Answers.Body: <p>You can easily find a counterexample. Suppose that there are three nodes <span class="math-container">$s$</span>, <span class="math-container">$p$</span>, and <span class="math-container">$goal$</span> such that <span class="math-container">$s \rightarrow p \rightarrow goal$</span>. The real cost of going from <span class="math-container">$s$</span> to <span class="math-container">$p$</span> is <span class="math-container">$c(s,p) = 10$</span> and <span class="math-container">$c(p, goal) = 10$</span>. Also, <span class="math-container">$h_1(s) = 18$</span>, <span class="math-container">$h_1(p) = 9$</span>, <span class="math-container">$h_1(goal) = 0$</span>, <span class="math-container">$h_2(s) = 17$</span>, <span class="math-container">$h_2(p)  = 1$</span>.On the other hand, <span class="math-container">$h^*(s) = 19$</span> and <span class="math-container">$h^*(p) = 10$</span>.</p>
  <p>Now, <span class="math-container">$h_1(s) \leqslant c(s,p) + h_1(p)$</span> and <span class="math-container">$h_1(s) \leqslant c(s, goal) + h_1(goal)$</span> that satisfies the consistency constraint of <span class="math-container">$h_1$</span>. Also, <span class="math-container">$h_2(s) \leqslant h^*(s)$</span> and <span class="math-container">$h_2(s) \leqslant h^*(p)$</span> that approve <span class="math-container">$h_2$</span> is admissible.</p>
  <p>However, for any node <span class="math-container">$n$</span>, if we define <span class="math-container">$h_3(n) = \min(h_1(n), h_2(n))$</span>:</p>
  <p><span class="math-container">$$h_3(s) = \min(h_1(s), h_2(s)) = 17 \not\leqslant c(s,p) + \min(h_1(p), h_2(p)) = 
  $$</span>
  <span class="math-container">$$c(s,p) + h_3(p) = 10 + 1 = 11. $$</span>
  It means that <span class="math-container">$h_3$</span> as a heuristic function is not consistent.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-05-08T01:44:01.713Z
Answers.LastActivityDate: 2021-05-09T13:54:18.283Z
Answers.LastEditDate: 2021-05-09T13:54:18.283Z
Answers.UserDisplayName: OmG
Body: <p>If <span class="math-container">$h_1(s)$</span> is a consistent heuristic and <span class="math-container">$h_2(s)$</span> is a admissible heuristic, is <span class="math-container">$\min(h_1(s),\ h_2(s))$</span> consistent?</p>
  
Tags: <search><proofs><admissible-heuristic><consistent-heuristic><heuristic-functions>
Title: Is $\min(h_1(s),\ h_2(s))$ consistent?
UserDisplayName: hello
ThreadText: # Is $\min(h_1(s),\ h_2(s))$ consistent?
  
  (Asked by: username_0 on 2021-05-06)
  
  If $h\_1(s)$ is a consistent heuristic and $h\_2(s)$ is a admissible heuristic, is $\min(h\_1(s),\ h\_2(s))$ consistent?
  
  **Tags:** search, proofs, admissible-heuristic, consistent-heuristic, heuristic-functions
  
  ---
  
  ## 1 answers
  
  ### Answer by username_1 on 2021-05-08. Score: 1
  
  You can easily find a counterexample. Suppose that there are three nodes $s$, $p$, and $goal$ such that $s \rightarrow p \rightarrow goal$. The real cost of going from $s$ to $p$ is $c(s,p) = 10$ and $c(p, goal) = 10$. Also, $h\_1(s) = 18$, $h\_1(p) = 9$, $h\_1(goal) = 0$, $h\_2(s) = 17$, $h\_2(p) = 1$.On the other hand, $h^\*(s) = 19$ and $h^\*(p) = 10$.
  
  Now, $h\_1(s) \leqslant c(s,p) + h\_1(p)$ and $h\_1(s) \leqslant c(s, goal) + h\_1(goal)$ that satisfies the consistency constraint of $h\_1$. Also, $h\_2(s) \leqslant h^\*(s)$ and $h\_2(s) \leqslant h^\*(p)$ that approve $h\_2$ is admissible.
  
  However, for any node $n$, if we define $h\_3(n) = \min(h\_1(n), h\_2(n))$:
  
  $$h\_3(s) = \min(h\_1(s), h\_2(s)) = 17 \not\leqslant c(s,p) + \min(h\_1(p), h\_2(p)) =
  $$
  $$c(s,p) + h\_3(p) = 10 + 1 = 11. $$
  It means that $h\_3$ as a heuristic function is not consistent.
  
  
  ---
  
  