Question

I have this massive array of ints from 0-4 in this triangle. I am trying to learn dynamic programming with Ruby and would like some assistance in calculating the number of paths in the triangle that meet three criterion:

1. You must start at one of the zero points in the row with 70 elements.
2. Your path can be directly above you one row (if there is a number directly above) or one row up heading diagonal to the left. One of these options is always available
3. The sum of the path you take to get to the zero on the first row must add up to 140.

Example, start at the second zero in the bottom row. You can move directly up to the one or diagonal left to the 4. In either case, the number you arrive at must be added to the running count of all the numbers you have visited. From the 1 you can travel to a 2 (running sum = 3) directly above or to the 0 (running sum = 1) diagonal to the left.

0   41   302   2413   13024   024130   4130241   30241302   241302413   1302413024   02413024130   413024130241   3024130241302   24130241302413   130241302413024   0241302413024130   41302413024130241   302413024130241302   2413024130241302413   13024130241302413024   024130241302413024130   4130241302413024130241   30241302413024130241302   241302413024130241302413   1302413024130241302413024   02413024130241302413024130   413024130241302413024130241   3024130241302413024130241302   24130241302413024130241302413   130241302413024130241302413024   0241302413024130241302413024130   41302413024130241302413024130241   302413024130241302413024130241302   2413024130241302413024130241302413   13024130241302413024130241302413024   024130241302413024130241302413024130   4130241302413024130241302413024130241   30241302413024130241302413024130241302   241302413024130241302413024130241302413   1302413024130241302413024130241302413024   02413024130241302413024130241302413024130   413024130241302413024130241302413024130241   3024130241302413024130241302413024130241302   24130241302413024130241302413024130241302413   130241302413024130241302413024130241302413024   0241302413024130241302413024130241302413024130   41302413024130241302413024130241302413024130241   302413024130241302413024130241302413024130241302   2413024130241302413024130241302413024130241302413   13024130241302413024130241302413024130241302413024   024130241302413024130241302413024130241302413024130   4130241302413024130241302413024130241302413024130241   30241302413024130241302413024130241302413024130241302   241302413024130241302413024130241302413024130241302413   1302413024130241302413024130241302413024130241302413024   02413024130241302413024130241302413024130241302413024130   413024130241302413024130241302413024130241302413024130241   3024130241302413024130241302413024130241302413024130241302   24130241302413024130241302413024130241302413024130241302413   130241302413024130241302413024130241302413024130241302413024   0241302413024130241302413024130241302413024130241302413024130   41302413024130241302413024130241302413024130241302413024130241   302413024130241302413024130241302413024130241302413024130241302   2413024130241302413024130241302413024130241302413024130241302413   13024130241302413024130241302413024130241302413024130241302413024   024130241302413024130241302413024130241302413024130241302413024130   4130241302413024130241302413024130241302413024130241302413024130241   30241302413024130241302413024130241302413024130241302413024130241302   241302413024130241302413024130241302413024130241302413024130241302413   1302413024130241302413024130241302413024130241302413024130241302413024   02413024130241302413024130241302413024130241302413024130241302413024130   

Answer 1

But I like homework 🙂

I find it easier to reason about the ‘paths’ problem when starting from the top, and following the rules the other way around.

This means:

• a partial path can be the top zero, or an extended partial path
• the extensions of a partial path Pr,c are, unless r is the last row, in which they’re complete, the union of
  • the extensions of Pr,c + P(r+1),c
  • the extensions of Pr,c + P(r+1),c+1

The ‘sum’ rule just selects certain of all complete paths.