I was recently working on a problem that need the idea of nim-multiplication. While nim-addition plays a nice role in all 1D combinatorial games, nim-multiplication takes its place in their 2D counterparts. An excellent read on the subject of Impartial Combinatorial Games can be found in Thomas S. Ferguson's Game Theory.

Now, back to nim-multiplication. While nim-addition, denoted by $\oplus$, is easy to calculate in a computer with a simple BitXOR operation, nim-multiplication, denoted by $\otimes$, is not as simple. Every source I found in the Internet listed the following two rules to simplify nim-multiplication.

To further simplify the task, we can make use of the following two additional rules.

Before solving for the nim-multiplication problem, let's just revisit some properties of the $\text{AND}$ product. It is easy that $\text{AND}$ satisfies the following three properties.

Using these properties, we can show that

Now the nim-product of arbitrary numbers can be split into several nim-products of powers-of-2 because

$x\otimes y=(2^{x_0}\oplus2^{x_1}\oplus2^{x_2}\oplus\cdots)\otimes(2^{y_0}\oplus2^{y_1}\oplus2^{y_2}\oplus\cdots)=\displaystyle\sum_{i,j \geq 0}2^{x_i}\otimes 2^{y_j}$

Now for the nim-product of two distinct powers-of-2,

\begin{align}

2^x\otimes 2^y & = (2^{x-(x\text{ AND }y)}\cdot 2^{x\text{ AND }y})\otimes (2^{y-(x\text{ AND }y)}\cdot 2^{x\text{ AND }y}) \\

& = (2^{x-(x\text{ AND }y)}\otimes 2^{x\text{ AND }y})\otimes (2^{y-(x\text{ AND }y)}\otimes 2^{x\text{ AND }y}) \\

& = (2^{x-(x\text{ AND }y)}\otimes 2^{y-(x\text{ AND }y)})\otimes (2^{x\text{ AND }y}\otimes 2^{x\text{ AND }y}) \\

& = 2^{x-(x\text{ AND }y)}\cdot 2^{y-(x\text{ AND }y)}\otimes (2^{x\text{ AND }y}\otimes 2^{x\text{ AND }y})\\

\end{align}

In the first equality, we have used both

The two results above together form the recursion that can be used to solve the nim-multiplication problem. Together with the facts that the nim-product of any number with 0 is 0 and the nim-product of a number with 1 is the number itself, it's not so hard to code the above for nim-multiplication.

UPDATE: My python implementation of the above algorithm.

def memoize(fn):

"""returns a memoized version of any function that can be called

with the same list of arguments.

Usage: foo = memoize(foo)"""

def handle_item(x):

if isinstance(x, dict):

return make_tuple(sorted(x.items()))

elif hasattr(x, '__iter__'):

return make_tuple(x)

else:

return x

def make_tuple(L):

return tuple(handle_item(x) for x in L)

def foo(*args, **kwargs):

items_cache = make_tuple(sorted(kwargs.items()))

args_cache = make_tuple(args)

if (args_cache, items_cache) not in foo.past_calls:

foo.past_calls[(args_cache, items_cache)] = fn(*args,**kwargs)

return foo.past_calls[(args_cache, items_cache)]

foo.past_calls = {}

foo.__name__ = 'memoized_' + fn.__name__

return foo

@memoize

def nimmult(x, y):

if x < y:

return nimmult(y, x)

elif y == 0:

return 0

elif y == 1:

return x

elif x & (x - 1):

return nimmult((x & - x), y) ^ nimmult(x - (x & - x), y)

elif y & (y - 1):

return nimmult(x, (y & - y)) ^ nimmult(x, y - (y & - y))

elif x in [2, 4, 16, 256, 65536, 4294967296]:

return 3 * x // 2 if x == y else x * y

elif x == y:

tempx, p = x, 0

while tempx > 1:

tempx >>= 1

p += 1

p = p & -p

p = 1 << p

return nimmult(3 * p // 2, nimmult(x // p, y // p))

else:

tempx, tempy, px, py = x, y, 0, 0

while tempx > 1:

tempx >>= 1

px += 1

while tempy > 1:

tempy >>= 1

py += 1

temp = px & py

temp = 1 << temp

return nimmult(x // temp * y // temp, nimmult(temp, temp))

I've included the memoize operator that I usually use in the code. In summary, Yes.. It's as simple as that :)

Until then

Yours Aye

Me

Now, back to nim-multiplication. While nim-addition, denoted by $\oplus$, is easy to calculate in a computer with a simple BitXOR operation, nim-multiplication, denoted by $\otimes$, is not as simple. Every source I found in the Internet listed the following two rules to simplify nim-multiplication.

**Rule 1**: The nim-product of a Fermat power and any smaller number is just their ordinary product.**Rule 2**: The nim-product of a Fermat power with itself is 3/2-th of the Fermat power.

These two rules, in addition to the fact that nim-product is distributive w.r.t nim-addition and the commutative property of nim-multiplication, certainly simplifies the task.

To further simplify the task, we can make use of the following two additional rules.

**Rule 3**: For any sequence of non-negative integers $p_0<p_1<p_2<\cdots$, we have $2^{2^{p_0}}\otimes2^{2^{p_1}}\otimes2^{2^{p_2}}\otimes\cdots=2^{2^{p_0}}\cdot2^{2^{p_1}}\cdot2^{2^{p_2}}\cdot\cdots$. This also shows that any power of 2 can be written as nim-product of distinct Fermat powers by writing the exponent as sums of powers-of -2.**Rule 4**:If $x\text{ AND }y=0$, then $x\oplus y=x + y$**Rule 5**: If $x\text{ AND }y=0$, then $2^x\otimes 2^y=2^x\cdot2^y$

**Rule 3**works because we can repeatedly apply

**Rule 1**for nim-products starting from the left and convert them to ordinary products. Note that any resulting term on the left will always be smaller than the term that comes after.

**Rule 4**works because the AND condition makes sure that no bits are identical in the binary representation of $x$ and $y$.

**Rule 5**works because the nim-product can be converted to nim-products of Fermat powers. The AND condition makes sure that no two Fermat powers are equal and hence we can sort the Fermat powers and apply

**Rule 3**.

Before solving for the nim-multiplication problem, let's just revisit some properties of the $\text{AND}$ product. It is easy that $\text{AND}$ satisfies the following three properties.

- $x\text{ AND }x=x$
- Associative: $x\text{ AND }(y \text{ AND }z)=(x\text{ AND }y)\text{ AND }z$
- Commutative: $x\text{ AND }y=y\text{ AND }x$

Using these properties, we can show that

**P1**: $(x\text{ AND }y)\text{ AND }(x-(x\text{ AND }y))=0$**P2**: $(x\text{ AND }y)\text{ AND }(y-(x\text{ AND }y))=0$**P3**: $ (x-(x\text{ AND }y))\text{ AND }(y-(x\text{ AND }y)) = 0$

Now the nim-product of arbitrary numbers can be split into several nim-products of powers-of-2 because

$x\otimes y=(2^{x_0}\oplus2^{x_1}\oplus2^{x_2}\oplus\cdots)\otimes(2^{y_0}\oplus2^{y_1}\oplus2^{y_2}\oplus\cdots)=\displaystyle\sum_{i,j \geq 0}2^{x_i}\otimes 2^{y_j}$

where the summation symbol should be understood as a nim-sum.

Let's first solve for nim-squares of powers-of-2. That is a nim-product of a power-of-2 with itself.

\begin{align}

2^x\otimes 2^x & = (2^{x-(x\text{ AND }-x)}\cdot 2^{x\text{ AND }-x})\otimes (2^{x-(x\text{ AND }-x)}\cdot 2^{x\text{ AND }-x}) \\

& = (2^{x-(x\text{ AND }-x)}\otimes 2^{x\text{ AND }-x})\otimes (2^{x-(x\text{ AND }-x)}\otimes 2^{x\text{ AND }-x}) \\

& = (2^{x\text{ AND }-x}\otimes 2^{x\text{ AND }-x})\otimes (2^{x-(x\text{ AND }-x)}\otimes 2^{x-(x\text{ AND }-x)}) \\

& = (3/2)2^{x\text{ AND }-x}\otimes(2^{x-(x\text{ AND }-x)}\otimes 2^{x-(x\text{ AND }-x)}) \\

\end{align}

We could convert ordinary product in the second equality into an nim-product because the AND product of the exponents can be clearly seen to be zero by using $y=-x$ in

\begin{align}

2^x\otimes 2^x & = (2^{x-(x\text{ AND }-x)}\cdot 2^{x\text{ AND }-x})\otimes (2^{x-(x\text{ AND }-x)}\cdot 2^{x\text{ AND }-x}) \\

& = (2^{x-(x\text{ AND }-x)}\otimes 2^{x\text{ AND }-x})\otimes (2^{x-(x\text{ AND }-x)}\otimes 2^{x\text{ AND }-x}) \\

& = (2^{x\text{ AND }-x}\otimes 2^{x\text{ AND }-x})\otimes (2^{x-(x\text{ AND }-x)}\otimes 2^{x-(x\text{ AND }-x)}) \\

& = (3/2)2^{x\text{ AND }-x}\otimes(2^{x-(x\text{ AND }-x)}\otimes 2^{x-(x\text{ AND }-x)}) \\

\end{align}

We could convert ordinary product in the second equality into an nim-product because the AND product of the exponents can be clearly seen to be zero by using $y=-x$ in

**P1**.Now for the nim-product of two distinct powers-of-2,

\begin{align}

2^x\otimes 2^y & = (2^{x-(x\text{ AND }y)}\cdot 2^{x\text{ AND }y})\otimes (2^{y-(x\text{ AND }y)}\cdot 2^{x\text{ AND }y}) \\

& = (2^{x-(x\text{ AND }y)}\otimes 2^{x\text{ AND }y})\otimes (2^{y-(x\text{ AND }y)}\otimes 2^{x\text{ AND }y}) \\

& = (2^{x-(x\text{ AND }y)}\otimes 2^{y-(x\text{ AND }y)})\otimes (2^{x\text{ AND }y}\otimes 2^{x\text{ AND }y}) \\

& = 2^{x-(x\text{ AND }y)}\cdot 2^{y-(x\text{ AND }y)}\otimes (2^{x\text{ AND }y}\otimes 2^{x\text{ AND }y})\\

\end{align}

In the first equality, we have used both

**P1**and

**P2**to convert from ordinary product to nim-product. In the last equality, we have used

**P3**to do the opposite.

The two results above together form the recursion that can be used to solve the nim-multiplication problem. Together with the facts that the nim-product of any number with 0 is 0 and the nim-product of a number with 1 is the number itself, it's not so hard to code the above for nim-multiplication.

UPDATE: My python implementation of the above algorithm.

def memoize(fn):

"""returns a memoized version of any function that can be called

with the same list of arguments.

Usage: foo = memoize(foo)"""

def handle_item(x):

if isinstance(x, dict):

return make_tuple(sorted(x.items()))

elif hasattr(x, '__iter__'):

return make_tuple(x)

else:

return x

def make_tuple(L):

return tuple(handle_item(x) for x in L)

def foo(*args, **kwargs):

items_cache = make_tuple(sorted(kwargs.items()))

args_cache = make_tuple(args)

if (args_cache, items_cache) not in foo.past_calls:

foo.past_calls[(args_cache, items_cache)] = fn(*args,**kwargs)

return foo.past_calls[(args_cache, items_cache)]

foo.past_calls = {}

foo.__name__ = 'memoized_' + fn.__name__

return foo

@memoize

def nimmult(x, y):

if x < y:

return nimmult(y, x)

elif y == 0:

return 0

elif y == 1:

return x

elif x & (x - 1):

return nimmult((x & - x), y) ^ nimmult(x - (x & - x), y)

elif y & (y - 1):

return nimmult(x, (y & - y)) ^ nimmult(x, y - (y & - y))

elif x in [2, 4, 16, 256, 65536, 4294967296]:

return 3 * x // 2 if x == y else x * y

elif x == y:

tempx, p = x, 0

while tempx > 1:

tempx >>= 1

p += 1

p = p & -p

p = 1 << p

return nimmult(3 * p // 2, nimmult(x // p, y // p))

else:

tempx, tempy, px, py = x, y, 0, 0

while tempx > 1:

tempx >>= 1

px += 1

while tempy > 1:

tempy >>= 1

py += 1

temp = px & py

temp = 1 << temp

return nimmult(x // temp * y // temp, nimmult(temp, temp))

I've included the memoize operator that I usually use in the code. In summary, Yes.. It's as simple as that :)

Until then

Yours Aye

Me