algorithm - Solving the similar recurrence: T(n) = 3T(n/3) + n/3 -


दिये गये .. 

  टी (0) = 3 n & lt; = 1 टी के लिए (एन) = 3 टी (एन / 3) + n / 3 n & gt; 1   
 तो इसका जवाब है कि  O (nlogn)  हो सकता है .. यह है कि मैंने यह कैसे किया और यह मुझे सही जवाब नहीं दे रहा है:  < पूर्व>  टी (एन) = 3 टी (एन / 3) + एन / 3 टी (एन / 3) = 3 टी (एन / 3 ^ 2) + एन / 3 ^ 2   < P> इसे टी (एन) में जमा कर देता है ..  
  टी (एन) = 3 (3 टी (n / 3 ^ 2) + n / 3 ^ 2) + n / 3 टी ( N / 3 ^ 2) = 3 (3 (3 टी (n / 3 ^ 3) + n / 3 ^ 3) + n / 3 ^ 2) + n / 3   
 आखिरकार 'जैसे दिखेंगे ..  <पूर्व>  टी (एन) = 3 ^ के (टी (एन / 3 ^ के)) + सीएन / 3 ^ के   <पी > सेट  k = lgn ..   
  टी (एन) = 3 ^ lgn * (टी (एन / 3 ^ lgn)) + cn / 3 ^ lgn टी (N) = n * T (0) + c T (n) = 3n + c   
 उत्तर का  ओ (n)  हालांकि ... गलत क्या है मेरे चरणों के साथ?  
 "post-text" itemprop = "text"> 
  टी (एन) = 3 टी (n / 3) + n / 3 टी (एन / 3) = 3 टी (एन / 9) + एन / 9 टी (एन) = 3 (3 टी (एन / 9) + एन / 9) + एन / 3 = 9 टी (एन / 9) + 2 * N / 3 // कथन 1 टी (एन / 9) = 3 टी (एन / 27) + एन / 27 टी (एन) = 9 (3 टी (एन / 27) + एन / 27) + 2 * एन / 3 // विवरण में टी (एन / 9) की जगह 1 = 27 टी (एन / 27) + 3 * (एन / 3) टी (एन) = 3 ^ के * टी (एन / 3 ^ के) + के * (एन / 3 ) // अंततः   
 लॉग को n के आधार पर 3 के साथ बदलें।  
  टी (एन) = एन टी (1) + (लॉग एन) (एन / 3); // टी (1) = 3 टी (एन) = 3 * एन + (लॉग एन) (एन / 3); इसलिए, हे (एन * लॉगन)

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