# BosonSampling and (BKS) Noise Sensitivity

Update (Nov 2014): Noise sensitivity of BosonSampling and computational complexity of noisy BosonSampling are studied in this paper by Guy Kindler and me. Some of my predictions from this post turned out to be false. In particular the noisy BosonSampling is not  flat and it does depend on the input matrix.  However when the noise level is a constant BosonSampling is in P, and when it is above 1 over the number of bosons, we cannot expect robust experimental outcomes.

—–

Following are some preliminary observations connecting BosonSampling, an interesting  computational task that quantum computers can perform (that we discussed in this post), and noise-sensitivity in the sense of Benjamini, Schramm, and myself (that we discussed here and here.)

## BosonSampling and computational-complexity hierarchy-collapse

Suppose that you start with n bosons each can have m locations. The i-th boson is in superposition and occupies the j-th location with complex weight $a_{ij}$. The bosons are indistinguishable which makes the weight for a certain occupation pattern proportional to the permanent of a certain n by n submatrix of the n by m matrix of weights.

Boson Sampling is a task that a quantum computer can perform. As a matter of fact, it only requires a “boson machine” which represents only a fragment of quantum computation. A boson machine is a quantum computer which only manipulates indistinguishable bosons with gated described by phaseshifters and beamsplitters.

BosonSampling and boson machines were studied in a recent paper The Computational Complexity of Linear Optics of Scott Aaronson and Alex Arkhipov (AA). They proved (Theorem 1 in the paper) that if (exact) BosonSampling can be performed by a classical computer then this implies a collapse of the computational-complexity polynomial hierarchy (PH, for short). This result adds to a similar result achieved independently by Michael J. Bremner, Richard Jozsa, and Dan J. Shepherd (BJS) in the paper entitled: “Classical simulation of commuting quantum computations implies collapse of the polynomial hierarchy,” and to older results by  Barbara Terhal and David DiVincenzo (TD) in the paper Adaptive quantum computation, constant depth quantum circuits and Arthur-Merlin games, Quant. Inf. Comp. 4, 134-145 (2004).

Since universal quantum computers can achieve BosonSampling (and the other related computational tasks considered by TD and BJS), this is a very strong indication for the computational complexity advantage of quantum computers which arguably brings us with quantum computers to the “cozy neighborhood” of NP-hardness.

Noisy quantum computers with quantum fault-tolerance are also capable of exact BosonSampling and this strong computational-complexity quantum-superiority applies to them as well.

## Realistic BosonSampling and Gaussian Permanent Estimation (GPE)

Aaronson an Arkhipov considered the following question that they referred to as Gaussian Permanent Approximation.

Problem (Problem 2 from AA’s paper): ($|GPE|_{\pm}^2$): Given as imput a matrix ${\cal N}(0,1)_{\bf C}^{n \times n}$ of i.i.d Gaussians,together with error bounds ε, δ > o, estimate to within additive error $\pm \epsilon n!$ with probability at leat 1-δ over X, in $poly(n,1/\epsilon,1/\delta)$ time.

They conjectured that such Gaussian Permanent Approximation is computationally hard and showed (Theorem 3) that this would imply that sampling w.r.t. states achievable by boson machines cannot even be approximated by classical computing (unless PH collapses). They regarded questions about approximation more realistic in the context of decoherence where we cannot expect exact sampling.

Scott Aaronson also expressed guarded optimism that even without quantum fault-tolerance BosonSampling can be demonstrated by boson machines for 20-30 bosons, leading to strong experimental evidence for computational advantage of quantum computers (or, if you wish, against the efficient Church-Turing thesis).

Is it so?

## More realistic BosonSampling and Noisy Gaussian Permanent Estimation (NGPE)

Let us consider the following variation that we refer to as Noisy Gaussian Permanent Estimation:

Problem 2′: ($|NGPE|_{\pm}^2$): Given as imput a matrix $M=$ ${\cal N}(0,1)_{\bf C}^{n \times n}$ of i.i.d Gaussians, and a parameter t>0 let NPER (M),  be the expected value of the permanent for $\sqrt {1-t^2}M+E$ where E= ${\cal N}(0,t)_{\bf C}^{n \times n}$.  Given the input matrix M together with error bounds ε, δ > o, estimate NPER(M) to within additive error $\pm \epsilon n!$ with probability at leat 1-δ over X, in $poly(n,1/\epsilon,1/\delta)$ time.

Problem 2′ seems more relevant for noisy boson machines (without fault-tolerance). The noisy state of the computer is based on every single boson  being slightly mixed, and the permanent computation will average these individual mixtures. We can consider the relevant value for t to be a small constant. Can we expect Problem 2′ to be hard?

The answer for Question 2′ is surprising. I expect that even when $t$ is very very tiny, namely $t=n^{-\beta}$ for $\beta <1$, the expected value of NPER(M) (essentially) does not depend at all on M!

## Noise Sensitivity a la Benjamini, Kalai and Schramm

Noise sensitivity for the sense described here for Boolean functions was studied in a paper by Benjamini Schramm and me in 1999.  (A related notion was studied by Tsirelson and Vershik.) Lectures on noise sensitivity and percolation is a new beautiful monograph by Christophe Garban and Jeff Steif which contains a description of noise sensitivity. The setting extends easily to the Gaussian case. See this paper by Kindler and O’donnell for the Gaussian case. In 2007, Ofer Zeituni and I studied the noise sensitivity in the Gaussian model of the maximal eigenvalue of random Gaussian matrices (but did not write it up).

Noise sensitivity depends on the degree of the support of the Fourier expansion. For determinants or permanents of an n by n matrices the basic formulas as sums of generalized diagonals describe the Fourier expansion,  so the Fourier coefficients are supported on degree-n monomials. This implies that the determinant and the permanent are very noise sensitive.

## Noisy Gaussian Permanent Estimation is easy

Noisy Gaussian Permanent Estimation is easy, even for very small amount of noise, because the outcome does not depend at all on the input. It is an interesting question what is the hardness of NGPE is when the noise is below the level of noise sensitivity.

Update (March, 2014) Exploring the connection between BosonSampling and BKS-noise sensitivity shows that the picture drawn here is incorrect. Indeed, the square of the permanent is not noise stable even when the noise is fairly small and this shows that the noisy distribution does not approximate the noiseless distribution. However the noisy distribution does depend on the input!

## AA’s protocol and experimental BosonSampling

Scott and Alex proposed a simple experiment described as follows : “An important motivation for our results is that they immediately suggest a linear-optics experiment, which would use simple optical elements (beamsplitters and phaseshifters) to induce a Haar-random $m \times m$ unitary transformation U on an input state of n photons, and would then check that the probabilities of various final states of the photons correspond to the permanents of $n \times n$ submatrices, as predicted by quantum mechanics.”

Recently, four groups carried out interesting BosonSampling experiments with 3 bosons (thus for permanents of 3 by 3 matrices.) (See this post on Scott’s blog.)

BKS-noise sensitivity is giving  simple predictions on how things will behave as a function of the number of bosons and this can be tested even with experiments with very small number of bosons. When you increase the number of bosons the distribution will quickly become uniform for the various final states. The correlation between the probabilities and the value corresponding to permanents will rapidly go to zero.

## Some follow-up questions

Here are a few interesting questions that deserve further study.

1) Does problem 2′ capture the general behavior of noisy boson machines? To what generality noise sensitivity applies for general functions described by Boson sampling distributions?

(There are several versions for photons-based quantum computers including even an important  model by Knill, Laflamme, and Milburn that support universal quantum computation. The relevance of BKS noise-sensitivity should be explored carefully for the various versions.)

2) Is the connection with noise sensitivity relevant to the possibility to have boson machines with fault tolerance?

3) What is the Gaussian-quantum analog for BKS’s theorem asserting that noise sensitivity is the law unless  we have substantial correlation with the majority function?

4) What can be said about noise-sensitivity of measurements for other quantum codes?

## A few more remarks:

### More regarding noisy boson machines and quantum fault tolerance

Noisy boson machines and BosonSampling are related to various other issues regarding quantum fault-tolerance. See my added recent remarks (about the issue of synchronization, and possible modeling using smoothed Lindblad evolutions) to my old post on AA’s work.

### Noise sensitivity and the special role of the majority function

The main result of Itai, Oded, and me was that a Boolean function which is not noise sensitive must have a substantial correlation with the majority function. Noise sensitivity and the special role of majority for it gave me some motivation to look at quantum fault-tolerance in 2005  (see also this post,) and this is briefly discussed in my first paper on the subject, but until now I didn’t find an actual connection between quantum fault-tolerance and BKS-noise-sensitivity.

### Censorship

It is an interesting question which bosonic states are realistic, and it came up in some of my papers and in the discussion with Aram Harrow and Steve Flammia (and their paper on my “Conjecture C”).

### A sort of conclusion

BosonSampling was offered as a way to prove that quantum mechanics allows a computational advantage without using the computational advantage for actual algorithmic purpose. If you wish, the AA’s protocol is offered as a sort of zero-knowledge proof that the efficient Church-Turing thesis is false.  It is a beautiful idea that attracted interest and good subsequent work from theoreticians and experimentalists. If the ideas described here are correct, BosonSampling and boson machines may give a clear understanding based on BKS noise-sensitivity for why quantum mechanics does not offer computational superiority (at least not without the magic of quantum fault-tolerance).

### Avi’s joke and common sense

Here is a quote from AA referring to a joke by Avi Wigderson: “Besides bosons, the other basic particles in the universe are fermions; these include matter particles such as quarks and electrons. Remarkably, the amplitudes for n-fermion processes are given not by permanents but by determinants of n×n matrices. Despite the similarity of their definitions, it is well-known that the permanent and determinant differ dramatically in their computational properties; the former is #P-complete while the latter is in P. In a lecture in 2000, Wigderson called attention to this striking connection between the boson and fermion dichotomy of physics and the permanent-determinant dichotomy of computer science. He joked that, between bosons and fermions, ‘the bosons got the harder job.’ One could view this paper as a formalization of Wigderson’s joke.”

While sharing the admiration to Avi in general and Avi’s jokes in particular, if we do want to take Avi’s joke seriously (as we always should), then the common-sense approach would be first to try to understand why is it that nature treats bosons and fermions quite equally and the dramatic computational distinction is not manifested at all. The answer is that a crucial ingredient for a computational model is the modeling of noise/errors, and that noise-sensitivity makes bosons and fermions quite similar physically and computationally.

### Eigenvalues, determinants, and Yuval Filmus

It is an interesting question (that I asked over Mathoverflow) to understand the Fourier expansion of the set of eigenvalues, the maximum eigenvalue and related functions. At a later point,  last May,  I was curious about the Fourier expansion of the determinant, and for the Boolean case I noticed remarkable properties of its Fourier expansion. So I decided to ask Yuval Filmus about it:

Dear Yuval

I am curious about the following. Let D be the function defined on {-1,1}^n^2
which associates to every +1/1 matrix its determinant.
What can be said about the Fourier transform of D? It looks to me that easy arguments shows that the Fourier transform is supported only on subsets of the entries
so that in every raw and columns there are odd number of entries. Likely there are even further restrictions that I would be interested to explore.
Do you know anything about it?
best Gil

Yuval’s answer came a couple of hours later like a cold shower:

Hi Gil,

The determinant is a sum of products of generalized diagonals.
Each generalized diagonal is just a Fourier character, and they are all different.

In other words, the usual formula for the determinant *is* its Fourier transform

This reminded me of a lovely story of how I introduced Moni Naor to himself that I should tell sometime.

### What else can a quantum computer sample?

The ability of quantum computers to sample (exactly) random complex Gaussian matrices according to the value of their permanents is truly amazing! If you are not too impressed by efficient factoring but still do not believe that QC can reach the neighborhood of NP-hard problems you may find this possibility too good to be true.

I am curious if sharp P reductions give us further results of this nature. For example,  can a QC sample random 3-SAT formulas (by a uniform distribution or by a certain other distribution coming from sharp-P reductions) according to the number of their satisfying assignments?

Can QC sample integer polytopes by their volume or by the number of integer points in them? Graphs by the number of 4-colorings?

# Lawler-Kozdron-Richards-Stroock’s combined Proof for the Matrix-Tree theorem and Wilson’s Theorem

David Wilson and a cover of Shlomo’s recent book “Curvature in mathematics and physics”

A few weeks ago, in David Kazhdan’s basic notion seminar, Shlomo Sternberg gave a lovely presentation Kirchho ff and Wilson via Kozdron and Stroock. The lecture is based on the work presented in the very recent paper by Michael J. Kozdron,  Larissa M. Richards, and Daniel W. Stroock: Determinants, their applications to Markov processes, and a random walk proof of Kirchhoff’s matrix tree theorem. Preprint, 2013. Available online at arXiv:1306.2059.

Here is the abstract:

Kirchhoff’s formula for the number of spanning trees in a connected graph  is over 150 years old. For example, it says that if $c_2, \dots, c_n$ are the nonzero  eigenvalues of the Laplacian, then the number k of spanning trees is $k= (1/n)c_2\cdots c_n.$ There are many proofs.  An algorithm due to Wilson via loop erased random walks produces such a tree, and Wilson’s theorem is that all spanning trees are produced by his algorithm with equal probability. Hence,  after the fact, we know that Wilson’s algorithm produces any given tree with probability 1/k.  A proof due to Lawler, using the Green’s function, shows directly that Wilson’s algorithm has the probability 1/k  of producing any given spanning tree, thus simultaneously proving Wilson’s theorem and Kirchhoff’s formula. Lawler’s proof has been considerably simplified by Kozdron and Stroock. I plan to explain their proof. The lecture will be completely self-contained, using only Cramer’s rule from linear algebra.

(Here are also lecture notes of the lecture by Ron Rosenthal.)

Here is some background.

## The matrix-tree theorem

The matrix tree theorem asserts that the number of rooted spanning trees of a connected graph G  is the product of the non-zero eigenvalues of L(G), the Laplacian of G.

Suppose that G has n vertices. The Laplacian of G is the matrix whose (i,i)-entry is the degree of the ith vertex, and its (i,j) entry for $i \ne j$ is 0 if the ith vertex is not adjacent to the jth vertex, and -1 if they are adjacent. So  L(G)=D-A(G) where A(G) is the adjacency matrix of G, and D is a diagonal matrix whose entries are the degrees of the vertices.  An equivalent formulation of the matrix-tree theorem is that the number of spanning trees is the determinant of a matrix obtained from the Laplacian by deleting the j th row and j th column.

We considered a high dimensional generalization of the matrix tree theorem in these posts (I, II, III, IV).

## How to generate a random spanning tree for a graph G?

### Using the matrix-tree theorem

Method A: Start with an edge $e \in G$, use the matrix-tree theorem to compute the probability $p_e$ that e belongs to a random spanning tree of G, take e with probability $p_e$. If e is taken consider the contraction $G/e$ and if G is not taken consider the deletion $G \backslash e$ and continue.

This is an efficient method to generate a random spanning tree according to the uniform probability distribution. You can extend it by assigning each edge a weight and chosing a tree with probability proportional to the product of its weights.

### Random weights and greedy

Method B: Assign each edge a random real number between 0 and 1 and chose the spanning tree which minimizes the sum of weights via the greedy algorithm.

This is a wonderful method but it leads to a different probability distribution on random spanning trees which is very interesting!

### The Aldous-Broder random walk method

Method C: The Aldous-Broder theorem. Start a simple random walk from a vertex of the graph until reaching all vertices, and take each edge that did not form a cycle with earlier edges. (Or, in other words, take every edge that reduced the number of connected components of the graph on the whole vertex set and visited edges.)

Amazingly, this leads to a random uniform spanning tree. The next method is also very amazing and important for many applications.

### David Wilson’s algorithm

Method D: Wilson’s algorithm. Fix a vertex as a root. (Later the root will be a whole set of vertices, and a tree on them.) Start from an arbitrary vertex u not in the root and take a simple random walk until you reach the root. Next, erase all edges in cycles of the path created by the random walk so you will left with a simple path from  u to the root. Add this path to the root and continue!

Here is a link to Wilson’s paper! Here is a nice presentation by Chatterji  and Gulwani.

# Oz’ Balls Problem: The Solution

A commentator named Oz proposed the following question: You have a box with n red balls and n blue balls. You take out each time a ball at random but, if the ball was red, you put it back in the box and take out a blue ball. If the ball was blue, you put it back in the box and take out a red ball.

You keep doing it until left only with balls of the same color. How many balls will be left (as a function of n)?

Peter Shor wrote in a comment “I’m fairly sure that there is not enough bias to get $cn$, but it intuitively seems far too much bias to still be $c \sqrt{n}$. I want to say $n^c$. At a wild guess, it’s either $c = \frac{2}{3}$or $c = \frac{3}{4}$, since those are the simplest exponents between $\frac{1}{2}$ and $1$.”  The comment followed by a heuristic argument of Kevin Kostelo and computer experiments by Lior Silberman that supported the answer $n^{3/4}$.

# Taking balls away: Oz’ Version

This post is based on a comment by Oz to our question about balls with two colors:

“There is an interesting (and more difficult) variation I once heard but can’t recall where:

You have a box with n red balls and n blue balls. You take out each time a ball at random as before. But, if the ball was red, you put it back in the box and take out a blue ball. If the ball was blue, you put it back in the box and take out a red ball.

You keep as before until left only with balls of the same color. How many balls will be left (as a function of n)?

1) Roughly  εn for some ε>0.

2) Roughly $\sqrt n$?

3) Roughly log n?

4) Roughly a constant?

5) Some other behavior

You have a box with n red balls and n blue balls. You take out balls one by one at random until left only with balls of the same color. How many balls will be left (as a function of n)?

1) Roughly  εn for some ε>0.

2) Roughly $\sqrt n$?

3) Roughly log n?

4) Roughly a constant?

Here is the collective intuition regarding this problem

# Itai Ashlagi, Yashodhan Kanoria, and Jacob Leshno: What a Difference an Additional Man makes?

We are considering the stable marriage theorem. Suppose that there are n men and n women. If the preferences are random and men are proposing, what is the likely average women’s rank of their husbands, and what is the likely average men’s rank of their wives?

Boris Pittel proved that on average a man will be matched to the woman in place log n on his list. (Place one is his most preferred woman.) A woman will be matched on average to a man ranked n/log n on her list.

We asked in the post “Test your intuition (19)”  what is the situation if there is one additional man, and men are still proposing. This question is based on a conversation with Jacob Leshno who told me about a remarkable paper Unbalanced random matching markets by Itai Ashlagi, Yash Kanoria, and Jacob Leshno. Continue reading

# Test Your Intuition (19): The Advantage of the Proposers in the Stable Matching Algorithm

### Stable mariage

The Gale-Shapley stable matching theorem and the algorithm.

GALE-SHAPLEY THEOREM Consider a society of n men and n women and suppose that every man [and every woman] have a preference (linear) relation on the women [men] he [she] knows. Then there is a stable marriage, namely a perfect matching between the men and the women so that there are no men and women which are not matched so that both of them prefer the other on their spouces.

Proof: Consider the following algorithm, on day 1 every man goes to the first woman on his list and every woman select the best man among those who come to her and reject the others. On the second day every rejected men go to the second woman on his list and every woman select one man from all man that comes to her (including the man she selected in the previous day if there was such a man) and rejects all others, and so on. This process will terminate after finitely many days and with a stable marriage! To see that the process terminate note that each day at least one man will come to a new women, or go back home after beeing rejected from every women (n+1 possibilities) and none of these possibilitie will ever repeat itself so after at most $n^2+n$ days things will stabilize. When it terminates we have a stable marriage because suppose women W and men M are not married at the end. If M is married to a women he prefers less then W or to no women at all it means that M visited W and she rejected him so she had a better men than M.  Sababa!
It turns out that the above algorithm where the men are proposing and being rejected is optimal for the men! If a man M is matched to a woman W then there is not a single stable marriage where M can be matched to a woman higher on his list. Similarly this algorithm is worst for the women. But by how much?

### Random independent preferences

Question 1:  There are n men and n women. If the preferences are random and men are proposing, what is the likely average women’s rank of their husbands, and what is the likely average men’s rank of their wives.

You can test your intuition, or look at the answer and for a follow up question after the fold.