Re: [Haskell-cafe] complex multiplication function in Haskell

On 27/04/2017, at 12:02 AM, Mahmoud Murad wrote:

Hi there, The number 354162 has 6 digits and contains all digits from 1 to 6, and it can be considered as a multiplication of 3 * 54 = 162. so what I need is a function that will take all the real number and return the number that can be split as above, 354162 .. 3 * 54 = 162 I have no idea how to do such a thing! any help!

Start by spelling out just what the conditions are. - Are you looking specifically for 6-digit numbers or could it be more? (Because you say "ALL the real number[s]". - Are you sure you want REAL numbers and not INTEGERS? One way to read what you have written is - find a permutation [a,b,c,d,e,f] of [1,2,3,4,5,6] - such that a*(b*10+c) == (d*10+e)*10+f. There are only 6! = 720 permutations of six digits, so not a lot to check. For a computer program, I'd just code that. We can restructure this as - find three different digits 1 <= a,b,c <= 6 - compute def = a*(b*10+c) - check that the digits of def are all 1..6 and all different and different from a,b,c. There are 120 cases to check here. As it happens, Data.List provides permutations Prelude> import Data.List Prelude Data.List> permutations [1..3] [[1,2,3],[2,1,3],[3,2,1],[2,3,1],[3,1,2],[1,3,2]] Prelude Data.List> length (permutations [1..6]) 720 You can combine this with a list comprehension to get the numbers you want.