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Comments:
<0> a few lets wouldn't hurt.
<1> yeah, you have that stray 'doInput list' at the top level there
<2> Cale: no main calls doInput
<1> then you need to indent the doInput
<3> What's input for Theta?
<4> As it's part of the definition of 'main'.
<0> @type sum
<5> forall a. (Num a) => [a] -> a
<3> Cale, what's the input for theta?
<1> Aleks: I don't understand the context of your question
<1> Manyfold: also, what's list?
<1> Manyfold: perhaps you want to p*** it the empty list?
<3> sin(theta) = 2/3 and pi<=(thetat)<=pi/2, find cot(theta)
<3> that's just (theta) not thetat
<3> understand now?
<6> Aleks: have you tried drawing a right triangle to solve that problem?
<7> back
<8> int-e annotated #21967 with "(for Manyfold) this could work" at http://paste.lisp.org/display/21967#1
<9> Manyfol1: yay
<7> Cale: it's excercise 3.10 in the tutorial you pointed me recently
<1> Manyfol1: yeah, I recognise it
<1> int-e: that still has indentation and typing problems
<3> <6> ...no... it would help though :x
<0> Cale: I guess the if then else has problems. what typo did I miss?
<1> int-e: not typos, but various types don't match
<0> oh yes. getLine is bad :)
<7> num is really a string?
<8> int-e annotated #21967 with "(for Manyfold) with style changes; this one actually works." at http://paste.lisp.org/display/21967#2
<7> so num = read numString peerhaps?
<0> you have to use let for declaring pure values within do blocks
<0> you can either use readLn (as I did) which is basically a getLine followed by a read.
<10> sum = foldr (+) 0
<10> product = foldr (*) 1
<0> there are really many ways to do it: a) x <- getLine; let y = read x b) x <- liftM read (getLine) c) x <- getLine >>= return . read d) x <- return . read =<< getLine (some people prefer this to c) )
<0> e) instead of let y = read x you could use y <- return (read x)
<10> x <- read `fmap` getLine
<0> then there's readLn and I think that covers the most reasonable versions.
<8> Cale annotated #21967 with "a solution to the problem" at http://paste.lisp.org/display/21967#3
<1> readLn
<1> @type readLn
<5> forall a. (Read a) => IO a
<1> oh, it was mentioned
<11> Hrm
<11> x^(ab) - x^a <- can this be reduced anymore?
<1> it can be factored
<10> x^a(x^b - 1)
<11> right eight
<0> xerox: not quite
<11> x^a*x^b - x^a
<10> No!
<11> no
<11> nono
<11> x^(ab) = x^a^b
<10> Right :)
<0> x^a (x^(ab-a) - 1) = x^a (x-1) (x^(ab-a-1) + ... + x + 1)
<0> CoffeeBuzz: I'd read that as x^(a^b) though, not (x^a)^b
<11> int-e, yeah, i ***umed you knew whay i meant ;)
<0> CoffeeBuzz: better add these parentheses.
<12> CoffeeBuzz: ^ is right ***ociative so a^b^c = a^(b^c).
<3> Thanks for all the help, cya!
<11> % Solve[x^a + 1 == 0]
<5> CoffeeBuzz: {{a -> (I*Pi)/Log[x]}}
<11> how do you tell it to solve for x?
<11> % Solve[x^a + 1 == 0, x]
<5> CoffeeBuzz: {{x -> (-1)^a^(-1)}}
<11> nevermind
<11> % Solve[ (e^(a*k^2*s^2) / (k^2*s^2) == 0, s]
<5> CoffeeBuzz: $Failed
<11> pfft
<11> oh
<11> % Solve[ e^(a*k^2*s^2) / (k^2*s^2) == e^a, s]
<5> CoffeeBuzz: {{s -> ((-I)*Sqrt[ProductLog[-((a*Log[e])/e^a)]])/(Sqrt[a]*k*Sqrt[Log[e]])}, {s -> (I*Sqrt[ProductLog[-((a*Log[e])/e^a)]])/(Sqrt[a]*k*Sqrt[Log[e]])}}
<13> CoffeeBuzz, e^(blah)/blah is never 0
<11> yeah just realized that ;)
<11> ok, time to type in this monster
<11> % Solve[ e^(a*k^2*s^2) / (k^2*s^2) - e^(a*r^2 + s^2) / (r^2 + s^2) == e^(a*r^2 + a*s^2*t^2) / (r^2 + s^2*t^2) - e^(a*r^2 + a*y^2) / (r^2 + y^2), y]
<5> CoffeeBuzz: {{y -> -(Sqrt[-(a*r^2*Log[e]) - ProductLog[(a*Log[e])/(e^(a*k^2*s^2)/(k^2*s^2) - e^(a*r^2 + s^2)/(r^2 + s^2) - e^(a*r^2 + a*s^2*t^2)/(r^2 + s^2*t^2))]]/(Sqrt[a]*Sqrt[Log[e]]))}, {y ->
<5> Sqrt[-(a*r^2*Log[e]) - ProductLog[(a*Log[e])/(e^(a*k^2*s^2)/(k^2*s^2) - e^(a*r^2 + s^2)/(r^2 + s^2) - e^(a*r^2 + a*s^2*t^2)/(r^2 + s^2*t^2))]]/(Sqrt[a]*Sqrt[Log[e]])}}
<11> % Solve[ e^(a*k^2*s^2) / (k^2*s^2) - e^(a*r^2 + s^2) / (r^2 + s^2) == e^(a*r^2 + a*s^2*(k^2 - 1)) / (r^2 + s^2*(k^2 - 1)) - e^(a*r^2 + a*y^2) / (r^2 + y^2), y]
<5> CoffeeBuzz: {{y -> -(Sqrt[-(a*r^2*Log[e]) - ProductLog[(a*Log[e])/(e^(a*k^2*s^2)/(k^2*s^2) - e^(a*r^2 + s^2)/(r^2 + s^2) - e^(a*r^2 + a*(-1 + k^2)*s^2)/(r^2 - s^2 + k^2*s^2))]]/(Sqrt[a]*Sqrt[Log[e]])
<5> )}, {y -> Sqrt[-(a*r^2*Log[e]) - ProductLog[(a*Log[e])/(e^(a*k^2*s^2)/(k^2*s^2) - e^(a*r^2 + s^2)/(r^2 + s^2) - e^(a*r^2 + a*(-1 + k^2)*s^2)/(r^2 - s^2 + k^2*s^2))]]/(Sqrt[a]*Sqrt[Log[e]])}}
<13> what the hell
<11> does it use e = exponential ?? or is there some f'n to do that
<11> exp?
<11> E?
<14> Use E
<11> % Solve[ E^(a*k^2*s^2) / (k^2*s^2) - E^(a*r^2 + s^2) / (r^2 + s^2) == E^(a*r^2 + a*s^2*(k^2 - 1)) / (r^2 + s^2*(k^2 - 1)) - E^(a*r^2 + a*y^2) / (r^2 + y^2), y]
<5> CoffeeBuzz: {{y -> -(Sqrt[-(a*r^2) - ProductLog[a/(E^(a*k^2*s^2)/(k^2*s^2) - E^(a*r^2 + s^2)/(r^2 + s^2) - E^(a*r^2 + a*(-1 + k^2)*s^2)/(r^2 - s^2 + k^2*s^2))]]/Sqrt[a])}, {y -> Sqrt[-(a*r^2) -
<5> ProductLog[a/(E^(a*k^2*s^2)/(k^2*s^2) - E^(a*r^2 + s^2)/(r^2 + s^2) - E^(a*r^2 + a*(-1 + k^2)*s^2)/(r^2 - s^2 + k^2*s^2))]]/Sqrt[a]}}
<11> thats not as pretty as i hoped it would be
<13> if you're trying to solve horrible equations then you probably need to rethink the problem
<11> well... i'm building a difference of gaussian image pyramid using incremental gaussian blurring (std dev = s)... but I pre-smooth with gaussian stddev = r.
<11> say I is my image, G(s) is Gu***ian blur kernel, stddev s. let L(s) = G(s)*I (* = convolution). Then I multiply s by some constant factor k, and subtract the two so D(s) = L(ks) - L(s)
<11> next level... D(ks) = L(k^2s) - L(ks) ... etc
<11> but ... i figured instead of using huge kernels as k gets raised to higher powers, you can do it incrementally. L(k^(p+1)s) = L(k^ps)*G(u) ... where u ends up = k^p*s*sqrt(k^2 - 1) ..
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