ITEC 120 2008spring ibarland, jdymacek

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lect12b
while loops
continued

Review

• Work through an example by hand. What “so-far” information do you accumulate? (Make a local-variable to store that info, which will change over time.)
• What corresponds to the new-information-looking-at-right-now? (Make a local-variable to store that info, which will change over time.)
• For each of these two pieces of info, how do they change from one iteration to the next? That is, how do you construct the new values out of the pieces available to you?

We finished last lecture having walked through

/** Announce a countdown, starting at 10.
 * @return A countdown starting at 10.  
 *   That is, "10...9...8... [etc] 2...1...Liftoff!"
 */
public String countdown() {
  String chantSoFar = "";  /* A local variable, to accumulate the answer. */
  int t = 10;         /* The next number to count down. */

  while (t >= 0) {
    chantSoFar = chantSoFar + (t + "...");
    t = t-1;
    }

  return (chantSoFar + "Liftoff!");
  }

• Fix the bug about including 0.
• Change the function so that it starts the countdown from a specified point, not just 10.

  style tip: Don't just make the parameter named t and then assign to it (but mention that approach). Keeping the parameter untouched segues well into the next problem.

• Challenge: Without changing the answer being computed, can you change how the answer is arrived at, by t counts from 1 up to (say) 10, and prepends that number to chant?

  teaching comment: this was a useful exercise to do in class; or at least, it wasn't obvious to people!

  
  To think about:
  • When first working through this by hand, what “so-far” information do you accumulate? (Make a local-variable to store that info, which will change over time.)
    In for-each loops, we still had to do this.
  • What corresponds to the case-looking-at-right-now? (Make a local-variable to store that info, which will change over time.)
    In for-each loops, this was the local variable you declared.
  • For each of these two pieces of info, how do they change from one iteration to the next? That is, how do you construct the new values out of the pieces available to you?
    In for-each loops, we still had to do this.
  • Are you sure that your loop-condition will eventually become false? (Your loop-body had better change something to eventually cause this!)
    This issue is new with while loops; we didn't have it with for-each loops.

  Solution; .jar

Exercise In bowling, pins are arranged in a triangle with four layers; there are a total of 10 pins. You might be inspired on the weekends to make up your own bowling formats, lining up empty bottles of Grape Nehi, and realizing that you aren't confined to only four rows; you might have 5 rows (requiring 5 more empties, for a total of 15), or 6 rows (requiring yet 6 more empties, for a total of 21).

Of course, such home-grown bowling is only interesting for so long, and then you start wondering, in general, how many pins are needed for bowling-variants with other numbers of rows?

      o         ( 1 pin  in a 1-high triangle)
     o o        ( 3 pins in a 2-high triangle)
    o o o       ( 6 pins in a 3-high triangle)
   o o o o      (10 pins in a 4-high triangle)
  o o o o o            etc.
 o o o o o o

Before launching into the code, let's compute some examples/test-cases by hand, paying attention to our own thought process; that will give us a clue as to how to compute the general case.

/** Return the nth triangular number.
 * @param  n The number of rows in a triangle.
 * @return The number of pins needed to set up a n rows of pins,
 *         each row containing one more pin than the previous.
 *         That is, return 1+2+3+...+(n-1)+n.
 * Test cases:
 * triangularNumber(4) = 10
 * triangularNumber(6) = 21
 *   ...other tests?...
 */

teaching note: Be sure to run a couple of live tests of this, to set up triangularPyramid, which will call this method.

Solution; .jar

Exercise Consider stacking oranges. This isn't two-dimensional like bowling pins sitting on the floor; oranges get stacked in pyramids. Every pyramid begins with a single orange at its peak, but there are several ways to extend the pyramid.

• You could have square pyramids (like the Egyptian and Mayan tombs), one peak orange (layer 1) is supported by four oranges (layer 2), which is supported by nine oranges (layer 3), which is supported by sixteen oranges (layer 4), etc.. The cross-section of the nth layer is an n×n square.
  Write a program to compute the number of oranges in a square pyramid of height n.
  
• If you want your stack to look tall, then you might instead use triangular pyramids (tetrahedra): The top layer has one orange, the next has three oranges, the next has a triangle of six oranges, the next has a triangle of ten oranges, etc.. What is the general pattern, for the number of oranges on the nth layer?
  Hint: if the general case involves a triangularNumber, then your code should call that previous function! (Unless you are clever enough to find a closed-form formula for tetrahedral numbers.)
  Write a program to compute the number of oranges in a triangular pyramid of height n, using a while-loop.
  

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©2008, Ian Barland, Radford University Last modified 2008.Jan.24 (Thu)

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