join =: 4 : 0
neq =. (- |. x) i.&0@:=&((x <.&# y)&{.) y NB. len of max prefix of y
that matches suffix of x
((-neq)}.x) , neq}.y
)
ddup =. (({. join&$: }.)~ <.@-:@#)^:(1<#)NB. split & recur
ddup 2 1 1 _1 2 _2 _1
2
Henry Rich
On 12/6/2019 6:26 PM, Louis de Forcrand wrote:
Not particularly J-ish, but (array-shuffling aside) linear solution:
s=: ,`(1}.])@.(= -@{.)/
s 1 _1 2 _2{~100?.@#4
2 1 2 1 _2 _1 2 1 1 2 2 2 _1 2 _1 2 1 2 2 2 1 1 _2 _1 2 1 _2 _1 _2 _2 _2 _2 _1
_2 _1 _1 _2 _2 1 1 1 2 1 _2 1 _2 _1 _1 _1 2
Since the reduced form of the input list is unique, we are free to perform
reductions in any order we please; in particular, we can start simplifying from
the back, which is what s does.
Might do strange stuff on an empty input list.
Cheers,
Louis
----Original Message----
From : [email protected]
Date : 06/12/2019 - 23:51 (CEST)
To : [email protected]
Subject : Re: [Jprogramming] Removing annihilating pairs
If the answer to Jimmy's question is no, then the uniqueness of the resulting
array has to do (surprisingly) with free groups
(https://en.wikipedia.org/wiki/Free_group).
Indeed we can view a vector of numbers as a word over the alphabet [0,∞) of
positive real numbers (where negative numbers / zero are inverses of letters in
our alphabet). The fact that all maximal reductions (reductions which contain
no adjacent inverses) of a given word are equal means exactly that our reduced
list is unique.
For a proof see:
https://math.stackexchange.com/a/2425147
I'll look into this problem, it's interesting!
Cheers,
Louis
----Original Message----
From : [email protected]
Date : 06/12/2019 - 23:36 (CEST)
To : [email protected]
Subject : Re: [Jprogramming] Removing annihilating pairs
Hi,
is the expected output of transforming 1 3 _3 3 5 to be 1 3 5 or 1 5 ?
On Fri, Dec 6, 2019 at 7:15 AM R.E. Boss <[email protected]> wrote:
Given an array with zero or more annihilating pairs, i.e., two subsequent
numbers which add up to zero, the question is to clean up the array by
deleting all annihilating pairs such that no such pairs are left.
I do have a solution that is both elegant and efficient (I believe), but I
am curious about other thoughts.
foo 2 1 1 _1 2 _2 _1
2
foo 1 _1 2 _2{~100?.@#4 NB. Notice (?.)
2 1 2 1 _2 _1 2 1 1 2 2 2 _1 2 _1 2 1 2 2 2 1 1 _2 _1 2 1 _2 _1 _2 _2 _2
_2 _1 _2 _1 _1 _2 _2 1 1 1 2 1 _2 1 _2 _1 _1 _1 2
R.E. Boss
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