changeset 121:87f69a87c491

Fix grammar.
author Oleksandr Gavenko <gavenkoa@gmail.com>
date Mon, 27 Mar 2017 15:44:35 +0300
parents 5cba87554532
children 2fea753a51ff
files bd51fb68-e005-11e6-8ee5-485b39c42d0f/index.rst
diffstat 1 files changed, 4 insertions(+), 4 deletions(-) [+]
line wrap: on
line diff
--- a/bd51fb68-e005-11e6-8ee5-485b39c42d0f/index.rst	Mon Mar 27 10:23:21 2017 +0300
+++ b/bd51fb68-e005-11e6-8ee5-485b39c42d0f/index.rst	Mon Mar 27 15:44:35 2017 +0300
@@ -19,7 +19,7 @@
 equal or fast one become ``None``. First incremented by one step, second incremented by two steps.
 They start from beginning of list with fast one step ahead.
 
-If there is loop with length ``len`` and loop begin in ``off`` offset at some step ``n`` we will
+If there is a loop with length ``len`` and loop begin from ``off`` offset at some step ``n`` we will
 have slow and fast pointer in same position::
 
   slow: off + {(n - off) mod len}
@@ -31,8 +31,8 @@
   1 + n = 0  (mod len)
   n = len - 1  (mod len)
 
-so with heppen after ``len-1`` or ``2*len-1`` or some ``X*len-1`` that should be greater then
-``off``.
+which will happen after ``len-1`` or ``2*len-1`` or some ``X*len-1`` step that should be greater
+then ``off``.
 
 Our lists may look like::
 
@@ -113,5 +113,5 @@
   True
 
 From above formula it take ``8`` steps as a first number of ``3*X - 1`` form that bigger then ``6``
-(this is happend when ``X=3``).
+(this is happened when ``X=3``).