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Unformatted text preview: 1 What Is Data Mining? Originally, data mining" was a statistician's term for overusing data to draw invalid inferences. Bonferroni's theorem warns us that if there are too many possible conclusions to draw, some will be true for purely statistical reasons, with no physical validity. Famous example: David Rhine, a parapsychologist" at Duke in the 1950's tested students for extrasensory perception" by asking them to guess 10 cards | red or black. He found about 1 1000 of them guessed all 10, and instead of realizing that that is what you'd expect from random guessing, declared them to have ESP. When he retested them, he found they did no better than average. His conclusion: telling people they have ESP causes them to lose it! Our de nition: discovery of useful summaries of data." 1.1 Applications Some examples of successes": 1. Decision trees constructed from bank-loan histories to produce algorithms to decide whether to grant a loan. 2. Patterns of traveler behavior mined to manage the sale of discounted seats on planes, rooms in hotels, etc. 3. Diapers and beer." Observation that customers who buy diapers are more likely to by beer than average allowed supermarkets to place beer and diapers nearby, knowing many customers would walk between them. Placing potato chips between increased sales of all three items. 4. Skycat and Sloan Sky Survey: clustering sky objects by their radiation levels in di erent bands allowed astromomers to distinguish between galaxies, nearby stars, and many other kinds of celestial objects. 5. Comparison of the genotype of people with without a condition allowed the discovery of a set of genes that together account for many cases of diabetes. This sort of mining will become much more important as the human genome is constructed. 1.2 The Data-Mining Communities 1. 2. 3. 4. 5. As data-mining has become recognized as a powerful tool, several di erent communities have laid claim to the subject: Statistics. AI, where it is called machine learning." Researchers in clustering algorithms. Visualization researchers. Databases. We'll be taking this approach, of course, concentrating on the challenges that appear when the data is large and the computations complex. In a sense, data mining can be thought of as algorithms for executing very complex queries on non-main-memory data. 1 1.3 Stages of the Data-Mining Process 1. Data gathering, e.g., data warehousing, Web crawling. 2. Data cleansing : eliminate errors and or bogus data, e.g., patient fever = 125. 3. Feature extraction : obtaining only the interesting attributes of the data, e.g., date acquired" is probably not useful for clustering celestial objects, as in Skycat. 4. Pattern extraction and discovery. This is the stage that is often thought of as data mining," and is where we shall concentrate our e ort. 5. Visualization of the data. 6. Evaluation of results; not every discovered fact is useful, or even true! Judgement is necessary before following your software's conclusions. 2 2 Association Rules and Frequent Itemsets The market-basket problem assumes we have some large number of items, e.g., bread," milk." Customers ll their market baskets with some subset of the items, and we get to know what items people buy together, even if we don't know who they are. Marketers use this information to position items, and control the way a typical customer traverses the store. In addition to the marketing application, the same sort of question has the following uses: 1. Baskets = documents; items = words. Words appearing frequently together in documents may represent phrases or linked concepts. Can be used for intelligence gathering. 2. Baskets = sentences, items = documents. Two documents with many of the same sentences could represent plagiarism or mirror sites on the Web. 2.1 Goals for Market-Basket Mining 1. Association rules are statements of the form fX1 ; X2 ; : : :; Xn g  Y , meaning that if we nd all of X1 ; X2; : : :; Xn in the market basket, then we have a good chance of nding Y . The probability of nding Y for us to accept this rule is called the con dence of the rule. We normally would search only for rules that had con dence above a certain threshold. We may also ask that the con dence be signi cantly higher than it would be if items were placed at random into baskets. For example, we might nd a rule like fmilk; butterg  bread simply because a lot of people buy bread. However, the beer diapers story asserts that the rule fdiapersg  beer holds with con dence sigini cantly greater than the fraction of baskets that contain beer. 2. Causality. Ideally, we would like to know that in an association rule the presence of X1 ; : : :; Xn actually causes" Y to be bought. However, causality" is an elusive concept. nevertheless, for market-basket data, the following test suggests what causality means. If we lower the price of diapers and raise the price of beer, we can lure diaper buyers, who are more likely to pick up beer while in the store, thus covering our losses on the diapers. That strategy works because diapers causes beer." However, working it the other way round, running a sale on beer and raising the price of diapers, will not result in beer buyers buying diapers in any great numbers, and we lose money. 3. Frequent itemsets. In many but not all situations, we only care about association rules or causalities involving sets of items that appear frequently in baskets. For example, we cannot run a good marketing strategy involving items that no one buys anyway. Thus, much data mining starts with the assumption that we only care about sets of items with high support; i.e., they appear together in many baskets. We then nd association rules or causalities only involving a high-support set of items i.e., fX1 ; : : :; Xn ; Y g must appear in at least a certain percent of the baskets, called the support threshold. 2.2 Framework for Frequent Itemset Mining We use the term frequent itemset for a set S that appears in at least fraction s of the baskets," where s is some chosen constant, typically 0.01 or 1. We assume data is too large to t in main memory. Either it is stored in a RDB, say as a relation BasketsBID; item or as a at le of records of the form BID; item1; item2; : : : ; itemn. When evaluating the running time of algorithms we: Count the number of passes through the data. Since the principal cost is often the time it takes to read data from disk, the number of times we need to read each datum is often the best measure of running time of the algorithm. There is a key principle, called monotonicity or the a-priori trick that helps us nd frequent itemsets: If a set of items S is frequent i.e., appears in at least fraction s of the baskets, then every subset of S is also frequent. 3 To nd frequent itemsets, we can: 1. Proceed levelwise, nding rst the frequent items sets of size 1, then the frequent pairs, the frequent triples, etc. In our discussion, we concentrate on nding frequent pairs because: a Often, pairs are enough. b In many data sets, the hardest part is nding the pairs; proceeding to higher levels takes less time than nding frequent pairs. Levelwise algorithms use one pass per level. 2. Find all maximal frequent itemsets i.e., sets S such that no proper superset of S is frequent in one pass or a few passes. 2.3 The A-Priori Algorithm This algorithm proceeds levelwise. 1. Given support threshold s, in the rst pass we nd the items that appear in at least fraction s of the baskets. This set is called L1 , the frequent items. Presumably there is enough main memory to count occurrences of each item, since a typical store sells no more than 100,000 di erent items. 2. Pairs of items in L1 become the candidate pairs C2 for the second pass. We hope that the size of C2 is not so large that there is not room for an integer count per candidate pair. The pairs in C2 whose count reaches s are the frequent pairs, L2. 3. The candidate triples, C3 are those sets fA; B; C g such that all of fA; B g, fA; C g, and fB; C g are in L2 . On the third pass, count the occurrences of triples in C3; those with a count of at least s are the frequent triples, L3 . 4. Proceed as far as you like or the sets become empty. Li is the frequent sets of size i; Ci+1 is the set of sets of size i + 1 such that each subset of size i is in Li . 2.4 Why A-Priori Helps Consider the following SQL on a BasketsBID; item relation with 108 tuples involving 107 baskets of 10 items each; assume 100,000 di erent items typical of Wal-Mart, e.g.. SELECT b1.item, b2.item, COUNT* FROM Baskets b1, Baskets b2 WHERE b1.BID = b2.BID AND b1.item GROUP BY b1.item, b2.item HAVING COUNT* = s; b2.item Note: s is the support threshold, and the second term of the WHERE clause is to prevent pairs of items that are really one item, and to prevent pairs from appearing ,twice.  In the join Baskets . Baskets, each basket contributes 10 = 45 pairs, so the join has 4:5  108 tuples. 2 A-priori pushes the HAVING down the expression tree," causing us rst to replace Baskets by the result of SELECT * FROM Baskets GROUP by item HAVING COUNT* = s; If s = 0:01, then at most 1000 items' groups can pass the HAVING condition. Reason: there are 108 item occurrences, and an item needs 0:01  107 = 105 of those to appear in 1 of the baskets. Although 99 of the items are thrown away by a-priori, we should not assume the resulting Baskets relation has only 106 tuples. In fact, all the tuples may be for the high-support items. However, in real situations, the shrinkage in Baskets is substantial, and the size of the join shrinks in proportion to the square of the shrinkage in Baskets. 4 2.5 Improvements to A-Priori Two types: 1. Cut down the size of the candidate sets Ci for i  2. This option is important, even for nding frequent pairs, since the number of candidates must be su ciently small that a count for each can t in main memory. 2. Merge the attempts to nd L1 ; L2 ; L3; : : : into one or two passes, rather than a pass per level. 2.6 PCY Algorithm Park, Chen, and Yu proposed using a hash table to determine on the rst pass while L1 is being determined that many pairs are not possibly frequent. Takes advantage of the fact that main memory is usualy much bigger than the number of items. During the two passes to nd L2, the main memory is laid out as in Fig. 1. Count items Frequent items Bitmap Hash table Counts for candidate pairs Pass 1 Pass 2 Figure 1: Two passes of the PCY algorithm Assume that data is stored as a at le, with records consisting of a basket ID and a list of its items. 1. Pass 1: a Count occurrences of all items. b For each bucket, consisting of items fi1 ; : : :; ik g, hash all pairs to a bucket of the hash table, and increment the count of the bucket by 1. c At the end of the pass, determine L1 , the items with counts at least s. d Also at the end, determine those buckets with counts at least s. Key point: a pair i; j  cannot be frequent unless it hashes to a frequent bucket, so pairs that hash to other buckets need not be candidates in C2. Replace the hash table by a bitmap, with one bit per bucket: 1 if the bucket was frequent, 0 if not. 2. Pass 2: a Main memory holds a list of all the frequent items, i.e. L1 . b Main memory also holds the bitmap summarizing the results of the hashing from pass 1. Key point: The buckets must use 16 or 32 bits for a count, but these are compressed to 1 bit. Thus, even if the hash table occupied almost the entire main memory on pass 1, its bitmap ocupies no more than 1 16 of main memory on pass 2. 5 c Finally, main memory also holds a table with all the candidate pairs and their counts. A pair i; j  can be a candidate in C2 only if all of the following are true: i. i is in L1 . ii. j is in L1 . iii. i; j  hashes to a frequent bucket. It is the last condition that distinguishes PCY from straight a-priori and reduces the requirements for memory in pass 2. d During pass 2, we consider each basket, and each pair of its items, making the test outlined above. If a pair meets all three conditions, add to its count in memory, or create an entry for it if one does not yet exist. When does PCY beat a-priori? When there are too many pairs of items from L1 to t a table of candidate pairs and their counts in main memory, yet the number of frequent buckets in the PCY algorithm is su ciently small that it reduces the size of C2 below what can t in memory even with 1 16 of it given over to the bitmap. When will most of the buckets be infrequent in PCY? When there are a few frequent pairs, but most pairs are so infrequent that even when the counts of all the pairs that hash to a given bucket are added, they still are unlikely to sum to s or more. 2.7 The Iceberg" Extensions to PCY 1. Multiple hash tables : share memory between two or more hash tables on pass 1, as in Fig. 2. On pass 2, a bitmap is stored for each hash table; note that the space needed for all these bitmaps is exactly the same as what is needed for the one bitmap in PCY, since the total number of buckets represented is the same. In order to be a candidate in C2, a pair must: a Consist of items from L1 , and b Hash to a frequent bucket in every hash table. Count items Hash table 1 Hash table 2 Counts for candidate pairs Frequent items Bitmaps Pass 1 Pass 2 Figure 2: Multiple hash tables memory utilization 2. Iterated hash tables Multistage : Instead of checking candidates in pass 2, we run another hash table di erent hash function! in pass 2, but we only hash those pairs that meet the test of PCY; i.e., they are both from L1 and hashed to a frequent bucket on pass 1. On the third pass, we keep bitmaps from both hash tables, and treat a pair as a candidate in C2 only if: a Both items are in L1 . 6 b The pair hashed to a frequent bucket on pass 1. c The pair also was hashed to a frequent bucket on pass 2. Figure 3 suggests the use of memory. This scheme could be extended to more passes, but there is a limit, because eventually the memory becomes full of bitmaps, and we can't count any candidates. Count items Frequent items Bitmap Hash table Another hash table Frequent items Bitmap Bitmap Counts for candidate pairs Pass 1 Pass 2 Pass 3 Figure 3: Multistage hash tables memory utilization When does multiple hash tables help? When most buckets on the rst pass of PCY have counts way below the threshold s. Then, we can double the counts in buckets and still have most buckets below threshold. When does multistage help? When the number of frequent buckets on the rst pass is high e.g., 50, but not all buckets. Then, a second hashing with some of the pairs ignored may reduce the number of frequent buckets signi cantly. The methods above are best when you only want frequent pairs, a common case. If we want all maximal frequent itemsets, including large sets, too many passes may be needed. There are several approaches to getting all frequent itemsets in two passes or less. They each rely on randomness of data in some way. 1. Simple approach : Taka a main-memory-sized sample of the data. Run a levelwise algorithm in main memory so you don't have to pay for disk I O, and hope that the sample will give you the truly frequent sets. Note that you must scale the threshold s back; e.g., if your sample is 1 of the data, use s=100 as your support threshold. You can make a complete pass through the data to verify that the frequent itemsets of the sample are truly frequent, but you will miss a set that is frequent in the whole data but not in the sample. To minimize false negatives, you can lower the threshold a bit in the sample, thus nding more candidates for the full pass through the data. Risk: you will have too many candidates to t in main memory. 2. SON95 Savasere, Omiecinski, and Navathe from 1995 VLDB; referenced by Toivonen. Read subsets of the data into main memory, and apply the simple approach" to discover candidate sets. Every basket is part of one such main-memory subset. On the second pass, a set is a candidate if it was identi ed as a candidate in any one or more of the subsets. 7 2.8 All Frequent Itemsets in Two Passes Key point: A set cannot be frequent in the entire data unless it is frequent in at least one subset. 3. Toivonen's Algorithm : a Take a sample that ts in main memory. Run the simple approach on this data, but with a threshold lowered so that we are unlikely to miss any truly frequent itemsets e.g., if sample is 1 of the data, use s=125 as the support threshold. b Add to the candidates of the sample the negative border: those sets of items S such that S is not identi ed as frequent in the sample, but every immediate subset of S is. For example, if ABCD is not frequent in the sample, but all of ABC; ABD; ACD, and BCD are frequent in the sample, then ABCD is in the negative border. c Make a pass over the data, counting all the candidate itemsets and the negative border. If no member of the negative border is frequent in the full data, then the frequent itemsets are exactly those candidates that are above threshold. d Unfortunately, if there is a member of the negative border that turns out to be frequent, then we don't know whether some of its supersets are also frequent, so the whole process needs to be repeated or we accept what we have and don't worry about a few false negatives. 8 3 Low-Support, High-Correlation Mining We continue to assume a market-basket" model for data, and we visualize the data as a boolean matrix, where rows = baskets and columns = items. Key assumptions: 1. Matrix is very sparse; almost all 0's. 2. The number of columns items is su ciently small that we can store something per column in main memory, but su ciently large that we cannot store something per pair of items in main memory same assumption we've made in all association-rule work so far. 3. The number of rows is so large that we cannot store the entire matrix in memory, even if we take advantage of sparseness and compress again, sames assumption as always. 4. We are not interested in high-support pairs or sets of columns; rather we want highly correlated pairs of columns. 3.1 Applications While marketing applications generally care only about high support it doesn't pay to try to market things that nobody buys anyway, there are several applications that meet the model above, especially the point about pairs of columns items with low support but high correlation being interesting: 1. Rows and columns are Web pages; r; c = 1 means that the page of row r links to the page of column c. Similar columns may be pages about the same topic. 2. Same as 1, but the page of column c links to the page of row r. Now, similar columns may represent mirror pages. 3. Rows = Web pages or documents; columns = words. Similar columns are words that appear almost always together, e.g., phrases." 4. Same as 3, but rows are sentences. Similar columns may indicate mirror pages or plagiarisms. 3.2 Similarity Example 3.1 : Think of a column as the set of rows in which the column has a 1. Then the similarity of two columns C1 and C2 is SimC1 ; C2 = jC1 C2 j=jC1 C2j. 0 1 1 0 1 0 1 0 1 = 2 5 = 40 similar 0 1 1 2 3.3 Signatures Key idea: map  hash" each column C to a small amount of data the signature, SigC  such that: 1. SigC  is small enough that a signature for each column can be t in main memory. 2. Columns C1 and C2 are highly similar if and only if SigC1  and SigC2  are highly similar. But note that we need to de ne similarity" for signatures. 9 An idea that doesn't work: Pick 100 rows at random, and make that string of 100 bits be the signature for each column. The reason is that the matrix is assumed sparse, so many columns will have an all-0 signature even if they are quite dissimilar. Useful convention: given two columns C1 and C2, we'll refer to rows as being of four types | a; b; c; d | depending on their bits in these columns, as follows: Type C1 C2 a 11 b 10 c 01 d 00 We'll also use a as the number of rows of type a," and so on. Note, SimC1 ; C2 = a=a + b + c. But since most rows are of type d, a selection of, say, 100 random rows will be all of type d, so the similarity of the columns in these 100 rows is not even de ned. 3.4 Min Hashing Imagine the rows permuted randomly in order. Hash" each column C to hC , the number of the rst row in which column C has a 1. The probability that hC1 = hC2  is a=a + b + c, since the hash values agree if the rst row with a 1 in either column is of type a, and they disagree if the rst such row is of type b or c. Note this probability is the same as SimC1 ; C2. If we repeat the experiment, with a new permutation of rows a large number of times, say 100, we get a signature consisting of 100 row numbers for each column. The similarity" of these lists fraction of positions in which they agree will be very close to the similarity of the columns. Important trick: we don't actually permute the rows, which would take many passes over the entire data. Rather, we read the rows in whatever order, and hash each row using say 100 di erent hash functions. For each column we maintain the lowest hash value of a row in which that column has a 1, independently for each of the 100 hash functions. After considering all rows, we shall have for each column the rst rows in which the column has 1, if the rows had been permuted in the orders given by each of the 100 hash functions. 3.5 Locality-Sensitive Hashing Problem: we've got signatures for all the columns in main memory, and similar signatures mean similar columns, with high probability, but there still may be so many columns that doing anything that is quadratic in the number of columns, even in main memory, is prohibitive. Locality-sensitive hashing LSH is a technique to be used in main memory for approximating the set of similar column-pairs with a lot less than quadratic work. The goal: in time proportional to the number of columns, eliminate as possible similar pairs the vast majority of the column pairs. 1. Think of the signatures as columns of integers. 2. Partition the rows of the signatures into bands, say l bands of r rows each. 3. Hash the columns in each band into buckets. A pair of columns is a candidate-pair if they hash to the same bucket in any band. 4. After identifying candidates, verify each candidate-pair Ci ; Cj  by examining SigCi  and SigCj  for similarity. 10 100 integers each. The signatures take 40Mb of memory, not too much by today's standards. Suppose we want pairs that are 80 similar. We'll look at the signatures, rather than the columns, so we are really identifying columns whose signatures are 80 similar | not quite the same thing. If two columns are 80 similar, then the probability that they are identical in any one band of 5 integers is 0:85 = 0:328. The probability that they are not similar in any of the 20 bands is 1 , :32820 = :00035. Thus, all but about 1 3000 of the pairs with 80-similar signatures will be identi ed as candidates. Now, suppose two columns are only 40 similar. Then the probability that they are identical in one band is 0:45 = :01, and the probability that they are similar in at least one of the 20 bands is no more than 0.2. Thus, we can skip at least 4 5 of the pairs that will turn out not to be candidates, if 40 is the typical similarity of columns. In fact, most pairs of columns will be a lot less than 40 similar, so we really eliminate a huge fraction of the dissimilar columns. 2 Example 3.2 : To see the e ect of LSH, consider data with 100,000 columns, and signatures consisting of Min hashing requires that we hash each row number k times, if we want a signature of k integers. With k-min hashing. In k-min hashing we instead hash each row once, and for each column, we take the k lowest-numbered rows in which that column has a 1 as the signature. To see why the similarity of these signatures is almost the same as the similarity of the columns from which they are derived, examine Fig. refkmin- g. This gure represents the signatures Sig1 and Sig2 for columns C1 and C2, respectively, as if the rows were permuted in the order of their hash values, and rows of type d neither column has 1 are omitted. Thus, we see only rows of types a, b, and c, and we indicate that a row is in the signature by a 1. Sig1 1 1 1 0 . . . 1 1 1 Sig2 1 0 1 1 . . . 1 0 1 3.6 k -Min Hashing 100 1’s 100 1’s Figure 4: Example of the signatures of two columns using k-min hashing Let us assume c  b, so the typical situation assuming k = 100 is as shown in Fig. 4: the top 100 rows in the rst column includes some rows that are not among the top 100 rows for the second column. Then an estimate of the similarity of Sig1 and Sig2 can be computed as follows: 100 jSig1 S ig2 j = a + a c because on average, the fraction of the 100 top rows of C2 that are also rows of C1 is a=a + c. Also: 100 jSig1 S ig2 j = 100 + a + c c 11 The argument is that all 100 rows of Sig1 are in the union. In addition, those rows of Sig2 that are not rows of Sig1 are in the union, and the latter set of rows is on average 100c=a + c rows. Thus, the similarity of Sig1 and Sig2 is: jSig1 S ig2 j = 100ca = a a+ jSig1 S ig2 j 100 + 100cc a + 2c a+ Note that if c is close to b, then the similarity of the signatures is close to the similarity of the columns, which is a=a + b + c. In fact, if the columns are very similar, then b and c are both small compared to a, and the similarities of the signatures and columns must be close. 3.7 Ampli cation of 1's Hamming LSH If columns are not sparse, but have about 50 1's, then we don't need min-hashing; a random collection of rows serves as a signature. Hamming LSH constructs a series of matrices, each with half as many rows as the previous, by OR-ing together two consecutive rows from the previous, as in Fig. 5. OR OR OR OR Figure 5: Construction of a series of exponentially smaller, denser matrices There are no more than log n matrices if n is the number of rows. The total number of rows in all matrices is 2n, and they can all be computed with one pass through the original matrix, storing the large ones on disk. In each matrix, produce as candidate pairs those columns that: 1. Have a medium density of 1's, say between 20 and 80, and 2. Are likely to be similar, based on an LSH test. Note that the density range 20 80 guarantees that any two columns that are at least 50 similar will be considered together in at least one matrix, unless by bad luck their relative densities change due to the OR operation combining two 1's into one. A second pass through the original data con rms which of the candidates are really similar. This method exploits an idea that can be useful elsewhere: similar columns have similar numbers of 1's, so there is no point ever comparing columns whose numbers of 1's are very di erent. 12 4 Query Flocks Goal: apply a-priori trick and other association-rule tricks to a more general class of complex queries. 4.1 Query Flock Notation A query ock is a generate-and-test system consisting of: 1. A query with parameters; we write the query in Datalog to simplify certain optimizations later. 2. A lter condition that says when the values of the parameters yields a query result that we accept. Note that the query ock is really a single query about its parameters; the parametrized-query component is not the real query. Example 4.1 : Frequent item pairs in a relation BasketsBID; item can be written as the query ock: Answerb - Basketsb,$1 AND Basketsb,$2 =s COUNTAnswer If we replace parameters $1 and $2 by values, e.g., diapers" and beer," respectively, then the query is asking for the set of basket ID's such that the basket contains both diapers and beer. The condition on the answer says that there must be at least s such baskets, where s is the support threshold. Thus, this query ock asks the usual question about the parameters $1 and $2: which pairs of items appear in at least s baskets?" 2 Example 4.2 : Here is a less usual example. It supposes relations: 1. Custname; attr; value. Tuple n; a; v means the customer with name n has value v for attribute a. For instance, Sue; age; 45 means that Sue is of age 45. 2. Buysname; prod tells what products each customer buys. 3. Typeprod; type tells the type of each product, e.g., product Coke" is of type soft drink." Here is the query ock that asks for values of some attribute that occur at least s times among buyers of a certain type of product: Answern - Custn,$a,$v AND Buysn,p AND Typep,$t =s COUNTAnswer 2 4.2 Execution Strategies The analog of a-priori is the observation that if we delete one or more subgoals from a Datalog query, the size of the set of answers can only increase. Our hope is that by computing some temporary relations using a subset of the subgoals, we can lter the sets of values for one or more parameters, using computations that are much less expensive than computing the entire query about the full set of parameters. We can describe the intermediate steps, as well as the nal computation of the parameter-values that pass the test by a sequence of steps of the form Relation := FILTER parameters , query , condition  The query is the ock query, with zero or more subgoals eliminated. A requirement is that this query be safe ; i.e., every variable appearing in the head appears in a nonnegated subgoal involving a relation i.e., not a subgoal involving an arithmetic comparison like a b. 13 The parameters are those appearing in the query. The condition is the same as the condition of the ock itself. Example 4.3 : The ock of Example 4.1 might be solved by using the rst subgoal to lter $1 and the second subgoal to lter $2. OK1$1 := FILTER $1 , Answerb - Basketsb,$1, COUNTAnswer = s OK2$2 := FILTER $2 , Answerb - Basketsb,$2, COUNTAnswer = s OK$1,$2 := FILTER $1,$2 , Answerb - Basketsb,$1 AND Basketsb,$2 AND OK1$1 AND OK2$2, COUNTAnswer = s Of course a clever ocks compiler recognizes that these two ltering steps are really the same and only computes one of OK1 and OK2. The reason a-priori often saves a lot of time is because the join of four relations at the last step computation of OK $1; $2 can be carried out in an order that reduces the size of intermediate relations, when compared with just joining Baskets with itself, as suggested by the ordering of Fig. 6. JOIN JOIN Baskets OK1 Baskets JOIN OK2 Figure 6: Preferred order for join in market-basket ock Notice that the ordering in Fig. 6 is not a left-deep ordering, which suggests that the typical commercial DBMS would not nd this order, and a query- ocks compiler needs to feed simpler queries to the DBMS so the right order of join is used by the DBMS. 2 Example 4.4 : Now let us consider how we might use lter steps to improve the running time of the nal join in Example 4.2. Using just the Cust subgoal is a lter on f$a; $vg, but there is no useful lter for just one of these parameters. We cannot use: Answern - Typep,$t to lter $t, because the query is not safe n appears in the head but not the body. However, Answern - Buysn,p AND Typep,$t is safe and may be used. A possible plan for optimizing this query ock is in Fig. 7. Figure 8 shows the preferred join order for the nal step. 2 14 OK1$a,$v := FILTER $a,$v , Answern - Custn,$a,$v, COUNTAnswer = s OK2$t := FILTER $t , Answern - Buysn,p AND Typep,$t, COUNTAnswer = s OK$a,$v,$t := FILTER $a,$v,$t , Answern - Custn,$a,$v, AND Buysn,p AND Typep,$t AND OK1$a,$v AND OK2$t, COUNTAnswer = s Figure 7: Query- ock plan for Example 4.2 JOIN JOIN Buys OK1 JOIN Cust Type JOIN OK2 Figure 8: Join order for nal step in Fig. 7 15 5 Web Search Outline: 1. Page rank, for discovering the most important" pages on the Web, as used in Google. 2. Hubs and authorities, a more detailed evaluation of the importance of Web pages using a variant of the eigenvector calculation used for Page rank. 5.1 Page Rank Intuitively, we solve the recursive de nition of importance": a page is important if important pages link to it. Create a stochastic matrix of the Web; that is: 1. Each page i corresponds to row i and column i of the matrix. 2. If page j has n successors links, then the ij th entry is 1=n if page i is one of these n successors of page j , and 0 otherwise. The intuition behind this matrix is: Imagine that initially each page has one unit of importance. At each round, each page shares whatever importance it has among its successors, and receives new importance from its predecessors. Eventually, the importance of each page reaches a limit, which happens to be its component in the principal eigenvector of this matrix. That importance is also the probability that a Web surfer, starting at a random page, and following random links from each page will be at the page in question after a long series of links. Example 5.1 : In 1839, the Web consisted on only three pages | Netscape, Microsoft, and Amazon. The links among these pages were as shown in Fig. 9. Ne MS Am Figure 9: The Web in 1839 Let n; m; a be the vector of importances for the three pages: Netscape, Microsoft, Amazon, in that order. Then the equation describing the asymptotic values of these three variables is: 2 32 32 3 n 1=2 0 1=2 n 4 m 5=4 0 0 1=2 5 4 m 5 a 1=2 1 0 a For example, the rst column of the matrix re ects the fact that Netscape divides its importance between itself and Amazon. The second column indicates that Microsoft gives all its importance to Amazon. We can solve equations like this one by starting with the assumption n = m = a = 1, and applying the matrix to the current estimate of these values repeatedly. The rst four iterations give the following estimates: 16 n=11 54 98 54 m = 1 1 2 3 4 1 2 11 16 a = 1 32 1 11 8 17 16 In the limit, the solution is n = a = 6=5; m = 3=5. That is, Netscape and Amazon each have the same importance, and twice the importance of Microsoft well this was 1839. 2 Note that we can never get absolute values of n, m, and a, just their ratios, since the initial assumption that they were each 1 was arbitrary. Since the matrix is stochastic sum of each column is 1, the above relaxation process converges to the principal eigenvector. 1. Dead ends : a page that has no successors has nowhere to send its importance. Eventually, all importance will leak out of" the Web. 2. Spider traps : a group of one or more pages that have no links out of the group will eventually accumulate all the importance of the Web. Example 5.2 : Suppose Microsoft tries to duck charges that it is a monopoly by removing all links from its site. The new Web is as shown in Fig. 10, and the matrix describing transitions is: 2 32 32 3 n 1=2 0 1=2 n 4 m 5=4 0 0 1=2 5 4 m 5 a 1=2 0 0 a 5.2 Problems With Real Web Graphs Ne MS Am Figure 10: Microsoft becomes a dead end The rst four steps of the iterative solution are: n=11 34 58 12 m = 1 1 2 1 4 1 4 3 16 a = 1 1 2 1 2 3 8 5 16 Eventually, each of n, m, and a become 0; i.e., all the importance leaked out. 2 Example 5.3 : Angered by the decision, Microsoft decides it will link only to itself from now on. Now, Microsoft has become a spider trap. The new Web is in Fig. 11, and the equation to solve is: 3 32 32 2 n 1=2 0 1=2 n 4 m 5=4 0 1 1=2 5 4 m 5 a 1=2 0 0 a The rst steps of the solution are: 17 Ne MS Am Figure 11: Microsoft becomes a spider trap n=11 34 58 12 m = 1 32 74 2 35 16 a = 1 1 2 1 2 3 8 5 16 Now, m converges to 3, and n = a = 0. 2 5.3 Google Solution to Dead Ends and Spider Traps Instead of applying the matrix directly, tax" each page some fraction of its current importance, and distribute the taxed importance equally among all pages. Example 5.4 : If we use a 20 tax, the equation of Example 5.3 becomes: 2 4 n 1=2 0 1=2 m 5 = 0:8 4 0 1 1=2 a 1=2 0 0 3 2 32 54 n 0:2 m 5 + 4 0:2 a 0:2 3 2 3 5 The solution to this equation is n = 7=11; m = 21=11; a = 5=11. Note that the sum of the three values is not 3, but there is a more reasonable distribution of importance than in Example 5.3. 2 5.4 Google Anti-Spam Devices Spamming" is the attempt by many Web sites to appear to be about a subject that will attract surfers, without truly being about that subject. Google, like other search engines, tries to match the words in your query to the words on the Web pages. However, Google, unlike other engines tends to believe what others say about you in their anchor text, making it harder for you to appear to be about something you are not. The use of Page rank to measure importance, rather than the more naive number of links into the page" also protects against spammers. The naive measure can be fooled by the spammer who creates 1000 pages that mutually link to one another, while Page rank recognizes that none of the pages have any real importance. 5.5 Hubs and Authorities Intuitively, we de ne hub" and authority" in a mutually recursive way: a hub links to many authorities, and an authority is linked to by many hubs. 18 Authorities turn out to be pages that o er information about a topic, e.g., the Quest home page about the IBM data-mining project. Hubs are pages that don't provide the information, but tell you where to nd the information, e.g., the CS345 home page. Uses a matrix formulation similar to that of Page rank, but without the stochastic restriction. We count each link as 1, regardless of how many successors or predecessors a page has. Repeated application of the matrix leads to divergence, but we can introduce scaling factors and keep the computed values of authority" and hubbiness" for each page within nite bounds. De ne a matrix A whose rows and columns correspond to Web pages, with entry Aij = 1 if page i links to page j , and 0 if not. Notice that AT , the transpose of A, looks like the matrix used for computing Page rank, but AT has 1's where the Page-rank matrix has fractions. Let ~ and ~ be vectors, whose ith component corresponds to the degrees of authority and hubbiness of the a h ith page. let  and  be suitable scaling factors to be determined later. Then we can state: 1. ~ = A~ . That is, the hubbiness of each page is the sum of the authorities of all the pages it links to, h a scaled by . 2. ~ = AT ~ . That is, the authority of each page is the sum of the hubbiness of all the pages that link a h to it, scaled by . We can derive from 1 and 2, using simple substitution, two equations that relate vectors ~ and ~ only a h to themselves: ~ = AT A~ ; ~ = AAT ~ a ah h As a result, we can compute ~ and ~ by relaxation, giving us the principal eigenvectors of the matrices h a AAT and AT A, respectively. Ne MS Am Figure 12: Web for Example 5.5 Example 5.5 : Consider the Web of Fig. 12. The relevant matrices are: 2 A=4 111 001 110 3 5 2 AT = 4 101 101 110 3 5 2 AAT = 4 312 110 202 3 5 2 AT A = 4 221 221 112 3 5 If we use  =  = 1 and assume that the vectors ~ = hn; hm ; ha and ~ = an; am; aa are each initially h a ~ are: 1; 1; 1 , the rst three iterations of the equations for ~ and h a 19 an = 1 5 24 114 am = 1 5 24 114 aa = 1 4 18 84 hn = 1 6 28 132 hm = 1 2 8 36 ha = 1 4 20 96 For instance, the vector ~ , properly scaled, will converge to a vector where an = am , and each of these is a p greater than aa in the ratio 1 + 3 : 2, or about 1.36. 2 20 6 Mining the Web Outline: 1. Dynamic itemset counting : Searching for interesting sets of items in a space too large ever to consider even each pair of items. 2. Books and authors" : Sergey Brin's intriguing experiment to mine the Web for relational data. 6.1 Finding Unusual Itemsets The problem is to nd sets of words that appear together unusually often" on the Web, e.g., New" and York" or f Dutchess", of", York"g. Unusually often" can be de ned in various ways, in order to capture the idea that the number of Web documents containing the set of words is much greater than what one would expect if words were sprinkled at random, each word with its own probability of occurrence in a document. One appropriate way is entropy per word in the set. Formally, the interest of a set of words S is log2 Q probS  w in S probw jS j Note that we divide by the size of S to avoid the Bonferroni e ect," where there are so many sets of a given size that some, by chance alone, will appear to be correlated. Example: If words a, b, and c each appear in 1 of all documents, and S = fa; b; cg appears in 0.1 , of documents, then the interestingness of S is log2 :001=:01  :01  :01 =3 = log21000=3 or about 3.3. Technical problem: interest is not monotone, or downwards closed," the way high support is. That is, we can have a set S with a high value of interest, yet some, or even all, of its immediate proper subsets are not interesting. In contrast, if S has high support, then all of its subsets have support at least as high. Technical problem: With more than 108 di erent words appearing in the Web, it is not possible even to consider all pairs of words. 6.2 The DICE Engine DICE dynamic itemset counting engine repeatedly visits the pages of the Web, in a round-robin fashion. At all times, it is counting occurrences of certain sets of words, and of the individual words in that set. The number of sets being counted is small enough that the counts t in main memory. From time to time, say every 5000 pages, DICE reconsiders the sets that it is counting. It throws away those sets that have the lowest interest, and replaces them with other sets. The choice of new sets is based on the heavy edge property, which is an experimentally justi ed observation that those words that appear in a high-interest set are more likely than others to appear in other high-interest sets. Thus, when selecting new sets to start counting, DICE is biased in favor of words that already appear in high-interest sets. However, it does not rely on those words exclusively, or else it could never nd high-interests sets composed of the many words it has never looked at. Some but not all of the constructions that DICE uses to create new sets are: 1. Two random words. This is the only rule that is independent of the heavy edge assumption, and helps new words get into the pool. 2. A word in one of the interesting sets and one random word. 21 3. Two words from two di erent interesting pairs. 4. The union of two interesting sets whose intersection is of size 2 or more. 5. fa; b; cg if all of fa; bg, fa; cg, and fb; cg are found to be interesting. Of course, there are generally too many options to do all of the above in all possible ways, so a random selection among options, giving some choices to each of the rules, is used. The general idea is to search the Web for facts of a given type, typically what might form the tuples of a relation such as Bookstitle; author. The computation is suggested by Fig. 13. Current patterns Sample data Find patterns Find data 6.3 Books and Authors Current data Figure 13: Extracting relations from the Web 1. Start with a sample of the tuples one would like to nd. In the example discussed in the Brin paper, ve examples of book titles and their authors were used. 2. Given a set of known examples, nd where that data appears on the Web. If a pattern is found that identi es several examples of known tuples, and is su ciently speci c that it is unlikely to identify too much, then accept this pattern. 3. Given a set of accepted patterns, nd the data that appears in these patterns, add it to the set of known data. 4. Repeat steps 2 and 3 several times. In the example cited, four rounds were used, leading to 15,000 tuples; about 95 were true title-author pairs. 6.4 What is a Pattern? The notion suggested consists of ve elements: 1. The order ; i.e., whether the title appears prior to the author in hte text, or vice-versa. In a more general case, where tuples have more than 2 components, the order would be the permutation of components. 2. The URL pre x. 3. The pre x of text, just prior to the rst of the title or author. 4. The middle : text appearing between the two data elements. 5. The su x of text following the second of the two data elements. Both the pre x and su x were limited to 10 characters. Example 6.1 : A possible pattern might consist of the following: 22 1. Order: title then author. 2. URL pre x: class 3. Pre x, middle, and su x of the following form: LI I title I by author P Here the pre x is LI I , the middle is I by including the blank after by", and the su x is P . The title is whatever appears between the pre x and middle; the author is whatever appears between the middle and su x. The intuition behind why this pattern might be good is that there are probably lots of reading lists among the class pages at Stanford. 2 To focus on patterns that are likely to be accurate, Brin used several constraints on patterns, as follows: Let the speci city of a pattern be the product of the lengths of the pre x, middle, su x, and URL pre x. Roughly, the speci city measures how likely we are to nd the pattern; the higher the speci city, the fewer occurrences we expect. Then a pattern must meet two conditions to be accepted: 1. There must be at least 2 known data items that appear in this pattern. 2. The product of the speci city of the pattern and the number of occurrences of data items in the pattern must exceed a certain threshold T not speci ed. 6.5 Data Occurrences An occurrence of a tuple is associated with a pattern in which it occurs; i.e., the same title and author might appear in several di erent patterns. Thus, a data occurrence consists of: 1. The particular title and author. 2. The complete URL, not just the pre x as for a pattern. 3. The order, pre x, middle, and su x of the pattern in which the title and author occurred. 6.6 Finding Data Occurrences Given Data If we have some known title-author pairs, our rst step in nding new patterns is to search the Web to see where these titles and authors occur. We assume that there is an index of the Web, so given a word, we can nd pointers to all the pages containing that word. The method used is essentially a-priori: 1. Find pointers to all those pages containing any known author. Since author names generally consist of 2 words, use the index for each rst name and last name, and check that the occurrences are consecutive in the document. 2. Find pointers to all those pages containing any known title. Start by nding pages with each word of a title, and then checking that the words appear in order on the page. 3. Intersect the sets of pages that have an author and a title on them. Only these pages need to be searched to nd the patterns in which a known title-author pair is found. For the pre x and su x, take the 10 surrounding characters, or fewer if there are not as many as 10. 23 6.7 Building Patterns from Data Occurrences 1. Group the data occurrences according to their order and middle. For example, one group in the group-by" might correspond to the order title-then-author" and the middle I by . 2. For each group, nd the longest common pre x, su x, and URL pre x. 3. If the speci city test for this pattern is met, then accept the pattern. 4. If the speci city test is not met, then try to split the group into two by extending the length of the URL pre x by one character, and repeat from step 2. If it is impossible to split the group because there is only one URL then we fail to produce a pattern from the group. class cs345 index.html class cs145 intro.html class cs140 readings.html Example 6.2 : Suppose our group contains the three URL's: The common pre x is class cs . If we have to split the group, then the next character, 3 versus 1, breaks the group into two, with those data occurrences in the rst page there could be many such occurrences going into one group, and those occurrences on the other two pages going into another. 2 6.8 Finding Occurrences Given Patterns 1. Find all URL's that match the URL pre x in at least one pattern. 2. For each of those pages, scan the text using a regular expression built from the pattern's pre x, middle, and su x. 3. Extract from each match the title and author, according the order speci ed in the pattern. 24 7 Clustering Given points in some space | often a high-dimensional space | group the points into a small number of clusters, each cluster consisting of points that are near" in some sense. Some applications: 1. Many years ago, during a cholera outbreak in London, a physician plotted the location of cases on a map, getting a plot that looked like Fig. 14. Properly visualized, the data indicated that cases clustered around certain intersections, where there were polluted wells, not only exposing the cause of cholera, but indicating what to do about the problem. Alas, not all data mining is this easy, often because the clusters are in so many dimensions that visualization is very hard. xx x xx x xx xxx xx x x xx xxx x Figure 14: Clusters of cholera cases indicated where the polluted wells were 2. Skycat clustered 2  109 sky objects into stars, galaxies, quasars, etc. Each object was a point in a space of 7 dimensions, with each dimension representing radiation in one band of the spectrum. The Sloan Sky Survey is a more ambitious attempt to catalog and cluster the entire visible universe. 3. Documents may be thought of as points in a high-dimensional space, where each dimension corresponds to one possible word. The position of a document in a dimension is the number of times the word occurs in the document or just 1 if it occurs, 0 if not. Clusters of documents in this space often correspond to groups of documents on the same topic. 7.1 Distance Measures To discuss whether a set of points is close enough to be considered a cluster, we need a distance measure Dx; y that tells how far points x and y are. The usual axioms for a distance measure D are: 1. Dx; x = 0. A point is distance 0 from itself. 2. Dx; y = Dy; x. Distance is symmetric. 3. Dx; y  Dx; z  + Dz; y. The triangle inequality. Often, our points may be thought to live in a k-dimensional Euclidean space, and the distance between any two points, say x = x1; x2; : : :; xk and y = y1; y2 ; : : :; yk is given in one of the usual manners: 1. Common distance  L2 norm": q Pk 2. Manhattan distance  L1 norm": 3. Max of dimensions  L1 norm": i=1 xi , yi 2 . Pk i=1 jxi , yi j. maxk=1 jxi , yi j. i When there is no Euclidean space in which to place the points, clustering becomes more di cult. Here are some examples where a distance measure without a Euclidean space makes sense. 25 dimension corresponds to one word. However, computation of distances would be prohibitive in a space this large. A better approach is to base the distance Dx; y on the dot product of vectors corresponding to x and y, since we then have to deal only with the words actually present in both x and y. We need to compute the lengths of the vectors involved, which is the square root of the sum of the squares of the numbers of occurrences of each word. The sum of the product of the occurrences of each word in each document is divided by each of the lengths to get a normalized dot product. We subtract this quantity from 1 to get the distance between x and y. For instance, suppose there were only 4 words of interest, and the vectors x = 2; 0; 3; 1 and y = 5; 3; 2; 0 represented the numbers of occurrences of these words in the twop documents. The dot product x:y is p 2  5 + 0  3 + 3p 2 + 1  0 = 16, the length of the rst vector is 22 + 02 + 32 + 12 = 14, and the length  p of the second is 52 + 32 + 22 + 02 = 38. Thus, p Dx; y = 1 , p 16 = 0:304 14 38 As another example, suppose document x has word-vector a1; a2; : : : , and y is two copies of x; i.e., y = 2a1; 2a2; : : : . Then P 2 ia =0 Dx; y = 1 , pP 2p2Pi i ai i2ai 2 That is, x and y are essentially the same document; surely they are on the same topic, and deserve to be clustered together. 2 Example 7.1 : We can think of Web pages as points in the roughly 10 -dimensional space where each 8 Example 7.2 : Character strings, such as DNA sequences may be similar even though there are some insertions and deletions as well as changes in some characters. For instance, abcde and bcdxye are rather similar, even though they don't have any positions in common, and don't even have the same length. Thus, instead of trying to construct a Euclidean space with one dimension for each position, we can de ne the distance function Dx; y = jxj + jyj , 2jLCS x; yj, where LCS stands for the longest common subsequence of x and y. In our example, LCS abcde; bcdxye is bcde, of length 4, so Dabcde; bcdxye = 5 + 6 , 2  4 = 3; i.e., the strings are fairly close. 2 7.2 The Curse of Dimensionality An unintuitive consequence of working in a high-dimensional space is that almost all pairs of points are about as far away as average. maximum possible distance, 2. However, suppose k is large, say 100, or 100,000. Regardless of which norm we use, L2 , L1 , or L1 , we know that Dx; y  maxi jxi , yi j, if x = x1; x2; : : : and y = y1 ; y2; : : : . For large k, it is very likely that there will be some dimension i such that xi and yi are almost as di erent as possible, even if x and y are very close in other dimensions. Thus, Dx; y is going to be very close to 1. 2 Another interesting consequence of high-dimensionality is that all vectors, such as x = x1; x2; : :,: and y = y1 ; y2; : : : are almost orthogonal. The reason is that if we project x and y onto any of the k planes 2 formed by two of the k axes, there is going to be one in which the projected vectors are almost orthogonal. Example 7.3 : Suppose we throw points at random into a k-dimensional unit cube. If k = 2, we expect that the points will spread out in the plane, with some very nearby points and some pairs at almost the p 7.3 Approaches to Clustering At a high level, we can divide clustering algorithms into two broad classes: 1. Centroid approaches. We guess the centroids or central point in each cluster, and assign points to the cluster of their nearest centroid. 26 2. Hierarchical approaches. We begin assuming that each point is a cluster by itself. We repeatedly merge nearby clusters, using some measure of how close two clusters are e.g., distance between their centroids, or how good a cluster the resulting group would be e.g., the average distance of points in the cluster from the resulting centroid. 7.4 Outline We shall consider the following algorithms for clustering; they di er in whether or not they assume a Euclidean distance, and in whether they use a centroid or hierarchical approach. 1. BFR: Centroid based; assumes Euclidean measure, with clusters formed by a Gaussian process in each dimension around the centroid. 2. Fastmap: Not really a clustering algorithm, but a way to construct a low-dimensional Euclidean space from an arbitrary distance measure. 3. GRGPF: Centroid-based, but uses only a distance measure, not a Euclidean space. 4. CURE: Hierarchical and Euclidean, this algorithm deals with odd-shaped clusters. Note that all the algorithms we consider are oriented toward clustering large amounts of data, in particular, data so large it does not t in main memory. Thus, we are especially concerned with the number of passes through the data that must be made, preferably one pass. 7.5 The k-Means Algorithm This algorithm is a popular main-memory algorithm, on which the BFR algorithm is based. k-means picks k cluster centroids and assigns points to the clusters by picking the closest centroid to the point in question. As points are assigned to clusters, the centroid of the cluster may migrate. Example 7.4 : For a very simple example of ve points in two dimensions, observe Fig. 15. Suppose we assign the points 1, 2, 3, 4, and 5 in that order, with k = 2. Then the points 1 and 2 are assigned to the two clusters, and become their centroids for the moment. 1 a 3 b 4 c 2 5 Figure 15: An example of the k-means algorithm When we consider point 3, suppose it is closer to 1, so 3 joins the cluster of 1, whose centroid moves to the point indicated as a. Suppose that when we assign 4, we nd that 4 is closer to 2 than to a, so 4 joins 2 in its cluster, whose center thus moves to b. Finally, 5 is closer to a than b, so it joins the cluster f1; 3g, whose centroid moves to c. 2 We can initialize the k centroids by picking points su ciently far away from any other centroid, until we have k. As computation progresses, we can decide to split one cluster and merge two, to keep the total at k. A test for whether to do so might be to ask whether so doing reduces the average distance from points to their centroids. 27 Having located the centroids of the k clusters, we can reassign all points, since some points that were assigned early may actually wind up closer to another centroid, as the centroids move about. If we are not sure of k, we can try di erent values of k until we nd the smallest k such that increasing k does not much decrease the average distance of points to their centroids. Example 7.5 illustrates this point. Example 7.5 : Consider the data suggested by Fig. 16. Clearly, k = 3 is the right number of clusters, but suppose we rst try k = 1. Then all points are in one cluster, and the average distance to the centroid will be high. xx xxx xx x x xx x x x xx xx x Average radius 1 2 k 3 4 Figure 16: Discovering that k = 3 is the right number of clusters Suppose we then try k = 2. One of the three clusters will be by itself and the other two will be forced into one cluster, as suggested by the dotted lines. The average distance of points to the centroid will thus shrink considerably. If k = 3, then each of the apparent clusters should be a cluster by itself, and the average distance from points to their centroids shrinks again, as indicated by the graph in Fig. 16. However, if we increase k to 4, then one of the true clusters will be arti cially partitioned into two nearby clusters, as suggested by the solid lines. The average distance to centroid will drop a bit, but not much. It is this failure to drop further that tips us o that k = 3 is right, even if the data is in so many dimensions that we cannot visualize the clusters. 2 7.6 The BFR Algorithm Based on k-means, this algorithm reads its data once, consuming a main-memory-full at a time. The algorithm works best if the clusters are normally distributed around a central point, perhaps with a di erent standard deviation in each dimension. Figure 17 suggests what the data belonging to a typical cluster in two-dimensions might look like. A centroid, marked by +, has points scattered around, with the standard deviation in the horizontal dimension being twice what it is in the vertical dimension. About 70 of the points will lie within the 1 ellipse; 95 will lie within 2 , 99.9 within 3 , and 99.9999 within 4 . 7.7 Representation of Clusters in BFR Having worked on Skycat, Usama Fayyad the F" in BFR probably thought of clusters as galaxies," as suggested in Fig. 18. A cluster consists of: 1. A central core, the Discard set DS. This set of points is considered certain to belong to the cluster. All the points in this set are replaced by some simple statistics, described below. Note: although called discarded" points, these points in truth have a signi cant e ect throughout the running of the algorithm, since they determine collectively where the centroid is and what the standard deviation of the cluster is in each dimension. 28 1σ 2σ Figure 17: Points in a cluster are assumed to have been generated by taking a centroid for the cluster and adding to it in each dimension a normally distributed random variable with mean 0 2. Surrounding subgalaxies, collectively called the Compression set CS. Each subcluster in the CS consists of a group of points that are su ciently close to each other that they can be replaced by their statistics, just like the DS for a cluster is. However, they are su ciently far away from any cluster's centroid, that we are not yet sure which cluster they belong to. 3. Individual stars that are not part of a galaxy or subgalaxy, the Retained set RS. These points can neither be assigned to any cluster nor can they be grouped into a subcluster of the CS. They are stored in main memory, as individual points, along with the statistics of the DS and CS. The statistics used to represent each cluster of the DS and each subcluster of the CS are: 1. The count of the number of points, N . 2. The vector of sums of the coordinates of the points in each dimension. The vector is called SUM, and the component in the ith dimension is SUMi . 3. The vector of sums of squares of the coordinates of the points in each dimension, called SUMSQ. The component in dimension i is SUMSQi . Note that these three pieces of information, totaling 2k + 1 numbers if there are k dimensions, are su cient to compute important statistics of a cluster or subcluster, and they are more convenient to maintain as points are added to clusters than would, say, be the mean and variance in each dimension. For instance: The coordinate i of the centroid of the cluster in dimension i is SUMi =N . The variance in dimension i is SUMSQi , SUMi 2 N N and the standard deviation i is the square root of that. 7.8 Processing a Main-Memory-Full of Points in BFR With the rst load of main memory, BFR selects the k cluster centroids, using some chosen main-memory algorithm, e.g., pick a sample of points, optimize the clusters exactly, and chose their centroids as the initial centroids. The entire main-memory full of points is then processed like any subsequent memory-load of points, as follows: 29 Compression set (CS) Retained set (RS) Discard set (DS) Figure 18: Examples of a cluster DS, subclusters CS, and individual points RS 1. Determine which points are su ciently close to a current centroid that they may be taken into the DS and their statistics N , SUM , SUMSQ combined with the prior statistics of the cluster. BFR suggests two ways to determine whether a point is close enough to a centroid to be taken into the DS: a Take all points whose Mahalanobis radius is below a certain threshold, say four times the standard deviation of the cluster. The Mahalanobis radius is essentially the distance from the centroid, scaled in each dimension by i , the standard deviation in that dimension. More precisely, if i is the mean in dimension i, then the radius of point y = y1; y2 ; : : : is s X yi , i 2 i i 2. 3. 4. 5. b Based on the number of points in the various clusters, ask whether it is likely say at the 95 level that the currently closest centroid will in the future move su ciently far away from the point y, and another centroid will move su ciently close to y that the latter is then closer to y than the former. If it is unlikely that the closest centroid for y will ever be di erent, then assign y to the DS, and place it in the cluster of the closest centroid. Adjust the statistics N , SUM and SUMSQ for each cluster to include the DS points just included. In main memory, attempt to cluster the points that have not yet been placed in the DS, including points of the RS from previous rounds. If we nd a cluster of points whose variance is below a chosen threshold, then we shall regard these points as a subcluster, replace them by their statistics, and consider them part of the CS. All other points are placed in the RS. Consider merging a new subcluster with a previous subcluster of the CS. The test for whether it is desirable to do so is that the combined set of points will have a variance below a threshold. Note that the statistics kept for CS subclusters is su cient to compute the variance of the combined set. If we are at the last round, i.e., there is no more data, then we can assign subclusters in CS and points in RS to their nearest cluster, even though they will of necessity be fairly far away from any cluster centroid. 30 8 More About Clustering We continue our discussion of large-scale clustering algorithms, covering: 1. Fastmap, and other ways to create a Euclidean space from an arbitrary distance measure. 2. The GRGPF algorithm for clustering without a Euclidean space. 3. The CURE algorithm for clustering odd-shaped clusters in a Euclidean space. 8.1 Simple Approaches to Building a Euclidean Space From a Distance Measure Any n points can be placed in n , 1-dimensional space, with distances preserved exactly. Example 8.1 : Figure 19 shows how we can place three points a, b, and c, with only a distance measure D given, in 2-dimensional space. Start by placing a and b distance Da; b apart. Draw a circle of radius Da; c around a and a circle of radius Db; c around b. By the triangle inequality, which D must obey, these circles intersect. Pick one of the two points of intersection as c. 2 c a D(a,c) D(a,b) b D(b,c) Figure 19: Placing three points in two dimensions However, we usually have far too many points to try to place them in one fewer dimension. A more brute-force approach to placing n points in a k-dimensional space, where k n, is called multidimensional scaling. 1. Start with the n points placed in k-dim space at random. 2. Take as the error" the energy of a system of springs, each of length Dx; y, that we imagine are strung between each pair of points x and y. The energy in a spring is the square of the di erence between its actual length and Dx; y. 3. Visit each point in turn, and try to place it in a position that minimizes the total energy of the springs. Since moving points around can a ect the optimal position of other points, we must visit points repeatedly, until no more improvements can be made. At this point, we are in a local optimum, which may or may not be the global optimum. Example 8.2 : Suppose we have three points in a 3-4-5 right triangle," as suggested in Fig. 20. If we try to place these points in one dimension, there must be some stretching and shrinking of springs. The optimum con guration, also shown in Fig. 20, is when the springs of length 3 and 4 are compressed to 7 3 and 10 3, while the spring of length 5 is stretched to 17 3. In this con guration, the total energy of the 7 system is 3 , 3 2 + 4 , 10 2 + 5 , 17 2 = 4=3. 2 3 3 31 b 5 a 3 c b 7/3 c 17/3 10/3 a 4 Figure 20: Optimal placement of three points in one dimension 8.2 Fastmap Problem: multidimensional scaling requires at least On2  time to handle n points, since we must compute all distances at least once | perhaps more than once. Fastmap is a method for creating k pseudo-axes that can serve to place n points in a k-dim space in Onk time. In outline, Fastmap picks k pairs of points ai; bi , each of which pairs serves as the ends" of one of the k axes of the k-dim space. Using the law of cosines, we can calculate the projection" x of any point c onto the line ab, using only the distances between points, not any assumed coordinates of these points in a plane. The diagram is shown in Fig. 21, and the formula is 2 D2 a; b 2 x = D a; c + 2Da; b , D b; c c D(a,c) D(b,c) a x D(a,b) b Figure 21: Computing the projection of point c onto line ab Having picked a pair of points a; b as an axis, part of the distance between any two points c and d is accounted for by the projections of c and d onto line ab, and the remainder of the distance is in other dimensions. If the projections of c and d are x and y, respectively, then in the future as we select other axes, the distance Dcurrentc; d should be related to the given distance function D by 2 Dcurrentc; d = D2 c; d , x , y2 The explanation is suggested by Fig. 22. Here, then, is the outline of the Fastmap algorithm. It computes for each point c, k projections, which we shall refer to as c1; c2; : : :; ck onto the k axes, which are determined by pairs of points a1 ; b1; a2 ; b2; : : :; ak ; bk. For i = 1; 2; : : :; k do the following: 1. Using the current distance Dcurrent, Pick ai and bi, as follows: a Pick a random point c. b Pick ai to be the point as far as possible from c, using distance Dcurrent. c Pick bi to be the point as far as possible from ai . 32 d D(c,d) D current (c,d) c a x y b Figure 22: Subtracting the distance along axis ab to get the current" distance between points c and d in other dimensions 2. For each point x, compute xi , using the law-of-cosines formula described above. 3. Change the de nition of Dcurrent to subtract the distance in the ith dimension as well as previous dimensions. That is s X Dcurrentx; y = D2 x; y , xj  , yj  2 Note that no computation is involved in this step; we just use this formula when we need to nd the current distance between speci c pairs of points in steps 1 and 2. j i 8.3 Ways to Use Fastmap in Clustering Algorithms 1. Map the data to k dimensions in Onk time. Cluster the points using any method that works for Euclidean spaces. Hope the resulting clusters are good unlikely, according to GRGPF. 2. Improvement: Map to k dimensions using Fastmap, but use the Euclidean space only to estimate the clustroids : points that are closest on average to the other members of their cluster like a centroid in a Euclidean space, but it has to be a particular point of the cluster, not a location that has no point. If there is a lot of data, we can use a sample for this stage. Then, assign points to clusters based on their true distance using the distance measure, not the Euclidean space to the clustroids. 8.4 Hierarchical Clustering A general technique that, if we are not careful, will take On2 time to cluster n points, is hierarchical clustering. 1. Start with each point in a cluster by itself. 2. Repeatedly select two clusters to merge. In general, we want to pick the two clusters that are closest, but there are various ways we could measure closeness." Some possibilities: a Distance between their centroids or if the space is not Euclidean, between their clustroids. b Minimum distance between nodes in the clusters. c Maximum distance between nodes in the clusters. d Average distance between nodes of the clusters. 3. End the merger process when we have few enough" clusters. Possibilities: 33 a Use a k-means approach | merge until only k clusters remain. b Stop merging clusters when the only clusters that can result from merging fail to meet some criterion of compactness, e.g., the average distance of nodes to their clustroid or centroid is too high. 8.5 The GRGPF Algorithm This algorithm assumes there is a distance measure D, but no Euclidean space. It also assumes that there is too much data to t in main memory. The data structure it uses to store clusters is like an R-tree. Nodes of the tree are disk blocks, and we store di erent things at leaf and interior nodes: In leaf blocks, we store cluster features that summarize a cluster in a manner similar to BFR. However, since there is no Euclidean space, the features are somewhat di erent, as follows: a The number of points in the cluster, N . b The clustroid: that point in the cluster that minimizes the rowsum, i.e., the sum of the squares of the distances to the other points of the cluster. ^ If C is a cluster, C will denote its clustroid. P ^ Thus, the rowsum of the clustroid is X in C DC; X . Notice that the rowsum of the clustroid is analogous to the statistic SUMSQ that was used in BFR. However, SUMSQ is relative to the origin of the Euclidean space, while GRGPF assumes no such space. The rowsum can be used to compute a statistic, the radius of the cluster that p analogous to the standard deviation of a cluster in BFR. is The formula is radius = rowsum=N . c The p points in the cluster that are closest to the clustroid and their rowsums, for some chosen constant p. d The p points in the cluster that are farthest from the clustroid. In interior nodes, we keep samples of the clustroids of the clusters represented by the descendants of this tree node. An e ort is made to keep the clusters in each subtree close. As in an R-tree, the interior nodes thus inform about the approximate region in which clusters at their descendants are found. When we need to insert a point into some cluster, we start at the root and proceed down the tree, choosing only those paths along which a reasonably close cluster might be found, judging from the samples at each interior node. 8.6 Maintenance of Cluster Features by GRGPF There is an initialization phase on a main-memory full of data, where the rst estimate of clusters is made, and the clusters themselves are arranged in a hierarchy corresponding to the R-tree." The need for this hierarchy was mentioned in item 8.5 above. We then examine all the other points and try to insert them into the existing clusters. The choice of clusters is reconsidered if there are too many to t the representations in main memory, or if a cluster gets too big too high a radius. There are many details about what happens when new points are added. Here are some of the key points: ^ a A new point X is placed in the cluster C such that DC; X  is a minimum. Use the samples at the interior nodes of the tree to avoid searching the entire tree and all the clusters. b If point X is added to cluster C , we add to each of the 2p + 1 rowsums maintained for that ^ cluster rowsums of C and the p closest and p furthest points the square of the distance from ^ that point to X . We also estimate the rowsum of X as Nr2 + NDC; X , where r is the radius of the cluster. The validity of this formula, as an approximation is, based on the property of the curse of dimensionality" we discussed in Section 7.2. That is, for each of the N points Y in the cluster, the distance between X and Y can be measured by going to the clustroid the term 34 ^ DC; X , and then from the clustroid to Y which is Y 's contribution to r. If we assume that ^ the lines in a hypothetical Euclidean space from C to X and Y are likely to be almost perpendicular, then the formula is justi ed by the Pythagorean theorem. c We must consider the possibility that after adding X , cluster C now hasa di erent clustroid, ^ which could be X or one of the p points closest to C . If, after accounting for X , one of these points has a lower rowsum, it becomes the clustroid. obviously, this process cannot be maintained inde nitely, since eventually the clustroid will migrate outside the p closest points. Thus, there may need to be periodic recomputation of the cluster features for a cluster, which is possible | at the cost of disk I O's | because all points in the cluster are stored on disk, even if they are not in the main-memory tree that stores the cluster features and samples. 8.7 Splitting and Merging Clusters in GRGPF Sometimes, the radius of the clustroid exceeds a given threshold, and the algorithm decides to split the cluster into two. This process requires bringing the entire cluster into main memory and performing some split algorithm on just these points in memory. An additional consequence is that the number of clusters in one leaf node increases by 1. If the node disk block over ows, then it must be split, as in a B-tree. That, in turn, may cause nodes up the tree to be split, and in the worst case, there is no room in main memory to hold the entire tree any more. In that case, the solution is to raise the threshold that the radius of a cluster may have, and consider merging clusters throughout the tree. Suppose we consider merging clusters Ci and Cj . We wish to avoid having to bring all of the points of these clusters into main memory, so we guess at the clustroid and the rowsums as follows: a Assume that the clustroid of the combined cluster will be one of the points furthest from the clustroid of either Ci or Cj . Remember that these points are all kept in the tree with their cluster features. b To decide on the new clustroid, we need to estimate the rowsum for each point X in either cluster. Suppose for example that X is in Ci. Then we estimate: , ^ ^ ^ ^ RowsumX  = Rowsumi X  + Nj D2 X; Ci  + D2 Ci; Cj  + Rowsumj Cj  Here, Nj is the number of points in Cj , and as always, hats indicate the clustroid of a cluster. The rationale for the above formula is that: i. The rst term represents the distances from X to all the nodes in its cluster. ii. The middle term represents part of the paths from X to each point in Cj . We assume that ^ ^ the path starts out going from X to Ci, the clustroid of Ci . From there, it goes to Cj . Note ^ that, by the curse of dimensionality," we may assume that the lines from X to Ci and from ^^ Ci to Cj are perpendicular" and combine them using the Pythagorean theorem. iii. The nal term represents the component of the paths from X to all the members of Cj that ^ continues after Cj is reached from X . Again, it is the curse of dimensionality" assumption ^ that lets us assume all the lines from Cj to members of Cj are orthogonal to the line from ^i to Cj . ^ C c Having decided on the new clustroid, compute the rowsums for all points that we retain in the cluster features of the new cluster using the same formula as in 2. 8.8 CURE We now return to clustering in a Euclidean space. The special problem solved by CURE is that when clusters are not neatly expressed as Gaussian noise around a central point as BFR assume many things can go wrong in a k-means approach. 35 Example 8.3 : Figure 23 suggests two possible problems. In a, even though k = 4 is the right number of clusters solid circles represent the extent of the clusters, an algorithm that tries to minimize distances to a centroid might well cluster points as suggested by the dotted lines, with the large cluster broken into three parts, and the three small clusters combined into one. In b, a long cluster could be interpreted as many round clusters, since that would minimize the average distance of points to the cluster centroids Notice, however, that by using the Mahalanobis radius, as in Section 7.8, a cluster that is long is one dimension is recognized as circular," and we would not expect BFR to make the mistake of Fig. 23b. 2 (a) 4 clusters are correct, but they’re the wrong 4! (b) One long cluster looks like many small ones! Figure 23: Problems with centroid-based clustering Here is an outline of the CURE algorithm: 1. Start with a main memory full of random points. Cluster these points using the hierarchical approach of Section 8.4. Note that hierarchical clustering tends to avoid the problems of Fig. 23, as long as the true clusters are dense with points throughout. 2. For each cluster, choose c sample" points for some constant c. These points are picked to be as dispersed as possible, then moved slightly closer to the mean, as follows: a Pick the rst sample point to be the point of the cluster farthest from the centroid. b Repeatedly pick additional sample points by choosing that point of the cluster whose minimum distance to an already chosen sample point is as great as possible. c When c sample points are chosen, move all the samples toward the centroid by some fractional distance, e.g., 20 of the way toward the centroid. As a result, the sample points need not be real points of the cluster, but that fact is unimportant. The net e ect is that the samples are typical" points, well dispersed around the cluster, no matter what the cluster's shape is. 3. Assign all points, including those involved in steps 1 and 2 to the nearest cluster, where nearest" means shortest distance to some sample point. Example 8.4 : Figure 24 suggests the process of picking c = 6 sample points from an elongated cluster, and then moving htem 20 of the way toward the centroid. 2 36 3 2 5 6 1 4 Figure 24: Selecting sample points in CURE 37 9 Sequence Matching Sequences are lists of values S = x1 ; x2; : : :; xk , although we shall often think of the same sequence as a continuous function de ned on the interval 0-to-1. That is, the sequence S can be thought of as sample values from a continuous function S t, with xi = S i=k. 9.1 Sequence-Matching Problems In the simplest case, we are given a collection of sequences fS1 ; S2 ; : : :; Sn g, and a query sequence Q, each of the same length. Our problem is to nd that sequence Si whose distance from Q is the , minimum, where R  distance" is de ned by the energy" of the di erence of the sequences; i.e., DS; T  = 01 S t , T t 2dt. For instance, the Si 's might be records of the prices of various stocks, and Q is the price of IBM stock, delayed by one day. If we found some Si that was very similar to Q, we could use the price of the stock Si to predict the price of IBM stock the next day, Notes: Do not try this at home. Anything easy to mine about stock prices is already being done, and the market has adjusted to whatever knowledge can be gleaned. Sequence matching is a great opportunity to violate the Bonferroni principal, since there has to be a closest sequence." For instance, a famous mistake was looking in the UN book of world statistics to nd the statistic that best predicted the Dow-Jones average. It was cotton production in Bangladesh." We could treat sequences of length k as points in a k-dim vector space, but doing so is not likely to be useful. Usually, k will be so high, that spacial index techniques like kd-trees or R trees will be useless. The trick adopted by Faloutsos and his colleagues is to map sequences to the rst few termsRof their Fourier transforms. Formally, the j th term of the Fourier Expansion of the function S t is 01 S te2jit dt. Recall that the imaginary exponential ei x is de ned to be sin x + i cos x. Thus, the real and imaginary parts of Xj tell how well S t matches sine and cosine functions that have j periods within the interval 0-to-1, i.e., sin 2jt and cos 2jt. Example 9.1 : Figure 25 suggests a simple function S t solid and compares it to the single-period sine function dotted and single-period cosine function dashed. The integral of the product S t sin 2t + iS t cos 2t is the complex number X1 . 2 9.2 Fourier Transforms as Indexes for Sequences Figure 25: Function S t matches the sine only slightly; the cosine better but not perfectly Key point: Parsival's Theorem states that the energy in a signal S t is the same as the energy in its P Fourier transform: i0 jXi j2. Note jX j is the magnitude of a complex number X . 38 Key application: If S t is actually the di erence of two sequences, then the distance between those R sequences, which is 01 S 2 tdt is the same as the sum of the squares of the magnitudes of the di erences of the Fourier coe cients of the two sequences. Since the square of a magnitude is always positive, we can nd all the sequences whose distance from a given query sequence Q is no more than 2 by nding all those sequences S the sum of whose magnitudes of the di erences of the rst m Fourier coe cients is no more than 2 . We shall retrieve some false drops" | sequences that are close to Q in their rst m coe cients, but di er greatly in their higher coe cients. There is some justi cation for the rule that all the energy is in the low-order Fourier coe cients," so retrieval based on the rst few terms is quite accurate in practice. 9.3 Matching Queries to Sequences of the Same Length As long as stored sequences and queries are of the same length, we can arrange the sequences in a simple, low-dimension data structure, as follows: 1. Compute the rst m Fourier coe cients of each sequence Si , for some small, xed m. It is often su cient to use m = 2 or m = 3. Since Fourier coe cients are complex, that is the equivalent of a 4or 6-dim space. 2. Place each sequence Si in this space according to the values of its rst m Fourier coe cients. The points themselves are stored in some suitable multidimensional data structure. 3. To search for all sequences whose distance from query Q is no more than 2, compute the rst m Fourier coe cients of Q, and look for points in the space no more distant than from the point corresponding to Q. 4. Since there are false drops, compare each retrieved sequence with Q to be sure that the distance is truly no greater than 2. In some applications, it may make sense to normalize" sequences and queries to make the 0th Fourier coe cient which is the average value 0 and perhaps to make the variance i.e., the energy" of each sequence the same. Example 9.2 : In Fig. 26a are two sequences that have a large di erence, but are in some sense the same sequence, shifted and scaled vertically. We could shift them vertically to make their averages be 0, as in Fig. 26b. However, that change still leaves one sequence always having twice the value of the other. If we also scaled them by multiplying the more slowly varying by 2, they would become the same sequence. 2 9.4 Queries That are Shorter than the Sequences Assume that queries are at least of length w. The sequential scan method for nding close within distance 2  matches to a query Q is as follows: 1. Store the rst m coe cients of the Fourier transform of each subsequence of length w for each sequence. As a result, if sequences are of length k, we store k , w + 1 points for each sequence. 2. Given a query Q, take the rst w values of Q and compute the Fourier transform of that sequence of length w. Retrieve each sequence S and position p that matches within . 3. For each such S and p, compare the entire query Q with sequence S starting at position p. If the distance is no more than 2, report the match. 39 (a) (b) Figure 26: Shifting and scaling can make two sequence that look rather di erent become the same 9.5 Trails The problem with sequential-scan is that the number of points that must be stored is almost the product of the number of sequences and the length of the sequences. The FRM paper takes advantage of the fact that as we slide a window of length w across a sequence S , the low-order Fourier coe cients will not vary too rapidly. We thus get a trail if we plot the points corresponding to consecutive windows of length w, as suggested in Fig. 27. Figure 27: Trials and the rectangles that bound them Instead of storing individual points, we store rectangles in the appropriate number of dimensions that bound the points in one segment of the trail, which is also suggested by Fig. 27. A rectangle can be stored using two opposite vertices, for example. Thus, if rectangles tend to represent many points, then the space ued to store the rectangles is much less than the space needed to store individual points. To retrieve the points whose distance from a query Q of length w is no more than , nd the rectangles that are within of Q. Match Q only against the sequences that correspond to at least one retrieved rectangle, and only at begining positions represented by those rectangles. Partitioning trails into rectangles is an opimization problem. FRM uses as the cost of storing a rectangle whose side in dimension i is Li as a fraction of the length of the total distance along the Q ith axis: i L + i + 0:5. For example, a rectangle of fractional sides 1=4 and 1=3 would have cost 40 3=45=6 = 5=8. Start from the beginning of the trail, and form rectangles by adding points as we treavel along the trail. When deciding whether or not to add another point to the current rectangle, make sure that the cost of the new rectangle per point covered decreases. If it increases, then start a new rectangle instead. 9.6 Matching Queries of Arbitrary Length If the query Q has length w, just nd the rectangles within distance from Q, as discussed in Section 9.5. Note that Q may actually be within a rectangle, and even so, Q may be distant from actual points on the trail, as suggested in Fig. 28. Q 1 Q 2 Figure 28: Two examples of queries near a rectangle; note that the query inside the rectangle is actually further from the points on the trail than the query outside the rectangle, but both need to be considered However, if the query Q has length greater than w, say pw for some constant p 1, there are several things we can do: 1. Search for just the rst w values of Q, retrieve the matching sequences and their positions, and then compare the entire Q with the sequence starting at that position. This method is like sequential scan, but it takes advantage of the storage e ciency of trails. 2. Search for the rectangles that are su ciently close within  to the rst w values of Q, the next w values of Q, and so on. Only a sequence that has at least one rectangle that is within of each of these subsequences of Q is a candidate match. Check the candidate matches. 3. Like 2, but check a sequence S only if it has a rectangle within distance =pp from at least one of 2 the p subsequences of Q. Note that the distance of sequences must be at most p, and if all p pp from any point on the trail of S , then p = p2 = 2 is a lower subsequences of Q are at least = bound on the distance between Q and any subsequence of S . 41 10 Mining Episodes In the episode model, the data is a history of events ; each event has a type and a time of occurrence. An example of event type might be: switch 34 became overloaded and had to drop a packet." It is probably too general to have an event time of the form some switch became overloaded." Several events may occur at the same time, and there is no guarantee that some event happens at every time unit. 10.1 Applications of Epsiode Mining Mining epsiodes," that is, sequences or sets of event types that occur within a short window, has been used to help predict outages in the Finnish power grid; the paper of MTV is an abstraction of this work. The same ideas could be used to monitor packet-switching or other communication networks, to develop rules for rerouting data in advance of congestion. It may also be possible to mine for rules that predict failures from combinations of manifestations in a variety of complex systems. A parallel episode is a set of event types, e.g., fA; B; C g. In diagrams, these episodes are represented by a vertical box with A, B , and C within. The intent of a parallel episode is that each of the events in the episode occurs within a window of time, but the order is not important. A serial episode is a list of event types, e.g., A; B; C . In diagrams, these events are shown in a horizontal box, in order. The intent is that within a window of time, these events occur in order. Note that a single event may be thought of as both a parallel and a serial episode. A composite episode is built recursively from events by serial and parallel composition. That is, a composite episode is either: An event, The serial composition of two or more events, or The parallel composition of two or more events. Thus. every serial and parallel episode is also a composite episode, but there are episodes that are composite, yet not serial or parallel. 10.2 Epsiodes Example 10.1 : Figure 29 shows a composite episode. It is the serial composition of three episodes. The rst is the single event A. Then comes the parallel epsiode fB; C; Dg. The third is a composite episode consisting of the parallel composition of the serial episodes E; F  and G; H . Two examples of orders of these 8 events that are consistent with this episode are ABCDEGFH and ACDBGHEF . 2 B A C D E G F H Figure 29: A composite episode 42 10.3 Monotonicity of Episodes and the A-Priori Algorithm Given a window length w an amount of time during which an episode may occur, an episode is frequent if it occurs in at least s windows, where s is the support threshold. Note that the same sequence of events may show up as an episode in several consecutive windows; we count one unit for each such window. However, no window can be credited with having the same episode occur more than once, even if we can construct the episode from several di erent sets of events in the same window. Like frequent itemsets, frequent episodes are monotone: if an episode E is frequent, then so is any episode formed by deleting some events from E . Thus, we can construct all frequent episodes levelwise," using a trick that is very much like a-priori. 1. Let Ci be the candidate episodes of size number of events i and let Li be the frequent episodes of size i. 2. For a basis, C1 is the set of all event types. 3. Construct Ci+1 from Li by putting in Ci+1 exactly those episodes E of size i + 1 such that deleting any one event from E yields an event from L + i. Example 10.2 : The epsiode of size 3 in Fig. 30a can be frequent i.e., in C3 only if the three episodes of Fig. 30b are frequent i.e., in L2 . 2 A C B A B A C B C (a) (b) Figure 30: Episode a is a candidate in C3 only if all three episodes b are in L2 10.4 Checking Parallel Episodes The big problem is converting Ci into Li by going once through the data and counting the support for each episode in Ci . A dumb algorithm will look at each window of length w in turn, and for each candidate episode check whether it is there; if so, add 1 to the support for the episode. The goal of MTV is to perform work that is only proportional to the sum over all events of the number of candidate episodes including the subepisodes of composite episodes that have that event. The following algorithm was developed in class; it seems less restrictive than the one given in the paper, since it does not assume the impossibility of an event type appearing several times in a window. We describe how the data is scanned once, to compute Li from Ci, considering each window in turn, from the beginning of the sequence. The data structure needed: 1. For each event type A: a A count A.count of the number of times A has been seen in the present window. b A linked list A.contains of all the episodes E that contain A. 2. For each candidate episode E : a A time E.startingTime that is the beginning of a consecutive sequence of windows in which E has always been present, up to the present window. 43 b An integer, the number of windows in which E has appeared, not including windows since startingTime. c An integer E.missing giving the number of events A of E that are not in the present window. Note that E is present in the window if and only if E.missing==0. The heart of the algorithm is what we do when we slide the window one time unit forward. We must consider what happens when an event A drops out of the beginning, and what happens when an event B is included at the front. If A drops out of the window: A.count--; ifA.count==0 * we just lost A * forall E on A.contains E.missing++; ifE.missing==1 * we just lost E * += E.startingTime - currentTime; If B enters the window: B.count++; ifB.count==1 * we just gained B * forall E on B.contains E.missing--; ifE.missing==0 * we just gained E * E.startingTime = currentTime; 10.5 Checking Serial Episodes To check serial episode A1 ; A2; : : :; An we simulate a nondeterministic nite automaton that recognizes the string :  A1 A2    An , as suggested by Fig. 31. As we scan the events in the data, we keep track of the set of states that the NFA is in. q0 A1 q1 A2 q2 ... An qn any other other any Figure 31: An NFA for recognizing a serial episode Here is a rough description of how the NFA's for the various events E are used, assuming a data structure similar to that described in Section 10.4 for parallel episodes. Note that the NFA stays in the same set of states it was in if the input symbol is an event that is not part of its episode. Thus, simulating this automaton requires 0 time unless its episode is on a list like A.contains for the current event A see the data structure in Section 10.4. As we simulate, we keep for each state the NFA is currently in the most recent point at which we could have started the NFA and still gotten to that state. This value is: 1. The current time if we enter state q1. 2. The time of qi,1 if we enter state qi by reading Ai . 44 3. The same time as previously, if we were already in qi and the next input is other than Ai . If the NFA for episode E enters the accepting state qn, but was not previously in that state, then set E.startingTime to the time currently associated with qn . If any time associated with a state is less that the current time minus w the window length, then delete that state from the set of states the NFA is in. If the accepting state is thus lost, add E.startingTime minus the current time to Example 10.3 : Suppose the event E is A; B; C . The NFA for E is as in Fig. 32. A 0 1 B 2 C 3 any other other any Figure 32: NFA for the episode A; B; C  Let the data include the following occurrences of A, B , and C , along with other event types, not shown: A; B; A; B; C ; all these events are within the current window. Then the states of the NFA after each of these events is as shown in Fig. 33. Solid lines show how each state is derived from the previous one by transitions of the NFA, and dashed lines indicate the associated time for each state. 2 A 0 0 1 B 0 1 2 A 0 1 2 B 0 1 2 C 0 1 2 3 Figure 33: Maintaining the set of states and their associated times 10.6 Counting Composite Events To extend the above ideas to composite episodes, we must keep a machine" of some sort for each subepisode. It is the job of each machine to report to the machines for the superepisodes of which it is a part whether or not it is present, and if so what is the most recent time at which it can be construed to have begun. If the subepisode is the parallel composition of events and or other subepisodes, we use a data structure like that of Section 10.4, but we replace the count for an event by the most recent beginning time for a composite episode. For serial episodes, we maintain an automaton, but the inputs are occurrences of its subepisodes. The starting time and support only need to be maintained for episodes in Ci, not for subepisodes. 45 ...
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This note was uploaded on 01/31/2011 for the course CS 345 taught by Professor Dunbar,a during the Fall '07 term at UC Davis.

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