python-openopt 0.38+svn1589-1 / usr / share / pyshared / openopt / solvers / HongKongOpt / qlcp.py

'''
Copyright (c) 2010 Enzo Michelangeli and IT Vision Ltd

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'''

from numpy import *
from scipy.linalg import lu_factor, lu_solve
from LCPSolve import LCPSolve

def qlcp(Q, e, A=None, b=None, Aeq=None, beq=None, lb=None, ub=None, QI=None):
    '''
    Minimizes e'x + 1/2 x'Q x subject to optional inequality, equality and
    box-bound (converted ti inequality) constraints.
    Note: x is NOT assumed to be non-negative by default.
    This quadratic solver works by converting the QP problem 
    into an LCP problem. It does well up to few hundred variables
    and dense problems (it doesn't take advantage of sparsity).
    If there are equality constraints, the problem may be feasible 
    even when Q is singular. If Q is not singular, it is possible to
    precompute its inverse and pass it as parameter QI (this is
    useful in SQP applications with approximation of the Hessian and
    its inverse, such as DFP or BFGS.
    Returns: x, the solution (or None in case of failure due to ray 
    termination in the LCP solver).
    '''
    nvars = Q.shape[0] # also e.shape[0]
    # convert lb and ub (if present) into Ax <=> b conditions, but
    # skip redundant rows: the ones where lb[i] == -Inf or ub[i] == Inf
    if lb != None:
        delmask = (lb != -Inf)
        addA = compress(delmask, eye(nvars), axis=0)
        addb = compress(delmask, lb, axis=0)
        A = vstack([A, -addA]) if A != None else -addA
        b = concatenate([b, -addb]) if b != None else -addb
    if ub != None:
        delmask = (ub != Inf)
        addA = compress(delmask, eye(nvars), axis=0)
        addb = compress(delmask, ub, axis=0)
        A = vstack([A, addA]) if A != None else addA
        b = concatenate([b, addb]) if b != None else addb

    n_ineq = A.shape[0] if A != None else 0
    #print "nett ineq cons:", n_ineq

    # if there are equality constraints, it's equiv to particular MLCP that 
    # can anyway be converted to LCP 
    '''
    The Karush-Kuhn-Tucker first-order conditions (being mu and lambda the
    KKT multipliers) are:

        (1.1) e + Q x + Aeq' mu + A' lambda = 0
        (1.2) Aeq x = beq
        (1.3) s = b - A x
        (1.4) s >= 0
        (1.5) lambda >= 0   
        (1.6) s' lambda = 0

    lambda are the multipliers of inequality constr., mu of equality constr.,
    and s are the slack variables for inequalities.

    This is a MLCP, where s and lambda are complementary. However, 
    we can re-write (1.1) and (1.2) as:

        | Q  Aeq'| * | x |  =  - | e + A' lambda |
        |-Aeq  0 |   |mu |       |      beq      |

    ...and, as long as 

        | Q  Aeq'|
        |-Aeq  0 |

    ...(in the program called B) is non-singular, we can solve:

        | x | = inv(| Q  Aeq'|) * | -e - A' lambda |
        |mu |       |-Aeq  0 |    |      -beq      |

    Then, if we define:

        M =     | A 0 | * inv( | Q  Aeq'| ) * | A'|
                               |-Aeq  0 |     | 0 |    

        q = b + | A 0 | * inv( | Q  Aeq'| ) * |  e  |
                               |-Aeq  0 |     | beq |

    ...(1.3) can be rewritten as an LCP problem in (s, lambda):

        s = M lambda + q

    (proof: replace M and q in the eq. above, and simplify remembering
    that | A'| lmbd == | A' lmbd |  , finally reobtaining  s = b - Ax )
         | 0 |         |    0    |

    Now, as we saw,    

        | Q  Aeq'| * | x |  = - | e + A' lambda | 
        |-Aeq  0 |   |mu |      |      beq      |

    ...so we can find

        | x | = inv( | Q  Aeq'| ) * - | e + A' lambda |
        |mu |        |-Aeq    |       |      beq      |

    ...and x will be in the first nvar elements of the solution.
    (we can also verify that b - A x == s)

    The advantage of having an LCP in lambda and s alone is also greater efficiency
    trough reduction of dimensionality (no mu's and x's are calculated by the Lemke
    solver). Also, x's are not necessarily positive (unlike s's and lambda's), 
    unless there are eplicit A,b conditions about it. However, the matrix inversion
    does cause a loss of accuracy(which could be estimated through B's condition
    number: should this value be returned with the status?).
    
    '''
    if Aeq != None:
        n_eq = Aeq.shape[0]
        B = vstack([
                hstack([ Q, Aeq.T]),
                hstack([-Aeq, zeros((n_eq, n_eq))])
             ])
        A0 = hstack([A, zeros((n_ineq, n_eq))]) if A != None else None
    else:
        B = Q
        A0 = A

    #print "det(B):", linalg.det(B)
    #print "B's log10(condition number):", log10(linalg.cond(B))

    ee = concatenate((e, beq)) if Aeq != None else e
    
    if A == None: # if no ineq constraints, no need of LCP: just solve a linear system
        xmu = linalg.solve(B, ee)
        x = xmu[:nvars]
    else:   # ve have to compute B's inverse, possibly using Q's inverse (if passed as parameter)
        if QI == None:
            # Even when Q is singular, B might not be, as long as the Eq. Constr. define a suitable subspace
            BI = linalg.inv(B)
        else:  # the inverse of Q was precomputed and passed by the caller (which requires Q not to be singular!)
            if Aeq == None:
                BI = QI
            else:
                # Use formula (1) at http://www.csd.uwo.ca/~watt/pub/reprints/2006-mc-bminv-poster.pdf
                # This is applicable only as long as Q and Aeq.T * Q * Aeq are not singular (i.e.,
                # Q not singular and Aeq full row rank).
                QIAeqT = dot(QI,Aeq.T)
                SQI = linalg.inv(dot(Aeq, QIAeqT))   # inverse of Shur's complement of Q in B
                QIAeqTSQI = dot(QIAeqT,SQI)
                BI = vstack([
                    hstack([QI-dot(dot(QIAeqTSQI,Aeq),QI), -QIAeqTSQI]),
                    hstack([dot(SQI,dot(Aeq,QI)),                 SQI]),
                    ])            
        
        A0BI = dot(A0, BI)
        M = dot(A0BI, A0.T)
        q = b + dot(A0BI, ee)

        # LCP: s = M lambda + q, s >= 0, lambda >= 0, s'lambda = 0
        # print "M is",M.shape,", q is ",q.shape
        s, lmbd, retcode = LCPSolve(M,q) 

        if retcode[0] == 1:
            # xmu = dot(BI, -concatenate([e + dot(A.T, lmbd), beq])) if Aeq != None else dot(BI, -(e + dot(A.T, lmbd)))
            kk = -concatenate([e + dot(A.T, lmbd), beq]) if Aeq != None else -(e + dot(A.T, lmbd))

            xmu = dot(BI, kk)

            x = xmu[:nvars]
        else:
            x = None

    return x