Answers.Body: <p>Define P as a CSP where X, Y are the variables, domain of both is {1,2,3,4} and conditions in normal form are:</p>
  
  <ol>
  <li>node-condition X&lt;4</li>
  <li>arc-condition X=Y</li>
  </ol>
  
  <p>P is 2-consistent (arc consistent) because for any X value it is possible to find a Y value that fulfills the arc-condition X=Y.</p>
  
  <p>However, P is not 1-consistent (node-consistent) because exist a X value (X=4) that can not fulfill the node condition x&lt;4.</p>
  
  <p>For these reasons, this problem is 2-consistent but not strongly 2-consistent.  </p>
  
  <p>Obviously, it is straightforward to convert this example in a strongly 2-consistent problem, just reducing the domain to {1,2,3}.</p>
  
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-18T16:42:19.630Z
Answers.LastActivityDate: 2018-02-18T19:00:54.843Z
Answers.LastEditDate: 2018-02-18T19:00:54.843Z
Answers.UserDisplayName: pasaba por aqui
Body: <p>From Russell-Norvig:</p>
  
  <blockquote>
    <p>A CSP is strongly k-consistent if it is k-consistent and is also (k − 1)-consistent, (k − 2)-consistent, . . . all the way down to 1-consistent.</p>
  </blockquote>
  
  <p>How can a CSP be k-consistent without being (k - 1)-consistent? I can't think of any counter example for this case. Any help would be appreciated.</p>
  
Tags: <linear-algebra>
Title: Does k consistency always imply (k - 1) consistency?
UserDisplayName: amad-person
ThreadText: # Does k consistency always imply (k - 1) consistency?
  
  (Asked by: username_0 on 2018-02-18)
  
  From Russell-Norvig:
  
  > A CSP is strongly k-consistent if it is k-consistent and is also (k − 1)-consistent, (k − 2)-consistent, . . . all the way down to 1-consistent.
  
  How can a CSP be k-consistent without being (k - 1)-consistent? I can't think of any counter example for this case. Any help would be appreciated.
  
  **Tags:** linear-algebra
  
  ---
  
  ## 1 answers
  
  ### Answer by username_1 on 2018-02-18. Score: 2
  
  Define P as a CSP where X, Y are the variables, domain of both is {1,2,3,4} and conditions in normal form are:
  
  1. node-condition X<4
  2. arc-condition X=Y
  
  P is 2-consistent (arc consistent) because for any X value it is possible to find a Y value that fulfills the arc-condition X=Y.
  
  However, P is not 1-consistent (node-consistent) because exist a X value (X=4) that can not fulfill the node condition x<4.
  
  For these reasons, this problem is 2-consistent but not strongly 2-consistent.
  
  Obviously, it is straightforward to convert this example in a strongly 2-consistent problem, just reducing the domain to {1,2,3}.
  
  
  ---
  
  