# Young’s lattice and the RSK algorithm¶

This section provides some examples on the RSK algorithm explained in Chapter 8 of Stanley’s book [Stanley2013].

We begin by creating the first few levels of Young’s lattice $$Y$$. For we need to define the elements and the order relation for the poset, which is containment of partitions:

sage: level = 6
sage: elements = [b for n in range(level) for b in Partitions(n)]
sage: ord = lambda x,y: y.contains(x)
sage: H = Y.hasse_diagram()
sage: view(H)      # optional


We can now define the up and down operators on $$\QQ Y$$. First we do so on partitions:

sage: QQY = CombinatorialFreeModule(QQ,elements)

sage: def up_operator_on_partitions(la):
....:     covers = Y.upper_covers(la)
....:     return sum(QQY(c) for c in covers)

sage: def down_operator_on_partitions(la):
....:     covers = Y.lower_covers(la)
....:     return sum(QQY(c) for c in covers)


Here is the result when we apply the operators to the partition $$(2,1)$$:

sage: la = Partition([2,1])
sage: up_operator_on_partitions(la)
B[[2, 1, 1]] + B[[2, 2]] + B[[3, 1]]
sage: down_operator_on_partitions(la)
B[[1, 1]] + B[[2]]


Now we define the up and down operator on $$\QQ Y$$:

sage: def up_operator(b):
....:     return sum(b.coefficient(p)*up_operator_on_partitions(p) for p in b.support())

sage: def down_operator(b):
....:     return sum(b.coefficient(p)*down_operator_on_partitions(p) for p in b.support())


We can check the identity $$D_{i+1} U_i - U_{i-1} D_i = I_i$$ explicitly on all partitions of 3 (so that $$i=3$$):

sage: for p in Partitions(3):
....:     b = QQY(p)
....:     assert down_operator(up_operator(b)) - up_operator(down_operator(b)) == b
....:


We can also check that the coefficient of $$\lambda \vdash n$$ in $$U^n(\emptyset)$$ is equal to the number of standard Young tableaux of shape $$\lambda$$:

sage: u = QQY(Partition([]))
sage: for i in range(4):
....:     u = up_operator(u)
....:
sage: u
B[[1, 1, 1, 1]] + 3*B[[2, 1, 1]] + 2*B[[2, 2]] + 3*B[[3, 1]] + B[[4]]


For example, the number of standard Young tableaux of shape $$(2,1,1)$$ is 3:

sage: StandardTableaux([2,1,1]).cardinality()
3


We can test this in general:

sage: for la in u.support():
....:     assert u[la] == StandardTableaux(la).cardinality()
....:


Let us now turn to the RSK algorithm. If we want to verify Example 8.12, we can do so as follows:

sage: p = Permutation([4,2,7,3,6,1,5])
sage: p.robinson_schensted()
[[[1, 3, 5], [2, 6], [4, 7]], [[1, 3, 5], [2, 4], [6, 7]]]


The tableaux can also be displayed as tableaux:

sage: P,Q = p.robinson_schensted()
sage: P.pp()
1  3  5
2  6
4  7
sage: Q.pp()
1  3  5
2  4
6  7


The inverse RSK algorithm is also implemented:

sage: RSK_inverse(P,Q, output='word')
word: 4273615


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