it is an algorithm in vector quantization
127353
a lot of big words there that I dont get but try the below link for a start maybe?
20294
The algorithm is designed to show your capability to code correctly. It requires you to have full understanding of the syntax of each operator and to be able to follow instructions.
163193
It has 2-bytes code, so you need to examine two consequent bytes. It is not C++ specific, it is PC-specific.
69367
you probably have the compression option selected ..
474686
omg its a calculus 1 optimization problem.
497625
It can be encoded with another 1*2 matrix through addition.\n\nFor example,\n\noriginal matrix = (x   y)\n\nencoding matrix = (a b)\n\nCoded matrix = (x+a    y+b)
653564
it is a type of Machinery.
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LZ77 and LZ78 are the names for the two lossless data compression algorithms published in papers by Abraham Lempel and Jacob Ziv in 1977 and 1978. These two algorithms form the basis for most of the LZ variations including LZW, LZSS and others. They are both dictionary coders, unlike minimum redundancy coders or run length coders. LZ77 is the "sliding window" compression algorithm, which was later shown to be equivalent to the explicit dictionary technique first given in LZ78.\n\nLZ77:\nThe LZ77 algorithm works by keeping a history window of the most recently seen data and comparing the current data being encoded with the data in the history window. What is actually placed into the compressed stream are references to the position in the history window, and the length of the match. If a match cannot be found the character itself is simply encoded into the stream after being flagged as a literal. As of 2004, the most popular LZ77 based compression method is called DEFLATE; it combines LZ77 with Huffman coding.\n\nLZ78:\nWhile the LZ77 algorithm works on past data, the LZ78 algorithm attempts to work on future data. It does this by forward scanning the input buffer and matching it against a dictionary it maintains. It will scan into the buffer until it cannot find a match in the dictionary. At this point it will output the location of the word in the dictionary, if one is available, the match length and the character that caused a match failure. The resulting word is then added to the dictionary.\nThough initially popular, the popularity of LZ78 later dampened, possibly because for the first few decades after it was introduced, parts of LZ78 were patent encumbered in the United States. The most popular form of LZ78 compression was the LZW algorithm, a modification of the LZ78 algorithm made by Terry Welch.\n\nLZ78 Example:\nThis example shows the LZ78 algorithm in action, showing the status of the output and the dictionary at every stage, both in encoding and decoding the message. In order to keep things clear, let us assume that we're dealing with a simple alphabet - capital letters only, and no punctuation or spaces. This example has been constructed to give reasonable compression on a very short message; when used on real data, repetition is generally less pronounced, and so the initial parts of a message will see little compression. As the message grows, however, the compression ratio tends asymptotically to the maximum.[1] A message to be sent might then look like the following:\n\nTOBEORNOTTOBEORTOBEORNOT#\n\nThe # is a marker used to show that the end of the message has been reached. Clearly, then, we have 27 symbols in our alphabet. A computer will render these as strings of bits; 5-bit strings are needed to give sufficient combinations to encompass the entire dictionary. As the dictionary grows, the strings will need to grow in length to accommodate the additional entries. A 5-bit string gives 25 = 32 possible combinations of bits, and so when the 33rd dictionary word is created, the algorithm will have to start using 6-bit strings. Note that since the all-zero string 00000 is used, and is labelled "0", the 33rd dictionary entry will be labelled 32. The initial dictionary, then, will consist of the following:\n\n# = 00000\nA = 00001\nB = 00010\nC = 00011\n.\n.\n.\nZ = 11010\n\n\n[edit]\n\nEncoding:\nIf we weren't using LZ78, and just sent the message as it stands (25 symbols at 5 bits each), it would require 125 bits. We will be able to compare this figure to the LZ78 output later. We are now in a position to apply LZ78 to the message.\n\nSymbol:   Bit Code:    New Dictionary Entry:\n          (= output)\n\nT      20 = 10100      28: TO\nO      15 = 01111      29: OB\nB       2 = 00010      30: BE\nE       5 = 00101      31: EO\nO      15 = 01111      32: OR <--- start using 6-bit strings\nR      18 = 010010     33: RN\nN      14 = 001110     34: NO\nO      15 = 001111     35: OT\nT      20
465157
A. Just multiply CF and then subract the resulting matrix from C. Is there something specific about the required operations you don't understand?
197608
Software programming.
242956
these are read with basic images.  rather than crossing a laser beam across it, a "photograph" of the pattern is taken, and a binary code can be read using the light and dark areas of the pattern.
507014
If you look at a picture (aka a graph) of the function, it will all become clear...\n\nhttp://upload.wikimedia.org/wikipedia/en/3/3e/1-over-x.png
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HI: I tried to answer your email but it said that your email address has not been confirmed and so the email didn't go through.\n\nalpha and beta are scalers for the vector space V.  This problem is actually so easy that people don't believe what they've done.  You really just need to show that Span S behaves like V.  \n\nSpan S must be a vector space itself.  Two elements from the span is \nu=a1*v1 + a2*v2...ak*vk\nand \nw=b1*v1 + b2*v2...bk*vk\n\nshow that alpha*u+beta*w is in Span S where alpha, beta are scalers for your vector space.\n\nalpha*u+beta*w = (alpha*a1+beta*b1)*v1 + ....\nis a linear combination of the vectors in Span S, thus it's in Span S.  \n\nSo, Span S is not empty and is closed, thus its a subspace.
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The singular value decomposition for a matrix A writes A as a product (hanger)(stretcher)(aligner). \n \n\nIt's an amazing and useful fact that every m x n matrix has a singular value decomposition. \n\n-------------------\nIn linear algebra, the singular value decomposition (SVD) is an important factorization of a rectangular real or complex matrix, with several applications in signal processing and statistics.\n\nThe SVD can be seen as a generalization of the spectral theorem to arbitrary, not necessarily square, matrices. The spectral theorem says that normal matrices can be unitarily diagonalized using a basis of eigenvectors.\n\n---------------------------------\nSingular value decomposition takes a rectangular matrix of gene expression data (defined as A, where A is a n x p matrix) in which the n rows represents the genes, and the p columns represents the experimental conditions.\n\n---------------------------------\nA very powerful set of techniques dealing with sets of equations or matrices that are either singular or numerically very close to singular is the so-called singular value decomposition (SVD). SVD allows one to diagnose the problems in a given matrix and provides numerical answer as well.\n\nAny m x n matrix a (m >= n) can be written as the product of an m x n column-orthogonal matrix u, an n x n diagonal matrix with positive or zero elements, and the transpose of an n x n orthogonal matrix v:\n\n\nI hope that all these are useful, good luck!!!!!!!!!!!!
125143
Sounds like you may need to read books by Han and Kamber, Witten and Frank for a better understanding of the data mining process and algorithms.
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http://cwx.prenhall.com/bookbind/pubbooks/hillchem3/medialib/media_portfolio/text_images/CH08/FG08_17.JPG
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It'd help to have  more detailed information. \nWhat does it mean that "pixels are independent?" \nAssuming that pixels are multivariate random variables, having n possible values for each color, the intensity of each component surely is correlated with that of adjacent pixels, I'd suppose.\nNow, what do you mean by "compression result"?\nPut more info. Your problem seems nice to tackle