Chip-Firing and Fractional Bases

# Chip-Firing and Fractional Bases

Matvey Borodin    Hannah Han    Kaylee Ji    Tanya Khovanova    Alexander Peng    David Sun    Isabel Tu    Jason Yang    William Yang    Kevin Zhang    Kevin Zhao
###### Abstract

We study a particular chip-firing process on an infinite path graph. At any time when there are at least chips at a vertex, chips fire to the left and chips fire to the right. We describe the final state of this process when we start with chips at the origin.

## 1 Introduction

A chip-firing game is a one-player game played on a graph. It was invented around 1983. Chips are distributed between vertices of a graph. At each point a vertex that have the number of chips that is at least its degree can ‘fire’ by sending a chip along each incident edge . The game was later expended to directed graphs, where a move consists of selecting a vertex with at least as many chips as its outdegree, and sending one chip along each outgoing edge to its neighbors .

We study a particular chip-firing game suggested by Prof. James Propp. This chip-firing game is played on an integer line. We consider an integer line as a graph with integers as nodes, and neighboring integers are connected by edges. At any point in the game there might be chips at nodes. We call our game an - firing game. If there are at least chips at a particular node, then the node ‘fires’: chips are moved to the left and chips are moved to the right. This game can be considered as a game on a directed graph if we replace an edge connecting two consecutive integers with edges directed from the larger to the smaller integer and edges directed the opposite way.

In this paper we describe the final state of the game when we start with chips at the origin.

We start with preliminaries and examples in Section 2. We also describe fractional bases as they are related to this game. In Section 3 we represent a state in the game as a series which immediately relate the chip-firing game to base . In Section 4 we reduce the case of not coprime and to the case of their coprime factors. In Section 5 we consider an example of 2-3 firing which illustrates what happens for any - firing.

To give a description of the final state, we start with describing the right side in Section 6 and end with describing the left side in Section 7.

We finish paper with an example of 1- firing in Section 8.

## 2 Preliminaries and Examples

Our graph is an infinite path marked with integers. We visualize it as a horizontal line with integers increasing to the right.

We start with chips at the node marked 0. We call the starting vertex the origin. At each moment a vertex has at least chips, it fires chips to the left and chips to the right. The order in which vertices fire does not matter. Given the infinity of the graph, the game terminates at a unique state, which we call the final state.

We represent each configuration with a string of integers, where we drop all zeros on the left and on the right and where we mark the origin with a dot to its right. We call this dot the radix. For example, consider 1-2 chip firing. If we start with seven chips in the origin, we get 22.12 as the final stage, see Figure 1. Figure 1: 1-2 Chip Firing starting with 7 chips

Given the final string representing an integer , the final string for can be found by simply increasing the origin node by 1, then firing if necessary. For example, to find the string for 8 in the 1-2 firing, we add 1 to the string for 7 resulting in 23.12. Then firing produces the final string 111.1112.

Every integer has a unique final state.

Our results will connect the left part of the final state to the fractional base . Here is a description of number in such a base. We start with in the unit’s place. If we subtract from and add to the ten’s place. The same rule works independently of placement and continues until all the digits are below . We can view this as a chip-firing process, where chips at one vertex are replaced with chips at a vertex to the left. Such fractional bases were introduced by Propp  and popularized by Tanton .

Suppose a number is written as in base . Then its value equals .

## 3 Series

We can also represent a state of the game as a series. If we have chips at node , the series are

 ∑mamtm.

One firing at node means adding the polynomial

 atk+1−(a+b)tk+btk−1

to the series.

We can factor the polynomial into

 (at−b)(t−1)tk−1.

It follows that the value of the series does not change for and . When , the series sum to the total number of chips. Not surprisingly, this value does not change. For , the series represent the value of the state interpreted in base . It follows that if we interpret the string corresponding to a state as a number in base , its value does not change after firing. This leads to the following proposition.

###### Proposition 3.1.

If we start with chips at the origin, the final state interpreted as a number in base equals .

More surprisingly the left/right part of the final state when interpreted in base is also an integer. Before proving it we introduce a name for the vertex to the right of the origin. We call it the origout.

###### Proposition 3.2.

If we start with chips at the origin, the part of the final state before the radix point interpreted as a number in base is an integer. The same is true for the part after the radix.

###### Proof.

The value of the left part before the radix interpreted in base can only change during the firing of nodes at the origin and origout. If the origin fires, the value of the left part before the radix decreases by . If the origout fires the value of the left part increases by . In any case it remains an integer.

The value of the right part in base 3/2 is the difference between the total value in base 3/2 and the value of the left part. The total value is equal to the starting number , so the right part is an integer too. ∎

We actually proved a stronger statement.

###### Corollary 3.3.

The value of the right side interprested in base is times the difference in the number of firings between the origin and the origout.

## 4 a-b Firing, when a and b are not coprime

Suppose and have their greatest common divisor . Then we can express the final result of - firing through - firing.

###### Theorem 4.1.

Suppose , where . The final state of an - firing of can be reconstructed from the final state of an - firing of in the following manner: Take the final state of an - firing of , multiply each digit by and add to the digit before the radix.

###### Proof.

We can group chips into bags of chips each and consider the bags to be new chips. ∎

For example, the final state of 5 in a 1-1 firing is 111.11, while the final state of 25 in a 5-5 firing is 555.55 and the final state of 26 in a 5-5 firing is 556.55.

For another example, the final state of 21 in a 2-3 firing is 442.2243, while the final state of 42 in a 4-6 firing is 884.4486.

Consider the - firing as an example.

As firing is symmetrical with respect to the origin, the final result is symmetrical too. The final string is easily described.

###### Lemma 4.2.

If we start with chips, the final state for the - firing has the remainder of divided by chips at the origin. It also has consecutive vertices on the each sides with chips.

From now on we assume that and are coprime. We also assume that , as the case can be described by symmetry.

## 5 2-3 Firing.

Consider a 2-3 firing. Here is the first several results for starting chips, where ranges from 0 to 28: 0., 1., 2., 3., 4., 20.3, 21.3, 22.3, 23.3, 24.3, 42.13, 43.13, 44.13, 213.43, 214.43, 232.413, 233.413, 234.413, 422.2413, 423.2413, 424.2413, 442.2243, 443.2243, 444.2243, 2332.222413, 2333.222413, 2334.222413, 4222.2222243.

We can see how transition from number to number works starting from . The number before the radix behaves similar to numbers in base 3/2: decrement last digit of at least 5 by 3, increment the digit just left to it by 2. On the right after the radix the pattern follows the following rule. If the left part of ends in several threes or fours, then for each three and four the following thing happens: 413 is replaced by 243, and 43 is replaced by 413.

Also, for , the left part before the radix, when interpreted in base 3/2 is equal to . Correspondingly, the right side interpreted in base 3/2 equals 2.

This example illustrates what happens for other coprime and , which we prove in the rest of the paper. Namely, starting from some , the transition from the final state of one integer to the next can be described as follows.

• The left part before the radix behaves similar to base while using digits between and inclusive (see Lemma 7.2). That is, if we get at least at one digit placement, we subtract and carry to the left.

• The right part consists of a repeated digit followed by patterns that cycle following each carry in the left part.

• The left part of number interpreted in base equals , where is a constant. Equivalently, the right part after the radix in base equals .

Now we are ready to describe the final state of - firing for coprime and , where . We start with the right side.

## 6 The Right Side

The value plays an important role in the coming formulae and proofs. We denote it by .

###### Lemma 6.1.

If , then . Also .

###### Proof.

First independently of the value of positive integer . If , then . Hence, . The second part follows from the first part: . ∎

This means that for , the values and do not fire, while the value fires.

Let us consider what happens on the right side after the origin fires.

The only times when the value of the right side interpreted in base changes is when the origin fires or when the origout fires.

We mark the origout value as bold when it is at least , that is, when it is about to fire. After the first firing of the origin the right side is:

 b.

The second time the origin fires, we get:

 b→(2b)→(b−a)b.

The third time the origin fires, the firing does not propagate to the origout:

 (b−a)b→(2b−a)b.

We continue with firing from the origin:

 (2b−a)b→(3b−a)b→(2b−a)(b−a)b.

And the next time the origout fires:

 (2b−a)(b−a)b→(3b−a)(b−a)b→(2b−2a)(2b−a)b.

One more firing, which does not propagate:

 (2b−2a)(2b−a)b→(3b−2a)(2b−a)b.

The pattern emerges. Notice that if the origin fires for the -th time, the origout fires too, unless is a triangular number.

Now we show what happens when the origin fires for the -th time:

 (kb−(k−1)a)…(3b−2a)(2b−a)b→((k+1)b−(k−1)a)…(2a−b)b
 →(kb−(k−1)a)…(3b−2a)(2b−a)(b−a)b.

This is followed by

 (kb−(k−1)a)…(3b−2a)(2b−a)(b−a)b→((kb−(k−1)a)…(3b−2a)(2b−2a)(2b−a)b.

Notice, that the value stopped the firing from propagating. In the next round the firing is stopped at the value . After several rounds we get the value:

 (kb−(k−1)a)…((m+1)b−ma)(mb−ma)(mb−(m−1)b)…(3b−2a)(2b−a)b.

This continues until we reach

 (kb−ka)(kb−(k−1)a)…(3b−2a)(2b−a)b.

On the next firing from the origin the firing does not propagate to the origout:

 ((k+1)b−ka)(kb−(k−1)a)…(3b−2a)(2b−a)b,

and the cycles starts again.

The pattern changes after origin firings when we reach

 (cb−ca)(cb−(c−1)a)…(3b−2a)(2b−a)b.

On the next firing from the origin we get:

 (cb−ca)(cb−(c−1)a)…(3b−2a)(2b−a)b→
 ((c+1)b−ca)(cb−(c−1)a)((c−1)b−(c−2)a)…(3b−2a)(2b−a)b
 →(cb−ca)(cb−ca)(cb−(c−1)a)((c−1)b−(c−2)a)…(3b−2a)(2b−a)(b−a)b.

From now on. every firing from the origin is matched by the firing of the origout. The result is adding one digit to the front of the right side, while the very end cycles the same way as before. Therefore, we have the following lemma.

###### Lemma 6.2.

Eventually the right side consists of repeated digit followed by the string which then cycles as described above. The eventual value of the right side interpreted in base is .

###### Proof.

The value of the right side is the difference of the number of firings between the origin and the origout times . There were firings from the origin that did not propagate to the origout. ∎

Notice that the cycle length of the very last digits is .

Example. Consider for example a 3-4 firing. In this case That means, the repeated digit is . The digits on the very right cycle through , , and . That is through 654, 6514, and 6254. The value of the right side interpreted in base 4/3 eventually becomes 9.

For some values of and we can exactly calculate the data we are interested in:

## 7 The Left Side

We showed in the previous section that eventually whenever the origin fires, the origout fires too. We can say: when the origin fires it sends chips to the left and gets chips. In other words: as soon as the origin has chips, of the chips explode, and chips appear to the left and another chips stay in place.

###### Lemma 7.1.

Eventually every significant digit on the left of the radix is at least . Moreover, the same is true at the intermediate stages of the firing process.

###### Proof.

We know that eventually each time the origin fires, the origout fires too. That means the origin is always at least after that. The next time the point to the left of the origin fires, the origin gets at least chips and fires again. That means the point to the left always gets chips back after firing. Continuing we see that eventually every significant digit to the left is at least . ∎

From here we can show that there is an equivalent description of the eventual behavior of the left side.

###### Lemma 7.2.

Eventually the left side behaves as the following chip-firing game: As soon as a point collects at least chips, it removes chips from its own pile and adds chips to the left.

###### Proof.

Suppose the value of the right side in base is stabilized and every digit on the left side is at least . We assume that when the origin fires, it sends chips to the left and then replaces the remaining chips with chips.

Consider a state on the left of the radix in the form , where ends at a digit not exceeding , and for and . If we add one chip to the origin, the origin fires:

 Xdkdk−1…d2(d1+a)a.

Then the digit next to the origin fires, causing the origin to fire again:

 Xdkdk−1…(d2+a)(d1−b)(a+b)Xdkdk−1…(d2+a)(d1−b+a)a.

Now the hundred’s digit fires:

 Xdkdk−1…(d3+a)(d2−b)(d1+a)a,

causing the ten’s digit and the origin to fire:

 Xdkdk−1…(d3+a)(d2−b+a)(d1−b)(a+b)Xdkdk−1…(d3+a)(d2−b+a)(d1−b+a)a,

Then the thousand’s digit fires, causing all the consecutive digits to fire ending in:

 Xdkdk−1…(d4+a)(d3−b+a)(d2−b+a)(d1−b+a)a.

After we reach and it fires, it causes all the digits to the right to fire ending in:

 (X+a)(dk−b+a)(dk−1−b+a)…(d3−b+a)(d2−b+a)(d1−b+a)a.

On the other hand, the second process we described ends in the same string. ∎

This is similar to base , except the carry happens not when a digit reaches , but rather when it reaches .

## 8 1-b Firing

The formulae are particularly simple for 1- firing. This case corresponds to the first line in the Table 1, where .

The repeated digit after the radix is , and the eventual value of the right side interpreted in base is . Correspondingly, for , the left part of 1- firing interpreted in base is equal . The value of the right side stabilizes after the first firing, that is after .

Also the cycle length is 1. That means the right side is always .

The number of digits to the right of the radix in the final state of integer is

 n−b−2∑i=1νb(i),

where the -adic valuation of . That is, , where is the largest power of that divides .

Now we wish to describe the left side. Consider a sequence that depends on an integer parameter . Each element of is a finite integer string consisting of digits 1 through . The sequence is arranged in an increasing order. In particular, , , and . Here is, for example, sequence :

 1, 2, 11, 12, 21, 22, 111, 112,….

We need this sequence to help us describe how the left part of the final state of 1- chip-firing changes from one number to the next.

Consider the final state of an in 1- firing, where . If , the final representation of it is . After that the left part before the radix follows sequence .

The left part is particularly easy to describe for . That is, for 1-2 firing. Here is the description of the final state for .

To get the left side we represent in binary, remove the first digit, and shift all the other digits up by 1. For example, 21 in binary is 10101. Remove the first digit to get 0101, then shift by 1 to get 1212.

## 9 Acknowledgements

We are grateful to Prof. James Propp for suggesting this project to us. We also want to thank PRIMES STEP for giving us the opportunity to do this research.

## References

•  A. Björner, L. Lovász, P. W. Shor, Chip-firing games on graphs. European Journal of Combinatorics archive, Volume 12 Issue 4, July 1991, pp 283–291, doi:10.1016/S0195-6698(13)80111-4.
•  A. Björner, L. Lovász, Chip-Firing Games on Directed Graphs. Journal of Algebraic Combinatorics, December 1992, Volume 1, Issue 4, pp 305–328, doi:10.1023/A:1022467132614
•  J. Propp, How do you write one hundred in base 3/2? https://mathenchant.wordpress.com/2017/09/17/how-do-you-write-one-hundred-in-base-32/, accessed in April 2018.
•  J. Tanton, Companion Guide to Exploding dots, James Tanton, 2013.
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