Let

Unable to parse markup [type=CF_TEX]

be a boolean function that tells us if we can reduce the substring

Unable to parse markup [type=CF_TEX]

. Also define function

Unable to parse markup [type=CF_TEX]

and

Unable to parse markup [type=CF_TEX]

, where

Unable to parse markup [type=CF_TEX]

the largest index

Unable to parse markup [type=CF_TEX]

such that

Unable to parse markup [type=CF_TEX]

, and

Unable to parse markup [type=CF_TEX]

as the lowest left index with the mentioned property.

Now here's the code for the recurence:

dp[i][j]=0;
if (r[i]>i) dp[i][j]=dp[r[i]+1][j];
for (k=r[i]+1; k<=j; k++)
    if (s[i]==s[k])
       dp[i][j]=dp[i][j] | (dp[r[i]+1][l[k]-1]] & dp[k+1][j]);

(There are also some cornes cases)

We observe that if

Unable to parse markup [type=CF_TEX]

then we can transition to state

Unable to parse markup [type=CF_TEX]

and see if we can reduce the substring

Unable to parse markup [type=CF_TEX]

. But also we can combine the substring

Unable to parse markup [type=CF_TEX]

with some other substring

Unable to parse markup [type=CF_TEX]

if all their characters are equal. So we are left to reduce

Unable to parse markup [type=CF_TEX]

and

Unable to parse markup [type=CF_TEX]

. We use

Unable to parse markup [type=CF_TEX]

and

Unable to parse markup [type=CF_TEX]

because it's always better to pick these indeces instead of fixing the right point of an interval and try to extend it to the left until we get to a different character.

I know that I worded it pretty badly, but try to simulate the recurrence on the examples and it should become clear why this kind of approach works. (but take notice, I'm not 100% sure that this solution is correct)