AE1APS Algorithmic Problem Solving
John Drake
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Coursework 3 due on Thursday at 1pm
(i.e. before the start of the tutorial).
Submit either hard-copy (including
name/student number), or e-mail soft-copy to
me.
Normal university regulations apply to late
submissions
Please let me know if you have any problems!
Office SEB325
John.drake@nottingham.edu.cn
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The position in a given sum game is (l, r).
A move affects one component (i.e one game)
An assignment can either be l := l’ or r :=r’
(i.e. the left or right component is affected)
Define two functions L and R on the left and
right positions such that;
(l, r) is a losing position exactly when L(l) = R(r)
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L and R must have equal values on end
positions
All moves from a losing position should
result in a winning position satisfying
L(l) != R(r)
Applying the winning strategy from a winning
position, should result in a losing position
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We require that:
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For end positions L(l) = R(r) = 0
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For every l’ such that there is a move from l
to l’ - L(l) != L(l’)
For any number m < R(r), it is possible to
move from r to r’ such that R(r’) = m (similar
for left).
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Fortunately for us such a function exists 
The MEX function, short for “minimal
excludant”, satisfies these conditions and is
therefore used as the functions L and R
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Let p be a position in a game. The mex value
of p (mex(p)), is defined as the smallest
natural number, n, such that
There is no move in the game from p to
position q satisfying mex(q) = n
For m < n, there is a move from p to q
satisfying mex(q) = m.
This is a recursive definition.
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Informally, mex(p) is the minimum number
that is excluded from the mex numbers of
positions q to which a move can be made
from p (i.e. q is a successor of p).
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The graphs do not have regular structure, so
the mex numbers must be calculated by hand
End positions have a mex number of 0
A mex number can be give to a node when all
its successors have been assigned a mex
number
The number is the smallest number that is
not included in the mex numbers of its
successors
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Suppose we start with “ok” (o,k) as our position
This is a winning position because the mex
numbers are different
The winning strategy is to move the right graph
to node i which has the same mex number as “O”
i.e. we are restoring the balance (R(r ) = L(l ))
The opponent will then disrupt the balance
(making this an inequality)
The first player the repeats the strategy of
making the mex numbers equal, moving in the
game with the highest mex number, until the
opponent can move no further
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Note that due to the lack of structure, the
mex number of the 15 left and 11 right
positions were calculated individually
In total we only did 15+11 (=24) calculations,
and not 15*11 (=165) calculations 
a)
(15)
b)
(20)