Module:Rational: Difference between revisions

← Older edit Newer edit →

@@ Line 112: / Line 112: @@
 function p.div(a, b)
 	return p.mul(a, p.inv(b))
+end
+-- compute a^b; b must be an integer
+function p.pow(a, b)
+	if type(a) == 'number' then
+		a = p.new(a)
+	end
+	if type(b) ~= 'number' then
+		return nil
+	end
+	if a.nan then
+		return { nan = true }
+	end
+	if a.inf then
+		if b == 0 then
+			return { nan = true }
+		elseif b > 0 then
+			return { inf = true, sign = math.pow(a.sign, b) }
+		else
+			return { zero = true, sign = math.pow(a.sign, b) }
+		end
+	end
+	if a.zero then
+		if b == 0 then
+			return p.new(1)
+		elseif b > 0 then
+			return { zero = true, sign = math.pow(a.sign, b) }
+		else
+			return { inf = true, sign = math.pow(a.sign, b) }
+		end
+	end
+	local c = p.new(1)
+	for i = 1, math.abs(b) do
+		if b > 0 then
+			c = p.mul(c, a)
+		else
+			c = p.div(c, a)
+		end
+	end
+	return c
+end
+-- compute a canonical representation of `a` modulo powers of `b`
+function p.modulo_mul(a, b)
+	if type(a) == 'number' then
+		a = p.new(a)
+	end
+	if type(b) == 'number' then
+		b = p.new(b)
+	end
+	if a.nan or b.nan or a.inf or b.inf or a.zero or b.zero then
+		return p.copy(a)
+	end
+	local neg_power = -1/0
+	local pos_power = 1/0
+	for factor, power in pairs(b) do
+		if type(factor) == 'number' then
+			if (power > 0 and (a[factor] or 0) >= 0) or (power < 0 and (a[factor] or 0) <= 0) then
+				pos_power = math.min(pos_power,
+					math.floor((a[factor] or 0) / power)
+				)
+			else
+				neg_power = math.max(neg_power,
+					-math.ceil(math.abs(a[factor] or 0) / math.abs(power))
+				)
+			end
+		end
+	end
+	local power = 0
+	if neg_power ~= neg_power + 1 and neg_power < 0 then
+		power = neg_power
+	end
+	if pos_power ~= pos_power + 1 and pos_power > 0 then
+		power = pos_power
+	end
+	return p.div(a, p.pow(b, power))
 end