Team Members
I received this data set from Elizabeth Jones, Raju Patel, and Corrie Collier.
Materials
This lab requires colored paper, scissors, and three index cards labeled A, B, and C.
Procedure
Cut out different colored squares of increasing size (we used five). The index cards A, B, and C represent 3 poles and the squares represent disks of increasing size. The disks are placed on pole A in ascending order with the largest on the bottom and the smallest on the top. The end goal is to get all of the disks in the same ascending arrangement onto pole B. However, only one disk may be moved at a time and a larger disk may never be placed on top of a smaller disk. Solve the problem for different numbers of disks and record the number of moves required to solve the problem.
Data Set
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Scatter Plot
Analysis of Data
Let x = the number of disks used. Let y = f(x) = the number of moves required to solve the puzzle for x disks. The domain of the function is the nonnegative integers (or as many as you have the time and energy to pursue; our choice of domain {1,2,3,4,5} was sufficient to reveal the pattern). The range of the function is also the nonnegative integers.
The sharply increasing y-values and rate of change shown by the scatter plot suggest an exponential function. In fact, a close at the data reveals that our data set satisfies the function y=f(x)=2^x-1.
Just out of curiosity, I would like to arrive at this nice closed form function from a regression equation. However, Microsoft Excel uses only base e exponential functions. Fortunately, the
TI -83 graphing calculator produces exponential regression equations of the form y=A*(B)^x.
My first attempt result in the following equation which is quite different from y=2^x — 1.
Y=(.4936582871)*(2.334368097)^x
Notice that the generic y=A*B^x equation has y-intercept of A. In the simplest exponential functions the y-intercept is 1. However, our y=2^x —1 is extremely simple yet has y-intercept 0. ( 0 disks can be moved in 0 moves.) This occurs because the function is shifted vertically down 1 unit; that is, all the y-values are decreased by 1 unit from a standard exponential function with y-intercept 1.
Since the calculator’s generic equation does not contain a constant for vertical shift, I will add 1 to each of my y-values. Then when the calculator produces an equation y=A*B^x for the new data, it will give us the corresponding equation y=A*(B)^x —1 for our original data. Finally ,we have presented the data in a way that allows the calculator to recognize the perfect fit. The constants returned from our regression equation are A=1 and B=2, so indeed y = 2^x -1.
Conclusion
Just as we observed with Now You’ve Got It, exponential functions exhibit increasingly large increases in y-value for relatively small increases in the x-value for larger x-values. The real life application is this: Don’t attempt to solve the Towers of Hanoi puzzle by hand for even 10 disks unless you have lots of free time since you would have to make ((2)^10)-1 = 1023 moves!