ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Mon, 08 Feb 2021 11:13:39 +0100- Finding an involution with Sagehttps://ask.sagemath.org/question/55622/finding-an-involution-with-sage/I have four lists L1,LL1 and L2, LL2 and a bijection G from LL1 to LL2. LL1 is a sublist of L1 and LL2 a sublist of L2.
For example they look like this:
L1=[ [ [ 2, 2, 2, 1 ], 4 ], [ [ 3, 2, 2, 1 ], 3 ], [ [ 2, 3, 2, 1 ], 3 ], [ [ 3, 3, 2, 1 ], 3 ], [ [ 4, 3, 2, 1 ], 2 ] ]
LL1=[ [ [ 2, 2, 2, 1 ], 4 ], [ [ 3, 2, 2, 1 ], 3 ], [ [ 2, 3, 2, 1 ], 3 ], [ [ 4, 3, 2, 1 ], 2 ] ]
and
L2=[ [ [ 2, 2, 2, 1 ], 2 ], [ [ 3, 2, 2, 1 ], 3 ], [ [ 2, 3, 2, 1 ], 3 ], [ [ 3, 3, 2, 1 ], 3 ], [ [ 4, 3, 2, 1 ], 4 ] ]
LL2=[ [ [ 2, 2, 2, 1 ], 2 ], [ [ 3, 2, 2, 1 ], 3 ], [ [ 2, 3, 2, 1 ], 3 ], [ [ 4, 3, 2, 1 ], 4 ] ]
The bijection G looks like this:
[ [ [ [ 2, 2, 2, 1 ], 4 ], [ [ 4, 3, 2, 1 ], 4 ] ], [ [ [ 3, 2, 2, 1 ], 3 ], [ [ 2, 3, 2, 1 ], 3 ] ], [ [ [ 2, 3, 2, 1 ], 3 ], [ [ 3, 2, 2, 1 ], 3 ] ],
[ [ [ 4, 3, 2, 1 ], 2 ], [ [ 2, 2, 2, 1 ], 2 ] ] ]
Thus an element in such a list is a tuple of the form [ [ 2, 2, 2, 1 ], 4 ] where [2,2,2,1] correponds to some algebraic object and the second entry is number. L1 and L2 will always have the same number of elements and LL1 and LL2 will have the same number of elements. The bijection G is given also as a list, where an entry such as
[ [ [ 2, 2, 2, 1 ], 4 ], [ [ 4, 3, 2, 1 ], 4 ] ]
means of course that G maps [ [ 2, 2, 2, 1 ], 4 ] to [ [ 4, 3, 2, 1 ], 4 ] .
Im searching for bijections F between L1 and L2 that satisfy the following three conditions:
1. F is an involution, that is $F^2=id$.
2. F preserves the number.
3. F restricted to LL1 maps to LL2 via the given involution G.
In the above example F is uniquely given as follows:
[ [ [ [ 2, 2, 2, 1 ], 4 ], [ [ 4, 3, 2, 1 ], 4 ] ], [ [ [ 3, 2, 2, 1 ], 3 ], [ [ 2, 3, 2, 1 ], 3 ] ], [ [ [ 2, 3, 2, 1 ], 3 ], [ [ 3, 2, 2, 1 ], 3 ] ],
[ [ [ 4, 3, 2, 1 ], 2 ], [ [ 2, 2, 2, 1 ], 2 ] ],[ [ [ 3, 3, 2, 1 ], 3 ], [ [ 3, 3, 2, 1 ], 3 ] ] ]
Here is another example where the lists are larger:
L1=[ [ [ 2, 2, 2, 2, 1 ], 5 ], [ [ 3, 2, 2, 2, 1 ], 4 ], [ [ 2, 3, 2, 2, 1 ], 3 ], [ [ 3, 3, 2, 2, 1 ], 4 ], [ [ 4, 3, 2, 2, 1 ], 3 ],
[ [ 2, 2, 3, 2, 1 ], 4 ], [ [ 3, 2, 3, 2, 1 ], 3 ], [ [ 2, 3, 3, 2, 1 ], 4 ], [ [ 3, 3, 3, 2, 1 ], 4 ], [ [ 4, 3, 3, 2, 1 ], 3 ], [ [ 2, 4, 3, 2, 1 ], 3 ], [
[ 3, 4, 3, 2, 1 ], 3 ],
[ [ 4, 4, 3, 2, 1 ], 3 ], [ [ 5, 4, 3, 2, 1 ], 2 ] ]
LL1=[ [ [ 2, 2, 2, 2, 1 ], 5 ], [ [ 3, 2, 2, 2, 1 ], 4 ], [ [ 2, 3, 2, 2, 1 ], 3 ], [ [ 4, 3, 2, 2, 1 ], 3 ], [ [ 2, 2, 3, 2, 1 ], 4 ],
[ [ 3, 2, 3, 2, 1 ], 3 ], [ [ 2, 4, 3, 2, 1 ], 3 ], [ [ 5, 4, 3, 2, 1 ], 2 ] ]
L2=[ [ [ 2, 2, 2, 2, 1 ], 2 ], [ [ 3, 2, 2, 2, 1 ], 3 ], [ [ 2, 3, 2, 2, 1 ], 3 ], [ [ 3, 3, 2, 2, 1 ], 3 ], [ [ 4, 3, 2, 2, 1 ], 4 ],
[ [ 2, 2, 3, 2, 1 ], 3 ], [ [ 3, 2, 3, 2, 1 ], 3 ], [ [ 2, 3, 3, 2, 1 ], 3 ], [ [ 3, 3, 3, 2, 1 ], 3 ], [ [ 4, 3, 3, 2, 1 ], 4 ], [ [ 2, 4, 3, 2, 1 ], 4 ], [
[ 3, 4, 3, 2, 1 ], 4 ],
[ [ 4, 4, 3, 2, 1 ], 4 ], [ [ 5, 4, 3, 2, 1 ], 5 ] ]
LL2=[ [ [ 2, 2, 2, 2, 1 ], 2 ], [ [ 3, 2, 2, 2, 1 ], 3 ], [ [ 2, 3, 2, 2, 1 ], 3 ], [ [ 4, 3, 2, 2, 1 ], 4 ], [ [ 2, 2, 3, 2, 1 ], 3 ],
[ [ 3, 2, 3, 2, 1 ], 3 ], [ [ 2, 4, 3, 2, 1 ], 4 ], [ [ 5, 4, 3, 2, 1 ], 5 ] ]
G=[ [ [ [ 2, 2, 2, 2, 1 ], 5 ], [ [ 5, 4, 3, 2, 1 ], 5 ] ], [ [ [ 3, 2, 2, 2, 1 ], 4 ], [ [ 2, 4, 3, 2, 1 ], 4 ] ], [ [ [ 2, 3, 2, 2, 1 ], 3 ], [ [ 3, 2, 3,
2, 1 ], 3 ] ],
[ [ [ 4, 3, 2, 2, 1 ], 3 ], [ [ 2, 2, 3, 2, 1 ], 3 ] ], [ [ [ 2, 2, 3, 2, 1 ], 4 ], [ [ 4, 3, 2, 2, 1 ], 4 ] ], [ [ [ 3, 2, 3, 2, 1 ], 3 ], [ [ 2, 3, 2, 2, 1
], 3 ] ],
[ [ [ 2, 4, 3, 2, 1 ], 3 ], [ [ 3, 2, 2, 2, 1 ], 3 ] ], [ [ [ 5, 4, 3, 2, 1 ], 2 ], [ [ 2, 2, 2, 2, 1 ], 2 ] ] ]
There might be many F satisfying the three conditions.
Thus my question: Is there an easy method to find all such bijections F satisfying the three conditions when L1,LL1,L2,LL2 and G are given?klaaaMon, 08 Feb 2021 11:13:39 +0100https://ask.sagemath.org/question/55622/