[SOLVED] Numbers to score and identify a situation
by lucmove from LinuxQuestions.org on (#5H9WF)
By "numbers to score and identify a situation" I mean, for example, file permissions. 4, 2, 1, and 0 allow very unique and specific sum combinations that identify each specific situation and can be translated into one single number.
0 (0+0+0) - No permission.
1 (0+0+1) - Only execute permission.
2 (0+2+0) - Only write permission.
3 (0+2+1) - Write and execute permissions.
4 (4+0+0) - Only read permission.
5 (4+0+1) - Read and execute permission.
6 (4+2+0) - Read and write permissions.
7 (4+2+1) - Read, write, and execute permission.
Each sum is unique.
I don't know if there is a name for that specifically in math. Is there?
Anyway, how far can I go and how am I supposed to find other numbers?
In my code, zero is not possible because I iterate through a list of conditions and add a value to a variable whenever a condition is true. So I'm using 1, 2, 3, and 7:
1+2=3
1+3=4
1+7=8
2+3=5
2+7=9
3+7=10 (*)
1+2+3=6
1+2+7=10 (*collision!)
2+3+7=12
1+3+7=11
So 7 is not good because it may cause a collision. Thinking "manually," I conclude that I should replace it with 13, merely because it is the smallest number that is not the sum of any two or three of the smaller numbers, i.e. it is out of reach of collisions.
But no.
1+2=3
1+3=4
1+13=14
2+3=5
2+13=15
3+13=16 (*)
1+2+3=6
1+2+13=16 (*collision!)
1+3+13=17
2+3+13=18
1+2+3+13=22
OK, I'm learning a lesson here. 7 is not the problem. The real problem is I can't have 1, 2, and 3 because 1+2=3 and that will always be trouble further ahead. One more try:
2, 3, 6, 10:
2+3=5
2+6=8
2+10=12
3+6=9
3+10=13
6+10=16
2+3+6=11
2+3+10=15
2+6+10=18
3+6+10=19
2+3+6+10=21
Good. No collisions.
But finding 6 and 10 was tedious. What is the correct logic for this? Is there some way to calculate it?
If I need more numbers (and I do), what would the next numbers be? How am I supposed to find them?
TIA
0 (0+0+0) - No permission.
1 (0+0+1) - Only execute permission.
2 (0+2+0) - Only write permission.
3 (0+2+1) - Write and execute permissions.
4 (4+0+0) - Only read permission.
5 (4+0+1) - Read and execute permission.
6 (4+2+0) - Read and write permissions.
7 (4+2+1) - Read, write, and execute permission.
Each sum is unique.
I don't know if there is a name for that specifically in math. Is there?
Anyway, how far can I go and how am I supposed to find other numbers?
In my code, zero is not possible because I iterate through a list of conditions and add a value to a variable whenever a condition is true. So I'm using 1, 2, 3, and 7:
1+2=3
1+3=4
1+7=8
2+3=5
2+7=9
3+7=10 (*)
1+2+3=6
1+2+7=10 (*collision!)
2+3+7=12
1+3+7=11
So 7 is not good because it may cause a collision. Thinking "manually," I conclude that I should replace it with 13, merely because it is the smallest number that is not the sum of any two or three of the smaller numbers, i.e. it is out of reach of collisions.
But no.
1+2=3
1+3=4
1+13=14
2+3=5
2+13=15
3+13=16 (*)
1+2+3=6
1+2+13=16 (*collision!)
1+3+13=17
2+3+13=18
1+2+3+13=22
OK, I'm learning a lesson here. 7 is not the problem. The real problem is I can't have 1, 2, and 3 because 1+2=3 and that will always be trouble further ahead. One more try:
2, 3, 6, 10:
2+3=5
2+6=8
2+10=12
3+6=9
3+10=13
6+10=16
2+3+6=11
2+3+10=15
2+6+10=18
3+6+10=19
2+3+6+10=21
Good. No collisions.
But finding 6 and 10 was tedious. What is the correct logic for this? Is there some way to calculate it?
If I need more numbers (and I do), what would the next numbers be? How am I supposed to find them?
TIA