About n-4 case in tiling problem

#1

after we get f(n-4) , we can arrange the last 4 tiles horizontally as well as vertically. So, there must be a total contribution of f(n-4) + 1 from the case of f(n-4)

#2

https://drive.google.com/file/d/145AxHBgqqGXBUUn65hY_7BNu3geUgbRe/view?usp=sharing

i have explained the problem with a pink pen in the image

#3

@bhavyarocks19 the question is concerned about the total no of ways of arrangement of brick this is not like that f(n-4)+1
case 1 if we place tile like vertically then left part is f(n-1) this is first way if you fill all tiles vertically.
case 2 if we place tile like horizontally then left par is f(n-4) this is second way if you fill tiles horizontally.
the total no of ways which is summation of vertical ways + horizontal ways so recurrence relation is nothing but f(n)=f(n-1)+f(n-4) if any queries come feel free to ask

#4

in f(n-4) case , we are just putting last 4 tiles vertically,

in f(n-1) case , we are just putting only last tile vertically and not caring of what pattern is there before that last tile , as it may also be horizontal placing of bricks before last tile.

#5

@bhavyarocks19 you don’t think like this way think recursively based on size left to allocate tile recursion will do there work
eq if n==5
1 way f(n-1)
you can either solve it with one 1 vertical tile and one horizontal tile or take all vertical tile this can be a way
2
if f(n-4)
then in this only one vertical tile or one horizontal tile will be placed
so all way will be calculate by recursion by summation of both these two

#6

Hey Bhavya,
As you are not responding to this thread, I am marking your doubt as Resolved for now. Re-open it if required.

Please mark your doubts as resolved in your course’s “ Ask Doubt ” section, when your doubt is resolved.