# Logic behind this problem

Hi, I have gone forward and took out the base cases from the problem, which are: if n==m, there are 2 ways and if m > n, there is one way. I am having trouble finding the recurrence relation. Pls help me out with understanding the logic of the recurrence relation of this problem.

Hi @akshatpriyansh,
In this problem you have figured the base cases correct that if n<m then return 1 and if n==m return 2. The recurrsion relation will be that if n>m then add the two recursions . One with same function but n as n-1 and the other will be same function but with n= n-m . Now this will the two cases one in which tile are placed vertically will be where n=n-1 and the one where tiles are placed horizontally will be where n=n-m . And also i will recommend that rather than forming a recurrence relation try to take a array of size n+1 and fill it from index i= 1 where size of n =1 till index i= n and where each index has the total possible way of that case on that index i. Recurrsion relation may give you time error.

How n =n-m finalised .?