How to construct a (3,6)NBIBD(v)

We can use the results for (3,6)GWhD(v)'s, OBIBD(v,3,2;2)'s and PMD(v,6,1)'s. For v = 6n, 6n+1 the (3,6)GWhD paper gives the initial results. Note that solutions for small designs may be found here or (in more detail, for v≤16 only) here. A list of known (3,6)INBIBD's is given here. The presentation given below uses only very easy and well-established methods.

v=6n+1

A (3,6)GWhD(6n+1) design exists for all n≥1, so the same is true for (3,6)NBIBD(6n+1)'s.

v=6n

It is not possible to construct a (3,6)GWhD(6), a (3,6)NBIBD(6) or, indeed, a RBIBD(6,3,2).

There are 73 values of v, v≡0 mod 6, for which the status of the (3,6)GWhD(v) is unknown.

Values of v, v≡0 mod 6, for which (3;6)GWhD(v )is unknown
18 42 48 54 60 66
72 90 102 108 114 120
126 132 138 150 174 216
222234 270 282 294 306
318 324 330 336 348 354
366384 396 402 414 420
426 450 462 480 504 510
522528 534 594 756 762
774 798 810 816 822 828
846858 882 894 906 912
918 924 942 954 960 966
972978 984 1026 1056 1098
1194     

Most of these can be derived from PMD(v,6,1)'s, so the remaining list of missing designs in this category is:

Values of v, v≡0 mod 6, for which (3,6)NBIBD(v)is required
18 48 54 60 72 90
102108114 132 138 150

Some of these can be solved quite easily.

  • 18 - a solution for (3,6)NBIBD(18) is given here
  • 90=10.9 DP: inflate the blocks of a (3,6)NBIBD(10) with TD(6,9)'s, and fill the holes of the resulting (3,6)HNBIBD with (3,6)NBIBD(9)'s.
  • 108=12.9 DP: inflate the blocks of a (3,6)NBIBD(12) with TD(6,9)'s, and fill the holes of the resulting (3,6)HNBIBD with (3,6)NBIBD(9)'s.
  • 138 - CRC III.2.17-18 states that if there exists a TD(m+1,m), with m+1=k+n, then there exists a {k-1,k+1}-GDD of type (m-1)kn1. (This is a result due to M.Greig, though I must admit I cannot find it in the reference given in the CRC.) Substitute k=8, n=10, m=17 to get a {7,9}-GDD of type 168101. Inflate the blocks and groups to get the required (3,6)NBIBD. Solutions for (3,6)NBIBDs for v∈{9,10,16} are given here.
  • 150=10.15 DP: inflate the blocks of a (3,6)NBIBD(10) with TD(6,15)'s, and fill the holes of the resulting (3,6)HNBIBD with (3,6)NBIBD(15)'s.

A (3,6)NBIBD(v), v≡0 mod 6, v≥6, exists except for v=6 and possibly for v∈{48,54,60,72,102,114,132}.

Preliminary observation

Note that a TD(10,s) exists, for s of the form 6t+1, for all s≥13, except for s ∈ {55,85,91,115,133,145,175,235,295,301,427,553,565,655,1141}

v=6n+3

A PMD(v,6,1), v≡3 mod 6 exists for all v > 657, and for v ∈ {141,147,189,201,225,231, 249,465} and the ranges [261,291], [303,369], [381,405], [417,429], [441,447] [477,483],[501,507], [525,603], [627,651].

Solutions for (3,6)NBIBD(v)'s for v ∈ {9,15,21,27} are given here.

Using the preliminary observations above, we can construct a table of values of v≤657 for which designs can be constructed using the dropping points method or from PMDs, as follows.

        s    
n6n+11560 168 121824 303642
213 117 129 141147  
319 171 183189 201  
425 225231237243249 261267
531273279285291297303309315321
637327333339345351357363369 
743381387393399405411417423429
849 441447453459465471477483
955  501507  525531537
1061543549555561567573579585591
1167597603 615621627633639645
1273651657663669675681687693699

The entries in orange are derived from PMDs.

The dropping points method is used to get a PBD(9(6n+1)+r,{9,10,6n+1,s},1) from a TD(10,6n+1), where n≥2, and s∈{156,0,168,12,18,24,30,36,42}; then inflate the blocks with (3,6)NBIBDs. These solutions are shown in blue. (No entries in the table by this method for s ∈ {156,168}: if solutions were known for v ∈ {48,60}, they would be used instead. No solutions are obtained for 6n+1=55, because no TD(10,55) is known.)

This leaves unsolved the 32 values of v below.

3339 45 51 57 63 697581
8793 99105 111123 135 153159
165 177195 207 213219255 375435
489495513 519609    

Some of these can be solved quite easily.

  • 57=7.8+1 SDP: inflate the blocks of a (3,6)NBIBD(7) with TD(6,8)'s to give a (3,6)HNBIBD of type 86. With 1 extra point, fill the holes with (3,6)INBIBD(9,1)'s and a (3,6)NBIBD(9).
  • 63=9.7 DP: inflate the blocks of a (3,6)NBIBD(9) with TD(6,7)'s to give a (3,6)HNBIBD of type 79. Fill the holes with (3,6)NBIBD(7)'s.
  • 81=9.9 DP: inflate the blocks of a (3,6)NBIBD(9) with TD(6,9)'s to give a (3,6)HNBIBD of type 99. Fill the holes with (3,6)NBIBD(9)'s.
  • 93=9.9+2.6 Weighting: starting with a TD(10,9), give 6 points of one group a weight of 2, the remainder of that group a weight of 0, and all other points a weight of 1. Inflate the blocks with (3,6)INBIBD(11,2)'s or (3,6)NBIBD(9)'s to give a (3,6)HNBIBD of type 99121. Fill the holes with (3,6)NBIBD(9)'s and a (3,6)NBIBD(12).
  • 99=9.9+2.9 Weighting: starting with a TD(10,9), give 9 points of one group a weight of 2, the remainder of that group a weight of 0, and all other points a weight of 1. Inflate the blocks with (3,6)INBIBD(11,2)'s or (3,6)NBIBD(9)'s to give a (3,6)HNBIBD of type 99181. Fill the holes with (3,6)NBIBD(9)'s and a (3,6)NBIBD(18).
  • 105=7.15 DP: inflate the blocks of a (3,6)NBIBD(15) with TD(6,7)'s to give a (3,6)HNBIBD of type 715. Fill the holes with (3,6)NBIBD(7)'s.
  • 111=10.11+1 SDP: inflate the blocks of a (3,6)NBIBD(10) with TD(6,11)'s to give a (3,6)HNBIBD of type 1110. With 1 extra point, fill the holes with (3,6)INBIBD(12,1)'s and a (3,6)NBIBD(12).
  • 123=7.17+4 SDP: inflate the blocks of a (3,6)NBIBD(7) with TD(6,17)'s to give a (3,6)HNBIBD of type 177. With 4 extra points, fill the holes with (3,6)INBIBD(21,4)'s and a (3,6)NBIBD(21).
  • 135=15.9 DP: inflate the blocks of a (3,6)NBIBD(15) with TD(6,9)'s to give a (3,6)HNBIBD of type 915. Fill the holes with (3,6)NBIBD(9)'s.
  • 159=9.16+15 Dropping points: from a TD(10,16) drop one point and inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {9,10,15,16}.
  • 165=12.13+9 Dropping points: from a TD(13,13) drop four points from one group and inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {9,12,13}.
  • 177=9.19+6 Spike: from a TD(15,19) drop all but one of the points from 6 groups, keeping all the points on one line of 15 points. Inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {9,10,15,19}.
  • 195=12.16+3 Spike: from a TD(15,16) drop all but one of the points from 3 groups, keeping all the points on one line of 15 points. Inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {12,13,15,16}.
  • 207=12.16+15 Dropping points: from a TD(13,16) drop one point and inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {12,13,15,16}.
  • 213=10.21+3 SDP: inflate the blocks of a (3,6)NBIBD(10) with TD(6,21)'s to give a (3,6)HNBIBD of type 2110. With 3 extra points, fill the holes with (3,6)INBIBD(24,3)'s and a (3,6)NBIBD(24).
  • 219=9.(24-1)+12 SIP: inflate the blocks of a (3,6)NBIBD(19) with (TD(6,24)- TD(6,1))s to give a (3,6)HNBIBD of type (24,1)9. With 3 extra points, fill the holes with (3,6)INBIBD(27,4)'s and a (3,6)NBIBD(12).
  • 255=9.27+12 Dropping points: from a TD(10,27) drop 15 points from 0ne group and inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {9,10,12,27}.
  • 435=7.62+1 SDP: inflate the blocks of a (3,6)NBIBD(7) with TD(6,62)'s to give a (3,6)HNBIBD of type 627. With 1 extra point, fill the holes with (3,6)INBIBD(63,1)'s and a (3,6)NBIBD(63).
  • 489=15.31+24 Dropping points: from a TD(16,31) drop 7 points from one group and inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {15,16,24,31}.
  • 495=15.31+30 Dropping points: from a TD(16,31) drop one point and inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {15,16,30,31}.
  • 513=57.9 DP: inflate the blocks of a (3,6)NBIBD(57) - see a bove - with TD(6,9)'s to give a (3,6)HNBIBD of type 957. Fill the holes with (3,6)NBIBD(9)'s.
  • 519=12.43+3 Spike: from a TD(15,43) drop all but one of the points from 3 groups, keeping all the points on one line of 15 points. Inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {12,13,15,43}.
  • 609=9.67+6 Spike: from a TD(15,67) drop all but one of the points from 6 groups, keeping all the points on one line of 15 points. Inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {9,10,15,67}.

This leaves the following values unsolved: {33,39,45,51,69,75,87}

v=6n+4

A PMD(v,6,1), v≡4 mod 6 exists for all v > 154, and for v ∈ {28,40,46}.

Solutions for (3,6)NBIBDs for v∈{10,16} are given here.

This leaves unsolved the 20 values of v below.

22 34 4652 58 6470 76 82
8894 100 106 112 118124 130 136
142 148       

Some of these have simple solutions.

  • 58 Abel, Bennett & Zhang give a HPMD(58,6,1) of type 7791; convert this into a (3,6)HNBIBD(58) of the same type, and fill in the holes.
  • 64=7.9+1 SDP: inflate the blocks of a (3,6)NBIBD(7) with TD(6,9)'s to give a (3,6)HNBIBD of type 97. With 1 extra point, fill the holes with (3,6)INBIBD(10,1)'s and a (3,6)NBIBD(10).
  • 70=10.7 DP: inflate the blocks of a (3,6)NBIBD(10) with TD(6,7)'s to give a (3,6)HNBIBD of type 710. Fill the holes with (3,6)NBIBD(7)'s.
  • 76 - an IPMD(76,15) exists; convert this to a (3,6)INBIBD(76,15), and fill the hole with a (3,6)NBIBD(15).
  • 82=9.9+1 Dropping points: from a TD(10,9) drop 8 points from one group and inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {9,10}.
  • 88=9.9+7 Dropping points: from a TD(10,9) drop 2 points from one group and inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {7,9,10}.
  • 94=7.13+3 SDP: inflate the blocks of a (3,6)NBIBD(7) with TD(6,13)'s to give a (3,6)HNBIBD of type 137. With 3 extra points, fill the holes with (3,6)INBIBD(16,3)'s and a (3,6)NBIBD(16).
  • 100=9.11+1 SDP: inflate the blocks of a (3,6)NBIBD(9) with TD(6,11)'s to give a (3,6)HNBIBD of type 119. With 1 extra point, fill the holes with (3,6)INBIBD(12,1)'s and a (3,6)NBIBD(12).
  • 106=7.15+1 SDP: inflate the blocks of a (3,6)NBIBD(7) with TD(6,15)'s to give a (3,6)HNBIBD of type 157. With 1 extra point, fill the holes with (3,6)INBIBD(16,1)'s and a (3,6)NBIBD(16).
  • 112=7.16 DP: inflate the blocks of a (3,6)NBIBD(7) with TD(6,16)'s to give a (3,6)HNBIBD of type 167. Fill the holes with (3,6)NBIBD(16)'s.
  • 118=9.13+1 Dropping points: from a TD(10,13) drop 12 points from one group and inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {9,10,13}.
  • 124=9.13+7 Dropping points: from a TD(10,13) drop 6 points from one group and inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {7,9,10,13}.
  • 130=10.13 DP: inflate the blocks of a (3,6)NBIBD(10) with TD(6,13)'s to give a (3,6)HNBIBD of type 1310. Fill the holes with (3,6)NBIBD(13)'s.
  • 136 - an IPMD(136,27) exists; convert this to a (3,6)INBIBD(136,27) and fill the hole with a (3,6)NBIBD(27).
  • 148=9.16+4 Spike: from a TD(13,16) drop all but one of the points from 4 groups, keeping all the points on one line of 13 points. Inflate the blocks of the resulting PBD with (3,6)NBIBD(v)s for v ∈ {9,10,13,16}.

A (3,6)NBIBD(v) exists for v≡4 mod 6, v>4, except possibly for v∈{22,34,52,142}.

This leaves 18 cases open for further investigation. Doubtless, I will still have missed a trick or two: I am particularly miffed by 22 and 48.