This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ; M 2 = M*M = [(0, 1), (1, 1)]* [(0, 1), (1, 1)] = [(1, 1), (1, 2)] ; M 3 = M 2 *M= [(1, 1), (1, 2)]* [(0, 1), (1, 1)] = [(1, 2), (2, 3)] ; M 4 =M 3 *M= [(1, 2), (2, 3)]* [(0, 1), (1, 1)] = [(2, 3), (3, 5)] ; M 5 =M 4 *M= [(2,3), (3,5)]* [(0, 1), (1, 1)] = [(3, 5), (5, 8)]; As can be seen from the above multiplication my formula holds true. It must be this way because the Fibonacci sequence uses the last two numbers computed to get the next number (ie: F N =F N2 + F N1 ) . Thus, since the matrix we are multiplying has two ones in the second column the two numbers in the second row of the matrix it is being multiplied by must be the F N2 and the F N1 numbers, based on the conventions of matrix multiplication. The pattern that I found was based on repeated multiplication of the original matrix M until a pattern was determined. (2) G N = c N1 (b) + c N2 d(a)+ (N2)c N3 d(b)+(N3)c N4 d 2 (a)...
View
Full Document
 Fall '07
 GABOW
 Algorithms, Multiplication, Boulder, Stephanie Pitts

Click to edit the document details