Recursive Back Tracking - Elder Laboratory

Recursive Back Tracking - Elder Laboratory

Dynamic Programming Credits: Many of these slides were originally authored by Jeff Edmonds, York University. Thanks Jeff! Optimization Problems For most, the best known algorithm runs in exponential time. Some have quick Greedy or Dynamic Programming algorithms. COSC 3101E 2 What is Dynamic Programming? Dynamic programming solves optimization problems by combining solutions to subproblems

Programming refers to a tabular method with a series of choices, not coding COSC 3101E 3 What is Dynamic Programming? A set of choices must be made to arrive at an optimal solution As choices are made, subproblems of the same form arise frequently The key is to store the solutions of subproblems to be reused in the future

COSC 3101E 4 Example 1 Fibonacci numbers are defined by: F0 0 F1 1 Fi Fi 1 Fi 2 for i 2. COSC 3101E 5 Fibonacci Example Time? COSC 3101E 6 Fibonacci Example

Time: Exponential COSC 3101E Waste time 7 redoing work Memoization Definition: An algorithmic technique which saves (memoizes) a computed answer for later reuse, rather than recomputing the answer. Memo functions were invented by Professor Donald Michie of Edinburgh University. The idea was further developed by Robin Popplestone in his Pop2 language. It was later integrated into LISP. This same principle is found at the hardware level in computer architectures which use a cache to store recently accessed memory locations. ["'Memo' functions: and machine learning", Donald Michie, Nature, 218, 19-22, 1968]. COSC 3101E 8 Memoization in Optimization

Remember the solutions for the subinstances If the same subinstance needs to be solved again, the same answer can be used. COSC 3101E 9 Memoization Memoization reduces the complexity from exponential to linear! COSC 3101E 10 From Memoization to Dynamic Programming Determine the set of subinstances that need to be solved. Instead of recursing from top to bottom, solve each of the required subinstances in smallest to largest order, storing results along the way. COSC 3101E

11 Dynamic Programming First determine the complete set of subinstances {100, 99, 98,, 0} Compute them in an order Smallest to largest such that no friend must wait. COSC 3101E 0 1 12 Dynamic Programming Fill out a table containing an optimal solution for each subinstance. 0 0, 1

1, COSC 3101E 1 2 2, 3, 3 5 2.191020 3.541020 4, 5, . 99, 100 0 1 13 Dynamic Programming Time Complexity? Linear! COSC 3101E

14 Dynamic Programming vs Divide-and-Conquer Recall the divide-and-conquer approach Partition the problem into independent subproblems Solve the subproblems recursively Combine solutions of subproblems e.g., mergesort, quicksort

This contrasts with the dynamic programming approach COSC 3101E 15 Dynamic Programming vs Divide-and-Conquer Dynamic programming is applicable when subproblems are not independent i.e., subproblems share subsubproblems Solve every subsubproblem only once and store the answer for use when it reappears A divide-and-conquer approach will do more work than

necessary COSC 3101E 16 A Sequence of 3 Steps A dynamic programming approach consists of a sequence of 3 steps 1. Characterize the structure of an optimal solution 2. Recursively define the value of an optimal solution 3. Compute the value of an optimal solution in a bottom-up fashion

COSC 3101E 17 Elements of Dynamic Programming For dynamic programming to be applicable, an optimization problem must have: 1. Optimal substructure 2. An optimal solution to the problem contains within it optimal solutions to subproblems (but this may also mean a greedy strategy applies) Overlapping subproblems COSC 3101E

The space of subproblems must be small; i.e., the same subproblems are encountered over and over 18 Elements of Dynamic Programming Dynamic programming uses optimal substructure from the bottom up: First find optimal solutions to subproblems Then choose which to use in optimal solution to problem. COSC 3101E 19 Example 2. Making Change Making Change To find the minimum number of Canadian coins to make any amount, the greedy method always works At each step, just choose the largest coin that does not overshoot the

desired amount The greedy method would not work if we did not have 5 coins For 31 cents, the greedy method gives seven coins (25+1+1+1+1+1+1), but we can do it with four (10+10+10+1) The greedy method also would not work if we had a 21 coin For 63 cents, the greedy method gives six coins (25+25+10+1+1+1), but we can do it with three (21+21+21) How can we find the minimum number of coins for any given set of denominations? COSC 3101E 21 Example

We assume coins in the following denominations: 1 5 10 21 25 Well use 63 as our goal COSC 3101E 22 A simple solution We always need a 1 coin, otherwise no solution exists for making one cent To make K cents: If there is a K-cent coin, then that one coin is the minimum Otherwise, for each value i < K, Find the minimum number of coins needed to make i cents Find the minimum number of coins needed to make K - i cents Choose the i that minimizes this sum This algorithm can be viewed as divide-and-conquer, or as brute force

This solution is very recursive It requires exponential work It is infeasible to solve for 63 COSC 3101E 23 Another solution We can reduce the problem recursively by choosing the first coin, and solving for the amount that is left For 63: One 1 coin plus the best solution for 62 One 5 coin plus the best solution for 58 One 10 coin plus the best solution for 53 One 21 coin plus the best solution for 42 One 25 coin plus the best solution for 38 Choose the best solution from among the 5 given above Instead of solving 62 recursive problems, we solve 5 This is still a very expensive algorithm COSC 3101E 24

End of Lecture 18 Nov 15, 2007 COSC 3101E 26 A dynamic programming solution Idea: Solve first for one cent, then two cents, then three cents, etc., up to the desired amount Save each answer in an array ! For each new amount N, compute all the possible pairs of previous answers which sum to N For example, to find the solution for 13, First, solve for all of 1, 2, 3, ..., 12 Next, choose the best solution among: Solution for 1 + solution for 12 Solution for 2 + solution for 11

Solution for 3 + solution for 10 Solution for 4 + solution for 9 Solution for 5 + solution for 8 Solution for 6 + solution for 7 COSC 3101E 27 An even better dynamic programming solution In fact, we can do a bit better than this, since the coins come in only a small number of denominations (1, 5, 10, 21, 25) For each new amount N, compute cost of solution based on smaller sum + one additional coin. For example, to find the solution for 13, First, solve for all of 1, 2, 3, ..., 12 Next, choose the best solution among: solution for 12 + 1 coin solution for 8 + 5 coin solution for 3 + 10 coin COSC 3101E

28 Making Change: Recurrence Relation Let sum value of change to re turn Let d [1...n ] denominations available Let mincoins(sum) minimum number of coins required to make change totalling sum. Let onecoin(sum) one coin in optimal set of coins to make change totalling sum. Then mincoins(sum) min(mincoins(sum - d ) 1) d sum onecoin(sum) argmin(mincoins(sum - d )) d sum COSC 3101E 29 function coins = makechange(d, sum) %precondition: d=set of denominations (must include penny), sum=change to be made %postcondition: coins = a minimal set of coins summing to sum

mincoins(0) = 0 mincoins(1sum) = for i = 1:sum %LI: mincoins(0...i-1) holds the min number of coins required to make change of (0i-1). % onecoin(1...i-1) holds the value of one coin in a minimal set of coins making the correct change. for j = 1:length(d) %try each denomination if d(j) <= i & mincoins(i-d(j)) + 1 < mincoins(i) mincoins(i) = mincoins(i-d(j)) + 1 %best solution so far onecoin(i) = d(j) ncoins = mincoins(sum) change = sum for i = 1:ncoins %recover coins in optimal set coins(i) = onecoin(change) change = change - coins(i) COSC 3101E 30 How good is the algorithm? The first algorithm is exponential, with a base proportional to sum (e.g., 63).

The second algorithm is much better exponential with a base proportional to the number of denominations (e.g., 5). The dynamic programming algorithm is O(sum number of denominations) COSC 3101E 31 Elements of Dynamic Programming For dynamic programming to be applicable, an optimization problem must have: 1. Optimal substructure COSC 3101E An optimal solution to the problem contains within it optimal solutions

to subproblems (but this may also mean a greedy strategy applies) 32 Elements of Dynamic Programming Dynamic programming uses optimal substructure from the bottom up: First find optimal solutions to subproblems Then choose which to use in optimal solution to problem. COSC 3101E 33 Example Proof of Optimal Substructure Consider the problem of making N with the fewest number of coins Either there is an N coin, or The set of coins n making up an optimal solution for N can be divided into two nonempty subsets, n1and n2, which make N1 and N2 change respectively, where N1 + N2 = N. If either N1 or N2 can be made with fewer coins, then clearly N can be made with fewer coins, hence solution was not optimal.

Thus each subset n1 and n2 must themselves be optimal solutions to the subproblems of making N1 and N2 change, respectively. COSC 3101E 34 Optimal Substructure Optimal substructure means that Every optimal solution to a problem contains... ...optimal solutions to subproblems Optimal substructure does not mean that If you have optimal solutions to all subproblems... ...then you can combine any of them to get an optimal solution to a larger problem.

Example: In Canadian coinage, The optimal solution to 7 is 5 + 1 + 1, and The optimal solution to 6 is 5 + 1, but The optimal solution to 13 is not 5 + 1 + 1 + 5 + 1 But there is some way of dividing up 13 into subsets with optimal solutions (say, 11 + 2) that will give an optimal solution for 13 Hence, the making change problem exhibits optimal substructure. COSC 3101E 35 Optimal Substructure Thus the step of choosing which subsolutions to combine is a key part of a dynamic programming algorithm. COSC 3101E 36

Dont all problems have this optimal substructure property? Longest simple path B 1 2 3 Consider the following graph: The longest simple path (path not containing a cycle) from A to D is A B C D However, the subpath A B is not the longest simple path from

A to B (A C B is longer) The principle of optimality is not satisfied for this problem Hence, the longest simple path problem cannot be solved by a dynamic programming approach COSC 3101E A 38 1 C 4 D

Example 2. Knapsack Problem Get as much value as you can into the knapsack COSC 3101E 39 The (General) 0-1 Knapsack Problem 0-1 knapsack problem: n items. Item i is worth $vi , weighs wi pounds. Find a most valuable subset of items with total weight W. vi, wi and W are all integers. Have to either take an item or not take it - cant take part of it. Is there a greedy solution to this problem? COSC 3101E 40

What are good greedy local choices? Select most valuable object? Select smallest object? Select object most valuable by weight? COSC 3101E 41 Some example problem instances Let W Capacity of knapsack = 10kg Problem Instance 1: v1 $60, w1 6kg v1 $60, w1 10kg Problem Instance 3: v1 $60, w1 6kg v2 $50, w2 5kg v2 $50, w2 9kg v2 $40, w2 5kg

Problem Instance 2: v3 $50, w3 5kg v3 $40, w3 5kg Select most valuable object? All Fail! Select smallest object? Select object most valuable by weight? COSC 3101E 42 Simplified 0-1 Knapsack Problem The general 0-1 knapsack problem cannot be solved by a greedy algorithm. What if we make the problem simpler: Suppose vi wi Can this simplified knapsack problem be solved by a greedy algorithm?

No! COSC 3101E 43 Some example problem instances Let W Capacity of knapsack = 10kg Problem Instance 1: Problem Instance 2: v1 w1 6 v1 w1 10 v2 w2 5 v2 w2 9 v3 w3 5 Select largest (most valuable) object? Both Fail!

Select smallest object? COSC 3101E 44 Approximate Greedy Solution For the simplified knapsack problem: the greedy solution (taking the most valuable object first) isnt that bad: 1 V V , where 2 V Total value of items selected by greedy algorithm V Total value of items selected by optimal algorithm COSC 3101E 45 End of Lecture 19 Nov 20, 2007 YORK UNIVERSITY

Office of the Dean FACULTY OF SCIENCE AND ENGINEERING MEMORANDUM TO: Distribution FROM: Walter P. Tholen, Associate Dean (Research & Faculty Affairs) DATE: November 15, 2007 SUBJECT: NSERC Undergraduate Student Research Awards (NSERC USRA) It is a pleasure to bring to your attention that for 2008-2009 York has been allocated 51 NSERC USRAs. Considering not all offers will be accepted, we should try to recruit at least 70 qualified applicants. Please bring these awards to the attention of qualified students as soon as possible. Further information on NSERC USRAs can be found on the NSERC website. The application form and instructions (Form 202) are now available only on the NSERC website and can be printed from there. Please read all instructions before completing the application form. I. NSERC USRA in Universities (a) The value of the award from NSERC is $4,500 for 16 consecutive weeks. Faculty members must pay the student at least 25% ($1,125) on top of this. If a student is selected from another university to hold their award at York, the supervisor will be responsible for paying at least $787.50 extra; i.e., 4% vacation pay and 10% for benefits. Travel allowance may also be available when applicable. (b) At York, the award can only be held during the summer session. (c) NSERC expects students to work the full term. NSERC may approve shorter tenure in exceptional circumstances (these circumstances must be approved by NSERC), so the departments must make the students aware that the awards are for the

full 16 weeks. (d) Students must return the completed applications to their departmental offices by January 25, 2008. Transcripts for York students can be printed by the departments as NSERC does not require that an official transcript be sent to them. Departments must submit their completed applications and transcripts, along with their rankings, to my 47 COSC 3101E office no later than February 8, 2008. Approximate Greedy Solution 1 Claim:V V 2 Proof: Let W capacity of knapsack. Let s value (weight) of object in optimal solution but not selected by greedy algorithm. 1 Suppose V V 2 1 Then s V (since object was not selected by greedy algorithm) 2

But since V s V W , object would be selected by greedy algorithm. Contradiction! And running time O( n log n) where n number of items COSC 3101E 48 Dynamic Programming Solution The General 0-1 Knapsack Problem can be solved by dynamic programming. Let W capacity of knapsack (kg) Let (v i ,w i ) value ($) and weight (kg) of item i [1...n] Let c[i ,w ] value of optimal solution for knapsack of capacity w and items drawn from [1...i ] 0 if i 0 or w 0 Then c i ,w c i 1, w if w i w max v c [i 1, w w ], c[i 1, w ] if i 0 and w w i

i i COSC 3101E 49 Correctness Let W capacity of knapsack (kg) Let (v i ,w i ) value ($) and weight (kg) of item i [1...n ] Let c[i ,w ] value of optimal solution for knapsack of capacity w and items drawn from [1...i ] 0 if i 0 or w 0 Then c i ,w c i 1, w if w i w max v c[ i 1, w w ], c[i 1, w ] if i 0 and w w i i i Idea: c[i 1,w ] value of optimal solution for capacity w and items drawn only from [1...i 1]

What happens when we are also allowed to consider item i ? Case 1. Optimal solution does not include item i . Total value is the same as before. One of these must be true! Case 2. Optimal solution does include item i. Total value is: Value of item i Value of optimal solution for remaining capacity of knapsack and allowable items COSC 3101E 50 Bottom-Up Computation Let W capacity of knapsack (kg) Let (v i ,w i ) value ($) and weight (kg) of item i [1...n ] Let c[i ,w ] value of optimal solution for knapsack of capacity w and items drawn from [1...i ] 0 if i 0 or w 0 Then c i ,w c i 1, w if w i w max v c[i 1, w w ], c[i 1, w ] if i 0 and w w

i i i Need only ensure that c[i 1,v ] has been computed v w c[i,w] i 0 1 2 i n COSC 3101E w Allowed Items 0 1 2 w W {} {1} {1 2}

{1 2 i} c[i,w] {1 2 n} 51 Integer Knapsack Solution Value=0 if no items or no knapsack. COSC 3101E 52 Integer Knapsack Solution Fill in table row-wise Recurrence relation COSC 3101E 53

Example Capacity W 6 0 if i 0 or w 0 Then c i ,w c i 1, w if w i w max v c[i 1, w w ], c[i 1, w ] if i 0 and w w i i i i v w 1 1 2 2 3 3 3 5 1 4 2 5 5 6 3 6 10 5

COSC 3101E c[i,w] i 0 1 2 3 4 5 6 Allowed Items {} {1} {1 2} {1 2 3} {1 2 3 4} {1 2 3 4 5} {1 2 3 4 5 6} w 0 0

0 0 0 0 0 0 54 1 0 0 0 5 5 5 5 2 0 1 1 5 5 5

5 3 0 1 3 6 6 6 6 4 0 1 3 8 8 11 11 5 0 1 4

8 8 11 11 6 0 1 4 9 9 12 15 Solving for the Items to Pack 0 if i 0 or w 0 Then c i ,w c i 1, w if w i w max v c[i 1, w w ], c[i 1, w ] if i 0 and w w i i i

i v w 1 1 2 2 3 3 3 5 1 4 2 5 5 6 3 6 10 5 COSC 3101E c[i,w] i 0 1 2 3 4 5 6 w

Allowed Items 0 {} 0 {1} 0 {1 2} 0 {1 2 3} 0 {1 2 3 4} 0 {1 2 3 4 5} 0 {1 2 3 4 5 6} 0 i n 1 0 0 0 5 5 5

5 2 0 1 1 5 5 5 5 3 0 1 3 6 6 6 6 55 4 0

1 3 8 8 11 11 5 0 1 4 8 8 11 11 6 0 1 4 9 9 12 15

w W items {} loop for i n downto 1 if c[i ,w ] c[i 1,w ] items items {i } w w w i Second Example Capacity W 6 0 if i 0 or w 0 Then c i ,w c i 1, w if w i w max v c[i 1, w w ], c[i 1, w ] if i 0 and w w i i i i

1 2 3 4 5 6 v w 1 2 4 3 2 1 5 4 4 3 2 3 COSC 3101E c[i,w] i 0 1 2 3 4

5 6 w Allowed Items 0 {} 0 {1} 0 {1 2} 0 {1 2 3} 0 {1 2 3 4} 0 {1 2 3 4 5} 0 {1 2 3 4 5 6} 0 56 1 0 0

0 2 2 2 2 2 0 1 1 2 2 2 2 3 0 1 4 4 4 4 4

4 0 1 4 6 6 6 6 5 0 1 5 6 7 7 7 6 0 1 5 7 7

8 8 Knapsack Problem: Running Time Running time (nW). (cf. Making change (dsum)). Not polynomial in input size! COSC 3101E 57 End of Lecture 20 Nov 22, 2007 Recall: Knapsack ProblemCapacity W 0 if i 0 or w 0 Then c i ,w c i 1, w if w i w max v c[i 1, w w ], c[i 1, w ] if i 0 and w w i i i

i 1 2 3 4 5 6 v w 1 2 4 3 2 1 5 4 4 3 2 3 COSC 3101E c[i,w] i 0 1

2 3 4 5 6 w Allowed Items 0 {} 0 {1} 0 {1 2} 0 {1 2 3} 0 {1 2 3 4} 0 {1 2 3 4 5} 0 {1 2 3 4 5 6} 0 59

1 0 0 0 2 2 2 2 2 0 1 1 2 2 2 2 3 0 1 4 4 4

4 4 4 0 1 4 6 6 6 6 5 0 1 5 6 7 7 7 6 0 1

5 7 7 8 8 6 Observation from Last Day (Jonathon): We could still implement this recurrence relation directly as a recursive program. 0 if i 0 or w 0 Then c i ,w c i 1, w if w i w max v c [i 1, w w ], c [i 1, w ] if i 0 and w w i i i c[i,w]

COSC 3101E w i Allowed Items 0 1 2 3 4 5 6 0

{} 0 0 0 0 0 0 0 1 {1} 0 0

1 1 1 1 1 2 {1 2} 0 0 1 4 4

5 5 3 {1 2 3 } 0 2 2 4 6 6 7 4

{1 2 3 4 } 0 2 2 4 6 7 7 5 {1 2 3 4 5 } 0 2

2 4 6 7 8 6 {1 2 3 4 5 6 0 2 2 4 6

7 8 } 60 Recall: Memoization in Optimization Remember the solutions for the subinstances If the same subinstance needs to be solved again, the same answer can be used. COSC 3101E 61 Memoization Memoization reduces the complexity from exponential to linear! COSC 3101E 62

From Memoization to Dynamic Programming Determine the set of subinstances that need to be solved. Instead of recursing from top to bottom, solve each of the required subinstances in smallest to largest order, storing results along the way. COSC 3101E 63 Dynamic Programming Examples 1. Fibonacci numbers 2. Making change 3. 0-1 Knapsack problem 4. Activity Scheduling with profits COSC 3101E 64 Recall: The Activity (Job/Event) Selection Problem Ingredients: Instances: Events with starting and finishing times

<,, ,>. Solutions: A set of events that do not overlap. Value of Solution: The number of events scheduled. Goal: Given a set of events, schedule as many as possible. COSC 3101E 65 From Previous Lecture: Problem can be solved by greedy algorithm Greedy Criteria: Motivation: COSC 3101E Earliest Finishing Time Schedule the event that will free up your room for someone else as soon as possible. 66

Works! But what if activities have different values? Activity Selection with Profits: Input: information ( si , fi , gi ) about n activities, where si start time of activity i f i finishing time of activity i gi value (profit) of activity i A feasible schedule is a set S {1, 2,..., n} such that i, j S , activities i and j do not conflict. Output: A feasible schedule S with maximum profit P ( S ) gi iS COSC 3101E 67 Will a greedy algorithm based on finishing time still work? g 2 10

g3 1 g1 1 COSC 3101E No! 68 Dynamic Programming Solution Precomputation: Time? 1. Sort activities according to finishing time: f1 f 2 f n (O(n log n)) 2. i {1,..., n}, compute H (i) max{l {1, 2,..., i 1}| f l si } (O( n log n)) i.e. H (i ) is the last event that ends before event i starts. COSC 3101E 69

Step 1. Define an array of values to compute i {0,..., n}, A(i ) largest profit attainable from the (feasible) scheduling of a subset of activities from {1, 2,..., i} Ultimately, we are interested in A(n) COSC 3101E 70 Step 2. Provide a Recurrent Solution 1. Sort activities according to finishing time: f1 f 2 f n (O(n log n)) 2. i {1,..., n}, compute H (i) max{l {1, 2,..., i 1} | f l si } (O(n log n)) i.e. H (i ) is the last event that ends before event i starts. A(0) 0 One of these must be true! A(i ) max{ A(i 1), gi A( H (i ))}, i {1,..., n} Decide not to schedule

activity i COSC 3101E Profit from scheduling activity i Optimal profit from scheduling activities that end before activity i begins 71 Step 3. Provide an Algorithm function A=actselwithp(g, H, n) % assumes inputs sorted by finishing time A(0)=0 for i=1:n A(i)=max(A(i-1), g(i)+A(H(i))) end COSC 3101E Running time? O(n)

72 Step 4. Compute Optimal Solution Invoke with: printasp(A,H,n,Activities to Schedule:) function actstring=printasp(A,H,i,actstring) if i=0 return Running time? O(n) if A(i)>A(i-1) actstring = printasp(A, H, H(i), actstring) actstring = [actstring, sprintf('%d ', i)] else actstring = printasp(A, H, i-1, actstring) COSC 3101E 73 Example Activity i

1 2 3 4 Start si 0 2 3 2 Finish fi 3 6

6 10 Profit gi 20 30 20 30 ? ? ? ? H(i)

COSC 3101E 74 Example Activity i 1 2 3 4 Start si 0 2 3 2

Finish fi 3 6 6 10 Profit gi 20 30 20 30 0 ?

? ? H(i) COSC 3101E 75 Example Activity i 1 2 3 4 Start si

0 2 3 2 Finish fi 3 6 6 10 Profit gi 20 30

20 30 0 0 ? ? H(i) COSC 3101E 76 Example Activity i 1 2

3 4 Start si 0 2 3 2 Finish fi 3 6 6 10

Profit gi 20 30 20 30 0 0 1 ? H(i) COSC 3101E 77

Example Activity i 1 2 3 4 Start si 0 2 3 2 Finish fi

3 6 6 10 Profit gi 20 A(0) 0 A(1) max{0, 20 A( H (1))} 20 A(2) max{20,30 A( H (2))} 30 30 20 30 A(3) max{30, 20 A( H (3))} 40 H(i)

0 0 1 0 A(4) max{40,30 A( H (4))} 40 COSC 3101E 78 Dynamic Programming Examples 1. Fibonacci numbers 2. Making change 3. 0-1 Knapsack problem 4. Activity scheduling with profits 5. Longest common subsequence COSC 3101E 79

Longest Common Subsequence Input: 2 sequences, X = x1, . . . , xm and Y = y1, . . . , yn. Output: a subsequence common to both whose length is longest. Note: A subsequence doesnt have to be consecutive, but it has to be in order. COSC 3101E 80 Example 3. Longest Common Subsequence COSC 3101E 81 Brute-force Algorithm For every subsequence of X , check whether it's a subsequence of Y . Time: (n 2m ). 2m subsequences of X to check.

Each subsequence takes ( n) time to check: scan Y for first letter, from there scan for second, and so on. COSC 3101E 82 Step 1. Define Data Structure Input: 2 sequences, X = x1, . . . , xm and Y = y1, . . . , yn. Let c(i , j ) length of LCS for X i and Y j Ultimately, we are interested in c( m, n). COSC 3101E 83 Step 2. Define Recurrence Case 1. Input sequence is empty COSC 3101E

84 Recurrence Case 2. Last elements match: must be part of an LCS X Xi 1 Yj 1 Y Z COSC 3101E __ __ __ __ m 85

Recurrence X Case 3. Last elements dont match: at most one of them is part of LCS Xi 1 Yj Y X Choose! Xi X Yj Yj 1 Y Y

COSC 3101E Xi 86 Step 3. Provide an Algorithm Running time? O(mn) COSC 3101E 90 Step 4. Compute Optimal Solution Running time? O(m+n) COSC 3101E 91 Example

COSC 3101E 92

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