1 What is combinatorics?

Transcription

1 1 What is combiatorics Combiatorics is the brach of mathematics dealig with thigs that are discrete, such as the itegers, or words created from a alphabet. This is i cotrast to aalysis, which deals with the properties of cotiuous systems, such as the real umber lie or a differetial equatio. May (but ot all) combiatorial problems also address systems which are fiite, so the systems beig ivestigated will have a specific umber of elemets: i fact, the questio how may elemets are i a particular system is a commo combiatorial query. The above loose defiitio of combiatorics is iclusive of a few other disciplies. Exploratios of the arithmetic properties of the itegers may be cosidered to fall uder the headig of umber theory; exploratios of multiplicatio-like ad additio-like operatios o discrete structures geerally falls uder the headig of algebra. The distictios betwee these disciplies ad combiatorics, especially as regards fiite structures, are somewhat amorphous. Broadly speakig, combiatorics teds to ask four types of questios: Eumerative questios How large is a particular set Existetial questios Is a set of coditios o a discrete system actually satisfiable Extremal questios How large does a discrete structure eed to become before certai substructures become ievitable Computatioal questios How log would it take a computer to aswer a questio posed about a discrete system To give some cotext to the above descriptios, here are a few fudametally combiatorial questios: How may umbers less tha cosist of distict digits appearig i ascedig order How may people eed to be attedig a party i order to guaratee that there are either 5 mutual acquaitaces or 5 mutual stragers If 20 people each choose a set of acceptable roommates, ca each perso be paired with a acceptable roommate How could a computer program be writte to fid such a pairig How may aagrams are there of the word MISSISSIPPI If we remove oe square from a 7 9 checkerboard, ca the remaiig squares be covered with domioes Does it matter which square we remove If we walk five blocks orth ad six blocks east, takig a radom route, what is the probability that we will pass by a ewsstad oe block to the orth ad oe block east of our startig positio After we have completed both semester of this course, you will kow the aswer to five of these questios as well as may others! Page 1 of 10 August 25, 2009

2 2 Itroductory Eumerative Methods The absolute simplest eumerative method is exhaustive eumeratio, also kow as brute force : simply list out every Questio 1: How may three-digit umbers are there whose digits add up to 6 Aswer 1: Cosider those umers startig with 1 ad with the remaiig digits addig up to 5: 105, 114, 123, 132, 141, ad 150. The those startig with 2, ad havig remaiig digits addig up to 4: 204, 213, 222, 231, 240; ow those startig with 3 : 303, 312, 321, 330. Fially, 4 yields 402, 411, ad 420, 5 gives 501 ad 510, ad 6 gives oly 600. We have listed 21 umbers here. Exhaustive eumeratio works, if the set beig explored is small eough ad the approach is methodical eough. But it s easy to get wrog if you re careless. It s better to use priciples to build up a cout o large sets from small sets. 2.1 A multiplicative priciple If the items beig couted ca be split ito two or more idepedetly coutable parts, the multiplicative priciple is a fudametal meas for coutig the items i questio: Propositio 1. If the elemets of a set X ca be decomposed ito pairs of elemets of A ad B such that each pair (a, b) correspods to oe ad oly oe elemet of X, the X = A B. So, for example: Questio 2: How may 2-letter strigs ca be costructed cosistig of a cosoat followed by a vowel Aswer 2: There are 21 cosoats ad 5 vowels. If we were to take A above to be the cosoatset ad B the vowel-set, there is a trivial associatio with our set of two-letter strigs via cocateatio, e.g. TO ( T, O ), so the size of our desired set is 21 5 = 105. We ca also use this to strig together groups of more tha two items at a time: Questio 3: A licese plate umber cosists of three letters followed by 3 umbers. To avoid cofusio, the letters I ad O are ot used, as are the umbers 1 ad 0. How may licese plates are possible Aswer 3: We have 24 possibilities for each letter, ad 8 possibilities for each umber. Thus, the umber of licese plates is the product of our umber of choices for each positio: = (24 3 )(8 3 ) = As log as the umber of choices for each part of the problem is idepedet of the choice made for the other parts of the problem, the multiplicative priciple holds, which allows us to use it eve i cases where there is apparetly ot idepedece of choice: Questio 4: How may 2-digit umbers are there with distict digits Aswer 4: The first digit ca be ay umber from 1 through 9; there are 9 possibilities for this digit. It seems that the possibilities for the secod digit deped o the choice of the first digit, but regardless of what first digit was chose, there are 9 possibilities for the secod digit (what they are depeds o the first digit, but how may of them there are is ot). Thus, there are 9 9 = 81 such umbers. Questio 5: How may orderigs are there of the letters ABCDE Page 2 of 10 August 25, 2009

3 Aswer 5: We ca choose ay of the 5 letters to be first; we ca choose ay of the 4 remaiig letters to be secod; ay of the 3 remaiig letters to be third, ad ay of the 2 remaiig letters to be fourth. At this poit, we have oly oe letter left, ad it must be fifth. Multiplyig together the umber of choices we have at each phase of the strig costructio, we have = 120. Beware, however, of situatios where the umber of possibilities for differet parts of the problem are ot idepedet! Questio 6: How may two-letter strigs are there with at most oe vowel Aswer 6: This is ot somethig we ca solve easily usig the multiplicative priciple! We might aïvely assert that there are 26 possibilities for the first letter, but we ru ito a problem: if the first letter is a cosoat, we have 26 possibilities for the secod letter; if it s a vowel, however, the secod letter would have to be a cosoat, so we d oly have 21 choices. 2.2 A additive priciple If we ca divide the set of objects beig couted ito two subsets, ad cout each oe idividually, we ca add up their sizes to fid the size of the origial set. Propositio 2. If set X ca be partitioed ito disjoit subsets A ad B, the X = A + B. This is useful for problems which behave differetly i differet cases: Questio 7: How may two-letter strigs are there with at most oe vowel Aswer 7: As metioed above, our choice of the first letter dictates the umber of possibilities for the secod letter. We ca thus divide the set we mea to cout ito two sets correspodig to our two cases: Case I: The first letter is a vowel (so the secod letter must be a cosoat). This case allows 5 choices for the first letter ad 21 for the secod, yieldig 5 21 = 105 two-letter strigs. Case II: The first letter is a cosoat (so the secod letter could be aythig). This case allows 21 choices for the first letter, ad 26 for the secod, yieldig = 546 two-letter strigs. We add these two distict cases to get the total umber of strigs, = 651. As see above, the additive priciple is frequetly used i tadem with other methods. I additio, it ca be used i place of other methods, if ecessary: Questio 8: How may 2-letter strigs ca be costructed cosistig of either a cosoat followed by a vowel, or a vowel followed by a cosoat Aswer 8: We ca divide this ito two cases: cosoat followed by a vowel has 21 5 = 105 possibilities, ad vowel followed by a coosoat has 5 21 = 105 possibilities, for a total of = 210 possibilities. However, we could also solve this usig solely the multiplicative priciple, by cosiderig three thigs to be set: a choice of vowel, a choice of cosoat, ad a order, e.g. TO ( O, T,C V), OF ( O, F,V C). There are 5 choices of vowel, 21 choices of cosoat, ad two orders, givig = 210 possibilities. Page 3 of 10 August 25, 2009

4 2.3 A Subtractive Priciple Sometimes, it s easier to overcout a set ad the exclude the thigs you did t mea to cout tha it is to try to exclude them from the start. Propositio 3. If set X is a subset of set A, ad B cosists of all elemets of A ot i X, the X = A B. Questio 9: How may two-letter strigs are there with at most oe vowel (have t we already asked this oe) Aswer 9: We ca cout all two-letter strigs easily: there are 26 choices of first letter, ad 26 choices of secod letter, for = 676. But this icludes two-letter strigs with two vowels, ad we emphatically do t wat to iclude those. How may of them are there Well, we d eed two vowels, so there would be 5 5 = 25 ways to get these uwated strigs. There are thus = 651 strigs with fewer tha 2 vowels. The subtractive priciple as preseted here is actually ot of use too ofte, but some modificatios o the same idea we ll fid to be extremely useful later. 2.4 Symmetry removal with divisio This priciple is extremely useful, but also dagerously easy to misuse: Propositio 4. If each elemet of set X ca be associated with elemets of set A such that each elemet of set A is associated with exactly oe elemet of X, X = A. The use of this method ca be see with a simple eumeratio which we will fid extremely useful later: Questio 10: How may 3-elemet subsets are there of the set {0, 1, 2,..., 9} Aswer 10: We will start with a questio aki to oe we ve already aswered: how may sequeces of three distict digits from 0 to 9 are there There are 10 possibilities for the first digit, ad ay of the ie uused possibilities ca be used for the secod digit, ad ay of the eight remaiig after that ca be used for the third. So there are = 720 possible sequeces. But sequeces are ot sets! The sequece 314 ad 134 both correspod to the set {1, 3, 4}. However, sequeces of distict terms correspod to sets i a ice predictable fashio: each set of three elemets ca have ay of the 3 elemets first, ay of the 2 remaiig elemets secod, ad the leftover elemet third, so each set correspods to 6 differet sequeces whe reordered. Thus, to fid the umber of sets, we simply cout groups of 6 sequeces, which would be = 120. Be wary, however, of situatios where a correspodece is ot cosistetly -to-oe for a particular value of : Questio 11: How may choices of 3 umbers from the set {0, 1, 2,..., 9}, allowig repetitios ad igorig order of selectio, are there Aswer 11: This seems quite similar to the above problem: why ot cout strigs of 3 umbers ad divide by the umber of arragemets which correspod to the same thig i differet orders But differet choices correspod to differet ubmers of strigs. While [1, 3, 4] correspods to 6 strigs as see above, [1, 1, 4] oly correspods to 3 (114, 141, ad 411), ad [1, 1, 1] oly correspods to oe. Page 4 of 10 August 25, 2009

5 You ca use the divisio method o this if you use the additive priciple to partitio the set of choices ito three classes based o how may of their elemets are the same: [a, a, a], [a, b, b], ad [a, b, c]. It is left as a exercise to the reader to get from this method to the correct aswer, which is Two Forms of Equivalece 3.1 Bijectios Betwee Coutable Structures Oe useful trick which we ll see more i comig days is showig that the same eumeratio ca be used to cout differet structures. For istace, we might cosider the ways to build a biary tree with 3 odes; there are 5 ways to do this: We might also cosider the umer of ways to triagulate a petago: Aother questio whose aswer is 5 is How may ways are there to est 3 pairs of paretheses. We ca do this i the followig ways: ((())), (()()), (())(), ()(()), ad ()()(). These might be coicidece! But if we look at 4-ode biary trees, triagulatios of the hexago, or estigs of 4 sets of paretheses, we see that there are 14 ways to do each of these. This may be more tha coicidece! I fact, we may show that all of these objects are coutable the same way by explicitly costructig a bijectio betwee them. For istace, we ca tur a parethesis structure ito a tree by the rule: (A)B becomes a ode whose left child is whatever tree is associated with A ad whose right child is whatever tree is associated with B ; coversely, we ca associate ay tree with a parethesis-system by associatig a tree with left-brach A ad right-brach B with the parethesis-system (A)B. Similarly, triagulatios could be associated with trees by idetifyig each tragle with a ode, which produces a atural biary tree. By associatig these systems with each other we show that they are eumerated the same way. This is useful i case oe of these structures is easier to cout tha the others (i this case, biary trees are particularly easy to cout). The umbers i this example are called the Catala umbers. Richard Staley, i Eumerative Combiatorics, Volume 2, idetifies 66 differet structures eumerated by the Catala umbers. 3.2 Multiple Coutig Methodologies For a Sigle Structure I additio to usig oe coutig method to eumerate multiple differet-seemig structure, oe powerful trick i our arseal is usig two differet methods to cout the same thig, ad i so Page 5 of 10 August 25, 2009

6 doig build a equality betwee two differet-seemig algebraic quatities. Example: Questio 7 ad Questio 9 really asked the exact same thig, but we aswered it two differet ways. From this, eve if we could t do simple arithmetic, we d kow that = That s a pretty silly idetity, sice it s just arithmetic. However, we could use the same techique, slightly geeralized, to prove a familiar algebraic idetity: Questio 12: Give a combiatorial proof that for itegers m 0, m 2 2 = (m+)(m ). Aswer 12: Let X = {a 1, a 2,..., a, b +1, b +2,..., b m }. We will ask the questio how may ways are there to select a ordered pair of elemets of X such that ot both of them are as This is a geeralizatio of our above argumet, with a i ad b j as stadis for vowels ad cosoats respectively. We may cout the umber of ways to get such a ordered pair usig the additive priciple. If the first term i the ordered pair is a a i (which could happe i ay of ways), the the secod term must be a b j (which ca happe i ay of m ways). We have (m ) possible ordered pairs begiig with a a. If we start with a b, however (whcih ca be doe i ay of m ways), we have free choice of ay of the m elemets of X for the secod term. Thus there are (m )m ordered pairs whose first term is a b. Takig these possibilities together, we see that there are ( m) + (m )m = ( m)( + m) ordered pairs i total. Alteratively, however, we could cout all the ordered pairs (which have m choices for each term, for a total of m 2 ) ad use the subtractive priciple to remove those of the form (a i, a j ), of which there ca be easily see to be 2. Thus, this same eumeratio ca be computed as m 2 2. We thus have a combiatorial presetatio that ( m)( m) = m 2 2. This too is a bit of a parlor trick: it avoids usig the distributive property, but it s otherwise difficult to see why this might be useful. The big beefits of demostratig equality this way came ito play whe we itroduce the combiatio statistic. Defiitio 1. The umber of possible k-elemet subsets of a -elemet set is the combiatio statistic or biomial coefficiet, deoted ( k), ad read choose k. This expressio is equal to! k!( k)!.! The demostratio that this statistic is actually used to solve Questio 11. k!( k)! is simply a geeralizatio of the argumet The ( biomial coefficiets are subject to a tremedous umber of idetities; oe of the simplest is ( k) = 1 ) ( k + 1 k 1). This ca be proved algebraically, but doig so is perfectly miserable, ad more importatly, uillumiatig: ( ) ( ) 1 1 ( 1)! + = k k 1 k!( 1 k)! + ( 1)! (k 1)!( k)! ( 1)!( k) = k![( k)( k 1)!] + ( 1)!(k) [k(k 1)!]( k)! ( 1)!( k) ( 1)!(k) = + = k!( k)! ( 1)! k!( k)! =! k!( k)! = k!( k)! ( ) k Page 6 of 10 August 25, 2009

7 Questio 13: osese How ca we prove that ( ) ( k = 1 ) ( k + 1 k 1) without all this awful factorial Aswer ) 13: We will cout a particular set i two differet ways! Let X = {1, 2, 3,..., }. The, clearly couts the k-elemet subsets of X. ( k We hope that we ca somehow use ( ) ( 1 k + 1 k 1) to cout the same thig, ad do so by artificially itroducig a additive-priciple decompositio of the k-elemet subsets of X ito two cases. Let X = {1, 2, 3,..., 1}. The, i costructig a subset S of X, we have two distict possibilities: Case I: S cotais. The, sice we kow oe elemet of S, we oly eed to choose k 1 others to brig it up to k 1 elemets. These elemets must be umbers less tha ; that is, elemets of X. There are thus ( 1 k 1) possible such k-elemet sets. Case I: S does ot cotai. Sice we have t assiged ay elemets to S, we eed to choose all k of them. These elemets must be umbers less tha, sice was specifically excluded. There are thus ( ) 1 k possible such k-elemet sets. Usig the additive priciple to assemble these, we see that there are ( ) ( 1 k ) k subsets of X with k elemets. Sice the biomial coefficiet has such a simple eumerative iterpretatio, it leds itself to may equivalecies of eumeratios: Questio 14: Prove that ( ) + 0 ( ) + 1 ( ) + 2 ( ) ( ) ( ) = 2 Aswer 14: The left side of this equatio is easily iterpreted as a cout related to subsets of a set with elemets. Let X = {1, 2,..., }. The, the terms of the sum o the left side cout the umber of zero-elemet subsets of X, the the umber of oe-elemet subsets, the the umber of two-elemet subsets, ad so forth up to the umber of -elemet subsets. Addig together the umber of subsets of each size gives a umber with a simple iterpretatio: it is the umber of subsets of X, regardless of size. Now, to form a equality with the right side of the equaito, we wat to fid a way to argue that the ubmer of subsets of X is also 2. To do so, we might cosider the followig costructio techique for a subset of X: for each elemet of X, we have two choices; it is either i the subset or it is out of the subset. Sice we make a choice amog two alteratives ad repeat this choice times, the umber of ways to build a subset is = 2. Almost all idetities ivolvig the combiatio statistic are ameable to a combiatorial proof, ad frequetly these proofs are much easier tha tryig to prove the same thig algebraically. 4 The Twelvefold Way The twelvefold way is a term developed by Richard Staley to describe 12 closely related eumeratio statistics; i the course of developig these statistics, we ll lear a great deal about eumerative methods. We ve see several differet sorts of ways to eumerate selectios so far. There were our strigs, where we cared about order but did t force ay relatioship amog the strig elemets; there were Page 7 of 10 August 25, 2009

8 multisets, where we allowed repetitios but did t care about order, there were strigs of distict elemets, where we cared about order ad were restrictig choices, ad there were sets, which did t regard order but had choice-restricito. It s clear that the questio how may differet ways ca we take thigs from a pool of k has more tha oe iterpretatio. I order to specify this questio, we eed to kow more about what distiguishes two ways, ad what restrictios we have o our choice. There are three properties of a selectio statistic which we ll fid of iterest: Selected-elemet restrictio Ca we select the same elemet multiple times Do we demad that each elemet be selected at least oce Order cosideratio Is the choice AAC differet from ACA Elemet cosideratio Do the idividual thigs chose have itrisic properties Do we thik of ABC merely as choosig 3 differet thigs ad idistiguishable from CAB or eve XYZ We would always thik of AAB ad ABC as differet, however; oe has a repetitio, ad the other does ot. All i all, we will have 12 differet statistics, based o our aswer to these questios: Choosig k el ts from a set of size Free choice No repeats Each el t at least oce Ordered choice, distiguished el ts (1) (2) (3) Uordered choice, distiguished el ts (4) (5) (6) Ordered choice, udistiguished el ts (7) (8) (9) Uordered choice, udistiguished el ts (10) (11) (12) We ve actually see examples of (1), (2), ad (5) already. There are two commo other ways of iterpretig these eumerative statistics, which may be useful, or easier to visualize. Gia-Carlo Rota preseted these statistics as a questio of puttig balls ito boxes. For istace, istead of selectig the three-elemet subset {1, 3, 7} from {1, 2,..., 10}, he would visualize this as a system of 10 umbered boxes, with three ulabeled balls placed i boxes 1, 3, ad 7. I cotrast, if he wated to describe the sequece (1, 7, 3), he would describe the same set of boxes, but umbered balls to idicate order of appearace, so that ball 1 would go i box 1, ball 2 i box 7, ad ball 3 i box 3. Thus, the same statistics could be represeted by the umber of cofiguratios of balls ito boxes, depedig o box capacity ad whether thigs were labeled. Puttig k balls ito boxes Free choice 1 ball per box 1 ball per box Labeled balls, labeled boxes (1) (2) (3) Ulabeled balls, labeled boxes (4) (5) (6) Labeled balls, ulabeled boxes (7) (8) (9) Ulabeled balls, ulabeled boxes (10) (11) (12) Aother popular iterpretatio of these statistics is that of Richard Staley, where he couts equivalece classes of fuctios from a set of size k to a set of size. The questio of equivalece depeds o whether two fuctios are idetical up to permutatio of the elemets of oe set; e.g. If X = {A, B, C} ad Y = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, the the fuctio {(A, 2), (B, 5), (C, 8)} would be cosidered as equivalet to {(A, 8), (B, 2), (C, 5)} uder a permutatio of the elemets of X. Page 8 of 10 August 25, 2009

9 This correspods to order-distiguishability, or ball-labelig i the two prior iterpretatios. I this iterpretatio, the restrictios o repeated choice or requiremet to choose all elemets ca easily be stated as ijectivity ad surjectivity: Fuctios from X to Y with X = k, Y = Ay fuctios Ijective fuctios Surjective fuctios I total (1) (2) (3) Equivalece classes uder permutig X (4) (5) (6) Equivalece classes uder permutig Y (7) (8) (9) Equivalece classes uder permutig Xad Y (10) (11) (12) Now, with may iterpretatios of these statistics, we are i a positio to start discussig the actual values i each cell. Cell (5) couts the umber of uordered choices of k distict elemets from a -elemet set. This! we kow to be k!( k)! = ( k). Cell (2) couts the umber of ordered choices of distict elemets; this is actually a precursor to our calculatio for cell (5), which we attaied via symmetry-divisio from the umber of ordered choices. The umber of ordered choices was ( 1)( 2)( 3) ( k + 1) =! ( k)! = k!( k). This is sometimes kow as the permutatio statistic, ad may be deoted P k. Cell (1) is also easy to cout usig the multiplicative priciple. We choose ay of values for our first choice; we choose ay of values for our secod choice, ad so forth k times over, which we ca do i = k ways. The ext easy cell to fill is the surprisigly esoteric (11) ad almost icidetally (8). I the ballsad-boxes paradigm, this would be the questio: how may ways ca we put k balls i ulabeled boxes such that each box cotais o more tha oe ball If k >, this is clearly impossible. If k, the we could put oe ball per box util we have k boxes cotaiig oe ball each, ad k boxes cotaiig o balls. If the boxes are ulabeled, all ways of doig this are idetical. Thus, statistics (8) ad (11) happe to always be either 1 or 0, depedig k. We thous have a pretty good start o fillig out our table: Choosig k el ts from a set of size Free choice No repeats Each el t at least oce Ordered choice, distiguished el ts k k! ( k) ( = P k Uordered choice, distiguished el ts { k) 1 if k Ordered choice, udistiguished el ts 0 if < k { 1 if k Uordered choice, udistiguished el ts 0 if < k 4.1 Cells (4) ad (6): quatity assigmets Cells (4) ad (6) of the twelvefold way ca easily be idetified as assigmets of quatities to each of differet etities. This is most evidet i the balls-i-boxes paradigm, uder which we are simply distributig k balls amog boxes, freely i the case of cell (4), freely after assigig oe Page 9 of 10 August 25, 2009

10 to each box i the case of cell (6). We might ask this questio for specific values of ad k to get a better idea of what we re lookig at: Questio 15: How may ways are there to distribute 10 idetical items (e.g. cois) amog four idividuals so that each idividual receives at least oe item Aswer 15: We culd list all the ways to do this out, but it would be astoishigly tedious to do so. A revisualizatio of the problem, though, allows us to rephrase it i terms of a quatity we kow how to fid. A divisio of 10 cois amog 4 idividuals could be represeted by layig out all 10 cois i a row, ad the placig 3 dividers, ad assigig each sectio to a idividual. So, for istace, we could represet a divisio i which Alice receives 3 cois, Bob 4, Carl 2, ad Diae 1 with the followig placemet of dividers: So, i geeral, we ca associate a divisio of 10 items amog people with a choice of three dividerpositios amog the 10 items. Sice the dividers go betwee the items, there are 9 possible dividerlocatios amog 10 items; as a result, there are ( 9 3) = 84 ways to do so. Geeralizig the above, we see that if we are assigig k items to idividuals such that each idividual gets at least oe, we ca view this as placig 1 dividers i the k 1 iterstices betwee the items, ad that there are ( k 1 1) ways to do so. We ca use a similar approach to fid a statistic for cell (4): we ca either bootstrap off of our kow value for (6), or subtly modify it. To bootstrap, ote that we ca form a easy bijectio betwee free asigmets of k items to etities ad assigmets of + k items to etities i which each etity gets at least oe item. The bijectio is simple: distribute the extra items givig oe to each idividual or reverse the process by takig oe item away from each idividual. Thus, all we eed to kow is how to calculate statistic (6) with parameters ad + k: this would give ( +k 1 1 ). Alteratively, we could fid (4) by simply cosiderig the placemet of items ad dividers as was doe for (6), but here placig multiple dividers adjacet is ot a problem, so istead of fixig the locatio of items ad puttig dividers amog them, we cosider + k 1 ope spaces, ad fill each with either oe of k items item or oe of 1 dividers. We are thus selectig 1 locatios from + k 1 possible lcoatios for our dividers, which ca be doe i ( ) +k 1 1 ways. Our table for the twelvefold way is rapidly fillig! Choosig k el ts from a set of size Free choice No repeats Each el t at least oce Ordered choice, distiguished el ts k k! ( ) k ( Uordered choice, distiguished el ts +k 1 ) ( ( k 1 ) 1 { k) 1 1 if k Ordered choice, udistiguished el ts 0 if < k { 1 if k Uordered choice, udistiguished el ts 0 if < k The five remaiig spaces will require more clever tools to fill i. Page 10 of 10 August 25, 2009