The standard inner product of two vectors has some nice geometric properties. Given two vectors $ x, y \in \mathbb{R}^n$, where by $ x_i$ I mean the $ i$-th coordinate of $ x$, the standard inner product (which I will interchangeably call the dot product) is defined by the formula
This formula, simple as it is, produces a lot of interesting geometry. An important such property, one which is discussed in machine learning circles more than pure math, is that it is a very convenient decision rule.
In particular, say we’re in the Euclidean plane, and we have a line $ L$ passing through the origin, with $ w$ being a unit vector perpendicular to $ L$ (“the normal” to the line).
If you take any vector $ x$, then the dot product $ \langle x, w \rangle$ is positive if $ x$ is on the same side of $ L$ as $ w$, and negative otherwise. The dot product is zero if and only if $ x$ is exactly on the line $ L$, including when $ x$ is the zero vector.
Left: the dot product of $ w$ and $ x$ is positive, meaning they are on the same side of $ w$. Right: The dot product is negative, and they are on opposite sides.
Here is an interactive demonstration of this property. Click the image below to go to the demo, and you can drag the vector arrowheads and see the decision rule change.
It’s always curious, at first, that multiplying and summing produces such geometry. Why should this seemingly trivial arithmetic do anything useful at all?
The core fact that makes it work, however, is that the dot product tells you how one vector projects onto another. When I say “projecting” a vector $ x$ onto another vector $ w$, I mean you take only the components of $ x$ that point in the direction of $ w$. The demo shows what the result looks like using the red (or green) vector.
In two dimensions this is easy to see, as you can draw the triangle which has $ x$ as the hypotenuse, with $ w$ spanning one of the two legs of the triangle as follows:
If we call $ a$ the (vector) leg of the triangle parallel to $ w$, while $ b$ is the dotted line (as a vector, parallel to $ L$), then as vectors $ x = a + b$. The projection of $ x$ onto $ w$ is just $ a$.
Another way to think of this is that the projection is $ x$, modified by removing any part of $ x$ that is perpendicular to $ w$. Using some colorful language: you put your hands on either side of $ x$ and $ w$, and then you squish $ x$ onto $ w$ along the line perpendicular to $ w$ (i.e., along $ b$).
And if $ w$ is a unit vector, then the length of $ a$—that is, the length of the projection of $ x$ onto $ w$—is exactly the inner product product $ \langle x, w \rangle$.
Moreover, if the angle between $ x$ and $ w$ is larger than 90 degrees, the projected vector will point in the opposite direction of $ w$, so it’s really a “signed” length.
Left: the projection points in the same direction as $ w$. Right: the projection points in the opposite direction.
And this is precisely why the decision rule works. This 90-degree boundary is the line perpendicular to $ w$.
More technically said: Let $ x, y \in \mathbb{R}^n$ be two vectors, and $ \langle x,y \rangle $ their dot product. Define by $ \| y \|$ the length of $ y$, specifically $ \sqrt{\langle y, y \rangle}$. Define by $ \text{proj}_{y}(x)$ by first letting $ y’ = \frac{y}{\| y \|}$, and then let $ \text{proj}_{y}(x) = \langle x,y’ \rangle y’$. In words, you scale $ y$ to a unit vector $ y’$, use the result to compute the inner product, and then scale $ y$ so that it’s length is $ \langle x, y’ \rangle$. Then
Theorem: Geometrically, $ \text{proj}_y(x)$ is the projection of $ x$ onto the line spanned by $ y$.
This theorem is true for any $ n$-dimensional vector space, since if you have two vectors you can simply apply the reasoning for 2-dimensions to the 2-dimensional plane containing $ x$ and $ y$. In that case, the decision boundary for a positive/negative output is the entire $ n-1$ dimensional hyperplane perpendicular to $ y$ (the projected vector).
In fact, the usual formula for the angle between two vectors, i.e. the formula $ \langle x, y \rangle = \|x \| \cdot \| y \| \cos \theta$, is a restatement of the projection theorem in terms of trigonometry. The $ \langle x, y’ \rangle$ part of the projection formula (how much you scale the output) is equal to $ \| x \| \cos \theta$. At the end of this post we have a proof of the cosine-angle formula above.
Part of why this decision rule property is so important is that this is a linear function, and linear functions can be optimized relatively easily. When I say that, I specifically mean that there are many known algorithms for optimizing linear functions, which don’t have obscene runtime or space requirements. This is a big reason why mathematicians and statisticians start the mathematical modeling process with linear functions. They’re inherently simpler.
In fact, there are many techniques in machine learning—a prominent one is the so-called Kernel Trick—that exist solely to take data that is not inherently linear in nature (cannot be fruitfully analyzed by linear methods) and transform it into a dataset that is. Using the Kernel Trick as an example to foreshadow some future posts on Support Vector Machines, the idea is to take data which cannot be separated by a line, and transform it (usually by adding new coordinates) so that it can. Then the decision rule, computed in the larger space, is just a dot product. Irene Papakonstantinou neatly demonstrates this with paper folding and scissors. The tradeoff is that the size of the ambient space increases, and it might increase so much that it makes computation intractable. Luckily, the Kernel Trick avoids this by remembering where the data came from, so that one can take advantage of the smaller space to compute what would be the inner product in the larger space.
Next time we’ll see how this decision rule shows up in an optimization problem: finding the “best” hyperplane that separates an input set of red and blue points into monochromatic regions (provided that is possible). Finding this separator is core subroutine of the Support Vector Machine technique, and therein lie interesting algorithms. After we see the core SVM algorithm, we’ll see how the Kernel Trick fits into the method to allow nonlinear decision boundaries.
Proof of the cosine angle formula
Theorem: The inner product $ \langle v, w \rangle$ is equal to $ \| v \| \| w \| \cos(\theta)$, where $ \theta$ is the angle between the two vectors.
Note that this angle is computed in the 2-dimensional subspace spanned by $ v, w$, viewed as a typical flat plane, and this is a 2-dimensional plane regardless of the dimension of $ v, w$.
Proof. If either $ v$ or $ w$ is zero, then both sides of the equation are zero and the theorem is trivial, so we may assume both are nonzero. Label a triangle with sides $ v,w$ and the third side $ v-w$. Now the length of each side is $ \| v \|, \| w\|,$ and $ \| v-w \|$, respectively. Assume for the moment that $ \theta$ is not 0 or 180 degrees, so that this triangle is not degenerate.
$ \displaystyle \| v – w \|^2 = \| v \|^2 + \| w \|^2 – 2 \| v \| \| w \| \cos(\theta)$
Moreover, The left hand side is the inner product of $ v-w$ with itself, i.e. $ \| v – w \|^2 = \langle v-w , v-w \rangle$. We’ll expand $ \langle v-w, v-w \rangle$ using two facts. The first is trivial from the formula, that inner product is symmetric: $ \langle v,w \rangle = \langle w, v \rangle$. Second is that the inner product is linear in each input. In particular for the first input: $ \langle x + y, z \rangle = \langle x, z \rangle + \langle y, z \rangle$ and $ \langle cx, z \rangle = c \langle x, z \rangle$. The same holds for the second input by symmetry of the two inputs. Hence we can split up $ \langle v-w, v-w \rangle$ as follows.
$ \displaystyle \begin{aligned} \langle v-w, v-w \rangle &= \langle v, v-w \rangle – \langle w, v-w \rangle \\ &= \langle v, v \rangle – \langle v, w \rangle – \langle w, v \rangle + \langle w, w \rangle \\ &= \| v \|^2 – 2 \langle v, w \rangle + \| w \|^2 \\ \end{aligned}$
Combining our two offset equations, we can subtract $ \| v \|^2 + \| w \|^2$ from each side and get
Which, after dividing by $ -2$, proves the theorem if $ \theta \not \in \{0, 180 \}$.
Now if $ \theta = 0$ or 180 degrees, the vectors are parallel, so we can write one as a scalar multiple of the other. Say $ w = cv$ for $ c \in \mathbb{R}$. In that case, $ \langle v, cv \rangle = c \| v \| \| v \|$. Now $ \| w \| = | c | \| v \|$, since a norm is a length and is hence non-negative (but $ c$ can be negative). Indeed, if $ v, w$ are parallel but pointing in opposite directions, then $ c < 0$, so $ \cos(\theta) = -1$, and $ c \| v \| = – \| w \|$. Otherwise $ c > 0$ and $ \cos(\theta) = 1$. This allows us to write $ c \| v \| \| v \| = \| w \| \| v \| \cos(\theta)$, and this completes the final case of the theorem.
The singular value decomposition (SVD) of a matrix is a fundamental tool in computer science, data analysis, and statistics. It’s used for all kinds of applications from regression to prediction, to finding approximate solutions to optimization problems. In this series of two posts we’ll motivate, define, compute, and use the singular value decomposition to analyze some data. (Jump to the second post)
I want to spend the first post entirely on motivation and background. As part of this, I think we need a little reminder about how linear algebra equivocates linear subspaces and matrices. I say “I think” because what I’m going to say seems rarely spelled out in detail. Indeed, I was confused myself when I first started to read about linear algebra applied to algorithms, machine learning, and data science, despite having a solid understanding of linear algebra from a mathematical perspective. The concern is the connection between matrices as transformations and matrices as a “convenient” way to organize data.
Data vs. maps
Linear algebra aficionados like to express deep facts via statements about matrix factorization. That is, they’ll say something opaque like (and this is the complete statement for SVD we’ll get to in the post):
The SVD of an $ m \times n$ matrix $ A$ with real values is a factorization of $ A$ as $ U\Sigma V^T$, where $ U$ is an $ m \times m$ orthogonal matrix, $ V$ is an $ n \times n$ orthogonal matrix, and $ \Sigma$ is a diagonal matrix with nonnegative real entries on the diagonal.
Okay, I can understand the words individually, but what does it mean in terms of the big picture? There are two seemingly conflicting interpretations of matrices that muddle our vision.
The first is that $ A$ is a linear map from some $ n$-dimensional vector space to an $ m$-dimensional one. Let’s work with real numbers and call the domain vector space $ \mathbb{R}^n$ and the codomain $ \mathbb{R}^m$. In this interpretation the factorization expresses a change of basis in the domain and codomain. Specifically, $ V$ expresses a change of basis from the usual basis of $ \mathbb{R}^n$ to some other basis, and $ U$ does the same for the co-domain $ \mathbb{R}^m$.
That’s fine and dandy, so long as the data that makes up $ A$ is the description of some linear map we’d like to learn more about. Years ago on this blog we did exactly this analysis of a linear map modeling a random walk through the internet, and we ended up with Google’s PageRank algorithm. However, in most linear algebra applications $ A$ actually contains data in its rows or columns. That’s the second interpretation. That is, each row of $ A$ is a data point in $ \mathbb{R}^n$, and there are $ m$ total data points, and they represent observations of some process happening in the world. The data points are like people rating movies or weather sensors measuring wind and temperature. If you’re not indoctrinated with the linear algebra world view, you might not naturally think of this as a mapping of vectors from one vector space to another.
Most of the time when people talk about linear algebra (even mathematicians), they’ll stick entirely to the linear map perspective or the data perspective, which is kind of frustrating when you’re learning it for the first time. It seems like the data perspective is just a tidy convenience, that it just “makes sense” to put some data in a table. In my experience the singular value decomposition is the first time that the two perspectives collide, and (at least in my case) it comes with cognitive dissonance.
The way these two ideas combine is that the data is thought of as the image of the basis vectors of $ \mathbb{R}^n$ under the linear map specified by $ A$. Here is an example to make this concrete. Let’s say I want to express people rating movies. Each row will correspond to the ratings of a movie, and each column will correspond to a person, and the $ i,j$ entry of the matrix $ A$ is the rating person $ j$ gives to movie $ i$.
In reality they’re rated on a scale from 1 to 5 stars, but to keep things simple we’ll just say that the ratings can be any real numbers (they just happened to pick integers). So this matrix represents a linear map. The domain is $ \mathbb{R}^3$, and the basis vectors are called people, and the codomain is $ \mathbb{R}^8$, whose basis vectors are movies.
Now the data set is represented by $ A(\vec e_{\textup{Aisha}}),A(\vec e_{\textup{Bob}}),A(\vec e_{\textup{Chandrika}})$, and by the definition of how a matrix represents a linear map, the entires of these vectors are exactly the columns of $ A$. If the codomain is really big, then the image of $ A$ is a small-dimensional linear subspace of the codomain. This is an important step, that we’ve increased our view from just the individual data points to all of their linear combinations as a subspace.
Why is this helpful at all? This is where we start to see the modeling assumptions of linear algebra show through. If we’re trying to use this matrix to say something about how people rate movies (maybe we want to predict how a new person will rate these movies), we would need to be able to represent that person as a linear combination of Aisha, Bob, and Chandrika. Likewise, if we had a new movie and we wanted to use this matrix to say anything about it, we’d have to represent the movie as a linear combination of the existing movies.
Of course, I don’t literally mean that a movie (as in, the bits comprising a file containing a movie) can be represented as a linear combination of other movies. I mean that we can represent a movie formally as a linear combination in some abstract vector space for the task at hand. In other words, we’re representing those features of the movie that influence its rating abstractly as a vector. We don’t have a legitimate mathematical way to understand that, so the vector is a proxy.
It’s totally unclear what this means in terms of real life, except that you can hope (or hypothesize, or verify), that if the rating process of movies is “linear” in nature then this formal representation will accurately reflect the real world. It’s like how physicists all secretly know that mathematics doesn’t literally dictate the laws of nature, because humans made up math in their heads and if you poke nature too hard the math breaks down, but it’s so damn convenient to describe hypotheses (and so damn accurate), that we can’t avoid using it to design airplanes. And we haven’t found anything better than math for this purpose.
Likewise, movie ratings aren’t literally a linear map, but if we pretend they are we can make algorithms that predict how people rate movies with pretty good accuracy. So if you know that Skyfall gets ratings 1,2, and 1 from Aisha, Bob, and Chandrika, respectively, then a new person would rate Skyfall based on a linear combination of how well they align with these three people. In other words, up to a linear combination, in this example Aisha, Bob, and Chandrika epitomize the process of rating movies.
And now we get to the key: factoring the matrix via SVD provides an alternative and more useful way to represent the process of people rating movies. By changing the basis of one or both vector spaces involved, we isolate the different (orthogonal) characteristics of the process. In the context of our movie example, “factorization” means the following:
Come up with a special list of vectors $ v_1, v_2, \dots, v_8$ so that every movie can be written as a linear combination of the $ v_i$.
Do the analogous thing for people to get $ p_1, p_2, p_3$.
Do (1) and (2) in such a way that the map $ A$ is diagonal with respect to both new bases simultaneously.
One might think of the $ v_i$ as “idealized movies” and the $ p_j$ as “idealized critics.” If you want to use this data to say things about the world, you’d be making the assumption that any person can be written as a linear combination of the $ p_j$ and any movie can be written as a linear combination of the $ v_i$. These are the rows/columns of $ U, V$ from the factorization. To reiterate, these linear combinations are only with respect to the task of rating movies. And they’re “special” because they make the matrix diagonal.
If the world was logical (and I’m not saying it is) then maybe $ v_1$ would correspond to some idealized notion of “action movie,” and $ p_1$ would correspond to some idealized notion of “action movie lover.” Then it makes sense why the mapping would be diagonal in this basis: an action movie lover only loves action movies, so $ p_1$ gives a rating of zero to everything except $ v_1$. A movie is represented by how it decomposes (linearly) into “idealized” movies. To make up some arbitrary numbers, maybe Skyfall is 2/3 action movie, 1/5 dystopian sci-fi, and -6/7 comedic romance. Likewise a person would be represented by how they decompose (via linear combination) into a action movie lover, rom-com lover, etc.
To be completely clear, the singular value decomposition does not find the ideal sci-fi movie. The “ideal”ness of the singular value decomposition is with respect to the inherent geometric structure of the data coupled with the assumptions of linearity. Whether this has anything at all to do with how humans classify movies is a separate question, and the answer is almost certainly no.
With this perspective we’re almost ready to talk about the singular value decomposition. I just want to take a moment to write down a list of the assumptions that we’d need if we want to ensure that, given a data set of movie ratings, we can use linear algebra to make exact claims about world.
All people rate movies via the same linear map.
Every person can be expressed (for the sole purpose of movie ratings) as linear combinations of “ideal” people. Likewise for movies.
The “idealized” movies and people can be expressed as linear combinations of the movies/people in our particular data set.
There are no errors in the ratings.
One could have a deep and interesting discussion about the philosophical (or ethical, or cultural) aspects of these assumptions. But since the internet prefers to watch respectful discourse burn, we’ll turn to algorithms instead.
Approximating subspaces
In our present context, the singular value decomposition (SVD) isn’t meant to be a complete description of a mapping in a new basis, as we said above. Rather, we want to use it to approximate the mapping $ A$ by low-dimensional linear things. When I say “low-dimensional linear things” I mean that given $ A$, we’d want to find another matrix $ B$ which is measurably similar to $ A$ in some way, and has low rank compared to $ A$.
How do we know that $ A$ isn’t already low rank? The reasons is that data with even the tiniest bit of noise is full rank with overwhelming probability. A concrete way to say this is that the space of low-rank matrices has small dimension (in the sense of a manifold) inside the space of all matrices. So perturbing even a single entry by an infinitesimally small amount would increase the rank.
We don’t need to understand manifolds to understand the SVD, though. For our example of people rating movies the full-rank property should be obvious. The noise and randomness and arbitrariness in human preferences certainly destroys any “perfect” linear structure we could hope to find, and in particular that means the data set itself, i.e. the image of $ A$, is a large-dimensional subspace of the codomain.
Finding a low-rank approximation can be thought of as “smoothing” the noise out of the data. And this works particularly well when the underlying process is close to a linear map. That is, when the data is close to being contained entirely in a single subspace of relatively low-dimension. One way to think of why this might be the case is that if the process you’re observing is truly linear, but the data you get is corrupted by small amounts of noise. Then $ A$ will be close to low rank in a measurable sense (to be defined mathematically in the sequel post) and the low-rank approximation $ B$ will be a more efficient, accurate, and generalizable surrogate for $ A$.
In terms of our earlier list of assumptions about when you can linear algebra to solve problems, for the SVD we can add “approximately” to the first three assumptions, and “not too many errors” to the fourth. If those assumptions hold, SVD will give us a matrix $ B$ which accurately represents the process being measured. Conversely, if SVD does well, then you have some evidence that the process is linear-esque.
To be more specific with notation, if $ A$ is a matrix representing some dataset via the image of $ A$ and you provide a small integer $ k$, then the singular value decomposition computes the rank $ k$ matrix $ B_k$ which best approximates $ A$. And since now we’re comfortable identifying a data matrix with the subspace defined by its image, this is the same thing as finding the $ k$-dimensional subspace of the image of $ A$ which is the best approximation of the data (i.e., the image of $ B_k$). We’ll quantify what we mean by “best approximation” in the next post.
That’s it, as far as intuitively understanding what the SVD is. I should add that the SVD doesn’t only allow one to compute a rank $ k$ approximation, it actually allows you to set $ k=n$ and get an exact representation of $ A$. We just won’t use it for that purpose in this series.
The second bit of intuition is the following. It’s only slightly closer to rigor, but somehow this little insight really made SVD click for me personally:
The SVD is what you get when you iteratively solve the greedy optimization problem of fitting data to a line.
By that I mean, you can compute the SVD by doing the following:
What’s the best line fitting my data?
Okay, ignoring that first line, what’s the next best line?
Okay, ignoring all the lines in the span of those first two lines, what’s the next best line?
Ignoring all the lines in the span of the first three lines, what’s the next best line?
(repeat)
It should be shocking that this works. For most problems, in math and in life, the greedy algorithm is far from optimal. When it happens, once every blue moon, that the greedy algorithm is the best solution to a natural problem (and not obviously so, or just approximately so), it’s our intellectual duty to stop what we’re doing, sit up straight, and really understand and appreciate it. These wonders transcend political squabbles and sports scores. And we’ll start the next post immediately by diving into this greedy optimization problem.
The geometric perspective
There are two other perspectives I want to discuss here, though it may be more appropriate for a reader who is not familiar with the SVD to wait to read this after the sequel to this post. I’m just going to relate my understanding (in terms of the greedy algorithm and data approximations) to the geometric and statistical perspectives on the SVD.
Michael Nielsen wrote a long and detailed article presenting some ideas about a “new medium” in which to think about mathematics. He demonstrates is framework by looking at the singular value decomposition for 2×2 matrices. His explanation for the intuition behind the SVD is that you can take any matrix (linear map) and break it up into three pieces: a rotation about the origin, a rescaling of each coordinate, followed by another rotation about the origin. While I have the utmost respect for Nielsen (his book on quantum mechanics is the best text in the field), this explanation never quite made SVD click for me personally. It seems like a restatement of the opaque SVD definition (as a matrix factorization) into geometric terms. Indeed, an orthogonal matrix is a rotation, and a diagonal matrix is a rescaling of each coordinate.
To me, the key that’s missing from this explanation is the emphasis on the approximation. What makes the SVD so magical isn’t that the factorization exists in the first place, but rather that the SVD has these layers of increasingly good approximation. Though the terminology will come in the next post, these layers are the (ordered) singular vectors and singular values. And moreover, that the algorithmic process of constructing these layers necessarily goes in order from strongest approximation to weakest.
Another geometric perspective that highlights this is that the rank-$ k$ approximation provided by the SVD is a geometric projection of a matrix onto the space of rank at-most-k matrices with respect to the “spectral norm” on matrices (the spectral norm of $ A$ is the largest eigenvalue of $ A^TA$). The change of basis described above makes this projection very easy: given a singular value decomposition you just take the top $ k$ singular vectors. Indeed, the Eckart-Young theorem formalizes this with the statement that the rank-$ k$ SVD $ B_k$ minimizes the distance (w.r.t. the spectral norm) between the original matrix $ A$ and any rank $ k$ matrix. So you can prove that SVD gives you the best rank $ k$ approximation of $ A$ by some reasonable measure.
Next time: algorithms
Next time we’ll connect all this to the formal definitions and rigor. We’ll study the greedy algorithm approach, and then we’ll implement the SVD and test it on some data.
Last time we left off with the tantalizing question: how do you do a quantum “AND” operation on two qubits? In this post we’ll see why the tensor product is the natural mathematical way to represent the joint state of multiple qubits. Then we’ll define some basic quantum gates, and present the definition of a quantum circuit.
Working with Multiple Qubits
In a classical system, if you have two bits with values $ b_1, b_2$, then the “joint state” of the two bits is given by the concatenated string $ b_1b_2$. But if we have two qubits $ v, w$, which are vectors in $ \mathbb{C}^2$, how do we represent their joint state?
There are seemingly infinitely many things we could try, but let’s entertain the simplest idea for the sake of exercising our linear algebra intuition. The simplest idea is to just “concatenate” the vectors as one does in linear algebra: represent the joint system as $ (v, w) \in \mathbb{C}^2 \oplus \mathbb{C}^2$. Recall that the direct sum of two vector spaces is just what you’d want out of “concatenation” of vectors. It treats the two components as completely independent of each other, and there’s an easy way to take any vector in the sum and decompose it into two vectors in the pieces.
Why does this fail to meet our requirements of qubits? Here’s one reason: $ (v, w)$ is not a unit vector when $ v$ and $ w$ are separately unit vectors. Indeed, $ \left \| (v,w) \right \|^2 = \left \| v \right \|^2 + \left \| w \right \|^2 = 2$. We could normalize everything, and that would work for a while, but we would still run into problems. A better reason is that direct sums screw up measurement. In particular, if you have two qubits (and they’re independent, in a sense we’ll make clear later), you should be able to measure one without affecting the other. But if we use the direct sum method for combining qubits, then measuring one qubit would collapse the other! There are times when we want this to happen, but we don’t always want it to happen. Alas, there should be better reasons out there (besides, “physics says so”) but I haven’t come across them yet.
So the nice mathematical alternative is to make the joint state of two qubits $ v,w$ the tensor product $ v \otimes w$. For a review of the basic properties of tensors and multilinear maps, see our post on the subject. Suffice it for now to remind the reader that the basis of the tensor space $ U \otimes V$ consists of all the tensors of the basis elements of the pieces $ U$ and $ V$: $ u_i \otimes v_j$. As such, the dimension of $ U \otimes V$ is the product of the dimensions $ \text{dim}(U) \text{dim}(V)$.
As a consequence of this and the fact that all $ \mathbb{C}$-vector spaces of the same dimension are the same (isomorphic), the state space of a set of $ n$ qubits can be identified with $ \mathbb{C}^{2^n}$. This is one way to see why quantum computing has the potential to be strictly more powerful than classical computing: $ n$ qubits provide a state space with $ 2^n$ coefficients, each of which is a complex number. With classical probabilistic computing we only get $ n$ “coefficients.” This isn’t a proof that quantum computing is more powerful, but a wink and a nudge that it could be.
While most of the time we’ll just write our states in terms of tensors (using the $ \otimes$ symbol), we could write out the vector representation of $ v \otimes w$ in terms of the vectors $ v = (v_1, v_2), w=(w_1, w_2)$. It’s just $ (v_1w_1, v_1w_2, v_2w_1, v_2w_2)$, with the obvious generalization to vectors of any dimension. This already fixes our earlier problem with norms: the norm of a tensor of two vectors is the product of the two norms. So tensors of unit vectors are unit vectors. Moreover, if you measure the first qubit, that just sets the $ v_1, v_2$ above to zero or one, leaving a joint state that is still a valid
Likewise, given two linear maps $ A, B$, we can describe the map $ A \otimes B$ on the tensor space both in terms of pure tensors ($ (A \otimes B)(v \otimes w) = Av \otimes Bw$) and in terms of a matrix. In the same vein as the representation for vectors, the matrix corresponding to $ A \otimes B$ is
One of the strange things about tensor products, which very visibly manifests itself in “strange quantum behavior,” is that not every vector in a tensor space can be represented as a single tensor product of some vectors. Let’s work with an example: $ \mathbb{C}^2 \otimes \mathbb{C}^2$, and denote by $ e_0, e_1$ the computational basis vectors (the same letters are used for each copy of $ \mathbb{C}^2$). Sometimes you’ll get a vector like
And if you’re lucky you’ll notice that this can be factored and written as $ \frac{1}{\sqrt{2}}(e_0 + e_1) \otimes e_0$. Other times, though, you’ll get a vector like
And it’s a deep fact that this cannot be factored into a tensor product of two vectors (prove it as an exercise). If a vector $ v$ in a tensor space can be written as a single tensor product of vectors, we call $ v$ a pure tensor. Otherwise, using some physics lingo, we call the state represented by $ v$ entangled. So if you did the exercise you proved that not all tensors are pure tensors, or equivalently that there exist entangled quantum states. The latter sounds so much more impressive. We’ll see in a future post why these entangled states are so important in quantum computing.
Now we need to explain how to extend gates and qubit measurements to state spaces with multiple qubits. The first is easy: just as we often restrict our classical gates to a few bits (like the AND of two bits), we restrict multi-qubit quantum gates to operate on at most three qubits.
Definition: A quantum gate $ G$ is a unitary map $ \mathbb{C}^{2^n} \to \mathbb{C}^{2^n}$ where $ n$ is at most 3, (recall, $ (\mathbb{C}^2)^{\otimes 3} = \mathbb{C}^{2^3}$ is the state space for 3 qubits).
Now let’s see how to implement AND and OR for two qubits. You might be wondering why we need three qubits in the definition above, and, perhaps surprisingly, we’ll see that AND and OR require us to work with three qubits.
Because how would one compute an AND of two qubits? Taking a naive approach from how we did the quantum NOT, we would label $ e_0$ as “false” and $ e_1$ as “true,” and we’d want to map $ e_1 \otimes e_1 \mapsto e_1$ and all other possibilities to $ e_0$. The main problem is that this is not an invertible function! Remember, all quantum operations are unitary matrices and all unitary matrices have inverses, so we have to model AND and OR as an invertible operation. We also have a “type error,” since the output is not even in the same vector space as the input, but any way to fix that would still run into the invertibility problem.
The way to deal with this is to add an extra “scratch work” qubit that is used for nothing else except to make the operation invertible. So now say we have three qubits $ a, b, c$, and we want to compute $ a$ AND $ b$ in the sensible way described above. What we do is map
$ \displaystyle a \otimes b \otimes c \mapsto a \otimes b \otimes (c \oplus (a \wedge b))$
Here $ a \wedge b$ is the usual AND (where we interpret, e.g., $ e_1 \wedge e_0 = e_0$), and $ \oplus$ is the exclusive or operation on bits. It’s clear that this mapping makes sense for “bits” (the true/false interpretation of basis vectors) and so we can extend it to a linear map by writing down the matrix.
This gate is often called the Toffoli gate by physicists, but we’ll just call it the (quantum) AND gate. Note that the column $ ijk$ represents the input $ e_i \otimes e_j \otimes e_k$, and the 1 in that column denotes the row whose label is the output. In particular, if we want to do an AND then we’ll ensure the “scratch work” qubit is $ e_0$, so we can ignore half the columns above where the third qubit is 1. The reader should write down the analogous construction for a quantum OR.
From now on, when we’re describing a basis state like $ e_1 \otimes e_0 \otimes e_1$, we’ll denote it as $ e_{101}$, and more generally when $ i$ is a nonnegative integer or a binary string we’ll denote the basis state as $ e_i$. We’re taking advantage of the correspondence between the $ 2^n$ binary strings and the $ 2^n$ basis states, and it compactifies notation.
Once we define a quantum circuit, it will be easy to show that using quantum AND’s, OR’s and NOT’s, we can achieve any computation that a classical circuit can.
We have one more issue we’d like to bring up before we define quantum circuits. We’re being a bit too slick when we say we’re working with “at most three qubits.” If we have ten qubits, potentially all entangled up in a weird way, how can we apply a mapping to only some of those qubits? Indeed, we only defined AND for $ \mathbb{C}^8$, so how can we extend that to an AND of three qubits sitting inside any $ \mathbb{C}^{2^n}$ we please? The answer is to apply the Kronecker product with the identity matrix appropriately. Let’s do a simple example of this to make everything stick.
Say I want to apply the quantum NOT gate to a qubit $ v$, and I have four other qubits $ w_1, w_2, w_3, w_4$ so that they’re all in the joint state $ x = v \otimes w_1 \otimes w_2 \otimes w_3 \otimes w_4$. I form the NOT gate, which I’ll call $ A$, and then I apply the gate $ A \otimes I_{2^4}$ to $ x$ (since there are 4 of the $ w_i$). This will compute the tensor $ Av \otimes I_2 w_1 \otimes I_2 w_2 \otimes I_2 w_3 \otimes I_2 w_4$, as desired.
In particular, you can represent a gate that depends on only 3 qubits by writing down the 3×3 matrix and the three indices it operates on. Note that this requires only 12 (possibly complex) numbers to write down, and so it takes “constant space” to represent a single gate.
Quantum Circuits
Here we are at the definition of a quantum circuit.
Definition: A quantum circuit is a list $ G_1, \dots, G_T$ of $ 2^m \times 2^m$ unitary matrices, such that each $ G_i$ depends on at most 3 qubits.
We’ll write down what it means to “compute” something with a quantum circuit, but for now we can imagine drawing it like a usual circuit. We write the input state as some unit vector $ x \in C^{2^n}$ (which may or may not be a pure tensor), each qubit making up the vector is associated to a “wire,” and at each step we pick three of the wires, send them to the next quantum gate $ G_i$, and use the three output wires for further computations. The final output is the matrix product applied to the input $ G_T \dots G_1x$. We imagine that each gate takes only one step to compute (recall, in our first post one “step” was a photon flying through a special material, so it’s not like we have to multiply these matrices by hand).
So now we have to say how a quantum circuit could solve a problem. At all levels of mathematical maturity we should have some idea how a regular circuit solves a problem: there is some distinguished output wire or set of wires containing the answer. For a quantum circuit it’s basically the same, except that at the end of the circuit we get a single quantum state (a tensor in this big vector space), and we just measure that state. Like the case of a single qubit, if the vector has coordinates $ x = (x_1, \dots, x_{2^n})$, they must satisfy $ \sum_i |x_i|^2 = 1$, and the probability of the measurement producing index $ j$ is $ |x_j|^2$. The result of that measurement is an integer (some classical bits) that represent our answer. As a side effect, the vector $ x$ is mutated into the basis state $ e_j$. As we’ve said we may need to repeat a quantum computation over and over to get a good answer with high probability, so we can imagine that a quantum circuit is used as some subroutine in a larger (otherwise classical) algorithm that allows for pre- and post-processing on the quantum part.
The final caveat is that we allow one to include as many scratchwork qubits as one needs in their circuit. This makes it possible already to simulate any classical circuit using a quantum circuit. Let’s prove it as a theorem.
Theorem: Given a classical circuit $ C$ with a single output bit, there is a quantum circuit $ D$ that computes the same function.
Proof. Let $ x$ be a binary string input to $ C$, and suppose that $ C$ has $ s$ gates $ g_1, \dots, g_s$, each being either AND, OR, or NOT, and with $ g_s$ being the output gate. To construct $ D$, we can replace every $ g_i$ with their quantum counterparts $ G_i$. Recall that this takes $ e_{b_1b_20} \mapsto e_{b_1b_2(g_i(b_1, b_2))}$. And so we need to add a single scratchwork qubit for each one (really we only need it for the ANDs and ORs, but who cares). This means that our start state is $ e_{x} \otimes e_{0^s} = e_{x0^s}$. Really, we need one of these gates $ G_i$ for each wire going out of the classical gate $ g_i$, but with some extra tricks one can do it with a single quantum gate that uses multiple scratchwork qubits. The crucial thing to note is that the state vector is always a basis vector!
If we call $ z$ the contents of all the scratchwork after the quantum circuit described above runs and $ z_0$ the initial state of the scratchwork, then what we did was extend the function $ x \mapsto C(x)$ to a function $ e_{xz_0} \mapsto e_{xz}$. In particular, one of the bits in the $ z$ part is the output of the last gate of $ C$, and everything is 0-1 valued. So we can measure the state vector, get the string $ xz$ and inspect the bit of $ z$ which corresponds to the output wire of the final gate of the original circuit $ C$. This is your answer.
$ \square$
It should be clear that the single output bit extends to the general case easily. We can split a circuit with lots of output bits into a bunch of circuits with single output bits in the obvious way and combine the quantum versions together.
Next time we’ll finally look at our first quantum algorithms. And along the way we’ll see some more significant quantum operations that make use of the properties that make the quantum world interesting. Until then!
In the last post in this series we saw some simple examples of linear programs, derived the concept of a dual linear program, and saw the duality theorem and the complementary slackness conditions which give a rough sketch of the stopping criterion for an algorithm. This time we’ll go ahead and write this algorithm for solving linear programs, and next time we’ll apply the algorithm to an industry-strength version of the nutrition problem we saw last time. The algorithm we’ll implement is called the simplex algorithm. It was the first algorithm for solving linear programs, invented in the 1940’s by George Dantzig, and it’s still the leading practical algorithm, and it was a key part of a Nobel Prize. It’s by far one of the most important algorithms ever devised.
The simplex algorithm can solve any kind of linear program, but it only accepts a special form of the program as input. So first we have to do some manipulations. Recall that the primal form of a linear program was the following minimization problem.
$ \min \left \langle c, x \right \rangle \\ \textup{s.t. } Ax \geq b, x \geq 0$
where the brackets mean “dot product.” And its dual is
$ \max \left \langle y, b \right \rangle \\ \textup{s.t. } A^Ty \leq c, y \geq 0$
The linear program can actually have more complicated constraints than just the ones above. In general, one might want to have “greater than” and “less than” constraints in the same problem. It turns out that this isn’t any harder, and moreover the simplex algorithm only uses equality constraints, and with some finicky algebra we can turn any set of inequality or equality constraints into a set of equality constraints.
We’ll call our goal the “standard form,” which is as follows:
$ \max \left \langle c, x \right \rangle \\ \textup{s.t. } Ax = b, x \geq 0$
It seems impossible to get the usual minimization/maximization problem into standard form until you realize there’s nothing stopping you from adding more variables to the problem. That is, say we’re given a constraint like:
$ \displaystyle x_7 + x_3 \leq 10,$
we can add a new variable $ \xi$, called a slack variable, so that we get an equality:
$ \displaystyle x_7 + x_3 + \xi = 10$
And now we can just impose that $ \xi \geq 0$. The idea is that $ \xi$ represents how much “slack” there is in the inequality, and you can always choose it to make the condition an equality. So if the equality holds and the variables are nonnegative, then the $ x_i$ will still satisfy their original inequality. For “greater than” constraints, we can do the same thing but subtract a nonnegative variable. Finally, if we have a minimization problem “$ \min z$” we can convert it to $ \max -z$.
So, to combine all of this together, if we have the following linear program with each kind of constraint,
We can add new variables $ \xi_1, \xi_2$, and write it as
By defining the vector variable $ x = (x_1, x_2, x_3, \xi_1, \xi_2)$ and $ c = (-1,-1,-1,0,0)$ and $ A$ to have $ -1, 0, 1$ as appropriately for the new variables, we see that the system is written in standard form.
This is the kind of tedious transformation we can automate with a program. Assuming there are $ n$ variables, the input consists of the vector $ c$ of length $ n$, and three matrix-vector pairs $ (A, b)$ representing the three kinds of constraints. It’s a bit annoying to describe, but the essential idea is that we compute a rectangular “identity” matrix whose diagonal entries are $ \pm 1$, and then join this with the original constraint matrix row-wise. The reader can see the full implementation in the Github repository for this post, though we won’t use this particular functionality in the algorithm that follows.
There are some other additional things we could do: for example there might be some variables that are completely unrestricted. What you do in this case is take an unrestricted variable $ z$ and replace it by the difference of two unrestricted variables $ z’ – z”$. For simplicity we’ll ignore this, but it would be a fruitful exercise for the reader to augment the function to account for these.
What happened to the slackness conditions?
The “standard form” of our linear program raises an obvious question: how can the complementary slackness conditions make sense if everything is an equality? It turns out that one can redo all the work one did for linear programs of the form we gave last time (minimize w.r.t. greater-than constraints) for programs in the new “standard form” above. We even get the same complementary slackness conditions! If you want to, you can do this entire routine quite a bit faster if you invoke the power of Lagrangians. We won’t do that here, but the tool shows up as a way to work with primal-dual conversions in many other parts of mathematics, so it’s a good buzzword to keep in mind.
In our case, the only difference with the complementary slackness conditions is that one of the two is trivial: $ \left \langle y^*, Ax^* – b \right \rangle = 0$. This is because if our candidate solution $ x^*$ is feasible, then it will have to satisfy $ Ax = b$ already. The other one, that $ \left \langle x^*, A^Ty^* – c \right \rangle = 0$, is the only one we need to worry about.
Again, the complementary slackness conditions give us inspiration here. Recall that, informally, they say that when a variable is used at all, it is used as much as it can be to fulfill its constraint (the corresponding dual constraint is tight). So a solution will correspond to a choice of some variables which are either used or not, and a choice of nonzero variables will correspond to a solution. We even saw this happen in the last post when we observed that broccoli trumps oranges. If we can get a good handle on how to navigate the set of these solutions, then we’ll have a nifty algorithm.
Let’s make this official and lay out our assumptions.
Extreme points and basic solutions
Remember that the graphical way to solve a linear program is to look at the line (or hyperplane) given by $ \langle c, x \rangle = q$ and keep increasing $ q$ (or decreasing it, if you are minimizing) until the very last moment when this line touches the region of feasible solutions. Also recall that the “feasible region” is just the set of all solutions to $ Ax = b$, that is the solutions that satisfy the constraints. We imagined this picture:
The constraints define a convex area of “feasible solutions.” Image source: Wikipedia.
With this geometric intuition it’s clear that there will always be an optimal solution on a vertex of the feasible region. These points are called extreme points of the feasible region. But because we will almost never work in the plane again (even introducing slack variables makes us relatively high dimensional!) we want an algebraic characterization of these extreme points.
If you have a little bit of practice with convex sets the correct definition is very natural. Recall that a set $ X$ is convex if for any two points $ x, y \in X$ every point on the line segment between $ x$ and $ y$ is also in $ X$. An algebraic way to say this (thinking of these points now as vectors) is that every point $ \delta x + (1-\delta) y \in X$ when $ 0 \leq \delta \leq 1$. Now an extreme point is just a point that isn’t on the inside of any such line, i.e. can’t be written this way for $ 0 < \delta < 1$. For example,
A convex set with extremal points in red. Image credit Wikipedia.
Another way to say this is that if $ z$ is an extreme point then whenever $ z$ can be written as $ \delta x + (1-\delta) y$ for some $ 0 < \delta < 1$, then actually $ x=y=z$. Now since our constraints are all linear (and there are a finite number of them) they won’t define a convex set with weird curves like the one above. This means that there are a finite number of extreme points that just correspond to the intersections of some of the constraints. So there are at most $ 2^n$ possibilities.
Indeed we want a characterization of extreme points that’s specific to linear programs in standard form, “$ \max \langle c, x \rangle \textup{ s.t. } Ax=b, x \geq 0$.” And here is one.
Definition: Let $ A$ be an $ m \times n$ matrix with $ n \geq m$. A solution $ x$ to $ Ax=b$ is called basic if at most $ m$ of its entries are nonzero.
The reason we call it “basic” is because, under some mild assumptions we describe below, a basic solution corresponds to a vector space basis of $ \mathbb{R}^m$. Which basis? The one given by the $ m$ columns of $ A$ used in the basic solution. We don’t need to talk about bases like this, though, so in the event of a headache just think of the basis as a set $ B \subset \{ 1, 2, \dots, n \}$ of size $ m$ corresponding to the nonzero entries of the basic solution.
Indeed, what we’re doing here is looking at the matrix $ A_B$ formed by taking the columns of $ A$ whose indices are in $ B$, and the vector $ x_B$ in the same way, and looking at the equation $ A_Bx_B = b$. If all the parts of $ x$ that we removed were zero then this will hold if and only if $ Ax=b$. One might worry that $ A_B$ is not invertible, so we’ll go ahead and assume it is. In fact, we’ll assume that every set of $ m$ columns of $ A$ forms a basis and that the rows of $ A$ are also linearly independent. This isn’t without loss of generality because if some rows or columns are not linearly independent, we can remove the offending constraints and variables without changing the set of solutions (this is why it’s so nice to work with the standard form).
Moreover, we’ll assume that every basic solution has exactly $ m$ nonzero variables. A basic solution which doesn’t satisfy this assumption is called degenerate, and they’ll essentially be special corner cases in the simplex algorithm. Finally, we call a basic solution feasible if (in addition to satisfying $ Ax=b$) it satisfies $ x \geq 0$. Now that we’ve made all these assumptions it’s easy to see that choosing $ m$ nonzero variables uniquely determines a basic feasible solution. Again calling the sub-matrix $ A_B$ for a basis $ B$, it’s just $ x_B = A_B^{-1}b$. Now to finish our characterization, we just have to show that under the same assumptions basic feasible solutions are exactly the extremal points of the feasible region.
Proposition: A vector $ x$ is a basic feasible solution if and only if it’s an extreme point of the set $ \{ x : Ax = b, x \geq 0 \}$.
Proof. For one direction, suppose you have a basic feasible solution $ x$, and say we write it as $ x = \delta y + (1-\delta) z$ for some $ 0 < \delta < 1$. We want to show that this implies $ y = z$. Since all of these points are in the feasible region, all of their coordinates are nonnegative. So whenever a coordinate $ x_i = 0$ it must be that both $ y_i = z_i = 0$. Since $ x$ has exactly $ n-m$ zero entries, it must be that $ y, z$ both have at least $ n-m$ zero entries, and hence $ y,z$ are both basic. By our non-degeneracy assumption they both then have exactly $ m$ nonzero entries. Let $ B$ be the set of the nonzero indices of $ x$. Because $ Ay = Az = b$, we have $ A(y-z) = 0$. Now $ y-z$ has all of its nonzero entries in $ B$, and because the columns of $ A_B$ are linearly independent, the fact that $ A_B(y-z) = 0$ implies $ y-z = 0$.
In the other direction, suppose that you have some extreme point $ x$ which is feasible but not basic. In other words, there are more than $ m$ nonzero entries of $ x$, and we’ll call the indices $ J = \{ j_1, \dots, j_t \}$ where $ t > m$. The columns of $ A_J$ are linearly dependent (since they’re $ t$ vectors in $ \mathbb{R}^m$), and so let $ \sum_{i=1}^t z_{j_i} A_{j_i}$ be a nontrivial linear combination of the columns of $ A$. Add zeros to make the $ z_{j_i}$ into a length $ n$ vector $ z$, so that $ Az = 0$. Now
And if we pick $ \varepsilon$ sufficiently small $ x \pm \varepsilon z$ will still be nonnegative, because the only entries we’re changing of $ x$ are the strictly positive ones. Then $ x = \delta (x + \varepsilon z) + (1 – \delta) \varepsilon z$ for $ \delta = 1/2$, but this is very embarrassing for $ x$ who was supposed to be an extreme point. $ \square$
Now that we know extreme points are the same as basic feasible solutions, we need to show that any linear program that has some solution has a basic feasible solution. This is clear geometrically: any time you have an optimum it has to either lie on a line or at a vertex, and if it lies on a line then you can slide it to a vertex without changing its value. Nevertheless, it is a useful exercise to go through the algebra.
Theorem. Whenever a linear program is feasible and bounded, it has a basic feasible solution.
Proof. Let $ x$ be an optimal solution to the LP. If $ x$ has at most $ m$ nonzero entries then it’s a basic solution and by the non-degeneracy assumption it must have exactly $ m$ nonzero entries. In this case there’s nothing to do, so suppose that $ x$ has $ r > m$ nonzero entries. It can’t be a basic feasible solution, and hence is not an extreme point of the set of feasible solutions (as proved by the last theorem). So write it as $ x = \delta y + (1-\delta) z$ for some feasible $ y \neq z$ and $ 0 < \delta < 1$.
The only thing we know about $ x$ is it’s optimal. Let $ c$ be the cost vector, and the optimality says that $ \langle c,x \rangle \geq \langle c,y \rangle$, and $ \langle c,x \rangle \geq \langle c,z \rangle$. We claim that in fact these are equal, that $ y, z$ are both optimal as well. Indeed, say $ y$ were not optimal, then
Which can be rearranged to show that $ \langle c,y \rangle < \langle c, z \rangle$. Unfortunately for $ x$, this implies that it was not optimal all along:
An identical argument works to show $ z$ is optimal, too. Now we claim we can use $ y,z$ to get a new solution that has fewer than $ r$ nonzero entries. Once we show this we’re done: inductively repeat the argument with the smaller solution until we get down to exactly $ m$ nonzero variables. As before we know that $ y,z$ must have at least as many zeros as $ x$. If they have more zeros we’re done. And if they have exactly as many zeros we can do the following trick. Write $ w = \gamma y + (1- \gamma)z$ for a $ \gamma \in \mathbb{R}$ we’ll choose later. Note that no matter the $ \gamma$, $ w$ is optimal. Rewriting $ w = z + \gamma (y-z)$, we just have to pick a $ \gamma$ that ensures one of the nonzero coefficients of $ z$ is zeroed out while maintaining nonnegativity. Indeed, we can just look at the index $ i$ which minimizes $ z_i / (y-z)_i$ and use $ \delta = – z_i / (y-z)_i$. $ \square$.
So we have an immediate (and inefficient) combinatorial algorithm: enumerate all subsets of size $ m$, compute the corresponding basic feasible solution $ x_B = A_B^{-1}b$, and see which gives the biggest objective value. The problem is that, even if we knew the value of $ m$, this would take time $ n^m$, and it’s not uncommon for $ m$ to be in the tens or hundreds (and if we don’t know $ m$ the trivial search is exponential).
So we have to be smarter, and this is where the simplex tableau comes in.
The simplex tableau
Now say you have any basis $ B$ and any feasible solution $ x$. For now $ x$ might not be a basic solution, and even if it is, its basis of nonzero entries might not be the same as $ B$. We can decompose the equation $ Ax = b$ into the basis part and the non basis part:
$ A_Bx_B + A_{B’} x_{B’} = b$
and solving the equation for $ x_B$ gives
$ x_B = A^{-1}_B(b – A_{B’} x_{B’})$
It may look like we’re making a wicked abuse of notation here, but both $ A_Bx_B$ and $ A_{B’}x_{B’}$ are vectors of length $ m$ so the dimensions actually do work out. Now our feasible solution $ x$ has to satisfy $ Ax = b$, and the entries of $ x$ are all nonnegative, so it must be that $ x_B \geq 0$ and $ x_{B’} \geq 0$, and by the equality above $ A^{-1}_B (b – A_{B’}x_{B’}) \geq 0$ as well. Now let’s write the maximization objective $ \langle c, x \rangle$ by expanding it first in terms of the $ x_B, x_{B’}$, and then expanding $ x_B$.
If we want to maximize the objective, we can just maximize this last line. There are two cases. In the first, the vector $ c_{B’} – (A^{-1}_B A_{B’})^T c_B \leq 0$ and $ A_B^{-1}b \geq 0$. In the above equation, this tells us that making any component of $ x_{B’}$ bigger will decrease the overall objective. In other words, $ \langle c, x \rangle \leq \langle c_B, A_B^{-1}b \rangle$. Picking $ x = A_B^{-1}b$ (with zeros in the non basis part) meets this bound and hence must be optimal. In other words, no matter what basis $ B$ we’ve chosen (i.e., no matter the candidate basic feasible solution), if the two conditions hold then we’re done.
Now the crux of the algorithm is the second case: if the conditions aren’t met, we can pick a positive index of $ c_{B’} – (A_B^{-1}A_{B’})^Tc_B$ and increase the corresponding value of $ x_{B’}$ to increase the objective value. As we do this, other variables in the solution will change as well (by decreasing), and we have to stop when one of them hits zero. In doing so, this changes the basis by removing one index and adding another. In reality, we’ll figure out how much to increase ahead of time, and the change will correspond to a single elementary row-operation in a matrix.
Indeed, the matrix we’ll use to represent all of this data is called a tableau in the literature. The columns of the tableau will correspond to variables, and the rows to constraints. The last row of the tableau will maintain a candidate solution $ y$ to the dual problem. Here’s a rough picture to keep the different parts clear while we go through the details.
But to make it work we do a slick trick, which is to “left-multiply everything” by $ A_B^{-1}$. In particular, if we have an LP given by $ c, A, b$, then for any basis it’s equivalent to the LP given by $ c, A_B^{-1}A, A_{B}^{-1} b$ (just multiply your solution to the new program by $ A_B$ to get a solution to the old one). And so the actual tableau will be of this form.
When we say it’s in this form, it’s really only true up to rearranging columns. This is because the chosen basis will always be represented by an identity matrix (as it is to start with), so to find the basis you can find the embedded identity sub-matrix. In fact, the beginning of the simplex algorithm will have the initial basis sitting in the last few columns of the tableau.
Let’s look a little bit closer at the last row. The first portion is zero because $ A_B^{-1}A_B$ is the identity. But furthermore with this $ A_B^{-1}$ trick the dual LP involves $ A_B^{-1}$ everywhere there’s a variable. In particular, joining all but the last column of the last row of the tableau, we have the vector $ c – A_B^T(A_B^{-1})^T c$, and setting $ y = A_B^{-1}c_B$ we get a candidate solution for the dual. What makes the trick even slicker is that $ A_B^{-1}b$ is already the candidate solution $ x_B$, since $ (A_B^{-1}A)_B^{-1}$ is the identity. So we’re implicitly keeping track of two solutions here, one for the primal LP, given by the last column of the tableau, and one for the dual, contained in the last row of the tableau.
I told you the last row was the dual solution, so why all the other crap there? This is the final slick in the trick: the last row further encodes the complementary slackness conditions. Now that we recognize the dual candidate sitting there, the complementary slackness conditions simply ask for the last row to be non-positive (this is just another way of saying what we said at the beginning of this section!). You should check this, but it gives us a stopping criterion: if the last row is non-positive then stop and output the last column.
The simplex algorithm
Now (finally!) we can describe and implement the simplex algorithm in its full glory. Recall that our informal setup has been:
Find an initial basic feasible solution, and set up the corresponding tableau.
Find a positive index of the last row, and increase the corresponding variable (adding it to the basis) just enough to make another variable from the basis zero (removing it from the basis).
Repeat step 2 until the last row is nonpositive.
Output the last column.
This is almost correct, except for some details about how increasing the corresponding variables works. What we’ll really do is represent the basis variables as pivots (ones in the tableau) and then the first 1 in each row will be the variable whose value is given by the entry in the last column of that row. So, for example, the last entry in the first row may be the optimal value for $ x_5$, if the fifth column is the first entry in row 1 to have a 1.
As we describe the algorithm, we’ll illustrate it running on a simple example. In doing this we’ll see what all the different parts of the tableau correspond to from the previous section in each step of the algorithm.
Spoiler alert: the optimum is $ x_1 = 2, x_2 = 1$ and the value of the max is 8.
So let’s be more programmatically formal about this. The main routine is essentially pseudocode, and the difficulty is in implementing the helper functions
def simplex(c, A, b):
tableau = initialTableau(c, A, b)
while canImprove(tableau):
pivot = findPivotIndex(tableau)
pivotAbout(tableau, pivot)
return primalSolution(tableau), objectiveValue(tableau)
Let’s start with the initial tableau. We’ll assume the user’s inputs already include the slack variables. In particular, our example data before adding slack is
c = [3, 2]
A = [[1, 2], [1, -1]]
b = [4, 1]
And after adding slack:
c = [3, 2, 0, 0]
A = [[1, 2, 1, 0],
[1, -1, 0, 1]]
b = [4, 1]
Now to set up the initial tableau we need an initial feasible solution in mind. The reader is recommended to work this part out with a pencil, since it’s much easier to write down than it is to explain. Since we introduced slack variables, our initial feasible solution (basis) $ B$ can just be $ (0,0,1,1)$. And so $ x_B$ is just the slack variables, $ c_B$ is the zero vector, and $ A_B$ is the 2×2 identity matrix. Now $ A_B^{-1}A_{B’} = A_{B’}$, which is just the original two columns of $ A$ we started with, and $ A_B^{-1}b = b$. For the last row, $ c_B$ is zero so the part under $ A_B^{-1}A_B$ is the zero vector. The part under $ A_B^{-1}A_{B’}$ is just $ c_{B’} = (3,2)$.
Rather than move columns around every time the basis $ B$ changes, we’ll keep the tableau columns in order of $ (x_1, \dots, x_n, \xi_1, \dots, \xi_m)$. In other words, for our example the initial tableau should look like this.
So implementing initialTableau is just a matter of putting the data in the right place.
def initialTableau(c, A, b):
tableau = [row[:] + [x] for row, x in zip(A, b)]
tableau.append(c[:] + [0])
return tableau
As an aside: in the event that we don’t start with the trivial basic feasible solution of “trivially use the slack variables,” we’d have to do a lot more work in this function. Next, the primalSolution() and objectiveValue() functions are simple, because they just extract the encoded information out from the tableau (some helper functions are omitted for brevity).
def primalSolution(tableau):
# the pivot columns denote which variables are used
columns = transpose(tableau)
indices = [j for j, col in enumerate(columns[:-1]) if isPivotCol(col)]
return list(zip(indices, columns[-1]))
def objectiveValue(tableau):
return -(tableau[-1][-1])
Similarly, the canImprove() function just checks if there’s a nonnegative entry in the last row
def canImprove(tableau):
lastRow = tableau[-1]
return any(x > 0 for x in lastRow[:-1])
Let’s run the first loop of our simplex algorithm. The first step is checking to see if anything can be improved (in our example it can). Then we have to find a pivot entry in the tableau. This part includes some edge-case checking, but if the edge cases aren’t a problem then the strategy is simple: find a positive entry corresponding to some entry $ j$ of $ B’$, and then pick an appropriate entry in that column to use as the pivot. Pivoting increases the value of $ x_j$ (from zero) to whatever is the largest we can make it without making some other variables become negative. As we’ve said before, we’ll stop increasing $ x_j$ when some other variable hits zero, and we can compute which will be the first to do so by looking at the current values of $ x_B = A_B^{-1}b$ (in the last column of the tableau), and seeing how pivoting will affect them. If you stare at it for long enough, it becomes clear that the first variable to hit zero will be the entry $ x_i$ of the basis for which $ x_i / A_{i,j}$ is minimal (and $ A_{i,j}$ has to be positve). This is because, in order to maintain the linear equalities, every entry of $ x_B$ will be decreased by that value during a pivot, and we can’t let any of the variables become negative.
All of this results in the following function, where we have left out the degeneracy/unboundedness checks.
[UPDATE 2018-04-21]: The pivot choices are not as simple as I thought at the time I wrote this. See the discussion on this issue, but the short story is that I was increasing the variable too much, and to fix it it’s easier to update the pivot column choice to be the smallest positive entry of the last row. The code on github is updated to reflect that, but this post will remain unchanged.
def findPivotIndex(tableau):
# pick first nonzero index of the last row
column = [i for i,x in enumerate(tableau[-1][:-1]) if x > 0][0]
quotients = [(i, r[-1] / r[column]) for i,r in enumerate(tableau[:-1]) if r[column] > 0]
# pick row index minimizing the quotient
row = min(quotients, key=lambda x: x[1])[0]
return row, column
For our example, the minimizer is the $ (1,0)$ entry (second row, first column). Pivoting is just doing the usual elementary row operations (we covered this in a primer a while back on row-reduction). The pivot function we use here is no different, and in particular mutates the list in place.
def pivotAbout(tableau, pivot):
i,j = pivot
pivotDenom = tableau[i][j]
tableau[i] = [x / pivotDenom for x in tableau[i]]
for k,row in enumerate(tableau):
if k != i:
pivotRowMultiple = [y * tableau[k][j] for y in tableau[i]]
tableau[k] = [x - y for x,y in zip(tableau[k], pivotRowMultiple)]
And in our example pivoting around the chosen entry gives the new tableau.
In particular, $ B$ is now $ (1,0,1,0)$, since our pivot removed the second slack variable $ \xi_2$ from the basis. Currently our solution has $ x_1 = 1, \xi_1 = 3$. Notice how the identity submatrix is still sitting in there, the columns are just swapped around.
There’s still a positive entry in the bottom row, so let’s continue. The next pivot is (0,1), and pivoting around that entry gives the following tableau:
And because all of the entries in the bottom row are negative, we’re done. We read off the solution as we described, so that the first variable is 2 and the second is 1, and the objective value is the opposite of the bottom right entry, 8.
To see all of the source code, including the edge-case-checking we left out of this post, see the Github repository for this post.
Obvious questions and sad answers
An obvious question is: what is the runtime of the simplex algorithm? Is it polynomial in the size of the tableau? Is it even guaranteed to stop at some point? The surprising truth is that nobody knows the answer to all of these questions! Originally (in the 1940’s) the simplex algorithm actually had an exponential runtime in the worst case, though this was not known until 1972. And indeed, to this day while some variations are known to terminate, no variation is known to have polynomial runtime in the worst case. Some of the choices we made in our implementation (for example, picking the first column with a positive entry in the bottom row) have the potential to cycle, i.e., variables leave and enter the basis without changing the objective at all. Doing something like picking a random positive column, or picking the column which will increase the objective value by the largest amount are alternatives. Unfortunately, every single pivot-picking rule is known to give rise to exponential-time simplex algorithms in the worst case (in fact, this was discovered as recently as 2011!). So it remains open whether there is a variant of the simplex method that runs in guaranteed polynomial time.
But then, in a stunning turn of events, Leonid Khachiyan proved in the 70’s that in fact linear programs can always be solved in polynomial time, via a completely different algorithm called the ellipsoid method. Following that was a method called the interior point method, which is significantly more efficient. Both of these algorithms generalize to problems that are harder than linear programming as well, so we will probably cover them in the distant future of this blog.
Despite the celebratory nature of these two results, people still use the simplex algorithm for industrial applications of linear programming. The reason is that it’s much faster in practice, and much simpler to implement and experiment with.
The next obvious question has to do with the poignant observation that whole numbers are great. That is, you often want the solution to your problem to involve integers, and not real numbers. But adding the constraint that the variables in a linear program need to be integer valued (even just 0-1 valued!) is NP-complete. This problem is called integer linear programming, or just integer programming (IP). So we can’t hope to solve IP, and rightly so: the reader can verify easily that boolean satisfiability instances can be written as linear programs where each clause corresponds to a constraint.
This brings up a very interesting theoretical issue: if we take an integer program and just remove the integrality constraints, and solve the resulting linear program, how far away are the two solutions? If they’re close, then we can hope to give a good approximation to the integer program by solving the linear program and somehow turning the resulting solution back into an integer solution. In fact this is a very popular technique called LP-rounding. We’ll also likely cover that on this blog at some point.
Oh there’s so much to do and so little time! Until next time.
