structured/0707.4565.structured.json

{"paper_meta":{"paper_id":"arxiv:0707.4565","title":"0707.4565","authors":[],"year":null,"published":null,"updated":null,"venue_or_source":"arxiv-papers-shard","primary_category":"cs","secondary_categories":[],"doi":null,"license":"unknown","source_type":"pdf+shard","source_url":null,"pdf_url":null},"abstract":{"raw":"","cleaned":"","offsets":[]},"full_paper_text":{"raw_ordered_text":"arXiv:0707.4565v3 [cs.CC] 16 Apr 2008\nOn the Complexity of the Interlace Polynomial∗\nMarkus Bl ̈aser, Christian Hoffmann\nNovember 16, 2021\nAbstract\nWe consider the two-variable interlace polynomial introduced by Arratia,\nBollob ́as and Sorkin (2004). We develop graph transformations which allow\nus to derive point-to-point reductions for the interlace polynomial. Exploiting\nthese reductions we obtain new results concerning the computational complex-\nity of evaluating the interlace polynomial at a fixed point. Regarding exact\nevaluation, we prove that the interlace polynomial is #P-hard to evaluate at\nevery point of the plane, except on one line, where it is trivially polynomial\ntime computable, and four lines, where the complexity is still open. This solves\na problem posed by Arratia, Bollob ́as and Sorkin (2004). In particular, three\nspecializations of the two-variable interlace polynomial, the vertex-nullity in-\nterlace polynomial, the vertex-rank interlace polynomial and the independent\nset polynomial, are almost everywhere #P-hard to evaluate, too. For the inde-\npendent set polynomial, our reductions allow us to prove that it is even hard\nto approximate at every point except at 0.\n1\nIntroduction\nThe number of Euler circuits in specific graphs and their interlacings turned out\nto be a central issue in the solution of a problem related to DNA sequencing by\nhybridization [ABCS00]. This led to the definition of a new graph polynomial, the\none-variable interlace polynomial [ABS04a].\nFurther research on this polynomial\ninspired the definition of a two-variable interlace polynomial q(G; x, y) containing\nas special cases the following graph polynomials: qN(G; y) = q(G; 2, y) is the origi-\nnal one-variable interlace polynomial which was renamed to “vertex-nullity interlace\n∗A preliminary version of this work has appeared in the proceedings of STACS 2008.\n1\n\npolynomial”, qR(G; x) = q(G; x, 2) is the new “vertex-rank interlace polynomial” and\nI(G; x) = q(G; 1, 1 + x) is the independent set polynomial1 [ABS04b].\nAlthough the interlace polynomial q(G; x, y) is a different object from the cele-\nbrated Tutte polynomial (also known as dichromatic polynomial, see, for instance,\n[Tut84]), they are also similar to each other. While the Tutte polynomial can be\ndefined recursively by a deletion-contraction identity on edges, the interlace polyno-\nmial satisfies recurrence relations involving several operations on vertices (deletion,\npivotization, complementation).\nBesides the deletion-contraction identity, the so called state expansion is a well-\nknown way to define the Tutte polynomial. Here the similarity to the two-variable\ninterlace polynomial is especially striking: while the interlace polynomial is defined\nas a sum over all vertex subsets of the graph using the rank of adjacency matrices\n(see (2.1)), the state expansion of the Tutte polynomial can be interpreted as a sum\nover all edge subsets of the graph using the rank of incidence matrices (see (4.1))\n[ABS04b, Section 1].\nReferences to further work on the interlace polynomial can be found in [ABS04b]\nand [EMS07].\n1.1\nPrevious work\nThe aim of this paper is to explore the computational complexity of evaluating2 the\ntwo-variable interlace polynomial q(G; x, y). For the Tutte polynomial this problem\nwas solved in [JVW90]: Evaluating the Tutte polynomial is #P-hard at any alge-\nbraical point of the plane, except on the hyperbola (x −1)(y −1) = 1 and at a\nfew special points, where the Tutte polynomial can be evaluated in polynomial time.\nFor the two-variable interlace polynomial q(G; x, y), only on a one-dimensional sub-\nset of the plane (on the lines x = 2 and x = 1) some results about the evaluation\ncomplexity are known.\nA connection between the vertex-nullity interlace polynomial and the Tutte poly-\nnomial of planar graphs [ABS04a, End of Section 7], [EMS07, Theorem 3.1] shows\nthat evaluating q is #P-hard almost everywhere on the line x = 2 (Corollary 4.4).\nIt has also been noticed that q(G; 1, 2) evaluates to the number of independent\nsets of G [ABS04b, Section 5], which is #P-hard to compute [Val79]. Recent work\non the matching generating polynomial [AM07] implies that evaluating q is #P-hard\nalmost everywhere on the line x = 1 (Corollary 4.10).\n1The independent set polynomial of a graph G is defined as I(G; x) = P\nj≥0 i(G; j)xj, where\ni(G; j) denotes the number of independent sets of cardinality j of G.\n2See Section 2.2 for a precise definition.\n2\n\nA key ingredient of [JVW90] is to apply graph transformations known as stretch-\ning and thickening of edges. For the Tutte polynomial, these graph transformations\nallow us to reduce the evaluation at one point to the evaluation at another point.\nFor the interlace polynomial no such graph transformations have been given so far.\n1.2\nOur results\nWe develop three graph transformations which are useful for the interlace polyno-\nmial: cloning of vertices and adding combs or cycles to the vertices. Applying these\ntransformations allows us to reduce the evaluation of the interlace polynomial at\nsome point to the evaluation of it at another point, see Theorem 3.3, Theorem 3.5\nand Theorem 3.7. We exploit this to obtain the following new results about the\ncomputational complexity of q(G; x, y).\nWe prove that the two-variable interlace polynomial q(G; x, y) is #P-hard to\nevaluate at almost every point of the plane, Theorem 4.11, see also Figure 1. Even\nthough there are some unknown (gray, in Figure 1) lines left on the complexity\nmap for q, this solves a challenge posed in [ABS04b, Section 5]. In particular we\nobtain the new result that evaluating the vertex-rank interlace polynomial qR(G; x)\nis #P-hard at almost every point (Corollary 4.12). Our techniques also give a new\nproof that the independent set polynomial is #P-hard to evaluate almost everywhere\n(Corollary 4.10).\nApart from these results on the computational complexity of evaluating the in-\nterlace polynomial exactly, we also show that the values of the independent set poly-\nnomial (which is the interlace polynomial q(G; x, y) on the line x = 1) are hard to\napproximate almost everywhere (Theorem 5.4).\n2\nPreliminaries\n2.1\nInterlace Polynomials\nWe consider undirected graphs without multiple edges but with self loops allowed.\nLet G = (V, E) be such a graph and A ⊆V . By G[A] we denote (A, {e|e ∈E, e ⊆\nA}), the subgraph of G induced by A. The adjacency matrix of G is the symmetric\nn × n-matrix M = (mij) over F2 = {0, 1} with mi,j = 1 iff{i, j} ∈E. The rank\nof this matrix is its rank over F2. Slightly abusing notation we write rk(G) for this\nrank. This allows us to define the two-variable interlace polynomial.\n3\n\nDefinition 2.1 ([ABS04b]). Let G = (V, E) be an undirected graph. The interlace\npolynomial q(G; x, y) of G is defined as\nq(G; x, y) =\nX\nA⊆V\n(x −1)rk(G[A])(y −1)|A|−rk(G[A]).\n(2.1)\nIn Section 3 we will introduce graph transformations which perform one and the\nsame operation (cloning one single vertex, adding a comb or a cycle to one single\nvertex, resp.) on every vertex of a graph. Instead of relating the interlace polynomial\nof the original graph directly to the interlace polynomial of the transformed graph,\nwe will analyze how, say, cloning one single vertex changes the interlace polynomial.\nTo express this, we must be able to treat the vertex being cloned in a particular\nway, differently from the other vertices. This becomes possible using a multivariate\nversion of the interlace polynomial, in which each vertex has its own variable. Once\nwe can express the effect of cloning one vertex by an appropriate substitution of\nthe vertex variable in the multivariate interlace polynomial, cloning all the vertices\namounts to a simple substitution of all vertex variables and brings us back to a\nbivariate interlace polynomial. This procedure has been applied successfully to the\nTutte polynomial [Sok05, BM06].\nWe choose the following multivariate interlace polynomial, which is similar to the\nmultivariate Tutte polynomial of Sokal [Sok05] and a specialization of the multivari-\nate interlace polynomial defined by Courcelle [Cou07].\nDefinition 2.2. Let G = (V, E) be an undirected graph. For each v ∈V let xv\nbe an indeterminate. Writing xA for Q\nv∈A xv, we define the following multivariate\ninterlace polynomial:\nP(G; u, x) =\nX\nA⊆V\nxAurk(G[A]).\nSubstituting each xv in P(G; u, x) by x, we obtain another bivariate interlace poly-\nnomial:\nP(G; u, x) =\nX\nA⊆V\nx|A|urk(G[A]).\nAn easy calculation proves that q and P are closely related:\nLemma 2.3. Let G be a graph. Then we have the polynomial identities q(G; x, y) =\nP(G; x−1\ny−1, y −1) and P(G; u, x) = q(G; ux + 1, x + 1).\n4\n\n2.2\nEvaluating Graph Polynomials\nGiven ξ, υ ∈Q we want to analyze the following computational problem:\nInput Graph G\nOutput q(G; ξ, υ)\nThis is what we mean by “evaluating the interlace polynomial q at the point (ξ, υ)”.\nAs an abbreviation for this computational problem we write\nq(ξ, υ),\nwhich should not be confused with the expression q(G; ξ, υ) denoting just a value in\nQ. Evaluating other graph polynomials such as P, qN, qR and I is defined accordingly.\nIf P1 and P2 are computational problems we use P1 ⪯T P2 (P1 ⪯m P2) to denote a\npolynomial time Turing reduction (polynomial time many-one reduction, resp.) from\nP1 to P2. For instance, Lemma 2.3 gives\nCorollary 2.4. For ξ, υ ∈ ̃Q, υ ̸= 1, we have q(ξ, υ) ⪯m P( ξ−1\nυ−1, υ −1). For μ, ξ ∈ ̃Q\nwe have P(μ, ξ) ⪯m q(μξ + 1, ξ + 1).\nHere ̃Q denotes some finite dimensional field extension Q ⊆ ̃Q ⊆R, which has\na discrete representation. As\n√\n2 will play an important role but we are not able to\nuse arbitrary real numbers as the input for a Turing machine, we use ̃Q instead of\nQ or R. We fix some ̃Q for the rest of this paper. This construction is done in the\nspirit of Jaeger, Vertigan, and Welsh [JVW90] who also propose to adjoin a finite\nnumber of points to Q in order to talk about the complexity at irrational points. To\nsome extent, this is an ad hoc construction, but it is sufficient for this work.\n3\nGraph Transformations for the Interlace Poly-\nnomial\nNow we describe our graph transformations, vertex cloning and adding combs or\ncycles to the vertices. The main results of this section are Theorem 3.3, Theorem 3.5\nand Theorem 3.7, which describe the effect of these graph transformations on the\ninterlace polynomial.\n3.1\nCloning\nCloning vertices in the graph yields our first graph transformation.\n5\n\nCloning one vertex\nLet G = (V, E) be a graph. Let a ∈V be some vertex (the one which will be cloned)\nand N the set of neighbors of a, V ′ = V \\ {a} and M = V ′ \\ N. The graph G\nwith a cloned, Gaa, is obtained out of G in the following way: Insert a new isolated\nvertex a′. Connect a′ to all vertices in N. If a does not have a self loop, we are done.\nOtherwise connect a and a′ and insert a self loop at a′. Thus, adjacency matrices of\nthe original (cloned, resp.) graph are\nB =\na\nN\nM\na\nb\n1\n0\nN\n1\nA11\nA12\nM\n0\nA21\nA22\nand\nBaa =\na′\na\nN\nM\na′\nb\nb\n1\n0\na\nb\nb\n1\n0\nN\n1\n1\nA11\nA12\nM\n0\n0\nA21\nA22\n,\nresp, (3.1)\nwhere b = 1 if a has a self loop and b = 0 otherwise. As the first column of Baa equals\nits second column, as well as the first row equals the second row, we can remove the\nfirst row and the first column of Baa without changing the rank. This also holds\nwhen we consider the adjacency matrices of G[A] (Gaa[A], resp.) instead of G (Gaa\nresp.) for A ⊆V ′. Thus we have for any A ⊆V ′\nrk(Gaa[A]) = rk(G[A]),\n(3.2)\nrk(Gaa[A ∪{a, a′}]) = rk(Gaa[A ∪{a}]) = rk(Gaa[A ∪{a′}]) = rk(G[A ∪{a}]).\n(3.3)\nLet x = (xv)v∈V (Gaa) be a labeling of the vertices of Gaa by indeterminates.\nDefine X to denote the following labeling of the vertices of G: Xv := xv for all\nv ∈V ′, Xa := (1 + xa)(1 + xa′) −1 = xa + xa′ + xaxa′. Then we have\nLemma 3.1. P(Gaa; u, x) = P(G; u, X).\nProof. On the one hand we have\nP(Gaa; u, x)\n=\nX\nA⊆V ′\nxA(urk(Gaa[A]) + xaurk(Gaa[A∪{a}]) + xa′urk(Gaa[A∪{a′}]) + xaxa′urk(Gaa[A∪{a,a′}]))\n=\nX\nA⊆V ′\nxA(urk(G[A]) + (xa + xa′ + xaxa′)urk(G[A∪{a}])) by (3.2), (3.3).\n6\n\nOn the other hand we have\nP(G; u, X) =\nX\nA⊆V ′\nXA(urk(G[A]) + Xaurk(G[A∪{a}]))\n=\nX\nA⊆V ′\nxA(urk(G[A]) + (xa + xa′ + xaxa′)urk(G[A∪{a}])).\nCloning all vertices\nFix some k. Given a graph G, the graph Gk is obtained by cloning each vertex of\nG exactly k −1 times. Note that the result of the cloning is independent of the\norder in which the different vertices are cloned. For a ∈V (G) let a1, . . . , ak be the\ncorresponding vertices in Gk. For a vertex labeling x of Gk we define the vertex\nlabeling X of G by Xa = (1 + xa1)(1 + xa2) · · · (1 + xak) −1 for a ∈V (G). Applying\nLemma 3.1 repeatedly we obtain\nLemma 3.2. P(Gk; u, x) = P(G; u, X).\nSubstitution of xv by x for all vertices v gives\nTheorem 3.3. Let G be a graph and Gk be obtained out of G by cloning each vertex\nof G exactly k −1 times. Then\nP(Gk; u, x) = P(G; u, (1 + x)k −1).\n(3.4)\nAs we will use it in the proof of Theorem 4.11, we note the following identity for\nq, which can be derived from Theorem 3.3 using Lemma 2.3:\nq(Gk; x, y) = q(G; (x −1)yk −1\ny −1 + 1, yk).\n(3.5)\nTheorem 3.3 also implies the following reduction for the interlace polynomial,\nwhich is the foundation for our results in Section 4.\nProposition 3.4. Let B2 = {0, −1, −2} and x be an indeterminate. For μ ∈ ̃Q, ξ ∈\n ̃Q \\ B2 we have P(μ, x) ⪯T P(μ, ξ). (For any μ ∈ ̃Q, we write P(μ, x) to denote\nthe following computational problem: given a graph G compute P(G; μ, x), which is\na polynomial in x with coefficients in ̃Q.)\n7\n\nProof. Let μ and ξ be given such that they fulfill the precondition of the proposition.\nGiven a graph G =: G1 with n vertices, we build G2, G3, . . . , Gn+1, where Gi is\nobtained out of G by cloning each vertex i −1 times.\nThis is possible in time\npolynomial in n. By Theorem 3.3, a call to an oracle for P(μ, ξ) with input Gi gives\nus P(G; μ, (1 + ξ)i −1) for i = 1, . . . , n + 1. The restriction on ξ guarantees that for\ni = 1, 2, 3, . . . the expression (1 + ξ)i −1 evaluates to pairwise different values. Thus,\nfor P(G; μ, x), which is a polynomial in x of degree ≤n, we have obtained the values\nat n + 1 distinct points. Using Lagrange interpolation we determine the coefficients\nof P(G; μ, x).\n3.2\nAdding Combs\nThe comb transformation sometimes helps, when cloning has not the desired effect.\nLet G = (V, E) be a graph and a ∈V some vertex. Then we define the k-comb of\nG at a as Ga,k = (V ∪{a1, . . . , ak}, E ∪{{a, a1}, . . . , {a, ak}}), with a1, . . . , ak being\nnew vertices.\nUsing similar arguments as with vertex cloning, adding combs to vertices yields\na point-to-point reduction for the interlace polynomial, too.\nTheorem 3.5. Let G be a graph and Gk be obtained out of G by performing a k-comb\noperation at every vertex. Then\nP(Gk; u, x) = p(k, u, x)|V (G)|P(G; u, x/p(k, u, x)),\n(3.6)\nwhere p(k, u, x) = (1 + x)k(xu2 + 1) −xu2.\nProof. The adjacency matrices of the original graph G (the graph Ga,k with a k-comb\nat a, resp.) are\na\nV ′\na\nb\nc\nV ′\ncT\nA\nand\na1\na2\n. . .\nak\na\nV ′\na1\n1\na2\n1\n...\n...\nak\n1\na\n1\n1\n. . .\n1\nb\nc\nV ′\ncT\nA11\n, resp.,\n(3.7)\nwith empty entries being zero. Consider A ⊆V (Ga,k). Let M := A ∩{a, a1, . . . , ak}.\nBy (3.7), the rank of Ga,k is related to the rank for G in the following way:\n8\n\n• If a ̸∈M, then rk(Ga,k[A]) = rk(G[A \\ M]).\n• If a ∈M and M ∩{a1, . . . , ak} ̸= ∅, then rk(Ga,k[A]) = rk(G[A \\ M]) + 2: Let\nw.l.o.g. a1 ∈M. Consider the adjacency matrix of Ga,k[A] and the following\noperations on it, which leave the rank unchanged. Using the first column we\nremove all 1s in the a-row, except the 1 in the first column. Using the first row\nwe remove all 1s in the a-column, except the 1 in the first row. The resulting\nmatrix B is a (k + |V |) × (k + |V |) matrix with 1s at positions (a, a1) and\n(a1, a), the submatrix of A11 induced by A \\ M in the lower right corner and\nzeros everywhere else. Thus rk(B) = rk(G[A \\ M]) + 2.\n• If M = {a}, then rk(Ga,k[A]) = rk(G[A]).\nLetting r(A) := rk(G[A]) and ra(A) := rk(G[A ∪{a}]) for A ⊆V ′, we see that\nP(Ga,k; u, x) equals\nX\nA⊆V ′\nxA\n \nur(A) X\n∅⊆S⊆{a1,...,ak}\nxS + xau2 ·\nX\n∅⊊S⊆{a1,...,ak}\nxS\n|\n{z\n}\n=:p(k,u,x)\n \n+ xaura(A) \nNote that p(k, u, x) does only depend on xa, xa1, . . . , xak, but not on xv for any v ∈V ′.\nAs we have\nP(G; u, X) =\nX\nA⊆V ′\nXA(ur(A) + Xaura(A)),\nwe conclude\nP(Ga,k; u, x) = p(k, u, x)P(G; u, X),\nwhere Xv = xv for v ∈V ′ and Xa =\nxa\np(k,u,x).\nWe can perform a k-comb operation at every a ∈V and call the result Gk.\nSubstituting x for xv, v ∈Gk, concludes the proof.\nThis yields\nProposition 3.6. Let p(k, u, x) = (1+x)k(xu2+1)−xu2. Let k be a positive integer\nand μ, ξ ∈ ̃Q. If p(k, μ, ξ) ̸= 0, we have P(μ, ξ/p(k, μ, ξ)) ⪯m P(μ, ξ).\n9\n\n3.3\nAdding Cycles\nLet G = (V, G) be a graph and a ∈V some vertex. Consider the graph Ga,k =\n(V ∪{1, 2, . . . , k −1}, E ∪{{a, 1}, {a, k −1}} ∪{{i −1, i} | 1 < i < k}), with\n1, 2, . . . , k −1 being new vertices. We say that Ga,k has been obtained out of G by\nadding a k-cycle to a.\nTheorem 3.7. Let G be a graph and Gk be obtained out of G by adding a k-cycle to\nevery vertex. Then P(Gk; u, x) = pk(u, x)P(G; u, qk(u, x)/pk(u, x)) for k = 3, 4 with\np3(u, x) = 1 + 2x + 3x2u2, q3(u, x) = x + x3u2, p4(u, x) = 1 + 3x + x2 + 2x2u2 + x3u2\nand q4(u, x) = x2 + 2x3u2 + x4u2.\nProposition 3.8. P(0, 1) ⪯m P(0, −1) and P(μ, −4) ⪯m P(μ, −2) for every μ ∈ ̃Q.\nProof. The first reduction follows from Theorem 3.7 adding 3-cycles, the second\nadding 4-cycles.\nProof of Theorem 3.7. We use the same idea as in the proof of Theorem 3.5. Consider\nthe case of a 3-cycle added at vertex a. Let V ′ = V \\ {a}. The adjacency matrix of\nGa,3 is\n1\n2\na\nV ′\n1\n1\n1\n2\n1\n1\na\n1\n1\nb\nc\nV ′\ncT\nA11\nwith empty entries being zero. Adding the second row to the first row and the second\ncolumn to the first column and subsequently the first row to the third row and the\nfirst column to the third column does not change the rank and gives\n1\n2\na\nV ′\n1\n1\n2\n1\na\nb\nc\nV ′\ncT\nA11\n.\nThis shows that rk(Ga,3[A] = rk(G[A\\{1, 2}])+2 for all A, {1, 2, a} ⊆A ⊆V (Ga,3).\nUsing arguments similar to this one and the ones in the proof of Theorem 3.5 we\nfind that\n• rk(Ga,3[A]) = rk(G[A ∩V ′] + 2 for all A, {a} ⊆A ⊆V (Ga,3) and either 1 ∈A\nor 2 ∈A,\n10\n\n• rk(Ga,3[A]) = rk(G[A] for all A, {a} ⊆A ⊆V (Ga,3) and {1, 2} ∩A = ∅,\n• rk(Ga,3[A]) = rk(G[A ∩V ′]) + rk(P2[A ∩V (P2)]) for all A ⊆V (Ga,3), a ̸∈A,\nwhere P2 is the the path with two vertices 1, 2.\nLetting again r(A) := rk(G[A]) and ra(A) := rk(G[A∪{a}]) for A ⊆V ′, we see that\nP(Ga,3; u, x) equals\nX\nA⊆V ′\nxA\n \nur(A) 1 + x1 + x2 + x1x2u2 + x1xau2 + x2xau2\n|\n{z\n}\n=:p3(u,x)\n \n+ ura(A)(xa + x1x2xau2\n|\n{z\n}\n=:q3(u,x)\n)\n \n,\nwhich equals p3(u, x)P(G; u, X) if we define X by Xv = xv for v ∈V ′ and Xa =\nq3(u, x)/p3(u, x).\nWe can use this identity for every vertex a and substitute xa,\na ∈V , by a single variable x. This gives the statement of the theorem concerning\n3-cycles. For 4-cycles we proceed in a similar fashion.\n4\nComplexity of evaluating the Interlace Polyno-\nmial exactly\nThe goal of this section is to uncover the complexity maps for P and q as indicated\nin Figure 1. While the left hand side (complexity map for P) is intended to follow\nthe arguments which prove the hardness, the right hand side (complexity map for q)\nfocuses on presenting the results.\nRemark 4.1. P(μ, 0) and P(1, ξ) are trivially solvable in polynomial time for any\nμ, ξ ∈ ̃Q, as P(G; μ, 0) = 1 and P(G; 1, ξ) = (1 + ξ)|V |.\nThus, on the thick black lines x = 0 and u = 1 in the left half of Figure 1, P can\nbe evaluated in polynomial time. By Lemma 2.3, these lines in the complexity map\nfor P correspond to the point (1, 1) and the line x = y, resp., in the complexity map\nfor q, see the right half of Figure 1.\n4.1\nIdentifying hard points\nWe want to establish Corollary 4.4 and Remark 4.5 which tell us, that P is #P-hard\nto evaluate almost everywhere on the dashed hyperbola in Figure 1 and at (0, 1). To\nthis end we collect known hardness results about the interlace polynomial.\n11\n\nFigure 1: Complexity of the interlace polynomials P and q. α =\n√\n2, β = 1/\n√\n2\nLet t(G; x, y) denote the Tutte polynomial of an undirected graph G = (V, E). It\nmay be defined by its state expansion as\nt(G; x, y) =\nX\nB⊆E(G)\n(x −1)r(E)−r(B)(y −1)|B|−r(B),\n(4.1)\nwhere r(B) is the F2-rank of the incidence matrix of G[B] = (V, B), the subgraph of\nG induced by B. (Note that r(B) equals the number of vertices of G[B] minus the\nnumber of components of G[B], which is the rank of B in the cycle matriod of G.)\nFor details about the Tutte polynomial we refer to standard literature [Tut84, BO92,\nWel93]. The complexity of the Tutte polynomial has been studied extensively. In\nparticular, the following result is known.\nTheorem 4.2 ([Ver05]). Evaluating the Tutte polynomial of planar graphs at (ξ, ξ)\nis #P-hard for all ξ ∈ ̃Q except for ξ ∈{0, 1, 2, 1 ±\n√\n2}.\nWe will profit from this by a connection between the interlace polynomial and\nthe Tutte polynomial of planar graphs. This connection is established via medial\ngraphs. For any planar graph G one can build the oriented medial graph ⃗Gm, find\nan Euler circuit C in ⃗Gm and obtain the circle graph H of C. The whole procedure\ncan be performed in polynomial time. For details we refer to [EMS07]. We will use\n12\n\nTheorem 4.3 ([ABS04a, End of Section 7]; [EMS07, Theorem 3.1]). Let G be a\nplanar graph, ⃗Gm be the oriented medial graph of G and H be the circle graph of\nsome Euler circuit C of ⃗Gm. Then q(H; 2, y) = t(G; y, y). Thus we have t(υ, υ) ⪯m\nP(\n1\nυ−1, υ −1), where t(υ, υ) denotes the problem of evaluating the Tutte polynomial\nof a planar graph at (υ, υ).\nProof. See the references for q(H; 2, y) = t(G; y, y) and use Lemma 2.3.\nWe set α =\n√\n2 and β = 1/\n√\n2.\nLet B1 = {±1, ±β, 0}. Theorem 4.2 and\nTheorem 4.3 yield\nCorollary 4.4. Evaluating the vertex-nullity interlace polynomial qN is #P-hard\nalmost everywhere. In particular, we have:\n• The problem qN(2) is trivially solvable in polynomial time.\n• For any υ ∈ ̃Q \\ {0, 1, 2, 1 ± α} the problem qN(υ) = q(2, υ) is #P-hard. Or,\nin other words, for any μ ∈ ̃Q \\ B1 the problem P(μ, 1/μ) is #P-hard.\nRemark 4.5. P(0, 1) is #P-hard, as P(G; 0, 1) equals the number of independent\nsets of G, which is #P-hard to compute [Val79].\n4.2\nReducing to hard points\nThe cloning reduction allows us to spread the collected hardness over almost the\nwhole plane: Combining Corollary 4.4 and Remark 4.5 with Proposition 3.4 we\nobtain\nProposition 4.6. Let B1 = {±1, ±β, 0} and B2 = {0, −1, −2} (as defined on\nPages 13 and 7, resp.). Let (μ, ξ) ∈(( ̃Q \\ B1) ∪{0}) × ( ̃Q \\ B2). Then P(μ, ξ)\nis #P-hard.\nThis tells us that P is #P-hard to evaluate at every point in the left half of\nFigure 1 not lying on one of the seven thick lines (three of which are solid gray ones,\ntwo of which are solid black ones, and two of which are dashed brown ones). Using\nthe comb and cycle reductions we are able to reveal the hardness of the interlace\npolynomial P on the lines x = −1 and x = −2:\nProposition 4.7. For μ ∈( ̃Q \\ B1) ∪{0} the problem P(μ, −1) is #P-hard.\n13\n\nProof. For μ = 0 we use Proposition 3.8 and Remark 4.5. If μ ̸= 0, we can use\nProposition 3.6, which yields P(μ, −1/μ2) ⪯m P(μ, −1). For μ = ±1 this reduces\n(±1, −1) to itself.\nFor μ = ±β this reduces (β, −2) to (β, −1) and (−β, −2) to\n(−β, −1). For other μ this gives a reduction of some point, which is already known\nas #P-hard by Proposition 4.6, to (μ, −1).\nProposition 4.8. For μ ∈( ̃Q \\ B1) ∪{0} the problem P(μ, −2) is #P-hard.\nProof. Use Proposition 3.8 and Proposition 4.6.\n4.3\nSumming up\nFirst we summarize our knowledge about P.\nTheorem 4.9. Let β = 1/\n√\n2.\n1. P(μ, ξ) is computable in polynomial time on the lines μ = 1 and ξ = 0.\n2. For (μ, ξ) ∈( ̃Q \\ {−1, −β, β, 1}) × ( ̃Q \\ {0}) the problem P(μ, ξ) is #P-hard.\nProof. Summary of Remark 4.1, Proposition 4.6, Proposition 4.7, Proposition 4.8.\nThe hardness of P(0, −1) follows from Corollary 4.10.\nIn particular we obtain the following corollary about the complexity of the inde-\npendent set polynomial, which also follows from [AM07].\nCorollary 4.10. Evaluating the independent set polynomial I(λ) = P(0, λ) = q(1, 1+\nλ) is #P-hard at all λ ∈ ̃Q except at λ = 0, where it is computable in polynomial\ntime.\nNow we turn to the complexity of q, see also the right half of Figure 1.\nTheorem 4.11. The two-variable interlace polynomial q is #P-hard to evaluate\nalmost everywhere. In particular, we have:\n1. q(ξ, υ) is computable in polynomial time on the line ξ = υ.\n2. Let ξ ∈ ̃Q\\{1} and x be an indeterminate. Then q(ξ, 1) is as hard as computing\nthe whole polynomial q(x, 1).\n3. q(ξ, υ) is #P-hard for all\n(ξ, υ) ∈{(ξ, υ) ∈ ̃Q2 | υ ̸= ±(ξ −1) + 1 and υ ̸= ±\n√\n2(ξ −1) + 1 and υ ̸= 1}.\n14\n\nProof of Theorem 4.11 (Sketch). (1) and (3) follow from Remark 4.1 and Theorem 4.9\nusing Lemma 2.3. For ξ ̸= 1, (3.5) gives q(Gk; ξ, 1) = q(G; k(ξ −1) + 1, 1), which\nyields enough points for interpolation in the same way as in Proposition 3.4 using\nk = 1, 2, 3, . . . This proves (2).\nTheorem 4.11 implies\nCorollary 4.12. Let β = 1/\n√\n2. Evaluating the vertex-rank interlace polynomial\nqR(G; x) is #P-hard at all ξ ∈ ̃Q except at ξ = 0, 1 −β, 1 + β (complexity open) and\nξ = 2 (computable in polynomial time).\n5\nInapproximability of the Independent Set Poly-\nnomial\nProvided we can evaluate the independent set polynomial at some fixed point, vertex\ncloning (adding combs, resp.) allows us to evaluate it at very large points. In this\nsection we exploit this to prove that the independent set polynomial is hard to\napproximate. Similar results are shown in [GJ07] for the Tutte polynomial.\nDefinition 5.1. Let λ ∈ ̃Q and ε > 0. By a randomized 2n1−ε-approximation algo-\nrithm for I(λ) we mean a randomized algorithm, that, given a graph G with n nodes,\nruns in time polynomial in n and returns ̃I(G; λ) ∈ ̃Q such that\nPr[2−n1−εI(G; λ) ≤ ̃I(G; λ) ≤2n1−εI(G; λ)] ≥3\n4.\nIn [GJ07], (non)approximability in the weaker sense of (not) admitting an FPRAS\nis considered.\nDefinition 5.2. Let λ ∈ ̃Q. A fully polynomial randomized approximation scheme\n(FPRAS) for I(λ) is a randomized algorithm, that given a graph G with n nodes and\nan error tolerance ε, 0 < ε < 1, runs in time polynomial in n and 1/ε and returns\n ̃I(G; λ) ∈ ̃Q such that\nPr[2−εI(G; λ) ≤ ̃I(G; λ) ≤2εI(G; λ)] ≥3\n4.\nLemma 5.3. For every λ ∈ ̃Q, 0 ̸= |1 + λ| ̸= 1, and every ε, 0 < ε < 1, there is no\nrandomized polynomial time 2n1−ε-approximation algorithm for I(λ) unless RP = NP.\n15\n\nTheorem 5.4. For every λ ∈ ̃Q\\{0} and every ε, 0 < ε < 1, there is no randomized\npolynomial time 2n1−ε-approximation algorithm (and thus also no FPRAS) for I(λ)\nunless RP = NP.\nProof. Lemma 5.3 gives the inapproximability at λ ∈ ̃Q \\ {−2, −1, 0}. By (3.6) we\ncould turn an approximation algorithm for I(−2) into an approximation algorithm for\nI(2) which would imply RP = NP by Lemma 5.3. For I(−1) we use Theorem 3.7.\nProof of Lemma 5.3. Fix λ ∈ ̃Q, 0 ̸= |1 + λ| ̸= 1, and ε, 0 < ε < 1. Assume we\nhave a randomized 2n1−ε-approximation algorithm A for I(λ). Given a graph G,\nTheorem 3.3 and Theorem 3.5, resp., will allow us to evaluate the independent set\npolynomial at a point ξ with |ξ| that large, that an approximation of I(G; ξ) can be\nused to recover the degree of I(G; x), which is the size of a maximum independent\nset of G. As computing this number is NP-hard, a randomized 2n1−ε-approximation\nalgorithm for I(G; λ) would yield an RP-algorithm for an NP-hard problem, which\nimplies RP = NP.\nLet G = (V, E) be a graph with |V | = n. We distinguish two cases. If |1+λ| > 1,\nwe choose a positive integer l such that (nl)1−ε ≥n2 and with ξ := (1 + λ)l −1 we\nhave\n|ξ| > 22(nl)1−ε+n+2.\n(5.1)\nAs λ and ε are constant, this can be achieved by choosing l = poly(n).\nIf 0 <\n|1 + λ| < 1, we choose a positive integer l such that with ξ :=\nλ\n(1+λ)l (5.1) holds.\nBy Theorem 3.3 (Theorem 3.5, resp.) we have I(G; ξ) = I(Gl; λ) (I(G; ξ) = (1 +\nλ)−l|V |I(Gl; λ), resp.).\nAlgorithm A returns on input Gl within time poly(nl) =\npoly(n) an approximation ̃I(Gl; λ), such that with ̃I(G; ξ) := ̃I(Gl; λ) ( ̃I(G; ξ) :=\n ̃I(Gl;λ)\n(1+λ)l|V |, resp.) we have\n2−(nl)1−εI(G; ξ) ≤ ̃I(G; ξ) ≤2(nl)1−εI(G; ξ)\n(5.2)\nwith high probability.\nLet c be the size of a maximum independent set of G, and let N be the number\nof independent sets of maximum size. We have\nI(G; x) = Nxc +\nX\n0≤j≤c−1\ni(G; j)xj\nand thus\n I(G; ξ)\nξc\n−N\n ≤\nX\n0≤j≤c−1\ni(G; j)|ξ|j−c\n≤c2n|ξ|−1 ≤2log n+n|ξ|−1 (5.1)\n< 1\n2.\n(5.3)\n16\n\nIf we could evaluate I(G; ξ) exactly, we could try all c ∈{1, . . . , n} to find the one for\nwhich I(G;ξ)\nξc\nis a good estimation for N, 1 ≤N ≤2n. This c is unique as |ξ| > 2n2.\nThe following calculation shows that this is also possible using the approximation\nalgorithm A.\nUsing A we compute ̃N( ̃c) :=\n ̃I(G;ξ)\nξ ̃c\nfor all ̃c ∈{1, . . . , n}. We claim that c is the\nunique ̃c with\n2−(nl)1−ε−1 ≤ ̃N( ̃c) ≤2(nl)1−ε+n+1.\n(5.4)\nLet us prove this claim. As 1 ≤N ≤2n and by (5.3), we know that\n1\n2 ≤I(G, ξ)\nξc\n≤2n+1.\n(5.5)\nThus, by (5.2), ̃c = c fulfills (5.4).\nOn the other hand, when ̃c ≤c −1 we have\n| ̃N( ̃c)|\n(5.2),(5.5)\n≥\n2−(nl)1−ε−1|ξ|\n(5.1)\n> 2(nl)1−ε+n+1.\nWhen ̃c ≥c + 1 we have | ̃N( ̃c)| < 2−(nl)1−ε−1 by similar arguments. This shows\nthat any integer ̃c, ̃c ̸= c, does not fulfill (5.4). Thus, c can be found in randomized\npolynomial time using A.\nAcknowledgement\nWe would like to thank Johann A. Makowsky for valuable comments on an earlier\nversion of this work and for drawing our attention to [AM07].\nReferences\n[ABCS00] Richard Arratia, B ́ela Bollob ́as, Don Coppersmith, and Gregory B.\nSorkin. Euler circuits and DNA sequencing by hybridization. Discrete\nApplied Mathematics, 104(1-3):63–96, 15 August 2000.\n[ABS04a] Richard Arratia, B ́ela Bollob ́as, and Gregory B. Sorkin. The interlace\npolynomial of a graph. J. Comb. Theory Ser. B, 92(2):199–233, 2004.\n[ABS04b] Richard Arratia, B ́ela Bollob ́as, and Gregory B. Sorkin. A two-variable\ninterlace polynomial. Combinatorica, 24(4):567–584, 2004.\n17\n\n[AM07]\nIlia Averbouch and J. A. Makowsky.\nThe complexity of multivariate\nmatching polynomials, February 2007. Preprint.\n[BM06]\nMarkus Bl ̈aser and Johann Makowsky. Hip hip hooray for Sokal, 2006.\nUnpublished note.\n[BO92]\nThomas Brylawski and James Oxley. The Tutte polynomial and its ap-\nplications. In Neil White, editor, Matroid Applications, Encyclopedia of\nMathematics and its Applications. Cambridge University Press, 1992.\n[Cou07]\nBruno Courcelle. A multivariate interlace polynomial, 2007. Preprint,\narXiv:cs.LO/0702016 v2.\n[EMS07]\nJoanna A. Ellis-Monaghan and Irasema Sarmiento. Distance hereditary\ngraphs and the interlace polynomial. Comb. Probab. Comput., 16(6):947–\n973, 2007.\n[GJ07]\nLeslie Ann Goldberg and Mark Jerrum. Inapproximability of the Tutte\npolynomial. In STOC ’07: Proceedings of the 39th Annual ACM Sym-\nposium on Theory of Computing, pages 459–468, New York, NY, USA,\n2007. ACM Press.\n[JVW90]\nF. Jaeger, D. L. Vertigan, and D. J. A. Welsh. On the computational com-\nplexity of the Jones and the Tutte polynomials. Math. Proc. Cambridge\nPhilos. Soc., 108:35–53, 1990.\n[Sok05]\nAlan D. Sokal. The multivariate Tutte polynomial (alias Potts model)\nfor graphs and matroids. In Bridget S. Webb, editor, Surveys in Combi-\nnatorics 2005. Cambridge University Press, 2005.\n[Tut84]\nW. T. Tutte. Graph Theory. Addison Wesley, 1984.\n[Val79]\nLeslie G. Valiant. The complexity of enumeration and reliability problems.\nSIAM Journal on Computing, 8(3):410–421, 1979.\n[Ver05]\nDirk Vertigan.\nThe computational complexity of Tutte invariants for\nplanar graphs. SIAM Journal on Computing, 35(3):690–712, 2005.\n[Wel93]\nD. J. A. Welsh. Complexity: knots, colourings and counting. Cambridge\nUniversity Press, New York, NY, USA, 1993.\n18","paragraphs":[{"paragraph_id":"p1","order":1,"text":"arXiv:0707.4565v3 [cs.CC] 16 Apr 2008\nOn the Complexity of the Interlace Polynomial∗\nMarkus Bl ̈aser, Christian Hoffmann\nNovember 16, 2021\nAbstract\nWe consider the two-variable interlace polynomial introduced by Arratia,\nBollob ́as and Sorkin (2004). We develop graph transformations which allow\nus to derive point-to-point reductions for the interlace polynomial. Exploiting\nthese reductions we obtain new results concerning the computational complex-\nity of evaluating the interlace polynomial at a fixed point. Regarding exact\nevaluation, we prove that the interlace polynomial is #P-hard to evaluate at\nevery point of the plane, except on one line, where it is trivially polynomial\ntime computable, and four lines, where the complexity is still open. This solves\na problem posed by Arratia, Bollob ́as and Sorkin (2004). In particular, three\nspecializations of the two-variable interlace polynomial, the vertex-nullity in-\nterlace polynomial, the vertex-rank interlace polynomial and the independent\nset polynomial, are almost everywhere #P-hard to evaluate, too. For the inde-\npendent set polynomial, our reductions allow us to prove that it is even hard\nto approximate at every point except at 0.\n1\nIntroduction\nThe number of Euler circuits in specific graphs and their interlacings turned out\nto be a central issue in the solution of a problem related to DNA sequencing by\nhybridization [ABCS00]. This led to the definition of a new graph polynomial, the\none-variable interlace polynomial [ABS04a].\nFurther research on this polynomial\ninspired the definition of a two-variable interlace polynomial q(G; x, y) containing\nas special cases the following graph polynomials: qN(G; y) = q(G; 2, y) is the origi-\nnal one-variable interlace polynomial which was renamed to “vertex-nullity interlace\n∗A preliminary version of this work has appeared in the proceedings of STACS 2008.\n1"},{"paragraph_id":"p2","order":2,"text":"polynomial”, qR(G; x) = q(G; x, 2) is the new “vertex-rank interlace polynomial” and\nI(G; x) = q(G; 1, 1 + x) is the independent set polynomial1 [ABS04b].\nAlthough the interlace polynomial q(G; x, y) is a different object from the cele-\nbrated Tutte polynomial (also known as dichromatic polynomial, see, for instance,\n[Tut84]), they are also similar to each other. While the Tutte polynomial can be\ndefined recursively by a deletion-contraction identity on edges, the interlace polyno-\nmial satisfies recurrence relations involving several operations on vertices (deletion,\npivotization, complementation).\nBesides the deletion-contraction identity, the so called state expansion is a well-\nknown way to define the Tutte polynomial. Here the similarity to the two-variable\ninterlace polynomial is especially striking: while the interlace polynomial is defined\nas a sum over all vertex subsets of the graph using the rank of adjacency matrices\n(see (2.1)), the state expansion of the Tutte polynomial can be interpreted as a sum\nover all edge subsets of the graph using the rank of incidence matrices (see (4.1))\n[ABS04b, Section 1].\nReferences to further work on the interlace polynomial can be found in [ABS04b]\nand [EMS07].\n1.1\nPrevious work\nThe aim of this paper is to explore the computational complexity of evaluating2 the\ntwo-variable interlace polynomial q(G; x, y). For the Tutte polynomial this problem\nwas solved in [JVW90]: Evaluating the Tutte polynomial is #P-hard at any alge-\nbraical point of the plane, except on the hyperbola (x −1)(y −1) = 1 and at a\nfew special points, where the Tutte polynomial can be evaluated in polynomial time.\nFor the two-variable interlace polynomial q(G; x, y), only on a one-dimensional sub-\nset of the plane (on the lines x = 2 and x = 1) some results about the evaluation\ncomplexity are known.\nA connection between the vertex-nullity interlace polynomial and the Tutte poly-\nnomial of planar graphs [ABS04a, End of Section 7], [EMS07, Theorem 3.1] shows\nthat evaluating q is #P-hard almost everywhere on the line x = 2 (Corollary 4.4).\nIt has also been noticed that q(G; 1, 2) evaluates to the number of independent\nsets of G [ABS04b, Section 5], which is #P-hard to compute [Val79]. Recent work\non the matching generating polynomial [AM07] implies that evaluating q is #P-hard\nalmost everywhere on the line x = 1 (Corollary 4.10).\n1The independent set polynomial of a graph G is defined as I(G; x) = P\nj≥0 i(G; j)xj, where\ni(G; j) denotes the number of independent sets of cardinality j of G.\n2See Section 2.2 for a precise definition.\n2"},{"paragraph_id":"p3","order":3,"text":"A key ingredient of [JVW90] is to apply graph transformations known as stretch-\ning and thickening of edges. For the Tutte polynomial, these graph transformations\nallow us to reduce the evaluation at one point to the evaluation at another point.\nFor the interlace polynomial no such graph transformations have been given so far.\n1.2\nOur results\nWe develop three graph transformations which are useful for the interlace polyno-\nmial: cloning of vertices and adding combs or cycles to the vertices. Applying these\ntransformations allows us to reduce the evaluation of the interlace polynomial at\nsome point to the evaluation of it at another point, see Theorem 3.3, Theorem 3.5\nand Theorem 3.7. We exploit this to obtain the following new results about the\ncomputational complexity of q(G; x, y).\nWe prove that the two-variable interlace polynomial q(G; x, y) is #P-hard to\nevaluate at almost every point of the plane, Theorem 4.11, see also Figure 1. Even\nthough there are some unknown (gray, in Figure 1) lines left on the complexity\nmap for q, this solves a challenge posed in [ABS04b, Section 5]. In particular we\nobtain the new result that evaluating the vertex-rank interlace polynomial qR(G; x)\nis #P-hard at almost every point (Corollary 4.12). Our techniques also give a new\nproof that the independent set polynomial is #P-hard to evaluate almost everywhere\n(Corollary 4.10).\nApart from these results on the computational complexity of evaluating the in-\nterlace polynomial exactly, we also show that the values of the independent set poly-\nnomial (which is the interlace polynomial q(G; x, y) on the line x = 1) are hard to\napproximate almost everywhere (Theorem 5.4).\n2\nPreliminaries\n2.1\nInterlace Polynomials\nWe consider undirected graphs without multiple edges but with self loops allowed.\nLet G = (V, E) be such a graph and A ⊆V . By G[A] we denote (A, {e|e ∈E, e ⊆\nA}), the subgraph of G induced by A. The adjacency matrix of G is the symmetric\nn × n-matrix M = (mij) over F2 = {0, 1} with mi,j = 1 iff{i, j} ∈E. The rank\nof this matrix is its rank over F2. Slightly abusing notation we write rk(G) for this\nrank. This allows us to define the two-variable interlace polynomial.\n3"},{"paragraph_id":"p4","order":4,"text":"Definition 2.1 ([ABS04b]). Let G = (V, E) be an undirected graph. The interlace\npolynomial q(G; x, y) of G is defined as\nq(G; x, y) =\nX\nA⊆V\n(x −1)rk(G[A])(y −1)|A|−rk(G[A]).\n(2.1)\nIn Section 3 we will introduce graph transformations which perform one and the\nsame operation (cloning one single vertex, adding a comb or a cycle to one single\nvertex, resp.) on every vertex of a graph. Instead of relating the interlace polynomial\nof the original graph directly to the interlace polynomial of the transformed graph,\nwe will analyze how, say, cloning one single vertex changes the interlace polynomial.\nTo express this, we must be able to treat the vertex being cloned in a particular\nway, differently from the other vertices. This becomes possible using a multivariate\nversion of the interlace polynomial, in which each vertex has its own variable. Once\nwe can express the effect of cloning one vertex by an appropriate substitution of\nthe vertex variable in the multivariate interlace polynomial, cloning all the vertices\namounts to a simple substitution of all vertex variables and brings us back to a\nbivariate interlace polynomial. This procedure has been applied successfully to the\nTutte polynomial [Sok05, BM06].\nWe choose the following multivariate interlace polynomial, which is similar to the\nmultivariate Tutte polynomial of Sokal [Sok05] and a specialization of the multivari-\nate interlace polynomial defined by Courcelle [Cou07].\nDefinition 2.2. Let G = (V, E) be an undirected graph. For each v ∈V let xv\nbe an indeterminate. Writing xA for Q\nv∈A xv, we define the following multivariate\ninterlace polynomial:\nP(G; u, x) =\nX\nA⊆V\nxAurk(G[A]).\nSubstituting each xv in P(G; u, x) by x, we obtain another bivariate interlace poly-\nnomial:\nP(G; u, x) =\nX\nA⊆V\nx|A|urk(G[A]).\nAn easy calculation proves that q and P are closely related:\nLemma 2.3. Let G be a graph. Then we have the polynomial identities q(G; x, y) =\nP(G; x−1\ny−1, y −1) and P(G; u, x) = q(G; ux + 1, x + 1).\n4"},{"paragraph_id":"p5","order":5,"text":"2.2\nEvaluating Graph Polynomials\nGiven ξ, υ ∈Q we want to analyze the following computational problem:\nInput Graph G\nOutput q(G; ξ, υ)\nThis is what we mean by “evaluating the interlace polynomial q at the point (ξ, υ)”.\nAs an abbreviation for this computational problem we write\nq(ξ, υ),\nwhich should not be confused with the expression q(G; ξ, υ) denoting just a value in\nQ. Evaluating other graph polynomials such as P, qN, qR and I is defined accordingly.\nIf P1 and P2 are computational problems we use P1 ⪯T P2 (P1 ⪯m P2) to denote a\npolynomial time Turing reduction (polynomial time many-one reduction, resp.) from\nP1 to P2. For instance, Lemma 2.3 gives\nCorollary 2.4. For ξ, υ ∈ ̃Q, υ ̸= 1, we have q(ξ, υ) ⪯m P( ξ−1\nυ−1, υ −1). For μ, ξ ∈ ̃Q\nwe have P(μ, ξ) ⪯m q(μξ + 1, ξ + 1).\nHere ̃Q denotes some finite dimensional field extension Q ⊆ ̃Q ⊆R, which has\na discrete representation. As\n√\n2 will play an important role but we are not able to\nuse arbitrary real numbers as the input for a Turing machine, we use ̃Q instead of\nQ or R. We fix some ̃Q for the rest of this paper. This construction is done in the\nspirit of Jaeger, Vertigan, and Welsh [JVW90] who also propose to adjoin a finite\nnumber of points to Q in order to talk about the complexity at irrational points. To\nsome extent, this is an ad hoc construction, but it is sufficient for this work.\n3\nGraph Transformations for the Interlace Poly-\nnomial\nNow we describe our graph transformations, vertex cloning and adding combs or\ncycles to the vertices. The main results of this section are Theorem 3.3, Theorem 3.5\nand Theorem 3.7, which describe the effect of these graph transformations on the\ninterlace polynomial.\n3.1\nCloning\nCloning vertices in the graph yields our first graph transformation.\n5"},{"paragraph_id":"p6","order":6,"text":"Cloning one vertex\nLet G = (V, E) be a graph. Let a ∈V be some vertex (the one which will be cloned)\nand N the set of neighbors of a, V ′ = V \\ {a} and M = V ′ \\ N. The graph G\nwith a cloned, Gaa, is obtained out of G in the following way: Insert a new isolated\nvertex a′. Connect a′ to all vertices in N. If a does not have a self loop, we are done.\nOtherwise connect a and a′ and insert a self loop at a′. Thus, adjacency matrices of\nthe original (cloned, resp.) graph are\nB =\na\nN\nM\na\nb\n1\n0\nN\n1\nA11\nA12\nM\n0\nA21\nA22\nand\nBaa =\na′\na\nN\nM\na′\nb\nb\n1\n0\na\nb\nb\n1\n0\nN\n1\n1\nA11\nA12\nM\n0\n0\nA21\nA22\n,\nresp, (3.1)\nwhere b = 1 if a has a self loop and b = 0 otherwise. As the first column of Baa equals\nits second column, as well as the first row equals the second row, we can remove the\nfirst row and the first column of Baa without changing the rank. This also holds\nwhen we consider the adjacency matrices of G[A] (Gaa[A], resp.) instead of G (Gaa\nresp.) for A ⊆V ′. Thus we have for any A ⊆V ′\nrk(Gaa[A]) = rk(G[A]),\n(3.2)\nrk(Gaa[A ∪{a, a′}]) = rk(Gaa[A ∪{a}]) = rk(Gaa[A ∪{a′}]) = rk(G[A ∪{a}]).\n(3.3)\nLet x = (xv)v∈V (Gaa) be a labeling of the vertices of Gaa by indeterminates.\nDefine X to denote the following labeling of the vertices of G: Xv := xv for all\nv ∈V ′, Xa := (1 + xa)(1 + xa′) −1 = xa + xa′ + xaxa′. Then we have\nLemma 3.1. P(Gaa; u, x) = P(G; u, X).\nProof. On the one hand we have\nP(Gaa; u, x)\n=\nX\nA⊆V ′\nxA(urk(Gaa[A]) + xaurk(Gaa[A∪{a}]) + xa′urk(Gaa[A∪{a′}]) + xaxa′urk(Gaa[A∪{a,a′}]))\n=\nX\nA⊆V ′\nxA(urk(G[A]) + (xa + xa′ + xaxa′)urk(G[A∪{a}])) by (3.2), (3.3).\n6"},{"paragraph_id":"p7","order":7,"text":"On the other hand we have\nP(G; u, X) =\nX\nA⊆V ′\nXA(urk(G[A]) + Xaurk(G[A∪{a}]))\n=\nX\nA⊆V ′\nxA(urk(G[A]) + (xa + xa′ + xaxa′)urk(G[A∪{a}])).\nCloning all vertices\nFix some k. Given a graph G, the graph Gk is obtained by cloning each vertex of\nG exactly k −1 times. Note that the result of the cloning is independent of the\norder in which the different vertices are cloned. For a ∈V (G) let a1, . . . , ak be the\ncorresponding vertices in Gk. For a vertex labeling x of Gk we define the vertex\nlabeling X of G by Xa = (1 + xa1)(1 + xa2) · · · (1 + xak) −1 for a ∈V (G). Applying\nLemma 3.1 repeatedly we obtain\nLemma 3.2. P(Gk; u, x) = P(G; u, X).\nSubstitution of xv by x for all vertices v gives\nTheorem 3.3. Let G be a graph and Gk be obtained out of G by cloning each vertex\nof G exactly k −1 times. Then\nP(Gk; u, x) = P(G; u, (1 + x)k −1).\n(3.4)\nAs we will use it in the proof of Theorem 4.11, we note the following identity for\nq, which can be derived from Theorem 3.3 using Lemma 2.3:\nq(Gk; x, y) = q(G; (x −1)yk −1\ny −1 + 1, yk).\n(3.5)\nTheorem 3.3 also implies the following reduction for the interlace polynomial,\nwhich is the foundation for our results in Section 4.\nProposition 3.4. Let B2 = {0, −1, −2} and x be an indeterminate. For μ ∈ ̃Q, ξ ∈\n ̃Q \\ B2 we have P(μ, x) ⪯T P(μ, ξ). (For any μ ∈ ̃Q, we write P(μ, x) to denote\nthe following computational problem: given a graph G compute P(G; μ, x), which is\na polynomial in x with coefficients in ̃Q.)\n7"},{"paragraph_id":"p8","order":8,"text":"Proof. Let μ and ξ be given such that they fulfill the precondition of the proposition.\nGiven a graph G =: G1 with n vertices, we build G2, G3, . . . , Gn+1, where Gi is\nobtained out of G by cloning each vertex i −1 times.\nThis is possible in time\npolynomial in n. By Theorem 3.3, a call to an oracle for P(μ, ξ) with input Gi gives\nus P(G; μ, (1 + ξ)i −1) for i = 1, . . . , n + 1. The restriction on ξ guarantees that for\ni = 1, 2, 3, . . . the expression (1 + ξ)i −1 evaluates to pairwise different values. Thus,\nfor P(G; μ, x), which is a polynomial in x of degree ≤n, we have obtained the values\nat n + 1 distinct points. Using Lagrange interpolation we determine the coefficients\nof P(G; μ, x).\n3.2\nAdding Combs\nThe comb transformation sometimes helps, when cloning has not the desired effect.\nLet G = (V, E) be a graph and a ∈V some vertex. Then we define the k-comb of\nG at a as Ga,k = (V ∪{a1, . . . , ak}, E ∪{{a, a1}, . . . , {a, ak}}), with a1, . . . , ak being\nnew vertices.\nUsing similar arguments as with vertex cloning, adding combs to vertices yields\na point-to-point reduction for the interlace polynomial, too.\nTheorem 3.5. Let G be a graph and Gk be obtained out of G by performing a k-comb\noperation at every vertex. Then\nP(Gk; u, x) = p(k, u, x)|V (G)|P(G; u, x/p(k, u, x)),\n(3.6)\nwhere p(k, u, x) = (1 + x)k(xu2 + 1) −xu2.\nProof. The adjacency matrices of the original graph G (the graph Ga,k with a k-comb\nat a, resp.) are\na\nV ′\na\nb\nc\nV ′\ncT\nA\nand\na1\na2\n. . .\nak\na\nV ′\na1\n1\na2\n1\n...\n...\nak\n1\na\n1\n1\n. . .\n1\nb\nc\nV ′\ncT\nA11\n, resp.,\n(3.7)\nwith empty entries being zero. Consider A ⊆V (Ga,k). Let M := A ∩{a, a1, . . . , ak}.\nBy (3.7), the rank of Ga,k is related to the rank for G in the following way:\n8"},{"paragraph_id":"p9","order":9,"text":"• If a ̸∈M, then rk(Ga,k[A]) = rk(G[A \\ M]).\n• If a ∈M and M ∩{a1, . . . , ak} ̸= ∅, then rk(Ga,k[A]) = rk(G[A \\ M]) + 2: Let\nw.l.o.g. a1 ∈M. Consider the adjacency matrix of Ga,k[A] and the following\noperations on it, which leave the rank unchanged. Using the first column we\nremove all 1s in the a-row, except the 1 in the first column. Using the first row\nwe remove all 1s in the a-column, except the 1 in the first row. The resulting\nmatrix B is a (k + |V |) × (k + |V |) matrix with 1s at positions (a, a1) and\n(a1, a), the submatrix of A11 induced by A \\ M in the lower right corner and\nzeros everywhere else. Thus rk(B) = rk(G[A \\ M]) + 2.\n• If M = {a}, then rk(Ga,k[A]) = rk(G[A]).\nLetting r(A) := rk(G[A]) and ra(A) := rk(G[A ∪{a}]) for A ⊆V ′, we see that\nP(Ga,k; u, x) equals\nX\nA⊆V ′\nxA"},{"paragraph_id":"p10","order":10,"text":"ur(A) X\n∅⊆S⊆{a1,...,ak}\nxS + xau2 ·\nX\n∅⊊S⊆{a1,...,ak}\nxS\n|\n{z\n}\n=:p(k,u,x)"},{"paragraph_id":"p11","order":11,"text":"+ xaura(A) \nNote that p(k, u, x) does only depend on xa, xa1, . . . , xak, but not on xv for any v ∈V ′.\nAs we have\nP(G; u, X) =\nX\nA⊆V ′\nXA(ur(A) + Xaura(A)),\nwe conclude\nP(Ga,k; u, x) = p(k, u, x)P(G; u, X),\nwhere Xv = xv for v ∈V ′ and Xa =\nxa\np(k,u,x).\nWe can perform a k-comb operation at every a ∈V and call the result Gk.\nSubstituting x for xv, v ∈Gk, concludes the proof.\nThis yields\nProposition 3.6. Let p(k, u, x) = (1+x)k(xu2+1)−xu2. Let k be a positive integer\nand μ, ξ ∈ ̃Q. If p(k, μ, ξ) ̸= 0, we have P(μ, ξ/p(k, μ, ξ)) ⪯m P(μ, ξ).\n9"},{"paragraph_id":"p12","order":12,"text":"3.3\nAdding Cycles\nLet G = (V, G) be a graph and a ∈V some vertex. Consider the graph Ga,k =\n(V ∪{1, 2, . . . , k −1}, E ∪{{a, 1}, {a, k −1}} ∪{{i −1, i} | 1 < i < k}), with\n1, 2, . . . , k −1 being new vertices. We say that Ga,k has been obtained out of G by\nadding a k-cycle to a.\nTheorem 3.7. Let G be a graph and Gk be obtained out of G by adding a k-cycle to\nevery vertex. Then P(Gk; u, x) = pk(u, x)P(G; u, qk(u, x)/pk(u, x)) for k = 3, 4 with\np3(u, x) = 1 + 2x + 3x2u2, q3(u, x) = x + x3u2, p4(u, x) = 1 + 3x + x2 + 2x2u2 + x3u2\nand q4(u, x) = x2 + 2x3u2 + x4u2.\nProposition 3.8. P(0, 1) ⪯m P(0, −1) and P(μ, −4) ⪯m P(μ, −2) for every μ ∈ ̃Q.\nProof. The first reduction follows from Theorem 3.7 adding 3-cycles, the second\nadding 4-cycles.\nProof of Theorem 3.7. We use the same idea as in the proof of Theorem 3.5. Consider\nthe case of a 3-cycle added at vertex a. Let V ′ = V \\ {a}. The adjacency matrix of\nGa,3 is\n1\n2\na\nV ′\n1\n1\n1\n2\n1\n1\na\n1\n1\nb\nc\nV ′\ncT\nA11\nwith empty entries being zero. Adding the second row to the first row and the second\ncolumn to the first column and subsequently the first row to the third row and the\nfirst column to the third column does not change the rank and gives\n1\n2\na\nV ′\n1\n1\n2\n1\na\nb\nc\nV ′\ncT\nA11\n.\nThis shows that rk(Ga,3[A] = rk(G[A\\{1, 2}])+2 for all A, {1, 2, a} ⊆A ⊆V (Ga,3).\nUsing arguments similar to this one and the ones in the proof of Theorem 3.5 we\nfind that\n• rk(Ga,3[A]) = rk(G[A ∩V ′] + 2 for all A, {a} ⊆A ⊆V (Ga,3) and either 1 ∈A\nor 2 ∈A,\n10"},{"paragraph_id":"p13","order":13,"text":"• rk(Ga,3[A]) = rk(G[A] for all A, {a} ⊆A ⊆V (Ga,3) and {1, 2} ∩A = ∅,\n• rk(Ga,3[A]) = rk(G[A ∩V ′]) + rk(P2[A ∩V (P2)]) for all A ⊆V (Ga,3), a ̸∈A,\nwhere P2 is the the path with two vertices 1, 2.\nLetting again r(A) := rk(G[A]) and ra(A) := rk(G[A∪{a}]) for A ⊆V ′, we see that\nP(Ga,3; u, x) equals\nX\nA⊆V ′\nxA"},{"paragraph_id":"p14","order":14,"text":"ur(A) 1 + x1 + x2 + x1x2u2 + x1xau2 + x2xau2\n|\n{z\n}\n=:p3(u,x)"},{"paragraph_id":"p15","order":15,"text":"+ ura(A)(xa + x1x2xau2\n|\n{z\n}\n=:q3(u,x)\n)"},{"paragraph_id":"p16","order":16,"text":",\nwhich equals p3(u, x)P(G; u, X) if we define X by Xv = xv for v ∈V ′ and Xa =\nq3(u, x)/p3(u, x).\nWe can use this identity for every vertex a and substitute xa,\na ∈V , by a single variable x. This gives the statement of the theorem concerning\n3-cycles. For 4-cycles we proceed in a similar fashion.\n4\nComplexity of evaluating the Interlace Polyno-\nmial exactly\nThe goal of this section is to uncover the complexity maps for P and q as indicated\nin Figure 1. While the left hand side (complexity map for P) is intended to follow\nthe arguments which prove the hardness, the right hand side (complexity map for q)\nfocuses on presenting the results.\nRemark 4.1. P(μ, 0) and P(1, ξ) are trivially solvable in polynomial time for any\nμ, ξ ∈ ̃Q, as P(G; μ, 0) = 1 and P(G; 1, ξ) = (1 + ξ)|V |.\nThus, on the thick black lines x = 0 and u = 1 in the left half of Figure 1, P can\nbe evaluated in polynomial time. By Lemma 2.3, these lines in the complexity map\nfor P correspond to the point (1, 1) and the line x = y, resp., in the complexity map\nfor q, see the right half of Figure 1.\n4.1\nIdentifying hard points\nWe want to establish Corollary 4.4 and Remark 4.5 which tell us, that P is #P-hard\nto evaluate almost everywhere on the dashed hyperbola in Figure 1 and at (0, 1). To\nthis end we collect known hardness results about the interlace polynomial.\n11"},{"paragraph_id":"p17","order":17,"text":"Figure 1: Complexity of the interlace polynomials P and q. α =\n√\n2, β = 1/\n√\n2\nLet t(G; x, y) denote the Tutte polynomial of an undirected graph G = (V, E). It\nmay be defined by its state expansion as\nt(G; x, y) =\nX\nB⊆E(G)\n(x −1)r(E)−r(B)(y −1)|B|−r(B),\n(4.1)\nwhere r(B) is the F2-rank of the incidence matrix of G[B] = (V, B), the subgraph of\nG induced by B. (Note that r(B) equals the number of vertices of G[B] minus the\nnumber of components of G[B], which is the rank of B in the cycle matriod of G.)\nFor details about the Tutte polynomial we refer to standard literature [Tut84, BO92,\nWel93]. The complexity of the Tutte polynomial has been studied extensively. In\nparticular, the following result is known.\nTheorem 4.2 ([Ver05]). Evaluating the Tutte polynomial of planar graphs at (ξ, ξ)\nis #P-hard for all ξ ∈ ̃Q except for ξ ∈{0, 1, 2, 1 ±\n√\n2}.\nWe will profit from this by a connection between the interlace polynomial and\nthe Tutte polynomial of planar graphs. This connection is established via medial\ngraphs. For any planar graph G one can build the oriented medial graph ⃗Gm, find\nan Euler circuit C in ⃗Gm and obtain the circle graph H of C. The whole procedure\ncan be performed in polynomial time. For details we refer to [EMS07]. We will use\n12"},{"paragraph_id":"p18","order":18,"text":"Theorem 4.3 ([ABS04a, End of Section 7]; [EMS07, Theorem 3.1]). Let G be a\nplanar graph, ⃗Gm be the oriented medial graph of G and H be the circle graph of\nsome Euler circuit C of ⃗Gm. Then q(H; 2, y) = t(G; y, y). Thus we have t(υ, υ) ⪯m\nP(\n1\nυ−1, υ −1), where t(υ, υ) denotes the problem of evaluating the Tutte polynomial\nof a planar graph at (υ, υ).\nProof. See the references for q(H; 2, y) = t(G; y, y) and use Lemma 2.3.\nWe set α =\n√\n2 and β = 1/\n√\n2.\nLet B1 = {±1, ±β, 0}. Theorem 4.2 and\nTheorem 4.3 yield\nCorollary 4.4. Evaluating the vertex-nullity interlace polynomial qN is #P-hard\nalmost everywhere. In particular, we have:\n• The problem qN(2) is trivially solvable in polynomial time.\n• For any υ ∈ ̃Q \\ {0, 1, 2, 1 ± α} the problem qN(υ) = q(2, υ) is #P-hard. Or,\nin other words, for any μ ∈ ̃Q \\ B1 the problem P(μ, 1/μ) is #P-hard.\nRemark 4.5. P(0, 1) is #P-hard, as P(G; 0, 1) equals the number of independent\nsets of G, which is #P-hard to compute [Val79].\n4.2\nReducing to hard points\nThe cloning reduction allows us to spread the collected hardness over almost the\nwhole plane: Combining Corollary 4.4 and Remark 4.5 with Proposition 3.4 we\nobtain\nProposition 4.6. Let B1 = {±1, ±β, 0} and B2 = {0, −1, −2} (as defined on\nPages 13 and 7, resp.). Let (μ, ξ) ∈(( ̃Q \\ B1) ∪{0}) × ( ̃Q \\ B2). Then P(μ, ξ)\nis #P-hard.\nThis tells us that P is #P-hard to evaluate at every point in the left half of\nFigure 1 not lying on one of the seven thick lines (three of which are solid gray ones,\ntwo of which are solid black ones, and two of which are dashed brown ones). Using\nthe comb and cycle reductions we are able to reveal the hardness of the interlace\npolynomial P on the lines x = −1 and x = −2:\nProposition 4.7. For μ ∈( ̃Q \\ B1) ∪{0} the problem P(μ, −1) is #P-hard.\n13"},{"paragraph_id":"p19","order":19,"text":"Proof. For μ = 0 we use Proposition 3.8 and Remark 4.5. If μ ̸= 0, we can use\nProposition 3.6, which yields P(μ, −1/μ2) ⪯m P(μ, −1). For μ = ±1 this reduces\n(±1, −1) to itself.\nFor μ = ±β this reduces (β, −2) to (β, −1) and (−β, −2) to\n(−β, −1). For other μ this gives a reduction of some point, which is already known\nas #P-hard by Proposition 4.6, to (μ, −1).\nProposition 4.8. For μ ∈( ̃Q \\ B1) ∪{0} the problem P(μ, −2) is #P-hard.\nProof. Use Proposition 3.8 and Proposition 4.6.\n4.3\nSumming up\nFirst we summarize our knowledge about P.\nTheorem 4.9. Let β = 1/\n√\n2.\n1. P(μ, ξ) is computable in polynomial time on the lines μ = 1 and ξ = 0.\n2. For (μ, ξ) ∈( ̃Q \\ {−1, −β, β, 1}) × ( ̃Q \\ {0}) the problem P(μ, ξ) is #P-hard.\nProof. Summary of Remark 4.1, Proposition 4.6, Proposition 4.7, Proposition 4.8.\nThe hardness of P(0, −1) follows from Corollary 4.10.\nIn particular we obtain the following corollary about the complexity of the inde-\npendent set polynomial, which also follows from [AM07].\nCorollary 4.10. Evaluating the independent set polynomial I(λ) = P(0, λ) = q(1, 1+\nλ) is #P-hard at all λ ∈ ̃Q except at λ = 0, where it is computable in polynomial\ntime.\nNow we turn to the complexity of q, see also the right half of Figure 1.\nTheorem 4.11. The two-variable interlace polynomial q is #P-hard to evaluate\nalmost everywhere. In particular, we have:\n1. q(ξ, υ) is computable in polynomial time on the line ξ = υ.\n2. Let ξ ∈ ̃Q\\{1} and x be an indeterminate. Then q(ξ, 1) is as hard as computing\nthe whole polynomial q(x, 1).\n3. q(ξ, υ) is #P-hard for all\n(ξ, υ) ∈{(ξ, υ) ∈ ̃Q2 | υ ̸= ±(ξ −1) + 1 and υ ̸= ±\n√\n2(ξ −1) + 1 and υ ̸= 1}.\n14"},{"paragraph_id":"p20","order":20,"text":"Proof of Theorem 4.11 (Sketch). (1) and (3) follow from Remark 4.1 and Theorem 4.9\nusing Lemma 2.3. For ξ ̸= 1, (3.5) gives q(Gk; ξ, 1) = q(G; k(ξ −1) + 1, 1), which\nyields enough points for interpolation in the same way as in Proposition 3.4 using\nk = 1, 2, 3, . . . This proves (2).\nTheorem 4.11 implies\nCorollary 4.12. Let β = 1/\n√\n2. Evaluating the vertex-rank interlace polynomial\nqR(G; x) is #P-hard at all ξ ∈ ̃Q except at ξ = 0, 1 −β, 1 + β (complexity open) and\nξ = 2 (computable in polynomial time).\n5\nInapproximability of the Independent Set Poly-\nnomial\nProvided we can evaluate the independent set polynomial at some fixed point, vertex\ncloning (adding combs, resp.) allows us to evaluate it at very large points. In this\nsection we exploit this to prove that the independent set polynomial is hard to\napproximate. Similar results are shown in [GJ07] for the Tutte polynomial.\nDefinition 5.1. Let λ ∈ ̃Q and ε > 0. By a randomized 2n1−ε-approximation algo-\nrithm for I(λ) we mean a randomized algorithm, that, given a graph G with n nodes,\nruns in time polynomial in n and returns ̃I(G; λ) ∈ ̃Q such that\nPr[2−n1−εI(G; λ) ≤ ̃I(G; λ) ≤2n1−εI(G; λ)] ≥3\n4.\nIn [GJ07], (non)approximability in the weaker sense of (not) admitting an FPRAS\nis considered.\nDefinition 5.2. Let λ ∈ ̃Q. A fully polynomial randomized approximation scheme\n(FPRAS) for I(λ) is a randomized algorithm, that given a graph G with n nodes and\nan error tolerance ε, 0 < ε < 1, runs in time polynomial in n and 1/ε and returns\n ̃I(G; λ) ∈ ̃Q such that\nPr[2−εI(G; λ) ≤ ̃I(G; λ) ≤2εI(G; λ)] ≥3\n4.\nLemma 5.3. For every λ ∈ ̃Q, 0 ̸= |1 + λ| ̸= 1, and every ε, 0 < ε < 1, there is no\nrandomized polynomial time 2n1−ε-approximation algorithm for I(λ) unless RP = NP.\n15"},{"paragraph_id":"p21","order":21,"text":"Theorem 5.4. For every λ ∈ ̃Q\\{0} and every ε, 0 < ε < 1, there is no randomized\npolynomial time 2n1−ε-approximation algorithm (and thus also no FPRAS) for I(λ)\nunless RP = NP.\nProof. Lemma 5.3 gives the inapproximability at λ ∈ ̃Q \\ {−2, −1, 0}. By (3.6) we\ncould turn an approximation algorithm for I(−2) into an approximation algorithm for\nI(2) which would imply RP = NP by Lemma 5.3. For I(−1) we use Theorem 3.7.\nProof of Lemma 5.3. Fix λ ∈ ̃Q, 0 ̸= |1 + λ| ̸= 1, and ε, 0 < ε < 1. Assume we\nhave a randomized 2n1−ε-approximation algorithm A for I(λ). Given a graph G,\nTheorem 3.3 and Theorem 3.5, resp., will allow us to evaluate the independent set\npolynomial at a point ξ with |ξ| that large, that an approximation of I(G; ξ) can be\nused to recover the degree of I(G; x), which is the size of a maximum independent\nset of G. As computing this number is NP-hard, a randomized 2n1−ε-approximation\nalgorithm for I(G; λ) would yield an RP-algorithm for an NP-hard problem, which\nimplies RP = NP.\nLet G = (V, E) be a graph with |V | = n. We distinguish two cases. If |1+λ| > 1,\nwe choose a positive integer l such that (nl)1−ε ≥n2 and with ξ := (1 + λ)l −1 we\nhave\n|ξ| > 22(nl)1−ε+n+2.\n(5.1)\nAs λ and ε are constant, this can be achieved by choosing l = poly(n).\nIf 0 <\n|1 + λ| < 1, we choose a positive integer l such that with ξ :=\nλ\n(1+λ)l (5.1) holds.\nBy Theorem 3.3 (Theorem 3.5, resp.) we have I(G; ξ) = I(Gl; λ) (I(G; ξ) = (1 +\nλ)−l|V |I(Gl; λ), resp.).\nAlgorithm A returns on input Gl within time poly(nl) =\npoly(n) an approximation ̃I(Gl; λ), such that with ̃I(G; ξ) := ̃I(Gl; λ) ( ̃I(G; ξ) :=\n ̃I(Gl;λ)\n(1+λ)l|V |, resp.) we have\n2−(nl)1−εI(G; ξ) ≤ ̃I(G; ξ) ≤2(nl)1−εI(G; ξ)\n(5.2)\nwith high probability.\nLet c be the size of a maximum independent set of G, and let N be the number\nof independent sets of maximum size. We have\nI(G; x) = Nxc +\nX\n0≤j≤c−1\ni(G; j)xj\nand thus\n I(G; ξ)\nξc\n−N\n ≤\nX\n0≤j≤c−1\ni(G; j)|ξ|j−c\n≤c2n|ξ|−1 ≤2log n+n|ξ|−1 (5.1)\n< 1\n2.\n(5.3)\n16"},{"paragraph_id":"p22","order":22,"text":"If we could evaluate I(G; ξ) exactly, we could try all c ∈{1, . . . , n} to find the one for\nwhich I(G;ξ)\nξc\nis a good estimation for N, 1 ≤N ≤2n. This c is unique as |ξ| > 2n2.\nThe following calculation shows that this is also possible using the approximation\nalgorithm A.\nUsing A we compute ̃N( ̃c) :=\n ̃I(G;ξ)\nξ ̃c\nfor all ̃c ∈{1, . . . , n}. We claim that c is the\nunique ̃c with\n2−(nl)1−ε−1 ≤ ̃N( ̃c) ≤2(nl)1−ε+n+1.\n(5.4)\nLet us prove this claim. As 1 ≤N ≤2n and by (5.3), we know that\n1\n2 ≤I(G, ξ)\nξc\n≤2n+1.\n(5.5)\nThus, by (5.2), ̃c = c fulfills (5.4).\nOn the other hand, when ̃c ≤c −1 we have\n| ̃N( ̃c)|\n(5.2),(5.5)\n≥\n2−(nl)1−ε−1|ξ|\n(5.1)\n> 2(nl)1−ε+n+1.\nWhen ̃c ≥c + 1 we have | ̃N( ̃c)| < 2−(nl)1−ε−1 by similar arguments. This shows\nthat any integer ̃c, ̃c ̸= c, does not fulfill (5.4). Thus, c can be found in randomized\npolynomial time using A.\nAcknowledgement\nWe would like to thank Johann A. Makowsky for valuable comments on an earlier\nversion of this work and for drawing our attention to [AM07].\nReferences\n[ABCS00] Richard Arratia, B ́ela Bollob ́as, Don Coppersmith, and Gregory B.\nSorkin. Euler circuits and DNA sequencing by hybridization. Discrete\nApplied Mathematics, 104(1-3):63–96, 15 August 2000.\n[ABS04a] Richard Arratia, B ́ela Bollob ́as, and Gregory B. Sorkin. The interlace\npolynomial of a graph. J. Comb. Theory Ser. B, 92(2):199–233, 2004.\n[ABS04b] Richard Arratia, B ́ela Bollob ́as, and Gregory B. Sorkin. A two-variable\ninterlace polynomial. Combinatorica, 24(4):567–584, 2004.\n17"},{"paragraph_id":"p23","order":23,"text":"[AM07]\nIlia Averbouch and J. A. Makowsky.\nThe complexity of multivariate\nmatching polynomials, February 2007. Preprint.\n[BM06]\nMarkus Bl ̈aser and Johann Makowsky. Hip hip hooray for Sokal, 2006.\nUnpublished note.\n[BO92]\nThomas Brylawski and James Oxley. The Tutte polynomial and its ap-\nplications. In Neil White, editor, Matroid Applications, Encyclopedia of\nMathematics and its Applications. Cambridge University Press, 1992.\n[Cou07]\nBruno Courcelle. A multivariate interlace polynomial, 2007. Preprint,\narXiv:cs.LO/0702016 v2.\n[EMS07]\nJoanna A. Ellis-Monaghan and Irasema Sarmiento. Distance hereditary\ngraphs and the interlace polynomial. Comb. Probab. Comput., 16(6):947–\n973, 2007.\n[GJ07]\nLeslie Ann Goldberg and Mark Jerrum. Inapproximability of the Tutte\npolynomial. In STOC ’07: Proceedings of the 39th Annual ACM Sym-\nposium on Theory of Computing, pages 459–468, New York, NY, USA,\n2007. ACM Press.\n[JVW90]\nF. Jaeger, D. L. Vertigan, and D. J. A. Welsh. On the computational com-\nplexity of the Jones and the Tutte polynomials. Math. Proc. Cambridge\nPhilos. Soc., 108:35–53, 1990.\n[Sok05]\nAlan D. Sokal. The multivariate Tutte polynomial (alias Potts model)\nfor graphs and matroids. In Bridget S. Webb, editor, Surveys in Combi-\nnatorics 2005. Cambridge University Press, 2005.\n[Tut84]\nW. T. Tutte. Graph Theory. Addison Wesley, 1984.\n[Val79]\nLeslie G. Valiant. The complexity of enumeration and reliability problems.\nSIAM Journal on Computing, 8(3):410–421, 1979.\n[Ver05]\nDirk Vertigan.\nThe computational complexity of Tutte invariants for\nplanar graphs. SIAM Journal on Computing, 35(3):690–712, 2005.\n[Wel93]\nD. J. A. Welsh. Complexity: knots, colourings and counting. Cambridge\nUniversity Press, New York, NY, USA, 1993.\n18"}],"pages":[{"page":1,"text":"arXiv:0707.4565v3 [cs.CC] 16 Apr 2008\nOn the Complexity of the Interlace Polynomial∗\nMarkus Bl ̈aser, Christian Hoffmann\nNovember 16, 2021\nAbstract\nWe consider the two-variable interlace polynomial introduced by Arratia,\nBollob ́as and Sorkin (2004). We develop graph transformations which allow\nus to derive point-to-point reductions for the interlace polynomial. Exploiting\nthese reductions we obtain new results concerning the computational complex-\nity of evaluating the interlace polynomial at a fixed point. Regarding exact\nevaluation, we prove that the interlace polynomial is #P-hard to evaluate at\nevery point of the plane, except on one line, where it is trivially polynomial\ntime computable, and four lines, where the complexity is still open. This solves\na problem posed by Arratia, Bollob ́as and Sorkin (2004). In particular, three\nspecializations of the two-variable interlace polynomial, the vertex-nullity in-\nterlace polynomial, the vertex-rank interlace polynomial and the independent\nset polynomial, are almost everywhere #P-hard to evaluate, too. For the inde-\npendent set polynomial, our reductions allow us to prove that it is even hard\nto approximate at every point except at 0.\n1\nIntroduction\nThe number of Euler circuits in specific graphs and their interlacings turned out\nto be a central issue in the solution of a problem related to DNA sequencing by\nhybridization [ABCS00]. This led to the definition of a new graph polynomial, the\none-variable interlace polynomial [ABS04a].\nFurther research on this polynomial\ninspired the definition of a two-variable interlace polynomial q(G; x, y) containing\nas special cases the following graph polynomials: qN(G; y) = q(G; 2, y) is the origi-\nnal one-variable interlace polynomial which was renamed to “vertex-nullity interlace\n∗A preliminary version of this work has appeared in the proceedings of STACS 2008.\n1"},{"page":2,"text":"polynomial”, qR(G; x) = q(G; x, 2) is the new “vertex-rank interlace polynomial” and\nI(G; x) = q(G; 1, 1 + x) is the independent set polynomial1 [ABS04b].\nAlthough the interlace polynomial q(G; x, y) is a different object from the cele-\nbrated Tutte polynomial (also known as dichromatic polynomial, see, for instance,\n[Tut84]), they are also similar to each other. While the Tutte polynomial can be\ndefined recursively by a deletion-contraction identity on edges, the interlace polyno-\nmial satisfies recurrence relations involving several operations on vertices (deletion,\npivotization, complementation).\nBesides the deletion-contraction identity, the so called state expansion is a well-\nknown way to define the Tutte polynomial. Here the similarity to the two-variable\ninterlace polynomial is especially striking: while the interlace polynomial is defined\nas a sum over all vertex subsets of the graph using the rank of adjacency matrices\n(see (2.1)), the state expansion of the Tutte polynomial can be interpreted as a sum\nover all edge subsets of the graph using the rank of incidence matrices (see (4.1))\n[ABS04b, Section 1].\nReferences to further work on the interlace polynomial can be found in [ABS04b]\nand [EMS07].\n1.1\nPrevious work\nThe aim of this paper is to explore the computational complexity of evaluating2 the\ntwo-variable interlace polynomial q(G; x, y). For the Tutte polynomial this problem\nwas solved in [JVW90]: Evaluating the Tutte polynomial is #P-hard at any alge-\nbraical point of the plane, except on the hyperbola (x −1)(y −1) = 1 and at a\nfew special points, where the Tutte polynomial can be evaluated in polynomial time.\nFor the two-variable interlace polynomial q(G; x, y), only on a one-dimensional sub-\nset of the plane (on the lines x = 2 and x = 1) some results about the evaluation\ncomplexity are known.\nA connection between the vertex-nullity interlace polynomial and the Tutte poly-\nnomial of planar graphs [ABS04a, End of Section 7], [EMS07, Theorem 3.1] shows\nthat evaluating q is #P-hard almost everywhere on the line x = 2 (Corollary 4.4).\nIt has also been noticed that q(G; 1, 2) evaluates to the number of independent\nsets of G [ABS04b, Section 5], which is #P-hard to compute [Val79]. Recent work\non the matching generating polynomial [AM07] implies that evaluating q is #P-hard\nalmost everywhere on the line x = 1 (Corollary 4.10).\n1The independent set polynomial of a graph G is defined as I(G; x) = P\nj≥0 i(G; j)xj, where\ni(G; j) denotes the number of independent sets of cardinality j of G.\n2See Section 2.2 for a precise definition.\n2"},{"page":3,"text":"A key ingredient of [JVW90] is to apply graph transformations known as stretch-\ning and thickening of edges. For the Tutte polynomial, these graph transformations\nallow us to reduce the evaluation at one point to the evaluation at another point.\nFor the interlace polynomial no such graph transformations have been given so far.\n1.2\nOur results\nWe develop three graph transformations which are useful for the interlace polyno-\nmial: cloning of vertices and adding combs or cycles to the vertices. Applying these\ntransformations allows us to reduce the evaluation of the interlace polynomial at\nsome point to the evaluation of it at another point, see Theorem 3.3, Theorem 3.5\nand Theorem 3.7. We exploit this to obtain the following new results about the\ncomputational complexity of q(G; x, y).\nWe prove that the two-variable interlace polynomial q(G; x, y) is #P-hard to\nevaluate at almost every point of the plane, Theorem 4.11, see also Figure 1. Even\nthough there are some unknown (gray, in Figure 1) lines left on the complexity\nmap for q, this solves a challenge posed in [ABS04b, Section 5]. In particular we\nobtain the new result that evaluating the vertex-rank interlace polynomial qR(G; x)\nis #P-hard at almost every point (Corollary 4.12). Our techniques also give a new\nproof that the independent set polynomial is #P-hard to evaluate almost everywhere\n(Corollary 4.10).\nApart from these results on the computational complexity of evaluating the in-\nterlace polynomial exactly, we also show that the values of the independent set poly-\nnomial (which is the interlace polynomial q(G; x, y) on the line x = 1) are hard to\napproximate almost everywhere (Theorem 5.4).\n2\nPreliminaries\n2.1\nInterlace Polynomials\nWe consider undirected graphs without multiple edges but with self loops allowed.\nLet G = (V, E) be such a graph and A ⊆V . By G[A] we denote (A, {e|e ∈E, e ⊆\nA}), the subgraph of G induced by A. The adjacency matrix of G is the symmetric\nn × n-matrix M = (mij) over F2 = {0, 1} with mi,j = 1 iff{i, j} ∈E. The rank\nof this matrix is its rank over F2. Slightly abusing notation we write rk(G) for this\nrank. This allows us to define the two-variable interlace polynomial.\n3"},{"page":4,"text":"Definition 2.1 ([ABS04b]). Let G = (V, E) be an undirected graph. The interlace\npolynomial q(G; x, y) of G is defined as\nq(G; x, y) =\nX\nA⊆V\n(x −1)rk(G[A])(y −1)|A|−rk(G[A]).\n(2.1)\nIn Section 3 we will introduce graph transformations which perform one and the\nsame operation (cloning one single vertex, adding a comb or a cycle to one single\nvertex, resp.) on every vertex of a graph. Instead of relating the interlace polynomial\nof the original graph directly to the interlace polynomial of the transformed graph,\nwe will analyze how, say, cloning one single vertex changes the interlace polynomial.\nTo express this, we must be able to treat the vertex being cloned in a particular\nway, differently from the other vertices. This becomes possible using a multivariate\nversion of the interlace polynomial, in which each vertex has its own variable. Once\nwe can express the effect of cloning one vertex by an appropriate substitution of\nthe vertex variable in the multivariate interlace polynomial, cloning all the vertices\namounts to a simple substitution of all vertex variables and brings us back to a\nbivariate interlace polynomial. This procedure has been applied successfully to the\nTutte polynomial [Sok05, BM06].\nWe choose the following multivariate interlace polynomial, which is similar to the\nmultivariate Tutte polynomial of Sokal [Sok05] and a specialization of the multivari-\nate interlace polynomial defined by Courcelle [Cou07].\nDefinition 2.2. Let G = (V, E) be an undirected graph. For each v ∈V let xv\nbe an indeterminate. Writing xA for Q\nv∈A xv, we define the following multivariate\ninterlace polynomial:\nP(G; u, x) =\nX\nA⊆V\nxAurk(G[A]).\nSubstituting each xv in P(G; u, x) by x, we obtain another bivariate interlace poly-\nnomial:\nP(G; u, x) =\nX\nA⊆V\nx|A|urk(G[A]).\nAn easy calculation proves that q and P are closely related:\nLemma 2.3. Let G be a graph. Then we have the polynomial identities q(G; x, y) =\nP(G; x−1\ny−1, y −1) and P(G; u, x) = q(G; ux + 1, x + 1).\n4"},{"page":5,"text":"2.2\nEvaluating Graph Polynomials\nGiven ξ, υ ∈Q we want to analyze the following computational problem:\nInput Graph G\nOutput q(G; ξ, υ)\nThis is what we mean by “evaluating the interlace polynomial q at the point (ξ, υ)”.\nAs an abbreviation for this computational problem we write\nq(ξ, υ),\nwhich should not be confused with the expression q(G; ξ, υ) denoting just a value in\nQ. Evaluating other graph polynomials such as P, qN, qR and I is defined accordingly.\nIf P1 and P2 are computational problems we use P1 ⪯T P2 (P1 ⪯m P2) to denote a\npolynomial time Turing reduction (polynomial time many-one reduction, resp.) from\nP1 to P2. For instance, Lemma 2.3 gives\nCorollary 2.4. For ξ, υ ∈ ̃Q, υ ̸= 1, we have q(ξ, υ) ⪯m P( ξ−1\nυ−1, υ −1). For μ, ξ ∈ ̃Q\nwe have P(μ, ξ) ⪯m q(μξ + 1, ξ + 1).\nHere ̃Q denotes some finite dimensional field extension Q ⊆ ̃Q ⊆R, which has\na discrete representation. As\n√\n2 will play an important role but we are not able to\nuse arbitrary real numbers as the input for a Turing machine, we use ̃Q instead of\nQ or R. We fix some ̃Q for the rest of this paper. This construction is done in the\nspirit of Jaeger, Vertigan, and Welsh [JVW90] who also propose to adjoin a finite\nnumber of points to Q in order to talk about the complexity at irrational points. To\nsome extent, this is an ad hoc construction, but it is sufficient for this work.\n3\nGraph Transformations for the Interlace Poly-\nnomial\nNow we describe our graph transformations, vertex cloning and adding combs or\ncycles to the vertices. The main results of this section are Theorem 3.3, Theorem 3.5\nand Theorem 3.7, which describe the effect of these graph transformations on the\ninterlace polynomial.\n3.1\nCloning\nCloning vertices in the graph yields our first graph transformation.\n5"},{"page":6,"text":"Cloning one vertex\nLet G = (V, E) be a graph. Let a ∈V be some vertex (the one which will be cloned)\nand N the set of neighbors of a, V ′ = V \\ {a} and M = V ′ \\ N. The graph G\nwith a cloned, Gaa, is obtained out of G in the following way: Insert a new isolated\nvertex a′. Connect a′ to all vertices in N. If a does not have a self loop, we are done.\nOtherwise connect a and a′ and insert a self loop at a′. Thus, adjacency matrices of\nthe original (cloned, resp.) graph are\nB =\na\nN\nM\na\nb\n1\n0\nN\n1\nA11\nA12\nM\n0\nA21\nA22\nand\nBaa =\na′\na\nN\nM\na′\nb\nb\n1\n0\na\nb\nb\n1\n0\nN\n1\n1\nA11\nA12\nM\n0\n0\nA21\nA22\n,\nresp, (3.1)\nwhere b = 1 if a has a self loop and b = 0 otherwise. As the first column of Baa equals\nits second column, as well as the first row equals the second row, we can remove the\nfirst row and the first column of Baa without changing the rank. This also holds\nwhen we consider the adjacency matrices of G[A] (Gaa[A], resp.) instead of G (Gaa\nresp.) for A ⊆V ′. Thus we have for any A ⊆V ′\nrk(Gaa[A]) = rk(G[A]),\n(3.2)\nrk(Gaa[A ∪{a, a′}]) = rk(Gaa[A ∪{a}]) = rk(Gaa[A ∪{a′}]) = rk(G[A ∪{a}]).\n(3.3)\nLet x = (xv)v∈V (Gaa) be a labeling of the vertices of Gaa by indeterminates.\nDefine X to denote the following labeling of the vertices of G: Xv := xv for all\nv ∈V ′, Xa := (1 + xa)(1 + xa′) −1 = xa + xa′ + xaxa′. Then we have\nLemma 3.1. P(Gaa; u, x) = P(G; u, X).\nProof. On the one hand we have\nP(Gaa; u, x)\n=\nX\nA⊆V ′\nxA(urk(Gaa[A]) + xaurk(Gaa[A∪{a}]) + xa′urk(Gaa[A∪{a′}]) + xaxa′urk(Gaa[A∪{a,a′}]))\n=\nX\nA⊆V ′\nxA(urk(G[A]) + (xa + xa′ + xaxa′)urk(G[A∪{a}])) by (3.2), (3.3).\n6"},{"page":7,"text":"On the other hand we have\nP(G; u, X) =\nX\nA⊆V ′\nXA(urk(G[A]) + Xaurk(G[A∪{a}]))\n=\nX\nA⊆V ′\nxA(urk(G[A]) + (xa + xa′ + xaxa′)urk(G[A∪{a}])).\nCloning all vertices\nFix some k. Given a graph G, the graph Gk is obtained by cloning each vertex of\nG exactly k −1 times. Note that the result of the cloning is independent of the\norder in which the different vertices are cloned. For a ∈V (G) let a1, . . . , ak be the\ncorresponding vertices in Gk. For a vertex labeling x of Gk we define the vertex\nlabeling X of G by Xa = (1 + xa1)(1 + xa2) · · · (1 + xak) −1 for a ∈V (G). Applying\nLemma 3.1 repeatedly we obtain\nLemma 3.2. P(Gk; u, x) = P(G; u, X).\nSubstitution of xv by x for all vertices v gives\nTheorem 3.3. Let G be a graph and Gk be obtained out of G by cloning each vertex\nof G exactly k −1 times. Then\nP(Gk; u, x) = P(G; u, (1 + x)k −1).\n(3.4)\nAs we will use it in the proof of Theorem 4.11, we note the following identity for\nq, which can be derived from Theorem 3.3 using Lemma 2.3:\nq(Gk; x, y) = q(G; (x −1)yk −1\ny −1 + 1, yk).\n(3.5)\nTheorem 3.3 also implies the following reduction for the interlace polynomial,\nwhich is the foundation for our results in Section 4.\nProposition 3.4. Let B2 = {0, −1, −2} and x be an indeterminate. For μ ∈ ̃Q, ξ ∈\n ̃Q \\ B2 we have P(μ, x) ⪯T P(μ, ξ). (For any μ ∈ ̃Q, we write P(μ, x) to denote\nthe following computational problem: given a graph G compute P(G; μ, x), which is\na polynomial in x with coefficients in ̃Q.)\n7"},{"page":8,"text":"Proof. Let μ and ξ be given such that they fulfill the precondition of the proposition.\nGiven a graph G =: G1 with n vertices, we build G2, G3, . . . , Gn+1, where Gi is\nobtained out of G by cloning each vertex i −1 times.\nThis is possible in time\npolynomial in n. By Theorem 3.3, a call to an oracle for P(μ, ξ) with input Gi gives\nus P(G; μ, (1 + ξ)i −1) for i = 1, . . . , n + 1. The restriction on ξ guarantees that for\ni = 1, 2, 3, . . . the expression (1 + ξ)i −1 evaluates to pairwise different values. Thus,\nfor P(G; μ, x), which is a polynomial in x of degree ≤n, we have obtained the values\nat n + 1 distinct points. Using Lagrange interpolation we determine the coefficients\nof P(G; μ, x).\n3.2\nAdding Combs\nThe comb transformation sometimes helps, when cloning has not the desired effect.\nLet G = (V, E) be a graph and a ∈V some vertex. Then we define the k-comb of\nG at a as Ga,k = (V ∪{a1, . . . , ak}, E ∪{{a, a1}, . . . , {a, ak}}), with a1, . . . , ak being\nnew vertices.\nUsing similar arguments as with vertex cloning, adding combs to vertices yields\na point-to-point reduction for the interlace polynomial, too.\nTheorem 3.5. Let G be a graph and Gk be obtained out of G by performing a k-comb\noperation at every vertex. Then\nP(Gk; u, x) = p(k, u, x)|V (G)|P(G; u, x/p(k, u, x)),\n(3.6)\nwhere p(k, u, x) = (1 + x)k(xu2 + 1) −xu2.\nProof. The adjacency matrices of the original graph G (the graph Ga,k with a k-comb\nat a, resp.) are\na\nV ′\na\nb\nc\nV ′\ncT\nA\nand\na1\na2\n. . .\nak\na\nV ′\na1\n1\na2\n1\n...\n...\nak\n1\na\n1\n1\n. . .\n1\nb\nc\nV ′\ncT\nA11\n, resp.,\n(3.7)\nwith empty entries being zero. Consider A ⊆V (Ga,k). Let M := A ∩{a, a1, . . . , ak}.\nBy (3.7), the rank of Ga,k is related to the rank for G in the following way:\n8"},{"page":9,"text":"• If a ̸∈M, then rk(Ga,k[A]) = rk(G[A \\ M]).\n• If a ∈M and M ∩{a1, . . . , ak} ̸= ∅, then rk(Ga,k[A]) = rk(G[A \\ M]) + 2: Let\nw.l.o.g. a1 ∈M. Consider the adjacency matrix of Ga,k[A] and the following\noperations on it, which leave the rank unchanged. Using the first column we\nremove all 1s in the a-row, except the 1 in the first column. Using the first row\nwe remove all 1s in the a-column, except the 1 in the first row. The resulting\nmatrix B is a (k + |V |) × (k + |V |) matrix with 1s at positions (a, a1) and\n(a1, a), the submatrix of A11 induced by A \\ M in the lower right corner and\nzeros everywhere else. Thus rk(B) = rk(G[A \\ M]) + 2.\n• If M = {a}, then rk(Ga,k[A]) = rk(G[A]).\nLetting r(A) := rk(G[A]) and ra(A) := rk(G[A ∪{a}]) for A ⊆V ′, we see that\nP(Ga,k; u, x) equals\nX\nA⊆V ′\nxA\n \nur(A) X\n∅⊆S⊆{a1,...,ak}\nxS + xau2 ·\nX\n∅⊊S⊆{a1,...,ak}\nxS\n|\n{z\n}\n=:p(k,u,x)\n \n+ xaura(A) \nNote that p(k, u, x) does only depend on xa, xa1, . . . , xak, but not on xv for any v ∈V ′.\nAs we have\nP(G; u, X) =\nX\nA⊆V ′\nXA(ur(A) + Xaura(A)),\nwe conclude\nP(Ga,k; u, x) = p(k, u, x)P(G; u, X),\nwhere Xv = xv for v ∈V ′ and Xa =\nxa\np(k,u,x).\nWe can perform a k-comb operation at every a ∈V and call the result Gk.\nSubstituting x for xv, v ∈Gk, concludes the proof.\nThis yields\nProposition 3.6. Let p(k, u, x) = (1+x)k(xu2+1)−xu2. Let k be a positive integer\nand μ, ξ ∈ ̃Q. If p(k, μ, ξ) ̸= 0, we have P(μ, ξ/p(k, μ, ξ)) ⪯m P(μ, ξ).\n9"},{"page":10,"text":"3.3\nAdding Cycles\nLet G = (V, G) be a graph and a ∈V some vertex. Consider the graph Ga,k =\n(V ∪{1, 2, . . . , k −1}, E ∪{{a, 1}, {a, k −1}} ∪{{i −1, i} | 1 < i < k}), with\n1, 2, . . . , k −1 being new vertices. We say that Ga,k has been obtained out of G by\nadding a k-cycle to a.\nTheorem 3.7. Let G be a graph and Gk be obtained out of G by adding a k-cycle to\nevery vertex. Then P(Gk; u, x) = pk(u, x)P(G; u, qk(u, x)/pk(u, x)) for k = 3, 4 with\np3(u, x) = 1 + 2x + 3x2u2, q3(u, x) = x + x3u2, p4(u, x) = 1 + 3x + x2 + 2x2u2 + x3u2\nand q4(u, x) = x2 + 2x3u2 + x4u2.\nProposition 3.8. P(0, 1) ⪯m P(0, −1) and P(μ, −4) ⪯m P(μ, −2) for every μ ∈ ̃Q.\nProof. The first reduction follows from Theorem 3.7 adding 3-cycles, the second\nadding 4-cycles.\nProof of Theorem 3.7. We use the same idea as in the proof of Theorem 3.5. Consider\nthe case of a 3-cycle added at vertex a. Let V ′ = V \\ {a}. The adjacency matrix of\nGa,3 is\n1\n2\na\nV ′\n1\n1\n1\n2\n1\n1\na\n1\n1\nb\nc\nV ′\ncT\nA11\nwith empty entries being zero. Adding the second row to the first row and the second\ncolumn to the first column and subsequently the first row to the third row and the\nfirst column to the third column does not change the rank and gives\n1\n2\na\nV ′\n1\n1\n2\n1\na\nb\nc\nV ′\ncT\nA11\n.\nThis shows that rk(Ga,3[A] = rk(G[A\\{1, 2}])+2 for all A, {1, 2, a} ⊆A ⊆V (Ga,3).\nUsing arguments similar to this one and the ones in the proof of Theorem 3.5 we\nfind that\n• rk(Ga,3[A]) = rk(G[A ∩V ′] + 2 for all A, {a} ⊆A ⊆V (Ga,3) and either 1 ∈A\nor 2 ∈A,\n10"},{"page":11,"text":"• rk(Ga,3[A]) = rk(G[A] for all A, {a} ⊆A ⊆V (Ga,3) and {1, 2} ∩A = ∅,\n• rk(Ga,3[A]) = rk(G[A ∩V ′]) + rk(P2[A ∩V (P2)]) for all A ⊆V (Ga,3), a ̸∈A,\nwhere P2 is the the path with two vertices 1, 2.\nLetting again r(A) := rk(G[A]) and ra(A) := rk(G[A∪{a}]) for A ⊆V ′, we see that\nP(Ga,3; u, x) equals\nX\nA⊆V ′\nxA\n \nur(A) 1 + x1 + x2 + x1x2u2 + x1xau2 + x2xau2\n|\n{z\n}\n=:p3(u,x)\n \n+ ura(A)(xa + x1x2xau2\n|\n{z\n}\n=:q3(u,x)\n)\n \n,\nwhich equals p3(u, x)P(G; u, X) if we define X by Xv = xv for v ∈V ′ and Xa =\nq3(u, x)/p3(u, x).\nWe can use this identity for every vertex a and substitute xa,\na ∈V , by a single variable x. This gives the statement of the theorem concerning\n3-cycles. For 4-cycles we proceed in a similar fashion.\n4\nComplexity of evaluating the Interlace Polyno-\nmial exactly\nThe goal of this section is to uncover the complexity maps for P and q as indicated\nin Figure 1. While the left hand side (complexity map for P) is intended to follow\nthe arguments which prove the hardness, the right hand side (complexity map for q)\nfocuses on presenting the results.\nRemark 4.1. P(μ, 0) and P(1, ξ) are trivially solvable in polynomial time for any\nμ, ξ ∈ ̃Q, as P(G; μ, 0) = 1 and P(G; 1, ξ) = (1 + ξ)|V |.\nThus, on the thick black lines x = 0 and u = 1 in the left half of Figure 1, P can\nbe evaluated in polynomial time. By Lemma 2.3, these lines in the complexity map\nfor P correspond to the point (1, 1) and the line x = y, resp., in the complexity map\nfor q, see the right half of Figure 1.\n4.1\nIdentifying hard points\nWe want to establish Corollary 4.4 and Remark 4.5 which tell us, that P is #P-hard\nto evaluate almost everywhere on the dashed hyperbola in Figure 1 and at (0, 1). To\nthis end we collect known hardness results about the interlace polynomial.\n11"},{"page":12,"text":"Figure 1: Complexity of the interlace polynomials P and q. α =\n√\n2, β = 1/\n√\n2\nLet t(G; x, y) denote the Tutte polynomial of an undirected graph G = (V, E). It\nmay be defined by its state expansion as\nt(G; x, y) =\nX\nB⊆E(G)\n(x −1)r(E)−r(B)(y −1)|B|−r(B),\n(4.1)\nwhere r(B) is the F2-rank of the incidence matrix of G[B] = (V, B), the subgraph of\nG induced by B. (Note that r(B) equals the number of vertices of G[B] minus the\nnumber of components of G[B], which is the rank of B in the cycle matriod of G.)\nFor details about the Tutte polynomial we refer to standard literature [Tut84, BO92,\nWel93]. The complexity of the Tutte polynomial has been studied extensively. In\nparticular, the following result is known.\nTheorem 4.2 ([Ver05]). Evaluating the Tutte polynomial of planar graphs at (ξ, ξ)\nis #P-hard for all ξ ∈ ̃Q except for ξ ∈{0, 1, 2, 1 ±\n√\n2}.\nWe will profit from this by a connection between the interlace polynomial and\nthe Tutte polynomial of planar graphs. This connection is established via medial\ngraphs. For any planar graph G one can build the oriented medial graph ⃗Gm, find\nan Euler circuit C in ⃗Gm and obtain the circle graph H of C. The whole procedure\ncan be performed in polynomial time. For details we refer to [EMS07]. We will use\n12"},{"page":13,"text":"Theorem 4.3 ([ABS04a, End of Section 7]; [EMS07, Theorem 3.1]). Let G be a\nplanar graph, ⃗Gm be the oriented medial graph of G and H be the circle graph of\nsome Euler circuit C of ⃗Gm. Then q(H; 2, y) = t(G; y, y). Thus we have t(υ, υ) ⪯m\nP(\n1\nυ−1, υ −1), where t(υ, υ) denotes the problem of evaluating the Tutte polynomial\nof a planar graph at (υ, υ).\nProof. See the references for q(H; 2, y) = t(G; y, y) and use Lemma 2.3.\nWe set α =\n√\n2 and β = 1/\n√\n2.\nLet B1 = {±1, ±β, 0}. Theorem 4.2 and\nTheorem 4.3 yield\nCorollary 4.4. Evaluating the vertex-nullity interlace polynomial qN is #P-hard\nalmost everywhere. In particular, we have:\n• The problem qN(2) is trivially solvable in polynomial time.\n• For any υ ∈ ̃Q \\ {0, 1, 2, 1 ± α} the problem qN(υ) = q(2, υ) is #P-hard. Or,\nin other words, for any μ ∈ ̃Q \\ B1 the problem P(μ, 1/μ) is #P-hard.\nRemark 4.5. P(0, 1) is #P-hard, as P(G; 0, 1) equals the number of independent\nsets of G, which is #P-hard to compute [Val79].\n4.2\nReducing to hard points\nThe cloning reduction allows us to spread the collected hardness over almost the\nwhole plane: Combining Corollary 4.4 and Remark 4.5 with Proposition 3.4 we\nobtain\nProposition 4.6. Let B1 = {±1, ±β, 0} and B2 = {0, −1, −2} (as defined on\nPages 13 and 7, resp.). Let (μ, ξ) ∈(( ̃Q \\ B1) ∪{0}) × ( ̃Q \\ B2). Then P(μ, ξ)\nis #P-hard.\nThis tells us that P is #P-hard to evaluate at every point in the left half of\nFigure 1 not lying on one of the seven thick lines (three of which are solid gray ones,\ntwo of which are solid black ones, and two of which are dashed brown ones). Using\nthe comb and cycle reductions we are able to reveal the hardness of the interlace\npolynomial P on the lines x = −1 and x = −2:\nProposition 4.7. For μ ∈( ̃Q \\ B1) ∪{0} the problem P(μ, −1) is #P-hard.\n13"},{"page":14,"text":"Proof. For μ = 0 we use Proposition 3.8 and Remark 4.5. If μ ̸= 0, we can use\nProposition 3.6, which yields P(μ, −1/μ2) ⪯m P(μ, −1). For μ = ±1 this reduces\n(±1, −1) to itself.\nFor μ = ±β this reduces (β, −2) to (β, −1) and (−β, −2) to\n(−β, −1). For other μ this gives a reduction of some point, which is already known\nas #P-hard by Proposition 4.6, to (μ, −1).\nProposition 4.8. For μ ∈( ̃Q \\ B1) ∪{0} the problem P(μ, −2) is #P-hard.\nProof. Use Proposition 3.8 and Proposition 4.6.\n4.3\nSumming up\nFirst we summarize our knowledge about P.\nTheorem 4.9. Let β = 1/\n√\n2.\n1. P(μ, ξ) is computable in polynomial time on the lines μ = 1 and ξ = 0.\n2. For (μ, ξ) ∈( ̃Q \\ {−1, −β, β, 1}) × ( ̃Q \\ {0}) the problem P(μ, ξ) is #P-hard.\nProof. Summary of Remark 4.1, Proposition 4.6, Proposition 4.7, Proposition 4.8.\nThe hardness of P(0, −1) follows from Corollary 4.10.\nIn particular we obtain the following corollary about the complexity of the inde-\npendent set polynomial, which also follows from [AM07].\nCorollary 4.10. Evaluating the independent set polynomial I(λ) = P(0, λ) = q(1, 1+\nλ) is #P-hard at all λ ∈ ̃Q except at λ = 0, where it is computable in polynomial\ntime.\nNow we turn to the complexity of q, see also the right half of Figure 1.\nTheorem 4.11. The two-variable interlace polynomial q is #P-hard to evaluate\nalmost everywhere. In particular, we have:\n1. q(ξ, υ) is computable in polynomial time on the line ξ = υ.\n2. Let ξ ∈ ̃Q\\{1} and x be an indeterminate. Then q(ξ, 1) is as hard as computing\nthe whole polynomial q(x, 1).\n3. q(ξ, υ) is #P-hard for all\n(ξ, υ) ∈{(ξ, υ) ∈ ̃Q2 | υ ̸= ±(ξ −1) + 1 and υ ̸= ±\n√\n2(ξ −1) + 1 and υ ̸= 1}.\n14"},{"page":15,"text":"Proof of Theorem 4.11 (Sketch). (1) and (3) follow from Remark 4.1 and Theorem 4.9\nusing Lemma 2.3. For ξ ̸= 1, (3.5) gives q(Gk; ξ, 1) = q(G; k(ξ −1) + 1, 1), which\nyields enough points for interpolation in the same way as in Proposition 3.4 using\nk = 1, 2, 3, . . . This proves (2).\nTheorem 4.11 implies\nCorollary 4.12. Let β = 1/\n√\n2. Evaluating the vertex-rank interlace polynomial\nqR(G; x) is #P-hard at all ξ ∈ ̃Q except at ξ = 0, 1 −β, 1 + β (complexity open) and\nξ = 2 (computable in polynomial time).\n5\nInapproximability of the Independent Set Poly-\nnomial\nProvided we can evaluate the independent set polynomial at some fixed point, vertex\ncloning (adding combs, resp.) allows us to evaluate it at very large points. In this\nsection we exploit this to prove that the independent set polynomial is hard to\napproximate. Similar results are shown in [GJ07] for the Tutte polynomial.\nDefinition 5.1. Let λ ∈ ̃Q and ε > 0. By a randomized 2n1−ε-approximation algo-\nrithm for I(λ) we mean a randomized algorithm, that, given a graph G with n nodes,\nruns in time polynomial in n and returns ̃I(G; λ) ∈ ̃Q such that\nPr[2−n1−εI(G; λ) ≤ ̃I(G; λ) ≤2n1−εI(G; λ)] ≥3\n4.\nIn [GJ07], (non)approximability in the weaker sense of (not) admitting an FPRAS\nis considered.\nDefinition 5.2. Let λ ∈ ̃Q. A fully polynomial randomized approximation scheme\n(FPRAS) for I(λ) is a randomized algorithm, that given a graph G with n nodes and\nan error tolerance ε, 0 < ε < 1, runs in time polynomial in n and 1/ε and returns\n ̃I(G; λ) ∈ ̃Q such that\nPr[2−εI(G; λ) ≤ ̃I(G; λ) ≤2εI(G; λ)] ≥3\n4.\nLemma 5.3. For every λ ∈ ̃Q, 0 ̸= |1 + λ| ̸= 1, and every ε, 0 < ε < 1, there is no\nrandomized polynomial time 2n1−ε-approximation algorithm for I(λ) unless RP = NP.\n15"},{"page":16,"text":"Theorem 5.4. For every λ ∈ ̃Q\\{0} and every ε, 0 < ε < 1, there is no randomized\npolynomial time 2n1−ε-approximation algorithm (and thus also no FPRAS) for I(λ)\nunless RP = NP.\nProof. Lemma 5.3 gives the inapproximability at λ ∈ ̃Q \\ {−2, −1, 0}. By (3.6) we\ncould turn an approximation algorithm for I(−2) into an approximation algorithm for\nI(2) which would imply RP = NP by Lemma 5.3. For I(−1) we use Theorem 3.7.\nProof of Lemma 5.3. Fix λ ∈ ̃Q, 0 ̸= |1 + λ| ̸= 1, and ε, 0 < ε < 1. Assume we\nhave a randomized 2n1−ε-approximation algorithm A for I(λ). Given a graph G,\nTheorem 3.3 and Theorem 3.5, resp., will allow us to evaluate the independent set\npolynomial at a point ξ with |ξ| that large, that an approximation of I(G; ξ) can be\nused to recover the degree of I(G; x), which is the size of a maximum independent\nset of G. As computing this number is NP-hard, a randomized 2n1−ε-approximation\nalgorithm for I(G; λ) would yield an RP-algorithm for an NP-hard problem, which\nimplies RP = NP.\nLet G = (V, E) be a graph with |V | = n. We distinguish two cases. If |1+λ| > 1,\nwe choose a positive integer l such that (nl)1−ε ≥n2 and with ξ := (1 + λ)l −1 we\nhave\n|ξ| > 22(nl)1−ε+n+2.\n(5.1)\nAs λ and ε are constant, this can be achieved by choosing l = poly(n).\nIf 0 <\n|1 + λ| < 1, we choose a positive integer l such that with ξ :=\nλ\n(1+λ)l (5.1) holds.\nBy Theorem 3.3 (Theorem 3.5, resp.) we have I(G; ξ) = I(Gl; λ) (I(G; ξ) = (1 +\nλ)−l|V |I(Gl; λ), resp.).\nAlgorithm A returns on input Gl within time poly(nl) =\npoly(n) an approximation ̃I(Gl; λ), such that with ̃I(G; ξ) := ̃I(Gl; λ) ( ̃I(G; ξ) :=\n ̃I(Gl;λ)\n(1+λ)l|V |, resp.) we have\n2−(nl)1−εI(G; ξ) ≤ ̃I(G; ξ) ≤2(nl)1−εI(G; ξ)\n(5.2)\nwith high probability.\nLet c be the size of a maximum independent set of G, and let N be the number\nof independent sets of maximum size. We have\nI(G; x) = Nxc +\nX\n0≤j≤c−1\ni(G; j)xj\nand thus\n I(G; ξ)\nξc\n−N\n ≤\nX\n0≤j≤c−1\ni(G; j)|ξ|j−c\n≤c2n|ξ|−1 ≤2log n+n|ξ|−1 (5.1)\n< 1\n2.\n(5.3)\n16"},{"page":17,"text":"If we could evaluate I(G; ξ) exactly, we could try all c ∈{1, . . . , n} to find the one for\nwhich I(G;ξ)\nξc\nis a good estimation for N, 1 ≤N ≤2n. This c is unique as |ξ| > 2n2.\nThe following calculation shows that this is also possible using the approximation\nalgorithm A.\nUsing A we compute ̃N( ̃c) :=\n ̃I(G;ξ)\nξ ̃c\nfor all ̃c ∈{1, . . . , n}. We claim that c is the\nunique ̃c with\n2−(nl)1−ε−1 ≤ ̃N( ̃c) ≤2(nl)1−ε+n+1.\n(5.4)\nLet us prove this claim. As 1 ≤N ≤2n and by (5.3), we know that\n1\n2 ≤I(G, ξ)\nξc\n≤2n+1.\n(5.5)\nThus, by (5.2), ̃c = c fulfills (5.4).\nOn the other hand, when ̃c ≤c −1 we have\n| ̃N( ̃c)|\n(5.2),(5.5)\n≥\n2−(nl)1−ε−1|ξ|\n(5.1)\n> 2(nl)1−ε+n+1.\nWhen ̃c ≥c + 1 we have | ̃N( ̃c)| < 2−(nl)1−ε−1 by similar arguments. This shows\nthat any integer ̃c, ̃c ̸= c, does not fulfill (5.4). Thus, c can be found in randomized\npolynomial time using A.\nAcknowledgement\nWe would like to thank Johann A. Makowsky for valuable comments on an earlier\nversion of this work and for drawing our attention to [AM07].\nReferences\n[ABCS00] Richard Arratia, B ́ela Bollob ́as, Don Coppersmith, and Gregory B.\nSorkin. Euler circuits and DNA sequencing by hybridization. Discrete\nApplied Mathematics, 104(1-3):63–96, 15 August 2000.\n[ABS04a] Richard Arratia, B ́ela Bollob ́as, and Gregory B. Sorkin. The interlace\npolynomial of a graph. J. Comb. Theory Ser. B, 92(2):199–233, 2004.\n[ABS04b] Richard Arratia, B ́ela Bollob ́as, and Gregory B. Sorkin. A two-variable\ninterlace polynomial. Combinatorica, 24(4):567–584, 2004.\n17"},{"page":18,"text":"[AM07]\nIlia Averbouch and J. A. Makowsky.\nThe complexity of multivariate\nmatching polynomials, February 2007. Preprint.\n[BM06]\nMarkus Bl ̈aser and Johann Makowsky. Hip hip hooray for Sokal, 2006.\nUnpublished note.\n[BO92]\nThomas Brylawski and James Oxley. The Tutte polynomial and its ap-\nplications. In Neil White, editor, Matroid Applications, Encyclopedia of\nMathematics and its Applications. Cambridge University Press, 1992.\n[Cou07]\nBruno Courcelle. A multivariate interlace polynomial, 2007. Preprint,\narXiv:cs.LO/0702016 v2.\n[EMS07]\nJoanna A. Ellis-Monaghan and Irasema Sarmiento. Distance hereditary\ngraphs and the interlace polynomial. Comb. Probab. Comput., 16(6):947–\n973, 2007.\n[GJ07]\nLeslie Ann Goldberg and Mark Jerrum. Inapproximability of the Tutte\npolynomial. In STOC ’07: Proceedings of the 39th Annual ACM Sym-\nposium on Theory of Computing, pages 459–468, New York, NY, USA,\n2007. ACM Press.\n[JVW90]\nF. Jaeger, D. L. Vertigan, and D. J. A. Welsh. On the computational com-\nplexity of the Jones and the Tutte polynomials. Math. Proc. Cambridge\nPhilos. Soc., 108:35–53, 1990.\n[Sok05]\nAlan D. Sokal. The multivariate Tutte polynomial (alias Potts model)\nfor graphs and matroids. In Bridget S. Webb, editor, Surveys in Combi-\nnatorics 2005. Cambridge University Press, 2005.\n[Tut84]\nW. T. Tutte. Graph Theory. Addison Wesley, 1984.\n[Val79]\nLeslie G. Valiant. The complexity of enumeration and reliability problems.\nSIAM Journal on Computing, 8(3):410–421, 1979.\n[Ver05]\nDirk Vertigan.\nThe computational complexity of Tutte invariants for\nplanar graphs. SIAM Journal on Computing, 35(3):690–712, 2005.\n[Wel93]\nD. J. A. Welsh. Complexity: knots, colourings and counting. Cambridge\nUniversity Press, New York, NY, USA, 1993.\n18"}]},"section_tree":[],"assets":{"figures":[],"tables":[],"images":[]},"math_expressions":{"inline":[],"block":[{"equation_id":"eq1","equation_number":null,"raw_text":"as special cases the following graph polynomials: qN(G; y) = q(G; 2, y) is the origi-","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq2","equation_number":null,"raw_text":"polynomial”, qR(G; x) = q(G; x, 2) is the new “vertex-rank interlace polynomial” and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq3","equation_number":null,"raw_text":"I(G; x) = q(G; 1, 1 + x) is the independent set polynomial1 [ABS04b].","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq4","equation_number":null,"raw_text":"braical point of the plane, except on the hyperbola (x −1)(y −1) = 1 and at a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq5","equation_number":null,"raw_text":"set of the plane (on the lines x = 2 and x = 1) some results about the evaluation","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq6","equation_number":null,"raw_text":"that evaluating q is #P-hard almost everywhere on the line x = 2 (Corollary 4.4).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq7","equation_number":null,"raw_text":"almost everywhere on the line x = 1 (Corollary 4.10).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq8","equation_number":null,"raw_text":"1The independent set polynomial of a graph G is defined as I(G; x) = P","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq9","equation_number":null,"raw_text":"nomial (which is the interlace polynomial q(G; x, y) on the line x = 1) are hard to","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq10","equation_number":null,"raw_text":"Let G = (V, E) be such a graph and A ⊆V . By G[A] we denote (A, {e|e ∈E, e ⊆","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq11","equation_number":null,"raw_text":"n × n-matrix M = (mij) over F2 = {0, 1} with mi,j = 1 iff{i, j} ∈E. The rank","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq12","equation_number":null,"raw_text":"Definition 2.1 ([ABS04b]). Let G = (V, E) be an undirected graph. The interlace","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq13","equation_number":null,"raw_text":"q(G; x, y) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq14","equation_number":null,"raw_text":"Definition 2.2. Let G = (V, E) be an undirected graph. For each v ∈V let xv","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq15","equation_number":null,"raw_text":"P(G; u, x) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq16","equation_number":null,"raw_text":"P(G; u, x) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq17","equation_number":null,"raw_text":"Lemma 2.3. Let G be a graph. Then we have the polynomial identities q(G; x, y) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq18","equation_number":null,"raw_text":"y−1, y −1) and P(G; u, x) = q(G; ux + 1, x + 1).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq19","equation_number":null,"raw_text":"Corollary 2.4. For ξ, υ ∈ ̃Q, υ ̸= 1, we have q(ξ, υ) ⪯m P( ξ−1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq20","equation_number":null,"raw_text":"Let G = (V, E) be a graph. Let a ∈V be some vertex (the one which will be cloned)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq21","equation_number":null,"raw_text":"and N the set of neighbors of a, V ′ = V \\ {a} and M = V ′ \\ N. The graph G","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq22","equation_number":null,"raw_text":"B =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq23","equation_number":null,"raw_text":"Baa =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq24","equation_number":null,"raw_text":"where b = 1 if a has a self loop and b = 0 otherwise. As the first column of Baa equals","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq25","equation_number":null,"raw_text":"rk(Gaa[A]) = rk(G[A]),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq26","equation_number":null,"raw_text":"rk(Gaa[A ∪{a, a′}]) = rk(Gaa[A ∪{a}]) = rk(Gaa[A ∪{a′}]) = rk(G[A ∪{a}]).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq27","equation_number":null,"raw_text":"Let x = (xv)v∈V (Gaa) be a labeling of the vertices of Gaa by indeterminates.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq28","equation_number":null,"raw_text":"Define X to denote the following labeling of the vertices of G: Xv := xv for all","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq29","equation_number":null,"raw_text":"v ∈V ′, Xa := (1 + xa)(1 + xa′) −1 = xa + xa′ + xaxa′. Then we have","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq30","equation_number":null,"raw_text":"Lemma 3.1. P(Gaa; u, x) = P(G; u, X).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq31","equation_number":null,"raw_text":"P(G; u, X) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq32","equation_number":null,"raw_text":"labeling X of G by Xa = (1 + xa1)(1 + xa2) · · · (1 + xak) −1 for a ∈V (G). Applying","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq33","equation_number":null,"raw_text":"Lemma 3.2. P(Gk; u, x) = P(G; u, X).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq34","equation_number":null,"raw_text":"P(Gk; u, x) = P(G; u, (1 + x)k −1).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq35","equation_number":null,"raw_text":"q(Gk; x, y) = q(G; (x −1)yk −1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq36","equation_number":null,"raw_text":"Proposition 3.4. Let B2 = {0, −1, −2} and x be an indeterminate. For μ ∈ ̃Q, ξ ∈","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq37","equation_number":null,"raw_text":"Given a graph G =: G1 with n vertices, we build G2, G3, . . . , Gn+1, where Gi is","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq38","equation_number":null,"raw_text":"us P(G; μ, (1 + ξ)i −1) for i = 1, . . . , n + 1. The restriction on ξ guarantees that for","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq39","equation_number":null,"raw_text":"i = 1, 2, 3, . . . the expression (1 + ξ)i −1 evaluates to pairwise different values. Thus,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq40","equation_number":null,"raw_text":"Let G = (V, E) be a graph and a ∈V some vertex. Then we define the k-comb of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq41","equation_number":null,"raw_text":"G at a as Ga,k = (V ∪{a1, . . . , ak}, E ∪{{a, a1}, . . . , {a, ak}}), with a1, . . . , ak being","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq42","equation_number":null,"raw_text":"P(Gk; u, x) = p(k, u, x)|V (G)|P(G; u, x/p(k, u, x)),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq43","equation_number":null,"raw_text":"where p(k, u, x) = (1 + x)k(xu2 + 1) −xu2.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq44","equation_number":null,"raw_text":"with empty entries being zero. Consider A ⊆V (Ga,k). Let M := A ∩{a, a1, . . . , ak}.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq45","equation_number":null,"raw_text":"• If a ̸∈M, then rk(Ga,k[A]) = rk(G[A \\ M]).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq46","equation_number":null,"raw_text":"• If a ∈M and M ∩{a1, . . . , ak} ̸= ∅, then rk(Ga,k[A]) = rk(G[A \\ M]) + 2: Let","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq47","equation_number":null,"raw_text":"zeros everywhere else. Thus rk(B) = rk(G[A \\ M]) + 2.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq48","equation_number":null,"raw_text":"• If M = {a}, then rk(Ga,k[A]) = rk(G[A]).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq49","equation_number":null,"raw_text":"Letting r(A) := rk(G[A]) and ra(A) := rk(G[A ∪{a}]) for A ⊆V ′, we see that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq50","equation_number":null,"raw_text":"=:p(k,u,x)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq51","equation_number":null,"raw_text":"P(G; u, X) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq52","equation_number":null,"raw_text":"P(Ga,k; u, x) = p(k, u, x)P(G; u, X),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq53","equation_number":null,"raw_text":"where Xv = xv for v ∈V ′ and Xa =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq54","equation_number":null,"raw_text":"Proposition 3.6. Let p(k, u, x) = (1+x)k(xu2+1)−xu2. Let k be a positive integer","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq55","equation_number":null,"raw_text":"and μ, ξ ∈ ̃Q. If p(k, μ, ξ) ̸= 0, we have P(μ, ξ/p(k, μ, ξ)) ⪯m P(μ, ξ).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq56","equation_number":null,"raw_text":"Let G = (V, G) be a graph and a ∈V some vertex. Consider the graph Ga,k =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq57","equation_number":null,"raw_text":"every vertex. Then P(Gk; u, x) = pk(u, x)P(G; u, qk(u, x)/pk(u, x)) for k = 3, 4 with","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq58","equation_number":null,"raw_text":"p3(u, x) = 1 + 2x + 3x2u2, q3(u, x) = x + x3u2, p4(u, x) = 1 + 3x + x2 + 2x2u2 + x3u2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq59","equation_number":null,"raw_text":"and q4(u, x) = x2 + 2x3u2 + x4u2.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq60","equation_number":null,"raw_text":"the case of a 3-cycle added at vertex a. Let V ′ = V \\ {a}. The adjacency matrix of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq61","equation_number":null,"raw_text":"This shows that rk(Ga,3[A] = rk(G[A\\{1, 2}])+2 for all A, {1, 2, a} ⊆A ⊆V (Ga,3).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq62","equation_number":null,"raw_text":"• rk(Ga,3[A]) = rk(G[A ∩V ′] + 2 for all A, {a} ⊆A ⊆V (Ga,3) and either 1 ∈A","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq63","equation_number":null,"raw_text":"• rk(Ga,3[A]) = rk(G[A] for all A, {a} ⊆A ⊆V (Ga,3) and {1, 2} ∩A = ∅,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq64","equation_number":null,"raw_text":"• rk(Ga,3[A]) = rk(G[A ∩V ′]) + rk(P2[A ∩V (P2)]) for all A ⊆V (Ga,3), a ̸∈A,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq65","equation_number":null,"raw_text":"Letting again r(A) := rk(G[A]) and ra(A) := rk(G[A∪{a}]) for A ⊆V ′, we see that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq66","equation_number":null,"raw_text":"=:p3(u,x)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq67","equation_number":null,"raw_text":"=:q3(u,x)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq68","equation_number":null,"raw_text":"which equals p3(u, x)P(G; u, X) if we define X by Xv = xv for v ∈V ′ and Xa =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq69","equation_number":null,"raw_text":"μ, ξ ∈ ̃Q, as P(G; μ, 0) = 1 and P(G; 1, ξ) = (1 + ξ)|V |.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq70","equation_number":null,"raw_text":"Thus, on the thick black lines x = 0 and u = 1 in the left half of Figure 1, P can","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq71","equation_number":null,"raw_text":"for P correspond to the point (1, 1) and the line x = y, resp., in the complexity map","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq72","equation_number":null,"raw_text":"Figure 1: Complexity of the interlace polynomials P and q. α =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq73","equation_number":null,"raw_text":"Let t(G; x, y) denote the Tutte polynomial of an undirected graph G = (V, E). It","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq74","equation_number":null,"raw_text":"t(G; x, y) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq75","equation_number":null,"raw_text":"where r(B) is the F2-rank of the incidence matrix of G[B] = (V, B), the subgraph of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq76","equation_number":null,"raw_text":"some Euler circuit C of ⃗Gm. Then q(H; 2, y) = t(G; y, y). Thus we have t(υ, υ) ⪯m","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq77","equation_number":null,"raw_text":"Proof. See the references for q(H; 2, y) = t(G; y, y) and use Lemma 2.3.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq78","equation_number":null,"raw_text":"We set α =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq79","equation_number":null,"raw_text":"2 and β = 1/","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq80","equation_number":null,"raw_text":"Let B1 = {±1, ±β, 0}. Theorem 4.2 and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq81","equation_number":null,"raw_text":"• For any υ ∈ ̃Q \\ {0, 1, 2, 1 ± α} the problem qN(υ) = q(2, υ) is #P-hard. Or,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq82","equation_number":null,"raw_text":"Proposition 4.6. Let B1 = {±1, ±β, 0} and B2 = {0, −1, −2} (as defined on","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq83","equation_number":null,"raw_text":"polynomial P on the lines x = −1 and x = −2:","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq84","equation_number":null,"raw_text":"Proof. For μ = 0 we use Proposition 3.8 and Remark 4.5. If μ ̸= 0, we can use","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq85","equation_number":null,"raw_text":"Proposition 3.6, which yields P(μ, −1/μ2) ⪯m P(μ, −1). For μ = ±1 this reduces","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq86","equation_number":null,"raw_text":"For μ = ±β this reduces (β, −2) to (β, −1) and (−β, −2) to","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq87","equation_number":null,"raw_text":"Theorem 4.9. Let β = 1/","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq88","equation_number":null,"raw_text":"1. P(μ, ξ) is computable in polynomial time on the lines μ = 1 and ξ = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq89","equation_number":null,"raw_text":"Corollary 4.10. Evaluating the independent set polynomial I(λ) = P(0, λ) = q(1, 1+","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq90","equation_number":null,"raw_text":"λ) is #P-hard at all λ ∈ ̃Q except at λ = 0, where it is computable in polynomial","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq91","equation_number":null,"raw_text":"1. q(ξ, υ) is computable in polynomial time on the line ξ = υ.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq92","equation_number":null,"raw_text":"(ξ, υ) ∈{(ξ, υ) ∈ ̃Q2 | υ ̸= ±(ξ −1) + 1 and υ ̸= ±","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq93","equation_number":null,"raw_text":"2(ξ −1) + 1 and υ ̸= 1}.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq94","equation_number":null,"raw_text":"using Lemma 2.3. For ξ ̸= 1, (3.5) gives q(Gk; ξ, 1) = q(G; k(ξ −1) + 1, 1), which","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq95","equation_number":null,"raw_text":"k = 1, 2, 3, . . . This proves (2).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq96","equation_number":null,"raw_text":"Corollary 4.12. Let β = 1/","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq97","equation_number":null,"raw_text":"qR(G; x) is #P-hard at all ξ ∈ ̃Q except at ξ = 0, 1 −β, 1 + β (complexity open) and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq98","equation_number":null,"raw_text":"ξ = 2 (computable in polynomial time).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq99","equation_number":null,"raw_text":"Lemma 5.3. For every λ ∈ ̃Q, 0 ̸= |1 + λ| ̸= 1, and every ε, 0 < ε < 1, there is no","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq100","equation_number":null,"raw_text":"randomized polynomial time 2n1−ε-approximation algorithm for I(λ) unless RP = NP.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq101","equation_number":null,"raw_text":"unless RP = NP.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq102","equation_number":null,"raw_text":"I(2) which would imply RP = NP by Lemma 5.3. For I(−1) we use Theorem 3.7.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq103","equation_number":null,"raw_text":"Proof of Lemma 5.3. Fix λ ∈ ̃Q, 0 ̸= |1 + λ| ̸= 1, and ε, 0 < ε < 1. Assume we","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq104","equation_number":null,"raw_text":"implies RP = NP.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq105","equation_number":null,"raw_text":"Let G = (V, E) be a graph with |V | = n. We distinguish two cases. If |1+λ| > 1,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq106","equation_number":null,"raw_text":"we choose a positive integer l such that (nl)1−ε ≥n2 and with ξ := (1 + λ)l −1 we","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq107","equation_number":null,"raw_text":"As λ and ε are constant, this can be achieved by choosing l = poly(n).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq108","equation_number":null,"raw_text":"|1 + λ| < 1, we choose a positive integer l such that with ξ :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq109","equation_number":null,"raw_text":"By Theorem 3.3 (Theorem 3.5, resp.) we have I(G; ξ) = I(Gl; λ) (I(G; ξ) = (1 +","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq110","equation_number":null,"raw_text":"Algorithm A returns on input Gl within time poly(nl) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq111","equation_number":null,"raw_text":"poly(n) an approximation ̃I(Gl; λ), such that with ̃I(G; ξ) := ̃I(Gl; λ) ( ̃I(G; ξ) :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq112","equation_number":null,"raw_text":"I(G; x) = Nxc +","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq113","equation_number":null,"raw_text":"Using A we compute ̃N( ̃c) :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq114","equation_number":null,"raw_text":"Thus, by (5.2), ̃c = c fulfills (5.4).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq115","equation_number":null,"raw_text":"that any integer ̃c, ̃c ̸= c, does not fulfill (5.4). Thus, c can be found in randomized","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"}],"equation_refs":[]},"symbol_table":[],"references":{"in_text_markers":[],"bibliography":[]},"claims":[],"flow_graph":{"nodes":[],"edges":[]},"quality":{"text_char_count":31556,"parse_confidence":0.5,"equation_parse_rate_proxy":1.0,"citation_resolution_rate_proxy":0.0,"structure_coverage_proxy":0.0,"asset_coverage_proxy":0.0,"accepted_for_training":true}}