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Jan 14th, 2020
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  1. \KwIn{A matrix $A$ of size $n \times m$}
  2. \KwOut{A column index $i$ representing the location of the tallest building.}
  3. \tcc{Initialize location variable.}
  4. $tallestLocation \gets 0$
  5.  
  6. $tallestHeight \gets 0$
  7.  
  8. \For{$i \in \{0,\ldots,m-1\}$}{
  9. \tcc{Use binary search to get the height of the column and compare it to the current tallest height}
  10. \tcc{Now choose an index p to partition around}
  11. $p \gets \lfloor n/2 \rfloor$
  12.  
  13. \While{$p \leq n-1$ \text{and} $p \geq 0$} {
  14. \If {$A[i, p] = 0$} {
  15. \tcc{Height found}
  16. \If {the number above = 1 OR $p = n-1$ OR $p=0$} {
  17. \If {$p \geq tallestHeight$} {
  18. $tallestHeight \gets j$
  19.  
  20. $tallestLocation \gets i$
  21. }
  22. break out of while loop
  23. }
  24. \Else {
  25. \tcc{Set new partition point}
  26. $p \gets midpoint of $
  27. }
  28. }
  29. }
  30. }
  31. \caption{\textsc{findTallest1} returns the location of the tallest building.}\label{alg:alg2}
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