diff --git a/edit_distance_l1_matching_v2.txt b/edit_distance_l1_matching_v2.txt
index e20a021..ad88f1c 100644
--- a/edit_distance_l1_matching_v2.txt
+++ b/edit_distance_l1_matching_v2.txt
@@ -59,7 +59,7 @@ form needed is their Theorem 7.
 there exists an embedding phi: {0,1}^d -> l1 with distortion at most
 2^{c sqrt(log d log log d)}."
 
-https://web.cs.ucla.edu/~rafail/PUBLIC/68.pdf
+./pdf/unknown-nd.pdf
 
 Khot and Naor obtained an Omega(sqrt(log d / log log d)) lower bound by
 studying cuts and Fourier analysis of noisy shifts. Krauthgamer and Rabani
@@ -68,11 +68,11 @@ Deza, Gupta, Indyk and Raskhodnikova constructed binary subsets whose L1
 distortion approaches 3/2. The recursive construction reaches the scale of the
 Ostrovsky-Rabani exponent.
 
-https://web.math.princeton.edu/~naor/homepage%20files/nonembed-final.pdf
+./pdf/unknown-nd.pdf
 
 https://doi.org/10.1137/060660126
 
-https://www.cs.columbia.edu/~andoni/papers/k2n2.pdf
+./pdf/unknown-nd.pdf
 
 An increasing matching between x and y is a list of pairs (i,j) such that
 x_i = y_j, and both i and j strictly increase from one pair to the next. Let
@@ -216,11 +216,11 @@ and use |exp(2 pi i theta)-1|^2 = 4 sin^2(pi theta). QED.
 
 Proof of the proposition. By the cut lemma, each displacement of H is a
 nonnegative linear combination of displacements of Boolean cut maps. Both
-sides of the desired inequality are linear in the cut weights, so it is
-enough to prove the inequality for H = 1_A for a subset A of G.
+sides of the desired ineuality are linear in the cut weights, so it is
+enough to prove the ineuality for H = 1_A for a subset A of G.
 
 Let K = <v_1> x ... x <v_b>, a subgroup of G. Every shift in the desired
-inequality preserves every coset of K. On a fixed coset choose a
+ineuality preserves every coset of K. On a fixed coset choose a
 representative c and identify T = prod_j Z/p_jZ with the coset by
 
     t -> c + sum_j t_j v_j.
@@ -284,7 +284,7 @@ The component terms on the right therefore have combined multiplier
 again bounding the left multiplier.
 
 Multiplying by |hat F(lambda)|^2 and summing over lambda proves the coset
-inequality. Averaging over the equal-size cosets gives the uniform average
+ineuality. Averaging over the equal-size cosets gives the uniform average
 over G. Summing the nonnegative cut weights proves the proposition for H.
 QED.
 
@@ -504,7 +504,7 @@ hence
     E >= T d q / (d+q).
 
 The latter is at least (T/2) min(d,q). This holds for every matching,
-including when d=0, and proves the displayed inequality. Finally,
+including when d=0, and proves the displayed ineuality. Finally,
 Delta(X,Y)=2 r_N(X,Y). QED.
 
 Apply the lemma to the words at z and at z+u v_{s,p}, where 1 <= u < p_1.
@@ -707,7 +707,7 @@ length:
       <= w Delta(x,y) + w ||x|-|y||
       <= 2w Delta(x,y).
 
-The last inequality follows because one insertion or deletion changes length
+The last ineuality follows because one insertion or deletion changes length
 by one. QED.
 
 The recurrence for Q_s is bounded first. For sufficiently large k, 2P <=
@@ -724,7 +724,7 @@ Then
       <= P^{-b/6}
       <= h^{-k}.
 
-For the last inequality, P=b^2, so
+For the last ineuality, P=b^2, so
 
     P^{b/6} = b^{b/3} = b^{(100/3)k} >= (b/10)^k = h^k.
 
@@ -853,8 +853,8 @@ on gaps between successive lengths N_k.
 
 Finally, choose two letters in any finite alphabet Sigma of size at least
 two. The strings over those two letters form an isometric copy of binary edit
-distance. One inequality follows because every binary edit script is also a
-Sigma-script. For the reverse inequality, project all other letters to one of
+distance. One ineuality follows because every binary edit script is also a
+Sigma-script. For the reverse ineuality, project all other letters to one of
 the chosen letters; each operation in a Sigma-script becomes at most one
 binary edit. Restricting an embedding to this subspace gives
 
@@ -880,7 +880,7 @@ where
 
 Here ED permits unit-cost substitutions as well as insertions and deletions.
 
-https://web.cs.ucla.edu/~rafail/PUBLIC/68.pdf
+./pdf/unknown-nd.pdf
 
 On the finite domain {0,1}^m, the minimum pairwise ratio of image distance to
 edit distance is positive. Dividing by that ratio makes the theorem