pan/lcra: Use Chaitin's spilling heuristic

Not much of a difference but slightly better and slightly less
arbitrary.

total instructions in shared programs: 3560 -> 3559 (-0.03%)
instructions in affected programs: 44 -> 43 (-2.27%)
helped: 1
HURT: 0

total bundles in shared programs: 1844 -> 1843 (-0.05%)
bundles in affected programs: 23 -> 22 (-4.35%)
helped: 1
HURT: 0

Signed-off-by: Alyssa Rosenzweig <alyssa.rosenzweig@collabora.com>

diff --git a/src/panfrost/midgard/lcra.c b/src/panfrost/midgard/lcra.c
index d86c35ffe09df1c3fe31725da4a2877e9da95458..f5e2ff2d8fec14e619d5cc4cdae3d334ab9dfc5a 100644 (file)
--- a/src/panfrost/midgard/lcra.c
+++ b/src/panfrost/midgard/lcra.c
@@ -201,7 +201,7 @@ lcra_count_constraints(struct lcra_state *l, unsigned i)
 signed
 lcra_get_best_spill_node(struct lcra_state *l)
 {
-        signed best_benefit = INT_MIN;
+        float best_benefit = -1.0;
         signed best_node = -1;
 
         for (unsigned i = 0; i < l->node_count; ++i) {
@@ -209,14 +209,10 @@ lcra_get_best_spill_node(struct lcra_state *l)
                 if (l->class[i] != l->spill_class) continue;
                 if (l->spill_cost[i] < 0) continue;
 
-                /* Compute the benefit of spilling a node as the number of
-                 * constraints on the node plus the number of slots it occupies
-                 * minus a multiple of the cost. TODO: Come up with a formula
-                 * whose use I can justify beyond "it makes my shaderdb look
-                 * happy" */
-
-                signed benefit = lcra_count_constraints(l, i);
-                benefit -= l->spill_cost[i] * 2;
+                /* Adapted from Chaitin's heuristic */
+                float constraints = lcra_count_constraints(l, i);
+                float cost = (l->spill_cost[i] + 1);
+                float benefit = constraints / cost;
 
                 if (benefit > best_benefit) {
                         best_benefit = benefit;