Assignment #12
Nicole Mosteller
EMAT 6680
Investigation #7: Problem: Place four numbers in the first row as follows
A B C D.
For each successive row replace the entries by the absolute value of the difference of the entry just above and the entry
just to the right in the previous row. In the fourth position use the absolute value of the difference of the fourth and the first (i. e. cycle)
|A - B| |B - C| |C - D| |D - A|.
To begin this investigation, I chose the most obvious entries for A, B, C, and D - 1, 2, 3, and 4 respectively.
Following the instructions, I achieved the matrix in Figure 1.
A |
B |
C |
D |
1 |
2 |
3 |
4 |
1 |
1 |
1 |
3 |
0 |
0 |
2 |
2 |
0 |
2 |
0 |
2 |
2 |
2 |
2 |
2 |
0 |
0 |
0 |
0 |
Figure 1.
This general case gives an idea that successive differences will eventually lead to a row of zeroes.
This case also allows us to anticipate the rows preceding the row of zeroes.
Let the row of zeroes be in the ith row, then the (i-1)th row must have the same value,
and in the (i-2)th row, the values in Column A = Column C while the values in Column B = Column D.
In the case were A = 1, B = 2, C = 3, and D = 4, we see that the zeroes appear in then 7th row.
To test this hypothesis, I decided to generate many tables with random values for A, B, C, and D.
Microsoft Excel makes finding random values as well as the recursive differences easy to find.
Below shows several of the tables generated with Excel.
A |
B |
C |
D |
11 |
961 |
360 |
301 |
950 |
601 |
59 |
290 |
349 |
542 |
231 |
660 |
193 |
311 |
429 |
311 |
118 |
118 |
118 |
118 |
0 |
0 |
0 |
0 |
Figure 2: The 6th row gives the row of zeroes. Notice that the cells in the 5th row have equal values,
and the 4th row has Cell A = Cell C and Cell B = Cell D.
A |
B |
C |
D |
40 |
656 |
516 |
369 |
616 |
140 |
147 |
329 |
476 |
7 |
182 |
287 |
469 |
175 |
105 |
189 |
294 |
70 |
84 |
280 |
224 |
14 |
196 |
14 |
210 |
182 |
182 |
210 |
28 |
0 |
28 |
0 |
28 |
28 |
28 |
28 |
0 |
0 |
0 |
0 |
Figure 3: The 10th row gives the row of zeroes. Notice that the cells in the 9th row have equal values,
and the 8th row has Cell A = Cell C and Cell B = Cell D.
A |
B |
C |
D |
217 |
951 |
744 |
796 |
734 |
207 |
52 |
579 |
527 |
155 |
527 |
155 |
372 |
372 |
372 |
372 |
0 |
0 |
0 |
0 |
Figure 4: The 5th row gives the row of zeroes. Notice that the cells in the 4th row have equal values,
and the 3th row has Cell A = Cell C and Cell B = Cell D.
To make your own investigation
Click Here! for the Random Numbers for A, B, C, and D on Excel.
From this investigation, I became interested to know if it was possible to arrive at the zero row after more than 10 recursions.
And after a paper and pencil investigation, I found that it is possible. The case that I found was just one case out of 4,294,967,296 cases.
Click Here! to see a the work into this investigation as well as an explanation.
After discussing this problem with another UGA Student - Thanks to Signe Kastberg! - since we are not
restricted in choosing A, B, C, and D from the positive intergers, we see that it is possible
to find cases where the zero row occurs after 10 recursions. See Figure 5 below.
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