We are given a natural number $x\in\mathbb{N}^+$, and $x-1$ adders. An adder in our case would be a component with two inputs $x_1,x_2$ and a single output $y$ such that $y=x_1+x_2$. Our goal is to design a system that has $x$ as its single input, and $x^2$ as its single output. No constants are available to use. Define $n:\mathbb{N}^+\to\mathbb{N}$ such that $\forall x\in\mathbb{N}^+$, $n(x)$ is the minimum amount of adders needed in order to implement such system that calculates $x^2$. What is $n(x)$ precisely?
For example: In order to calculate $7^2$ efficiently:
- First step: we only have $7$, so we must do $7+7=14$. (1 adder)
- Second step: Now we also have $14$, so we can calculate $7+14=21$ and $14+14=28$ (2 adders; 3 overall)
- Final step: Now we have $21$ and $28$, so we can calculate $21+28=49$ (1 adder; 4 overall) and we're done.
It needs to be proven now that $7^2$ can't be calculated using $3$ adders or less (pretty intuitive), and then it can be concluded that $n(7)=4$.
It is clear that $n(x)\leq x-1$. I was able to show that $n(x)$ is not increasing (for instance - $n(7)=4$ while $n(8)=3$), but I think I did manage to find a tighter upper bound - which is $n(x)\leq2\lfloor\log_2(x)\rfloor$. I'm having trouble with finding the exact expression for $n(x)$.
This was my way of thinking. For every $m\in\mathbb{N}^+$, define the group of natural numbers $A_m$ s.t.:
$$A_m\triangleq \{x\in\mathbb{N}^+|\lceil\log_2(x)\rceil=m\}$$
So for example, $A_3=\{5,6,7,8\}$. It is clear that $x_m\equiv2^m-1\in A_m$. This demands proof, but I tend to believe that out of all the integers of $A_m$, $x_m$ requires the greatest amount of adders. Pay attention that:
$$x_m^2=x_m(2^{m-1}+2^{m-1}-1)=2^{m-1}x_m+\sum_{k=0}^{m-2}2^kx_m$$
You can efficiently calculate $2^{m-1}x_m$ using $m-1$ adders. When you do that, you have the input $2^kx_m$ available, for every $k\in[1,m-1]_\mathbb{N}$. So $\sum_{k=0}^{m-2}2^kx_m$ can be calculated with $m-2$ adders. Then, a final adder is needed in order to calculate $x_m^2$. So overall:
$$n(2^m-1)=2m-2$$
Given that $x_m^2$ cannot be calculated using $2m-3$ adders or less. Pay attention that for $m=3$, we get $n(7)=4$, as before.
I tried to do the same thing with different patterns, but I didn't manage to conclude the general form. Some help would be appreciated :)
Note: This question can be rephrased in a clearer way. Suppose you initially have a set $S=\{1\}$, each time you can pick up two (can be duplicate) numbers in $S$ and put the sum of those two numbers in $S$. What is the least number of steps, or $|S|$, that is needed such that $n\in S$? (The question is not the same, but the answers are (I believe) the same, since all my equations can simply be divided by $x_m$). (Thank you @JetfiRex for this)
