cosc 3101nj. elder announcements midterms are marked assignment 2: –still analyzing
DESCRIPTION
COSC 3101NJ. Elder What is a loop invariant? An assertion about the state of one or more variables used in the loop. When the exit condition is met, the LI leads naturally to the postcondition (the goal of the algorithm). Thus the LI must be a statement about the variables that store the results of the computation.TRANSCRIPT
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COSC 3101N J. Elder
Announcements• Midterms are marked
• Assignment 2: – Still analyzing
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COSC 3101N J. Elder
Loop Invariants (Revisited)• Question 5(a): Design an iterative
algorithm for Parity:
p=0for i = 1 to n do p = xor(p,s[i])return p
Loop Invariant?
LI: After i iterations are performed, p=Parity(s[1…i])
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COSC 3101N J. Elder
What is a loop invariant?
• An assertion about the state of one or more variables used in the loop.
• When the exit condition is met, the LI leads naturally to the postcondition (the goal of the algorithm).
• Thus the LI must be a statement about the variables that store the results of the computation.
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COSC 3101N J. Elder
Know What an LI Is Not
``the LI is NOT...''– code
– The steps taken by the algorithm
– A statement about the range of values assumed by the loop index.
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COSC 3101N J. Elder
Dynamic Programming: Recurrence
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COSC 3101N J. Elder
Dynamic programming
• Step 1: Describe an array of values you want to compute.
• Step 2: Give a recurrence for computing later values from earlier (bottom-up).
• Step 3: Give a high-level program.
• Step 4: Show how to use values in the array to compute an optimal solution.
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COSC 3101N J. Elder
Example 1. Rock climbing
At every step our climber can reach exactly three handholds: above, above and to the right and above and to the left.
There is a table of “danger ratings” provided. The “Danger” of a path is the sum of danger ratings of all handholds on the path.
5 3
4
2
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COSC 3101N J. Elder
For every handhold, there is only one “path” rating. Once we have reached a hold, we don’t need to know how we got there to move to the next level.
This is called an “optimal substructure” property. Once we know optimal solutions to
subproblems, we can compute an optimal solution to the problem itself.
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COSC 3101N J. Elder
Step 2. Define a RecurrenceLet C(i,j) represent the danger of hold (i,j)
Let A(i,j) represent the cumulative danger of the safest path from the bottom to hold (i,j)
ThenA(i,j) = C(i,j)+min{A(i-1,j-1),A(i-1,j),A(i-1,j+1)}
i.e., the safest path to hold (i,j) subsumes the safest path to holds at level i-1 from which hold (i,j) can be reached.
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COSC 3101N J. Elder
Example 2. Activity Scheduling with Profits
1 1g 2 10g
3 1g
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COSC 3101N J. Elder
Step 2. Provide a Recurrent Solution
(0) 0( ) max{ ( 1), ( ( ))}, {1,..., }i
AA i A i g A H i i n
Decide not to schedule activity i
Profit from scheduling activity i
Optimal profit from scheduling activities that end before activity i begins
1 21. Sort activities according to fini ( (shing lo ti g ))me: nf f O nf n
2. {1,..., }, compute ( ) max{ {1,2,..., 1 ( ( log )} } )| l ii n H Oi l i f s n n
. . ( ) is the last event that ends before event starts.i e H i i
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COSC 3101N J. Elder
Example 3: Scheduling Jobs with Deadlines, Profits and Durations
information ( , , ) about activities, whereinteger deadline for job integer duration of job real-valued pr
Input
ofit of job
:
i i i
i
i
i
d t g nd it ig i
A C is a sequence { (1), (2),..., ( )} such that: ( ) scheduled start time of job ( ) 1 if job is not scheduled
schedule C C C C nC i iC i i
A feasible schedule C with maximum profit: (Output ): ii C
P C g
A schedule C is if each scheduled job finishes by its deadlineand no two scheduled jobs over
feasilap
bleped.
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COSC 3101N J. Elder
Dynamic Programming Solution
Precomputation:
1 2Sort activities according to de ( ( log ))adline: nd Od n nd
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COSC 3101N J. Elder
Step 2. Provide a Recurrent Solution(0, ) 0 {0,..., }
For {1,..., }, {0,..., }, define t min{ , }( is the latest time that we can schedule job so that itends both by its deadline and by time .)Then:
0 ( , ) ( 1, )
i i
A t t d
i n t d t d tt i
t
t A i t A i tt
0 ( , ) max{ ( 1, ), ( 1, )}iA i t A i t g A i t
Decide not to schedule job i
Profit from job i Profit from scheduling activities that end before job i begins
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COSC 3101N J. Elder
Step 2. (cntd…) Proving the Recurrent Solution
We effectively schedule job i at the latest possible time.
This leaves the largest and earliest contiguous block of time for scheduling jobs with earlier deadlines.
For {1,..., }, {0,..., }, define t min{ , }( is the latest time that we can schedule job so that itends both by its deadline and by time .)
i ii n t d t d tt i
t
Then:0 ( , ) ( 1, )t A i t A i t
0 ( , ) max{ ( 1, ), ( 1, )}it A i t A i t g A i t
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COSC 3101N J. Elder
2d 1id id1d … t
it
t
event iCase 1
2d 1id id1d … t
it
t
event iCase 2
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COSC 3101N J. Elder
Example 3. Longest Common Subsequence
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COSC 3101N J. Elder
Optimal Substructure• Input: 2 sequences, X = x1, . . . , xm and Y =
y1, . . . , yn.
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COSC 3101N J. Elder
Proof of Optimal Substructure Part 1
1 1
1 1
1
{ ,..., }; { ,..., }{ ,..., }; { ,..., }
{ ,..., } is an LCS of an
Notati
on:
d .
m n
i i i i
k
X x x Y y yX x x Y y yZ z z X Y
1 1 11. If , then and is an LCS of and . :
m n k m n k m n
Theorex y z x y Z X
mY
Proof that :k m nz x y Suppose . Then could append to common subsequenceof and of length 1 contradiction.
k m m nz x x y ZX Y k
1 1 1Proof that is an LCS of and :k m nZ X Y
1 1 1 is a length-( 1) common subsequence of and .k m nZ k X Y
1 1Suppose there is a longer common subsequence of and .m nW X Y
Appending to common subsequence of length >k contradictionm nx y W
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COSC 3101N J. Elder
Proof of Optimal Substructure Part 2
1 1
1 1
1
{ ,..., }; { ,..., }{ ,..., }; { ,..., }
{ ,..., } is an LCS of an
Notati
on:
d .
m n
i i i i
k
X x x Y y yX x x Y y yZ z z X Y
12. If and , then is an LCS of :
an d .m n k m mx y z x ZTheor
Ym
Xe
Proof :
1 is a common subsequence of and .k m mz x Z X Y
1Suppose there is a common subsequence of and mW X Y
of length also a common subsequence of and contradictionk W X Y
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COSC 3101N J. Elder
Proof of Optimal Substructure Part 3
1 1
1 1
1
{ ,..., }; { ,..., }{ ,..., }; { ,..., }
{ ,..., } is an LCS of an
Notati
on:
d .
m n
i i i i
k
X x x Y y yX x x Y y yZ z z X Y
12. If and , then is an LCS of a:
nd .m n k n nx y zTheor
y Z X Yem
Proof :
As for Part 2.
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COSC 3101N J. Elder
Step 2. Provide a Recurrent Solution
Input sequences are empty
Last elements match: must be part of LCS
Last elements don’t match: at most one of them is part of LCS
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COSC 3101N J. Elder
Example 6: Longest Increasing Subsequence
• Input: 1 sequence, X = x1, . . . , xn.
• Output: the longest increasing subsequence of X.
• Note: A subsequence doesn’t have to be consecutive, but it has to be in order.
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COSC 3101N J. Elder
Step 1. Define an array of values to compute
{0,..., }, length of LIS of endiA( ng in .) iii n X x
Ultimately, we are intere maxsted { ( }in .) |1A i i n
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COSC 3101N J. Elder
Step 2. Provide a Recurrent Solution
for 1 ,( ) 1 max{ ( ) |1 and }j i
i nA i A j j i x x
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COSC 3101N J. Elder
Step 3. Provide an Algorithm
Running time? O(n2)function A=LIS(X)
for i=1:length(X)
m=0;
for j=1:i-1
if X(j) < X(i) & A(j) > m
m=A(j);
end
end
A(i)=m+1;
end
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COSC 3101N J. Elder
Step 4. Compute Optimal Solution
Running time? O(n)
function lis=printLIS(X, A)[m,mi]=max(A);lis=printLISm(X,A,mi,'LIS: ');lis=[lis,sprintf('%d', X(mi))];
function lis=printLISm(X, A, mi, lis)if A(mi) > 1 i=mi-1; while ~(X(i) < X(mi) & A(i) == A(mi)-1) i=i-1; end lis=printLISm(X, A, i, lis); lis=[lis, sprintf('%d ', X(i))];end
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COSC 3101N J. Elder
LIS Example
X = 96 24 61 49 90 77 46 2 83 45
A = 1 1 2 2 3 3 2 1 4 2
> printLIS(X,A)
> LIS: 24 49 77 83
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COSC 3101N J. Elder
Example 6: Optimal Binary Search Trees
1 1 2
Sequence { ,..., } of distinct keys,
probability
Inp
that a search is for key
ut:
n n
i i
K k k n k k kp k
BST with minimum expected searcOutp
h ut:
cost
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COSC 3101N J. Elder
Expected Search Cost
Which BST is more efficient?
Cost = # of items examined.
For key , cost = depth ( ) 1i T ik k
1
[search cost in ]
(depth ( ) 1)n
T i ii
E T
k p
1
1 depth ( )n
T i ii
k p
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COSC 3101N J. Elder
Observations
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COSC 3101N J. Elder
Optimal Substructure
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COSC 3101N J. Elder
Optimal Substructure (cntd…)
TLet e expected search cost in subtree , starting from root( )T T
depth ( )j
l T ll i
p k
TLet e expected search cost in subtree , starting from root( )T T
(depth (root( )) depth ( ))j
l T T ll i
p T k
depth (root( )) depth ( )j j
T l l T ll i l i
T p p k
Tdepth (root( )) ej
T ll i
T p
T
T
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COSC 3101N J. Elder
Recursive Solution
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COSC 3101N J. Elder
Recursive Solution (cntd…)
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COSC 3101N J. Elder
Step 2. Provide a Recurrent Solution
Expected cost of search for left subtree
Added cost when subtrees embedded under root
Expected cost of search for right subtree
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COSC 3101N J. Elder
n)
Step 3. Provide an Algorithm
Running time? O(n3)
work on subtrees of increasing size l
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COSC 3101N J. Elder
Example
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COSC 3101N J. Elder
Example (cntd…)
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COSC 3101N J. Elder
Step 4. Compute Optimal Solution
Running time? O(n)
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COSC 3101N J. Elder
Elements of Dynamic Programming
• Optimal substructure: – an optimal solution to the problem contains
within it optimal solutions to subproblems.
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COSC 3101N J. Elder
Elements of Dynamic Programming
• Cut and paste: prove optimal substructure by contradiction:– assume an optimal solution to a problem with suboptimal solution to
subproblem
– cut out the suboptimal solution to the subproblem.
– paste in the optimal solution to the subproblem.
– show that this results in a better solution to the original problem.
– This contradicts our assertion that our original solution is optimal.
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COSC 3101N J. Elder
• 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.
Elements of Dynamic Programming
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COSC 3101N J. Elder
Section V. Graph Algorithms
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COSC 3101N J. Elder
(c) The subgraph of the graph in part (a) induced by the vertex set {1,2,3,6}.
(a) A directed graph G = (V, E), where V = {1,2,3,4,5,6} and E = {(1,2), (2,2), (2,4), (2,5), (4,1), (4,5), (5,4), (6,3)}. The edge (2,2) is a self-loop.
(b) An undirected graph G = (V,E), where V = {1,2,3,4,5,6} and E = {(1,2), (1,5), (2,5), (3,6)}. The vertex 4 is isolated.
Directed and Undirected Graphs
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COSC 3101N J. Elder
Graph Isomorphism
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COSC 3101N J. Elder
Trees
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COSC 3101N J. Elder
Representations: Undirected Graphs
Adjacency List Adjacency Matrix
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COSC 3101N J. Elder
Representations: Directed Graphs
Adjacency List Adjacency Matrix