partitioning a list

[Checco7]

how can I partition a list into a list?

example sublists who have sum 4:

list [1,3,4,2,2] produces [[1,3], [4], [2,2]]

I tried a lot but I did not succeed.

Thanks if you help me

I would like to take an L1 list, a list of L2 lists. true if:

L1 is a list of number

L2 is a list of list where the sum is a certain number, for example 20, from 0 to i, and the list i + 1 is the remainder.

example:

partition20 ([12, 11, 7,3,18,1,2], [[12,7,1], [18,2], [11,3]])

my test:

 sum([],0).

sum([X],X).

sum([X,Y|Xs],S):-

   sum([Y|Xs],S1),

    S is X + S1.

rest([], [], []).

rest([A|Ra], [A|Rb], Z) :-

   rest(Ra, Rb, Z).

rest([A|Ra], Rb, [A|Z]) :-

   rest(Ra, Rb, Z).

   

partition([], []).

partition([A|Ra], [[A|Rb]|Rest]) :-

   rest(Ra, Rb, Z),

   partition(Z, Rest),

   sum([A|Rb],20).

[Barb Knox]

On 27/06/2018, at 11:05, Checco7 <pefs...@gmail.com> wrote:

how can I partition a list into a list?

example sublists who have sum 4:

list [1,3,4,2,2] produces [[1,3], [4], [2,2]]

I tried a lot but I did not succeed.

If you have tried a lot, then please show us a couple of your attempts.  This will provide a basis for giving you useful help.

Thanks if you help me

And the Evil Eye if we don’t?

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|  BBB                b    \    Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |   ,008015L080180,022036,029037
|  B  B  a  a   r     b  b  |   ,047045,L014114L4.
|  BBB    aa a  r     bbb   |
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[Volker Wysk]

Am Dienstag, 26. Juni 2018, 16:05:11 CEST schrieb Checco7:
> how can I partition a list into a list?
> example sublists who have sum 4:
> list [1,3,4,2,2] produces [[1,3], [4], [2,2]]

This doesn't explain, what you have in mind...

Volker

[Checco7]

some help in the right solution?

[Kuniaki Mukai]

Hi, 

Did you find a solution for that lovely exercise ?  I hope you did,

but here is my solution based on select/3 just in case, 

though I am not  sure it is a solution to  your problem.

sublist(0, X, X, Y, Y):-!.

sublist(N, X, Y, Z, U):- N>0, select(A, X, X0),

N0 is N-A,

sublist(N0, X0, Y, [A|Z], U).

?- sublist(3, [1,3,2], R).

%@ R = [[2, 1], [3]] .

sublist(_, [], []).

sublist(N, X, [Z|Y]):-sublist(N, X, X0, [], Z),

sublist(N, X0, Y).

?- sublist(4, [1,3,4,2,2], R).

%@ R = [[3, 1], [4], [2, 2]] .

Hope this help,

Kuniaki Mukai

On Jun 28, 2018, at 3:13, Checco7 <pefs...@gmail.com> wrote:

I would like to take an L1 list, a list of L2 lists. true if:

L1 is a list of number

L2 is a list of list where the sum is a certain number, for example 20, from 0 to i, and the list i + 1 is the remainder.

example:

partition20 ([12, 11, 7,3,18,1,2], [[12,7,1], [18,2], [11,3]])

my test:

 sum([],0).

sum([X],X).

sum([X,Y|Xs],S):-

   sum([Y|Xs],S1),

    S is X + S1.

resto([], [], []).

rest([A|Ra], [A|Rb], Z) :-

   rest(Ra, Rb, Z).

rest([A|Ra], Rb, [A|Z]) :-

   rest(Ra, Rb, Z).

   

partition([], []).

partition([A|Ra], [[A|Rb]|Rest]) :-

   rest(Ra, Rb, Z),

   partition(Z, Rest),

   sum([A|Rb],20).

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[Kuniaki Mukai]

Hi,

I have changed my codes in the previous post a little bit in order to

keep the ordering for each sublist with that of the original one

by using difference lists, which are ubiquitous in Prolog

programming.

sublist(0, X, X, Y, Y):-!.

sublist(N, X, Y, [A|Z], U):- N>0, select(A, X, X0),

N0 is N-A,

sublist(N0, X0, Y, Z, U).

%

partition_by_weight(_, [], []).

partition_by_weight(N, X, [Z|Y]):-sublist(N, X, X0, Z, []),

partition_by_weight(N, X0, Y).

% ?- partition_by_weight(4, [1,3,4,2,2], R).

%@ R = [[1, 3], [4], [2, 2]] .

If I am helping you too much, I am very sorry. Your problem

was interesting for me too. 

Thanks,

Kuniaki Mukai

On Jun 28, 2018, at 3:16, Checco7 <pefs...@gmail.com> wrote:

[Kuniaki Mukai]

Hi,

I noticed that my previous codes searches unnecessary “solutions”.

Here is a revised version in that point, though I only hope so

without confidence. Now I am a person who would like to know

an  "official" answer to the problem if any.

% ?- partition_by_weight(4, [1,3,4,2,2], R).

%@ R = [[1, 3], [4], [2, 2]] ;

%@ false.

% ?- partition_by_weight(5, [1,3,4,2,3,2], R).

%@ R = [[1, 4], [3, 2], [3, 2]] ;

%@ R = [[1, 4], [3, 2], [2, 3]] ;

%@ false.

partition_by_weight(_, [], []).

partition_by_weight(N, X, [Z|Y]):- sublist(N, X, Z, X0),

partition_by_weight(N, X0, Y).

% ?- sublist(4, [1,3,4,2,2], R, X).

sublist(N, [A|X], [A|Z], X0):- N0 is N-A,

sublist_aux(N0, X, X0, Z, []).

%

sublist_aux(0, X, X, Y, Y):-!.

sublist_aux(N, X, Y, [A|Z], U):- N>0,

select(A, X, X0),

N0 is N-A,

sublist_aux(N0, X0, Y, Z, U).

Kuniaki Mukai