It is always a good idea to find a recursive algorithm first and then turn it into a table.

1. $f(C,n)$
2. $~~$if(C = $\emptyset$) return 0;
3. $~~$else
4. $~~~~$opt = 0;
5. $~~~~$for each $c\in C$ do
6. $~~~~~~D=\{d\in C:d<c\}$
7. $~~~~~~E=\{e-c:e\in D,e>c\}$
8. $~~~~~~opt = min\{opt,f(D,c)+f(E,n-c)\}$
9. $~~~~$return $opt+n$;

So you may ask: isn't there too many subsets of C to be put in a table? Observe that only 'consecutive' subsets are needed. And there are only $n \choose 2$ of them.(why?) Another problem is: some entries will change value in $E$. We can walk around this by indicating start and end in each $f$ rather than just specifying the length.

It is always a good idea to find a recursive algorithm first and then turn it into a table.

1. $f(C,n)$
2. $~~$if(C = $\emptyset$) return 0;
3. $~~$else
4. $~~~~$opt = infinity;
5. $~~~~$for each $c\in C$ do
6. $~~~~~~D=\{d\in C:d<c\}$
7. $~~~~~~E=\{e-c:e\in D,e>c\}$
8. $~~~~~~opt = min\{opt,f(D,c)+f(E,n-c)\}$
9. $~~~~$return $opt+n$;

So you may ask: isn't there too many subsets of C to be put in a table? Observe that only 'consecutive' subsets are needed. And there are only $n \choose 2$ of them.(why?) Another problem is: some entries will change value in $E$. We can walk around this by indicating start and end in each $f$ rather than just specifying the length.