# How can we express value of cosine similarity of two documents into percentage?

We were doing project work for plagiarism checking. For this purpose, we have taken a term frequency vector of two documents and measured the similarity using a cosine similarity measure. The value of cosine similarity is limited between 0 and 1. We know that the value of cosine similarity will be 1 if two documents exactly match with one another. In this case, we can say 100% match. Moreover, the value will be 0 for no match i.e. 0 % match. Furthermore, if the value is 0.65, then how do we find the percentage from this score? The definition and formula of cosine similarity are shown in the following figure.

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– D.W.
Nov 9, 2021 at 4:37

Suppose that your two vectors $$x,y$$ belong to $$\{\pm1\}^n$$. The formula of cosine similarity in this case gives the cosine similarity as $$\frac{1}{n} \sum_{i=1}^n x_i y_i = \frac{\#\{i : x_i = y_i\} - \#\{i : x_i \neq y_i\}}{n} = 1 - 2\Pr_i[x_i \neq y_i] = 2\Pr_i[x_i=y_i] - 1.$$ As can be seen, the range of cosine similarity is actually $$[-1,1]$$.
Sometimes other proxies of cosine similarity are used. For example, cosine distance is one minus cosine similarity; this ranges over $$[0,2]$$. Similarly, "half cosine distance" is half the cosine distance, which ranges over $$[0,1]$$. In the case above, half cosine distance is exactly $$\Pr_i[x_i \neq y_i]$$.
Starting with the actual cosine similarity $$S$$, you can extract $$\Pr_i[x_i = y_i]$$ as $$\frac{1+S}{2}$$. In the case of two vectors $$x,y \in \{\pm 1\}^n$$, this is the percentage of indices in which $$x_i$$ equals $$y_i$$. For general vectors $$x,y$$, the quantity $$\frac{1+S}{2}$$ no longer has this interpretation. It is just a distance measure.