Skip to content
Fuzzy text search

Li, Bo (normalisation)

disnorm(x,y)=2⋅dis(x,y)∣x∣+∣y∣+dis(x,y)dis_{norm}(x, y) = \frac{2 \cdot dis(x,y)}{|x| + |y| + dis(x, y)}

Open question 🤔

disnorm(a,b)=2⋅disLev(a,b)∣a∣+∣b∣+disLev(a,b)=23dis_{norm}(a, b) = \frac{2 \cdot dis_{Lev}(a, b)}{|a| + |b| + dis_{Lev}(a, b)} = \frac{2}{3}

a and b are two different strings, shouldn’t distance be 1. For comparison Levy et al. (normalisation) is 1 in this case.

Reading