# AE1APS Algorithmic Problem Solving John Drake.  Coursework 3 due on Thursday at 1pm  (i.e. before the start of the tutorial).  Submit either hard-copy.

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AE1APS Algorithmic Problem Solving John Drake

 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

 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)

 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

 We require that:  For end positions L(l) = R(r) = 0  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).

 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

 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.

 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).

 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

 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

 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

b) a) b) (15)(20)

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