Calculate the \(L^{p}\) scores.

## Usage

```
LPScores(powerRelation, elements = powerRelation$elements)
LPRanking(powerRelation)
lexcelPScores(powerRelation, elements = powerRelation$elements)
lexcelPRanking(powerRelation)
```

## Arguments

- powerRelation
A

`PowerRelation`

object created by`PowerRelation()`

or`as.PowerRelation()`

- elements
Vector of elements of which to calculate their scores. By default, the scores of all elements in

`powerRelation$elements`

are considered.

## Value

Score function returns a list of type `LPScores`

and length of `powerRelation$elements`

(unless parameter `elements`

is specified).
Each index contains a vector of length `length(powerRelation$elements)`

.

Ranking function returns corresponding `SocialRanking`

object.

## Details

Let \(N\) be a set of elements, \(\succsim \in \mathcal{T}(\mathcal{P})\) a power relation, and \(\Sigma_1 \succ \Sigma_2 \succ \dots \succ \Sigma_m\) its corresponding quotient order.

For an element \(i \in N\), construct a matrix \(M^\succsim_i\) with \(m\) columns and \(|N|\) rows. Whereas each column \(q\) represents an equivalence class, each row \(p\) corresponds to the coalition size.

$$(M^\succsim_i)_{p,q} = |\lbrace S \in \Sigma_q: |S| = p \text{ and } i \in S\rbrace|$$

For \(i, j \in N\), the social ranking solution \(L^p\) then ranks \(i\) strictly above \(j\) if one of the following conditions hold:

\(\lbrace i \rbrace \succ \lbrace j \rbrace\);

\(\lbrace i \rbrace, \lbrace j \rbrace \in \Sigma_k\) and there exists a row \(p_0 \in \lbrace 2, \dots, |N|\rbrace\) such that: $$\sum_{q < k} (M^\succsim_i)_{p,q} = \sum_{q < k} (M^\succsim_j)_{p,q}\quad \forall p < p_0,\text{ and}$$ $$\sum_{q < k} (M^\succsim_i)_{p_0,q} > \sum_{q < k} (M^\succsim_j)_{p_0,q}.$$

In `R`

, given two matrices `M_i`

and `M_j`

, this comparison could be expressed as

```
# function that returns TRUE if i should be ranked strictly above j
k_i <- which(M_i[1,] == 1)
k_j <- which(M_j[1,] == 1)
if(k_i != k_j) return(k_i < k_j)
if(k_i == 1) return(FALSE)
# get sum for each row
# removing the first row implies that we start in row 2
sums_i <- apply(M_i[-1,seq(k_i-1)], 1, sum)
sums_j <- apply(M_j[-1,seq(k_j-1)], 1, sum)
# apply lexcel comparison
i <- which(a != b)
return(length(i) > 0 && a[i[1]] > b[i[1]])
```

## Example

Let \(\succsim: (123 \sim 12 \sim 2) \succ (13 \sim 23) \succ (1 \sim 3 \sim \{\})\). From this, we get the following three matrices:

$$ M^\succsim_1 = \begin{bmatrix} 0 & 0 & 1\\ 1 & 1 & 0\\ 1 & 0 & 0 \end{bmatrix} M^\succsim_2 = \begin{bmatrix} 1 & 0 & 0\\ 1 & 0 & 1\\ 1 & 0 & 0 \end{bmatrix} M^\succsim_3 = \begin{bmatrix} 0 & 0 & 1\\ 0 & 2 & 0\\ 1 & 0 & 0 \end{bmatrix} $$

\((M^\succsim_2)_{2,3}\) in this context refers to the value in the second row and third column of element 2, in this case \(1\).

In the example, \(2\) will be immediately put above \(1\) and \(3\) because \(\lbrace 2 \rbrace \succ \lbrace 1 \rbrace\) and \(\lbrace 2 \rbrace \succ \lbrace 3 \rbrace\). Since \(\lbrace 1 \rbrace \sim \lbrace 3 \rbrace\), we next consider the coalitions of size 2. Here, it turns out that \((M^\succsim_1)_{2,1} + (M^\succsim_1)_{2,2} = 1 + 1\) is equal to \((M^\succsim_3)_{2,1} + (M^\succsim_3)_{2,2} = 0 + 2\). For obvious reasons the grand coalition does not have to be considered, thus \(1\) and \(3\) are considered equally powerful by the \(L^p\) solution.

\(L^{p}\) is a social ranking solution belonging to the family of lexicographical ranking functions.
While related to `L1Ranking()`

, it incorporates the property of "standardness", stating that if the
singleton coalition \(\lbrace i\rbrace \succ \lbrace j\rbrace\), then the ranking solution
should also prefer \(i\) over \(j\).

If \(\lbrace i\rbrace \sim \lbrace j\rbrace\), then all coalitions from size 2 and upward are inspected, giving higher precedence to coalitions with a lower number of elements. While this preference is similar to the \(L^{(1)}\), it differs in two notable ways:

If \(\lbrace i\rbrace, \lbrace j\rbrace \in \Sigma_k\), then only coalitions \(S \succsim (\lbrace i \rbrace \sim \lbrace j \rbrace)\) are considered,

From this subset of coalitions, consider the total number of coalitions \(i\) (or \(j\)) belongs to, given each coalition size. This may ignore information about the distribution of these coalitions within the different equivalence classes, which \(L^{(1)}\) and the slight variation \(L^{p^*}\) of the \(L^p\) solution take into account.

## Alterations

The matrices as described above and in Béal S, Rémila E, Solal P (2022).
“Lexicographic solutions for coalitional rankings based on individual and collective performances.”
*Journal of Mathematical Economics*, **102**, 102738.
can be investigated with the `L1Scores()`

function.

For efficiency, `LPScores()`

discards much of the redundant information.
Instead of a matrix for each element, it returns a vector of size \(|N|\).

Given a score vector `v`

for an element `i`

, `v[1]`

is the position of the singleton coalition `{i}`

.
This implies that if `v[1] < w[1]`

, where `w`

is the score vector of an element `j`

, then `i`

is ranked strictly above `j`

.

`v[2]`

, `v[3]`

, ..., `v[n]`

then indicates the number of coalitions of size `2`

, `3`

, ..., `n`

that the element `i`

appears in.

## Aliases

For better discoverability, `lexcelPScores()`

and `lexcelPRanking()`

serve as aliases for `LPScores()`

and `LPRanking()`

, respectively.

## References

Béal S, Rémila E, Solal P (2022).
“Lexicographic solutions for coalitional rankings based on individual and collective performances.”
*Journal of Mathematical Economics*, **102**, 102738.

## See also

Other ranking solution functions:
`L1Scores()`

,
`L2Scores()`

,
`LPSScores()`

,
`copelandScores()`

,
`cumulativeScores()`

,
`kramerSimpsonScores()`

,
`lexcelScores()`

,
`ordinalBanzhafScores()`

## Examples

```
pr <- as.PowerRelation("(123 ~ 13 ~ 2) > (12 ~ 1 ~ 3) > (23 ~ {})")
scores <- LPScores(pr)
scores$`2`
#> [1] 1 0 0
# [1] 1 0 0
LPRanking(pr)
#> 2 > 1 ~ 3
# 2 > 1 ~ 3
# Since L^(1) also the relation {1,2}, which ranks above {2,3}, it will place 1 above 3
L1Ranking(pr)
#> 2 > 1 > 3
# 2 > 1 > 3
```