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hi, i think there could be written algos based on grid concept for the stored values, n a small square 4 data, also, that possibly work in O(N) time n O(N^(2/3)) mem, for example or even less O(N^(1/2)), O((log(N)), or evn O(1) mem for O(N) time, so i dont think really its need to trade time for mem . for example O(1) mem algo first seek for target without filling then intelligent border keeping also condition of conexivitty (local test) 4 the remaining zone... only if there is no bordering n need to spilt, then proceed splitting... O(N^1/2) of O(N^2/3) are evn more simple they use O(N) time. it some cautious terms N is the maximum number of pixel of a possible image to b filled... — Preceding unsigned comment added by 93.118.212.93 (talk) 10:29, 18 January 2013 (UTC) <span style="font-size: smaller;" class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/93.118.212.93|93.118.212.93]] ([[User talk:93.118.212.93|talk]]) </span><!-- Template:Unsigned IP --> <!--Autosigned by SineBot-->
== a lil-big difference in K. algo idea ==
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)
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