How far can Nim in disguise be stretched?

A move in the game of nim consists of taking any positive number of tokens from a single pile. Suppose we add the class of moves of taking a nonnegative number of tokens jointly from all the piles. We give a complete answer to the question which moves in the class can be adjoined without changing the winning strategy of nim. The results apply to other combinatorial games with unbounded Sprague-Grundy function values. We formulate two weakened conditions of the notion of nim-sum 0 for proving the results.Comment: To appear in J. Combinatorial Theory (A

C1C^1-actions of Baumslag-Solitar groups on S1S^1

Let $BS(1, n)=$ be the solvable Baumslag-Solitar group, where $n\geq 2$. It is known that B(1, n) is isomorphic to the group generated by the two affine maps of the line : $f_0(x) = x + 1$ and $h_0(x) = nx$. The action on S^1 = \RR \cup {\infty} generated by these two affine maps $f_0$ and $h_0$ is called the standard affine one. We prove that any representation of BS(1,n) into $Diff^1(S^1)$ is (up to a finite index subgroup) semiconjugated to the standard affine action