Response: explained in the explanation section
Explanation:
Given that:
Assume D(k) =║ true if [1::: k] is a valid sequence of words, or false otherwise
the sub problem s[1::: k] is a valid sequence of words IFF s[1::: 1] is valid and s[ 1 + 1::: k] is a valid word.
Thus, we derive that D(k) is defined by the following recurrence relation:
D(k) = ║ false max(d[l] ∧ DICT(s[1 + 1::: k]) otherwise
Algorithm:
Valid sentence (s,k)
D [1::: k] ∦ array of boolean variables.
for a ← 1 to m
do;
d(0) ← false
for b ← 0 to a - j
for b ← 0 to a - j
do;
if D[b] ∧ DICT s([b + 1::: a])
d (a) ← True
(b). Algorithm Output
if D[k] == True
stack = temp stack ∦stack assists in displaying the strings in order
c = k
while C > 0
stack push (s [w(c)]::: C] // w(p) denotes the index in s[1::: k] of the valid word // at position c
P = W (p) - 1
output stack
= 0 =
cheers, I hope this aids you!!!
