By Michael Huber

By reason of the category of the finite easy teams, it hasbeen attainable in recent times to symbolize Steiner t-designs, that's t -(v, okay, 1) designs,mainly for t = 2, admitting teams of automorphisms with sufficiently strongsymmetry homes. even though, regardless of the finite easy staff type, forSteiner t-designs with t > 2 each one of these characterizations have remained longstandingchallenging difficulties. specially, the choice of all flag-transitiveSteiner t-designs with three ≤ t ≤ 6 is of specific curiosity and has been open for about40 years (cf. Delandtsheer (Geom. Dedicata forty-one, p. 147, 1992 and guide of IncidenceGeometry, Elsevier technological know-how, Amsterdam, 1995, p. 273), yet possibly datingback to 1965).The current paper maintains the author's paintings (see Huber (J. Comb. conception Ser.A ninety four, 180-190, 2001; Adv. Geom. five, 195-221, 2005; J. Algebr. Comb., 2007, toappear)) of classifying all flag-transitive Steiner 3-designs and 4-designs. We supply acomplete category of all flag-transitive Steiner 5-designs and end up furthermorethat there are not any non-trivial flag-transitive Steiner 6-designs. either effects depend upon theclassification of the finite 3-homogeneous permutation teams. in addition, we surveysome of the main basic effects on hugely symmetric Steiner t-designs.

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**Extra resources for A census of highly symmetric combinatorial designs**

**Sample text**

Starting with an ordered quadruple of positive integers, a generalized Euclidean algorithm is applied successively as follows: if the numbers are x, y, u, v and x > y, then the quadruple is replaced by x − y, y, u + v, v. Otherwise, it is replaced by x, y − x, u, v + u. The algorithm stops when the numbers in the first pair become equal (in which case they are equal to the greatest common divisor of x and y). Assume that we start with m, n, m, n. Prove that when the algorithm ends, the arithmetic mean of the numbers in the second pair equals the least common multiple of m and n.

Verify the equality 3 √ 20 + 14 2 + 3 √ 20 − 14 2 = 4. Solution. Apparently, this problem has nothing to do with polynomials. But let us denote the complicated irrational expression by x and analyze its properties. Because of the cube roots, it becomes natural to raise x to the third power: √ √ x 3 = 20 + 14 2 + 20 − 14 2 √ √ √ √ 3 3 3 20 + 14 2 + 20 − 14 2 + 3 (20 + 14 2)(20 − 14 2) √ 3 = 40 + 3x 400 − 392 = 40 + 6x. And now we see that x satisfies the polynomial equation x 3 − 6x − 40 = 0. We have already been told that 4 is a root of this equation.

It is good to know that the quest for invariants dominated twentieth-century geometry. That being said, let us return to the realm of elementary mathematics with a short list problem from the 46th International Mathematical Olympiad. Example. There are n markers, each with one side white and the other side black, aligned in a row with their white sides up. At each step, if possible, we choose a marker with the white side up (but not one of the outermost markers), remove it, and reverse the two neighboring markers.

### A census of highly symmetric combinatorial designs by Michael Huber

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