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== Possible linear algorithm simple n efficient just like K idea ==
(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
once we got first result from these 3 there might b an easy way to compute the other two, like O(N)time complexity which is the algo main idea, n honestly idk if its working, but if it does i wish credits :))
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