# Trouble In Triplicate

## Kevin Wald

The game described is like the card game Set, except that:

(1) Cards differ only in color, pattern, and shape, not count.

(2) The rule for forming triples is different. The 3x3 arrangement of contestants hints at what is going on:

• The three in the diagonal are ENO, NEO, and EON, "dancing"; they are all "1" (and are chanting to that effect).
• The three below the diagonal are Zero, Love, and a goose egg; they are all "0".
• The three above the diagonal contain examples of color (Gray), pattern (Paisley), and shape (Sphere), the three qualities a card can have.

So we have:

 1 shape color 0 1 pattern 0 0 1

that is, a 3x3 upper triangular matrix with unit diagonal. If we take the three values of each quality to represent 0, 1, and 2 (with addition and multiplication mod 3), these will form a (nonabelian) group with one element for each card. A triple, as Vanna implies, consists of a set of cards that multiply together into the identity element:

 1 0 0 0 1 0 0 0 1

In particular, if you multiply three matrices of this form, you get:

 1 s1 c1 1 s2 c2 1 s3 c3 1 (s1+s2+s3) (c1+c2+c3+s1p2+s1p3+s2p3) 0 1 p1 * 0 1 p2 * 0 1 p3 = 0 1 (p1+p2+p3) 0 0 1 0 0 1 0 0 1 0 0 1

So for shapes and patterns, the corresponding numbers just have to add up to zero, which means that shapes and patterns work the same as in regular Set: they have to be all the same or all different. For colors, however, there is an additional term of s1p2 + s1p3 + s2p3. From the examples given, we know that:

(i) squiggle * hollow + squiggle * striped + diamond * striped = 0
(ii) squiggle * hollow + squiggle * striped + oval * striped = 0
(iii) diamond * solid + diamond * striped + squiggle * striped = (0 - red - red - green) = red - green

From (i) and (ii), striped * (diamond - oval) = 0, so since diamond != oval, striped = 0; then (from (i) or (ii)) squiggle * hollow = 0, so since hollow != striped = 0, squiggle = 0.

Thus, solid and hollow are 1 and 2 in some order, and diamond and oval are 1 and 2 in some order; also, from (iii), red = green + diamond * solid. We can't pin down exact values any further than that, but that is enough to determine all the correct triples entirely; any arbitrary assignment obeying these rules, such as:

0 1 2 squiggle blue striped diamond green solid oval red hollow

will give the same triples as any other.

We can then fill in all the triples in the illustration; the result is (where sCp indicates shape, color, and pattern):

```2B1 0R1 1B1     1B1 0G1 2B1     2B2 0G1 1B0     1B1 0G1 2B1
1G2-----2R1-----0R0             2R2-----0R2-----1B2
0B0 2B0 1B0  |  1B0 0B2 2B1     2B0 1B0 0B0  |  2B2 2R0 2B1
0G1#############1R2#############2B0
0B1 2R0 1B2  |  1B1 0G0 2B2  #  1B2 1R1 1B0  |  0B1 2G1 1B1
1R2-----1R1-----1G0      #      2B2-----1R1-----0R0
1B1 0G1 2B1     0B2 2G0 1B1  #  1B2 0R0 2B1     2B1 1R2 0B0
0G0
1B1 0G2 2B0     1B2 0R2 2B2  #  2B1 0R2 1B0     2B1 2R2 2B0
2B2-----2R1-----2R0      #      2G2-----2G1-----2R0
1B1 1G2 1B0  |  0B2 1B1 2B0  #  1B0 1R2 1B1  |  0B1 2G1 1B1
0R2#############2R1#############1R0
0B2 2B1 1B0  |  0B2 2R2 1B2     1B2 0R1 2B0  |  2B1 0R2 1B0
0G2-----1G0-----2B1             1R2-----0B2-----2B2
1B2 1G0 1B1     0B1 2R0 1B2     2B2 2R0 2B1     0B1 1B2 2B0
```

If you take the enclosed vertical triples, in order by the letters labeling them, and translate the colors of each triple into the letter whose background has the same three colors, you get LOOKER TURNED SET TO CRAP, which is a cryptic clue for the answer, SPECTATOR.