A few things figuring out helped to solve this problem.

  • for n+2 gon there are maximum n*(n+1)/2 -1  (summation of n natural numbers -1 = S) diagonals and minimum S-(n-1) diagonals, that means for S-(n-1) number of diagonals we need minimum n+2 polygon.
  • Example : there is no diagonal for triangle , 2 diagonals for Quadrilaterals 5 for pentagons. max number of diagonal can be found simply by subtracting edge number from C(n,r) = n! / ( r! (n – r)! ).


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