diff 2b337de..45db8d1 trace_reconstruction_anchor_induction_v2.txt
diff --git a/trace_reconstruction_anchor_induction_v2.txt b/trace_reconstruction_anchor_induction_v2.txt
index 2dbe208..6b2f7fa 100644
--- a/trace_reconstruction_anchor_induction_v2.txt
+++ b/trace_reconstruction_anchor_induction_v2.txt
@@ -536,3 +536,658 @@ ordinary polynomial.
This is the whole point of the latest-anchor convention. The latest
index is either at the mismatch or to the right of it; anything to the
left has already been decided and cancels.
+
+========================================================================
+7. THE INDUCTION GOAL
+========================================================================
+
+At stage j, with 0 <= j <= b, we want a probe with the following
+properties.
+
+ (G1) Order and blocks.
+ 1 <= k <= 2 P_0 4^j and 1 <= A <= 4^j.
+
+ (G2) Masks. For every level h < j there are at most 4^{j-h-1} masks of
+ that level, and every loose edge is covered by at least one of
+ their differences.
+
+ (G3) A pivot. There is a real theta with |theta| <= 5 a_j such that
+
+ |H( exp(-a_j + i theta) )| >= exp( - B_j / f(theta/a_j) ).
+
+Stop as soon as a_j n <= .001. This happens by j = b, because
+a_b n < .0001. At the moment of stopping, the radius parameter a_j is at
+most .001/n, i.e. very far inside the unit disk.
+
+Interpretation of (G3). The point exp(-a_j + i theta) sits just inside
+the unit circle, at distance about a_j from it. The factor
+1/f(theta/a_j) is between exp(-25/20) and 1 when |theta| <= 5 a_j, so it
+is a mild penalty for choosing a direction theta away from the positive
+real axis.
+
+Size of the sums. Every probe satisfying (G1) and (G2) has
+
+ sup_m |s_x(m)| <= Z_j, sup_m |s_y(m)| <= Z_j,
+
+where
+
+ Z_j = 2^{4^j} * ( 2 / (epsilon a_j) )^{4^j}.
+
+Why. Each loose edge contributes a decay factor at least
+exp(-epsilon a_j Delta), where Delta is its gap, because every level
+h < j has t_h >= a_j and the coverage condition forces at least one
+applicable mask. Summing over Delta >= 1 gives at most
+1/(1 - exp(-epsilon a_j)) <= 2/(epsilon a_j). There are at most
+A <= 4^j loose edges. Bit predicates are at most 1. Each mask factor has
+modulus at most 2 by (M1), and there are at most 4^j of them, giving
+2^{4^j}. Exact edges are fixed at distance 1 and contribute 1.
+
+Consequence, used many times below: the auxiliary magnitudes are tiny
+compared to the accuracy budget. For 0 <= j < b,
+
+ log Z_j + log Z_{j+1} + log(1/a_j) + log(1/t_j)
+ <= 10^21 ell_* 4^j (b + 2^j)
+ <= 10^{-60} B_j.
+
+Proof: each term is at most about 4^j log(1/a_j), and
+log(1/a_h) <= 10^18 (b + 2^h) ell_*. So the left side is at most
+10^18 * 4^j * (b + 2^j) * ell_* times a small constant. And
+
+ 4^j (b + 2^j) <= 8^j (b + 1) <= 8^j (b + 1)^3.
+
+Since B_j contains 10^100 * 8^j * (b + 1)^3 * ell_*, there is enormous
+room.
+
+
+========================================================================
+8. THE BASE CASE: A PERIOD-AVOIDING WORD
+========================================================================
+
+We must produce a probe at stage 0. It will have all edges exact, no
+masks, and order k = 2 P_0.
+
+8.1 Finding the word.
+
+By the padding, d >= 2 P_0 for all sufficiently large n. Let
+
+ w' = x_{d-2P_0+1} ... x_{d-1} = y_{d-2P_0+1} ... y_{d-1}
+
+be the common last 2 P_0 - 1 symbols before the mismatch.
+
+CLAIM. At least one of the two words w'0 and w'1 has no period strictly
+smaller than P_0.
+
+Proof. Suppose w'0 has period p_1 < P_0 and w'1 has period p_2 < P_0.
+Then w', being a prefix of each, has both periods. Since
+
+ |w'| = 2 P_0 - 1 >= p_1 + p_2 - gcd(p_1, p_2)
+
+(because p_1, p_2 <= P_0 - 1), Fine and Wilf gives that w' has period
+g = gcd(p_1, p_2).
+
+Look at positions 2 P_0 - p_1 and 2 P_0 - p_2 inside w'. They differ by
+p_2 - p_1, which is a multiple of g, so they carry the same symbol.
+
+But period p_1 of w'0 forces (w'0)_{2P_0 - p_1} to equal (w'0)_{2P_0},
+which is 0. And period p_2 of w'1 forces (w'1)_{2P_0 - p_2} to equal
+(w'1)_{2P_0}, which is 1. So the common symbol would be both 0 and 1.
+Contradiction.
+
+Choose w in {w'0, w'1} with no period below P_0.
+
+8.2 Building the base probe.
+
+Let k = 2 P_0. Take all edges exact. Take predicates chi_e that test
+the e-th symbol of w. There are no masks, so A = 1.
+
+The number of occurrences of w ending exactly at d differs between x
+and y: by construction one of them has w there and the other does not.
+Hence s_x(0) - s_y(0) is +1 or -1, so the two strings already disagree
+at offset 0.
+
+8.3 Occurrences are far apart.
+
+Suppose w occurs in x ending at d + m_1 and again ending at d + m_2,
+with 0 < m_2 - m_1 < P_0. Call delta = m_2 - m_1.
+
+The two occurrences have length 2 P_0 and are shifted by delta. At every
+position covered by both, the symbol is determined twice: once from the
+first occurrence and once from the second. Reading off the two
+determinations, we get w_j = w_{j+delta} whenever both indices lie in
+[1, 2 P_0]. That is exactly the statement that w has period delta.
+
+Since delta < P_0, this contradicts the choice of w. The same argument
+applies in y. So occurrence endings in each individual string are
+P_0-separated.
+
+8.4 The shape of H.
+
+Putting the two previous paragraphs together, up to an overall sign,
+
+ H(z) = 1 - eta z^u + sum_{m >= P_0} c_m z^m
+
+with
+
+ 1 <= u < P_0, eta in {0,1}, |c_m| <= 1,
+
+and if eta = 0 then u is arbitrary, since the term -eta z^u is then
+absent for every choice of u.
+
+Explanation. s_x(0) - s_y(0) = +-1 gives the constant term. The other
+string has its nearest occurrence ending at d + u for some 1 <= u < P_0,
+or none at all, in which case eta = 0; this gives the -eta z^u term.
+Every further occurrence, in either string, ends at least P_0 away from
+d, giving the tail.
+
+8.5 The pivot by the Poisson inequality.
+
+Let v = 10 log(P_0) / P_0. At the positive real point z = exp(-v):
+
+ The first two terms have modulus at least
+
+ 1 - exp(-u v) >= 1 - exp(-v) >= v/2,
+
+ since v is tiny.
+
+ The tail is at most
+
+ sum_{m >= P_0} exp(-v m) = exp(-v P_0)/(1 - exp(-v))
+ <= 2 P_0^{-10} / v <= v/6,
+
+ because exp(-v P_0) = P_0^{-10} and 1 - exp(-v) >= v/2.
+
+Therefore |H(exp(-v))| >= v/2 - v/6 = v/3.
+
+On the circle |z| = exp(-a_0) the crude bound |H(z)| <= 2/a_0 holds.
+There are at most about 1/a_0 contributing offsets, each coefficient at
+most 2; since 1/a_0 is astronomically large, this is trivial.
+
+The Poisson inequality on the disk of radius exp(-a_0) centred at 0,
+evaluated at the interior point exp(-v) which is much closer to 1, then
+gives
+
+ max_{|theta| <= a_0} |H(exp(-a_0 + i theta))|
+ >= exp( - O( (v/a_0) log(1/a_0) ) ).
+
+Indeed, the harmonic weight of that arc is at least a constant times
+a_0/v (because v >= 2 a_0 and v < .01), and log|H| is bounded above by
+log(2/a_0) elsewhere. Boundary zeros are removed by the usual
+regularization log(|H| + delta), delta -> 0.
+
+Size of the exponent. We have
+
+ log(1/a_0) = 10^16 + log(D L_0) <= 10^16 + 35 + b,
+ v/a_0 <= 10 log(2 P_0) <= 10 (10^16 + 36 + b),
+
+so the exponent is at most about 10 (10^16 + 36 + b)^2, which is under
+1.1 * 10^33 for any conceivable n. On the other hand B_0 is at least
+10^100. So the exponent is at most 10^{-66} B_0, comfortably below the
+requirement B_0 / f(theta/a_0) >= 0.94 B_0.
+
+This gives (G1), (G2), and (G3) at j = 0: order 2 P_0 <= 2 P_0 4^0,
+block count 1 <= 4^0, no masks, and |theta| <= a_0 <= 5 a_0.
+
+
+========================================================================
+9. THE STEP: A GAUSSIAN WINDOW AND AN ARC TEST
+========================================================================
+
+Suppose (G1), (G2), (G3) hold at stage j < b. Write
+
+ a = a_j, t = a_{j+1}, B = B_j, E = B / f(theta/a_j).
+
+Recall t/a <= v_0, far below .01; a <= a_0, far below .001; and, during
+a genuine step (one not at the stopping point), a > .001/n.
+
+9.1 Step one: a window.
+
+Define the window
+
+ sigma^2 = 2.4 B / t^2
+ ell = floor( 10^{-10} E )
+ v(m) = exp(-y^2/2) * sum_{k=0}^{ell} (y^2/2)^k / k!,
+
+with y = m/sigma throughout. The function v is a polynomial in m^2 of
+degree 2 ell; it satisfies 0 <= v <= 1; and it is a smooth window, equal
+to 1 inside a window and 0 far outside.
+
+CLAIM 9.1. For 0 <= m <= 3 E/a we have 1 - v(m) <= exp(-2E).
+
+Proof. On this range,
+
+ y^2/2 <= 2 E (t/a)^2 <= 2 E exp(-2 * 10^16),
+
+because t/a <= exp(-10^16). So w = y^2/2 is astronomically small, and
+the Taylor remainder of the exponential series after ell terms,
+multiplied by exp(-w), is at most (e w/(ell+1))^{ell+1}, which is far
+below exp(-2E) because ell is a positive fraction of E.
+
+Interpretation: the window radius 3E/a is astronomically smaller than
+the Gaussian scale sigma, so within the window the correction is
+negligible.
+
+Project to real or imaginary part. Choose c(s) = Re(s) or c(s) = Im(s),
+whichever keeps at least half of the pivot. Since H = Re H + i Im H,
+one of the two has modulus at least half of |H|.
+
+Define, for 0 <= m with d + m < n, and zero elsewhere,
+
+ f_x(m) = ( 2 + c( s_x(m)/Z_j ) ) v(m) / Y
+ f_y(m) = ( 2 + c( s_y(m)/Z_j ) ) v(m) / Y
+ Y = 3 sum_m v(m).
+
+Two immediate facts. Since |s_x|, |s_y| <= Z_j, we have
+0 <= 2 + c(s/Z_j) <= 3, so 0 <= f_x, f_y and sum_m f_x(m), sum_m f_y(m)
+<= 1. Also Y >= 3, because the term m = 0 has v(0) = 1; so there is no
+division by zero.
+
+The normalizing factor is harmless: log Y <= 10^{-8} E. To see this,
+split at H_* = ceil(1000 B / t). The sum of v within that range is at
+most about 2 H_*, and outside it v(m) is at most
+2^ell exp(-m^2/(4 sigma^2)), which decays like a Gaussian. Using
+log(1/t) <= 10^{-60} B, B <= 4E, and log B = o(E), the claim follows.
+
+9.2 Step two: a polynomial on the unit circle.
+
+Define
+
+ J(z) = sum_m ( f_x(m) - f_y(m) ) z^m.
+
+Two bounds.
+
+ On the whole unit circle, sup_{|z| = 1} |J(z)| <= 2, because
+
+ sum_m |f_x - f_y| <= sum_m f_x + sum_m f_y <= 2.
+
+ At the pivot z_0 = exp(-a + i theta),
+
+ |J(z_0)| >= exp( -1.0005 E ).
+
+ Reason: masking costs at most exp(-2E) on the window (Claim 9.1) and
+ is exponentially small beyond it; 1/Y costs at most exp(-10^{-8} E);
+ log Z_j costs at most 10^{-60} B. Adding these up gives the stated
+ loss.
+
+The truncation of J to |m| <= H_* is denoted J_diamond. Because all
+negative powers of z vanish in the difference, and the omitted tail has
+l_1 norm below exp(-200E), we also get
+
+ |J_diamond(z_0)| >= exp( -1.0006 E ).
+
+9.3 Step three: a discrete grid on the arc.
+
+Put
+
+ N = ceil( (1000 B / t)^6 ),
+ h = 2 pi / (N a),
+ v_0^* = theta / a,
+ gamma = floor( 10^4 / h ),
+ F(l) = sum_m f_x(m) exp( 2 pi i l m / N ),
+ G(l) = sum_m f_y(m) exp( 2 pi i l m / N ).
+
+The grid points are z_l = exp(-a) exp(2 pi i l / N) = exp(-a + i a l h),
+for |l| <= gamma. Note that |F|, |G| <= 1.
+
+CLAIM 9.2 (phase test). Put Lambda = exp(-6B). Then there exist
+r in {2,3,4} and integers l_1,...,l_r with
+
+ |l_i| <= gamma and |l_1 + ... + l_r| <= 1,
+
+such that
+
+ | prod_i F(l_i) - prod_i G(l_i) | > Lambda.
+
+Sketch of proof. Introduce the deficits
+
+ m_l = max( |F(l)|, |G(l)| ) (so 0 <= m_l <= 1)
+ g_l = min( 50, E^{-1} log(1/m_l) )
+ q_l = min( 50, E^{-1} log(2 m_l / |F(l) - G(l)|) ),
+
+with the conventions g_l = 50 when m_l = 0, and q_l = 50 when the
+denominator vanishes. Both are nonnegative and symmetric, because F and
+G are real on the grid by construction.
+
+The grid weights are
+
+ k_l = h / ( pi ( 1 + (l h - v_0^*)^2 ) ) for |l| <= gamma,
+
+and we define
+
+ G_cal = sum_l k_l g_l, Q = sum_l k_l q_l.
+
+The Poisson inequality, applied to the pivot bound on J_diamond, implies
+the deficit budget
+
+ G_cal + Q <= 1.003.
+
+The proof is a Riemann-sum approximation to the Poisson integral.
+Concretely, for each l,
+
+ g_l + q_l <= min( 100, E^{-1} log( 2 / |J(exp(-a + i a l h))| ) )
+
+up to the truncation correction, and the Poisson integral of that capped
+quantity is at most 1.0025. The rest is bookkeeping. The Poisson density
+in the rescaled variable x = l h, for the disk of radius exp(-a), is at
+least (1 - 2a)/(pi (1 + (x - v_0^*)^2)), and the grid sum differs from
+the integral by o(1) uniformly, by layer cake and bounded variation.
+
+Now the combinatorial core. For every integer j'shift with
+1 <= j'shift <= S/h, every fractional vertex cover of the graph on the
+integer points in [-gamma, gamma] with edges (l, l + j'shift) has total
+k_l weight at least M. Here S and M are the numbers in the table below.
+
+Proof. Restrict to one residue class mod j'shift: it is a chain. Extend
+the weights to the full Cauchy kernel on the whole line; the tail costs
+less than .0001. A fractional cover assigns a number in [0,1] to each
+vertex, with endpoint sums at least 1. It can be rounded to a random
+binary cover without increasing expected weight: pick one uniform
+threshold U_0 in [0,1], use "vertex number >= U_0" on even vertices and
+"vertex number >= 1 - U_0" on odd vertices. So it is enough to bound
+binary covers.
+
+In each chain, take the central vertex l nearest to v_0^*/h. The
+weights decrease outward along both arms. A binary cover either omits
+the centre, in which case both first neighbours are forced and the
+disjoint edges (2,3), (4,5), ... force the odd weights on both arms; or
+it includes the centre, in which case the disjoint edges (1,2), (3,4),
+... force the even weights. So the minimum is the lesser of the two
+parity masses.
+
+The parity masses have closed forms. With d' = j'shift * h,
+x = l h - v_0^*, and K_Cauchy(x) = 1/(pi (1 + x^2)),
+
+ Xi(x) = sum_{nu in Z} K_Cauchy(x + 2 nu d')
+ = (1/(2d')) * (1 + 2 sum_{m>=1} exp(-m pi/d') cos(m pi x/d')),
+
+using the Fourier transform identity
+int K_Cauchy(x) exp(i xi x) dx = exp(-|xi|) and summing over periods.
+The even mass is h Xi(x). The odd mass is h Xi(x + d'), which is never
+larger: write Xi in closed form with r_0 = exp(-pi/d'),
+
+ Xi(x) = (1 - r_0^2) / (2 d' (1 + r_0^2 - 2 r_0 cos(pi x/d'))),
+
+and cos is maximal at x = 0.
+
+Summing the odd masses over the j'shift central representatives, up to
+O(h) error, gives the integral
+
+ int_{-d'/2}^{d'/2} Xi(x + d') dx
+ = 1/2 - (2/pi) arctan( exp(-pi/d') ).
+
+This is smallest at the largest possible d', which is S. So the cover
+weight is at least M.
+
+Second, directly by Riemann sum,
+
+ sum_{|l| <= S/h} k_l > J_*.
+
+Table. Round |v_0^*| up to w in {1,2,3,4,5}. Set
+
+ w 1 2 3 4 5
+ S 1.5 2.1 2.8 3.6 4.4
+ M .419 .357 .297 .246 .208
+ J_* .524 .453 .380 .335 .292
+
+These are conservative, in the sense of the four inequalities
+
+ M < 1/2 - (2/pi) arctan( exp(-pi/S) ) - .0002
+ J_* < (1/pi)( arctan(S+w) + arctan(S-w) ) - .0002
+ T_lim(v) = 6 f(v) - .1
+ T_lim(w-1) > max{ 2.006/M,
+ 1.005 / (J_* - (1 + 2 J_*/M)^2 M/16) }.
+
+All of them can be checked with rational arithmetic, without any
+sensitive numerical integration:
+
+ use 3.14159 < pi < 3.14160;
+ bound exp(-pi/S) from above by a truncated exponential series;
+ write arctan(S+w) = pi/2 - arctan(1/(S+w));
+ the remaining arctangent arguments have modulus at most .6, so
+ evaluating their alternating series to six terms costs at most
+ .6^{15}/15;
+ lower-bound exp(u) in f(w-1) by 1 + u + u^2/2.
+
+The resulting worst-endpoint values are
+
+ w 1 2 3 4 5
+ max{...} 4.96 5.79 6.94 8.26 9.71
+
+while
+
+ T_lim(w-1) = 6 exp((w-1)^2/20) - .1
+
+equals about 5.90, 6.52, 7.41, 8.73, 10.51. Each exceeds the
+corresponding maximum.
+
+Now the contradiction. Put R = G_cal/M and T_0 = T_lim(v_0^*), which is
+at least T_lim(w-1). Then 2R < T_0, and there exists an integer s with
+
+ |s| <= S/h, 2R + q_s < T_0, 2g_s + q_s < T_0.
+
+Otherwise, pointwise on the whole interval,
+
+ q_s >= (T_0 - 2R) max( 0, 1 - 2 g_s/T_0 ),
+
+so by the cover bound
+
+ Q >= (T_0 - 2 G_cal/M)(J_* - 2 G_cal/T_0),
+
+and minimizing this quadratic in G_cal gives
+
+ G_cal + Q >= T_0 ( J_* - (1 + 2 J_*/M)^2 M/16 ) > 1.004,
+
+contradicting G_cal + Q <= 1.003. Choose the witness with s >= 0 by
+symmetry.
+
+For each 0 <= j'shift <= s, choose a reference pair (e_j, e_j +
+j'shift) whose end deficits sum to at most R. Such a pair exists: if
+the minimum were R' > R, the clipped deficits min(1, g_l/R') would be a
+fractional cover of cost G_cal/R' < M, contradicting the cover bound.
+For j'shift = 0, minimize 2 g_l and use the total kernel mass > .999
+together with M < .499.
+
+Suppose the claim fails. At every pair (l, -l), the failure bounds the
+modulus of F(l) - G(l) by Lambda/m_l. At all relevant indices
+2 g_l < T_0 < 50, so the cap is inactive and m_l = exp(-g_l E). Write
+
+ U(l) = arg F(l) - arg G(l) modulo 2 pi.
+
+By reality of F and G, U(-l) = -U(l). Set
+
+ d_j = U(e_j + j) - U(e_j), so d_0 = 0.
+
+The failed quadruple product at frequencies
+
+ (e_j + j, -e_j, -e_{j-1} - j + 1, e_{j-1})
+
+sums to 1, so it bounds d_j - d_{j-1} mod 2 pi by 100 Lambda exp(2 R E):
+both products have modulus at least exp(-2RE)/16, and two complex
+numbers differing by at most Lambda, both of modulus at least v, have
+angular distance at most pi Lambda/v.
+
+Summing over j gives
+
+ |d_s| <= 100 N Lambda exp(2 R E) modulo 2 pi.
+
+Similarly, the failed triple product at (e_s + s, -e_s, -s) sums to 1
+and gives
+
+ |U(s) - d_s| <= 100 Lambda exp( (R + g_s) E ).
+
+But R + g_s + q_s <= max(2R + q_s, 2g_s + q_s) < T_0. Therefore
+
+ |F(s) - G(s)| / m_s
+ <= Lambda/m_s^2
+ + 100 Lambda ( exp((R+g_s)E) + N exp(2RE) )
+ < exp(-q_s E),
+
+using log N <= 10^{-8} E and R + g_s + q_s < T_0 < 50. That contradicts
+the uncapped definition of q_s, which was supposed to stay below 50. The
+claim holds.
+
+Consequence. We now have r in {2,3,4} and angles theta_1,...,theta_r
+with
+
+ |theta_s| <= 10^4 a and |theta_1 + ... + theta_r| <= .01 t,
+
+and a product discrepancy larger than Lambda.
+
+
+========================================================================
+10. SHUFFLE: MERGING THE COPIES
+========================================================================
+
+The discrepancy is a difference of products. We expand it.
+
+10.1 Components.
+
+Each array copy contributing a factor (one F or one G) is a bounded
+combination of three component types:
+
+ (C1) a constant "latest valid" probe with a single wildcard position
+ (value 1),
+
+ (C2) the stage-j probe,
+
+ (C3) the complex conjugate of the stage-j probe (which also
+ conjugates the masks).
+
+Each copy also carries the Gaussian weight v(m_s), the phase
+exp(i theta_s m_s), and a scalar coefficient of modulus at most 1 (from
+the floor, from 1/Y, from 1/Z_j, and from the projection c).
+
+10.2 Categories.
+
+Group the expanded products into categories. A category records
+
+ for each copy, its type (C1), (C2), or (C3);
+
+ the total weak ordering of all selected source positions across all
+ copies, including all equalities, i.e. all ties.
+
+The number P of nonempty categories satisfies log P = o(B), uniformly.
+
+Why. Once the types are fixed and all exact adjacencies inside each copy
+are honoured, each copy contributes at most 4^{j+1} contiguous blocks.
+Specify the weak order of the block starts: at most m! 2^m possibilities
+for m <= 4^{j+1} blocks. Then process blocks in start order. A new start
+either lies strictly beyond the maximum end index seen so far, or its
+offset from that maximum is an integer between 0 and k-1. This
+determines the entire weak ordering. The total number of categories is
+at most 81 (4k+1)^{4^{j+1}}, so
+
+ log P = O( 4^j log( 4k + 4^{j+1} + 2 ) ) = O(4^j b) = o(B),
+
+including the case k = 1.
+
+10.3 Merged tuples.
+
+Fix a category. All of its exact adjacency constraints act across
+adjacent distinct slots of the merged order: if an old exact edge
+spanned an intervening distinct slot, the category would be empty.
+Collect these as the merged exact edges, and declare the rest loose.
+
+The merged order has k' <= 4k slots and at most as many blocks as the
+sum of the old block counts (each old block is contained in one merged
+block, and each merged block contains at least one old block). Tie
+bit-predicates are multiplied. Retain every old mask unchanged; each
+contributes at most 4 copies here. The wildcard component contributes
+no mask.
+
+Coverage. Every loose merged edge is covered either by an inherited
+mask or by the new mask we are about to add. Consider a copy with a slot
+at the left endpoint of a loose merged edge. If that slot is the copy's
+latest, it is not the overall latest (since the copy is not the
+component chosen last), so it contributes a new covering interval. If it
+is not the copy's latest, then the copy has a consecutive pair
+straddling the edge; that pair was loose in the copy, so an old mask
+already covered it.
+
+Latest component and differences. Choose the component whose latest
+selected index is last in the category, with ties broken
+deterministically, and relabel it as s = r. Let Y_0 be the overall
+latest offset. Put
+
+ g_r = 0,
+ g_s = Y_0 - m_s >= 0 for s < r,
+ g_bar = (g_1 + ... + g_r)/r,
+ Q(g) = sum_s (g_s - g_bar)^2.
+
+10.4 Phase bookkeeping.
+
+The product of phases factorizes:
+
+ prod_{s=1}^r exp(i theta_s m_s)
+ = exp( i (theta_1 + ... + theta_r) Y_0 )
+ * prod_{s<r} exp( -i theta_s g_s ).
+
+So the outer phase carries the tiny sum theta_1 + ... + theta_r, and the
+inner phases live in the differences g_s. The inner phases become a mask
+of level j. This is the only place where the relaxation of the angle
+budget pays off.
+
+
+========================================================================
+11. CENTER FOURIER EXPANSION: AN ADMISSIBLE MASK
+========================================================================
+
+The Gaussian weight is not yet a mask in the sense of (M1) and (M2). We
+convert it.
+
+Momentarily vary Y_0 as a real variable and define
+
+ W_g(Y_0) = exp(t Y_0) * prod_{s=1}^r v(Y_0 - g_s).
+
+CLAIM 11.1. There is a centred Gaussian probability nu(d omega) with
+variance r / ((1 - zeta) sigma^2), where zeta = 10^{-5}, such that
+
+ W_g(Y_0) = exp(1.203 B) int_R exp(i omega Y_0) K(g, omega) nu(d omega),
+
+where
+
+ K(g, omega) = exp(-1.203 B) * W_g_hat(omega) / (d nu / d omega),
+
+and W_g_hat is the Fourier transform
+
+ W_g_hat(omega) = (2 pi)^{-1} int_R W_g(Y_0) exp(-i omega Y_0) dY_0.
+
+Moreover, for all real omega and all g_s >= 0,
+
+ |K(g, omega)| <= exp( - epsilon t (g_1 + ... + g_r) ). (11.1)
+
+Sketch of proof. Scale:
+
+ z = (Y_0 - g_bar) sqrt(r) / sigma
+ rho = t sigma / sqrt(r)
+ u' = omega sigma / sqrt(r)
+ d_s = (g_s - g_bar) / sigma
+ p_ell(x) = sum_{k=0}^{ell} (x^2/2)^k / k!.
+
+Shifting the contour by -i u', which is legal because the integrand is a
+polynomial times a Gaussian, gives
+
+ W_g_hat = (2 pi)^{-1/2}
+ * exp( (t - i omega) g_bar - Q(g)/(2 sigma^2)
+ + rho^2/2 - (u')^2/2 - i rho u' )
+ * E_{X ~ N(rho,1)} prod_s p_ell( (X - i u')/sqrt(r) - d_s ).
+
+The product of polynomials is at most zeta^{-r ell} times
+exp( (zeta/2)( X^2 + (u')^2 + Q(g)/sigma^2 ) ). Integrating X gives
+
+ |W_g_hat(omega)| / (density of nu at omega)
+ <= zeta^{-r ell}/(1 - zeta)
+ * exp( t g_bar - (1 - zeta) Q(g)/(2 sigma^2)
+ + 2.4 B / (2 r (1 - zeta)) ).
+
+Now Q(g) >= r g_bar^2/(r-1), because g_r = 0. So the exponent above,
+plus the epsilon t sum_s g_s term, which equals epsilon t r g_bar, is at
+most
+
+ (1 + r epsilon)^2 (2.4 B) (r-1) / (2 (1 - zeta) r).
+
+For r = 4 this is at most .9001 B. Add 2.4B/(2r(1-zeta)) <= .3001 B and
+the negligible log term; the total is below 1.203 B. Hence (11.1).
+
+Interpretation. K is a Gaussian bump in g. For now it decays like
+exp(-epsilon t |g|_1), which is the (M1) behaviour, before we make it
+compact.