Example.

Hence byNine ¼(19)(3B)(5D)(7F) = [(08)(2A)][(08)(19)(2A)(3B)(5D)(7F)], where the first bracket requires 23 steps, and the second is the elementary rule 1000 (only the leftmost cell is active). The difference between the two realizations (sequences sð Þ <sup>1</sup>;…;<sup>24</sup> in hex), where the outer parentheses are inverses of each other and the inner part is self-inverse, and

0110 0010 1100 … 1001 0011 0001 0x62C478E0BF5DA931

Active cells FED...210 ! 62C...931 some patterns in binary All values F...0 in hexadecimal 1111 1110 1101 … 0010 0001 0000 0xFEDCBA9876543210 1110 1111 1101 … 0011 0000 0001 0xEFDCAB9867452301 1100 0101 0111 … 1011 1010 1001 0xC57E03124D6F8BA9 1000 0000 0110 … 1110 1111 1101 0x806B5243197ACEFD 0010 1010 0100 … 1100 0101 0111 0x2A43F86B91D0EC57 0011 1111 0001 … 1000 0000 0110 0x3F12AC7ED495B806 1011 0111 1001 … 0000 1000 0110 0xB79A2CFE541D3086 1111 0111 1101 … 0100 1100 0110 0xF7DE28BA105934C6 0111 1111 0101 … 0100 1100 0110 0x7F5EA03298D1B4C6 0110 1110 0100 … 0101 1100 0111 0x6E4FB12398D0A5C7 0110 1110 0100 … 1101 1100 1111 0x6E4739AB10582DCF 0110 1010 0000 … 1001 1000 1011 0x6A073DEF541C298B 0100 0000 1010 … 0001 0010 0011 0x40ADB7C5F69E8123 0000 0100 1110 … 0101 0010 0011 0x04E9F781B6DAC523 1010 0110 1100 … 1111 1000 1011 0xA6C15D293470EF8B 1011 0111 1100 … 1110 1000 1010 0xB7C04D392561FE8A 0011 1111 1100 … 1110 0000 0010 0x3FC845B1AD697E02 0010 1010 1000 … 1011 0101 0011 0x2A8C10E4F97D6B53 1010 0010 0000 … 0011 1101 1011 0xA20C98E471F563DB 1110 0010 0100 … 0011 1001 1111 0xE248DCA075B1639F 1110 1010 0100 … 1011 0001 0111 0xEA405C28FD396B17 1010 1110 0000 … 1111 0101 0111 0xAE04182CB93D6F57 0000 1100 1010 … 0101 1111 1101 0x0CA6928E31B745FD 0100 1000 1110 … 0001 1011 1001 0x48E6D2CA35F701B9

194 From Natural to Artificial Intelligence - Algorithms and Applications

their difference:

1.: (5A581248)1(A5A5A)4(842185A5)8

2.: (48181A52)1(A5A5A)4(25A18184)8

Diff 124–81A——-A18–421-

Table 3. Exponentiation x ↦ 3<sup>x</sup> in F17.

We start with the effect of AS <sup>i</sup> = "�" for the locally bijective ECA-57. We consider four adjacent cells and the effect of � between the two middle cells, which are thus updated simultaneously, s ¼ 0110.


We obtain the patterns 0011 and 0101 twice, while missing 0001 and 0111. Hence the image is smaller than the full 24 by 2 or by a factor 7/8 in general.

We have the following in-degree distribution (loss pattern 4 of [[10], Table 8]):


Twelve patterns map bijectively (1,1) to 12 patterns, 4 patterns map 2:1 to 2 patterns, while 2 patterns of the range are not met at all (0,1).

We have #(domain) = in-degree�#(range) for every in-degree. Also, the overall sum is

$$\sum \# (\text{domain}) = \sum \text{ in-degree} \times \# (\text{range}) = 2^n.$$

Any additional simultaneous firing may increase the losses. Additional in-degree distributions (loss patterns) for ECAs and QCAs may be found in ([10], Table 8).

We restate Theorem 3 (i) from [14]:

#### Theorem 2.

We assume an ECA that generates at least the alternating group A2<sup>n</sup> , when using temporal rules from f g <; > <sup>n</sup> \ <sup>&</sup>lt;n, <sup>&</sup>gt;<sup>n</sup> f g. Let <sup>f</sup> : <sup>F</sup><sup>n</sup> <sup>2</sup> ! <sup>F</sup><sup>n</sup> <sup>2</sup> be any non-bijective function on n symbols, where we require n ≥ 4 for ECA.

Let # wð Þ¼ ∣f g vjf vð Þ¼ w ∣ be the number of configurations v leading to configuration w. Then f is clocking-computable by an ECA with in-degree distribution φð Þ¼ 0; 1; 2 ð Þ 12; 2; 2 or ð Þ 24; 4; 4 , if and only if

$$\sum\_{w \in \mathbb{F}\_2^n} \lfloor \#(w)/2 \rfloor \ge \wp(2) \cdot 2^n / \sum\_{k=0}^2 \wp(k) \cdot 1$$

Proof. See [14], Theorem 3. □

#### 5.2. Algorithm for non-bijective functions

The computation of a non-bijective function can be decomposed into 3 steps:

Step I. We start in the middle: Shrink the 2<sup>n</sup> singletons of the domain to the desired distribution on the range. Find a sequence of elementary steps, necessarily including non-bijective ones, which generates the same distribution of counts in the image space as for the original function. Usually there are more than one of these sequences with the same complexity. Values/patterns as such do not yet play a role; therefore the requirements are usually easy to meet in a variety of ways, and the sequence of necessary elementary steps is short.

Step II. Bijectively map the input of the original function to the input of any of the results of Step I in such a way that the desired image counts are matched. Only the occurence counts must be observed, while, within this restriction, pattern values can be permuted.

Step III. Bijectively map the output from Step I to the output of the desired function, considering the flow within Steps II and I in that order. Now all values matter and cannot be permuted. However, this step takes place on a subset of size <sup>∣</sup>Im fð Þ<sup>∣</sup> instead of 2<sup>n</sup>.

In Steps II and III we use the meet-in-the-middle approach, applying update rules starting both from the input (identity) and the desired function values to take advantage of the birthday paradox: In this way, 2 � ffiffi # <sup>p</sup> patterns are sufficient to generate ffiffi # <sup>p</sup> � ffiffi # <sup>p</sup> <sup>¼</sup> # potential matches in the middle (here # ≈ ∣A2<sup>n</sup> ∣), a considerable saving in both space and time.

5.3. Example of a non-bijective function

output, lies within the set 0f g ; 1; 2; 3; 4; 6; 9 .

5.3.1. Step I

5.3.2. Step II

different possibilities for Step II.

(0001,0100,1001).

multiplication up to 3 � 3 ¼ 9 on the torus of size n ¼ 4

The 4-bit input is interpreted as a pair of numbers from the set 0f g ; 1; 2; 3 , whose product, the

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

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

197

In Out In Out In Out In Out 00.00 0000 01.00 0000 10.00 0000 11.00 0000 00.01 0000 01.01 0001 10.01 0010 11.01 0011 00.10 0000 01.10 0010 10.10 0100 11.10 0110 00.11 0000 01.11 0011 10.11 0110 11.11 1001

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

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

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

0123.4567.89AB.CDEF <sup>0011</sup> 160000 3012.7654.A99A.EFDC <sup>1101</sup> 142000 ADCB.E781.7557.B218 <sup>0110</sup> 5 4100 CBAF.85E5.5335.F05E <sup>0111</sup> 5 3010 AED8.E292.2222.8729 3 3001

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!

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

1 2357

within each set. Finally, the three remaining input patterns stay on their own.

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.

All patterns in hex Active cells Occurrence counts
