Hybrid probabilistic logic programs extends probabilistic logic programs by adding the possibility to manage continuous random variables. Despite the maturity of the field, a semantics that unifies discrete and continuous random variables and function symbols was still missing. In this paper, we summarize the main concepts behind a new proposed semantics for hybrid probabilistic logic programs with function symbols.
1. Contribution
This paper is a technical summary of [
pick(0,hearts) :- \+ spades(0), \+ clubs(0).
This program describes a game of card with only 1 round (identified with 0) and 3 cards. A player draws a card that can be either of spades, clubs, or hearts. These have all the same probability. Note that, to describe three possible cards we only need 2 (probabilistic) facts, since the third can be represented with the negation of the other two (see line 5). The predicate pick/2 describes the three possible outcomes.
The probability of a query asked on the program of Example 1 is easy to compute. For example, the probability of pick(0,hearts) is (1 − 1/3) · (1 − 1/2) = 1/3. However, we can make the program more interesting by adding multiple rounds. To do this, we introduce a function symbol s/1 to the program of Example 1.
Example 2. Cards with multiple rounds. We extend Example 1 by considering multiple rounds. We introduce the function symbol s/1 to indicate a round and with s(X) we indicate the round after the round X. So, starting from 0 (first round), we have s(0) for the second round, s(s(0)) for the third round and so on. We introduce an additional rule: the game stops when the player picks a card of hearts. The program thus became:
A possible question could be: “what is the probability that the player picks at least one time spades?” This probability can be computed by asking the query ___.
To compute the probability of the query ___ from Example 2, we need to consider pairwise incompatible covering set of explanations [4]. If we replace spades with 1 and clubs with 2 and we use 0 and 1 to indicate respectively not selected and selected, we get a pairwise incompatible covering set of explanations = { 0, 1, . . .} with 0 = {(1, {/0}, 1)} 1 = {(1, {/0}, 0), (2, {/0}, 1), (1, {/(0)}, 1)}
. . . = {(1, {/0}, 0), (2, {/0}, 1), . . . , (1, {/− 1(0)}, 0),
(2, {/− 1(0)}, 1), (1, {/(0)}, 1)} . . .
From here, we can compute the probability of the query = ___ as () = 1 1 (︂ 2 1 )︂ 3 + 3 · 3 · 2 since we have a sum of a geometric series.
We can even further extend the previous example by also considering continuous random variables. To represent these, we use the syntax
: where is an atom and is a special atom that denotes its probability density. Example 3. Cards multiple rounds and continuous random variables. We extend Example 2 by adding another rule: the player still draws a card but, in addition, he/she need to spin a wheel. If the axis of the wheel is between 0 and 180 degrees the game stops. This scenario can be encoded with:
In line 1 we have a continuous probabilistic fact angle/2 where its argument X follows a uniform distribution between 0 and 360.
We may be still interested in computing the probability of the query ___ from Example 3. Diferently from Example 2, we now need to consider a mutually disjoint covering set of worlds . First, we partition the random variables in two sets: a countable set X of continuous random variables (identified by 0, (0), . . . , where each element has a range [0, 360]) and a countable set Y of discrete random variables (where each element can be true or false). The set = 0 ∪ 1 . . . is such that: 0 = {(X, Y) | X = (0, 1, . . .), Y = (0, 0, 1, 1, . . .),
0 ∈]180, 360], 0 = 1} 1 = {(X, Y) | X = (0, 1, . . .), Y = (0, 0, 1, 1, . . .),
0 ∈]180, 360], 0 = 0, 0 = 1, 1 ∈]180, 360], 1 = 1} . . .
In other words: for 0, spades was selected at round 0 (0 = 1) and the wheel (0) in the same round was in the range ]180, 360]; for 1, spades was not selected at round 0 (0 = 0), clubs was selected at round 0 (0 = 1), the wheel (0) was in the range ]180, 360] at round 0, spades was selected at round (0) (1 = 1) and the wheel (1) was in the range ]180, 360] at round (0), and so on.
The probability for each can be computed by multiplying the discrete and continuous components. For 0 (the process is similar for all the ∈ ) we have: (0) = =
Y({(0, 0, 1, 1, . . .) | 0 = 1}) X where 13 is the contribution of the discrete random variable (spades) and 3610 is the contribution of the continuous one (angle). By considering all the values obtained for all the , we get 13 · 21 · ∑︀∞=0( 23 · 12 · 21 ) = 16 · ∑︀∞=0( 16 ) = 16 · 56 = 15 as probability for the query at_least_once_spades.
In [