sprague-grundy : what if player allowed to take half of the coins? In sprague-grundy , I've seen that the answer is the XOR of all the numbers of coins in each pile. But what if a player is only allowed to remove at least one and at most half of the coins from each pile? Suppose if a pile is contain 7 coins than at least 1 and at most 3 ($\lfloor\frac72\rfloor$) coins can be removed. If the number is 6 then 1 to 3.
If the number of coins is 1 in a pile than only 1 coin can be removed(not $\lfloor\frac12\rfloor$), otherwise the above rule is applied.
 A: It sounds like you're asking about a variant of Nim in which the legal moves on a heap of $n$ beans are to either remove $1$ bean, or to remove no more than half the beans.  As you already know, for Nim one defines the nim-sum of two heap sizes as their binary XOR; the nim-value of a position is then the nim-sum of all the heap sizes; a winning position is just one with a nonzero nim-value; and a winning move is a move that leaves the nim-value at zero.  The most fundamental result in the theory of impartial games is that any position has an associated nim-value that is the nim-sum of the nim-values of its components.  This is not always useful, because the components are not always easy to identify, but in your case each heap is still a single component.  The only difference is that the nim-value of a single heap in your game will be $f(n)$ instead of $n$, for a function $f(n)$ given by
$$
f(n)=\text{mex}\{f(n-k):k=1\vee 2k\le n\},$$
where the "mex" of a set of natural numbers is the minimum excluded value.  So
$$
\begin{eqnarray}
f(0)&=&\text{mex}\{\}=0 \\
f(1)&=&\text{mex}\{0\}=1 \\
f(2)&=&\text{mex}\{1\}=0 \\
f(3)&=&\text{mex}\{0 \}=1 \\
f(4)&=&\text{mex}\{0,1 \}=2 \\
f(5)&=&\text{mex}\{1,2\}=0 \\
f(6)&=&\text{mex}\{1,2,0\}=3 \\
f(7)&=&\text{mex}\{2,0,3\}=1 \\
f(8)&=&\text{mex}\{2,0,3,1\}=4 \\
f(9)&=&\text{mex}\{0,3,1,4\}=2 \\
f(10)&=&\text{mex}\{0,3,1,4,2\}=5 \\
f(11)&=&\text{mex}\{3,1,4,2,5\}=0 \\
f(12)&=&\text{mex}\{3,1,4,2,5,0\}=6
\end{eqnarray}
$$
and so on.  The losing single-heap sizes are those with nim-value zero: $2,5,11,...$; each is double the previous size plus one.

Update: Looking further at the sequence of nim-values reveals the pattern.  At each odd $n=2k+1$, the size $k$ becomes unreachable, and as a consequence $f(2k+1)=f(k)$.  At each even $n=2k$, a new value is attained: $f(2k)=k$ for $k>1$.  This recursion makes it easy to determine the nim-value for any $n$:
$f(1)=1$, $f(0)=f(2)=0$, $f(2k+1)=f(k)$ for $k>1$, and $f(2k)=k$ for $k>1$.
