CSE 101. Algorithm Design and Analysis Miles Jones Office 4208 CSE Building Lecture 20: Dynamic Programming
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1 CSE 101 Algorithm Design and Analysis Miles Jones Office 4208 CSE Building Lecture 20: Dynamic Programming
2 DYNAMIC PROGRAMMING Dynamic programming is an algorithmic paradigm in which a problem is solved by identifying a collection of subproblems and tackling them one by one, smallest first, using the answers to small problems to help figure out larger ones, until they are all solved. Examples:
3 DYNAMIC PROGRAMMING Dynamic programming is an algorithmic paradigm in which a problem is solved by identifying a collection of subproblems and tackling them one by one, smallest first, using the answers to small problems to help figure out larger ones, until they are all solved. Examples: findmax, findmin, fib2,
4 DYNAMIC PROGRAMMING (SHORTEST PATH IN DAGS)
5 DYNAMIC PROGRAMMING (SHORTEST PATH IN DAGS)
6 DYNAMIC PROGRAMMING (SHORTEST PATH IN DAGS) Shortest distance from D to another node x will be denoted dist(x). Notice that the shortest distance from D to C is dist C =
7 DYNAMIC PROGRAMMING (SHORTEST PATH IN DAGS) Shortest distance from D to another node x will be denoted dist(x). Notice that the shortest distance from D to C is dist C = min(dist E + 5, dist B + 2)
8 DYNAMIC PROGRAMMING (SHORTEST PATH IN DAGS) Shortest distance from D to another node x will be denoted dist(x). Notice that the shortest distance from D to C is dist C = min(dist E + 5, dist B + 2) This kind of relation can be written for every node. Since it s a DAG, the arrows only go to the right so by the time we get to node x, we have all the information needed!!
9 DYNAMIC PROGRAMMING (SHORTEST PATH IN DAGS) Step1: Define the subproblems: Step 2: Base Case: Step 3: express recursively: Step 4: order the subproblems
10 DYNAMIC PROGRAMMING (SHORTEST PATH IN DAGS) Step1: Define the subproblems: the distance to the ith vertex Step 2: Base Case: the distance to the first vertex to itself is 0 Step 3: express recursively: dist(v) = min (u,v) E dist u + l u, v Step 4: order the subproblems linearized order
11 DYNAMIC PROGRAMMING (SHORTEST PATH IN DAGS) initialize all dist(.) values to infinity dist(s):=0 for each v V\{s} in linearized order dist(v)= min (u,v) E dist u + l u, v Like D/C, this algorithm solves a family of subproblems. We start with dist(s)=0 and we get to the larger subproblems in linearized order by using the smaller subproblems.
12 DYNAMIC PROGRAMMING (SHORTEST PATH IN DAGS)
13 DP (LONGEST INCREASING SUBSEQUENCE) Given a sequence of distinct positive integers a[1],,a[n] An increasing subsequence is a sequence a[i_1],,a[i_k] such that i_1< <i_k and a[i_1]< <a[i_k]. For Example: 15, 18, 8, 11, 5, 12, 16, 2, 20, 9, 10, 4 5, 16, 20 is an increasing subsequence. How long is the longest increasing subsequence?
14 DP (LONGEST INCREASING SUBSEQUENCE) Let s make a DAG out of our example:
15 DP (LONGEST INCREASING SUBSEQUENCE) Let s make a DAG out of our example: Now, instead of finding the longest increasing subsequence of a list of integers, we are finding the longest path in a DAG!!!!
16 DYNAMIC PROGRAMMING (LONGEST INCREASING SUBSEQUENCE) Step1: Define the subproblems: Step 2: Base Case: Step 3: express recursively: Step 4: order the subproblems
17 DYNAMIC PROGRAMMING (LONGEST INCREASING SUBSEQUENCE) Step1: Define the subproblems: L(k) will be the length of the longest increasing subsequence ending exactly at position k Step 2: Base Case: L(1) = 0 Step 3: express recursively: L(k) = 1+max({L[i]:(i,j) is an edge}) Step 4: order the subproblems from left to right
18 DP (LONGEST INCREASING SUBSEQUENCE) Finding longest path in a DAG: L[1]:=0 for j=1 n L[j]=1+max({L[i]:(i,j) is an edge}) prev(j)=i return max({l[j]})
19 DP (LONGEST INCREASING SUBSEQUENCE) Let s make a DAG out of our example:
20 DP (LONGEST INCREASING SUBSEQUENCE) Let s make a DAG out of our example:
21 DP (LONGEST INCREASING SUBSEQUENCE) Finding longest path in a DAG: L[1]:=0 for j=1 n L[j]=max({L[i]:(i,j) is an edge}) prev(j)=i return max({l[j]}) How long does this take?
22 DP (LONGEST INCREASING SUBSEQUENCE) Finding longest path in a DAG: L[1]:=0 for j=1 n L[j]=max({L[i]:(i,j) is an edge}) prev(j)=i return max({l[j]}) How long does this take? To solve L[j]=max({L[i]:(i,j) is an edge}), we need to know L[i] for each edge (i,j) in E. This is equal to the indegree of j. So we sum over all vertices we get that j V so the runtime is O( E ). d in (j) = E
23 DP (LONGEST INCREASING SUBSEQUENCE) The runtime is dependent on the number of edges in the DAG. Note that if the sequence is increasing If the sequenece is decreasing then
24 DP (LONGEST INCREASING SUBSEQUENCE) The runtime is dependent on the number of edges in the DAG. What are the maximum and minimum number of edges?
25 DP (LONGEST INCREASING SUBSEQUENCE) The runtime is dependent on the number of edges in the DAG. Note that if the sequence is increasing then E = n If the sequenece is decreasing then E =
26 DP (LONGEST INCREASING SUBSEQUENCE) What is the expected number of edges?
27 DP (STRING RECONSTRUCTION) Given a string of letters with no spaces or punctuation, how would you figure out how to separate the words? Example: THESEARETHERULES
28 DP (STRING RECONSTRUCTION) Given a string of letters with no spaces or punctuation, how would you figure out how to separate the words? Example: THESEARETHERULES greedy approach: Find the first real word, remove it from the string and repeat on the remaining string. If it doesn t work try a different way.
29 DP (STRING RECONSTRUCTION) THESEARETHERULES
30 DYNAMIC PROGRAMMING (STRING RECONSTRUCTION) Step1: Define the subproblems: Step 2: Base Case: Step 3: express recursively: Step 4: order the subproblems
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