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The Lightest Higgs Boson Mass in the Minimal Supersymmetric Standard Model

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The lightest Higgs boson mass in the Minimal Supersymmetric Standard Model ?J.A.Casas 1,2,J.R.Espinosa 2?,M.Quir′o s 2and A.Riotto 2?1CERN,TH Division,CH-1211Geneva 23,Switzerland 2Instituto de Estructura de la Materia,CSIC Serrano 123,28006-Madrid,Spain Abstract We compute the upper bound on the mass of the lightest Higgs boson in the Minimal Supersymmetric Standard Model in a model-independent way,includ-ing leading (one-loop)and next-to-leading order (two-loop)radiative corrections.We ?nd that (contrary to some recent claims)the two-loop corrections are nega-tive with respect to the one-loop result and relatively small (<～3%).After de?ning physical (pole)top quark mass M t ,by including QCD self-energies,and physical Higgs mass M H ,by including the electroweak self-energies Π M 2H ?Π(0),we obtain the upper limit on M H as a function of supersymmetric parameters.We include as supersymmetric parameters the scale of supersymmetry breaking M S ,

the value of tan βand the mixing between stops X t =A t +μcot β(which is re-sponsible for the threshold correction on the Higgs quartic coupling).Our results do not depend on further details of the supersymmetric model.In particular,for M S ≤1TeV,maximal threshold e?ect X 2t =6M 2S and any value of tan β,we ?nd M H

≤140GeV for M t ≤190GeV.In the particular scenario where the top is in its infrared ?xed point we ?nd M H ≤86GeV for M t =170GeV.

CERN–TH.7334/94

July 1994

CERN–TH.7334/94IEM–FT–87/94hep-ph/9407389

1Introduction

There are good reasons to believe that the Standard Model(SM)is not the ultimate theory since it is unable to answer many fundamental questions.One of them,why and how the electroweak and the Planck scales are so hierarchically separated,has mo-tivated the proposal of the Minimal Supersymmetric extension of the Standard Model (MSSM)as the underlying theory at scales of order1TeV.Indeed,the supersymmet-ric scale cannot be very large not to spoil the solution to the hierarchy problem,but the ambiguity about its maximal allowable size makes it very di?cult to give model–independent predictions testable in accelerators.

Fortunately,the supersymmetric predictions for the upper bound on the lightest Higgs boson mass represent a(perhaps unique)exception to this rule.Therefore,they are crucial for the experimental veri?cation of supersymmetry.This importance is reinforced by the fact that uncovering the Higgs boson of the SM is one of the main challenges for present(LEP,Tevatron)and future(LEP-200,LHC)accelerators.Much work has been recently devoted to this subject[1–10],but still there is a substantial disagreement on the?nal results(see e.g.[5,9,10]),especially concerning the two–loop corrections(which,as we will see,are crucial).

The aim of this paper is to evaluate the two-loop predictions of the MSSM on the lightest Higgs mass in a consistent and model-independent way.In fact,our results will not depend on the details of the MSSM(e.g.the assumption or not of universality of the soft breaking terms,gauge and Yukawa uni?cation,etc.).Actually,when we will refer to the MSSM,we will simply mean the supersymmetric version of the SM with minimal particle content.The consistency of the approach will allow us to show up the reasons of the disagreement between previous results.We also give the predictions on the Higgs mass for a particularly appealing scenario,namely the assumption that the recently detected top quark[11]has the Yukawa coupling in its infrared?xed point.It will turn out that in this case the Higgs boson should be just around the corner.

Let us brie?y review the status of the subject and the most recent contributions on it.The MSSM has an extended Higgs sector with two Higgs doublets with opposite hypercharges:H1,responsible for the mass of the charged leptons and the down-type quarks,and H2,which gives a mass to the up-type quarks.After the Higgs mechanism there remain three physical scalars,two CP-even and one CP-odd Higgs bosons.In particular,the lightest CP-even Higgs boson mass satis?es the tree-level bound

m2H≤M2Z cos22β,(1.1) where tanβ=v2/v1is the ratio of the Vacuum Expectation Values(VEV’s)of the neutral components of the two Higgs?elds H2and H1,respectively.The relation (1.1)implies that m2H

(RG)techniques[4].All methods found excellent agreement with each other and large

radiative corrections,mainly controlled by the top Yukawa coupling,which could make

the lightest CP-even Higgs boson to escape experimental detection at LEP-200.In particular,the RG approach(which will be followed in the present paper)is based on

the fact that supersymmetry decouples and below the scale of supersymmetry breaking M S the e?ective theory is the SM.Assuming M2Z?M2S the tree-level bound(1.1)is saturated at the scale M S and the e?ective SM at scales between M Z and M S contains

the Higgs doublet

H=H1cosβ+iσ2H?2sinβ,(1.2) with a quartic coupling taking,at the scale M S,the(tree level)value

λ=

1Strictly speaking this has been proven[7]for a theory with a single mass scale.

on the validity of the perturbative expansion for V.In particular,although the whole e?ective potential is scale-invariant,the one-loop approximation is not,thus one has to be careful about the choice of the renormalization scale.As we will see this fact is at the origin of some misunderstandings in previous works.Furthermore,since the Higgs mass is computed to one-loop order there are a number of one-loop e?ects that need to be considered for the consistency of the procedure:

i)As it is explained in Appendix A,the tree level quartic coupling

(1.3)

receives

one-loop threshold contributions at the M S scale.These are given by

?λ=3h4t

M2S 2?X2t

MS renormalization scheme.We also comment on the introduction of the Higgs sector both in the e?ective potential and in the Higgs boson self-energy.

2The one-loop e?ective potential

Our starting point in this section will be the e?ective potential of the Standard Model. This can be written in the’t Hooft-Landau gauge as[8]

V eff(μ(t),λi(t);φ(t))≡V0+V1+···,(2.1) whereλi≡(g,g′,λ,h t,m2)runs over all dimensionless and dimensionful couplings and V0,V1are respectively the tree level potential and the one-loop correction,namely

V0=?1

λ(t)φ4(t),(2.2)

V1=1

μ2(t)?

MS renormalization scheme[12].The parametersλ(t)and

m(t)are the Standard Model quartic coupling and mass,running with the RGE,while the running Higgs?eld is

φ(t)=ξ(t)φc,(2.4)φc being the classical?eld and

ξ(t)=e? t0γ(t′)dt′,(2.5) whereγ(t)is the Higgs?eld anomalous dimension.V1in eq.(2.3)contains the radiative corrections where only the top quark and the W,Z gauge bosons are propagating in the loop.For the moment we will disregard those coming from the Higgs and Goldstone boson propagation2.They can be easily introduced,as shown in Appendix B,without altering the numerical results that are obtained in this paper.The mass parameters in (2.3)are given by

m2W(t)=

[g2(t)+g′2(t)]φ2(t),

m2t(t)=

2In particular,the Goldstone boson contribution to the e?ective potential generates,on the running Higgs mass,an infrared logarithmic divergence.This divergence is cancelled when the physical Higgs mass is considered,as will be shown in Appendix B.

where g,g′and h t are the SU(2),U(1)and top Yukawa coupling,respectively.Finally the scaleμ(t)is related to the running parameter t by

μ(t)=μe t,(2.7) whereμis a?xed scale,that we will take equal to the physical Z mass,M Z.

The complete e?ective potential V eff(μ(t),λi(t);φ(t))and its n?th derivative are scale-independent(see e.g.ref.[8]),i.e.

dV(n)

eff

V eff(μ(t),λi(t);φ(t)).(2.9)

?φ(t)n

The above property allows in principle to?x a di?erent scale for each value of the classical?eldφc,i.e.μ(t)=f(φc)or,equivalently,t=t(φc)=log(f(φc)/μ).In that case V eff becomes V eff(φc)≡V eff(f(φc),λi(t(φc)).Then,from(2.8)and(2.9)one can readily prove that

V(n)

eff t=t(φc)=d n V eff(φc)

3E.g.in studies of the e?ective potential stability[8]to control large logarithms that can arise for large?eld values.It has also been used in ref.[9],but as we will see,it is an unnatural complication in this context.

iv)The last,but not the least,drawback is that once one has chosen a particular function t(φc),one looses track of scale invariance and so there is no way of checking how good the approximation is at the minimum of the e?ective potential. For the above mentioned reasons we will keep t andφc as independent variables.We will minimize the e?ective potential(2.1),truncated at one-loop,at some?xed scale t=t?,i.e.

?V eff

φ(t)

ξ(t?)

.(2.13) Accordingly,we must impose

v= φ(t Z) = φ(t?)

ξ(t Z)

2Gμ)?1/2=246.22GeV,(2.15) is the“measured”VEV for the Higgs?eld5[16].

We can trade φ(t?) by m2(t?)from the condition of minimum(2.11),which trans-lates,using(2.14),into the boundary condition for m2(t):

m2(t?)

2λ(t?)ξ2(t?)+

g4(t?) log g2(t?)ξ2(t?)v23

4μ2(t?)?

2μ2(t?)?1

(2.16)

We can de?ne now m H (t ?)as the second derivative of the e?ective potential at the minimum,evaluated at the scale t ?,i.e.

m 2H (t ?)=?2V eff

64π2ξ2(t ?)v 2 g 4(t ?) log g

(t ?)ξ2(t ?)v 2

4μ2(t ?)+2

2μ2(t ?) ,(2.17)

where we have used (2.16).At an arbitrary scale t we can use the scale independence property of the whole potential,see (2.9),and write

m 2H (t )=m 2H (t ?)ξ

(t ?)

?φ(t )2

φ(t )= φ(t ) =?m 2(t )+3

8π2 14μ2(t )+1

4μ2(t )+1 ?h 4t (t ) 3log h 2t (t )ξ2

(t )v 2

a)The one-loop case,whereβandγ–functions are considered to one-loop and the

e?ective potential is approximated by the tree level term(2.2).In this case the leading logarithms are resummed to all-loop in the e?ective potential.The Higgs mass includes then all-loop leading logarithm contributions[6,7].

b)The two-loop case,whereβandγ–functions in the RGE are considered to two-loop

order and the e?ective potential is considered in the one-loop approximation,(2.2) and(2.3).In this case the leading and next-to-leading logarithms are resummed to all-loop in the e?ective potential.The Higgs mass includes then all-loop leading and next-to-leading logarithm contributions[7].

We will impose as boundary conditions:

?For the gauge couplings:

g(M Z)=0.650,

g′(M Z)=0.355,

g3(M Z)=1.23.

(3.1)

?For the top Yukawa coupling:

h t(m t)=√

,(3.2)

where m t is the running mass:m t(μ(t)=m t)=m t.?For the quartic coupling:

λ(M S)=1

16π2 2X2t6M4S

.(3.3)

?For the mass m2(t)we will take as boundary condition the value determined by eq.(2.16)at the scale t?.

Note that the previous boundary conditions depend on the values of m t,t?(i.e.the minimization scale)and the supersymmetric parameters M S,X t and tanβ.

Our main task now will be to determine the optimal minimization scale t?as a function of m t.Our criterion will be that t?be in the region where the e?ective potential is more scale independent.We estimate this in the following way.For?xed t?and m t,all boundary conditions(3.1)–(3.3)are determined and the e?ective potential (2.1)can be computed for any value of t.Thus the corresponding value ofφ(t),say φmin(t),that minimizes the one-loop potential can be numerically evaluated.Had we considered the whole e?ective potential,its scale independence property(2.12)would

imply thatφmin(t)/ξ(t)= φ(t) /ξ(t),where φ(t) has been de?ned in(2.13),is t–independent.Since the one-loop approximation is not exactly scale independent the most appropriate scale is the scale t s for whichφmin(t)/ξ(t)has a stationary point,i.e.

ξ(t) t=t s=0.(3.4) Of course the scale at which(3.4)occurs depends on t?and m t,i.e.t s≡t s(t?,m t). Therefore,for a given value of m t the optimal scale t?is de?ned as

t s≡t s(t?,m t)=t?,(3.5) i.e.it is the scale which simultaneously de?nes the boundary condition(2.16)and agrees with the extremal of the functionφmin(t)/ξ(t).

This procedure is illustrated in Fig.1,where we plotφmin(t)/ξ(t)vs.μ(t)for m t=160GeV and supersymmetric parameters M S=1TeV,X t=0,cos22β=1. The solid curve shows the two-loop result whose stationary point satis?es eq.(3.5) and therefore de?nes the optimal scale t?.Note that there is no?ne tuning in the choice(3.5)since any point near the stationary point would be equally appropriate because the curveφmin(t)/ξ(t)is very?at in that region.We have also shown in Fig.1 the curveφmin(t)/ξ(t)in the one-loop approximation(dashed curve)withμ(t?)?xed by the two-loop result.We can see that the one-loop curve is much steeper than the two-loop curve,which means that the one-loop result,i.e.the tree level potential(2.2) improved by the one-loop RGE,is far from being scale independent at any scale.This feature has also been observed for the MSSM one-loop e?ective potential[17].

We plot in Figs.2a,b,c,d,μ(t?)as a function of m t for the di?erent values of su-persymmetric parameters M S=1,10TeV;X2t=0,6M2S.In all plots the solid curve corresponds to cos22β=1and the dashed curve to cos22β=0.

Once we have determinedμ(t?)for?xed values of m t and all supersymmetric pa-rameters,the Higgs running mass m H(t)is given by(2.18),while the mass de?ned as the second derivative of the e?ective potential m H,der(t)is given by(2.19).By de?nition both masses coincide at t?

m H(t?)=m H,der(t?),(3.6) and the ratio m H,der(t)/m H(t)should be equal to one for an exactly scale independent e?ective potential.Consistency of our procedure requires the curve m H,der(t)/m H(t)to be?at in the region where we minimize,i.e.at t?.In Fig.3we plot m H,der(t)/m H(t) for m t=160GeV and the values of supersymmetric parameters as in Fig.1,M S=1 TeV,X t=0,cos22β=1.We see that at the pointμ(t?)the curve is in its?at region, thoughμ(t?)does not exactly coincide with the extremum of m H,der(t)/m H(t).

Neither m H nor m t are physical masses.They are computed from the e?ective po-tential,i.e.at zero external momentum,and need to be corrected by the corresponding polarizations to obtain the propagator pole physical masses.This will be done in sec-tion4.For the time being,and just to compare with other results in the literature,we

will neglect the shift to the physical poles and de?ne the top mass by the boundary condition(3.2)and the Higgs mass by the usual condition

m H(μ(t)=m H)=m H.(3.7) In this way the Higgs mass m H can be unambiguously determined.We plot in Fig.4 m H as a function of m t for the values of supersymmetric parameters M S=1TeV, X t=0,cos22β=1.The thin solid line corresponds to the two-loop result and the dashed-line corresponds to the one-loop result.(We will disregard for the moment the thick solid line,which corresponds to shifting the running masses to propagator poles.) The main feature that arises from Fig.4is that the two-loop corrections are negative with respect to the one-loop result.This feature is in qualitative agreement with our previous two-loop result[5](where the one-loop corrections to the e?ective potential where not considered)and with others from di?erent authors[10].This comparison will be done in some detail in section5.

4Numerical Results

We will present,in this section,the numerical results on the Higgs mass,evaluated in the next-to-leading logarithm approximation,as a function of the top quark mass. The running top quark mass m t that we have been using in the previous section was evaluated at the scaleμ(t)=m t,i.e.m t(m t)=m t(see eq.(3.2)).However the running mass does not coincide with the gauge invariant pole of the top quark propagator M t. In the Landau gauge the relationship between the running m t and the physical(pole) mass M t is given by[14]

M t= 1+4π m t(M t).(4.1)

On the other hand,the running Higgs mass,m H(t),given by eq.(2.18),has a scale variation

dm2H(t)

The calculation ofΠ(M2H)?Π(0)is presented in Appendix B.We note that the scale dependence(at the one-loop level)of m H(t),as provided by(4.2),is cancelled by the scale dependence ofΠ(M2H)?Π(0).The only remaining scale dependence of M H comes from two-loop contributions.This feature is exhibited in Fig.5where we plot M H in the range ofμ(t)between M S and M Z for m t=160GeV and the supersymmetric parameters M S=1TeV,X t=0,cos22β=1.We can see that M H has a negligible variation(～0.5GeV)between M S and M Z.For the sake of comparison we have also plotted the running mass m H(t)whose variation is more appreciable(～2.5GeV).This e?ect is more accentuated for larger top masses.

We can see from(4.1)and from Fig.4that the e?ect of considering the pole mass M t is negative on the two-loop corrections to the Higgs mass,while the e?ect ofΠ(M2H)?Π(0)is positive(～2GeV for m t=160GeV and～5GeV for m t=215GeV)thus partially compensating the e?ect of(4.1).The global e?ect is negative and small as can be seen in Fig.4.The thick solid line indicates M H as a function of M t.It is below the thin solid line which was the two-loop evaluation of m H as a function of m t.The comparison with the one-loop result(dashed line)shows that two-loop corrections are negative with respect to the one-loop result.Numerically they are small(～1GeV) for M t=120GeV and larger(～6?7GeV)for M t=220GeV.

In Fig.6a,b,c,d we plot M H as a function of M t for values of supersymmetric parameters M S=1,10TeV;X2t=0,6M2S;cos22β=0,1.In all cases the solid curve corresponds to cos22β=1and the dashed curve to cos22β=0.Notice that the dependence on the mixing parameter X t is sizeable.In all the?gures we have used the lower limit on the top quark mass,M t>120GeV at95%C.L.,from the CDF dilepton channel[11].If we use the recent evidence for the top quark production at CDF with

a mass M t=174±10+13

?12GeV[11]and the bounds for supersymmetric parameters

M S≤1TeV and maximal threshold correction X2t=6M2S,we obtain the absolute upper bound M H<140GeV.

The dependence of M H on tanβis exhibited in Fig.7where we?x M t=170GeV and M S=1TeV.The solid curve corresponds to the absolute upper bound for the mixing X2t=6M2S and the dashed curve to the case of zero mixing.Concerning the dependence of M H on the stop mixing X t parameter(or equivalently the stop splitting), we have found it to be sizeable,as can be seen from Figs.6.

5Connection with other approaches

Some papers have recently appeared aiming to estimate the mass of the lightest Higgs boson up to next-to-leading order in the MSSM,and with apparently contradictory results.In this section we will comment on those papers in the context of our formalism and will exhibit the origin of the disagreements.

In ref.[9]the RG approach was used to estimate the mass of the lightest Higgs boson up to next-to-leading order.The relevant points of their calculation are the following:i) They considered the one-loop correction to the e?ective potential in the approximation

g=g′=0,ii)They neglected the wave-function renormalization leading to physical masses for the top quark and Higgs boson,iii)They minimized the e?ective potential at the scaleμ(t?)=v.In addition they only consider the case of zero stop mixing, X t=0.

Contrary to our results,they found that the two-loop correction is positive and sizeable with respect to the one-loop result.For instance in the typical case M S=1 TeV,X t=0,cos22β=1,they?nd for m t=170GeV the two-loop result m H～117 GeV,while we?nd for M t=170GeV the two-loop result M H～111GeV.We have

been able to trace back the di?erence between the two results to the points i)–iii) above.In particular,the authors of ref.[9]?nd for the same values of the parameters a positive two-loop correction with respect to the one-loop result～7GeV while our two-loop result is smaller than the one-loop result by～3GeV.To understand the origin of this discrepancy we have plotted in Fig.8the physical Higgs mass M H as a function of the minimization scaleμ(t?)for M t=170GeV,M S=1TeV and X t=0.Thick lines correspond to cos22β=1and thin lines to cos22β=0.Solid lines represent M H evaluated in the two-loop approximation and dashed lines in the one-loop approximation.Our chosen value ofμ(t?)is indicated in the?gure with an open diamond and that chosen in ref.[9],μ(t?)=v,with an open square.We can see that M H evaluated at two-loop is very stable againstμ(t?):it varies～5GeV in the whole interval M Z≤μ(t?)≤M S and～2GeV between the diamond and the square.However the value of M H evaluated at one-loop is very unstable.In fact the two-loop correction changes from negative atμ(t)=M Z(>～10%)to positive at μ(t)=M S(>～20%).In the region of our chosen value ofμ(t?)it is negative and～3 GeV while atμ(t?)=v it is positive and greater～7GeV.Fig.8shows that though our choice ofμ(t?)is more in agreement with perturbation theory thanμ(t?)=v the di?erence is however not important when plotting the physical Higgs mass(which was not considered in ref.[9]).

In ref.[10]the diagrammatic and e?ective potential approaches were used to evaluate the lightest Higgs mass at two-loop order in the framework of the MSSM.Various approximations,like g=g′=0in the e?ective potential,were used,and only the case of zero stop mixing,X t=0,was considered.Our results agree with those of ref.[10]within less than～3GeV.In fact for M t=150GeV,X t=0,cos22β=1 and M S=1TeV(M S=10TeV)ref.[10]?nds M H～107GeV(M H～110GeV) while we?nd from Figs.6a and6b M H～103GeV(M H～110GeV).We consider this agreement as satisfactory given the approximations used in ref.[10].The fact that two-loop corrections found in ref.[10]are sizeable with respect to the one-loop result can be explained as a consequence of the approximation used there to estimate the one-loop result.Taking M S=10TeV and M t=150GeV,they found m H=138GeV at one-loop using a simple approximation that takes into account only the leading part(～M4t log(M2S/M2t))of the corrections,as it is common practice in some phenomenological analysis.Actually,a slightly more so?sticated approximation(as the one labelled1βin their paper)or the one-loop result obtained by a numerical integration of the RGE(as

in our approach)gives m H～115GeV(both methods agree up to a di?erence～2?3 GeV due to subleading e?ects,such as those provided by considering gauge couplings in the e?ective potential)for the above mentioned values of the parameters.This has to be compared with the two-loop result M H～110GeV and shows that the net e?ect of two-loop corrections is indeed negative and small6.

Finally bounds on the lightest Higgs mass have been recently analyzed in ref.[18] in the context of models with Yukawa uni?cation.The results of ref.[18]are presented in the one-loop approximation.They choose M Z as the minimization scale and?nd larger bounds than in previous estimates.For instance it is found,for M t=170 GeV and any value of the supersymmetric parameters such that M S<1TeV,that M H<～102GeV.We have tried to reproduce their results using our formalism.In fact, their solution for small tanβis close to the?xed point solution where tanβand m t are related through[19]

m t=(196GeV)[1+2(α3(M Z)?0.12)]sinβ.(5.1) Now,?xing M S=1TeV and maximal mixing7X2t=6M2S we?nd,for M t=170GeV that tanβ=1.74from(5.1)and M H～99GeV if we?xμ(t?)=M Z,in agreement with the result of ref.[18].However this large value of M H is a clear consequence of the chosen minimization scale where two-loop corrections are very large.In fact forμ(t?)=M Z our two-loop result gives M H～85GeV.Moving to the region where perturbation theory is more reliable,as we have done along this paper,we would obtain the?nal two-loop result,M H～86GeV.We have plotted in Fig.9M H as a function of M t for M S=1TeV and tanβ,determined from(5.1),corresponding to the?xed point solution.The solid(dashed)line corresponds to the case of maximal mixing, X2t=6M2S(zero mixing,X t=0).

6Conclusions

We have computed in this paper the upper bounds on the mass of the lightest Higgs boson in the MSSM including leading and next-to-leading logarithm radiative correc-tions.We have used an RG approach by means of a careful treatment of the e?ective potential in the SM,assuming that supersymmetry is decoupled from the SM,which we have shown to be an excellent approximation for M S>～1TeV or even much less. Our results have covered the whole parameter space of supersymmetric parameters;in

fact,when we refer to the MSSM,we simply mean the supersymmetric version of the

SM with minimal particle content.We have also included QCD radiative corrections

to the top quark mass,which gives a negative contribution to the Higgs mass as a function of the physical(pole)top mass M t,and electroweak radiative corrections to

the Higgs mass M H,Π(M2H)?Π(0),providing a positive contribution to M H and par-tially cancelling the former ones.Therefore the balance of including the contribution of

QCD and electroweak self-energies to the quark and Higgs masses is negative and small

(<～2%for M t<200GeV).We also made a reliable estimate of the total uncertainty in the?nal results,which turns out to be quite small(～2GeV).

Concerning the numerical results,we have found in particular that for M t≤190 GeV,M S≤1TeV and any value of tanβwe have M H<120GeV for X t=0,and M H<140GeV for X2t=6M2S(maximal threshold e?ect).We have also applied our

calculation to a particularly appealing scenario,namely the assumption that the top has the Yukawa coupling in its infrared?xed point.Then,the bounds are much more stringent. E.g.for M t=170GeV,M S≤1TeV and X2t=6M2S,we?nd M H≤86 GeV.

Two papers have recently tried to incorporate radiative corrections to the Higgs

mass up to the next-to-leading order and with qualitatively di?erent results.In ref.[9], using the RG approach,positive,and large next-to-leading corrections with respect to the one-loop results were found.In ref.[10],using diagrammatic and e?ective potential methods in a particular MSSM as well as various approximations,it was found that two-loop corrections are also sizeable,but negative with respect to the one-loop result!

Using the RG approach,as ref.[9],we have found(unlike in ref.[9])that two-loop

corrections are negative with respect to the one-loop result.We have traced back the origin of this disagreement in their choice of the minimization scale.Furthermore the authors of ref.[9]neglected various e?ects(as the contribution of gauge bosons to the one–loop e?ective potential,or the wave function renormalization of top quark and Higgs boson)and considered only the case with zero stop mixing.

On the other hand,we have found that the abnormal size of the two-loop corrections obtained in[10]is a consequence of an excesively rough estimate of the one-loop result, but we are in agreement with their?nal two-loop result.In fact our two-loop results di?er from those of ref.[10]by less than3%.Also our results show a large sensitivity of the Higgs mass to the stop mixing parameter.

Finally we would like to comment brie?y on the generality of our results.As was already stated,we are assuming average squark masses M2S?M2Z,and that all supersymmetric particle masses are>～M S.If we relax the last assumption,i.e.if some supersymmetric particles were much lighter,the value of the quartic coupling at M S (see eq.(1.3))would be slightly increased and,correspondingly,our bounds would be slightly relaxed.We have made an estimate of this e?ect.Assuming an extreme case where all gauginos,higgsinos and sleptons have masses～M Z,we have found for M S=1TeV and cos22β=1an increase of the Higgs mass～2%.For values of tanβclose to one(as those appearing in infrared?xed point scenarios)the corresponding e?ect is negligible.On the other hand,our numerical results have been computed

for M S≥1TeV.For values of M S≤1TeV the bounds on the lightest Higgs mass are lowered.Hence,in this sense,all our results can be considered as absolute upper bounds.

Acknowledgements

We thank R.Hemp?ing and S.Peris for illuminating discussions.We also thank M. Carena,G.Kane,M.Mangano,N.Polonsky,S.Pokorski,A.Santamar′?a,C.Wagner and F.Zwirner for useful discussions and comments.

Appendix A

We comment here on the origin and size of threshold contributions to the quartic coupling of the SM,λ,at the supersymmetric scale.We also evaluate the e?ective D=6operators that are relevant for the Higgs potential,showing that for M S≥1TeV they are negligible.

As it is well known,the one-loop correction to the MSSM e?ective potential is dominated by the stop contribution

V MSSM 1=

Q2?

32π2 2X2t6M2S

h4t H4,(A.3)

where have rede?ned h t to be the usual top Yukawa coupling in the SM.This gives a threshold contribution to the SM quartic coupling

?λ=3h4t

M2S 2?X2t

Alternatively,eq.(A.4)can be obtained diagrammatically from two kinds of dia-grams exchanging stops(see the?rst paper of ref.[4]).Both methods give the same result.

It is worth noticing that(A.4)has a maximum for X2t=6M2S,which means that the threshold e?ect cannot be arbitrarily large.Note also that X2t=6M2S is barely consistent with the bound from color conserving minimum[13],so the case of maximal threshold represents an extreme situation.

Analogously,we can obtain from(A.1)the relevant e?ective D=6operators,i.e. those proportional to H6.These turn out to be

3M2S ? X2t3M6

? X6t

(?φ0)2?1

(?φR)2?12δZ H(?φR)2?1

whereΠ0(p2)is the(unrenormalized)self-energy of the Higgs boson?eld and in the last equality we have neglected higher order terms in the perturbative expansion.The renormalized self-energy is de?ned as

ΠR(p2)=Π0(p2)+δm2?δZ H p2.(B.3) Thus

ΓR(p2)=p2? m2R+ΠR(p2) .(B.4) Consequently,the physical(pole)mass,M2H,satis?es the relation

M2H=m2R+ΠR(p2=M2H).(B.5) On the other hand,the running mass m2H,de?ned as the second derivative of the renormalized e?ective potential(see eqs.(2.17)and(2.18)),is given by

?2V eff

m2H=

d log Z H

=?2γM2H(B.10)

d logμ

which cancels the scale dependence of m2H(see eq.(4.2)).

We now want to give the complete expression for the physical mass M2H.The quantity?Πin the Landau gauge is given by the sum of the following terms ?Π=?Πtt(top contribution)

+?ΠW±W?+?ΠZ0Z0

(gauge and Goldstone bosons contribution) +?ΠW±χ?+?ΠZ0χ

+?Πχ±χ?+?Πχ

3χ3

+?ΠHH(pure scalar bosons contribution).(B.11)

In eq.(B.11)we have taken into account only the contribution from the heaviest fermion,the top,and indicated by χ±and χ3the charged and the neutral Goldstone bosons,respectively.

The complete expression for the di?erent contributions to ?Πcalculated in the

8π2 ?2M 2

t Z M 2

t 2M 2H log M 2

t M 2H ?2 .

(B.12)

ii )Gauge bosons and Goldstone bosons contribution:

?ΠW ±W ?+?ΠZ 0Z 0+?ΠW ±χ?+?ΠZ 0χ3

=g 2M 2

W 2I (3)

W ?M 2H I (4)

+M 2

H 2I (6)

W ?1

16π2 I (7)

W (μ2)+I (8)

W (μ2)

+116π2I (9)W +

M 2W x (1?x )?i?

=Z M 2

W M 2W y (1?y )

12?1

M 2

H +1

M 2W Z M 2W 6(M 2

H ?M 2W )3

M 2W ,

I 3W = 10 10dxdy y 361

W 3M 2H M 2H M 2H +1

M 6H ?1 log 1?M 2

H M 2W (1?xy )?M 2

H y (1?y )?i?=1M 2H +

M 2H +1

M 4H ?1 log 1?M 2H

I 5W = 1

0 10

10dxdydz

z (1?z )316M 2W

4?M 2H M 2H +(M 2H ?M 2W )3

M 2W ?1M 4W log M 2H 6

M 2H [M 2W (1?y ?z (x ?y ))?M 2H z (1?z )?i?]2

=?2

M 4H +1M 2W

Z M 2W 3M 6H M 4W

log 1?

M 2H 3M 4W log 1?M 2H 3

M 2W M 2W +1M 2H +iπM 4W ,I 7W = 1

0dx (1+2x )log

M 2W x ?M 2H x (1?x )?i?μ2+M 2W M 4H ?3M 2W

M 2W ,

I 8W =

10dx log M 2W x ?i?μ2,I 9W = 10 10dxdy

M 2W ?1

M 2W +1M 2H +iπ

8π2 ?3+5

M 2W +1M 2W +M 4H M 2H ?M 4H M 2W

高难度的数独技巧窍门

-! 1 2 □ 4 D fi 1 8 9 A 8 1 3 B 7 6 1 Ei 9 A C 9 1 8 7 5 D 1 7 8 4 3 9 E 3 8 9 1 4 F 5 4 9 i 6 8 1 G 9 3 H 1 L 3 8 1 如左图，观察行 B ,我们发现除了 B3 单元格以外其余的八个单元格已经填入了 1、2、4、5、6、7、8、9,还有3没有填写，所以3就应该填入B3单元格。这是行唯一解法。 1 2 3 4 S 6 ? S 如左图，观察D7-F9这个九宫格，我们发现除了 E7单元格以外其余的八个单元格已经填入了 1、2、3、4、6、7、 & 9，还有5没有填写，所以5就应该填入E7单元格。这是九宫格唯一解法。 A 1 J R c D E F G E T fl 3 1. 5 B 2 41 1 3 1 ti 1 7 8 5 S 2 3 9 3 8 g 1 4 T 5 4 g T 2 3 0 a 1 2 3 1 6 1 3 e 1

! -单元唯一法在解题初期应用的几率并不高，而在解题后期，随着越来越多的单元格填上了数字, 使得应用这一方法的条件也逐渐得以满足。 △基础摒除法基础摒除法是直观法中最常用的方法，也是在平常解决数独谜题时使用最频繁的方法。单元排除法使用得当的话，甚至可以单独处理中等难度的谜题。使用单元排除法的目的就是要在某一单元（即行，列或区块）中找到能填入某一数字的唯一位置，换句话说，就是把单元中其他的空白位置都排除掉。那么要如何排除其余的空格呢？当然还是不能忘了游戏规则，由于1-9的数字在每一行、每一列、每一个九宫格都要出现且只能出现一次，所以：如果某行中已经有了某一数字，则该行中的其他位置不可能再出现这一数字；如果某列中已经有了某一数字，则该列中的其他位置不可能再出现这一数字；如果某区块中已经有了某一数字，则该区块中的其他位置不可能再出现这一数字。基础摒除法可以分为行摒除、列摒除和九宫格摒除。如左图，观察D1-F3这个九宫格。由于11 格有数字9, 所以第1列其它所有单元格都不能填入9;由于B2格有数字 9,所以第2列其它所有单元格都不能填入9 ；由于D8格有数字 9，所以行D其它所有单元格都不能填入9。这样，D1-F3这个九宫格内只有E3单元格能够填入数字9。所以E3单元格的答案就是9。如左图，观察行H。由于C3格有数字4, 所以第3列其他所有单元格不能填入数字4; 由于E8格有数字4，所以第8列其他所有单元格不能填入数字4;由于I4格有数字4，所以G4-I6这个九宫格内其他所有单元格不能填入数字4。这样行H中能够填入数字4的单元格只有H9。所以H9单元格的答案就是4。

The way常见用法

The way 的用法 Ⅰ常见用法： 1)the way+ that 2)the way + in which（最为正式的用法） 3)the way + 省略（最为自然的用法）举例：I like the way in which he talks. I like the way that he talks. I like the way he talks. Ⅱ习惯用法：在当代美国英语中，the way用作为副词的对格，“the way+ 从句”实际上相当于一个状语从句来修饰整个句子。 1)The way =as I am talking to you just the way I’d talk to my own child. He did not do it the way his friends did. Most fruits are naturally sweet and we can eat them just the way they are—all we have to do is to clean and peel them. 2）The way= according to the way/ judging from the way The way you answer the question, you are an excellent student. The way most people look at you, you’d think trash man is a monster. 3）The way =how/ how much No one can imagine the way he missed her. 4）The way =because

数独入门你必须掌握的那些规则和技巧

数独入门：你必须掌握的那些规则和技巧数独的规则在空格内填入数字1-9，使得每行、每列和每个宫内数字都不重复。注意：数独题目满足条件的答案是唯一的。

数独的元素数独的元素主要包括行、列和宫。这三者划分出数独有三种不同形态的区域，而数独规则就是要求在这些区域内出现的数字都为1~9。元素坐标图：行：数独盘面内横向一组九格的区域，用字母表示其位置；列：数独盘面内纵向一组九格的区域，用数字表示其位置；宫：数独盘面内3×3格被粗线划分的区域，用中文数字表示其位置。格的坐标：利用表示行位置的字母和表示列位置的数字定位数独盘面内每个格子的具体位置，如A3格，F8格等。数独技巧 1.?宫内排除法排除法就是利用数独中行、列和宫内不能填入相同数字的规则，利用已出现的数字对同行、同列和同宫内其他格进行排斥相同数字的方法。宫内排除法就是将一个宫作为目标，用某个数字对它进行排除，最终得到这个宫内只有一格出现该数字的方法。技巧示意图：

宫内排除法如上图所示，A2、B4和F7三格内的1都对三宫进行排除，这时三宫内只有C9格可以填入1，本图例就是对三宫运用的排除法。 2.?行列排除法行列排除法就是将一行或一列作为目标，用某个数字对它进行排除，最终得到这个行列内只有一格出现该数字的方法。技巧示意图：

行列排除法如上图所示，D2和B8两格内的6都对F行进行排除，这时F行内只有F5格可以填入6，本图例就是对F行运用的排除法。 3.?区块排除法区块排除法就是先利用宫内排除法在某个宫内形成一个区块，利用该区块的排除再结合其他已知数共同确定某宫内只有一格出现该数字的方法。技巧示意图：区块排除法如上图所示，B4格的7对五宫进行排除，在五宫内形成了一个含数字7的区块。无论该区块中F5格是7还是F6格是7，都可以对F行其他格的7进行排除。再结合H7格的7同时对六宫进行排除，得到六宫内只有D8格可以填7。 4.?宫内数对占位法数对占位法指的是在某个区域中使得某两数只能出现在某两格内，这时虽然无法判断这两个数字的位置，但可以利用两数的占位排斥掉其他数字出现在这两格，再结合排除法就可以间接填出下个数字。技巧示意图：

The way的用法及其含义(二)

The way的用法及其含义（二）二、the way在句中的语法作用 the way在句中可以作主语、宾语或表语： 1.作主语 The way you are doing it is completely crazy.你这个干法简直发疯。 The way she puts on that accent really irritates me. 她故意操那种口音的样子实在令我恼火。The way she behaved towards him was utterly ruthless. 她对待他真是无情至极。 Words are important, but the way a person stands, folds his or her arms or moves his or her hands can also give us information about his or her feelings. 言语固然重要,但人的站姿,抱臂的方式和手势也回告诉我们他(她)的情感。 2.作宾语 I hate the way she stared at me.我讨厌她盯我看的样子。 We like the way that her hair hangs down.我们喜欢她的头发笔直地垂下来。 You could tell she was foreign by the way she was dressed. 从她的穿著就可以看出她是外国人。 She could not hide her amusement at the way he was dancing. 她见他跳舞的姿势，忍俊不禁。 3.作表语 This is the way the accident happened.这就是事故如何发生的。 Believe it or not, that's the way it is. 信不信由你, 反正事情就是这样。 That's the way I look at it, too. 我也是这么想。 That was the way minority nationalities were treated in old China. 那就是少数民族在旧中

数独教案完整版

数独教案基本项目课程名称：感受数独魅力授课对象：三到六年级学生课程类型：逻辑思维课，选修课教学材料：自编纲要教学时间：一学期，每周1课时，共18课时具体教学方案一、指导思想数学是神奇的世界，肯定有不少学生产生了浓厚的兴趣。为此，训练学生的思维活动是重中之重。数学思维活动在数学教学课堂中探求问题的思考、推理、论证的过程等一系列数学活动都是数学教学中实施思维训练的理论依据之一。因此，开展校本数独课程，一是能更好的促进学生数学思维能力的发展，符合课改的要求；二是填补了我们课改中的弱项。二、教学目标 1、尊重学生的主体地位和主体人格，培养学生自主性、主动性，引导学生在掌握数学思维成果的过程中学会学习、学会创造。 2、将数学知识寓于游戏之中，教师适当穿针引线，把单调的数

学过程变为艺术性的游戏活动，让学生在游戏中学习在玩中收获。 3、课堂上围绕“趣”字，把数学知识容于活动中，使学生在好奇中，在追求答案的过程中提高自己的观察能力，想象能力，分析能力和逻辑推理能力。力求体现我们的智慧秘诀：“做数学，玩数学，学数学”。三、教学措施 1、结合教材，精选小学数学的教学内容，以适应社会发展和进一步学习的需要。力求题材内容生活化，形式多样化，解题思路方程化，教学活动实践化。 2、教学内容的选编体现教与学的辨证统一。教学内容呈现以心理学的知识为基础，符合儿童认知性和连续性的统一，使数学知识和技能的掌握与儿童思维发展能力相一致。 3、教学内容形式生动活泼，符合学生年龄特点，赋予启发性，趣味性和全面性，可以扩大学生的学习数学的积极性。 4、每次数学思维训练课都有中心，有讨论有交流有准备。有阶段性总结和反思。四、教学内容

(完整版)the的用法

定冠词the的用法：定冠词the与指示代词this ,that同源,有“那(这)个”的意思,但较弱,可以和一个名词连用,来表示某个或某些特定的人或东西. (1)特指双方都明白的人或物 Take the medicine.把药吃了. (2)上文提到过的人或事 He bought a house.他买了幢房子. I've been to the house.我去过那幢房子. (3)指世界上独一无二的事物 the sun ,the sky ,the moon, the earth (4)单数名词连用表示一类事物 the dollar 美元 the fox 狐狸或与形容词或分词连用,表示一类人 the rich 富人 the living 生者 (5)用在序数词和形容词最高级,及形容词等前面 Where do you live?你住在哪? I live on the second floor.我住在二楼. That's the very thing I've been looking for.那正是我要找的东西. (6)与复数名词连用,指整个群体 They are the teachers of this school.(指全体教师) They are teachers of this school.(指部分教师) (7)表示所有,相当于物主代词,用在表示身体部位的名词前 She caught me by the arm.她抓住了我的手臂. (8)用在某些有普通名词构成的国家名称,机关团体,阶级等专有名词前 the People's Republic of China 中华人民共和国 the United States 美国 (9)用在表示乐器的名词前 She plays the piano.她会弹钢琴. (10)用在姓氏的复数名词之前,表示一家人 the Greens 格林一家人(或格林夫妇) (11)用在惯用语中 in the day, in the morning... the day before yesterday, the next morning... in the sky... in the dark... in the end... on the whole, by the way...

数独入门：你必须掌握的那些规则和技巧

数独入门：你必须掌握的那些规则和技巧标准化文件发布号：（9312-EUATWW-MWUB-WUNN-INNUL-DDQTY-KII

数独的元素数独的元素主要包括行、列和宫。这三者划分出数独有三种不同形态的区域，而数独规则就是要求在这些区域内出现的数字都为1~9。元素坐标图：行：数独盘面内横向一组九格的区域，用字母表示其位置；列：数独盘面内纵向一组九格的区域，用数字表示其位置；宫：数独盘面内3×3格被粗线划分的区域，用中文数字表示其位置。格的坐标：利用表示行位置的字母和表示列位置的数字定位数独盘面内每个格子的具体位置，如A3格，F8格等。数独技巧 1. 宫内排除法排除法就是利用数独中行、列和宫内不能填入相同数字的规则，利用已出现的数字对同行、同列和同宫内其他格进行排斥相同数字的方法。宫内排除法就是将一个宫作为目标，用某个数字对它进行排除，最终得到这个宫内只有一格出现该数字的方法。技巧示意图：

宫内排除法如上图所示，A2、B4和F7三格内的1都对三宫进行排除，这时三宫内只有C9格可以填入1，本图例就是对三宫运用的排除法。 2. 行列排除法行列排除法就是将一行或一列作为目标，用某个数字对它进行排除，最终得到这个行列内只有一格出现该数字的方法。技巧示意图：

行列排除法如上图所示，D2和B8两格内的6都对F行进行排除，这时F行内只有F5格可以填入6，本图例就是对F行运用的排除法。 3. 区块排除法区块排除法就是先利用宫内排除法在某个宫内形成一个区块，利用该区块的排除再结合其他已知数共同确定某宫内只有一格出现该数字的方法。技巧示意图：区块排除法如上图所示，B4格的7对五宫进行排除，在五宫内形成了一个含数字7的区块。无论该区块中F5格是7还是F6格是7，都可以对F行其他格的7进行排除。再结合H7格的7同时对六宫进行排除，得到六宫内只有D8格可以填7。 4. 宫内数对占位法数对占位法指的是在某个区域中使得某两数只能出现在某两格内，这时虽然无法判断这两个数字的位置，但可以利用两数的占位排斥掉其他数字出现在这两格，再结合排除法就可以间接填出下个数字。技巧示意图：

“the way+从句”结构的意义及用法

“tｈｅwａｙ+从句”结构的意义及用法首先让我们来看下面这个句子: Ｒead the foｌloｗｉnｇpassagｅａnd talkａbout ｉt wi ｔh your classmaｔes．Try tｏｔeｌl whaｔyｏu thiｎk ｏf Tom aｎd ｏｆｔhe ｗay the chiｌｄrenｔrｅated ｈim. 在这个句子中，ｔhe waｙ是先行词,后面是省略了关系副词that或in whｉch的定语从句。下面我们将叙述“the ｗaｙ+从句”结构的用法。 1.tｈe wａy之后，引导定语从句的关系词是tｈat而不是hoｗ,因此，<<现代英语惯用法词典＞＞中所给出的下面两个句子是错误的：Ｔhｉs is ｔｈeｗayｈowｉtｈａppened. This is the way how he ａlwａｙs treats me. 2．在正式语体中,thaｔ可被in which所代替;在非正式语体中，ｔhat则往往省略。由此我们得到theｗay后接定语从句时的三种模式：1) tｈe ｗaｙ＋ｔhａt-从句２）the way +ｉn whiｃh－从句3) the ｗay +从句例如：Ｔｈe way(in whiｃh ，thａt) ｔhｅｓｅｃomｒade ｓloｏｋａｔｐrobｌems is wrong．这些同志看问题的方法

不对。Ｔｈｅｗａy(that ,ｉn ｗhiｃh)yoｕ’ｒe doinｇit is comple ｔeｌy crａzy.你这么个干法,简直发疯。 Wｅａdmiｒｅd him for thｅｗay ｉｎｗhｉch he fａｃeｓdiｆficultieｓ． Waｌlａｃe ａｎd Ｄarwiｎｇreed ｏn the way ｉｎwhi ｃh ｄｉfｆｅrｅnt forms ｏf life had ｂegｕn.华莱士和达尔文对不同类型的生物是如何起源的持相同的观点。 Thｉs is ｔｈe ｗaｙ（tｈａｔ) hｅdｉd it. I lｉkeｄｔhe waｙ(ｔhat) ｓｈeｏｒganized ｔhe ｍeetｉng． 3.theｗay(tｈat)有时可以与hｏw(作“如何”解）通用。例如： That’s tｈe ｗaｙ(tｈaｔ) shｅspoke. = Tｈat’s how shｅspoke.

数独高级技巧入门链的逻辑及

[数独高级技巧入门]链的逻辑及AIC 这个帖子主要想阐述链是什么，怎么使用链，以及链的逻辑过程，帮助大家首先了解原理，那么以后关于chain、wing之类的按照这个思路都非常容易理解。首先我想说明下什么是“强”关系，什么是“弱”关系强关系是说A与B两个事件，假如A不成立，则B一定成立。弱关系是说A与B两个事件，假如A成立，则B 一定不成立。举一个简单的例子帮助大家体会：（图中被划短横线的格表示不含候选数1）这是一个数独的宫，根据数独规则一个宫内出现数字1-9各一次，可以做出以下两点推断：1.左上格不是1，则右中格一定是1；2.左上格是1，则右中格一定不是1。第一种推断得到这两格的1是强关系，所以可以说两格之间形成一条强链，强链我们通常以双横线表示（==）；第二种推断得到这两格的1是弱关系，所以可以说两格之间形成一条弱链，弱链我们通常以单横线表示（——）。再举一个例子：（图中被划短横线的格表示不含候选数1）上图可以做出三大点推断：1.左上格是1，则中上格及右中格一定不是1；2.中上格是1，则左上格及右中格一定不是1；3.右中格是1，则左上格及中上格一定不是1。这个例子里，存在着3条弱链，分别是（左上--中上）、（左上--右中）、（中上--右中）。上面说的是同一数字的强弱关系，当然强弱关系可以不局限于一个数字，下面用例子来说明：（图中被短横线划掉的格说明未知其候选数情况）根据右上格的候选数仅有1与2可以做出以下推断：1.如果该格不能是1，则一定为2；2.如果该格是1，则一定不是2。推断一说明数字1与2之间是强关系，形成强链；推断二说明其为弱关系，形成弱链。（图中被短横线划掉的格说明未知其候选数情况）右上格有3个候选数，我们可以做出以下推断：1.如果这格为1，则不能为2或3；2.如果这格为2，则不能为1或3；3.如果这格为3，则不能为1或2。数字1与2、2与3、1与3之间分别为一条弱链。像第二张图这样的关系推断，大家可能会不以为意，但是这是理解强弱关系的一个很好的例子，对于后面将要叙述的内容也会有所帮助。相信通过上面的说明大家已经了解了强弱链是什么，接下来我们将强弱链连接起来。第一种情况：A==B--C==D由A的真假情况可以做出以下BCD关系的枚举。再次请大家注意本文开头所提到的强弱关系本质1.强关系是说A与B 两个事件，假如A不成立，则B一定成立。2.弱关系是说A与B两个事件，假如A成立，则B一定不成立。（图中红色部分表示根据上一个的真假情况必然是这样的推导）可见A与D不全为假，即A与D一定有一个为真。当A 与D有等位群格位的交集时，即可做出相应删减。（图示技巧名为Skyscraper）根据强弱关系，我们找到了一条符合 A==B--C==D的强弱链组：r3c1(2)==r3c7(2)--r9c7(2)==r9c2(2)。根据上文提到的逻辑关系，可以得到r3c1=2与r9c2=2至少有一个成立，所以可以删去它们等位群格位的交集（即橙色区域）的候选数2。补充说明：发现很多人对于第七列的画法存在疑问，为什么不标双线（强链），因为这里运用的是“是A非B”的弱关系，所以只能是标单线（弱链）的，关于“强强强” 的链接我们在后文提到是无法得到任何结论的。我们可以从强弱关系的逻辑把上述这条链走一遍，共有以下两种情形：1）r3c1=2；2）r3c1<>2->r3c7=2

way 用法

表示“方式”、“方法”，注意以下用法： 1.表示用某种方法或按某种方式，通常用介词in(此介词有时可省略)。如： Do it (in) your own way. 按你自己的方法做吧。 Please do not talk (in) that way. 请不要那样说。 2.表示做某事的方式或方法，其后可接不定式或of doing sth。如： It’s the best way of studying [to study] English. 这是学习英语的最好方法。 There are different ways to do [of doing] it. 做这事有不同的办法。 3.其后通常可直接跟一个定语从句(不用任何引导词)，也可跟由that 或in which 引导的定语从句，但是其后的从句不能由how 来引导。如：我不喜欢他说话的态度。正：I don’t like the way he spoke. 正：I don’t like the way that he spoke. 正：I don’t like the way in which he spoke. 误：I don’t like the way how he spoke. 4.注意以下各句the way 的用法： That’s the way (=how) he spoke. 那就是他说话的方式。 Nobody else loves you the way(=as) I do. 没有人像我这样爱你。 The way (=According as) you are studying now, you won’tmake much progress. 根据你现在学习情况来看，你不会有多大的进步。 2007年陕西省高考英语中有这样一道单项填空题： ——I think he is taking an active part insocial work. ——I agree with you_____. A、in a way B、on the way C、by the way D、in the way 此题答案选A。要想弄清为什么选A，而不选其他几项，则要弄清选项中含way的四个短语的不同意义和用法，下面我们就对此作一归纳和小结。一、in a way的用法表示：在一定程度上，从某方面说。如： In a way he was right.在某种程度上他是对的。注：in a way也可说成in one way。二、on the way的用法 1、表示：即将来(去)，就要来(去)。如： Spring is on the way.春天快到了。 I'd better be on my way soon.我最好还是快点儿走。 Radio forecasts said a sixth-grade wind was on the way.无线电预报说将有六级大风。 2、表示：在路上，在行进中。如： He stopped for breakfast on the way.他中途停下吃早点。 We had some good laughs on the way.我们在路上好好笑了一阵子。 3、表示：(婴儿)尚未出生。如： She has two children with another one on the way.她有两个孩子，现在还怀着一个。 She's got five children，and another one is on the way.她已经有5个孩子了，另一个又快生了。三、by the way的用法

数独技巧3

X翼删减法、剑鱼删减法 X翼删减法：两列只有两格可以填入6，且这4 剑鱼删减法与X翼删减法道理相同，由2列拓剑鱼删减法除了以上标准型（3-3-3，3列都有 3个候选数），还由一些

X翼删减法实例：剑鱼删减法实例：

Turbot Fish 删减法 1楼 Turbot Fish介绍之前做个简单的铺垫，简单介绍一下强弱链的关系。单链分为强链和弱链。强链：某行、列或宫只存在2个某候选数，这两个数就构成强链，两数非真即假。这里用红线连接表示。弱链：某行、列或宫存在3个或3格以上某候选数，这些数就构成弱链，其中一个为真则其余为假；其中一个为假则不能判断其余的真假。这里用蓝线连接表示。根据强链两端数字，一个为真另一个为假的特性可以引申出某些三条连续单链组有排除候选数的情况。 “强-强-强链”和“强-弱-强”链都可以导致“长链”两端数字交叉处格中的该数被删除。

下边给出两种“三连链”的图:（两图中“长链”形状可以互换）说明： “强-强-强链”由于链两端数非真即假的特性，标成红蓝两组，红为真则蓝为假，反之亦然。“长链”两端也为一红一蓝，肯定有一个是真，所以排除掉共同区域格（橙色格）中的x。 “强-弱-强链”虽然不像“三强”中数字真假那么分明，但注意弱链的两端，弱链一端为真另一端也为假，这两端的数字分别连接强链，所以导致“长链”两端数同样是一真一假。如果弱链两端均为假，则长链两端数都为真。综上：同样排除掉共同区域格（橙色格）中的x。所以，可以看出“强-强-强链”与“强-弱-强链”在排除两端数字交叉区域数字的效果上是“等价”的。

The way的用法及其含义(一)

The way的用法及其含义（一）有这样一个句子：In 1770 the room was completed the way she wanted. 1770年，这间琥珀屋按照她的要求完成了。 the way在句中的语法作用是什么？其意义如何？在阅读时，学生经常会碰到一些含有the way 的句子，如：No one knows the way he invented the machine. He did not do the experiment the way his teacher told him.等等。他们对the way 的用法和含义比较模糊。在这几个句子中，the way之后的部分都是定语从句。第一句的意思是，“没人知道他是怎样发明这台机器的。”the way的意思相当于how；第二句的意思是，“他没有按照老师说的那样做实验。”the way 的意思相当于as。在In 1770 the room was completed the way she wanted.这句话中，the way也是as的含义。随着现代英语的发展，the way的用法已越来越普遍了。下面，我们从the way的语法作用和意义等方面做一考查和分析：一、the way作先行词，后接定语从句以下3种表达都是正确的。例如：“我喜欢她笑的样子。” 1. the way+ in which +从句 I like the way in which she smiles. 2. the way+ that +从句 I like the way that she smiles. 3. the way + 从句(省略了in which或that) I like the way she smiles. 又如：“火灾如何发生的，有好几种说法。” 1. There were several theories about the way in which the fire started. 2. There were several theories about the way that the fire started.

数独的解法与技巧

数独的直观式解题技巧直观法概说前言数独这个数字解谜游戏，完全不必要用到算术！会用到的只是推理与逻辑。刚开始接触数独时，即使是只须用到"基础摒除法"及"唯一解法"技巧的简易级谜题，就已可让我们焦头烂额了，但是随着我们深陷数独的迷人世界之后，这类简易级的数独谜题必定在短时间内难再使我们获得征服的满足。于是，当我们逐步深入、进阶到更难的游戏后，我们将会需要发展出更多的解谜技巧。虽然最好的技巧便是我们自己发现的窍门，这样我们很容易就能记住它们，运用自如，不需要别人来耳提面命。但是如果完全不去观摩学习他人发展出来的技巧，而全靠自己摸索，那将是一个非常坚苦的挑战，也不是正确的学习之道！所以让我们一齐来探讨数独的解谜方法吧！数独的解谜技巧，刚开始发展时，以直观法为主，对于初入门的玩家来说，这也是一般人较容易理解、接受的方法，对于一般报章杂志及大众化网站上的数独谜题而言，如果能灵活直观法的各项法则，通常已游刃有余。直观法详说直观法的特性： 1.不需任何辅助工具就可应用。所以要玩报章杂志上的数独谜题时，只要有一枝笔就可以开始了，有人会说：可能需要橡皮擦吧答案是：不用！只要你把握数独游戏的填制原则：绝不猜测。灵活运用本站所介绍的直观填制法，确实可以不必使用橡皮擦。

2.从接到数独谜题的那一刻起就可以立即开始解题。 3.初学者或没有计算机辅助时的首要解题方法。 4.相对而言，能解出的谜题较简单。直观法的主要的技巧： 1.基础摒除法。 2.唯一解法。 3.区块摒除法。 4.唯余解法。 5.单元摒除法。 6.矩形摒除法。 7.余数测试法。基础摒除法前言对第一次接触数独游戏，接受了 1 ～ 9 的数字在每一行、每一列、每一个九宫格都只能出现一次的规则后，开始要解题的玩家来说，基础摒除法绝对是他第一个想到及使用的方法，十分的自然、也十分的简易。如果能够细心、系统化的运用基础摒除法，一般报章杂志或较大众化的数独网站上的数独谜题几乎全部可解出来。只不过大部分的玩家都不知如何系统化的运用基础摒除法罢了！基础摒除法虽然简单，但在实际应用时，仍然可分成三个部分：

数独入门教程

数独入门教程数独是一种填数的小游戏，从出现到现在已有几十年的历史了，从最初刊登到报纸和书籍上，现在搬到电脑上，玩起来更加方便了。这篇数独游戏的入门篇，对于初学者有很大帮助。一、数独（SuDoku）介绍数独是一种源自18世纪末的瑞士，后在美国发展、并在日本得以发扬光大的数学智力拼图游戏。拼图是九宫格（即3格宽×3格高）的正方形状，每一格又细分为一个九宫格。在每一个小九宫格中，分别填上1至9的数字，让整个大九宫格每一列、每一行的数字都不重复。数独的玩法逻辑简单，数字排列方式千变万化，数独是锻炼脑筋的好方法。历史如今数独的雏型首先于1970年代由美国的一家数学逻辑游戏杂志发表，当时名为Number Place。现今流行的数独于1984年由日本游戏杂志《パズル通信ニコリ》发表并得了现时的名称。数独本是“独立的数字”的省略，因为每一个方格都填上一个个位数。数独冲出日本成为英国当下的流行游戏，多得曾任香港高等法院法官的高乐德（Wayne Gould）。2004年，他在日本旅行的时候，发现杂志的这款游戏，便带回伦敦向《泰晤士报》推介并获得接纳。英国《每日邮报》也于三日后开始连载，使数独在英国正式掀起热潮。其他国家和地区受其影响也开始连载数独。数独术语要理解如何对一个数独题求解，我们先来介绍一些在本网站中使用的术语。单元格和值一个数独谜题通常包含有9x9=81个单元格，每个单元格仅能填写一个值。对一个未完成的数独题，有些单元格中已经填入了值，另外的单元格则为空，等待解题者来完成。行和列习惯上，横为行，纵为列，在这里也不例外。行由横向的9个单元格组成，而列由纵向的9 个单元格组成。很明显，整个谜题由9行和9列组成。为了避免混淆，这里用大写英文字母和数字分别表示行和列。例如，单元格[G6]指的是行G和第6列交界处的单元格，它已填入了值7。

way 的用法

way 的用法【语境展示】 1. Now I’ll show you how to do the experiment in a different way. 下面我来演示如何用一种不同的方法做这个实验。 2. The teacher had a strange way to make his classes lively and interesting. 这位老师有种奇怪的办法让他的课生动有趣。 3. Can you tell me the best way of working out this problem? 你能告诉我算出这道题的最好方法吗？ 4. I don’t know the way (that / in which) he helped her out. 我不知道他用什么方法帮助她摆脱困境的。 5. The way (that / which) he talked about to solve the problem was difficult to understand. 他所谈到的解决这个问题的方法难以理解。 6. I don’t like the way that / which is being widely used for saving water. 我不喜欢这种正在被广泛使用的节水方法。 7. They did not do it the way we do now. 他们以前的做法和我们现在不一样。【归纳总结】 ●way作“方法，方式”讲时，如表示“以……方式”，前面常加介词in。如例1； ●way作“方法，方式”讲时，其后可接不定式to do sth.，也可接of doing sth. 作定语，表示做某事的方法。如例2，例3；

(完整版)数独解题的基本技巧完整篇

数独解题的基本技巧完整篇-----由浅入深的学习以前已经写过类似的文章，不过好像太偏向于高难度的技巧，像是X-Wing， Y-Wing，Swordfish等等，说实在的真要用到它们，技巧上可还难的很，而且能够运用到的场合也并不多。现在我选择了以下十三个图形范例，说明技巧的运用，应该算是由浅入深的方法，如果读者能够确实了解使得思路开通，自然能成为各类数独的解题高手了。(尤其是9-13项) 例题-1基本交叉排除法（Cross Elimination）说明：利用同一排的三个九宫内，两个相同数字找出另一个相同数字的位置。（数字5）例题-2三连数空格的利用(Blank Triples)

说明：正中央的九宫内有一整排的三个空格，称为三连空格。位在同一排其他两个九宫内的数字，应该会在本九宫内的其他位置。（数字4与7）例题-3三连数满格的利用（Full Triples）说明：中下位置的九宫内，上排已全有数字，针对右侧九宫的数字4，只能在本九宫的下排位置，以及左侧九宫的上排位置。

例题-4基本交叉排除法（Cross Elimination）说明：有时候利用两个位置的交叉排除，也能得到答案。（数字8的位置）例题-5单排数字的交叉排除（Straight Line）

说明：中间横排数字2的位置只能在最右侧。（由于没有相同两数的交叉，很容易被忽略）例题-6三连空格的利用(Blank Triples) 说明：本题同样是三连空格，但是不同的应用。正中央九宫内的其他数字，应该要出现在其他九宫内与三连空格同一排的位置。（数字2与3应该在另外两个红筐位置，因而这三连空格的数字为4，6，9，蓝筐内为4。）例题-7双位交互排除法----这是很多难题的唯一破解方法(第3点定位)

the-way-的用法讲解学习

t h e-w a y-的用法

The way 的用法 "the way+从句"结构在英语教科书中出现的频率较高, the way 是先行词, 其后是定语从句.它有三种表达形式:1) the way+that 2)the way+ in which 3)the way + 从句(省略了that或in which),在通常情况下, 用in which 引导的定语从句最为正式,用that的次之,而省略了关系代词that 或 in which 的, 反而显得更自然,最为常用.如下面三句话所示,其意义相同. I like the way in which he talks. I like the way that he talks. I like the way he talks. 一.在当代美国英语中,the way用作为副词的对格,"the way+从句"实际上相当于一个状语从句来修饰全句. the way=as 1)I'm talking to you just the way I'd talk to a boy of my own. 我和你说话就象和自己孩子说话一样. 2)He did not do it the way his friend did. 他没有象他朋友那样去做此事. 3)Most fruits are naturally sweet and we can eat them just the way they are ----all we have to do is clean or peel them . 大部分水果天然甜润,可以直接食用,我们只需要把他们清洗一下或去皮.

数独的7种解法

数独解法七种解法：前言数独这个数字解谜游戏，完全不必要用到算术！会用到的只是推理与逻辑。刚开始接触数独时，即使是只须用到"唯一解"技巧的简易级谜题，就已可让我们焦头烂额了，但是随着我们深陷数独的迷人世界之后，这类简易级的数独谜题必定在短时间内难再使我们获得征服的满足。于是，当我们逐步深入、进阶到更难的游戏后，我们将会需要发展龈?多的解谜技巧。虽然最好的技巧便是我们自己发现的窍门，这样我们很容易??能记住它们，运用自如，不需要别人来耳提面命。但是如果完全不去观摩学习他人发展出来的技巧，而全靠自己摸索，那将是一个非常坚苦的挑战，也不是正确的学习之道！所以让我们一齐来探讨数独的解谜方法吧！数独的解谜技巧，刚开始发展时，以直观式的唯一解及摒除法为主，对于初入门的玩家来说，这也是一般人较容易理解、接受的方法，对于一般简易级或中级的数独谜题，如果能灵活运用此二法则，通常已游刃有余。

1.唯一解法当数独谜题中的某一个宫格因为所处的列、行或九宫格已出现过的数字已达8 个，那么这个宫格所能填入的数字就剩下这个还没出现过的数字了。 <图1> (9, 8)出现唯一解了 <图1>是最明显的唯一解出现时机，请看第8 行，由(1,8) ～(8,8) 都已填入数字了，只剩(9,8)还是空白，此时(9,8)中应填入的数字，当然就是第8 行中还没出现过的数字了！请一个个数字核对一下，哦！是数字8 还没出现过，所以(9,8) 中该填入的数字就是数字 8 了。 <图2> (8, 9)出现唯一解了 <图2>是另一个明显出现唯一解的情形，请看第8 列，由(8,1) ～(8,8) 都已填入数

way的用法总结大全

way的用法总结大全 way的用法你知道多少，今天给大家带来way的用法，希望能够帮助到大家，下面就和大家分享，来欣赏一下吧。 way的用法总结大全 way的意思 n. 道路,方法,方向,某方面 adv. 远远地，大大地 way用法 way可以用作名词 way的基本意思是“路,道,街,径”,一般用来指具体的“路,道路”,也可指通向某地的“方向”“路线”或做某事所采用的手段,即“方式,方法”。way还可指“习俗,作风”“距离”“附近,周围”“某方面”等。 way作“方法,方式,手段”解时,前面常加介词in。如果way前有this, that等限定词,介词可省略,但如果放在句首,介词则不可省略。

way作“方式,方法”解时,其后可接of v -ing或to- v 作定语,也可接定语从句,引导从句的关系代词或关系副词常可省略。 way用作名词的用法例句 I am on my way to the grocery store.我正在去杂货店的路上。 We lost the way in the dark.我们在黑夜中迷路了。 He asked me the way to London.他问我去伦敦的路。 way可以用作副词 way用作副词时意思是“远远地,大大地”,通常指在程度或距离上有一定的差距。 way back表示“很久以前”。 way用作副词的用法例句 It seems like Im always way too busy with work.我工作总是太忙了。 His ideas were way ahead of his time.他的思想远远超越了他那个时代。 She finished the race way ahead of the other runners.她第一个跑到终点,远远领先于其他选手。 way用法例句

标准数独技巧整理

标准数独技巧 ?唯一数Last Value ?适用情况：当某行、某列或某宫中已经出现八个不同数字时，最后一格即剩下还未出现过的第九个数。图中这一行已经出现数字1、2、3、4、5、6、7、8，所以余下的星号格为9。 ?实际应用：宫摒除Hidden Single in Box

?适用情况：观察某一个数字A，根据数独规则，在同行、列、宫内无重复数字，若一格是A，则其所在行、列、宫都不会再有A，若以此得出某一宫内数字A仅剩一个可能位置，则可以判断这格就是A。图中对于第一宫，由于四个A的影响，第一宫只有一个地方可能填A，即星号处。 ?实际应用：

?行列摒除Hidden Single in Row/Column ?适用情况：观察某一个数字A，根据数独规则，在同行、列、宫内无重复数字，若一格是A，则其所在行、列、宫都不会再有A，若以此得出某一行或列内数字A仅剩一个可能位置，则可以判断这格就是A。图中对于第一行，由于四格受A的影响，第一行只有一个地方可能填A，即星号处。 ?实际应用：

?唯一余数Naked Single ?适用情况：观察某一格，根据数独规则，一格与其所在的行列宫没有重复数字，点算这格所在行列宫已经出现过的数字，若已经出现8个不同的数字，则这格就是第9个没有出现过的数。对于星号格，其所在行（第一行）已经出现2346，所在列（第五列）已经出现15，所在宫（第二宫）已经出现2678，即 12345678均出现了，故星号格为9。

?实际应用： ?宫摒除区块Pointing ?适用情况：在进行宫摒除时，发现某数在某宫可能位置不止一个，但是可能位置处在同行或同列，则可以排除相应行或列中除他们外其他格的该数。数字5对第二宫摒除发现第二宫5的可能位置是2个星号格，虽然目前不能确定是哪一格，但可以确定的是第三行除了星号格外其他格（用短横线标示）一定不是5。如下图所示：

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