The (Sub/Sup) Majority Learning

Take a bitstring a, and an integer $\theta$. The oracle will respond 0 if $d(a,x)\leq \theta$  if queried with x and 1 otherwise. In other terms this oracle reply yes if two bitstrings agree on at least $\theta$ bit.

Here are the probability with a simple membership oracle (used with the usual phase kickback trick). Only even n gives an invertible matrix for the use of impatient learning.

For all these value, the optimum is achieved when the phase kickback takes value in {-1 ; 1}.

We have a function $f : \mathbb{Z}_2^n\rightarrow \mathbb{Z}_2$ that is True for only one $a \in \mathbb{Z}_2^n$. We also have a quantum oracle that returns the value of the function f. Classicaly, finding the value of $a$ will take n-1 steps. With the quantum version (Grover's Algorithm) it takes only $\sqrt{n}$ steps.

The idea (without explanations) is to take an equally distributed weights vector as input $x$, and to iterate a sequence of unitary operations during a certain amount of time.  After this given number of steps a measurement on the input vector is made, and is has been proved that we will observed the correct answer with probability 1. The number of steps is proportional to $\sqrt{n}$.

Bellow I put the picture representing the evolution (step after step) of the probability to observe the correct answer. You can see the oscillations that are well explained by geometrical description of this algorithm.

Python Code

Bellow you'll find the code associated to this example. You'll also be able to visualize the animation of the probabilities distribution according to the time.  The code is written in Python and requires Numpy to work. The associated functions classes are not useful for this simple case, but you'll understand in the coming post why I need them.

• Grover Algorithm - Python (run python grover.py)
  

Interesting fact:

If you use more than one correct possible answer (as in the case of perceptron learning) and a different matrix (namely with two -1 instead of only one), the behaviour could be close to the Grover's Algorithm. Since I was studying perceptron, I implemented a function that output an oracle given a certain threshold. If this threshold is 0, then only one correct answer exist. If it's one, then n+1 correct answers are possible ... but they are all in relation with the initial teacher.

If we fix the teacher, select a threshold of 1, and use the matrix H.diag(-1,1,...,1,-1).H and apply the same algorithm we get the following probability distribution for the teacher :

If you want to use my code, you have to put teachers=mat_teacher_bin(N,1) on line 36 and uncomment line 69

Explainations? For the moment none. But I plan to read the following paper : Improved bond in Oracle identification

]]> http://jice.lavocat.name/blog/2009/08/grovers-search-algorithm-in-python/feed/ 0 http://jice.lavocat.name/blog/2009/07/tout-sur-hadamard/ http://jice.lavocat.name/blog/2009/07/tout-sur-hadamard/#comments Mon, 20 Jul 2009 01:37:33 +0000 Jice http://jice.lavocat.name/blog/?p=712 Qui est donc ce fameux Hadamard dont tout le monde parle en informatique quantique? Jacques Hadamard est un mathématicien français (1865-1963) à qui l'on doit les célèbres matrices (encore un normalien).

Je vous conseille la biographie suivante sur Jacques Hadamard : Jacques Hadamard un mathématicien universel

Ainsi que ce site regroupant les propriétés des-dites matrices : http://chaos.if.uj.edu.pl/~karol/hadamard/index.php

The majority problem is equivalent to the perceptron learning. For each $a \in \mathbb{Z}_n$ define a function $m_a : \mathbb{Z}_2^N \rightarrow \mathbb{Z}_2$ :

$m_a(x)= \begin{cases} 1 & \text{ if } wt(a-x)\leq n/2 \\0 & \text{ otherwise } \end{cases}$

Where wt is the weight of a bit-string (number of 1).

Alternatively we can write : $m_a(x) = \Theta (n/2 - wt(x-a) )$.

The problem is : determine a given an access to answers from $m_a$.

We saw in the presentation (link available soon) that the classical complexity was $O(n)$ while the quantum complexity was $O(\sqrt{n})$. In the following we show the classical bound.

The classical algorithm

For the classical case we will use a dichotomic process as in the binary chop. Here is the algorithm :

• At the first step you look the result for 0...0. If it's 1 you take all the possible weights that could have the same result ($2^n / 2 = 2^{n-1}$ weights possible). If it's 0 you take the complementary set of bits (the same number of weights if n is even).
• After that, you choose a new vector in your possible set, with the biggest Hamming distance with the previous tested bit. Then you look the result. After that step you would be able to remove another time : $2^{n-1} / 2$ remaining weights .
• ... you repeat these steps n-1 times at all, in the end you have $2^n/2^{n-1} = 2$ possible states.
• In the end, you may have two bits that you will separate by choosing a special bit b.  This bit b will depend on the two possible states and differentiate them.
• And you are done in n steps.

Example for n=3

For n=3, let's take the example of the following target bit : w = 001.

• I first try 000. I get 1. So the possible weights are : 000 / 001 / 010 / 100
• Then I test 001. I get 1. So the remaining states are 001 / 000 ( 011 / 101 are not possible because they are not in the previous set)
• Finally I test 101. I get 1. So the only possible result is 001.

Here is a list of the ones I found interesting to read :

First a general definition of the context :  Jorge Castro -  How Many Query Superpositions Are Needed to Learn?

A  paper about lots of bounds : Rocco Servedio & Steven Gortler - Equivalences and separations between quantum and classical learnability

Chapter 6 of Ashley Montanaro - Structure randomness ...

A recent paper that tries to avoid quantum oracle queries : Alp Atici & Rocco Servedio - Quantum algorithms for learning and testing juntas

And finally a well-known chapter about technics for computing quantum complexities : Ethan Bernstein, Umesh Vazirani - Quantum complexity theory

$i\hbar\frac{\partial}{\partial t}\left|\Psi(t)\right>=H\left|\Psi(t)\right>$

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