Haskell Hamming Sequence Logic
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It was suggested I place this code here. I had posted it on StackOverflow for kind of review.
I found in Excel that a candidate number when evenly divided by one of [2,3,5] in can be found in previously generated hamming numbers. The problem is, I thought this would be a little faster. It seems terribly slow. It is, first-of-all correct.
The function takes two parameters, a seed list of Hamming numbers less than 10 and a candidate list of any size. I prefer to us a list of [2,3,5] multiples I call base
base = scanl (b a -> a+b) 2 $ cycle [1,1,1,1,2,1,1,2,2,1,1,2,2,1,1,2,1,1,1,1,2,2]
hamx ls (x:xs)
| even x && elem (div x 2) ls = hamx (x:ls) xs
| mod x 3 == 0 && elem (div x 3) ls = hamx (x:ls) xs
| mod x 5 == 0 && elem (div x 5) ls = hamx (x:ls) xs
| null xs = ls
| otherwise = hamx ls xs
I tried filter and any and others in place of elem. The list generates in reverse. The elem finds Hamming matches faster from the bottom and new Hamming numbers appear also at the bottom. So, the bottom is the top.
haskell
asked 10 mins ago
fp_mora
1012
add a comment |
up vote
0
down vote
favorite
It was suggested I place this code here. I had posted it on StackOverflow for kind of review.
I found in Excel that a candidate number when evenly divided by one of [2,3,5] in can be found in previously generated hamming numbers. The problem is, I thought this would be a little faster. It seems terribly slow. It is, first-of-all correct.
The function takes two parameters, a seed list of Hamming numbers less than 10 and a candidate list of any size. I prefer to us a list of [2,3,5] multiples I call base
base = scanl (b a -> a+b) 2 $ cycle [1,1,1,1,2,1,1,2,2,1,1,2,2,1,1,2,1,1,1,1,2,2]
hamx ls (x:xs)
| even x && elem (div x 2) ls = hamx (x:ls) xs
| mod x 3 == 0 && elem (div x 3) ls = hamx (x:ls) xs
| mod x 5 == 0 && elem (div x 5) ls = hamx (x:ls) xs
| null xs = ls
| otherwise = hamx ls xs
I tried filter and any and others in place of elem. The list generates in reverse. The elem finds Hamming matches faster from the bottom and new Hamming numbers appear also at the bottom. So, the bottom is the top.
haskell
asked 10 mins ago
fp_mora
1012
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
It was suggested I place this code here. I had posted it on StackOverflow for kind of review.
I found in Excel that a candidate number when evenly divided by one of [2,3,5] in can be found in previously generated hamming numbers. The problem is, I thought this would be a little faster. It seems terribly slow. It is, first-of-all correct.
The function takes two parameters, a seed list of Hamming numbers less than 10 and a candidate list of any size. I prefer to us a list of [2,3,5] multiples I call base
base = scanl (b a -> a+b) 2 $ cycle [1,1,1,1,2,1,1,2,2,1,1,2,2,1,1,2,1,1,1,1,2,2]
hamx ls (x:xs)
| even x && elem (div x 2) ls = hamx (x:ls) xs
| mod x 3 == 0 && elem (div x 3) ls = hamx (x:ls) xs
| mod x 5 == 0 && elem (div x 5) ls = hamx (x:ls) xs
| null xs = ls
| otherwise = hamx ls xs
I tried filter and any and others in place of elem. The list generates in reverse. The elem finds Hamming matches faster from the bottom and new Hamming numbers appear also at the bottom. So, the bottom is the top.
haskell
asked 10 mins ago
fp_mora
1012
It was suggested I place this code here. I had posted it on StackOverflow for kind of review.
I found in Excel that a candidate number when evenly divided by one of [2,3,5] in can be found in previously generated hamming numbers. The problem is, I thought this would be a little faster. It seems terribly slow. It is, first-of-all correct.
The function takes two parameters, a seed list of Hamming numbers less than 10 and a candidate list of any size. I prefer to us a list of [2,3,5] multiples I call base
base = scanl (b a -> a+b) 2 $ cycle [1,1,1,1,2,1,1,2,2,1,1,2,2,1,1,2,1,1,1,1,2,2]
hamx ls (x:xs)
| even x && elem (div x 2) ls = hamx (x:ls) xs
| mod x 3 == 0 && elem (div x 3) ls = hamx (x:ls) xs
| mod x 5 == 0 && elem (div x 5) ls = hamx (x:ls) xs
| null xs = ls
| otherwise = hamx ls xs
I tried filter and any and others in place of elem. The list generates in reverse. The elem finds Hamming matches faster from the bottom and new Hamming numbers appear also at the bottom. So, the bottom is the top.
haskell
haskell
asked 10 mins ago
fp_mora
1012
asked 10 mins ago
fp_mora
1012
asked 10 mins ago
fp_mora
1012
asked 10 mins ago
fp_mora
1012
asked 10 mins ago
fp_mora
1012
1012
add a comment |
add a comment |
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