Question

4. Suppose you will travel on a railway and there are n ticket offices to buy a ticket along the railway. You will change trains during your journey. At the beginning, you are given for each 1 ≤ i < j ≤ n and pi,j is the price of a ticket for a train that goes from ticket office i to ticket office j. These prices are arbitrary. For instance, it is possible to see a price combination as p1,3 = 15 and p1,5 = 10. You start at ticket office 1 and must finish at the ticket office n (by using the trains).

(a) Suggest an efficient dynamic programming algorithm that minimizes the money you spent in your journey.

(b) What is the complexity of your algorithm?

Answer 1

(a) Suggest an efficient dynamic programming algorithm that minimizes the money you spent in your journey.

Explanation of programming approach

Here the dynamic programming is represented in more generalized way.

Here the starting ticket office number is customized. So user can start from any ticket office and it will show from source ticket office to all possible ticket offices’ minimum fares and the intermediate journey ticket office route.

As the source ticket office number is user defined, so if user take console input 1, it means journey starts from ticket office 1. It will give minimum fare for all destination ticket offices. Here all the destination ticket office are nth ticket office for a individual source to destination ticket office with the intermediate ticket office.

/* program to calculate minimum ticket fare from source ticket office to destination ticket office*/

#include<stdio.h>
#include<conio.h>
#define infy_value 9999999

void tickrt_price(int ticket_price[100][100],int n,int start_station);

int main()
{
   int ticket_price[100][100],i,j,n,st;
   printf("How many stations are there:");
   scanf("%d",&n);
   printf("\nEnter the stations to stations ticket fare:\n");
  
   for(i=1;i<=n;i++)
   {
       for(j=1;j<=n;j++)
       {
       if (i==j){
          printf("\nStation: %d to Station: %d (Rs.) ", i,j);
          ticket_price[i][j]=0;
          printf("%d",ticket_price[i][j]);
       }
       else{
       printf("\nStation: %d to Station: %d (Rs.) ", i,j);
       scanf("%d",&ticket_price[i][j]);
   }
    }
   }
   printf("\n Station to station ticket fare details is (Rs./-):\n\n");
   for(i=1;i<=n;i++)
   {
       for(j=1;j<=n;j++)
       {
       printf("%d\t ",ticket_price[i][j]);
    }
    printf("\n");
   }
   printf("\nEnter the journey starting stastion number:-> ");
   scanf("%d",&st);
  
   tickrt_price(ticket_price,n,st);
  
   return 0;
}

void tickrt_price(int ticket_price[100][100],int n,int start_station)
{

   int ticket_fare[100][100],min_fare[100],pd[100];
   int travel_stn[100],c,min_fare_value,next_station,i,j;

   for(i=1;i<=n;i++)
   {
       for(j=1;j<=n;j++)
       {
           if(ticket_price[i][j]==0)
               ticket_fare[i][j]=infy_value;
           else
               ticket_fare[i][j]=ticket_price[i][j];
   }
}   
  
   for(i=1;i<=n;i++)
   {
       min_fare[i]=ticket_fare[start_station][i];
       pd[i]=start_station;
       travel_stn[i]=0;
   }
  
   min_fare[start_station]=0;
   travel_stn[start_station]=1;
   c=1;
  
   while(c<=n-1)
   {
       min_fare_value=infy_value;
      
       for(i=1;i<=n;i++)
           if(min_fare[i]<min_fare_value && !travel_stn[i])
           {
               min_fare_value=min_fare[i];
               next_station=i;
           }
          
           travel_stn[next_station]=1;
           for(i=1;i<=n;i++)
               if(!travel_stn[i])
                   if(min_fare_value+ticket_fare[next_station][i]<min_fare[i])
                   {
                       min_fare[i]=min_fare_value+ticket_fare[next_station][i];
                       pd[i]=next_station;
                   }
       c++;
   }

  
   for(i=1;i<=n;i++)
   {
       if(i!=start_station)
       {
           printf("\n");
           printf("\nMinimum Train Fare from Station: %d to Station: %d = Rs.%d /-",start_station,i,min_fare[i]);
           printf("\nJourney path of station : Stn:%d",i);
          
           j=i;
           do
           {
               j=pd[j];
               printf(" <===stn:%d ",j);
           }while(j!=start_station);
   }
}   
}

/* minimize the money in journey function part */

void tickrt_price(int ticket_price[100][100],int n,int start_station)
{

   int ticket_fare[100][100],min_fare[100],pd[100];
   int travel_stn[100],c,min_fare_value,next_station,i,j;

   for(i=1;i<=n;i++)
   {
       for(j=1;j<=n;j++)
       {
           if(ticket_price[i][j]==0)
               ticket_fare[i][j]=infy_value;
           else
               ticket_fare[i][j]=ticket_price[i][j];
   }
}   
  
   for(i=1;i<=n;i++)
   {
       min_fare[i]=ticket_fare[start_station][i];
       pd[i]=start_station;
       travel_stn[i]=0;
   }
  
   min_fare[start_station]=0;
   travel_stn[start_station]=1;
   c=1;
  
   while(c<=n-1)
   {
       min_fare_value=infy_value;
      
       for(i=1;i<=n;i++)
           if(min_fare[i]<min_fare_value && !travel_stn[i])
           {
               min_fare_value=min_fare[i];
               next_station=i;
           }
          
           travel_stn[next_station]=1;
           for(i=1;i<=n;i++)
               if(!travel_stn[i])
                   if(min_fare_value+ticket_fare[next_station][i]<min_fare[i])
                   {
                       min_fare[i]=min_fare_value+ticket_fare[next_station][i];
                       pd[i]=next_station;
                   }
       c++;
   }

  
   for(i=1;i<=n;i++)
   {
       if(i!=start_station)
       {
           printf("\n");
           printf("\nMinimum Train Fare from Station: %d to Station: %d = Rs.%d /-",start_station,i,min_fare[i]);
           printf("\nJourney path of station : Stn:%d",i);
          
           j=i;
           do
           {
               j=pd[j];
               printf(" <===stn:%d ",j);
           }while(j!=start_station);
   }
}   
}

output ( As the output screen is large so the overall output is divided in two parts)

(b) What is the complexity of your algorithm?

Here the algorithm is designed with a adjacency matrix representation where the ticket office to ticket office fares are expressed through matrix form with n x n matrix.

Now to generate fare from source ticket office to destination ticket office every pair of ticket offices are compared with other pair of ticket office. Here we implement nested loop structure where each loop has n iterations (Where total number of ticket offices are n).So total time complexity will be in n² form and it can be represented as O(n2).