A Simplex-Type Voronoi Algorithm Based on Short Vector Computations of Copositive Quadratic Forms

Simons workshop Lattices Geometry, Algorithms and Hardness Berkeley, February 21st 2020 A simplex-type Voronoi algorithm based on short vector computations of copositive quadratic forms

Achill Schürmann (Universität Rostock)

based on work with Mathieu Dutour Sikirić and Frank Vallentin ( for Q n positive definite ) Perfect Forms 2 S>0

DEF: min(Q)= min Q[x] is the arithmetical minimum • x n 0 2Z \{ } Q is uniquely determined by min(Q) and Q perfect • , n MinQ = x Z : Q[x]=min(Q) { 2 }

V(Q)=cone xxt : x MinQ is Voronoi cone of Q • { 2 } (Voronoi cones are full dimensional if and only if Q is perfect!)

THM: Voronoi cones give a polyhedral tessellation of n S>0 and there are only ﬁnitely many up to GL n ( Z ) -equivalence. Voronoi’s Reduction Theory

n t GLn(Z) acts on >0 by Q U QU S 7! Georgy Voronoi (1868 – 1908) Task of a reduction theory is to provide a fundamental domain

Voronoi’s algorithm gives a recipe for the construction of a complete list of such polyhedral cones up to GL n ( Z ) -equivalence Ryshkov Polyhedron

The set of all positive deﬁnite quadratic forms / matrices with arithmeticalRyshkov minimum Polyhedra at least 1 is called Ryshkov polyhedron

= Q n : Q[x] 1 for all x Zn 0 R 2 S>0 2 \{ } is a locally ﬁnite polyhedron • R is a locally finite polyhedron • R Vertices of are perfect forms Vertices of are• perfect R • R

1 n n ↵ (det(Q + ↵Q0)) is strictly concave on • 7! S>0 Voronoi’s algorithm Voronoi’s Algorithm

Start with a perfect form Q

1. SVP: Compute Min Q and describing inequalities of the polyhedral cone

n (Q)= Q0 : Q0[x] 1 for all x Min Q P { 2 S 2 } 2. PolyRepConv: Enumerate extreme rays R ,...,R of (Q) 1 k P 3. SVPs: Determine contiguous perfect forms Qi = Q + ↵Ri, i =1,...,k

4. ISOMs: Test if Qi is arithmetically equivalent to a known form 5. Repeat steps 1.–4. for new perfect forms

( graph traversal search on edge graph of Ryshkov polyhedron ) 2M.DutourSikiri´c,A.Sch¨urmann,andF.Vallentin

From the algorithmic side some successful ideas have been proposed by Jarre A SIMPLEXand Schmallowsky ALGORITHM [14], Nie [17], FOR Sponsel RATIONAL and Dür [20], Groetzner and Dür [11], and Anstreicher,CP-FACTORIZATION Burer, and Dickinson [4, Section 3.3]. In this paper we describe a new procedure that works for all matrices in the rational closure MATHIEU DUTOUR SIKIRIC,´ ACHILL SCHURMANN,¨ ˜ AND FRANKT VALLENTINn n = cone xx : x Q 0 2M.DutourSikirić,A.Schürmann,andF.VallentinCP { 2 } of the cone of completely positive matrices. Moreover, it also computes certificates Abstract. Wein case describeA and analyze. an algorithmic procedure, similar to the From the algorithmic62 CPn side some successful ideas have been proposed by Jarre simplex algorithmand SchmallowskyIn for contrast linear programming, to [14], most Nie of the [17], which other Sponsel tests approaches whether and Dür mentioned or [20], not a Groetzner given above (the and exception Dür [11], being symmetric matrix is completely positive. For matrices in the rational closure andthe Anstreicher, approach Burer, by Anstreicher, and Dickinson Burer [4, and Section Dickinson 3.3]. whose factorization method is of the cone ofbased completely on the positive ellipsoid matrices method), our procedure our algorithm computes is a exact rational in the sense that it uses cp-factorization.In this For paper matrices we which describe are not a new completely procedure positive that a works certificate for all matrices in the rational numbers only if the input matrix is rational and so does not have to cope in form ofrational a separating closure hyperplane is computed, whenever our procedure ter- with numerical instabilities.˜ In particularT it findsn a rational cp-factorization if it minates. We conjecture that there existsn a= pivot cone rulexx so: thatx ourQ 0 procedure CP { 2 } always terminatesof theexists. cone for However, of rational completely input questions matrices. positive of convergence matrices. Moreover, remain, in it also particular computes for all certificatesA which are not contained in the “irrational boundary part” (bd n) ˜ n. in case A n. CP \ CP In contrast62 CP to most of the other approaches mentioned above (the exception being 2. The copositive minimum and copositive perfect matrices the approach by Anstreicher, Burer and Dickinson whose factorization method is basedBy on then we ellipsoid denote method), the Euclidean our algorithm vector space is exact of symmetric in the sensen thatn matrices it uses with S n ⇥ rationalinner numbers product1. onlyA,Introduction B if the= Trace( input matrixAB)= is rationalA andB .Withrespecttothisinner so does not have to cope h i i,j=1 ij ij Copositive programmingwithproduct numericalGeneralization we gives instabilities. have a common the following In framework particular duality to it relations formulatefinds a rational… between manyand cp-factorizationapplication! di thecult cone of copositive if it exists. However, questions of convergence remain,P in particular for all A which are optimization problemsmatrices as convex and and conic the cone ones. of completely In fact, many positive NP-hard matrices problems not containedIDEA: in Generalize the “irrational Voronoi’s boundary theory part”n to (bd n) ˜ n. are known to have such reformulationsn =( (seen)⇤ for= exampleB : theA, B surveys0 for [2, all 8]).A Alln , otherCOP convexCP cones and{ 2theirS dualsh iCP \ CP 2 CP } the diculty of theseand2. problemsThe copositive appears(Opgenorth, minimum to be 2001) “converted” and copositive into the perfect diculty matrices of understanding the conen of copositive matrices n whichn =( consistsn)⇤. of all symmetric By n we denoteT the EuclideanCOP vectorCPn spaceCOP of symmetric n n matrices with n n matrices B So,S inwithIn order particularx toBx show to0 that forthe all acompletely givenx R symmetric0 positive.n Its dual matrix cone cone A isis notthe⇥ completely cone positive, it ⇥ inner2 S product A, B = Trace(AB2)= AijBij.Withrespecttothisinner h i i,j=1 productsuces we to have find the a copositive followingT matrix dualitynB relationsn betweenwith B,A the< cone0. We of copositive call B a sepa- n = cone xx : x R P0 2andCOP its dual, theh copositivei cone matricesrating and witnessCP and thefor coneA { of completelyn in2 this case,} positive because matrices the linear hyperplane orthogonal to B separates A and62 CP . of completely positive n n matrices. Therefore,n itn seems no surprise that many Using⇥ then notation=( nBCP)⇤[x=] forB xTBx:=A,B,xx B T0, for we all obtainA n , basic questions about this coneCOP are stillCP open{ and2 appearS h h toi be dii cult. 2 CP } and n n One important problem is to find an algorithmicn = B test: B[x deciding] 0 for all whetherx R or0 . not COP { 2n S=( n)⇤. 2 } a given symmetric matrixDefinitionA is 2.1. completelyFor a symmetric positive.CP matrix IfCOP possibleB n onewe define would the likecopositive to minimum So, in order to show that a given symmetric matrix2 S A is not completely positive, it obtain a certificateas for either A n or A n n. In terms of the definitions the suces to find a2 copositiveCPn matrix62 >CP0 B n withn B,A < 0. We call B a sepa- most natural certificate for A CPn is giving⇢ S a cp-factorization2⇢COPCOP h n i rating witness for A minin this(B)=inf case, becauseB[v]: thev linearZ 0 hyperplane0 , orthogonal 2 CP62 CPn COP 2 \{ } to Bandseparates we denotem A and the setn of. vectors attaining it by A, B = (A B) n h i T CP· T nT S UsingA = the notationxixi Bwith[x] forxx1,...,xBx =mB,xxnR 0, we obtain Min (B)= v h Z2 0 : iB[v]=min (B) . i=1 COP n 2 COPn X n = B : B[x] 0 for all x R 0 . The followingCOP proposition{ 2 S shows that matrices2 in the } interior of the cone of and for A Definitionn givingcopositive a 2.1. separating matricesFor a symmetric attain hyperplanes their matrix copositive definedB minimum. byn weB define thencopositiveso that minimum B,A < 0.62 DickinsonCP and Gijben [5] showed that this (strong)2 S membership2 COP prob- h i as lem is NP-hard. Lemma 2.2. Let B be a matrix in the interiorn of the cone of copositive matrices. min (B)=inf B[v]:v Z 0 0 , Then, the copositiveCOP minimum of B is strictly2 positive\{ } and it is attained by only andfinitely we denote many the vectors. set of vectors attaining it by Date:April15,2018. n Min (B)= v Z 0 : B[v]=min (B) . 2010 Mathematics SubjectProof. Classification.Since B isCOP copositive,90C20. we2 have the inequalityCOP min (B) 0. Suppose that Key words and phrases. copositive programming, complete positivity, matrixn factorization,COP Themin following(B) = proposition 0. Then there shows is a sequence that matricesvi Z in0 the0 interiorof pairwise of the distinct cone oflattice copositive minimum. COP 2 \{ } copositivepoints matrices such that attainB[vi]tendstozerowhen their copositive minimum.i tends to infinity. From the sequence The first author wasv supportedwe construct by the a newHumboldt sequence foundation.u of points The third on the author unit is sphere partiallySn 1 by setting supported by the SFB/TRRLemmai 2.2. 191 “SymplecticLet B be a Structures matrix in in the Geometry,i interior of Algebra the cone and of Dynamics”, copositive matrices. funded by the DFG.Then, the copositive minimum of B is strictly positive and it is attained by only finitely many vectors. 1 Proof. Since B is copositive, we have the inequality min (B) 0. Suppose that n COP min (B) = 0. Then there is a sequence vi Z 0 0 of pairwise distinct lattice COP 2 \{ } points such that B[vi]tendstozerowheni tends to infinity. From the sequence n 1 vi we construct a new sequence ui of points on the unit sphere S by setting Application: Copositive Optimization

• Copositive optimization problems are convex conic problems

min C, Q such that Q, A = b , i = 1,...,m h i h ii i and Q CONE 2

n CONE = R 0 CONE = or Linear Programming (LP) CPn COPn n Copositive Programming (CP) CONE = 0 Semideﬁnite ProgrammingS (SDP)

˜ n Task: Certify or disprove Q n = cone xx> : x Q 0 2 CP 2 ( due to duality theory we can give certiﬁcates for solutions of convex conic problems ) Copositive minimum (COP-SVP)

DEF: min Q =minQ[x] is the copositive minimum n COP x Z 0 0 2 \{ } Difﬁcult to compute!?

THM: (Bundfuss and Dür, 2008) For Q int we can construct a family of simplices k 2 COPn n in the standard simplex = x R 0 : x1 + ...xn = 1 2 k such that each has vertices v1,...vn with vi>Qvj > 0

A ﬁrst naive algorithm: ”Fincke-Pohst strategy” to compute min Q in each cone k COP Generalized Ryshkov polyhedron

The set of all copositive quadratic forms / matrices with copositive minimum at least 1 is called Ryshkov polyhedron n = Q n : Q[x] 1 for all x Z 0 0 R 2 COP 2 \{ } DEF: Q int is called -perfect if and only if 2 COPn COP Q is uniquely determined by min Q and COP n Min Q = x Z 0 : Q[x]=min Q COP 2 COP is a locally finite polyhedron (with dead-ends / rays) • R Vertices of are -perfect • R COP 4M.DutourSikirić,A.Schürmann,andF.Vallentin 4M.DutourSikirić,A.Schürmann,andF.Vallentin

Definition 2.3. A copositive matrix B int( n) is called -perfect if it is Definition 2.3. A copositive2 COP matrix B int(COP ) is called -perfect if it is uniquely determined by its copositive minimum min B2andCOP the setn Min COPB attaining it. uniquely determined by its copositiveCOP minimum min BCOPand the set Min B attaining it. COP COP In other words, B int( n)is -perfect if and only if it is the unique solution of the system2 ofIn linear otherCOP equations words, COPB int( n)is -perfect if and only if it is the unique solution of the system2 of linearCOP equationsCOP T B,vv =min B, forT all v Min B. h i COP B,vv =min2 COPB, for all v Min B. h i COP 2 COP In particular all verticesIn particular of are all vertices-perfect. of are -perfect. R COP R COP Lemma 2.4. -perfectA copositive matrices exist in all starting dimensions (dimension pointn =1being COP Lemma 2.4. -perfect matrices exist in all dimensions (dimension n =1being trivial): For dimensiontrivial):n 2 Forthe dimension followingCOP n matrix2 the following matrix 2 10... 20 10... 0 ...... 0 12. 0. 12. 1 . . . 1 . . . . . . THM:(4) . Q.A = . . . . (4) QAn = B 0 . .n B. 0 0 C. is . -perfect. 0 C B B C COP C B . . . B . C.. .. C B . .. .. B2 . 1C. . 2 1C B B C C B 0 ... 0 B120 C... 0 12C B B C C 1 @ A 1 is -perfect;@ Q is a vertex of A. is -perfect; Q is a vertex of 2 .An COP 2 An COP R R Proof. Matrix QAn is positive definite since Proof. Matrix QAn is positive definite since n 1 n 1 2 2 2 2 QAn [x]=x21 + 2 (xi xi+1) + xn for x R. QAn [x]=x + (xi xi+1) + x for x R. 2 1 in=1 2 i=1 X In particularX it lies in the interior of the copositive cone. Furthermore, In particular it lies in the interior of the copositive cone. Furthermore, k k min QA =2 with Min QA = ej :1 j k n , COP n COP n 8    9 min QA =2 with Min QA = ej :1 j k i=jn , COP n COP n 8   Gram matrix of the root⇤ lattice An,avery COP important lattice in the theory of sphere packings, see Conway and Sloane [3]. The matrix QAn is also known as a Gram matrix of the root lattice An,avery important lattice in the theory of sphere packings,3. seeThe Conway algorithm and Sloane [3].

In this section3. The we algorithm show how one can solve the linear program (2). Our algorithm is similar to the simplex algorithm for linear programming, as it walks along a In this section we showpath of how subsequently one can solve constructed the linear program-perfect (2). matrices, Our algorithm which are vertices of the COP is similar to the simplexpolyhedral algorithm set forand linear which programming, are connected by as edges. it walks along a R path of subsequently constructed -perfect matrices, which are vertices of the polyhedral set and which are connectedCOP by edges. R Voronoi-type simplex algorithm Input: A n 2 S>0 COP-SVPs: Obtain an initial -perfect matrix B COP P 1. BP, A < 0 A n (with witness BP) h i 62 CP 2. LP: A cone xx> : x Min BP A ˜ n 2 2 COP 2 CP 3. COP-SVP: Compute Min BP and the polyhedral cone COP n (BP)= B : B[x] 1 for all x Min BP P { 2 S 2 COP }

4. PolyRepConv: Determine a generator R of an extreme ray of (BP) P with A, R < 0. ( flexible ”pivot-rule” ) h i 5. SimplexDiv: R n A n (with witness R) 2 COP 62 CP 6. COP-SVPs: Determine the contiguous -perfect matrix COP BN := BP + R with > 0 and min BN = 1 COP 7. Set BP := BN and 1. Interior cases (algorithm terminates)

63 EX: A = 32 ✓ ◆

Starting with QA2 one iteration of the algorithm finds 1 3/2 the -perfect matrix BP = and COP 3/23 ✓ ◆ 1 1 > 1 1 > 2 2 > A = + + 0 0 1 1 1 1 ✓ ◆✓ ◆ ✓ ◆✓ ◆ ✓ ◆✓ ◆ 12 M.DutourSikiri´c,A.Sch¨urmann,andF.Vallentin

Note that there is always a unique choice in Step 4 in case A is a 2 2 rank-1 matrix. Note also that the vectors represent fractions that converge to ⇥p2. Every second vector corresponds to a convergent of the continued fraction of p2. For instance,

1 99/70 = 1 + . 1 2+ 1 2+ 1 2+ 1 2+ 2

The -perfect matrix after ten iterations of the algorithm is COP 4756 6726 B(10) = . P 6726 9512 ✓ ◆

(i) It can be shown that the matrices BP converge to a multiple of

1 p2 B = satisfying A, B = 0 and X, B 0 for all X 2. p22 h i h i 2 CP ✓ ◆

(i) However, every one of the inﬁnitely many perfect matrices BP satisﬁes

X, B(i) > 0 for all X . h P i 2 CP2

5. Computational Experiments We implemented our algorithm. The source code, written in C++, is available on TODO: github [?]. In this section we report how it performs on several examples, most of them were previously discussed in the literature. TODO: The ﬁrst example is the matrix

A1 = ... Boundary cases from ˜ n which lies in the interior of 5. Asimplexalgorithmforrationalcp-factorization 13 The second example is from(algorithmCP [11] and terminates lies in the boundarywith a suitable of pivot-rule)CP CP5 Starting from QA5 our algorithm does five steps to find the factorization A2 = 10 T 85115i=1 vivi with 58511 0P 1 v1 =(0, 0, 0, 1, 1) EX:A2 = 15851 from Groetzner, Dür (2018) B11585C v2 =(0, 0, 1, 1, 0) Asimplexalgorithmforrationalcp-factorizationB C 13 B51158C v =(0, 0, 1, 2, 1) B C 3 @ A Starting from QA5 our algorithm does five steps to find thev4 factorization=(0, 1, 1, 0A, 0)2 = 10 T Starting with QA , our algorithm finds a cp-factorization after 5 iterations i=1 vivi with 5 v5 =(0, 1, 2, 1, 0)

P v1 =(0, 0, 0, 1, 1) v6 =(1, 0, 0, 0, 1)

v2 =(0, 0, 1, 1, 0) v7 =(1, 0, 0, 1, 2)

v3 =(0, 0, 1, 2, 1) v8 =(1, 1, 0, 0, 0)

v4 =(0, 1, 1, 0, 0) v9 =(1, 2, 1, 0, 0)

v5 =(0, 1, 2, 1, 0) v10 =(2, 1, 0, 0, 1) giving va6 certiﬁcate=(1, 0 ,for0, 0 the, 1) matrix to be completely positive Thev7 third=(1 example, 0, 0, 1, 2) is from [17, Example 6.2] and it positive semideﬁnite, but not completelyv8 =(1 positive, 1, 0, 0, 0)

v9 =(1, 2, 1, 0, 0) 11001 12100 v10 =(2, 1, 0, 0, 1) 0 1 A3 = 01210 B00121C B C B10016C The third example is from [17, Example 6.2] and it positive semidefinite,B but notC @ A completely positive Starting from QA5 our algorithm does 18 steps to find the copositive matrix 11001 363/5 2126/35 2879/70 608/21 4519/210 12100 0 2126/351 1787/35 347/10 1025/42 253/14 A = 012100 1 3 B = 2879/70 347/10 829/35 1748/105 371/30 B3 00121C B B 608/21C 1025/42 1748/105 1237/105 601/70 C B10016B C C B B 4519/210C 253/14 371/30 601/70 671/105 C @ B A C Starting from QA5 our algorithm does 18@ steps to find the copositive matrix A with A3,B3 = 2/5 showing that A3 5. 363/5 2126The/h35 fourth 2879i and/70 last example 608/21 is the family451962 CP/ of210 completely positive matrices TODO 2126/35 1787/35 347/10 1025/42 253/14 0 1 B = 2879/70 347/10 829/35 1748/105 371/30 3 B 608/21 1025/42 1748/105 1237/105 601/70 C B from [14]. C B 4519/210 253/14 371/30 601/70 671/105 C B TODO: Some words about the runningC time. @ A with A3,B3 = 2/5 showing that A3 5. Theh fourthi and last example is the family62 CP of completely positiveAcknowledgement matrices TODO The second author likes to thank Je↵ Lagarias for a helpful discussion about Farey sequences. from [14]. This project has received funding from the European Union’s Horizon 2020 re- TODO: Some words aboutsearch the and running innovation time. programme under the Marie Sklodowska-Curie agreement No 764759. Acknowledgement The second author likes to thank Je↵ Lagarias for a helpfulReferences discussion about Farey sequences. [1] S. Bundfuss and M. Dür, Algorithmic copositivity detection by simplicial partition,Linear This project has received fundingAlgebra from Appl. the428 European(2008), 1511–1523. Union’s Horizon 2020 re- search and innovation programme under the Marie Sklodowska-Curie agreement No 764759.

References [1] S. Bundfuss and M. D¨ur, Algorithmic copositivity detection by simplicial partition,Linear Algebra Appl. 428 (2008), 1511–1523. Asimplexalgorithmforrationalcp-factorization 13

Starting from QA5 our algorithm does ﬁveAsimplexalgorithmforrationalcp-factorization steps to ﬁnd the factorization A2 = 13 10 T i=1 vivi with

Starting from QA5 our algorithm does five steps to find the factorization A2 = P 10 Tv1 =(0, 0, 0, 1, 1) i=1 vivi with v2 =(0, 0, 1, 1, 0) P v1 =(0, 0, 0, 1, 1) v3 =(0, 0, 1, 2, 1) v2 =(0, 0, 1, 1, 0) v4 =(0, 1, 1, 0, 0) v3 =(0, 0, 1, 2, 1) v5 =(0, 1, 2, 1, 0) v4 =(0, 1, 1, 0, 0) v6 =(1, 0, 0, 0, 1) v5 =(0, 1, 2, 1, 0) v7 =(1, 0, 0, 1, 2) v6 =(1, 0, 0, 0, 1) v8 =(1, 1, 0, 0, 0) v7 =(1, 0, 0, 1, 2) v9 =(1, 2, 1, 0, 0) v8 =(1, 1, 0, 0, 0) v10 =(2, 1, 0, 0, 1) Exterior casesv9 =(1, 2, 1, 0, 0) v10 =(2, 1, 0, 0, 1) The third example is from(algorithm [17, Example conjectured 6.2] and it positiveto terminate) semidefinite, but not completely positive The third example11001 is from [17, Example 6.2] and it positive semidefinite, but not completely positive12100 0 1 11001 EX:A3 = 01210 from Nie (2014) 12100 B00121C 0 1 B AC= 01210 B10016C3 B C B00121C Starting from Q our algorithm does@ 18 steps to findA theB copositive matrixC A5 B10016C Starting with Q , after 18 iterationsB our algorithmC finds the -perfect 363/5 Starting2126 from/35Q 2879ourA/5 algorithm70 608 does@/21 18 steps4519 to/ find210A the copositive matrix A5 COP 2126/35 1787/35 347/10 1025/42 253/14 0 363/5 2126/35 2879/70 6081/21 4519/210 B3 = 2879/70 347/10 829/35 1748/105 371/30 2126/35 1787/35 347/10 1025/42 253/14 B 608/21 1025/042 1748/105 1237/105 601/70 C 1 B B3 = 2879/70 347/10 829/35 1748C /105 371/30 B 4519/210 253/14 371/30 601/70 671/105 C B B 608/21 1025/42 1748/105 1237C/105 601/70 C B C with A ,B@ = 2/5 showing thatB 4519A /210. 253/14 371/30 601A /70 671/105 C 3 3 B 3 5 C Theh fourthi and last example is@ the family62 CP of completely positive matrices TODO A with A3,B3 = 2/5 showing that A3 5. Theh fourthgivingi and a last certificate example for is the the family matrix62 CP of completelynot to be completely positive matrices positiveTODO from [14]. TODO: Some wordsfrom about [14]. the running time. TODO: Some words about the running time. Acknowledgement The second author likes to thank Je↵ LagariasAcknowledgement for a helpful discussion about Farey sequences. The second author likes to thank Je↵ Lagarias for a helpful discussion about This project has receivedFarey sequences. funding from the European Union’s Horizon 2020 re- search and innovation programmeThis project under has received the Marie funding Sklodowska-Curie from the European agreement Union’s Horizon 2020 re- No 764759. search and innovation programme under the Marie Sklodowska-Curie agreement No 764759. References [1] S. Bundfuss and M. Dür, Algorithmic copositivity detectionReferences by simplicial partition,Linear Algebra Appl. 428 (2008),[1] S. 1511–1523. Bundfuss and M. Dür, Algorithmic copositivity detection by simplicial partition,Linear Algebra Appl. 428 (2008), 1511–1523. Open Questions / TODOs

• Find suitable / good pivot rules for boundary cases

• Prove termination of algorithm for exterior cases

• Improve computations in practice

• … in particular: ﬁnd a better algorithm to compute min and the set of its representatives Min COP COP ( COP-SVPs ) References

• Achill Schürmann, Computational Geometry of Positive Deﬁnite Quadratic Forms, University Lecture Series, AMS, Providence, RI, 2009.

• Mathieu Dutour Sikirić, Achill Schürmann and Frank Vallentin, Rational factorizations of completely positive matrices, Linear Algebra and its Applications, 523 (2017), 46–51.

• Mathieu Dutour Sikirić, Achill Schürmann and Frank Vallentin, A simplex algorithm for rational cp-factorization, Math. Prog. A, 2020, online ﬁrst

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