From e56b32e5a8df18a931b83a0cd3af5b7f188d40a4 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=C3=89mile=20Nadeau?= <nadeau.emile@gmail.com>
Date: Tue, 27 Aug 2019 18:16:06 +0000
Subject: [PATCH] update readme to show the option to count objects

---
 README.rst | 19 +++++++++++++++++++
 1 file changed, 19 insertions(+)

diff --git a/README.rst b/README.rst
index 830a2e87e..a2906110c 100644
--- a/README.rst
+++ b/README.rst
@@ -177,6 +177,20 @@ the ``from_dict`` function is written in such a way that for any class
    ...                    data['alphabet'],
    ...                    bool(int(data['just_prefix'])))
 
+We also add some methods that we will need to get the enumerations of the
+objects.
+
+.. code:: python
+
+   ...     def is_epsilon(self):
+   ...         return self.prefix == "" and self.just_prefix
+   ...
+   ...     def is_atom(self):
+   ...         return len(self.prefix) == 1 and self.just_prefix
+   ...
+   ...     def is_positive(self):
+   ...         return len(self.prefix) > 0
+
 Our ``CombinatorialClass`` is now ready. What is left to do is create
 the strategies that the ``CombinatorialSpecificationSearcher`` will use
 for performing combinatorial exploration. This is given in the form of a
@@ -394,5 +408,10 @@ we see that the minimum polynomial satisfied by the generating function
 and moreover
 ``F = -(x**7 + x**5 + x**4 + x**3 + x**2 + 1)/(x**6 + x**3 - x**2 + 2*x - 1)``.
 
+.. code:: python
+
+   >>> tree.count_objects_of_length(10)
+   331
+
 You can now try this yourself using the file ``example.py``, which can
 count any set of words avoiding consecutive patterns.