Jwj ( 3 4 2 N 2 ) 2 N01 (2 N 2 ) 2 N01 < 2 2 N N 2 0n K =n 2

^ d jw0j (w0) + ^ d jw1j (w1) 2 + E(jwj + 1) = d 0 (w0) + d 0 (w1) 2 so d 0 obeys the average law (without conservation). Proof.[of Theorem 19] Note the d 0 constructed in Theorem 18 is generally not conservative. We will parallel Theorem 4, by constructing a slothful two-sided quasipolynomially precise cover. The construction of a conservative cover from a slothful cover preserves two-sided quasipolynomially precision. We assume powers of two are in the dependency sets. Paralleling Theorem 4, put i = max G ^ d;djwje;djwje \ f0; : : : ; jwjg: We note that except on a set of polylog size, ^ d jwj (w) makes changes of size at most 1=(jwj 2 + 1): In Theorem 4 these small changes were absorbed by adding 3 1 2 0i to ^ d djwje (w[0::i]); here we add E(i): 16 to conguration C: Dene \y at x allows C" to mean that there is some w 2 W such that the string at position xn 2 in w is y. a contradiction. Let A be the set of x's such that 3=4 of the y's at x allow C: Consider the x's in increasing order. If x = 2 A then reset S Sy; for y along the path of decreasing d: If x 2 A; then if there's a setting y of ![2 1 3 n + xn 2 ; 2 1 3 n + (x + 1)n 2 0 1] with d(CSy) < d(S) 0 2 30n D; reset S Sy: After 2 1 2 n03 of the x's in A; either the value of d has decreased to zero or for all y we have d(CSy) d(CS) 0 2 30n D: Then, by Markov, at least 3=4 the y make d(CSy) < d(CS) +2 50n : Since x 2 A; 3=4 of the y's at x allow C; so 1=2 of the y's satisfy both Setting [3 n + xn 2 ; 3 n + (x + 1)n 2) to y allows C d(CSy) < d(CS) + 2 50n : Fix one of these y's, extend S Sy; nd a setting of [3 n ; 2 13 n) witnessing that y at x allows C; and nally extend S along the path of decreasing d: We have d(![0::3 n+1 0 1]) d(C) + 2 50n (1 + O(1)=n …