# Circular sets and powers of two

by G. Lathoud, May 2017

## Summary

This paper investigates a non-uniform way to visit a circular set without repetition. Interestingly, it turns out that only circular sets of $2^q$ elements permit this.

## Definition: circular set

Take $N$ holes arranged in a circle:

...pick one as the first hole #0, chose a direction (e.g. counterclockwise), and name the following holes accordingly #1, #2, ..., #($N-1$):

## Filling the set: an example ($N=8$)

At this point all holes have been filled, so we stop. The order in which we filled the holes: $[ #0, #1, #3, #6, #2, #7, #5, #4 ]$ can be seen as a permutation of the first $N=8$ non-negative integers.

## Success and failure

Repetitions are not accepted, i.e., whenever we land on a hole that has already been filled, we call that a failure. Example: $N=3$:

Looking at the first few values of $N$:


N=2     success
N=3              failure
N=4     success
N=5              failure
N=6              failure
N=7              failure
N=8     success
N=9              failure
...              failure
N=15             failure
N=16    success
N=17             failure


Successes seem to correspond to powers of two: $N=2^q$ where $q \in \mathbb{N}_+^*$

## Main result

A circular set of $N$ elements can be visited in the above–described non–uniform way if and only if $N$ is a power of two ($N=2^q$).

Formally: we define the property:

$P_{N} \triangleq$ "for the circular set of $N$ holes, for each step $i = 1 \dots N$, we land on an empty hole (an thus after the $N$ steps all holes are filled)".

The main result of the present paper is:

$$N\ is\ a\ power\ of\ two\ \Leftrightarrow\ P_N\ true$$

where "$N$ is a power of two" means $log_2(N) \in \mathbb{N}_+$ or equivalently: $\exists\ q\ \in \mathbb{N}_+\ s.t.\ N=2^q$.

Appendix (A1) demonstrates that $N\ power\ of\ 2 \Rightarrow\ P_N\ true$.

Appendix (A2) demonstrates that $N\ not\ a\ power\ of\ 2 \Rightarrow\ P_N\ false$.

# Openings

## (O1): Filling, seen as a permutation

The order in which we fill the holes, e.g. for $N=8$: $[ #0, #1, #3, #6, #2, #7, #5, #4 ]$ can be seen as a permutation of the first $N=8$ non-negative integers.

What happens if we repeat this permutation?


N=8 (2^3)

step  0  current  0,1,2,3,4,5,6,7
step  1  current  0,1,3,6,2,7,5,4
step  2  current  0,1,6,5,3,4,7,2
step  3  current  0,1,5,7,6,2,4,3
step  4  current  0,1,7,4,5,3,2,6
step  5  current  0,1,4,2,7,6,3,5
step  6  current  0,1,2,3,4,5,6,7

=> period: 6


So we can observe a periodicity. What about other values of $N=2^q$?


N=   4 (2^2)     period:       2
N=   8 (2^3)     period:       6
N=  16 (2^4)     period:      14
N=  32 (2^5)     period:      30
N=  64 (2^6)     period:    2280
N= 128 (2^7)     period:   18480
N= 256 (2^8)     period:    2964
N= 512 (2^9)     period:   10248
N=1024 (2^10)    period: 6036022


These results were obtained using this JavaScript code, running it in a browser console.

More results, thanks to James Waldby:

N=2048     (2^11)  period: 2013788
N=4096     (2^12)  period: 182700
N=8192     (2^13)  period: 23591700149148
N=16384    (2^14)  period: 19958999294520
N=32768    (2^15)  period: 6959639094969183480
N=65536    (2^16)  period: 26192250762550684725571680
N=131072   (2^17)  period: 3643607285677121265513086436360
N=262144   (2^18)  period: 552425176567480329867600
N=524288   (2^19)  period: 1104886883792843540368115115103595714814480
N=1048576  (2^20)  period: 598516403943830902612507349753400
N=2097152  (2^21)  period: 145205898084376835266284046071840
N=4194304  (2^22)  period: 2218238870433506133002290060782067400445862406820
N=8388608  (2^23)  period: 1284879737068585718960296765948422911523733198261847333520
N=16777216 (2^24)  period: 44942096361329939149161891089213686455000
N=33554432 (2^25)  period: 29792203744656340656203106393819329353621249230
N=67108864 (2^26)  period: 100060465407260592657061740391450002788223169808280

Open questions: For powers of two $N=2^q$, can someone derive a formula giving the period as a function of $N$, i.e. the series (2,6,14,30,2280,...)?

## (O2): About other filling methods

This paper investigated the particular filling method, where at each step $i$ we move $j=i$ holes forward, thus defining the series: $$(j)_i = (1,2,3,\dots,N-1)$$ Besides the obvious "uniform" filling method, where we move 1 hole forward each time: $$(j)_i = (1,1,1,\dots,1)$$ ...are there other "non-uniform" filling methods without repetition, at least for $N$ being a power of two?

For example, for $N=2^2=4$ the answer is yes. Besides the filling method investigated so far: $$(j)_i = (1,2,3)$$ there is also: $$(j)_i = (3,2,1)$$ which is equivalent to invert the direction. If we additionally restrict $(j)_i$ being itself a permutation of $(1,2,3,...,N-1)$, these are the only two possibilities for $(j)_i$ for $N=4$.

Open question: Are there other methods $(j)_i$ to fill without repetition, which work for all $N$ powers of two? Especially when we restrict the series $(j)_i$ being itself a permutation of $(i)_i = (1,2,3,...,N-1)$ ?

# Appendices

## (A1) Show that $N\ power\ of\ 2 \Rightarrow\ P_N\ true$

Let us assume $H_1$ and $H_2$:

$H_1$
$N$ is a power of two: $\exists\ q\ \in \mathbb{N}_+^*\ s.t.\ N=2^q$
$H_2$
$P_N\ false$, i.e. in at least one of the $N$ steps $i=1 \dots N$, we land on a hole that has already been filled: $$\exists (a,b)\in\mathbb{N}^2\ s.t.\ 0 \le a \lt b \lt N\ and\ V_a \equiv V_b\ [N]$$

where:

• $V_i$ is, at step $i$, the total number of holes we've been moving since the beginning: $$V_i \triangleq \sum_{j=1}^{i} j = \frac{i(i+1)}{2}$$
• $V_a \equiv V_b\ [N]$ means congruence modulo $N$: $$\exists k \in \mathbb{N}\ s.t.\ V_b - V_a = k \cdot N$$

$H_2$ implies: $$\exists k \in \mathbb{N}\ s.t.\ \frac{b(b+1)}{2} - \frac{a(a+1)}{2} = k \cdot N$$

which we can rewrite: $$b^2 - a^2 + b - a = 2 \cdot k \cdot N$$ $$(b - a) \cdot (b + a + 1) = 2 \cdot k \cdot N \hspace{2.5em} (\alpha)$$

Observations: $b-a$ and $b+a$ have same parities, hence $b-a$ and $b+a+1$ have opposite parities, i.e. one is odd and the other one is even.

On the right hand side, $2 \cdot N = 2^{q+1}$ is a power of two, thus $k$ must be odd. Moreover, since $(b - a)$, $(b+a+1)$ and $2 \cdot N$ are all non-negative, we have $k > 0$.

Summarized: $k \ge 1$, and $k$ is the only odd term on the right hand side.

Let us assume $k = 1$. Because the two terms on the left hand side have opposite parities, then either $a+b+1=1$ (impossible), or $b-a=1$ i.e. $b=a+1$ so $(\alpha)$ can be written: $1 \cdot 2 \cdot b = 1 \cdot 2 \cdot N$, and thus $b=N$, which is impossible as well.

Therefore: $k$ is odd and $k \ge 3$.

### (A1.a) Let us assume $b-a=k$

$(\alpha)$ can be rewritten: $$k \cdot (1+2a+k) = 2 \cdot k \cdot N$$ $$1+2a+k = 2\cdot N$$ $$a = N - \frac{k+1}{2}$$ $$b = a + k = N + \frac{2k - k - 1}{ 2 }$$ $$b = N + \frac{k-1}{2}$$ Since $k \ge 3$, we have $b \gt N$, which is impossible.

### (A1.b) Let us assume $a+b+1=k$

i.e. $$b-a = b-(k - (b+1)) = 2b - k + 1$$ $(\alpha)$ can be rewritten: $$(2b - k + 1) \cdot k = 2 \cdot k \cdot N$$ $$2b - k + 1 = 2 N$$ $$b = N + \frac{k-1}{2}$$ Since $k \ge 3$, we have $b \gt N$, which is impossible.

### Conclusion of (A1)

For $q \in \mathbb{N}_+^*$ and $N=2^q$, assuming $P_N$ false leads to a contradiction, therefore $P_N$ is true. $P_{2^0}=P_1$ is obvious, therefore:

$$\forall q \in \mathbb{N}_+\ P_{2^q}\ true$$

## (A2) Show that $N\ not\ a\ power\ of\ 2 \Rightarrow\ P_N\ false$

Formally: we want to show that:

$$\forall N \in \mathbb{N}_+^* \ s.t.\ log_2 N \notin \mathbb{N}$$ $$\exists (a,b) \in \mathbb{N}^2 \hspace{1em} 0 \le a \lt b \lt N \hspace{1em} s.t. \hspace{1em} V_a \equiv V_b \ [N]$$

where $V_i \triangleq \frac{i(i+1)}{2}$ is the number of holes we've moved since the beginning.

For $(a,b) \in \mathbb{N}^2$, we define the property:

$$T_{a,b} \hspace{1em} \triangleq \hspace{1em} 0 \le a \lt b \lt N \hspace{0.75em} and\hspace{0.75em} V_a \equiv V_b \ [N]$$

To prove the result, we need to find at least one value of $(a,b)$ that verifies $T_{a,b}$

### (A2.1) Case: $N=2p+1$ where $p \in \mathbb{N}_+^*$

Since $V_i \triangleq \frac{i(i+1)}{2}$ we can write: $$V_{p+1} - V_{p-1} = p+1 \ + \ p = N$$ Thus, $a = p - 1$ and $b = p + 1$ verify $T_{a,b}$.

### Transition

It remains to find $(a,b)$ verifying $T_{a,b}$ when $N=2p$ is not a power of two.

### (A2.2) Case: $N$ even but not a power of two

i.e. $$\exists (p,q) \in (\mathbb{N}_+^*)^2 \hspace{1.5em} N=2^q \cdot (2p+1)$$ e.g.


N =  6 = 2*3   q:1   p:1
N = 10 = 2*5   q:1   p:2
N = 12 = 4*3   q:2   p:1
N = 14 = 2*7   q:1   p:3


#### (A2.2.1) When $p \ge 2^q$

Let $a = p - 2^q$ and $b = p + 2^q$.

We have $0 \le a$ and $a \lt b$ and $N - b = 2^q \cdot 2p - p = p (2^{q+1} - 1) > 0$

therefore we have $0 \le a \lt b \lt N$.

Besides,

$$V_b - V_a = \sum_{j=1}^{p+2^q} j \hspace{1em} - \hspace{1em} \sum_{j=1}^{p-2^q} j$$ $$= \sum_{j=p-2^q}^{p+2^q} j \hspace{1em} - \hspace{1em} (p - 2^q)$$ $$= p \cdot (2 \cdot 2^q + 1) - (p - 2^q)$$ $$= 2^q \cdot (2p + 1)$$ $$= N$$

$(a,b)$ verify $T_{a,b}$.

#### (A2.2.2) When $p \lt 2^q$

Let $a = 2^q - p -1$ and $b = 2^q + p$.

We can show, as in (A2.2.1), that $0 \le a \lt b \lt N$.

Besides,

$$V_b - V_a = \sum_{j=1}^{2^q + p} j \hspace{1em} - \hspace{1em} \sum_{j=1}^{2^q-p-1} j$$ $$= \sum_{j=2^q -p}^{2^q + p} j = 2^q \cdot (2p+1)$$ $$= N$$

Thus $(a,b)$ verify $T_{a,b}$.