<sup>p</sup> � ffiffi #

<sup>p</sup> <sup>¼</sup> # potential

must be observed, while, within this restriction, pattern values can be permuted.

X 2

φð Þk :

k¼0

<sup>2</sup> be any non-bijective function on n symbols,

<sup>2</sup> ! <sup>F</sup><sup>n</sup>

The image consists of 7 patterns with occurence counts 7 (0000), 2 (0010,0011,0110), and 1 (0001,0100,1001).

In Step I, we thus have to shrink the set of patterns present from 16 to 7 in such a way that (any) 7 patterns are mapped to the same one, and additionally 3 sets of two patterns are joined within each set. Finally, the three remaining input patterns stay on their own.

An exhaustive search over sequences with 4 update rules, starting and ending with a nonbijective one, yields 72 such sequences with the desired shrinking factor, one example is the sequence <sup>s</sup>ð Þ <sup>1</sup>;…;<sup>4</sup> <sup>¼</sup> <sup>0011</sup>; <sup>1101</sup>; <sup>0110</sup>; 0111 of active cells.


#### 5.3.2. Step II

After Step I, the output pattern 2 has frequency 7. Therefore, 2 has to match 0 in Step III, while 0,1,2,3,4,8,C are matched to 5,7,8,9,A,B,E in any order in Step II. Thus, we already have 7! different possibilities for Step II.

Similarly, we can maps 0110 and 1001 to either 1f g ; 4 , 3f g ;C , or 6f g ; F and so forth. Let c run through the occurrence counts, here c ¼ 7, 2 and 1 are actually taken, and let then p cð Þ be the number of output patterns with in-degree c, here pð Þ¼ 7 1, pð Þ¼ 2 pð Þ¼ 1 3. We get

$$\prod\_{c} c!^{p(c)} p(c)! = 7!^1 1! \cdot (2!)^3 3! \cdot (1!)^3 3! = 1451520$$

16. Again, meet-in-the-middle yields a match. Be aware that the necessary effort does not drop

In total, 13 + 4 + 15 = 32 steps are required to compute this multiplication function (Table 5).

In the case of unfair bijective functions, any subset of cells may fire simultaneously, provided

• No cell is active during two consecutive time steps (these two would cancel each other).

Global efficiency—which shortest update sequence generates a certain function/permutation is beyond the scope of this paper (it is dealt with implicitly by brute force in a breadth-first

ð Þt

<sup>k</sup> ¼ 1 ðcell ck active in step tÞ implies that.

<sup>k</sup> could be removed, since each single-location action is an

<sup>k</sup> could be moved to s

ð Þ t�1

<sup>k</sup> , maintaining

ð Þt

bijectivity. Hence ð Þ iii is required. □ Using this lemma, we now can compute an upper bound on the number of bijective functions realizable with t update steps: We define a matrix A with rows and columns indexed by the bijective elementary update rules, here named sð Þ1 , sð Þ2 , …. Set aij ≔ 1, if rule s ið Þ followed by

<sup>2</sup><sup>n</sup> ð Þ �jIm fð Þj ! in general.

Hard, firm, soft … Etherealware: Computing by Temporal Order of Clocking

http://dx.doi.org/10.5772/intechopen.80432

199

ð Þ <sup>16</sup>�<sup>7</sup> ! <sup>≈</sup> <sup>11</sup>! or <sup>2</sup>n!

We define local efficiency of a bijective update sequence by two properties:

• No active cell can be moved to the previous time step.

<sup>k</sup>þ<sup>1</sup> <sup>¼</sup> <sup>0</sup> <sup>ð</sup>both inactiveÞ,

ð Þ t�1

ð Þt

<sup>k</sup>þ<sup>1</sup> were both inactive, the active cell <sup>s</sup>

<sup>k</sup>þ<sup>1</sup> equals 1, active.

from 16! to 7! but only to <sup>16</sup>!

that no adjacent cells are contained in the set.

A sequence of rules is bijective and efficient, if s

ð Þ t�1 <sup>k</sup>�<sup>1</sup> and <sup>s</sup>

ð Þi is required for and then ensures bijectivity.

ð Þ t�1 <sup>k</sup> and s

ð Þt

<sup>k</sup> ¼ 0 ðinactiveÞ, and.

6. Efficiency

manner). Lemma 3.

i. s ð Þt

ii. s

Proof.

If s ð Þ t�1

If s ð Þ t�1 <sup>k</sup>�<sup>1</sup> and <sup>s</sup>

ð Þ t�1

iii. at least one of s

<sup>k</sup>�<sup>1</sup> <sup>¼</sup> 0 and <sup>s</sup>

<sup>k</sup> was active, both s

involution. Thus ð Þii is required.

rule s jð Þ is efficient, aij ≔ 0 otherwise.

Let λð Þ n be the largest eigenvalue of A.

ð Þ t�1

bijective functions consistent with the count distribution to choose from in Step II. One of these is given in the left column of Table 4.

#### 5.3.3. Step III

The outcome of Steps I + II completely fixes the necessary permutation for Step III. However, we do not have to deal with 2<sup>n</sup> values, but only <sup>∣</sup>Imð Þ<sup>f</sup> <sup>∣</sup> are relevant, in our example 7 instead of


Table 4. Steps II, I, and III for multiplication up to 3 � 3 ¼ 9.


Table 5. Steps II, I, and III by output patterns.

16. Again, meet-in-the-middle yields a match. Be aware that the necessary effort does not drop from 16! to 7! but only to <sup>16</sup>! ð Þ <sup>16</sup>�<sup>7</sup> ! <sup>≈</sup> <sup>11</sup>! or <sup>2</sup>n! <sup>2</sup><sup>n</sup> ð Þ �jIm fð Þj ! in general.

In total, 13 + 4 + 15 = 32 steps are required to compute this multiplication function (Table 5).
