ASKSAGE: Sage Q&A Forum - Latest question feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Sun, 05 Jul 2020 13:36:40 -0500Combination under constrained situation with a conditionhttps://ask.sagemath.org/question/52345/combination-under-constrained-situation-with-a-condition/Step 1: I want to take a number `n` as input from user
Step 2: We form the set `S` consisting of elements from `0` to `n*(2^{n-1})`
Step 3: Now I pick each possible two-element subsets of `S` and store it in `P`.
Step 4: Now I need to pick `n*(2^{n-1})` two-element subsets from `P`
such that each number that occurs in that set occurs exactly `n` times
neither less nor more and put them all in a list.
Example
n = 2
n*(2^{n-1}) = 2*(2^{2-1}) = 4
S = {0,1,2,3,4}
p = {(0,1),(0,2),(0,3),(0,4),(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}
Step 4 One element which satisfies condition of step 4; HERE `n = 2`.
`{(0,1),(1,2),(2,3),(0,3)}` which has `2*(2^{n-1}) = 2*(2^{2-1}) = 4` elements.
Now see in the above set
- `0` occurred `n=2` times only in `(0,1)` and `(0,3)` only
- `1` occurred `n=2` times only in `(0,1)` and `(1,2)` only
- `2` occurred `n=2` times only in `(1,2)` and `(2,3)` only
- `3` occurred `n=2` times only in `(2,3)` and `(0,3)` only
Similarly we may get for `{(0,1),(1,4),(4,2),(2,0)}` we can easily verify like above.
Now based on `n` the number of elements size etc will vary.
Kindly help if possible any one.sriramSun, 05 Jul 2020 13:36:40 -0500https://ask.sagemath.org/question/52345/