New answers tagged

0

If you take the initial conditions $T(0) = 0$ and $T(1) = 1$ then you get A018819 (up to shift), which is essentially A000123. The asymptotics are known to be $T(n) = n^{\Theta(\log n)}$. The references under the second entry contain more accurate asymptotic information. More explicitly, A018819 satisfies the recurrence $a(2m+1) = a(2m) = a(2m-1) + a(m)$, ...


0

Your algorithm is not correct. Here is a counterexample: $D$ contains the words $aa,bb,ac,bc$; $v=aa$; and $w=bb$. A transformation is possible $(aa \to ac \to bc \to bb)$ but your algorithm won't find it.


0

The optimal solution will always have the following form: You will change the types of the stones t times, at a total cost of tx. And for each type, you buy the stone when it is cheapest. Let's say t = 3, so you changed the types three times. To get a stone of type 7, for example, you could have bought the 7th stone immediately, the 6th stone after one ...


2

Usually when trying to figure out a dynamic programming solution's time complexity, you need to find two values: The number of elements that will be calculated (or a bound on this number. Notice that "elements that would be calculated" refers to the elements you save with the memoization technique) The time it takes to process an item, assuming ...


2

The DPLL algorithm doesn't try to satisfy as many clauses as possible. If there is a satisfying assignment, DPLL will find it. Otherwise DPLL will try a series of partial assignments until it runs out of possibilities, and then it will declare the formula unsatisfiable. But in the case of an unsatisfiable formula, there is no guarantee that DPLL will ...


Top 50 recent answers are included