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POBinary chop: if list[middle] == key case
 

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  1. POBinary chop: if list[middle] == key case
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    <p>I'm revising algorithms for my exams and I was trying to solve this exercise but I couldn't come up with a solution.</p> <p>This is the pseudo-code.</p> <pre><code>1. int search (int [] a, int x) { 2. // Pre: ∃i:Nat (0≤i&lt;a.length ∧ a[i]=x) ∧ a is in ascending order 3. // Post: 0≤ r≤ a.length ∧ 4. // ∀i:int.(0 ≤ i &lt; r → a[i] &lt; x) ∧ a[r]=x 5. int left, middle; int right = a.length; 6. if (a[0]&gt;=x) return 0; else left=0; //a[left]&lt;x (a) 7. while ((right-left&gt;1){ (b) 8. // invariant: (I1) 0≤ left &lt; right ≤ a.length ∧ 9. // (I2) ∀i:int(0 ≤ i ≤ left → a[i] &lt; x) ∧ 10. // (I3) ∀i:int(right ≤ i &lt; a.length → a[i] &gt; x) 11. // (note: a[i]&gt;x not a[i]≥x) 12. // Variant: right-left-1 13. middle = (left+right) / 2; // left &lt; middle &lt; right (*) 14. if ( a[middle]== x) return middle; 15. else {if a[middle)&lt;x) left = middle ; 16. else right=middle;} (c) 17. } 18. } // left+1=right } (d) </code></pre> <p>So, the code as it is now doesn't satisfy the post condition because for certain inputs (e.g. x = 1 and a={0,1,1,1,1}) the value returned by line 14 doesn't satisfy the post-condition on line 4. The exercise is asking to : "Replace "return middle;" on line 14. by a while loop and return code so that it satisfies the postcondition. Include variant and invariant strong enough to imply the postcondition on return. Hint: Make sure you include to state what doesn't change."</p> <p>I have not been able to find a solution. Can anyone help me?</p> <p>Thanks in advance, VJ</p> <p>EDIT: Ok, thank you both of you for your help.</p> <pre><code>while(middle &gt; 0 &amp;&amp; a[middle] == x){ middle--; </code></pre> <p>} return middle;</p> <p>I chose the variant to be middle. And the invariant to be :</p> <p>0x</p> <p>Do You reckon this is correct?</p>
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