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GENERATE AND TEST FRAMEWORKNattee Niparnan
OPTIMIZATION EXAMPLE: FINDING MAX VALUE IN AN ARRAY
25
2 34
43 4 9 0 -5 87
0 5 6 1
There are N possible answersThe first elementThe second element3rd, 4th …
Try all of themRemember the best one
GENERATE AND TEST FRAMEWORK
Define the set of admissible solution Generate all of them
(generating) For each generated solution
Test whether it is the one we want By the “evaluation” function (testing)
for optimization: remember the best one so far For decision: report when we found the “correct”
one
COMBINATION AND PERMUTATION
In many case, the set of the admissible solutions is a set of “combination” or “permutation” of something
We need to knows how to generate all permutations and combinations
GENERATING ALL POSSIBLE ANSWER
COMBINATION
Given N things Generate all possible selections of K things
from N things
Ex. N = 3, k = 2
COMBINATION WITH REPLACEMENT
Given N things Generate all possible selections of K things
from N things When something is selected, we are permit to
select that things again (we replace the selected thing in the pool)
Ex. N = 3, k = 2
BREAKING THE PADLOCK
BREAKING THE PADLOCK Assuming we have four rings Assuming each ring has following mark
We try
…. Backtracking
Undone the second step,
switch to another value
KEY IDEA
A problem consists of several similar steps Choosing a things from the pool
We need to remember the things we done so far
GENERAL FRAMEWORK
Storage
Step that have been done
Engine
Initial Step
1. Get a step that is not complete
2. Try any possible next step
3. Store each newly generated next step
COMBINATION
Generate all combinations of lock key Represent key by int
=0=1=2=3
Step = sequence of selected symbols
E.g., ’01’ ‘0003’
GENERAL SEARCH Storage s S ‘’ While s is not empty
Curr Storage.get If Curr is the last step
evaluate Else
Generate all next step from Curr Push them to S
For all symbol i New = curr+I Storage.push(new)
If length(curr) == 4
SEARCH SPACE
Set of all admissible solutions E.g., Combination Padlock
Search space = 0000 4444 For a task with distinct steps it is easy to
enumerate the search space with “backtracking”
for most of the case, we can do “backtracking”
BACKTRACKING
A tool to enumerate the search space Usually using the “stack” to implement the
storage Employ the processor stack i.e., using the “recursive” paradigm
COMBINATION EXAMPLE BY BACKTRACKING
The process automatically remember the step
void backtracking(int step,int *sol) { if (step < num_step) { for (int i = 0; i < num_symbol; i++) {
sol[step] = i; backtracking(step + 1,sol);}
} else { check(sol); }}
SEARCH TREE Sols in each step
0 00 000 0000 check 000 0001 check 000 0002 check 000 0003 check 000 00 001 0010 check
SEARCH TREE
0 1 2 3
01 02 03 04 …
… … … …… … … …… … … …
8-QUEEN
Given a chess board 8 queens
X X X
X X X
X X X X
X X X
X X X
8-QUEENS PROBLEM
Try to place the queens so that they don’t get in the others’ ways
Q
Q
Q
Q
Q
Q
Q
Q
SOLVING THE PROBLEM
Define the search space What is the search space of this problem? How large it is?
Choose an appropriate representation
EXAMPLE
Every possible placement of queens Size: 648
Representation: a set of queens position E.g., (1,1) (1,2) (2,5) (4,1) (1,2) (3,4) (8,8) (7,6)
This includes overlapping placement!!!
EXAMPLE
Another representation Try to exclude overlapping
Use combination without replacement This is a combination
Selecting 8 positions out of 64 positions Size: (64)! / (64 – 8)! * 8!
Implementation: in the “generating next step”, check for overlapping
COMBINATION WITHOUT REPLACEMENT
We go over all position For each position, we either “choose” or “skip”
that position for the queen
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
COMBINATION WITHOUT REPLACEMENT
void e_queen(int step,int *mark_on_board) { if (step < 64) { mark_on_board[step] = 0; e_queen(step + 1,mark_on_board); mark_on_board[step] = 1; e_queen(step + 1,mark_on_board); } else { check(mark_on_board); }}
Also has to check whether
we mark exactly 8 spots
Mark_on_board is a binary string indicate that whether
the position is selected
COMBINATION WITHOUT REPLACEMENT
The generated mark_on_board includes 000000000000 select no position (obviously
not the answer) 111111000000 select 6 positions (obviously
not the answer)
We must limit our selection to be exactly 4
COMBINATION WITHOUT REPLACEMENT
void e_queen(int step,int *mark_on_board,int chosen) { if (step < 64) { if ((64 – 8) – (step – chosen) > 0) { mark_on_board[step] = 0; e_queen(step + 1,mark_on_board,chosen); } if (8 - chosen > 0) { mark_on_board[step] = 1; e_queen(step + 1,mark_on_board,chosen+1); } } else { check(mark_on_board); }}
Number of possible 0
Number of possible 1
EXAMPLE
Any better way? For each row, there should be only one queen
Size: 88
Representation: sequence of columns E.g., (1,2,3,4,5,6,7,8)
USING BACKTRACKING?
The problem consists of 8 step Placing each queen
We never sure whether the queen we place would lead to a solution
Backtracking is an appropriate way
SOLVING THE PROBLEM: ENUMERATING SEARCH SPACE There are eight possible ways in each step There are eight steps Very similar to the combination problem
void e_queen(int step,int *queen_pos) { if (step < 8) { for (int i = 0; i < 8; i++) {
queen_pos[step] = i; e_queen(step + 1, queen_pos);}
} else { check(queen_pos); }}
8-QUEEN BY PERMUTATION
Queen should not be in the same column The solution should never have any column
repeated E.g.,
(1,2,3,4,5,6,7,1) is bad (column collision (1,1,3,4,5,6,7,5) is bad as well….
(1,2,3,4,5,6,7,8) is good There should be no duplicate column index!!!
PERMUTATION Given N symbols A permutation is the element arrange in any
order E.g., 1 2 3 4 Shows
1 2 3 4 1 2 4 3 1 3 2 4 1 3 4 2 … 4 3 2 1
For each step, we have to known which one is used
PERMUTATION BY BACKTRACKING The problem consists of several similar steps Special condition
Symbols never repeat How to do?
Easy way: Generate all combination (as done before)
Check for ones that symbols do not repeat Better way:
Remember what symbols are used
PERMUTATIONvoid backtracking(int step,int *sol) { if (step < num_step) { for (int i = 0; i < num_symbol; i++) { if not_used(sol,i,step) {
sol[step] = i; backtracking(step,sol);
}}
} else { check(sol); }}
Bool not_used(int *sol,int value,int step) { for (int i = 0;i < step; i++) { if (sol[i] == value) return false; } return true;}
PERMUTATION
More proper ways
void backtracking(int step,int *sol,bool *used) { if (step < num_step) { for (int i = 0; i < num_symbol; i++) { if (!used[i]) {
used[i] = true;sol[step] = i;
backtracking(step,sol,used); used[i] = false; }
} } else { check(sol); }}
BACKTRACKING PROBLEMS
Given N Find any sequence of (a1,a2,a3,…) such that
a1 +a2 + a3 +… + ak = N ai > 0 ai <= aj for all i < j ai is an integer
EXAMPLE
N = 4 1 + 1 + 1 + 1 1 + 1 + 2 1 + 3 2 + 2 4
SOLVING WITH BACKTRACKING
Representation Array of ai
Step Choosing the next value for ai
BRANCH & BOUNDTechnique to reduce enumeration
MAIN IDEA
We should not enumerate solution that will never produce a solution
We have done that!!! 8-queens By naïve combination, we will have to do all 648
But, by each improvement, we further reduce what we have to do
PERMUTATION
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PERMUTATION
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10 11 12 20 21 2200 01 02
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010
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KEY
If we know, at any step, that the solution is not feasible Then, it is futile to further search along that path
OPTIMIZATION PROBLEM
All previous examples are decision problems Asking whether the solution satisfy the criteria
Now, we consider broader set of problem, the optimization problem
Example For all students, find one with maximum height
EVALUATION
Representation: x = id of student “goodness evaluation” evaluate(x)
For all x in the search space, we have to find one with maximum evaluate(x)
SOLVING OPTIMIZATION PROBLEM
Enumerate all possible solution Calculate its value
Remember the max
void backtracking(int step,int *sol) { if (step < num_step) { for (int i = 0; i < num_symbol; i++) {
sol[step] = i; backtracking(step,sol);}
} else { value = evaluate(sol); if (value > max) remember(value,sol); }}
BRANCH & BOUND IN OPTIMIZATION PROBLEM For many problems, it is possible to assert its
goodness even the solution is not complete If we can predict the best value for the
remaining step, then we can use that value to “bound” our search
EXAMPLE
Assuming that we have 10 steps At step 7, the goodness of the partial solution
is X Assuming that we know that the remaining
step could not produce a solution better than Y If we have found a solution better than X+Y
We can simply “bound” the search
KEYS
We must know the so-called “upper bound” of the remaining step It should be compute easily
EXAMPLE
23 35 2
Let value at this point be 10
If we know that this path never bet higher than 13 (which make
10 + 13 < 35)We can neglect it
KNAPSACK PROBLEM
KNAPSACK PROBLEM
Given a sack, able to hold K kg Given a list of objects
Each has a weight and a value Try to pack the object in the sack so that the
total value is maximized
VARIATION Rational Knapsack
Object is like a gold bar, we can cut it in to piece with the same value/weight
0-1 Knapsack Object cannot be broken, we have to choose to
take (1) or leave (0) the object E.g.
K = 50 Objects = (60,10) (100,20) (120,30) Best solution = choose second and third
RATIONAL KNAPSACK
Can be solved by greedy Sort object according to value/weight ratio Pick objects by that ratio
If object is larger than the remaining capacity, just divide it
0-1 KNAPSACK WITH B&B
0-1 knapsack is very suitable for B&B We can calculate the goodness of the partial
solution Just sum the value of the selected objects
We have fast, good upper bounds (several one) The sum of remaining unselected objects
The sum of remaining unselected object that don’t exceed the capacity
The solution of the “rational knapsack” of the remaining objects with the remaining capacity
TASK
Implement 0-1 Knapsack
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