I'm trying to understand _why_ an alternate formula for finding the _best_ $p$ items in a 0/1 knapsack with $n$ items isn't working. The formula was proposed by [@Carlos Linares López](http://cs.stackexchange.com/users/1337/carlos-linares-l%C3%B3pez) in this answer: [Variant of the Knapsack Problem](http://cs.stackexchange.com/a/18529/37728). 

Please read the previous answer for context; also I'm positive that the formula is still correct, but either my use or understanding of it is incorrect.

Let me illustrate the issue:

- Item 1: weight=1, value=1
- item 2: weight=1, value=4
- item 3: weight=1, value=3

The calculated _dp_ table (_total items visited denoted in superscript_): 
$$
\begin{array}{c|ccc}
3 & 1^1 & 5^2 & 8^3 \\
2 & 1^1 & 5^2 & 7^2 \\
1 & 1^1 & 4^1 & 4^1 \\
\hline
 & 1 & 2 & 3
\end{array}
$$

Now to find the _best_ two items out of the three ($p=2$), we use Carlos _alternate_ formula $\max\limits_{p=0,j-1}\{T_{i, p}\}$. The _best 2_ dp table looks as follows:
$$
\begin{array}{c|ccc}
3 & 1^1 & 5^2 & 4^2 \\
2 & 1^1 & 5^2 & 7^2 \\
1 & 1^1 & 4^1 & 4^1 \\
\hline
 & 1 & 2 & 3
\end{array}
$$
You can clearly see this is wrong--look at the values of $T_{3,3}$ and $T_{2,3}$. Anyway this is how I calculated the table: the first two columns ($j$-items) are identical. The third column ($j=3$) is where we start using the formula.

- $T_{1,3} = T_{1,2}$
- $T_{2,3} = T_{i-w_j,j-1,k-1}+v_j = T_{1,j-1,k-1}+v_j = T_{1,j-1,k-1}+v_j = \max\limits_{p=0,j-1}\{T_{i, p}\} + v_j = \max\limits_{p=0,1}\{T_{1, p}\} + v_j = \max\{T_{1,1}, T_{1,2}\} + v_j = \max\{1,4\} + 3 = 4 + 3 = 7$
- $T_{3,3} = T_{i-w_j,j-1,k-1}+v_j = T_{2,j-1,k-1}+v_j = T_{2,j-1,k-1}+v_j = \max\limits_{p=0,j-1}\{T_{i, p}\} + v_j = \max\limits_{p=0,1}\{T_{2, p}\} + v_j = \max\{T_{2,1}, \require{enclose}\enclose{updiagonalstrike}{T_{2,2}}\} + v_j = \max\{1,\enclose{updiagonalstrike}{5}\} + 3 = 1 + 3 = 4$

Now the reason the $T_{2,2} (5)$ is crossed-out because the _total items visited_ count for that entry is already $2$, so I'm _assuming_ we just skip it? But (un?)fortunately the two best items are $j=2$ and $j=3$.

For reference, I implemented using the _k-dimension_ and that one gives the correct results. However, I'm still curious as to why this specific formula isn't working; I'm sure I'm doing something silly--please help me find it =)

_EDIT: Dp tables for $k=1,2,3$_


K=1
$
\begin{array}{c|ccc}
3 & 1^1 & 4^1 & 4^1 \\
2 & 1^1 & 4^1 & 4^1 \\
1 & 1^1 & 4^1 & 4^1 \\
\hline
 & 1 & 2 & 3
\end{array}
$

K=2
$
\begin{array}{c|ccc}
3 & 1^1 & 5^2 & 7^2 \\
2 & 1^1 & 5^2 & 7^2 \\
1 & 1^1 & 4^1 & 4^1 \\
\hline
 & 1 & 2 & 3
\end{array}
$

K=3
$
\begin{array}{c|ccc}
3 & 1^1 & 5^2 & 8^3 \\
2 & 1^1 & 5^2 & 7^2 \\
1 & 1^1 & 4^1 & 4^1 \\
\hline
 & 1 & 2 & 3
\end{array}
$