The reason I redefined Eq is:
When doing `union` over two list of MyTuple is just base on its first
element.
Basically it means: [(1,2), (2,2)] `union` [(1,0), (2,0), (0,0)]
produce [(1,2), (2,2), (0,0)]
rather than [(1,2),(2,2),(1,0),(2,0),(0,0)] by default.
-Haisheng
On Sun, Feb 5, 2012
*d = [sum $ map (a !!) [i | i - b, j - c, i + j 3, i + j == dIndex] |
dIndex - [0..3]] *
This is cool.
-Simon
On Sun, Feb 5, 2012 at 5:07 PM, L Corbijn aspergesoe...@gmail.com wrote:
On Sun, Feb 5, 2012 at 7:28 AM, Haisheng Wu fre...@gmail.com wrote:
a = [1,1,1,1]
b = [0,1,2,3]
d =
Okay.
But that's misleading, as normally x == y = True *iff* compare x y = EQ,
but there this is not verified, as you don't redefined
A function like unionBy doesn't exist, so IMHO to limit ambiguity, it
would be a good idea to use MyTuple *only *at the specific place where you
need its Eq
How about this:
import Array
a = [1,1,1,1]
b = [0,1,2,3]
c = [0,2]
d = elems $ accumArray (+) 0 (0,3) [(i+j,a!!i) | i-b, j-c, i+j3]
--
Thomas H
On 2012-02-06 12:01 , Haisheng Wu wrote:
*d = [sum $ map (a !!) [i | i - b, j - c, i + j 3, i + j == dIndex] |
dIndex
- [0..3]] *
* Haisheng Wu fre...@gmail.com [2012-02-05 14:28:10+0800]
a = [1,1,1,1]
b = [0,1,2,3]
d = [0,0,0,0]
for i in b:
for j in c:
if (i+j)3:
d[i+j] += a[i]
Do you have any cool solution in FP way?
You can use IntMap as a replacement for arrays:
(I didn't understand your
Haisheng Wu fre...@gmail.com writes:
a = [1,1,1,1]
b = [0,1,2,3]
d = [0,0,0,0]
for i in b:
for j in c:
if (i+j)3:
d[i+j] += a[i]
Do you have any cool solution in FP way?
I find the above sufficiently alien that I can’t work out what
it’s meant to do (what is it actually
Sorry there is a mistake in the problem description.
Here it is in Python:
a = [1,1,1,1] b = [0,1,2,3] c = [0,2] d = [0,0,0,0]
for i in b:
for j in c:
if (i+j)3:
d[i+j] += a[i]
-Haisheng
On Sun, Feb 5, 2012 at 2:28 PM, Haisheng Wu fre...@gmail.com wrote:
a =
Concerning your first solution, I don't understand why you redefine Eq but
not Ord instance. Ord will still work by comparing the tuples and not the
first elements of said tuples.
Plus the good news is you don't have to do this: just use regular tuples
and use sort*By *or group*By *functions from
For instance your Eq instance could have been written
x == y = (==) `on` (fst . getTuple)
Sorry, wrong arity:
(==) = (==) `on` (fst . getTuple)
Okay for the imperative code.
2012/2/5 Yves Parès yves.pa...@gmail.com
Concerning your first solution, I don't understand why you redefine Eq but
+= a[i] is the same as +=1, isn't it?
(i accidentally didn't reply to the list on my first try. sorry.)
Am 05.02.2012 16:36, schrieb Haisheng Wu:
Sorry there is a mistake in the problem description.
Here it is in Python:
a = [1,1,1,1] b = [0,1,2,3] c = [0,2] d = [0,0,0,0]
for i in b:
Can you write it as a Python function? Another way of asking: is the
goal to mutate d or is it to produce the list?
On 2012-02-05 23.36.28 +0800, Haisheng Wu wrote:
Sorry there is a mistake in the problem description.
Here it is in Python:
a = [1,1,1,1] b = [0,1,2,3] c = [0,2] d =
On Sun, Feb 5, 2012 at 2:28 PM, Haisheng Wu fre...@gmail.com wrote:
for i in b:
for j in c:
if (i+j)3:
d[i+j] += a[i]
Do you have any cool solution in FP way?
Not sure whether this is cool, but here it is nonetheless:
a = repeat 1;
b = [0..3];
c = [0,2];
d = map (sum ∘ map ((a
a = [1,1,1,1]
b = [0,1,2,3]
d = [0,0,0,0]
for i in b:
for j in c:
if (i+j)3:
d[i+j] += a[i]
My just work implementation in Haskell
http://hpaste.org/57452
Another people implementation in Haskell with Monad and it turns out
complex and very imperatively.
http://hpaste.org/57358
Do
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