1. Edit from old slide since 2000 P. Chongstitvatana 26 Nov 2010
Backtrack: 8-queen
Edit from old slide since 2000
P. Chongstitvatana 26 Nov 2010

2. Backtracking Eight queens problem
1 try all possible C64 8 = 4,426,165,368
2 never put more than one queen on a given row,
vector representation : each row specify which column (3,1,6,2,8,6,4,7)

3. X X X X X X X X
(3,1,6,2,8,6,4,7)

4. Num. Of positions = 8 8 = 16,777,216 (first soln after 1,299,852 )
Queen1
for i1 = 1 to 8 do
for i2 = 1 to 8 do
for i3 = 1 to 8 do
sol = [i1, i2, i8 ]
if solution ( sol ) then write sol stop
write “there is no solution”
Num. Of positions = 8 8 = 16,777,216
(first soln after 1,299,852 )

5. 3 Never put queen on the same row (different numbers on soln vector)
sol = initial-permutation
while sol != final-permutation and not solution(sol) do
sol = next-permutation
if solution(sol) then write sol
else write “there is no solution”

6. T[1 . . n] is a global array initialize to [1,2,. . n]
Permutation
T[1 . . n] is a global array initialize to [1,2,. . n]
initial call perm(1)
Perm(i)
if i = n then use T
else for j = i to n do exchange T[i] and T[j]
perm(i+1)
exchange T[i] and T[j]
Number of positions 8! = 40,320 (first soln after 2830)

7. 8-queen as tree search
a vector V[1. .k] of integers between 1 and 8 is k-promising, if none of the k queens threatens any of the others.
A vector V is k-promising if, for every pair of integers i and j between 1 and k with i != j, we have V[i] - V[j] is-not-in {i-j, 0, j-i}.
Solutions to the 8-queen correspond to vectors that are 8-promising.

8. V is (k+1)-promising, and U[i] = V[i] for every i in [1..k]
Let N be the set of k-promising vectors, k: Let G = (N,A) be the directed graph such that (U,V) is-in A iff there exists an integer k, k:0..8 , such that
U is k-promising
V is (k+1)-promising, and
U[i] = V[i] for every i in [1..k]
k = 0
. . .
Number of node < 8!
(node 2057, first soln after 114 )
k = 8

9. General Template for backtracking
Backtrack ( v[1..k] ) // v is k-promising vector
if solution ( v ) then write v
else for each (k+1)-promising vector w
such that w[1..k] = v[1..k]
do backtrack( w[1.. k+1] )