Question

Help with Pascal’s Triangle: Paths and Binary Strings

Suppose you want to create a path between each number on Pascal's Triangle. For this exercise, suppose the only moves allowed are to go down one row either to the left or to the right. We will code the path by using bit strings. In particular, a O will be used for each move downward to the left, and a 1 for each move downward to the right. So, for example, consider the first five rows of Pascal's Triangle below, and the path shown between the top number 1 (labelled START) and the left-most 3. START Row Row Row 001 Row 10 Row This path involves starting at the top 1 labelled START and first going down and to the left (code with a 0), then down to the left again (code with another O), and finally down to the right (code with a 1). Hence, this path would be coded with binary string 001. This code is then recorded at the ending location on the triangle. For Option #1, complete the following tasks based on the coding scheme described above: a. Determine if there are additional paths between the START (number 1 at the top) and end point (leftmost 3). If so, describe them in words, by tracing the path on the triangle, and as binary strings. Then record all such binary strings at the ending location. b. Find and record at least 5 paths using binary strings between the START location and numbers in rows 4 and 5. Compare the binary strings for each number and discuss why those similarities and differences might exist. Also, explain any connections you notice between the number of possible paths ending at each location and the corresponding entry of Pascal's Triangle. c. Discuss what information about an endpoint and the paths leading to it can be gathered from the length of its binary strings and the number of 1s in them. d. Explain the addition rule of Pascal's Triangle in your own words in terms of the path coding scheme you worked with in this assignment. e. Note the type of symmetry that Pascal's Triangle has and explain it in terms of the paths.

Answer 1

a. The additional paths between the starting point (number 1 at the top) and

the ending point (leftmost 3) as shown in the given diagram are

1. Going down left, then down right and finally down left. The corresponding binary string is 010.

.

Binary string: START olo O O O O 5 10 10

2. Going down right, then down left and again down left. The corresponding binary string is 100.

Binary string: START 100 o O O 5 10 10

b. Consider the leftmost 4 in the 4th row. The paths are

.

START Binary string: No. of paths 0001, 0010, 0100, 1000 O O 5 10 10

Consider the middle 6 in the 4th row. The paths are

.

START No. of paths Binary string : 0011, 0101, 010, 1010, 6 1100 ,1001 O 5 10 10

Consider the rightmost 4 in the 4th row. The paths are

.

Binary string: START No. of paths -4 1110, 1110, 1011, 0111 ☺ O 5 10 10

Consider the leftmost 5 in the 5th row. The paths are

.

Binary string: START No. of paths 00001, 00010, - 5 G 00100, 01000, 10000 O 10 10

Consider the rightmost 5 in the 5th row. The paths are

.

Binary string: START No. of paths 11110, 11101, - 5 11011, 10ill, 01111 O O 5 10 10

Similarities and differences:

The number of digits in a binary string of n-th row is n. If we ignore the border line 1s of the triangle and begin counting from the left, then number of 1s in the binary string in the m-th number of a row is m.

Also, the number of paths for a particular entry is k.

c. As already stated in (b) , the length of a binary string of a number in n-th row is n. The number of 1s in the binary string in the m-th number of a row is m.

d. Consider two entries and in a particular row. Adding them, one will get the entry in the next row.

Hence number of paths leading to the entry in a particular row is the sum of the numbers of paths leading to the two entries and in the previous row.