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Elliott Wave: count hypotheses, rules and invalidation

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• The common impulse illustration has five waves followed by an A–B–C correction. Treat this as a classification hypothesis.
• Under a standard non-diagonal impulse convention, wave 2 does not retrace beyond the start of wave 1, wave 3 is not the shortest among 1/3/5, and wave 4 does not overlap wave 1 territory.
• Diagonal and corrective variants use different rules. State the allowed pattern family before counting.
• A rule such as “wave 3 is not shortest” cannot be finally checked until wave 5 is observed. Do not pretend the finished count was known at wave 3.
• Build pivots using a fixed causal reversal threshold or delayed confirmation. Save every provisional count and its invalidation time.
• Compare competing counts and a simple swing/trend baseline. An LLM may describe candidates, but numerical code should check their constraints.
• Measure predictive value from the timestamp the candidate was available, not from the earliest labeled turning point.

Use the rule

• Fix the definition, units and information timestamp.
• Compute the example and inspect the graph.
• Compare with a simple baseline on untouched periods after costs.

Before moving on

Write the exact rule, its availability time, an invalidation condition and a fair out-of-sample test.

Research checkpoint · reviewed 12 September 2026

• Status: accepted ICAART 2025 conference paper; arXiv abstract and bibliographic metadata reviewed only.
• Scope: combines wave analysis with language models, retrieval and reinforcement learning on historical major US-company data. Exact dates, split design and executable costs were not verified from the abstract.
• Use as an architecture research lead. Benchmark any implementation against fixed numerical swing rules and audit pretraining, retrospective wave labels and available information. No reported performance is adopted here.

Wawer & Chudziak · ElliottAgents (ICAART 2025) ↗

def wave_three_not_shortest(pivots):
    if len(pivots)!=6: raise ValueError("Six completed pivots required")
    p=pivots
    lengths=[abs(p[1]-p[0]),abs(p[3]-p[2]),abs(p[5]-p[4])]
    return lengths,lengths[1]>=min(lengths[0],lengths[2])

print(wave_three_not_shortest([100,110,105,123,115,127]))