Geometric Algebra Model of Distributed Representations 7 Table 1 Contents of the clean-up memory used in tests described throughout this paper.

| Number of Contents blades 1 A total of 42 atomic objects: 19 fillers, 7 single-feature roles and 8 double-feature roles 2 (1a) biteagt ∗Fido+biteob j ∗Pat (2a) f leeagt ∗Pat + f leeob j ∗Fido 3 (3a) seeagt ∗ John+seeob j∗(1a) (PSmith) name ∗Pat +sex ∗male+age ∗ 66 4 (1b) biteagt ∗Fido+biteob j ∗PSmith (2c) f leeagt ∗PSmith+ f leeob j ∗Fido (4a) causeagt∗(1a)+causeob j∗(2a) 5 (3b) seeagt ∗ John+seeob j∗(1b) (5a) seeagt ∗ John+seeob j∗(4a) 6 (4c) causeagt∗(1b)+causeob j∗(2a) 7 (DogFido) class ∗ animal +type ∗ dog+taste ∗ chickenlike +name ∗Fido+age ∗ 7+sex ∗male+occupation ∗ pet 8 (1c) biteagt ∗DogFido+biteob j ∗Pat (2b) f leeagt ∗Pat + f leeob j ∗DogFido (4b) causeagt∗(1b)+causeob j∗(2c) 9 (3c) seeagt ∗ John+seeob j∗(1c) (5b) seeagt ∗ John+seeob j∗(4b) 10 (1d) biteagt ∗DogFido+biteob j ∗PSmith (2d) f leeagt ∗PSmith+ f leeob j ∗DogFido 11 (3d) seeagt ∗ John+seeob j∗(1d) (∗) Each sentence carries a number (e.g. "(3a)") to make further equations more readable.   |
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## 3.1 Right-Hand-Side Questions

In the previous Section we commented on the use of reversions and the choice of the side of a statement that a question should be asked on. The argument for righthand-side (generally: fixed-hand-side) questions without a reversion was that rules of decomposition of a statement should be clear and unchangeable. However, the use of right-hand-side questions poses a problem best described by the following example. Let the clean-up memory contain seven roles and fillers

seeagt = c00101, John = c00101, seeob j = c01010, Pat = c10000, biteagt = c10110, Fido = c10001, biteob j = c00001, (33)
$$(34)^{\frac{1}{2}}$$
$$\mathrm{T}^{1}=\mathrm{T}^{10001},$$
$$(35)$$
and two sentences mentioned in Table 1

(1a) biteagt ∗Fido+biteob j ∗Pat = c00111 −c10001, (34)
(3a) $see_{agt}*John+see_{obj}*(1\text{a})=-c_{00000}-c_{01101}-c_{11011}$.  
Let us now ask a question (3a) ] see*ob j* = (3a) ∗ see*ob j*. The decoded answer