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THE CONSCIOUSNESS ATLAS
Read. Experiment. Follow the evidence.

ConsciousnessLearn · Explore · Joseph Anady
12 minute lessonReady to learn

Read the normalized MIP objective

Separate predictive information loss, its normalizer, and the selected partition.

Course lessons · 4 of 5
Your goal

Evaluate candidate ratios and explain the conditions of the recovered benchmark.

Useful background: Enumerate graph bipartitions

01 · Understand the idea

Read the numerator term by term

The benchmark’s stochastic interaction adds the next-state uncertainty of each part conditioned on its own current state, then subtracts the whole system’s next-state uncertainty conditioned on the whole current state. It measures a specific observational loss of predictive integration.

The denominator K is the smaller present-state differential entropy of the two parts. The saved benchmark uses natural logarithms, fixed innovation scale, and a positive denominator. The ratio J is what the search minimizes.

J(S)=H(Xt+1,S∣Xt,S)+H(Xt+1,Sˉ∣Xt,Sˉ)−H(Xt+1∣Xt)min⁡{H(Xt,S),H(Xt,Sˉ)}J(S)=\frac{H(X_{t+1,S}\mid X_{t,S})+H(X_{t+1,\bar S}\mid X_{t,\bar S})-H(X_{t+1}\mid X_t)}{\min\{H(X_{t,S}),H(X_{t,\bar S})\}}

Preserve the model behind the numbers

The recovered study uses small synthetic linear stochastic networks with a stationary population covariance. Its exact reference enumerates the finite bipartition set under a recorded floating-point tolerance. It is not a claim of exact rational arithmetic or a full IIT 4.0 calculation.

Gaussian differential entropy depends on coordinate scale. Changing variable units can change the normalizer and selected split. A reproducible comparison therefore keeps the generator, covariance definition, units, and candidate set fixed.

02 · Work through an example

Score a hypothetical candidate

1Compute raw loss

Suppose the two partwise conditional entropies are four and five nats, and the whole-system conditional entropy is seven nats.

ΦSI=4+5−7=2\Phi_{\rm SI}=4+5-7=2
2Compute the normalizer

If the two present-state entropies are six and four nats, the smaller one is four.

K=min⁡(6,4)=4K=\min(6,4)=4
3Form the normalized score

The ratio is one half. Store both the ratio and the raw two-nat loss so the two quantities remain distinguishable.

J=2/4=0.5J=2/4=0.5
03 · Change something

Make the idea respond

Use this graph lab to compare how a raw cut and a degree-volume normalization rank candidate splits. It computes NCut = cut/vol(A) + cut/vol(B); it does not compute the conditional Gaussian entropies, entropy normalizer K, or J from this lesson. The worked numerical example and source links supply the information-objective calculation. Keep those two normalization rules labeled when comparing their behavior.

Change one input at a time. Predict what should happen, then compare the computed result with your prediction.

Teaching model · computed in your browser

Find every graph partition

Change a tiny graph and examine every possible split. The minimum is computed by complete enumeration.

NCut⁡(A,B)=cut⁡(A,B)vol⁡(A)+cut⁡(A,B)vol⁡(B)\operatorname{NCut}(A,B)=\frac{\operatorname{cut}(A,B)}{\operatorname{vol}(A)}+\frac{\operatorname{cut}(A,B)}{\operatorname{vol}(B)}

A and B are nonempty complementary node sets. cut(A, B) is the sum of edge weights crossing the split, counted once. vol(A) is the sum of weighted degrees of nodes in A, including both ends of internal edges; vol(B) is defined likewise. A zero volume makes this objective undefined. This is a normalized-cut teaching example, not the manuscript’s normalized minimum information partition (MIP). It does not calculate integrated information or benchmark Queyranne, annealing, or QAOA.

Try this: lower the bridge weight to zero. Inspect both the minimum and any undefined partitions. Then switch from two clusters to a cycle.

Experiment controls

The default worked example is shown below.

Distinct bipartitions312^(6 − 1) − 1
Smallest defined NCut0.0645
Selected NCut0.0645This split attains the minimum
Minimizing splits1
Selected graph partitionNumbers on edges are weights. Letters inside circles identify nodes; the group label beneath each node identifies its part. All exact data appears in the tables below.1110.2111AGroup ABGroup ACGroup ADGroup BEGroup BFGroup B
Group A: A, B, C. Group B: D, E, F. Dashed edges cross the split; solid edges stay within a part. Edge numbers are weights.

For this split: cut(A, B) = 0.2; vol(A) = 6.2; vol(B) = 6.2. NCut = 0.2 / 6.2 + 0.2 / 6.2 = 0.0645. Every unordered, nonempty bipartition appears once because node A is fixed in the first part.

Every distinct bipartition — complete enumeration
Group AGroup BCutvol(A)vol(B)NCut
AB, C, D, E, F2210.41.1923
A, BC, D, E, F248.40.7381
A, CB, D, E, F2.24.28.20.7921
A, B, CD, E, F0.26.26.20.0645 · minimum
A, DB, C, E, F4.24.28.21.5122
A, B, DC, E, F4.26.26.21.3548
A, C, DB, E, F46.461.2917
A, B, C, DE, F28.440.7381
A, EB, C, D, F448.41.4762
A, B, EC, D, F466.41.2917
A, C, EB, D, F4.26.26.21.3548
A, B, C, ED, F2.28.24.20.7921
A, D, EB, C, F4.26.26.21.3548
A, B, D, EC, F4.28.24.21.5122
A, C, D, EB, F48.441.4762
A, B, C, D, EF210.421.1923
A, FB, C, D, E448.41.4762
A, B, FC, D, E466.41.2917
A, C, FB, D, E4.26.26.21.3548
A, B, C, FD, E2.28.24.20.7921
A, D, FB, C, E4.26.26.21.3548
A, B, D, FC, E4.28.24.21.5122
A, C, D, FB, E48.441.4762
A, B, C, D, FE210.421.1923
A, E, FB, C, D466.41.2917
A, B, E, FC, D484.41.4091
A, C, E, FB, D4.28.24.21.5122
A, B, C, E, FD2.210.22.21.2157
A, D, E, FB, C2.28.24.20.7921
A, B, D, E, FC2.210.22.21.2157
A, C, D, E, FB210.421.1923
Inspect the exact adjacency matrix
Symmetric edge-weight matrix; diagonal entries are zero
NodeABCDEF
A011000
B101000
C1100.200
D000.2011
E000101
F000110
Open the full lab
04 · Check your understanding

Your turn.

Try an answer before opening a hint. Feedback will explain the result, and you can practice again.

Interactive practice is loading. The lesson, worked example, and printable questions remain available below.

View printable practice and solutions

Use these questions when studying without interactive controls. Opening a written solution does not earn progress.

  1. Partwise conditional entropies are 3 and 4 nats; whole-system conditional entropy is 5 nats. What is the raw stochastic interaction?

    Read the solution

    The raw value is 3 + 4 − 5 = 2 nats.

  2. That raw value is 2 nats and the two normalizing entropies are 8 and 4 nats. What is J?

    Read the solution

    K = 4 nats, so J = 2/4 = 0.5.

  3. Which value determines the minimizing partition in this normalized benchmark?

    • The raw numerator alone.
    • The largest partition index.
    • The number of times a result was displayed.
    • J, the normalized ratio.
    Read the solution

    The benchmark minimizes the ratio J over valid candidate splits.

  4. Why does the benchmark fix variable scale?

    • Differential entropy can change under rescaling.
    • Natural logarithms prevent all scale effects.
    • Scale only changes page layout.
    • A smaller unit always proves greater integration.
    Read the solution

    Fixing units and innovation scale keeps the objective comparable.

05 · Apply it to the archive

Follow the original record

Use the definitions you just practiced to read the project’s own notation, assumptions, and evidence.

Normalized minimum stochastic-interaction partitionGaussian stochastic interaction and marginal differential entropyExact partitions and measured search performance
Take it with you

The ideas to keep

  • Minimize the defined ratio.
  • Store raw and normalized results separately.
  • Exact finite search is specific to its model and tolerance.

Reading completion is separate from your practice evidence.

Progress is kept in this browser. How learning works

Search equations, builds, manuscript text, and recovered sources.