Add comments about the choice of memory size for sort managers

src/phase2.hpp

@@ -147,6 +147,11 @@ Phase2Results RunPhase2(
         // * update (pos, offset) to remain valid after table_index-1 has been
         //   compacted.
         // * sort by pos
+        //
+        // As we have to sort two adjacent tables at the same time in phase 3,
+        // we can use only a half of memory_size for SortManager. However,
+        // table 1 is already sorted, so we can use all memory for sorting
+        // table 2.
 
         auto sort_manager = std::make_unique<SortManager>(
             table_index == 2 ? memory_size : memory_size / 2,

src/phase3.hpp

@@ -190,6 +190,8 @@ Phase3Results RunPhase3(
             L_sort_manager->FreeMemory();
         }
 
+        // We read only from this SortManager during the second pass, so all
+        // memory is available
         R_sort_manager = std::make_unique<SortManager>(
             memory_size,
             num_buckets,
@@ -363,8 +365,12 @@ Phase3Results RunPhase3(
             L_sort_manager.reset();
         }
 
-        // L sort manager will be used for the writer, and R sort manager will be used for the
-        // reader
+        // In the second pass we read from R sort manager and write to L sort
+        // manager, and they both handle table (table_index + 1)'s data.
+        // For tables below 6 we can only use a half of memory_size since it
+        // will be sorted in the first pass of the next iteration together with
+        // the next table, which will use the other half of memory_size.
+        // Tables 6 and 7 will be sorted alone, so we use all memory for them.
         R_sort_manager->FreeMemory();
         L_sort_manager = std::make_unique<SortManager>(
             (table_index >= 5) ? memory_size : (memory_size / 2),