logic.py

               ]))

# ______________________________________________________________________________

# Example application (not in the book).
# You can use the Expr class to do symbolic differentiation.  This used to be
# a part of AI; now it is considered a separate field, Symbolic Algebra.


def diff(y, x):
    """Return the symbolic derivative, dy/dx, as an Expr.
    However, you probably want to simplify the results with simp.
    >>> diff(x * x, x)
    ((x * 1) + (x * 1))
    """
    if y == x:
        return 1
    elif not y.args:
        return 0
    else:
        u, op, v = y.args[0], y.op, y.args[-1]
        if op == '+':
            return diff(u, x) + diff(v, x)
        elif op == '-' and len(y.args) == 1:
            return -diff(u, x)
        elif op == '-':
            return diff(u, x) - diff(v, x)
        elif op == '*':
            return u * diff(v, x) + v * diff(u, x)
        elif op == '/':
            return (v * diff(u, x) - u * diff(v, x)) / (v * v)
        elif op == '**' and isnumber(x.op):
            return (v * u ** (v - 1) * diff(u, x))
        elif op == '**':
            return (v * u ** (v - 1) * diff(u, x) +
                    u ** v * Expr('log')(u) * diff(v, x))
        elif op == 'log':
            return diff(u, x) / u
        else:
            raise ValueError("Unknown op: {} in diff({}, {})".format(op, y, x))


def simp(x):
    """Simplify the expression x."""
    if isnumber(x) or not x.args:
        return x
    args = list(map(simp, x.args))
    u, op, v = args[0], x.op, args[-1]
    if op == '+':
        if v == 0:
            return u
        if u == 0:
            return v
        if u == v:
            return 2 * u
        if u == -v or v == -u:
            return 0
    elif op == '-' and len(args) == 1:
        if u.op == '-' and len(u.args) == 1:
            return u.args[0]  # --y ==> y
    elif op == '-':
        if v == 0:
            return u
        if u == 0:
            return -v
        if u == v:
            return 0
        if u == -v or v == -u:
            return 0
    elif op == '*':
        if u == 0 or v == 0:
            return 0
        if u == 1:
            return v
        if v == 1:
            return u
        if u == v:
            return u ** 2
    elif op == '/':
        if u == 0:
            return 0
        if v == 0:
            return Expr('Undefined')
        if u == v:
            return 1
        if u == -v or v == -u:
            return 0
    elif op == '**':
        if u == 0:
            return 0
        if v == 0:
            return 1
        if u == 1:
            return 1
        if v == 1:
            return u
    elif op == 'log':
        if u == 1:
            return 0
    else:
        raise ValueError("Unknown op: " + op)
    # If we fall through to here, we can not simplify further
    return Expr(op, *args)


def d(y, x):
    """Differentiate and then simplify."""
    return simp(diff(y, x))