Resource Estimator - T states per rotation formula

[bdg221]

Describe the bug

The formula used by the Resource Estimator for number of T states to implement an arbitrary rotation does not match the cited paper.

The formula used by the Resource Estimator is:
T ( \epsilon ) = 0.53 log2 ( 1 / \epsilon ) + 5.3

In practice, the number of T states per rotation is:

                (NUM_TS_PER_ROTATION_A_COEFFICIENT
                    * ((self.rotation_count as f64) / eps_synthesis).log2()
                    + NUM_TS_PER_ROTATION_B_COEFFICIENT)
                    .ceil() as _,

(see logical_counts.rs)
where:

/// A coefficient in Ts per rotation
pub const NUM_TS_PER_ROTATION_A_COEFFICIENT: f64 = 0.53;

/// A coefficient in Ts per rotation
pub const NUM_TS_PER_ROTATION_B_COEFFICIENT: f64 = 5.3;

(see constants.rs)

This information is seen by the user in the "Resource estimates breakdown" section of the estimate report. For example, the double-factorized chemistry sample shows:

The number of T states to implement a rotation with an arbitrary angle is ⌈0.53log2(11,988,044/0.0033333333333333335)+5.3⌉ [arXiv:2203.10064]. For simplicity, we use this formula for all single-qubit arbitrary angle rotations, and do not distinguish between best, worst, and average cases.

However, the linked paper arXiv:2203.10064 does not include this formula.

The table on page 8 of the paper lists a "Mixed fallback" approximation protocol for Clifford+T (T-count) as 0.53 log2 (1/ \epsilon ) + 4.86 and the "Mix fallback" approximation protocol for Clifford+sqrt(T) (T count) as 0.56 log2 (1 /\epsilon) + 5.30.

To Reproduce
N/A

Expected behavior

Either the formula should match the paper or an explanation of the formula should be provided.

Screenshots

System information
N/A

Additional context
N/A

[msoeken]

Thanks for catching this @bdg221. We will change 5.3 to 4.86 to match the mixed fallback mean formula for the Clifford+T gate set.