Divide and conquer approach for Karatsuba algorithm









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I'm trying to write Karatsuba Algorithm using the divide and conquer approach in Haskell. I did it with merge sort algorithm but can't figure it out here and it's a little bit embarrassing at this point.



My main function looks like this:



dz test end divide combine x = 
if test x then end x
else combine(map(dz test end divide combine) (divide x))


I test it for values 1234 and 5678: dz test end divide combine [1234, 5678,2].



So I have to write test, end, divide and combine functions.



lengthNumb x = length . show $ x
test (x:x1:xs) = (lengthNumb x) < 4 || (lengthNumb x1) < 4
end (x:y:z:xs) = [x * y, z]


This is pretty straightforward. I just check if both numbers that I want to multiply are at least 4 digits long. If not I just use simple multiplication and carry m value. I know that Karatsuba works better for bigger numbers but this is just for testing purposes.



divide = 
divide (x:x1:x2:xs) =
let y1 = x `div` 10^x2
y2 = x `mod` 10^x2
y3 = x2 `div` 2
z1 = x1 `div` 10^x2
z2 = x1 `mod` 10^x2
in [[y1,y2,y3],[z1,z2,y3],[y1+y2, z1+z2, y3]]

combine [[x, x1],[y,y1],[z,z1]] = x * 10^(2*x1) + y + (z - x - y) * 10^x1


I was told that combine function should only do the final multiplication. So I guess it should get three numbers as an input (each with their m value) and then also do the necessary subtraction ( z-x-y ) because I couldn't do it in divide.



But this is wrong. I get an error:



Occurs check: cannot construct the infinite type: a ~ [a]
Expected type: [[a]] -> [a]

Actual type: [[[a]]] -> [a]


I think it is a problem with the parameters of combine function but I don't know how to fix it. I also think that there could be a problem with how combine and divide work together because in one of the previous iterations the final result of multiplication was wrong.










share|improve this question

















  • 6




    Write type signatures! They will show you everything.
    – luqui
    Nov 9 at 18:29






  • 5




    To elaborate on @luqui's comment: in general, if you add a type signature to each of your functions, the error messages will be closer to the code that needs to change. Often they will also be more concrete & easier to interpret.
    – bergey
    Nov 9 at 18:34










  • @luqui I am currently learning how to do it by myself but I am not there yet. I know that I can check signatures using :t command but that doesn't seem to help much. divide has [a] -> [[a]] and combine has [[b]] -> b which looks alright to me.
    – SuperM4n
    Nov 9 at 18:36






  • 1




    "I am not there yet." Just write divide :: [a] -> [[a]] above your divide definition, and combine :: [[b]] -> b above your combine definition (and similar signatures above other definitions). That tells the type checker what types you expect those functions to be, so it can figure out whether the type error is in the caller or the callee. It helps a lot with error clarity.
    – DarthFennec
    Nov 9 at 19:00










  • @DarthFennec I'm sorry, I misunderstood above comments. I wrote "I am not there yet" because I thought that guys above were talking about the derivation of types (I don't know if this is the correct term in English.). I did as you asked and now indeed I have a more detailed error message. I'll try to improve my code, thank you.
    – SuperM4n
    Nov 9 at 19:11














up vote
0
down vote

favorite












I'm trying to write Karatsuba Algorithm using the divide and conquer approach in Haskell. I did it with merge sort algorithm but can't figure it out here and it's a little bit embarrassing at this point.



My main function looks like this:



dz test end divide combine x = 
if test x then end x
else combine(map(dz test end divide combine) (divide x))


I test it for values 1234 and 5678: dz test end divide combine [1234, 5678,2].



So I have to write test, end, divide and combine functions.



lengthNumb x = length . show $ x
test (x:x1:xs) = (lengthNumb x) < 4 || (lengthNumb x1) < 4
end (x:y:z:xs) = [x * y, z]


This is pretty straightforward. I just check if both numbers that I want to multiply are at least 4 digits long. If not I just use simple multiplication and carry m value. I know that Karatsuba works better for bigger numbers but this is just for testing purposes.



divide = 
divide (x:x1:x2:xs) =
let y1 = x `div` 10^x2
y2 = x `mod` 10^x2
y3 = x2 `div` 2
z1 = x1 `div` 10^x2
z2 = x1 `mod` 10^x2
in [[y1,y2,y3],[z1,z2,y3],[y1+y2, z1+z2, y3]]

combine [[x, x1],[y,y1],[z,z1]] = x * 10^(2*x1) + y + (z - x - y) * 10^x1


I was told that combine function should only do the final multiplication. So I guess it should get three numbers as an input (each with their m value) and then also do the necessary subtraction ( z-x-y ) because I couldn't do it in divide.



But this is wrong. I get an error:



Occurs check: cannot construct the infinite type: a ~ [a]
Expected type: [[a]] -> [a]

Actual type: [[[a]]] -> [a]


I think it is a problem with the parameters of combine function but I don't know how to fix it. I also think that there could be a problem with how combine and divide work together because in one of the previous iterations the final result of multiplication was wrong.










share|improve this question

















  • 6




    Write type signatures! They will show you everything.
    – luqui
    Nov 9 at 18:29






  • 5




    To elaborate on @luqui's comment: in general, if you add a type signature to each of your functions, the error messages will be closer to the code that needs to change. Often they will also be more concrete & easier to interpret.
    – bergey
    Nov 9 at 18:34










  • @luqui I am currently learning how to do it by myself but I am not there yet. I know that I can check signatures using :t command but that doesn't seem to help much. divide has [a] -> [[a]] and combine has [[b]] -> b which looks alright to me.
    – SuperM4n
    Nov 9 at 18:36






  • 1




    "I am not there yet." Just write divide :: [a] -> [[a]] above your divide definition, and combine :: [[b]] -> b above your combine definition (and similar signatures above other definitions). That tells the type checker what types you expect those functions to be, so it can figure out whether the type error is in the caller or the callee. It helps a lot with error clarity.
    – DarthFennec
    Nov 9 at 19:00










  • @DarthFennec I'm sorry, I misunderstood above comments. I wrote "I am not there yet" because I thought that guys above were talking about the derivation of types (I don't know if this is the correct term in English.). I did as you asked and now indeed I have a more detailed error message. I'll try to improve my code, thank you.
    – SuperM4n
    Nov 9 at 19:11












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm trying to write Karatsuba Algorithm using the divide and conquer approach in Haskell. I did it with merge sort algorithm but can't figure it out here and it's a little bit embarrassing at this point.



My main function looks like this:



dz test end divide combine x = 
if test x then end x
else combine(map(dz test end divide combine) (divide x))


I test it for values 1234 and 5678: dz test end divide combine [1234, 5678,2].



So I have to write test, end, divide and combine functions.



lengthNumb x = length . show $ x
test (x:x1:xs) = (lengthNumb x) < 4 || (lengthNumb x1) < 4
end (x:y:z:xs) = [x * y, z]


This is pretty straightforward. I just check if both numbers that I want to multiply are at least 4 digits long. If not I just use simple multiplication and carry m value. I know that Karatsuba works better for bigger numbers but this is just for testing purposes.



divide = 
divide (x:x1:x2:xs) =
let y1 = x `div` 10^x2
y2 = x `mod` 10^x2
y3 = x2 `div` 2
z1 = x1 `div` 10^x2
z2 = x1 `mod` 10^x2
in [[y1,y2,y3],[z1,z2,y3],[y1+y2, z1+z2, y3]]

combine [[x, x1],[y,y1],[z,z1]] = x * 10^(2*x1) + y + (z - x - y) * 10^x1


I was told that combine function should only do the final multiplication. So I guess it should get three numbers as an input (each with their m value) and then also do the necessary subtraction ( z-x-y ) because I couldn't do it in divide.



But this is wrong. I get an error:



Occurs check: cannot construct the infinite type: a ~ [a]
Expected type: [[a]] -> [a]

Actual type: [[[a]]] -> [a]


I think it is a problem with the parameters of combine function but I don't know how to fix it. I also think that there could be a problem with how combine and divide work together because in one of the previous iterations the final result of multiplication was wrong.










share|improve this question













I'm trying to write Karatsuba Algorithm using the divide and conquer approach in Haskell. I did it with merge sort algorithm but can't figure it out here and it's a little bit embarrassing at this point.



My main function looks like this:



dz test end divide combine x = 
if test x then end x
else combine(map(dz test end divide combine) (divide x))


I test it for values 1234 and 5678: dz test end divide combine [1234, 5678,2].



So I have to write test, end, divide and combine functions.



lengthNumb x = length . show $ x
test (x:x1:xs) = (lengthNumb x) < 4 || (lengthNumb x1) < 4
end (x:y:z:xs) = [x * y, z]


This is pretty straightforward. I just check if both numbers that I want to multiply are at least 4 digits long. If not I just use simple multiplication and carry m value. I know that Karatsuba works better for bigger numbers but this is just for testing purposes.



divide = 
divide (x:x1:x2:xs) =
let y1 = x `div` 10^x2
y2 = x `mod` 10^x2
y3 = x2 `div` 2
z1 = x1 `div` 10^x2
z2 = x1 `mod` 10^x2
in [[y1,y2,y3],[z1,z2,y3],[y1+y2, z1+z2, y3]]

combine [[x, x1],[y,y1],[z,z1]] = x * 10^(2*x1) + y + (z - x - y) * 10^x1


I was told that combine function should only do the final multiplication. So I guess it should get three numbers as an input (each with their m value) and then also do the necessary subtraction ( z-x-y ) because I couldn't do it in divide.



But this is wrong. I get an error:



Occurs check: cannot construct the infinite type: a ~ [a]
Expected type: [[a]] -> [a]

Actual type: [[[a]]] -> [a]


I think it is a problem with the parameters of combine function but I don't know how to fix it. I also think that there could be a problem with how combine and divide work together because in one of the previous iterations the final result of multiplication was wrong.







haskell divide-and-conquer karatsuba






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share|improve this question











share|improve this question




share|improve this question










asked Nov 9 at 18:18









SuperM4n

9611




9611







  • 6




    Write type signatures! They will show you everything.
    – luqui
    Nov 9 at 18:29






  • 5




    To elaborate on @luqui's comment: in general, if you add a type signature to each of your functions, the error messages will be closer to the code that needs to change. Often they will also be more concrete & easier to interpret.
    – bergey
    Nov 9 at 18:34










  • @luqui I am currently learning how to do it by myself but I am not there yet. I know that I can check signatures using :t command but that doesn't seem to help much. divide has [a] -> [[a]] and combine has [[b]] -> b which looks alright to me.
    – SuperM4n
    Nov 9 at 18:36






  • 1




    "I am not there yet." Just write divide :: [a] -> [[a]] above your divide definition, and combine :: [[b]] -> b above your combine definition (and similar signatures above other definitions). That tells the type checker what types you expect those functions to be, so it can figure out whether the type error is in the caller or the callee. It helps a lot with error clarity.
    – DarthFennec
    Nov 9 at 19:00










  • @DarthFennec I'm sorry, I misunderstood above comments. I wrote "I am not there yet" because I thought that guys above were talking about the derivation of types (I don't know if this is the correct term in English.). I did as you asked and now indeed I have a more detailed error message. I'll try to improve my code, thank you.
    – SuperM4n
    Nov 9 at 19:11












  • 6




    Write type signatures! They will show you everything.
    – luqui
    Nov 9 at 18:29






  • 5




    To elaborate on @luqui's comment: in general, if you add a type signature to each of your functions, the error messages will be closer to the code that needs to change. Often they will also be more concrete & easier to interpret.
    – bergey
    Nov 9 at 18:34










  • @luqui I am currently learning how to do it by myself but I am not there yet. I know that I can check signatures using :t command but that doesn't seem to help much. divide has [a] -> [[a]] and combine has [[b]] -> b which looks alright to me.
    – SuperM4n
    Nov 9 at 18:36






  • 1




    "I am not there yet." Just write divide :: [a] -> [[a]] above your divide definition, and combine :: [[b]] -> b above your combine definition (and similar signatures above other definitions). That tells the type checker what types you expect those functions to be, so it can figure out whether the type error is in the caller or the callee. It helps a lot with error clarity.
    – DarthFennec
    Nov 9 at 19:00










  • @DarthFennec I'm sorry, I misunderstood above comments. I wrote "I am not there yet" because I thought that guys above were talking about the derivation of types (I don't know if this is the correct term in English.). I did as you asked and now indeed I have a more detailed error message. I'll try to improve my code, thank you.
    – SuperM4n
    Nov 9 at 19:11







6




6




Write type signatures! They will show you everything.
– luqui
Nov 9 at 18:29




Write type signatures! They will show you everything.
– luqui
Nov 9 at 18:29




5




5




To elaborate on @luqui's comment: in general, if you add a type signature to each of your functions, the error messages will be closer to the code that needs to change. Often they will also be more concrete & easier to interpret.
– bergey
Nov 9 at 18:34




To elaborate on @luqui's comment: in general, if you add a type signature to each of your functions, the error messages will be closer to the code that needs to change. Often they will also be more concrete & easier to interpret.
– bergey
Nov 9 at 18:34












@luqui I am currently learning how to do it by myself but I am not there yet. I know that I can check signatures using :t command but that doesn't seem to help much. divide has [a] -> [[a]] and combine has [[b]] -> b which looks alright to me.
– SuperM4n
Nov 9 at 18:36




@luqui I am currently learning how to do it by myself but I am not there yet. I know that I can check signatures using :t command but that doesn't seem to help much. divide has [a] -> [[a]] and combine has [[b]] -> b which looks alright to me.
– SuperM4n
Nov 9 at 18:36




1




1




"I am not there yet." Just write divide :: [a] -> [[a]] above your divide definition, and combine :: [[b]] -> b above your combine definition (and similar signatures above other definitions). That tells the type checker what types you expect those functions to be, so it can figure out whether the type error is in the caller or the callee. It helps a lot with error clarity.
– DarthFennec
Nov 9 at 19:00




"I am not there yet." Just write divide :: [a] -> [[a]] above your divide definition, and combine :: [[b]] -> b above your combine definition (and similar signatures above other definitions). That tells the type checker what types you expect those functions to be, so it can figure out whether the type error is in the caller or the callee. It helps a lot with error clarity.
– DarthFennec
Nov 9 at 19:00












@DarthFennec I'm sorry, I misunderstood above comments. I wrote "I am not there yet" because I thought that guys above were talking about the derivation of types (I don't know if this is the correct term in English.). I did as you asked and now indeed I have a more detailed error message. I'll try to improve my code, thank you.
– SuperM4n
Nov 9 at 19:11




@DarthFennec I'm sorry, I misunderstood above comments. I wrote "I am not there yet" because I thought that guys above were talking about the derivation of types (I don't know if this is the correct term in English.). I did as you asked and now indeed I have a more detailed error message. I'll try to improve my code, thank you.
– SuperM4n
Nov 9 at 19:11

















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