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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