dependent-bases.md
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Dependent bases
- 2026-09-14 [Proved] The collapse theorem: for one root
r >= 2, basesbase_i = r^(e_i)withi = 1..m, any dimensiondim >= 1and any digit setsA_iinside{0,...,base_i - 1}^dimeach containing0, putM = lcm(e_1,...,e_m)andB = r^M; thencap_i F(base_i, A_i) = F(B, A)exactly, whereA = (cap_i F(base_i, A_i)) cap [0, B)^dimis the joint set's own bottom block, the count law is exact atcard {n in [0, B^level)^dim : n in F(B, A)} = (card A)^levelfor everylevel >= 0, and the attractor of{x -> (x + a)/B : a in A}has Hausdorff and box dimensionlog card A / (M log r). Proof in four steps:e_idividingMmakes everye_i-group of base-rpositions sit inside oneM-group, so no group straddles andlcmis forced rather than chosen; each digit at basebase_iis then a function of one base-Bdigit, so the constraint is per base-Bdigit with nothing carried between them; the top base-Bdigit is read with its fullM / e_ibase-base_idigits while the base-base_iexpansion ofnstops at its top nonzero digit, so the two readings agree exactly when0 in A_i; and membership one digit at a time gives the count with no error term while the maps send the unit cube to boxes with disjoint interiors, so the open set condition gives the dimension. No step usesdim = 1and no step asksA_ito be a product across thedimaxes, sodim >= 2is covered with no extra hypothesis, compound designs included. Across two dependence classes the theorem says nothing. Witness: lab/py/base-collapse verbsblocks,dim,check;bases.md:92,bases.md:101-107. - 2026-09-14 [Proved] Multiplicative dependence is an equivalence relation on bases
>= 2and each class is the set of integer powers of its least member: reading a base by its vector of prime exponents turnsp^a = q^baseinto parallel vectors, so a class is the set of integer points on one ray through the origin, its least memberris the primitive vector on that ray and every member isr^efor one integere >= 1; transitivity is one line,p^a = q^baseandq^c = s^dgivep^(ac) = s^(bd). The half of the rule that follows is collapse first: partition the bases into dependence classes and replace each class by the single design the collapse theorem gives it, with its own baser^M, its own digit set and its exact dimension; a transversality budget inside a class is not allowed, by the Refuted row below. Witness: lab/py/base-collapse verbblocks;bases.md:88,bases.md:126. - 2026-09-14 [Verified] Four dependent cells rebuilt from their digit sets alone, root, exponents,
M, block set and exact dimension: base4on{0,1}with base8on{0,1,2,3}givesr = 2,M = 6,A = {0,1,16,17}and dimension exactly1/3; base4on{0,1}with base16on{0,1,4,5}givesM = 4,A = {0,1,4,5}and exactly1/2; base9on{0,1,2}with base27on{0,...,8}givesr = 3,M = 6, nine blocks and exactly1/3; bases4, 8, 16on{0,1},{0,1,2,3},{0,...,7}giveM = 12, sixteen blocks and exactly1/3. The block set is built twice per cell and asserted equal, once by sieving allr^Mbase-rwords against the per-group constraint and once by testing every integer belowr^Mfor membership in each original design. Witness: lab/py/base-collapse verbsblocks,dim;bases.md:115-120. - 2026-09-14 [Verified] The exact count law survives brute force to
10^13on all four cells, three checks each: every element ofF(B, A)below10^13passes a digit test in each original base; the joint count from enumerating the lowest-dimension original design (2^22 - 1elements at three cells,3^14 - 1at the base-9 cell) and filtering it by the others equals the count ofF(B, A)below10^13, which with the first check gives set equality, the four counts reading32767,4194303,19682,32767; and the joint count belowB^levelis exactly(card A)^levelat everylevelwithB^level <= 10^13, reachinglevel = 7and16384,level = 10and1048576,level = 4and6561,level = 3and4096. Witness: lab/py/base-collapse verbcheck;bases.md:124. - 2026-09-14 [Verified] A dependence class can give a block count that is not a power of its root, so the exact dimension can be irrational: base
4on{0,1,2}with base16on the full digit set givesr = 2, exponents2, 4,M = 4andcard A = 9, dimensionlog_2(9) / 4 = log_2(3) / 2. The dimension islog_r(card A) / Malways, and for the least baserof the class it is rational exactly whencard Ais a power ofr, since a primitiveris no perfect power. Witness: lab/py/base-collapse verbdimcellI;bases.md:107. - 2026-09-14 [Verified] The hypothesis
0 in A_iis sharp and not decoration: atr = 2withbase_1 = 2onA_1 = {1}andbase_2 = 4on the full digit set,M = 2and the bottom block isA = {0, 1, 3}, yetF(4, A)holds4,5,12and13, whose base-2 words carry a digit outsideA_1, so the collapse strictly over-counts the joint set. Witness: lab/py/base-collapse verbblocks;bases.md:105. - 2026-09-14 [Conjecture] Budgeting across the collapsed classes alone, with
mthe number of classes and not the number of bases, is the right second step: nothing here shows a collapsed class behaves in a cross-class budget like an ordinary design of the same dimension, so the step is assumed. Witness:bases.md:126; no proof and no lab verb. - 2026-09-14 [Refuted] The naive budget
sum_i dim A_i - (m - 1)is not an upper bound on a multiplicatively dependent cell and is low on every one tested: base4on{0,1}with base8on{0,1,2,3}has exact dimension1/3against a budget of1/2 + 2/3 - 1 = 1/6; base4on{0,1}with base16on{0,1,4,5}has exact dimension1/2against a budget of1/2 + 1/2 - 1 = 0; base9on{0,1,2}with base27on{0,...,8}reads1/3against1/6; and the three-base cell4, 8, 16reads1/3againstmax(0, 1/2 + 2/3 + 3/4 - 2) = 0, the budget being read at zero because no dimension is negative and the raw sum there is-1/12. The second cell is the clean failure:F(16, {0,1,4,5}) = F(4, {0,1})as sets, since a base-16 digit lies in{0,1,4,5}exactly when both of its base-4 digits lie in{0,1}, so the budget prices the intersection of a set with itself at dimension0while the truth is that set, of dimension1/2. A budget built on transversality cannot be applied inside a dependence class. Witness: lab/py/base-collapse verbsdim,check;bases.md:113,bases.md:115-120.