Iwerlipse: Difference between revisions

The '''Iwerlipse''' (pronounced ee-ver-lipse /ˈivɜlɪps/) is the range of positions Mario can occupy while he is in the air and cannot turn, assuming that drag is constant. The boundary of the Iwerlipse is reached by '''Arctan Straining''' - the optimal way of air straining that gives the greatest distance along a chosen angle ''ω'' for a given number of frames ''T''. For a single frame ''T'' = 1, the Iwerlipse is an ellipse with half-width 10 (in the sideways direction) and half-height 1.5 (in the forwards direction). For general ''T'', it resembles an ellipse with half-width 10''T'' and half-height 3''T''(''T''+1)/4, but bulges at the corners, covering more area than an ellipse with the same dimensions. As shown by ''Grassdigger'', the shape is well approximated by a generalised superellipse, and thus ''Pannenkoek2012'' proposed calling this shape the Iwerlipse - a pun on the name of ''Iwer Sonsch'', who first discoverered that Arctan Straining was optimal in April 2018.

Arctan Straining gives the maximum possible distance along a target angle ''ω'' that differs from the facing angle ''θ'', over a finite number of frames ''T''. It is performed by choosing straining angles 𝜙_t (intendedYaw) such that

with ''t'' being the frame number between 1 and ''T''. (Usually only the top row of this equation is given, but since the arctan function only gives results between ±90°, we must add 180° to the result if the target angle is behind Mario.)

* Forward velocity accelerates gradually over time, while sideways speed is reset every frame.

==Update Equations==

When Mario is in the air and cannot turn, the function <code>update_air_without_turn</code> is responsible for updating Mario's velocity variables from one frame, ''t''-1, to the next, ''t''. Without loss of generality, we can choose coordinates so that Mario's facing angle is zero. In other words, he is facing along the z-axis, while his sideways speed is directed along the x-axis. Mario's forward velocity and position coordinates are then updated via

z_t &= z_{t-1} + v_t,\\

x_t &= x_{t-1} + 10\sin(\phi_t),

where 𝜙_t is the straining angle (intendedYaw) on frame ''t''. The total effect of drag is determined by

where ''D''_cap is a soft velocity cap equal to 48 for long jumps and 32 otherwise. In practice, these drag boundaries are far enough apart that the drag is often constant over large periods of time. In this case we can write Mario's position after ''T'' frames as

z_T &= z_0 + v_0 T - \frac{DT(T+1)}{2} + 1.5 \sum_{t=1}^T (T+1-t)\cos(\phi_t).

In deriving the expression for ''z_T'' we have used that the sum of the first ''T'' integers is equal to the triangle number ''T''(''T''+1)/2, and we have grouped together terms of cos(𝜙_t). Notice that, since forward straining produces an acceleration that is not reset between frames, earlier frames have a stronger effect on ''z_T'' than later frames. The effective strength is 1.5(''T''+1-''t''), i.e. 1.5 multiplied by the number of frames remaining (including the current frame), since the forward velocity gained on one frame produces a displacement also on all remaining frames.

==Arctan Straining Proof==

The displacement vector after ''T'' frames is '''R''' = (''x_T'' − ''x₀'') '''x̂''' + (''z_T'' − ''z₀'') '''ẑ'''. Taking the dot product of this with the unit vector along the target angle ''ω'' gives the distance moved along this angle

S &= \left(v_0 T - \frac{DT(T+1)}{2}\right)\cos(\omega)\\

&+ \sum_{t=1}^T \left[10\sin(\phi_t)\sin(\omega) + 1.5(T+1-t)\cos(\phi_t)\cos(\omega)\right],\\

where the first line is a constant displacement due to our initial velocity and drag, and the second line gives the effective contribution from straining on each frame. Thanks to this separation, we can maximize the total distance by individually maximizing each contribution, d''S''/d𝜙_t = 0, yielding

<math> \tan(\phi_t - \theta) = \frac{10\tan(\omega - \theta)}{1.5(T+1-t)},</math>

where we have also rotated the coordinate system back to allow for general facing angles θ. Since tan(''x'') = tan(''x''±180°), this equation has two unique solutions for 𝜙_t which are given above.

== Iwerlipse Equations ==

By writing sin(𝜙_t) and cos(𝜙_t) in terms of tan(𝜙_t), and using the optimal straining relation, we obtain exact expressions for the boundary of the Iwerlipse after ''T'' frames

where ''β'' = 1.5/10tan(''ω''), and ⟨''z''⟩ = ''z''₀ + ''v''₀''T'' - ''DT''(''T''+1)/2. We can restrict 0 < ''ω'' < 90°, with the four combinations of plus and minus signs giving the remaining quadrants. The Iwerlipse's half-width is ''x''_max = 10''T'', while its half-height is ''z''_max = 3''T''(''T''+1)/4.

If ''T'' is large, the exact expressions involve large summation terms, so it may be useful to have approximate expressions that are easier to compute.

By replacing the summations from ''t'' = 1 to ''t'' = ''T'' with an integral from ''t'' = 1/2 to ''t'' = ''T'' + 1/2 (essentially performing the midpoint rule of numerical integration in reverse), we obtain

with the notation [F(t)]_a^b = F(b) − F(a). This approximation converges to the true result very quickly - even for ''T'' = 2, the RMS error (normalized by ''x''_max and ''z''_max) is 4.3×10⁻³, while for ''T > 6'' it is less than 10⁻³. Similar expressions obtained using the Euler–Maclaurin formula converge more slowly, even with first-order correction terms.

<math> \left(\frac{x_T - x_0}{x_\mathrm{max}}\right)^{m(T)} + \left(\frac{z_T - \langle z \rangle}{x_\mathrm{max}}\right)^{n(T)} = 1, </math>

== Quarterstep Penalty ==

If a number of quartersteps, ''q'' ∈ {0,1,2,3}, are lost on the final frame, the forward velocity gained on frame ''t'' no longer accumulates forward distance over ''T''+1-''t'' frames, but instead over ''T'' - ''q''/4 + 1 - ''t'' frames. The optimal straining relation then becomes:

The optimal straining angle for the final frame is not affected, since in this case the ellipse shrinks in both directions rather than just vertically, i.e. a factor (1-''q''/4) cancels in the numerator and denominator.

Iwer states the optimal straining equation for the first time in the following form

This is equivalent to our expression above, since arccot(''x'') = arctan(1/''x''),   

with, ''r'' = 1/tan(''ω''), and, ''t''' = ''T'' + 1 - ''t''.   

(If using this expression, we need to add 180° if the target angle is to the right of Mario, since arccot only gives results between 0° and 180°.)

==See Also==