Number of Ways to Walk Between Corners of Blocks

In summary, if you want to walk from the corner of a block to the corner of another block (I depart from the corner of coordinates (0,0) and want to get to the corner with coordinates (X,Y), with X and Y being whole numbers.), then there are N*N possible ways to do so, where N is a fixed number.
  • #1
matiasmorant
39
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suppose that I want to walk from the corner of a block to the corner of another block (I depart from the corner of coordinates (0,0) and want to get to the corner with coordinates (X,Y), with X and Y being whole numbers. ); and I want to do so by walking N block sides (I can walk over the same side more than once).

how many ways are there of doing so?
 
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  • #2
hi matiasmorant! :smile:
matiasmorant said:
(I can walk over the same side more than once).

i don't get it … surely the number of possible ways is infinite? :confused:

(i can just keep going back for my keys! :biggrin:)
 
  • #3
I believe you might be thinking of a grid of say n*n square blocks. How many ways can you traverse the grid from one end of a diagonal to the other (following the orthogonal block sides) with the caveat that you must be closer to your objective with each transit of a block side. Show a general solution for any n>1.
 
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  • #4
tiny-tim said:
hi matiasmorant! :smile:


i don't get it … surely the number of possible ways is infinite? :confused:

(i can just keep going back for my keys! :biggrin:)

yes, sure, but only N times!

I can only walk N squares sides, if I keep on coming back for my keys, then I won't have enough permitted sides left to get to my destination :-D

To start with, you can take the case where N is the least number of sides in which I can reach both corners. For example, if I want to reach the corner with coordinates (1,Y) then there Y possible routes, if I want to reach the corner with coordinates (2,Y) then there are (Y^3 - Y)/6 possible routes (that's, with N being the least). what about (X,Y) ?

However, if N is not the least... let's say N=3 and I want to get to (1,1). Then, it can't be done, because, I could get there by walking 2 sides, 4 sides (coming back for my keys once), 6 sides (coming back for my keys twice), but not 3 sides.
That's it.

Some advice in the case where N is the least is also appreciated :-)
 
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  • #5
SW VandeCarr said:
I believe you might be thinking of a grid of say n*n square blocks. How many ways can you traverse the grid from one end of a diagonal to the other (following the orthogonal block sides) with the caveat that you must be closer to your objective with each transit of a block side. Show a general solution for any n>1.

That's almost what I meant :-). but withOUT the caveat that you must be closer to your objective with each transit of a block side. I have to walk N block sides before getting to my destination. I might want to get to a corner which is just a block away (0,1), but I have to walk, say, 3 sides (so I have to come back sometimes). the answer then, would be 9 possible routes, if I didn't count bad. In your problem the answer would be just 1. Besides, I couldn't get to (1,1) by walking 3 sides :-). I'm very interested in the answer to the case you explained to.
 
  • #6
matiasmorant said:
That's almost what I meant :-)m the answer would be just 1. Besides, I couldn't get to (1,1) by walking 3 sides :-). I'm very interested in the answer to the case you explained to.

In my problem, every block would be walked on all four sides total for all paths. Also, another hint: every path is the same length (measured in block sides). Btw, I don't see it as a combinatorial problem and it's not really very hard.

EDIT: If I understand your problem correctly, my problem becomes your problem if you take all paths in both directions. That is, the solution to your problem is just twice the number of paths as mine (assuming you're defining the grid space the same way I am, as n x n square blocks with the ends of one of the diagonals defining the start and finish for all paths.)
 
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  • #7
matiasmorant said:
yes, sure, but only N times!
…
However, if N is not the least

ah, so the question is …

how many ways of going from (0,0) to (x,y) in x + y + 2N steps, where N is a fixed number ≥ 0

EDIT: the way to solve this would be to deal with the case of x + 2n steps sideways and y + 2(N - n) steps up, for each n ≤ N

each case would then be the problem of arranging x + 2n items in a total of x + y + 2N :wink:​
 
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What is the "Number of Ways to Walk Between Corners of Blocks" problem?

The "Number of Ways to Walk Between Corners of Blocks" problem is a mathematical concept that involves finding the number of ways a person can walk from one corner of a grid of blocks to the opposite corner, while only being able to move in certain directions (usually up or right).

Why is the "Number of Ways to Walk Between Corners of Blocks" problem important?

This problem is important because it has real-world applications in fields such as computer science and engineering. It can help determine the efficiency of algorithms and the shortest paths in a grid-like environment.

How do you solve the "Number of Ways to Walk Between Corners of Blocks" problem?

To solve this problem, you can use a mathematical formula called the binomial coefficient, also known as the "choose" function. This formula allows you to calculate the number of possible combinations or paths in a grid of blocks.

What factors affect the number of ways to walk between corners of blocks?

The number of ways to walk between corners of blocks is affected by the size of the grid, the starting and ending positions, and the direction of movement allowed. Generally, a larger grid and more restricted movement will result in a higher number of possible paths.

Are there any limitations to the "Number of Ways to Walk Between Corners of Blocks" problem?

Yes, there are some limitations to this problem. It assumes that the person can only move in certain directions and cannot backtrack or move diagonally. In reality, there may be other factors that affect a person's movement, making the problem more complex.

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