tests(eviction/lfu): add tie-break tests to assert LRU-on-ties behavior (insert order and access order)

Author: hyp3rd

pkg/eviction/lfu_test.go

@@ -0,0 +1,61 @@
+package eviction
+
+import "testing"
+
+// Test that when two items have equal frequency, the older (least-recent) is evicted first.
+func TestLFU_EvictsOldestOnTie_InsertOrder(t *testing.T) {
+	lfu, err := NewLFUAlgorithm(2)
+	if err != nil {
+		t.Fatalf("NewLFUAlgorithm error: %v", err)
+	}
+
+	lfu.Set("a", 1)
+	lfu.Set("b", 2)
+
+	// Both have count=1, but "a" is older (inserted first), so it should be evicted first.
+	key, ok := lfu.Evict()
+	if !ok {
+		t.Fatalf("expected an eviction, got none")
+	}
+	if key != "a" {
+		t.Fatalf("expected 'a' to be evicted first on tie, got %q", key)
+	}
+
+	// Next eviction should evict the remaining one
+	key, ok = lfu.Evict()
+	if !ok {
+		t.Fatalf("expected a second eviction, got none")
+	}
+	if key != "b" {
+		t.Fatalf("expected 'b' to be evicted second, got %q", key)
+	}
+}
+
+// Test that recency is used to break ties after accesses with equalized frequency.
+func TestLFU_EvictsOldestOnTie_AccessOrder(t *testing.T) {
+	lfu, err := NewLFUAlgorithm(2)
+	if err != nil {
+		t.Fatalf("NewLFUAlgorithm error: %v", err)
+	}
+
+	lfu.Set("a", 1)
+	lfu.Set("b", 2)
+
+	// Make both have the same frequency (2) but different recency orders.
+	// Access sequence: bump "b" first, then "a" -> both count=2, and "a" is more recent.
+	if _, ok := lfu.Get("b"); !ok {
+		t.Fatalf("expected to get 'b'")
+	}
+	if _, ok := lfu.Get("a"); !ok {
+		t.Fatalf("expected to get 'a'")
+	}
+
+	// On tie (count=2), the older (least-recent) is "b" now, so it should be evicted first.
+	key, ok := lfu.Evict()
+	if !ok {
+		t.Fatalf("expected an eviction, got none")
+	}
+	if key != "b" {
+		t.Fatalf("expected 'b' to be evicted first on tie after accesses, got %q", key)
+	}
+}