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this idea could make the difference between O(N^1.59) n O(N*log(N)) by simply redisn the plan of recursive calls of this good idea: basically instead of planning 3 muls of T(n/2) we plan (and that is reffering to the muls of sums) the T(n/2) sum n then the two remaining two T(n/4) sums. i agree there is some terms let 4 autosolve but this doesent yet charge the data stack (hypotetical data stack of some implementention of the algo) n those remaining unconsidered ("yet") terms are expected to become smaller n smaller. practically if the tack grows with N/2+N/4+N/4=N then this small idea could lead us to an O(n*log(n)) time complexity algo. i agree my "demonstration" here is far to b complete, take it as a Russian style problem , if u wish :) Florin Matei, Romania [[Special:Contributions/93.118.212.93|93.118.212.93]] ([[User talk:93.118.212.93|talk]]) 13:12, 14 April 2013 (UTC)
:No, russian style is to leave the solution of a problem to the reader. What you want is to leave the formulation of a problem to the reader. This is crank style. Come back when you have an implementation with demonstrable runtimes.--[[User:LutzL|LutzL]] ([[User talk:LutzL|talk]]) 12:13, 15 April 2013 (UTC)
well, another pure creative idea is that observing/remarking first that if we manage to write the master method equivalent for finding O() of Karatsuba algo idea in such way that being possible the following kinda write T(N)=3*(1-p)*T(N/2), p a small percentage of 1 meaning the problem solved first from each of those 3 planned, we might find better O() for K. algo idea... basically insted of planning a mul of T(N/2) we solve a small percentage of that n planning only the remaining (1-p)*T(N/2) claiming we already talk abt a better O()... as a rule of thumb the lil mul n the remaining mul should b equalize in time wasting in order to find a most representative O()
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