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Algorithm cost equals (1+1/c)(1 + 1/c)(1+1/c) times its dual objective

Proved
PrimalDualOnline.SkiRental.alg_cost_eq_ratio_mul_dual

by moutei · Sep 16, 2026 · Mathlib 0df444a (Lean v4.33.1)

competitive-analysislinear-programmingonline-algorithmsprimal-dualski-rental

Let B≥1B \ge 1B≥1 be the purchase price, k≥0k \ge 0k≥0 the number of ski days, and run the fractional primal-dual algorithm with rate c∗(B)=(1+1/B)B−1c^{*}(B) = (1 + 1/B)^{B} - 1c∗(B)=(1+1/B)B−1. Write ALG(B,k)=Bxk+∑j<kzj\mathrm{ALG}(B,k) = B x_k + \sum_{j<k} z_jALG(B,k)=Bxk​+∑j<k​zj​ for the primal objective of the algorithm's solution and DUAL(B,k)=∑j<kyj\mathrm{DUAL}(B,k) = \sum_{j<k} y_jDUAL(B,k)=∑j<k​yj​ for the dual objective it has accumulated. Then

ALG(B,k)  =  (1+1c∗(B))⋅DUAL(B,k).\mathrm{ALG}(B,k) \;=\; \left(1 + \frac{1}{c^{*}(B)}\right) \cdot \mathrm{DUAL}(B,k) .ALG(B,k)=(1+c∗(B)1​)⋅DUAL(B,k).

This is an exact equality in both directions, with no error term. Combinatorially, on each active day the primal objective grows by 1+1/c1 + 1/c1+1/c while the dual grows by 111, and on each day after the buy variable has reached 111 neither grows; summing over the kkk days turns the per-day relation into an identity between the two totals, each equal to (1+1/c∗(B))min⁡(k,B)\left(1 + 1/c^{*}(B)\right)\min(k, B)(1+1/c∗(B))min(k,B).

This statement is the pivot of the mission. It converts a question about the algorithm's cost, which depends on the whole request sequence, into a question about the dual objective, which weak duality makes a lower bound on the offline optimum.

Source correspondence — read this before attacking the target.

  • What the thesis states (p. 18, claim (ii)): "in each iteration, the ratio between the change in the primal and dual objective functions is bounded by (1+1/c)(1+1/c)(1+1/c)." The headline claim is an inequality, and it is stated per iteration, not in aggregate.
  • What the thesis's proof computes (p. 19): on an active day, "the increase in the primal objective function is BΔx+zj=x+1/c+1−x=1+1/cB\Delta x + z_j = x + 1/c + 1 - x = 1 + 1/cBΔx+zj​=x+1/c+1−x=1+1/c" — an exact per-day equality, and the dual increase is exactly 111.
  • What this Lean theorem states: an exact equality between the aggregate primal cost and (1+1/c)\left(1 + 1/c\right)(1+1/c) times the aggregate dual objective.
  • Introduced by the formalization: this is a derived strengthening, not a verbatim transcription. It strengthens the source's "bounded by" to an equality, and it aggregates what the source states per iteration. Both steps are licensed by the exact computation in the source's own proof, but neither is how the source phrases the claim. A reader comparing the mission against Chapter 3 should expect to find the per-day equality and the per-iteration bound there, and to find the aggregate equality only here.

Formalization Note. The primal objective is read at the final buy level xkx_kxk​ together with the rent variables at indices 0,…,k−10, \dots, k-10,…,k−1; index kkk contributes a buy term but no rent term. Both totals vanish at k=0k = 0k=0 and are constant in kkk once k≥Bk \ge Bk≥B.

Preamble
import Definitions.Def_PrimalDualOnline_SkiRentalLP
import Definitions.Def_PrimalDualOnline_SkiRentalAlg
import Mathlib.Tactic

open PrimalDualOnline.SkiRental
Formal statement
theorem PrimalDualOnline.SkiRental.alg_cost_eq_ratio_mul_dual
    (B : ℕ) (hB : 0 < B) (k : ℕ) :
    algCost B (cOpt B) k = (1 + 1 / cOpt B) * algDualValue B (cOpt B) k := by sorry
Source
Niv Buchbinder, "Designing Competitive Online Algorithms via a Primal-Dual Approach", PhD thesis, Tel Aviv University, 2008, https://www.tau.ac.il/~nivb/download/phd-thsis.pdf, Chapter 3, p. 18 (PDF p. 34): claim (ii) of the analysis ("in each iteration, the ratio between the change in the primal and dual objective functions is bounded by (1+1/c)"), in aggregate form.
Read-back

What the Lean code literally says, in plain math · claude-opus-5

Binders. The statement quantifies over exactly two variables, both explicit: natural numbers BBB and kkk. It carries exactly one hypothesis, 0<B0 < B0<B. There are no implicit arguments, no typeclass assumptions, no other side conditions — nothing constrains kkk relative to BBB, and nothing constrains the free parameter ccc, because the statement instantiates it at one specific value.

The constant. c∗(B)=(1+1B)B−1c^*(B) = \left(1 + \frac{1}{B}\right)^{B} - 1c∗(B)=(1+B1​)B−1, a real base raised to a natural-number power. The scalar on the right-hand side is

1+1c∗(B)  =  (1+1B)B(1+1B)B−1.1 + \frac{1}{c^*(B)} \;=\; \frac{\left(1+\frac{1}{B}\right)^{B}}{\left(1+\frac{1}{B}\right)^{B} - 1}.1+c∗(B)1​=(1+B1​)B−1(1+B1​)B​.

For B≥1B \ge 1B≥1 this is 222 when B=1B = 1B=1 and lies in (ee−1, 2]\left(\tfrac{e}{e-1},\,2\right](e−1e​,2] in general; the statement does not name, bound, or characterize it beyond writing it as that expression.

The sequences. With c=c∗(B)c = c^*(B)c=c∗(B) fixed throughout:

x0=0,xj+1={xj(1+1B)+1c B,xj<1,xj,xj≥1,x_0 = 0, \qquad x_{j+1} = \begin{cases} x_j\left(1 + \dfrac{1}{B}\right) + \dfrac{1}{c\,B}, & x_j < 1,\\[2mm] x_j, & x_j \ge 1, \end{cases}x0​=0,xj+1​=⎩⎨⎧​xj​(1+B1​)+cB1​,xj​,​xj​<1,xj​≥1,​

and, each looking at the same index jjj (not j+1j+1j+1), zj=1−xjz_j = 1 - x_jzj​=1−xj​ if xj<1x_j < 1xj​<1 else 000, and yj=1y_j = 1yj​=1 if xj<1x_j < 1xj​<1 else 000.

The two sides, expanded. The left-hand side is the "primal cost" of the pair (xk,(zj)j<k)(x_k, (z_j)_{j<k})(xk​,(zj​)j<k​): LHS=B⋅xk+∑j=0k−1zj\text{LHS} = B \cdot x_k + \sum_{j=0}^{k-1} z_jLHS=B⋅xk​+∑j=0k−1​zj​. The right-hand side is the scalar times a plain sum: RHS=(1+1c∗(B))∑j=0k−1yj\text{RHS} = \left(1 + \frac{1}{c^*(B)}\right)\sum_{j=0}^{k-1} y_jRHS=(1+c∗(B)1​)∑j=0k−1​yj​. Both sums run over the kkk indices 0,…,k−10,\dots,k-10,…,k−1 and are empty when k=0k = 0k=0.

What each side evaluates to combinatorially. Since yjy_jyj​ indicates xj<1x_j < 1xj​<1, the dual sum counts indices: ∑j<kyj=#{j<k:xj<1}\sum_{j<k} y_j = \#\{j < k : x_j < 1\}∑j<k​yj​=#{j<k:xj​<1}. Unwinding the recursion with c=c∗(B)c = c^*(B)c=c∗(B), as long as the guard holds xj=(1+1/B)j−1c∗(B)x_j = \frac{(1+1/B)^j - 1}{c^*(B)}xj​=c∗(B)(1+1/B)j−1​, so xj<1x_j < 1xj​<1 exactly for j<Bj < Bj<B, xB=1x_B = 1xB​=1 exactly, and xj=1x_j = 1xj​=1 for j≥Bj \ge Bj≥B; hence the count is min⁡(k,B)\min(k,B)min(k,B). The primal side collapses correspondingly, and both sides equal

(1+1(1+1B)B−1)⋅min⁡(k,B).\left(1 + \frac{1}{\left(1+\frac{1}{B}\right)^{B} - 1}\right)\cdot \min(k, B).(1+(1+B1​)B−11​)⋅min(k,B).

Exactness. The claim is an equality of real numbers, not an inequality in either direction and not an asymptotic or approximate relation: no error term, no limit, no "for sufficiently large kkk". The factor is asserted to be attained on the nose, for every kkk and every B≥1B \ge 1B≥1.

Index asymmetry. The two sides do not range over the same indices. The left side evaluates xxx at index kkk — one past the last index appearing in either sum — and adds the kkk terms z0,…,zk−1z_0, \dots, z_{k-1}z0​,…,zk−1​; the right side involves only y0,…,yk−1y_0, \dots, y_{k-1}y0​,…,yk−1​ and has no term at index kkk. Moreover zjz_jzj​ and yjy_jyj​ are both computed from xjx_jxj​, the value before the jjj-th update, whereas B⋅xkB \cdot x_kB⋅xk​ uses the value after kkk updates.

Degenerate cases. k=0k = 0k=0: both sums empty, x0=0x_0 = 0x0​=0, so the claim reads 0=00 = 00=0. k=Bk = Bk=B: xB=1x_B = 1xB​=1 exactly and both sides equal B(1+1/c∗(B))B\left(1 + 1/c^*(B)\right)B(1+1/c∗(B)); yB=0y_B = 0yB​=0 but index BBB is outside the summation range anyway. k>Bk > Bk>B: the sequence is frozen at xk=1x_k = 1xk​=1 and all summands with index ≥B\ge B≥B vanish, so both sides are constant in kkk; the equality for large kkk carries no more information than at k=Bk = Bk=B. B=1B = 1B=1: c∗(1)=1c^*(1) = 1c∗(1)=1, factor exactly 222, and for k≥1k \ge 1k≥1 the claim reads 1⋅1+(1−0)=2⋅11 \cdot 1 + (1 - 0) = 2 \cdot 11⋅1+(1−0)=2⋅1. B=0B = 0B=0 is excluded, though all expressions remain well-formed under 1/0=01/0 = 01/0=0.

What is NOT asserted. No optimality, minimality, or competitive-ratio claim: no offline optimum, adversary, or comparison quantity appears, so nothing says the ratio compares the cost to anything other than ∑j<kyj\sum_{j<k} y_j∑j<k​yj​. Nothing asserts c∗(B)c^*(B)c∗(B) is the best, unique, or only constant with this property, and nothing for any other ccc. No feasibility claim: not that xk+zj≥1x_k + z_j \ge 1xk​+zj​≥1, not that yj≤1y_j \le 1yj​≤1, not that (yj)(y_j)(yj​) is feasible for any dual program; the words "primal" and "dual" in the definition names carry no asserted linear-programming content, and no weak- or strong-duality statement is present. No structural facts about the sequence are asserted, even though they follow from the definitions: not xj≤1x_j \le 1xj​≤1, not monotonicity, not zj≥0z_j \ge 0zj​≥0, not the closed form. Neither side is asserted to equal min⁡(k,B)\min(k,B)min(k,B) or the scaled version — that identification is a consequence of unfolding, not part of the statement. Nothing about limits as B→∞B \to \inftyB→∞ or about e/(e−1)e/(e-1)e/(e−1), nothing about any interpretation as days, prices, or decisions, and nothing about any algorithm's behaviour beyond the literal arithmetic identity.

Human review
  • Endorsed by Shuze Chen · Sep 16, 2026

  • Endorsed by moutei · Sep 16, 2026

    Confirmed by the mission captain (proposal self-audit).

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