Dynamic Programming. CS 374: Algorithms & Models of Computation, Fall Lecture 11. October 1, 2015
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1 CS 374: Algorithms & Models of Computation, Fall 2015 Dynamic Programming Lecture 11 October 1, 2015 Chandra & Manoj (UIUC) CS374 1 Fall / 32
2 Dynamic Programming Dynamic Programming is smart recursion plus memoization Chandra & Manoj (UIUC) CS374 2 Fall / 32
3 Dynamic Programming Dynamic Programming is smart recursion plus memoization Question: Suppose we have a recursive program foo(x) that takes an input x. On input of size n the number of distinct sub-problems that foo(x) generates is at most A(n) foo(x) spends at most B(n) time not counting the time for its recursive calls. What is an upper bound on the running time of memoized version of foo(x) if x = n? Chandra & Manoj (UIUC) CS374 2 Fall / 32
4 Dynamic Programming Dynamic Programming is smart recursion plus memoization Question: Suppose we have a recursive program foo(x) that takes an input x. On input of size n the number of distinct sub-problems that foo(x) generates is at most A(n) foo(x) spends at most B(n) time not counting the time for its recursive calls. What is an upper bound on the running time of memoized version of foo(x) if x = n? O(A(n)B(n)). Chandra & Manoj (UIUC) CS374 2 Fall / 32
5 Part I Longest Increasing Subsequence Chandra & Manoj (UIUC) CS374 3 Fall / 32
6 Sequences Definition Sequence: an ordered list a 1, a 2,..., a n. Length of a sequence is number of elements in the list. Definition a i1,..., a ik is a subsequence of a 1,..., a n if 1 i 1 < i 2 <... < i k n. Definition A sequence is increasing if a 1 < a 2 <... < a n. It is non-decreasing if a 1 a 2... a n. Similarly decreasing and non-increasing. Chandra & Manoj (UIUC) CS374 4 Fall / 32
7 Sequences Example... Example 1 Sequence: 6, 3, 5, 2, 7, 8, 1, 9 2 Subsequence of above sequence: 5, 2, 1 3 Increasing sequence: 3, 5, 9, 17, 54 4 Decreasing sequence: 34, 21, 7, 5, 1 5 Increasing subsequence of the first sequence: 2, 7, 9. Chandra & Manoj (UIUC) CS374 5 Fall / 32
8 Longest Increasing Subsequence Problem Input A sequence of numbers a 1, a 2,..., a n Goal Find an increasing subsequence a i1, a i2,..., a ik maximum length of Chandra & Manoj (UIUC) CS374 6 Fall / 32
9 Longest Increasing Subsequence Problem Input A sequence of numbers a 1, a 2,..., a n Goal Find an increasing subsequence a i1, a i2,..., a ik maximum length of Example 1 Sequence: 6, 3, 5, 2, 7, 8, 1 2 Increasing subsequences: 6, 7, 8 and 3, 5, 7, 8 and 2, 7 etc 3 Longest increasing subsequence: 3, 5, 7, 8 Chandra & Manoj (UIUC) CS374 6 Fall / 32
10 Recursive Approach: Take 1 LIS: Longest increasing subsequence Can we find a recursive algorithm for LIS? LIS(A[1..n]): Chandra & Manoj (UIUC) CS374 7 Fall / 32
11 Recursive Approach: Take 1 LIS: Longest increasing subsequence Can we find a recursive algorithm for LIS? LIS(A[1..n]): 1 Case 1: Does not contain A[n] in which case LIS(A[1..n]) = LIS(A[1..(n 1)]) 2 Case 2: contains A[n] in which case LIS(A[1..n]) is not so clear. Observation For second case we want to find a subsequence in A[1..(n 1)] that is restricted to numbers less than A[n]. This suggests that a more general problem is LIS smaller(a[1..n], x) which gives the longest increasing subsequence in A where each number in the sequence is less than x. Chandra & Manoj (UIUC) CS374 7 Fall / 32
12 Recursive Approach LIS(A[1..n]): the length of longest increasing subsequence in A LIS smaller(a[1..n], x): length of longest increasing subsequence in A[1..n] with all numbers in subsequence less than x LIS smaller(a[1..n], x): if (n = 0) then return 0 m = LIS smaller(a[1..(n 1)], x) if (A[n] < x) then m = max(m, 1 + LIS smaller(a[1..(n 1)], A[n])) Output m LIS(A[1..n]): return LIS smaller(a[1..n], ) Chandra & Manoj (UIUC) CS374 8 Fall / 32
13 Example Sequence: A[1..8] = 6, 3, 5, 2, 7, 8, 1, 9 Chandra & Manoj (UIUC) CS374 9 Fall / 32
14 Recursive Approach LIS smaller(a[1..n], x): if (n = 0) then return 0 m = LIS smaller(a[1..(n 1)], x) if (A[n] < x) then m = max(m, 1 + LIS smaller(a[1..(n 1)], A[n])) Output m LIS(A[1..n]): return LIS smaller(a[1..n], ) How many distinct sub-problems will LIS smaller(a[1..n], ) generate? Chandra & Manoj (UIUC) CS Fall / 32
15 Recursive Approach LIS smaller(a[1..n], x): if (n = 0) then return 0 m = LIS smaller(a[1..(n 1)], x) if (A[n] < x) then m = max(m, 1 + LIS smaller(a[1..(n 1)], A[n])) Output m LIS(A[1..n]): return LIS smaller(a[1..n], ) How many distinct sub-problems will LIS smaller(a[1..n], ) generate? O(n 2 ) Chandra & Manoj (UIUC) CS Fall / 32
16 Recursive Approach LIS smaller(a[1..n], x): if (n = 0) then return 0 m = LIS smaller(a[1..(n 1)], x) if (A[n] < x) then m = max(m, 1 + LIS smaller(a[1..(n 1)], A[n])) Output m LIS(A[1..n]): return LIS smaller(a[1..n], ) How many distinct sub-problems will LIS smaller(a[1..n], ) generate? O(n 2 ) What is the running time if we memoize recursion? Chandra & Manoj (UIUC) CS Fall / 32
17 Recursive Approach LIS smaller(a[1..n], x): if (n = 0) then return 0 m = LIS smaller(a[1..(n 1)], x) if (A[n] < x) then m = max(m, 1 + LIS smaller(a[1..(n 1)], A[n])) Output m LIS(A[1..n]): return LIS smaller(a[1..n], ) How many distinct sub-problems will LIS smaller(a[1..n], ) generate? O(n 2 ) What is the running time if we memoize recursion? O(n 2 ) since each call takes O(1) time to assemble the answers from to recursive calls and no other computation. Chandra & Manoj (UIUC) CS Fall / 32
18 Recursive Approach LIS smaller(a[1..n], x): if (n = 0) then return 0 m = LIS smaller(a[1..(n 1)], x) if (A[n] < x) then m = max(m, 1 + LIS smaller(a[1..(n 1)], A[n])) Output m LIS(A[1..n]): return LIS smaller(a[1..n], ) How many distinct sub-problems will LIS smaller(a[1..n], ) generate? O(n 2 ) What is the running time if we memoize recursion? O(n 2 ) since each call takes O(1) time to assemble the answers from to recursive calls and no other computation. How much space for memoization? Chandra & Manoj (UIUC) CS Fall / 32
19 Recursive Approach LIS smaller(a[1..n], x): if (n = 0) then return 0 m = LIS smaller(a[1..(n 1)], x) if (A[n] < x) then m = max(m, 1 + LIS smaller(a[1..(n 1)], A[n])) Output m LIS(A[1..n]): return LIS smaller(a[1..n], ) How many distinct sub-problems will LIS smaller(a[1..n], ) generate? O(n 2 ) What is the running time if we memoize recursion? O(n 2 ) since each call takes O(1) time to assemble the answers from to recursive calls and no other computation. How much space for memoization? O(n 2 ) Chandra & Manoj (UIUC) CS Fall / 32
20 Recursive Algorithm: Take 2 Definition LISEnding(A[1..n]): length of longest increasing sub-sequence that ends in A[n]. Question: can we obtain a recursive expression? Chandra & Manoj (UIUC) CS Fall / 32
21 Recursive Algorithm: Take 2 Definition LISEnding(A[1..n]): length of longest increasing sub-sequence that ends in A[n]. Question: can we obtain a recursive expression? LISEnding(A[1..n]) = ( ) max 1 + LISEnding(A[1..i]) i:a[i]<a[n] Chandra & Manoj (UIUC) CS Fall / 32
22 Example Sequence: A[1..8] = 6, 3, 5, 2, 7, 8, 1, 9 Chandra & Manoj (UIUC) CS Fall / 32
23 Recursive Algorithm: Take 2 LIS ending alg(a[1..n]): if (n = 0) return 0 m = 1 for i = 1 to n 1 do if (A[i] < A[n]) ( then ) m = max m, 1 + LIS ending alg(a[1..i]) return m LIS(A[1..n]): return max n i=1lis ending alg(a[1... i]) Chandra & Manoj (UIUC) CS Fall / 32
24 Recursive Algorithm: Take 2 LIS ending alg(a[1..n]): if (n = 0) return 0 m = 1 for i = 1 to n 1 do if (A[i] < A[n]) ( then ) m = max m, 1 + LIS ending alg(a[1..i]) return m LIS(A[1..n]): return max n i=1lis ending alg(a[1... i]) How many distinct sub-problems will LIS ending alg(a[1..n]) generate? Chandra & Manoj (UIUC) CS Fall / 32
25 Recursive Algorithm: Take 2 LIS ending alg(a[1..n]): if (n = 0) return 0 m = 1 for i = 1 to n 1 do if (A[i] < A[n]) ( then ) m = max m, 1 + LIS ending alg(a[1..i]) return m LIS(A[1..n]): return max n i=1lis ending alg(a[1... i]) How many distinct sub-problems will LIS ending alg(a[1..n]) generate? O(n) Chandra & Manoj (UIUC) CS Fall / 32
26 Recursive Algorithm: Take 2 LIS ending alg(a[1..n]): if (n = 0) return 0 m = 1 for i = 1 to n 1 do if (A[i] < A[n]) ( then ) m = max m, 1 + LIS ending alg(a[1..i]) return m LIS(A[1..n]): return max n i=1lis ending alg(a[1... i]) How many distinct sub-problems will LIS ending alg(a[1..n]) generate? O(n) What is the running time if we memoize recursion? Chandra & Manoj (UIUC) CS Fall / 32
27 Recursive Algorithm: Take 2 LIS ending alg(a[1..n]): if (n = 0) return 0 m = 1 for i = 1 to n 1 do if (A[i] < A[n]) ( then ) m = max m, 1 + LIS ending alg(a[1..i]) return m LIS(A[1..n]): return max n i=1lis ending alg(a[1... i]) How many distinct sub-problems will LIS ending alg(a[1..n]) generate? O(n) What is the running time if we memoize recursion? O(n 2 ) since each call takes O(n) time Chandra & Manoj (UIUC) CS Fall / 32
28 Recursive Algorithm: Take 2 LIS ending alg(a[1..n]): if (n = 0) return 0 m = 1 for i = 1 to n 1 do if (A[i] < A[n]) ( then ) m = max m, 1 + LIS ending alg(a[1..i]) return m LIS(A[1..n]): return max n i=1lis ending alg(a[1... i]) How many distinct sub-problems will LIS ending alg(a[1..n]) generate? O(n) What is the running time if we memoize recursion? O(n 2 ) since each call takes O(n) time How much space for memoization? Chandra & Manoj (UIUC) CS Fall / 32
29 Recursive Algorithm: Take 2 LIS ending alg(a[1..n]): if (n = 0) return 0 m = 1 for i = 1 to n 1 do if (A[i] < A[n]) ( then ) m = max m, 1 + LIS ending alg(a[1..i]) return m LIS(A[1..n]): return max n i=1lis ending alg(a[1... i]) How many distinct sub-problems will LIS ending alg(a[1..n]) generate? O(n) What is the running time if we memoize recursion? O(n 2 ) since each call takes O(n) time How much space for memoization? O(n) Chandra & Manoj (UIUC) CS Fall / 32
30 Removing recursion to obtain iterative algorithm Typically, after finding a dynamic programming recursion, we often convert the recursive algorithm into an iterative algorithm via explicit memoization and bottom up computation. Why? Chandra & Manoj (UIUC) CS Fall / 32
31 Removing recursion to obtain iterative algorithm Typically, after finding a dynamic programming recursion, we often convert the recursive algorithm into an iterative algorithm via explicit memoization and bottom up computation. Why? Mainly for further optimization of running time and space. Chandra & Manoj (UIUC) CS Fall / 32
32 Removing recursion to obtain iterative algorithm Typically, after finding a dynamic programming recursion, we often convert the recursive algorithm into an iterative algorithm via explicit memoization and bottom up computation. Why? Mainly for further optimization of running time and space. How? First, allocate a data structure (usually an array or a multi-dimensional array that can hold values for each of the subproblems) Figure out a way to order the computation of the sub-problems starting from the base case. Chandra & Manoj (UIUC) CS Fall / 32
33 Removing recursion to obtain iterative algorithm Caveat: Dynamic programming is not about filling tables. It is about finding a smart recursion. First, find the correct recursion. Chandra & Manoj (UIUC) CS Fall / 32 Typically, after finding a dynamic programming recursion, we often convert the recursive algorithm into an iterative algorithm via explicit memoization and bottom up computation. Why? Mainly for further optimization of running time and space. How? First, allocate a data structure (usually an array or a multi-dimensional array that can hold values for each of the subproblems) Figure out a way to order the computation of the sub-problems starting from the base case.
34 Iterative Algorithm via Memoization Compute the values LIS ending alg(a[1..i]) iteratively in a bottom up fashion. LIS ending alg(a[1..n]): Array L[1..n] (* L[i] = value of LIS ending alg(a[1..i]) *) for i = 1 to n do L[i] = 1 for j = 1 to i 1 do if (A[j] < A[i]) do L[i] = max(l[i], 1 + L[j]) return L LIS(A[1..n]): L = LIS ending alg(a[1..n]) return the maximum value in L Chandra & Manoj (UIUC) CS Fall / 32
35 Iterative Algorithm via Memoization Simplifying: LIS(A[1..n]): Array L[1..n] (* L[i] stores the value LISEnding(A[1..i]) *) m = 0 for i = 1 to n do L[i] = 1 for j = 1 to i 1 do if (A[j] < A[i]) do L[i] = max(l[i], 1 + L[j]) m = max(m, L[i]) return m Chandra & Manoj (UIUC) CS Fall / 32
36 Iterative Algorithm via Memoization Simplifying: LIS(A[1..n]): Array L[1..n] (* L[i] stores the value LISEnding(A[1..i]) *) m = 0 for i = 1 to n do L[i] = 1 for j = 1 to i 1 do if (A[j] < A[i]) do L[i] = max(l[i], 1 + L[j]) m = max(m, L[i]) return m Correctness: Via induction following the recursion Running time: Chandra & Manoj (UIUC) CS Fall / 32
37 Iterative Algorithm via Memoization Simplifying: LIS(A[1..n]): Array L[1..n] (* L[i] stores the value LISEnding(A[1..i]) *) m = 0 for i = 1 to n do L[i] = 1 for j = 1 to i 1 do if (A[j] < A[i]) do L[i] = max(l[i], 1 + L[j]) m = max(m, L[i]) return m Correctness: Via induction following the recursion Running time: O(n 2 ) Space: Chandra & Manoj (UIUC) CS Fall / 32
38 Iterative Algorithm via Memoization Simplifying: LIS(A[1..n]): Array L[1..n] (* L[i] stores the value LISEnding(A[1..i]) *) m = 0 for i = 1 to n do L[i] = 1 for j = 1 to i 1 do if (A[j] < A[i]) do L[i] = max(l[i], 1 + L[j]) m = max(m, L[i]) return m Correctness: Via induction following the recursion Running time: O(n 2 ) Space: Θ(n) Chandra & Manoj (UIUC) CS Fall / 32
39 Iterative Algorithm via Memoization Simplifying: LIS(A[1..n]): Array L[1..n] (* L[i] stores the value LISEnding(A[1..i]) *) m = 0 for i = 1 to n do L[i] = 1 for j = 1 to i 1 do if (A[j] < A[i]) do L[i] = max(l[i], 1 + L[j]) m = max(m, L[i]) return m Correctness: Via induction following the recursion Running time: O(n 2 ) Space: Θ(n) O(n log n) run-time achievable via better data structures. Chandra & Manoj (UIUC) CS Fall / 32
40 Example Example 1 Sequence: 6, 3, 5, 2, 7, 8, 1 2 Longest increasing subsequence: 3, 5, 7, 8 Chandra & Manoj (UIUC) CS Fall / 32
41 Example Example 1 Sequence: 6, 3, 5, 2, 7, 8, 1 2 Longest increasing subsequence: 3, 5, 7, 8 1 L[i] is value of longest increasing subsequence ending in A[i] 2 Recursive algorithm computes L[i] from L[1] to L[i 1] 3 Iterative algorithm builds up the values from L[1] to L[n] Chandra & Manoj (UIUC) CS Fall / 32
42 Computing Solutions 1 Memoization + Recursion/Iteration allows one to compute the optimal value. What about the actual sub-sequence? 2 Two methods 1 Explicit: For each subproblem find an optimum solution for that subproblem while computing the optimum value for that subproblem. Typically slow but automatic. 2 Implicit: For each subproblem keep track of sufficient information (decision) on how optimum solution for subproblem was computed. Reconstruct optimum solution later via stored information. Typically much more efficient but requires more thought. Chandra & Manoj (UIUC) CS Fall / 32
43 Computing Solution: Explicit method for LIS LIS(A[1..n]): Array L[1..n] (* L[i] stores the value LISEnding(A[1..i]) *) Array S[1..n] (* S[i] stores the sequence achieving L[i] *) m = 0 h = 0 for i = 1 to n do L[i] = 1 S[i] = [i] for j = 1 to i 1 do if (A[j] < A[i]) and (L[i] < 1 + L[j]) do L[i] = 1 + L[j] S[i] = concat(s[j], [i]) if (m < L[i]) m = L[i], h = i return m, S[h] Chandra & Manoj (UIUC) CS Fall / 32
44 Computing Solution: Explicit method for LIS LIS(A[1..n]): Array L[1..n] (* L[i] stores the value LISEnding(A[1..i]) *) Array S[1..n] (* S[i] stores the sequence achieving L[i] *) m = 0 h = 0 for i = 1 to n do L[i] = 1 S[i] = [i] for j = 1 to i 1 do if (A[j] < A[i]) and (L[i] < 1 + L[j]) do L[i] = 1 + L[j] S[i] = concat(s[j], [i]) if (m < L[i]) m = L[i], h = i return m, S[h] Running time: O(n 3 ) Space: O(n 2 ). Extra time/space to store, copy Chandra & Manoj (UIUC) CS Fall / 32
45 Computing Solution: Implicit method for LIS LIS(A[1..n]): Array L[1..n] (* L[i] stores the value LISEnding(A[1..i]) *) Array D[1..n] (* D[i] stores how L[i] was computed *) m = 0 h = 0 for i = 1 to n do L[i] = 1 D[i] = i for j = 1 to i 1 do if (A[j] < A[i]) and (L[i] < 1 + L[j]) do L[i] = 1 + L[j] D[i] = j if (m < L[i]) m = L[i], h = i m = L[h] is optimum value Chandra & Manoj (UIUC) CS Fall / 32
46 Computing Solution: Implicit method for LIS LIS(A[1..n]): Array L[1..n] (* L[i] stores the value LISEnding(A[1..i]) *) Array D[1..n] (* D[i] stores how L[i] was computed *) m = 0 h = 0 for i = 1 to n do L[i] = 1 D[i] = i for j = 1 to i 1 do if (A[j] < A[i]) and (L[i] < 1 + L[j]) do L[i] = 1 + L[j] D[i] = j if (m < L[i]) m = L[i], h = i m = L[h] is optimum value Question: Can we obtain solution from stored D values and h? Chandra & Manoj (UIUC) CS Fall / 32
47 Computing Solution: Implicit method for LIS LIS(A[1..n]): Array L[1..n] (* L[i] stores the value LISEnding(A[1..i]) *) Array D[1..n] (* D[i] stores how L[i] was computed *) m = 0, h = 0 for i = 1 to n do L[i] = 1 D[i] = 0 for j = 1 to i 1 do if (A[j] < A[i]) and (L[i] < 1 + L[j]) do L[i] = 1 + L[j], D[i] = j if (m < L[i]) m = L[i], h = i S = empty sequence while (h > 0) do add L[h] to front of S h = D[h] Output optimum value m, and an optimum subsequence S Chandra & Manoj (UIUC) CS Fall / 32
48 Computing Solution: Implicit method for LIS LIS(A[1..n]): Array L[1..n] (* L[i] stores the value LISEnding(A[1..i]) *) Array D[1..n] (* D[i] stores how L[i] was computed *) m = 0, h = 0 for i = 1 to n do L[i] = 1 D[i] = 0 for j = 1 to i 1 do if (A[j] < A[i]) and (L[i] < 1 + L[j]) do L[i] = 1 + L[j], D[i] = j if (m < L[i]) m = L[i], h = i S = empty sequence while (h > 0) do add L[h] to front of S h = D[h] Output optimum value m, and an optimum subsequence S Running time: O(n 2 ) Space: O(n). Chandra & Manoj (UIUC) CS Fall / 32
49 Dynamic Programming 1 Find a smart recursion for the problem in which the number of distinct subproblems is small; polynomial in the original problem size. 2 Estimate the number of subproblems, the time to evaluate each subproblem and the space needed to store the value. This gives an upper bound on the total running time if we use automatic memoization. 3 Eliminate recursion and find an iterative algorithm to compute the problems bottom up by storing the intermediate values in an appropriate data structure; need to find the right way or order the subproblem evaluation. This leads to an explicit algorithm. 4 Optimize the resulting algorithm further Chandra & Manoj (UIUC) CS Fall / 32
50 Part II Checking if string in L Chandra & Manoj (UIUC) CS Fall / 32
51 Problem Input A string w Σ and access to a language L Σ via function IsStringinL(string x) that decides whether x is in L Goal Decide if w L using IsStringinL(string x) as a black box sub-routine Example Suppose L is English and we have a procedure to check whether a string/word is in the English dictionary. Is the string isthisanenglishsentence in English? Is stampstamp in English? Is zibzzzad in English? Chandra & Manoj (UIUC) CS Fall / 32
52 Recursive Solution When is w L? Chandra & Manoj (UIUC) CS Fall / 32
53 Recursive Solution When is w L? w L if w L or if w = uv where u L and v L Chandra & Manoj (UIUC) CS Fall / 32
54 Recursive Solution When is w L? w L if w L or if w = uv where u L and v L Assume w is stored in array A[1..n] IsStringinLstar(A[1..n]): If (IsStringinL(A[1..n])) Output YES Else For (i = 1 to n 1) do If (IsStringinL(A[1..i]) and IsStringinLstar(A[i + 1..n])) Output YES Output NO Chandra & Manoj (UIUC) CS Fall / 32
55 Recursive Solution Assume w is stored in array A[1..n] IsStringinLstar(A[1..n]): If (IsStringinL(A[1..n])) Output YES Else For (i = 1 to n 1) do If (IsStringinL(A[1..i]) and IsStringinLstar(A[i + 1..n])) Output YES Output NO How many distinct sub-problems does IsStringinLstar(A[1..n]) generate? Chandra & Manoj (UIUC) CS Fall / 32
56 Recursive Solution Assume w is stored in array A[1..n] IsStringinLstar(A[1..n]): If (IsStringinL(A[1..n])) Output YES Else For (i = 1 to n 1) do If (IsStringinL(A[1..i]) and IsStringinLstar(A[i + 1..n])) Output YES Output NO How many distinct sub-problems does IsStringinLstar(A[1..n]) generate? O(n) Chandra & Manoj (UIUC) CS Fall / 32
57 Recursive Solution Assume w is stored in array A[1..n] IsStringinLstar(A[1..n]): If (IsStringinL(A[1..n])) Output YES Else For (i = 1 to n 1) do If (IsStringinL(A[1..i]) and IsStringinLstar(A[i + 1..n])) Output YES Output NO How many distinct sub-problems does IsStringinLstar(A[1..n]) generate? O(n) What is running time of memoized version of IsStringinLstar(A[1..n])? Chandra & Manoj (UIUC) CS Fall / 32
58 Recursive Solution Assume w is stored in array A[1..n] IsStringinLstar(A[1..n]): If (IsStringinL(A[1..n])) Output YES Else For (i = 1 to n 1) do If (IsStringinL(A[1..i]) and IsStringinLstar(A[i + 1..n])) Output YES Output NO How many distinct sub-problems does IsStringinLstar(A[1..n]) generate? O(n) What is running time of memoized version of IsStringinLstar(A[1..n])? O(n 2 ) Chandra & Manoj (UIUC) CS Fall / 32
59 Recursive Solution Assume w is stored in array A[1..n] IsStringinLstar(A[1..n]): If (IsStringinL(A[1..n])) Output YES Else For (i = 1 to n 1) do If (IsStringinL(A[1..i]) and IsStringinLstar(A[i + 1..n])) Output YES Output NO How many distinct sub-problems does IsStringinLstar(A[1..n]) generate? O(n) What is running time of memoized version of IsStringinLstar(A[1..n])? O(n 2 ) What is space requirement of memoized version of IsStringinLstar(A[1..n])? Chandra & Manoj (UIUC) CS Fall / 32
60 Recursive Solution Assume w is stored in array A[1..n] IsStringinLstar(A[1..n]): If (IsStringinL(A[1..n])) Output YES Else For (i = 1 to n 1) do If (IsStringinL(A[1..i]) and IsStringinLstar(A[i + 1..n])) Output YES Output NO How many distinct sub-problems does IsStringinLstar(A[1..n]) generate? O(n) What is running time of memoized version of IsStringinLstar(A[1..n])? O(n 2 ) What is space requirement of memoized version of IsStringinLstar(A[1..n])? O(n) Chandra & Manoj (UIUC) CS Fall / 32
61 A variation Input A string w Σ and access to a language L Σ via function IsStringinL(string x) that decides whether x is in L, and non-negative integer k Goal Decide if w L k using IsStringinL(string x) as a black box sub-routine Example Suppose L is English and we have a procedure to check whether a string/word is in the English dictionary. Is the string isthisanenglishsentence in English 5? Is the string isthisanenglishsentence in English 4? Is asinineat in English 2? Is asinineat in English 4? Is zibzzzad in English 1? Chandra & Manoj (UIUC) CS Fall / 32
62 Recursive Solution When is w L k? Chandra & Manoj (UIUC) CS Fall / 32
63 Recursive Solution When is w L k? k = 0: w L k iff w = ɛ k = 1: w L k iff w L k > 1: w L k if w = uv with u L and v L k 1 Chandra & Manoj (UIUC) CS Fall / 32
64 Recursive Solution Output NO Chandra & Manoj (UIUC) CS Fall / 32 When is w L k? k = 0: w L k iff w = ɛ k = 1: w L k iff w L k > 1: w L k if w = uv with u L and v L k 1 Assume w is stored in array A[1..n] IsStringinLk(A[1..n], k): If (k = 0) If (n = 0) Output YES Else Ouput NO If (k = 1) Output IsStringinL(A[1..n]) Else For (i = 1 to n 1) do If (IsStringinL(A[1..i]) and IsStringinLk(A[i + 1..n], k 1)) Output YES
65 Analysis IsStringinLk(A[1..n], k): If (k = 0) If (n = 0) Output YES Else Ouput NO If (k = 1) Output IsStringinL(A[1..n]) Else For (i = 1 to n 1) do If (IsStringinL(A[1..i]) and IsStringinLk(A[i + 1..n], k 1)) Output YES Output NO How many distinct sub-problems are generated by IsStringinLk(A[1..n], k)? Chandra & Manoj (UIUC) CS Fall / 32
66 Analysis IsStringinLk(A[1..n], k): If (k = 0) If (n = 0) Output YES Else Ouput NO If (k = 1) Output IsStringinL(A[1..n]) Else For (i = 1 to n 1) do If (IsStringinL(A[1..i]) and IsStringinLk(A[i + 1..n], k 1)) Output YES Output NO How many distinct sub-problems are generated by IsStringinLk(A[1..n], k)? O(nk) Chandra & Manoj (UIUC) CS Fall / 32
67 Analysis IsStringinLk(A[1..n], k): If (k = 0) If (n = 0) Output YES Else Ouput NO If (k = 1) Output IsStringinL(A[1..n]) Else For (i = 1 to n 1) do If (IsStringinL(A[1..i]) and IsStringinLk(A[i + 1..n], k 1)) Output YES Output NO How many distinct sub-problems are generated by IsStringinLk(A[1..n], k)? O(nk) How much space? Chandra & Manoj (UIUC) CS Fall / 32
68 Analysis IsStringinLk(A[1..n], k): If (k = 0) If (n = 0) Output YES Else Ouput NO If (k = 1) Output IsStringinL(A[1..n]) Else For (i = 1 to n 1) do If (IsStringinL(A[1..i]) and IsStringinLk(A[i + 1..n], k 1)) Output YES Output NO How many distinct sub-problems are generated by IsStringinLk(A[1..n], k)? O(nk) How much space? O(nk) pause Running time? Chandra & Manoj (UIUC) CS Fall / 32
69 Analysis IsStringinLk(A[1..n], k): If (k = 0) If (n = 0) Output YES Else Ouput NO If (k = 1) Output IsStringinL(A[1..n]) Else For (i = 1 to n 1) do If (IsStringinL(A[1..i]) and IsStringinLk(A[i + 1..n], k 1)) Output YES Output NO How many distinct sub-problems are generated by IsStringinLk(A[1..n], k)? O(nk) How much space? O(nk) pause Running time? O(n 2 k) Chandra & Manoj (UIUC) CS Fall / 32
70 Another variant Question: What if we want to check if w L i for some 0 i k? That is, is w k i=0 Li? Chandra & Manoj (UIUC) CS Fall / 32
71 Exercise Definition A string is a palindrome if w = w R. Examples: I, RACECAR, MALAYALAM, DOOFFOOD Chandra & Manoj (UIUC) CS Fall / 32
72 Exercise Definition A string is a palindrome if w = w R. Examples: I, RACECAR, MALAYALAM, DOOFFOOD Problem: Given a string w find the longest subsequence of w that is a palindrome. Example MAHDYNAMICPROGRAMZLETMESHOWYOUTHEM has MHYMRORMYHM as a palindromic subsequence Chandra & Manoj (UIUC) CS Fall / 32
73 Exercise Assume w is stored in an array A[1..n] LPS(A[1..n]): length of longest palindromic subsequence of A. Recursive expression/code? Chandra & Manoj (UIUC) CS Fall / 32
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