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✓ 7 results · all verified · 3 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 4 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Revised simplex linear programming and certificates

1 · Prerequisites

2 · Summary

Objective

This page develops the revised simplex method for finite linear programs in standard form. It defines bases, primal and dual revised-simplex steps, the ratio rules, Phase I with redundant-row removal, and exact primal, dual, unboundedness, and infeasibility certificates. Bland's least-index rule handles degenerate pivots and gives finite termination in exact arithmetic over finitely many bases; this is not a polynomial-time claim. A finite-pivot-budget dual run may return inconclusive. The residual audit reports exact residuals and a primal-dual objective difference for the stored finite data; it does not certify uncertainty in data before storage. Sparse pricing counts apply to the explicitly stated stored-entry stream and make no wall-time or bandwidth claim.

The finite-precision basis service recomputes canonical NA02 QR of each admitted current basis and its transpose, reuses the current factors for primal, multiplier and requested direction solves, and discards them after an established basis replacement. A finite budget and explicit numerical-failure or pivot-inconclusive statuses prevent unvalidated floating signs or ratios from becoming exact pivots. Independent rounded residual certificates and a finite rational upper-bound option accompany the existing exact stored-data audit and objective-gap bound. Only that audit’s exact feasibility, dual-sign and zero-gap tests certify stored-data optimality; small residuals alone certify neither LP optimality nor infeasibility. The exact primal, dual and PhaseI methods retain their existing claims.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Revised simplex standard-form linear program

Definition

Let m,n be nonnegative integers, A∈Rm×n, b∈Rm, and c∈Rn. The standard-form minimization linear program with data (A,b,c) is min⁡{cTx:Ax=b, x≥0}. Its feasible set is P(A,b)={x∈Rn:Ax=b, xj≥0 for every j}, and a feasible x is optimal if cTx≤cTz for every z∈P(A,b). The notation x≥0 is componentwise. The revised-simplex basis algorithms on this page require a full-row-rank matrix, meaning rank⁡(A)=m; a rank-deficient raw system must first be reduced by exact row operations, with inconsistent rows reported as infeasible. For m=0, the empty system has its usual zero-dimensional interpretation.

A vector q∈Rm satisfying ATq≥0 and bTq<0 is called a Farkas infeasibility certificate for this sign convention. Indeed, any feasible x≥0 would give bTq=xTATq≥0.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Basis dictionaries and reduced costs for revised simplex

Definition

Use the standard-form problem of Revised simplex standard-form linear program. Assume A has full row rank m. An ordered basis is a list B=(B1,…,Bm) of distinct column indices for which the square matrix AB=[aB1 ⋯ aBm] is invertible. Write N={1,…,n}∖{B1,…,Bm} and cB=(cB1,…,cBm)T. The basic vector x(B) is defined by xB(B)=AB−1b and xj(B)=0 for j∈N; the basis is primal feasible, and x(B) is a basic feasible solution, exactly when AB−1b≥0.

The basis multiplier is y(B)=AB−TcB. For every column j, its reduced cost is cˉj=cj−ajTy(B). In particular, cˉBi=0 for each basic index. The basis is dual feasible when cˉj≥0 for every j∈N. When m=0, take the empty basis and the empty multiplier; these formulas remain meaningful.

A primal-feasible basis is degenerate if some basic coordinate is zero. A simplex pivot is degenerate if its ratio-test step length is zero.

AlgorithmAlgorithm: AI-adaptedProof: Not applicableOpen item page →

Exact residual audit of finite-precision simplex candidates

Definition

Input a standard-form problem as in Revised simplex standard-form linear program, whose stored matrix, right-hand side, and cost entries are finite binary floating-point values, together with finite stored candidate vectors x^∈Rn and y^∈Rm. A finite binary value is interpreted as its exact dyadic rational (an integer times an integer power of 2). If a dimension is wrong or any value is NaN or infinite, return invalid-input. Otherwise use exact integer/rational arithmetic to compute rp=Ax^−b, rd=c−ATy^, g=cTx^−bTy^, ρp=∥rp∥∞, and ρd=∥rd∥∞, where ∥z∥∞=max⁡i∣zi∣ and the empty-vector value is zero. Return these exact values and ∣g∣ as the exact primal-dual objective difference for the stored data. Return verified-optimal-for-stored-data exactly when x^≥0, rp=0, rd≥0, and g=0. In every other finite-input case return candidate-only with the residuals and gap; a nonzero or small residual is never rounded to zero and does not certify infeasibility.

The exact arithmetic is a verification procedure for a candidate produced by a floating-point solver; it does not define a basis factorization or bound uncertainty in data before they were stored. If a statement about pre-rounded data is required, separate certified input intervals must be supplied and propagated. The procedure makes no operation-count or runtime claim.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Basic feasible solutions, extreme points, and basis existence

Statement

Let A∈Rm×n have full row rank and let P(A,b) be the feasible set of Revised simplex standard-form linear program. For a feasible x, define its support by S(x)={j:xj>0}. A set C is convex if tu+(1−t)v∈C for all u,v∈C and 0<t<1. A point x∈C is an extreme point if every representation x=tu+(1−t)v with u,v∈C and 0<t<1 forces u=v=x. Then x is an extreme point of P(A,b) if and only if the columns {aj:j∈S(x)} are linearly independent. This is equivalent to x being a basic feasible solution for Basis dictionaries and reduced costs for revised simplex. In particular, every nonempty P(A,b) has a basic feasible solution. The statement includes m=0 using the empty-basis convention.

Facts & Assumptions

Given: A full-row-rank matrix A, a vector b, and a feasible point x∈P(A,b).

[F1]

Revised simplex standard-form linear program: Its feasible set is P(A,b)={x∈Rn:Ax=b, xj≥0 for every j}.

[F2]

Basis dictionaries and reduced costs for revised simplex: The basic vector x(B) is defined by xB(B)=AB−1b and xj(B)=0 for j∈N; the basis is primal feasible, and x(B) is a basic feasible solution, exactly when AB−1b≥0.

Proof

1.1givenF1chooseconstructalgebra

If the support columns indexed by S(x) are dependent, choose a nonzero vector d supported on S(x) with Ad=0. Since each xj>0 on this finite support, choose ε>0 smaller than xj/∣dj∣ for every nonzero dj. Then x+εd and x−εd are distinct nonnegative solutions of Az=b, and their midpoint is x; hence x is not extreme.

1.2givenF1algebra

Suppose the support columns are independent and x=tu+(1−t)v for feasible u,v and 0<t<1. For j∉S(x), nonnegativity and xj=0 force uj=vj=0. Thus u−v is supported on S(x) and A(u−v)=0; independence forces u=v, and the convex-combination identity then gives u=v=x. This proves extremality under its stated definition.

2.1givenF1step 1.1choosealgebra

If the support columns of a feasible point are dependent, take d as in step 1.1 and choose its sign so some dj<0. With t=min⁡dj<0xj/(−dj)>0, the point x+td remains feasible, remains nonnegative, and has strictly fewer positive coordinates. Repeating this support reduction at most n times gives a feasible point whose support columns are independent.

3.1F1F2step 2.1chooseconstruct

Because A has rank m, its columns span Rm. Extend the independent support columns from step 2.1 to m independent columns of A by repeatedly taking the least-index column outside their current span; assigning zero to each added coordinate makes the reduced point the basic vector of that basis. It is therefore a basic feasible solution. When m=0, support reduction gives the zero vector and the empty list is the required basis.

4.1F1F2step 1.1step 1.2step 3.1∎

A basic feasible solution has support contained in an independent basis, so its support columns are independent and step 1.2 makes it extreme. Conversely, an extreme feasible point has independent support by step 1.1 and extends to a basis by step 3.1. Thus the three stated conditions are equivalent, and step 3.1 proves existence whenever P(A,b) is nonempty.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Primal revised-simplex pivot identities and certificates

Statement

Let B be a primal-feasible basis for the standard-form problem of Revised simplex standard-form linear program and Basis dictionaries and reduced costs for revised simplex, and let j be a nonbasic index with reduced cost cˉj<0. Put d=AB−1aj. If di≤0 for every basic row i, then rj=1, rB=−d, and rk=0 for the other nonbasic indices define a feasible ray with Ar=0 and cTr=cˉj<0; the LP is unbounded below. If some di>0, the step θ=min⁡di>0(xBi/di) is finite and nonnegative, every minimizing row gives a valid pivot, the new basic vector is feasible, and the objective change is θcˉj≤0, strictly negative exactly when θ>0. Finally, if all nonbasic reduced costs are nonnegative at a primal-feasible basis, its basic vector is optimal.

Facts & Assumptions

Given: The standard-form data, a primal-feasible basis B, and a nonbasic j with cˉj<0 when considering a pivot.

[F1]

Revised simplex standard-form linear program: Its feasible set is P(A,b)={x∈Rn:Ax=b, xj≥0 for every j}, and a feasible x is optimal if cTx≤cTz for every z∈P(A,b).

[F2]

Basis dictionaries and reduced costs for revised simplex: For every column j, its reduced cost is cˉj=cj−ajTy(B), and each basic reduced cost is zero.

[F3]

Inequality constrained programs, tangent cones, constraint qualifications, and KKT: The KKT system requires primal feasibility, nonnegative inequality multipliers, complementary products λigi(x)=0, and zero Lagrangian gradient.

[F4]

Convex KKT conditions certify a global minimum and an outward normal: On an open convex domain, a differentiable convex objective and inequalities with a feasible KKT tuple and nonnegative inequality multipliers certify a global minimum.

Proof

1.1F1F2algebra

Define rj=1, rB=−d, and all other nonbasic components of r to be zero. Since ABd=aj, we have Ar=aj−ABd=0; also cTr=cj−cBTAB−1aj=cˉj.

2.1givenF1step 1.1algebra

If every di≤0, then r≥0, so x+λr is feasible for every λ≥0. Its cost is cTx+λcˉj, which tends to −∞ as λ→∞; this proves the ray branch.

2.2givenF1step 1.1choosealgebra

If some di>0, set θ=min⁡di>0xBi/di. Each ratio is nonnegative, the finite nonempty index set has a minimum, and xB−θd≥0: rows with di>0 satisfy the ratio bound, while rows with di≤0 do not decrease. The entering component is θ≥0, so x+θr is feasible.

3.1F2step 2.2algebra

For any minimizing row i, replacing column Bi by aj gives a basis: in coordinates of AB, the replacement column is d, whose ith entry is positive, so expansion along that column gives a nonzero determinant. The unique basic vector of the new basis is x+θr, because that vector satisfies the constraints and has zero on the remaining nonbasic variables.

3.2step 1.1step 2.2algebra

The objective change is cT(x+θr)−cTx=θcTr=θcˉj. Since θ≥0 and cˉj<0, it is nonpositive and is zero exactly when θ=0, the degenerate-pivot case.

4.1givenF1F2F3F4algebra∎

If n=0, the feasible set contains only the empty vector, so its feasible basic vector is optimal. Suppose n≥1. Set U=Rn, f(z)=cTz, h(z)=Az−b, gj(z)=−zj, ν=−y(B), and λ=c−ATy(B). The open convex domain is U=Rn, and the affine functions are differentiable and convex. The vector λ is the reduced-cost vector, including zero basic coordinates by [F2], so it is nonnegative; x is feasible by hypothesis. For each j, either j is nonbasic and xj=0, or j is basic and λj=0, so λjgj(x)=0. Since Dg=−I, the stationarity expression is ∇f(x)+ATν+Dg(x)Tλ=c−ATy(B)−λ=0. These are precisely the KKT conditions of [F3], and [F4] proves that x globally minimizes cTz over P(A,b).

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

Primal revised simplex with Bland's rule

Definition

Input the exact standard-form data of Revised simplex standard-form linear program with full-row-rank A and a primal-feasible basis B as in Basis dictionaries and reduced costs for revised simplex. Use exact arithmetic. At each iteration compute xB=AB−1b, y=AB−TcB, and all nonbasic reduced costs cˉj=cj−ajTy.

  1. If every nonbasic cˉj≥0, return optimal with x(B) and y.
  2. Otherwise choose the least index j∉B with cˉj<0 and solve ABd=aj.
  3. If no component di is positive, return unbounded with the ray whose jth component is 1, whose basic components are −d, and whose other nonbasic components are 0.
  4. Otherwise compute θ=min⁡di>0xBi/di. Among minimizing rows, leave the basic variable of least index. Replace its basis column by aj and repeat.

The algorithm assumes an exact initial feasible basis. It makes no iteration bound polynomial in m or n.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Dual revised simplex with a finite pivot budget

Definition

Input the exact standard-form data and a basis B for which every nonbasic reduced cost is nonnegative, as in Revised simplex standard-form linear program and Basis dictionaries and reduced costs for revised simplex. Also input a nonnegative integer pivot budget K. Set k=0 and repeat:

  1. Compute xB=AB−1b and y=AB−TcB. If xB≥0, return optimal with the primal basic vector and y.
  2. Otherwise choose the least-index basic variable Bi with xBi<0. For each nonbasic j compute dij=(AB−1aj)i. If no dij<0, return infeasible with q=AB−Tei.
  3. If k=K, return pivot-budget-exhausted with the current basis and multiplier y=AB−TcB.
  4. Among j with dij<0, choose one minimizing cˉj/(−dij), with least variable index among ties. Replace Bi by j, set k←k+1, and repeat.

The pivot-budget-exhausted status is inconclusive. No finite-success or anti-cycling claim is part of this definition.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Sparse reduced-cost pricing operation count

Statement

Let A∈Rm×n be represented by a valid CSR record with row pointers r, column indices ct, values vt, and s stored entries, under Compressed sparse row matrices with stored-entry semantics. Matrix and vector coordinates here are zero-based: row i and column j correspond to row i+1 and column j+1 in Basis dictionaries and reduced costs for revised simplex, with the same relabelling for y,c,w and basis-column indices. For each 0≤t<s, let i(t) be the unique row satisfying ri(t)≤t<ri(t)+1. For a cost vector c∈Rn and given y∈Rm, initialize w∈Rn to zero and, for t=0,…,s−1, update wct←wct+vtyi(t). In exact real arithmetic the resulting vector is ATy. This loop uses exactly s multiplications and s additions, preceded by n zero initializations; forming c−w uses n subtractions and gives the reduced-cost vector of Basis dictionaries and reduced costs for revised simplex under the stated relabelling when y=y(B). Every stored entry is counted, including repeated column indices and explicitly stored zeros. The claim is an arithmetic-operation count; it gives no runtime, bandwidth, or factorization bound.

Facts & Assumptions

Given: A valid CSR record, y∈Rm, and exact real arithmetic for the stated updates.

[F1]

Basis dictionaries and reduced costs for revised simplex: For every one-based column index ℓ, its reduced cost is cˉℓ=cℓ−aℓTy(B); this item's zero-based column j corresponds to ℓ=j+1.

[F2]

Compressed sparse row matrices with stored-entry semantics: The stored entries in row i have indices ri≤t<ri+1, and the represented coordinate is Aij=∑ri≤t<ri+11ct=jvt.

[A1]

The loop processes every stored index exactly once in its increasing array order, using one exact real multiplication and addition per index.

Proof

1.1givenF2A1induction

The intervals [ri,ri+1) are disjoint and, because r0=0 and rm=s, their union is [0,s). Thus each stored index has exactly one row i(t), including the empty-stream case s=0. After processing the first k stored indices, the accumulator satisfies wj=∑0≤t<k, ct=jvtyi(t): this holds at k=0, and the next update adds exactly its one summand to component ck.

2.1F1F2step 1.1algebra

At k=s, for each column j the CSR reconstruction in [F2] gives (ATy)j=∑i=0m−1Aijyi=∑i=0m−1∑ri≤t<ri+11ct=jvtyi=∑0≤t<s, ct=jvtyi(t)=wj. The finite row intervals partition the stored indices, so the reindexing counts every stored term once; real multiplication distributes over finite sums. By [F1], c−w is the reduced-cost vector when y=y(B).

3.1A1step 2.1algebra∎

Each of the s updates performs one multiplication and one addition, and initialization and final subtraction each touch exactly n coordinates. The count follows the stated loop, so duplicate indices and zero-valued entries still contribute one multiplication and one addition each; no timing, bandwidth, or factorization conclusion follows.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Primal-dual residual gap bound for a simplex candidate

Statement

For a real standard-form problem with A∈Rm×n and candidate vectors x∈Rn, y∈Rm, define rp=Ax−b, rd=c−ATy, and g=cTx−bTy. When the data and candidates are finite stored binary floating-point values, Exact residual audit of finite-precision simplex candidates computes precisely these quantities, interpreting the stored values as exact dyadic rationals. For all the stated real data, g=xTrd+yTrp, and consequently ∣g∣≤∥x∥1∥rd∥∞+∥y∥1∥rp∥∞. Here ∥z∥1=∑i∣zi∣ and ∥z∥∞=max⁡i∣zi∣, with both norms equal to zero on the empty vector. This bounds the primal-dual objective difference; when either vector is infeasible it does not bound distance to an optimal value. If Ax=b, x≥0, c−ATy≥0, and g=0, then x is optimal.

Facts & Assumptions

Given: A real standard-form problem, dimension-matched real vectors x,y, and the residuals and gap defined in the Statement.

[F1]

Revised simplex standard-form linear program: Its feasible set is P(A,b)={x∈Rn:Ax=b, xj≥0 for every j}, and a feasible x is optimal if cTx≤cTz for every z∈P(A,b).

[F2]

Exact residual audit of finite-precision simplex candidates: For dimension-matched finite stored binary data and candidates, exact integer/rational arithmetic computes rp=Ax^−b, rd=c−ATy^, and g=cTx^−bTy^. The status verified-optimal-for-stored-data is returned exactly when x^≥0, rp=0, rd≥0, and g=0.

Proof

1.1givenalgebra

Expanding the right-hand side and cancelling the two equal scalar products gives xT(c−ATy)+yT(Ax−b)=cTx−bTy=g.

2.1givenstep 1.1algebra

The triangle inequality and ∣xTrd∣≤∑i∣xi∣∥rd∥∞ give ∣g∣≤∥x∥1∥rd∥∞+∥y∥1∥rp∥∞. The same calculation holds for empty vectors with the stated zero norms.

3.1F1F2givenstep 1.1algebra∎

If x is primal feasible and y has ATy≤c, then every feasible z satisfies cTz≥yTAz=yTb. The stated g=0 makes cTx=yTb, so x attains this lower bound and is optimal. In particular, every verified-optimal-for-stored-data output of the audit certifies optimality for its stored data, because its exact tests are these hypotheses.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Bland-rule finite termination and optimal-basis existence

Statement

For exact arithmetic and a finite standard-form LP with full-row-rank A, the primal algorithm of Primal revised simplex with Bland's rule, started from any primal-feasible basis, terminates after finitely many pivots at an optimal basis or returns a feasible ray of strictly negative cost. If the feasible set is nonempty and the objective is bounded below, an optimal basis exists and is returned. The number of pivots is at most the number of possible bases, hence at most (nm); this is a finiteness bound, not a polynomial-time claim. Every pivot may be degenerate, so the objective need not decrease strictly at each step.

Facts & Assumptions

Given: Exact standard-form data with A of full row rank, a primal-feasible basis, and the deterministic least-index entering and leaving rules.

[F1]

Basis dictionaries and reduced costs for revised simplex: In particular, cˉBi=0 for each basic index.

[F2]

Primal revised-simplex pivot identities and certificates: The objective change is θcˉj≤0, strictly negative exactly when θ>0.

[F3]

Primal revised simplex with Bland's rule: Otherwise choose the least index j∉B with cˉj<0 and solve ABd=aj.

[F4]

Basic feasible solutions, extreme points, and basis existence: In particular, every nonempty P(A,b) has a basic feasible solution.

Proof

1.1F2F3algebra

Every pivot in [F3] uses a negative reduced cost and a nonnegative ratio, so [F2] makes the objective nonincreasing. If a basis repeats, the deterministic rule repeats the same future pivots and produces a cycle; the objective returns to its initial value, forcing every pivot decrease to be zero. Since each entering reduced cost is strictly negative, every pivot in the cycle has θ=0, so the represented feasible vector stays fixed throughout the cycle.

2.1F1F3step 1.1

Let T be the indices that enter in this cycle and let q=max⁡T. Because the basis returns, q also leaves. Choose a tableau P where q enters and a tableau Q where q leaves. At P, its reduced cost rqP<0 and rjP≥0 for every j<q: a basic reduced cost is zero, and a lower-index eligible nonbasic variable would have been chosen instead. At Q, the entering variable t is in T, is not the basic variable q, and therefore satisfies t<q.

3.1F1F2F3step 2.1algebra

At Q, put d=ABQ−1at and define the entering direction zt=1, zBQ=−d, with other nonbasic coordinates zero. Then Az=0, cTz=rtQ<0, and zq<0 because q leaves through a row with positive pivot coefficient. If rP=c−ATyP is the reduced-cost vector at P, then rP⋅z=cTz<0 since Az=0.

4.1F1step 1.1step 2.1step 3.1algebra

For every j>q, rjPzj=0: if j is basic at P then rjP=0, while if it is nonbasic at P, maximality of q means it never enters during the cycle and it is nonbasic at Q, where zj=0. Also rqPzq>0. Thus the negative sum rP⋅z forces some j<q with rjPzj<0; because rjP≥0, this gives rjP>0 and zj<0. Such a j is nonbasic at P, basic at Q, and has dj>0. Its value is zero at P and stays zero during the degenerate cycle, so at Q it ties with q at ratio zero.

5.1F3step 4.1

Since j<q is an eligible leaving variable with the same minimum ratio as q, [F3] would choose j to leave at Q, contradicting the choice of q. Therefore no basis repeats.

6.1F2F4step 5.1algebra∎

There are at most (nm) possible column bases, so the algorithm terminates after at most that many visited bases. At termination [F2] gives either an optimal basis or an unbounded ray. If the objective is bounded below, the ray branch is impossible. If the feasible set is nonempty, [F4] supplies a starting basis; hence a finite optimum is attained at an optimal basis found by the algorithm. The combinatorial bound does not imply polynomial-time complexity.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Dual-simplex invariant and Farkas infeasibility certificate

Statement

Let B be a basis of the minimization problem for which all nonbasic reduced costs are nonnegative. At an iteration of Dual revised simplex with a finite pivot budget, a pivot chosen from a row i with xBi<0 and a column j with dij=(AB−1aj)i<0 minimizing cˉj/(−dij) preserves dual feasibility. If there is no nonbasic j with dij<0, then q=AB−Tei satisfies ATq≥0 and bTq<0, so q is a Farkas infeasibility certificate as defined in Revised simplex standard-form linear program and the primal problem is infeasible. If the current basis is both primal and dual feasible, it is optimal. A finite pivot-budget exhaustion is inconclusive and does not assert infeasibility or convergence.

Facts & Assumptions

Given: Exact standard-form data, a basis B, a dual-feasible multiplier y=AB−TcB, and the dual-simplex row and ratio choices.

[F1]

Basis dictionaries and reduced costs for revised simplex: For every column j, its reduced cost is cˉj=cj−ajTy(B).

[F2]

Dual revised simplex with a finite pivot budget: Among j with dij<0, choose one minimizing cˉj/(−dij), with least variable index among ties.

[F3]

Revised simplex standard-form linear program: A vector q∈Rm satisfying ATq≥0 and bTq<0 is called a Farkas infeasibility certificate for this sign convention.

Proof

1.1F1F2algebra

Fix an eligible pivot (i,j) and put q=AB−Tei. With α=cˉj/dij≤0, the vector y′=y+αq satisfies ajTy′=cj, so the entering column has new reduced cost zero. The leaving basic column aBi has qTaBi=1, hence its new reduced cost is −α≥0.

1.2F1F2F3algebra

If no nonbasic column has dij<0, then ATq≥0: the basic columns give the coordinates of ei, and every nonbasic column has nonnegative ith transformed entry. Also bTq=eiTAB−1b=xBi<0. By [F3], q is a Farkas infeasibility certificate; any feasible x≥0 would give bTq=xTATq≥0, a contradiction.

2.1F1F2step 1.1algebra

For any other nonbasic k, write dik=qTak. Its new reduced cost is cˉk′=cˉk−αdik. If dik≥0, this is nonnegative because α≤0. If dik<0, the ratio rule gives cˉj/(−dij)≤cˉk/(−dik), so cˉk′=(−dik)(cˉk/(−dik)−cˉj/(−dij))≥0. Thus every new nonbasic reduced cost is nonnegative.

3.1F1F2algebra∎

If the basis is primal feasible, its basic vector x and dual-feasible y satisfy cTx=bTy, because cBTAB−1b=yTb. For every feasible z, ATy≤c and z≥0 imply cTz≥bTy=cTx, so x is optimal. The algorithm's budget-exhausted branch asserts no further mathematical conclusion.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Phase I revised simplex and rank reduction

Definition

Input any exact equality system Ax=b with x≥0, including a rank-deficient raw system. Let D be diagonal with Dii=−1 when bi<0 and Dii=1 otherwise; set A~=DA and b~=Db≥0. Add artificial variables u∈Rm and solve the Phase I problem min⁡1Tu subject to A~x+u=b~, x,u≥0, starting at x=0,u=b~ with artificial basis Im. Use the exact primal algorithm of Primal revised simplex with Bland's rule and its finite-termination result Bland-rule finite termination and optimal-basis existence.

At a terminal optimum, if 1Tu>0, return infeasible and the basis multiplier y for the Phase I objective. If the optimum is zero, set every artificial variable to zero and remove artificial basic variables as follows. For a zero-valued artificial basic variable, inspect its row in B−1A~. If some original nonbasic column has a nonzero coefficient there, pivot the least-index such column into the basis; the pivot may have either sign because the leaving basic value is zero. If no such coefficient exists, and the transformed right-hand-side entry is zero, delete that redundant equation and its zero artificial basic variable. Repeat until all remaining basic columns are original; return the row-reduced equivalent system and its original feasible basis. A zero row with nonzero transformed right-hand side returns inconsistent-row. All tests are exact; a finite-precision near-zero objective is not treated as zero.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Phase I feasible-basis extraction and infeasibility certificate

Statement

For any exact system Ax=b, x≥0, the Phase I procedure of Phase I revised simplex and rank reduction terminates. Its minimum artificial-variable sum is zero if and only if the original system is feasible. If the minimum is positive, its returned vector yields a Farkas infeasibility certificate, as defined in Revised simplex standard-form linear program, with ATq≥0 and bTq<0. If the minimum is zero, the extraction procedure returns an equivalent full-row-rank reduced system with a basic feasible solution using only original columns; zero-valued artificial basic variables are either pivoted out or their redundant equations are removed.

Facts & Assumptions

Given: Finite exact matrices and vectors A,b, with nonnegative original variables.

[F1]

Phase I revised simplex and rank reduction: Phase I row-signs the equations, uses [DA I] with initial x=0,u=Db, minimizes 1Tu, and extracts zero artificial basics by pivots or zero-row deletion.

[F2]

Bland-rule finite termination and optimal-basis existence: Under its exact-arithmetic, finite-standard-form, full-row-rank hypotheses, the primal run terminates at an optimal basis or returns a strictly negative-cost feasible ray.

[F3]

Basis dictionaries and reduced costs for revised simplex: The basis multiplier is y(B)=AB−TcB.

[F4]

Revised simplex standard-form linear program: A vector q∈Rm satisfying ATq≥0 and bTq<0 is called a Farkas infeasibility certificate for this sign convention.

[F5]

Farkas alternative for a finite linear system: Exactly one of x≥0,Ax=b and existence of q with ATq≥0,bTq<0 holds for every finite real A,b.

Proof

1.1givenF1algebra

Let Dii=−1 when bi<0 and Dii=1 otherwise. Since D is invertible and D−1=D, Ax=b is equivalent to DAx=Db, and Db≥0. The augmented matrix [DA I] has full row rank because it contains the identity columns; x=0,u=Db is feasible for the Phase I problem and has objective 1TDb≥0.

1.2F1F2algebra

By [F2] the Phase I primal algorithm terminates at an optimum or an unbounded ray. Its objective is ∑iui≥0 at every feasible point, so a negative-cost ray is impossible; it terminates at a minimum with nonnegative reduced costs.

2.1F1F2F3step 1.2algebra

Suppose the minimum w∗>0. Any feasible original x would give a Phase I feasible point (x,u=0) of objective zero, contradicting minimality. Let y be the terminal basis multiplier for the Phase I objective (0,1). Nonnegative reduced costs on the original columns, using [F3], give (DA)Ty≤0, and the basic-vector identity gives (Db)Ty=w∗>0.

2.2F1step 1.2algebra

Suppose instead that w∗=0. Since every artificial variable is nonnegative and their sum is zero, each artificial variable has value zero, and the original coordinates satisfy DAx=Db. If an artificial variable is basic at tableau row k and some original nonbasic column has a nonzero coefficient in row k of B−1DA, pivot that column in. The pivot is nonzero, and the entering variable and leaving artificial variable both have value zero, so the represented feasible point is unchanged and one artificial basic variable is removed.

3.1F4step 2.1algebra

Set q=−DTy. Then ATq=−(DA)Ty≥0 and bTq=−(Db)Ty<0, so [F4] identifies it as a Farkas certificate. The finite-system alternative [F5] says exactly one of feasibility and such a certificate holds; because this q satisfies the certificate branch, the original system is infeasible.

3.2F1step 2.2algebra

If no original nonbasic column has a nonzero coefficient in row k, every original basic column has a unit column there at a different row, so the entire row k of B−1DA is zero. Every nonbasic artificial variable is zero, every other basic artificial column has zero in row k, and the basic artificial value at row k is zero; hence row k of B−1Db is also zero. The transformed equation is 0=0 and may be deleted without changing the feasible set; deleting that row and its artificial basis column leaves an invertible identity submatrix for the remaining basis.

4.1F1step 2.2step 3.2construct∎

Each pivot or row deletion removes one artificial basic variable, so after at most m such operations no artificial variable remains basic. The remaining basis consists of original columns and is invertible for the retained row-reduced system; its basic vector is feasible because the represented point never changed. The retained equations are equivalent to the original system by the invertible row operations and deletion only of 0=0 rows. This proves the zero-optimum extraction claim, including redundant-row removal.

AlgorithmAlgorithm: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Revised-simplex basis QR recomputation with finite residual and pivot statuses

Definition

Input finite stored binary data for the standard-form LP of Basis dictionaries and reduced costs for revised simplex, with full-row-rank A∈Rm×n, a candidate ordered basis B of m distinct in-range indices, and a nonnegative integer budget K of basis evaluations. Stored values represent their exact dyadic rationals. Invalid dimensions, nonfinite inputs or repeated/out-of-range basis indices return invalid-input. Nonsingularity of C=AB must be established as part of an admitted basis input; a numerical rank threshold is not such an establishment. A basis whose validity is unavailable returns basis-validity-inconclusive.

A caller may request a primal or dual simplex transition after an audit, or only an audit. It may supply a finite sequence of transition records, one for each requested basis replacement, from Primal revised simplex with Bland's rule or Dual revised simplex with a finite pivot budget. A transition record has fields mode (primal or dual), basis_before (the ordered current basis, with its validity witness), leaving_position i, and entering_index j; it is checked rather than trusted as already verified. The initial basis and each record's basis_before must carry the validity witness described here. After a replacement, its checked nonzero pivot coefficient establishes the new basis; any next transition record must carry its own checked validity witness.

An admitted basis input includes either an exact rational inverse witness G with CG=GC=Im, checked exactly, or a rational matrix H and rational q with 0≤q<1 and ∥Im−HC∥∞≤q, also checked exactly. For matrices with at least one row, ∥M∥∞=max⁡i∑j∣Mij∣; the empty matrix has norm 0. For vectors it is the maximum absolute coordinate, with empty value 0. For m=0, the empty inverse witness satisfies the identities. If neither validity certificate is verified, return basis-validity-inconclusive before factorization. A numerical rank threshold is never a validity certificate.

With an exact inverse witness, first compare basis_before with the current ordered basis exactly. Then check a transition by exact rational arithmetic: compute xB=Gb, y=GTcB, dk=Gak for every nonbasic k, and cˉk=ck−akTy. A primal record is admissible exactly when xB≥0, j is the least-index nonbasic column with cˉj<0, some drj>0, and i minimizes xBr/drj over rows with drj>0, with ties resolved by least basic-variable index. A dual record is admissible exactly when every nonbasic cˉk≥0, Bi is the least-index basic variable with xBi<0, dij<0, and j minimizes cˉk/(−dik) over nonbasic k with dik<0, with ties resolved by least variable index. In either case, the checked nonzero pivot coefficient proves the replacement basis nonsingular.

Certified rational enclosures may instead establish a transition. For each current D=C and D=CT, provide rational HD,qD,MD with 0≤qD<1, ∥Im−HDD∥∞≤qD, and MD≥∥HD∥∞/(1−qD); verify these inequalities exactly. For each solve Dz=r, combine its returned candidate z^ and a rational upper bound η for its Step 3 residual bound to form the coordinate intervals [z^ℓ−MDη,z^ℓ+MDη]. For closed rational intervals I=[a,b] and J=[c,d], use I+J=[a+c,b+d], I−J=[a−d,b−c], IJ=[min⁡{ac,ad,bc,bd},max⁡{ac,ad,bc,bd}], and I/J=I⋅[min⁡{1/c,1/d},max⁡{1/c,1/d}] when 0∉J; division is undefined otherwise. Evaluate dot products, reduced costs, and ratio expressions with these exact rational endpoint operations. First compare basis_before with the current ordered basis exactly. An interval proves a sign or order only when its endpoints imply that relation. If a ratio comparison is not decided by interval endpoints, return pivot-inconclusive unless an exact inverse witness G is supplied for the current basis; with G, compute the exact rational ratios and compare them exactly, resolving equality by the least-index rule. Admit the record only when the required signs and comparisons, including all least-index rules, are proved. If a needed sign, ratio comparison, or basis match is not established, return pivot-inconclusive with the current basis and its last completed audit, without changing the basis.

For m≥1, use the operationwise arithmetic of Operationwise real and complex floating point arithmetic with rounded square roots and fused multiply-add, with 0<u<1, mu<1, (m+3)u<1, and γm+3<1/8. All nonzero exact intermediates must be normal. Norms and certificate expressions below are evaluated exactly or by certified upward bounds; failure to obtain a finite valid bound is a numerical failure. Process the following loop:

  1. Maintain an attempt counter t, initially 0. If t=K, return budget-exhausted with the current basis, the last completed audit if any, and the basis to which that audit belongs. Otherwise set t←t+1 before starting this basis evaluation. Every budget, numerical-failure, basis-validity-inconclusive, or pivot-inconclusive return includes the current basis and identifies the basis associated with any retained audit; state explicitly when no audit exists.

  2. Separately for D=C=AB and D=CT, copy the current stored entries into a working matrix W. For k=1,…,m, set r=m−k+1, take the active column x=Wk:m,k and trailing block T=Wk:m,k+1:m, and call Scaled Householder reflector formation and application once on (x,T). On success, save its reflector data and replace the active block Wk:m,k:m by its returned output, which writes the new diagonal entry and exact zeros below it; leave earlier rows and columns unchanged. If x=0, save an identity marker and leave the trailing block unchanged. Stop with numerical-failure at the first failed call, recording the factorization, column k, and failure stage. The final W is the computed upper-triangular factor R^D.

    To apply these stored reflectors to any right-hand side s, start with z=s and visit k=1,…,m in that same order. Skip an identity marker. Otherwise, on the active suffix zk:m compute the sequential rounded dot d=fl⁡(vkTzk:m), then p=fl⁡(τ^kvk), u=fl⁡(p d) componentwise, and zk:m=fl⁡(zk:m−u) componentwise, using the operation order of Scaled Householder reflector formation and application without a column overwrite. Stop at the first failure and record the right-hand side, reflector index, and failure stage. For Dz=s, apply this replay to obtain the transformed right-hand side and solve R^Dz^=z by descending substitution: for i=m,…,1, sequentially compute ti=fl⁡(∑j=i+1m(R^D)ijz^j), hi=fl⁡(zi−ti), and z^i=fl⁡(hi/(R^D)ii). A zero diagonal or any failed operation returns numerical-failure with the solve and row index. Reuse the C factorization for x^B and every nonbasic direction d^j; reuse the CT factorization for y^. Never substitute a rounded dense orthogonal product for the exact orthogonal representative in the proof. After every admitted basis replacement discard both factorizations and recompute; no rank-one factorization-update implementation is claimed.

  3. Independently certify each returned solve for its actual current matrix D=C or CT and right-hand side r. Compute v^=fl⁡(Dz^) and w^=fl⁡(v^−r) by the sequential products/subtraction of Rounded matrix products give certified linear and normal-equation residual allowances. Put gm=1+6u1−(m−1)u−1,W=gm∥D∥F∥z^∥2+u(∥v^∥2+∥r∥2). With the local coefficient κm of The scaled computed Householder step satisfies the local backward error premise, put β=(1+κm)m−1,βD=β+γm+3(1+β), and report the valid upper bound η(D,r,z^)=min⁡{∥w^∥2+W, βD∥D∥F∥z^∥2+β∥r∥2}. A finite concrete certificate option uses exact rational arithmetic on the stored dyadic entries and a rational upper bound for u satisfying all displayed admissibility inequalities. Use ∥v∥2≤∑i∣vi∣ and ∥D∥F≤∑i,j∣Dij∣ as rational upper bounds for every norm in the two nonnegative expressions. For the local coefficient, substitute eˉm=(1+u)m+3−1 for em=(1+u)m/2+3−1 in the source formula: put dˉm=eˉm+u(2+eˉm), am=(1+u)/(1−u)m−1, gmQR=mu/(1−mu), cm=2((1+am)(1+u)2(1+gmQR)−1), and κˉm=max⁡{eˉm+8dˉm,(1+u)cm+3u}. These are rational upper bounds because m+3≥m/2+3 and all constituent nonnegative formulas increase in their arguments. Then βˉ=(1+κˉm)m−1 and βˉD=βˉ+γm+3(1+βˉ) are rational upper bounds for the QR coefficients. Compute both candidate bounds by finite exact rational sums/products/powers and take their minimum as the returned η. This implements the certificate without a norm oracle; other finite certified upper-bound procedures are allowed. Any factorization, substitution, dot, subtraction, range or certificate failure returns numerical-failure, with its stage and the last completed audit; it returns no certificate for the failed solve.

  4. Embed x^B in x^∈Rn with zero nonbasic entries, and call Exact residual audit of finite-precision simplex candidates on these actual stored x^,y^. Retain its exact rp=Ax^−b, rd=c−ATy^, objective difference g, and audit status together with all solve residual bounds. basis-solves-residual-certified means only that the returned bounds certify those linear equations; any requested thresholds must also be checked before declaring them met. verified-optimal-for-stored-data is issued only by the exact audit's primal/sign/zero-gap checks. Other completed candidates remain candidate-only; small residuals or a failed/ambiguous pivot never certify infeasibility.

  5. Return if audit-only mode or the caller has no further transition. Otherwise compare the next record with the current basis and apply the exact or enclosure admission predicate above. A failed or unresolved predicate returns pivot-inconclusive or basis-validity-inconclusive, retaining the current basis and last completed audit. On admission, replace position i by j, discard both factorizations, and continue at Step 1. Exact driver invariants remain those of the existing driver, not consequences of successful floating solves.

For m=0, skip QR and substitution: the basis and multiplier are empty, every direction is the empty vector, and their basis-equation residuals and solve bounds η are zero. Evaluate the existing exact audit with x^=0; no basis replacement is possible. The budget and invalid-input branches remain in force.

Remarks

No exact-arithmetic simplex method, Bland termination theorem, dual invariant, PhaseI certificate or existing residual audit is replaced. The procedure recomputes canonical current-basis QR data, checks caller-supplied transitions, and reports computed candidates with independently certified residuals. A useful forward-accuracy interpretation requires a conditioning bound; signs and ratios are used only when exact checks or certified enclosures establish them. Pre-rounded input uncertainty requires separate data enclosures. No numerical-success, anti-cycling or polynomial-time guarantee is added.

TheoremStatement: AI-adaptedProof: AI-generatedOpen item page →

Finite QR basis recomputation has sound residual and stored-data audit statuses

Statement

For admitted inputs of Revised-simplex basis QR recomputation with finite residual and pivot statuses, the procedure terminates after at most K basis-evaluation attempts with its defined output or failure status. Each completed solve certificate satisfies ∥Dz^−r∥2≤η(D,r,z^) for its actual current D=AB or ABT and right-hand side. If z∗=D−1r denotes the exact solution, then also ∥z^−z∗∥2≤∥D−1∥2η(D,r,z^). This conditional inequality does not compute or bound ∥D−1∥2. In particular, its returned x^B,y^,d^j have the respective certified primal-basis, multiplier and direction residuals. Its exact stored-data audit obeys g=x^Trd+y^Trp,∣g∣≤∥x^∥1∥rd∥∞+∥y^∥1∥rp∥∞. An admitted exact inverse or H,q basis witness proves the current basis nonsingular; every checked MD bounds the corresponding ∥D−1∥∞. A transition admitted by either exact rational checks or certified intervals satisfies the exact primal or dual pivot rule in the algorithm, and its replacement basis is nonsingular. Only the exact audit's feasibility, dual-sign and zero-gap checks issue verified-optimal-for-stored-data. Every unestablished transition or unresolved sign/ratio comparison stops without changing the basis. The theorem grants neither forward accuracy without conditioning nor correctness of unvalidated floating pivot choices, an infeasibility certificate from near-zero data, or numerical termination at an optimum.

Facts & Assumptions

Given: The admitted inputs, finite budget, arithmetic and certificate hypotheses of the algorithm.

[F1]

Revised-simplex basis QR recomputation with finite residual and pivot statuses specifies fresh current-basis QR recomputation, bounded attempts, first failures, established transitions, exact stored-data audit and all return statuses.

[F2]

Computed reduced QR least squares is backward stable for full column rank supplies a square exactly orthogonal representative Q^, triangular R^, and perturbations D+ΔQ=Q^R^,(R^+ΔR)z^=Q^T(r+δr), with ∥ΔQ∥F≤β∥D∥F, ∥δr∥2≤β∥r∥2, and ∥ΔR∥F≤γm+3∥R^∥F on every successful branch. Its additional perturbed-least-squares uniqueness premise is not needed for these equations.

[F3]

Rounded matrix products give certified linear and normal-equation residual allowances bounds ∥w^−(Dz^−r)∥2≤W for the specified successful residual evaluation and certified norms.

[F4]

Primal-dual residual gap bound for a simplex candidate gives the exact audit identity, norm bound and conditional optimality for the stored data.

[A1]

Every claimed successful call obeys the stated source arithmetic/range hypotheses, and bound arithmetic is exact or certified upward.

Proof

1.1F1givencasesalgebra

If m=0, the empty basis is nonsingular on the zero space. For m≥1, an exact inverse witness G gives C−1=G. Otherwise let E=I−HC with ∥E∥∞≤q<1. The row-sum norm inequality gives (1−q)∥v∥∞≤∥HCv∥∞ for every v, so both HC and C have trivial kernel and, being square, are nonsingular. If M≥∥H∥∞/(1−q) is supplied, applying this inequality to v=(HC)−1w gives ∥(HC)−1w∥∞≤∥w∥∞/(1−q), and C−1=(HC)−1H yields ∥C−1∥∞≤M. The same argument applies to each separately checked D=C or CT witness in interval mode.

2.1A1F1F2step 1.1casesalgebra

If m=0, the empty solve residuals are zero and only the finite audit and budget branches remain. If m≥1, step 1.1 makes each square D nonsingular and hence full column rank. Apply [F2] under [A1]; since the exact representative is square, Q^Q^T=I. Multiplication of its triangular equation therefore gives (D+ΔD)z^=r+δr, where ΔD=ΔQ+Q^ΔR. No nonsingularity of R^+ΔR or exact orthogonality of a rounded dense output is used.

3.1A1F2step 2.1algebra

Orthogonality and [F2] imply ∥R^∥F=∥D+ΔQ∥F≤(1+β)∥D∥F. Hence ∥ΔD∥F≤[β+γm+3(1+β)]∥D∥F=βD∥D∥F. Rearranging step 2.1 yields ∥Dz^−r∥2≤∥δr∥2+∥ΔD∥F∥z^∥2≤β∥r∥2+βD∥D∥F∥z^∥2.

4.1A1F1F3step 1.1step 3.1algebra

The triangle inequality and [F3] give the independent upper bound ∥Dz^−r∥2≤∥w^∥2+W. Both quantities in the reported minimum are upper bounds, so their minimum is an upper bound. Applying this to each current primal, transpose-multiplier and direction solve proves all residual conclusions. The same conclusion holds when the algorithm uses rational or other certified upper bounds, since both evaluated candidates remain upper bounds. Finally D(z^−z∗)=Dz^−r, so the defining induced-norm inequality gives ∥z^−z∗∥2≤∥D−1∥2∥Dz^−r∥2≤∥D−1∥2η. This is conditional on the actual inverse norm, not an assertion that the procedure computes it. For interval mode, [F1] forms each coordinate interval with radius MDη; step 1.1 gives ∥D−1∥∞≤MD, while step 3.1's residual bound and ∥z∥∞≤∥z∥2 imply ∥z^−D−1r∥∞≤MDη. Thus the intervals contain the exact solution.

5.1F1step 1.1step 4.1algebra

The interval addition and subtraction endpoints bound the corresponding sums and differences; the four endpoint products bound every product, and reciprocal endpoints bound reciprocals on either interval that avoids zero. Induction therefore encloses every exact reduced cost and ratio evaluated by [F1]. Any endpoint comparison that admits a sign or order is true of the enclosed exact value; any unresolved comparison either is checked by an exact inverse witness or returns inconclusive. Hence every admitted exact-witness or interval record satisfies the corresponding primal or dual pivot rule. If position i is replaced by column j, then C−1AB′ is the identity with column i replaced by d=C−1aj, so det⁡(AB′)=det⁡(C)di≠0 by the checked pivot sign. The replacement basis is nonsingular.

5.2F1F4step 4.1algebra

By [F1], the completed audit uses the actual stored candidate vectors and original stored LP data. Apply [F4] to obtain the displayed identity, norm bound and exactly conditional optimality statement. Since [F1] labels no other branch verified-optimal and rejects unestablished transitions, no small residual, unresolved sign/ratio comparison or numerical failure is converted into a pivot, optimality or infeasibility certificate.

6.1F1step 1.1step 5.2givenalgebra∎

Each canonical factorization, vector application, substitution and residual loop has finite dimensions; each elementary call either completes or returns its first failure by [F1]. Verifying a transition record uses only finitely many exact rational matrix products or certified interval operations over the finite basis and column sets; unresolved comparisons return inconclusive. The dyadic audit uses finitely many exact integer/rational operations. Before each attempt the counter increases, and [F1] allows at most K attempts and no unbounded internal transition search. Every admitted basis replacement discards old factors before computing new certificates. Consequently the procedure terminates with only the stated current-basis outputs or failure/inconclusive status, proving the remaining claims.

5 · Examples, counterexamples and false statements

None yet.

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