How can this assert() ever trigger?

[Joseph Martinot-Lagarde]

Le 10/05/2014 17:24, Albert van der Horst a écrit :
> I have the following code for calculating the determinant of
> a matrix. It works inasfar that it gives the same result as an
> octave program on a same matrix.
>
> / ----------------------------------------------------------------
>
> def determinant( mat ):
>      ''' Return the determinant of the n by n matrix mat
>          i row j column
>          Destroys mat ! '''
>      #print "getting determinat of", mat
>      n=len(mat)
>      nom = 1.
>      if n == 1: return mat[0][0]
>      lastr = mat.pop()
>      jx=-1
>      for j in xrange(n):
>         if lastr[j]:
>             jx=j
>             break
>      if jx==-1: return 0.
>      result = lastr[jx]
>      assert(result<>0.)
>      # Make column jx zero by subtracting a multiple of the last row.
>      for i in xrange(n-1):
>          pivot = mat[i][jx]
>          if 0. == pivot: continue
>          assert(result<>0.)
>          nom *= result   # Compenstate for multiplying a row.
>          for j in xrange(n):
>              mat[i][j] *= result
>          for j in xrange(n):
>              mat[i][j] -= pivot*lastr[j]
>      # Remove colunm jx
>      for i in xrange(n-1):
>         x= mat[i].pop(jx)
>         assert( x==0 )
>
>      if (n-1+jx)%2<>0: result = -result
>      det = determinant( mat )
>      assert(nom<>0.)
>      return result*det/nom
>
> /-----------------------------------------
>
> Now on some matrices the assert triggers, meaning that nom is zero.
> How can that ever happen? mon start out as 1. and gets multiplied
> with a number that is asserted to be not zero.
>
> Any hints appreciated.
>
> Groetjes Albert
>
I know it's not the question, but if you want a replacement for octave 
did you try numpy (and scipy) ? The determinant would be computer faster 
and with less memory than with your function.

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