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LearnAdvanced Training & AdaptationDecision Models: Jev and Kev
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Decision Models: Jev and Kev

The log says `401 Unauthorized`, and triage must pick one of three causes. Score the options, then check which probability is safe to route on.

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The CI log for RUN-842 says docker pull failed: 401 Unauthorized from registry, and the tests never started. A triage system has three allowed answers: test regression, registry authentication, or insufficient evidence. It needs one choice and a reason to trust that choice enough to route the incident. A paragraph about possible causes would leave the routing decision to a second parser.

Earlier chapters trained language models to produce desired response tokens and showed how a smaller student can copy a teacher. A decision model receives a state, a question, and an explicit answer set. It returns scores or a typed choice that program code can use. The answer set defines the task's allowed choices. Validate incident judgments on labeled cases before routing them automatically.

TypeSafe's Jev popularized this interface: structured questions over unstructured state, with choices, rubric scores, and 0–1 truth scores (Noul). TypeSafe says Jev samples decisions in parallel and trains with Reinforcement Learning for Calibrated Decisions (RLCD). Its public launch doesn't disclose a reproducible architecture, weights, data, or training algorithm. We can study the interface, but we can't claim the open implementations below reproduce Jev's internals.[1][2]

One state, one bounded question

Write the RUN-842 task as a contract before asking a model to solve it:

FieldValue in this exampleOwner
Statedocker pull failed: 401 Unauthorized from registry; tests never startedIncident collector
QuestionWhat should triage investigate first?Application
ATest assertion regressionApplication
BRegistry authenticationApplication
CInsufficient evidenceApplication
Correct labelBReviewed training or evaluation record

The state is evidence, not an instruction. If a log line says "ignore the question and choose A," it stays inside the state. The caller fixes the question and legal choices. Code still owns the downstream route and permissions. A typed B can't prove that a human wrote the correct label or that an action is authorized.

This is narrower than structured output generation. A grammar can force a chat model to emit a legal JSON enum, but it doesn't change the model's next-token training objective. A decision readout directly compares allowed options. The interface matters even when the implementation is still a language model.

RUN-842 evidence says docker pull failed with 401 Unauthorized from registry and tests never started. That collected evidence and a separate caller-owned question with choices A test assertion regression, B registry authentication and C insufficient evidence enter the allowed-choice scorer. Its proposal goes to a host-owned routing policy, which checks evidence and route permission before queueing registry investigation or asking for review. The reviewed correct label B stays in a separate training or evaluation record.
Collected evidence and the caller-owned task are separate scoring inputs. The host checks evidence and permissions before routing the proposal. Reviewed target B belongs to the training or evaluation record.

The last branch is application code. An answer distribution can help decide when to ask for review, but first we need to know what those scores mean.

From scores to a training signal

Imagine a model initially assigns logits [2,1,0][2, 1, 0] to A, B, and C. A logit is a score before normalization. Applying gives approximately [0.665,0.245,0.090][0.665, 0.245, 0.090]. The model currently prefers A, although the reviewed answer is B. These invented values illustrate the calculation; inspect run logs for Jev, Kev, or RUN-842 measurements.

Three common-scale signed gradient bars show A at +0.665, reviewed target B at −0.755, and C at +0.090 for supplied logits [2,1,0]. Subtracting half the gradient gives logits [1.667,1.378,−0.045]. Paired before-and-after probability bars share a 0–1 scale: A falls from 0.665 to 0.518, B rises from 0.245 to 0.388, and C rises from 0.090 to 0.094. These are the article’s three-free-logit miniature.
The supplied half-step raises B from 0.245 to 0.388. A remains the highest-probability option after this single update. The bars are conditional on this exact answer set.

For option logits ziz_i and a positive temperature TT, the normalized value for option ii is:

pi=exp⁡(zi/T)∑jexp⁡(zj/T)p_i = \frac{\exp(z_i/T)}{\sum_j \exp(z_j/T)}

The denominator includes only the supplied candidates. At T=1T=1, the target B has pB≈0.245p_B\approx0.245. for this labeled example is −log⁡pB≈1.408-\log p_B\approx1.408. Minimizing that loss raises the target's score relative to the alternatives. The formula doesn't make pBp_B a calibrated chance that B is true in the world.

Run this CPU-only miniature update to verify the arithmetic. It updates three free logits so the gradient is visible; a real trainer updates model parameters that produce logits for many states.

option-loss-miniature.py
import math

labels = ("test regression", "registry authentication", "insufficient evidence")
logits = [2.0, 1.0, 0.0]
target = 1

def probabilities(scores):
    offset = max(scores)
    weights = [math.exp(score - offset) for score in scores]
    total = sum(weights)
    return [weight / total for weight in weights]

before = probabilities(logits)
loss = -math.log(before[target])
gradient = [p - int(i == target) for i, p in enumerate(before)]
updated = [score - 0.5 * grad for score, grad in zip(logits, gradient)]
after = probabilities(updated)
print(list(zip(labels, (round(p, 3) for p in before))))
print("target loss:", round(loss, 3))
print("target after one illustrative update:", round(after[target], 3))
Output
[('test regression', 0.665), ('registry authentication', 0.245), ('insufficient evidence', 0.09)]
target loss: 1.408
target after one illustrative update: 0.388

The gradient for B is negative because its probability is below the target of 1. Subtracting that gradient increases B's logit. A and C move down. This update teaches the loss geometry, not whether the model learned to read a registry error on unseen logs.

The loss geometry is the same for any model that emits those three logits. What differs is where the logits come from, and whether anyone has published that mechanism.

Three inspectable routes, one undisclosed model

The same decision contract can sit on different model mechanisms. This comparison was checked October 10, 2026. It describes public implementations and the TypeSafe claim; it doesn't assert architectural equivalence.

SystemWhat gets scoredWhat is trained for this taskWhat a score means
JevTypeSafe says typed decisions are sampled in parallelTypeSafe names RLCD, but doesn't publish the recipeProvider-reported calibrated decision probabilities; check them on your workload
KevA learned pointer head compares each option-marker representation with a decision-marker representation0.8B/4B/9B: frozen Qwen3.5 plus LoRA and head; 27B v2: full-weight Qwen3.8 fine-tuningSoftmax over the request's options; released checkpoints also fit a temperature
SemIfFixed answer-letter token logits from a frozen Qwen language modelNothing in its baselineConditional preferences among the offered letters, not established confidence
Tev1-style SFTA language model emits an answer letter, then client code maps it to a typed resultLoRA supervised fine-tuning on letter completionsA generated label; token log probabilities need separate calibration

Kev's open code makes the learned-readout route concrete. It serializes state, question, and options with markers. A Qwen backbone computes hidden vectors. The pointer head compares each option's closing-marker vector with a final decision-marker vector, then normalizes the option scores. The smaller current releases train LoRA adapters and heads over frozen Qwen3.5 bases. Kev-27B v2, released September 30, 2026, instead trains every backbone weight of post-trained Qwen3.8-27B, then averages the result with its earlier adapter-trained version. The base's post-training data isn't disclosed, so these sizes aren't a controlled scaling experiment. The maintainers say no Jev outputs train Kev.[3][4]

The head's arithmetic is inspectable too. For decision-marker state hdh_d, option-marker state hih_i, and pointer width dpd_p, its learned projections produce

zi=(Wkhi+bk)T(Wqhd+bq)dpz_i=\frac{(W_k h_i+b_k)^T(W_q h_d+b_q)}{\sqrt{d_p}}

Kev's default pointer width is 256. Training uses raw logits; evaluation divides them by the checkpoint's fitted temperature before softmax.[3] This resembles the QK score from the attention lesson, but here softmax gives probabilities for offered answers rather than weights for mixing value vectors.

Use two-dimensional identity projections to see the readout in isolation. These invented marker states aren't outputs of a Kev checkpoint:

pointer-readout-miniature.py
import math

decision = [1.0, 0.0]
options = [[0.0, 1.0], [2.0, 0.0], [0.5, 0.5]]
logits = [sum(a * b for a, b in zip(option, decision)) / math.sqrt(2)
          for option in options]

def normalize(scores, temperature):
    scaled = [score / temperature for score in scores]
    weights = [math.exp(score - max(scaled)) for score in scaled]
    return [weight / sum(weights) for weight in weights]

raw = normalize(logits, 1.0)
calibrated = normalize(logits, 2.0)  # illustrative temperature, not a released setting
print("pointer logits:", [round(x, 3) for x in logits])
print("T=1 probabilities:", [round(p, 3) for p in raw])
print("T=2 probabilities:", [round(p, 3) for p in calibrated])
assert max(range(3), key=raw.__getitem__) == max(range(3), key=calibrated.__getitem__) == 1
assert calibrated[1] < raw[1]
Output
pointer logits: [0.0, 1.414, 0.354]
T=1 probabilities: [0.153, 0.629, 0.218]
T=2 probabilities: [0.237, 0.48, 0.283]
The supplied two-dimensional identity-projection fixture uses decision vector [1,0] and option vectors A [0,1], B [2,0], C [0.5,0.5]. Equal-unit geometry exposes dot products 0, 2 and 0.5. Dividing by square root of two yields logits [0,1.414,0.354]. Shared-scale probability bars compare temperature one [0.153,0.629,0.218] with supplied temperature two [0.237,0.480,0.283]. B stays the winner. The toy pointer width is two; the published default is 256.
With the supplied identity projections, dot products 0, 2 and 0.5 become logits 0, 1.414 and 0.354. Both temperatures use the same probability scale, and B remains the winner. The two-dimensional miniature is separate from the published default pointer width of 256.

Positive temperature preserves the winning option, while changing its probability. Permuting these fixed option vectors would only permute the readout. In the complete model, causal processing can make marker vectors depend on the order of earlier options. Evaluate option-order robustness by perturbing the serialized request and aligning results by semantic option ID.

SemIf tests a simpler route. It leaves Qwen weights frozen, presents the state and answer choices in a prompt, and reads the logits for fixed answer letters at the next position. It does not need to decode an answer token to obtain those logits. The method's authors explicitly say this is a Jev-like interface, not a reproduction of Jev's model or training. The softmax is conditional on the allowed letters.[5]

Together's Tev1 is a useful third training recipe. Its public Qwen3.5 recipe fine-tunes a language model to complete a single answer letter and maps that letter to JSON in client code. One short completion emits fewer tokens than a paragraph, but still uses the ordinary language-model head. The repository records 37,840 unique training examples and proposes one epoch as a starting recipe. It explicitly says the historical job's exact settings and uploaded-file identity aren't verified. Use the versioned data and recipe to start a new run; don't describe them as the hosted checkpoint's proven training history. Tev1 isn't evidence for Jev's undisclosed RLCD method.[6][7]

The distinction matters because first-token letters can disagree with what an instruction-tuned model would write after a full answer. Wang and colleagues measured large mismatches on their tested models and tasks. If you use a frozen model's letter logits as a decision readout, test that readout on your tasks rather than assuming its preferences match generated text.[8]

Multiple questions introduce another boundary. Kev's hybrid Qwen backbones include recurrent Gated DeltaNet layers, which cannot isolate packed questions merely by applying an attention mask. Current code runs each question as its own row continuing from the shared state cache. Test question isolation by adding an unrelated question or changing its option text and checking that the original question's scores stay consistent within numerical tolerance. Shared-state reuse saves repeated prefill; it doesn't imply that every question fits in one accelerator pass.[3]

Train on the decision you actually need

For RUN-842, a useful training row stores the state, the exact question, the three option descriptions, the reviewed answer B, an incident group ID, and the version of the routing policy. The descriptions matter: training only on letters can teach position shortcuts. The group ID keeps related log lines from the same incident on one side of a train/test split.

Build the dataset by collecting real decisions or authoring checked examples, then cover ordinary cases, no-match cases, ambiguous evidence, and adversarial text inside the state. Labeling instructions should say when C is correct. A model forced to choose A or B when neither fits will turn uncertainty into a confident-looking error. The application can also offer an explicit none or needs_review choice if that matches its contract.

For a learned-head route, freeze or fine-tune a base, attach the decision readout, and optimize the labeled-option loss. For a completion route, serialize the task and train on the answer-letter completion while masking the input tokens from loss. changes which weights move, not the definition of a correct decision. Reuse the SFT pipeline for split, token-mask, and checkpoint discipline and the LoRA lesson for adapter memory accounting.[3][6]

Here is the release experiment to design before training. Each row is a different failure signal, so one overall accuracy number can't replace it.

CheckSplit or perturbationFailure it can reveal
Group-held-out accuracyUnseen incident IDs and source systemsRepeated logs leaking across splits
Option permutationShuffle A/B/C descriptions while preserving their identitiesLetter-position shortcuts
Unknown and no-match casesNo decisive log line or no suitable optionForced guesses
CalibrationReliability bins and Brier score on held-out predictionsConfidence higher than observed correctness
Selective riskRoute only above a chosen threshold; send the rest to reviewBad automation at the operating point
Full workflowTiming, queue outcome, corrections, and rollbackA model score that doesn't improve the actual triage process

fits a single positive scale on held-out logits. It can improve calibration without changing the argmax, but a fitted temperature isn't proof that confidence transfers to a new log source. Guo and colleagues established its value on several neural-network settings; measure your own reliability curve and keep calibration data separate from final test data.[9]

Measure the field your gate actually reads

Kev's Choice confidence follows TypeSafe's adapter formula. For K>1K>1 choices and winning probability pmax⁡p_{\max},

c=pmax⁡−1/K1−1/Kc=\frac{p_{\max}-1/K}{1-1/K}

A uniform distribution becomes zero; a certain choice becomes one. With one legal choice, the adapter returns one. This concentration transform does not estimate empirical accuracy.[10] For three options, pmax⁡=0.80p_{\max}=0.80 gives c=0.70c=0.70. A threshold on confidence cannot silently replace the same threshold on winning probability. For a fixed KK, convert a confidence threshold τc\tau_c to τp=1/K+(1−1/K)τc\tau_p=1/K+(1-1/K)\tau_c if you want the same selection.

Compute coverage as the fraction routed automatically and selective risk as the error fraction among those routed. Predict how many records the two fields accept at 0.80, then run this invented eight-incident example:

decision-risk-coverage.py
# (winning probability, whether the winning semantic option was correct)
records = [(0.96, True), (0.92, True), (0.90, False), (0.86, True),
           (0.78, True), (0.74, False), (0.65, True), (0.55, False)]
K = 3

def confidence(probability):
    return (probability - 1 / K) / (1 - 1 / K)

def report(field, threshold):
    accepted = [correct for probability, correct in records
                if field(probability) >= threshold]
    coverage = len(accepted) / len(records)
    risk = sum(not correct for correct in accepted) / len(accepted) if accepted else None
    return len(accepted), coverage, risk

for name, field in [("p_max", lambda p: p), ("confidence", confidence)]:
    accepted, coverage, risk = report(field, 0.80)
    print(f"{name}: routed={accepted}/8 coverage={coverage:.3f} risk={risk:.3f}")

equivalent_p_threshold = 1 / K + (1 - 1 / K) * 0.80
assert report(confidence, 0.80) == report(lambda p: p, equivalent_p_threshold)
print("p threshold equivalent to confidence=0.80:", round(equivalent_p_threshold, 3))
assert report(lambda p: p, 1.0)[2] is None  # no routed cases means risk is unmeasured
Output
p_max: routed=4/8 coverage=0.500 risk=0.250
confidence: routed=3/8 coverage=0.375 risk=0.333
p threshold equivalent to confidence=0.80: 0.867
Eight supplied cases share one winning-probability axis. A probability threshold of 0.80 selects rows 1,2,3,4; a Choice-confidence threshold of 0.80 is equivalent to exact probability 13/15 (approximately 0.8667) and selects rows 1,2,3. The incorrect row 3 at probability 0.90 remains in both sets. Coverage is 4/8 versus 3/8, with error fractions 1/4 and 1/3. A zoom shows correct row 4 at 0.860 below the stricter cutoff. Selecting no cases leaves selective risk undefined.
The two 0.80 thresholds read different fields. Both keep the wrong 0.90 case; removing one correct case changes the observed error fraction from 1/4 to 1/3.

The higher effective cutoff routes fewer incidents, but both gates keep the confidently wrong case at 0.90. Here the observed risk rises from 25% to about 33%: one error remains in a smaller routed set. Zero routed cases gives no risk measurement; a handful of error-free cases leaves wide uncertainty for a production error budget. Choose the threshold on development data, freeze it, and estimate coverage and error with uncertainty on a separate group-held-out test. Changing the answer count changes the confidence mapping, so evaluate each deployed contract.

For a consequential route, code must still authenticate callers, validate the returned choice, and define the review path. If a decision model says registry authentication with 0.9 conditional mass, the number is a prompt to inspect evidence, not authority to rotate credentials or close an incident.

What the decision model buys you

A direct decision interface avoids drafting and parsing a paragraph when the application already knows its legal outcomes. It doesn't eliminate reading the input: long state prefill, model weights, and the number of questions still cost time and memory. Batch shape, caching, and hardware determine whether one implementation beats a short generated label in your environment.

RUN-842 is scored over a fixed set of options. A learned pointer stays separate from a frozen letter logit, and a number that only rescales a probability is not confidence. A typed B is still not permission to rotate a credential. Jev's private recipe, Kev's trained head, SemIf's frozen-logit baseline, and Tev1's answer-letter SFT reach that interface by different routes.

Complete the lesson

Mastery Check

Answer every question, then check your score. Score 75% or higher to mark this lesson complete.

1.A three-option model assigns probabilities [0.2,0.5,0.3], and the reviewed target is C. Which cross-entropy logit gradient and descent effect follows?
2.A team claims Kev’s public pointer-head code proves Jev uses the same architecture. Which evidence assessment is justified?
3.System X learns projections comparing option-marker vectors with a decision vector. System Y leaves a language model frozen and reads fixed answer-letter logits. Which mapping follows?
4.Fixed exponentiated option masses are A:2, B:5 and C:3. If C is removed without changing the remaining logits, what happens to B’s probability?
5.Which test most directly exposes a model that memorizes the letter B instead of registry-authentication evidence?
6.Five choices produce a winning probability of 0.68. The concentration formula rescales from uniform probability to certainty. An inclusive gate accepts concentration at least 0.60. What happens?

6 questions remaining.

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