I tied to simplify the task as much as possible, so I could apply it to my algorithm.
And here is the challenge for mathematicians and programmers:
I need to create a method where I pass parameter int n:
public void optionality_generator(int n){
//some kind of loops, or recursions...to make it workable
System.out.println("current combination: ...");
}
The output should show all possible combinations of true's and false's.
Here is examples where N=1; N=2; N=3; N=4; N=5 where x=false and 0=true; Please note, empty break lines is just for you to recognise easier the patterns. Hopefully, I included all possible combinations):
Combination of 1:
0
x
Combination of 2:
00
x0
0x
xx
Combination of 3:
000
X00
0X0
00X
XX0
0XX
XXX
Combination of 4:
0000
X000
0X00
00X0
000X
XX00
X0X0
X00X
0XX0
0X0X
00XX
XXX0
XX0X
X0XX
0XXX
XXXX
Combination of 5:
00000
X0000
0X000
00X00
000X0
0000X
XX000
X0X00
X00X0
X000X
X0X00
X00X0
X000X
0XX00
0X0X0
0X00X
00XX0
00X0X
000XX
XXX00
XX0X0
XX00X
X0XX0
X0X0X
X00XX
0XXX0
0XX0X
00XXX
XXXX0
XXX0X
XX0XX
X0XXX
0XXXX
XXXXX
Also, If you see the output, here is the pattern I recognized, that all combinations are inverted on half (e.g first combination is 00000 last one will be XXXXX, second one X0000, one before the last one will be 0XXXX etc..). Maybe, this pattern will help to make the whole algorithm more efficient, not sure about this.
Thank you in advance!
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