In this note I prove that the muffin bound S(r) is continuous on several intervals, including [4/3,8/5]-{3/2} and [9/5,15/7]-{2}. I first prove this for optimal "binary" solutions, which are those in which every muffin splits in exactly 2 pieces, then address the general optimum. Let S'(r) denote the optimum binary solution value. Consider example r = 11/6 in interval [9/5,2). The first level decomposition of a solution to 11/6 is: r = m/p = 11/6, 22 muffin pieces [3] x 2 Z parts, 6 pieces [4] x 4 A parts, 16 pieces 5 excess muffins (10 pieces) in A There are fewer Z pieces (6) than A pieces (16), so if a solution with all Z pieces the same size (11/18) and no piece smaller than a Z piece complement (7/18) is feasible, it will be optimal. The 6 Z piece complements must be A pieces. Pack them in the 6 A parts as evenly as possible, accounting for 6 of the 11 muffins. The remaining 5 muffins (10 pieces) all pack in A parts. Call these "m" pieces. This gives a tentative solution [a,z]*6, [m*2]*5; [z*3]*2; [a*2,m*2]*2, [a,m*3]*2 The "fill set" is this decomposition is the portions of the A parts that contain m pieces. This fill set has 2 2-parts, the [m*2] subparts of the [a*2,m*2] 4-parts, and 2 3-parts. The key idea of the proof is that we can reduce problems r in [9/5,2) to smaller muffin problems whose first level structure is r's fill set. Then an affine transform of the reduced problem solution realizes a solution for the fill set: fill set for 11/6 has 2 2-parts and 2 3-parts fill set reduces to muffin problem 5/4 r = m/p = 5/4, 10 muffin pieces [2] x 2 Z parts, 4 pieces [3] x 2 A parts, 6 pieces 1 excess muffins (2 pieces) in A with lexical optimum [z*2]*2; [a*2,m]*2 = [5/8*2]*2; [3/8*2,1/2]*2 The 2-parts and 3-parts in the reduced problem have average piece sizes (5/4)/2 = 15/24 and (5/4)/3 = 10/24 respectively. The 2-parts in the 11/6's fill set have average piece size (11/6-7/18*2)/2 = = 57/108. The fill set 3-parts have average piece size (11/6-7/18)/3 = 52/108. The affine map from the 5/4 solution to an 11/6 fill set solution thus maps 15/24 |-> 57/108 and 10/24 |-> 52/108, which shifts piece sizes closer to 1/2 by a factor of 9/2. The solved 11/6 A-parts from this map are: 1/108 * { [42*2,57*2]*2, [42,52*3]*2 } Theorem: S(r) = 1-r/3 on [9/5,2). Proof: S(r) <= 1-r/3 is a long known lower bound. Here we realize solutions for all r in [9/5,2), proving the bound is tight. There is a function Q which reduces each instance r in [9/5,2) to a smaller problem Q(r) matching r's fill set, constructed as in the above example for r = 11/6. Q turns out to be a Mobius transform. It is uniquely determined by the image of 3 of its points, e.g, Q(11/6) = 5/4 Q(9/5) = 1 Q(15/8) = 3/2 and we have Q(r) = (7r-12)/(2r-3). Q maps [9/5,2) to [1,2). For reduced problem t = Q(r), we have numerator(t) < numerator(r) and t's total piece count divides r's fill set piece count, i.e., 2k * numerator(t) = r's fill set piece count for some k>0. The affine map g from solution(t) to solution(fill_set(r)) shifts each piece size x toward 1/2 with some scale factor g_s: g(x) = (x-1/2)*g_s + 1/2 A straightforward calculation finds g_s = g_s(r) = 2r/3-1. Let S_2(r) = g(S'(Q(r))), the size of the smallest piece in r's fill set. Here's a direct check that S_2(r) > S(r) on [9/5,2): For r in [9/5,15/8], 1 <= t = Q(r) <= 3/2, S'(t) > 1/4, and 1/5 <= g_s(r) <= 1/4. Thus S_2(r) = g(S'(t)) > g(1/4) >= 7/16 > 2/5 >= S(r). For r in [15/8,2), 3/2 <= t = Q(r) < 2, S'(t) > 1/3, and 1/4 <= g_s(r) < 1/3. Thus S_2(r) = g(S'(t)) >= g(1/3) >= 4/9 > 3/8 >= S(r). Since S_2(r) is the smallest fill piece size and S_2(r) > S(r), we've just proved this reduction finds the optimal second largest distinct piece size for r in [9/5,2). Note finally that g defines a bijection between binary solutions to Q(r) and binary solutions to r's fill set which preserves lexical optimality. QED. We have S'(r) = S(r) for all nonintegral r >= 3/2. The S'/S bounds and reduction functions for intervals [4/3,8/5] and [9/5,15/7] are: S'(r) = 1-r/2 on [4/3,3/2), Q<3,1>(r) = (5r-6)/(2r-2) Q: [4/3,3/2) |-> [1,3/2) S(r) = r/4 on (3/2,8/5], Q<4>(r) = (23-14r)/(26-16r) Q: [8/5,3/2) |-> [1,3/2) S(r) = 1-r/3 on [9/5,2), Q<4,1>(r) = (7r-12)/(2r-3) Q: [9/5,2) |-> [1,2) S(r) = r/5 on (2,15/7], Q<5>(r) = (20-5r)/(9-2r) Q: [15/7,2) |-> [1,2) I haven't yet calculated the affine scale factor function for 3 of these cases, though it's easy to find for any individual instance. Note that 1/4 < S'(r) <= 1/3 on [4/3,3/2) for the binary solution optimum, but the general optimum is S(r) = 1/3. This general optimum feasibility is also provable by a Mobius reduction to binary solutions of smaller instances. S(r) = 1/3 for r = m/p in [4/3,3/2) is achieved by splitting muffins into either 2 pieces or exact thirds so there are exactly 3p pieces and all persons get 3 pieces. The "fill sets" in this range are the person pieces left after removing all exact 1/3 pieces, a set of 2-parts and 3-parts. r = 4/3 has all 2-parts for its fill set, and r approaching 3/2 approaches a fill set of all 3-parts. The reduction which maps these fill sets to smaller problems is Q_3(r) = 3r-3 For example, consider r = 22/15 in [4/3,3/2). We achieve S(r) = 1/3 with 1 muffin [1/3*3] and 21 2 piece muffins [m*2]*21. The solution for persons has the form [1/3,m*2]*3, [m*3]*12 Removing the [1/3] pieces leaves a fill set of 3 2-parts [m*2] and 12 3-parts [m*3]. Q_3(r) = 3r-3 reduces this to Q_3(22/15) = 7/5 with binary (not general) muffin decomposition r = m/p = 7/5, 14 muffin pieces [2] x 1 Z parts, 2 pieces [3] x 4 A parts, 12 pieces 5 excess muffins (10 pieces) in A and optimal binary solution [z*2]; [a,m*2]*2, [m*3]*2; 7/5's fill set reduces (by Q_2(r) = Q(r) = (5r-6)/(2r-2)) to Q(7/5) = 5/4 with previously seen optimal solution [z*2]*2; [a*2,m]*2 = [5/8*2]*2; [3/8*2,1/2]*2 This maps affinely back to optimal binary solution 1/20 * { [14*2]; [6,11*2]*2, [9*2,10]*2 } for 7/5. Replicated 3 times, this maps affinely into the fill set for 22/15, giving the optimal person solution 1/60 * { [20,34*2]*3; [26,31*2]*6, [29*2,30]*6 } for 22/15. Note that "ternary" optimal solutions don't map affinely into fill sets in these reductions. They are a top level exception, but only binary solutions (in this note) are targets in the recursive reduction to smaller problems. There are two exceptions S'(r) < S(r) where the binary optimum is not the general optimum. One exception is ternary optima where S'(r) < 1/3 = S(r). I've just proved continuity and optimality of both the general and the binary optimum on [4/3,3/2), by Mobius reductions that prove fill set solution feasibility. The other exception is S'(r) = 1/2 < 1 = S(r), which only occurs at integral r. This case and other optimal solutions that use unsplit muffins are simply handled by rejoining any combinable pairs that a binary solution might have.