Design and Analysis of Algorithm
Unit-V
Backtracking
The General method
The 8-queens problem
Sum of subsets
Graph coloring
Hamiltonian cycles
Knapsack problem
The General Method
•
The name backtrack was first coined by D.H. Lehmer in the 1950s
•
Backtracking is one of the most general techniques used to search the state space
tree systematically to find solution to some problem.
•
The backtracking technique is similar to Depth first search in a directed graph.
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Design and Analysis of Algorithm
•
The solution in backtracking must be expressed as an n-tuple solution vector
(x1,x2,…,xn) where Xi are chosen from some finite set Si
•
The solution vector must satisfy the criterion function P(x1,………,xn) (sometimes
called bounding function).
•
For any problem we solve using backtracking require to satisfy complex set of
constraints. These constraints can be divided into two categories:
Explicit constraints
Implicit constraints
•
Explicit constraints:
Explicit constraints are rules that restrict each xi value to be chosen from a given set
Si .
e.g.
xi
}
or
xi = 0 or 1 or
Si = {a : li
•
0
a
Si = { all nonnegative real numbers
Si = { 0, 1 }
li
xi
ui
or
ui }
Implicit constraints:
The implicit constraints are rules that determine which of the tuple in the solution
space satisfy the criterion function.
e.g 8-Queen problem
No two queens are on the same row
No two queens are on the same column
No two queens are on the same diagonal
Recursive backtracking algorithm
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Design and Analysis of Algorithm
Iterative backtracking algorithm
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Design and Analysis of Algorithm
Terminologies:
The tree organization of the solution space is state space tree.
Each node in the state space tree defines problem state
All paths from the root node to other nodes defines the state space of the problem
Solution states are those problem states s for which the path from the root to s defines
a tuple in the solution space.
Answer states are those solution states s for which the path from root to s defines a
tuple that is a member of solutions (it satisfies the implicit constraints).
A node which has been generated and all of whose children have not yet been generated
is called a live node
The live node whose children are currently being generated is called E-node
A dead node is a generated node which is not to be expanded further or all of whose
children have been generated
Tree organization that are problem instance dependent are called dynamic tree.
Bounding functions are used to kill live nodes without generating all their children
Tree organization of 4-Queens problem (permutation tree)
•
In 4-Queens problem, 4-Queens are to be placed on 4x4 chessboard so that
No two queens are on the same row
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Design and Analysis of Algorithm
No two queens are on the same column
No two queens on the same diagonal
•
The possible tree organization of 4-Queens problem is called a permutation tree.
•
In permutation tree, the solution space consists of 4! Permutation of 4-tuple solution
vector (x1, x2, x3, x4)
•
The edges in a tree are labeled by possible value xi. Edges from level1 to level2 nodes
specify the value for x1, edges from level2 to level3 nodes specify the value for x2 and
so on.
•
The solution space is defined by all paths from root node to a leaf node.
The 8-Queens problem
•
The problem is to place 8-Queens on 8x8 chessboard so that they don’t attach
each other by being in the same row, same column and same diagonal.
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Design and Analysis of Algorithm
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We can have solution (x1,x2,…,x8) in which each xi is the column of ith row where
ith queen is placed.
•
Two Queens on the same diagonal that run from upper left to the lower right has
the same (row - column) value. Also two Queens on the same diagonal that goes
from the upper right to lower left has the same (row + column) value.
•
Suppose two queens are at position (i,j) and (k,l).Then they are on the same
diagonal iff.
i-j=k-l
………(1)
or
i+j=k+l
………(2)
The first equation implies:
j-l=i-k
and the second equation implies:
j-l=k-i
So, two Queens placed on the same diagonal iff
|j-l| = |i-k|
Example:
4-Queens problem
Q1→
Q2→
Q3→
Q4→
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Design and Analysis of Algorithm
The portion of State-space tree of 4-queens problem is
8-queen problem
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The solution is (4, 6, 8, 2, 7, 1, 3, 5)
Algorithm NQueens(k, n)
// Using backtracking, this procedure prints all possible placements of n queens on
// an nXn chessboard so that they are nonattacking.
{
for (int i=1; i<=n; i++) {
if (Place(k, i)) {
x[k] = i;
if (k==n)
then write (x[1…n]);
else NQueens(k+1, n);
}
}
}
Algorithm Place( k, i)
//Returns true if a queen can be placed in kth row and ith column. Otherwise it
// returns false. x[] is a global array whose first (k-1) values have been set. abs(r) //
returns the absolute value of r.
{
for (int j=1; j < k; j++)
if ((x[j] == i) // Two in the same column
|| (abs(x[j]-i) == abs(j-k))) // or in the same diagonal
return(false);
return(true);
}
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Design and Analysis of Algorithm
Sum of subset
•
In a sum of subset problem given a set of n distinct positive numbers(usually called
weights) and a positive integer m
•
The problem is to find all subset of the given set whose sum is equal to a positive
integer m
•
We could formulate this problem using fixed-size or variable-sized tuple.
•
We consider a backtracking solution using fixed-size tuple strategy. In this case the
element xi of the solution vector is either one or zero depending on whether the
weight wi is included or not.
•
The root of tree represents the starting point. The children of any node are easily
generated. For a node at level i the left child corresponds to xi=1 and right to xi=0
•
The simple choice for the bounding function Bk(x1,..,xk)=true iff
k
n
wi xi
wi m
i 1
i k 1
•
Clearly x1,..,xk can’t lead to an answer node if this condition is not satisfy.
•
If we assume that wi’s are initially in nondecreasing order, (x1,..,xk) cannot lead to
an answer node if
k
wi xi wk
1
m
i 1
•
So, the bounding function is
k
Bk (x1,...,xk ) trueiff wi xi
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wi
m
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Design and Analysis of Algorithm
i 1 i k 1
and
wi xi wk
1
k
m
i 1
Algorithm SumOfSub(s, k, r)
// Find all subsets of w[1:n] that sum to m. The values of x[j], 1<=j<k, have already
//been determined. s= ∑j=1,k-1 w[j]*x[j] and r= ∑j=k,n w[j]. The w[j]'s are in //
nondecreasing order. It is assumed that w[1]<=m and ∑i=1n w[i]>=m.
{
// Generate left child. Note that s+w[k] <= m sine Bk-1 is true.
x[k] = 1;
;// Subset found
if (s+w[k] == m) then write (x[1:k])
else if (s+w[k]+w[k+1] <= m)
SumOfSub(s+w[k], k+1, r-w[k]);
// Generate right child and evaluate Bk-1.
if ((s+r-w[k] >= m) && (s+w[k+1] <= m)) {
x[k] = 0;
SumOfSub(s, k+1, r-w[k]);
}
}
Example1:
Given n=6, w[1:6]={5,10,12,13,15,18}. Find all possible subset whose sums are
m=30 using sum of subset algorithm. Draw portion of state space tree that is
generated
2011 10M
Example2:
Given n=4, (w1, w2, w3, w4)=(7,11,13, 24). Find all possible subset whose
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sums are m=31 using sum of subset algorithm.
2015
5M
Example2:
Let w=[w1, w2, w3, w4,w5,w6]=(5, 7,10,12, 15, 18, 20]. Find all possible subset
whose sums are m=35 using sum of subset algorithm. Draw portion of state space
tree that is generated
2014 10M
Solution to Example1:
Given n=6, w[1:6]={5,10,12,13,15,18}, m=30.
All possible subset whose sum=30 are:
A = (5, 10, 15 ) i.e. ( 1, 1, 0 ,0, 1, 0)
B = (5, 12, 13 ) i.e. ( 1, 0, 1, 1, 0, 0)
C = (12, 18 )
i.e. ( 0, 0, 1, 0, 0, 1)
Portion of state space tree is:
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Design and Analysis of Algorithm
Graph Coloring
•
In this problem, for any given graph G and a positive integer m. We will have to color
each of the vertices in G using at most m different colors in such a way that no two
adjacent vertices have the same color.
•
This is called m-colorability decision problem.
•
The m-colorability problem is optimized when for the smaller integer m the graph
G can be colored.
•
The smaller integer m for which the graph G can be colored is called the chromatic
number of the graph.
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For example.
The minimum number of colors required to color the graph shown below is 3. Hence
we can say that graph’s chromatic number is 3.
•
Consider a simple graph containing four nodes. The graph can be colored using
atmost 3 colors.
•
The possible choices to color each node using three colors are:
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Design and Analysis of Algorithm
Planar Graph
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A graph is said to be planar iff it can be drawn in a plane in such a way that no two
edges cross each other.
•
Consider a Map with different regions be colored in such a way that no two adjacent
regions have the same color.
•
The map can easily be transformed into a graph. Each region of the map becomes a
node, and if two regions are adjacent, then the corresponding nodes are joined by an
edge.
•
The following figure shows a map with five regions and its corresponding graph.
This map requires four colors.
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Hamiltonian Cycles
•
Let G = (V, E) be a connected graph with n vertices. A Hamiltonian cycle is a path
that visits each vertex exactly once and return to the starting vertex.
•
In other words, if a Hamiltonian cycle begins at some vertex v1Є G and the vertices
of G are visited in the order v1, v2, ………… vn+1, then the edges (vi, vi+1) are in the E, 1
≤ i ≤ n and the vi are distinct except for v1 and vn+1.
Example 1:
The graph contains the Hamiltonian cycle 1, 2, 8, 7, 6, 5, 4, 3, 1
Example 2:
There is no way to visit all the vertices exactly once. Hence The graph
Contains no Hamiltonian cycle
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