Abstract: Corewar is a programming game played on a board simulating computer memory. This board is called the core, and consists of N cells, each containing an assembly language command. Programs run simultaneously, with the objective of killing other programs in core and thus becoming the sole owner of the core. Addition, subtraction, and multiplication give the core a ℤ/Nℤ ring structure. N is called the coresize. Popular values of N are 8000, 8192, and 55440.

In this article, we examine ways to optimize simple bombing or scanning step sizes. We give formulas to measure ‘optimization’ of a step for a given coresize, and prove that imp-pairs have the same optima score.

## Contents

1. Terminology
2. Euclid's Algorithm and Continued Fractions
3. Some Models for Bombing Core (incomplete)
4. A Better Formula for the Length of Intervals
5. The Optima Step Formula

## Terminology

We use the following symbols:

• ℤ⁺ : The set of all positive integers.
• (A,B) : The greatest common factor of positive integers A,B.
• x⌋ : The integral part of a non-negative real number x.
• A|B : The integer A divides the integer B evenly (that is, without remainder).

## Euclid's Algorithm and Continued Fractions

In this section, we build the mathematical framework for the results of this article. We explore the relationship betwwen Euclid's Algorithm and continued fractions, and prove a classic result of Euler, who seems to have anticipated Corewar in 1764.

Recall that Euclid's Algorithm computes (A,B), with integers A>B>0, as follows:

Let X0 = A and X1 = B.

Inductively define

Xi+1 = Xi-1 - Xi*Yi, (1)

with

Yi=⌊Xi-1/Xi⌋. (2)

Clearly, 0 ≤ Xi+1 < Xi. If Xi+1=0, our induction must stop (because Yi+1 is undefined). In this case, define k to be the largest i for which Xi is non-zero. Since Xi+1<Xi for all i, the method of least descent (aka Noetherian Induction) assures that such a k exists.

If N|Xi-1 and N|Xi, then N|Xi+1 by definition of Xi+1. The same reasoning gives us a result in the opposite direction: If N|Xi+1 and N|Xi, then N|Xi-1. Using induction, we see that N|A and N|B if and only if N|Xk (and N|Xk+1, which is always true) for any integer N>0. Hence, Xk and (A,B) have the same factors, so Xk=(A,B).

Definition: Given A_1,A_2,...,A_n in Z+, we define the continued fraction generated by the A_i's as

<A_1,A_2,...,A_n> = A_1+1/(A_2+1/...(A_{n-1}+1/A_n)...) (3) We define the length of <A_1,...,A_n> to be n. Clearly, <A_1,...,A_n> is always rational. We define the numerator of <A_1,...,A_n> to be p, where p/q is the reduced fraction with p/q=<A_1,...,A_n> and denote it by p=[A_1,...,A_n]. We allow "empty" numerators, defining []=1. Propositon 1: Given A,B in Z+ with A>B, we have A/B = <Y_1,...,Y_k> where the Y_i's and k are defined by Euclid's algorithm using (1) and (2). Proof: We the notations of Euclid's algorithm above and induction on k. When k=1, we have X_2=0 (by definition of k), so we know by (1) with i=1 that 0 = X_0-Y_1*X_1. Hence, <Y_1> = Y_1 = X_0/X_1. Now, suppose k>1. Note that X_1>X_2>0, so we can use Euclid's algorthm on A'= X_1, B'= X_2 to find each Y'_i, and it is easy to see that Y'_i = Y_{i+1}. Hence, for induction we may assume that X_1 / X_2 = <Y_2,Y_3,...,Y_k>. We then have <Y_1,Y_2,...,Y_k>= Y_1+1/<Y_2,Y_3,...,Y_k> = Y_1+1/(X_1/X_2) = (Y_1 * X_1 + X_2)/X_1 = X_0/X_1 = A/B.

Corollary 2: If we are given the Y_i's and m=(A,B), we can recover A and B. Proof: <Y_1,...,Y_k> is rational, hence can be uniquely written as a reduced fraction a/b. Existence: Since this fraction is reduced, we have (a,b)=1. Define A= ma, B= mb. Then A/B=a/b=<Y_1,...,Y_k> and (A,B)=m*(a,b)=1. Uniqueness: If A/B=A'/B'=<Y_1,...,Y_k> and (A,B)=(A',B')=m then A/m,B/m are in Z+ and (A/m,B/m)=1, so (A/m)/(B/m) is the reduced fraction a/b. Hence A/m=a, B/m=b. Similary, A'/m=a, B'/m=b. Thus A=A' and B=B'.

Corollary 3: Given A,B in Z+ with A>B and (A,B)=1, we have A = [Y_1,...,Y_k], B = [Y_2,...,Y_k] where the Y_i's and k are defined by Euclid's algorithm using (1) and (2). Proof: From Proposition 1, we know A/B=<Y_1,...,Y_k>. This fraction is reduced when (A,B)=1, so A = [Y_1,...,Y_k] by definition. If k=1, then B=1=[]. Now if k>1 and C/D is the reduced fraction <Y_2,...,Y_k>, then we have A/B= Y_1 + 1/(C/D) = (Y_1*C + D)/C. The fractions on both sides of this equation are reduced, hence B=C. But C = [Y_2,...,Y_k] by definition.

Corollary 4: For any continued fraction <A_1,...,A_n> with n>=2, the following identity holds: [A_1,...,A_n] = A_1*[A_2,...,A_n] + [A_3,...,A_n]. (4) Proof: If A/B=<A_1,...,A_n> and C/D=<A_2,...,A_n> are both reduced fractions, then A/B= (A_1*C + D)/C, as in the proof of Corollary 3. Multiplying both sides by B=C, we obtain: A = A_1*C + D. From Corollary 3, we have A = [A_1,...,A_n], C = [A_2,...,A_n], and D = [A_3,...,A_n].

Examples: 43/10 = <4,3,3>. Also, 43/10 = <4,3,2,1>, even though these are not the Y_i defined in (2). 43/30 = <1,2,3,4>. If (A,B)=1 and A/B = <1,1,...,1> (k times), then A=F_{k+1} and B=F_k, where F_i is the i'th term in the Fibonacci sequence {1,1,2,3,5,8,13,...}. We now come to the main result of this section, proved by Euler in 1764. Theorem 5: For any continued fraction <A_1,...,A_n>, the following relations hold: [A_1,...,A_n]=[A_n,...,A_1] (5) [A_2,...,A_n] * [A_1,...,A_{n-1}] = [A_1,...,A_n] * [A_2,...,A_{n-1}] - (-1)^n (n>1) (6) Proof: We use induction on n. When n=0 or n=1, (5) is tautological. And for n=2, we have, by (4), [A_1,A_2]= A_1 * [A_2]+ [] = A_1 * A_2 + 1. Hence, [A_2] * [A_1] = A_2 * A_1 = (A_1 * A_2 + 1)*1 -1 = [A_1,A_2]*[]-(-1)^2, so (6) holds. For n>=2, we use (4) and induction on (5) repeatedly to get: [A_1,...,A_2] = A_1 * [A_2,...,A_n] + [A_3,...,A_n] = A_1 * [A_n,...,A_2] + [A_n,...,A_3] = A_1 * (A_n * [A_{n-1},...,A_2] + [A_{n-2},...,A_2]) + A_n * [A_{n-1},...,A_3] + [A_{n-2},...,A_3] = A_1 * A_n * [A_2,...,A_{n-1}] + A_1 * [A_2,...,A_{n-2}] + A_n * [A_3,...,A_{n-1}] + [A_3,...,A_{n-2}]. If we replace (A_1,...,A_n) with (A_n,...,A_1), the first and fourth terms of the right-hand expression stay the same, and the second and third terms get swapped. Thus, [A_1,...,A_n] = [A_n,...,A_1] since they both equal the same expression. This proves (5) for all n>0. Finally, for n>=3, we can assume by induction that (6) holds for <A_2,...,A_n>. In other words, [A_2,...,A_{n-1}]*[A_3,...,A_n}] = [A_3,...,A_{n-1}]*[A_2,...,A_n] -(-1)^{n-1}. Thus, [A_1,...,A_n]*[A_2,...,A_{n-1}] = (A_1*[A_2,...,A_n]+[A_3,...,A_n]) *[A_2,...,A_{n-1}] = A_1*[A_2,...,A_n]*[A_2,...,A_{n-1}] +[A_3,...,A_n]*[A_2,...,A_{n-1}] = A_1*[A_2,...,A_n]*[A_2,...,A_{n-1}] +[A_3,...,A_{n-1}]*[A_2,...,A_n]+(-1)^n = [A_2,...,A_n]*(A_1*[A_2,...,A_{n-1}]+[A_3,...,A_{n-1}])+(-1)^n = [A_2,...,A_n]*[A_1,...,A_{n-1}]+(-1)^n. This is (6) for <A_1,...,A_n>, so by induction, (6) holds for all n>1.

Corollary 6: Given positive integers A,B, with A/B = <Y_1,Y_2,...,Y_k>, define positive integers A',B' by A'/B' = <Y_k,Y_{k-1},...,Y_1> (7) where (A',B')=(A,B) (8) Then the following relationships hold: A' = A; (9) B'*B = -(-1)^k*(A,B)^2 (mod A). (10) Proof: Let m=(A,B). Then (A/m)/(B/m) is the reduced form of A/B, whence A = m*[Y_1,...,Y_k], B = m*[Y_2,...,Y_k]. Similarly, A' = m*[Y_k,...,Y_1], B' = m*[Y_{k-1},...,Y_1]. Then (9) follows from (5), and (6) can be written as: (B/m)*(B'/m)=(A/m)*[Y_2,...,Y_{k-1}]-(-1)^k. We multiply both sides by m^2 to get B*B'=A*m*[Y_2,...,Y_{k-1}]-(-1)^k*m^2. Then (10) follows from passing to congruence (mod A).

## Some Models for Bombing Core

Let N be the coresize and let K<N be a step size. The mathematical model for core is Z/NZ, and there is a natural map f:Z->Z/NZ given by f(n)=n+N. Rather than mess with cosets, we will refer to elements of core by representatives in Z. For a<b in Z, we define the open interval (a,b) in Z/NZ to be the the image of (a,b) in Z under f. Similarly with half-open intervals [a,b).

Examples: Suppose N=8000. Then (1,4) is an open interval containing the images of 2 and 3. (K-N,K) is an interval containing everything in Z/NZ except for the image of K. And (5,6) is the empty interval. Note that [a,b)=(a-1,b) for all intervals.

Definition: The length of an interval (a,b) is the number of elements in this interval, given by

||(a,b)||= max( b-a-1, N ).

Similarly,

||[a,b)||= max ( b-a, N ).

The Bombing Model: The first bomb is dropped at K=K-N, leaving the largest unbombed interval (K-N,K). The second bomb is dropped at 2*K, and in general, the j'th bomb is dropped at j*K. The last bomb will be dropped at 0, completing the bombing run. If the gcf (K,N)=m, we say that K bombs core mod m, because only each m'th location will get bombed with this pattern. We see that N/m bombs are dropped in a bombing run.

The Partiton Model: If we consider the bombs as merely partitions of the core into intervals, we can imagine the j'th bomb being dropped between the j*K'th core location and its predecessor. Thus, after the first bomb, we would be left with the interval [K-N,K). Except for the type of interval, this model is very similar to the bombing model. In particular, the last bomb will be dropped between -1 and 0, partitioning core into N/m intervals of m elements each.

We use the partiton model to keep the math simple, but similar results hold for the bombing model.

Definition: For j>0, let U_j be the interval containing 0 in the partition after j bombs.

Theorem 7: For all j with 0<j<N/m, the j+1'st bomb partitions an interval of length ||U_j|| into two intervals of lengths ||[0,b)|| and ||[a,0)||, where U_j=[a,b).

Proof: We first make two observations: If, after j bombs, a bomb has been dropped at (i.e. before) a location c but no bomb has been dropped at c+K, then the bomb dropped at c was the j'th bomb, hence c=j*K (mod N). Similarly, if no bomb has been dropped at c but a bomb has been dropped at c+K, then the bomb dropped at c+K was the first bomb, hence c=0.

Now, if [d,e) is the interval in the partition in which the j+1'st bomb will be dropped, we must first show that ||[a,b)||=||[d,e)||. If [d,e) contains 0 then this is tautological, so we may assume that 0 is not in [d,e).

We claim that [d+K,e+K) is in the partition. If d+K is unbombed, then the first observation shows us that d is the j'th bomb. Thus, the j+1'st bomb will be dropped at d+K, so d+K is in the interval [d,e). Then e-K is also in this interval, and as it is unbombed, the second observation tells us it is 0, contrary to assumption. Similarly, e+K has been bombed (although the notation turns a bit weird). The second observation also assures us that no bombs are between d+K and e+K. This proves the claim.

By induction, either there is an interval in the partition of the form [d+g*K,e+g*K) which contains 0, or every interval of the form [d+g*K,e+g*K) with g>=0 is in the partition. In the first case, we have

||[a,b)||=||[d+g*K,e+G*K)||=e-d=||[d,e)||,

so we are done. Otherwise, since d is in the bombing run, we know that d=l*K for some l with 1<l<N/m, so with g=N/m-l, we get

d+g*K=(N/m)*K=N*(K/m)=0 (mod N),

whence [d+g*K,e+g*K) is an interval in the partition containing 0, and we have the first case after all.

We must show that...

Corollary 8: For all j<N/m, there are at most three different interval sizes in the partition after j bombs.

Proof: ...

Corollary 9: ||U_j|| is the maximum size of all intervals after j bombs.

Proof: We know that U_j is the largest interval of the three in corollary 8.

## A Better Formula for the Length of Intervals

Here is some terminology for the next result.

Let X_0=N, X_1=K, and define X_i and Y_i using Euclid's algorithm (that is, equations (1) and (2)). Further, define Z_i inductively by Z_0=0, Z_1=1, and

Z_{i+1}=Z_i*Y_i+Z_{i-1}. (11)

By proposition 1, we have X_0/X_1=(X_0/m)/(X_1/m)=<Y_1,...,Y_k>. Applying Corollary 3 to X_0/m,X_1/m, we get

X_0= m*[Y_1,...,Y_k]

X_1= m*[Y_2,...,Y_k]

and inductively using Corollary 4 with equation (1), we get

X_i= m*[Y_{i+1},...,Y_k] (0<=i<=k). (12)

Inductively using Corollary 4 with equation (11), we also get

Z_i= [Y_{i-1},...,Y_1] (1<=i<=k+1). (13)

Now we can describe the bombing sequence as a series of bombing runs by looking at the intervals U_i containing 0.

Definition: The i'th bombing run for i<k consists of all bombs dropped starting with Z_i+Z_{i-1}'th bomb and ending with Z_{i+1}+Z_i-1'th bomb. The k'th bombing run consists of all bombs dropped starting with the Z_k+z_{k-1}'th bomb and ending with the Z_{k+1}-1'th bomb.

Theorem 9: Every bomb (except for the last one) belongs to exactly one bombing run. if the j'th bomb is in the i'th bombing run, then

U_j = [-X_{i-1}+p*X_i,X_i) (for i odd) (14)

U_j = [-X_i,X_{i-1}-p*X_i) (for I even)

where p=[(j-Z_i-Z_{i-1})/Z_i]+1. Note that p, hence U_j, changes values every Z_i bombs.

Proof: We note that Z_1+Z_0=1, so the first bomb belongs to the first bombing run. We also see that the first bomb in the i+1'st bombing run is the bomb directly after the last bomb in the i'th bombing run. Finally, the last bomb of the k'th bombing run is at Z_{k+1}-1=N/m-1. Thus, all of the bombs except the N/m'th one are in some bombing interval.

We now prove (14) inductively on i. When i=1, we have p=j. According to the partition model, U_1=[-N+K,K), and as long as j*K<=N, we know that U_j=[-N+j*K,K). But j*K<=N if and only if j<=Y_1. This range of j is precisely the first bombing run, since Z_2=Y_1 and Z_1=1. Now, (14) follows for i=1. Note also that the last U_j of the first bombing run is

U_{Z_2}=[-X_0+Y_1*X_1 , X_1)

=[-X_2 , X_1),

and the next bomb is dropped at X_1-X_2.

We now prove (14) for i even. The proof of i odd is obtained by negating all bomb locations and changing intervals [-a,b) to [-b,a). Inductively, we assume that

The Z_h'th bomb dropped at X_h, for all h with 0<=h<=i. The Z_i+Z_{i-1}'th bomb dropped at X_{i-1}-X_i. We ask two questions: What values do the intervals U_j take during the i'th bombing run? And for what j do the U_j shrink? The first question is answered by corollary 8: For any j, if U_{j-1}=[-a,b) with a<b, then either U_j=U_{j-1} or U_j=[0,b) or U_j=[-a,b-a). The first case gives us no new values, and the second case only occurs when 0 is bombed, at the end of the bombing pattern, which we can ignore (since U_j isn't even defined for j=N/m). Thus, the U_j after the start of the i'th bombing run have the form [-X_i, X_{i-1}-p*X_i) with p>=1 and X_{i-1}-p*X_i>= 0, whch means p<=Y_i. We have by hypothesis that the first bomb in the i'th bombing run is dropped at X_{i-1}-X_i and shrinks U_j. So, what is the gap in time between a bombs being dropped at X_{i-1}-(p-1)*X_i and X_{i-1}-p*X_i? Since all bombs are evenly spaced, this gap will be the same for all p, and with p=1, we know that the Z_{i-1}'th bomb dropped at X_{i-1} and the Z_{i-1}+Z_i'th bomb drops at X_{i-1}-X_i. Thus the value of p increases every Z_i bombs. We know that the value of p starts at one with the Z_i+Z_{i-1}'th bomb, so we can verify that the p given in the statement of the theorem is correct. Next, we must verify that the bombs we just dropped were the i'th bombing run. We started with bomb Z_i+Z_{i-1}, and dropped Z_i bombs for each p between 1 and Y_i (except for the last bombing run, where p only ranged between 1 and Y_i-1). Thus, the last bomb in this bombing was the Z_i+Z_{i-1}+Y_i*Z_i-1=Z_i+Z_{i+1}-1'th bomb (or Z_{k+1}-1=N/m-1 for the last run). From the last two paragraphs, with p=Y_i+1, we see that the first bomb of the next bombing run is dropped at X_{i-1}-(p+1)*X_i= X_{i+1}-X_i. And with p=Y_i, we have the Z_{i+1}'th bomb is the Z_i*Y_i+Z_{i-1}'th bomb, which drops at X_{i-1}-X_i*Y_i= X_{i+1}. This finishes the induction step.

---

## The Optima Step Formula

In Corewar, the step size of a bombing or scanning pattern is the distance between successive bombing or scanning locations. Chosing a good step size is critical to the success of a warrior, so we need a way of measuring the effectiveness of a step size.

What makes one step size better than another? Here is one train of thought: Larger enemies can usually be found faster than small ones. If your step takes the same time to find small enemies as large ones, it probably needs to be optimized against larger enemies. On the other hand, if your step size is too coarse, small programs can slip through the cracks. For instance, assuming 8000 instructions in core, a step size of 4000 is bad. The first two bombs are spaced well, but the pattern only hits two core locations. An ideal pattern will divide the core into successively smaller chunks, until it has checked all locations in core (except where your code resides).

We can measure the success of this method by adding the lengths of the largest untouched segments during different stages of bombing. This sum is the optima function of a coresize and step. A lower function value between step-sizes with the same greatest common factor indicates a more optimal pattern.

Definition: The optima function is defined to be:

O(N,K)= sumj( ||Uj|| ).

Corollary 8 assures that this is the sum we want. Proposition 10: O(N,K) = O(N,N-K). Proof: In coresize N, we can consider a step-size of N-K as a step-size of K bombing in the other direction. If the U_j are the intervals containing 0 for step-size K, as above, then, by symmetry, the intervals U'_j containing 0 for step-size N-K will just be the U_j reversed; that is, if for some j, U_j=[-a,b), then U'_j=[-b,a). Note that ||U_j|| = a+b = ||U'_j||. We sum over all j to get the result.

Theorem 11: O(N,K)= sum_{i=1}^k(X_{i-1}*Z{i+1}-X_i*Z_i-X_i*Z_i*(Y_i*(Y_i-1)/2) Proof: We write O(N,K)= sum_{i=1}^k O_i(N,K), where each O_i is the sum of the ||U_j|| from the i'th bombing run. In the i'th bombing run, we have (by theorem 9) Z_i intervals of length X_{i-1}-(p-1)*X_i, for p ranging from 1 to Y_i. Adding these up, we think that O_i(N,K)= sum_{p=1}^{Y_i} Z_i*(X_{i-1}-(p-1)*X_i) = sum_{p=1}^{Y_i}(Z_i*X_{i-1}) - sum_{p=1}^{Y_i}(Z_i*X_i*(p-1)) = Y_i*Z_i*X_{i-1} - Z_i*X_i*(Y_i*(Y_i-1)/2) Well, not quite. We forgot that the last bombing run only uses the values of p from 1 to Y_k-1. Thus we must add the correction term -Z_k*(X_{k-1}-(Y_k-1)*X_k)= -Z_k*(X_{k-1}-Y_k*X_k+X_k) = -Z_k*(X_{k+1}+X_k) = -Z_k*X_k since X_{k+1}=0. We thus have O(N,K)= (15) sum_{i=1}^k( Y_i*Z_i*X_{i-1} - Z_i*X_i*(Y_i*(Y_i-1)/2) ) -Z_k*X_k. We also have the following telescoping sum: sum_{i=1}^k( X_i*Z_i - X_{i-1}*Z_{i-1}) = X_k*Z_k - X_0*Z_0 = X_k*Z_k since Z_0=0. Substituting this into (15), we get O(N,K)= (16) sum_{i=1}^k( Y_i*Z_i*X_{i-1} - Z_i*X_i*(Y_i*(Y_i-1)/2) - X_i*Z_i + X_{i-1}*Z_{i-1}). and the first and fourth terms of the sum combine thusly: Y_i*Z_i*X_{i-1} + X_{i-1}*Z_{i-1} = X_{i-1}*(Y_i*Z_i+Z_{i-1}) = X_{i-1}*Z{i+1}. to yield the formula we wished to prove.

Corollary 12: If {K,K''} is an imp-pair for coresize N, then O(N,K)=O(N,K''). Proof: Recall that {K,K''} is an imp-pair if and only if K*K'' = 1 (mod N). For any given K,N, we know that K'' exists if and only if (K,N)=1, and we know that K'' is unique. Construct K' and N' as in corollary 6. Then we have N=[Y_1,...,Y_k], K=[Y_2,...,Y_k], N'=[Y_k,...,Y_1], K'=[Y_{k-1},...,Y_1]. If we define X'_i,Y'_i, and Z'_i by X'_0=N',X'_1=K' and the recurrence relations in section 3, we get the follwing for all applicable i: Y'_i=Y_{k+1-i} X'_i=Z_{k+1-i} Z'_i=X_{k+1-i}.

If we apply theorem 11 and reverse the order of summation, we get

O(N',K')= sum_{i=1}^k(X'_{i-1}*Z'{i+1} -X'_i*Z'_i -X'_i*Z'_i*(Y'_i*(Y'_i-1)/2) = sum_{i=1}^k(X'_{k+1-(i-1)}*Z'_{k+1-(i+1)} -X'_{k+1-i}*Z'_{k+1-i} -X'_{k+1-i}*Z'_{k+1-i}*(Y'_{k+1-i}*(Y'_{k+1-i}-1)/2) = sum_{i=1}^k(Z_{i+1}*X_{i-1} -Z_i*X_i -Z_i*X_i*(Y_i*(Y_i-1)/2) = O(N,K).

Now, by corollary 6, we have N'=N and K*K'= +/- 1 (mod N), so by uniqueness of K'', we have either K''=K' (mod N) or K''=-K' (mod N). In the first case, O(N,K'')=O(N',K'), and in the second case, O(N,K'')= O(N,N-K')=O(N',K') by proposition 10.