This is the definition of theta notation:

The function $T(N)$ is $\Theta(g(N))$ if there exist positive constants $C_1$, $C_2$ and $N_0$ such that $C_1g(N) \le T(N) \le C_2g(N)$ for all $N \ge N_0$.

My calculation is below. When I solved the inequality, $N \ge 0$. Am I making a mistake or am I on the right track?

Assuming 1 time unit equals 1 memory read or write, I have counted the time units to get to the expression $T(N) = 4KN + 37N + 11K + 22$.

This is a sorting algorithm that uses hashing. The inputs are an array $A$ of non-negative integers of length $N$, $k$ is the largest integer value in $A$, and $a$ and $b$ are positive integers. I also implemented this in javascript and the output is wrong so I think whoever made this made a mistake.

function Sort(A,N,k,a,b)
    C=new array(N) of zeroes    // 4 time units
    R=new array(N) of zeroes   // 4 time units
    pos=0 // 3
    for 0 <= j < N          // 7N+5 time units
        C[(a*A[j]+b)%N]=C[(a*A[j]+b)%N]+1 // 20N time units
    for 0 <= i < N          // 7N+5 time units
        for pos <= r < pos+C[(a*i+b)%N] // 11K time units 
            R[r]=i     // 4KN time units
        pos=r =       // 3N time units
    return R         // 1 time unit
end function

The function $T(N)$ is $\Theta(g(N))$ if there exist positive constants $C_1$, $C_2$ and $N_0$ such that $C_1g(N) \le T(N) \le C_2g(N)$ for all $N \ge N_0$.

We can prove that $T(N) = 4KN + 37N + 11K + 22 = \Theta(g(KN))$.

$$(T(N) = 4KN + 37N + 11K + 22) \le (g(N) =4KN + 37N + 11K + 22 + KN)$$

$$(T(N) = 4KN + 37N + 11K + 22) \le (g(N) = 5KN + 37N + 11K + 22)$$

Therefore, the value of $C_2$ becomes 5. If we solve the above inequality for $N$, $N \ge 0$.

  • $\begingroup$ Answer to title: yes, why not? Look at $f(n)\in\Theta(f(n))$. $\endgroup$
    – zkutch
    Commented Mar 17, 2022 at 1:01

1 Answer 1


It doesn't matter whether you allow $N_0 = 0$ or not (though per your definition, it is not allowed). If $C_1 g(N) \le T(N) \le C_2 g(N)$ holds for all $N \ge 0$, then in particular, it holds for all $N \geq 1$, or for all $N \geq 1000$, for that matter.

The condition "$C_1 g(N) \le T(N) \le C_2 g(N)$ for all $N \ge N_0$" only requires that if $N \ge N_0$ then $C_1 g(N) \le T(N) \le C_2 g(N)$. There is not requirement in the other direction. It's not an if-and-only-if.


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