Re: [algogeeks] coloring problem Dynamic programming

2013-12-15 Thread kumar raja
Saurabh your solutions seems right , but can u explain me how did u arrive
at the time and space complexity with some proof /pseudocode/ explanation?


On 15 December 2013 09:47, Saurabh Paliwal saurabh.paliwa...@gmail.comwrote:

 what is the issue with the usual recurrence :-

 Let answer[i][j] represent minimum cost of coloring i houses with ith
 house colored with jth color.

 answer[i][j] = min(answer[i-1][t]) + colorvalue[t] where t varies from 1
 to k except j.

 initial setting :-  answer[1][j] = colorvalue[j]

 final answer is min(answer[n][j]) for all j from 1 to k.

 Time complexity = O(n*k*k)
 Space complexity = O(k) if only 2 rows are taken ( effectively only 2 are
 required ).


 On Sun, Dec 15, 2013 at 12:44 AM, kumar raja rajkumar.cs...@gmail.comwrote:

 You have 'n' houses and 'k' colors. The cost of coloring a house with one
 of the 'k' colors is different from coloring another house with same color.
  The cost of coloring the house with different colors are different. So now
 how to colour the 'n' houses such that no two adjcent houses will get the
 same color and the total coloring cost for 'n' houses is minimized.



 Regards,
 Kumar Raja.

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  -Saurabh Paliwal

B-Tech. Comp. Science and Engg.

IIT ROORKEE

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Re: [algogeeks] coloring problem Dynamic programming

2013-12-15 Thread kumar raja
No need for the explanation ,got it man thanks.


On 15 December 2013 16:20, kumar raja rajkumar.cs...@gmail.com wrote:

 Saurabh your solutions seems right , but can u explain me how did u arrive
 at the time and space complexity with some proof /pseudocode/ explanation?


 On 15 December 2013 09:47, Saurabh Paliwal saurabh.paliwa...@gmail.comwrote:

 what is the issue with the usual recurrence :-

 Let answer[i][j] represent minimum cost of coloring i houses with ith
 house colored with jth color.

 answer[i][j] = min(answer[i-1][t]) + colorvalue[t] where t varies from 1
 to k except j.

 initial setting :-  answer[1][j] = colorvalue[j]

 final answer is min(answer[n][j]) for all j from 1 to k.

 Time complexity = O(n*k*k)
 Space complexity = O(k) if only 2 rows are taken ( effectively only 2 are
 required ).


 On Sun, Dec 15, 2013 at 12:44 AM, kumar raja rajkumar.cs...@gmail.comwrote:

 You have 'n' houses and 'k' colors. The cost of coloring a house with
 one of the 'k' colors is different from coloring another house with same
 color.  The cost of coloring the house with different colors are different.
 So now how to colour the 'n' houses such that no two adjcent houses will
 get the same color and the total coloring cost for 'n' houses is minimized.



 Regards,
 Kumar Raja.

 --
 You received this message because you are subscribed to the Google
 Groups Algorithm Geeks group.
 To unsubscribe from this group and stop receiving emails from it, send
 an email to algogeeks+unsubscr...@googlegroups.com.




 --
  -Saurabh Paliwal

B-Tech. Comp. Science and Engg.

IIT ROORKEE

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Re: [algogeeks] coloring problem Dynamic programming

2013-12-15 Thread kumar raja
What can be the recurrence relation if the condition is changed to No 3
adjacent houses should get same color?


On 15 December 2013 16:26, kumar raja rajkumar.cs...@gmail.com wrote:

 No need for the explanation ,got it man thanks.


 On 15 December 2013 16:20, kumar raja rajkumar.cs...@gmail.com wrote:

 Saurabh your solutions seems right , but can u explain me how did u
 arrive at the time and space complexity with some proof /pseudocode/
 explanation?


 On 15 December 2013 09:47, Saurabh Paliwal 
 saurabh.paliwa...@gmail.comwrote:

 what is the issue with the usual recurrence :-

 Let answer[i][j] represent minimum cost of coloring i houses with ith
 house colored with jth color.

 answer[i][j] = min(answer[i-1][t]) + colorvalue[t] where t varies from 1
 to k except j.

 initial setting :-  answer[1][j] = colorvalue[j]

 final answer is min(answer[n][j]) for all j from 1 to k.

 Time complexity = O(n*k*k)
 Space complexity = O(k) if only 2 rows are taken ( effectively only 2
 are required ).


 On Sun, Dec 15, 2013 at 12:44 AM, kumar raja 
 rajkumar.cs...@gmail.comwrote:

 You have 'n' houses and 'k' colors. The cost of coloring a house with
 one of the 'k' colors is different from coloring another house with same
 color.  The cost of coloring the house with different colors are different.
 So now how to colour the 'n' houses such that no two adjcent houses will
 get the same color and the total coloring cost for 'n' houses is minimized.



 Regards,
 Kumar Raja.

 --
 You received this message because you are subscribed to the Google
 Groups Algorithm Geeks group.
 To unsubscribe from this group and stop receiving emails from it, send
 an email to algogeeks+unsubscr...@googlegroups.com.




 --
  -Saurabh Paliwal

B-Tech. Comp. Science and Engg.

IIT ROORKEE

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 You received this message because you are subscribed to the Google
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