Combinatorics problem about determining number of paths

In summary, the conversation discusses a variation of the problem where the number of paths on a discrete grid is determined. The variation involves allowing a specified number of reverse movements in addition to the movements in the proper direction. The use of the inclusion-exclusion principle is suggested to address the issue of preventing paths with adjacent x's and w's. To solve the simpler case where only reverse horizontal movements are allowed, the paths are divided into two groups and the principle is used to count the number of paths with no adjacent x's and w's. Generalizing this approach to the case with reverse movements in both dimensions may require further considerations.
  • #1
Sprooth
17
0
Hi,

I'm trying to figure out a variation of the problem where you determine the number of paths on a discrete grid. For example, for the paths from (0,0) to (a,b), you can consider it to be a rearrangement of a word with a x's and b y's, so the number of paths is (a+b)! / a!b!.

The variation I would like to figure out would allow a specified number of reverse movements in addition to the movements in the proper direction. I am representing a movement to the right by "x", a movement to the left by "w", a movement upward by "y", and a movement downward by "z". If I am finding a path from (0,0) to (a,b) with c reverse horizontal movements and d reverse vertical movements, then the number of paths I have so far would be:

(a + b + 2c + 2d)! / (a+c)!(b+d)!c!d!

since the paths could be considered as a rearrangement of a word made of a+c x's, b+d y's, c w's and d z's. (a+c x's since there must be an additional movement to the right for every allowed movement to the left.

I believe this is correct so far, but now I want to prevent paths where for example it goes right, then left, then right with a net result of just going right. To do this I want to find rearrangements of the word where there are no adjacent x's and w's and no adjacent y's and z's. I began by simplifying the problem to one where only reverse horizontal movements were allowed and I would then attempt to generalize to the case with reverse movements in both dimensions. However, I got stuck even in the simpler case.

Would someone be able to help me through the simpler case (where there would be x's for going right, y's for going up, and w's for going left)? Then I will try to generalize to the other case.

Thanks for your help.
 
Physics news on Phys.org
  • #2


Hello,

Thank you for your inquiry. This is an interesting variation of the problem and it seems like you have made some good progress in your approach. To address the issue of preventing paths with adjacent x's and w's, you could consider using a technique called "inclusion-exclusion principle". This principle can be used to count the number of objects that satisfy certain conditions, while excluding those that do not.

In this case, you could consider the paths that have adjacent x's and w's as one group, and the paths that do not have adjacent x's and w's as another group. Using the principle, you can then subtract the number of paths with adjacent x's and w's from the total number of paths, to get the number of paths without adjacent x's and w's.

For the simpler case where only reverse horizontal movements are allowed, you could approach it by considering the paths that start with a reverse movement (w) and those that do not. You could then use the inclusion-exclusion principle to count the number of paths with no adjacent x's and w's. Generalizing this approach to the case with reverse movements in both dimensions may require some additional considerations, but it could be a good starting point.

I hope this helps. Let me know if you have any further questions or if you need any clarification. Good luck with your research!
 

1. How do you determine the number of paths in a combinatorics problem?

To determine the number of paths in a combinatorics problem, you need to first understand the problem and identify the variables involved. Then, you can use the fundamental principle of counting or various combinatorial formulas such as permutations and combinations to calculate the number of paths.

2. What is the difference between permutations and combinations in combinatorics?

Permutations refer to the arrangement of objects in a specific order, while combinations refer to the selection of objects without considering the order. In combinatorics problems, permutations are used when order matters, while combinations are used when order does not matter.

3. Can you provide an example of a combinatorics problem about determining the number of paths?

One example of a combinatorics problem about determining the number of paths is the "lattice path problem." In this problem, a person starts at the bottom left corner of a grid and can only move to the right or up to reach the top right corner. The question is how many different paths can the person take to reach the top right corner?

4. How do you approach a complex combinatorics problem about determining the number of paths?

To approach a complex combinatorics problem about determining the number of paths, it is important to break down the problem into smaller, simpler parts. Then, you can use the principles of counting and combinatorial formulas to solve each part separately and combine the solutions to get the final answer.

5. Are there any real-world applications of combinatorics problems about determining the number of paths?

Yes, combinatorics problems about determining the number of paths have various real-world applications, such as in transportation planning, network routing, and designing computer algorithms. These problems help in finding the most efficient and cost-effective routes or solutions for different scenarios.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
288
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
453
  • Calculus and Beyond Homework Help
Replies
3
Views
238
  • Calculus and Beyond Homework Help
Replies
2
Views
587
  • Calculus and Beyond Homework Help
Replies
2
Views
492
  • Calculus and Beyond Homework Help
Replies
14
Views
630
  • Calculus and Beyond Homework Help
Replies
2
Views
249
  • Calculus and Beyond Homework Help
Replies
15
Views
2K
Back
Top