Special Issue (2003), S245–S253 Advances in Geometry ( de Gruyter 2003 Blocking sets in PG

Special Issue(2003),S245–S253Advances in Geometry(de Gruyter2003 Blocking sets in PG(2,p)for small p,and partial spreadsin PG(3,7)Aart Blokhuis,Andries E.Brouwer and Henny A.WilbrinkDedicated to Professor Adriano Barlotti on the occasion of his80th birthdayAbstract.Wefind all minimal blocking sets of size32ðpþ1Þin PGð2;pÞfor p<41.There isone new sporadic example,for p¼13.Wefind all maximal partial spreads of size45in PGð3;7Þ.1Minimal nontrivial blocking sets in PG(2,p)A blocking set in a projective plane is a set of points meeting all lines.It is called nontrivial when it does not contain a line.An m-secant of a set is a line meeting the set in precisely m points.Blokhuis[2]shows that in a Desarguesian projective plane PGð2;pÞof primeorder p,a nontrivial blocking set has size at least32ðpþ1Þ,and,moreover,thatin case of equality each point of the blocking set lies on precisely12ðpÀ1Þtangents(1-secants).Nontrivial blocking sets of size3ðpþ1Þexist for all p.Indeed,an example is given by the projective triangle:the set consisting of the pointsð0;1;Às2Þ,ð1;Às2;0Þ,ðÀs2;0;1Þwith s A F p.No nontrivial blocking set of size qþm in PGð2;qÞcan have a k-secant for k>m, and in particular such a set of size3ðpþ1Þin PGð2;pÞcannot have a k-secant withk>12ðpþ3Þ.The triangle has three12ðpþ3Þ-secants.Conversely,Lova´sz andSchrijver[10]show that any nontrivial blocking set of size3ðpþ1Þwith a1ðpþ3Þ-secant must be projectively equivalent to the triangle.(They put the given secant at infinity and show that the remaining p a‰ne points can be taken to be the points ða;aðpþ1Þ=2Þfor a A F p.)A blocking set S in PGð2;qÞis called of Re´dei type when there is a line L such that j S n L j¼q.Thus,we know the blocking sets of Re´dei type meeting the Blokhuis bound in PGð2;pÞ,p prime.Let us call a nontrivial blocking set in PGð2;pÞthat meets the Blokhuis bound sporadic if it is not of Re´dei type.A single sporadic blocking set (in PGð2;7Þ)was known.Here wefind a second sporadic blocking set(in PGð2;13Þ) and show that no other sporadic blocking sets exist in PGð2;pÞ,p<41.2The Blokhuis boundTheorem 2.1([2]).Let S be a nontrivial blocking set in PG ð2;p Þ,p prime .Then j S j d 32ðp þ1Þ.If equality holds ,then each point of S lies on precisely 12ðp À1Þtangents .Proof.Let S ¼fða i ;b i ;c i Þj i ¼1;...;q þm g be a minimal blocking set in PG ð2;q Þ,where q is a power of the prime p .The polynomial F ðX ;Y ;Z Þ¼Q i ða i X þb i Y þc i Z Þvanishes in all points ðx ;y ;z Þ,hence can be written asF ðX ;Y ;Z Þ¼A ðX ;Y ;Z ÞðX q ÀX ÞþB ðX ;Y ;Z ÞðY q ÀY ÞþC ðX ;Y ;Z ÞðZ q ÀZ Þ.Since F ðX ;Y ;Z Þis homogeneous,all low degree terms cancel,and we have F ðX ;Y ;Z Þ¼A 0ðX ;Y ;Z ÞX q þB 0ðX ;Y ;Z ÞY q þC 0ðX ;Y ;Z ÞZ q ,where F has de-gree q þm and A 0;B 0;C 0have degree m .Assume that j S j <2q ,so that no cancella-tion takes place between the terms on the right hand side.Let the line Z ¼0contain l points of S ,and assume that ð1;0;0ÞA S .Now divide by X and substitute X ¼0,Y ¼1to get f ðZ Þ¼b ðZ Þþc ðZ ÞZ q where f has degree q þm Àl and factors completely,and c has degree m Àl and b has degree at most m À1.Write f ðZ Þ¼s ðZ ÞÁr ðZ Þwhere s contains every irreducible factor of f just once,and r contains the repeated factors.Then s j ðb þcZ q Þand s j ðZ q ÀZ Þso s j ðb þcZ Þ.And r j f 0¼b 0þc 0Z q ,so that f ¼rs j ðb þcZ Þðb 0þc 0Z q Þ,and hence f j ðb þcZ Þðb 0c Àbc 0Þ.If the factors on the right are nonzero,it follows that q þm Àl c 2ðm À1Þþm Àl À1that is,m d ðq þ3Þ=2.And in case of equality the degree of s equals the degree of b þcZ so that ð1;0;0Þlies on precisely ðq À1Þ=2tangents.If b þcZ ¼0then f ¼c ÁðZ q ÀZ Þand it follows that ð1;0;0Þdoes not lie on a tangent,i.e.,S is not minimal,contradiction.If b 0c Àbc 0¼0then b and c di¤er by a p -th power.In the particular case q ¼p (and m <q )it follows that they di¤er by a constant factor,say b ðZ Þ¼a Ác ðZ Þ,and f ðZ Þ¼c ðZ ÞÁða þZ Þq so that S contains (and hence is)a line.3Lacunary polynomialsWe see that the blocking set problem leads one to search for polynomials f ðx Þ,g ðx Þ,h ðx Þ,where f factors completely into linear factors and g and h have degree at most 1ðq þ1Þsuch that f ¼x q g þh .(Indeed,in the proof above we found such an f given a small blocking set S ,a point P inside,and a line L passing through that point.An e -fold linear factor of f corresponds to a line on P distinct from L meeting S in e þ1points.The line L meets S in j S j Àdegree ðf Þpoints.Below we take j S j ¼32ðq þ1Þ.)This equation has solutions that need not correspond to blocking sets.We give a few examples.a)(For odd q ,say q ¼2r þ1.)Take f ðx Þ¼x Q ðx Àa Þ3where the product is over the nonzero squares a .Then f satisfies f ðx Þ¼x ðx r À1Þ3¼x q g þh with g ðx Þ¼x r À3,h ðx Þ¼3x r þ1Àx .This would correspond to line intersections (with frequen-Aart Blokhuis,Andries E.Brouwer and Henny A.WilbrinkS246cies written as exponents)1r 224r .For q ¼7this is the function for the blocking set ð1;0;0Þ,ð0;1;0Þ,ð0;0;1Þ,ða ;b ;1Þwith a ;b A f 1;2;4g .b)(For q ¼4t þ1.)Take f ðx Þ¼x Q ðx Àa ÞQ ðx Àb Þ4where the product is over the nonzero squares a and fourth powers b .Here f ðx Þ¼x ðx 2t À1Þðx t À1Þ4¼x q g þh with g ðx Þ¼x 2t À4x t þ5and h ðx Þ¼À5x 2t þ1þ4x t þ1Àx .This would cor-respond to line intersections 12t 2t þ26t .c)(For q ¼4t þ1.)Take f ðx Þ¼x t þ1Q ðx Àa ÞQ ðx Àb Þ2where the product is over the nonzero squares a and fourth powers b .Here f ðx Þ¼x t þ1ðx 2t À1Þðx t À1Þ2¼x q g þh with g ðx Þ¼x t À2and h ðx Þ¼2x 2t þ1Àx t þ1.This would correspond to line intersections 12t 2t 4t ðt þ2Þ2.For q ¼13this is a function for the blocking set ð1;0;0Þ,ð0;1;0Þ,ð0;0;1Þ,ð1;a ;0Þ,ð0;1;a Þ,ða ;0;1Þ,ðb ;c ;1Þwith a 3¼À1,b 3¼c 3¼1.d)(For q ¼13.)Take f ðx Þ¼x Q ðx Àa Þ4Q x À12a ÀÁwhere the product is over all a with a 3¼1.Here f ðx Þ¼x ðx 3À1Þ4x 3À1ÀÁ¼x q g þh with g ðx Þ¼x 3þ4and h ðx Þ¼5x 7À5x 4À5x .This would correspond to line intersections 162454,and indeed this occurs.These lacunary polynomials are just weighted subsets of the projective line,andin particular PGL ð2;q Þacts.For example,x !1x sends x q g þh to x q ~h þ~g where ~kðx Þ¼x ðq þ1Þ=2k ðx À1Þ.For completeness we describe the lacunary polynomials that correspond to the Re ´dei type blocking set:e)Take f ðx Þ¼x q Àx ðq þ1Þ=2¼x ðq þ1Þ=2Q ðx Àa Þwhere the product is over the nonzero squares a .f )Take f ðx Þ¼x q À2x ðq þ1Þ=2þx ¼x Q ðx Àa Þ2where the product is over the nonzero squares a .4Search setupWe search for lacunary polynomials as described above over the prime field F p by exploiting the equationf ¼x pg þh ¼a ðxg þh Þðgh 0Àg 0h Þfor some constant a ,where f factors into linear factors,and xg þh factors into dis-tinct linear factors,and g and h have degree at most 12ðp þ1Þ.If we guess xg þh and the constant of proportionality a and the constant term of gthen this relation gives a recurrence that allows us to compute all other coe‰cients of g ,and thus to find f .If we take l ¼1,then xg þh is a product of m ¼ðp þ3Þ=2distinct linear factors,and there are Àp m Ápossible choices for the set of roots of xg þh .We tried all possibilities for p <41,where PGL ð2;p Þwas used to divide the computation time by roughly p 3.This yields all possibilities for f ,and in partic-ular the multiplicities of the roots of f ,so that we know the sizes of the intersection of lines on some arbitrary point ð1;0;0Þwith S .This su‰ces to classify the pos-sible solutions.In fact,except for example d)in the previous section we only find solutions if xg þh ¼x ðq þ1Þ=2Àx .In a seperate section we will completely classify this special case.Blocking sets in PG ð2;p Þfor small p ,and partial spreads in PG ð3;7ÞS247Looking at p ¼31took 80minutes CPU time on an old Pentium running Linux,and p ¼37took four days.5ResultsThe results are as follows.First of all there are possibilities with a factor of multi-plicity 12ðp þ1Þ,i.e.,a 12ðp þ3Þ-secant,and we have a Re ´dei example,unique by Lova´sz and Schrijver.For the primes p ¼7;11;19;23;31there is a unique non-Re´dei intersection pattern,namely 1ðp À1Þ=2224ðp À1Þ=2(corresponding to the lacunary polynomial found under a)above).Counting the total number of lines on these points we see that this can be a blocking set only for p ¼7.It remains to investigate the cases p ¼7;13;17;29;37.5.1p F 7.For p ¼7there is a unique intersection pattern 132243(and no computer search is required to see that).It gives rise to a unique sporadic blocking set (of size12)(see also [4]).It arises as follows.The a‰ne plane AG ð2;3Þcan be embedded into PG ð2;q Þif and only if q ¼0;1ðmod 3Þ,as one easily checks by assigning coordinates to the 9points of AG ð2;3Þ(for more details see [9]and [1]).This embedding is unique up to isomorphism.The three lines in a parallel class of AG ð2;3Þare concurrent in PG ð2;q Þif and only if q ¼0ðmod 3Þ.For q ¼1ðmod 3Þthis 9-set can be found as the set of inflections of a nondegenerate cubic.Dualizing we find a dual a‰ne plane DAG ð2;3Þwith 12points,94-lines (3on each point)and 122-lines (2on each point)embedded in PG ð2;q Þfor q ¼1ðmod 3Þ.It has ðq 2þq þ1ÞÀ12ðq þ1À5ÞÀ9À12¼ðq À4Þðq À7Þ0-secants,and hence is a blocking set for q ¼4;7and for q ¼4even a 2-fold blocking set.The projective triangle in PG ð2;7Þcan also be viewed as a modification of AG ð2;3Þ:it arises by taking the 9points of AG ð2;3Þand adding the 3points of intersection of the lines of one parallel class.There are no other possibilities:Suppose the blocking set S has n i i -secants,1c i c 4.Then P n i ¼57,P in i ¼96,P Ài 2Án i ¼66by standard counting.And n 1¼36since we have equality in the Blokhuis bound.Hence n 2¼12,n 3¼0,n 4¼9.If there are m i i -secants on a fixed point s A S ,then P m i ¼8,P ði À1Þm i ¼11,m 1¼3so that m 2¼2,m 4¼3.This yields the DAG ð2;3Þstructure.More generally,Ga ´cs et al.showed in [6]that if a nontrivial blocking set S of size 3ðp þ1Þin PG ð2;p Þhas a k -secant for k d 1ðp þ1Þthen it is of Re ´dei type,unless p ¼7and we have this dual a‰ne plane.5.2p F 11.We already saw that for p ¼11nothing of interest happens.More gen-erally,Ga ´cs [5]showed that a k -secant with k ¼12ðp À1Þonly occurs for sets of Re ´dei type,and simple counting then shows that for p ¼11the set S must be of Re´dei type.5.3p F 13.For p ¼13there is a nice example again that is not of Re´dei type.Let q ¼1ðmod 3Þand take in PG ð2;q Þthe 9points of an embedded AG ð2;3Þtogether with the 12points of intersection of lines that are parallel in AG ð2;3Þ.This yields aAart Blokhuis,Andries E.Brouwer and Henny A.WilbrinkS248self-dual configuration.Indeed,these21points together with the21lines that contain more than two of the points have a structure that is that of PGð2;4Þin which the incidences between the9points of a unital(AGð2;3Þ)and the tangent at these points has been removed.There are125-secants,94-secants,362-secants,21ðqþ1À8Þ1-secants andðq2þqþ1ÞÀ21ðqþ1À8ÞÀ36À9À12¼ðqÀ7ÞðqÀ13Þ0-secants, so that this is a blocking set for q¼7;13,and for q¼7even a2-fold blocking set. For p¼13we have j S j¼21.The search shows that there are four possible inter-section patterns:a)162246,b)162563,c)16234352,d)162454.Let there be N a points of type a,etc.,and n i i-secants.If N b>0,then there is a6-secant,and it meets another126-secants,so13c n6¼3N b=6and N b>j S j,contradiction.So N b¼0.If also N a¼0then N cþN d¼21,n1¼126,n2¼32N cþ2N d,n4¼34N c,n5¼25N cþ45N d,Pn i¼132þ13þ1¼183,with unique solution N c¼12,N d¼9,n2¼36,n4¼9,n5¼12.Each4-secant meets the remaining eight,that is,the4-secants meet pairwise(in points of type c)),and the points of type c)form a DAGð2;3Þ.A5-secant meets the DAGð2;3Þin at most two points,so has at least three points of type d),and the points of type d)together with the5-secants form an AGð2;3Þ.Now everything is determined,and this indeed yields a solution.If N a>0then at most two points do not lie on a4-secant,so N d c2.If N c¼N d¼0,then N a¼21and n4¼6N a=4is not integral.Contradiction.So,n5¼2 5N cþ45N d>0.We have n4¼64N aþ34N c,so N c is even,and4j n5.Each5-secantmeets at leastfive more,so n5d8,i.e.,N cþ2N d d20,N cþN d d18,N a c3.If n5d12then N cþ2N d d30,N cþN d d28,contradiction.So n5¼8.Now n4¼6 4N aþ34N c¼32ðN aþN cþN dÞÀ34ðN cþ2N dÞ¼32Á21À34Á20is not integral.Con-tradiction.So,up to isomorphism there is a unique minimal blocking set in PGð2;13Þof size 21that is not of Re´dei type.5.4p F17.For p¼17we have j S j¼27.There are three possible intersection pat-terns:a)182248,b)182664,c)18244462.We have N aþN bþN c¼27and n1¼8Á27¼216,and n2¼N aþ3N bþ2N c, n4¼2N aþN c,so n2þn4¼3Á27¼81and n6¼172þ17þ1À216À81¼10. 2N bþN c¼3n6¼30,so N b d3.Now three points of type b)see twelve6-secants,but there are only ten,so there is a6-secant with at least two points of type b).But such a 6-secant meets at least3þ3þ1þ1þ1þ1¼10other6-secants,contradiction. So,no non-Re´dei sets occur for p¼17.5.5p F29.For p¼29we have j S j¼45.There are three possible intersection pat-terns:a)11422414,b)1142967,c)114274792.If type c)occurs then there are9-secants,and each9-secant meets another nine,so 10c n9¼2N c=9and N c d45so that all points are of type c).But then n4¼7N c=4 is not integral.Contradiction.So N c¼0.There are14N a=44-secants,so N a is even.There are7N b=66-secants, so N b is even.But N aþN b¼45.Contradiction.So,no non-Re´dei sets occur for p¼29.Blocking sets in PGð2;pÞfor small p,and partial spreads in PGð3;7ÞS2495.6p F 37and larger p .For p ¼37we have j S j ¼57.There are three possible intersection patterns:a)11822418,b)11821169,c)1182949112,and as before no non-Re´dei set can exist.Let us prove more generally that no sporadic blocking set exists in PG ð2;p Þ,p ¼4t þ1d 37,when only the three patterns a)12t 2242t ,b)12t 2t þ26t and c)12t 2t 4t ðt þ2Þ2do occur.We have j S j ¼6t þ3.If type c)occurs then there are ðt þ2Þ-secants,and each meets t þ2more,so t þ3c n t þ2¼2N c =ðt þ2Þc 2j S j =ðt þ2Þ<12,contradiction.So N c ¼0.Now N a þN b ¼j S j and n 1þn 2þn 4þn 6¼p 2þp þ1determines all values:N a ¼12,N b ¼6t À9,n 1¼12t 2þ6t ,n 2¼3t 2þ32t þ3,n 4¼6t ,n 6¼t 2À32t .Now a 4-line meets 4ð2t À1Þother 4-lines,contradicting n 4¼6t .So,for a new sporadic blocking set we need a new factorizing lacunary polynomial.6The special case xg B h F x (p B 1)/2C xIn this section we consider the modular di¤erential equationx p g þh ¼a ðxg þh Þðg 0h Àh 0g Þ;where xg þh factors into distinct linear factors,and g ;h A F p ½x are both of degree at most ðp þ1Þ=2,not both zero,and a is a nonzero constant.Write s :¼xg þh and t :¼ðx p Àx Þ=s .Then h ¼s Àxg and s 0t þst 0¼À1.Rewrite the original equation asðx p Àx Þg ¼s ðag 0s Àags 0þag 2À1Þ:Division by s givestg ¼ag 0s Àags 0þag 2À1¼ag 0s Àags 0þag 2þst 0þs 0t :This may be rewritten ass ðag 0þt 0Þ¼Àðag Àt Þðg Às 0Þ:We now consider the special case s ¼x n þ1Àx ,where n :¼ðp À1Þ=2.Then t ¼x n þ1,and our equation simplifies toðx n À1Þxag 0À12x n ¼ðx n þ1Àag Þg þ1À12x n :If u is a square in F Ãp (so that u n À1¼0)then g ðu ÞA À1;2a ÈÉ.Comparing degrees we see that g has degree at most n .Modulo x n this equation reduces further toxg 0¼g À1a ðg þ1Þmod x n :Aart Blokhuis,Andries E.Brouwer and Henny A.WilbrinkS250Note that g ð0ÞÀ1a ÀÁðg ð0Þþ1Þ¼0.Consider more generally the equation xg 0¼ðg Àb Þðg Àc Þmod x n ,say with g ð0Þ¼b .Then we get g ¼c þðb Àc Þ=ð1ÀCx b Àc Þmod x n for some constant C .(Indeed,the equation xg 0¼ðg Àb Þðg Àc Þsu‰ces to determine all coe‰cients of g in terms of earlier coe‰cients,except the coe‰cient of x i where i ¼b Àc .)In the above,1=ð1ÀCx d Þwas to be interpreted in F ½½x .We get a solution in polynomial form by replacing it by ð1ÀC m x dm Þ=ð1ÀCx d Þ,for some m such that dm d n .Thus,in our case,g ¼c þd 1ÀC m x dm1ÀCxd þex n ;for certain constants c ;d ;e ,where either d ¼0and the middle term is absent,or C 00,0<d <p ,m d 2,d ðm À1Þ<n c dm .Since g takes at most two values on nonzero squares,the same holds for 1ÀC m x dm 1ÀCx (when d 00).Thus,there are constants A ;B such that x n À1divides ð1ÀC m x dm ÀA ð1ÀCx d ÞÞð1ÀC m x dm ÀB ð1ÀCx d ÞÞ.This remains true if we re-place x dm by x dm Àn ,so either n c 2d ,m ¼2,d ¼n =2,or the right hand side vanishes and A ¼0,dm ¼n ,C m ¼1.In the former case we have (with new constants)g ¼c þdx n =2þex n with c ¼À1or c ¼1=a .Substitution and comparison of coe‰cients gives ða ;c ;d ;e Þ¼ðÀ2;À1;0;1=2Þor ða ;c ;d ;e Þ¼ðÀ2;À1;0;0Þor ða ;c ;d ;e Þ¼ðÀ4=3;À1;G 1=2;0Þor ða ;c ;d ;e Þ¼ðÀ2;À1=2;0;0Þor ða ;c ;d ;e Þ¼ðÀ4=3;À3=4;0;1=4Þor ða ;c ;d ;e Þ¼ðÀ4=5;À5=4;G 1;À1=4Þ,and these correspond to the examples f ),e),c),f ),a),b),respectively.In the latter case we have g ¼c þd 1Àx n 1ÀCx d þex n ,where n ¼dm ,C m ¼1and with-out loss of generality m d 3.The two values taken by g on the set of nonzero squaresdi¤er by 2þ1¼G n ¼H 1,so that a ¼À2and c þe ¼À1=paring leadingcoe‰cients we find e A f 0;À1=4g .Comparing constants we find c þd A fÀ1;À1=2g .The four possible values of d turn out to be 0;n =2;n ;3n =2,and we already handled those.Altogether the conclusion is that if x p g þh ¼a ðxg þh Þðg 0h Àh 0g Þand xg þh ¼x n þ1Àx ,with g ;h both of degree at most n þ1,then we have one of the examples from Section 3.7Partial spreads in PG(3,7)A spread in a point-line geometry is a partition of the point set into lines.A partial spread is a collection of pairwise disjoint lines.Given a partial spread in a point-line geometry,we shall call a point not covered by one of its lines a hole .Hirschfeld [8](Section 17.6)shows that PG ð3;q Þhas a maximal partial spread of size q 2Àq þ2for q >3(and a maximal partial spread of size 7for q ¼3).No larger maximal partial spreads (that are not spreads)are known,except for q ¼7,where Heden [7]constructed a maximal partial spread of size 45.Blocking sets in PG ð2;p Þfor small p ,and partial spreads in PG ð3;7ÞS251The relation with blocking sets in PG ð2;q Þis as follows:Given a maximal partial spread of size q 2þ1Àd in PG ð3;q Þ,where d >0,we find a nontrivial blocking set of size q þd in PG ð2;q Þ.(Indeed,we find such a blocking set by taking the set of holes in a plane that does not contain a line of the partial spread.)Since nontrivial blocking sets in PG ð2;7Þhave size at least 12,it follows that a partial spread in PG ð3;7Þthat is not a full spread has at most 45lines,that is,has at least 40holes.We did a complete search for partial spreads with 40holes and find that there are precisely 879nonisomorphic such partial spreads.The table below gives group order,number of isomorphism classes and total number of partial spreads.order#total 117445100802383496368037604804175********351512008391263601092332812408640020112962411118806014321204864total 87911070000Soicher [11]had already determined the partial spreads with 40holes and an auto-morphism group of order 5.The geometry of the set H of 40holes (complement of the union of a maximal partial spread S of size 45)is uniquely determined,as was already remarked by Heden.Indeed,each plane must meet H in either 5or 12points (depending on whether it contains a line of S or not),and the holes form a blocking set in each plane p with 12holes.(Otherwise there would be a line L in p disjoint from H ,and looking at the 8planes on L they must all have precisely 5points of H ,contradiction.)Thus,the planes with 12holes are either of the triangle or of the DAG ð2;3Þtype.Now all planes with 12holes must be of the same type.Indeed,let an m-line be a line with m holes.A plane of triangle type does not have 4-lines,while a plane of DAG ð2;3Þtype does not have 5-lines.In particular,a 4-line cannot meet a 5-line.Each hole in a plane of DAG ð2;3Þtype is on some 4-line,so no such hole can be on a 5-line.On a 4-line there are 8planes,four of DAG ð2;3Þtype,and we find at least 36holes on a 4-line,no room for a 5-line.Not all planes can be of triangle type.Indeed,suppose this is the case.Each 3-line is on three planes with 12holes and in each of these planes each of the three holes ofAart Blokhuis,Andries E.Brouwer and Henny A.WilbrinkS252Blocking sets in PGð2;pÞfor small p,and partial spreads in PGð3;7ÞS253 the3-line lies on a unique5-line.It follows that each hole is on precisely three5-lines (so that there are245-lines in all).On the other hand,the projective transformations thatfix the set of non-holes on a5-line have two orbits on the5holes,so that the two ‘corners’on that line in a triangle do not depend on the choice of triangle,so that these corners would be on six5-lines,contradiction.Thus,all planes are dual a‰ne planes.We have a geometry with points and4-lines, where two intersecting4-lines determine a plane,and each plane is dual a‰ne of order 3.By Cuypers[3]this is the geometry of points and hyperbolic lines and dual a‰ne planes of the Spð4;3Þgeometry.This is again a self-dual configuration that lives in PGð3;qÞfor all prime powers q¼1ðmod3Þ.(For example,in PGð3;4Þit lives as the nonisotropic points of a Uð4;2Þgeometry.)Explicit coordinates:take the4points ð1;0;0;0Þand the36pointsð0;1;a;ÀbÞwhere a3¼b3¼1and the coordinates may be permuted cyclically.References[1]A.Bichara,G.Korchma´ros,n2-sets in a projective plane which determine exactly n2þnlines.J.Geom.15(1980),175–181.MR82j:51016Zbl0459.51007[2]A.Blokhuis,On the size of a blocking set in PGð2;pÞ.Combinatorica14(1994),111–114.MR96b:51010Zbl0803.05011[3]H.Cuypers,Symplectic geometries,transvection groups,and bin.TheorySer.A65(1994),39–59.MR94m:51003Zbl0824.51003[4]J.W.Di Paola,On minimum blocking coalitions in small projective plane games.SIAMJ.Appl.Math.17(1969),378–392.MR40#1140Zbl0191.49601[5]A.Ga´cs,A remark on blocking sets of almost Re´dei type.J.Geom.60(1997),65–73.MR99e:51009Zbl0897.51004[6]A.Ga´cs,P.Sziklai,T.Szo˝nyi,Two remarks on blocking sets and nuclei in planes of primeorder.Des.Codes Cryptogr.10(1997),29–39.MR97j:51015Zbl0874.51002[7]O.Heden,A maximal partial spread of size45in PGð3;7Þ.Des.Codes Cryptogr.22(2001),331–334.MR2002a:51007Zbl0982.51005[8]J.W.P.Hirschfeld,Finite projective spaces of three dimensions.Oxford Univ.Press1985.MR87j:51013Zbl0574.51001[9]L.M.Kelly,S.Nwankpa,A‰ne embeddings of Sylvester-Gallai bin.Theory Ser.A14(1973),422–438.MR47#3207Zbl0282.05018[10]L.Lova´sz,A.Schrijver,Remarks on a theorem of Re´dei.Studia Sci.Math.Hungar.16(1983),449–454.MR85e:51017Zbl0535.51009[11]L.Soicher,Computation of partial spreads./~leonard/partialspreadsReceived7January,2003;revised5April,2003A.Blokhuis,A.E.Brouwer,H.A.Wilbrink,Dept.of Math.,Techn.Univ.Eindhoven,P.O.Box513,5600MB Eindhoven,NetherlandsEmail:{aartb,aeb,wsdwhw}@win.tue.nl。