feat: reveal multiple rangesets with one rangeset.

[yuroitaki]

Addresses #655

[codecov]

Codecov Report

Attention: Patch coverage is 86.99187% with 32 lines in your changes missing coverage. Please review.

Project coverage is 56.25%. Comparing base (65299d7) to head (47ff4ce).

Files with missing lines	Patch %	Lines
crates/core/src/transcript/proof.rs	87.15%	28 Missing ⚠️
crates/core/src/transcript.rs	50.00%	3 Missing ⚠️
crates/core/src/index.rs	94.73%	1 Missing ⚠️

Additional details and impacted files

@@            Coverage Diff             @@
##              dev     #664      +/-   ##
==========================================
+ Coverage   55.76%   56.25%   +0.48%     
==========================================
  Files         189      189              
  Lines       20886    20869      -17     
==========================================
+ Hits        11648    11740      +92     
+ Misses       9238     9129     -109     

☔ View full report in Codecov by Sentry.
📢 Have feedback on the report? Share it here.

[sinui0]

This doesn't address the issue, we're not looking for a method to pass in a slice. Instead, we want to address the problem that the user has to specify each range individually which corresponds to a commitment. When the user provides a range set to reveal, we should look into the encoding tree/hash commitments and determine if we can satisfy it with one or more of the available commitments.

[yuroitaki]

This doesn't address the issue, we're not looking for a method to pass in a slice. Instead, we want to address the problem that the user has to specify each range individually which corresponds to a commitment. When the user provides a range set to reveal, we should look into the encoding tree/hash commitments and determine if we can satisfy it with one or more of the available commitments.

Yeap! Got the same feedback from @themighty1 , and am working on it

[yuroitaki]

Just realised there is one more polishing to do, will notify when it's ready to be reviewed

[themighty1]

Very nicely done!!!

[themighty1]

I would avoid putting here the rest of the comment starting from NOTE:
The way that the reveal API works now is the actual expected behaviour.
Talking about "unnecessary computation time" will cause the user to wonder about such small details. But we know that compute overhead is so miniscule, it's not worth drawing attention to.

Additionally, we encourage to use reveal() only once but then immediately a few lines below call it multiple times in this example. This will make the user think the example is wrong/the comment is wrong/the user missed something.

I feel we can allow the user to be able to both call reveal one time or multiple times, depending on their code ergonomics needs.
I feel that the user should be exposed to as few inconsequential details as possible. But feel free to keep the NOTE: if you still think it is valuable.

[yuroitaki]

I would avoid putting here the rest of the comment starting from NOTE: The way that the reveal API works now is the actual expected behaviour. Talking about "unnecessary computation time" will cause the user to wonder about such small details. But we know that compute overhead is so miniscule, it's not worth drawing attention to.

Additionally, we encourage to use reveal() only once but then immediately a few lines below call it multiple times in this example. This will make the user think the example is wrong/the comment is wrong/the user missed something.

I feel we can allow the user to be able to both call reveal one time or multiple times, depending on their code ergonomics needs. I feel that the user should be exposed to as few inconsequential details as possible. But feel free to keep the NOTE: if you still think it is valuable.

Ah i see, i guess the point i was trying to convey wasn't clear: that for superset ranges, they should call reveal once after union-ing all of them; but for exact ranges, they can call multiple times. I was mainly worrying about the compute overhead when the following is executed N times (the number of times user call reveal with superset ranges):

 for committed_dir_idx in encoding_tree
      .transcript_indices()
          .into_iter()
          .filter(|(dir, _)| *dir == dir_idx.0)
      {
          if committed_dir_idx.1.is_subset(&dir_idx.1) {
              staged_subsets.push(committed_dir_idx);
          }
      }

Because if N is big, then this block might cause some noticeable overhead, since is_subset itself contains an inner loop. But rationally users wouldn't reveal such large N — let me try to change the wording to make it clearer and less confusing.

[yuroitaki]

@themighty1 just updated the example and new comments, let me know what you think :)