diff --git a/research/hidden-window-attainment-budgets/PROOF.md b/research/hidden-window-attainment-budgets/PROOF.md new file mode 100644 index 0000000000..f4667317d7 --- /dev/null +++ b/research/hidden-window-attainment-budgets/PROOF.md @@ -0,0 +1,131 @@ +# Exactly which budgets attain the hidden-window bound + +This scoped proof was produced by Nick Petalas. Apache-2.0. + +## Model and predecessor + +Fix integers h>=1 and N>=1. On ranks 2^j for 0<=j<=2h+2, let A_j,B_j +be nonnegative integer call multiplicities, at most N calls per profile. +Set Q(x)=sum_j(B_j-A_j)x^j and F(p)=Q(2^p). Require endpoint zeros of +multiplicity at least h at p=0,1 and two distinct interior zeros alpha= log_2((Z+1)/(Z-1)). +``` + +W counts calls on each side after cancelling common calls. It is not the +combined count. Eligible profiles have Z>1. The source supplies the bound, +one infinite attaining family and one further example. Its Open next +question 1 asks for the complete attainment classification. Our result +answers that question; no source executable or census is replayed. + +## Classification + +The bound is attainable at budget N if and only if L belongs to + +```text +{4n^2-1, 4n^2 : n>=3} union {2n(n+1) : n>=4}. +``` + +All attaining residuals are as follows, with the positive leading coefficient +convention above. Include both when both tests hold. + +```text +L=4n^2-1 or 4n^2, n>=3: + (a,z,c)=(n(n-1), 2n^2-1, n(n+1)), D=4n^2-1; + +L=2n(n+1), n>=4: + (a,z,c)=(n(n-1)/2, n^2+n-1, (n+1)(n+2)/2), D=L. +``` + +This is an existence statement at a budget. Other profiles at that budget +may have strictly wider windows. Elsewhere the bound is unattainable; this +does not identify the true optimum or imply that no eligible profile exists. + +## Necessity + +The quadratic formula gives s/r=(z+sqrt(Delta))/(z-sqrt(Delta)). Its +denominator is positive since c>0. Put k=R(1)=a+c-z. Strict interior +roots imply that k is a positive integer. Since D=2z+k<=L, we have 0=1/Z. Strict monotonicity of (1+t)/(1-t) on 0=1 and 01, hence u=c-a is a positive integer. Using a+c=z+k yields + +```text +u^2=(a+c)^2-4ac=2kz+k^2+1. +``` + +For k=1, u^2=2z+2 is even. Write u=2n, n>=1. Then + +```text +z=2n^2-1, a=n(n-1), c=n(n+1), D=4n^2-1. +r=(n+1)/n, s=n/(n-1). +``` + +Positivity and s<2 force n>=3. From z=Z and D<=L, exactly +L=4n^2-1 and L=4n^2 remain. + +For k=2, u^2=4z+5 is odd. Write u=2n+1, n>=0. Then + +```text +z=n^2+n-1, a=n(n-1)/2, c=(n+1)(n+2)/2, +D=2n(n+1)=2z+2, r=(n+2)/n, s=(n+1)/(n-1). +``` + +Positivity and s<2 force n>=4. Now D<=L leaves only L=2n(n+1). +These cases exhaust k and prove necessity. + +## Sufficiency and every profile realization + +Each listed triple has positive integer coefficients, Delta=1, strict +interior roots, z=Z and SD<=N. Form Q using the endpoint factors above. +All its roots are positive, so its coefficients alternate strictly in sign. +Let A be its negative coefficient part and B its positive coefficient part. +Then Q(1)=Q(2)=0 gives equal counts and rank masses, and + +```text +sum A_j = sum B_j = |Q(-1)|/2 = S(a+z+c) = SD. +``` + +The orientation makes Q positive on (r,s) and negative on (1,r) and (s,2). +Thus these profiles satisfy the premises and attain the bound. + +Every profile pair giving this Q is obtained by adding the same nonnegative +integer vector H to both disjoint coefficient parts. It is permitted exactly +when sum H_j<=N-SD. Writing N=SL+rho with 0<=rho