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1 | import exceptions |
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2 | class VectorShapeError(exceptions.Exception): pass |
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3 | |
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4 | |
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5 | def conjugate_gradient(A,b,x0=None,imax=10000,tol=1.0e-8,iprint=0): |
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6 | """ |
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7 | Try to solve linear equation Ax = b using |
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8 | conjugate gradient method |
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9 | |
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10 | Input |
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11 | A: matrix or function which applies a matrix, assumed symmetric |
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12 | A can be either dense or sparse |
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13 | b: right hand side |
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14 | x0: inital guess (default the 0 vector) |
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15 | imax: max number of iterations |
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16 | tol: tolerance used for residual |
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17 | |
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18 | Output |
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19 | x: approximate solution |
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20 | """ |
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21 | |
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22 | from Numeric import dot, array, Float, zeros |
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23 | |
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24 | b = array(b, typecode=Float) |
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25 | if len(b.shape) != 1 : |
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26 | raise VectorShapeError, 'input vector should consist of only one column' |
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27 | |
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28 | if x0 is None: |
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29 | x0 = zeros(b.shape, typecode=Float) |
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30 | else: |
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31 | x0 = array(x0, typecode=Float) |
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32 | |
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33 | |
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34 | #FIXME: Should test using None |
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35 | if iprint == 0: |
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36 | iprint = imax |
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37 | |
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38 | i=1 |
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39 | x = x0 |
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40 | r = b - A*x |
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41 | d = r |
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42 | rTr = dot(r,r) |
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43 | rTr0 = rTr |
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44 | |
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45 | while (i<imax and rTr>tol**2*rTr0): |
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46 | q = A*d |
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47 | alpha = rTr/dot(d,q) |
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48 | x = x + alpha*d |
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49 | if i%50 : |
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50 | r = b - A*x |
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51 | else: |
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52 | r = r - alpha*q |
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53 | rTrOld = rTr |
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54 | rTr = dot(r,r) |
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55 | bt = rTr/rTrOld |
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56 | |
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57 | d = r + bt*d |
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58 | i = i+1 |
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59 | if i%iprint == 0 : |
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60 | print 'i = %g rTr = %20.15e'% (i,rTr) |
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61 | |
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62 | #FIXME: Should this raise an exception? |
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63 | if i==imax: |
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64 | print 'max number of iterations attained' |
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65 | |
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66 | return x |
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67 | |
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68 | |
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