## Description

- Write a Fortran integer function CONVERT(N,T,M)which converts an integer argument N into a sequence of decimal digits (i.e. characters) in the string t. The function returns the number of characters stored in t, so, if N is equal to 123, the value returned is 3, and the first three characters of T are ’1’, ’2’ and ’3’. m is the (original) length of T and it cannot be exceeded; if the length of t is insuﬃcient for n, all characters of t should be set to ”*” and the value returned is m.

Write also a Fortran program which reads several integer numbers, invokes function CONVERT and prints the results.

- Finding the shortest way from one city to another, possibly through a number of other cities, is known as the shortest path problem. There are many algorithms for this problem; the approach outlined below is based on the Dijkstra’s algorithm.

Let a quadratic (integer) array A[1 : n, 1 : n] describe the connections among a set of cities (denoted “1”to “n”), where A[i, j] = A[j, i] is the distance from city “i” to city “j” if there is a direct connection between “i” and “j”; if there is no direct connection, a[i, j] = a[j, i] = 0. The iterative algorithm finding the shortest path from m to k uses an auxiliary array B[1 : n] which, at each step of iteration, indicates the shortest paths from m to each other node that has been reached so far. Initially, all elements of B are set to a large value, for example, 1 plus the sum of all elements of A, and B[m] is set to zero. In each iteration, the new value of B[i], i = m, is the smallest value of the sum B[j] + A[j, i] for all cities “j” which are directly connected to “i”:

B[i] = min (B[j] + A[j, i]), 1 ≤ i ≤ n, i = m,

1≤J≤N

A[j,i]=06

and the iteration continues as long as there are changes in B; when the values of B do not change, the algorithm ends and B[k] is the length of the shortest path from m to k.

Example: For a set of four cities, let:

0 | 10 | 5 | 0 | |||

A = | 10 | 0 | 3 | 2 | ||

5 | 3 | 0 | 6 | |||

0 | 2 | 6 | 0 |

The shortest path from “1” to “4” is found in the following way. Initially, B = [0, 53, 53, 53] (52 is the sum of all elements of A). Then, since “2” is connected to “1”, “3” and “4”, the new value of B[2] is min(0+10,53+3,53+2) = 10. The new value of B[3] is min(0+5,10+3,53+6) = 5 and the new value of B[4] is 11, so, after the first iteration B = [0, 10, 5, 11]. The second iteration creates B = [0, 8, 5, 10], and the third iteration does not change anything, so the process ends with B[4] = 10; the length of the shortest path from “1” to “4” is 10, and the path is (“1”,“3”,“2”,“4”) (finding this path, when its length is known, is another interesting problem). ✷

Write a Fortran function SHORTESTPATH (A,N,M,K) which returns the length of the shortest path from m to k if such a path exists, otherwise -1 should be returned. A is the array of distances. Write also a simple program which reads the data, calls the function and prints the result.