# CHALLENGE: WALK ALONG THE GRID

A classic Math Olympiad question: How many ways are there to go through an m-by-n grid by walking only up and right? Let's implement that in code. Given that there is a grid of size m-by-n, print the number of ways to go from the bottom left to the top right corner by only moving up and right. But we are not done yet. The challenge is too easy. Let us include a list of forbidden points you cannot visit, with coordinates (x,y) relative to the start point. Good luck!

11/4/2017 3:55:38 PM

👑 Prometheus 🇸🇬28 Answers

New Answer🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰🔰 //btw sorry vengat ... no time for making code ... but i would like to give my mathematical answer ... a general answer 👉1) suppose u want to go from A (x,y) to B (l,m) ... //only move up amd right then , l>=x & m>=y //obviously ☺ u need to move (l-x) ways to go right by 1 step & (m-y) steps to go down therefore 👉👉👉 {(l+m)-(x+y)} ! / {(l-x)! . (m-y)!} //basic method of bijection , //combinations ☺ 👉2)A (x,y) to B (l,m) without C (e,f) // check condition in 1) and also check whether C lies b/w way from A & B or not 👉👉👉answer in 1) - ((ways of going from A to C). (ways of going from C to B) //use formula derived in 1) 👉3) A (x,y) to B (l,m) through pt. P (s,k) 👉👉👉3)(no. of ways from A to P ). (no. of ways from P to B) //use formula derived in 1) //🙋🙋🙋for more than 1 forbideen points 👍👍👍 for more than 1 forbidden points ... u need to substract ways of passing through these points example ::: A to B & pt. C & D are forbidden points then , (total no. of ways from pt. A to pt. B ) - { (ways from pt. A to pt. C)(ways from pt. C to pt. B) + (same for pt. D) + (ways from pt.A to pt. C )(ways from pt.C to pt.D). (ways from pt. D to pt. B)} //hope u like the soln... , & happy coding☺☺☺ //bye bye tata 🌠

guys //just see my answer , no confusion be happy ☺ //& what u mean by 7×5 path ... might u mean from origin(@one corner) to opposite diagonal corner answer will be 12! / 7!.5! without obstacles ☺ ie 792 //thats correct

@kartikey 4C2 ie 6 ways moving right and upward will only be possible if respective coordinates of pt.A are smaller or equal to respective coordinates of B @yash welcome // now everyone can code the challenge ☺ ... hahaha

haan cbse //thats agrawal ... waise agarwal hi thaa but 10th ma agrawal chala gaya thaa toh ab ma agrawal use karta hun 😂

i will give it a try //wait ☺ //but its too easy make it , for making it more good for example ::: we need to walk from A(x,y) to pt. B(i,j) & we need to find ... 1) find total no. ways of going from pt. A to pt. B (ie no restriction) 2)find total no. of ways of going from A to B without going to coordinates p (l,m) 3)find total no. of ways of going from A to B through pt. p (l,m) //how about this ☺ @vengat

@yash , bhai yarr p, d block toh puche hi mat ... 1st term ma unhone band bajadi aur ab 2nd term ma organic chemistry aa gayi 😑 but i am finding it some interesting ... but rxns learn karne ma meri band baj gayi ha //par karni toh padegi kaise bhi ... best of luck 👍 to u ... 👉study chemistry hard

@ vengat... this code gives u no. of path to go from a start-point to an end-point..(only up / right) bonus...** give any starting and ending point ** give more than 1 forbidden points ** default starting point is left bottom, default ending point is top right ** default list forbidden points is set to empty ## pass arguments in function as u want ## it results -1 if starting point is in left side or in bottom of end point.. https://code.sololearn.com/c7pui8yMCwUR/?ref=app

An interesting problem. (There seems to be a slight difference of understanding as to whether an m x n grid has m rows and n columns from (1, 1) to (m, n) or m+1 rows and n+1 columns from (0, 0 to (m, n). I assume the latter.) If there are no forbidden points the answer is simply the number of permutations of m rightward moves and n upward moves which is factorial(m+n)/(factorial(m)*factorial(n)). If there is just one forbidden point the answer can be obtained by subtracting the number of routes through the forbidden point. If there is more than one forbidden point the answer becomes very messy as you have to consider routes through all combinations of the forbidden points. The alternative is to solve the problem for each point by first solving the problem for the point below and the point to the left and then add them which my program does. If the input to https://code.sololearn.com/cNwQxBj30TSc/#java is: 7 5 2 3 6 4 4 1 the output is: In a 7x5 grid without forbidden points there are 792 ways but with 3 forbidden points at (2, 3) (4, 1) (6, 4) there are just 187 ways.

@Ga Yes but, Pegasus said that we can move only in right and up, so I just need an example.

@ Gaurav @ Vcc 7 * 5 grid... {(7-1)*(5-1)} ! ---------------- = 210 (7-1) ! * (5-1) ! i implemented this thinking a matrix...thats why subtracted 1##### ## its corrected now...thanks to u two [email protected] Vcc.. suppose its 1 * 1 grid.. means just like a box... for forbidden points ony 0,0.. ur code shud give 1 way...of 2 ways... its giving 0

Here is my code. I'm not confident this is a right answer. https://code.sololearn.com/cXD0Sl2skBE1/?ref=app