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A full cogood-addition path from the empty partition to (3,1,1) at p = 3

Example

At p=3 the p-restricted partition (3,1,1) is reached from the empty partition by the five cogood additions ∅→ 0 (1)→ 2 (1,1)→ 1 (1,1,1)→ 1 (2,1,1)→ 2 (3,1,1), that is, (1)=f~0∅, (1,1)=f~2(1), (1,1,1)=f~1(1,1), (2,1,1)=f~1(1,1,1), (3,1,1)=f~2(2,1,1), where the labels record the residues of the added nodes (Good-node deletion and cogood-node addition operators). Reversing the path gives the good deletions (3,1,1)→ 2 (2,1,1)→ 1 (1,1,1)→ 1 (1,1)→ 2 (1)→ 0 ∅. The relevant reduced signatures are: (1) at i=0: (−); (1,1) at i=2: (−); (1,1,1) at i=1: (+,−); (2,1,1) at i=1: (−,−); (3,1,1) at i=2: (−); and (1,1) at i=1: (+,+), (2,1,1) at i=2: (+). The word (−,−) for (2,1,1) has two normal nodes of the same residue, while (+,−) for (1,1,1) shows that a reduced signature may contain both signs.

Facts & Assumptions

Given: The prime p=3, the residues of a node computed as c−r mod 3, and the partitions ∅, (1), (1,1), (1,1,1), (2,1,1), (3,1,1).

[F1]

The reduced i-signature of a partition is obtained by listing the addable and removable i-nodes top to bottom, marking them + and −, and cancelling −+ pairs; the good i-node is the removable node of the first surviving −, the cogood i-node is the addable node of the last surviving +, and e~i,f~i delete the good node and add the cogood node (Good-node deletion and cogood-node addition operators, Residue signatures, normal, good, conormal and cogood nodes).

[F2]

For p-restricted μ,ν one has μ=f~iν if and only if ν=e~iμ, with both sides defined simultaneously (Good-node deletion and cogood-node addition operators).

[F3]

The reachability proposition guarantees that a path of cogood additions from ∅ to every p-restricted partition exists; the present example exhibits one explicitly and checks it node by node (Every p-restricted partition is reached from the empty partition by cogood additions).

Verification

technique · direct
1.1givenF1

The 0-nodes of ∅ and (1). The empty partition has the single addable node (1,1) of residue c−r=0, so its residue-0 word is (+) and the cogood node is (1,1), whence f~0∅=(1); the partition (1) has the single removable node (1,1) of residue 0, so its residue-0 word is (−) with good node (1,1) and e~0(1)=∅.

1.2F1F2

The 1-nodes of (1,1) and (1,1,1). In (1,1) the residue-1 nodes are the addable nodes (1,2) (residue 2−1=1) and (3,1) (residue 1−3=−2≡1), ordered (1,2),(3,1) top to bottom; the word is (+,+), the cogood node is the last surviving +, namely (3,1), and f~1(1,1)=(1,1,1), while adding (1,2) would lead to (2,1). In (1,1,1) the residue-1 nodes are the addable node (1,2) and the removable node (3,1); the word (+,−) is already reduced, so the surviving + is (1,2) and the surviving − is (3,1): the cogood node is (1,2) with f~1(1,1,1)=(2,1,1), and the good node is (3,1) with e~1(1,1,1)=(1,1), in agreement with the inverse property of [F2].

2.1step 1.1F1

The 2-nodes of (1) and (1,1). The residue-2 nodes of (1): the addable node (2,1) has residue 1−2=−1≡2, and there is no other, so the word is (+) with cogood node (2,1) and f~2(1)=(1,1). For (1,1) the residue-2 node is the removable node (2,1) (residue 1−2≡2), the word is (−), the good node is (2,1) and e~2(1,1)=(1).

2.2step 1.2F1F2

The 1-nodes of (2,1,1) and the 2-nodes of (2,1,1) and (3,1,1). In (2,1,1) the residue-1 nodes are the removable nodes (1,2) (residue 2−1=1) and (3,1) (residue 1−3≡1), listed (1,2),(3,1): the word (−,−) is reduced, so the good node is the first surviving −, namely (1,2), and e~1(2,1,1)=(1,1,1). The residue-2 nodes of (2,1,1) are the single addable node (1,3) (residue 3−1=2): the word is (+) with cogood node (1,3) and f~2(2,1,1)=(3,1,1). In (3,1,1) the only residue-2 boundary node is the removable node (1,3); the interior node (2,1) has the same residue but does not enter the signature; the word is (−) with good node (1,3) and e~2(3,1,1)=(2,1,1).

3.1step 1.1step 2.1step 1.2step 2.2F2

The path and its reverse. Step 1.1, step 2.1, step 1.2 and step 2.2 exhibit, successively, ∅→(1)→(1,1)→(1,1,1)→(2,1,1)→(3,1,1) as cogood additions with residues 0,2,1,1,2, and each reverse step as the deletion of the good node of the same residue; by [F2] this is exactly the assertion that each forward step is a cogood addition and each reverse step a good deletion, and the five residue labels match the displayed path.

4.1step 1.1step 2.1step 1.2step 2.2F3∎

All reduced signatures claimed in the example are the ones computed: (−) for (1) at i=0 in step 1.1, (−) for (1,1) at i=2 in step 2.1, (+,+) for (1,1) and (+,−) for (1,1,1) at i=1 in step 1.2, and (−,−) for (2,1,1) at i=1, (+) for (2,1,1) at i=2 and (−) for (3,1,1) at i=2 in step 2.2; in particular (−,−) exhibits two normal nodes of one residue and (+,−) exhibits both signs in a reduced word, and the path still exists as [F3] predicts.

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