## In Search of a 4-by-11 Matrix

**IMPORTANT UPDATE [January 30, 2014]:** I have managed to solve the 4-by-11 case: there is no such matrix! Details of the computation that led to this result, as well as several other related results, are given in [4]. See Table 3 in that paper for an updated list of which cases still remain open (the smallest open cases are now 5-by-11 and 6-by-10).

After spinning my wheels on a problem for far too long, I’ve decided that it’s time to enlist the help of the mathematical and programming geniuses of the world wide web. The problem I’m interested in asks for a 4-by-11 matrix whose columns satisfy certain relationships. While the conditions are relatively easy to state, the problem size seems to be just slightly too large for me to solve myself.

### The Problem

The question I’m interested in (for reasons that are explained later in this blog post) is, given positive integers p and s, whether or not there exists a p-by-s matrix M with the following three properties:

- Every entry of M is a nonzero integer;
- The sum of any two columns of M contains a 0 entry; and
- There is no way to append a (s+1)th column to M so that M still has property 2.

In particular, I’m interested in whether or not such a matrix M exists when p = 4 and s = 11. But to help illustrate the above three properties, let’s consider the p = 3, s = 4 case first, where one such matrix M is:

The fact that M satisfies condition 2 can be checked by hand easily enough. For example, the sum of the first two columns of M is [0, -1, 3]^{T} which contains a 0 entry, and it is similarly straightforward to check that the other 5 sums of two columns of M each contain a 0 entry as well.

Checking property 3 is slightly more technical (NP-hard, even), but is still doable in small cases such as this one. For the above example, suppose that we could add a 5th column (which we will call z = [z_{1}, z_{2}, z_{3}]^{T}) to M such that its sum with any of the first 4 columns has a 0 entry. By looking at M’s first column, we see that one of z’s entries must be -1 (and by the cyclic symmetry of the entries of the last 3 columns of M, we can assume without loss of generality that z_{1} = -1). By looking at the last 3 columns of M, we then see that either z_{2} = 2 or z_{3} = -2, either z_{2} = 1 or z_{3} = 2, and either z_{2} = -2 or z_{3} = 1. Since there is no way to simultaneously satisfy all 3 of these requirements, no such column z exists.

### What’s Known (and What Isn’t)

As I mentioned earlier, the instance of this problem that I’m really interested in is when p = 4 and s = 11. Let’s first back up and briefly discuss what is known for different values of p and s:

**If s ≤ p then M does not exist.**To see this, simply note that property 3 can never be satisfied since you can always append one more column. If we denote the (i,j)-entry of M by m_{ij}and the i-th entry of the new column z by z_{i}, then you can choose z_{i}= -m_{ii}for i = 1, 2, …, s.**Given p, the smallest value of s for which M exists is:**(a) s = p+1 if p is odd, (b) s = p+2 if p = 4 or p ≡ 2 (mod 4), (c) s = p+3 if p = 8, and (d) s = p+4 otherwise. This result was proved in [1] (the connection between that paper and this blog post will be explained in the “Motivation” section below).**If s > 2**In this case, there is no way to satisfy property 2. This fact is trivial when p = 1 and can be proved for all p by induction (an exercise left to the reader?).^{p}then M does not exist.**If s = 2**To see this claim, let the columns of M be the 2^{p}then M exists.^{p}different columns consisting only of the entries 1 and -1. To see that property 2 is satisfied, simply notice that each column is different, so for any pair of columns, there is a row in which one column is 1 and the other column is -1. To see that property 3 is satisfied, observe that any new column must also consist entirely of 1′s and -1′s. However, every such column is already a column of M itself, and the sum of a column with itself will not have any 0 entries.**If s = 2**There is an inductive construction (with the p = 3, s = 4 example from the previous section as the base case) that works here. More specifically, if we let M^{p}– 4 (and p ≥ 3) then M exists._{p}denote a matrix M that works for a given value of p and s = 2^{p}– 4, we let B_{p}be the matrix from the s = 2^{p}case above, and 1_{k}denotes the row vector with k ones, then

is a solution to the problem for p’ = p+1 and s’ = 2^{p+1}– 4.**If 2**This is a non-trivial result that follows from [2].^{p}– 3 ≤ s ≤ 2^{p}– 1 then M does not exist.

Given p, the above results essentially tell us the largest and smallest values of s for which a solution M to the problem exists. However, we still don’t really know much about when solutions exist for intermediate values of s – we just have scattered results that say a solution does or does not exist in certain specific cases, without really illuminating what is going on. The following table summarizes what we know about when solutions do and do not exist for small values of p and s (a check mark ✓ means that a solution exists, a dash - means no solution exists, and ? means we don’t know).

s \ p | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|

1 | - | - | - | - | - |

2 | ✓ | - | - | - | - |

3 | - | - | - | - | - |

4 | - | ✓ | ✓ | - | - |

5 | - | - | - | - | - |

6 | - | - | - | ✓ | ✓ |

7 | - | - | - | ✓ | - |

8 | - | - | ✓ | ✓ | ✓ |

9 | - | - | - | ✓ | ? |

10 | - | - | - | ✓ | ? |

11 | - | - | - | ? | ? |

12 | - | - | - | ✓ | ✓ |

13 | - | - | - | - | ✓ |

14 | - | - | - | - | ✓ |

15 | - | - | - | - | ✓ |

16 | - | - | - | ✓ | ✓ |

17 – 26 | - | - | - | - | ✓ |

27 | - | - | - | - | ? |

28 | - | - | - | - | ✓ |

29 | - | - | - | - | - |

30 | - | - | - | - | - |

31 | - | - | - | - | - |

32 | - | - | - | - | ✓ |

The table above shows why I am interested in the p = 4, s = 11 case: it is the only case when p ≤ 4 whose solution still is not known. The other unknown cases (i.e., p = 5 and s ∈ {9,10,11,27}, and far too many to list when p ≥ 6) would be interesting to solve as well, but are a bit lower-priority.

### Some Simplifications

Some assumptions about the matrix M can be made without loss of generality, in order to reduce the search space a little bit. For example, since the values of the entries of M don’t really matter (other than the fact that they come in positive/negative pairs), the first column of M can always be chosen to consist entirely of ones (or any other value). Similarly, permuting the rows or columns of M does not affect whether or not it satisfies the three desired properties, so you can assume (for example) that the first row is in non-decreasing order.

Finally, since there is no advantage to having the integer k present in M unless -k is also present somewhere in M (i.e., if M does not contain any -k entries, you could always just replace every instance of k by 1 without affecting any of the three properties we want), we can assume that the entries of M are between -floor(s/2) and floor(s/2), inclusive.

### Motivation

The given problem arises from *unextendible product bases (UPBs)* in quantum information theory. A set of pure quantum states forms a UPB if and only if the following three properties hold:

- (product) Each state is a
*product state*(i.e., can be written in the form , where for all i); - (basis) The states are mutually orthogonal (i.e., for all i ≠ j); and
- (unextendible) There does not exist a product state with the property that for all j.

UPBs are useful because they can be used to construct quantum states with very strange entanglement properties [3], but their mathematical structure still isn’t very well-understood. While we can’t really expect an answer to the question of what sizes of UPBs are possible when the local dimensions are arbitrary (even just the *minimum* size of a UPB is still not known in full generality!), we might be able to hope for an answer if we focus on multi-qubit systems (i.e., the case when ).

In this case, the 3 properties above are isomorphic in a sense to the 3 properties listed at the start of this post. We associate each state with the j-th column of the matrix M. To each state in the product state decomposition of , we associate a unique integer in such a way that orthogonal states are associated with negatives of each other. The fact that for all i ≠ j is then equivalent to the requirement that te sum of any two columns of M has a 0 entry, and unextendibility of the product basis corresponds to not being able to add a new column to M without destroying property 2.

Thus this blog post is really asking whether or not there exists an 11-state UPB on 4 qubits. In order to illustrate this connection more explicitly, we return to the p = 3, s = 4 example from earlier. If we associate the matrix entries 1 and -1 with the orthogonal standard basis states and the entries 2 and -2 with the orthogonal states , then the matrix M corresponds to the following set of s = 4 product states in :

The fact that these states form a UPB is well-known – this is the “Shifts” UPB from [3], and was one of the first UPBs found.

**References**

- N. Johnston. The minimum size of qubit unextendible product bases. In
*Proceedings of the 8th Conference on the Theory of Quantum Computation, Communication and Cryptography (TQC)*, 2013. E-print: arXiv:1302.1604 [quant-ph], 2013. - L. Chen and D. Ž. Ðjoković. Separability problem for multipartite states of rank at most four.
*J. Phys. A: Math. Theor.*, 46:275304, 2013. E-print: arXiv:1301.2372 [quant-ph] - C. H. Bennett, D. P. DiVincenzo, T. Mor, P. W. Shor, J. A. Smolin, and B. M. Terhal. Unextendible product bases and bound entanglement.
*Phys. Rev. Lett.*, 82:5385–5388, 1999. E-print: arXiv:quant-ph/9808030 - N. Johnston. The structure of qubit unextendible product bases. E-print: arXiv:1401.7920 [quant-ph], 2014.

Great post, I agree.

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