Dynamic Programming design technique is one of the fundamental algorithm design techniques, and possibly one of the ones that are hardest to master for those who did not study it formally. In these slides (which are continuation of part 1 slides), we cover two problems: maximum value contiguous subarray, and maximum increasing subsequence.
Dynamic Programming is an algorithmic paradigm that solves a given complex problem by breaking it into subproblems and stores the results of subproblems to avoid computing the same results again.
Dynamic Programming design technique is one of the fundamental algorithm design techniques, and possibly one of the ones that are hardest to master for those who did not study it formally. In these slides (which are continuation of part 1 slides), we cover two problems: maximum value contiguous subarray, and maximum increasing subsequence.
Dynamic Programming is an algorithmic paradigm that solves a given complex problem by breaking it into subproblems and stores the results of subproblems to avoid computing the same results again.
it contains the detail information about Dynamic programming, Knapsack problem, Forward / backward knapsack, Optimal Binary Search Tree (OBST), Traveling sales person problem(TSP) using dynamic programming
In computer science, divide and conquer is an algorithm design paradigm based on multi-branched recursion. A divide-and-conquer algorithm works by recursively breaking down a problem into two or more sub-problems of the same or related type until these become simple enough to be solved directly.
Divide and Conquer Algorithms - D&C forms a distinct algorithm design technique in computer science, wherein a problem is solved by repeatedly invoking the algorithm on smaller occurrences of the same problem. Binary search, merge sort, Euclid's algorithm can all be formulated as examples of divide and conquer algorithms. Strassen's algorithm and Nearest Neighbor algorithm are two other examples.
it contains the detail information about Dynamic programming, Knapsack problem, Forward / backward knapsack, Optimal Binary Search Tree (OBST), Traveling sales person problem(TSP) using dynamic programming
In computer science, divide and conquer is an algorithm design paradigm based on multi-branched recursion. A divide-and-conquer algorithm works by recursively breaking down a problem into two or more sub-problems of the same or related type until these become simple enough to be solved directly.
Divide and Conquer Algorithms - D&C forms a distinct algorithm design technique in computer science, wherein a problem is solved by repeatedly invoking the algorithm on smaller occurrences of the same problem. Binary search, merge sort, Euclid's algorithm can all be formulated as examples of divide and conquer algorithms. Strassen's algorithm and Nearest Neighbor algorithm are two other examples.
This file contains the contents about dynamic programming, greedy approach, graph algorithm, spanning tree concepts, backtracking and branch and bound approach.
Dynamic Programming is one of the most interesting design techniques. The concise idea is to avoid recomputations. Matrix Chain Multiplication and All Pairs Shortest Paths are two interesting applications of this design technique
https://hasgeek.com/rootconf/data-privacy-conference/sub/synthetic-data-generation-VN92QpTzvTSAeepCW8YRMU
Synthetic data generation for relational data
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AI Pilot Review: The World’s First Virtual Assistant Marketing SuiteGoogle
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2. What will be covered
•Time complexity
•Dynamic Programming
• Backtracking
3. Time complexity of For loop
For (idx = 0; idx < N ; idx = idx + 1 )
{
print(idx)
}
Time complexity
is O(N)
4. Time complexity of For loop
For (idx = 1; idx < N ; idx = idx * 2)
{
print(idx)
}
Time complexity
is O(log2
(N))
idx = 1, 2, 4, 8, 16, 32…
N = 8, number of steps = 3
N = 32, number of steps = 5
5. Time complexity of For loops
For (idx = 1; idx < N ; idx = idx + 1)
{
aList.Append(idx)
}
Time complexity
is O(N2
)
1 2 N-5 N
Elements before you = 1 + 2 + 3 + 4 + … + N = N(N-1)/2
6. Time complexity of For loops
For (idx = 1; idx < N ; idx = idx + 1)
{
aList.Append(idx)
}
Time complexity
is O(N)
1 2 N-5 N
If you had a tail pointer ? Then 1 + 1 + 1 = N
Tail Pointer
7. Time complexity of For loops
For (idx = 1; idx < N ; idx = idx + 1)
{
binaryTree.Insert(idx)
}
Time complexity
is O(N . log2
(N))
A
K
Z S
T
E
Cost of one insert into Binary tree
= go down the depth of binary Tree
= log2
(N)
8. Time complexity of two For loops
For (i = 0; i < N ; i = i + 1)
for (j = 0; j < N ; j = j + 1)
{
print(i + j)
}
Time complexity
is O(N2
)
9. Time complexity of two For loops
for (i = 0; i < N ; i = i + 1)
for (j = 0; j < i ; j = j + 1)
{
print(i + j)
}
for (i = 0; i < N ; i = i + 1)
for (j = i; j < N ; j = j + 1)
{
print(i + j)
}
Left loop = 1 + 2 + 3 + 4 + … + N
= N(N-1)/2
Time complexity
is O(N2
)
Right loop = N + (N-1) + … + 2 + 1
10. Mark each algorithm on this chart
N2
N N3
exp(N)1
1
N
N.log2
(N)
N.log2
(N)
Exp(N)
TIME
SPAC
E
Double
For Loop
Single For
Loop
Tree
insert
List insert
List insert
with Tail
pointer
11. Many ways of solving a given problem
N2
N N3
exp(N)1
1
N
N.log2
(N)
N.log2
(N)
Exp(N)
TIME
SPAC
E
Time
efficient
Space
efficient
Space and
time
optimal
Cannot do better than Lower bound.
This depends on constraints of problem
12. Four ways to get home
Backtracking : Explore every point from current position. If you
don’t reach home, return to previous position and retry.
Divide and Conquer : Find way to get to Metro, then Majestic,
then home.
Dynamic Programming : Go home at lowest cost using Bus,
Metro, Cab and Rickshaw
Greedy : Find best next point in home-ward direction
14. Get home via Dynamic Programming
Office Home
Jayanagar
Majestic
Cubbon
Park
TravelCost (Home) =
Min ( TravelCost(Majestic), TravelCost(Cubbon Park) )
TravelCost (Cubbon Park) =
Min ( TravelCost (Direct), TravelCost(Jayanagar) )
Silk Board Jn
15. Dynamic Programming - subset sum
4 7
7
5
3
3
5
3
3
Either add this element OR not
4 7 5 3
Find subset whose sum is 14
5
3
16. DP - Subset Sum – recursive
Boolean hasSubsetSum(int curIdx, int curSum)
{
// either you add this element OR not add it
return hasSubsetSum(curIdx - 1, curSum)
|| hasSubsetSum(curIdx - 1, curSum - array[curIdx - 1])
}
17. DP - Subset Sum – recursive
Boolean hasSubsetSum(int curIdx, int curSum)
{
if (curSum == 0) return True
// either you add this element OR not add it
return hasSubsetSum(curIdx - 1, curSum)
|| hasSubsetSum(curIdx - 1, curSum - array[curIdx - 1])
}
18. DP - Subset Sum – recursive
Boolean hasSubsetSum(int curIdx, int curSum)
{
if (curSum == 0) return True
if (curIdx == 0) || (curSum != 0) return False
// either you add this element OR not add it
return hasSubsetSum(curIdx - 1, curSum)
|| hasSubsetSum(curIdx - 1, curSum - array[curIdx - 1])
}
19. DP - Subset Sum – recursive to iterative
• REMEMBER : When the Recursion is a function of 2 variables, your iterative solution
requires a 2-dim matrix
….hasSubsetSum(curIdx - 1, curSum)
|| hasSubsetSum(curIdx - 1, curSum - array[curIdx - 1])
0
1
…
last array element n-1
0 1 expectedSum
20. DP - Subset sum – iterative
Boolean hasSubsetSum(int expectedSum)
{
for (i = 1; i < n; i ++)
for (j = 1; j < expectedSum; j ++ )
{
subset[ i ][ j ] = subset[i - 1][ j ]
|| subset[i -1][ j – array[curIdx]]
}
}
22. How to get home via backtracking
Boolean PathToHome(current position)
{
if (reachedHome(current position))
return True
for (all available Lanes starting from current position)
{
gotHome = PathToHome(current position + lane)
if (gotHome = false)
Go back to previous position
}
return False
}
26. Backtracking – subset sum - Complexity
•Each element is either added or Not
•Number of such subsets = 2 * 2 * 2... (N times) = 2N
•Each subset has to be summed = sum of (roughly) N elements
Hence, time complexity
= number of elements in each subset x number of subsets
= O(N . 2N
)
27. Knights tour
Position = 2, 2
0,1 1,0 0,3 1,4 3,0 4,1 3,4 4,3
Each cell has 8 options
Depth of tree = 64
Check for all 64 cells
Time = O(864
)
30. Recap : Four ways to get home
Backtracking : Explore every point from current position. If you
don’t reach home, return to previous position and retry.
Divide and Conquer : Find way to get to Baiyappanahalli Metro,
then Majestic, then home.
Dynamic Programming : Go home at lowest cost using Bus,
Metro, Cab and Rickshaw
Greedy : Find best next point in home-ward direction
31. Difference between algorithm types
•“Greedy Algorithm” first makes choice and then solves sub-problem
•“Divide and conquer” generates non-overlapping sub-problems
•“Dynamic programming” finds optimal solution to sub-problem; but it
can generate overlapping sub-problems – hence “memoization”
•“Backtracking” can reach an invalid solution – that’s why its called
backtracking
32. Conclusion
To write good code…
•You have to read good code
•Work through many problems
•Memorize elegant patterns & tricks
33. Dynamic Programming - Rabbits
If rabbits die after 4 days, only those younger than 4 days are alive
Int getNumRabbits(int numDays)
{
if (n == 0) return 1
alive = [3 x getNumRabbits(numOfDays – 1) ]
- getNumOfRabbits[numOfDays – 4]
}
34. Invariants in an algorithm
•Invariants ensure that each step in your algorithm is correct
•Convert invariants into Assertions
•https://yourbasic.org/algorithms/loop-invariants-explained/
•Example : Take 2-3 algorithms and define one invariant in them
35. Think of the simplest cases
•Given a complicated problem
•E.g. In array of 100 elements, rearrange in positive & negative
Too hard to think of all combinations !
1. Lets reduce it to simplest possible examples
2. Take an array with 3 entries
3. Find a technique to solve this
4. Then expand it to solve larger problem
36. Think of the simplest cases
-3 -3 -3 4 4 4
-3 5 6
5 -3 6
5 -36
-3 -5 6
-3 -56
-3 -56
USE
SYMMETRY