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Find the number of paths on the Cartesian plane from (0, 0) to (n, n), which never raises above the y = x line. It is possible to make three types of moves along the path:

move up, i.e. from (i, j) to (i, j + 1);
move to the right, i.e. from (i, j) to (i + 1, j);
the right-up move, i.e. from (i, j) to (i + 1, j + 1)
amruth polavarapu
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  • Sorry, this isn't a site for doing your homework for you. I've you've got a specific problem we'll try to help you out, but dumping your homework assignment with no attempt shown to solve it or even a real question is frowned upon. – beaker Apr 11 '16 at 19:55
  • @beaker, This is not my homework problem. I could not approach these kind of problems. So all I wanted to know is how do I start solving these. I don't want answer. I want to know the approach – amruth polavarapu Apr 12 '16 at 09:29
  • Check [Catalan numbers](https://en.m.wikipedia.org/wiki/Catalan_number) – Ante Apr 12 '16 at 14:24
  • @Ante It's not Catalan Numbers since it have `(x, y) => (x + 1, y + 1)` – Sayakiss Apr 13 '16 at 08:01
  • @Sayakiss It is not Catalan, but Catalan numbers can be used to find the solution. Choosing positions of diagonal moves gives number of paths as sum( binomial(n,i) * catalan(i), for i in {0,...,n}). WolfranAlpha gives this [solution](https://www.wolframalpha.com/input/?i=sum(+binomial(n,i)+*+CatalanNumber(i),+i,+0,+n)). That can be computed by changing second parameter in 2F1. For [n=10](https://www.wolframalpha.com/input/?i=Hypergeometric2F1(1%2F2,+-10,+2,+-4)) states that number of paths is 223191. – Ante Apr 16 '16 at 21:19

2 Answers2

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Path count 101

First, we solve a simpler problem:

Find the number of paths on the Cartesian plane from (0, 0) to (n, n) with:

  • move up, i.e. from (i, j) to (i, j + 1);
  • move to the right, i.e. from (i, j) to (i + 1, j);

and we can go to grid which x < y.

How to solve it? Too Hard? Okay, we try to find the number of paths from (0, 0) to (2, 2) first. We could draw all paths in a grid:

enter image description here

We define 

f(x,y) => the number of paths from (0, 0) to (x, y)

You can see the path to (2, 2) from either (1, 2) or (1, 2), so we can get:

f(2, 2) = f(2, 1) + f(1, 2)

And then you will notice for point(x, y), its path from either (x, y - 1) or (x - 1, y). That's very natural, since we have only two possible moves:

  • move up, i.e. from (i, j) to (i, j + 1);
  • move to the right, i.e. from (i, j) to (i + 1, j);

I draw a larger illustration for you, and you can check our conclusion:

enter image description here

So we can get that:

f(x, y) = f(x, y - 1) + f(x - 1, y)

Wait... What if x = 0 or y = 0? That's quite direct:

if x = 0 => f(x, y) = f(x, y - 1)
if y = 0 => f(x, y) = f(x - 1, y)

The last... How about f(0, 0)? We define:

f(0, 0) = 1

since there just 1 path from (0,0) to (1,0) and (0, 1).

OK, summarise:

f(x, y) = f(x, y - 1) + f(x - 1, y)
if x = 0 => f(x, y) = f(x, y - 1)
if y = 0 => f(x, y) = f(x - 1, y)
f(0, 0) = 1

And by recursion, we can solve that problem.

Your problem

Now let's discuss your original problem, just modify our equations a little bit:

f(x, y) = f(x, y - 1) + f(x - 1, y) + f(x - 1, y - 1)
if x = 0 => f(x, y) = f(x, y - 1)
if y = 0 => f(x, y) = f(x - 1, y)
if x < y => f(x, y) = 0
f(0, 0) = 1

and it will result my code.

The last thing I add to my code is Memoization. In short, Memoization can eliminate the repeat calculation -- if we have calculated f(x,y) already, just store it in a dictionary and never calculate it again. You can read the wiki page for a further learning.

So, that's all of my code. If you still get some questions, you can leave a comment here, and I will reply it as soon as possible.

Code:

d = {}  # Memoization


def find(x, y):
    if x == 0 and y == 0:
        return 1
    if x < y:
        return 0
    if d.get((x, y)) is not None:
        return d.get((x, y))
    ret = 0
    if x > 0:
        ret += find(x - 1, y)
    if y > 0:
        ret += find(x, y - 1)
    if x > 0 and y > 0:
        ret += find(x - 1, y - 1)
    d[(x, y)] = ret
    return ret


print find(2, 1) # 4
Sayakiss
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  • Can you teach me how to approach these kinds of problems and in specific how to think while solving this. I don't want the code. I want to know your approach. – amruth polavarapu Apr 12 '16 at 09:31
  • @amruthvenkat Okay... Now I'm at work, I will explain my solution in details hours later... – Sayakiss Apr 12 '16 at 09:34
  • @amruthvenkat So you can vote my answer up. If there is no more further question about my answer, you may set my answer as accepted. – Sayakiss Apr 13 '16 at 05:19
  • I have voted your answer but since my reputation is not 15 it won't be displayed here... – amruth polavarapu Apr 13 '16 at 06:57
  • @amruthvenkat Okay... By the time you have at least 15 reputation, please do not forget to vote up for me... :) – Sayakiss Apr 13 '16 at 07:27
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For additional ideas for solving problems such as this one, there is a mathematical curiosity that is in 1-1 correspondence with not only the sequences produced by lattice walks where one's steps must remain below the line x = y, but a veritable plethora, of fantastical mathematical beasts that cut a wide swath of applicability to problem solving and research.

What are these curiosities? Catalan Numbers:

C_n = 1/(n+1)*(2n)Choose(n), n >= 0, where if n = 0, C_0 = 1.

They also count:

  1. The number of expressions containing $n$ pairs of parentheses
    • eg. n = 3: ((())), (()()), (())(), ()(()), ()()()
  2. Number of plane trees with n + 1 vertices
  3. Number of Triangulations of a convex (n + 2)-gon
  4. Monomials from the product: p(x1, ..., xn) = x1(x1 + x2)(x1 + x2 + x3) ... (x1 + ... + xn)
  5. bipartite vertices of rooted planar trees
  6. And soooo many more things

These object appear in a lot of active research in mathematical physics, for instance, which is an active area of algorithms research due to the enormous data sets.

So you never know what seemingly far flung concepts are intimately linked in some deep dark mathematical recess.

joshua fiddler
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