> The points should be aligned on a log-log plot to be a power function.
> As Robert Kern stated before, this problem should be not worse than
> O(n**3) - how have you implemented it?
>
Sure enough, the complete equation is t = 5e-05exp(1.1n), or t = 5e-05
X 3**n.
As for the implementation, it's pretty much brute force. Here is the
code itself - the cacheValue decorator memoizes the calculated inverse
for the given Matrix.
@cacheValue
def det(self):
"Function to return the determinant of the matrix."
if self.isSquare():
if self.numRows() > 2:
multiplier = 1
firstRow = self[1]
tmp = self._rows[1:]
rangelentmp = range(len(tmp))
col = 0
detsum = 0
for val in firstRow:
if val:
#~ tmp2 = Matrix([
RowVector(t[0:col]+t[col+1:]) for t in tmp ])
tmp2 = self.getCachedMatrix([
RowVector(t[0:col]+t[col+1:]) for t in tmp ])
detsum += ( multiplier * val * tmp2.det() )
multiplier = -multiplier
col += 1
return detsum
if self.numRows() == 2:
return self[1][1]*self[2][2]-self[1][2]*self[2][1]
if self.numRows() == 1:
return self[1][1]
if self.numRows() == 0:
return 0
else:
raise MatrixException("can only compute det for square
matrices")
def cofactor(self,i,j):
i-=1
j-=1
#~ tmp = Matrix([ RowVector(r[:i]+r[i+1:]) for r in
(self._rows[:j]+self._rows[j+1:]) ])
tmp = self.getCachedMatrix([ RowVector(r[:i]+r[i+1:]) for r in
(self._rows[:j]+self._rows[j+1:]) ])
if (i+j)%2:
return -tmp.det()
else:
return tmp.det()
#~ return (-1) ** (i+j) * tmp.det()
@cacheValue
def inverse(self):
if self.isSquare():
if self.det() != 0:
ret = Matrix( [ RowVector( [ self.cofactor(i,j) for j
in self.colrange() ] )
for i in self.rowrange() ] )
ret *= (1.0/self.det())
return ret
else:
raise MatrixException("cannot compute inverse for
singular matrices")
else:
raise MatrixException("can only compute inverse for square
matrices")
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