Combinatorial Problems and Algorithms in Comparative Genomics. Max Alekseyev

Size: px
Start display at page:

Download "Combinatorial Problems and Algorithms in Comparative Genomics. Max Alekseyev"

Transcription

1 Comintoril Prolems n Algorithms in Comprtive Genomis Mx Alekseyev Dept. Computer Siene, University of South Crolin, Columi, SC, U.S.A. Algorithmi Biology L, Aemi University, Sint Petersurg, Russi 2011

2 Genome Rerrngements Mouse X hromosome Unknown nestor ~ 80 M yers go Humn X hromosome

3 Genome Rerrngements: Evolutionry Senrios Unknown nestor ~ 80 M yers go Wht is the evolutionry senrio for trnsforming one genome into the other? Wht is the orgniztion of the nestrl genome? Are there ny rerrngement hotspots in mmmlin genomes? Reversl (inversion) flips segment of hromosome

4 Genome Rerrngements: Anestrl Reonstrution Wht is the evolutionry senrio for trnsforming one genome into the other? Wht is the orgniztion of the nestrl genome? Are there ny rerrngement hotspots in mmmlin genomes?

5 Genome Rerrngements: Evolutionry Erthqukes Wht is the evolutionry senrio for trnsforming one genome into the other? Wht is the orgniztion of the nestrl genome? Are there ny rerrngement hotspots in mmmlin genomes? (ontroversy in )

6 Genome Rerrngements: Evolutionry Erthqukes Wht is the evolutionry senrio for trnsforming one genome into the other? Wht is the orgniztion of the nestrl genome? Where re the rerrngement hotspots in mmmlin genomes?

7 Rerrngement Hotspots in Tumor Genomes Rerrngements my isrupt genes n lter gene regultion. Exmple: rerrngement in leukemi yiels Philelphi hromosome: Chr 9 Chr 22 promoter ABL gene promoter BCR gene promoter BCR gene promoter -1 onogene Thousns of iniviul rerrngements hotspots known for ifferent tumors.

8 Biologil Prolem: Who re evolutionry loser to humns: mie or ogs?

9 Who is Closer to Us: Mouse or Dog?

10 Primte Roent Crnivore Split roent-rnivore split time primte-roent split primte-rnivore split

11 Primte Roent Crnivore Split roent-rnivore split time primte-roent split primte-rnivore split

12 Primte Roent vs. Primte Crnivore Split July 2007 n up new ppers supporting the primte-rnivore split April 2007 Lunter et l., PLoS CB 2007 refute Cnnrozzi et l. rguments Jnury 2007 Cnnrozzi et. l., PLoS CB 2007 rgue for the primte-rnivore split 2001 Murphy et. l., Siene 2001 set new ominnt view: the primte-roent split efore 2001 most iologists elieve in the primte-rnivore split primte-roent split primte-rnivore split

13 Reonstrution of Anestrl Genomes: Humn / Mouse / Rt

14 Reonstrution of MANY Anestrl Genomes: Cn It Be Done?

15 Algorithmi Bkgroun: Genome Rerrngements n Brekpoint Grphs

16 Unihromosoml Genomes: Reversl Distne A reversl flips segment of hromosome. For given genomes P n Q, the numer of reversls in shortest series, trnsforming one genome into the other, is lle the reversl istne etween P n Q. Hnnenhlli n Pevzner (FOCS 1995) gve polynomil-time lgorithm for omputing the reversl istne.

17 Prefix Reversls A prefix reversl flips prefix permuttion. Pnke Flipping Prolem: sort given stk (permuttion) of pnkes of ifferent sizes with the minimum numer of flips of ny numer of top pnkes.

18 Multihromosoml Genomes: Genomi Distne Genomi Distne etween two genomes is the minimum numer of reversls, trnslotions, fusions, n fissions require to trnsform one genome into the other. Hnnenhlli n Pevzner (STOC 1995) extene their lgorithm for omputing the reversl istne to omputing the genomi istne. These lgorithms were followe y mny improvements: Kpln et l. 1999, Ber et l. 2001, Tesler 2002, Ozery-Flto & Shmir 2003, Tnnier & Sgot 2004, Bergeron , et.

19 HP Theory Is Rther Complite: Is There Simpler Alterntive? HP theory is key tool in most genome rerrngement stuies. However, it is rther omplite tht mkes it iffiult to pply in omplex setups. To stuy genome rerrngements in multiple genomes, we use 2-rek rerrngements, lso known s DCJ (Ynopoulus et l., Bioinformtis 2005).

20 Simplifying HP Theory: Swith from Liner to Cirulr Chromosomes A hromosome n e represente s yle with irete re n unirete lk eges, where: re eges enoe loks n their iretions; jent loks re onnete with lk eges.

21 Reversls on Cirulr Chromosomes reversl Reversls reple two lk eges with two other lk eges

22 Fissions fission Fissions split single yle (hromosome) into two. Fissions reple two lk eges with two other lk eges.

23 Trnslotions / Fusions fusion Trnslotions/Fusions trnsform two yles (hromosomes) into single one. They lso reple two lk eges with two other lk eges.

24 2-Breks 2-rek 2-Brek reples ny pir of lk eges with nother pir forming mthing on the sme 4 verties. Reversls, trnslotions, fusions, n fissions represent ll possile types of 2-reks.

25 2-Brek Distne The 2-Brek istne ist(p,q) etween genomes P n Q is the minimum numer of 2reks require to trnsform P into Q. In ontrst to the genomi istne, the 2-rek istne is esy to ompute.

26 Two Genomes s Blk-Re n Green-Re Cyles P Q

27 Rerrnging P in the Q orer P Q

28 Brekpoint Grph = Superposition of Genome Grphs: Gluing Re Eges with the Sme Lels P Brekpoint Grph (Bfn & Pevzner, FOCS 1994) G(P,Q) Q

29 Blk-Green Cyles Blk n green eges represent perfet mthings in the rekpoint grph. Therefore, together these eges form olletion of lk-green lternting yles (where the olor of eges lternte). The numer of lk-green yles yles(p,q) in the rekpoint grph G(P,Q) plys entrl role in omputing the 2-rek istne etween P n Q.

30 Rerrngements Chnge Cyles Trnsforming genome P into genome Q y 2-reks orrespons to trnsforming the rekpoint grph G(P,Q) into the rekpoint grph G(Q,Q). G(P,Q) yles(p,q) = 2 G(P',Q) yles(p',q) = 3 G(Q,Q) trivil yles yles(q,q) = 4 = loks(p,q)

31 Trnsforming P into Q y 2-reks 2-reks P=P0 P1... P= Q G(P,Q) G(P1,Q)... G(Q,Q) yles(p,q) yles... loks(p,q) yles # of lk-green yles inrese y loks(p,q) - yles(p,q) How muh eh 2-rek n ontriute to this inrese?

32 2-Brek Distne Any 2-Brek inreses the numer of yles y t most one (Δyles 1) 1 Any non-trivil yle n e split into two yles with 2-rek (Δyles = 1) 1 Every sorting y 2-rek must inrese the numer of yles y loks(p,q) yles(p,q) The 2-Brek Distne etween genomes P n Q: ist(p,q) = loks(p,q) yles(p,q) (p. Ynopoulos et l., 2005, Bergeron et l., 2006)

33 Complex Rerrngements: Trnspositions Snkoff, PLoS CB 2006, suggeste tht rerrngement hotspots oserve in the Pevzner-Tesler nlysis my e result of omplex rerrngement opertions suh s trnspositions. Sorting y Trnspositions Prolem: Given two genomes, fin the shortest sequene of trnspositions trnsforming one genome into the other. First 1.5-pproximtion lgorithm ws given y Bfn n Pevzner, SODA Reent hievement: pproximtion lgorithm ue to Elis n Hrtmn, WABI The omplexity sttus ws unknown for long time. Only in 2011 it ws prove to e NP-omplete.

34 Complex Rerrngements: Trnspositions (с)

35 Sorting y Trnspositions Sorting y Trnspositions Prolem: Given two genomes, fin the shortest sequene of trnspositions trnsforming one genome into the other. First 1.5-pproximtion lgorithm ws given y Bfn n Pevzner (SODA 1995). Best hievement: pproximtion lgorithm ue to Elis n Hrtmn (WABI 2005). Prove to e NP-omplete y Bulteu, Fertin, n Rusu (ICALP 2011).

36 Trnspositions trnsposition Trnspositions ut off segment of one hromosome n insert it t some position in the sme or nother hromosome

37 Trnspositions Are 3-Breks 3-rek 3-Brek reples ny triple of lk eges with nother triple forming mthing on the sme 6 verties. Trnspositions re 3-Breks.

38 3-Brek Distne: Fous on O Cyles A 3-rek n inrese the numer of o yles (i.e., yles with o numer of lk eges) y t most 2 (Δyleso 2) A non-trivil o yle n e split into three o yles with 3-rek (Δyleso = 2) An even yle n e split into two o yles with 3-rek (Δyleso = 2) The 3-rek Distne etween genomes P n Q is: 3(P,Q) = ( loks(p,q) yleso(p,q) ) / 2

39 Multi-Brek Rerrngements The stnr rerrngement opertions (reversls, trnslotions, fusions, n fissions) mke 2 rekges in genome n glue the resulting piees in new orer. k-brek rerrngement opertion mkes k rekges in genome n glues the resulting piees in new orer. Rerrngements re rre evolutionry events n iologists elieve tht krek rerrngements re unlikely for k>3 n reltively rre for k=3 (t lest in the mmmlin evolution). Also, in rition iology, hromosome errtions for k>2 (initive of hromosome mge rther thn evolutionry vile vritions) my e more ommon, e.g., omplex rerrngements in irrite humn lymphoytes (Shs et l., 2004; Levy et l., 2004).

40 Multi-Brek Rerrngements We propose ext formuls for the k-rek istne etween multi-hromosoml irulr genomes s well s liner-time lgorithm for omputing it. (MA & PP, Theor. Comput. Si. 2008) The ext formuls for k(p,q) eomes omplex s k grows, e.g.: The formul for 20(P,Q) is estimte to ontin over 1,500 terms.

41 Brekpoints Are Verties in Non-trivil Cyles Brekpoints orrespon to regions in the genome tht were roken y some rerrngement(s). In the rekpoint grph, rekpoints orrespon to verties hving two neighors (while verties with just one neighor represent ommon jenies etween loks in the genomes). All verties in non-trivil yles in the rekpoint grph represent rekpoints.

42 Brekpoint Uses n Reuses Eh 2-rek uses four verties (the enpoints of the ffete eges). A vertex (rekpoint) is reuse if it is use y t lest two ifferent 2-reks (i.e., the numer of uses > 1). 1 Numer of uses:

43 Algorithmi Prolem: Reonstrution of Anestrl Genomes

44 Anestrl Genomes Reonstrution in Nutshell Given set of genomes, reonstrut genomes of their ommon nestors. The evolutionry tree of these genomes my e known or unknown.

45 Existing Tools for Anestrl Genomes Reonstrution GRAPPA: J. Tng, B. Moret et l. (2001) MGR: G. Bourque n P. Pevzner (2002) InferCARs: J. M, D. Hussler et l. (2006) EMRAE: H. Zho n G. Bourque (2007) MGRA: M. Alekseyev n P. Pevzner (2009)

46 Anestrl Genomes Reonstrution Prolem (with known phylogeny) Input: set of k genomes n phylogeneti tree T Output: genomes t the internl noes of the tree T Ojetive: minimize the totl sum of the genomi istnes long the rnhes of T NP-omplete in the simplest se of k=3. Wht mkes it hr?

47 Brekpoints Are Footprints of Rerrngements on the Groun of Genomes Input: set of k genomes n phylogeneti tree T Output: genomes t the internl noes of the phylogeneti tree T Ojetive funtion: the sum of genomi istnes long the rnhes of T (ssuming the most prsimonious rerrngement senrio) NP-omplete in the simplest se of k=3. Wht mkes it hr? BREAKPOINTS RE-USES (resulting in messy footprints )! Anestrl Genome Reonstrutions of MANY Genomes (i.e., for lrge k) my e esier to solve.

48 Solution: Multiple Brekpoint Grphs n MGRA Algorithm

49 How to Construt Brekpoint Grph for Multiple Genomes? P Q R

50 Construting Multiple Brekpoint Grph: rerrnging P in the Q orer P Q R

51 Construting Multiple Brekpoint Grph: rerrnging R in the Q orer P Q R

52 Multiple Brekpoint Grph: Still Gluing Re Eges with the Sme Lels P Q R Multiple Brekpoint Grph G(P,Q,R)

53 Prolem: Reonstrut Anestors of Six Mmmlin Genomes time Mouse Rt Dog mque Humn Chimpnzee

54 Multiple Brekpoint Grph of 6 Genomes Multiple Brekpoint Grph G(M,R,D,Q,H,C) of the Mouse, Rt, Dog, mque, Humn, n Chimpnzee genomes.

55 k=2 Genomes: Two Wys of Sorting y 2-Breks Trnsforming P into Q with lk 2-reks: 2-reks P = P0 P1... P-1 P = Q G(P,Q) G(P1,Q)... G(P,Q) = G(Q,Q) Trnsforming Q into P with green 2-reks: 2-reks Q = Q0 Q1... Q = P G(P,Q) G(P,Q1)... G(P,Q) = G(P,P) Let's omine these two wys...

56 Sorting By 2-Breks: Meet In The Mile Let X e ny genome on pth from P to Q: P = P0 P1... Pm = X = Q-m... Q1 Q0 = Q 2-Breks t the left n right hn sies of X re inepenent! Sorting By 2-Breks Prolem is equivlent to fining shortest trnsformtion of G(P,Q) into set of trivil yles G(X,X) (n ientity rekpoint grph of priori unknown genome X): G(P0,Q0) G(P1,Q0) G(P1,Q1) G(P1,Q2)... G(X,X) The lk n green 2-reks my ritrrily lternte.

57 MGRA: Trnsformtion into n Ientity Brekpoint Grph We fin trnsformtion of the multiple rekpoint grph G(P1,P2,...,Pk) with relile rerrngements (reognize from their footprints ) into some ( priori unknown!) ientity multiple rekpoint grph G(X,X,...,X): G(P1,P2,...,Pk)... G(X,X,...,X) Eh rerrngement is onsistent with the given tree T n thus is ssigne to some rnh of T. Rerrngements re pplie in ritrry orer tht ielly (when there re no extensive rekpoint re-uses) oes not ffet the result. Previously pplie rerrngements my revel footprints of new ones.

58 Brekpoints s Rerrngements' Footprints Q D H C M R In this yle there four rekpoints of type MR+DQHC. It n e trnsforme into trivil multi-yles: y two 2-reks: one in Mouse, Mouse one in Rt; Rt or y four 2-reks: one in eh of Dog, Dog mque, mque Humn, Humn n Chimpnzee; pnzee or y single 2-rek operting in the Mouse-Rt nestor (i.e., simultneously in Mouse n Rt)

59 Tree-Consistent Rerrngements Eh rnh of the given tree T efines two omplementry groups of genomes, to eh of whih the sme rerrngements my e pplie simultneously. For exmple, the rnh lele MR+DQHC efines groups {M, R} (Mouse n Rt) n {D,Q,H,C} (Dog, mque, Humn, Chimpnzee). But there re no groups like {M,C} or {R,D,H}. So, we n pply the sme rerrngement to M n R simultneously, viewing it s hppening in their ommon nestor (enote MR) long the MR+DQHC rnh.

60 When All Relile 2-Breks Are Ientifie n Unone The multiple rekpoint grph is reue rmtilly! The remining (non-trivil) omponents n e proesse mnully in the se-y-se fshion.

61 MGRA: Reonstrution of the Anestrl Genomes The resulting ientity rekpoint grph G(X,X,...,X) efines its unerlying genome X. The reverse trnsformtion is pplie to the genome X to trnsform it into eh of the originl genomes P1, P2,..., Pk. This trnsformtion trverses ll internl noes of T n thus efines the nestrl genome t every noe.

62 Reonstrute X Chromosomes The Mouse, Rt, Dog, mque, Humn, Chimpnzee genomes n their reonstrute nestors:

63 If The Evolutionry Tree Is Not Known For the set of 7 mmmlin genomes: Mouse, Rt, Dog, mque, Humn, Chimpnzee, n Opossum, the evolutionry tree T is not known. Depening on the primte roent rnivore split, three topologies re possile (only two of them re vile). However, these three topologies shre mny ommon rnhes in T (onfient rnhes). We n restrit the trnsformtion only to suh rnhes in orer to simplify the rekpoint grph, not reking n eviene for either of the topologies.

64 Resolving The Primte-Roent-Crnivore Split Controversy We reue the multiple rekpoint grph G(M,D,Q,O) (of representtives of eh fmily n n outgroup) with relile 2-reks on the onfient groups of genomes. M M M Q Q O Q D O D O D Wht woul e n eviene for one topology over the others?

65 Rerrngement Eviene For The PrimteCrnivore Split Eh of the three topologies hs n unique rnh in the tree. A single rerrngement ssigne to suh rnh woul orrespon to lest two rerrngements if this rnh is sent in the tree. M O M M 12 Q 19 Q 26 D O D O D We oserve the prevlene of rerrngements' footprints speifi to the primte rnivore split. Q

66 Aknowlegments Pvel Pevzner, UC Sn Diego Glenn Tesler, UC Sn Diego Jin M, University of Illinois t Urn-Chmpign

67 Biologil Prolem: Why n Where Genome Rerrngements Hppen?

68 Chromosome Brekge Moels Chromosome Brekge Moels speify how hromosomes re roken y rerrngements. While the ext mehnism of rerrngements is not known, suh moels try to explin s mny s possile sttistil hrteristis oserve in rel genomes. The more hrteristis re pture y moel, the etter is this moel. The hoie of moel is prtiulrly importnt in simultions tht im retion of simulte genomes whose hrteristis shoul mth those of rel genomes.

69 Testing Moels Given hrteristi oserve in rel genomes n hromosome rekge moel, we n test whether the moel explins this hrteristi. Test: Simulte genomes using the moel n hek if the simulte genomes possess the require hrteristi. As soon s new hrteristi in rel genomes is isovere, the existing moels n e teste ginst it. If they fil, this lls for new moel tht woul explin ll previously known hrteristis s well s the new one.

70 Susumu Ohno: Rerrngements our rnomly Ohno, 1970, 1973 Rnom Brekge Hypothesis: Genomi rhitetures re shpe y rerrngements tht our rnomly.

71 Rnom Brekge Moel (RBM) The rnom rekge hypothesis ws emre y iologists n hs eome e fto theory of hromosome evolution. Neu & Tylor, Pro. Nt'l A. Sienes 1984 First onvining rguments in fvor of the Rnom Brekge Moel (RBM) RBM implies tht there is no rerrngement hotspots RBM ws re-iterte in hunres of ppers

72 Frgile Brekge Moel (FBM) Pevzner & Tesler, PNAS 2003 rgue tht every evolutionry senrio for trnsforming Mouse into Humn genome must result in lrge numer of rekpoint re-uses, ontrition to the RBM. propose the Frgile Brekge Moel (FBM) tht postultes existene of rerrngement hotspots n vst rekpoint re-use FBM implies tht the humn genome is mosi of soli n frgile regions

73 Reuttl of the Reuttl Snkoff & Trinh, J. Comput. Biol. 2004, presente rguments ginst the Frgile Brekge Moel: we hve shown tht rekpoint re use of the sme mgnitue s foun in Pevzner n Tesler, 2003 my very well e rtifts in ontext where NO re use tully ourre.

74 Reuttl of the Reuttl of the Reuttl Snkoff & Trinh, J. Comput. Biol. 2004, presente rguments ginst the Frgile Brekge Moel: we hve shown tht rekpoint re use of the sme mgnitue s foun in Pevzner n Tesler, 2003 my very well e rti fts in ontext where NO re use tully ourre. Peng et l., PLoS Comput. Biol. 2006, foun n error in the Snkoff Trinh rguments. Snkoff, PLoS Comput. Biol. 2006, knowlege the error: Not only i we foist hstily oneive n inor retly exeute simultion on n overworke RECOMB on ferene progrm ommittee, ut worse nostr mxim ulp we olige tem of high powere reserhers to len up fter us!

75 All Reent Stuies Support FBM Kikut et l., Genome Res. 2007:... the Neu n Tylor hypothesis is not possile for the explntion of synteny in rt.

76 With One Influentil Exeption M et l., Genome Res. 2006: Simultions suggest tht this frequeny of rekpoint re use is pproximtely wht one woul expet if rekge ws eqully likely for every genomi position reful nlysis [of the RBM vs. FBM ontroversy] is eyon the sope of this stuy.

77 Our Contriution We reonile the eviene for limite rekpoint reuse in M et l., 2006 with the Frgile Brekge Moel n revel rmpnt ut elusive rekpoint reuse. We provie eviene for the irth n eth of the frgile regions, implying tht they move to ifferent lotions in ifferent lineges, explining why M et l., 2006, foun limite rekpoint reuse etween ifferent rnhes of the evolutionry tree. We introue the Turnover Frgile Brekge Moel (TFBM) tht ounts for the irth n eth of the frgile regions n shes light on possile reltionship etween rerrngements n Mthing Segmentl Duplitions. Duplitions TFBM points to lotions of the urrently frgile regions in the humn genome.

78 Tests vs. Moels Why iologists elieve in RBM? Beuse RBM implies the exponentil istriution of the sizes of the synteny loks oserve in rel genomes. A flw in this logi: RBM is not the only moel tht omplies with the exponentil istriution test. Why Pevzner n Tesler refute RBM? Beuse RBM oes not omply with the rekpoint reuse test: RBM implies low reuse ut rel genomes revel high reuse. FBM omplies with oth the exponentil istriution n rekpoint reuse tests. But is there test tht oth RBM n FBM fil? Test Moel RBM FBM Exponentil Brekpoint istriution reuse YES NO YES YES

79 Tests vs. Moels Why iologists elieve in RBM? Beuse RBM implies the exponentil istriution of the sizes of the synteny loks oserve in rel genomes. A flw in this logi: RBM is not the only moel tht omplies with the exponentil istriution test. Why Pevzner n Tesler refute RBM? Beuse RBM oes not omply with the rekpoint reuse test: RBM implies low reuse ut rel genomes revel high reuse. FBM omplies with oth the exponentil istriution n rekpoint reuse tests. RBM n FBM fil the Multispeies Brekpoint Reuse (MBR) test. Test Moel RBM FBM Exponentil Brekpoint istriution reuse MBR YES NO NO YES YES NO

80 Tests vs. Moels Why iologists elieve in RBM? Beuse RBM implies the exponentil istriution of the sizes of the synteny loks oserve in rel genomes. A flw in this logi: RBM is not the only moel tht omplies with the exponentil istriution test. Why Pevzner n Tesler refute RBM? Beuse RBM oes not omply with the rekpoint reuse test: RBM implies low reuse ut rel genomes revel high reuse. FBM omplies with oth the exponentil istriution n rekpoint reuse tests. TFBM psses ll three tests. Test Moel RBM FBM TFBM Exponentil Brekpoint istriution reuse MBR YES NO NO YES YES NO YES YES YES

81 Algorithmi Prolem: Brekpoint Re-use Anlysis

82 Brekpoints Are Verties in Non-trivil Cyles Brekpoints orrespon to regions in the genome tht were roken y some rerrngement(s). In the rekpoint grph, rekpoints orrespon to verties hving two neighors (while verties with just one neighor represent ommon jenies etween synteny loks). All verties in non-trivil yles in the rekpoint grph represent rekpoints.

83 Brekpoint Uses n Reuses Eh 2-rek uses four verties (the enpoints of the ffete eges). A vertex (rekpoint) is reuse if it is use y t lest two ifferent 2-reks (i.e., the numer of uses > 1). 1 Numer of uses:

84 Intr- n Inter- Reuses For n evolutionry tree with known rerrngement senrios, rekpoint is intr-reuse on some rnh if it is use y t lest two ifferent 2-reks long this rnh. Similrly, rekpoint is inter-reuse ross two rnhes if it is use on oth these rnhes.

85 Rerrngement Senrios Remin Amiguous In mmmlin evolution we know only genomes of existing speies ut o not know the nestrl genomes. While nestrl genomes n e relily reonstrute, the ext rerrngement senrios etween them remin miguous. Cn we ompute the numer of rekpoint intrn inter- reuses without knowing rerrngement senrios?

86 Numer of Intr-Reuses (Lower Boun) For rerrngement senrio etween genomes P n Q: The numer of 2-reks is t lest ist(p,q) Eh 2-rek uses 4 rekpoints The numer of rekpoints is 2 loks(p,q) Hene the totl numer of intr-reuses is: 4 ist(p,q) 2 loks(p,q)

87 Numer of Inter-Reuses (Lower Boun) For two rnhes (P,Q) n (P',Q') in the tree: Set V of the verties in non-trivil yles in G(P,Q) represents the rekpoints etween genomes P n Q Set V' of the verties in non-trivil yles in G(P',Q') represents the rekpoints etween genomes P' n Q' Hene, the numer of inter-reuses is size of the intersetion of V n V'

88 Surprising Irregulrities in Brekpoint Reuse Aross Vrious Pirs of Brnhes Sttistis of rekpoint intr- n inter-reuses etween the rnhes of the tree of six mmmlin genomes: Colors represent the istne etween pir of rnhes: re = jent rnhes; green = rnhes seprte y one other rnh; yellow = rnhes seprte y two other rnhes. Wht is surprising out this Tle?

89 Solution: Turnover Frgile Brekge Moel n Multispeie Brekpoint Reuse Test

90 Turnover Frgile Brekge Moel (TFBM) The M et l. oservtion n the sttistis of inter-reuses inites: Brekpoint inter-reuses mostly hppen ross jent rnhes of the evolutionry tree. Turnover Frgile Brekge Moel (TFBM): Frgile regions re sujet to irth n eth proess n thus hve limite lifespn.

91 Simplest TFBM: Fixe Turnover Rte for Frgile Regions TFBM(m,n,x): TFBM(m,n,x) genomes hve m frgile regions n (out of m) frgile regions re tive eh 2-rek is pplie to 2 (out of n) rnomly hosen tive frgile regions fter eh 2-rek, x tive frgile regions (out of n) ie n x new tive frgile regions (out of m-n) m-n re orn FBM is prtiulr se of TFBM with x=0 RBM is prtiulr se of TFBM with x=0 n n=m

92 Reognizing the Birth n Deth Given n evolutionry tree with known rerrngement senrios, how one woul etermine whether they followe TFBM with x = 0 (tht is, FBM/RBM) or x > 0? Compring rekpoint inter-reuse ross ifferent pirs of rnhes woul help, ut it lso epens on the rnh lengths tht my iffer signifintly ross the tree.

93 Sle Brekpoint Reuse The numer of rekpoint intr- n inter- reuses epens on the length of rnhes. To eliminte this epeneny, we efine the sle intr- n inter- reuse: We efine n expresse nlytilly: E(t) = the expete numer of intr-reuses long rnh of length t; E(t1,t2) = the expete numer of inter-reuses ross rnhes of length t1 n t2. Sle intr- n inter-reuse is the numer of reuses ivie y E(t) or E(t1,t2) respetively.

94 Sle Inter-Reuse in Colore Cells (Simulte Genomes with Vrile Brnh Length) FBM TFBM TFBM TFBM Simultions for the se when n=900 out of m=2000 frgile regions re tive n vrious turnover rte x=0..4.

95 Mesuring Reuse in the Whole Evolutionry Tree TFBM suggests tht on verge the numer rekpoint reuses r(r1,r2) for 2-reks r1 n r2 epens on the istne (in the evolutionry tree) etween them. The lrger is the istne, the smller is r(r1,r2). Our gol is to efine single mesure for the whole tree tht woul esrie this tren n llow one to test whether the rerrngement proess follow the TFBM with x>0.

96 Multispeies Brekpoint Reuse The multispeies rekpoint reuse is funtion R(L) expressing verge rekpoint reuse etween pirs of rerrngements seprte y L other rerrngements in the given tree. It n e expliitly efine s: R(L) = Σ r(r1,r2) / Σ 1 where oth sums re tken over ll pirs of rerrngements r1 n r2 t istne L in the tree.

97 Multispeies Brekpoint Reuse Test For RBM/FBM, R(L) is onstnt. For TFBM with x > 0, R(L) is eresing funtion. MBR Test: ompute R(L), n hek if it is eresing. (A stronger vrint: etermine x n hek if x>0.)

98 Multispeies Brekpoint Reuse in TFBM (theoreti urve) For TFBM with prmeters m, n, x, we erive n nlyti formul: R(L) = 8(m-n)/(mn) * ( 1 xm/(n(m-n)) )L + 8/m For smll L, R(L) is pproximte y stright line: 8/n 8x/n2 L whih oes not epen on m. Given R(L), the prmeters n n x n e etermine from the vlue n slope of R(L) t L=0.

99 Multispeies Brekpoint Reuse in TFBM (theoreti vs. empiri urve) Simultions for the se when n=160 out of m=800 frgile regions re tive n vrious turnover rte x

100 From Simulte to Rel Genomes: Complitions It is esy to ompute R(L) for simulte genomes, whose rerrngement history is efine y simultions. For rel genomes, while we n relily reonstrut the nestrl genomes, the ext evolutionry senrios etween them remin miguous.

101 From Simulte to Rel Genomes: Complitions It is esy to ompute R(L) for simulte genomes, whose rerrngement history is efine y simultions. For rel genomes, while we n relily reonstrut the nestrl genomes, the ext evolutionry senrios etween them remin miguous. We n smple rnom senrios inste.

102 Multispeies Reuse etween Mmmlin Genomes Best fit: m 4017 n 196 x 1.12

103 Implitions: How will the Humn Genome Evolve in the Next Million Yers?

104 Preition Power of TFBM Cn we etermine urrently tive regions in the humn genome H from omprison with other mmmlin genomes? RBM provies no lue FBM suggests to onsier the rekpoints etween H n ny other genome TFBM suggests to onsier the losest genome suh s the mque-humn nestor QH. QH Brekpoints in G(QH,H) re likely to e reuse in the future rerrngements of H.

105 Vlition of Preitions for the Mque-Humn Anestor (QH) Preition of frgile regions on (QH,H) se on the mouse, rt, n og genomes: Using mouse genome M s proxy: ury 34 / 552 6% Using mouse-rt-og nestor genome MRD: MRD ury 18 / % Using mque genome Q: ury 10 / 68 16% (using synteny loks lrger thn 500K)

106 Puttive Ative Frgile Regions in the Humn Genome

107 Unsolve Mystery: Wht Cuses Frgility? Zho n Bourque, Genome Res. 2009, suggeste tht frgility is promote y Mthing Segmentl Duplitions, pir of long similr regions lote within rekpoint regions flnking rerrngement. TFBM is onsistent with this hypothesis sine the similrity etween MSDs eteriortes with time, implying tht MSDs re lso sujet to irth n eth proess.

108 Aknowlegments Pvel Pevzner, UC Sn Diego Glenn Tesler, UC Sn Diego Jin M, University of Illinois t Urn-Chmpign

Recent advances in analysis of evolutionary transpositions

Recent advances in analysis of evolutionary transpositions Reent vnes in nlysis of evolutionry trnspositions Mx Alekseyev Deprtment of Mthemtis / Computtionl Biology Institute George Wshington University 2015 Genome Rerrngements Mouse X hromosome unknown nestor

More information

Genome Rearrangements:

Genome Rearrangements: Genome Rerrngements: from Biologicl Prolem to Comintoril Algorithms (nd ck) Pvel Pevzner Deprtment of Computer Science, University of Cliforni t Sn Diego Genome Rerrngements Mouse X chromosome Unknown

More information

CS 491G Combinatorial Optimization Lecture Notes

CS 491G Combinatorial Optimization Lecture Notes CS 491G Comintoril Optimiztion Leture Notes Dvi Owen July 30, August 1 1 Mthings Figure 1: two possile mthings in simple grph. Definition 1 Given grph G = V, E, mthing is olletion of eges M suh tht e i,

More information

Counting Paths Between Vertices. Isomorphism of Graphs. Isomorphism of Graphs. Isomorphism of Graphs. Isomorphism of Graphs. Isomorphism of Graphs

Counting Paths Between Vertices. Isomorphism of Graphs. Isomorphism of Graphs. Isomorphism of Graphs. Isomorphism of Graphs. Isomorphism of Graphs Isomorphism of Grphs Definition The simple grphs G 1 = (V 1, E 1 ) n G = (V, E ) re isomorphi if there is ijetion (n oneto-one n onto funtion) f from V 1 to V with the property tht n re jent in G 1 if

More information

Solutions for HW9. Bipartite: put the red vertices in V 1 and the black in V 2. Not bipartite!

Solutions for HW9. Bipartite: put the red vertices in V 1 and the black in V 2. Not bipartite! Solutions for HW9 Exerise 28. () Drw C 6, W 6 K 6, n K 5,3. C 6 : W 6 : K 6 : K 5,3 : () Whih of the following re iprtite? Justify your nswer. Biprtite: put the re verties in V 1 n the lk in V 2. Biprtite:

More information

Mid-Term Examination - Spring 2014 Mathematical Programming with Applications to Economics Total Score: 45; Time: 3 hours

Mid-Term Examination - Spring 2014 Mathematical Programming with Applications to Economics Total Score: 45; Time: 3 hours Mi-Term Exmintion - Spring 0 Mthemtil Progrmming with Applitions to Eonomis Totl Sore: 5; Time: hours. Let G = (N, E) e irete grph. Define the inegree of vertex i N s the numer of eges tht re oming into

More information

CIT 596 Theory of Computation 1. Graphs and Digraphs

CIT 596 Theory of Computation 1. Graphs and Digraphs CIT 596 Theory of Computtion 1 A grph G = (V (G), E(G)) onsists of two finite sets: V (G), the vertex set of the grph, often enote y just V, whih is nonempty set of elements lle verties, n E(G), the ege

More information

Numbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point

Numbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point GCSE C Emple 7 Work out 9 Give your nswer in its simplest form Numers n inies Reiprote mens invert or turn upsie own The reiprol of is 9 9 Mke sure you only invert the frtion you re iviing y 7 You multiply

More information

22: Union Find. CS 473u - Algorithms - Spring April 14, We want to maintain a collection of sets, under the operations of:

22: Union Find. CS 473u - Algorithms - Spring April 14, We want to maintain a collection of sets, under the operations of: 22: Union Fin CS 473u - Algorithms - Spring 2005 April 14, 2005 1 Union-Fin We wnt to mintin olletion of sets, uner the opertions of: 1. MkeSet(x) - rete set tht ontins the single element x. 2. Fin(x)

More information

Factorising FACTORISING.

Factorising FACTORISING. Ftorising FACTORISING www.mthletis.om.u Ftorising FACTORISING Ftorising is the opposite of expning. It is the proess of putting expressions into rkets rther thn expning them out. In this setion you will

More information

I 3 2 = I I 4 = 2A

I 3 2 = I I 4 = 2A ECE 210 Eletril Ciruit Anlysis University of llinois t Chigo 2.13 We re ske to use KCL to fin urrents 1 4. The key point in pplying KCL in this prolem is to strt with noe where only one of the urrents

More information

for all x in [a,b], then the area of the region bounded by the graphs of f and g and the vertical lines x = a and x = b is b [ ( ) ( )] A= f x g x dx

for all x in [a,b], then the area of the region bounded by the graphs of f and g and the vertical lines x = a and x = b is b [ ( ) ( )] A= f x g x dx Applitions of Integrtion Are of Region Between Two Curves Ojetive: Fin the re of region etween two urves using integrtion. Fin the re of region etween interseting urves using integrtion. Desrie integrtion

More information

CSE 332. Sorting. Data Abstractions. CSE 332: Data Abstractions. QuickSort Cutoff 1. Where We Are 2. Bounding The MAXIMUM Problem 4

CSE 332. Sorting. Data Abstractions. CSE 332: Data Abstractions. QuickSort Cutoff 1. Where We Are 2. Bounding The MAXIMUM Problem 4 Am Blnk Leture 13 Winter 2016 CSE 332 CSE 332: Dt Astrtions Sorting Dt Astrtions QuikSort Cutoff 1 Where We Are 2 For smll n, the reursion is wste. The onstnts on quik/merge sort re higher thn the ones

More information

Lecture 6: Coding theory

Lecture 6: Coding theory Leture 6: Coing theory Biology 429 Crl Bergstrom Ferury 4, 2008 Soures: This leture loosely follows Cover n Thoms Chpter 5 n Yeung Chpter 3. As usul, some of the text n equtions re tken iretly from those

More information

CS 2204 DIGITAL LOGIC & STATE MACHINE DESIGN SPRING 2014

CS 2204 DIGITAL LOGIC & STATE MACHINE DESIGN SPRING 2014 S 224 DIGITAL LOGI & STATE MAHINE DESIGN SPRING 214 DUE : Mrh 27, 214 HOMEWORK III READ : Relte portions of hpters VII n VIII ASSIGNMENT : There re three questions. Solve ll homework n exm prolems s shown

More information

1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. 1 PYTHAGORAS THEOREM 1 1 Pythgors Theorem In this setion we will present geometri proof of the fmous theorem of Pythgors. Given right ngled tringle, the squre of the hypotenuse is equl to the sum of the

More information

Monochromatic Plane Matchings in Bicolored Point Set

Monochromatic Plane Matchings in Bicolored Point Set CCCG 2017, Ottw, Ontrio, July 26 28, 2017 Monohromti Plne Mthings in Biolore Point Set A. Krim Au-Affsh Sujoy Bhore Pz Crmi Astrt Motivte y networks interply, we stuy the prolem of omputing monohromti

More information

Project 6: Minigoals Towards Simplifying and Rewriting Expressions

Project 6: Minigoals Towards Simplifying and Rewriting Expressions MAT 51 Wldis Projet 6: Minigols Towrds Simplifying nd Rewriting Expressions The distriutive property nd like terms You hve proly lerned in previous lsses out dding like terms ut one prolem with the wy

More information

Section 1.3 Triangles

Section 1.3 Triangles Se 1.3 Tringles 21 Setion 1.3 Tringles LELING TRINGLE The line segments tht form tringle re lled the sides of the tringle. Eh pir of sides forms n ngle, lled n interior ngle, nd eh tringle hs three interior

More information

Common intervals of genomes. Mathieu Raffinot CNRS LIAFA

Common intervals of genomes. Mathieu Raffinot CNRS LIAFA Common intervls of genomes Mthieu Rffinot CNRS LIF Context: omprtive genomis. set of genomes prtilly/totlly nnotte Informtive group of genes or omins? Ex: COG tse Mny iffiulties! iology Wht re two similr

More information

Lecture 2: Cayley Graphs

Lecture 2: Cayley Graphs Mth 137B Professor: Pri Brtlett Leture 2: Cyley Grphs Week 3 UCSB 2014 (Relevnt soure mteril: Setion VIII.1 of Bollos s Moern Grph Theory; 3.7 of Gosil n Royle s Algeri Grph Theory; vrious ppers I ve re

More information

Necessary and sucient conditions for some two. Abstract. Further we show that the necessary conditions for the existence of an OD(44 s 1 s 2 )

Necessary and sucient conditions for some two. Abstract. Further we show that the necessary conditions for the existence of an OD(44 s 1 s 2 ) Neessry n suient onitions for some two vrile orthogonl esigns in orer 44 C. Koukouvinos, M. Mitrouli y, n Jennifer Seerry z Deite to Professor Anne Penfol Street Astrt We give new lgorithm whih llows us

More information

Now we must transform the original model so we can use the new parameters. = S max. Recruits

Now we must transform the original model so we can use the new parameters. = S max. Recruits MODEL FOR VARIABLE RECRUITMENT (ontinue) Alterntive Prmeteriztions of the pwner-reruit Moels We n write ny moel in numerous ifferent ut equivlent forms. Uner ertin irumstnes it is onvenient to work with

More information

Eigenvectors and Eigenvalues

Eigenvectors and Eigenvalues MTB 050 1 ORIGIN 1 Eigenvets n Eigenvlues This wksheet esries the lger use to lulte "prinipl" "hrteristi" iretions lle Eigenvets n the "prinipl" "hrteristi" vlues lle Eigenvlues ssoite with these iretions.

More information

Comparing the Pre-image and Image of a Dilation

Comparing the Pre-image and Image of a Dilation hpter Summry Key Terms Postultes nd Theorems similr tringles (.1) inluded ngle (.2) inluded side (.2) geometri men (.) indiret mesurement (.6) ngle-ngle Similrity Theorem (.2) Side-Side-Side Similrity

More information

CARLETON UNIVERSITY. 1.0 Problems and Most Solutions, Sect B, 2005

CARLETON UNIVERSITY. 1.0 Problems and Most Solutions, Sect B, 2005 RLETON UNIVERSIT eprtment of Eletronis ELE 2607 Swithing iruits erury 28, 05; 0 pm.0 Prolems n Most Solutions, Set, 2005 Jn. 2, #8 n #0; Simplify, Prove Prolem. #8 Simplify + + + Reue to four letters (literls).

More information

2.4 Theoretical Foundations

2.4 Theoretical Foundations 2 Progrmming Lnguge Syntx 2.4 Theoretil Fountions As note in the min text, snners n prsers re se on the finite utomt n pushown utomt tht form the ottom two levels of the Chomsky lnguge hierrhy. At eh level

More information

Lecture 4: Graph Theory and the Four-Color Theorem

Lecture 4: Graph Theory and the Four-Color Theorem CCS Disrete II Professor: Pri Brtlett Leture 4: Grph Theory n the Four-Color Theorem Week 4 UCSB 2015 Through the rest of this lss, we re going to refer frequently to things lle grphs! If you hen t seen

More information

Arrow s Impossibility Theorem

Arrow s Impossibility Theorem Rep Voting Prdoxes Properties Arrow s Theorem Arrow s Impossiility Theorem Leture 12 Arrow s Impossiility Theorem Leture 12, Slide 1 Rep Voting Prdoxes Properties Arrow s Theorem Leture Overview 1 Rep

More information

Probability. b a b. a b 32.

Probability. b a b. a b 32. Proility If n event n hppen in '' wys nd fil in '' wys, nd eh of these wys is eqully likely, then proility or the hne, or its hppening is, nd tht of its filing is eg, If in lottery there re prizes nd lnks,

More information

Surds and Indices. Surds and Indices. Curriculum Ready ACMNA: 233,

Surds and Indices. Surds and Indices. Curriculum Ready ACMNA: 233, Surs n Inies Surs n Inies Curriulum Rey ACMNA:, 6 www.mthletis.om Surs SURDS & & Inies INDICES Inies n surs re very losely relte. A numer uner (squre root sign) is lle sur if the squre root n t e simplifie.

More information

CS 360 Exam 2 Fall 2014 Name

CS 360 Exam 2 Fall 2014 Name CS 360 Exm 2 Fll 2014 Nme 1. The lsses shown elow efine singly-linke list n stk. Write three ifferent O(n)-time versions of the reverse_print metho s speifie elow. Eh version of the metho shoul output

More information

Lecture 8: Abstract Algebra

Lecture 8: Abstract Algebra Mth 94 Professor: Pri Brtlett Leture 8: Astrt Alger Week 8 UCSB 2015 This is the eighth week of the Mthemtis Sujet Test GRE prep ourse; here, we run very rough-n-tumle review of strt lger! As lwys, this

More information

Linear Algebra Introduction

Linear Algebra Introduction Introdution Wht is Liner Alger out? Liner Alger is rnh of mthemtis whih emerged yers k nd ws one of the pioneer rnhes of mthemtis Though, initilly it strted with solving of the simple liner eqution x +

More information

SIMPLE NONLINEAR GRAPHS

SIMPLE NONLINEAR GRAPHS S i m p l e N o n l i n e r G r p h s SIMPLE NONLINEAR GRAPHS www.mthletis.om.u Simple SIMPLE Nonliner NONLINEAR Grphs GRAPHS Liner equtions hve the form = m+ where the power of (n ) is lws. The re lle

More information

Data Structures LECTURE 10. Huffman coding. Example. Coding: problem definition

Data Structures LECTURE 10. Huffman coding. Example. Coding: problem definition Dt Strutures, Spring 24 L. Joskowiz Dt Strutures LEURE Humn oing Motivtion Uniquel eipherle oes Prei oes Humn oe onstrution Etensions n pplitions hpter 6.3 pp 385 392 in tetook Motivtion Suppose we wnt

More information

On a Class of Planar Graphs with Straight-Line Grid Drawings on Linear Area

On a Class of Planar Graphs with Straight-Line Grid Drawings on Linear Area Journl of Grph Algorithms n Applitions http://jg.info/ vol. 13, no. 2, pp. 153 177 (2009) On Clss of Plnr Grphs with Stright-Line Gri Drwings on Liner Are M. Rezul Krim 1,2 M. Siur Rhmn 1 1 Deprtment of

More information

Lecture 11 Binary Decision Diagrams (BDDs)

Lecture 11 Binary Decision Diagrams (BDDs) C 474A/57A Computer-Aie Logi Design Leture Binry Deision Digrms (BDDs) C 474/575 Susn Lyseky o 3 Boolen Logi untions Representtions untion n e represente in ierent wys ruth tle, eqution, K-mp, iruit, et

More information

Lesson 2: The Pythagorean Theorem and Similar Triangles. A Brief Review of the Pythagorean Theorem.

Lesson 2: The Pythagorean Theorem and Similar Triangles. A Brief Review of the Pythagorean Theorem. 27 Lesson 2: The Pythgoren Theorem nd Similr Tringles A Brief Review of the Pythgoren Theorem. Rell tht n ngle whih mesures 90º is lled right ngle. If one of the ngles of tringle is right ngle, then we

More information

18.06 Problem Set 4 Due Wednesday, Oct. 11, 2006 at 4:00 p.m. in 2-106

18.06 Problem Set 4 Due Wednesday, Oct. 11, 2006 at 4:00 p.m. in 2-106 8. Problem Set Due Wenesy, Ot., t : p.m. in - Problem Mony / Consier the eight vetors 5, 5, 5,..., () List ll of the one-element, linerly epenent sets forme from these. (b) Wht re the two-element, linerly

More information

Coding Techniques. Manjunatha. P. Professor Dept. of ECE. June 28, J.N.N. College of Engineering, Shimoga.

Coding Techniques. Manjunatha. P. Professor Dept. of ECE. June 28, J.N.N. College of Engineering, Shimoga. Coing Tehniques Mnjunth. P mnjup.jnne@gmil.om Professor Dept. of ECE J.N.N. College of Engineering, Shimog June 8, 3 Overview Convolutionl Enoing Mnjunth. P (JNNCE) Coing Tehniques June 8, 3 / 8 Overview

More information

Electronic Circuits I Revision after midterm

Electronic Circuits I Revision after midterm Eletroni Ciruits I Revision fter miterm Dr. Ahme ElShfee, ACU : Fll 2018, Eletroni Ciruits I -1 / 14 - MCQ1 # Question If the frequeny of the input voltge in Figure 2 36 is inrese, the output voltge will

More information

Let s divide up the interval [ ab, ] into n subintervals with the same length, so we have

Let s divide up the interval [ ab, ] into n subintervals with the same length, so we have III. INTEGRATION Eonomists seem muh more intereste in mrginl effets n ifferentition thn in integrtion. Integrtion is importnt for fining the epete vlue n vrine of rnom vriles, whih is use in eonometris

More information

Lecture Notes No. 10

Lecture Notes No. 10 2.6 System Identifition, Estimtion, nd Lerning Leture otes o. Mrh 3, 26 6 Model Struture of Liner ime Invrint Systems 6. Model Struture In representing dynmil system, the first step is to find n pproprite

More information

Metaheuristics for the Asymmetric Hamiltonian Path Problem

Metaheuristics for the Asymmetric Hamiltonian Path Problem Metheuristis for the Asymmetri Hmiltonin Pth Prolem João Pero PEDROSO INESC - Porto n DCC - Fule e Ciênis, Universie o Porto, Portugl jpp@f.up.pt Astrt. One of the most importnt pplitions of the Asymmetri

More information

CS 573 Automata Theory and Formal Languages

CS 573 Automata Theory and Formal Languages Non-determinism Automt Theory nd Forml Lnguges Professor Leslie Lnder Leture # 3 Septemer 6, 2 To hieve our gol, we need the onept of Non-deterministi Finite Automton with -moves (NFA) An NFA is tuple

More information

Momentum and Energy Review

Momentum and Energy Review Momentum n Energy Review Nme: Dte: 1. A 0.0600-kilogrm ll trveling t 60.0 meters per seon hits onrete wll. Wht spee must 0.0100-kilogrm ullet hve in orer to hit the wll with the sme mgnitue of momentum

More information

Equivalent fractions have the same value but they have different denominators. This means they have been divided into a different number of parts.

Equivalent fractions have the same value but they have different denominators. This means they have been divided into a different number of parts. Frtions equivlent frtions Equivlent frtions hve the sme vlue ut they hve ifferent enomintors. This mens they hve een ivie into ifferent numer of prts. Use the wll to fin the equivlent frtions: Wht frtions

More information

ANALYSIS AND MODELLING OF RAINFALL EVENTS

ANALYSIS AND MODELLING OF RAINFALL EVENTS Proeedings of the 14 th Interntionl Conferene on Environmentl Siene nd Tehnology Athens, Greee, 3-5 Septemer 215 ANALYSIS AND MODELLING OF RAINFALL EVENTS IOANNIDIS K., KARAGRIGORIOU A. nd LEKKAS D.F.

More information

Green s Theorem. (2x e y ) da. (2x e y ) dx dy. x 2 xe y. (1 e y ) dy. y=1. = y e y. y=0. = 2 e

Green s Theorem. (2x e y ) da. (2x e y ) dx dy. x 2 xe y. (1 e y ) dy. y=1. = y e y. y=0. = 2 e Green s Theorem. Let be the boundry of the unit squre, y, oriented ounterlokwise, nd let F be the vetor field F, y e y +, 2 y. Find F d r. Solution. Let s write P, y e y + nd Q, y 2 y, so tht F P, Q. Let

More information

MCH T 111 Handout Triangle Review Page 1 of 3

MCH T 111 Handout Triangle Review Page 1 of 3 Hnout Tringle Review Pge of 3 In the stuy of sttis, it is importnt tht you e le to solve lgeri equtions n tringle prolems using trigonometry. The following is review of trigonometry sis. Right Tringle:

More information

On the Spectra of Bipartite Directed Subgraphs of K 4

On the Spectra of Bipartite Directed Subgraphs of K 4 On the Spetr of Biprtite Direte Sugrphs of K 4 R. C. Bunge, 1 S. I. El-Znti, 1, H. J. Fry, 1 K. S. Kruss, 2 D. P. Roerts, 3 C. A. Sullivn, 4 A. A. Unsiker, 5 N. E. Witt 6 1 Illinois Stte University, Norml,

More information

arxiv: v2 [math.co] 31 Oct 2016

arxiv: v2 [math.co] 31 Oct 2016 On exlue minors of onnetivity 2 for the lss of frme mtrois rxiv:1502.06896v2 [mth.co] 31 Ot 2016 Mtt DeVos Dryl Funk Irene Pivotto Astrt We investigte the set of exlue minors of onnetivity 2 for the lss

More information

Algebra 2 Semester 1 Practice Final

Algebra 2 Semester 1 Practice Final Alger 2 Semester Prtie Finl Multiple Choie Ientify the hoie tht est ompletes the sttement or nswers the question. To whih set of numers oes the numer elong?. 2 5 integers rtionl numers irrtionl numers

More information

Graph Algorithms. Vertex set = { a,b,c,d } Edge set = { {a,c}, {b,c}, {c,d}, {b,d}} Figure 1: An example for a simple graph

Graph Algorithms. Vertex set = { a,b,c,d } Edge set = { {a,c}, {b,c}, {c,d}, {b,d}} Figure 1: An example for a simple graph Inin Institute of Informtion Tehnology Design n Mnufturing, Knheepurm, Chenni 00, Ini An Autonomous Institute uner MHRD, Govt of Ini http://www.iiitm..in COM 0T Design n Anlysis of Algorithms -Leture Notes

More information

Welcome. Balanced search trees. Balanced Search Trees. Inge Li Gørtz

Welcome. Balanced search trees. Balanced Search Trees. Inge Li Gørtz Welome nge Li Gørt. everse tehing n isussion of exerises: 02110 nge Li Gørt 3 tehing ssistnts 8.00-9.15 Group work 9.15-9.45 isussions of your solutions in lss 10.00-11.15 Leture 11.15-11.45 Work on exerises

More information

Algorithm Design and Analysis

Algorithm Design and Analysis Algorithm Design nd Anlysis LECTURE 8 Mx. lteness ont d Optiml Ching Adm Smith 9/12/2008 A. Smith; sed on slides y E. Demine, C. Leiserson, S. Rskhodnikov, K. Wyne Sheduling to Minimizing Lteness Minimizing

More information

Arrow s Impossibility Theorem

Arrow s Impossibility Theorem Rep Fun Gme Properties Arrow s Theorem Arrow s Impossiility Theorem Leture 12 Arrow s Impossiility Theorem Leture 12, Slide 1 Rep Fun Gme Properties Arrow s Theorem Leture Overview 1 Rep 2 Fun Gme 3 Properties

More information

Part I: Study the theorem statement.

Part I: Study the theorem statement. Nme 1 Nme 2 Nme 3 A STUDY OF PYTHAGORAS THEOREM Instrutions: Together in groups of 2 or 3, fill out the following worksheet. You my lift nswers from the reding, or nswer on your own. Turn in one pket for

More information

Graph Theory. Simple Graph G = (V, E). V={a,b,c,d,e,f,g,h,k} E={(a,b),(a,g),( a,h),(a,k),(b,c),(b,k),...,(h,k)}

Graph Theory. Simple Graph G = (V, E). V={a,b,c,d,e,f,g,h,k} E={(a,b),(a,g),( a,h),(a,k),(b,c),(b,k),...,(h,k)} Grph Theory Simple Grph G = (V, E). V ={verties}, E={eges}. h k g f e V={,,,,e,f,g,h,k} E={(,),(,g),(,h),(,k),(,),(,k),...,(h,k)} E =16. 1 Grph or Multi-Grph We llow loops n multiple eges. G = (V, E.ψ)

More information

6.5 Improper integrals

6.5 Improper integrals Eerpt from "Clulus" 3 AoPS In. www.rtofprolemsolving.om 6.5. IMPROPER INTEGRALS 6.5 Improper integrls As we ve seen, we use the definite integrl R f to ompute the re of the region under the grph of y =

More information

Phylogenies via Quartets

Phylogenies via Quartets Phylogenies vi Qurtets Dvi Brynt rynt@mth.mgill. LIRMM, Frne CRM, U. e M. U. Cnterury MGill University Bite-size trees There is only one unroote tree for one, two or three tx... But there re four unroote

More information

Review of Gaussian Quadrature method

Review of Gaussian Quadrature method Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge

More information

arxiv: v1 [cs.dm] 24 Jul 2017

arxiv: v1 [cs.dm] 24 Jul 2017 Some lsses of grphs tht re not PCGs 1 rxiv:1707.07436v1 [s.dm] 24 Jul 2017 Pierluigi Biohi Angelo Monti Tizin Clmoneri Rossell Petreshi Computer Siene Deprtment, Spienz University of Rome, Itly pierluigi.iohi@gmil.om,

More information

The DOACROSS statement

The DOACROSS statement The DOACROSS sttement Is prllel loop similr to DOALL, ut it llows prouer-onsumer type of synhroniztion. Synhroniztion is llowe from lower to higher itertions sine it is ssume tht lower itertions re selete

More information

I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3

I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3 2 The Prllel Circuit Electric Circuits: Figure 2- elow show ttery nd multiple resistors rrnged in prllel. Ech resistor receives portion of the current from the ttery sed on its resistnce. The split is

More information

Probability The Language of Chance P(A) Mathletics Instant Workbooks. Copyright

Probability The Language of Chance P(A) Mathletics Instant Workbooks. Copyright Proility The Lnguge of Chne Stuent Book - Series L-1 P(A) Mthletis Instnt Workooks Copyright Proility The Lnguge of Chne Stuent Book - Series L Contents Topis Topi 1 - Lnguge of proility Topi 2 - Smple

More information

Area and Perimeter. Area and Perimeter. Solutions. Curriculum Ready.

Area and Perimeter. Area and Perimeter. Solutions. Curriculum Ready. Are n Perimeter Are n Perimeter Solutions Curriulum Rey www.mthletis.om How oes it work? Solutions Are n Perimeter Pge questions Are using unit squres Are = whole squres Are = 6 whole squres = units =

More information

Graph width-parameters and algorithms

Graph width-parameters and algorithms Grph width-prmeters nd lgorithms Jisu Jeong (KAIST) joint work with Sigve Hortemo Sæther nd Jn Arne Telle (University of Bergen) 2015 KMS Annul Meeting 2015.10.24. YONSEI UNIVERSITY Grph width-prmeters

More information

COMPUTING THE QUARTET DISTANCE BETWEEN EVOLUTIONARY TREES OF BOUNDED DEGREE

COMPUTING THE QUARTET DISTANCE BETWEEN EVOLUTIONARY TREES OF BOUNDED DEGREE COMPUTING THE QUARTET DISTANCE BETWEEN EVOLUTIONARY TREES OF BOUNDED DEGREE M. STISSING, C. N. S. PEDERSEN, T. MAILUND AND G. S. BRODAL Bioinformtis Reserh Center, n Dept. of Computer Siene, University

More information

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES PAIR OF LINEAR EQUATIONS IN TWO VARIABLES. Two liner equtions in the sme two vriles re lled pir of liner equtions in two vriles. The most generl form of pir of liner equtions is x + y + 0 x + y + 0 where,,,,,,

More information

Discrete Structures Lecture 11

Discrete Structures Lecture 11 Introdution Good morning. In this setion we study funtions. A funtion is mpping from one set to nother set or, perhps, from one set to itself. We study the properties of funtions. A mpping my not e funtion.

More information

Technology Mapping Method for Low Power Consumption and High Performance in General-Synchronous Framework

Technology Mapping Method for Low Power Consumption and High Performance in General-Synchronous Framework R-17 SASIMI 015 Proeeings Tehnology Mpping Metho for Low Power Consumption n High Performne in Generl-Synhronous Frmework Junki Kwguhi Yukihie Kohir Shool of Computer Siene, the University of Aizu Aizu-Wkmtsu

More information

Outline. Theory-based Bayesian framework for property induction Causal structure induction

Outline. Theory-based Bayesian framework for property induction Causal structure induction Outline Theory-sed Byesin frmework for property indution Cusl struture indution Constrint-sed (ottom-up) lerning Theory-sed Byesin lerning The origins of usl knowledge Question: how do people relily ome

More information

Exercise sheet 6: Solutions

Exercise sheet 6: Solutions Eerise sheet 6: Solutions Cvet emptor: These re merel etended hints, rther thn omplete solutions. 1. If grph G hs hromti numer k > 1, prove tht its verte set n e prtitioned into two nonempt sets V 1 nd

More information

Analysis of Temporal Interactions with Link Streams and Stream Graphs

Analysis of Temporal Interactions with Link Streams and Stream Graphs Anlysis of Temporl Intertions with n Strem Grphs, Tiphine Vir, Clémene Mgnien http:// ltpy@ LIP6 CNRS n Soronne Université Pris, Frne 1/23 intertions over time 0 2 4 6 8,,, n for 10 time units time 2/23

More information

Activities. 4.1 Pythagoras' Theorem 4.2 Spirals 4.3 Clinometers 4.4 Radar 4.5 Posting Parcels 4.6 Interlocking Pipes 4.7 Sine Rule Notes and Solutions

Activities. 4.1 Pythagoras' Theorem 4.2 Spirals 4.3 Clinometers 4.4 Radar 4.5 Posting Parcels 4.6 Interlocking Pipes 4.7 Sine Rule Notes and Solutions MEP: Demonstrtion Projet UNIT 4: Trigonometry UNIT 4 Trigonometry tivities tivities 4. Pythgors' Theorem 4.2 Spirls 4.3 linometers 4.4 Rdr 4.5 Posting Prels 4.6 Interloking Pipes 4.7 Sine Rule Notes nd

More information

F / x everywhere in some domain containing R. Then, + ). (10.4.1)

F / x everywhere in some domain containing R. Then, + ). (10.4.1) 0.4 Green's theorem in the plne Double integrls over plne region my be trnsforme into line integrls over the bounry of the region n onversely. This is of prtil interest beuse it my simplify the evlution

More information

Global alignment. Genome Rearrangements Finding preserved genes. Lecture 18

Global alignment. Genome Rearrangements Finding preserved genes. Lecture 18 Computt onl Biology Leture 18 Genome Rerrngements Finding preserved genes We hve seen before how to rerrnge genome to obtin nother one bsed on: Reversls Knowledge of preserved bloks (or genes) Now we re

More information

Tutorial Worksheet. 1. Find all solutions to the linear system by following the given steps. x + 2y + 3z = 2 2x + 3y + z = 4.

Tutorial Worksheet. 1. Find all solutions to the linear system by following the given steps. x + 2y + 3z = 2 2x + 3y + z = 4. Mth 5 Tutoril Week 1 - Jnury 1 1 Nme Setion Tutoril Worksheet 1. Find ll solutions to the liner system by following the given steps x + y + z = x + y + z = 4. y + z = Step 1. Write down the rgumented mtrix

More information

Lecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations.

Lecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations. Lecture 3 3 Solving liner equtions In this lecture we will discuss lgorithms for solving systems of liner equtions Multiplictive identity Let us restrict ourselves to considering squre mtrices since one

More information

(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P.

(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P. Chpter 7: The Riemnn Integrl When the derivtive is introdued, it is not hrd to see tht the it of the differene quotient should be equl to the slope of the tngent line, or when the horizontl xis is time

More information

SECTION A STUDENT MATERIAL. Part 1. What and Why.?

SECTION A STUDENT MATERIAL. Part 1. What and Why.? SECTION A STUDENT MATERIAL Prt Wht nd Wh.? Student Mteril Prt Prolem n > 0 n > 0 Is the onverse true? Prolem If n is even then n is even. If n is even then n is even. Wht nd Wh? Eploring Pure Mths Are

More information

Subsequence Automata with Default Transitions

Subsequence Automata with Default Transitions Susequene Automt with Defult Trnsitions Philip Bille, Inge Li Gørtz, n Freerik Rye Skjoljensen Tehnil University of Denmrk {phi,inge,fskj}@tu.k Astrt. Let S e string of length n with hrters from n lphet

More information

Applied. Grade 9 Assessment of Mathematics. Multiple-Choice Items. Winter 2005

Applied. Grade 9 Assessment of Mathematics. Multiple-Choice Items. Winter 2005 Applie Gre 9 Assessment of Mthemtis Multiple-Choie Items Winter 2005 Plese note: The formt of these ooklets is slightly ifferent from tht use for the ssessment. The items themselves remin the sme. . Multiple-Choie

More information

QUADRATIC EQUATION. Contents

QUADRATIC EQUATION. Contents QUADRATIC EQUATION Contents Topi Pge No. Theory 0-04 Exerise - 05-09 Exerise - 09-3 Exerise - 3 4-5 Exerise - 4 6 Answer Key 7-8 Syllus Qudrti equtions with rel oeffiients, reltions etween roots nd oeffiients,

More information

Connectivity in Graphs. CS311H: Discrete Mathematics. Graph Theory II. Example. Paths. Connectedness. Example

Connectivity in Graphs. CS311H: Discrete Mathematics. Graph Theory II. Example. Paths. Connectedness. Example Connetiit in Grphs CSH: Disrete Mthemtis Grph Theor II Instrtor: Işıl Dillig Tpil qestion: Is it possile to get from some noe to nother noe? Emple: Trin netork if there is pth from to, possile to tke trin

More information

Algorithm Design and Analysis

Algorithm Design and Analysis Algorithm Design nd Anlysis LECTURE 5 Supplement Greedy Algorithms Cont d Minimizing lteness Ching (NOT overed in leture) Adm Smith 9/8/10 A. Smith; sed on slides y E. Demine, C. Leiserson, S. Rskhodnikov,

More information

4.5 THE FUNDAMENTAL THEOREM OF CALCULUS

4.5 THE FUNDAMENTAL THEOREM OF CALCULUS 4.5 The Funmentl Theorem of Clculus Contemporry Clculus 4.5 THE FUNDAMENTAL THEOREM OF CALCULUS This section contins the most importnt n most use theorem of clculus, THE Funmentl Theorem of Clculus. Discovere

More information

Section 2.1 Special Right Triangles

Section 2.1 Special Right Triangles Se..1 Speil Rigt Tringles 49 Te --90 Tringle Setion.1 Speil Rigt Tringles Te --90 tringle (or just 0-60-90) is so nme euse of its ngle mesures. Te lengts of te sies, toug, ve very speifi pttern to tem

More information

HS Pre-Algebra Notes Unit 9: Roots, Real Numbers and The Pythagorean Theorem

HS Pre-Algebra Notes Unit 9: Roots, Real Numbers and The Pythagorean Theorem HS Pre-Alger Notes Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Roots nd Cue Roots Syllus Ojetive 5.4: The student will find or pproximte squre roots of numers to 4. CCSS 8.EE.-: Evlute squre roots

More information

SOME INTEGRAL INEQUALITIES FOR HARMONICALLY CONVEX STOCHASTIC PROCESSES ON THE CO-ORDINATES

SOME INTEGRAL INEQUALITIES FOR HARMONICALLY CONVEX STOCHASTIC PROCESSES ON THE CO-ORDINATES Avne Mth Moels & Applitions Vol3 No 8 pp63-75 SOME INTEGRAL INEQUALITIES FOR HARMONICALLY CONVE STOCHASTIC PROCESSES ON THE CO-ORDINATES Nurgül Okur * Imt Işn Yusuf Ust 3 3 Giresun University Deprtment

More information

Part 4. Integration (with Proofs)

Part 4. Integration (with Proofs) Prt 4. Integrtion (with Proofs) 4.1 Definition Definition A prtition P of [, b] is finite set of points {x 0, x 1,..., x n } with = x 0 < x 1

More information

arxiv: v1 [cs.cg] 28 Apr 2009

arxiv: v1 [cs.cg] 28 Apr 2009 Orienttion-Constrine Retngulr Lyouts Dvi Eppstein 1 n Elen Mumfor 2 1 Deprtment of Computer Siene, University of Cliforni, Irvine, USA 2 Deprtment of Mthemtis n Computer Siene, TU Einhoven, The Netherlns

More information

Surface maps into free groups

Surface maps into free groups Surfce mps into free groups lden Wlker Novemer 10, 2014 Free groups wedge X of two circles: Set F = π 1 (X ) =,. We write cpitl letters for inverse, so = 1. e.g. () 1 = Commuttors Let x nd y e loops. The

More information

Trigonometry and Constructive Geometry

Trigonometry and Constructive Geometry Trigonometry nd Construtive Geometry Trining prolems for M2 2018 term 1 Ted Szylowie tedszy@gmil.om 1 Leling geometril figures 1. Prtie writing Greek letters. αβγδɛθλµπψ 2. Lel the sides, ngles nd verties

More information

April 8, 2017 Math 9. Geometry. Solving vector problems. Problem. Prove that if vectors and satisfy, then.

April 8, 2017 Math 9. Geometry. Solving vector problems. Problem. Prove that if vectors and satisfy, then. pril 8, 2017 Mth 9 Geometry Solving vetor prolems Prolem Prove tht if vetors nd stisfy, then Solution 1 onsider the vetor ddition prllelogrm shown in the Figure Sine its digonls hve equl length,, the prllelogrm

More information

If the numbering is a,b,c,d 1,2,3,4, then the matrix representation is as follows:

If the numbering is a,b,c,d 1,2,3,4, then the matrix representation is as follows: Reltions. Solutions 1. ) true; ) true; ) flse; ) true; e) flse; f) true; g) flse; h) true; 2. 2 A B 3. Consier ll reltions tht o not inlue the given pir s n element. Oviously, the rest of the reltions

More information

Solutions to Problem Set #1

Solutions to Problem Set #1 CSE 233 Spring, 2016 Solutions to Prolem Set #1 1. The movie tse onsists of the following two reltions movie: title, iretor, tor sheule: theter, title The first reltion provies titles, iretors, n tors

More information

Technische Universität München Winter term 2009/10 I7 Prof. J. Esparza / J. Křetínský / M. Luttenberger 11. Februar Solution

Technische Universität München Winter term 2009/10 I7 Prof. J. Esparza / J. Křetínský / M. Luttenberger 11. Februar Solution Tehnishe Universität Münhen Winter term 29/ I7 Prof. J. Esprz / J. Křetínský / M. Luttenerger. Ferur 2 Solution Automt nd Forml Lnguges Homework 2 Due 5..29. Exerise 2. Let A e the following finite utomton:

More information