Using the basic operations of relational algebra (RA), it is possible to obtain the largest value assigned to a given attribute of a relation. This post shows how this can be done.

To start, consider the following operators of RA (below, $R$ and $S$ are relations):

$\sigma_{\theta}(R)$ | select tuples (rows) from $R$ which satisfy the condition $\theta$, where $\theta$ consists of comparisons of attributes from $R$ using the following binary operators: $\lt$, $\leq$, $=$, $\geq$ and $\gt$ |

$\Pi_{a_1,\ldots,a_n}(R)$ | extract attributes (columns) $a_1,\ldots,a_n$ from $R$ (duplicate tuples are discarded so each tuple in the resulting relation is unique) |

$\rho_{a/b}(R)$ | rename attribute $b$ from $R$ to $a$ |

$R \bowtie_{\theta} S$ | $\theta$-join of $R$ and $S$: computes all combinations of tuples from $R$ and $S$ that satisfy the condition $\theta$ ($R \bowtie_{\theta} S = \sigma_{\theta}(R \times S)$, where $R \times S$ is the Cartesian product of $R$ and $S$) |

Consider now the relation $P$ below:

Name | Age |
---|---|

Peter | 21 |

Bob | 25 |

Alice | 32 |

John | 27 |

The maximum age of the people listed in $P$ can be retrieved as follows: $$ \max_{P}(\textrm{Age}) := \Pi_{\textrm{Age}} P - \Pi_{\textrm{Age}}\left[ P \underset{\textrm{Age} \lt \textrm{Age2}}{\bowtie} \left(\rho_{\textrm{Name2/Name}}\rho_{\textrm{Age2/Age}} P\right)\right] $$ In other words, we first obtain a relation $\Pi_{\textrm{Age}} P$ which contains a single column with all ages and subtract from it the set of all ages which are smaller than some other age. To clarify the second part, notice that: $$ P \underset{\textrm{Age} \lt \textrm{Age2}}{\bowtie} \left(\rho_{\textrm{Name2/Name}}\rho_{\textrm{Age2/Age}} P\right) $$ is a relation containing four columns ($\textrm{Name}$, $\textrm{Age}$, $\textrm{Name2}$, $\textrm{Age2}$) with each of its tuples satisfying $\textrm{Age} \lt \textrm{Age2}$. Applying $\Pi_{\textrm{Age}}$ to this relation gives us another relation with a single column ($\textrm{Age}$) containing all original age values from $P$ which are smaller than some other age value in $P$. Therefore, the relation $$ \Pi_{\textrm{Age}}\left[ P \underset{\textrm{Age} \lt \textrm{Age2}}{\bowtie} \left(\rho_{\textrm{Name2/Name}}\rho_{\textrm{Age2/Age}} P\right)\right] $$ contains all ages except the largest one, so removing these values from $\Pi_{\textrm{Age}} P$ yields a relation with a single age value: the largest one.

## Comments

If this does not seem obvious, notice that: $$ \Pi_{\textrm{Name}, \textrm{Age}}\left[ P \underset{\textrm{Age} \lt \textrm{Age2}}{\bowtie} \left(\rho_{\textrm{Name2/Name}}\rho_{\textrm{Age2/Age}} P\right)\right] $$ is a relation containing all people who are younger than someone else. By subtracting this relation from $\Pi_{\textrm{Name}, \textrm{Age}} P$, you obtain a relation containing only the people who are the oldest.

If you have additional attributes on your original relation $P$, change the projections accordingly to select the attributes you wish to keep at the end.