generate and test framework

GENERATE AND TEST FRAMEWORKNattee Niparnan

OPTIMIZATION EXAMPLE: FINDING MAX VALUE IN AN ARRAY25

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

QQ

QQ

QQ

QQ

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

0 1 2

10 11 12 20 21 2200 01 02

000001002

010011012

020021022

100101102

110111112

120121122

200201202

210211212

220221222

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 10If 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

generate and test framework

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