Can anyone explain me in below algorithm how the "ISZERO" function checking whether the polynomial is zero or not. Here "REM(P,e)" function removes all the values with exponent "e".
what i don't able to understand is the significance of "if COEF(P,e) = - c". And also what is this "SMULT" function.
structure POLYNOMIAL
declare ZERO( ) poly; ISZERO(poly) Boolean
COEF(poly,exp) coef;
ATTACH(poly,coef,exp) poly
REM(poly,exp) poly
SMULT(poly,coef,exp) poly
ADD(poly,poly) poly; MULT(poly,poly) poly;
for all P,Q, poly c,d, coef e,f exp let
REM(ZERO,f) :: = ZERO
REM(ATTACH(P,c,e),f) :: =
if e = f then REM(P,f) else ATTACH(REM(P,f),c,e)
***ISZERO(ZERO) :: = true
ISZERO(ATTACH(P,c,e)):: =
if COEF(P,e) = - c then ISZERO(REM(P,e)) else false***
COEF(ZERO,e) :: = 0
COEF(ATTACH(P,c,e),f) :: =
if e = f then c + COEF(P,f) else COEF(P,f)
SMULT(ZERO,d,f) :: = ZERO
SMULT(ATTACH(P,c,e),d,f) :: =
ATTACH(SMULT(P,d,f),c d,e + f)
ADD(P,ZERO):: = P
ADD(P,ATTACH(Q,d,f)) :: = ATTACH(ADD(P,Q),d,f)
MULT(P,ZERO) :: = ZERO
MULT(P,ATTACH(Q,d,f)) :: =
ADD(MULT(P,Q),SMULT(P,d,f))
end
end POLYNOMIAL
Without knowing what language this is, it looks like this line
ISZERO(ATTACH(P,c,e)):: =
if COEF(P,e) = - c then ISZERO(REM(P,e)) else false
is specifying ISZERO recursively. We are trying to determine whether ATTACH(P, c, e), otherwise known as P(x) + cx^e, is zero. It first checks whether the x^e coefficient of P is -c. If not, then P(x) + cx^e is definitely not zero, and you can return false immediately. Otherwise, P(x) + cx^e = REM(P, e), so you have to check ISZERO(REM(P, e)).
I believe SMULT is multiplication, so SMULT(P, a, b) is equivalent to a * x^b * P(x).
Related
Just as a silly example, say that I wish to solve for the following nonlinear equation x^2 - F(c)=0, where c can take different values between zero and one and F is a standard normal CDF. If I wish to solve for one particular value of c, I would use the following code:
c = linspace(0,1,100);
L = length(c);
x0 = c;
function Y = eq(x)
Y = x^2 - cdfnor("PQ",x-c(1),0,1)
endfunction
xres = fsolve(x0(1),eq);
My question is: Is there a way to solve for the equation for each value of c (and not only c(1))? Specifically, if I can use a loop over fsolve? If so, how?
Just modify your script like this:
c = linspace(0,1,100);
L = length(c);
x0 = c;
function Y = eq(x)
Y = x^2 - cdfnor("PQ",x-c,zeros(c),ones(c))
endfunction
xres = fsolve(x0,eq);
We want to compare a^b to c^d, and tell if the first is smaller, greater, or equal (where ^ denotes exponentiation).
Obviously, for very large numbers, we cannot explicitely compute these values.
The most common approach in this situation is to apply log on both sides and compare b * log(a) to d * log(c). The issue here is that logs are floating-point operations, and as such we cannot trust our answer with 100% confidence (there might be some values which are incredibly close, and because of floating-point error we get a wrong answer).
Is there an algorithm for solving this problem? I've been scouring the intrernet for this, but I can only find solutions which work for particular cases only (e.g. in which one exponent is a multiple of another), or which use floating point in some way (logarithms, division) etc.
This is sort of two questions in one:
Are they equal?
If not, which one is greater?
As Peter O. observes, it's easiest to build in a language that provides an arbitrary-precision fraction type. I'll use Python 3.
Let's assume without loss of generality that a ≤ c (swap if necessary) and b is relatively prime to d (divide both by the greatest common divisor).
To get at the core of the question, I'm going to assume that a, c > 0 and b, d ≥ 0. Removing this assumption is tedious but not difficult.
Equality test
There are some easy cases where a = 1 or b = 0 or c = 1 or d = 0.
Separately, necessary conditions for a^b = c^d are
i. b ≥ d, since otherwise b < d, which together with a ≤ c implies a^b < c^d;
ii. a is a divisor of c, since we know from (i) that a^b = c^d is a divisor of c^b = c^(b−d) c^d.
When these conditions hold, we can divide through by a^d to reduce the problem to testing whether a^(b−d) = (c/a)^d.
In Python 3:
def equal_powers(a, b, c, d):
while True:
lhs_is_one = a == 1 or b == 0
rhs_is_one = c == 1 or d == 0
if lhs_is_one or rhs_is_one:
return lhs_is_one and rhs_is_one
if a > c:
a, b, c, d = c, d, a, b
if b < d:
return False
q, r = divmod(c, a)
if r != 0:
return False
b -= d
c = q
def test_equal_powers():
for a in range(1, 25):
for b in range(25):
for c in range(1, 25):
for d in range(25):
assert equal_powers(a, b, c, d) == (a ** b == c ** d)
test_equal_powers()
Inequality test
Once we've established that the two quantities are not equal, it's time to figure out which one is greater. (Without the equality test, the code here could run forever.)
If you're doing this for real, you should consult an actual reference on computing elementary functions. I'm just going to try to do the simplest thing that works.
Time for a calculus refresher. We have the Taylor series
−log x = (1−x) + (1−x)^2/2 + (1−x)^3/3 + (1−x)^4/4 + ...
To get a lower bound, truncate the series. To get an upper bound, we can truncate but replace the final term (1−x)^n/n with (1−x)^n/n (1/x), since
(1−x)^n/n (1/x)
= (1−x)^n/n (1 + (1−x) + (1−x)^2 + ...)
= (1−x)^n/n + (1−x)^(n+1)/n + (1−x)^(n+2)/n + ...
> (1−x)^n/n + (1−x)^(n+1)/(n+1) + (1−x)^(n+2)/(n+2) + ...
To get a good convergence rate, we're going to want 0.5 ≤ x < 1, which we can achieve by dividing x by a power of two.
In Python, we'll represent a real number as an infinite generator of shrinking intervals that contain the true value. Once the intervals for b log a and d log c are disjoint, we can determine how they compare.
import fractions
def minus(x, y):
while True:
x_lo, x_hi = next(x)
y_lo, y_hi = next(y)
yield x_lo - y_hi, x_hi - y_lo
def times(b, x):
for lo, hi in x:
yield b * lo, b * hi
def restricted_log(a):
series = 0
n = 0
numerator = 1
while True:
n += 1
numerator *= 1 - a
series += fractions.Fraction(numerator, n)
yield -(series + fractions.Fraction(numerator * (1 - a), (n + 1) * a)), -series
def log(a):
n = 0
while a >= 1:
a = fractions.Fraction(a, 2)
n += 1
return minus(restricted_log(a), times(n, restricted_log(fractions.Fraction(1, 2))))
def less_powers(a, b, c, d):
lhs = times(b, log(a))
rhs = times(d, log(c))
while True:
lhs_lo, lhs_hi = next(lhs)
rhs_lo, rhs_hi = next(rhs)
if lhs_hi < rhs_lo:
return True
if rhs_hi < lhs_lo:
return False
def test_less_powers():
for a in range(1, 10):
for b in range(10):
for c in range(1, 10):
for d in range(10):
if a ** b != c ** d:
assert less_powers(a, b, c, d) == (a ** b < c ** d)
test_less_powers()
[] = always
O = next
! = negation
<> = eventually
Wondering is it []<> is that equivalent to just []?
Also having a hard time understanding how to distribute temporal logic.
[][] (a OR !b)
!<>(!a AND b)
[]([] a ==> <> b)
I'll use the following notations:
F = eventually
G = always
X = next
U = until
In my model-checking course, we defined LTL the following way:
LTL: p | φ ∩ ψ | ¬φ | Xφ | φ U ψ
With F being a syntactic sugar for :
F (future)
Fφ = True U φ
and G:
G (global)
Gφ = ¬F¬φ
With that, your question is :
Is it true that : Gφ ?= GFφ
GFφ <=> G (True U φ)
Knowing that :
P ⊧ φ U ψ <=> exists i >= 0: P_(>= i) ⊧ ψ AND forall 0 <= j < i : P_(<= j) ⊧ φ
From that, we can clearly see that GFφ indicates that it must always be true that φ will be always be verified after some time i, and before that (j before i) True must be verified (trivial).
But Gφ indicates that φ must always be true, "from now to forever" and not "from i to forever".
G p indicates that at all times p holds. GF p indidcates that at all times, eventually p will hold. So while the infinite trace pppppp... satisfies both of the specifications, an infinite trace of the form p(!p)(!p!)p(!p)p... satisfies only GF p but not G p.
To be clear, both these example traces need to contain infinitely many locations, where p holds. But in the case of GF p, and only in this case, it is acceptable that there be locations in between, where p does not hold.
So the short answer to the above question by counterexample is: no, those two specifications aren't the same.
I was reading on perceptrons and trying to implement one in haskell. The algorithm seems to be working as far as I can test. I'm going to rewrite the code entirely at some point, but before doing so I thought of asking a few questions that have arosen while coding this.
The neuron can be trained when returning the complete neuron. let neuron = train set [1,1] works, but if I change the train function to return an incomplete neuron without the inputs, or try to pattern match and create only an incomplete neuron, the code falls into neverending loop.
tl;dr when returning complete neuron everything works, but when returning curryable neuron, the code falls into a loop.
module Main where
import System.Random
type Inputs = [Float]
type Weights = [Float]
type Threshold = Float
type Output = Float
type Trainingset = [(Inputs, Output)]
data Neuron = Neuron Threshold Weights Inputs deriving Show
output :: Neuron -> Output
output (Neuron threshold weights inputs) =
if total >= threshold then 1 else 0
where total = sum $ zipWith (*) weights inputs
rate :: Float -> Float -> Float
rate t o = 0.1 * (t - o)
newweight :: Float -> Float -> Weights -> Inputs -> Weights
newweight t o weight input = zipWith nw weight input
where nw w x = w + (rate t o) * x
learn :: Neuron -> Float -> Neuron
learn on#(Neuron tr w i) t =
let o = output on
in Neuron tr (newweight t o w i) i
converged :: (Inputs -> Neuron) -> Trainingset -> Bool
converged n set = not $ any (\(i,o) -> output (n i) /= o) set
train :: Weights -> Trainingset -> Neuron
train w s = train' s (Neuron 1 w)
train' :: Trainingset -> (Inputs -> Neuron) -> Neuron
train' s n | not $ converged n set
= let (Neuron t w i) = train'' s n
in train' s (Neuron t w)
| otherwise = n $ fst $ head s
train'' :: Trainingset -> (Inputs -> Neuron) -> Neuron
train'' ((a,b):[]) n = learn (n a) b
train'' ((a,b):xs) n = let
(Neuron t w i) = learn (n a) b
in
train'' xs (Neuron t w)
set :: Trainingset
set = [
([1,0], 0),
([1,1], 1),
([0,1], 0),
([0,0], 0)
]
randomWeights :: Int -> IO [Float]
randomWeights n =
do
g <- newStdGen
return $ take n $ randomRs (-1, 1) g
main = do
w <- randomWeights 2
let (Neuron t w i) = train w set
print $ output $ (Neuron t w [1,1])
return ()
Edit: As per comments, specifying a little more.
Running with the code above, I get:
perceptron: <<loop>>
But by editing the main method to:
main = do
w <- randomWeights 2
let neuron = train w set
print $ neuron
return ()
(Notice the let neuron, and print rows), everything works and the output is:
Neuron 1.0 [0.71345896,0.33792675] [1.0,0.0]
Perhaps I am missing something, but I boiled your test case down to this program:
module Main where
data Foo a = Foo a
main = do
x ← getLine
let (Foo x) = Foo x
putStrLn x
This further simplifies to:
main = do
x ← getLine
let x = x
putStrLn x
The problem is that binding (Foo x) to something that depends on x
is a cyclic dependency. To evaluate x, we need to know the value of
x. OK, so we just need to calculate x. To calculate x, we need to
know the value of x. That's fine, we'll just calculate x. And so on.
This isn't C, remember: it's binding, not assignment, and the binding
is evaluated lazily.
Use better variable names, and it all works:
module Main where
data Foo a = Foo a
main = do
line ← getLine
let (Foo x) = Foo line
putStrLn x
(The variable in question, in your case, is w.)
This is a common mistake in Haskell. You cannot say things like:
let x = 0
let x = x + 1
And have it mean what it would in a language with assignment, or even nonrecursive binding. The first line is irrelevant, it gets shadowed by the second line, which defines x as x+1, that is, it defines recursively x = ((((...)+1)+1)+1)+1, which will loop upon evaluation.
I am working on the exercise questions of book The Lambda calculus. One of the questions that I am stuck is proving the following:
Show that the application is not associative; in fact, x(yz) not equals (xy)z
Here is what I have worked on so far:
Let x = λa.λb. ab
Let y = λb.λc. bc
Let z = λa.λc. ac
(xy)z => ((λa.λb. ab) (λb.λc. bc)) (λa.λc. ac)
=> (λb. (λb.λc. bc) b) (λa.λc. ac)
=> (λb.λc. bc) (λa.λc. ac)
=> (λc. (λa.λc. ac) c)
x(yz) => (λa.λb. ab) ((λb.λc. bc) (λa.λc. ac))
=> (λb. ((λb.λc. bc) (λa.λc. ac)) b)
=> (λb. (λc. (λa.λc. ac) c) b)
Is this correct? Please help me understand.
I also think that your counter-example is correct.
You can probably get a simpler counter-example like this:
let x = λa.n and y, z variables then:
(xy)z => ((λa.n) y) z => n z
x(yz) => (λa.n) (y z) => n
The derivations seem fine, at a glance.
Conceptually, just think that x, y, and z can represent any computable functions, and clearly, some of those functions are not associative. Say, x is 'subtract 2', y is 'divide by 2', and z is 'double'. For this example, x(yz) = 'subtract 2' and (xy)z = 'substract 1'.
It seems ok, but for simplicity, how about prove by contradiction?
Assume (xy)z = x(yz), and let
x = λa.λb. a # x = const
y = λa. -a # y = negate
z = 1 # z = 1
and show that ((xy)z) 0 ≠ (x(yz)) 0.
The book you mention by Barendregt is extremely formal and precise (a great book), so it would be nice to have the precise statement of the exercise.
I guess the actual goal was to find instantiations for x, y and z such that
x (y z) reduces to the boolean true = \xy.x and (x y) z reduces to the boolean false = \xy.y
Then, you can take e.g. x = \z.true and z = I = \z.z (y arbitrary).
But how can we prove that true is not convertible with false? You have no way to prove it inside the calculus, since you have no negation: you can only prove equalities and not inequalities. However, let us observe that if true=false then all terms are equal.
Indeed, for any M and N, if true = false then
true M N = false M N
but true M N reduces to M, while false M N reduces to N, so
M = N
Hence, if true = false all terms would be equal, and the calculus would be trivial. Since we can find not trivial models of the lambda calculus, no such
model may equate true and false (more generally may equate terms with different normal forms, that would require us to talk about the bohm-out technique).