# compute the similarity between two vectors [on hold]

Euclidean distance is a measure that may be used to compute the similarity between two vectors. Given a query $q$ and documents $d_1, \ldots, d_n$, we may rank the documents $\mathcal{D} = \{d_k\}_{k=1}^n$ in the increasing order of Euclidean distance from $q$.

Prove that if $q$ and the document vectors $\mathcal{D}$ are all normalized to unit vectors, then the rank ordering produced by Euclidean distance is identical to that produced by cosine similarity.

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## put on hold as off-topic by Thursday, Mark Fantini, Claude Leibovici, studiosus, Jean-Claude ArbautAug 27 at 5:32

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What is a cosine similarity? Computing $\dfrac{\vec{x}\cdot \vec{y}}{\|\vec{x}\| \|\vec{y}\|}$? –  gt6989b Mar 20 '13 at 17:10
–  benji_r Mar 20 '13 at 17:11

Doesn't seem true to me. Take $\mathcal{D}=\{(1,0), (0,1)\}$ and $q = (a,b)$ on the unit circle. Then, the cosine similarity would result in $q \cdot d_1 = a$ and $q \cdot d_2 = b$. Euclidean distance (we consider the square of the actual distance, it doesn't alter comparisons), however, would have $(a-1)^2+b^2$ and $a^2 + (b-1)^2$.

Note that $$\begin{split} (a-1)^2+b^2 &< a^2 + (b-1)^2 \\ a^2 -2a + 1+b^2 &< a^2 + b^2 -2b +1 \\ -2a &< -2b \\ b &< a \end{split}$$

So the distance ranking orders $d_1 < d_2$ iff the cosine ranking orders $d_2 < d_1$.

It seems, this is the counterexample, since they produce a different ordering. Or am I not understanding something?

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