Dynamic Programming. Coin Row Problem - How its Recurrsive relation is developed - dynamic-programming

Question

Problem: A set of n coins is placed in a row. The coins have positive values which need not be distinct. Find the maximum amount that can be collected given the constraint that no two adjacent coins can be picked up.
Its recursive relation is
F(n) = max{cn + F(n − 2), F(n − 1)} for n > 1,
F(0) = 0, F(1) = c1.
My question is how this recursive relation is developed. Please someone explain this to me.

Answer 1

First, envision a line of coins, with the value of each depicted by the variable ci:
c1 c2 c3 c4 c5 ... cn
If there are no coins, than obviously the max amount that can be made is 0. Likewise, if there is only 1 coin, the max amount is the value of that coin, c1. This accounts for the base case.
For the recursive case of the max value for n coins, start at cn, which is the right-most coin. Since the constraint is that you cannot select adjacent coins, the max value that you can achieve is either the right most coin plus the max achieved from 2 slots to the left (this accounts for the f(n - 2), or the max achieved by selecting the coin immediately to the left (accounting for the f(n - 1) case) and discarding the rightmost coin cn.
Considering the following line of coins again:
c1 c2 c3 c4 c5 c6
The f(6) case would look at c6 + the greatest amount from coins c1 - c4, OR the greatest amount from coins c1 - c5 (and excludes c6).
f(4), likewise, returns c4 + the greatest amount from coins c1 - c2, OR the great amount from coins c1 - c3 (again excluding c4).
f(2) returns c2 + c0 or the greatest amount from c1 (effectively c1) The first equates to c2, since c0 is 0 by the base case, and the second equates to c1 (again by the base case). So f(2) is really just the max of c1 or c2.
Note, too, that the f(n - 2) and f(n - 1) may be the same, since in the n - 1 case it might be beneficial to select the coin to the left (which is the f(n - 2) case). But that is why the first half is not merely f(n - 2), but also adds to it cn

Answer 2

Let's start at the end.
Let's denote the answer to the problem of n coins as F(n)
If you have zero coins, the amount is zero. so F(0) = 0.
If you have a single coin, the amount is that coin's value, so F(1)=c1
Now suppose someone told you the values of F(n-1), F(n-2). How can you use them to find F(n)?
If you have n coins, you have two possible moves:
Pick the nth coin, skip the adjacent one ((n-1)th coin, that's the rule!) and resume solving from there.
Skip the nth coin, and resume solving from the adjacent (n-1)th.
How do you express the notions of 1 and 2 with the tools you have?
If you pick the nth coin, it's value is Cn. Now you have to skip the (n-1)th coin, and continue solving from (n-2) coin. This is Cn + F(n-2).
If you skip the nth coin it contributes 0 to the solution, and now you resume solving from the (n-1)th coin. That's F(n-1).
Which one of either case 1 or case 2 is larger? You don't know. But you can express it as
max(Cn + F(n-2), F(n-1)),
which is saying "I don't which one is larger, but one of them is so return it please".
I hope this helps!

Answer 3

Let F(n) be the maximum amount that can be picked up from the row of n coins.
To derive a recurrence for F(n), we partition all the allowed coin selections into two groups:
those that include the last coin and those without it.
The largest amount we can get from the first group is equal to cn + F(n − 2)—
the value of the nth coin plus the maximum amount we can pick up from the first n − 2 coins.
The maximum amount we can get from the second group is equal to F(n − 1) by the definition of F(n).