(1) Lemma 7.1
  L = 1: maximum of c_1 over all binary cycles with n <= 16: 1 (first at (2, 'AB')); 2/(L+1) = 1
  L = 2: maximum of c_1 over all binary cycles with n <= 16: 2/3 (first at (3, 'ABB')); 2/(L+1) = 2/3
  L = 3: maximum of c_1 over all binary cycles with n <= 16: 1/2 (first at (4, 'ABBB')); 2/(L+1) = 1/2
  L = 4: maximum of c_1 over all binary cycles with n <= 16: 2/5 (first at (5, 'ABBBB')); 2/(L+1) = 2/5
(2) Lemma 7.2 on random grids
  grids tested: 170  inequalities (12) per grid: ABB d=3: 6, d=4: 10; ABBB d=3: 5, d=4: 8
  largest lhs/rhs seen: 1  (1 means attained)
(3) Corollary 1.8
  d = 4: c_4 = [Fraction(4, 1), Fraction(12, 1), Fraction(16, 1), Fraction(8, 1)]  combination = [Fraction(7, 1), Fraction(12, 1), Fraction(16, 1), Fraction(8, 1)]  difference = [Fraction(3, 1), Fraction(0, 1), Fraction(0, 1), Fraction(0, 1)] >= 0; bound 62/3
  d = 5: c_5 = [Fraction(5, 1), Fraction(20, 1), Fraction(40, 1), Fraction(40, 1), Fraction(16, 1)]  combination = [Fraction(12, 1), Fraction(20, 1), Fraction(40, 1), Fraction(40, 1), Fraction(16, 1)]  difference = [Fraction(7, 1), Fraction(0, 1), Fraction(0, 1), Fraction(0, 1), Fraction(0, 1)] >= 0; bound 64
  d = 3, ABBB: c_3 = [Fraction(3, 1), Fraction(6, 1), Fraction(4, 1)]  combination = [Fraction(3, 1), Fraction(6, 1), Fraction(4, 1)]  difference = [Fraction(0, 1), Fraction(0, 1), Fraction(0, 1)] ; bound 11/2
  lower bounds by stacking: 3*6 = 18 , 9*6 = 54 , 3*(8/5) = 24/5
(4) Remark 7.3: exact linear programs
  ABB, d = 4, k <= 3: optimal value 62/3 (note: 62/3); optimal D = ['0', '5/9', '2/3', '5/12']
     dual solution (inequality: multiplier): [((3,), '4'), ((1, 1, 1), '1'), ((1, 1, 2), '2')]
  ABB, d = 5, k <= 3: optimal value 64 (note: 64); optimal D = ['0', '2/9', '2/3', '2/3', '7/18']
     dual solution (inequality: multiplier): [((3,), '8'), ((4,), '16'), ((1, 1, 2), '4'), ((1, 2, 2), '4')]
  ABB, d = 6, k <= 3: optimal value 596/3 (note: 596/3); optimal D = ['0', '0', '1/2', '2/3', '2/3', '11/24']
     dual solution (inequality: multiplier): [((4,), '64'), ((1, 4), '12'), ((1, 2, 2), '14'), ((1, 2, 3), '8')]
  ABBB, d = 3, k <= 2: optimal value 11/2 (note: 11/2); optimal D = ['3/10', '1/2', '2/5']
     dual solution (inequality: multiplier): [((2,), '3'), ((1, 1), '1/2'), ((1, 2), '2')]
  D = (0, 5/9, 2/3, 5/12) is feasible for d = 4 with c_4 = 62/3
  the constant sequence D_j = 6/13 is feasible for d = 4..12 with value (6/13)(3^d-1)/2 > 2*3^(d-2): True
  upper bounds of (1): (3^d-1)/2 * C_1 = 80/3 242/3 13/2  and (26/2)(2/3) = 26/3
RESULT PASS
