Answers.Body: <p>No, it will not necessary be consistent or admissible. Consider this example, where <span class="math-container">$s$</span> is the start, <span class="math-container">$g$</span> is the goal, and the distance between them is 1.</p>
  
  <pre>
  s --1-- g
  </pre>
  
  <p>Assume that <span class="math-container">$h_0$</span> and <span class="math-container">$h_1$</span> are perfect heuristics. Then <span class="math-container">$h_0(s) = 1$</span> and <span class="math-container">$h_1(s) = 1$</span>. In this case the heuristic is inadmissible because <span class="math-container">$h_0(s)+h_1(s) = 2 &gt; d(s, g)$</span>. Similarly, as an undirected graph the heuristic will be inconsistent because <span class="math-container">$|h(s)-h(g)| &gt; d(s, g)$</span>.</p>
  
  <p>If you'd like to understand the conditions for the sum of heuristics to be consistent and admissible, I would look at the work on <a href="http://www.bgu.ac.il/~felner/2004/jairpdb.pdf" rel="noreferrer">additive PDB heuristics</a>.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2020-02-28T16:46:51.143Z
Answers.Text: For a more extreme version of this answer, consider taking a single admissible, consistent heuristic, and then adding up an infinite number of copies of them. For any base heuristic value $> 0$, this sum is going to end up being $\infty$, which is generally not admissible.
Answers.UserDisplayName: Dennis Soemers
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2020-02-28T16:33:33.900Z
Answers.LastActivityDate: 2020-02-28T16:41:38.610Z
Answers.LastEditDate: 2020-02-28T16:41:38.610Z
Answers.UserDisplayName: Nathan S.
Body: <p>Imagine that we have a set of heuristic functions <span class="math-container">$\{h_i\}_{i=1}^N$</span>, where each <span class="math-container">$h_i$</span> is both admissible and consistent (monotonic). Is <span class="math-container">$\sum_{i=1}^N h_i$</span> still consistent or not?</p>
  
  <p>Is there any proof or counterexample to show the contradiction?</p>
  
Tags: <search><proofs><heuristics><admissible-heuristic><consistent-heuristic>
Title: Is the summation of consistent heuristic functions also consistent?
UserDisplayName: Mostafa Ghadimi
ThreadText: # Is the summation of consistent heuristic functions also consistent?
  
  (Asked by: username_0 on 2020-02-28)
  
  Imagine that we have a set of heuristic functions $\{h\_i\}\_{i=1}^N$, where each $h\_i$ is both admissible and consistent (monotonic). Is $\sum\_{i=1}^N h\_i$ still consistent or not?
  
  Is there any proof or counterexample to show the contradiction?
  
  **Tags:** search, proofs, heuristics, admissible-heuristic, consistent-heuristic
  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_1 on 2020-02-28. Score: 5
  
  No, it will not necessary be consistent or admissible. Consider this example, where $s$ is the start, $g$ is the goal, and the distance between them is 1.
  
  ```
  
  s --1-- g
  
  ```
  
  Assume that $h\_0$ and $h\_1$ are perfect heuristics. Then $h\_0(s) = 1$ and $h\_1(s) = 1$. In this case the heuristic is inadmissible because $h\_0(s)+h\_1(s) = 2 > d(s, g)$. Similarly, as an undirected graph the heuristic will be inconsistent because $|h(s)-h(g)| > d(s, g)$.
  
  If you'd like to understand the conditions for the sum of heuristics to be consistent and admissible, I would look at the work on [additive PDB heuristics](http://www.bgu.ac.il/~felner/2004/jairpdb.pdf).
  
  > **Comments:**
  > **username_2**: For a more extreme version of this answer, consider taking a single admissible, consistent heuristic, and then adding up an infinite number of copies of them. For any base heuristic value $> 0$, this sum is going to end up being $\infty$, which is generally not admissible.  
  
  ---
  
  