Partizan Splittles G. A. MESDAL III

Transcription

1 Games of No Chance 3 MSRI Publications Volume 56, 009 Partizan Splittles G. A. MESDAL III ABSTRACT. Splittles is a nim variant in which a move consists of removing tokens from one heap, and (optionally) splitting the remaining heap into two. The possible numbers of tokens that can legally be removed are fixed, but the two players might have different subtraction sets. The nature of the game, and the analysis techniques employed, vary dramatically depending on the subtraction sets.. Introduction Partizan Splittles is a game played by two players, conventionally called Left and Right. A position in the game consists of a number of heaps of tokens and a move requires a player to choose a heap, remove some positive number, s, of tokens from the heap and optionally to split the remaining heap (if there are two or more tokens remaining) into two heaps. Two sets of positive integers S L and S R are fixed in advance, and there is an additional restriction that when Left moves she must choose s S L, while Right must choose s S R at his turn. The sets S L and S R are called the subtraction sets of Left and Right respectively. It is sometimes convenient to represent a position pictorially by one-dimensional blocks of boxes rather than heaps of tokens. A move is to remove a contiguous block of boxes; moves in the middle of a block are tantamount to splitting a heap. In this paper, we address several possible restrictions on S L and S R, each of which yields a game whose analysis requires different techniques from combinatorial game theory. For some choices of S L and S R, canonical forms are This work was conducted at the third meeting of Games at Dalhousie in Halifax, Nova Scotia, in June of 004. The authors are Michael Albert, Elwyn Berlekamp, William Fraser, J. P. Grossman, Richard Guy, Matt Herron, Lionel Levine, Richard Nowakowski, Paul Ottaway, Aaron Siegel, Angela Siegel, and David Wolfe. 447

2 448 G. A. MESDAL III readily available, while for others, canonical forms are complex and uninformative, while temperature theory and the relatively new techniques of reduced canonical forms reveal a great deal of information. While impartial octal games [BCG0] are well-studied, there has been surprisingly little work on partizan variants. In a partizan octal game, the two players have different octal codes indicating their legal moves. Each code is a sequence of octal digits, d 0 :d d d 3 : : :, where each d i includes a,, and/or 4, indicating whether it is legal to remove i tokens and leave 0,, and/or heaps, respectively. In Partizan Splittles, the octal codes consist entirely of 0s and 7s. For instance, if S L D f; 4g and S R D f; 5g, then the game is versus Thane Plambeck [Pla95], as well as Calistrate and Wolfe, have unpublished results in the game where players cannot split into two heaps; the octal codes for these games consist entirely of 0s and 3s. Our first example is simple.. f; oddsg versus f; oddsg THEOREM. If is an element of both subtraction sets and all the elements of both subtraction sets are odd numbers, then 0 if n is even, G n D if n is odd. PROOF. Each move changes the parity of the total number of tokens in all the heaps, and in the final position, which has zero tokens, this total is even. Thus, the game is she-loves-me she-loves-me-not. 3. fg versus fkg THEOREM. If S L D fg and S R D fkg, then G n is arithmetic-periodic with period k and saltus fk j 0g. In particular, n if n < k, G n D fk j 0g C G n k if n k. We can write G n more naturally with period k and saltus k 8 < n if n < k, G n D fn j n kg if k n < k, : k C G n k if n k. as PROOF. The proof follows in part from the fact that the conjectured saltus is exactly G k. So the theorem asserts that one can treat a single heap as a collection

3 PARTIZAN SPLITTLES 449 of heaps of size k and (possibly) a single remaining heap of size less than k without changing its value. For instance: D ƒ ƒ k k Clearly, G a D a for 0 a < k, since only Left can move from such a position. Likewise, G k D fk j 0g. In a general position, it suffices to show that any move that straddles a period boundary is dominated by one that does not, for then the game reduces to its decomposed form. Left s moves never straddle a boundary. As for Right s moves, assume inductively that shorter positions achieve their conjectured values and decompose at period boundaries. Observe that if acb DkCc for 0a; b; c <k, then G a C G b D a C b D k C c exceeds G k C G c D fk j 0g C c. Hence, Right prefers the latter move, avoiding a boundary. 4. f; kg versus f; k C g In this case, too, we can find exact values for all G n. The sequence is arithmetic-periodic with period 4k and saltus "!. (Note that "! D "C" D f" j g, and that " D f0 j #g is positive and infinitesimal with respect to ".) THEOREM 3. If S L D f; kg and S R D f; k C g then 8 0 C j:" ˆ<! if i is even and 0 i < k, C j:" G 4jkCi D! if i is odd and 0 i < k, " C j:" ˆ:! if i is even and k i < 4k, " C j:"! if i is odd and k i < 4k. That is, the values are given by k ƒ 0 0 : : : 0 " " " " : : : " " Period 4k, saltus "! PROOF. When n < 4k, it is easy to confirm that the proposed values of G n are correct. It thus suffices to show that G nc4k G n D "!. Assume, inductively, that the conjectured values are correct for heap sizes less than n C 4k. First, moves by either player that split G n into G a G b can be countered by splitting G nc4k into G ac4k C G b, leaving zero by induction.

4 450 G. A. MESDAL III G nc4k G n "! G n CG k G n "! G nck CG k G n "! G nc4k G n G n CG k G n " G n CG k CG k G n "! D G n C0C" G n "! < 0 G n CG k G n G nc4k G n "! G n CG k G n "! G nc4k G n " G n CG k G n G n CG k G n " Figure. Diagrams showing G nc4k D G n C "!. The first tree shows Right s winning responses to Left s moves, while second shows how Left wins when Right moves first. Except in one case, the response leaves a game equal to 0. Next, observe that Left s moves from G nc4k that leave a heap of size k are at least as good as her other moves. Similarly, Right s dominant splitting moves from G nc4k remove k C tokens, leaving a heap of size k. (When convincing yourself of these last assertions, it helps to keep in mind that all G a C G b for fixed a C b have the same -parity, and alternate rows add " and ".) Figure summarizes how the second player wins in response to Left s (respectively, Right s) options not yet dispensed with. COROLLARY 4. The values in the last theorem remain unchanged when Left has additional odd options, and/or Left has additional options exceeding k, and/or Right has additional odd options between and k C. PROOF. In all cases these new options are dominated. 5. f; othersg versus f; 3; 5; : : : ; k C g While the actual values might be quite complex and depend on the specific choice for S L, we can describe a few properties of the sequence G n. THEOREM 5. Suppose S L, and either S R D f; 3; 5; : : : ; k C g for some k or S R D f; 3; 5; : : :g. The following relations hold:

5 PARTIZAN SPLITTLES 45 G n G nc (5-) G nc D G n C (5-) G n G i C G j.for S R finite and n i j k even/ (5-3) PROOF. For (5-), in most sequences of play, Left wins G nc G n by matching options naturally, removing the same number of tokens as Right did from the opposite heap. Left can then leave a position of the form G ac CG b G a G b, for a; b 0, which is 0 by induction. The only exception is if Right removes n C from the first heap. In this case, Left removes n from the second, leaving G G D 0. We show (5-) and (5-) in tandem by induction. In particular, we assume that (5-) holds for n 0 n when proving (5-), but that (5-) holds for n 0 < n when proving (5-). For (5-), we wish to show the second player wins on the difference game We can depict this game as Left s moves G nc G n D G nc G n G : Right s moves f; othersg f; 3; 5; : : : ; k C g f; 3; 5; : : : ; k C g f; othersg The roles of the players are reversed in the second row, it being the negative of the game G n C G. So in the second row, Left removes elements from S R while Right removes elements from S L. If either player removes r boxes from the top row, leaving a block of length i odd (and, perhaps a second block of either parity), the other player can counter symmetrically by removing r boxes from the bottom, leaving a block of length i. The resulting position is 0 by induction. The reverse is also true; the second player can respond to moves on the bottom row that leave an even-length block. Pictorially, moves A on top leaving one end odd match up with moves A 0 on bottom leaving the same end even and one shorter. A B ƒ odd ƒ even A 0 B 0 Also shown are moves B which take a single box from one end of the top row, which match up with the move B 0 taking the lone box on the bottom row. So we are left with cases that split the top row into two even-length heaps or that split the bottom row into two odd-length heaps. Only Right can do the

6 45 G. A. MESDAL III latter, for it requires removing an even number. Left s responses parallel Right s moves as below. If Right removes C R on the top, Left s reply of C RL wins by application of (5-) then multiple applications of (5-). Similarly Left wins after Right s D R and Left s D RL. C R ƒ D RL ƒ E L ƒ ƒ C RL ƒ D R We are now left with the single case when Left removes an odd number from the top row, splitting it into two even-sized heaps, as in E L above. Right responds by removing as large an (odd) number as possible from one of these two even-sized heaps. If one of the heaps is of size k C (or S R is infinite), Right leaves that heap a singleton, canceling the single box on the second row, and wins by (5-). Otherwise, he has taken away k C, and wins by (5-) and (5-). Lastly, to prove (5-), we show that Left wins moving second on G n G i G j : The gap in the bottom row is of even length and at least k. Left can respond to moves that fail to straddle the gap as below: A ƒ B ƒ R B RL ƒ A 0 In particular moves outside the gap match up with moves in the other row, winning by induction. Left responds to moves inside the gap by responding on the odd side: Since the gap was of even length, and Right can remove only odd numbers, the gap is split into an even length and an odd length. Left then wins by application of (5-) to both sides. A Right move that straddles the gap can only straddle one side. Left responds by removing a like number from the side below Right s move: C R ƒ ƒ C RL Since the parity of the number of boxes in each row is preserved, each segment can be shortened to an even length by an even number of applications of (5-),

7 PARTIZAN SPLITTLES 453 which, since C D 0 and D, leave the game value unchanged. Left then proceeds to win by (5-) applied to both sides. 6. Odd versus Even In this section we consider the partizan splitting game where S L is the set of all (positive) odd integers and S R is the set of all (positive) even integers. A salient feature is that the endgame is overwhelmingly favorable to Left: G D, but there are no positions with negative right stop, since Left has a move from every nonterminal position. One would expect, therefore, that Left should prefer to split each position into as many components as possible, preferably odd in size, while Right should aim to annihilate each component as quickly as possible. Since Right will naturally give preference to destroying the largest heaps, one might also expect that Left would prefer to split each heap as evenly as possible. As is so often the case, the canonical forms of Odd versus Even are a mess, but the orthodox moves as defined by Berlekamp [Ber96], Definition 0 realize these intuitions precisely. Left s orthodox strategy is to split as evenly as possible; Right s is to consume the largest available heap. Furthermore, we will see that from positions of the form G k where it is most crucial that Left split evenly these are the unique orthodox options (Theorem ). The game also exhibits a fascinating logarithmic behavior: if Left and Right play orthodox strategies, with Left splitting evenly and Right consuming what he can, then the game will last for O.log n/ moves. Furthermore, from positions of the form G k where it is most crucial that Left split evenly Left s only orthodox move is the even split. By contrast, we note that, G 3 has seven canonical Left options. The main result is the following theorem, which gives the mean, m.g n /, and temperature, t.g n /, of every single-heap Odd versus Even position. THEOREM 6. Fix n and let k be such that k n < kc. Then m.g n / D n and t.g n / D n, where n D bn=c C C k n if n is even, I k n D n C if n is odd. To prove Theorem 6, we will define H n to be the auxiliary game where S L D fg and S R is the set of even integers. We will first show that m.h n / D n and t.h n / D n, and then argue that the means and temperatures do not change when Left s subtraction set includes other odd integers. We will need several lemmas. The first shows that if n is odd, then H n D H n C (canonically). This reduces Theorem 6 to the case where n is even.

8 454 G. A. MESDAL III LEMMA 7. Let n be odd. Then PROOF. We proceed by induction on n. To prove (6-), consider H nc H n < ; (6-) H n H n D : (6-) C H n H nc : Left can win by moving immediately to C H n ; this value is positive since R 0.H n / 0. If Right moves to C H a C H b H nc, Left counters to C H a CH b H ac H b. This is a winning move by induction on (6-) or (6-), depending on whether a is odd or even, respectively. If Right moves to C H n H a H b, then since n C is even, a C b is odd and hence one of a; b (say a) must be odd. Left counters to C H a C H b H a H b, which is 0 by induction. To prove (6-) we show that H n H n is a second-player win. If Right moves to H a C H b H n, then since n is odd, acb is also odd and hence one of a; b (say a) must be odd. Left counters to H a CH b H a H b, which by induction is equal to 0. Likewise, if Right moves to H n H a H b, then since n is even, a C b is odd and hence one of a; b (say a) must be even. Left counters to H ac C H b H a H b. Finally, if Right moves to H n H n, Left simply responds with H n H n. Conversely, if Left moves to H a C H b H n, then since n > we can assume without loss of generality that a>0. Right counters to H a CH b H a H b. By induction, this is 0 if a is odd, and negative if a is even. If instead Left moves to H n H a H b, then since n is even, acb is odd and hence one of a; b (say a) must be odd. Right counters to H ac C H b H a H b, which is negative by induction on (6-). The rather dry arithmetic of the n and n is described in the next two lemmas. LEMMA 8. Fix n > and let k be such that k n < kc. Then n n D k : PROOF. We may assume that n is even, for if n 0 D n C is odd, we have k n; n 0 < kc and n 0 n 0 D n n D k :

9 PARTIZAN SPLITTLES 455 We separate cases into n D k and n k. If n D k, k k D k C k C k. k /C k C k D C k C k C k D k : If n k, n=c n n D k C k D n kc C k C k.n /=C k C k n kc C k D k : Lemma 8 shows that the n are (nonstrictly) increasing; and therefore, up to parity, so are the n. Furthermore, up to parity, the rate of increase is decreasing. This fact will be critical in the proof of Theorem 6, since it quantifies the intuition that Left prefers to split as evenly as possible. LEMMA 9. Fix n > and let k be such that k n < kc. Then n C n D n C k C k : PROOF. Again, we separate into the same two cases. If n D k, k C k D k C k C k. k /C C k C k D C k C k C C k D k C k: Since n= k D, this yields the desired equality. When n k, notice that exactly one of n, n is odd, and in either case bn=c C b.n /=c D n. So, bn=c C n C n D C k b.n /=c C C C k k k.n /C D k C k D nc k C k : Since exactly one of n, n needed. is odd, we have n C n D n C n C, as PROOF. (of Theorem 6) As noted in the exposition, we first show that m.h n / D n and t.h n / D n, and then generalize to the G n. The proof is by induction on n. The base cases H D and H D f j 0g are easily verified. At odd stages of

10 456 G. A. MESDAL III the induction, the result is an immediate corollary of Lemma 7, so fix an even n >. Left has a move to H L n D H n= C H n=. By induction, we know that and since k m.h L n / D m.h n=/ C m.h n= / D n= C n= n= < k, Lemma 9 implies that n= C n= D n= C C.k / D bn=c C C k : k k Furthermore, t.hn L / maxf n=; n= g D n= : Now Right can remove the entire heap, moving to H R n m.h L n / m.h R n / D m.h L n / Now certainly n > n=. Therefore m.h L n / m.h R n / D bn=c C k C k > maxft.h L n /; t.h R n /g: D 0. Therefore D n : If H L n and H R n were the only options of H n, then by an elementary thermographic argument, we would have t.h n / D n and m.h n / D m.h n L/ C m.h n R/ D m.h n L/ D n D n : Certainly both players have other options available, so we conclude the proof by showing that Hn L and H n R are thermally optimal at all temperatures t n=. Since n= is an upper bound for t.hn L /, it suffices to show that, for any other options Hn L0, H n R0 L0, we have m.hn / m.h n L R0 / and m.hn / m.h n R/. This is trivial in the case of Right options, since no Odd versus Even position can have negative mean. Therefore, consider some arbitrary Left option H a C H b, with a > b and a C b D n. We necessarily have a n= and n= b, with a n= D.n= / b. Now since exactly one of n=, n= is odd, repeated applications of Lemma 8 imply that It follows that and hence, since a C b and n= C.n= a n= n= b : a C b < n= C n= / are both odd, a C b < n= C n= :

11 PARTIZAN SPLITTLES 457 This completes the proof for the H n. To conclude, we show (again by induction on n) that Left s additional options in G n convey no thermographic advantage. For suppose Left has a move from G n to G a CG b, where acb D n c for some c 0. By induction we may assume that m.g i / D i and t.g i / D i for all i < n. But just as before, we have a C b acc C b n= C n= so that G a C G b is thermally dominated at temperatures t n=. We conclude with a neat little theorem on orthodox moves. DEFINITION 0. Let G be a game and fix t 0. A Left option G L is said to be orthodox at temperature t if R t.g L / D L t.g L / C t. Likewise, a Right option G R is orthodox at temperature t if L t.g R / D R t.g/ t. We say that an option is orthodox if it is orthodox at temperature t.g/. That is, an orthodox move is one that achieves the best possible score at temperature t.g/. THEOREM. Let k > and n D k to Gn L D G k C G k.. Left s only orthodox move from G n is PROOF. Since n is odd, Left must split G n into two heaps that are either both even or both odd. It is easily seen that those options with both heaps even are badly dominated, so it suffices to show that Gn L is strictly optimal among those options with both heaps odd. By Lemma 8, Therefore, k k 3 D k ; but k C k D k : k C C k 3 < k C k : Repeated application of Lemma 8 also shows that a C b k C C k 3 for every other choice of a; b both odd with a C b < n. Therefore Gn L has the strictly highest mean among the Left options of G n. But we also know that t.g k C G k / k < n so G L n is the unique optimal move at temperature n.

12 458 G. A. MESDAL III 7. f; oddsg versus f; 4g Suppose S L is any set of odd numbers containing, and S R D f; 4g. The values G n for these games can be quite complex. For example, when S L Df; 3g the canonical form of G 4 contains 6 stops! Furthermore, the exact value of G n depends strongly on the specific set S L (if S L D fg then the canonical form of G 4 has only 6 stops). Although it is not practical to solve for G n exactly, we can find a very good approximation for G n. In particular, let f.n/ be the arithmetic-periodic sequence with period 4 and saltus 3/4 defined by 8 0 if n D 0, ˆ< if n D, f.n/ D = if n D, 3= if n D 3, ˆ: f.n 4/ C 3=4 if n 4. The main theorem of this section is that G n is infinitesimally close to f.n/ for any choice of S L that contains and zero or more other odd numbers. We begin by briefly reviewing some definitions and results that will be required. Infinitesimals. Write L 0.G/ and R 0.G/ for the Left and Right stops of G, respectively. A game G is infinitesimal if L 0.G/ D R 0.G/ D 0. Write G Inf H when G and H differ by an infinitesimal; we also say that G is H -ish. If G Inf H, then L 0.G/ D L 0.H/ and R 0.G/ D R 0.H/. The converse is in general not true, but if x is a number and L 0.G/ D R 0.G/ D x, then it is true that G Inf x, in which case we say that G is numberish. Write G Inf H if there is some infinitesimal " such that G H C ", and similarly for G Inf H. A Left option G L of G is Inf-dominated if G L0 Inf G L for some other Left option G L0, and similarly for Right options. In [GS07] it is shown that if G Inf H, then L 0.G/ L 0.H/ and R 0.G/ R 0.H/. More importantly, they show: PROPOSITION. If G is not a number and G 0 is obtained from G by repeatedly (i) eliminating Inf-dominated options, and (ii) replacing any option H with H 0 Inf H, then G 0 Inf G. Norton multiplication. Fix a game U > 0. The Norton product GU is defined by 8 G times G times ˆ< ƒ ƒ GU D 0 or U C U C C U or U U U if G is an integer, ˆ: GL U C.U C I/ ˇˇ GR U.U C I/ otherwise.

13 PARTIZAN SPLITTLES 459 where I ranges over all Left and Right incentives of G. We will use the following properties of Norton multiplication, which are proved in [BCG0]. PROPOSITION 3. Let U be any positive game. Then: (i) If G D H, then G U D H U (independence of form). (ii) G H if and only if G U H U (monotonicity). (iii).g C H/ U D G U C H U (distributivity). For our purposes, we take U D. Since the only Left or Right incentive of is, we have G D G L C j G R when G is not an integer. We note that G is equal to G overheated from to, an operation defined in [BCG0]. It is easy to verify by induction that if x is a number then L 0.x/ D dxe and R 0.x/ D bxc. LEMMA 4. If x D a=4 for some integer a, then f.x =4/ C j.x C =4/ g D x PROOF. If x is not an integer, then x Dfx =4 j x C =4g, so the result follows from the definition of Norton multiplication and Proposition 3(i). Otherwise, by symmetry, it suffices to show that Right has no winning move from f.x =4/ C j.x C =4/ g x: If Right moves in the first component, then the resulting game is.x C =4/ x D.=4/ which we can verify is 0, so Left wins. If Right moves in the second component, which has the effect of subtracting, then Left responds in the first, and the resulting game is.x =4/ C x D. =4/ C D.3=4/ which we can verify is 0, so again Left wins. PROOF OF MAIN RESULT. We will now show that G n Inf f.n/. Our proof is by induction. Suppose the result holds for all m < n. It is convenient to assume that n 4; for n < 4 we can easily validate the result by hand. We begin by showing that in the game G n, there are only one Left and one Right option that need to be considered. LEMMA 5. The Left options of G n are Inf-dominated by G n 4 C G 3 Inf.f.n/ C 3=4/:

14 460 G. A. MESDAL III The Right options are Inf-dominated by G n 4 Inf.f.n/ 3=4/: PROOF. The Left options of G n are G n k a C G k with a S L. Since f.m/ < f.m C / for all m and G m Inf f.m/ for m < n, G n k a C G k Inf G n k C G k so we may assume that a D. Next, since f is arithmeticperiodic with period 4, G n k C G k Inf G n kc3 C G k 4 for k 4, so we may assume that k < 4. This leaves us with four options to consider, which are infinitesimally close to: f.n /;.f.n / C /;.f.n 3/ C =/;.f.n 4/ C 3=/ It is easy to verify that for all m we have f.m/c3= f.mc3/; f.m/c= f.mc/; f.m/c f.mc/; from which it follows that.f.n 4/C3=/ D.f.n/C3=4/ Inf-dominates the others. The proof for the Right options is similar. Since f.m/ < f.m C / and f is arithmetic-periodic with period 4, we need only consider the four options G n k 4 C G k with k < 4, which are infinitesimally close to f.n 4/;.f.n 5/ C /;.f.n 6/ C =/;.f.n 7/ C 3=/ The same three inequalities as before show that f.n 4/ D.f.n/ 3=4/ Inf-dominates the others (note that for n D 4; 5; 6, not all the other options exist, but this does not affect the result). Next we show that G n has the same Left and Right stops as f.n/. LEMMA 6. L 0.G n / D df.n/e and R 0.G n / D bf.n/c. PROOF. First we compute maxfr 0.Gn L/g and minfl 0.Gn R /g. By Lemma 5, maxfr 0.G L n /g D R 0..f.n/ C 3=4// D bf.n/ C 3=4c D df.n/e: The last equality follows from the fact that f.n/ is of the form a=4 for some integer a. Similarly, minfl 0.G R n /g D L 0..f.n/ 3=4// D df.n/ 3=4e D bf.n/c: If G n is not a number then we are done, as then L 0.G n / D maxfr 0.Gn L /g and R 0.G n / D minfl 0.Gn R/g. If G n is a number then G n D L 0.G n / D R 0.G n /, but L 0.G n / maxfr 0.G L n /g D df.n/e bf.n/c D minfl 0.G R n /g R 0.G n / so in fact we must have equality throughout, which means that f.n/ is also an integer, and again we are done.

15 PARTIZAN SPLITTLES 46 From Lemma 6 it follows immediately that when f.n/ is an integer, G n Inf f.n/ Inf f.n/. Finally, if f.n/ is not an integer, then by Lemma 6, G n is not numberish. So by Lemma 5, Proposition and Lemma 4, G n Inf f.f.n/ C 3=4/ j.f.n/ 3=4/g Inf f.f.n/ =4/ C j.f.n/ C =4/ g D f.n/: References [BCG0] Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy. Winning Ways for Your Mathematical Plays. A K Peters, Ltd., Natick, Massachusetts, nd edition, 00. First edition published in 98 by Academic Press. [Ber96] Elwyn Berlekamp. The economist s view of combinatorial games. In Richard Nowakowski, editor, Games of No Chance: Combinatorial Games at MSRI, 994, pages 0 0. Cambridge University Press, Mathematical Sciences Research Institute Publications 9, 996. [GS07] J. P. Grossman and A. N. Siegel. Reductions of partisan games. In this volume. [Pla95] Thane Plambeck. Partisan subtraction games. Working notes, February G. A. MESDAL III malbert@cs.otago.ac.nz berlek@math.berkeley.edu bfraser@alumni.ucsd.edu jpg@ai.mit.edu rkg@cpsc.ucalgary.ca rjn@mscs.dal.ca ottaway@mathstat.dal.ca aaron.n.siegel@gmail.com siegel@mathstat.dal.ca wolfe@gustavus.edu

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