again

bibook/pairwise/needleman.md

@@ -173,7 +173,7 @@ In the same way memory usage also scales with $ \mathcal{O}(MN)$, as the scoring
 
 Big-O notation serves as a quick and effective tool for comparing different algorithms. For example, it allows us to see at a glance how the Needleman-Wunsch algorithm compares to other sequence alignment algorithms in terms of efficiency.
 
-A useful comparison is the complexity of our initial proposition, to enumerate and calculate the scores for all possible alignments of two sequences. This can be done by calculating the number of alignments with $k$ matches/mis-matches between the two sequences which is ${M \choose k}{N \choose k}$. If we asume that $N>M$ and sum this for all possible values of $k$, we get $\sum_{k=0}^M{M \choose k}{N \choose k}=\sum_{k=0}^M{M \choose M-k}{N \choose k}={N+M \choose M}=\frac{(M+N)!}{M!*N!}$ number of different aligments. This can be shown to follow $ \mathcal{O}(2^{2N}/\sqrt{N})$ {cite}`lange2002mathematical, eddy2004dynamic`.
+A useful comparison is the complexity of our initial proposition, to enumerate and calculate the scores for all possible alignments of two sequences. This can be done by calculating the number of alignments with $k$ matches/mis-matches between the two sequences which is ${M \choose k}{N \choose k}$. If we asume that $N>M$ and sum this for all possible values of $k$, we get $\sum_{k=0}^M{M \choose k}{N \choose k}=\sum_{k=0}^M{M \choose M-k}{N \choose k}={N+M \choose M}=\frac{(M+N)!}{M!*N!}$ number of different aligments. This can be [shown](https://math.stackexchange.com/a/4134185) to follow $\mathcal{O}((\frac{e(N+M)}{M})^M)$ {cite}`lange2002mathematical, eddy2004dynamic`.
 
 ```{bibliography}
 :filter: docname in docnames