epsilon vol02 2015april

197
Người viết cho rằng hầu hết khái niệm tưởng như trừu tượng đều có cội nguồn ở những thao tác toán học cụ thể và thông dụng. Người viết cũng cho rằng chỉ khi được nâng lên tầm khái niệm, các thao tác toán học mới trở thành những công cụ tư duy thật sự mạnh mẽ Phương trình đại số một ẩn số Ngô Bảo Châu Phương trình đại số 1 ẩn số Ngô Bảo Châu Bài toán Frobenius về những đồng xu Trần Nam Dũng Bài toán đội nón Đặng Nguyễn Đức Tiến Khám phá toán học thông qua các định lý hình học. Đào Thanh Oai Laurent Schwartz Hà Huy Khoái Giới thiệu đề Vietnam TST 2015, bình luận sơ bộ và tóm tắt cách giải. Trần Nam Dũng VÀ CÁC CHUYÊN MỤC KHÁC 13 apr 2015 Đại số tuyến tính VẠN TUẾ Nghịch đảo Möbius Ngô Quang Hưng Tạp chí online của cộng đồng những người yêu Toán T/p chí online cıa cºng đng nhœng ngưi yêu Toán

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  • Ngi vit cho rng hu ht khi nim tng nh tru tng u c ci ngun nhng thao tc ton hc c th v thng dng. Ngi vit cng cho rng ch khi c nng ln tm khi nim, cc thao tc ton hc mi tr thnh nhng cng c t duy tht s mnh m

    Phng trnh i s mt n sNg Bo Chu

    Phng trnh i s 1 n sNg Bo Chu

    Bi ton Frobenius v nhng ng xu Trn Nam Dng

    Bi ton i nn ng Nguyn c Tin

    Khm ph ton hc thng qua cc nh l hnh hc.o Thanh Oai

    Laurent SchwartzH Huy Khoi

    Gii thiu Vietnam TST 2015, bnh lun s b v tm tt cch gii. Trn Nam Dng

    V CC CHUYN MC KHC

    13 apr 2015

    i s tuyn tnh VN TUNghch o MbiusNg Quang Hng

    T p c h online ca cng ng nhng n g i y u T o n

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    Tp ch online ca cng ng nhng ngi yu Ton

    EPSILONCh bin: abc TRN NAM DNG

    Bin tp vin: NG NGUYN C TIN

    Bin tp vin: V QUC B CN

    Bin tp vin: TRN QUANG HNG

    Bin tp vin: L PHC L

    S 2, ngy 13 thng 04 nm 2015

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    Ton LI NG CHO EPSILON S 2

    Ban bin tp Epsilon

    Epsilon s 1 ra mt c n nhn mt cch nng nhit cabn c. l mt nim ng vin ln lao dnh cho Ban bintp, gip chng ti c thm nng lng, nhit huyt bctip trn con ng m khng hn ch c hoa hng.

    Epsilon nhn c mt vn tc ban u v mt gia tc. Bnhng dng. Vi s ng gp ca cng ng, hy vng Epsilons gi c nhp v u t ra mt vo ngy 13 cc thng chn phc v cng ng, em n mt mn n tinh thn v trongmt cuc sng ang vn rt nhiu nhng mn n.

    Epsilon mong mun s l mt nhp cu kt ni nhng itng vn cn xa cch nhau: l thuyt v thc tin, ton hc vcc mn khoa hc khc, gio vin v hc sinh, cc nh ton hcchuyn nghip v nhng ngi lm ton nghip d, ton hnlm v ton gii tr, ton cao cp v ton s cp. V th, Epsilons c s ha quyn ca nhng bi vit vi ni dung v phongcch rt khc nhau. Ban bin tp s tn trng cch hnh vnca cc tc gi m khng p t kin ca mnh, ch chnh sa cho bi tt hn, chnh xc hn.

    Epsilon s 2 m cc bn cm trn tay s c 12 bi vit cacc tc gi n t nhiu quc gia, nhiu thnh phn v cp chuyn nghip. K t s ny, Epsilon s dnh nhng trang vitca mnh gii thiu v tiu s cc nh ton hc ni tingth gii, ln ny s l bi vit ca GS H Huy Khoi v LoranSchwarz, nhn k nim 100 nm ngy sinh ca ng v bi vitv thin ti on mnh Evariste Galois song hnh vi bi vit vPhng trnh i s ca GS Ng Bo Chu. Cu ni gia tonhc v khoa hc my tnh trong s ny s c th hin bngbi vit ca GS Ng Quang Hng, H Buffalo, M v nghch oMobius. Hnh hc s cp, mn hc vn c coi l c xa v gici nht s nh li ti mi di gc nhn ca mt ngi yu

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    ton nghip d, k s o Thanh Oai trong bi Phng phpm rng v sng to cc nh l hnh hc c in.

    Cc bn hc sinh yu ton chc chn s tm c nhiu iub ch qua cc bi vit v cc bi ton thi chn HSG quc gia(VMO 2015) v chn i tuyn d IMO 2015 (Vietnam TST 2015)ca cc tc gi Trn Nam Dng, Nguyn Tt Thu, Trn QuangHng. Bi bnh lun ca Nguyn Vn Li (Budapest) v NguynHng Sn (Warsaw) v TST cng s gip cc bn hiu r hnv cc bi ton trong thi. c bit trong s ny s c bivit Inequalities, A Journey into Fibonacci and Lucas numbersca hai tc gi nc ngoi l Vandanjav Adiyasuren v BoldSanchir n t H QG Mng C. Nhng ai yu ton hc giitr s c tip tc cuc phiu lu k th vo vng quc canhng chic nn mu sc vi ngi hng dn vin ngNguyn c Tin (Trento, Italy). Mt nh o thut c o khcl Nguyn Quc Khnh s ra mt bn c mt chuyn mc lth v b ch: Gii thiu sch.

    Hy vng vi nhng bi vit nh th, mi c gi u c th tmc t nht l 10% iu mnh yu thch trong s ny. Nhth, Ban bin tp cm thy tht mn nguyn v cho rngnhim v hon thnh.

    V li nng lng, nhit huyt tip tc bc i.

    i nhiu ngi, bn s i rt xa ...

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    MC LC

    1 Li ng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

    2 Nhn 100 nm ngy sinh Laurent Schwartz

    H Huy Khoi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

    3 Phng trnh i s mt n sNg Bo Chu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

    4 variste GaloisLu Trng Lun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

    5 Nghch o Mobius

    Ng Quang Hng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

    6 Cc bi ton i nnng Nguyn c Tin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

    7 Bi ton Frobenius v nhng ng xu

    Trn Nam Dng - Nguyn Tt Thu . . . . . . . . . . . . . . . . . . . . . 73

    8 Vit Nam TST 2015

    Trn Nam Dng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

    9 Li gii v bnh lun hai bi hnh thi chn ituyn Vit Nam nm 2015

    Trn Quang Hng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

    10 Cc vn c in v hin i

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    Trn Nam Dng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

    11 Bt ng thc Shapiro . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

    12 Phng php m rng v sng to cc nh l hnhhc c in

    o Thanh Oai . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

    13 A journey into Fibonacci and Lucas numbers

    V. Adiyasuren - B. Sanchir . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

    14 Ton hc trong mt ai

    Nguyn Quc Khnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

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    Ton NHN 100 NM NGY SINH

    LAURENT SCHWARTZ

    H Huy Khoi

    H Ni

    Ti l nh ton hc. Ton hc y p cuc i ti.

    Laurent Schwartz 1 vit nh vy trong li m u cun hik ca ng. ng cng ni rng, ngoi ton hc, ng ginh rtnhiu thi gian ca i mnh cho cuc u tranh v quyn conngi, v quyn ca cc dn tc, ban u th nh mt ngiTroskit, sau th ng ngoi tt c cc ng phi! Vit Namchim mt v tr quan trng trong cc hot ng ca ng.Trong nhiu nm, ng lun ng hng u trong i ng nhngtr thc ln ca Phng Ty u tranh ng h cuc khng chinca nhn dn Vit Nam. Trong cun hi k dy 500 trang cang, c th tm thy khong 100 trang c nhc n Vit Nam.

    Laurent Schwartz sinh ngy 5 thng 3 nm 1915 ti Paris. Chang l mt bc s phu thut, m ng l ngi yu thin nhin,nh ng ni, sut ngy ch quanh qun vi mnh vn v baa con. Tui th ca ng tri qua m m lng quAutouillet, m ng gi mt cch tru mn trong hi k ca mnhl Khu vn Eden. Mi sau ny, ng vn thng xuyn tr vkhu vn , v nh ng k li, nhng nh l hay nht cang c tm thy ti khu vn Eden.

    Ngay t khi cn nh, Laurent Schwartz bc l thin hngnghin cu. Nu nh hu ht tr em hi lng vi nhng li giithch s lc ca b m khi chng hi ti sao?, th cu bLaurent khng nh vy. Cu lun i hi nhng li gii thchcn k, m t khi c tho mn. M cu rt lng tng trcnhng cu hi: Ti sao khi cm ci gy vo nc th thy ncong, ti sao trong cng mt nhit m khng kh lc th

    1Nh ton hc ngi Php (05/03/1915 - 04/07/2002)

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    lnh hn, lc th nng hn nc, ti sao khi lt p ci tha cph th khng bao gi ht c ph, m cn mt t dnh li tha,v cn rt nhiu nhng cu hi khc.

    cc lp tiu hc, Laurent Schwartz khng phi l hc sinhgii mn ton. ng rt nh li thy Thoridenet, ngi dy ngmn vn nm lp 5 tng ni vi m ng: Ti cha c hc sinhno gii nh vy v mn ting Latinh, nhng v ting Php, ngnng v ton th cu ta km hn mt cht. Tuy vy, cho d ngita ni vi b th no i na, cu ta s tr thnh nh ton hc!.Laurent Schwartz ni rng, nu khng c li khuyn ca ngthy dy vn th c l ng tr thnh nh ngn ng hc,ch khng phi nh ton hc! May mn na cho Laurent l cugp mt thy gio dy ton y nhit tm, thy Julien. ng gii thch cho hc sinh mt cch rt vui v v n gin nhngiu k diu ca mn hnh hc, m ra cho h mt th gii tonhc m trc h cha c bit n. Laurent Schwartz k,sau khi suy ngh vi ba tun, ng quyt nh tr thnh nhton hc. Theo ng, thin hng c sn trong con nging, nhng tr thnh hin thc nh thy gio. V th ngcho rng, vai tr ca ngi thy i vi tng lai hc sinh l c ngha quyt nh.

    Laurent Schwartz thi vo trng Ecole Normale Suprieure(Paris) nm 1934. Ecole Normale, ng c hc vi nhng gios ni ting nht thi by gi: Frchet, Montel, Borel, Denjoy,Julia, Elie Cartan, Lebesgue v Hadamard. Trong kho , ngcng vi Choquet, Marot l ba ngi xut sc nht.

    Nm 1937, ng tt nghip i hc Ecole Normale, lm nghincu sinh ti trng i hc Strasbourg v bo v lun n Tins nm 1943. Gio s hng dn lun n ca ng l Valiron, mttrong nhng nh ton hc ni ting nht thi v l thuythm. Vi nm sau, Valiron cng l ngi hng dn ca gios L Vn Thim.

    Trong cc nm 1944 1945 ng ging dy ti khoa Khoa hc Grenoble, sau chuyn v Nancy, nhn mt chc gio s khoa Khoa hc. Chnh trong thi gian ny, ng sng to ra cngtrnh ni ting v l thuyt cc hm suy rng.

    Nm 1953 Laurent Schwartz tr v Paris, lm gio s cho n1959. ng ging dy ti trng Ecole Polytechnique t 1959 n

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    1980, ri lm vic trng i hc Paris 7 ba nm, cho n ngyngh hu nm 1983.

    Cng hin ln nht cho ton hc ca Laurent Schwartz l cccng trnh ca ng v l thuyt phn b, c vit vo nhngnm 40. Nhng t tng ca ng theo hng ny c trnhby ln u tin nm 1948 trong bi M rng khi nim hm,o hm, bin i Fourier v cc ng dng ton hc, vt l.

    L thuyt phn b l s m rng ng k php tnh tch phn vvi phn. Do nhu cu ca Vt l hc, Heaviside v Dirac mrng php tnh vi cc ng dng c bit. Tuy nhin, phngphp ca h, cng nh nhng phng php tng t v ccphp tnh hnh thc khng c xy dng trn mt nn tngton hc cht ch. nhng nghin cu ca h c th trthnh mt l thuyt mi thc s ca vt l hc, cn trang bcho n mt c s ton hc vng chc. Chnh Dirac c ln ni:Khi bn nh xy dng mt l thuyt mi no trong vt l, ci duynht m bn c th tin tng l ton hc. Laurent Schwartz pht trin mt l thuyt lm c s cho cc phng php tnh tonnu trn trong vt l, lm cho nhng phng php tm cng dng ht sc rng ri trong nhng lnh vc khc nhau.

    Francois Treves ni v cng trnh ca Laurent Schwartz nhsau: T tng ca Laurent Schwartz cho mt cch l giithng nht tt c cc hm suy rng thm nhp trong gii tchnh l nhng phim hm tuyn tnh lin tc trn khng gian cchm kh vi v hn trit tiu ngoi mt tp compc. ng chomt cch m t c h thng v cht ch, hon ton da trn giitch hm tru tng v l thuyt i ngu. Cng cn nhc lirng, mt cch l gii nh vy c trc y trong cng trnhca Andr Weil v tch phn cc nhm compc a phng ... Dos i hi ca tnh kh vi trong l thuyt phn b, khng gian cchm th v i ngu ca chng i khi rt phc tp. iu nydn n nhng nghin cu si ni v cc khng gian vector topokhng thuc cc phm tr quen thuc nh khng gian Hilbert vkhng gian Banach. Nhng nghin cu ny, n lt mnh, chiuri nhng nh sng mi ln nhiu lnh vc ca Gii tch thun tu,nh Phng trnh o hm ring, hoc Hm s bin s phc.

    Nhng t tng ca Laurent Schwartz c th p dng chonhiu khng gian hm th khc nhau, nh chnh ng v nhiungi khc ch r ... Herald Bohr, ngi gii thiu cng trnh

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    ca Laurent Schwartz trong bui trao Gii thng Fields ngy30 thng 8 nm 1950 ti Harvard m t cc cng trnh caLaurent Schwartz vit nm 1948 nh sau: Chng chc chn str thnh nhng cng trnh kinh in ca ton hc thi i chngta ... Ti ngh rng, nhng ngi trch dn cng trnh ca ng,cng ging nh ti, s phi km nn mt nim phn khch dchu, nhn thy s hi ho tuyt vi ca mt cu trc tnh tonm l thuyt ny dn chng ta n, v hiu tm quan trng vu vit ca chng i vi nhiu phn ca gii tch cao cp, nhL thuyt ph, L thuyt th v, v ton b l thuyt phng trnho hm ring.

    Ngoi gii thng Fields, Laurent Schwartz cn nhn c giithng ca Vin hn lm khoa hc Paris cc nm 1955, 1964, 1972.Nm 1972 ng c bu lm Vin s Vin hn lm Php. ngc phong tin s danh d ca nhiu trng i hc, trong c Humboldt (1960), Brussels (1962), Lund (1981), Tel-Aviv (1981),Montreal (1985) v Athens (1993).

    Khng ch l nh ton hc ni ting, Laurent Schwartz cn cbit n nh l mt trong nhng tr thc ln sut i u tranhv t do ca cc dn tc. Laurent Schwartz ni rng, nhngnm Ecole Normale xc nh hon ton khuynh hngchnh tr ca ng: Chng chin tranh v bo v nhng gi trca con ngi. Cun sch ng Dng cp cu (IndochineSOS) ca Andre Viollis cho ng thy r ti c ca ch nghathc dn Php ng Dng. Quan im chnh tr ca ng thhin r nht trong phong tro chng chin tranh xm lc ca quc M Vit Nam. ng xng khu hiu Mt trn dntc gii phng s chin thng thay cho khu hiu m ng chol m h ca phong tro chng chin tranh Vit Nam Phpthi Ho bnh Vit Nam. Hot ng ca U ban quc giaVit Nam do ng sng lp gy c ting vang ln. ng htsc t ho khi vo khong l Noel nm 1966, nhn c bcin cm n v chc mng ca Ch tch H Ch Minh. ng nVit Nam nhiu ln trong thi k cn chin tranh, vi t cch lthnh vin trong To n quc t xt x ti c chin tranh caM Vit Nam (mt t chc quc t do nh ton hc, nh trithc ni ting ngi Anh, gii thng Nobel v vn hc nm 1950,hun tc Bertrand Russell sng lp). Nhng chuyn i v cclng qu Vit Nam lm cho ng thy yu mn c bit tnc v con ngi Vit Nam. Khng g c th ni y hn

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    tnh cm ca ng vi Vit Nam bng chnh nhng li ng vittrong hi k ca mnh: Vit Nam ghi du n trong cuc iti. Ti tng bit n ng Dng thuc a, qua cun sch caAndr Viollis vit nm 1931, m ti c nm 1935. Lc ti vatrn 20 tui. Cuc u tranh ca ti cho t do ca t nc nyl cuc u tranh di nht ca cuc i ti. Ti yu, v mimi yu Vit Nam, nhng phong cnh, nhng con ngi tuyt vi,nhng chic xe p. Trong ti, c mt cht no l ngi VitNam. Gp ngi Vit Nam, nghe ting h ni chuyn vi nhautrong xe but (m tt nhin l ti khng hiu), ti cm thy mtnim hnh phc khng ct ngha c. Si giy tnh cm nilin ti vi t nc ny.

    Nm 1998, khi Vin Ton hc t chc Hi ngh quc t nhn 80nm ngy sinh ca Gio s L Vn Thim, Laurent Schwartz rt xc ng thng bo cho Ban t chc rng ng rt munsang Vit Nam thm mt ln na, nhng tic l sc kho khngcho php. Khi ng qua i nm 2002, t Thng tin ton hc caHi ton hc Vit Nam c ng mt bi vit tng nh ng.Dng nh ng bit trc iu , nn vit trong hi kca mnh: Les Vietnamiens ne moublient pas (Ngi Vit Namkhng qun ti).

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    PHNG TRNH I SMT N S

    Ng Bo Chu

    i hc Chicago, M

    Tm tt

    T th k 20 trc Cng nguyn, ngi dn thnh Babylon bit gii phng trnh bc hai. Nhng phi n th k16 sau Cng nguyn, cc nh ton hc ca thi Phc hng:Tartaglia, Cardano, Ferrari, mi tm ra li gii cho phngtrnh bc ba v bc bn. u th k 19, Abel v Galois,hai thin ti ton hc bc mnh, chng minh nghim caphng trnh i s tng qut bc t nm tr i, khng thbiu din c nh mt biu thc i s vi cn thc nhtrong trng hp a thc bc khng qu bn. Cng trnhca Galois, vit ra nh li trng tri trc gi u sng,sau c xem nh mc khai sinh ca i s hin i.

    L thuyt Galois hin i c pht biu trn c s cckhi nim m rng trng v nhm Galois. Nhng khinim ny khng d nm bt. Mc ch ca bi vit ny lgip nhng ngi mi hc nm bt nhng khi nim ,thng qua vic tm hiu m thc m chng xut hin trongqu trnh tm nghim ca nhng phng trnh i s cth.

    Ngi vit cho rng hu ht khi nim tng nh trutng u c ci ngun nhng thao tc ton hc c thv thng dng. Ngi vit cng cho rng ch khi c nngln tm khi nim, cc thao tc ton hc mi tr thnhnhng cng c t duy tht s mnh m. Cu chuyn spk v thuyt Galois c th xem nh mt minh chng.

    hiu bi vit ny, ngi c cn mt s kin thc cbn v i s tuyn tnh, trong c bit quan trng lkhi nim chiu ca khng gian vector.

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    1. Lch s ca bi ton

    Vo th k th by trc cng nguyn, li gii cho phng trnhbc hai tng qut

    x2 + ax+ b = 0, (3.1)

    c nh ton hc Brahmagupta, ngi n , trnh bymt cch tng minh dng

    x =a?d

    2, (3.2)

    vi d = a2 4b l bit thc ca phng trnh bc hai.Trc , t khong th k 20 trc cng nguyn, ngi Babylon tm li gii hnh hc cho bi ton tng ng tm hai cnhca hnh ch nht bit trc chu vi v din tch ca n. Du vtca nhng phng php hnh hc khc nhau gii phngtrnh bc hai c pht hin trong hu ht cc nn vn minhc i t Babylon, Ai cp, Hy lp, n , Trung Hoa ...

    Phng trnh bc ba tng qut cng c ngi Babylon nghincu. Ngi Hy lp c i th xy dng nghim phng trnhbc ba bng thc k v compa nhng khng thnh cng.

    Nh ton hc Trung Hoa Wang Xiaotong a ra li gii cho 27phng trnh bc ba khc nhau, nhng khng a ra phngphp gii phng trnh bc ba tng qut.

    ng k nht l pht hin ca nh th ngi Ba t OmarKhayyam sng vo th mi mt. ng chng minh rng nghimc th xy dng nghim phng trnh bc ba bng cch ly giaohai ng conic. Ngoi ra, ng pht biu rng khng th xydng nghim phng trnh bc ba ch bng thc k v compa.Omar Khayyam khng a ra mt cng thc cho nghim caphng trnh bc ba ging nh cng thc (3.2) cho phngtrnh bc hai.

    Phi ch n thi k phc hng, nh ton hc Tartaglia, sng vo th k th mi su, mi a ra cng thc tng qutu tin cho nghim ca phng tnh bc ba

    ax3 + bx2 + cx+ d = 0, (3.3)

    dng

    x = 13a

    (b+ C+

    0

    C

    ), (3.4)

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    trong

    C =3

    d1 +

    a21 4302

    , (3.5)

    vi 0, 1 l cc a thc tng minh vi bin s a, b, c, d.

    Li gii cho phng trnh bc ba qu l rc ri, nhng li giicho phng trnh bc bn ca Ferrari cn rc ri hn nhiu.

    Nh ton hc Joseph Lagrange, ngi , l ngi a ra mtphng php chung gii c phng trnh bc ba v bc bn.Phng php ca Lagrange d trn khi nim gii thc mchng ta s xem xt k lng.

    Ruffini nghin cu phng php ca Lagrange v nhn thyrng n khng th m rng ra cho phng trnh c bc nm vbc cao hn na.

    Abel l ngi u tin a ra chng minh cht ch v khngnh phng trnh bc nm tng qut khng th gii c bngcn thc. nh l Abel-Ruffini cng c Galois, mt nh tonhc ngi Php, chng minh mt cch c lp. Nhng ng ixa hn Abel v a ra mt khi nim c tnh cht cch mng, l nhm Galois.

    2. V pht biu ca bi ton

    Bi ton ta quan tm chnh l vic biu din nghim ca phngtrnh a thc

    a0xn + a1x

    n1 + = 0, (3.6)di dng mt biu thc vi bin s a0, a1, . . . , m trong tac quyn dng bn php ton thng thng v cn thc.

    hiu r th no l biu din c di dng mt biu thcnh th, ta s cn khi nim trng v m rng trng. V dnh cc biu thc vi bin s a0, a1, . . . ,an m ch dng bnphp ton thng thng v vi h s hu t, l trng sinh rabi a0, a1, . . . ,an.

    Cu hi biu din nghim bng cn thc thc ra vn khngchun. Tht vy phng trnh bc n c th c ti n nghim chonn ht mp m cn lm r ta mun biu din nghim notrong s n nghim . D nhin trong cng thc (3.2), du cho

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    php ta biu din c nghim ca (3.1). Trong khi , cng thcca Tartaglia (3.5) dng nh cho ta su nghim khc nhauca phng trnh bc ba, ci r rng l khng th.

    Thc ra ta khng c cch no chn mt trong n nghim caphng trnh (3.6). Khi nim nhm Galois sinh ta chnh l lng ho s mp m ny. Ngc li, nh ta s phn tch, cutrc ca nhm Galois s quyt nh vic phng trnh (3.6) cth gii c bng cn thc hay khng.

    3. M rng bc hai

    gii phng trnh bc hai (3.1), ta thc hin php i biny = x + a

    2. Sau khi i bin, phng trnh (3.1) quy v dng

    n gin hny2 d = 0. (3.7)

    Ta c th coi y l mt ci mo quy phng trnh bc haitng qut (3.1) v phng trnh bc hai rt gn (3.7).

    Ta cng c th thay i quan im: Khng quan tm n victm ra dng chnh xc (3.2) ca nghim na, m ch quan tmn vic nghim c th biu din di dng biu thc i sca

    ?d. Lp lun c th s phc tp hn, nhng s m ra cho

    ta mt tm nhn mi.

    lm n gin vn , gi s cc h s a, b l s hu t. Tabit rng trong C, phng trnh (3.1) c hai nghim. Ta s khiu 1 P C l mt trong hai nghim ca n. Gi s 1 R Q, khi tp cc s phc c dng

    L = tm+ n1 | m, n P Qu,l mt khng gian vector hai chiu trn Q. T ng thc

    21 = (a1 + b),ta suy ra rng nu u, v P L th uv P L. Ta cng c th chngminh rng nu u P L t0u, th u1 P L. Nh vy L l mt trngcon ca C. Nu xem nh khng gian vector trn Q, n c chiubng 2. V th ta ni rng L l mt m rng bc hai ca Q.Ta thy nghim cn li, k hiu l 2, ca a thc

    P = x2 + ax+ b,

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    cng nm trong L. Tht vy, a thc bc hai P c mt nghim1 P L, nghim cn li 2 cng phi nm trong L v cng khngl s hu t. Ni cch khc, m rng bc hai sinh bi 1, trngvi m rng bc hai sinh bi 2

    tm+ n2 | m, n P Qu.Suy t (3.2) ra th c m rng bc hai sinh bi 1 hay 2 utrng vi m rng bc hai sinh bi cn bc hai ca bit thc

    Q[?d] = tm+ n?d | m, n P Qu. (3.8)

    y cng l mt cch din t vic c 1 v 2 u c thvit c di dng c dng m+

    ?d vi m, n P Q.

    nh l 3.1. Cho P P Q[x] l mt a thc bc hai bt kh quy,L l m rng bc hai ca Q sinh bi mt trong cc nghim ca P.Khi L = Q[

    ?d] vi d = a2 4b.

    D thy rng, nu L l m rng bc hai ca Q, khi mi phnt P L Q l nghim ca mt phng trnh bt kh quy bchai no . V th ta c th pht biu li nh l trn dng cng hn:

    nh l 3.2. Mi m rng bc hai ca Q u c dng L = Q[?d]

    vi d l mt s hu t no .

    M rng ra phng trnh bc cao hn, ta c th nh ngha rnhrt khi nim phng trnh gii c bng cn thc.

    4. Phng trnh gii c bng cn thc

    T nay tr i, ta s thay trng cc s hu t bi mt trng Kbt k. Thay cho trng cc s phc, ta cho trc mt trngng i s cha K. Xin nhc li rng trng K c gi l ngi s nu mi a thc P P K[x] bc n u c ng n nghimtrong K, nu ta m c bi. Ta s ch xt ti cc m rng ca Kcha trong K.

    a thc bc n

    P = xn + a1xn1 + + an P K[x],

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    c gi l bt kh quy nu n khng th phn tch c thnhtch ca hai a thc c bc nh hn. Gi s P l mt a thcbc n bt kh quy. Vi mi nghim P K ca P, ta t

    K[] = tm0 +m1+ +mn1n1 | m0, . . . ,mn1 P Ku. (3.9)S dng ng thc n = (a1n1+ +an), ta chng minh crng nu u, v P K[] th uv P K[]. Ngoi ra, nu u P K[] t0uth u1 P K[]. Ni cch khc, K[] l mt trng con ca K. Sdng gi thit P l a thc bt kh quy, ta chng minh crng K[], xem nh khng gian vector trn trng K, c chiubng n. Ni cch khc, K[] l mt m rng bc n ca K.

    Ta ni nghim c th biu din c bng biu thc i svi cn thc nu tn ti mt chui m rng trng lin tip

    K = K0 K1 K2 Kr, (3.10)sao cho vi mi i P t1, 2, . . . , ru, Ki l mt m rng bc ni caKi1 c dng

    Ki Ki1[x]/(xni i),v sao cho K[] Kr.Khi nim m rng trng cho php ta pht biu rnh rtcu hi liu nghim ca P c th biu din di dng t hpi s v cn thc hay khng. N cn cho php ta t ra nhngcu hi khc, su sc hn, v nghim ca a thc.

    5. Ph thuc i s gia cc nghim

    Nh trn, ta vn k hiu P P K[x] l mt a thc bt kh quybc n, v l mt nghim ca P trong K, K[] l m rng bc nca K bao gm cc t hp i s ca nh (3.9).

    Khc vi trng hp bc 2, khi n 3, nu 1 v 2 l hai nghimkhc nhau ca P, cc trng con K[1] v K[2] ca K, c th lkhc nhau, nh ta thy trong v d sau y.

    Xt trng hp K = Q v a thc P = x3 2. Nu P K l mtnghim ca P th hai nghim cn li s l j v j2. y ta sdng k hiu

    j = cos2pi

    3+ i sin

    2pi

    3, (3.11)

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    l cn bc ba nguyn s ca n v. D thy Q[] Q[j] v nudu bng xy ra th ta s c j P Q[]. Mt khc, m rng Q[j] lm rng bc 2 ca Q v j l nghim ca a thc bc hai x2+x+1,cho nn n khng th nm trong mt m rng bc ba. Tht vy,nu Q[j] Q[] th Q[] s l mt khng gian vector trn trngQ[j], cho nn chiu ca n nh khng gian vector trn Q phil mt s chn. Trong trng hp ny, cc m rng bc ba ngvi 3 nghim ca P = x3 2 l i mt khc nhau:

    Q[] Q[j] Q[j2]. (3.12)Khi K = Q[j], P = x3 2 vn l a thc bc ba bt kh quy trongK[x]. Nhng khi cc m rng bc ba ca K ng vi 3 nghimca P = x3 2 l trng nhau:

    K[] = K[j] = K[j2]. (3.13)

    Ta nhn thy trng hp u, v j khng ph thuc i svi nhau so vi trng c s K = Q. Ni cch khc j khng thbiu din c nh t hp i s ca vi h s hu t. Tuyvy, nu ta m rng trng c s thnh K = Q[j], th v j trnn ph thuc i s.

    V d ny a ta n vi khi nim trng phn r ca mta thc bt kh quy. Trng phn r ca mt a thc l cngc o s ph thuc i s gia cc nghim ca n. Trngphn r s ln nu cc nghim c t quan h i s, trngphn r s nh nu cc nghim c nhiu quan h i s.

    a thc bt kh quy P P K[x] bc n c gi l tch c nun c n nghim i mt khc nhau trong K. Trong trng hpc s khng, mi a thc bt kh quy u tch c. Trongtrng hp c s p 0, c nhng a thc bt kh quy nhngkhng tch c. Trong bi ny, ta s ch xt n nhng athc bt kh quy tch c.

    Cho P P K[x] l mt a thc bt kh quy bc n tch c c hs u bng mt. Ta k hiu 1, 2, . . . ,n P K l cc nghimca P, v gi trng phn r ca K l trng con ca K sinhbi 1, 2, . . . ,n. Trong vnh a thc L[x], a thc P phn rhon ton

    P = (x 1) . . . (x n),thnh tch cc tha s bc mt.

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    lm r ny, ta thc hin mt kho st. K hiu L l trngphn r ca P, khi L l trng con cc tiu cha tt c cctrng con K[1], K[2] . . . , K[n], cn gi l compositum caK[1], K[2] . . . , K[n]. K hiu Li l compositum ca K[1], K[2]. . . , K[n], khi ta c chui m rng trng lin tip

    K = L0 L1 Ln = L.K hiu li l bc ca m rng Li/Li1. Trng phn r L l mrng bc l1l2 . . . ln ca K. Cc s nguyn l1, l2, . . . , ln phn nhmc ph thuc i s gia cc nghim 1, 2, . . . , n. Ta cth kho st chng tun t nh sau

    V P l mt a thc bt kh quy bc n cho nn L1 l mrng bc l1 = n ca L0.

    Xt m rng tip theo L2/L1. a thc P xem nh phn tca L1[x] khng cn bt kh quy na, m c th phn tchc thnh

    P = (x 1)Q.Thnh phn Q c th l a thc bt kh quy, c th khng.

    Nu Q l mt a thc bt kh quy, bc n 1, th L2 s lmt m rng bc l2 = n 1 ca L1. Trong trng hp ny,1 v 2 khng c quan h i s g vi nhau.

    Nu Q P L1[x] khng phi a thc bt kh quy, ta c thphn tch n thnh Q = Q2Q3 vi Q2 l a thc bt khquy c nghim l 2. Khi L2 l m rng ca L1 c bc l2bng vi bc ca a thc Q2. Trong trng hp ny 1 v2 c ph thuc i s.

    Tip tc vi m rng L3/L2 . . .Qua kho st ny ta thy l1 l2 l mt dy s nguyngim tht s v t suy ra l1 ln n!. Dy s ny o mc ph thuc i s gia cc nghim 1, 2, . . . , n ca P.

    6. Nhm Galois ca mt a thc

    Cho P P K[x] l mt a thc bt kh quy bc n tch c.Nghim ca n trong K l 1, 1, . . . , n i mt khc nhau.

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    Ta k hiu L l trng phn r, trng con ca K sinh ra bi1, 1, . . . , n.

    Nhm Galois L/K l nhm cc t ng cu m rng L/K : Cct ng cu ca m rng L/K l cc song nh : L L bo toncu trc vnh v cu trc khng gian vector ca L trn trngK. Nu khng c nguy c nhm ln, ta vit gin lc ch s vngm hiu = L/K. nhn mnh s ph thuc vo P, chngta cng s gi l nhm Galois ca a thc P.

    Cho P v P L l mt nghim ca P, khi () cng l mtnghim ca P. V vy nhm Galois tc ng ln tp hp ccnghim ca P. Vi k hiu chn t1, . . . ,nu, cc nghim caP c nh s, tc ng ca ln chng c cho bi ngcu nhm

    P : nvo trong nhm cc hon v cp n.

    nh l 6.1. ng cu P : n l n nh. Tc ng ca ln tp t1, 2, . . . , nu l tc ng bc cu. S phn t ca ngbng vi bc ca m rng L/K.

    Chng minh khng nh th nht khng kh. Nu nm tronghch ca P, th (i) = i vi mi nghim ca P. Trong honcnh ny, tc ng tm thng ln ton b L v L c sinhra bi 1, 1, . . . , n.

    Chng minh khng nh th hai v th ba kh hn mt cht.Trc ht ta chng minh rng mi ng cu K - i s : L Ku c nh l L. Tht vy mi ng cu nh vy u bo tontp t1, 1, . . . , nu, m tp ny sinh ra L, cho nn (L) = L. Nhvy ta chng minh rng

    AutK(L) = HomK(L,K

    )(3.14)

    chng minh khng nh th ba, ta ch cn cn chng minhrng

    |HomK(L,K

    ) | = degK(L). (3.15)Mt m rng hu hn L ca K gi l m rng tch c nun tho mn tnh cht ny.

    Nu L = K[x]/P vi P P K[x] l mt a thc bt kh quy bc ntch c th L l m rng tch c. Tht vy, trong trng

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    hp ny, cho mt ng cu : L K tng ng vi cho (x)l mt nghim ca P trong K v c ng n nghim khc nhaunh vy.

    C th chng minh c rng compositum ca m rng tchc K1, K2, . . . , Kn vi K Ki K l m rng tch c. 1 Vth, trng phn r ca mt a thc tch c l mt m rngtch c.

    Ta quay li chng minh khng nh th hai: Tc ng ca ln tp cc nghim t1, 2, . . . , nu l tc ng bc cu. Ta schng minh tn ti P AutK(L) sao cho (1) = 2. Trc ht tac ng cu : K[1] K[2] cho bi 1 2 v 1 v 2 c cnga thc cc tiu l P. Ta cn chng minh rng c th thctrin thnh mt t ng cu : L L. kt thc chng minh, ta cn s dng thm mt tnh chtna ca m rng tch c: Nu L/K l mt m rng tch c,th tn ti P L l phn t sinh ca L. Ni cch khc tn ti P L sao cho L = K[]. 2M rng L/K[1] cng l m rng tch c, cho nn tn ti1 P L sao cho L = K[1][1]. Ta c th pht trin

    : K[1] K[2],

    thnh mt ng cu c dng

    : K[1][1] K[2][2],

    vi 2 P K c la chn thch hp. S dng (3.14), ta c

    K[2][2] = L,

    v l mt t ng cu ca L nh ta mong mun.

    Ta c th tm tt cc thng tin trong mc ny nh sau. Mi athc bt kh quy P ng mt nhm Galois . Nu cc nghim caP trong K c nh s th c th coi nh mt nhm con canhm i xng cp n, c tc ng bc cu ln tp t1, 2, . . . , nu.S phn t ca ng bng bc ca trng phn r.

    1Xem chng minh trong sch Algebra ca Lang.2Xem chng minh trong sch Algebra ca Lang.

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    7. Tng ng Galois

    Nh ta thy trong mc trc, nu L l trng phn r caa thc bt kh quy tch c P P K[x], th m rng L/K c tnhcht s phn t ca nhm AutK(L) ng bng vi bc ca mrng. M rng L/K c gi l m rng Galois nu tnh cht nyc tho mn. Trng phn r ca mt a thc bt kh quylun lun l m rng Galois.

    M rng L/K gi l m rng tch c nu tn ti mt a thcbt kh quy P P K[x] tch c sao cho L K[x]/(P), tt nhin cth c nhiu a thc P nh th. D thy nu L = K[x]/(P) l mrng tch c nh trn v l m rng Galois, th L l trngphn r ca P. Ni cch khc m rng Galois l trng phnr ca mt a thc no . Mt khc, n c th ng thi ltrng phn r ca nhiu a thc khc nhau.

    Khng phi m rng no cng l m rng Galois. Quay li vd (3.12) m rng bc ba Q[] ca Q, vi l mt nghim cax3 2. Nu P AutQ(Q[]) th () cng phi l mt nghim cax3 2. Theo (3.12) th khng th l mt t ng cu ca Q[]tr trng hp = 1.

    Ni cch khc Q[] khng phi biu din Galois.Ngc li, theo (3.13) th K[]/K l m rng Galois vi K = Q[j]vi j l cn nguyn s bc ba ca n v (3.11). Tng qut hn

    nh l 7.1. Nu xn a P K[x] l mt a thc bt kh quy tchc, khi nu K cha cn nguyn s bc n ca n v, th vimi nghim P K ca a thc xn a, m rng L = K[] ca K lm rng Galois.

    Cc m rng trung gian gia K v L c th c phn loi davo . y thng c coi l mnh quan trng nht trongl thuyt Galois, v c gi tng ng Galois. Vn ci gquan trng nht lun lun c th bn ci.

    nh l 7.2. Mi m rng trung gian K L 1 L ng vi mt nhmcon 1 ca bao gm cc phn t P tc ng ln L nh mtnh x L 1-tuyn tnh. Tng ng ngc li c cho bi L 1 = L 1 .Hn na, L 1 l m rng Galois ca K khi v ch khi 1 l nhm conchun tc ca . 3

    3Nhm con 1 c gi l nhm con chun tc nu vi mi P , tac 11 = 1.

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    Trong trng hp , ta c L 1/K = / 1.

    Trong mc 5, ta kho st trng phn r L ca mt a thcbt kh quy P P K[x] thng qua chui m rng lin tip

    K = L0 L1 L2 Ln = L,vi Li l compositum ca K[1], K[2], . . . , K[i]. Chui m rngtng ng vi chui cc nhm con ca

    = 0 1 n = 0. (3.16)C th chng minh c rng i = Xni vi mi i = 1, 2, . . . , n,trong ni l nhm con cc phn t ca n c nh cc phnt 1, 2, . . . , i trong tp t1, 2, . . . , nu.Cc nhm con 1, 2, . . . ni chung khng phi l nhm conchun tc ca , v L1, L2, . . . khng phi l m rng Galoisca K. Trong mt mc sau, mc 8, chng ta s kho st ktrng hp chui m rng Galois.

    8. Tiu chun gii c phng trnhbng cn thc

    Xt v d cho bi chui cc m rng cn thc nh (3.10)

    K = K0 K1 Kr, (3.17)vi Ki Ki1[x]/(xni i). Gi thit rng K cha cc cn nguyns ca n v bc n1, n2, . . . , nr. Khi vi mi i, Ki l m rngGalois ca Ki1, v t ta c th suy ra Kr l m rng Galoisca K. K hiu i = Kr/Ki , ta s c chui cc nhm con chuntc nh sau:

    = 0 1 2 vi

    i1/i = Ki/Ki1 Z/niZ.Ni cch khc, nhm l nhm gii c.

    nh l 8.1. Cho P P K[x] l mt a thc bt kh qui bc n tchc vi h s trong trng K c cn nguyn s ca n v micp r n. iu kin cn v nghim ca P c th biu dinnh mt biu thc i s vi cn thc l nhm Galois ca P lnhm gii c.

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    Cho P K l mt nghim ca P. Nh phn tch trong mc 4, biu din c di dng biu thc i s c cn thc tngng vi vic tn ti mt chui m rng lin tip nh (3.17),vi Ki Ki1[x]/(xni i), v K[] Kr. V trng phn r L caP l m rng Galois nh nht cha K[] cho nn:

    K L Kr.iu ny ko theo P = L/K l thng ca . Vi iu kintrng c s cha cn nguyn s ca n v, ta ch ra trn l l nhm gii c. V vy nhm Galois P ca a thcP cng l nhm gii c. P l nhm gii c l iu kin cn biu din c nh mt biu thc i s vi cn thc.

    chng minh iu kin ny cng l iu kin , ta cnchng minh rng nu K l mt trng c nghim nguyn s cpn ca n v v nu L/K l m rng Galois c nhm Galois ngcu vi Z/nZ, th tn ti P K sao cho

    L K[x]/(xn ).Gi s m rng L l trng phn r ca mt a thc bt khqui P P K[x] c bc n. Nu 1, 1, . . . , n l cc nghim ca P trong K th L l m rng ca K sinh bi cc nghim ny.

    Chn mt phn t sinh ca nhm Galois L/K = Z/nZ. V tcng ca L/K ln tp 1, , 1, . . . , n l tc ng bc cu, chonn tng ng vi mt hon v n - chu trnh. Ta c th gi s

    (1) = 2, (2) = 3, . . . ,(n) = 1.

    Chn P K l mt cn nguyn s cp n ca n v v thit lpgii thc Lagrange:

    = 1 + 2 + + n1n.Lp lun nh trong trng hp n = 3, ta thy n = l mtphn t ca K v t suy ra rng

    L K[x]/(xn ). bit mt phng trnh c th gii c bng cn thc haykhng, ta ch cn bit xem nhm Galois ca n c th giic hay khng. Tnh nhm Galois ca mt a thc P tu

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    hon ton khng d. Quy trnh kho st trnh by trong mc 5,cho ta mt s thng tin v nhm Galois, trong c mt chuinhm con (3.16), nhng thng l nhng nhm con khngchun tc. Trong trng hp ca a thc tng qut

    P = xn + a1xn1 + + an P K[x], (3.18)

    vi K = k(a1, a2, . . . , an) l trng cc phn thc vi bin sa1, a2, . . . , an, lp lun nh trong 9, ta thy P = n. Nh vyi vi phng trnh tng qut, ta cn tm hiu vi n no thnhm i xng n l nhm gii c. Ta s ch ra rng ccnhm 3,4 l gii c v nh c th tm ra cng thc biudin nghim ca phng trnh bc ba v bc bn. Trong khi n l nhm khng gii c vi mi n 5.

    9. Phng trnh bc ba

    Li gii phng trnh bc ba ca Tartaglia kh phc tp v nging nh mt ci g t trn tri ri xung. Li gii phng trnhbc bn ca Ferrari cn phc tp hn na. Sau ny, Lagrangea ra phng php rt p tm ra phng php chung choli gii ca Tartaglia, Cardano v Ferrari. ng sng to ra mtcng c mi, gi l resolvent, m chng ta s tm chuyn ngthnh gii thc.

    minh ho cho cng dng ca phng php Lagrange, ta strnh by li gii phng trnh bc ba tng qut theo phngphp ny vi cht t h tr ca l thuyt Galois. Mc d phngphp gii thc ca Lagrange c th din gii d dng bng lthuyt Galois, ta cng nn lu l nhm Galois sinh ra sauphng php gii thc Lagrange.

    Trong i s hin i, ta c th gn cho ch tng qut mtngha chnh xc: Ta chn trng c s

    K = k(a,b, c) R = k[a,b, c],

    l trng cc biu thc hu t vi bin s a, b, c v h s nmtrong mt trng k no . a thc bc ba tng qut l a thc

    P = x3 + ax2 + bx+ c P R[x].

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    Ta s gi s rng trng k c cn nguyn s bc ba ca n v.Ni cch khc a thc x3 1 P k[x] phn r hon ton thnh:

    x3 1 = (x 1)(x j)(x j2),vi j P k l cn nguyn s bc ba ca n v. Ly v d, ta c thchn k = Q[j] vi j nh trong (3.11).

    Cho K l mt trng ng i s cha K, v gi 1, 2,3 l ccnghim ca P trong K. Trng phn r ca K :

    L = k(1,2,3) S = k[1,2,3],l trng cc thng ca cc a thc c bin s 1, 2, 3 tho

    a = 1 + 2 + 3b = 12 + 23 + 31

    c = 123C th chng minh c rng S l module t do cp su trnvnh R v L l khng gian vector c chiu bng su trn trngK. Nh vy nhm Galois ca P c su phn t v v vy nch c th l ton b nhm i xng 3. gii phng trnh

    bc ba tng qut, ta s kho st nhm i xng 3. Ta s thhin mt hon v cp ba nh trong v d sau: (2, 3, 1) l hon v1 2, 2 3, 3 1. Cc phn t ca 3 c th c phn loinh sau

    Phn t n v (1, 2, 3). Phn t c 2-chu trnh (2, 1, 3), (1, 3, 2), (3, 2, 1). Phn t 3-chu trnh (2, 3, 1), (3, 1, 2).

    ng cu du sgn : 3 t1u gn cho ba phn t c 2-chutrnh du 1 v gn cho ba phn t cn li du +1. Hch cang cu du sgn : 3 t1u l nhm lun phin 3. Nhmny c 3 phn t:

    3 = t(1, 2, 3), (2, 3, 1), (3, 1, 2)uv ng cu vi nhm xch Z/3Z. Tm li ta c dy khp

    0 3 3 Z/2Z 0 (3.19)

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    vi 3 Z/3Z. Nhm 3 l nhm gii c.Nhm con 3 tng ng vi m rng trung gian K+ = L3 baogm cc phn t ca L c nh di tc ng ca 3. Ta cchui cc m rng lin tip

    K K+ L.vi K+/K l m rng Galois cp hai c nhm Galois Z/2Z, vL/K+ l m rng Galois cp ba c nhm Galois Z/3Z.

    Qu trnh biu din nghim ca phng trnh bc ba ph qunh biu thc i s c cn thc c th chia thnh hai bc:

    S dng dy khp (3.19) xy dng m rng trung gian.K K+ L vi K+/K l m rng Galois cp hai c nhmGalois Z/2Z, v L/K+ l m rng Galois cp ba c nhmGalois Z/3Z.

    Biu din cc m rng trung gian K+/K di dng

    K+ = K[x]/(x2 1),

    v L/K+ di dng L = K+[x]/(x3 2).

    Ta c th vit tng minh mt phn t ca K+ :

    = (1 2)(2 3)(3 1).D thy l n nh di tc ng ca 3. Ta cng ngaythy 2 = d vi

    d =ij

    (i j),

    l bit thc ca P. Bit thc d l mt phn t ca K v ta c

    K+ K[x]/(x2 + d). (3.20)Tht vy, ta c ng cu trng:

    K[x]/(x2 + d) K+,xc nh bi x . V c hai v u l khng gian vector haichiu trn K, v ng cu trng lun l n nh, cho nn nbt buc phi l song nh.

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    Ni cch khc, mi phn t ca K+ u c th biu din mtcch duy nht di dng m+ n vi m, n P K. chng t phng trnh tng qut bc ba c th gii cbng cn thc, ta ch cn cn chng minh rng m rng L/K+c th biu din c di dng phng trnh

    L = K+[x]/(x3 ),

    vi mt phn t P L+ no . K hiu = (2, 3, 1) l hon vtc ng ln cc nghim ca P nh sau: (1) = 2, (2) = 3v (3) = 1. Gii thc Lagrange c dng:

    = 1 + j2 + j23.

    Ta nhn thy rng:

    () = j1 v 2() = j2,

    v t suy ra = ()2() = 3.

    Phn t biu din nh tch ()2() hin nhin c tnh nnh di tc ng ca . V vy P K+.Ta khng nh rng

    L K+[x]/(x3 ). (3.21)Tht vy ta c ng cu vnh

    : K+[x]/(x3 ) L,

    xc nh bi x v 3 = . V () , cho nn R K, v nhca ng cu l mt m rng trung gian

    K im() Lvi K im(). Ta nhn thy chiu ca im() nh khng gianvector trn K phi bng ba v m rng bc ba L khng th chatrong n mt m rng bc hai.

    Ni cch khc, ng cu l ton nh. V khng gian ngunv ch c cng s chiu, l ng cu. Nh vy mi phn tca L, trong c 1, 2, 3, c th biu din mt cch duy nhtdi dng

    m0 +m1+m22, vi m0, m1, m2 P K+.

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    Bn thn tho m phng trnh 3 = vi P K+. Mi phnt ca K+ u c th biu din mt cch duy nht di dngm + n vi m, n P K v 2 = d. Chu kh tng minh ho cc hs m, n, m0, m1, m2, ta s tm lo c cng thc Tartaglia (3.5)cho nghim ca phng trnh bc ba tng qut

    x3 + ax2 + bx+ c = 0.

    10. Phng trnh bc bn

    nh l 10.1. Nhm 4 l nhm gii c.

    K hiu S l tp t1, 2, 3, 4u v 2(S) l tp cc tp con ca S cng hai phn t. Tp ny c 6 phn t chia thnh ba cp cctp con i nhau:

    t1, 2u, t3, 4ut1, 3u, t2, 4ut1, 4u, t2, 3u

    v gi 2(S) l tp cc cp tp con i nhau nh trn. Ta cnh x 2-1

    2(S) 2(S). (3.22)Nhm 4 cc hon v ca S, tc ng mt cch tng thch ln2(S) v 2(S). Tc ng ca 4 ln 2(S) cho ta mt ng cu4 3. Kho st k hn tc ng 4 ln 2(S) v 2(S) ta cdy khp

    0 (Z/2Z)2 4 3 0 (3.23)v t suy ra rng nhm 4 l nhm gii c.

    Bn c c th dng gii thc Lagrange tm ra biu thccn thc cho nghim phng trnh bc bn tng qut, gingnh trng hp phng trnh bc ba trnh by mc 9.

    11. Phng trnh bc nm tr ln

    nh l 11.1. Vi mi n 5, n khng phi l nhm gii c.Nhm i xng n c ng cu du vi hch l nhm lunphin n

    0 n n t1u 0 (3.24)

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    C th chng minh rng vi mi n 5, nhm lun phin nl nhm n, tc l mt nhm khng c nhm con chun tc,ngoi nhm tm thng v chnh n.

    Nhm con chun tc l hp ca mt s lp lin hp v s phnt ca nhm con bng vi tng s phn t ca mt s lp linhp. Mt khc, tng s ny phi l c ca s phn t ca 5.Bng cch lit tt c cc lp lin hp ca 5 v lc lng cachng, ta nhn ra rng tng lc lng ca mt s lp lin hpca 5 khng th ng bng mt c thc s ca 60. V thnhm 5 khng c nhm con chun tc no ngoi chnh n vnhm tm thng.

    Bn c c th chng minh bng qui np rng n l nhm nvi mi n 5 xut pht t trng hp n = 5.

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    Ton variste Galois

    Ngi dch: Lu Trng Lun 1

    i hc FPT

    variste Galois 2 sinh ti Bourg La Reine (gn Paris) l con ngNicholas Gabriel Galois v b Adelaide Marie Demante. Cha mGalois u l nhng tr thc c gio dc k v trit hc, vnhc c in v tn gio. Tuy nhin, khng ai trong gia nhGalois bc l kh nng ton hc. M ca Galois l ngi thyduy nht dy d ng cho n nm 12 tui. B dy ng ting HyLp, La tinh v tn gio, cng l lc b bt u gieo t tnghoi nghi ca mnh cho con. Cha ca Galois l ngi c nhhng trong cng ng v nm 1815 c bu lm th trngBourg-la-Reine.

    Thi im bt u nhng s kin lch s ng vai tr quantrng trong cuc i Galois chnh l cuc nh chim nh tBastille ngy 14/07/1789. T lc , vng triu Louis 16 b lunglay d i khi ng o ngi dn Php on kt li nhm lt s cai tr c quyn ca gio hi v nh nc.

    Bt chp nhng n lc tha hip, vua Louis 16 b mang raxt x sau khi tm cch trn khi t nc. Sau khi nh vuab hnh hnh vo ngy 21/01/1973, nc Php ri vo tnh trngkinh hong vi nhiu phin xt x chnh tr. n cui nm 1793,c ti 4595 t nhn chnh tr b giam gi Paris. Tuy nhin, tnhhnh t nc cng bt u sng sa hn sau khi qun inc ny, di s ch huy ca Napoleon ginh ht chinthng ny n chin thng khc.

    Napoleon tr thnh nht Tng ti vo nm 1800 ri ln ngiHong vo nm 1804. Qun i Php tip tc chinh phcchu u trong khi quyn lc ca Napoleon ngy cng c cngc. Nm 1811, Napoleon t n nh cao quyn lc. Nhng n

    1Ngun: www-history.mcs.st-andrews.ac.uk/Biographies/Galois.html2Nh ton hc ngi Php (25/10/1811 - 31/05/1832)

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    nm 1815 th quyn lc sp . Cuc tn cng Nga nm 1812tht bi ko theo nhng cuc bi trn khc. Qun lin minhtin vo Paris ngy 31/03/1814. Napoleon thoi v ngy 6/4 vLouis XVIII c lin minh a ln lm vua. Nm 1815 chngkin 100 ngy ni ting. Napoleon tin vo Paris ngy 20/3, bnh bi ti Waterloo ngy 18/6 v thoi v ln th hai ngy22/6. Louis XVIII c khi phc ngi vua nhng mt vo thng9/1824. Charles X tr thnh v vua mi.

    Galois lc ny ang i hc. ng vo hc lp 4 ni tr trngLyce of Louis-le-Grand ngy 06/10/1823. Trong hc k u tin,mt cuc ni lon nh n ra v 40 hc sinh b ui hc. Galoiskhng lin quan n s vic ny. Nm hc 1824 1825, ng thc lc gii v nhn nhiu gii thng. Tuy nhin, nm 1826,Galois phi hc li v mn hng bin khng t yu cu.

    Thng 2/1827 l thi im mang tnh bc ngoc trong cuc iGalois. ng vo lp ton u tin ca mnh, lp ca M. Vernier.Ngay lp tc, ng b ton hc cun ht v gio vin hngdn nhn xt:

    Nim am m ton hc chi phi cu y, ti ngh tt nht l bam cu y nn cho php cu khng hc bt k mn g khc ngoitr mn ny. Cu y ang lng ph thi gian y, lm phincc gio vin v chuc ly nhiu hnh pht.

    Hc b ca Galois bt u xut hin cc t c bit, k d, lpd v hng ni. C l y l nh ton hc lp d nht ca mithi i. Thy gio M. Vernier nhn xt:

    Thng minh, tin b ng k nhng cha phng php.

    Nm 1828, Galois thi vo trng Bch khoa Paris nhng trt.y l trng i hc danh ting Paris v Galois mun thivo trng thun li cho vic hc. ng thi ng cng munvo trng v lc ny phong tro chnh tr trong gii sinh vin y ang din ra rt mnh m trong khi Galois mun theo bccha m ng tr thnh ngi tch cc ng h phe cng ha.

    Tr v li Louis-le-Grand, Galois ghi danh vo lp chuyn tonca Louis Richard. Tuy nhin ng li tp trung ngy cng nhiuvo nhng nghin cu ca ring mnh v t n vic hcti trng hn. ng nghin cu mn hnh hc ca Legendre vcc lun n ca Lagrange. Richard nhn xt:

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    Cu sinh vin ny ch thch tp trung vo nhng vn nh caoca ton hc.

    Thng 4/1829 Galois cng b cng trnh ton hc u tin vlin phn s trn Annales de mathmatiques. Ngy 25/5 v 1/6ng gi cc bi vit v phng php gii cc phng trnh i scho Vin hn lm Khoa hc. Cauchy l ngi c phn cngnh gi cng trnh ny.

    Bi kch p n vi Galois vo ngy 02/07/1829 khi cha ng tt. Linh mc x Bourg-la-Reine gi mo tn ca th trngGalois trn nhng bi chm bim nhm vo nhng ngi thnca gia nh Galois. Cha ca Galois l ngi tt v v scandal vt qu sc chu ng ca ng. ng treo c t st trongcn h ca mnh Paris, cch Louis-le-Grand, ni con mnhang hc ch vi bc chn. Galois b tc ng nghim trngbi ci cht ca cha mnh v n nh hng ln n hngi sau ny ca ng.

    Vi tun sau ci cht ca cha mnh, Galois tip tc ng kthi vo trng Bch khoa Paris ln th hai. Ln ny ng litrt, c l mt phn do n ri vo thi im ti t nht ngaysau ci cht ca cha ng, mt phn do ng cha bao gi giivic din t nhng tng ton hc su sc ca mnh. V thGalois nh phi vo hc ti cole Normale, mt nhnh catrng Louis-le-Grand. vo c y, ng phi d k thit ti m nu vo c trng Bch khoa Paris th Galois khng cn n n. ng thi u v nhn bng tt nghip ngy29/12/1829. Ngi nh gi mn ton nhn xt:

    Sinh vin ny i khi din t kin kh hiu nhng cu ta thngminh v bc l tinh thn nghin cu c bit.

    Gim kho mn vn th nhn xt:

    y l sinh vin duy nht c kt qu rt t, cu ta tuyt i chngbit g. H ni vi ti cu ta c kh nng ton phi thng. Thtngc nhin v sau bui thi ny, ti tin l cu ta ch c mt chtthng minh.

    Galois gi thm cho Cauchy bi nghin cu v l thuyt phngtrnh nhng sau nhn ra n trng vi mt phn trong cngtrnh c ng sau khi mt ca Abel. Sau , Galois theo likhuyn ca Cauchy gi mt bi nghin cu mi v iu kinphng trnh gii c bng cn thc vo thng 2/1830. Bi

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    bo c gi cho Fourier, th k ca Vin hn lm Paris xtduyt Gii thng ln v ton hc. Fourier mt thng 4/1830 vcng trnh ny ca Galois khng bao gi c tm thy na.

    Galois, sau khi c cng trnh ca Abel v Jacobi bt tayvo vic nghin cu l thuyt hm eliptic v tch phn Abel. Vis h tr ca Jacques Sturm, ng cng b 3 bi bo trnBulletin de Frussac vo thng 4/1830. Tuy nhin, thng 6, ngbit c rng gii thng ca hc vin s c ng trao choAbel (sau khi mt) v Jacobi v cng trnh ca ng khng hc xem xt.

    Thng 7/1830 n ra cuc cch mng. Vua Charles X phi trnkhi nc Php. Bo ng din ra trn cc ng ph Parisv gim c trng cole Normale, M. Guigniault, ng cngtrng ngn sinh vin ra ngoi tham gia bo ng. Galoistm cch tro tng tham gia nhng khng thnh. Thng12/1830 M. Guigniault vit mt s bi bo ch trch sinh vinv Galois vit th p tr trn Gazette des coles, ch trch M.Guigniault v hnh ng giam sinh vin bn trong trng. V lth ny m Galois b ui hc v gia nhp i pho binh ca Vbinh quc gia, mt nhnh dn qun t v theo phe cng ha.Ngy 31/12/1830, i pho binh ca V binh quc gia b honggia ra lnh gii tn v vua mi Louis-Phillipe lo ngi y l mtmi e da i vi ngai vng.

    Hai bi cng b nh, mt bn tm tt ng trn trn Annalesde Gergonne (12/1830) v mt l th v vic ging dy khoa hctrn Gazette des coles (2/1/1831) l nhng n phm cui cngtrong i ng. Thng 1/1831 Galois n lc quay tr li vi tonhc. ng t chc vi lp v i s cao cp thu ht 40 sinh vinn d bui u tin nhng sau con s ny nhanh chnggim xung. ng c Poisson mi gi phin bn th ba cacng trnh ca mnh v phng trnh cho Vin hn lm v ng thc hin ngy 17/1.

    Ngy 18/4, Sophie Germain vit mt l th cho bn c, nh tonhc Libri k v tnh hnh ca Galois

    ... ci cht ca M. Fourier, l mt mt qu ln cho sinh vin Galoisngi m cho d tnh tnh k quc, bc l thin hng thngminh. Nhng bt hnh ny qu ln n n khin Galois bui khi cole Normale. Anh ta khng tin bc. H ni anh ta sb in hon ton. Ti s iu ny l tht...

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    Cui nm 1830, tt c 19 s quan thuc i pho binh ca Vbinh quc gia b bt v b kt ti m mu lt chnh quyn.H c tuyn trng n v ngy 09/05/1831, c 200 ngi theophe cng ha tp trung n ti mng vic ny. Trong ba tic,Galois nng ly trong khi tay cm con dao gm ang m ra nhv hm da nh vua, Louis-Phillipe. Sau ba n ti, Galois bbt v giam ti nh t Sainte-Plagie. ng c th ra sauphin x ngy 15/6.

    Ngy 14/7, ngy din ra cuc tn cng nh t Bastille v Galoisb bt tr li v mc ng phc ca i pho binh ca V binhquc gia vn b gii tn. Lc ng cng ang mang trongngi mt khu sng trng np n, vi khu sng ngnv mt dao gm. Galois b a tr li nh t Sainte-Plagie.Thi gian ny, ng nhn c tin cng trnh ca mnh b bcb. Poisson nhn xt:

    Lp lun ca Galois cha r rng v cha c pht trin y cho php chng ti nh gi c tnh chnh xc ca n.

    Tuy nhin, ng khuyn khch Galois cng b mt bn tm tty hn cng trnh ca mnh. trong nh t Sainte-Plagie,Galois c dng dao t t nhng nhng bn t khc ngnng li. Khi say ru trong t, ng bc l tm hn mnh:

    Bn c bit ti thiu g khng? Ti ch tit l vi bn thi: lngi m ti ch c th yu thng v yu thng trong tm khm.Ti mt cha ti v khng ai c th thay th c ng, bn chiu khng?

    Thng 3/1832, mt trn dch t qut qua Paris v nhng tnhn, trong c Galois, c chuyn n tri Sieur Fault-rier. Ni y, dng nh ng phi lng Stephanie-Felice duMotel, con gi mt bc s a phng. Sau khi ra t ngy 29/4,Galois vit th qua li vi Stephanie v r rng l c ny tmcch n trnh cuc tnh.

    Ci tn Stephanie xut hin nhiu ln bn l cc bn ghi chpca Galois. Galois u sng vi Perscheux dHerbinville ngy30/5 m l do khng ai r nhng chc chn vic ny c linquan n Stephanie. Mt trong s nhng ghi chp ny c cu:

    C g cn b sung trong chng minh ny. Nhng ti li khngc thi gian.

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    Chnh iu ny dn n huyn thoi rng ng dnh mcui cng vit li tt c nhng g ng bit v l thuyt nhm.Cu chuyn ny c v nh c phng i ln.

    Galois b thng trong cuc u sng v b dHerbinville vnhng ngi i theo b mc cho n khi mt nng dn tm thy.ng mt ti bnh vin Cochin ngy 31/5 v m tang c tchc ngy 2/6. N tr thnh tm im ca cuc ni lon caphe cng ha ko di nhiu ngy.

    Em trai Galois v bn ng, Chevalier, sao chp li cc bivit lin quan n ton hc ca ng v gi cho Gauss, Jacobicng nhng ngi khc. Galois cng tng m c c Jacobiv Gauss nhn xt cng trnh ca mnh. Ngi ta khng tmthy nhn xt ca nhng ngi ny. Tuy nhin, cc bi bo n tay Liouville, ngi vo thng 9/1843 cng b trc vin hnlm rng ng tm thy trong cc bi bo ca Galois mt ligii chnh xc.

    ... va chnh xc va su sc cho bi ton tuyt vi sau: Cho mtphng trnh bt kh quy vi bc nguyn t, xt xem n c giic bng cn thc khng?

    Liouville xut bn nhng bi bo ca Galois trn tp ch camnh nm 1846. L thuyt m Galois phc tho trong nhng bibo ny ngy nay c gi l l thuyt Galois.

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    Ton NGHCH O MOBIUS

    Ng Quang Hng (i hc Buffalo, M)

    Php nghch o Mobius khi nguyn l mt cng thc trong lthuyt s. n nhng nm 1960 th Gio s Gian-Carlo Rotacho chng ta thy cng thc trong l thuyt s l mt trnghp c bit ca mt cng thc p dng trn cc tp th t bnphn (poset). Cng thc Mobius tng qut c nhiu ng dngtrong Ton v My Tnh. Trong bi ny ta ro qua chng minhca php nghch o Mobius trn cc tp th t bn phn vmt vi ng dng ca n.

    1. Ba v d

    1.1. Ton t hp

    Cng thc inclusion-exclusion ni rng, m tng s nhc tc Ch Pho l b hoc th N l m, th ta cng s con ca chPho vi s con ca th N tr i s con chung. Ni cch khc,cho n tp hp hu hn A1, ,An th ta c th tnh lc lngca hi ca chng bng cng thc:n

    i=1Ai

    =ni=1 |Ai|

    1ijn |Ai XAj|+

    1ijkn |Ai XAj XAk| + (1)n1 |A1 X XAn|Cng thc ny mt s sch ni l ca Abraham de Moivre;nhng c v n xut hin nm 1854 t mt bi bo ca Danielda Silva, v ln na nm 1883 trong mt bi bo ca JosephSylvester [11].

    Bi tp 1.1. Nm 1891, Francois douard Anatole Lucas (cha bi ton thp H Ni) t cu hi sau y: cho mt ci bntrn v m cp v chng, c bao nhiu cch xp h ngi namn xem k sao cho khng cp v chng no ngi k nhau?" Tac th dng cng thc IE tr li cu hi ca Lucas.

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    1.2. L thuyt s

    Trong l thuyt s c mt cng thc gi l cng thc nghch oMobius [10], xinh hn hoa hu! Cng thc ny pht biu nhsau: Cho 2 hm s f,g bt k trn min s nguyn dng, ta c

    f(n) =d|ng(d), @n 1

    tng ng vi

    g(n) =d|n(d)f(n/d), @n 1

    trong (d) l hm Mobius nh ngha nh sau

    (d) =

    $&%1 d l tch ca mt s chn cc s nguyn t khc nhau1 d l tch ca mt s l cc s nguyn t khc nhau0 d c c s l bnh phng ca mt s nguyn t

    August Ferdinand Mobius l mt nh thin vn ngi c, tngl tr l ca Gauss; ng cng l tc gi ca ci bng Mobiuslng danh trong hnh hc T-p.

    1.3. Hnh t-p

    Cng thc a din Euler pht biu rng v e + f = 2, trong v, e, f l tng s nh, cnh, v mt ca mt khi a din bachiu. Euler khm ph ra cng thc ny nm 1752, nhng cv nh Descartes cng bit n t 1640. Trm nm sau, nm1852, Schlafli pht biu cng thc tng qut cho cc a dinli trong khng gian n-chiu, nhng chng minh ng phi chn ngi khng l Henry Poincar (1893, [4]).Cng thc Euler tng qut, cng gi l cng thc Euler-Poincar,pht biu nh sau. Gi Fi l tng s mt i-chiu ca a dinn chiu (mt 0-chiu l nh, mt 1-chiu l cnh, vn vn).Ta quy c Fn = 1 v F1 = 1 vit cho tin. Th ta c cng thcEuler-Poincar

    ni=1

    (1)iFi = 0.

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    1.4. Gian-Carlo Rota

    Nm 1964, trong bi u tin ca mt chui bi bo kinh int nn mng cho l thuyt t hp i s [5], Gian-Carlo Rotacho chng ta bit c ba cng thc trn chng qua l trnghp c bit ca phng php tnh nghch o Mobius trn cctp hp th t mt phn (partially ordered set, hay poset). Mphng php nghch o Mobius trn posets th chng qua chl pht biu sau y: nu A l mt ma trn vung kh nghch,th x = Ay tng ng vi y = A1x. i s tuyn tnh munnm! Rota c quyn sch rt th v c nhiu giai thoi ni tingtrong gii chuyn mn tn l Indiscrete Thoughts [6].Di y chng ta duyt qua phng php ca Rota, chngminh c ba cng thc trn, v chng minh b Sauer-Shelah t thng cng.

    2. Nghch o Mobius trn posets

    2.1. Tp hp th t bn phn (Poset)

    Poset i khi l mt tp hp m ta c th so snh ln nhgia mt s cp phn t nhng khng nht thit l so c ttc cc cp. Th t ln nh ny c tnh bc cu (transitive) vkhng to ra th t lun qun.C th hn, mt poset (tp th t bn phn) l mt cp (P,)trong P l mt tp hp v l mt quan h nh phn (hayquan h hai ngi) gia cc phn t ca P tha mn 3 tnh cht

    1. x y v y z suy ra x z, vi mi x,y, z P P (tnh bc cu transitive)

    2. x x, @x P P (tnh phn x reflexive)

    3. x y v y x suy ra x = y (tnh phn xng antisymmet-ric)

    V d 2.1. P = Bn l tp tt c cc tp con ca [n] v quanh nh phn l , ngha l X Y nu v ch nu X Y. Ciposet ny gi l i s Bool (Boolean algebra). Xem v d trnHnh 5.1.

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    H

    t2ut1u t3u

    t1, 3ut1, 2u t2, 3u

    t1, 2, 3u

    Hnh 5.1: i s Bool B3

    V d 2.2. P = Dn l tp tt c cc c s dng ca n, quanh nh phn l quan h chia ht, ngha l i j nu v ch nui|j. K hiu i|j ngha l j chia ht cho i (hay i chia ht j). Xem vd trn Hnh 5.2.

    1

    2 3 5

    4 6 10 15

    12 20 30

    60

    Hnh 5.2: Poset cc c s ca 60

    V d 2.3. P l tp tt c cc mt (faces) ca mt a in(polytope) trong khng gian n chiu; v x y nu mt x chatrong mt y. Mt rng cng l mt mt vi chiu 1, v ton

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    b a din l mt mt vi s chiu bng n. Poset ny cn gi lface lattice ca polytope. Xem v d trn Hnh 5.2.

    a

    b

    e

    c

    d

    H

    cea d b

    ebabadea ec ed cb cd

    ecbeab ead ecd abcd

    abcde

    Hnh 5.3: Face lattice ca hnh Pyramid

    2.2. Hm Mobius ca poset

    Nhng iu ta vit sau y ng cho mt trng K ty h v ccposets v hn (min l n hu hn a phng1). cho ngin, ta pht biu cc kt qu vi K = C v cc posets hu hnthi.Gi (P,) l mt poset hu hn. Ta xt cc ma trn kch thc|P| |P| sao cho (x,y) = 0 nu x y. Khi x y th (x,y) P Cty h. Tp cc ma trn ny gi l i s k (incidence algebra)ca P, k hiu l I(P). Trong i s k th ma trn nh nghabng

    (x,y) =

    #1 x = y

    0 x yl ma trn n v.

    nh l 2.4. Cho trc poset (P,) trong P hu hn. Xt mtma trn P I(P) ty th kh nghch nu v ch nu (x, x) 0, @x P P.

    1Ngha l s cc thnh vin nm gia mt cp x v y l hu hn vi micp x,y.

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    Chng minh. Nu ta v th" ca P bng cch xem P nh tpcc nh v v mt mi tn t x n y nu x y (nh trongHnh 5.1 v 5.2) th ta c mt th c hng nhng khng cvng trn (directed acyclic graph). Do , tn ti mt cch litk tt c cc phn t ca P t tri sang phi sao cho tt c ccmi tn u tr sang phi hoc tr vo chnh n (loop trong th). Th t ny gi l trt t t-p (topological ordering) ca th, l mt bi tp c bn khi hc cc thut ton duyt th.Nu ta vit cc ma trn P I(P) m cc hng v ct nh chs theo th t ny th ta c cc ma trn tam gic trn (upper-triangular). Do kh nghch nu v ch nu (x, x) 0, @x,ngha l cc phn t trn ng cho khc khng. Lu rngma trn nghch o cng l ma trn tam gic trn, v do cng thuc v i s k.

    Mt thnh vin quan trng ca i s k I(P) l ma trn , gil hm zeta ca P, nh ngha bng

    (x,y) =

    #1 x y0 x y

    nh ngha 2.5 (Hm Mobius ca mt poset). Hm Mobius caposet (P,), k hiu l , chnh l ma trn nghch o ca hmzeta . (Theo nh l 2.4 th kh nghch.)

    K n ta m t mt cng thc quy tnh hm Mobius camt poset. T nh ngha ca php nhn ma trn, vi , P I(P)bt k ta c

    ()(x,y) =zPP(x, z)(z,y) =

    xzy

    (x, z)(z,y),

    ti v nu x z th (x, z) = 0, cn nu z y th (z,y) = 0. Do, t = ta suy ra

    (x,y) =

    xzy(x, z)(z,y) =

    xzy

    (x, z).

    Hay vit c th hn th vi mi x,y P P ta c

    xzy(x,y) =

    #1 nu x = y0 nu x y. (5.1)

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    ng thc (5.1) suy ra cng thc quy np tnh (x,y):

    (x,y) =

    $&%1 x = y

    xzy (x, z) x y0 x y

    T cng thc ny ta suy ra gi tr hm Mobius cho ba posets trn. Hai ng thc u th d (lm bi tp), ci th ba th kh.

    1. Nu P = Bn l tp tt c cc tp con ca [n] (i s Bool),th

    (A,B) =

    #(1)|B||A| A B0 A B

    2. Nu P = Dn l tp tt c cc c s ca n, th

    (x,y) =

    #(1)r nu y/x l tch ca r s nguyn t khc nhau0 nu khng phi th.

    3. Nu P l face-lattice ca mt a in n chiu th

    (A,B) =

    #(1)dim(B)dim(A) if A B0 nu khng.

    (5.2)

    2.3. Nghch o Mobius

    Xt hai hm s f,g : P C bt k. Ta c th xem chng nh haivectors trong khng gian C|P|. Cng thc nghch o Mobiustrn poset ni hai iu rt n gin:

    f = g g = f, (5.3)v, xoay ngang cc vectors ra th

    f = g g = f. (5.4) hiu ngha t hp ca s tng ng ny, ta vit r rnghn mt cht v ta bit (x,y) v (x,y) bng 0 nu x y. Quanh (5.3) ni rng:

    f(x) =xy

    g(y), @x P P g(x) =xy

    (x,y)f(y), @y P P. (5.5)

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    ng thc ny ta hiu nh sau. Gi s ta c hm g gn mtcon s g(y) vo mi thnh vin y P P, v f gn vo mi x P P mtcon s l tng ca cc g(y) sao cho x y, th v phi ca (5.3)cho ta cch tnh g da trn f.i ngu li, quan h (5.4) ni rng:

    f(x) =xy

    g(y), @x P P g(x) =xy

    (y, x)f(y), @y P P. (5.6)

    V d 2.6. c cng thc Euler-Poincar, ta p dng (5.5)trong g(y) = 1 vi y = P v g(y) = 0 vi mi y cn li trong P.Khi , r rng l tt c cc f(x) u bng 1. Dng (5.2), ta c

    0 = g(H) =

    mt B

    (1)dim(B)dim(H)f(B) =

    mt B

    (1)dim(B)+1 = n

    i=1(1)iFi.

    V d 2.7. p dng (5.6) cho poset P = Dn, ta c ngay cng thcnghch o Mobius c in trong l thuyt s trn.

    V d 2.8. Cn cng thc inclusion-exclusion th sao? Cchhiu sau y s hu dng trong nhiu trng hp. Gi s ta cmt tp bi ve" U = A1Y YAn. Mi vin bi c nhiu mu. Ccmu c nh s t 1 n n. Gi Ai l tp cc vin bi c mui. Vi X [n] ty , gi g(X) l tp tt c cc vin bi ch c ngcc mu trong X m thi. Khi ,

    f(X) =XY

    g(Y)

    chnh l s cc vin bi m mi vin c t nht cc mu trong X,v f(H) = |U|. Do ,

    f(X) =

    iPXAi

    .

    p dng (5.5) cho poset P = Bn ta kt lun

    0 = g(H) =Y[n]

    (1)|Y|iPYAi

    .

    Chuyn f(H) = |U| sang mt v l ta c cng thc inclusion-exclusion.

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    3. B SauerShelah

    Chng ta t thng cng bng cch chng minh mt b thp quan trng gi l b Sauer-Shelah [7, 8]. B ny cng dng su sc trong l thuyt hc my, c th l l thuytchiu Vapnik-Chervonenkis (VC dimension) [13, 12].Gi F l mt b cc tp con ca [n]. Vi S [n] bt k, nhngha hnh chiu ca F ln S l tp

    F(S) = tFX S | F P Fu.Ta ni F bm nt S nu F(S) = 2|S|.

    B 3.1 (B Sauer-Shelah). Cho trc F l mt b cc tpcon ca [n]. Gi d l kch thc ln nht ca mt tp S [n] b Fbm nt, th

    |F| d(n) =di=0

    (n

    i

    ).

    Chng minh. Ta chng minh b ny bng phng phpchiu (i S tuyn tnh van tu!). Gi

    ([n]d)l tp tt c cc tp

    con ca [n] vi kch thc b hn hoc bng d. Vi mi F P F,nh ngha mt hm s hF :

    ([n]d) R nh sau:

    hF(X) =

    #1 X F0 X F .

    Cc hm hF l cc vectors trong khng gian Rd(n). C tt c |F|vectors hF, do nu chng c lp tuyn tnh th |F| d(n).Gi s chng khng c lp tuyn tnh, ngha l tn ti cc hs F sao cho

    FPFFhF = 0 (5.7)

    v cc h s ny khng cng bng 0. cho tin, ta m rngnh ngha v gn X = 0 vi mi X P 2[n]zF.T (5.7), vi X P ([n]d) bt k ta c FPF FhF(X) = 0, hay nicch khc vi X P ([n]d) ty ta c XY Y = 0. nh nghaX =

    XY g(Y), th ta va thy rng X = 0, @X P

    ([n]d).

    Gi Y l tp con nh nht ca [n] sao cho Y 0. (Nu ta ly tpF P F c kch thc ln nht sao cho F 0 th F = F 0, do

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    tn ti tp Y nh nht nh nh ngha.) D nhin |Y| d+ 1.Ta chng minh rng Y b F bm nt, t dn n iu v l. chng minh Y b bm nt th ta cn chng minh, vi Z Yty , tn ti F P F sao cho F X Y = Z. chng minh iu nyth ch cn chng minh

    A[n],AXY=ZA 0.

    l xong, ti v A = 0, @A R F. n y ta xt poset Bm gm ttc cc tp con ca Y Z (t m = |Y Z|). Poset ny l i sBool bc m. Vi mi phn t W Y Z, nh ngha

    g(W) =

    X:XXY=ZYWX.

    V nh ngha, vi mi V Y Z,

    f(V) =

    VWYZg(W).

    (Lu rng ta s dng dng (5.5) ca nghch o Mobius.) Dthy rng

    f(V) = ZYV , @V P Bm.Do Y l tp nh nht vi Y = 0, ta c f(V) = 0, @V Y Z, vf(Y Z) = Y 0. Theo nghch o Mobius ta c

    A[n],AXY=ZA = g(H) =

    VYZ

    (1)|V |f(V) = (1)|YZ|Y 0.

    4. Ch thch

    B sch ca Stanley [10, 9] l tham kho quan trng nht choton t hp m bao gm nghch o Mobius, t hp t-p. Schca Vapnik [12] ni v l thuyt hc my xc sut v l thuytchiu Vapnik-Chervonenkis. Mt tham kho tuyt vi khc choton T hp l quyn sch ca van Lint v Wilson [11]. Trongngnh my tnh, nghch o Mobius c ng dng trong c sd liu [2], thut ton [3, 1].

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    Ti liu tham kho

    [1] BJORKLUND, A., HUSFELDT, T., KASKI, P., AND KOIVISTO,M. Trimmed moebius inversion and graphs of boundeddegree. Theory Comput. Syst. 47, 3 (2010), 637654.

    [2] DALVI, N. N., AND SUCIU, D. The dichotomy of probabilisticinference for unions of conjunctive queries. J. ACM 59, 6(2012), 30.

    [3] NEDERLOF, J. Fast polynomial-space algorithms usingmobius inversion: Improving on steiner tree and relatedproblems. In Automata, Languages and Programming,36th International Colloquium, ICALP 2009, Rhodes, Greece,July 5-12, 2009, Proceedings, Part I (2009), S. Albers,A. Marchetti-Spaccamela, Y. Matias, S. E. Nikoletseas, andW. Thomas, Eds., vol. 5555 of Lecture Notes in ComputerScience, Springer, pp. 713725.

    [4] POINCAR, H. Sur la gnralisation dun thorme dEulerrelatif aux polydres. Comptes rendus hebdomadaires delAcadmie des sciences de Paris 117 (1893), 144145.

    [5] ROTA, G.-C. On the foundations of combinatorial theory. I.Theory of Mobius functions. Z. Wahrscheinlichkeitstheorieund Verw. Gebiete 2 (1964), 340368 (1964).

    [6] ROTA, G. C., AND PALOMBI, F. Indiscrete thoughts.Birkhauser, 1996.

    [7] SAUER, N. On the density of families of sets. J. Combinato-rial Theory Ser. A 13 (1972), 145147.

    [8] SHELAH, S. A combinatorial problem; stability and orderfor models and theories in infinitary languages. Pacific J.Math. 41 (1972), 247261.

    [9] STANLEY, R. P. Enumerative combinatorics. Vol. 2, vol. 62 ofCambridge Studies in Advanced Mathematics. CambridgeUniversity Press, Cambridge, 1999. With a foreword byGian-Carlo Rota and appendix 1 by Sergey Fomin.

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    [10] STANLEY, R. P. Enumerative combinatorics. Volume 1, sec-ond ed., vol. 49 of Cambridge Studies in Advanced Mathe-matics. Cambridge University Press, Cambridge, 2012.

    [11] VAN LINT, J. H., AND WILSON, R. M. A course in combina-torics, second ed. Cambridge University Press, Cambridge,2001.

    [12] VAPNIK, V. N. The Nature of Statistical Learning Theory.Springer-Verlag New York, Inc., New York, NY, USA, 1995.

    [13] VAPNIK, V. N., AND CHERVONENKIS, A. Y. Theory of uniformconvergence of frequencies of events to their probabilitiesand problems of search for an optimal solution from em-pirical data. Avtomat. i Telemeh., 2 (1971), 4253.

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    Ton CC BI TON I NN

    ng Nguyn c Tin (i hc Trento, Italy)

    Trong s 1 ca Epsilon, chng ti gii thiu vi c gi 15bi ton i nn c su tp trong hn 50 nm (t 1961 n2013), phn b nhng mc kh v th loi khc nhau. tip tc cuc hnh trnh, trong chuyn mc phn ny chngti chn lc v trnh by nhng li gii p cho cc bi ton c trnh by. Chng ti cng gii thiu vi c gi nhng ligii tng qut cng nh nhng ng dng thc t cho cc vn t nhng bi ton gii tr ny.

    1. Bi ton 3 chic nnNhc li : C 3 ngi chi,mi ngi c i ngunhin mt nn c mu hoc xanh dng. H nhnthy mu nn ca 2 bn mnhnhng khng bit mu camnh. Mi ngi cn phion ra mu nn ca mnh,hoc chn b qua nu khngon c. Nu t nht mtngi on ng mu nn vnhng ngi cn li khngon sai, h thng tr chi.H s thua nu c ngi onsai hoc c 3 cng chn bqua. H c trao i chinthut vi nhau trc khi chinhng trong khi tham gia thkhng c trao i bt cthng tin g. Tm chin thutc xc sut thng cao nht.

    bt u cuc hnh trnh,chng ti mi qu c gi cngquay li nhm 3 bi ton i nns 4, s 5, v s 6.Chng ta bt u bi bi tons i nn s 4 trc. Theo lutchi, ta c th thy rng nu nhc 3 u chn b qua th hs thua trong mi tnh hung,v vy, trong cc cch tr li dthy phi c t nht mt ngigi ln mu nn. Xt trng hpnu c 3 ngi chi u cngchn mt mu, hoc cng chnngu nhin m khng b qua thxc sut c 3 cng thng l1/8, tc 12.5% v c tt c 8 khnng khc nhau cho 3 ngi. Mtchin thut tt hn c th d dng c tm thy l 1 ngichn ngu nhin, v 2 ngi chn b qua. Lc ny ta c xc

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    sut thng l 1/2 = 50%. Liu y c phi l chin thut ttnht?

    Nh gi ca bi ton i nn s 6, chin thut ny cha phil chin thut tt nht. Li gii sau cho ra xc sut thng licao hn: nu mt ngi thy 2 ngi i nn khc mu nhau,h s chn b qua" v nu thy 2 ngi i nn trng munhau, s chn mu ngc li.Vi cch ny, h s lun on ng 6/8 trng hp v ch saitrong 2/8 trng hp khi c 3 nn cng mu nhau (xem v d Hnh 6.1).

    Hnh 6.1: Mi ngi nu thy 2 nn khc mu s chn b qua vnu thy 2 nn cng mu, s chn mu ngc li. Trong v d ny,ngi i nn s on mu (do thy 2 nn xanh) v 2 ngi inn xanh s chn b qua (do h thy 2 nn khc mu nhau).

    Li gii ny cho kh nng thng li l 6/8 = 75% nu nh nnc i tht s l ngu nhin. Tuy nhin, trong thc t (nhm t trong bi ton i nn s 6), ngi dn tr s nhanhchng nhn ra chin thut ca h v s i nn cho h cngmu nhau vi tn sut nhiu hn, hoc thm ch lun luni cng mu. Khi kh nng chin thng t 75% s gimxung thnh 0%! Vy ngi chi phi i ph th no vi ngidn tr xo quyt?

    Mt i sch c th ngh ra cho chin thut chi di hn l khingi qun tr i qua i nn cng mu, khi thy 2 nn gingmu nhau, h s chn mu . Nh vy vi trng hp ny hs thng 100%. V tt nhin, ngi qun tr li hc c cchny v quay li phn b vi 6 trng hp kia hoc quay li viphn b ngu nhin nn. Xc sut chin thng khi s thayi 75% > 0% > 100% > 75% > ... lp i lp li nh mt

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    cuc chin gia nhng ngi chi v ngi dn tr. Trung bnhchin thng do vy l 58%. Liu c cch lm tt hn 58% iph vi ngi dn tr xo quyt?

    Rt th v l chng ta c th nng cp chin thut c boton kh nng chin thng 75%. Chin thut ny nh sau:Chng ta c 8 cch i nn khc nhau v cc cch ny c thghp thnh tng cp i xng" nhau, i vi xanh. V d 3nn -- , s i vi 3 nn xanh-xanh-xanh XXX, hocX s i vi XX. Tng tng trn mt hnh lp phng vita cc nh tng ng ln lt vi cc cch i nn, v 2nh ni vi nhau khi ch c 1 nn khc mu nhau. Cc cuhnh i nn i xng" chnh l cc nh i xng nhau trnhnh lp phng ny.

    X

    X

    X

    XX

    XX

    XX

    XXX

    Chin thut chn hi sinhcp nh ng mu.

    X

    X

    X

    XX

    XX

    XX

    XXX

    Chin thut chn hi sinh2 nh XX v X.

    Hnh 6.2: Chin thut i ph vi ngi dn tr xo quyt".

    Vi cch th hin ny, nhim v ca ngi chi l xc nhnhm mnh nh no trn hnh lp phng! Vi mi ngichi, khi thy c 2 mu nn ca ngi chi i din, nginy s xc nh c mnh ang cnh no v nhim v lhoc b qua, hoc on xem mnh nh no. Vi cu hnhny, nu chp nhn hi sinh" mt cp nh i xng bt k,ta s lun xc nh c nh cn tm bng cch lun tr lihng v cc nh mu xanh khi tnh trng ang mt cnhni gia nh xanh" v nh xm", ngc li s chn b qua.Chin thut ny ch sai khi nhm ngi chi c nh ri vo cpnh c chn hi sinh", lc c 3 ngi u l 3 cnhni ra t cc nh xm" ny. Nh vy, cch chi theo chin

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    thut cu hi 1 l chn cp nh XXX v v c th ddng b bt bi" bi ngi dn tr chi (Hnh 6.2). Nhng ta cth chn cc cp nh nh (XX v X), hay (XX v X)...v u c xc sut thng li l 75% cho trng hp i nnngu nhin. Lc ny, ngi dn chng trnh khng th "btbi" v nhm ngi chi lun linh ng thay i c cp nhhi sinh" ngm nh, v buc ngi dn tr phi i nn mtcch ngu nhin cho h, v chin thng n vi xc sut 75%!

    Mt chin thut v d cho trng hp nu chn cp XX vX (xem Hnh 6.2 bn phi) nh sau:

    Ngi 1: nu gp xanh - s chn xanh, iu ny tngng vi cnh X - XX (mu trn hnh), v v cpnh hi sinh c cha X nn ta chn xanh. Nu gp - xanh s chn , iu ny tng ng vi cnh muxanh l cy trn hnh. Ngi 1 chn b qua nu thy 2 nncng mu (v cc cnh ny ni cc nh khng thuc tphi sinh").

    Ngi 2: nu gp nn cng mu, s chn mu , ngcli s chn b qua.

    Ngi 3: nu gp xanh - s chn ; nu gp - xanhs chn xanh, v b qua nu cng mu.

    Xt c th 8 cch i nn c th c, ta c kt qu nh sau:

    Nn 1 Nn 2 Nn 3 on 1 on 2 on 3 Kt quXanh Xanh Xanh B qua Xanh B qua ngXanh Xanh Xanh B qua B qua ngXanh Xanh Xanh SaiXanh B qua B qua ng Xanh Xanh B qua B qua Xanh ng Xanh Xanh Xanh Sai Xanh B qua B qua ng B qua B qua ng

    Vi 2 trng hp cn li (XX - X v X - XX) cng khngkh tm ra chin thut tng t vi kh nng thng 6/8. Nhvy, vi cch lm ngu nhin chn cp nh "hi sinh" ngi chi buc ngi dn tr phi i nn ngu nhin, v do vy, xcsut thng li bo ton vi t l 75%.

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    Tng qut ha cho 2n 1 ngi chiTip theo, chng ti xin mi c gi cng xem li gii m rngcho bi ton vi trng hp N = 2n 1 ngi chi, v y cngchnh l cu tr li cho bi ton i nn s 5 khi thay n = 4.

    Lu rng hin nay li gii p ch mi c tm thy trnghp tng qut nhng c bit ny, tc N = 2n1. Vi cc trnghp khc, li gii vn cn ang b ng. Chng ti s trnh bycch lm c th cho trng hp ny v sau s lin kt cchlm ny vi m Hamming, mt m sa li tuyn tnh ni tingv c s dng rng ri trong vin thng.

    Cch lm nh sau: gn th t cho ngi chi t 1 n N, vi cc s ny sang nh phn. Chng ta s s dng php snhkhc (XOR)1 trn cc chui nh phn ny. c tnh quan trngca php XOR l vi s k bt k, k k = 0, bi v k khp vichnh n trn mi bit. Gi T l tng (XOR) ca tt c ngi chic nn mu . Chin thut ca ngi chi l on sao cho Tkhc 0 v khi , h s thng nu T tht s khc 0 (tt nhin,h s sai khi T = 0).

    C th chin thut nh sau: Gi tk l tng XOR cc ngi inn m ngi th k thy c.

    Nu tk = 0, anh ta s on " (v nu anh ta on xanh,T s khng cp nht v s bng 0).

    Nu tk = k, anh ta s on xanh (v nu anh ta on th T = tk k = k k = 0). Ngc li, nu tk 0 v tk k, anh ta chn b qua.

    Lc ny nh ta thy, nu T tht s bng 0, tt c u s onsai. Nu T = k 0, ngi th k s on ng v tt c nhngngi khc u chn b qua, nn h s chin thng. V T c gitr t 0 n N nn xc sut thng li l N/(N+ 1).

    Liu chin thut ny c th nng cp ln i ph vi ngidn tr xo quyt"? Cu tr li l c vi cch lm tng t nhphn trnh by trn.

    1Php snh khc (thut ng ting Anh l Exclusive OR", hay cn gi lphp OR c loi tr, v cng cn gi l tng NIM), k hiu l , thc hin"cng" 2 chui nh phn bng cch so snh cc bit t phi sang tri. ngvi mi cp bit s cho kt qu l 1 nu 2 bit ny khc nhau v kt qu l 0nu 2 bit ny ging nhau. V d: 0110010 1010111 = 1100101.

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    M HAMMING V BI TON I NN S 5

    Gi s ti mun gi mt thng ip cho bn v thng ipny c m ha thnh cc bit nh phn (k s 0 hoc 1).Tuy nhin gia ng truyn c k can thip, v thay imt bit no trc khi bn nhn c thng ip. V vy,ti phi tm ra cch thm vo mt s thng tin b sung chothng ip sao cho khi nhn c, bn c th pht hinthng ip b thay i v hn na, bn c th khi phcli thng ip gc. Mt trong nhng phng php c bn phng chng vic sa i ny l s dng m ti din: tis gi mt ln 3 bit ging nhau cho cng 1 bit m ti mungi cho bn: 000 hoc 111. Lc ny, d k can thip c thayi bt k bit no, ti cng s khi phc li c. Cch lmny tuy t mc ch, nhng hiu qu khng cao, v nidung tha chim n 2 ln ni dung thng tin cn gi.

    M Hamminga a ra cch tt hn cho vic sa li ny, trong c 2k1 bit thng tin gi i ch c k bit b sung, cn li2kk1 bit l d liu. V d vi k = 4, khi gi mt thng ipc chiu di 15 bit th trong 11 bit l d liu (so vi ch 5bit l d liu phng php s dng m ti din). Chi titcch hot ng ca m Hamming c gi c th d dng tmthy ti Wikipedia. y, chng ti ch ch ra mi quan hgia Hamming v bi ton i nn s 5.

    Mt c tnh ng ch ca m Hamming l bt k chui(c chiu di 2k 1) no cng hoc l mt thng ip ng,hoc ch b sai mt bit. iu ny khc bit vi m ti dinv m ti din c mt s thng ip s khng bao gi xuthin. V d, bn s khng bao gi nhn c m 001011 v y c n 2 bit b thay i. m Hamming, mi chuiu c kh nng xut hin.

    By gi, gi s ta m ha nhng ngi i nn xanh l 0 vnn l 1, khi mt cch i nn cho N ngi tr thnhmt "thng ip" N bit. Lc ny, mi ngi s c th thyton b nhng bit khc, tr bit ca anh ta, v tt nhin chc 2 kh nng hoc anh ta l bit 0 (xanh), hoc l bit 1 ().Ngi chi s dng chin thut l lun gi nh chui N bit

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    hin ti KHNG PHI l mt "thng ip" ng. C th: numt ngi chi thy rng nu bit tng ng vi anh ta cgi tr 0 s bin chui thnh mt thng ip ng, anh tas chn "" (ng vi bit 1), v ngc li. Nu anh ta thyrng c 0 v 1 u c th lm thnh mt "thng ip" ng,anh ta chn b qua. Chin thut ny ch gp tht bi khi"thng ip" ban u ng v thnh cng nu "thng ip"sai. V c mi thng ip ng s c tng ng N thng ipsai, nn xc sut thng li l N/(N+ 1), tng t nh chinthut nu bi ton i nn s 5.

    aM Hamming c t theo tn ca nh ton hc ngi M, RichardWesley Hamming (1915 - 1998).

    2. Cc bi ton ca Konstantin Knop

    Chng ti mi c gi tip tc i n phn li gii ca 2 biton tuyt p do Konstantin Knop xut: bi ton i nn s7 v bi ton i nn s 9.

    Nh cp n Epsilon s 1, y l nhm cc bi tonm ngi chi xp thnh hng nn s lng nn m ngi chiquan st c gp gii hn. iu ny khng xy ra nhm biton i nn trn l mi ngi chi u thy c nn cami ngi chi khc (tr nn ca mnh). Tuy nhin, nhmbi ton trc, ngi chi khng bit c nhng ngi khcs on mu g, cn nhm bi ton ny, th t nu ln munn cng thuc v chin thut chi. tin theo di, chng tinhc li ca bi ton i nn s 7 cho trng hp 3 mu:

    C 100 ngi c xp thnh mt hng, mi ngi c i mtnn chn ngu nhin trong 3 mu cho trc. Mi ngi ch nhnthy mu nn ca nhng ngi ng trc mnh m khng thynn ca mnh v nhng ngi ng sau. Ln lt mi ngi sphi on mu nn ca mnh v h to cho nhng ngi khcnghe. Ngi ng cui cng (l ngi thy mu nn ca ton b99 ngi trc) l ngi bt u phi on. Ngi chi khng ctrao i bt k thng tin g vi nhau ngoi tr lng nghe mu nnt ngi on trc. ng sai cng ch c bit khi ngi cuicng on xong. Hy tm chin chut sao cho s ngi onsai l t nht.

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    y, sau khi quen thuc vi nhm bi ton phn 1, tathy y cng l mt dng m sa sai. C th trong bi tonny, chin thut s dng modulo c p dng. Chng ti trnhby y chin thut tng qut cho bi ton vi M ngi chiv N mu nn, chin thut nh sau:

    Gn mi mu tng ng vi mt con s t 0 n N 1. Ngi u tin tnh tng tt c nhng mu nn m anh taquan st c, ly modulo theoN v h mu nn tng ng.

    Nhng ngi tip theo, cn c vo li h ca nhng ngitrc , v tng ca nhng ngi ng trc mnh son ng mu nn ca mnh.

    Nh vy, cch lm ny ch c ngi u tin l c kh nngon sai, v v c N mu khc nhau, nn kh nng tt c mingi u on ng ch ph thuc vo kh nng ngi uon ng, v kh nng ny l 1/N.

    Hy th xt mt v d n gin vi 5 ngi chi, 3 mu nn vmi ngi ln lt i cc nn l (1, 2, 0, 2, 1):

    Ngi u tin, quan st c trc mnh c cc nn(2, 0, 2, 1), c tng bng 2 + 0 + 2 + 1 2 (mod 3), ngi nys h mnh c mu 2 (v sai).

    Ngi tip theo, quan st thy trc mnh c (0, 2, 1) vcn c vo li on ca ngi trc 2, nn s on l2 0 2 1 2 (mod 3) (ng). Ngi th 3, quan st thy (2, 1) v cn c vo li onca 2 ngi trc (2, 2), nn on l 2221 0 (mod 3)(ng).

    Ngi th 4, quan st thy (1) v cn c vo li on ca3 ngi trc (2, 2, 0), nn on l 2 2 0 1 2 (mod 3)(ng).

    Ngi cui cng, cn c vo li on ca 4 ngi trc(2, 2, 0, 2), nn on l 2 2 0 2 1 (mod 3) (ng).

    Kt qu l 4 ngi sau u lun lun on ng v ngi utin xc sut ng l 1/3.

    Tip theo, chng ti trnh by li gii cho bi ton i nn s 9. tin theo di, chng ti nu li y.

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    C 100 ngi xp thnh hng, mi ngi c i mt nn trongs 101 nn khc mu nhau. Mi ngi ch thy nn ca nhngngi ng trc mnh v khng thy nn ca mnh cng nhnhng ngi ng sau. Ln lt mi ngi t sau ra trc phion mu nn ca mnh v h to cho mi ngi cng nghe. Muno h ri s khng c h li na. Ngi chi khng ctrao i thng tin vi nhau. Tm chin thut sao cho kh nng ttc u on ng l cao nht.

    Thot nhn, 2 bi ton ny kh ging nhau, nhng khi tht sgii chng ta s thy y l mt bi khc, v mi bi c v pring bit. Trc khi i vo phn tch li gii cho bi ny, chngti xin nhc li hai khc bit c bn gia bi ton i nn s 9so vi bi ton s 7:

    N ngi chi, mi ngi i mt nn trong s N + 1 nnkhc mu nhau. iu ny c ngha l mu nn mi ngiu khc nhau. bi s 7, ta khng c rng buc gias ngi chi v s mu nn, hn na h c th i nnging mu nhau.

    Mu h ri khng c h li. bi s 7, iu kin nykhng c.

    Nh phn tch trn, ta thy rng c th p dng chinthut c vi modulo 101, nhng v iu kin khng cho phpmu h ri c dng li nn chin thut gp tr ngi! V ngichi u tin (ng cui hng) sau khi chn 1 mu nn(theo modulo), th kh nng trng vi mu ca mt trong nhngngi ng trc l rt cao, ln n 99/101, nn c th ni rnggn nh s c mt ngi tuy bit r mu nn ca mnh, nhngkhng th h! V chng ta gp b tc y!

    Konstantin Knop, tc gi ca bi ton, a ra gi nh sau:Hy ngi chi khng may mn h ln mu nn ca ngixp u hng (ngi khng thy nn ca ai). Lc ny, tt cnhng ngi chi cha on mu nn ca mnh nu thy cngi no (tr ngi u tin) on mu ca ngi uhng, s bit rng l k "khng may mn". Nhng ngi tiptheo, cn c vo s tnh c mu nn ca mnh (v khngtrng li na), v ch c ngi ng u hng (ngi on cuicng) l on sai, do mu nn ca anh ta b s dng.

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    Nh vy vi chin thut ny, khng qu 3 ngi phi hi sinh cu mi ngi khc. Trong s ny, khng phi ai cng l"anh hng", v ngi cui hng (ngi on u tin) bit munn ca mnh vi xc sut 1/2. Bng cch thng bo nhng gmnh thy, anh ta hi sinh c hi ca mnh. Hai ngi cn li"ra i" trc c khi h phi on.

    Liu phng n trnh by trn l ti u? Cu tr li l cha.Bn c hn s t cu hi v sao chng ti (v c tc gi cabi ton) bn v phng n ny, mc d n cha phi li giicn tm? n gin l v n p v n c th gii quyt tng qutvi M > N nn.

    i n n li gii ti u cho bi ton ny, chng ti xin mic gi ri xa modulo m chuyn sang mt phng thc khc:s dng hon v!

    Hy tng tng, ta c thm mt ngi chi "o", i chic nncn li v ng sau ngi cui hng. Nh vy, ton b cch inn tr thnh hon v ca 101 s!

    Lm th no ngi chi (khng o) u tin c th truynti thng tin cho cc ngi khc on ra hon v ca tt c bnh? Trc mt anh ta l 99 ngi, nn ch c 2 hon v c thc khi quan st t v tr ny. Vi ch mt bit thng tin (ng vi2 hon v) nh vy, lm sao h c th chin thng? Cu tr lil anh ta s chn mu sao cho hon v tng ng l mt honv chn (even permutation)2. Cn c vo nhng ngi chitip theo s on ra mu nn ca mnh, nm trong s 2 mum h cha nghe thy hoc cha quan st thy. Nh vo honv chn, h s chn ng mu nn ca mnh! Mt li gii ngin v tuyt p!

    Chng ta hy cng xt v d c th nh sau: Gi s c 3 ngichi v 4 nn:

    Gi s ngi ng cui hng thy nn ngi ng utin l 2 v ngi tip theo l 4, ngha l cc hon v cth c s c dng [* * 4 2]. C th, c 2 hon v c th c

    2Hon v chn (even permutation) l mt hon v c s lng nghch th(inversion), tc s cp m s ln xut hin trc s nh, l chn. V d[1 3 4 2] c 2 nghch th l (3, 2) v (4, 2) nn l hon v chn.

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    l [1 3 4 2] v [3 1 4 2]. V [1 3 4 2] l hon v chn ([3 1 42] l hon v l), nn ngi u tin s h mnh c mu 3(v gn 1 cho ngi "o").

    Lc ny, ngi tip theo c thng tin cho anh ta l [* 3 *2], nn c 2 hon v tng ng l [1 3 4 2] v [4 3 1 2], nnanh ta s chn [1 3 4 2] do [1 3 4 2] l hon v chn v [4 31 2] l hon v l (c 5 nghch th). V vy, anh ta h mu4 cho mnh. V y l kt qu ng.

    Ngi cui cng, lng nghe thng tin t 2 ngi trc, nnc [* 3 4 *] v c 2 hon v anh ta c th chn l [1 3 42] v [2 3 4 1]. y mt ln na anh ta s chn [1 3 42] v l hon v chn ([2 3 4 1] c 3 nghch th, nn lhon v l) nn s h mu 2 cho mnh.

    Vy theo chin thut ny, chc chn N 1 ngi sau s onng mu nn ca mnh v ngi u tin c 2 kh nng chn nn kh nng chin thng cho tt c mi ngi l 1/2.

    Tip theo, kt thc cho chuyn mc "Cc bi ton i nn",chng ti trnh by phn li gii cho mt trng hp rt kh vibi ton v hn nn, trong phn ln phi s dng n s htr ca my tnh.

    3. Bi ton v hn nn

    Trong phn ny, chng ti tham kho v gn nh dch li phnln nhng ghi ch t bi ton #1179 trong chuyn mc Cu hng tun ca nh ton hc Stan Wagon (i hc Macalester).Nh 2 phn trc, chng ti ng li bi y c gitin theo di.Bi ton ny c ra bi Lionel Levine (i hc Cornell) vonm 2011 nh sau: Bn ngi cng tham gia mt tr chi onnn nh sau: Ngi dn tr s i v hn cc nn c mu trnghoc en ln u mi ngi vi xc sut nn trng v en l nhnhau v bng 12 . Cc nn ca mi ngi c nh s ln lt1, 2, . . . Mi ngi chi ch thy c ton b nn ca 3 ngikhc nhng nn ca mnh th h khng thy.

    Mi ngi s c pht mt t giy v h c php ghi vo mt con s, ng vi ch s ca nn ca h m h on l muen. Sau khi nhn tr li, ngi dn tr s kim tra s c

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    ghi trn giy ca mi ngi.

    Nu c 4 ngi cng on ng (tc l 4 ngi u ghi c con sng vi nn mu en ca mnh), h thng tr chi, ngc li, chcn mt ngi on khng ng, h thua. Bn ngi c tholun trc chin thut trc khi chi v khng c bt k trao ino sau . H cng khng bit c thi im m nhng ngikhc a giy cho ngi dn tr. Hy tm chin thut xc sutthng l cao nht.

    V d: h u ghi s 2015 vo cc mnh giy. Khi , c hi chinthng s l 1/16 v xc sut nn th 2015 ca mi ngi l 1/2.

    3.1. Chin thut New Zealand

    Li gii u tin trnh by y c Stan Wagon gi l chinthut New Zealand (v i bng bu dc ca New Zealand ctn v biu tng l ALL-BLACKS). Chin thut nh sau:

    Ngi th j s tm ra ch s #j nh nht sao cho ton b nn th#j ca tt c nhng ngi quan st c u c mu en. Nginy s on mu nn th #j ca mnh cng l mu en.

    Chin thut ny s thnh cng nu ti v tr no tt c ui nn en v tht bi nu nh c 1 ngi i nn trng. Vc n kh nng c 1 ngi i nn trng v 1 kh nng tt cu i nn en, nn xc sut thnh cng ca chin thut sl 1/(n+ 1).

    Hy th phn tch cho trng hp n gin vi 2 ngi chi:ta thy nu cc nn u tin c mu l en-en, h s thnhcng. Nu c mu Trng-Trng h s i ln cp tip theo, nu

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    l en-Trng hoc Trng-en, h s tht bi. Nh vy xc sut chin thng l 1/3.

    Stan Wagon, sau nhn ra rng nu nh chin thut l tmra ch s #j nh nht trong c ng k nn trng cng sthnh cng vi xc sut 1/(n + 1) vi k n bt k. Chin thutNew Zealand vi m rng ny ng ngha vi k = 0. T tngny kh gn vi phn 1 ca bi trong chin thut i ph vingi dn tr "xo quyt".

    y c phi l chin thut tt nht? Cu tr li l cha phi,nhng y l mt chin thut n gin c hiu qu rt tt. TheoStan Wagon, vi n = 2 v n = 3, hiu qu ca chin thut c thso snh vi cc chin thut phc tp hn. Tuy nhin vi n tngqut, chin thut c a ra bi Peter Winkler s thnh cngvi xc sut xp x 1/log(n), khi n ln s vt tri chin thutNew Zealand.

    3.2. Chin thut Winkler (2011)

    Chin thut ny nh sau: tt c ngi chi s thng nht vinhau mt s t v mi ngi s on da trn 2 tiu ch sau y:

    Tiu ch (A). Trong s t nn u tin ca mi ngi uc t nht mt nn en.

    Tiu ch (B). Tng cc ch s ca nn en u tin trnu mi ngi chia ht cho t.

    Ngi chi th j quan st thy ton b n 1 nn en u tinca nhng ngi chi khc nn khi c t v p dng 2 tiu chtrn, anh ta s nhanh chng tnh ra c ch s nn en utin ca mnh. V l con s duy nht trong khong t 1 n t.

    Xt tiu ch (A) vi n = 1000 ngi chi v gi tr t = 13, khi xc sut trong 13 nn u c nn en xut hin ca mt ngil 1 (1/2)13 v do vy xc sut c n ngi u c nn en trongs t nn u s l (1 (1/2)13)1000 = 0.885, ngha l xp x 88.5%.Xc sut ny s tin dn n xp x 100% khi t tng.

    Xt tiu ch (B), xc sut tng tt c cc ch s ca nn enu tin chia ht cho t l 1/t.

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    Nh vy, chin thut Winkler ra s thnh cng khi c (A) v(B) u ng, ngha l xc sut thnh cng s l:(

    1 12t

    )n1

    t

    nu ta xt (A) v (B) nh 2 bin c c lp (mc d thc t (A)v (B) khng hon ton c lp).

    tm cc tr cho (1 12t)n 1t , ta xt o hm:

    BBt

    ((1 1

    2t)n

    t

    )=(1 2t)n(ntlog(2) 2t + 1)

    (2t 1)t2

    V t 0 v n 0 nn o hm bng 0 khi:ntlog(2) 2t + 1 = 0 2t = ntlog(2) + 1

    Ta c khai trin MacLaurin ca 2t l:

    2t =n=0

    tnlogn(2)

    n!

    Gi tr t tha mn s xp x t log2(2n 2).

    Cu hi t ra l vichin thut nh vy,xc sut chin thng lbao nhiu? V gi tr tcn chn l bao nhiu? chin thut Winklerthnh cng vi xc sutcao nht, cn tm ra gitr t sao cho xc sut cac (A) v (B) ng xy rat gi tr cc i.

    y, tiu ch (B) cn quan tm cao hn v c kh nng gi trti u s t c khi tng cc ch s u tin c gi tr khc0 (mod t), ngha l xc sut ti u c th xy ra khi tng nyng d vi s (mod t) 0 no . Chin thut Winkler v vy trthnh tm cp gi tr (t, s) sao cho gi tr xc sut t cc ivi u vo n cho trc. Ta c th tm ra gi tr ny thng quathut ton nh sau:

    Gi F(n, t, s) l s cch i nn cho n ngi vi t nn utin sao cho c 2 tiu ch (A) v (B) u tha. Khi , xcsut thnh cng chnh l F(n, t, s)/2tn.F c th xy dng quy nh sau:

    F[1, t, 0] = 1, lu , y cn hiu l s = t, v vi n = 1th ch c 1 kh nng l nn v tr t l nn c u tinc mu en.

    F[1, t, s] = 2ts, v ch cn nn v tr s l nn u tinmu en, cc nn sau c th c mu ty .

    F[n, t, s] = tk=1 2(tk) F[n - 1, t, modulo(s - k, s)], y66

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    ch l quy da trn kt qu bc trc.

    Mt s kt qu tnh ton vi chin thut Winkler:

    n = 2 v n = 3: chin thut Winkler khng tt hn chinthut New Zealand. C th xc sut ti u cho n = 2 lF(2, 2, 0)/16 = 5/16 1/3 v n = 3 l F(3, 3, 0)/512 = 121/512 1/4.

    n = 4, chin thut Winkler nhm ti s = 0 khng nhnhhn New Zealand, nhng vi kt qu ti u t = 4 v s = 1,xc sut l 421/2048, hay 0.2056, khong 3% nhiu hn1/5.

    Vi n ln, chin thut Winkler vt tri chin thut NewZealand. Vi n = 1000, (t, s) ti u l (13, 10) xc sut tnhc xp x 0.068, tc l 68 ln tt hn xc sut 1/1001ca chin thut New Zealand.

    3.3. Nhng tng mi

    Chin thut Winkler mc d cho ra kt qu rt tt vi n ln, tuynhin, y vn cha phi l chin thut tt nht. Nm 2014,Larry Carter, J-C Reyes, v Joel Rosenberg (San Diego) xut ra hai chin thut mi nh sau:

    Chin thut sa sai3. tng ca chin thut ny nh sau:nu nh mt ngi nhn ra rng c ng i no ca mnhs tht bi nu tun theo chin thut c bn ca Winkler, lc ngi ny s nhn ra rng h s tht bi nu cng nhctun theo chin thut gc, v vy thay v p dng trong on t1 n t chin thut Winkler, ngi ny s sa sai bng cch pdng chin thut trong on t t+1 n 2t. V nh vy, nu mingi u nhn ra khng th p dng chin thut trong on1 n t, h s sa sai bng cch nhy ln mt on. Chinthut ny vn s tht bi trong trng hp c ngi chi khngth tm thy li. V d nh ngi A c ton b t nn u umu trng, lc A s khng th nhn ra l cn phi sa li.Kim tra bng my tnh, chin thut ny tht s tt hn chinthut Winkler, vi (n = 4, t = 4, s = 1), chin thut nng xc sutthnh cng t 0.205566 ln 0.210090.

    3Thut ng gc l The reset strategy, y chng ti gi l sa sai datrn tng ca chin thut.

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    Chin thut hon v. tng c bn l thay v tun theo chinthut (B) tnh tng ton b nhng nn en u tin ca ngi th h s tnh cc tng t cc hon v khc nhau. iu nybuc phi s dng thm gi nh l ngi chi ngm nh mingi c th t khc bit. S dng tr gip ca my tnh, StanWagon cho ra kt qu vi n = 3, t = 4 v hon v (342), kt qut c l 263/1024 = 0.256, cao hn 1/4 ca chin thut NewZealand v 121/512 = 0.236 ca chin thut Winkler. Vi chinthut sa sai, kt qu l 0.259.

    J-C Reyes v Joel Rosenberg tip tc xut mt chin thutmi t kt qu tt hn cho trng hp n = 3. Stan Wagon cngkhng gii thch c v sao phng php ny c th t hiuqu cao nh vy! Tuy nhin, kim chng bng my tnh chothy kt qu tht s tt hn. C th l xc sut vi n = 3 l0.2617, cao hn ton b cc chin thut c gii thiu tnhn y.Phng php ny da trn 3 ma trn kch thc 8 8 c xydng trc:

    2, 1, 4, 3, 5, 6,8, 7, 1, 3, 2, 5,4, 7, 6, 8, 3, 2,1, 4, 6, 5, 7, 4,5, 4, 6, 2, 1, 3,1, 2, 4, 5, 7, 1,3, 2, 8, 6, 7, 6, 5,8, 2, 1, 4, 3, 6, 8,3, 7, 1, 4, 1, 2, 8,7, 2, 6, 5, 6, 2, 1.

    Ma trn A

    2, 1, 3, 5, 4, 6,7, 8, 1, 3, 2, 4,6, 5, 8, 7, 3, 2,1, 1, 5, 4, 6, 6,5, 4, 1, 3, 2, 7,6, 5, 4, 5, 6, 2,1, 3, 7, 7, 7, 6, 4,8, 3, 1, 2, 5, 6, 8,5, 7, 2, 2, 3, 1, 8,7, 8, 6, 1, 3, 1, 2.

    Ma trn B

    2, 1, 3, 5, 4, 6,7, 8, 1, 3, 2, 4,6, 5, 8, 7, 4, 2,1, 6, 5, 8, 8, 6,4, 5, 4, 2, 1, 7,6, 7, 5, 4, 6, 1,2, 3, 6, 1, 6, 7, 5,3, 8, 2, 1, 4, 8, 6,7, 6, 3, 1, 4, 2, 7,8, 4, 5, 2, 4, 2, 1.

    Ma trn C

    Chin thut p dng nh sau: ngi A s tm nn en nh nhtca B v C, sau s chn gi tr v tr tng ng ma trn A.V d nh A thy ngi B c nn u tin l mu en v ngiC c nn u tin mu en l nn th 5. Khi , A s chn s v tr (1, 5) ma trn A, ng vi gi tr 5. B v C cng lm tngt vi ma trn ca mnh. Chin thut s chc chn sai khi cmt ngi c nn en u tin nm ngoi 8 v tr u tin. Lu rng cc s trn dng v ct cc ma trn ny khng phi lhon v, nn tng kh nng ca cc ma trn l 8364 = 10173.

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    Chin thut Winkler vi (n = 3, t = 3, s = 0) tng ng vi chinthut ny vi ma