Based on this alternative version, the Levenshtein distance between the strings, word and work is ______ (Answer in integer).
To calculate the alternative Levenshtein distance between the strings "word" and "work" with each insertion and deletion costing 1 and no substitutions allowed, we proceed as follows:
- Compare the strings character by character:
- "word": w, o, r, d
- "work": w, o, r, k
- The first three characters (w, o, r) are identical in both strings, requiring no edits.
- The 4th character differs: "d" in "word" vs "k" in "work". Since substitutions are not allowed, we need to perform an insertion followed by a deletion.
- Delete "d" from "word" (cost: 1)
- Insert "k" into "word" (cost: 1)
- The total cost for completing the transformation from "word" to "work" is 1 + 1 = 2.
In conclusion, the alternative Levenshtein distance is 2, which fits within the specified range of (2,2).
Important Questions from Numerical Computation
$A$ is an ($n \times n$) matrix. Consider the following two statements
Statement 1: Columns of matrix $A$ are linearly independent
Statement 2: Inverse of matrix $A$ exists
Which one of the following statements is TRUE?
- Consider the function $G(x, y, z) = 0$. This function allows us to implicitly define each of three variables as a function of the other two variables. Assume that all partial derivatives of the function $G(x, y, z)$ exist everywhere. Then, the value of $\left(\frac{\partial z}{\partial x} \times \frac{\partial x}{\partial y} \times \frac{\partial y}{\partial z}\right)$ is _________ (in integer)
- In a given collection of documents, let $N$ be the total number of documents and let $d$ be the number of documents in which the term $t$ occurs.
Which ONE of the following fractions is used to define the inverse document frequency (idf) of the term $t$? - The total number of unique character bigrams in the string thoughtsoftheghosts (which contains 19 characters) is ______.
A bacterium can be approximated as a cylinder with a hemisphere at each end, as shown in the figure. The cylinder has a height of $1 \mu\text{m}$ and diameter of $1 \mu\text{m}$. Assuming that the density of the bacterium is equal to that of water, what is the approximate mass of this bacterium?
Given: density of water = $10^3 \text{ kg/m}^3$; $1 \mu\text{m} = 10^{-6}\text{ m}$
Volume of a cylinder = $\pi r^2 h$, where $r$ is the radius and $h$ is the height of the cylinder
Volume of a sphere = $\frac{4}{3} \pi r^3$, where $r$ is the radius of the sphere
