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== Possible linear algorithm simple n efficient just like K idea ==
ok, once n last 4 everybody that wanna challenge this idea: my challenge 2 u is that using bit
aritmetics , to easily compute , 4 example (A+4*B)*(C+4*D), O(N) time, when we already got (A+2*B)*(C+2*D), letters ar very long integers, lets say. i think there is a chance 4 this to work so give it a try, y not ?
(yeah, i think there is a chance 4 this to work, around of computed kernel which questions is if might work in recurrence in order to rightly compute the desired 2 similar values)
lets try to multiply m=AB with n=CD, A most significant part of m, etc... we could do that by computing first (A+2*B)*(C+2*D) , (A+4*B)*(C+4*D), n (A+8*B)*(C+8*D) , after that solving the system
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