Heads Up
Dan Baczkowski
Sarah Brown
Rose Kneen
Alison Reichert
Sponsor: Professor J. Neuzil
Kent State University
1. INTRODUCTION
In this paper we examine an impartial combinatorial game we
call Heads Up. This game is played using a string of coins where
two players alternately choose a heads up coin, remove it and flip
over any adjacent coins. The game ends when there are not any
heads up coins left. We will be using the normal play rule as
defined in, Winning Ways for your mathematical plays, [Berlekanp
et.al. 2001] “the player that is unable to move loses.” For
example, a simplified version of a Heads Up game is played as
follows:
Let H denote a heads up coin and T denote a coin facing tails up.
Given the string of coins THTH:
Player one has two moves:
 Taking the first heads up coin yields the result H + HH. (See
Figure 1)
1
(Notice the given string is now divided into two
smaller strings. We will denote this break in the
string with the addition symbol.)
 Player two now has three possible moves. Suppose
player two chooses the first H. The result is now
just HH.
 Then player one moves say, choosing the first H.
The result is now just T.
 It is now player two’s turn. Since there not any
heads up coins remaining and by the normal play
rule, player two loses the game.
 Taking the second heads up coin yields the result THH. (See
Figure 1)
 Player two has two possible moves. Suppose
player two chooses the second H. This results in
TT. The game is over and player one loses by the
normal play rule.
 Suppose player two chose the first H. This results
in H + T. Player one will take the H and win the
game by the normal play rule.
Figure 1
THTH
H + HH
0 + HH
T
THH
H+T
0
H+T
TT
0
From this illustration it is now clear that different choices of
moves generate different results. Our objective is to calculate the
2
values of strings of coins in an effort to decide upon the best
possible move, and ultimately a way to win the game.
So, how can you win this game? What are some winning
moves? What is the value of a particular string of coins? We
develope some theorems to help calculate the answers to these
questions for some unique and rather complex strings of coins.
These theorems are stated and explained in Sections 3, 4, and 5.
This impartial coin game originated from a solitaire coin
game called Solitaire on a Circle. This game can be found on the
website, www.cut-the-knot.com/SimpleGames/Csolitare.shtml.
We took this one player game that is won by removing all coins
and manufactured it into a different type of game with different
rules and values. We decided to name our new game Heads Up
because a play is made on a heads up coin. In addition to Heads
Up, we also developed a variation of the Heads Up game. This
variation of the game is explained in Section 6.
Other coin games can be found on the cut-the-knot website
and in the book Fair Game[Guy 1989] on pages 59 to 65. More
impartial and take and break games can be found in the book
Winning Ways for your mathematical plays on pages 53 through
117.
2. Preliminaries
Throughout this paper we will be using some of the basics of
impartial games. We would like to briefly explain some of the
basics. If you would like to examine these theories further, please
refer to Fair Game by Richard K. Guy.
All impartial games can be given a value, which is called a
NIM value (*n). Let G be an impartial game if the NIM value of
G is *n, then n is called the nimber. Note that nimbers are always
3
non-negative integers. It is important to note that for any impartial
game which is not equal to *0, there is a winning move for the
next player.
Bouton’s theorem [Bouton 1901] says that if we have a
game G that is equal to the sum of two smaller games G1 and G2
then we can add the NIM values of G1 and G2 to obtain the NIM
value of G. We can do NIM addition via a binary method. First
we write the nimbers in binary. We then add the place columns
modulo 2. The result is the nimber x and the NIM value of G is
*x. See the following illustration.
*5 + *7
5 in binary: 101
7 in binary: 111
101
+ 111
010
2 in binary is 010 so *5 + *7 = *2.
It is easy to see that *n + *n = *0 for all n. See chart 1 for a
sample of nimber sums.
4
Chart 1: Sum of Nimbers
*1
*2
*3
*4
*5
*6
*7
*1
*0
*3
*2
*5
*4
*7
*6
*2
*3
*0
*1
*6
*7
*4
*5
*3
*2
*1
*0
*7
*6
*5
*4
*4
*5
*6
*7
*0
*1
*2
*3
*5
*4
*7
*6
*1
*0
*3
*2
*6
*7
*4
*5
*2
*3
*0
*1
*7
*6
*5
*4
*3
*2
*1
*0
To figure out the NIM value of most games we use the
Sprague-Grundy theorem [Grundy 1939, Sprague 1935-36]. First
we write down all of the possible plays from the game in set
notation. We then find the value of each individual play. Finally,
the NIM value of the whole game is the mex of the set, which is
the least nimber, which is not in the set. See the following
illustration.
mex{ *0, *2, *4} = *1
mex{ *1, *2, *4} = *0
Now that we have looked at the basics of game theory, we
can now see how they apply to our game, Heads Up.
5
Chart 2: NIM Values for Heads Up
GAME
POSITION
PLAYABLE POSITION(S)
MEX
T
H
{}
{T} = {*0}
*0
*1
HH
HT
TT
{TH} = {*0}
{H} = {*1}
{}
*1
*0
*0
HHH
THH
HTH
THT
TTH
TTT
{TH, T+T} = {*0, *0}
{H+T, TT} = {*1, *0}
{HH} = {*1}
{H+H} = {*0}
{TH} = {*0}
{}
*1
*2
*0
*1
*1
*0
HHHH
THHH
HTHH
TTHH
THHT
HTTH
HTHT
HTTT
THTT
TTTT
{THH, T+TH} = {*2, *0}
{H+TH, TT+T, THT} = {*0, *0, *1}
{HHH, HH+T, HTT} = {*1, *1, *1}
{TH+T, TTT} = {*0, *0}
{H+TT} = {*1}
{HTH} = {*0}
{HHT, HH+H} = {*2, *0}
{HTT} = {*1}
{H+HT} = {*1}
{}
*1
*2
*0
*1
*0
*1
*1
*0
*0
*0
HHHHH
HHHHT
HHHTH
HHTHH
HHHTT
HTHTH
THTHT
HHTHT
HHTTT
HTTTH
HTTHT
THHTH
THHHT
HHTTH
HTTTT
THTTT
THHTT
TTHTT
HTHTT
TTTTT
{THHH, T+THH, HT+TH} = {*2, *2, *0}
{THHT, T+THT, HT+TT, HHT+H} = {*0, *1, *0, *3}
{THTH, T+TTH, HT+HH, HHHH} = {*1, *1, *1, *1}
{TTHH, T+HHH} = {*1, *1}
{THTT, T+TTT, HT+TH} = {*0, *0, *0}
{HHTH, HH+HH} = {*0, *0}
{H+HHT} = {*3}
{TTHT, T+HHT, HHH+H} = {*0, *2, *0}
{TTTT, T+HTT} = {*0, *1}
{HTTH} = {*1}
{HTHT, HTH+H} = {*1, *1}
{H+TTH, TT+HH, THHH} = {*0, *1, *2}
{H+THT, TT+TT} = {*0, *0}
{TTTH, T+HTH, HHTH} = {*0, *0, *0}
{HTTT} = {*0}
{H+HTT} = {*0}
{H+TTT, TT+HT} = {*1, *0}
{TH+HT} = {*0}
{HHTT, HH+HT} = {*1, *1}
{}
*1
*2
*0
*0
*1
*1
*0
*1
*2
*0
*0
*3
*1
*1
*1
*1
*2
*1
*0
*0
6
Theorems of “Lonely” H’s
From looking at chart 2 you may wonder if there is are certain
types of strings that follow a pattern of values. The easiest strings
to find the NIM values of are the ones that have only one H.
Consider the following theorem.
Theorem 1
The NIM value of each of the following (and the mirror image)
is *0 if n is odd and *1 if n is even.
a) HTn
b) HTnH
c) H2TnH
d) H2TnH2
in addition,
e) Tn = * 0 for all n and
f) H2Tn = *1 if n is even and
= *2 if n is odd.
We will prove (a) by induction. For the base step, notice that the
value of HT0=H=*1 and HT1=HT=*0 (see chart 2). For the
inductive step assume that the theorem holds for all positive
values of n less than k. Notice that the NIM value of HTk is
mex{HTk-1}. If k is odd, then k-1 is even, so by the induction
assumption HTk-1=*1 and HTk=mex{*1}=*0. If k is even, then k1 is odd, so by the induction assumption HTk-1=*0 and HTk=mex
{*0}=*1.
Exactly the same proof works for (b)
7
Part (c) is also proved by induction. H2T0H = H3=*1 and H2T1H
= H2TH=*0 (see chart 2) For the inductive step assume that the
theorem holds for all positive values of n less than k. There are
three plays from H2TkH Notice that the NIM value of H2TkH is
mex{ Tk+1H, T+HTk-1H, H2Tk-1H }
For k even, the first two have value *0 by parts (a) and (b) and the
third has value *0 by the induction assumption. Thus the mex is
*1. For k odd,
All three values are are 1 and the mex is 0.
The proof of part (d) is essentially the same as that of (c).
For part (f), there are two plays from H2Tn so
H2Tn = mex{ Tn+1, T+H Tn+1} =mex{*0,*1} = *2,
if n is even and is mex{*0,*0} = *1, if n is odd This completes the
proof of theorem 1.
Now that we have a theorem for XTnY with X and Y equal
to one of H0, H1 or H2 we need to consider what happens when
we add more H’s and T’s. We would like to be able to find the
value of any string combination of H’s and T’s. Our next theorem
generalizes the previous theorem and allows us to find the value
for more types of strings. First we need to define what type of
strings we are looking at.
Definition
In a string of H’s and T’s, an H is “lonely” if on both sides
of the H there is a T.
8
Our next theorem is about strings in which all the H’s are lonely
except possibly, an H or H2 at the beginning and/or the end
Theorem 2
Let G be the string XTm1HTm2H Tm3Tmk-1HTmkY
where there is at least one lonely H between the X and Y and
both X and Y are H0, H1, or H2. If all of the H’s between X and
Y are lonely (in other words
mi ≥1 for all i), then the NIM value of G is:
*0 if the number of T’s in the string is odd
and the value of G is
*1 if the number of T’s in the string is even.
Proof: In this proof we let ∑(G) denote the number of T’s in the
string, i.e. ∑(G) = m1 + m2 ++mk The proof is by induction on
the total number of symbols in the string. For the base case, the
shortest string with at least one lonely H is THT which has vaule
*1 (chart 2) and has an even number of T’s.
For the inductive step, let G = XTm1HTm2HHTmkY with X
and Y both equal to one of H0, H1, or H2. Assume G has at least
one lonely H and assume the result is true for strings that have the
correct form and are shorter than G. Note that every play from G
yields a string shorter than G or a sum of two strings, each of
which is shorter than G. However, some plays will increase ∑(G).
I. First Case: ∑(G). is odd .
It suffices to show all the plays from G have value *1 or *3. (
Then the mex will be *0)
A) Playing from X or Y with m1>1.
9
If X = H then G plays to
A1 = HTm1-1HTm2HHtmkY. If X = HH then G
plays to either A1 = Tm1+1HTm2HHTmkY or
A1 =T + HTm1-1HTm2HHTmkY Since m1>1, each
possibility for A1 has at least one lonely H and
∑(A1) = ∑(G)1 therefore, by the induction
assumption, A1 = *1. Note that this argument still
works if X = HH and and the first H is played even if
m1=1
B) Playing from X or Y with m1=1.
If X = H then G plays to A1 = HHTm2HHTmkY
If X = HH and the second H is played the result is
T+A1 = T+HHTm2HHTmkY
If A1 has a lonely H then then induction assumption
implies G = *1, if not then A1 = HHTm2Y with m2
even so G = A1 = *1 by the appropriate case of
Theorem 1
C)
Playing the first lonely H (or last, by symmetry). In
this case G plays to A1 + A2 =
m -1
m -1
m
m
XT 1 H + HT 2 HT 3HHT kY Then
∑(A1)+∑(A2) = ∑(G) - 2 which is
odd, so ∑(A1) = m1 –1 .and ∑(A2). Have opposite
parity.
i) If m1 –1 is odd and ∑(A2).is even then A1 = *0
( theorem 1) and A2= *1 by the induction
10
assumption if A2 has a lonely H and by theorem 1 if
A2 doesn’t.
ii) If = m1 –1 is even and ∑(A2).is odd then A1 = *1
(theorem 1) and A2= *0 by the induction assumption
if A2 has a lonely H and A2= *0 or *2 if A2. has no
lonely H (theorem 1) Therefore A1 + A2 = *1
or *3. This last possibility occurs if
m
m
G = XT 1HTHT 3 which plays to
XTm1-1+HHTm3
D) Playing a lonely H which is neither the first nor the
last. In this case G plays to A1 + A2 =
XTm1HTm2HHTmj –1 H + HTmj+1 –1 HTmkY
As above, ∑(A1). + ∑(A2).= ∑(G) - 2, so ∑(A1). and
∑(A2) have opposite parity. If both A1 and A2 have a
lonely H the induction hypothesis implies A1 = *0 and
A2 = *1, or vice versa. In either case A1 + A2 = *1
If the second lonely H is the play and m2 = 1 then
A1 won’t have a lonely H and G plays to
m
2
m –1
m
A1 + A2= XT 1H + HT 3 HHT kY
i) If A2 has a lonely H and X is not empty then, by
Theorem 1 for A1 the induction assumption for A2
the parity argument implies A1 + A2 = *0 + *1
(or vice versa)
ii) If A2 has a lonely H and X is empty then,
A1 + A2 = *1 + *0 = * 1 (m1 even) or
A1 + A2 = *2 + *1 = * 3 (m1 odd)
11
iii) If neither A1 nor A2 has a lonely H then
G = XTm1HTHTHTm4Y and G plays to
A1 + A2 = XTm1H2 + H2 Tm4Y Using Theorem 1
we get that one of A1 or A2 has value *1 and the
other is either *0 or *2. Hence the sum is *1 or *3
Second Case: ∑(G). is even
It suffices to show all the plays from G have value *0 or
*2 and at least one play has value *0 ( Then the mex
will be *1)
A) Playing from X or Y with m1>1 or
B) Playing from X or Y with m1=1
The proofs of these cases are the same as the proofs
of (I-A) and (I-B) with “odd” replaced by “even”
and the value of A1 is *0.
C) Playing the first lonely H (or last, by symmetry). In
this case G plays to
A1 + A2 = XTm1-1H + HTm2-1HTm3HHTmkY Then
∑(A1)+∑(A2) = ∑(G) - 2 which is even, so ∑(A1)
.and ∑(A2). have the same parity.
i) If ∑(A1) = m1 –1 and ∑(A2).are both even then
A1 = *1 ( theorem 1) and A2= *1 by the induction
assumption if A2 has a lonely H and by theorem 1 if
A2 doesn’t.
ii) If ∑(A1).= m1–1 and ∑(A2).are both odd then
A1 = *0 (theorem 1) and A2= *0 by the induction
12
assumption if A2 has a lonely H and A2= *0 or *2
if A2. has no lonely H (theorem 1) Therefore
A1 + A2 = *1 or *3. This last possibility occurs
XTm1HTHTm3 which plays to
XTm1-1+HHTm3 In this case, when the second H is
played we get XTm1HH+HTm3-1 which has value
*1+*1 = *0.
D) Playing a lonely H which is neither the first nor the
last. In this case G plays to A1+ A2 =
XTm1HTm2HHTmj –1 H + HTmj+1 –1 HTmkY
As above, ∑(A1). + ∑(A2).= ∑(G) - 2, so ∑(A1). and
∑(A2) have the same parity. If both A1 and A2 have a
lonely H the induction hypothesis implies A1 = A2
and hence A1 + A2 = *0 If the second lonely H is
the play and = m2 = 1 then A1 won’t have a lonely H
and G plays to
A1 + A2= XTm1H2 + HTm3 –1 HHTmkY
i) If A2 has a lonely H and X is not empty then, by
Theorem 1 for A1 the induction assumption for A2
the parity argument implies A1 and A2 are equal and so
A1 + A2 = *0
ii) If A2 has a lonely H and X is empty then,
A1 + A2 = *1 + *1 = * 0 (m1 even) or
A1 + A2 = *2 + *0 = * 2 (m1 odd).
When the value is *2 we must find another play on the
string that yields value *0, in order to make the mex
13
if G =
come out to be *1. In this case
G = Tm1HTH Tm3HTmkY and playing the first H
yields Tm1-1H + HH Tm3HTmkY which has value
1 *1 + *1 = *0. (Using the parity)
iii) If neither A1 nor A2 has a lonely H then
G = XTm1HTHTHTm4Y and G plays to
m 2
2 m
A1 + A2 = XT 1H + H T 4Y. As above, A1 and A2
have the same parity ansd, by Theorem 1, have the
same value and add to *0 unless X is empty and Y is
not (or vise versa). If X is empty and Y isn’t, then
playing the first H yields
A1 + A2 = Tm1-1H+HHTHTm4Y
Using Theorem 1 on A1 and the induction
assumption on A2, A1 and A2 have the same value and
add to *0. This completes the proof of Theorem 2. 
4. Theorems About H’s in Blocks
Lemma
The string of coins THkTn=
(a.).*(k mod 2) if n is odd
(b.) *(2[(k+1)/3]) if n is even
for k≥0 and n≥1. We use [ ] to denote the greatest integer function Note
that the case where n=0 is also true and is proven in Theorem 4.
Proof: First, notice that we are claiming if n is odd, then the value
is always *0 or *1. Also, if n is even, then we claim the values are
*(2[(k+1)/3]) which are 0,0,2,2,2,4,4,4,…,2m,2m,2m,… for k≥0.
14
So, if k equals 0 or 1 and n is even, we get 0. Also, every three
consecutive values of k after k equals 0 and 1 are the same value.
We start by evaluating all of the possible moves of THkTn. So, we
get
H + T Hk-2Tn by playing the first H,
T2 + T Hk-3Tn by playing the second H,
THT + T Hk-4Tn by playing the third H,
TH2T + THk-5Tn by playing the fourth H,
…
THpT + THqTn by playing the (p+2)nd H where p+q=k-3,
…
THk-4T + THTn by playing the (k-2)nd, ie. the third last H,
THk-3T + Tn+1 by playing the (k-1)st, ie. the second last H, and
THk-2T + HTn-1 by playing the kth, ie. the last H
=mex{H + THk-2Tn,…, THpT + THqTn,…, THk-2T + HTn-1}
where THpT + THqTn represents all middle plays for p+q=k-3 for
p,q≥0.
Now, we calculate the values of these different positions and
proceed by induction on k. The base case, THTn, was proved in
Theorem 2. For the inductive step, we assume the lemma is true
for all values less than k and prove that it holds for k. We now
calculate the values of the different plays when n is odd and n is
even.
(I.) So, for n odd, the values are
15
position
H + T Hk-2Tn
T2 + T Hk-3Tn
THT + T Hk-4Tn
TH2T + THk-5Tn
…
THpT + THqTn
…
THk-4T + THTn
THk-3T + Tn+1
THk-2T + HTn-1
mex
values assumed inductively
*1 + *(k-2) mod 2
*0 + *(k-3) mod 2
*1 + *(k-4) mod 2
*0 + *(k-5) mod 2
*p mod 2 + *q mod 2
*(k-4) mod 2 + *1
*(k-3) mod 2 + *0
*(k-2) mod 2 + *1
if k is odd
*1 + *1
*0 + *0
*1 + *1
*0 + *0
if k is even
*1 + *0
*0 + *1
*1 + *0
*0 + *1
?
?
*1 + *1
*0 + *0
*1 + *1
*1
*0 + *1
*1 + *0
*0 + *1
*0
Note that when k=2, we only use the first play, THk-2Tn, and the
last play, Hk-2T + HTn-1.
Note for the positions in the middle, THpT + THqTn where
p+q=k-3, that if k is odd, then k-3 is even. This implies that p and
q are both odd or both even, in other words, they are *1+*1 or
*0+*0 which both equal *0. Now, if k is even, then k-3 is odd.
This implies that one of p and q is odd and the other is even, in
other words, *1+*0 which equals *1. So, if k is odd, then
THpT + THqTn=*0, and if k is even, then THpT + THqTn=*1.
So the mex is *0 if k is even and *1 if k is odd.
(II.) So, for n even
Again, we break up into two cases depending on whether k is
even or odd. First, the value of the middle plays, THpT + THqTn,
is *p mod 2 + *2[(q+1)/3]. The following chart gives the values
for specific p and q.
(1.) k is even
16
P
0
1
2
…
k-9
k-8
k-7
k-6
k-5
k-4
k-3
Q *p mod 2
k-3
0
k-4
1
k-5
0
…
6
1
5
0
4
1
3
0
2
1
1
0
0
1
*2[(q+1)/3]
sum
2[(k-2)/3]
2[(k-2)/3]
2[(k-3)/3]
1 + 2[(k-3)/3]
2[(k-4)/3]
2[(k-4)/3]
4
4
2
2
2
0
0
5
4
3
2
3
0
1
This alternating pattern shows the values for the middle
plays. It appears that by the alternating values of *p mod 2 and by
the groups of three *2[(q+1)/3] that the sums of the middle plays
for k even will be every value starting from 0 to 2[(k-2)/3. Now,
consider the plays of taking the first and last H,
H + THk-2Tn and THk-2T + HTn-1. These values are
*1+*2[(k-1)/3] and *0+*0, respectively. So, the values in the set
of possible plays will consist of:
*0, *1, *2,…, *2[(k-4)/3], *1+*2[(k-3)/3], *2[(k-2)/3], and
*1+*2[(k-1)/3] which is the highest. We will consider the three
possible cases for k.
(i.) For k=3j, we use the rules for NIM addition to obtain
1+2[(k-1)/3]=1+2[(3j-1)/3]=1+2(j-1)=2j-1. This implies the
next integer, 2j, is the mex. We check and see
2[(k+1)/3] =2[(3j+1)/3]=2j, too.
(ii.) For k=3j+1, again we use the rules for NIM addition. We
note that this is a special case. It is true that 1+2[(k-1)/3] is
the highest value in the set of plays, but there are terms only
missing between it and the others. We look at the last four
values in the set of plays. We have
17
*2[(k-4)/3], *1+*2[(k-3)/3], *2[(k-2)/3], and *1+*2[(k-1)/3].
Note that for this case, k=3j+1, that
2[(k-4)/3]=2[(k-3)/3]=2[(k-2)/3]=2(j-1)=2j-2.
Then, we have the values 2j-2, 2j-1, and 2j-2 for
*2[(k-4)/3], *1+*2[(k-3)/3], *2[(k-2)/3], respectively.
Notice 1+2[(k-1)/3]=2j+1. Then, we are missing the integer
2j in our set of values. Thus, 2j should be our mex, and
observe it is by 2[(k+1)/3]= 2[(3j+2)/3]=2j.
(iii.) For k=3j+2, using rules for NIM addition,
1+2[(k-1)/3]=1+2[(3j+1)/3]=1+2j. This implies the next
integer is 2+2j which is the mex. Note that 2[(k+1)/3]=2j+2.
We have just proved that for the case where k is even, the mex is
*2[(k+1)/3].
(2.) k is odd
p
0
1
2
…
k-9
k-8
k-7
k-6
k-5
k-4
k-3
Q *p mod 2
k-3
1
k-4
0
k-5
1
…
6
0
5
1
4
0
3
1
2
0
1
1
0
0
*2[(q+1)/3]
Sum
2[(k-2)/3]
1 + 2[(k-2)/3]
2[(k-3)/3]
2[(k-3)/3]
2[(k-4)/3]
1 + 2[(k-4)/3]
4
4
2
2
2
0
0
4
5
2
3
2
1
0
The set of possible plays consists of *0, *1, *2,…, *1+*2[(k-4)/3],
*2[(k-3)/3], *1+*2[(k-2)/3], and *1+*2[(k-1)/3]. We will consider
the three possible cases for k.
18
(i.) For k=3j, we get 1+2[(k-1)/3]= 1+2[(3j-1)/3]=1+2(j-1)=2j-1.
This implies the next integer, 2j, is the mex. So, 2[(k+1)/3]=
2[(3j+1)/3]=2j, too.
(ii.) For k=3j+1, again we use the rules for NIM addition. We
note that this is a special case. It is true that 1+2[(k-1)/3] is
the highest value in the set of plays, but there are terms only
missing between it and the others. We look at the last four
values in the set of plays. We have *1+*2[(k-4)/3], *2[(k3)/3], *1+*2[(k-2)/3], and *1+*2[(k-1)/3]. Now, 1+2[(k4)/3]= 1+2[(k-2)/3]=2j-1 and 2[(k-3)/3]=2j-2. Then, we have
the values 2j-2 and 2j-1. Note that 1+2[(k-1)/3]=2j+1. Then,
we are missing the integer 2j in our set of values. Thus 2j
should be our mex, and observe it is by 2[(k+1)/3]=
2[(3j+2)/3]=2j.
(iii.) For k=3j+2, using rules for NIM addition, 1+2[(k-1)/3]=
1+2[(3j+1)/3]=1+2j. This implies the next integer is 2+2j
which is the mex. Note that 2[(k+1)/3]= 2[(3j+3)/3]=2j+2.
We have just proved for case (1.) k is even that the mex is
*2[(k+1)/3]. Thus, the lemma has been proved.
‫ڤ‬
Theorem 4
The string of coins HkTn=
(a.) *0 if k≡0 mod 6 and n≡0 mod 2
(b.) *1 if k≠ mod 6 and n≡0 mod 2
(c.) *2[(k+1)/3] if n is odd
where we use [ ] to denote the greatest integer function and for
k≥0, n≥1.
Proof: We start by evaluating all of the possible moves of HkTn.
So, we get
THk-2Tn by playing the first H,
T + THk-3Tn by playing the second H,
19
HT + T Hk-4Tn by playing the third H,
H2T + THk-5Tn by playing the fourth H,
…
HpT + THqTn by playing the (p+2)nd H where p+q=k-3,
…
Hk-4T + THTn by playing the (k-2)nd, ie. the third last H,
Hk-3T + Tn+1 by playing the (k-1)st, ie. the second last H, and
Hk-2T + HTn-1 by playing the kth, ie. the last H
=mex{THk-2Tn,…, HpT + THqTn,…, Hk-2T + HTn-1}
where HpT + THqTn represents all middle plays for p+q=k-3 and
p,q≥0.
Now, we calculate the values of these different positions and
prove by induction on k. Clearly if k=0, then there are no H’s,
therefore a string of T’s equals *0. The base case, HTn, was
proven in Theorem 1. For the inductive step, we assume the
theorem is true for all values less than k and prove that it holds for
k. We now calculate the values of the different plays when n is
odd and n is even.
(I.) So, for n odd, we look at the case where k is odd.
position
THk-2Tn
T + THk-3Tn
HT + THk-4Tn
H2T + THk-5Tn
H3T + THk-6Tn
H4T + THk-7Tn
…
HpT + THqTn
…
Hk-7T + TH4Tn
values assumed inductively
*2[(k-1)/3]
*0 + *0
*0 + *1
*2 + *0
*2 + *1
*2 + *0
Sum
*2[(k-1)/3]
*0
*1
*2
*3
*2
*2[(p+1)/3]+ *q mod 2
…
*2[(k-6)/3] + *0
*2[(k-6)/3]
20
Hk-6T + TH3Tn
Hk-5T + TH2Tn
Hk-4T + THTn
Hk-3T + Tn+1
Hk-2T + HTn-1
mex
*2[(k-5)/3] + *1
*2[(k-4)/3] + *0
*2[(k-3)/3] + *1
*2[(k-2)/3] + *0
*2[(k-1)/3] + *1
*2[(k-5)/3] + *1
*2[(k-4)/3]
*2[(k-3)/3] + *1
*2[(k-2)/3]
*2[(k-1)/3] + *1
*2[(k+1)/3]
Note that when k=2, we only use the first play, THk-2Tn, and the
last play, Hk-2T + HTn-1. We can see by the values of the sums
that for k odd, the set of possible plays will consist of *0, *1,
*2,…, *2[(k-4)/3]), *2[(k-3)/3]+*1, *2[(k-2)/3], *2[(k-1)/3]+*1
which is the highest. Just as proven in the lemma with three sub
cases, we obtain the mex to be *2[(k+1)/3]. This is done similarly
for the case when k is even. Furthermore, the proof for HkTn
where n is odd is the same for THkTn where n is even.
(II.) For n even,
position
THk-2Tn
T + THk-3Tn
HT + THk-4Tn
H2T + THk-5Tn
H3T + THk-6Tn
H4T + THk-7Tn
…
HpT + THqTn
…
Hk-7T + TH4Tn
Hk-6T + TH3Tn
Hk-5T + TH2Tn
Hk-4T + THTn
Hk-3T + Tn+1
Hk-2T + HTn-1
mex
How to obtain
Take first H
Take second H
Take third H
Take fourth H
Take fifth H
Take sixth H
values assumed inductively
2[(k-1)/3]
0 + 2[(k-2)/3]
0 + 2[(k-3)/3]
2 + 2[(k-4)/3]
2 + 2[(k-5)/3]
2 + 2[(k-6)/3]
Take (p+2)nd H
2[(p+1)/3] + 2[(q+1)/3]
Take (k-5)th H
Take (k-4)th H
Take (k-3)rd H
Take (k-2)nd H
Take (k-1)st H
Take last H
2[(k-6)/3] + 2
2[(k-5)/3] + 2
2[(k-4)/3] + 2
2[(k-3)/3] + 0
2[(k-2)/3] + 0
2[(k-1)/3] + 0
?
21
It appears that for these set of plays, that every value equals
*2i + *2j for some nonnegative integers i and j. Recall that in
adding nimbers this equals an even nimber or *0 if i=j. So the
next question, is it possible to get *2i + *2i = *0? We must look
somewhere in the middle of the chart to find if this is possible.
Really what we need to observe is the values of HpT + THqTn,
*2[(p+1)/3]+*2[(q+1)/3], because the first and last plays will
always be the highest values. Looking at the middle plays, we
will show what happens for all possible k.
(i.) k≡1 mod 6, k≡3 mod 6, k≡5 mod 6
In other words, we are looking at the case where k is odd.
When k is odd that implies that k-3=p+q is even. Well, if
p+q is even, then there exists a number when p=q. Then,
there exists a p and q such that *2[(p+1)/3] = *2[(q+1)/3].
This implies there is a *0 in the set of plays; therefore, the
mex of HkTn for n even and k odd is *1.
(ii.) k≡2 mod 6 (ie. k=6j+2) and k≡4 mod 6 (ie. k=6j+4)
So, let p=(k-2)/2 and let q=(k-2)/2 – 1. Then, we see that
*2[(p+1)/3]=2[k/6] and *2[(q+1)/3]=2[(k-2)/6]. Now for
k=6j+2 and k=6j+4, *2[(p+1)/3]=2j=*2[(q+1)/3]. So, there
always exist some values in the center of the chart that give
us *0. Thus we see for k≡2 mod 6 and k≡4 mod 6, the mex
of HkTn for n even is *1.
(iii.) k≡0 mod 6 (ie. k=6j)
Note that if there does exist a *0, then it would occur at the
center play or plays depending on whether there is an even or odd
number of plays. This is true, because the values of HpT alternate
0,0,2,2,2,4,4,4,… and the values of THqTn alternate the reverse
way.
Since k is divisible by 6, k is even which implies that p+q is
odd. Since p and q are integers, one of them is even and the other
22
is odd. For simplicity, let p be even and again let p=(k-2)/2, and
let q=(k-2)/2 – 1 which is odd. Simple induction shows that these
choices for p and q are the center values for HpT + THqTn, the
middle plays. Now, it is suffice to show that if there does not exist
a *0 for these particular p and q, then there does not exist a *0 in
the set of possible plays.
Our choice of p and q for k=6j tell us that *2[(p+1)/3]=2j and
*2[(q+1)/3]=2(j-1). This implies that the two values in the center
play are different even nimbers, thus the value equals a non-zero
even nimber. Therefore, there is no *0 in the set of We have now
proved possible plays which implies that our mex equals *0 for k≡
mod
We have now proved the theorem holds for all possible cases.
Theorem 4
: The NIM value for the string of coins THTHnT =
(a)
(b)
(c)
*0 when n = 0, 1, 2
*2 when n is odd n > 3
*3 when n is even n > 4
We will be using Theorem 3 and the Lemma to prove this
theorem. For referencing convenience, these are the values of the
two theorems being used and the one to be proved in chart format.
Proof:
23
We will prove the theorem for the case when n = 0, 1, and 2
directly.
Chart 2:
n=
0
1
2
3
4
5
6
7
8
9
10
Theorem 3
HnT
*0
*0
*2
*2
*2
*4
*4
*4
*6
*6
*6
Lemma
THnT
*0
*1
*0
*1
*0
*1
*0
*1
*0
*1
*0
THTHnT *0
*0
*0
*2
*3
*2
*3
*2
*3
*2
*3
Theorem 4
n = 0 THTT
H + HT from Chart 2 we get: *1 + *0 = *1 mex = *0
n = 1 THTHT
H + HHT from Chart 2 we get: *1 + *2 = *3 mex = *0
n = 2 THTHHT
H + HHHT from Chart 2 we get: *1 + *2 = *3
or THH + TT
*2 + *0 = *2
or THTT + H
*0 + *1 = *1 mex = *0
x
We will prove the rest of the values by induction on n.
24
(L) Base step
When N=3, the mex of THTHHHT equals *2.
MOVES
Taking the first H: H +
HHHHT
Second H: THH + THT
Third H: THTT + TT
Fourth H: THTHT + H
VALUES
*1 + *2 =
*3
*2 + *1 =
*3
*0 + *0 =
*0
*0 + *1 =
*1
mex { *3, *0, *1 } = *2.
When N=4, the mex of THTHHHHT equals *3.
MOVES
Taking the first H: H +
HHHHHT
Second H: THH + THHT
Third H: THTT + THT
Fourth H: THTHT + TT
Fifth H: THTHHT + H
VALUES
*1 + *4 =
*5
*2 + *0 =
*2
*0 + *1 =
*1
*0 + *0 =
*0
*0 + *1 =
*1
mex { *5, *2, *1, *0 } = *3.
25
(I.) Inductive step
Let D(n) stand for the values of the Theorem 3. (See above
chart)
Let A(P) stand for the values of the conjectured theorem.
(See above chart)
Assume the theorem holds for all values of n > 3.
When n is odd we must show that *0, *1, *3, and nimbers greater
than *3 are the only attainable values from the possible moves
forcing the mex to be *2.
(**Notice that taking an H surrounded by H’s in the middle of the
string of coins will create a total loss of three heads up coins from
the string. Hence, in line four, P + Q = n – 3.)
PLAY
1. Take the first
H
2. Second H
3. Last H
4. Any H in the
middle
RESULT
H + Hn+1T
VALUE
*1 + D(n+1)
SOLUTIONS
*5, *7, *9…
THH + THn-2T
THTHn-2T + H
THTHPT +
THQT
*2 + *1
A(n-2) + *1
A(P) + *0 or *1
*3
*3
*0, *1, *3
When n is even, we must show that *0, *1, *2, and nimbers
greater than *3 are the only attainable values from the possible
moves forcing the mex to be *3.
26
PLAY
1. Take the first
H
2. Second H
3. Last H
4. Any H in the
middle
RESULT
H + Hn+1T
VALUE
*1 + D(n+1)
SOLUTIONS
*5, *7, *9…
THH + THn-2T
THTHn-2T + H
THTHPT +
THQT
*2 + *0
*1 + A(n-2)
A(P) + *0 or *1
*2
*2
*0, *1, *2,
This completes the proof for all possible cases.
‫ڤ‬
Winning Strategies:
Given a string of coins that might be seen in a typical game of
Heads Up:
THTHHHHHHT
A winning move would be to take the fourth H and create a zero
game.
1. Given THTHHHHHHHHHHHT
If the first player takes the fifth H, and assuming both players
continue making all their best possible moves, who will win the
game?
6 A Variation of Heads Up
In this variation of Heads Up, each player must turn over exactly
one adjacent coin. For example, in the simple game H, there are
no moves since there are no adjacent coins, thus H = *0. In a
27
more complicated example, the possible moves from THTTH are
as follows:
H + TTH from removing the first H and flipping the first T;
T + HTH from removing the first H and flipping the second T;
THTH
from removing the last H and flipping the last T.
The value of this game is *1 (see Chart 3). The following table
provides the NIM values for several game positions.
Table 6.1
n
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
TH n
*0
*1
*1
*2
*3
*1
*1
*4
*3
*2
*4
*3
*4
*5
*2
Hn
*0
*0
*1
*2
*0
*3
*1
*0
*3
*3
*2
*4
*0
*5
*6
TH nT
*0
*1
*2
*0
*3
*1
*0
*3
*3
*2
*4
*0
*5
*6
*7
Based on the values from the Table 7.1, we can derive the
following theorem.
Theorem 5 THnT = Hn+1
Proof:We prove by induction on n.
(I) Base case, n = 0.
TH0T = TT = mex{} = *0 and,
H0+1 = H = mex{} = *0
(II) We assume the values are true for all values ≤ n-1 and prove
that it holds for n. So, the possible moves from THnT are as
follows:
H+ THn-1
from removing the first H and flipping the
adjacent T;
n-2
T+ TH T
from removing the first H and flipping the
adjacent H;
TT + THn-2 from removing the second H and flipping the
H on the left;
28
TH + THn-3T from removing the second H and flipping the
H on the right;
THT + THn-3 from removing the third H and flipping the H
on the left;
THH+THn-4T from removing the third H and flipping the H
on the right;
.
.
.
THi + THn-2-iT
from removing the ith H for i ≤ n-2.
And the possible moves from Hn+1 are as follows:
THn-1
T + Hn-1
H+THn-2
HT+Hn-2
HH+THn-2
from removing the first H and flipping the H on the
right;
from removing the second H and flipping the H on
the left;
from removing the second H and flipping the H on
the right;
from removing the third H and flipping the H on
the left;
from removing the third H and flipping the H on
the right;
.
.
.
THi + Hn-1-i
from removing the ith H for i ≤ n-1.
Obviously, THi = THi for all i, and THn-1 = H + THn-1 since
H = *0. Further, by the inductive step, Hn-1-i = THn-2-i, since
n-1-i > n-2-i. Therefore, T HnT = Hn+1 •
29
References
[Berlekanp et.al. 2001] E. R. Berlekamp, J. H. Conway, R. K. Guy
Winning Ways for Your Mathematical Plays A. K. Peters, Natick
MA, 2001
[Bouton 1901] Charles Bouton, Nim, a game with a complete mathematical theory,
Ann. Of Math., Princeton(2),3,p 35-39.
[Guy 1989] R. K. Guy, Fair Game: How to Play Impartial Combinatorial Games,
COMAP, Arlington, MA, 1989
[Grundy 1939] P. M. Grundy, “Mathematics and Games” Eureka 2 (1939) p 6-8,
Reprinted in Eureka 27 (1964) p 9-11
[Sprague 1935-36] R. Sprague, “Uber Mathematische Kampfspeile” , Tohoku Math
J. (1935-36) ,p 438-444
30