:py:mod:`nonlocal_games.quantum_hedging`
========================================

.. py:module:: nonlocal_games.quantum_hedging

.. autoapi-nested-parse::

   Semidefinite programs for obtaining values of quantum hedging scenarios.



Module Contents
---------------

Classes
~~~~~~~

.. autoapisummary::

   nonlocal_games.quantum_hedging.QuantumHedging




.. py:class:: QuantumHedging(q_a, num_reps)


   Calculate optimal winning probabilities for hedging scenarios.

   Calculate the maximal and minimal winning probabilities for quantum
   hedging to occur in certain two-party scenarios :cite:`Arunachalam_2017_QuantumHedging, Molina_2012_Hedging`.

   .. rubric:: Examples

   This example illustrates the initial example of perfect hedging when Alice
   and Bob play two repetitions of the game where Alice prepares the maximally
   entangled state:

   .. math::
       u = \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle,

   and Alice applies the measurement operator defined by vector

   .. math::
       v = \cos(\pi/8)|00\rangle + \sin(\pi/8)|11\rangle.

   As was illustrated in :cite:`Molina_2012_Hedging`, the hedging value of the above scenario is
   :math:`\cos(\pi/8)^2 \approx 0.8536`

   >>> from numpy import kron, cos, sin, pi, sqrt, isclose
   >>> from toqito.states import basis
   >>> from toqito.nonlocal_games.quantum_hedging import QuantumHedging
   >>>
   >>> e_0, e_1 = basis(2, 0), basis(2, 1)
   >>> e_00, e_01 = kron(e_0, e_0), kron(e_0, e_1)
   >>> e_10, e_11 = kron(e_1, e_0), kron(e_1, e_1)
   >>>
   >>> alpha = 1 / sqrt(2)
   >>> theta = pi / 8
   >>> w_var = alpha * cos(theta) * e_00 + sqrt(1 - alpha ** 2) * sin(theta) * e_11
   >>>
   >>> l_1 = -alpha * sin(theta) * e_00 + sqrt(1 - alpha ** 2) * cos(theta) * e_11
   >>> l_2 = alpha * sin(theta) * e_10
   >>> l_3 = sqrt(1 - alpha ** 2) * cos(theta) * e_01
   >>>
   >>> q_1 = w_var * w_var.conj().T
   >>> q_0 = l_1 * l_1.conj().T + l_2 * l_2.conj().T + l_3 * l_3.conj().T
   >>> molina_watrous = QuantumHedging(q_0, 1)
   >>>
   >>> # cos(pi/8)**2 \approx 0.8536
   >>> '%.2f' % molina_watrous.max_prob_outcome_a_primal()
   '0.85'

   .. note::
       You do not need to use `'%.2f' %` when you use this function.

       We use this to format our output such that `doctest` compares the calculated output to the
       expected output upto two decimal points only. The accuracy of the solvers can calculate the
       `float` output to a certain amount of precision such that the value deviates after a few digits
       of accuracy.

   .. rubric:: References

   .. bibliography::
       :filter: docname in docnames

   .. py:method:: max_prob_outcome_a_primal()

      Compute the maximal probability for calculating outcome "a".

      The primal problem for the maximal probability of "a" is given as:

      .. math::

          \begin{equation}
              \begin{aligned}
                  \text{maximize:} \quad & \langle Q_{a_1} \otimes \ldots
                                           \otimes Q_{a_n}, X \rangle \\
              \text{subject to:} \quad & \text{Tr}_{\mathcal{Y}_1 \otimes
                                          \ldots \otimes \mathcal{Y}_n}(X) =
                                          I_{\mathcal{X}_1 \otimes \ldots
                                          \otimes \mathcal{X}_n},\\
                                          & X \in \text{Pos}(\mathcal{Y}_1
                                          \otimes \mathcal{X}_1 \otimes \ldots
                                          \otimes \mathcal{Y}_n \otimes
                                          \mathcal{X}_n)
              \end{aligned}
          \end{equation}

      :return: The optimal maximal probability for obtaining outcome "a".


   .. py:method:: max_prob_outcome_a_dual()

      Compute the maximal probability for calculating outcome "a".

      The dual problem for the maximal probability of "a" is given as:

      .. math::

          \begin{equation}
              \begin{aligned}
                  \text{minimize:} \quad & \text{Tr}(Y) \\
                  \text{subject to:} \quad & \pi \left(I_{\mathcal{Y}_1
                  \otimes \ldots \otimes \mathcal{Y}_n} \otimes Y \right)
                  \pi^* \geq Q_{a_1} \otimes \ldots \otimes Q_{a_n}, \\
                  & Y \in \text{Herm} \left(\mathcal{X} \otimes \ldots \otimes
                  \mathcal{X}_n \right)
              \end{aligned}
          \end{equation}

      :return: The optimal maximal probability for obtaining outcome "a".


   .. py:method:: min_prob_outcome_a_primal()

      Compute the minimal probability for calculating outcome "a".

      The primal problem for the minimal probability of "a" is given as:

      .. math::

          \begin{equation}
              \begin{aligned}
                  \text{minimize:} \quad & \langle Q_{a_1} \otimes \ldots
                                           \otimes Q_{a_n}, X \rangle \\
              \text{subject to:} \quad & \text{Tr}_{\mathcal{Y}_1 \otimes
                                          \ldots \otimes \mathcal{Y}_n}(X) =
                                          I_{\mathcal{X}_1 \otimes \ldots
                                          \otimes \mathcal{X}_n},\\
                                          & X \in \text{Pos}(\mathcal{Y}_1
                                          \otimes \mathcal{X}_1 \otimes \ldots
                                          \otimes \mathcal{Y}_n \otimes
                                          \mathcal{X}_n)
              \end{aligned}
          \end{equation}

      :return: The optimal minimal probability for obtaining outcome "a".


   .. py:method:: min_prob_outcome_a_dual()

      Compute the minimal probability for calculating outcome "a".

      The dual problem for the minimal probability of "a" is given as:

      .. math::

          \begin{equation}
              \begin{aligned}
                  \text{maximize:} \quad & \text{Tr}(Y) \\
                  \text{subject to:} \quad & \pi \left(I_{\mathcal{Y}_1
                  \otimes \ldots \otimes \mathcal{Y}_n} \otimes Y \right)
                  \pi^* \leq Q_{a_1} \otimes \ldots \otimes Q_{a_n}, \\
                  & Y \in \text{Herm} \left(\mathcal{X} \otimes \ldots \otimes
                  \mathcal{X}_n \right)
              \end{aligned}
          \end{equation}

      :return: The optimal minimal probability for obtaining outcome "a".