The Substring Search Problem

Transcription

1 The Substing Seach Poblem One algoithm which is used in a vaiety of applications is the family of substing seach algoithms. These algoithms allow a use to detemine if, given two chaacte stings, one is contained within the othe. Stated moe fomally: Let A be an alphabet, a set of chaactes, of size. If we assume that each chaacte is equally likely, the pobability of andomly encounteing any chaacte is. Let S be a lage sting of N chaactes fom A. Let T be a smalle sting of m chaactes fom A. Assume m N. Poblem: Detemine if T is contained in S, and if so, at what index in S does T begin? The substing seach has applications in a wide vaiety of poblems, such as DNA compaison, plagaism detection tools, file seach tools, and moe. In addessing this poblem, and algoithms designed to solve it, we become awae that seach stings have impotant fundamental popeties. Most notably, the lengths of the elevant stings, and the chaacte set used in compising the stings. We wite a simple algoithm, below, to cay out ou substing seach. This is eflected in the figue below, in which it becomes appaent that thee ae N m + positions whee T might occu within S. We theefoe need to compae two stings of size m at most N m + times. int seach (sting, N, substing, m) fo i fom 0 to N - m fo j fom 0 to m- if substing[j]!= sting[i+j] beak else if j == m- etun i etun -

2 Main Sting Substing At fist glance, we might assume that the un time of this algoithm is descibed by T (n) = O(N m). Yet a close look indicates that the un time is heavily influenced by whethe o not the sting is eve found, and if so, how soon it s found. Clealy, if the substing matches the main sting at the fist position, we stop looking afte a single compaison of m chaactes. If the substing is neve matched, it equies N m chaacte compaes. This wide vaiance is not just a popety of the algoithm, but moeso the data. Since we see such vaiation on un time, T ( ), based on whethe o not the substing is matched ealy, late, o neve, it makes sense to analyze how likely each of these possibilities is. We might ask, fist, if thee ae othe factos which detemine the likelihood of finding a substing on any paticula compaison. To addess this question, let s pesent some examples of substing seach poblems.. Is the binay sting 0 within the lage sting Is the DNA sting T CAG within the lage sting AGCT T ACGGACAT CAGAC 3. Is the decimal sting 207 within the lage sting Is the English Chaacte sting woof within the lage sting I eally like dogs In looking at the fou cases above, it seems intuitive that the binay sting has the best chance of a match. Looking futhe, we might (and should) conclude that the DNA sting match is the next most likely case fo a match, followed by the decimal sting, and finally the English sting. The eason fo this is clea: having access to moe chaactes ceates moe stings (of any length), and makes a match less likely. We call the list of allowable chaactes the alphabet of a set of stings. The paticula list of chaactes in an alphabet doesn t eally matte in ou analysis... the size of the alphabet does. In the cases above, we utilize alphabets: {0, }, {A, C, G, T }, {0,, 2, 3, 4, 5, 6, 7, 8, 9}, {a...z}, with sizes 2, 4, 0, 26 (o 52 is we decide that ou stings ae case-sensitive). 2

3 Note: Some might not be familia with the DNA example. We will not descibe this in detail hee, except to identify the fou chaactes: A, C, G, T. It is undestood that DNA (Deoxyibonucleic Acid) is composed of suga, phosphate molecules, and fou nucleobases: Adenine, Cytosine, Guanine, Thiamine. It is these nucleobases, and the ode in which they occu in long DNA molecules, that identifies genetic taits and justifies it s application of the substing seach algoithm. Getting back to the elative likelihood of finding the andom substings of size 4 in lage stings of size 8, we fist stat with the simple question: what is the likelihood matching two andom stings of size 4, whee the stings come fom an alphabet of size 2, o 4, o 0, o 26? Since this equies two chaactes to match 4 times, we have the pobabilities: P = 4 = , 4 = , 4 = 0.000, 4 = , espectively. In geneal, the pobability of matching two stings of size m fom an alphabet of size is given by: P (m, ) = ( ) m In this fomula, we easily see the enomous vaiation in pobability given small diffeences in both and m. In pactice, is usually easonably small, but m can be vey lage, such as in the DNA example. Of couse, compaing two stings of the same size is just the fist step in ou algoithm, and aleady we see that the pobability of a match vaies widely with the paametes. Now, we incopoate this opeation into the lage algoithm, and see that the pobability of success is athe complicated. In fact, befoe we begin, we need to identify the useful mathematical fomulas involved in ou analysis. I. Some Relevant Seies Fomulas: Fist, a bief etun to the wondes of Calculus II. In paticula, let s ecall some finite seies fomulas. + x + x x k = xk+ + 2x + 3x kx k = kxk+ (k+)x k + x (x ) 2 You should ecognize the fist fomula as the geometic seies. Notice that if x <, we have: lim + x + k x x k x = lim k+ = k x x It also beas pointing out that the second fomula is the deivative with espect to x of the fist. These fomulae ae vey useful in woking with the geometic pobability distibution. In what follows, x will epesent a pobability, and thus we will geneally see that x. II. The Geometic Pobability Model In shot, suppose we have an expeiment 3

4 fo which the only possible two outcomes can be descibed as success and failue. We epeat this expeiment seveal times, and each time the pobability of success is the same. We define p as this pobability of success. Of couse this means that we can define the pobability of failue as q = p. The question hee is: given a paticula expeiment, how many times must we conduct ou simple expeiment befoe achieving the fist success? This numbe, say n, diffes fom expeiment to expeiment. But what we CAN ask, (and answe) ae questions such as i) what is the pobability that ou fist success will happen ON the nth tial, and ii) what is the pobability that ou fist success will happen BY the nth tial. Once we know how to answe these questions, we can then move on to ask: what is the AVERAGE numbe of tials needed to ealize the fist success. We answe all of these questions hee: Pobability of the fist success ON and BY the n th tial: P (fist success ON tial n) = q n p To undestand the fomula above, think of it as: if the fist success occus on the n-th tial, the fist n tials must be failues. The pobability of the fist n failues is given by q n, which is then followed by a success, with pobability p. Pobability of the fist success BY the n th tial: P (fist success BY tial n) = P (on ) + P (on 2) + P (on ( 3) +... ) + P (on n) = p + qp + q 2 p+,,, +q n p = p = q n The Aveage Value of a Vaiable given a specific Pobability Distibution: Suppose that we have an expeiment which poduces one of a finite set of numeical outcomes. We call this set the Sample Set and might epesent this as: X = {x, x 2, x 3,...x R }. Each outcome has a pobability given by P (x i ) = p i. Clealy, we have 0 p i and R p i =. We define the aveage, o expected value of outcomes as: k= E[X] = X = R x i p i k= The expected value gives us the aveage of all outcomes in an expeiment whee we pefom many tials. Example: Given a single six sided die, ou sample space is X = {, 2, 3, 4, 5, 6}. The pobability of each oll is given by p i =. The expected value, then, is given by: 6 q n q E[X] = = 2 6 =

5 Of couse, we will neve actually oll a 3.5, but if we oll a single die 000 times, the aveage oll will be vey close to 3.5. Pobability Analysis of the Substing Seach Algoithm: Hee, we wish to ask two questions, and detemine the answes in tems of the thee paametes, N, m,. The two questions ae i) What is the pobability that we will find the substing, T, of size m within the lage sting, S, of size N? ii) What ae the aveage numbe of sting-to-sting compaes that will occu in ou attempt to find substing T? By this point, ou emaining computations ae simple. To simplify the computations to follow, we define:. Pobability of a successful sting-to-sting match: p = ( ) m 2. Pobability of a failed sting-to-sting match:, and q = p = ( 3. The numbe of positions whee a substing match might occu: W = N m + ) m Pobability that a substing will be found at all: P = P (substing match BY N m + tial) = q W = (( ) m ) N m+ Clealy, this implies that the pobability of NO match at all is given by: P = P (No match BY N m + tial) = q W = (( ) m ) N m+ Both of these will be useful in computing the aveage numbe of sting-to-sting compaes in any substing seach: E[X] = W k P (match ON kth compae) + W P (no match) k= = W = ( k q k p + W q W = p (W q (W +) (W +)q W +) ) k= ( W q (W +) (W +)q W ++( q)w q W ( q) = + W q W ) ( = m (( ) m ) N m+ ) ) ( q) ( 2 q W q Back to ou fou cases... It is instuctive to demonstate the vaiation in pobabilities given small changes in alphabets. Conside the impotant computations applied to ou fou examples above, ecalling that in this examples, we have: N = 8, m = 4, W = N m + = 5. 5

6 p = ( ) 4 6 = 2 = 4 = 0 = q = ( ) P (found) = q W E[compaes] = qw p To summaize, let s make some obsevations. Fist, unless and m ae vey small, N will need to be VERY lage to allow a a chance of substing match. We notice that fo the last thee cases, the expected value of needed compaes was vey close to the total numbe of possible compaes. We might descibe this as the aveage cases is vey close to the wost case. We began by identifying the best case as a match on the fist compae, and found the complexity as T (N, m, ) = O(m). Likewise, we identified the wost case as having to pefom N m + compaisons of stings of m chaactes, and descibed this as T (N, m, ) = O(Nm m 2 + m) = O(N m). The aveage case would be found using the expected (aveage) numbe of sting-to-sting compaes, of m chaactes. That is, T (N, m, ) = m ( m (( ) m ) ) N m+. A pefectly appopiate question to ask at this point would be: can we do anything to speed this up? That is, to find a moe efficient vaiation of the substing seach algoithm. An idea that makes sense to pusue is to seek ways to make the individual sting-to-sting compaes moe efficient. Indeed thee has been pogess in this diection. Moe to follow... 6