games

Seating the first few thousand Knights

Feb 3rd

The Knight-seating problems asks for a consistent placing of n-th Knight at an odd root of unity, compatible with the two different realizations of the algebraic closure of the field with two elements. The first identifies the multiplicative group of its non-zero elements with the group of all odd complex roots of unity, under complex multiplication. The second uses Conway’s ‘simplicity rules’ to define an addition and multiplication on the set of all ordinal numbers.

The odd Knights of the round table-problem asks for a specific one-to-one correspondence between two realizations of ‘the’ algebraic closure $\overline{\mathbb{F}_2}$ of the field of two elements.

The first identifies the multiplicative group of its non-zero elements with the group of all odd complex roots of unity, under complex multiplication. The addition on $\overline{\mathbb{F}_2}$ is then recovered by inducing an involution on the odd roots, pairing the one corresponding to x to the one corresponding to x+1.

The second uses Conway’s ‘simplicity rules’ to define an addition and multiplication on the set of all ordinal numbers. Conway proves in ONAG that this becomes an algebraically closed field of characteristic two and that $\overline{\mathbb{F}_2}$ is the subfield of all ordinals smaller than $\omega^{\omega^{\omega}}$ . The finite ordinals (the natural numbers) form the quadratic closure of $\mathbb{F}_2$ .

On the natural numbers the Conway-addition is binary addition without carrying and Conway-multiplication is defined by the properties that two different Fermat-powers $N=2^{2^i}$ multiply as they do in the natural numbers, and, Fermat-powers square to its sesquimultiple, that is $N^2=\frac{3}{2}N$ . Moreover, all natural numbers smaller than $N=2^{2^{i}}$ form a finite field $\mathbb{F}_{2^{2^i}}$ . Using distributivity, one can write down a multiplication table for all 2-powers.

The Knight-seating problems asks for a consistent placing of n-th Knight K_n at an odd root of unity, compatible with the two different realizations of $\overline{\mathbb{F}_2}$ . Last time, we were able to place the first 15 Knights as below, and asked where you would seat $K_{16}$

K_4 was placed at $e^{2\pi i/15}$ as 4 was the smallest number generating the ‘Fermat’-field $\mathbb{F}_{2^{2^2}}$ (with multiplicative group of order 15) subject to the compatibility relation with the generator 2 of the smaller Fermat-field $\mathbb{F}_2$ (with group of order 15) that 4^5=2 .

To include the next Fermat-field $\mathbb{F}_{2^{2^3}}$ (with multiplicative group of order 255) consistently, we need to find the smallest number n generating the multiplicative group and satisfying the compatibility condition $n^{17}=4$ . Let’s first concentrate on finding the smallest generator : as 2 is a generator for 1st Fermat-field $\mathbb{F}_{2^{2^1}}$ and 4 a generator for the 2-nd Fermat-field $\mathbb{F}_{2^{2^2}}$ a natural conjecture might be that 16 is a generator for the 3-rd Fermat-field $\mathbb{F}_{2^{2^3}}$ and, more generally, that $2^{2^i}$ would be a generator for the next field $\mathbb{F}_{2^{2^{i+1}}}$ .

However, an “exercise” in the 1978-paper by Hendrik Lenstra Nim multiplication asks : “Prove that $2^{2^i}$ is a primitive root in the field $\mathbb{F}_{2^{2^{i+1}}}$ if and only if i=0 or 1.”

I’ve struggled with several of the ‘exercises’ in Lenstra’s paper to the extend I feared Alzheimer was setting in, only to find out, after taking pen and paper and spending a considerable amount of time calculating, that they are indeed merely exercises, when looked at properly… (Spoiler-warning : stop reading now if you want to go through this exercise yourself).

In the picture above I’ve added in red the number x(x+1)=x^2+1 to each of the involutions. Clearly, for each pair these numbers are all distinct and we see that for the indicated pairing they make up all numbers strictly less than 8.

By Conway’s simplicity rules (or by checking) the pair (16,17) gives the number 8. In other words, the equation x^2+x+8 is an irreducible polynomial over $\mathbb{F}_{16}$ having as its roots in $\mathbb{F}_{256}$ the numbers 16 and 17. But then, 16 and 17 are conjugated under the Galois-involution (the Frobenius $y \mapsto y^{16}$ ). That is, we have $16^{16}=17$ and $17^{16}=16$ and hence $16^{17}=8$ . Now, use the multiplication table in $\mathbb{F}_{16}$ given in the previous post (or compute!) to see that 8 is of order 5 (and NOT a generator). As a consequence, the multiplicative order of 16 is 5×17=85 and so 16 cannot be a generator in $\mathbb{F}_{256}$ . For general i one uses the fact that $2^{2^i}$ and $2^{2^i}+1$ are the roots of the polynomial $x^2+x+\prod_{j<i} 2^{2^j}$ over $\mathbb{F}_{2^{2^i}}$ and argues as before.

Right, but then what is the minimal generator satisfying $n^{17}=4$ ? By computing we see that the pairings of all numbers in the range 16…31 give us all numbers in the range 8…15 and by the above argument this implies that the 17-th powers of all numbers smaller than 32 must be different from 4. But then, the smallest candidate is 32 and one verifies that indeed $32^{17}=4$ (use the multiplication table given before).

Hence, we must place Knight $K_{32}$ at root $e^{2 \pi i/255}$ and place the other Knights prior to the 256-th at the corresponding power of 32. I forgot the argument I used to find-by-hand the requested place for Knight 16, but one can verify that $32^{171}=16$ so we seat $K_{16}$ at root $e^{342 \pi i/255}$ .

But what about Knight $K_{256}$ ? Well, by this time I was quite good at squaring and binary representations of integers, but also rather tired, and decided to leave that task to the computer.

If we denote Nim-addition and multiplication by $\oplus$ and $\otimes$ , then Conway’s simplicity results in ONAG establish a field-isomorphism between $~(\mathbb{N},\oplus,\otimes)$ and the field $\mathbb{F}_2(x_0,x_1,x_2,\hdots )$ where the x_i satisfy the Artin-Schreier equations

$x_i^2+x_i+\prod_{j < i} x_j = 0$

and the i-th Fermat-field $\mathbb{F}_{2^{2^i}}$ corresponds to $\mathbb{F}_2(x_0,x_1,\hdots,x_{i-1})$ . The correspondence between numbers and elements from these fields is given by taking $x_i \mapsto 2^{2^i}$ . But then, wecan write every 2-power as a product of the x_i and use the binary representation of numbers to perform all Nim-calculations with numbers in these fields.

Therefore, a quick and dirty way (and by no means the most efficient) to do Nim-calculations in the next Fermat-field consisting of all numbers smaller than 65536, is to use sage and set up the field $\mathbb{F}_2(x_0,x_1,x_2,x_3)$ by

R.< x,y,z,t > =GF(2)[]
S.< a,b,c,d >=R.quotient((x^2+x+1,y^2+y+x,z^2+z+x*y,t^2+t+x*y*z))

To find the smallest number generating the multiplicative group and satisfying the additional compatibility condition $n^{257}=32$ we have to find the smallest binary number $i_1i_2 \hdots i_{16}$ (larger than 255) satisfying

(i1*a*b*c*t+i2*b*c*t+i3*a*c*t+i4*c*t+i5*a*b*t+i6*b*t+
i7*a*t+i8*t+i9*a*b*c+i10*b*c+i11*a*c+i12*c+i13*a*b+
i14*b+i15*a+i16)^257=a*c

It takes a 2.4GHz 2Gb-RAM MacBook not that long to decide that the requested generator is 1051 (killing another optimistic conjecture that these generators might be 2-powers). So, we seat Knight $K_{1051}$ at root $e^{2 \pi i/65535}$ and can then arrange seatings for all Knight queued up until we reach the 65536-th! In particular, the first Knight we couldn’t place before, that is Knight $K_{256}$ , will be seated at root $e^{65826 \pi i/65535}$ .

If you’re lucky enough to own a computer with more RAM, or have the patience to make the search more efficient and get the seating arrangement for the next Fermat-field, please drop a comment.

I’ll leave you with another Lenstra-exercise which shouldn’t be too difficult for you to solve now : “Prove that $x^3=2^{2^i}$ has three solutions in $\mathbb{N}$ for each $i \geq 2$ .”

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Conway, fun, Galois, games, ONAG, ordinals

The odd knights of the round table

Jan 28th

Posted by lievenlb in games

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Here’s a tiny problem illustrating our limited knowledge of finite fields : “Imagine an infinite queue of Knights $\{ K_1,K_2,K_3,\hdots \}$ , waiting to be seated at the unit-circular table. The master of ceremony (that is, you) must give Knights K_a and K_b a place at an odd root of unity, say $\omega_a$ and $\omega_b$ , such that the seat at the odd root of unity $\omega_a \times \omega_b$ must be given to the Knight $K_{a \otimes b}$ , where $a \otimes b$ is the Nim-multiplication of and . Which place would you offer to Knight $K_{16}$ , or Knight K_n , or, if you’re into ordinals, Knight $K_{\omega}$ ?”

What does this have to do with finite fields? Well, consider the simplest of all finite field $\mathbb{F}_2 = \{ 0,1 \}$ and consider its algebraic closure $\overline{\mathbb{F}_2}$ . Last year, we’ve run a series starting here, identifying the field $\overline{\mathbb{F}_2}$ , following John H. Conway in ONAG, with the set of all ordinals smaller than $\omega^{\omega^{\omega}}$ , given the Nim addition and multiplication. I know that ordinal numbers may be intimidating at first, so let’s just restrict to ordinary natural numbers for now. The Nim-addition of two numbers $n \oplus m$ can be calculated by writing the numbers n and m in binary form and add them without carrying. For example, $9 \oplus 1 = 1001+1 = 1000 = 8$ . Nim-multiplication is slightly more complicated and is best expressed using the so-called Fermat-powers $F_n = 2^{2^n}$ . We then demand that $F_n \otimes m = F_n \times m$ whenever m < F_n and $F_n \otimes F_n = \frac{3}{2}F_n$ . Distributivity wrt. $\oplus$ can then be used to calculate arbitrary Nim-products. For example, $8 \otimes 3 = (4 \otimes 2) \otimes (2 \oplus 1) = (4 \otimes 3) \oplus (4 \otimes 2) = 12 \oplus 8 = 4$ . Conway’s remarkable result asserts that the ordinal numbers, equipped with Nim addition and multiplication, form an algebraically closed field of characteristic two. The closure $\overline{\mathbb{F}_2}$ is identified with the subfield of all ordinals smaller than $\omega^{\omega^{\omega}}$ . For those of you who don’t feel like going transfinite, the subfield $~(\mathbb{N},\oplus,\otimes)$ is identified with the quadratic closure of $\mathbb{F}_2$ .

The connection between $\overline{\mathbb{F}_2}$ and the odd roots of unity has been advocated by Alain Connes in his talk before a general public at the IHES : “L’ange de la géométrie, le diable de l’algèbre et le corps à un élément” (the angel of geometry, the devil of algebra and the field with one element). He describes its content briefly in this YouTube-video

At first it was unclear to me which ‘coupling-problem’ Alain meant, but this has been clarified in his paper together with Caterina Consani Characteristic one, entropy and the absolute point. The non-zero elements of $\overline{\mathbb{F}_2}$ can be identified with the set of all odd roots of unity. For, if x is such a unit, it belongs to a finite subfield of the form $\mathbb{F}_{2^n}$ for some n, and, as the group of units of any finite field is cyclic, x is an element of order 2^n-1 . Hence, $\mathbb{F}_{2^n}- \{ 0 \}$ can be identified with the set of 2^n-1 -roots of unity, with $e^{2 \pi i/n}$ corresponding to a generator of the unit-group. So, all elements of $\overline{\mathbb{F}_2}$ correspond to an odd root of unity. The observation that we get indeed all odd roots of unity may take you a couple of seconds¹.

Assuming we succeed in fixing a one-to-one correspondence between the non-zero elements of $\overline{\mathbb{F}_2}$ and the odd roots of unity $\mu_{odd}$ respecting multiplication, how can we recover the addition on $\overline{\mathbb{F}_2}$ ? Well, here’s Alain’s coupling function, he ties up an element x of the algebraic closure to the element s(x)=x+1 (and as we are in characteristic two, this is an involution, so also the element tied up to x+1 is s(x+1)=(x+1)+1=x. The clue being that multiplication together with the coupling map s allows us to compute any sum of two elements as $x+y=x \times s(\frac{y}{x}) = x \times (\frac{y}{x}+1)$ . For example, all information about the finite field $\mathbb{F}_{2^4}$ is encoded in this identification with the 15-th roots of unity, together with the pairing s depicted as

Okay, we now have two identifications of the algebraic closure $\overline{\mathbb{F}_2}$ : the smaller ordinals equipped with Nim addition and Nim multiplication and the odd roots of unity with complex-multiplication and the Connes-coupling s. The question we started from asks for a general recipe to identify these two approaches.

To those of you who are convinced that finite fields (LOL, even characteristic two!) are objects far too trivial to bother thinking about : as far as I know, NOBODY knows how to do this explicitly, even restricting the ordinals to merely the natural numbers!

Please feel challenged! To get you started, I’ll show you how to place the first 15 Knights and give you a procedure (though far from explicit) to continue. Here’s the Nim-picture compatible with that above

To verify this, and to illustrate the general strategy, I’d better hand you the Nim-tables of the first 16 numbers. Here they are

It is known that the finite subfields of $~(\mathbb{N},\oplus,\otimes)$ are precisely the sets of numbers smaller than the Fermat-powers F_n . So, the first one is all numbers smaller than F_1=4 (check!). The smallest generator of the multiplicative group (of order 3) is 2, so we take this to correspond to the unit-root $e^{2 \pi i/3}$ . The next subfield are all numbers smaller than F_2 = 16 and its multiplicative group has order 15. Now, choose the smallest integer k which generates this group, compatible with the condition that $k^{\otimes 5}=2$ . Verify that this number is 4 and that this forces the identification and coupling given above.

The next finite subfield would consist of all natural numbers smaller than F_3=256 . Hence, in this field we are looking for the smallest number k generating the multiplicative group of order 255 satisfying the extra condition that $k^{\otimes 17}=4$ which would fix an identification at that level. Then, the next level would be all numbers smaller than F_4=65536 and again we would like to find the smallest number generating the multiplicative group and such that the appropriate power is equal to the aforementioned k, etc. etc.

Can you give explicit (even inductive) formulae to achieve this? I guess even the problem of placing Knight 16 will give you a couple of hours to think about… (to be continued).

If m is odd, then (2,m)=1 and so 2 is a unit in the finite cyclic group $~(\mathbb{Z}/m\mathbb{Z})^*$ whence , so the m-roots of unity lie within those of order [↩]

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arxiv, Connes, Conway, fun, Galois, games, geometry, ONAG, ordinals, simples

On2 : extending Lenstra’s list

Jan 27th

Posted by lievenlb in games

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We have seen that John Conway defined a nim-addition and nim-multiplication on the ordinal numbers in such a way that the subfield $[\omega^{\omega^{\omega}}] \simeq \overline{\mathbb{F}}_2$ is the algebraic closure of the field on two elements. We’ve also seen how to do actual calculations in that field provided we can determine the mystery elements $\alpha_p$ , which are the smallest ordinals not being a p-th power of ordinals lesser than $[\omega^{\omega^{k-1}}]$ if is the k+1 -th prime number.

Hendrik Lenstra came up with an effective method to compute these elements $\alpha_p$ requiring a few computations in certain finite fields. I’ll give a rundown of his method and refer to his 1977-paper “On the algebraic closure of two” for full details.

For any ordinal $\alpha < \omega^{\omega^{\omega}}$ define its degree $d(\alpha)$ to be the degree of minimal polynomial for $\alpha$ over $\mathbb{F}_2 = [2]$ and for each prime number let f(p) be the smallest number such that is a divisor of 2^h-1 (clearly f(p) is a divisor of p-1 ).

In the previous post we have already defined ordinals $\kappa_{p^k}=[\omega^{\omega^{k-1}.p^{n-1}}]$ for prime-power indices, but we now need to extend this definition to allow for all indices. So. let be a natural number, the smallest prime number dividing and the highest power of dividing . Let g=[h/q] , then Lenstra defines

$\kappa_h = \begin{cases} \kappa_q~\text{if q divides}~d(\kappa_q)~\text{ and} \\ \kappa_g + \kappa_q = [\kappa_g + \kappa_q]~\text{otherwise} \end{cases}$

With these notations, the main result asserts the existence of natural numbers m,m' such that

$\alpha_p = [\kappa_{f(p)} + m] = [\kappa_{f(p)}] + m'$

Now, assume by induction that we have already determined the mystery numbers $\alpha_r$ for all odd primes r < p , then by teh argument of last time we can effectively compute in the field $[\kappa_p]$ . In particular, we can compute for every element its multiplicative order $ord(\beta)$ and therefore also its degree $d(\beta)$ which has to be the smallest number such that $ord(\beta)$ divides [2^h-1] .

Then, by the main result we only have to determine the smallest number m such that $\beta = [\kappa_{f(p)} +m]$ is not a p-th power in $\kappa_p$ which is equivalent to the condition that

$\beta^{(2^{d(\beta)}-1)/p} \not= 1$ if divides $[2^{d(\beta)}-1]$

All these conditions can be verified within suitable finite fields and hence are effective. In this manner, Lenstra could extend Conway’s calculations (probably using a home-made finite field program running on a slow 1977 machine) :

$\begin{array}{c|c|c} p & f(p) & \alpha_p \\ \hline 3 & 2 & [2] \\ 5 & 4 & [4] \\ 7 & 3 & [\omega]+1 \\ 11 & 10 & [\omega^{\omega}]+1 \\ 13 & 12 & [\omega]+4 \\ 17 & 8 & [16] \\ 19 & 18 & [\omega^3]+4 \\ 23 & 11 & [\omega^{\omega^3}]+1 \\ 29 & 28 & [\omega^{\omega^2}]+4 \\ 31 & 5 & [\omega^{\omega}]+1 \\ 37 & 36 & [\omega^3]+4 \\ 41 & 20 & [\omega^{\omega}]+1 \\ 43 & 14 & [\omega^{\omega^2}]+ 1 \end{array}$

Right, so let’s try the case p=47 . To begin, f(47)=23 whence we have to determine the smallest field containg $\kappa_{23}$ . By induction (Lenstra’s tabel) we know already that

$\kappa_{23}^{23} = \kappa_{11} + 1 = [\omega^{\omega^3}]+1$ and $\kappa_{11}^{11} = \kappa_5 + 1 = [\omega^{\omega}]+1$ and $\kappa_5^5=[4]$

Because the smallest field containg is $[16]=\mathbb{F}_{2^4}$ we have that $\mathbb{F}_2(4,\kappa_5,\kappa_{11}) \simeq \mathbb{F}_{2^{220}}$ . We can construct this finite field, together with a generator of its multiplicative group in Sage via

sage: f1.< a >=GF(2^220)

In this field we have to pinpoint the elements $4,\kappa_5$ and $\kappa_{11}$ . As has order in $\mathbb{F}_{2^4}$ we know that $\kappa_5$ has order . Hence we can take $\kappa_5 = a^{(2^{220}-1)/75}$ and then $4=\kappa_5^5$ .

If we denote $\kappa_5$ by x5 we can obtain $\kappa_{11}$ as x11 by the following sage-commands

sage: c=x5+1

sage: x11=c.nth_root(11)

It takes about 7 minutes to find x11 on a 2.4 GHz MacBook. Next, we have to set up the field extension determined by $\kappa_{23}$ (which we will call x in sage). This is done as follows

sage: p1.=PolynomialRing(f1)



sage: f=x^23-x11-1

sage: F2=f1.extension(f,'u')

The MacBook needed 8 minutes to set up this field which is isomorphic to $\mathbb{F}_{2^{5060}}$ . The relevant number is therefore $n=\frac{2^{5060}-1}{47}$ which is the gruesome

34648162040462867047623719793206539850507437636617898959901744136581
259144476645069239839415478030722644334257066691823210120548345667203443
317743531975748823386990680394012962375061822291120459167399032726669613
442804392429947890878007964213600720766879334103454250982141991553270171
938532417844211304203805934829097913753132491802446697429102630902307815
301045433019807776921086247690468136447620036910689177286910624860871748
150613285530830034500671245400628768674394130880959338197158054296625733
206509650361461537510912269982522844517989399782602216622257291361930850
885916974186835958466930689748400561295128553674118498999873244045842040
080195019701984054428846798610542372150816780493166669821114184374697446
637066566831036116390063418916814141753876530004881539570659100352197393
997895251223633176404672792711603439161147155163219282934597310848529360
118189507461132290706604796116111868096099527077437183219418195396666836
014856037176421475300935193266597196833361131333604528218621261753883518
667866835204501888103795022437662796445008236823338104580840186181111557
498232520943552183185687638366809541685702608288630073248626226874916669
186372183233071573318563658579214650042598011275864591248749957431967297
975078011358342282941831582626985121760847852546207377440873367589369439
085660784239080183415569559585998884824991911321095149718147110882474280
968166266224151511519773175933506503369761671964823112231808283557885030
984081329986188655169245595411930535264918359325712373064120338963742590
76555755141425

Remains ‘only’ to take x,x+1,etc. to the n-th power and verify which is the first to be unequal to 1. For this it is best to implement the usual powering trick (via digital expression of the exponent) in the field F2, something like

sage: def power(e,n): ...: le=n.bits() ...: v=n.digits() ...: mn=F2(e) ...: out=F2(1) ...: i=0 ...: while i< le : ...: if v[i]==1 : out=F2(outmn) ...: m=F2(mnmn) ...: mn=F2(m) ...: i=i+1 ...: return(out) ...:

then it takes about 20 seconds to verify that power(x,n)=1 but that power(x+1,n) is NOT! That is, we just checked that $\alpha_{47}=\kappa_{11}+1$ .

It turns out that 47 is the hardest nut to crack, the following primes are easier. Here’s the data (if I didn’t make mistakes…)

$\begin{array}{c|c|c} p & f(p) & \alpha_p \\ \hline 47 & 23 & [\omega^{\omega^{7}}]+1 \\ 53 & 52 & [\omega^{\omega^4}]+1 \\ 59 & 58 & [\omega^{\omega^8}]+1 \\ 61 & 60 & [\omega^{\omega}]+[\omega] \\ 67 & 66 & [\omega^{\omega^3}]+[\omega] \end{array}$

It seems that Magma is substantially better at finite field arithmetic, so if you are lucky enough to have it you’ll have no problem finding $\alpha_p$ for all primes less than 100 by the end of the day. If you do, please drop a comment with the results…

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Conway, mac

On2 : Conway’s nim-arithmetics

Jan 26th

Posted by lievenlb in games

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Last time we did recall Cantor’s addition and multiplication on ordinal numbers. Note that we can identify an ordinal number $\alpha$ with (the order type of) the set of all strictly smaller ordinals, that is, $\alpha = \{ \alpha'~:~\alpha' < \alpha \}$ . Given two ordinals $\alpha$ and $\beta$ we will denote their Cantor-sums and products as $[ \alpha + \beta]$ and $[\alpha . \beta]$ .

The reason for these square brackets is that John Conway constructed a well behaved nim-addition and nim-multiplication on all ordinals $\wis{On}_2$ by imposing the ‘simplest’ rules which make $\wis{On}_2$ into a field. By this we mean that, in order to define the addition $\alpha + \beta$ we must have constructed before all sums $\alpha' + \beta$ and $\alpha + \beta'$ with $\alpha' < \alpha$ and $\beta' < \beta$ . If + is going to be a well-defined addition on $\wis{On}_2$ clearly $\alpha + \beta$ cannot be equal to one of these previously constructed sums and the ‘simplicity rule’ asserts that we should take $\alpha+\beta$ the least ordinal different from all these sums $\alpha'+\beta$ and $\alpha+\beta'$ . In symbols, we define

$\alpha+ \beta = \wis{mex} \{ \alpha'+\beta,\alpha+ \beta'~|~\alpha' < \alpha, \beta' < \beta \}$

where $\wis{mex}$ stands for ‘minimal excluded value’. If you’d ever played the game of Nim you will recognize this as the Nim-addition, at least when $\alpha$ and $\beta$ are finite ordinals (that is, natural numbers) (to nim-add two numbers n and m write them out in binary digits and add without carrying). Alternatively, the nim-sum n+m can be found applying the following two rules :

the nim-sum of a number of distinct 2-powers is their ordinary sum (e.g. , and,
the nim-sum of two equal numbers is 0.

So, all we have to do is to write numbers n and m as sums of two powers, scratch equal terms and add normally. For example, 13+7=(8+4+1)+(4+2+1)=8+2=10 (of course this is just digital sum without carry in disguise).

Here’s the beginning of the nim-addition table on ordinals. For example, to define 13+7 we have to look at all values in the first 7 entries of the row of 13 (that is, $\{ 13,12,15,14,9,8,11 \}$ ) and the first 13 entries in the column of 7 (that is, $\{ 7,6,5,4,3,2,1,0,15,14,13,12,11 \}$ ) and find the first number not included in these two sets (which is indeed ).

In fact, the above two rules allow us to compute the nim-sum of any two ordinals. Recall from last time that every ordinal can be written uniquely as as a finite sum of (ordinal) 2-powers : $\alpha = [2^{\alpha_0} + 2^{\alpha_1} + \hdots + 2^{\alpha_k}]$ , so to determine the nim-sum $\alpha+\beta$ we write both ordinals as sums of ordinal 2-powers, delete powers appearing twice and take the Cantor ordinal sum of the remaining sum.

Nim-multiplication of ordinals is a bit more complicated. Here’s the definition as a minimal excluded value

$\alpha.\beta = \wis{mex} \{ \alpha'.\beta + \alpha.\beta' – \alpha'.\beta' \}$

for all $\alpha' < \alpha, \beta' < \beta$ . The rationale behind this being that both $\alpha-\alpha'$ and $\beta – \beta'$ are non-zero elements, so if $\wis{On}_2$ is going to be a field under nim-multiplication, their product should be non-zero (and hence strictly greater than 0), that is, $~(\alpha-\alpha').(\beta-\beta') > 0$ . Rewriting this we get $\alpha.\beta > \alpha'.\beta+\alpha.\beta'-\alpha'.\beta'$ and again the ‘simplicity rule’ asserts that $\alpha.\beta$ should be the least ordinal satisfying all these inequalities, leading to the $\wis{mex}$ -definition above. The table gives the beginning of the nim-multiplication table for ordinals. For finite ordinals n and m there is a simple 2 line procedure to compute their nim-product, similar to the addition-rules mentioned before :

the nim-product of a number of distinct Fermat 2-powers (that is, numbers of the form $2^{2^n}$ ) is their ordinary product (for example, ), and,
the square of a Fermat 2-power is its sesquimultiple (that is, the number obtained by multiplying with $1\frac{1}{2}$ in the ordinary sense). That is,

Using these rules, associativity and distributivity and our addition rules it is now easy to work out the nim-multiplication n.m : write out n and m as sums of (multiplications by 2-powers) of Fermat 2-powers and apply the rules. Here’s an example

5.9=(4+1).(4.2+1)=4^2.2+4.2+4+1=6.2+8+4+1=(4+2).2+13=4.2+2^2+13=8+3+13=6

Clearly, we’d love to have a similar procedure to calculate the nim-product $\alpha.\beta$ of arbitrary ordinals, or at least those smaller than $\omega^{\omega^{\omega}}$ (recall that Conway proved that this ordinal is isomorphic to the algebraic closure $\overline{\mathbb{F}}_2$ of the field of two elements). From now on we restrict to such ‘small’ ordinals and we introduce the following special elements :

$\kappa_{2^n} = [2^{2^{n-1}}]$ (these are the Fermat 2-powers) and for all primes p > 2 we define $\kappa_{p^n} = [\omega^{\omega^{k-1}.p^{n-1}}]$ where is the number of primes strictly smaller than (that is, for p=3 we have k=1, for p=5, k=2 etc.).

Again by associativity and distributivity we will be able to multiply two ordinals $< \omega^{\omega^{\omega}}$ if we know how to multiply a product

$[\omega^{\alpha}.2^{n_0}].[\omega^{\beta}.2^{m_0}]$ with $\alpha,\beta < [\omega^{\omega}]$ and $n_0,m_0 \in \N$ .

Now, $\alpha$ can be written uniquely as $[\omega^t.n_t+\omega^{t-1}.n_{t-1}+\hdots+\omega.n_2 + n_1]$ with t and all n_i natural numbers. Write each n_k in base where is the k+1 -th prime number, that is, we have for $n_0,n_1,\hdots,n_t$ an expression

$n_k=[\sum_j p^j.m(j,k)]$ with $0 \leq m(j,k) < p$

The point of all this is that any of the special elements we want to multiply can be written as a unique expression as a decreasing product

$[\omega^{\alpha}.2^{n_0}] = [ \prod_q \kappa_q^m(q) ]$

where runs over all prime powers. The crucial fact now is that for this decreasing product we have a rule similar to addition of 2-powers, that is Conway-products coincide with the Cantor-products

$[ \prod_q \kappa_q^m(q) ] = \prod_q \kappa_q^m(q)$

But then, using associativity and commutativity of the Conway-product we can ‘nearly’ describe all products $[\omega^{\alpha}.2^{n_0}].[\omega^{\beta}.2^{m_0}]$ . The remaining problem being that it may happen that for some q we will end up with an exponent m(q)+m(q')>p . But this can be solved if we know how to take p-powers. The rules for this are as follows

$~(\kappa_{2^n})^2 = \kappa_{2^n} + \prod_{1 \leq i < n} \kappa_{2^i}$ , for 2-powers, and,

$~(\kappa_{p^n})^p = \kappa_{p^{n-1}}$ for a prime p > 2 and for $n \geq 2$ , and finally

$~(\kappa_p)^p = \alpha_p$ for a prime p > 2 , where $\alpha_p$ is the smallest ordinal $< \kappa_p$ which cannot be written as a p-power $\beta^p$ with $\beta < \kappa_p$ . Summarizing : if we will be able to find these mysterious elements $\alpha_p$ for all prime numbers p, we are able to multiply in $[\omega^{\omega^{\omega}}]=\overline{\mathbb{F}}_2$ .

Let us determine the first one. We have that $\kappa_3 = \omega$ so we are looking for the smallest natural number $n < \omega$ which cannot be written in num-multiplication as n=m^3 for $m < \omega$ (that is, also a natural number). Clearly 1=1^3 but what about 2? Can 2 be a third root of a natural number wrt. nim-multiplication? From the tabel above we see that 2 has order 3 whence its cube root must be an element of order 9. Now, the only finite ordinals that are subfields of $\wis{On}_2$ are precisely the Fermat 2-powers, so if there is a finite cube root of 2, it must be contained in one of the finite fields $[2^{2^n}]$ (of which the mutiplicative group has order $2^{2^n}-1$ and one easily shows that 9 cannot be a divisor of any of the numbers $2^{2^n}-1$ , that is, 2 doesn’t have a finte 3-th root in nim! Phrased differently, we found our first mystery number $\alpha_3 = 2$ . That is, we have the marvelous identity in nim-arithmetic

$\omega^3 = 2$

Okay, so what is $\alpha_5$ ? Well, we have $\kappa_5 = [\omega^{\omega}]$ and we have to look for the smallest ordinal which cannot be written as a 5-th root. By inspection of the finite nim-table we see that 1,2 and 3 have 5-th roots in $\omega$ but 4 does not! The reason being that 4 has order 15 (check in the finite field [16]) and 25 cannot divide any number of the form $2^{2^n}-1$ . That is, $\alpha_5=4$ giving another crazy nim-identity

$~(\omega^{\omega})^5 = 4$

And, surprises continue to pop up… Conway showed that $\alpha_7 = \omega+1$ giving the nim-identity $~(\omega^{\omega^2})^7 = \omega+1$ . The proof of this already uses some clever finite field arguments. Because 7 doesn’t divide any number $2^{2^n}-1$ , none of the finite subfields $[2^{2^n}]$ contains a 7-th root of unity, so the 7-power map is injective whence surjective, so all finite ordinal have finite 7-th roots! That is, $\alpha_7 \geq \omega$ . Because $\omega$ lies in a cubic extension of the finite field [4], the field generated by $\omega$ has 64 elements and so its multiplicative group is cyclic of order 63 and as $\omega$ has order 9, it must be a 7-th power in this field. But, as the only 7th powers in that field are precisely the powers of $\omega$ and by inspection $\omega+1$ is not a 7-th power in that field (and hence also not in any field extension obtained by adjoining square, cube and fifth roots) so $\alpha_7=\omega +1$ .

Conway did stop at $\alpha_7$ but I’ve always been intrigued by that one line in ONAG p.61 : “Hendrik Lenstra has computed $\alpha_p$ for $p \leq 43$ “. Next time we will see how Lenstra managed to do this and we will use sage to extend his list a bit further, including the first open case : $\alpha_{47}= \omega^{\omega^7}+1$ .

For an enjoyable video on all of this, see Conway’s MSRI lecture on Infinite Games. The nim-arithmetic part is towards the end of the lecture but watching the whole video is a genuine treat!

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Conway, games, simples

On2 : transfinite number hacking

Jan 8th

Posted by lievenlb in games

1 comment

In ONAG, John Conway proves that the symmetric version of his recursive definition of addition and multiplcation on the surreal numbers make the class On of all Cantor’s ordinal numbers into an algebraically closed Field of characteristic two : On2 (pronounced ‘Onto’), and, in particular, he identifies a subfield with the algebraic closure of the field of two elements. What makes all of this somewhat confusing is that Cantor had already defined a (badly behaving) addition, multiplication and exponentiation on ordinal numbers.

Over the last week I’ve been playing a bit with sage to prove a few exotic identities involving ordinal numbers. Here’s one of them ( $\omega$ is the first infinite ordinal number, that is, $\omega=\{ 0,1,2,\hdots \}$ ),

$~(\omega^{\omega^{13}})^{47} = \omega^{\omega^7} + 1$

answering a question in Hendrik Lenstra’s paper Nim multiplication.

However, it will take us a couple of posts before we get there. Let’s begin by trying to explain what brought this on. On september 24th 2008 there was a meeting, intended for a general public, called a la rencontre des dechiffeurs, celebrating the 50th birthday of the IHES.

One of the speakers was Alain Connes and the official title of his talk was “L’ange de la géométrie, le diable de l’algèbre et le corps à un élément” (the angel of geometry, the devil of algebra and the field with one element). Instead, he talked about a seemingly trivial problem : what is the algebraic closure of $\mathbb{F}_2$ , the field with two elements? My only information about the actual content of the talk comes from the following YouTube-blurb

Alain argues that we do not have a satisfactory description of $\overline{\mathbb{F}}_2$ , the algebraic closure of $\mathbb{F}_2$ . Naturally, it is the union (or rather, limit) of all finite fields $\mathbb{F}_{2^n}$ , but, there are too many non-canonical choices to make here.

Recall that $\mathbb{F}_{2^k}$ is a subfield of $\mathbb{F}_{2^l}$ if and only if is a divisor of and so we would have to take the direct limit over the integers with respect to the divisibility relation… Of course, we can replace this by an increasing sequence of a selection of cofinal fields such as

$\mathbb{F}_{2^{1!}} \subset \mathbb{F}_{2^{2!}} \subset \mathbb{F}_{2^{3!}} \subset \hdots$

But then, there are several such suitable sequences! Another ambiguity comes from the description of $\mathbb{F}_{2^n}$ . Clearly it is of the form $\mathbb{F}_2[x]/(f(x))$ where f(x) is a monic irreducible polynomial of degree , but again, there are several such polynomials. An attempt to make a canonical choice of polynomial is to take the ‘first’ suitable one with respect to some natural ordering on the polynomials. This leads to the so called Conway polynomials.

Conway polynomials for the prime have only been determined up to degree 400-something, so in the increasing sequence above we would already be stuck at the sixth term $\mathbb{F}_{2^{6!}}$ …

So, what Alain Connes sets as a problem is to find another, more canonical, description of $\overline{\mathbb{F}}_2$ . The problem is not without real-life interest as most finite fields appearing in cryptography or coding theory are subfields of $\overline{\mathbb{F}}_2$ .

(My guess is that Alain originally wanted to talk about the action of the Galois group on the roots of unity, which would be the corresponding problem over the field with one element and would explain the title of the talk, but decided against it. If anyone knows what ‘coupling-problem’ he is referring to, please drop a comment.)

Surely, Connes is aware of the fact that there exists a nice canonical recursive construction of $\overline{\mathbb{F}}_2$ due to John Conway, using Georg Cantor’s ordinal numbers.

In fact, in chapter 6 of his book On Numbers And Games, John Conway proves that the symmetric version of his recursive definition of addition and multiplcation on the surreal numbers make the class $\wis{On}$ of all Cantor’s ordinal numbers into an algebraically closed Field of characteristic two : $\wis{On}_2$ (pronounced ‘Onto’), and, in particular, he identifies a subfield

$\overline{\mathbb{F}}_2 \simeq [ \omega^{\omega^{\omega}} ]$

with the algebraic closure of $\mathbb{F}_2$ . What makes all of this somewhat confusing is that Cantor had already defined a (badly behaving) addition, multiplication and exponentiation on ordinal numbers. To distinguish between the Cantor/Conway arithmetics, Conway (and later Lenstra) adopt the convention that any expression between square brackets refers to Cantor-arithmetic and un-squared ones to Conway’s. So, in the description of the algebraic closure just given $[ \omega^{\omega^{\omega}} ]$ is the ordinal defined by Cantor-exponentiation, whereas the exotic identity we started out with refers to Conway’s arithmetic on ordinal numbers.

Let’s recall briefly Cantor’s ordinal arithmetic. An ordinal number $\alpha$ is the order-type of a totally ordered set, that is, if there is an order preserving bijection between two totally ordered sets then they have the same ordinal number (or you might view $\alpha$ itself as a totally ordered set, namely the set of all strictly smaller ordinal numbers, so e.g. $0= \emptyset,1= \{ 0 \},2=\{ 0,1 \},\hdots$ ).

For two ordinals $\alpha$ and $\beta$ , the addition $[\alpha + \beta ]$ is the order-type of the totally ordered set $\alpha \sqcup \beta$ (the disjoint union) ordered compatible with the total orders in $\alpha$ and $\beta$ and such that every element of $\beta$ is strictly greater than any element from $\alpha$ . Observe that this definition depends on the order of the two factors. For example, $[1 + \omega] = \omega$ as there is an order preserving bijection $\{ \tilde{0},0,1,2,\hdots \} \rightarrow \{ 0,1,2,3,\hdots \}$ by $\tilde{0} \mapsto 0,n \mapsto n+1$ . However, $\omega \not= [\omega + 1]$ as there can be no order preserving bijection $\{ 0,1,2,\hdots \} \rightarrow \{ 0,1,2,\hdots,0_{max} \}$ as the first set has no maximal element whereas the second one does. So, Cantor’s addition has the bad property that it may be that $[\alpha + \beta] \not= [\beta + \alpha]$ .

The Cantor-multiplication $\alpha . \beta$ is the order-type of the product-set $\alpha \times \beta$ ordered via the last differing coordinate. Again, this product has the bad property that it may happen that $[\alpha . \beta] \not= [\beta . \alpha]$ (for example $[2 . \omega ] \not=[ \omega . 2 ]$ ). Finally, the exponential $\beta^{\alpha}$ is the order type of the set of all maps $f~:~\alpha \rightarrow \beta$ such that $f(a) \not=0$ for only finitely many $a \in \alpha$ , and ordered via the last differing function-value.

Cantor’s arithmetic allows normal-forms for ordinal numbers. More precisely, with respect to any ordinal number $\gamma \geq 2$ , every ordinal number $\alpha \geq 1$ has a unique expression as

$\alpha = [ \gamma^{\alpha_0}.\eta_0 + \gamma^{\alpha_1}.\eta_1 + \hdots + \gamma^{\alpha_m}.\eta_m]$

for some natural number and such that $\alpha \geq \alpha_0 > \alpha_1 > \hdots > \alpha_m \geq 0$ and all $1 \leq \eta_i < \gamma$ . In particular, taking the special cases $\gamma = 2$ and $\gamma = \omega$ , we have the following two canonical forms for any ordinal number $\alpha$

$[ 2^{\alpha_0} + 2^{\alpha_1} + \hdots + 2^{\alpha_m}] = \alpha = [ \omega^{\beta_0}.n_0 + \omega^{\beta_1}.n_1 + \hdots + \omega^{\beta_k}.n_k]$

with m,k,n_i natural numbers and $\alpha \geq \alpha_0 > \alpha_1 > \hdots > \alpha_m \geq 0$ and $\alpha \geq \beta_0 > \beta_1 > \hdots > \beta_k \geq 0$ . Both canonical forms will be important when we consider the (better behaved) Conway-arithmetic on $\wis{On}_2$ , next time.

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Connes, Conway, cryptography, Galois, games, geometry

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