← Back to context Comment by karamazov 4 years ago Or use the closed-form solution. 2 comments karamazov Reply jacobolus 4 years ago The "closed-form solution" is slower than standard method. It just uses arbitrary-precision fractional arithmetic (square root, exponentiation, division) instead of arbitrary-precision integer arithmetic (exponentiation of a 2x2 integer matrix). singhrac 4 years ago The best solution is to use the O(log n) time exponentiation of a matrix, which is fast enough to be constant.
jacobolus 4 years ago The "closed-form solution" is slower than standard method. It just uses arbitrary-precision fractional arithmetic (square root, exponentiation, division) instead of arbitrary-precision integer arithmetic (exponentiation of a 2x2 integer matrix). singhrac 4 years ago The best solution is to use the O(log n) time exponentiation of a matrix, which is fast enough to be constant.
singhrac 4 years ago The best solution is to use the O(log n) time exponentiation of a matrix, which is fast enough to be constant.
The "closed-form solution" is slower than standard method. It just uses arbitrary-precision fractional arithmetic (square root, exponentiation, division) instead of arbitrary-precision integer arithmetic (exponentiation of a 2x2 integer matrix).
The best solution is to use the O(log n) time exponentiation of a matrix, which is fast enough to be constant.