Computation of H values of some distributions of eight particles between eight boxes

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We will start with the distribution 8 0 0 0 0 0 0 0 (eight particles in one box, in the other boxes zero particles).

At equilibrium P_eq[A] = P_eq[B] = P_eq[C] =, etc. is equal to 1/8 .

When we compute the H values belonging to the corresponding distributions, we again follow the formula (where k generally signifies a box) :

that is to say we compute :
P[X,t] (where X is either A, or B, or, C, etc.),
P_eq[X] (which is 1/8),
P[X,t] / P_eq[X],
log ( P[X,t] / P_eq[X] ), and
P[X,t] . log ( P[X,t] / P_eq[X] ). This result is to be obtained for each box. The eight results will then be added together, yielding the H value for time t .
(When a point (.) is used, it means "times" (x), except where it is evidently a decimal point).
The distributions are assumed to be ordered in time, that is to say to each distribution a time t is associated. We do not consider the different values of t .

Distribution 8 0 0 0 0 0 0 0 .

Box A.
P[A,t] = 8/8 = 1 .
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = 1 / (1/8) = 8 .
log ( P[A,t] / P_eq[A] ) = log 8 = 2.08 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = 1 x 2.08 = 2.08 .

Box B.
P[B,t] = 0/8 = 0, so
P[B,t] . { log ( P[B,t] / P_eq[B] ) } = 0 .

The same goes for all the remaining boxes ( P[C,t] = 0/8, P[D,t] = 0/8, etc.).
So the H value for this distribution is
2.08 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 2.08 .

Distribution 7 1 0 0 0 0 0 0 .

Box A.
P[A,t] = 7/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = 7/8 / (1/8) = 7 .
log ( P[A,t] / P_eq[A] ) = log 7 = 1.95 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (7/8)(1.95) = 1.70 .

Box B.
P[B,t] = 1/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (1/8) / (1/8) = 1 .
log ( P[B,t] / P_eq[B] ) = log 1 = 0 .
P[B,t] . { log ( P[B,t] / P_eq[B] ) } = (1/8)(0) = 0 .

Box C.
P[C,t] = 0/8 = 0, so
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = 0 .

The same goes for all the remaining boxes ( P[D,t] = 0/8, P[E,t] = 0/8, etc.).
So the H value for this distribution is
1.70 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 1.70 .

Distribution 6 2 0 0 0 0 0 0 .

Box A.
P[A,t] = 6/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = 6/8 / (1/8) = 6 .
log ( P[A,t] / P_eq[A] ) = log 6 = 1.79 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (6/8)(1.79) = 1.34 .

Box B.
P[B,t] = 2/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (2/8) / (1/8) = 2 .
log ( P[A,t] / P_eq[A] ) = log 2 = 0.69 .
P[B,t] . { log ( P[A,t] / P_eq[A] ) } = (2/8)(0.69) = 0.17 .

Box C.
P[C,t] = 0/8 = 0, so
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = 0 .

The same goes for all the remaining boxes ( P[D,t] = 0/8, P[E,t] = 0/8, etc.).
So the H value for this distribution is
1.34 + 0.17 + 0 + 0 + 0 + 0 + 0 + 0 = 1.51 .

Distribution 4 4 0 0 0 0 0 0 .

Box A.
P[A,t] = 4/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = 4/8 / (1/8) = 4 .
log ( P[A,t] / P_eq[A] ) = log 4 = 1.39 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (4/8)(1.39) = 0.69 .

Box B.
P[B,t] = 4/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (4/8) / (1/8) = 4 .
log ( P[A,t] / P_eq[A] ) = log 4 = 1.39 .
P[B,t] . { log ( P[A,t] / P_eq[A] ) } = (4/8)(1.39) = 0.69 .

Box C.
P[C,t] = 0/8 = 0, so
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = 0 .

The same goes for all the remaining boxes ( P[D,t] = 0/8, P[E,t] = 0/8, etc.).
So the H value for this distribution is
0.69 + 0.69 + 0 + 0 + 0 + 0 + 0 + 0 = 1.39 .

Distribution 6 1 1 0 0 0 0 0 .

Box A.
P[A,t] = 6/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = 6/8 / (1/8) = 6 .
log ( P[A,t] / P_eq[A] ) = log 6 = 1.79 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (6/8)(1.79) = 1.34 .

Box B.
P[B,t] = 1/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (1/8) / (1/8) = 1 .
log ( P[A,t] / P_eq[A] ) = log 1 = 0 .
P[B,t] . { log ( P[A,t] / P_eq[A] ) } = (1/8)(0) = 0 .

Box C.
P[C,t] = 1/8.
P_eq[B] = 1/8 .
P[C,t] / P_eq[C] = (1/8) / (1/8) = 1 .
log ( P[C,t] / P_eq[C] ) = log 1 = 0 .
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = (1/8)(0) = 0 .

Box D.
P[D,t] = 0/8 = 0, so
P[D,t] . { log ( P[D,t] / P_eq[D] ) } = 0 .

The same goes for all the remaining boxes ( P[E,t] = 0/8, P[F,t] = 0/8, etc.).
So the H value for this distribution is
1.34 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 1.34 .

Distribution 5 2 1 0 0 0 0 0 .

Box A.
P[A,t] = 5/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = 5/8 / (1/8) = 5 .
log ( P[A,t] / P_eq[A] ) = log 5 = 1.61 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (5/8)(1.61) = 1.01 .

Box B.
P[B,t] = 2/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (2/8) / (1/8) = 2 .
log ( P[A,t] / P_eq[A] ) = log 2 = 0.69 .
P[B,t] . { log ( P[A,t] / P_eq[A] ) } = (2/8)(0.69) = 0.17 .

Box C.
P[C,t] = 1/8.
P_eq[B] = 1/8 .
P[C,t] / P_eq[C] = (1/8) / (1/8) = 1 .
log ( P[C,t] / P_eq[C] ) = log 1 = 0 .
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = (1/8)(0) = 0 .

Box D.
P[D,t] = 0/8 = 0, so
P[D,t] . { log ( P[D,t] / P_eq[D] ) } = 0 .

The same goes for all the remaining boxes ( P[E,t] = 0/8, P[F,t] = 0/8, etc.).
So the H value for this distribution is
1.01 + 0.17 + 0 + 0 + 0 + 0 + 0 + 0 = 1.18 .

Distribution 4 3 1 0 0 0 0 0 .

Box A.
P[A,t] = 4/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = 4/8 / (1/8) = 4 .
log ( P[A,t] / P_eq[A] ) = log 4 = 1.39 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (4/8)(1.39) = 0.69 .

Box B.
P[B,t] = 3/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (3/8) / (1/8) = 3 .
log ( P[B,t] / P_eq[B] ) = log 3 = 1.10 .
P[B,t] . { log ( P[B,t] / P_eq[B] ) } = (3/8)(1.10) = 0.41 .

Box C.
P[C,t] = 1/8.
P_eq[B] = 1/8 .
P[C,t] / P_eq[C] = (1/8) / (1/8) = 1 .
log ( P[C,t] / P_eq[C] ) = log 1 = 0 .
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = (1/8)(0) = 0 .

Box D.
P[D,t] = 0/8 = 0, so
P[D,t] . { log ( P[D,t] / P_eq[D] ) } = 0 .

The same goes for all the remaining boxes ( P[E,t] = 0/8, P[F,t] = 0/8, etc.).
So the H value for this distribution is
0.69 + 0.41 + 0 + 0 + 0 + 0 + 0 + 0 = 1.10 .

Distribution 5 1 1 1 0 0 0 0 .

Box A.
P[A,t] = 5/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = 5/8 / (1/8) = 5 .
log ( P[A,t] / P_eq[A] ) = log 5 = 1.61 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (5/8)(1.61) = 1.01 .

Box B.
P[B,t] = 1/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (1/8) / (1/8) = 1 .
log ( P[B,t] / P_eq[B] ) = log 1 = 0 .
P[B,t] . { log ( P[B,t] / P_eq[B] ) } = (1/8)(0) = 0 .

Box C.
P[C,t] = 1/8.
P_eq[B] = 1/8 .
P[C,t] / P_eq[C] = (1/8) / (1/8) = 1 .
log ( P[C,t] / P_eq[C] ) = log 1 = 0 .
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = (1/8)(0) = 0 .

Box D.
P[D,t] = 1/8.
P_eq[D] = 1/8 .
P[D,t] / P_eq[D] = (1/8) / (1/8) = 1 .
log ( P[D,t] / P_eq[D] ) = log 1 = 0 .
P[D,t] . { log ( P[D,t] / P_eq[D] ) } = (1/8)(0) = 0 .

Box E.
P[E,t] = 0/8 = 0, so
P[E,t] . { log ( P[E,t] / P_eq[E] ) } = 0 .

The same goes for all the remaining boxes ( P[F,t] = 0/8, P[G,t] = 0/8, etc.).
So the H value for this distribution is
1.01 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 1.01 .

Distribution 3 3 2 0 0 0 0 0 .

Box A.
P[A,t] = 3/8.
P_eq[B] = 1/8 .
P[A,t] / P_eq[A] = (3/8) / (1/8) = 3 .
log ( P[A,t] / P_eq[A] ) = log 3 = 1.10 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (3/8)(1.10) = 0.41 .

Box B.
P[B,t] = 3/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (3/8) / (1/8) = 3 .
log ( P[B,t] / P_eq[B] ) = log 3 = 1.10 .
P[B,t] . { log ( P[B,t] / P_eq[B] ) } = (3/8)(1.10) = 0.41 .

Box C.
P[C,t] = 2/8.
P_eq[C] = 1/8 .
P[C,t] / P_eq[C] = (2/8) / (1/8) = 2 .
log ( P[C,t] / P_eq[C] ) = log 2 = 0.69 .
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = (2/8)(0.69) = 0.17 .

Box D.
P[D,t] = 0/8 = 0, so
P[D,t] . { log ( P[D,t] / P_eq[D] ) } = 0 .

The same goes for all the remaining boxes ( P[E,t] = 0/8, P[F,t] = 0/8, etc.).
So the H value for this distribution is
0.41 + 0.41 + 0.17 + 0 + 0 + 0 + 0 + 0 = 0.99 .

Distribution 4 2 1 1 0 0 0 0 .

Box A.
P[A,t] = 4/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = 4/8 / (1/8) = 4 .
log ( P[A,t] / P_eq[A] ) = log 4 = 1.39 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (4/8)(1.39) = 0.69 .

Box B.
P[B,t] = 2/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (2/8) / (1/8) = 2 .
log ( P[B,t] / P_eq[B] ) = log 2 = 0.69 .
P[B,t] . { log ( P[B,t] / P_eq[B] ) } = (2/8)(0.69) = 0.17 .

Box C.
P[C,t] = 1/8.
P_eq[C] = 1/8 .
P[C,t] / P_eq[C] = (1/8) / (1/8) = 1 .
log ( P[C,t] / P_eq[C] ) = log 1 = 0 .
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = (1/8)(0) = 0 .

Box D.
P[D,t] = 1/8.
P_eq[D] = 1/8 .
P[D,t] / P_eq[D] = (1/8) / (1/8) = 1 .
log ( P[D,t] / P_eq[D] ) = log 1 = 0 .
P[D,t] . { log ( P[D,t] / P_eq[D] ) } = (1/8)(0) = 0 .

Box E.
P[E,t] = 0/8 = 0, so
P[E,t] . { log ( P[E,t] / P_eq[E] ) } = 0 .

The same goes for all the remaining boxes ( P[F,t] = 0/8, P[G,t] = 0/8, etc.).
So the H value for this distribution is
0.69 + 0.17 + 0 + 0 + 0 + 0 + 0 + 0 = 0.86 .

Distribution 3 2 2 1 0 0 0 0 .

Box A.
P[A,t] = 3/8.
P_eq[B] = 1/8 .
P[A,t] / P_eq[A] = (3/8) / (1/8) = 3 .
log ( P[A,t] / P_eq[A] ) = log 3 = 1.10 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (3/8)(1.10) = 0.41 .

Box B.
P[B,t] = 2/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (2/8) / (1/8) = 2 .
log ( P[A,t] / P_eq[A] ) = log 2 = 0.69 .
P[B,t] . { log ( P[A,t] / P_eq[A] ) } = (2/8)(0.69) = 0.17 .

Box C.
P[C,t] = 2/8.
P_eq[C] = 1/8 .
P[C,t] / P_eq[C] = (2/8) / (1/8) = 2 .
log ( P[C,t] / P_eq[C] ) = log 2 = 0.69 .
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = (2/8)(0.69) = 0.17 .

Box D.
P[D,t] = 1/8.
P_eq[D] = 1/8 .
P[D,t] / P_eq[D] = (1/8) / (1/8) = 1 .
log ( P[D,t] / P_eq[D] ) = log 1 = 0 .
P[D,t] . { log ( P[D,t] / P_eq[D] ) } = (1/8)(0) = 0 .

Box E.
P[E,t] = 0/8 = 0, so
P[E,t] . { log ( P[E,t] / P_eq[E] ) } = 0 .

The same goes for all the remaining boxes ( P[F,t] = 0/8, P[G,t] = 0/8, etc.).
So the H value for this distribution is
0.41 + 0.17 + 0.17 + 0 + 0 + 0 + 0 + 0 = 0.75 .

Distribution 4 1 1 1 0 0 0 0 .

Box A.
P[A,t] = 4/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = 4/8 / (1/8) = 4 .
log ( P[A,t] / P_eq[A] ) = log 4 = 1.39 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (4/8)(1.39) = 0.69 .

Box B.
P[B,t] = 1/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (1/8) / (1/8) = 1 .
log ( P[B,t] / P_eq[B] ) = log 1 = 0 .
P[B,t] . { log ( P[B,t] / P_eq[B] ) } = (1/8)(0) = 0 .

Box C.
P[C,t] = 1/8.
P_eq[C] = 1/8 .
P[C,t] / P_eq[C] = (1/8) / (1/8) = 1 .
log ( P[C,t] / P_eq[C] ) = log 1 = 0 .
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = (1/8)(0) = 0 .

Box D.
P[D,t] = 1/8.
P_eq[D] = 1/8 .
P[D,t] / P_eq[D] = (1/8) / (1/8) = 1 .
log ( P[D,t] / P_eq[D] ) = log 1 = 0 .
P[D,t] . { log ( P[D,t] / P_eq[D] ) } = (1/8)(0) = 0 .

Box E.
P[E,t] = 0/8 = 0, so
P[E,t] . { log ( P[E,t] / P_eq[E] ) } = 0 .

The same goes for all the remaining boxes ( P[F,t] = 0/8, P[G,t] = 0/8, etc.).
So the H value for this distribution is
0.69 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 0.69 .

Distribution 2 2 2 2 0 0 0 0 .

Box A.
P[A,t] = 2/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = (2/8) / (1/8) = 2 .
log ( P[A,t] / P_eq[A] ) = log 2 = 0.69 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (2/8)(0.69) = 0.17 .

Box B.
P[B,t] = 2/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (2/8) / (1/8) = 2 .
log ( P[B,t] / P_eq[B] ) = log 2 = 0.69 .
P[B,t] . { log ( P[B,t] / P_eq[B] ) } = (2/8)(0.69) = 0.17 .

Box C.
P[C,t] = 2/8.
P_eq[C] = 1/8 .
P[C,t] / P_eq[C] = (2/8) / (1/8) = 2 .
log ( P[C,t] / P_eq[C] ) = log 2 = 0.69 .
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = (2/8)(0.69) = 0.17 .

Box D.
P[D,t] = 2/8.
P_eq[D] = 1/8 .
P[D,t] / P_eq[D] = (2/8) / (1/8) = 2 .
log ( P[D,t] / P_eq[D] ) = log 2 = 0.69 .
P[D,t] . { log ( P[D,t] / P_eq[D] ) } = (2/8)(0.69) = 0.17 .

Box E.
P[E,t] = 0/8 = 0, so
P[E,t] . { log ( P[E,t] / P_eq[E] ) } = 0 .

The same goes for all the remaining boxes ( P[F,t] = 0/8, P[G,t] = 0/8, etc.).
So the H value for this distribution is
0.17 + 0.17 + 0.17 + 0.17 + 0 + 0 + 0 + 0 = 0.68 .

Reflecting on this latter result, see NOTE 1 .

Distribution 2 2 2 1 1 0 0 0 .

Box A.
P[A,t] = 2/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = (2/8) / (1/8) = 2 .
log ( P[A,t] / P_eq[A] ) = log 2 = 0.69 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (2/8)(0.69) = 0.17 .

Box B.
P[B,t] = 2/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (2/8) / (1/8) = 2 .
log ( P[B,t] / P_eq[B] ) = log 2 = 0.69 .
P[B,t] . { log ( P[B,t] / P_eq[B] ) } = (2/8)(0.69) = 0.17 .

Box C.
P[C,t] = 2/8.
P_eq[C] = 1/8 .
P[C,t] / P_eq[C] = (2/8) / (1/8) = 2 .
log ( P[C,t] / P_eq[C] ) = log 2 = 0.69 .
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = (2/8)(0.69) = 0.17 .

Box D.
P[D,t] = 1/8.
P_eq[D] = 1/8 .
P[D,t] / P_eq[D] = (1/8) / (1/8) = 1 .
log ( P[D,t] / P_eq[D] ) = log 1 = 0 .
P[D,t] . { log ( P[D,t] / P_eq[D] ) } = (1/8)(0) = 0 .

Box E.
P[E,t] = 1/8.
P_eq[E] = 1/8 .
P[E,t] / P_eq[E] = (1/8) / (1/8) = 1 .
log ( P[E,t] / P_eq[E] ) = log 1 = 0 .
P[E,t] . { log ( P[E,t] / P_eq[E] ) } = (1/8)(0) = 0 .

Box F.
P[F,t] = 0/8 = 0, so
P[F,t] . { log ( P[F,t] / P_eq[F] ) } = 0 .

The same goes for all the remaining boxes ( P[G,t] = 0/8, P[H,t] = 0/8).
So the H value for this distribution is
0.17 + 0.17 + 0.17 + 0 + 0 + 0 + 0 + 0 = 0.51 .

Distribution 3 1 1 1 1 1 0 0 .

Box A.
P[A,t] = 3/8.
P_eq[B] = 1/8 .
P[A,t] / P_eq[A] = (3/8) / (1/8) = 3 .
log ( P[A,t] / P_eq[A] ) = log 3 = 1.10 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (3/8)(1.10) = 0.41 .

Box B.
P[B,t] = 1/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (1/8) / (1/8) = 1 .
log ( P[B,t] / P_eq[B] ) = log 1 = 0 .
P[B,t] . { log ( P[B,t] / P_eq[B] ) } = (1/8)(0) = 0 .

Box C.
P[C,t] = 1/8.
P_eq[C] = 1/8 .
P[C,t] / P_eq[C] = (1/8) / (1/8) = 1 .
log ( P[C,t] / P_eq[C] ) = log 1 = 0 .
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = (1/8)(0) = 0 .

Box D.
P[D,t] = 1/8.
P_eq[D] = 1/8 .
P[D,t] / P_eq[D] = (1/8) / (1/8) = 1 .
log ( P[D,t] / P_eq[D] ) = log 1 = 0 .
P[D,t] . { log ( P[D,t] / P_eq[D] ) } = (1/8)(0) = 0 .

Box E.
P[E,t] = 1/8.
P_eq[E] = 1/8 .
P[E,t] / P_eq[E] = (1/8) / (1/8) = 1 .
log ( P[E,t] / P_eq[E] ) = log 1 = 0 .
P[E,t] . { log ( P[E,t] / P_eq[E] ) } = (1/8)(0) = 0 .

Box F.
P[F,t] = 1/8.
P_eq[F] = 1/8 .
P[F,t] / P_eq[F] = (1/8) / (1/8) = 1 .
log ( P[F,t] / P_eq[F] ) = log 1 = 0 .
P[F,t] . { log ( P[F,t] / P_eq[F] ) } = (1/8)(0) = 0 .

Box G.
P[G,t] = 0/8 = 0, so
P[G,t] . { log ( P[G,t] / P_eq[G] ) } = 0 .

The same goes for the remaining box ( P[H,t] = 0/8).
So the H value for this distribution is
0.41 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 0.41 .

Distribution 2 2 1 1 1 1 0 0 .

Box A.
P[A,t] = 2/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = (2/8) / (1/8) = 2 .
log ( P[A,t] / P_eq[A] ) = log 2 = 0.69 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (2/8)(0.69) = 0.17 .

Box B.
P[B,t] = 2/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (2/8) / (1/8) = 2 .
log ( P[B,t] / P_eq[B] ) = log 2 = 0.69 .
P[B,t] . { log ( P[B,t] / P_eq[B] ) } = (2/8)(0.69) = 0.17 .

Box C.
P[C,t] = 1/8.
P_eq[C] = 1/8 .
P[C,t] / P_eq[C] = (1/8) / (1/8) = 1 .
log ( P[C,t] / P_eq[C] ) = log 1 = 0 .
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = (1/8)(0) = 0 .

Box D.
P[D,t] = 1/8.
P_eq[D] = 1/8 .
P[D,t] / P_eq[D] = (1/8) / (1/8) = 1 .
log ( P[D,t] / P_eq[D] ) = log 1 = 0 .
P[D,t] . { log ( P[D,t] / P_eq[D] ) } = (1/8)(0) = 0 .

Box E.
P[E,t] = 1/8.
P_eq[E] = 1/8 .
P[E,t] / P_eq[E] = (1/8) / (1/8) = 1 .
log ( P[E,t] / P_eq[E] ) = log 1 = 0 .
P[E,t] . { log ( P[E,t] / P_eq[E] ) } = (1/8)(0) = 0 .

Box F.
P[F,t] = 1/8.
P_eq[F] = 1/8 .
P[F,t] / P_eq[F] = (1/8) / (1/8) = 1 .
log ( P[F,t] / P_eq[F] ) = log 1 = 0 .
P[F,t] . { log ( P[F,t] / P_eq[F] ) } = (1/8)(0) = 0 .

Box G.
P[G,t] = 0/8 = 0, so
P[G,t] . { log ( P[G,t] / P_eq[G] ) } = 0 .

The same goes for the remaining box ( P[H,t] = 0/8).
So the H value for this distribution is
0.17 + 0.17 + 0 + 0 + 0 + 0 + 0 + 0 = 0.34 .

Distribution 2 1 1 1 1 1 1 0 .

Box A.
P[A,t] = 2/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = (2/8) / (1/8) = 2 .
log ( P[A,t] / P_eq[A] ) = log 2 = 0.69 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (2/8)(0.69) = 0.17 .

Box B.
P[B,t] = 1/8.
P_eq[B] = 1/8 .
P[B,t] / P_eq[B] = (1/8) / (1/8) = 1 .
log ( P[B,t] / P_eq[B] ) = log 1 = 0 .
P[B,t] . { log ( P[B,t] / P_eq[B] ) } = (1/8)(0) = 0 .

Box C.
P[C,t] = 1/8.
P_eq[C] = 1/8 .
P[C,t] / P_eq[C] = (1/8) / (1/8) = 1 .
log ( P[C,t] / P_eq[C] ) = log 1 = 0 .
P[C,t] . { log ( P[C,t] / P_eq[C] ) } = (1/8)(0) = 0 .

Box D.
P[D,t] = 1/8.
P_eq[D] = 1/8 .
P[D,t] / P_eq[D] = (1/8) / (1/8) = 1 .
log ( P[D,t] / P_eq[D] ) = log 1 = 0 .
P[D,t] . { log ( P[D,t] / P_eq[D] ) } = (1/8)(0) = 0 .

Box E.
P[E,t] = 1/8.
P_eq[E] = 1/8 .
P[E,t] / P_eq[E] = (1/8) / (1/8) = 1 .
log ( P[E,t] / P_eq[E] ) = log 1 = 0 .
P[E,t] . { log ( P[E,t] / P_eq[E] ) } = (1/8)(0) = 0 .

Box F.
P[F,t] = 1/8.
P_eq[F] = 1/8 .
P[F,t] / P_eq[F] = (1/8) / (1/8) = 1 .
log ( P[F,t] / P_eq[F] ) = log 1 = 0 .
P[F,t] . { log ( P[F,t] / P_eq[F] ) } = (1/8)(0) = 0 .

Box G.
P[G,t] = 1/8.
P_eq[G] = 1/8 .
P[G,t] / P_eq[G] = (1/8) / (1/8) = 1 .
log ( P[G,t] / P_eq[G] ) = log 1 = 0 .
P[G,t] . { log ( P[G,t] / P_eq[G] ) } = (1/8)(0) = 0 .

Box H.
P[H,t] = 0/8 = 0, so
P[H,t] . { log ( P[H,t] / P_eq[H] ) } = 0 .

So the H value for this distribution is
0.17 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 0.17 .

Finally we consider the distribution 1 1 1 1 1 1 1 1 .

This is the equilibrium distribution.

Box A.
P[A,t] = 1/8.
P_eq[A] = 1/8 .
P[A,t] / P_eq[A] = (1/8) / (1/8) = 1 .
log ( P[A,t] / P_eq[A] ) = log 1 = 0 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (1/8)(0) = 0 .

The same goes for all the remaining boxes, B, C, D, E, F, G, and H.
So the H value for this distribution is
0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 0 .

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