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6.2. ACCELERATION-LEVEL CONSTRAINTS 61
Iteration
The algorithm loops over i, at iteration i it computes λi and ai while maintaining the complementarity
conditions on λj and aj for j < i. Indices j > i are ignored at the i-th iteration. Let
us look at the i-th iteration more closely.
When the i-th iteration is started, λi = 0 and ai = Ai,C∪NC · λC∪NC + bi. If (λi, ai) already
lies on a line of the complementarity diagram due to constraint i then (6.18) already holds for
i, index i is assigned to the corresponding index set and the algorithm can proceed to the next
iteration. Let us assume that (λi, ai) is not on a line.
Our goal is to “drive” i to NH, C or NL. We will do so in multiple steps. At each step we
compute a vector ∆ λ, the adjustments to λ, that will make (λi, ai) “move towards” a line in the
diagram. This will be looped until (λi, ai) ends up on a line. Index i will be then assigned to the
corresponding index set and the iteration will finish.
To ensure that conditions for constraints j < i are not violated while taking a step during
iteration i, we will require that clamped indices remain clamped and non-clamped indices remain
non-clamped or a “singular point” is reached. If (λj, aj) reaches a “singular point”, j is transferred
to the neighboring index set. We will now derive what ∆ λ should look like.
• Since we are not interested in constraints j > i at the i-th iteration, ∆ λj>i = 0.
• If ai > 0 we would like to decrease λi so that i would eventually reach NH, hence ∆λi
should be a negative non-zero value. If ai < 0 we would like to increase λi so that i would
eventually reach NL, hence ∆λi should be a positive non-zero value.
However we do not know what the exact value of ∆λi should be. To tackle this problem
we will write λdir = ∆λi = ∓1 (a λ “direction”) with a note that we will actually add
∆ λ · scale to λ and scale > 0 (a scaling factor) will be determined later.
• If ∆ λ · scale was added to λ, a = A · λ + b would change by ∆a · scale = A · ∆ λ · scale.
From the requirement that clamped indices remain clamped or reach “singular points” and
are transferred to NH or NL we get a condition ∆aC = 0 and from the requirement that
non-clamped indices remain non-clamped or reach “singular points” and are transferred
to C we get ∆ λNC = 0. Putting these two requirements together we get that ∆ λC =
−A −1
CC · AC,i · ∆λi.
This is easy to show. Since we work with matrix slices we can assume (without loss of
generality) that C = {1, . . . , k} and NC = {k + 1, . . . , i − 1}. Constraints j > i are ignored
at this iteration and to simplify writing we can assume that this is the last iteration. Then,
utilizing the requirement ∆λNC = 0, ∆λi = λdir and letting ∆λC = x,
⎛
A = ⎝
ACC AC,NC AC,i
⎞
⎠ ∆ ⎛
λ = ⎝
x
0
⎞
⎠
ANC,C ANC,NC ANC,i
Ai,C Ai,NC Aii
⎛
λdir
⎞
∆a = A · ∆ ACC · x + AC,i · λdir
λ = ⎝ ANC,C · x + ANC,i · λdir
⎠
Ai,C · x + Aii · λdir
and since it is required that ∆aC = 0 we get that x = ∆λC = −A −1
CC · AC,i · λdir.
We thus have
∆ λC = −A −1
CC · AC,i · λdir, ∆ λNC = 0, ∆ λi = λdir, ∆ λj>i = 0 (6.19)