A Stronger Connection between the Asymptotic Rank Conjecture and the Set Cover Conjecture

We give a short proof that Strassen’s asymptotic rank conjecture implies that for every ε > 0 there exists a (3/22/3 + ε)n-time algorithm for set cover on a universe of size n with sets of bounded size. This strengthens and simplifies a recent result of Bj'orklund and Kaski that Strassen’s asymptotic rank conjecture implies that the set cover conjecture is false. From another perspective, we show that the set cover conjecture implies that a particular family of tensors Tn ∈ ℂN ⊗ ℂN ⊗ ℂN has asymptotic rank greater than N1.08. Furthermore, if one could improve a known upper bound of 8n on the tensor rank of Tn to 2/9 · n8n for any n, then the set cover conjecture is false.


Introduction
In a recent preprint [BK23], Björklund and Kaski have shown that the set cover conjecture [CDL + 16, KT19] from the area of exact exponential algorithms, and Strassen's asymptotic rank conjecture [Str94], a broad generalization of the conjecture that the exponent of matrix multiplication ω equals 2, are inconsistent with each other.The purpose of this note is to give a simpler and quantitatively informative proof of this result.
We begin by recalling the relevant problems.
Problem 1.1 (s-Set Cover).Given t ∈ N and F ⊆ 2 [n] where each X ∈ F has size at most s, decide if there exist at most t sets in F whose union equals [n].
Via standard reductions we will be interested in the following algorithmic problem.
This can be solved in time 8 n ⋅ poly(n) as follows.Let f i ∶ Z 3n 2 → {0, 1} be the indicator function of F i .Note that (f 1 * f 2 * f 3 )(1 3n ) > 0 if and only if the answer to Problem 1.3 is "yes" (here (f * g)(x) ∶= ∑ y∈Z 3n 2 f (y)g(x − y) denotes the convolution of f and g).Using Fourier inversion in Z 3n 2 , we can compute this quantity in 8 n ⋅ poly(n) time.We will see that improving significantly on this simple algorithm would refute Conjecture 1.2 (this fact is essentially shown in [BK23] as well).
Clearly Problem 1.1 requires at least 3n n ≈ 6.75 n time.We show that one can nearly achieve this (that is, solve Problem 1.1 in time (6.75 + ε) n for any ε > 0) if the asymptotic rank conjecture holds.This improves on the main result in [BK23], which roughly showed that Problem 1.3 can be solved in time 7.999 n under this assumption.Similar to [BK23], we will obtain algorithms for Problem 1.3 from algorithms for evaluating certain trilinear forms, or tensors.These will in turn arise from hypothetical bounds on the asymptotic rank of certain "base" tensors.Our simplification is due to the fact that the trilinear forms we will be interested in are more symmetric than those in [BK23].This is also related to the fact that the main algorithmic problem studied in [BK23] is an "almost balanced" tripartitioning problem, rather than the "exactly balanced" Problem 1.3.

Background on asymptotic tensor rank
We refer the reader to [Blä13] and Chapters 14 and 15 of [BCS13] for background on tensor rank and its applications, but we quickly recap the relevant notions here.Throughout F denotes an arbitrary field.For our purposes one may assume without loss of generality that F = Q or F = Z p (this will be due to [BCS13, Section 15.3]).We assume access to a poly(n)-time algorithm for arithmetic with n-bit elements in the prime field of F. By a tensor we mean a trilinear form T ∶ F n × F n × F n → F, i.e., a cubic set-multilinear polynomial in three disjoint sets of n variables.We say that T has dimension n.The support of T is the set of monomials appearing with nonzero coefficient (with respect to some implicitly chosen basis).We say that T is concise if for all choices of bases, every variable is contained in some monomial in the support of T .We say that T is tight if for some choice of bases of F n , there exist injective functions f, g, h The following tensors are key.
X S Y T Z U .
Definition 1.5.Let T be a tensor.We say that T has rank one if The tensor rank of T , denoted R(T ), is the minimum number r such that T can be expressed as an F-linear combination of r rank-one tensors.
To orient the reader, it is not difficult to show that every tensor has rank at most O(n 2 ), and that "most" tensors meet this bound (see [BCS13,Theorem 20.9] for precise bounds).It is also We use the shorthand T ⊗r to denote the r-fold Kronecker product of T with itself.
As a result, the following is well-defined.
The significance of asymptotic rank is that it that it characterizes the asymptotic complexity of computing high Kronecker powers of T .More formally, if T is concise then R ̃(T ) ≤ x implies that for every ε > 0, one can compute T ⊗r using O((x + ε) r ) arithmetic operations.2This can be understood as a generalization of the relevance of tensor rank to fast matrix multiplication, and follows from a recursive algorithm analogous to Strassen's algorithm [Str69] and all subsequent improvements thereon.Furthermore, this is optimal, as the computation of a trilinear form T requires Ω(R(T )) multiplications [BS83, Corollary 6]. 3onjecture 1.8 (Asymptotic Rank Conjecture [Str94]).For every tensor That is, this posits that tensors of potentially high rank "simplify" as much as is possible under taking Kronecker powers.This conjecture is probably difficult, as a positive answer would imply that ω = 2 and a negative answer would necessarily give explicit high-rank tensors4 , a longstanding challenge in algebraic complexity.To illustrate how poorly understood the behavior of tensor rank under powering is, as far as we are aware it is consistent with the current state of knowledge that for all n-dimensional tensors, R(T ⊗2 ) ≤ 5n 2 -in other words, we do not even know if tensors simplify as much as is possible once they are squared!More embarrassingly yet, we are not aware of any efficiently computable map from the space of n-dimensional tensors to the space of n 2 -dimensional tensors whose image contains tensors of superlinear (i.e.ω(n 2 )) rank.Finally, we are unaware of any example of tensor with (border) rank greater than n but with asymptotic rank n.If no such example exists, then the moral opposite of the asymptotic rank conjecture is true!

Results
Our main result is the following: Theorem 1.9.For any k ∈ N and ε > 0, there is a randomized algorithm for balanced tripartitioning with runtime .
By a reduction from Problem 1.3 to Problem 1.1 similar to that of [BK23, Section 4], we then obtain the following.
Corollary 1.10.If the asymptotic rank conjecture is true, then for every fixed ε > 0 and s ∈ N, there is a randomized algorithm for s-set cover with runtime (3 For context, [BK23] showed that if the asymptotic rank conjecture is true, then for any s ∈ N there is a randomized algorithm for s-set cover with runtime 1.99999 n (see [BK23, Item 5 of Section 1.2]).
We now highlight some more-or-less immediate consequences of Theorem 1.9.
Corollary 1.12.If the set cover conjecture is true, then for every k, Before proceeding to the proofs of Theorem 1.9 and corollary 1.10, we pause to make some comments.
1. Assuming Conjecture 1.8, k = 11 is the smallest value for which Theorem 1.9 would beat the 8 n ⋅ poly(n)-time algorithm for Problem 1.3 sketched earlier.The resulting base tensor is of modest dimension 33 11 > 10 8 .The improvement over the trivial algorithm in [BK23] followed from a base tensor of dimension 7.

The tensor T 1 arose in work of Coppersmith and Winograd [CW87, Section 11], where it
was noted that if R ̃(T 1 ) = 3, then ω = 2.Note that Theorem 1.9 does not conflict with this possibility.Moreover, key to [CW87] is the fact that the induced matching number of the supporting hypergraph of T k is nearly maximal.See [FK14, Appendix A] for an exposition of this fact.In the language of [CKSU05], these induced matchings are called uniquely solvable puzzles.
3. Corollary 1.11 shows that the set cover conjecture implies the existence of an N -dimensional tensor (that is, T k where N = 3k k ) with asymptotic rank greater than N 1.08 .On the other hand, it is known that the asymptotic rank of any N -dimensional tensor is at most N 2ω 3 [Str88, Proposition 3.6].As 1.09 < 4 3, this does not imply that ω = 2 is inconsistent with Conjecture 1.2.4. The best upper bound we know on R(T k ) is 8 k 2, valid for any field with char(F) ≠ 2.
Corollary 1.12 says that a tantalizingly small improvement on this would violate Conjecture 1.2.This rank upper bound is closely related to the Fourier algorithm sketched earlier, and is realized as follows.Let G be an abelian group.Suppose that there is a function f ∶ [3k]   k → G and x ∈ G such that f (S) + f (T ) + f (U ) = x if and only if S, T, U are disjoint.We could then obtain the upper bound R(T k ) ≤ G by zeroing-out and relabeling variables in the tensor ∑ a,b,c∈G,a+b+c=x X a Y b Z c , which has rank G when char(F) ≠ 2. The most obvious way to instantiate this idea is to take G = Z 3k 2 , let f be the indicator vector of S, and let x be the all-ones vector.But we can do a little better by taking G = Z 3k−1 2 , letting f (S) be the the indicator vector ∑ s∈S e s if 1 ∈ S, and otherwise f (S) = 1 3k−1 + ∑ s≠1∈S e i , and letting x be the all-ones vector.
5. The proof of Theorem 1.9 will imply that all of these results hold for any tensor with the same support as T k .5 2 Proofs of Theorem 1.9 and Corollary 1.10 We first note that the tensors T k satisfy the condition in Conjecture 1.8.
Proposition 2.1.T k is concise and tight.
Proof.Conciseness of T k is equivalent to the saying that the bilinear form i=1 4 i witnesses the tightness of T k .
Proof of Theorem 1.9 .Given an instance of Problem 1.3 on a universe of size 3n 0 and with families F 10 , F 20 , F 30 , let n ∶= k⌈n 0 k⌉ be the smallest multiple of k larger than n 0 , and set r ∶= n k.Pick This yields an instance of Problem 1.3 which has a solution if and only if the original one did, which is constructible in time poly(n) (remembering k is a constant), and on a negligibly larger universe (since n < n 0 + k).
Note that if we set all variables X a , Y b , Z c where a ∉ F ′ 1 , b ∉ F ′ 2 , c ∉ F ′ 3 to zero, the resulting restriction of T ⊗r k is identically zero if and only if {F ′ i } contained a tripartition.So let Y be a subset of the prime field of F with Y = 4 (taking an extension if char(F) ∈ {2, 3}).Set X a = 0 if a ∉ F ′ 1 and let X a be a uniformly random element of Y otherwise, and similarly for the Y and Z variables.By what we just said this is always zero on "no" instances, and by the Schwartz-Zippel lemma this is nonzero with probability at least 3 4 on "yes" instances. 6By assumption, this evaluation can be done using O((R ̃(T k ) + ε 2) r ) field operations.Because asymptotic rank is invariant under field extension [BCS13,Proposition 15.17], we may assume that these field operations only involve elements in Y and a constant-sized subset of the prime field of F arising in a rank decomposition of a fixed power of T k .So this evaluation takes (R ̃(T k ) + ε 2) r ⋅ poly(n) time.
Repeat this test 1 p times and output "no" just when all tests fail.This always rejects no instances and accepts yes instances with probability at least 1 − (1 − 3p 4) 1 p > e −3 4 .The total time taken is Proof of Corollary 1.10 .Given set family F, first construct its downwards closure F ′ ∶= ∪ X∈F 2 X .This is done in time F ⋅ poly(n).
This is all done naïvely in time n n 3 ⋅ poly(n).If there were sets in F ′ t 1 , F ′ t 2 , F ′ t 3 partitioning [n], then there exists an S such that F ′′ t 1 ,S , F ′′ t 2 ,S , F ′′ t 3 ,S contains a balanced tripartition of [n] ∖ S, and conversely.The sets F ′′ t i ,S have equal size ⌊n 3⌋ − s by construction.We then call the balanced tripartitioning algorithm on the universe [n] − S with set families F ′′ t 1 ,S , F ′′ t 2 ,S , F ′′ t 3 ,S .The number of calls made is bounded by the number of triples (t 1 , t 2 , t 3 ) and the number of choices of S, which are both poly(n).If the asymptotic rank conjecture is true, then by Theorem 1.9, for any fixed k and δ > 0, each call takes (( 3k k + δ)27 k ( 3k k 2k k )) n (3k) ⋅ poly(n) time.As the first factor approaches (27 4) n 3 as k → ∞ and δ → 0, by choosing k sufficiently large and δ sufficiently small, we conclude.
with X i ≤ s, there exists a partition A ⊔ B ⊔ C = [m] such that ⌊n 3⌋ − s ≤ ⊔ a∈A X a ≤ ⌊n 3⌋ + s, and similarly with the sets indexed by B and C.Next, for each m ≤ t, compute all unions of all pairwise disjoint collections of m subsets in F ′ , having size at most n 3 + s.This can be done with dynamic programming in time n n 3 ⋅ poly(n).Next remove from the resulting set family all sets of size less than ⌊n 3⌋ − s.Let F ′ 1 , ..., F ′ t be the resulting set families.By the first paragraph, it now suffices to check for every (t 1 , t 2 , t 3 ) witht 1 + t 2 + t 3 ≤ t if there exist X ∈ F ′ t 1 , Y ∈ F ′ t 2 , Z ∈ F ′ t 3 partitioning [n].For every S ⊂ [n] of size 3s, and for every partition S = S 1 ⊔ S 2 ⊔ S 3 , for i ∈ [3] construct Note that F contains at most t sets covering [n] if and only if F ′ contains at most t sets partitioning [n].Moreover, notice that for any partition X 1 ⊔ ⋯ ⊔ X m = [n]