ? though the Choleski factorization H = LLt is available. Note that3 H ? 1 = L? t L? 1 , so the choice 3 It is traditional to write L? t for the transpose of L? 1 , which also is the inverse of Lt . 3 A = L? t works. Computing X = L? t Y is the same as solving for X in the equation Y = Lt X , which is the process of ba ck substitution in numerical linear algebra. In some applications one knows the eigenvectors of C (which also are the eigenvectors of H ), and the corresponding eigenvalues. These (either the eigenvectors or the eigenvectors and eigenvalues) sometimes are called principal com2 ponents.Let qj be the eigenvectors, normalized to be orthonormal, and ? j the corresponding eigenvalues of C , so that 2 Cqj = ? j qj , t qj qk = ? jk . t Denote the qj component of X by Zj = qj X . This is a linear function of X and t therefore Gaussian with mean zero. It’s variance (note: Zj = Zj = X t qj ) is 2 t t t 2 E [Zj ] = E [Zj  · Zj ] = qj E [XX t ]qj = qj Cqj = ? j . A similar calculation shows that Zj and Zk are uncorrelated and hence (as components of a multivariate normal) independent. Therefore, we can generate Yj as independent standard normals and sample the Zj using Zj = ? j Yj . (2) After that, we can get an X using Zj qj . X= (3) j =1 We restate this in matrix terms. Let Q be the orthogonal matrix whose columns are the orthonormal eigenvectors of C , and let ? 2 be the diagonal ma2 trix with ? j in the (j, j ) diagonal position. The eigenvalue/eigenvector relations are CQ = Q? 2 , Qt Q = I = QQt . (4) The multivariate normal vector Z = Qt X then has covariance matrix E [ZZ t ] = E [Qt XX t Q] = Qt CQ = ? 2 . This says that the Zj , the components of Z , are 2 independent univariate normals with variances ? j . Therefore, we may sample Z by choosing its components by (14) and then reconstruct X by X = QZ , which s the same as (15). Alternatively, we can calculate, using (17) that t C = Q? 2 Qt = Q Qt = (Q? ) (Q? ) . Therefore A = Q? satis? es AAt = C and X = AY = Q? Y = QZ has covariance C if the components of Y are independent standard univariate normals or 2 the components of Z are independent univariate normals with variance ? j . 3 Brownian motion examples We illustrate these ideas for various kids of Brownian motion. Let X (t) be a Brownian motion path. Choose a ? nal time t and a time step ? t = T /n. The 4 observation times will be tj = j ? t and the observations (or observation values) will be Xj = X (tj ).These observations may be assembled into a vector X = (X1 , . . . , Xn )t . We seek to generate sample observation vectors (or observation paths). How we do this depends on the boundary conditions. The simplest case is standard Brownian motion. Specifying X (0) = 0 is a Dirichlet boundary condition at t = 0. Saying nothing about X (T ) is a free (or Neumann) condition at t = T . The joint probability density for the observation vector, f (x) = f (x1 , . . . , xn ), is found by multiplying the conditional densities. Given Xk = X (tk ), the next observation Xk+1 = X (tk + ? ) is Gaussian with mean Xk and variance ? t, so its conditional density is v 2 1 e? (xk+1 ? Xk ) /2? t . 2? ?t Multiply these together and use X0 = 0 and you ? nd (with the convention x0 = 0) f (x1 , . . . , xn ) = 3. 1 1 2? ?t n/2 exp ?1 2 ? Deltat n? 1 (xk+ 1 ? xk )2 . (5) k=0 The random walk method The simplest and possibly best way to generate a sample observation path, X , comes from the derivation of (1). First generate X1 = X (? t) as a mean zero v univariate normal with mean zero and variance ? t, i. e. X1 = ? tY1 . Given X1 , X2 is a univariate normal with mean X1 and variance ? , so we may v take X2 = X1 + ? tY2 , and so on. This is the random walk method. If you just want to make standard Brownian motion paths, stop here. We push on for pedigogical purposes and to develop strategies that apply to other types of Brownian motion. We describe the random walk method in terms of the matrices above, starting by identifying the matrices C and H . Examining (1) leads to ? 2 ? 1 0  ·Ã‚ ·Ã‚ · ? ? ? 1 2 ? 1 0  ·Ã‚ ·Ã‚ · ? ? .. .. .. . . . 1 ? 0 ? 1 ? H= ?. .. ?t ? . . 2 ? 1 ?. ? .. ? . ? 1 2 0  ·Ã‚ ·Ã‚ · 0 ? 1 ? 0 .? .? .? ? ? ? ? 0? ? ? ?1 ? 1 This is a tridiagonal matrix with pattern ? 1, 2, ? except at the bottom right corner. O ne can calculate the covariances Cjk from the random walk representation v Xk = ? t (Y1 +  ·  ·  · + Yk ) . 5 Since the Yj are independent, we have Ckk = var(Xk ) = ? t  · k  · var(Yj ) = tk , and, supposing j < k , Cjk = E [Xj Xk ] = ? tE [((Y1 +  ·  ·  · + Yj ) + (Yj +1 +  ·  ·  · + Yk ))  · (Y1 +  ·  ·  · + Yj )] = 2 ?tE (Y1 +  ·  ·  · + Yj ) = tj . These combine into the familiar formula Cjk = cov(X (tj ), X (tk )) = min(tj , tk ) . This is the same as saying that the ? 1 ?1 ? ?. ?. C = ? t ? . ? ? ? 1 matrix C is 1  ·Ã‚ ·Ã‚ · 2 2  ·Ã‚ ·Ã‚ · 2 . . . 3  ·Ã‚ ·Ã‚ · . . . 2 3  ·Ã‚ ·Ã‚ · ? 1 2? ? ? 3? .? .? .? .. . (6) The random walk method for generating X may be expresses as ? ? ? Y ? X1 1 1 0  ·Ã‚ ·Ã‚ · 01 ? ? ? ?1 1 0  ·Ã‚ ·Ã‚ · 0 ? ? . ? ?.? ?.? v? ? . ? ?.? 1 0 . . ? . .? ? . ? = ? t ? 1 1 ? ? ? ? ?. . .. ? ? ? ?. . . .. ? ? ? ? 11 1  ·Ã‚ ·Ã‚ · 1 Yn Xn Thus, X = AY with ? ? 1 0  ·Ã‚ ·Ã‚ · 01 ?1 1 0  ·Ã‚ ·Ã‚ · 0 ? ? ? v? .? .? . ?1 1 1 0 .? A = ? t ? ?. . ? .. .. ?. . ? . 11 1  ·Ã‚ ·Ã‚ · 1 (7) The reader should do the matrix multiplication to check that indeed C = AAt for (6) and (7). Notice that H is a sparse matrix indicating short range interactions while C is full indicating long range correlations.This is true of in great number of physical applications, though it is rare to have an explicit formula for C . 6 We also can calculate the Choleski factorization of H . The reader can convince herself or himself that the Choleski factor, L, is bidiagonal, with nonzeros only on or immediately below the diagonal. However, the formulas are simpler if we reverse the order of the coordinates. Therefore we de? ne the coordinate reversed observation vector t X = (Xn , xn? 1 , . . . , Xn ) and whose covariance matrix is ? tn ? tn? 1 ? C=? . ?. . t1 tn? 1 tn? 1  ·  ·  · t1 t1 .. .  ·Ã‚ ·Ã‚ · ? ? ? , ? t1 and energy matrix ? 1 ? 1  ·Ã‚ ·Ã‚ · 0 ? 0 .? .? .? ? ? ?. ? 0? ? ? ?1 ? 2 ? ? ? 1 2 ? 1 0  ·Ã‚ ·Ã‚ · ? ? .. .. .. . . . 1 ? 0 ? 1 ? H= .. ?t ? . . ?. . 2 ? 1 ? ? .. ? . ? 1 2 0  ·Ã‚ ·Ã‚ · 0 ? 1 We seek the Choleski factorization H = LLt ? l1 0 ? m2 l2 1? L= v ? m3 ?t ? 0 ? . .. . . . with bidiagonal ?  ·Ã‚ ·Ã‚ · ? 0 ? ?. .. ? . ? .. . Multiplying out H = LLt leads to equations that successively determine the lk and mk : 2 l1 l 1 m2 2 2 l1 + l 2 l 2 m3 = 1 =? l1 = 1 , = ? 1 =? m2 = ? 1 , = 2 =? l2 = 1 , = 1 =? m3 = ? 1 , etc. , The result is H = LLt with L simply ? 1 0  ·Ã‚ ·Ã‚ · ? ? 1 10 1? .. L= v ? . ?t ? ? 1 ? . .. .. . . . . 7 ? ? ? ?. ? ? The sampling algorithm using this Y = Lt X : ? ? ? 1 Yn ? Yn? 1 ? ? ? ? ?0 ? ? 1? ? ? ? ? . ?= v ? ?.? ?t ? ?.? ?. ? ? ?. . Y1 0 information is to ? nd X from Y by solving ?1 0 1 .. . ?1 .. .  ·Ã‚ ·Ã‚ ·  ·Ã‚ ·Ã‚ · .. . 0 0 Xn . ? ? Xn? 1 . . . 0 . . ?1 X1 1 ? ? ? ? ? ? ? ? ? Solving from the bottom up (back substitution), we have Y1 = Y2 = v 1 v X1 =? X1 = ? tY1 , ?t v 1 v (X2 ? X1 ) =? X2 = X1 + ? tY2 , etc. ? t This whole process turns out to give the same random walk sampling method. Had we not gone to the time reversed (X , etc. variables, we could have calculated the bidiagonal Choleski factor L numerically. This works for any problem with a tridiagonal energy matrix H and has a name in the control theory/estimation literature that escapes me. In particular, it will allow to ? nd sample Brownian motion paths with other boundary conditions. 3. 2 The Brownian bridge construction The Brownian bridge construction is useful in the mathematical theory of Brownian motion. It also is the basis for the success of quasi Monte Carlo methods in ? nance. Suppose n is a power of 2: n = 2L . We will construct the observation path X through a sequence of L re? ements. First, notice that Xn is a univariate normal with mean zero and variance T , so we may take (with Yk,l being independent standard normals) v Xn = T Y1,1 . Given the value of Xn , the midoint observation, Xn/2 , is a univariate normal4 w ith mean 1 Xn and variance T /4, so we may take 2 Xn 2 v 1 T = Xn + Y2,1 . 2 2 At the ? rst level, we chose the endpoint value for X . We could draw a ? rst level path by connenting Xn to zero with a straight line. At the second level, or ? rst re? nement, we created a midpoint value. The second level path could be piecewise linear, connecting 0 to X n to Xn . 4 We assign this and related claims below as exercises for the student. 8 The second re? nement level creates values for the â€Å"quarter points†. Given n X n , X n is a normal with mean 1 X n and variance 1 T . Similarly, X 34 is a 2 42 2 4 2 1 1T normal with mean 2 (X n + Xn ) and variance 4 2 . Therefore, we may take 2 Xn = 4 1 1 Xn + 22 2 T Y3,1 2 and n X 34 = 1 1 (X n + Xn ) + 2 2 2 T Y3,2 . 2 1 The level three path would be piecewise linear with breakpoints at 1 , 2 , and 3 . 4 4 Note that in each case we add a mean zero normal of the appropriate variance to the linear interpolation value.In the general step, we go from the level k ? 1 path to the level k paths by creating values for the midpoints of the level k ? 1 intervals. The level k observations are X j . The values with even j are known from the previous 2k? 1 level, so we need values for odd j . That is, we want to interpolate between the j = 2m value and the j = 2m + 2 value and add a mean zero normal of the appropriate variance: X (2m+1)n = 2k? 1 1 2 mn X 2k? 1 + X (2m+2)n 2 2k? 1 + 1 2(k? 2)/2 T Ym,k . 2 The reader should check that the vector of standard normals Y = (Y1,1 , Y2,1 , Y3,1 , Y3,2 , . . . t indeed has n = 2L components. The value of this method for quasi Monte Carlo comes from the fact that the most important values that determine the large scale structure of X are the ? rst components of Y . As we will see, the components of the Y vectors of quasi Monte Carlo have uneven quality, with the ? rst components being the best. 3. 3 Principle components The principle component eigenvalues and eigenvectors for many types of Brownian motion are known in closed form. In many of these cases, the Fast Fourier Transform (FFT) algorithm leads to a reasonably fast sampling method.These FFT based methods are slower than random walk or Brownian bridge sampling for standard random walk, but they sometimes are the most e? cient for fractional Brownian motion. They may be better than Brownian bridge sampling with quasi Monte Carlo (I’m not sure about this). The eigenvectors of H are known5 to have components (qj,k is the k th component of eigenvector qj . ) qj,k = const  · sin(? j tk ) . 5 See e. g. Numerical Analysis by Eugene Isaacson and Herbert Keller. 9 (8) The n eigenvectors and eigenvalues then are determined by the allowed values of ? j , which, in turn, are determined throught the boundary conditions.We 2 2 can ? nd ? j in terms of ? j using the eigenvalue equation Hqj = ? j qj evaluated at any of the interior components 1 < k < n: 1 2 [? sin(? j (tk ? ?t)) + 2 sin(? j tk ) ? sin(? j (tk + ? t)) ] = ? j sin(? j tk ) . ?t Doing the math shown that the eigenvalue equation is satis? ed and that 2 ?j = 2 1 ? cos(? j ? t) . ?t (9) The eigenvalue equation also is satis? ed at k = 1 because the form (8) automatically satis? es the boundary condition qj,0 = 0. This is why we used the sine and not the cosine. Only special values ? j give qj,k that satisfy the eigenvalue equation at the right boundary point k = n. 10

Master and Margarita by Mikhail Bulgakov: two novels in one

â€Å"Master and Margarita† by Mikhail Bulgakov is probably the most famous and the most unusual novel ever created in the USSR. Bulgakov artistically investigated the theme of epic struggle of good and evil relating biblical events to the realities of Moscow in the mid-thirties, having demonstrated that people have not changed in the last two thousand of years except that â€Å"only the housing problem has   corrupted them†[1].The narrative of â€Å"Master and Margarita† is pretty sophisticated, including at least three plot lines: adventures of Woland and his suit in Moscow, story of Master and Margarita themselves, both taking place in Moscow, and a tale of Yeshua Ha-Nozri – a mendicant prophet before Pontius Pilate and an obvious allegory of Jesus Christ. This paper aims to investigate ties between modern and biblical events in the â€Å"Master and Margarita† and show how Bulgakov paralleled biblical characters and events with his contemporari es.Bulgakov’s mission in â€Å"Master and Margarita† was more than complicated because he devoted his novel to such sempiternal topics as love and fear, good and evil, God and Satan. He obviously demonstrated that those topics are timeless, for Master is interested in the same questions as Pilate was 1900 years before. At the end of the novel Moscow and Jerusalem seem to be united in a metaphysical oneness, and two plots turn out to be one. Jerusalem in the case symbolizes immortal ever-being world and Moscow is the world of the earth. At that Woland acts as a figure which ties the worlds, for he â€Å"was   on Pontius Pilate's   balcony, and in the garden when   he   talked with Kaifa, and on the platform, only   secretly, incognito†[2] and then visited Moscow and talked to Master. It is hard to believe, that Woland is a real Satan, he rather looks like exactly part of that power which eternally wills evil and eternally works good.The story opens by a discussion at Patriarch's Ponds and nothing unreal happens at the beginning, except for s strange presentiment of Berlioz. In order to continue his narration, Bulgakov needed to introduce the reader into fantastic multi-world reality. He accepts that the reader is so skeptic about existence of God and Satan, that Woland expects to notice an atheist in every window. In this scene Bulgakov puts the words into the mouths of Woland:   â€Å"Bear in mind that Jesus did exist†[3]. Woland says this to Berlioz, but it is Bulgakov who says to the reader: bear in mind that Jesus did exist.After that Bulgakov’s story of Jesus is perceived by the reader as actual, and the biblical world is easily connected with earthly world. Moreover, the existence of the divine reality is proved even by its fiercest enemies – Berlioz and Ivanushka. The latter has blamed Jesus so desperately, â€Å"his Jesus came out, well, completely alive, the once-existing   Jesus, though,   tru e,   a Jesus   furnished   with   all negative features†[4]. Berlioz tells Ivanushka that the rhyme has to be rewritten, but it is then rewritten not by Ivanushka but by Woland. So, concludes Bulgakov, whatever we mean about Jesus in this life, divine reality will remain unchanged.In order to stress the biblical chapters and keep the style unified Bulgakov showed Jerusalem as a vision of Woland or a dream of Ivanushka or a story told by Master or read by Margarita. His tone changes from satiric or sympathetic to unimpassioned speech in the evangelical passages. In contrast, when Bulgakov turns back to Moscow, his characters are   no longer legendary and obtain everyday features. Thusly he achieves to basic aims – inseparably integrates biblical chapters into the novel and creates an illusion of parallel between the two worlds.Events are separated almost by two thousand years of time, but analogies are easy to trace. Both stories happen in May in the days befor e Easter, both in the mid-thirties of I and XX century AD. The weather and temporary changes are almost same. And surely the crowd is the same, whether it is a crowd which gathers to stare at Yeshua’s execution or a crowd in the vaudeville house.  The tale of both worlds ends with the same conclusive phrase â€Å"the cruel fifth procurator of Judea, the equestrian Pontius Pilate†[5], and so Bulgakov shows that the outcome of the novel is justification and recognition of truth. When Pilate, a biblical symbol of a man who yields before evil, shrives, he enters into the moonlight, being â€Å"forgiven on the eve of Sunday†. While Pilate walks with Yeshua by the moonlight beam Ivanushka stands at the beginning of this beam, symbolizing those whose choice is still not sure.The most recognizable â€Å"double-character† is Yeshua and Master – an analogy of Christ and Bulgakov himself. Yeshua does not act as a Saviour, he is rather a creator who faces in comprehension and repudiation. Same happens to Master, whose novel is welcomed by angry critics and who is finally symbolically executed, because he is unable to create any more and falls into insanity. However, his novel is not lost, and later it is read by Yeshua-Jesus himself, so Master’s novel is a story of biblical events and at the same time a link between those events and our world.  In contrast to all other human characters, even Margarita, Master and Yeshua are the only ones who struggle for truth and sincerity to the end. Pilate had a heart with Yeshua and hinted, that only a slight lie could save his life. Yeshua refused and became a martyr. Pilate could not overcome his fear and sentenced himself to everlasting spiritual unrest. Surprisingly, Pilate’s vis-à  -vis in the modern world is Margarita, who leaves Master in a critical moment and has a feeling of guilt for long years. But at the end she is saved by Woland and Pilate is forgiven by Christ.Virtual ly every character in the novel has to face a choice situation and the choice he or she makes in the usual world predetermines his or her future existence in the spiritual reality. Master chooses to forget about his novel and about Margarita and so he becomes Pilate’s accomplice in execution of Christ. Therefore, he is not worthy of light and he never meets Jesus. Divine will is passed to him by Woland, who, in turn, receives it from Matthew Levi. Characters of the paralleled novels never meet together and their ties are limited to communication via messengers and joining into one at the end.As regards the other characters, their choice is rather comic, than dramatic. Berlioz’s uncle chooses to live in Moscow, and the only thing he is interested in when he receives a message about death of his nephew is to receive ownership of an apartment. Nikanor Ivanovich, chairman of the tenants' association', chooses to take money from a suspicious foreigner.The crowd of women cho oses to change their dress at a show arranged by Fagot and Behemoth. Their punishments are ridiculous but this does not mean that their actions will not be judged later in Ewigkeit. Fate of Berlioz is the most frightening warning. Woland appeared to be an adept of a theory â€Å"which holds that it will be given to each according to his faith.†[6] In the earthly life Berlioz has chosen to believe that there is nothing in the afterlife, and Woland has put him into nonentity in the spiritual world.The final scenes of the novel show the idea, that examples of the great masters are never lost, for â€Å"manuscripts don't burn†[7]. They become a part of eternity, and their creators obtain eternal life. So they are able to cross the border between two worlds. The fate of Bulgakov’s novel is the best illustration for this thesis, for it has been published only decades after his death and immediately gained global recognition. Perhaps this was in a way a fourth plot lin e of the novel – Bulgakov’s own life became a parallel for lives of Yeshua and Master.References:Mikhail Bulgakov. The Master and Margarita. Penguin Books Ltd; New Ed edition, 2004. 432 pages.[1] Mikhail Bulgakov. The Master and Margarita. Penguin Books Ltd; New Ed edition, 2004.   p. 124[2] Ibid, p. 42 [3] Ibid, p. 18[4] Ibid, p. 8 [5] Ibid, p.380 [6] Ibid, p.275  [7] Ibid, p.369

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