sk
algorithm
0 Answer(s)      5 years and 2 months ago
Posted in : Java Beginners

Hi all,

i have a task which is create an algorithm from a java code.and i need help from u guys to check on my algo put a comment on what should i change or why it is wrong.

I really appreciate if u could help me on this matter...

thanks a lot.

here is my algorithm:

```>
Input:
input configuration set,S (number of value for p1(v1), number of value for p2(v2),.....number of value for pn (vn))
number of interaction strength, t
number of parameter, p
Output:
test case suite , A

begin
initialize binary_list as empty binary set;
choose binary sets (in parameter order p1,p2,... pn) where number of bit '1' equal to t;
push binary sets into binary_list;

while binary_list not empty
for each binary_list: b = binary set
//count number of value combination base on b;
initialize count = 1;
for (i <= p)
if b(i) = '1';
v(i);
count = count * v(i);
end if
end for

initialize repetition = 1;
for (i<= p)
if b(i) = '1'
initialize current_combination = 0;
initialize temp as an empty set of temporary test set;
while (current_combination < count)
initialize v_o_p = 0;
choose value where parameter (i);
insert or update v_o_p into temp;
repeat same value for repetition time;
end while

repetition = repetition * v(i);

end if

else if b(i) = '0'
choose value from parameter (i) randomly;
update temp;

end for

end for each
end while

//write candidates into final test set

while temp is not empty
pop 1st test case in temp and push into A;

for each of test case in temp: tc = test case

compare tc to test case in A;

if (compare == 0)
insert tc into A;
else
remove tc in temp;
end for each
end while

end

the brief explanation by example:

user input:
set of parameter: A:a1,a2,a3; B:b1,b2,b3; C:c1,c2,c3

number of parameter= 3

the value of each parameter = 3 for A, 3 for B, 3 for C

number of strength,t = 2

the set of binary in this case is = 3 (base on parameter orientation ABC)

>
1- 011
>
2- 101
>
3- 110

operates for each binary set:
example for binary set no 1= 011;
starts from last bit which is '1';
so in empty set named temp for first bit should have:

>

xxc1

xxc2

xxc3

xxc1

xxc2

xxc3

xxc1

xxc2

xxc3

2nd bit, temp should have (updated temp):

xb1c1

xb1c2

xb1c3

xb2c1

xb2c2

xb2c3

xb3c1

xb3c2

xb3c3

last round for 1st bit is '0', so should select randomly from parameter A's values:

a1b1c1

a3b1c2

a2b1c3

a1b2c1

a3b2c2

a3b2c3

a2b3c1

a3b3c2

a1b3c3

this repeated until all binary sets operated with same operation.

then all data in temp will be compared to each other if no duplicate data, push into final_set.
```

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