**2.2 The algorithm for finding the minimum number of transitions in a coverage state**

• *S y*<sup>1</sup>

**Figure 3.**

• *S y*<sup>2</sup>

• *S y*<sup>3</sup>

*T*<sup>∗</sup> *y*<sup>3</sup> � � � � �

> *t* 0 <sup>2</sup> ¼ *t*3.

� � <sup>¼</sup> <sup>1</sup> *S y*<sup>4</sup>

*The petri net reachable tree.*

� � <sup>¼</sup> <sup>2</sup> *S y*<sup>5</sup>

� � <sup>¼</sup> <sup>1</sup> *S y*<sup>6</sup>

� � <sup>¼</sup> <sup>2</sup>

� � <sup>¼</sup> <sup>2</sup>

� � <sup>¼</sup> <sup>3</sup>

use the algorithm. Let it be a covering peak.

example *t*2, *t*<sup>3</sup> let us mark *t*

2.For each chosen transitions, *t*

• If for the same *t*

*ai* ¼ *μ*½ � *x <sup>j</sup>* in which case *t*

then *ai* ¼ max *μ*½ � *x <sup>j</sup>*

0

will correspond *ai* with *σ* and *ai* ¼ *μ*ð Þ *x <sup>j</sup>*

case there is no related position for *t*

0

*<sup>i</sup>* transition for 1≤*j*≤j j *P* in the way that *δ μ y*<sup>0</sup> ½ �, *t*

following way:

• If for *t* 0

Moreover, for *t*

*t* 0 *<sup>i</sup> σ* ¼ *t* 0 *<sup>i</sup>*<sup>1</sup> … *t* 0 *ik t* 0 *ik* ¼ *t* 0 *i*

no *t* 0

**119**

We found out that in minimum number, *S y*<sup>1</sup>

*The Possibilities of Modeling Petri Nets and Their Extensions*

*DOI: http://dx.doi.org/10.5772/intechopen.90275*

0

0

0

, *μ*½ � *x <sup>k</sup>* n o. � � <sup>¼</sup> *S y*<sup>3</sup>

*<sup>i</sup>* <sup>¼</sup> *tj*,1<sup>≤</sup> *<sup>i</sup>*<sup>≤</sup> *<sup>T</sup>*<sup>∗</sup> j j, and *tj* <sup>∈</sup>*T*<sup>∗</sup> . We get *<sup>t</sup>*

*<sup>i</sup>* is corresponded with *ai* numbers in the

*<sup>i</sup>* transition corresponds with *Pj* position.

*<sup>i</sup>* transition ∃1≤*k* 6¼ *j*≤j j *P* in the way that *δ μ y*<sup>0</sup> ½ �, *t*

*<sup>i</sup>* transition we will correspond *Pj* and *Pk* positions. If instead of

0 *i* � �

� � for <sup>∃</sup>1<sup>≤</sup> *<sup>j</sup>*≤j j *<sup>P</sup>* in the way that *δ μ <sup>y</sup>*<sup>0</sup> ð Þ ½ �, *<sup>σ</sup> <sup>j</sup>* <sup>¼</sup> *<sup>ω</sup>*, *<sup>y</sup>*<sup>0</sup> <sup>∈</sup> *<sup>G</sup>*, then we

� ¼ 3 ) need to take a *y*<sup>1</sup> peak. Choosing the appropriate peak coverage, we

1.We choose the path connecting the root of the tree with*y* and *T* <sup>∗</sup> ; for our

*<sup>i</sup>* transition ∃1≤*j*≤j j *P* in the way that*δ μ y*<sup>0</sup> ½ �, *t*

.

0 *i* .

In this case, we will correspond *σ* with *Pj* position. In the opposite case, if there is

� �, *<sup>T</sup>*<sup>∗</sup> *<sup>y</sup>*<sup>1</sup> � � � � �

0 *i* � � � ¼ 2, and

0

*<sup>j</sup>* ¼ *ω*, *y*<sup>0</sup> ∈ *G* then

*<sup>j</sup>* ¼ *ω*, then *ai* ¼ 1, in which

0 *i* � �

*<sup>k</sup>* ¼ *ω*,

<sup>1</sup> ¼ *t*2, and

Consider the Petri net in **Figure 1** and the corresponding reachable tree TT (see **Figure 3**).

We note the set of states in Petri nets with *P*. Let T\* denote the succession of transitions from the root *TT* to *y*, the transition sequence with *G* the succession of the peaks in T\*.

Consider μ½ �¼ x ð Þ 0, 1, 15, 13 state. Let us find the *y* peak of this reachable tree for which the following inequality holds: *μ*½ � *y* ≥*μ*½ � *x* .

Assume that such peaks are *y*1, … , *ym*. Let us choose one peak among the peaks on which we will use the algorithm.

For every yi peak, we profile μ yi � �.

Suppose in μ yi � � there is *<sup>ω</sup>* in *<sup>μ</sup> yi* � � *i*1 , … , *μ yi* � � *ik*. For each μ yi � � we count <sup>S</sup> <sup>¼</sup> <sup>P</sup>ik zjð Þt , where.

j¼i1 zjðÞ¼ t # Pj,I tð Þ<sup>k</sup> � �, ∀tk ∈T<sup>∗</sup> :

We take the yi for which the S is the minimum. If for any peak, these numbers are equal, then we take the yi in which T<sup>∗</sup> height is the minimum.

For example, μ½ � x , we will cover the following peak:

$$\bullet \ y\_1 \,\,\mu\left[\boldsymbol{\uprho}\_1\right] = (\mathbf{1}, \mathbf{1}, \boldsymbol{\alpha}, \boldsymbol{\alpha}), \,\, T^\* = \{t\_2, t\_3\}.$$

$$\bullet \ y\_2 \,\,\mu \left[ \mathcal{y}\_2 \right] = (\mathbf{0}, \mathbf{2}, \boldsymbol{\alpha}, \boldsymbol{\alpha}), \,\, T^\* = \{t\_2, t\_3, t\_1\}.$$


*The Possibilities of Modeling Petri Nets and Their Extensions DOI: http://dx.doi.org/10.5772/intechopen.90275*

**Figure 3.** *The petri net reachable tree.*

same state that has already been developed, then we will call the new peak repeated

repeat, and the states grow. Let us introduce the idea of infinite much as ω:

Let us give several definitions, which will be used in the entire work. **Definition 2**. A peak is called a boundary if it is in a processing state. **Definition 3**. The peak is called a terminal if it does not contain a subtree. **Definition 4**. The peak is called internal if it has already been processed. **Definition 5**. The boundary peak is repeated if there is an internal peak with the

If there is a path like / \*\* / in the tree, then the path through the second peak may

ω≥ω, ω þ a ¼ ω, and ω � a ¼ ω, where a ¼ const: for example, instead of (5,⋯), we will write ð*ω*, ⋯). In this case the tree will become as finite, and we will have loss

A description of the structure of the reachability tree algorithm can be seen in an

With the help of this algorithm, we will build (as in **Figure 1**) the Petri net

**2.2 The algorithm for finding the minimum number of transitions in a coverage**

Consider the Petri net in **Figure 1** and the corresponding reachable tree TT (see

Consider μ½ �¼ x ð Þ 0, 1, 15, 13 state. Let us find the *y* peak of this reachable tree for

Assume that such peaks are *y*1, … , *ym*. Let us choose one peak among the peaks

, … , *μ yi* � �

We take the yi for which the S is the minimum. If for any peak, these numbers

*ik*. For each μ yi

� � we count

� �.

� � *i*1

are equal, then we take the yi in which T<sup>∗</sup> height is the minimum.

For example, μ½ � x , we will cover the following peak:

� � <sup>¼</sup> ð Þ 1, 1,*ω*, *<sup>ω</sup>* , *<sup>T</sup>*<sup>∗</sup> <sup>¼</sup> f g *<sup>t</sup>*2, *<sup>t</sup>*<sup>3</sup> .

� � <sup>¼</sup> ð Þ 0, 2, *<sup>ω</sup>*,*<sup>ω</sup>* , *<sup>T</sup>* <sup>∗</sup> <sup>¼</sup> f g *<sup>t</sup>*2, *<sup>t</sup>*3, *<sup>t</sup>*<sup>1</sup> .

� � <sup>¼</sup> ð Þ 1, 1,*ω*, *<sup>ω</sup>* , *<sup>T</sup>* <sup>∗</sup> <sup>¼</sup> f g *<sup>t</sup>*2, *<sup>t</sup>*3, *<sup>t</sup>*<sup>2</sup> .

� � <sup>¼</sup> ð Þ 1, 1,*ω*,*<sup>ω</sup>* , *<sup>T</sup>* <sup>∗</sup> <sup>¼</sup> f g *<sup>t</sup>*2, *<sup>t</sup>*3, *<sup>t</sup>*<sup>3</sup> .

� � <sup>¼</sup> ð Þ 1, 1,*ω*, *<sup>ω</sup>* , *<sup>T</sup>*<sup>∗</sup> <sup>¼</sup> f g *<sup>t</sup>*3, *<sup>t</sup>*2, *<sup>t</sup>*<sup>3</sup> .

� � <sup>¼</sup> ð Þ 0, 2, *<sup>ω</sup>*,*<sup>ω</sup>* , *<sup>T</sup>*<sup>∗</sup> <sup>¼</sup> f g *<sup>t</sup>*3, *<sup>t</sup>*1, *<sup>t</sup>*2, *<sup>t</sup>*<sup>3</sup> .

We note the set of states in Petri nets with *P*. Let T\* denote the succession of transitions from the root *TT* to *y*, the transition sequence with *G* the succession of

and will not develop it.

*Numerical Modeling and Computer Simulation*

of information [2].

earlier published article [8].

reachable tree (see **Figure 3**).

which the following inequality holds: *μ*½ � *y* ≥*μ*½ � *x* .

� � there is *<sup>ω</sup>* in *<sup>μ</sup> yi*

� �, ∀tk ∈T<sup>∗</sup> :

on which we will use the algorithm. For every yi peak, we profile μ yi

zjð Þt , where.

same state.

**state**

**Figure 3**).

<sup>S</sup> <sup>¼</sup> <sup>P</sup>ik j¼i1

• *y*<sup>1</sup> *μ y*<sup>1</sup>

• *y*<sup>2</sup> *μ y*<sup>2</sup>

• *y*<sup>3</sup> *μ y*<sup>3</sup>

• *y*<sup>4</sup> *μ y*<sup>4</sup>

• *y*<sup>5</sup> *μ y*<sup>5</sup>

• *y*<sup>6</sup> *μ y*<sup>6</sup>

**118**

the peaks in T\*.

Suppose in μ yi

zjðÞ¼ t # Pj,I tð Þ<sup>k</sup>


We found out that in minimum number, *S y*<sup>1</sup> � � <sup>¼</sup> *S y*<sup>3</sup> � �, *<sup>T</sup>*<sup>∗</sup> *<sup>y</sup>*<sup>1</sup> � � � � � � ¼ 2, and *T*<sup>∗</sup> *y*<sup>3</sup> � � � � � � ¼ 3 ) need to take a *y*<sup>1</sup> peak. Choosing the appropriate peak coverage, we use the algorithm. Let it be a covering peak.

	- If for *t* 0 *<sup>i</sup>* transition ∃1≤*j*≤j j *P* in the way that*δ μ y*<sup>0</sup> ½ �, *t* 0 *i* � � *<sup>j</sup>* ¼ *ω*, *y*<sup>0</sup> ∈ *G* then *ai* ¼ *μ*½ � *x <sup>j</sup>* in which case *t* 0 *<sup>i</sup>* transition corresponds with *Pj* position.
	- If for the same *t* 0 *<sup>i</sup>* transition ∃1≤*k* 6¼ *j*≤j j *P* in the way that *δ μ y*<sup>0</sup> ½ �, *t* 0 *i* � � *<sup>k</sup>* ¼ *ω*, then *ai* ¼ max *μ*½ � *x <sup>j</sup>* , *μ*½ � *x <sup>k</sup>* n o.

Moreover, for *t* 0 *<sup>i</sup>* transition we will correspond *Pj* and *Pk* positions. If instead of *t* 0 *<sup>i</sup> σ* ¼ *t* 0 *<sup>i</sup>*<sup>1</sup> … *t* 0 *ik t* 0 *ik* ¼ *t* 0 *i* � � for <sup>∃</sup>1<sup>≤</sup> *<sup>j</sup>*≤j j *<sup>P</sup>* in the way that *δ μ <sup>y</sup>*<sup>0</sup> ð Þ ½ �, *<sup>σ</sup> <sup>j</sup>* <sup>¼</sup> *<sup>ω</sup>*, *<sup>y</sup>*<sup>0</sup> <sup>∈</sup> *<sup>G</sup>*, then we will correspond *ai* with *σ* and *ai* ¼ *μ*ð Þ *x <sup>j</sup>* .

In this case, we will correspond *σ* with *Pj* position. In the opposite case, if there is no *t* 0 *<sup>i</sup>* transition for 1≤*j*≤j j *P* in the way that *δ μ y*<sup>0</sup> ½ �, *t* 0 *i* � � *<sup>j</sup>* ¼ *ω*, then *ai* ¼ 1, in which case there is no related position for *t* 0 *i* .

For example:


Now, we will define the following action for *ai*:

$$a\_i = \begin{cases} \frac{a\_i - n}{m}, & \text{if } (a\_i - n) \text{modm} = \mathbf{0} \\\left[\frac{a\_i - n}{m}\right] + \mathbf{1}, & \text{if } (a\_i - n) \text{modm} \neq \mathbf{0} \end{cases}$$

Given our denote, we get that state *μ*½ �¼ *x* ð Þ 0, 1, 15, 13 covers state *μ*½ �¼ *y* ð Þ 1, 1,*ω*,*ω :* To achieve the goal, we need to implement the *t*<sup>2</sup> transition 27 times and

� �≤*ts <sup>y</sup>*<sup>2</sup>

� �, ∀*tk* ∈*T*<sup>∗</sup> .

*<sup>i</sup>* <sup>¼</sup> *<sup>d</sup>*<sup>0</sup> <sup>þ</sup> <sup>X</sup>

� �.

s are equal to 1. Without breaking the sense we will

*<sup>μ</sup>*½ � *<sup>x</sup> <sup>l</sup>* <sup>þ</sup> *<sup>m</sup>*<sup>0</sup> � *<sup>b</sup>*<sup>0</sup> � �

*μ*½ � *x <sup>l</sup>* þ *n*<sup>0</sup>

*<sup>i</sup>* � *<sup>b</sup>*<sup>0</sup> � � <sup>¼</sup> *ts <sup>y</sup>*<sup>1</sup>

*<sup>d</sup>*<sup>00</sup> <sup>þ</sup> <sup>X</sup> *k*0

> 0 1, ⋯, *t* 0

*y*<sup>0</sup> covering peak has less number of transitions than

*i*¼*d*00þ1

, as *S y*<sup>1</sup>

� � <sup>¼</sup> *S y*<sup>2</sup>

*μ*½ � *x <sup>l</sup>* þ *n*<sup>00</sup> *<sup>i</sup>* � *<sup>b</sup>*<sup>00</sup> � �

*<sup>k</sup>* the succession of

*<sup>k</sup>* move is in minimal state.

� �<*S y*<sup>2</sup>

� �& *T* <sup>∗</sup> 1 � � � �

� �

� �,

s is equal to 1. We will get

*k*

*i*¼*d*<sup>0</sup> þ1

*k*

*i*¼*d*<sup>0</sup> þ1

> *d*0 <*d*<sup>00</sup>

0 1, ⋯, *t* 0

� � � � � �.

<sup>1</sup> transition must be implemented for 27 times and *t*

, *<sup>l</sup>* <sup>¼</sup> *<sup>T</sup>* <sup>∗</sup> j j ð Þ*<sup>y</sup>* . Which means *ts*ð Þ*<sup>y</sup>* is *<sup>y</sup>* the number of

0 2

0

*The Possibilities of Modeling Petri Nets and Their Extensions*

*DOI: http://dx.doi.org/10.5772/intechopen.90275*

*<sup>i</sup>*¼1*<sup>b</sup> j i*

� �≥*μ*½ � *<sup>x</sup>* . In that case *ts <sup>y</sup>*<sup>1</sup>

*zj*ð Þ*t* , where *zj*ðÞ¼ *t* # *Pj*,*I t*ð Þ*<sup>k</sup>*

� � � � � �.

*i*

*<sup>i</sup>* , where *<sup>k</sup>*<sup>0</sup> <sup>¼</sup> *<sup>T</sup>*<sup>∗</sup> *<sup>y</sup>*<sup>2</sup>

*<sup>i</sup>* � *<sup>b</sup>*<sup>00</sup> � �≥*d*<sup>0</sup> <sup>þ</sup> <sup>X</sup>

**Lemma 2.** Suppose *y*<sup>1</sup> and *y*<sup>2</sup> are covering peaks. There is *S y*<sup>1</sup>

� �.

*k*

*i*¼*d*<sup>0</sup> þ1

Proof: Let *y* be the covering peak in our algorithm and *t*

�

transitions. It must be shown that the number of *t*

*<sup>i</sup>* s, number of *d*<sup>00</sup> is equal to 1. Moreover *d*<sup>00</sup> ≤*d*<sup>0</sup>

*μ*½ � *x <sup>l</sup>* þ *n*<sup>0</sup> *<sup>i</sup>* � *<sup>b</sup>*<sup>0</sup> � � <sup>&</sup>lt;

**Theorem.** Through the abovementioned number of covering state, transition

*k*

*i*¼*d*<sup>0</sup> þ1 *b*0

, where *<sup>k</sup>* <sup>¼</sup> *<sup>T</sup>*<sup>∗</sup> *<sup>y</sup>*<sup>1</sup>

*i*

� � <sup>¼</sup> *<sup>d</sup>*<sup>0</sup> <sup>þ</sup> <sup>X</sup>

*<sup>i</sup>*¼<sup>1</sup>*b*<sup>00</sup>

*μ*½ � *x <sup>l</sup>* þ *n*<sup>00</sup>

� �<*ts <sup>y</sup>*<sup>2</sup>

*<sup>i</sup>* <sup>¼</sup> *<sup>d</sup>* <sup>þ</sup> <sup>X</sup>

**Lemma 1**. Suppose there are *y*<sup>1</sup> and *y*<sup>2</sup> peaks in the way that

The above shows that the *t*

the *t*<sup>3</sup> transition for 15 times. Let us assign *ts*ð Þ¼ *<sup>y</sup>* <sup>P</sup>*<sup>l</sup>*

In this number some *b*<sup>0</sup>

suppose that the first *d*<sup>0</sup> number of *b*<sup>0</sup>

*ts y*<sup>1</sup>

� � <sup>¼</sup> <sup>P</sup>*<sup>k</sup>*<sup>0</sup>

*k*0

*i*¼*d*00þ1

� �≤*ts <sup>y</sup>*<sup>2</sup> � �

transition for 15 times.

enabled transitions.

� �<sup>≥</sup> *<sup>μ</sup>*½ � *<sup>x</sup>* &*<sup>μ</sup> <sup>y</sup>*<sup>2</sup>

*S y*ð Þ¼ <sup>P</sup> *i k j*¼*i*<sup>1</sup>

*ts*ð Þ¼ *<sup>y</sup>* <sup>P</sup> *k i*¼1 *b*0 *i*

We have *ts y*<sup>2</sup>

Suppose for *b*<sup>00</sup>

� � <sup>¼</sup> *<sup>d</sup>*<sup>00</sup> <sup>þ</sup> <sup>X</sup>

) *ts y*<sup>1</sup>

The lemma is proven.

� : in this case *ts y*<sup>1</sup>

*k*

*i*¼*d*<sup>0</sup> þ1 *b*0

� � <sup>¼</sup> *<sup>d</sup>*<sup>0</sup> <sup>þ</sup> <sup>X</sup>

¼ *ts y*<sup>2</sup> � �*:*

The lemma is proven.

algorithm is in its minimal state.

For this we need to show that ∃

0 1, ⋯, *t* 0 *k*.

*ts y*<sup>2</sup>

< *T* <sup>∗</sup> 2 � � �

Proof:

*ts y*<sup>1</sup>

the number of *t*

**121**

Proof: We have.

*μ y*<sup>1</sup>

where *n* to *t* 0 *<sup>i</sup>* or in *σ* corresponding *Pj* position, the number of tokens are in their first position, and *m* from *t* 0 *<sup>i</sup>* or *σ* to the number of the arrows in the state: # *Pj*, *O t*<sup>0</sup> *i* � � � � .

If for *t* 0 *<sup>i</sup>* transition *P*1, ⋯, *Pk* positions correspond, then we will take the *P*<sup>1</sup> position for which *μ*½ � *x* <sup>1</sup> ¼ *ai*. In this case *n* ¼ *μ*0½ � *P*<sup>1</sup> , and *m* ¼ # *Pj*, *O t*<sup>0</sup> *i* � � � � .

If there is no corresponding position for *t* 0 *<sup>i</sup>* transition, then we will leave *ai* to remain the same.

For example:

$$\begin{aligned} a\_1 &= (a\_1 - 1)/\mathbf{1} = (\mathbf{13} - \mathbf{1})/\mathbf{1} = \mathbf{12} \\ a\_2 &= (a\_2 - \mathbf{0})/\mathbf{1} = (\mathbf{15} - \mathbf{0})/\mathbf{1} = \mathbf{15}. \end{aligned}$$

Let us mark *b*<sup>1</sup> *<sup>i</sup>* ¼ *ai*. For example:

$$\begin{aligned} b\_1^1 &= a\_1 = \mathbf{12} \\ b\_2^1 &= a\_2 = \mathbf{15}. \end{aligned}$$

#### *2.2.1 Cumulative move*

We will take *T* <sup>∗</sup> last transition or the succession of transition, fix it and mark as *tα*.

*t<sup>α</sup>* corresponding *b*<sup>1</sup> *<sup>i</sup>* is marked as *α* which we also fix. The fixed *b j <sup>i</sup>* does not change in the next moves.

We consider all *T* <sup>∗</sup> items from the right to left, starting from *tα*.

Suppose *t* 0 *<sup>k</sup>* is the considered transition or the transition succession and *P*<sup>1</sup> is the corresponding position of *t* 0 *k*.

If *P*<sup>1</sup> ∈*I t*ð Þ*<sup>α</sup>* , then *t* 0 *<sup>k</sup>* corresponding *b j <sup>i</sup>* in the next move will get the following value: *b j*þ1 *<sup>i</sup>* ¼ *b j <sup>i</sup>* þ *α* � *l*, where *<sup>l</sup>* from *P*<sup>1</sup> position *t<sup>α</sup>* is the number of arrows.

Suppose *t* 0 *<sup>k</sup>* corresponds with *P*1, ⋯, *Pl* positions. If ∃1≤*j*≤*l* in the way that*Pj* ∈*I t*ð Þ*<sup>α</sup>* , then *b j*þ1 *<sup>i</sup>* ¼ *b j <sup>i</sup>* þ *α* � *l*, where *l* ¼ # *Pj*,*I t*ð Þ*<sup>α</sup>* � �. In the opposite case, *b j*þ1 *<sup>i</sup>* ¼ *b j i* .

After that, we fix *t<sup>α</sup>* the previous transition action and denote it as *tα*.

We denote the new *t<sup>α</sup>* corresponding to *b j <sup>i</sup>* as *α* and go to the second step again. It follows that for each transition or sequence of transitions, there will be a correspondingly fixed number *b j i* , which will mark the number of implementation of the transition or sequence of transitions. For example:

*The Possibilities of Modeling Petri Nets and Their Extensions DOI: http://dx.doi.org/10.5772/intechopen.90275*

$$\begin{array}{cc} t\_1' & t\_2' \\ \mathbf{12} & \mathbf{15} \\ \mathbf{27} & \mathbf{15} \end{array}$$

The above shows that the *t* 0 <sup>1</sup> transition must be implemented for 27 times and *t* 0 2 transition for 15 times.

Given our denote, we get that state *μ*½ �¼ *x* ð Þ 0, 1, 15, 13 covers state *μ*½ �¼ *y* ð Þ 1, 1,*ω*,*ω :* To achieve the goal, we need to implement the *t*<sup>2</sup> transition 27 times and the *t*<sup>3</sup> transition for 15 times.

Let us assign *ts*ð Þ¼ *<sup>y</sup>* <sup>P</sup>*<sup>l</sup> <sup>i</sup>*¼1*<sup>b</sup> j i* , *<sup>l</sup>* <sup>¼</sup> *<sup>T</sup>* <sup>∗</sup> j j ð Þ*<sup>y</sup>* . Which means *ts*ð Þ*<sup>y</sup>* is *<sup>y</sup>* the number of enabled transitions.

**Lemma 1**. Suppose there are *y*<sup>1</sup> and *y*<sup>2</sup> peaks in the way that *μ y*<sup>1</sup> � �<sup>≥</sup> *<sup>μ</sup>*½ � *<sup>x</sup>* &*<sup>μ</sup> <sup>y</sup>*<sup>2</sup> � �≥*μ*½ � *<sup>x</sup>* . In that case *ts <sup>y</sup>*<sup>1</sup> � �≤*ts <sup>y</sup>*<sup>2</sup> � �.

Proof: We have.

For example:

where *n* to *t*

remain the same. For example:

Let us mark *b*<sup>1</sup>

*2.2.1 Cumulative move*

*t<sup>α</sup>* corresponding *b*<sup>1</sup>

change in the next moves.

0

corresponding position of *t*

If *P*<sup>1</sup> ∈*I t*ð Þ*<sup>α</sup>* , then *t*

0

correspondingly fixed number *b*

Suppose *t*

*j*þ1 *<sup>i</sup>* ¼ *b j*

that*Pj* ∈*I t*ð Þ*<sup>α</sup>* , then *b*

Suppose *t*

as *tα*.

value: *b*

*b j*þ1 *<sup>i</sup>* ¼ *b j i* .

**120**
