1. Specker Derivative Game Karl Lieberherr Alex Dubreuil Spring 2009

2. Mega moves in classic and secret SDG White-black mega move –white: offer derivatives –black: buy derivatives or reoffer –if bought then repeat r times for each bought derivative: –white: deliver raw material with witness quality(S) of secret finished product S –black: deliver finished product FP –white: reveal secret S –black: check secret S against witness quality(S) –win »classic SDG: satisfaction ratio sr(FP) wrt all. win if sr(FP) >= price * 1. »secret SDG: satisfaction ratio sr(FP) wrt secret S (think of secret S as the maximum): win if sr(FP) >= price * quality(S). –pay for performance in raw material finishing: aggregate wins

3. derivative: (CSP predicate)

4. SDG Game Versions T Ball (one relation) Softball –Slow Pitch (recognizing noise) one implication chain of any number of relations. –Fast Pitch any number of relations –Level k Independent (k independent relations with no implication relationship). Note: Level 1 Independent = T Ball –Level k Reduced (any number of relations that can be reduced to Level k Independent.) Note: Slow Pitch is a special case of Level 1 Reduced. Baseball –Classic and Secret CSP Any Combinatorial Maximization Problem T Ball and Softball are based on CSP

5. SDG Game Versions T Ball Slow Pitch Fast Pitch Level k Independent Level k Reduced T Ball = Fast Pitch Level 1 Independent Slow Pitch = Special case of Fast Pitch Level 1 Reduced Softball

6. 12481248 35961012 7111314 Level 0 Level 1 odd Level 2 Level 1 even Level 3 15 (Implied by all) Null level0 (Implies all)

7. 12481248 35961012 7111314 Level 0 Level 1 odd Level 2 Level 1 even Level 3 15 (Implied by all) Null level0 (Implies all) Derivative: (1 2 3 10 12 7 11 13 14)ARITY 2

8. Interpretation: Arity 2 One node per relation. Edges: implication between relations. There are choose(4,2) independent relations for arity 2. choose(4,k), k =0…4, the maximum is at k=4/2=2. choose(n,k), k = 0…n, the maximum is at k = n/2.

9. Arity 3 choose(n,k), k = 0…n, the maximum is at k = n/2. n=8, choose(8,4)=8*7*6*5/(1*2*3*4)=70 Note: sum[k=0..n] choose(n,k) = 2 n