1 files changed, 0 insertions, 107 deletions
diff --git a/app-arch/lbzip2/files/lbzip2-2.2-assertion.patch b/app-arch/lbzip2/files/lbzip2-2.2-assertion.patch
deleted file mode 100644
index b4946175b947..000000000000
--- a/app-arch/lbzip2/files/lbzip2-2.2-assertion.patch
+++ /dev/null
@@ -1,107 +0,0 @@
-From 57eeee36927f8a40ece1ca06c674e0bd56d0f21f Mon Sep 17 00:00:00 2001
-Message-Id: <57eeee36927f8a40ece1ca06c674e0bd56d0f21f.1358019732.git.jlec@gentoo.org>
-From: Mikolaj Izdebski <zurgunt@gmail.com>
-Date: Sat, 20 Oct 2012 18:37:17 +0200
-Subject: [PATCH] Fix assertion failure, closes #8
-
-src/encode.c (generate_initial_trees): Rewrite from scratch.
----
- src/encode.c | 61 +++++++++++++++++++++++++++++++++++++++++++++++-------------
- 1 file changed, 48 insertions(+), 13 deletions(-)
-
-diff --git a/src/encode.c b/src/encode.c
-index 09cfacc..00a78dc 100644
---- a/src/encode.c
-+++ b/src/encode.c
-@@ -763,40 +763,75 @@ assign_codes(uint32_t C[], uint32_t L[], uint8_t B[], uint32_t n)
- 
- /* Create initial mapping of symbols to trees.
- 
--   The goal is to divide all as symbols [0,as) into nt equivalence classes
-+   The goal is to divide all as symbols [0,as) into nt equivalence classes (EC)
-    [0,nt) such that standard deviation of symbol frequencies in classes is
-    minimal. We use a kind of a heuristic to achieve that. There might exist a
-    better way to achieve that, but this one seems to be good (and fast) enough.
- 
-    If the symbol v belongs to the equivalence class t then set s->length[t][v]
-    to zero. Otherwise set it to 1.
--
--   TODO: This piece of code really needs some R&D...
- */
- static void
- generate_initial_trees(struct encoder_state *s, unsigned nm, unsigned nt)
- {
--  unsigned a, b, c, t;
--
--  /* Equivalence classes are empty. */
-+  unsigned a, b;   /* range [a,b) of symbols forming current EC */
-+  unsigned freq;   /* symbol frequency */
-+  unsigned cum;    /* cumulative frequency */
-+  unsigned as;     /* effective alphabet size (alphabet size minus number
-+                      of symbols with frequency equal to zero) */
-+  unsigned t;      /* current tree */
-+
-+  /* Equivalence classes are initially empty. */
-   memset(s->length, 1, sizeof(s->length));
- 
-+  /* Determine effective alphabet size. */
-+  as = 0;
-+  for (a = 0, cum = 0; cum < nm; a++) {
-+    freq = s->lookup[0][a];
-+    cum += freq;
-+    as += min(freq, 1);
-+  }
-+  assert(cum == nm);
-+
-+  /* Bound number of EC by number of symbols. Each EC is non-empty, so number
-+     of symbol EC must be <= number of symbols. */
-+  nt = min(nt, as);
-+
-   /* For each equivalence class: */
--  for (a = 0, t = 0; t < nt; t++) {
-+  a = 0;
-+  for (t = 0; nt > 0; t++, nt--) {
-+    assert(nm > 0);
-+    assert(as >= nt);
-+
-     /* Find a range of symbols which total count is roughly proportional to one
-        nt-th of all values. */
--    for (c = 0, b = a; c * (nt-t) < nm; b++)
--      c += s->lookup[0][b];
--    assert(a < b);
--    if (a < b-1 && (2*c - s->lookup[0][b-1]) * (nt-t) > 2*nm) {
--      c -= s->lookup[0][--b];
-+    freq = s->lookup[0][a];
-+    cum = freq;
-+    as -= min(freq, 1);
-+    b = a+1;
-+    while (as > nt-1 && cum * nt < nm) {
-+      freq = s->lookup[0][b];
-+      cum += freq;
-+      as -= min(freq, 1);
-+      b++;
-     }
--    nm -= c;
-+    if (cum > freq && (2*cum - freq) * nt > 2*nm) {
-+      cum -= freq;
-+      as += min(freq, 1);
-+      b--;
-+    }
-+    assert(a < b);
-+    assert(cum > 0);
-+    assert(cum <= nm);
-+    assert(as >= nt-1);
-+    Trace(("Tree %u: EC=[%3u,%3u), |EC|=%3u, cum=%6u", t, a, b, b-a, cum));
- 
-     /* Now [a,b) is our range -- assign it to equivalence class t. */
-     bzero(&s->length[t][a], b - a);
-     a = b;
-+    nm -= cum;
-   }
-+  assert(as == 0);
-   assert(nm == 0);
- }
- 
--- 
-1.8.1
-