compiler: Split up Cluster.used_dimensions to fix Lift

Author: FabioLuporini

devito/ir/clusters/cluster.py

@@ -172,21 +172,34 @@ def dimensions(self):
         return set().union(*[i._defines for i in self.ispace.dimensions])
 
     @cached_property
-    def used_dimensions(self):
+    def exprs_dimensions(self):
         """
-        The Dimensions that *actually* appear among the expressions in ``self``.
-        These do not necessarily coincide the IterationSpace Dimensions; for
-        example, reduction or redundant (i.e., invariant) Dimensions won't
-        appear in an expression.
+        The Dimensions that appear explicitly in the Cluster expressions.
         """
-        dims_exprs = {i for i in self.free_symbols if i.is_Dimension}
-
+        dims_explicit = {i for i in self.free_symbols if i.is_Dimension}
         dims_implicit = set().union(*[set(e.implicit_dims) for e in self.exprs])
+        return dims_explicit | dims_implicit
 
+    @cached_property
+    def guards_dimensions(self):
+        """
+        The Dimensions that appear explicitly in the Cluster guards.
+        """
         syms_guards = set().union(*[e.free_symbols for e in self.guards.values()])
         dims_guards = {i for i in syms_guards if i.is_Dimension}
+        return dims_guards
 
-        return dims_exprs | dims_implicit | dims_guards
+    @cached_property
+    def used_dimensions(self):
+        """
+        All the Dimensions that appear explicitly either within the expressions
+        or the guards.
+
+        Note that, in some cases, some of the IterationSpace Dimensions might
+        not appear here among the used Dimensions -- for example, reduction or
+        redundant (i.e., invariant) Dimensions.
+        """
+        return self.exprs_dimensions | self.guards_dimensions
 
     @cached_property
     def dist_dimensions(self):

devito/passes/clusters/misc.py

@@ -55,7 +55,7 @@ def callback(self, clusters, prefix):
                 continue
 
             # Is `c` a real candidate -- is there at least one invariant Dimension?
-            if any(d._defines & hope_invariant for d in c.used_dimensions):
+            if any(d._defines & hope_invariant for d in c.exprs_dimensions):
                 processed.append(c)
                 continue
 
@@ -69,16 +69,16 @@ def callback(self, clusters, prefix):
             # All of the inner Dimensions must appear in the write-to region
             # otherwise we would violate data dependencies. Consider
             #
-            # 1)                 2)                          3)
-            # for i              for i                       for i
-            #   for x              for x                       for x
-            #     r = f(a[x])        for y                       for y
-            #                          r[x] = f(a[x, y])           r[x, y] = f(a[x, y])
+            # 1)                2)                        3)
+            # for i             for i                     for i
+            #   for x             for x                     for x
+            #     r = f(a[x])       for y                     for y
+            #                         r[x] = f(a[x, y])         r[x, y] = f(a[x, y])
             #
             # In 1) and 2) lifting is infeasible; in 3) the statement can
             # be lifted outside the `i` loop as `r`'s write-to region contains
             # both `x` and `y`
-            xed = {d._defines for d in c.used_dimensions if d not in outer}
+            xed = {d._defines for d in c.exprs_dimensions if d not in outer}
             if not all(i & set(w.dimensions) for i, w in product(xed, c.scope.writes)):
                 processed.append(c)
                 continue