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I have an algorithm which depends on three variables an where the running time is in $\mathcal{O}(m+2 m\cdot n\cdot p+p\cdot(n+m))$ and I would like to simplified it. I proceeded as follows :

\begin{eqnarray} \mathcal{O}(3m+2 m\cdot n\cdot p+p\cdot(n+m)) &=& \mathcal{O}(m\cdot(3+2n\cdot p)+p\cdot(n+m))\\ &=& \mathcal{O}(2m\cdot n\cdot p+n\cdot p + m\cdot p)\\ &=& \mathcal{O}(n\cdot p \cdot(2m+1) + m\cdot p)\\ &=& \mathcal{O}(2n\cdot m \cdot p + m\cdot p)\\ &=& \mathcal{O}(m \cdot p\cdot(2n+1))\\ &=& \mathcal{O}(2n\cdot m \cdot p)\\ &=& \mathcal{O}(m\cdot n \cdot p)\\ \end{eqnarray}

I successively replace $3+2n\cdot p$ by $2n\cdot p$, $2m+1$ by $2m$ and $2n+1$ by $2n$ since they are asymptotically equivalent. Is it correct?

More generally can I always replace any subexpression with an asymptotically equivalent one within the big-O expression?

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All is correct, but - it can be done in one step after you represent this expression as a sum of terms: $$\mathcal{O}(3m+2m⋅n⋅p+p⋅n+p⋅m)=\mathcal{O}(m⋅n⋅p)$$ because:

  • The term $(m⋅n⋅p)$ dominates all the other terms - $(3m)$, $(p⋅n)$, and $(p⋅m)$ - in this sum.
  • Constants are eliminated according to the $\mathcal{O}$ definition.
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