Abstract
Analogy-making is at the core of human and artificial intelligence and creativity with applications to such diverse tasks as proving mathematical theorems and building mathematical theories, common sense reasoning, learning, language acquisition, and story telling. This paper introduces from first principles an abstract algebraic framework of analogical proportions of the form ‘a is to b what c is to d’ in the general setting of universal algebra. This enables us to compare mathematical objects possibly across different domains in a uniform way which is crucial for AI-systems. It turns out that our notion of analogical proportions has appealing mathematical properties. As we construct our model from first principles using only elementary concepts of universal algebra, and since our model questions some basic properties of analogical proportions presupposed in the literature, to convince the reader of the plausibility of our model we show that it can be naturally embedded into first-order logic via model-theoretic types and prove from that perspective that analogical proportions are compatible with structure-preserving mappings. This provides conceptual evidence for its applicability. In a broader sense, this paper is a first step towards a theory of analogical reasoning and learning systems with potential applications to fundamental AI-problems like common sense reasoning and computational learning and creativity.
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20 July 2022
Springer Nature’s version of this paper was updated to reflect the Funding information: Open access funding provided by Austrian Science Fund (FWF)
Notes
See Convention 15.
We omit here the superscript from notation, that is, we write s(e1) instead of \(s^{\mathbb {M}}(\mathbf e_{1})\) et cetera.
To be more succinct, we have summarized here two diagrams into one separated by semicolons.
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I thank the anonymous reviewers whose comments and suggestions helped improve and clarify this paper. This work has been supported by the Austrian Science Fund (FWF) project P31063-N35.
Appendix
Appendix
Example 1
Explicating Example 19, we wish to compute all solutions to the analogical equation in the multiplicative algebra
given by
Recall that \(\mathbb {N}_{2}=\{2,3,\ldots \}\) consists of the natural numbers not containing 0 and 1—this will allow us to bound the number of generalizations of a given number by the number of its prime factors. Following Fact 16, we first compute all solutions to the arrow equation in \(\mathbb {M}\) given by
The justification z → 4 is unique to x = 4, which immediately entails that it is a solution to (2). Notice that in the algebra \((\mathbb Q,\cdot )\), the rewrite rule \(z\to \frac z 5\) is a justification of 20 → 4 : ⋅ 30 → d iff d = 6, which yields the solution \(x=\frac {30} 5=6\). Unfortunately, this justification is not available in \(\mathbb {M}\) and to prove that 6 is a solution to 20 → 4 : ⋅ 30 → x, we need to show that either \(Jus_{\mathbb {M}}(20,4)\cup Jus_{\mathbb {M}}(30,6)\) contains only trivial justifications, or that \(Jus_{\mathbb {M}}(20\to 4~{:\!\cdot }~ 30\to 6)\) is a non-empty maximal set of justifications with respect to 6. Here a natural question arises: Are x = 4, 6 the only solutions to the arrow equation (2) in \(\mathbb {M}\)? The answer is ‘no’ as we can show that, unexpectedly, x = 9 is another justifiable solution!
We begin by computing all \(\mathbb {M}\)-generalizations of 20, 4, and 30. For this, it will be convenient to first compute their unique prime decompositions:
These numbers are similar in the following sense:
-
1.
The only difference between 20 and 30 is the second prime factor; both numbers are divisible by 2 and 5.
-
2.
The only difference between 20 and 4 is that 4 is not divisible by 5; both numbers are divisible by 2 and 4.
-
3.
The numbers 4 and 30 have the first prime factor 2 in common.
These similarities are reflected in the computation of their \(\mathbb {M}\)-generalizations:
This yields
We now compute all non-empty sets of justifications \(Jus_{\mathbb {M}}(20\to 4~{:\!\cdot }~ 30\to d)\), for all \(d\in \mathbb {N}_{2}\), with the following procedure (cf. Pseudocode ??): for each \(\mathbb {M}\)-generalization \(s(\mathbf z)\in gen_{\mathbb {M}}(20,30)\) and each witness (e1,e2) satisfyingFootnote 1,Footnote 2
and for each \(t(\mathbf z)\in gen_{\mathbb {M}}(4)\) satisfying
we add s → t to \(Jus_{\mathbb {M}}(20\to 4~{:\!\cdot }~ 30\to t(\mathbf e_{2}))\).
First of all, notice that z → 4 is (only) in \(Jus_{\mathbb {M}}(20\to 4~{:\!\cdot }~ 30\to 4)\), which immediately yields the solution x = 4 to (2) (cf. Remark 25).
-
1.
s(z) = 10z: The only witnesses \(e_{1},e_{2}\in \mathbb {M}\) satisfying (4) are e1 = 2 and e2 = 3, and the only \(t_{1},t_{2}\in gen_{\mathbb {M}}(4)\) satisfying 4 = t(2) are t1(z) = 2z and t2(z) = z2. We therefore have \(10z\to 2z\in Jus_{\mathbb {M}}(20\to 4~{:\!\cdot }~ 30\to t_{1}(3))=Jus_{\mathbb {M}}(20\to 4~{:\!\cdot }~ 30\to 6)\) and \(10z\to z^{2}\in Jus_{\mathbb {M}}(20\to 4~{:\!\cdot }~ 30\to t_{2}(3))=Jus_{\mathbb {M}}(20\to 4~{:\!\cdot }~ 30\to 9)\). This can be depicted as follows:
Since 3 is a unique witness satisfying s(3) = 30, in case 10z → 2z is a justification of 20 → 4 : ⋅ 30 → d, we must have d = 2 ⋅ 3 = 6, which shows that \(Jus_{\mathbb {M}}(20\to 4~{:\!\cdot }~ 30\to 6)\) is non-empty and subset maximal with respect to 6 (see the Uniqueness Lemma 23). We have thus derived
$$ \begin{array}{@{}rcl@{}} \mathbb{M}\models 20\to 4~{:\!\cdot}~ 30\to 6. \end{array} $$Notice that the justification 10z → 2z emulates the justification \(z\to \frac z 5\) from above. The same line of reasoning with 10z → z2 instead of 10z → 2z shows
$$ \begin{array}{@{}rcl@{}} \mathbb{M}\models 20\to 4~{:\!\cdot}~ 30\to 9. \end{array} $$ -
2.
s(z1,z2) = 5z1z2: The only \(\mathbf e_{1}\in \mathbb {M}^{2}\) satisfying 20 = s(e1) is given by (2,2). There are two \(\mathbf e_{2}^{(1)},\mathbf e_{2}^{(2)}\in \mathbb {M}^{2}\) satisfying \(30=s(\mathbf e_{2}^{(1)})=s(\mathbf e_{2}^{(2)})\) given by \(\mathbf e_{2}^{(1)}=(2,3)\) and \(\mathbf e_{2}^{(2)}=(3,2)\). Moreover, we have five \(t_{1}(z_{1},z_{2}),t_{2}(z_{1},z_{2}),t_{3}(z_{1},z_{2}),t_{4}(z_{1},z_{2}),t_{5}(z_{1},z_{2})\in gen_{\mathbb {M}}(4)\) satisfying 4 = ti(2, 2), 1 ≤ i ≤ 5, given by
$$ \begin{array}{@{}rcl@{}} &&t_{1}(z_{1},z_{2})=2z_{1},\quad t_{2}(z_{1},z_{2})=2z_{2},\quad t_{3}(z_{1},z_{2})={z_{1}^{2}},\quad t_{4}(z_{1},z_{2})={z_{2}^{2}},\quad\\ &&t_{5}(z_{1},z_{2})=z_{1}z_{2}. \end{array} $$We therefore have the following justifications:Footnote 3
Notice the symmetries between the first and second, and between the third and fourth diagrams above; in what follows, we will only mention one of multiple symmetric cases.
-
3.
s(z1,z2) = 2z1z2: In the remaining cases, we present only the self-explaining diagrams enumerating all valid arrow proportions:
We see here that 20 → 4 : ⋅ 30 → 10 has the justification 2z1z2 → 2z1 in \(\mathbb {M}\), which is also a justification of 20 → 4 : ⋅ 30 → 6; on the other hand, we have seen above that 5z1z2 → 2z1 and 5z1z2 → 2z2 are justifications of 20 → 4 : ⋅ 30 → 6, but not of 20 → 4 : ⋅ 30 → 10, which shows that, up to this point, \(Jus_{\mathbb {M}}(20\to 4~{:\!\cdot }~ 30\to 10)\) is strictly contained in \(Jus_{\mathbb {M}}(20\to 4~{:\!\cdot }~ 30\to 6)\) and therefore not subset maximal with respect to 10. The case t(z1,z2) = 2z2 is analogous. We further have the arrow proportion
The only remaining case \(t(z_{1},z_{2})={z_{2}^{2}}\) is analogous.
-
4.
s(z1,z2,z3) = z1z2z3:
Here it is sufficient to analyze the three cases (2, 3, 5); (2, 5, 3); (3, 2, 5), where the second argument varies, instead of all six permutations of (2, 3, 5) since for 2z2 only the second argument is relevant. The cases t(z1,z2,z3) = 2z1 and t(z1,z2,z3) = 2z3 are analogous. We further have the arrow proportions
The cases \(t(z_{1},z_{2},z_{3})={z_{1}^{2}}\) and \(t(z_{1},z_{2},z_{3})={z_{3}^{2}}\) are analogous.
-
5.
s(z1,z2) = z1z2:
The case t(z1,z2) = z2 is analogous. We further have the arrow proportion
The case \(t(z_{1},z_{2})={z_{2}^{2}}\) is analogous. The next case is:
The case t(z1,z2) = 2z2 is analogous.
-
6.
s(z) = 5z:
Observe that transforming 20 into 4 means removing the prime factor 5 from 20 in (3)—transforming 30 ‘in the same way’ therefore means here to remove the prime factor 5 from 30 in (3), yielding the solution x = 6. This shows that 5z → z emulates the justification \(z\to \frac z 5\) mentioned at the beginning of the example not available in \(\mathbb {M}\).
-
7.
s(z) = 2z: There is a unique e1 = 10 with 20 = s(e1)—there is no \(t(z)\in gen_{\mathbb {M}}(4)\) satisfying 4 = t(10). This shows that there is no justification 2z → t(z) of 20 → 4 : ⋅ 30 → d, for any \(d\in \mathbb {N}_{2}\).
-
8.
s(z) = z: There is a unique e1 = 20 with 20 = s(e1)—there is no \(t(z)\in gen_{\mathbb {M}}(4)\) satisfying 4 = t(20). This shows that there is no justification z → t(z) of 20 → 4 : ⋅ 30 → d, for any \(d\in \mathbb {N}_{2}\).
To summarize, we have the following non-empty sets of justifications:
We see that only x = 4, 6, 9 have maximal non-empty sets of justifications, proving
The solution x = 4 follows easily with z → 4 and Remark 25. The solution x = 6 is intuitive and has, among others, the natural justifications 10z → 2z and 5z → z resembling \(z\to \frac z 5\) in \((\mathbb Q,\cdot )\). Lastly, the solution x = 9 has the unique justification 10z → z2.
We now want to compute all solutions in \(\mathbb {M}\) to the directed analogical equation
Recall from Fact 16 that \(d\in \mathbb {M}\) is a solution to 20 : 4 : ⋅ 30 : x iff d is a solution to 20 → 4 : ⋅ 30 → x and 30 is a solution to 4 → 20 : ⋅ d → x in \(\mathbb {M}\). We already know that 4, 6, and 9 are the only solutions to 20 → 4 : ⋅ 30 → x in \(\mathbb {M}\). We check whether 4, 6, and 9 are solutions to (5) separately:
-
1.
There can be no justification of 4 → 20 : ⋅ 4 → 30 in \(\mathbb {M}\)—for example, given the \(\mathbb {M}\)-generalization 2z of 4 in \(\mathbb {M}\), there is no \(\mathbb {M}\)-generalization t of 20 and 30 such that t(2) = 20 and t(2) = 30, which can be depicted as follows:
This shows
$$ \begin{array}{@{}rcl@{}} \emptyset=Jus_{\mathbb{M}}(4\to 20~{:\!\cdot}~ 4\to 30)\subsetneq Jus_{\mathbb{M}}(4\to 20~{:\!\cdot}~ 4\to 20)=\{4\to 20,\ldots\}, \end{array} $$which means that 30 is not a solution to 4 → 20 : ⋅ 4 → x and, hence, 4 is not a solution to (5):
$$ \begin{array}{@{}rcl@{}} \mathbb{M}\not\models 20:4~{:\!\cdot}~ 30:4. \end{array} $$ -
2.
On the other hand, the justifications
and
show that 30 is a solution to 4 → 20 : ⋅ 6 → x and 4 → 20 : ⋅ 9 → x in \(\mathbb {M}\) (Uniqueness Lemma 23), thus proving
$$ \begin{array}{@{}rcl@{}} \mathbb{M}\models 20:4~{:\!\cdot}~ 30:6 \quad\text{and}\quad \mathbb{M}\models 20:4~{:\!\cdot}~ 30:9. \end{array} $$This shows that 6 and 9 are the only solutions to 20 : 4 : ⋅ 30 : x in \(\mathbb {M}\), that is,
$$ \begin{array}{@{}rcl@{}} Sol_{\mathbb{M}}(20:4~{:\!\cdot}~ 30:x)=\{6,9\}. \end{array} $$
It remains to check whether 6 and 9 are solutions to 20 : 4 :: 30 : x in (1):
-
1.
We first verify that 6 is a solution by showing the remaining relation
$$ \begin{array}{@{}rcl@{}} \mathbb{M}\models 30:6~{:\!\cdot}~ 20:4. \end{array} $$For this, we prove
$$ \begin{array}{@{}rcl@{}} \mathbb{M}\models 30\to 6~{:\!\cdot}~ 20\to 4 \quad\text{and}\quad \mathbb{M}\models 6\to 30~{:\!\cdot}~ 4\to 20. \end{array} $$The first relation is justified by 5z → z, that is, 5z → z is a justification of 30 → 6 : ⋅ 20 → d in \(\mathbb {M}\), \(d\in \mathbb {M}\), iff there are \(e_{1},e_{2}\in \mathbb {M}\) such that
$$ \begin{array}{@{}rcl@{}} 30=5e_{1} \quad\text{and}\quad 6=e_{1} \quad\text{and}\quad 20=5e_{2} \quad\text{and}\quad d=e_{2}, \end{array} $$which is equivalent to d = 4. An analogous argument using the justification z → 5z proves the second relation.
-
2.
Finally, we show that 9 is a solution by showing the remaining relation
$$ \begin{array}{@{}rcl@{}} \mathbb{M}\models 30:9~{:\!\cdot}~ 20:4. \end{array} $$For this, we prove
$$ \begin{array}{@{}rcl@{}} \mathbb{M}\models 30\to 9~{:\!\cdot}~ 20\to 4 \quad\text{and}\quad \mathbb{M}\models 9\to 30~{:\!\cdot}~ 4\to 20. \end{array} $$The first relation is justified by 10z → z2 and the second by z2 → 10z by a similar argument as for 6 in the previous item.
We have thus shown
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Antić, C. Analogical proportions. Ann Math Artif Intell 90, 595–644 (2022). https://doi.org/10.1007/s10472-022-09798-y
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DOI: https://doi.org/10.1007/s10472-022-09798-y