First what wikipedia states: wikipedia says that sadi has the following symmetries: [3

^{+},4,3] (i.e. diminished 24-cell symmetry), [(3,3)

^{+},4] (i.e. diminished 16-cell symmetry) and [3

^{1,1,1}]

^{+}(i.e. chiral demitesseractic symmetry). Now I think that apart from these symmetries, sadi also has full demicubic symmetry.

The normal construction of sadi from demicubic symmetry is as a snub, i.e. you take the 24-cell: o3x3o*b3o. Truncate it: x3x3x*b3x, and finally snub it: s3s3s*b3s.

A funny property of the 24-cell is that it's self-dual. I think this property can be exploited to give it full demitesseractic symmetry.

when you take the 24-cell: o3x3o*b3o, and first dual it, before you apply the alternation, you get the following:

24-cell: o3x3o*b3o

dual 24-cell: ooq3ooo3oqo3*b3qoo&#zx <- still in perfect demitesseractic symmetry, but oriented differently

truncated dual 24-cell: qAooAq3oooooo3AqqAoo*b3ooAqqA&#zx <- also perfect demitesseractic symmetry, although a bit unwieldy ( A=2sqrt(2) )

This buildup of the truncated dual 24-cell is nicely grouped in pairs: You have qA3oo3Aq*b3oo + oo3oo3qA*b3Aq + Aq3oo3oo*b3qA. When you draw these as diagrams, you can see the pairs look like rotors that have their nodes swapped. You can say one of every pair is "left-turning", and the other one is "right turning" The alternation process is done by taking all things that are turning one way, so you get: q3oo3A*b3o + o3o3q*b3A + A3o3o*b3q. This is still in full demitesseractic symmetry. now the relaxation is done by changing every q in an x and every A in an f. then you get the compound I found in the D4.11/D4.12 threat.

Sadi can't get full 16-cell symmetry this way, nor full 24-cell symmetry, but do you think it is possible that it does have full demitesseractic symmetry?