Compe buffer source interferes with Ultisnips tabstops

Question

Compe buffer source interferes with Ultisnips tabstops

physicophilic opened this issue 3 years ago · 6 comments

Checkhealth

health#compe#check
========================================================================
## compe:snippet
  - OK: snippet engine detected.

## compe:mapping
  - INFO: `compe#complete` is not mapped
  - INFO: `compe#confirm` is not mapped
  - INFO: `compe#close` is not mapped
  - INFO: `compe#scroll` is not mapped

Describe the bug

I like using buffer source in markdown. But after writing a lot, my snippet tabstops start to fail. I can't tab through them. See the example below.

Add the snippets to /.config/nvim/UltiSnips/markdown.snippets

snippet td "to the ... power" iA
^{$1}$0
endsnippet

and

snippet mk "Inline math" wA
$$1$$0
endsnippet

To Reproduce

Steps to reproduce the behavior with the minimal config:

Take this minimal config

if has('vim_starting')
  set encoding=utf-8
endif
scriptencoding utf-8

if &compatible
  set nocompatible
endif
let s:plug_dir = expand('/tmp/plugged/vim-plug')
if !isdirectory(s:plug_dir)
  execute printf('!curl -fLo %s/autoload/plug.vim --create-dirs https://raw.githubusercontent.com/junegunn/vim-plug/master/plug.vim', s:plug_dir)
end

execute 'set runtimepath+=' . s:plug_dir
call plug#begin(s:plug_dir)
Plug 'SirVer/ultisnips'
Plug 'hrsh7th/nvim-compe'
call plug#end()
PlugInstall | quit

" Options or settings here.

set completeopt=menuone,noinsert,noselect " better completion
let g:compe = {}
let g:compe.enabled = v:true
let g:compe.autocomplete = v:true
let g:compe.debug = v:false
let g:compe.min_length = 1
let g:compe.preselect = 'enable'
let g:compe.throttle_time = 80
let g:compe.source_timeout = 200
let g:compe.resolve_timeout = 800
let g:compe.incomplete_delay = 400
let g:compe.max_abbr_width = 100
let g:compe.max_kind_width = 100
let g:compe.max_menu_width = 100
let g:compe.documentation = v:true

let g:compe.source = {}
let g:compe.source.path = v:true
let g:compe.source.buffer = v:true
let g:compe.source.calc = v:true
let g:compe.source.nvim_lsp = v:true
let g:compe.source.nvim_lua = v:true
let g:compe.source.vsnip = v:true
let g:compe.source.ultisnips = v:true
let g:compe.source.luasnip = v:true
let g:compe.source.emoji = v:true

let g:UltiSnipsExpandTrigger = '<Tab>'
let g:UltiSnipsJumpForwardTrigger = '<Tab>'
let g:UltiSnipsJumpBackwardTrigger = '<S-Tab>'
let g:UltiSnipsSnippetsDir = $HOME.'.config/nvim/UltiSnips'
let g:UltiSnipsSnippetDirectories = [$HOME.'/.config/nvim/UltiSnips']

Open a markdown buffer, and enter this content (just a sample which makes it fail for me). Sorry for adding the whole thing, cause I don't know what really triggers this.


$\circ$ Proposal of observable correspondence - CFT correlation functions $\langle \Phi(y_1) \Phi(y_2) \cdot  \cdot \cdot  \Phi(y_n)\rangle$ arise from
  dependence of supergravity action $S_G(y, r)$ on asymptotic boundary conditions, where $y$ is used for coordinates on boundary. 


 
  That makes sense, $S_G(y, r\to  \infty)$ ought to do the job.
 

 


 - Masses of particles in supergravity give (scaling) dimension of CFT operators
   as a special case (it appears).


### Introduction

Understanding gauge theory with group $SU(N)$ for large $N$ has importance in
solving confinement problem, which has led Witten to this perhaps, because
Maldacena's proposal links conformally invariant 'cousins' of $SU(N)$ (for large $N$) with super gravity - although I don't see who's a cousin of who. 

$\mathcal{N} = 4$ super Yang Mills in 4D has $SU(N)$ gauge group and is also a CFT. I don't know what's $\mathcal{N}$. But this theory is important example, cited everywhere. Let's do a map between this and IIB superstring theory to which Maldacena declares it is dual. 

| super YM  |   superstring |
|:-----------------:|:-----:|
| $g_{YM}$ | $g_{st} \sim g_{YM}^2 \cdot \phi \cdot R_s$ |
| $SU(N)$| ?|
| $M^{4}$| $AdS_5 \times S^{5}$ |

The spheres in IIB have radius $R_s = (g_{YM}^2 N)^{1/4}$ and $\phi$ is some
kind of flux on $S^{5}$. The bulk theory is supergravity if $N\to \infty$ and
$g^2_{YM} N \to \infty$ as well. First is related to degrees of freedom of QFT ($c_{eff} = N^2 -1$) and $g^2_{YM} N$ is a modified coupling constant $\lambda$.  If this is true, and $g_{st}$ is small in this case, $\phi$ better be non-trivial.


#### AdS space 

* There's a "boundary" at spatial infinity. This makes quantization difficult. What does it mean? Hawking's book's reference is given.  See later.
* Boundary of $AdS_{d+1}$ is $M_{d}$: minkowski space - almost!
* $SO(2,d)$ acts on AdS bulk as 'ordinary symmetry' group (isometries) but same
  group acts on $M_d$ as conformal group. On $M_d$, one sees 'singleton
  representations', basically seen as free field theories - what we did. Much references.


The way he puts it is interesting. 



Other things in intro:

- Minkowski space as boundary of AdS - exercise. 
- AdS description using coordinates similar to boundary coordinates. General
  behaviour of AdS a bit more - exercise.

- Holography in general. Gives examples of other kinds. AdS-CFT has 'covariance'
  - bulk theory is relativistic - under $SO(2,d)$. Others may or may not have this, and the way covariance is achieved may be different.

### Boundary behaviour  

* Euclidean $AdS_{d+1}$

What does he mean? Surely not Euclidean space $\mathbb{R}^{d+1}$,  as
that's Euclidean "Minkowski space", not Euclidean "AdS". So we try being
smarter. AdS space comes from a hypersurface of Minkowski space in 1 higher
dimension, so no surprise that Euclidean does similar. However, he makes this
claim:

- Boundary of Euclidean AdS = Minkowski = boundary of true AdS

Let $\sum_{i=0}^{d} y_{i}^2<1$. Open unit ball in $d+1$ dimensional Euclidean
space $(y_{0}, .., y_{d})$,  written $B_{d+1}. $Then? 

$$ 
ds^2 = \frac{4}{(1 - |y|^2)^2} \sum_{i=0}^{d} d y_{i}^2
$$
is the metric of Euclidean AdS. Here $|y|^2 = y_{0}^2 + y_{1}^2 + \cdot \cdot +
y_{d}^2$. So the metric blows up where ball ends. But hey. He says $B_{d+1}$ is
$AdS_{d+1}$ itself! So metric isn't singular. I think I have seen this form
elsewhere. It's the Poincare patch I think. 

Compactification. 

* Boundary of $B_{d+1}$ is $\sum_i y_{i}^2 = 1$, which is $\mathbb{S}^{d}$. This
  can be included in space, but the metric will break down. To do this right,
  pick $f$ defined on $\overline{B}_{d+1}$, which vanishes (with first order zero) on $\mathbb{S}^{d}$,
  like $f = 1 - |y|^2$,  then set 
$$ 
d \widetilde{s}^2 = f^2 ds ^2 
$$
That modifies metric inside, but that's okay for now.. for some reason. 
Let $f \to fe^{w}$ for some real function $w$ on $\overline{B}_{d+1}$. This would ensure $d \widetilde{s}^2$ is line element for whole $AdS$.  Then, 

$$ 
d \widetilde{s}^2 = f^2 e^{2 w} ds ^2 
$$
which on the sphere would kill $f^2$, and the boundary metric would be left
conformally invariant because of $e^{2w}$ factor.  


He gives 2 other representations. One is akin to global coordinates and another
to Poincare patch again. This is kind of weird business. The global coordinates
define $r = \tanh (y/2)$.  Is this really AdS space? I should look at this.

At the end of the buffer, try doing this. First mk ---> $|$ then $atd$ --> $a^{|}$ ---> $a^{d |}$ . Here | is the cursor. Hitting tab twice will reveal the error. It happens 90% of the times

Actual vs expected behaviour

1 tab should give $a^{d}|$ which it does. 2 tabs should give $a^{d}$|, which it does not 90% of times.

Try deleting all text from buffer and try it again; then it works 100% times.

Additional context (optional)

I think some time variable can be adjusted to prevent this behaviour?

And thank you for the plugin!

Edit: I forgot to add that

On removing buffer source from minimal.vim problem vanishes.
it happens for other kinds of snippets too, after a lot of text has been added. Is it possible that buffer source only tries to complete 'words'? After all I don't want buffer to complete anything with {,}, (, etc..

Answer 1 · 2021-07-21T17:16:24.000Z

Seems like #167 describes the solution? I tried adding default_pattern=[[\w*]], which seems to fix the issue. That restricts the completion options from buffer to only alpha-numerics. Is this the optimal way?

Answer 2 · 2021-07-23T04:52:09.000Z

Hm.... Sorry. this issue is a bit difficult with my English skills...

I can't think nvim-compe breaks ultisnips tab expansion behavior... it's very weird.
(I can believe if you claims nvim-compe doesn't show ultinsips completion)

Answer 3 · 2021-07-23T08:38:26.000Z

It’s okay! No problem! I know some Japanese so let me try that.

Tashkani ultisnips snippets o complete koto ga dekimasu. Mondai sore ja nai. Buffer source o tsukatte toki ultisnips no tab ga chotte dake kowarimasu. Sore dake desu.

Tatoeba snippet ga buffer no words to onaji nattara, snippet no execution jikan ni nvim-compe no pop up ga ikimasu. Sono ato tab wa kikanai.

Sono tame ni buffer ‘default_pattern’ o [[\w\+]] set kuretanda, soshite ‘tab’ wa ima ok desu. #167 to onaji desu.

chotto shitsumon wa kore deshita: kono pattern wa best desu ka? Anata no default pattern wa osoraku ‘{,},^..}’ o recognise kuretanda. Kore wa ‘[A-Za-z0-9]’ dake o recognise ga dekimasu. Sore ni, buffer no kotoba to betsu nani mono o complete koto no hitsuyo wa arimasu ka?

Nihongo de ayamachi o yurushtekudasai demo kono ato mo romaji de hanasu koto no hitsuyo nara oshiete kudasai.

Arigatou! Hopefully I have conveyed it to you!

Answer 4 · 2021-07-23T17:56:38.000Z

@physicophilic I'm very surprised by your Japanese!!! Your Japanese is very well and thank you for your effort.
But someone might see this issue for finding information so I think English is better.

The nvim-compe's default pattern is the following.

[['\%(-\?\d\+\%(\.\d\+\)\?\|\h\w*\%(-\w*\)*\)']]

I think it recognize word, saneke_case, kebab-case, 0.1, -0.1, 1 and -1.

The word's pattern is \h\w*\%(-\w*\)*.

I think [\h\w*\%(-\w*\)* is very similar to \w\+ .... it's weird...

Answer 5 · 2021-07-27T12:46:34.000Z

Thank you @hrsh7th !
I learnt most of it from anime. Once took an N5 course at college to watch more anime without subtitles 😄

I think you're right. I have been trying to work but the problem is still there. The reason is something else. I don't know what I should do.

Answer 6 · 2021-09-25T14:26:06.000Z

This issue has been automatically marked as stale because it has not had recent activity. It will be closed if no further activity occurs. Thank you for your contributions.