You have $n$ sticks of arbitrary lengths, not necessarily integral.
By cutting some sticks (one cut cuts one stick, but we can cut as often as we want), you want to get $k<n$ sticks such that:
- All these $k$ sticks have the same length;
- All $k$ sticks are at least as long as all other sticks.
Note that we obtain $n + C$ sticks after performing $C$ cuts.
What algorithm would you use such that the number of necessary cuts is minimal? And what is that number?
As an example, take $k=2$ and any $n\geq 2$. The following algorithm can be used:
- Order the sticks by descending order of length such that $L_1\geq L_2 \geq \ldots \geq L_n$.
- If $L_1\geq 2 L_2$ then cut stick #1 to two equal pieces. There are now two sticks of length $L_1 / 2$, which are at least as long as the remaining sticks $2 \ldots n$.
- Otherwise ($L_1 < 2 L_2$), cut stick #1 to two unequal pieces of sizes $L_2$ and $L_1-L_2$. There are now two sticks of length $L_2$, which is longer than $L_1-L_2$ and the other sticks $3 \ldots n$.
In both cases, a single cut is sufficient.
I tried to generalize this to larger $k$, but there seem to be a lot of cases to consider. Can you find an elegant solution?
Asked By : Erel Segal-Halevi
Answered By : Raphael
The first core observation to solving this problem is that the feasibility of a cutting length $l$,
$\qquad\displaystyle \operatorname{Feasible}(l) = \biggl[\, \sum_{i=1}^n \Bigl\lfloor\frac{L_i}{l} \Bigr\rfloor \geq k \,\biggr]$,
is piecewise-constant, left-continuous and non-increasing in $l$. Since the number of necessary cuts behaves similarly, finding the optimal length is just
$\qquad\displaystyle l^{\star} = \max \{ l \mid \operatorname{Feasible}(l) \}$.
Furthermore, as the other answers have proposed, all jump discontinuities have the form $L_i/j$. This leaves us with a discrete, one-dimensional search problem amenable to binary search (after sorting a finite set of candidates).
Note furthermore that we only need to consider the $L_i$ that are shorter than the $k$-largest one, since that one is always feasible.
Then, different bounds on $j$ lead to algorithms of different efficiency.
- $1 \leq j \leq k$ results in a search space of quadratic size (in $k$),
- $1 \leq j \leq \lceil k/i \rceil$ in a linearithmic one (assuming the $L_i$ are sorted by decreasing size), and
- slightly more involved bounds in a linear one.
Using this, we can solve the proposed problem in time $\Theta(n + k \log k)$ and space $\Theta(n + k)$.
One further observation is that the sum in $\mathrm{Feasible}$ grows in $l$ by $1$ for each candidate $L_i/j$ "passed", counting duplicates. Using this, we can use rank selection instead of binar search and obtain an algorithm that runs in time and space $\Theta(n)$, which is optimal.
Find the details in our article Efficient Algorithms for Envy-Free Stick Division With Fewest Cuts (by Reitzig and Wild, 2015).
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Question Source : http://cs.stackexchange.com/questions/30073
