Foundations to FrontierPart IX: The graduate coreBezhig

GB: Cumulative learning, data, and research assessment

6 exercises, each with a complete solution.

On this page
  1. Smallest useful hints
  2. Complete solutions: GB
  3. How to use an assessment result

Prerequisites: G11–G16 and the relevant project chapters. These are fresh synthesis tasks, not another count of earlier questions. A passing response should include assumptions and verification, not just arithmetic. Optional extra experiments are unassigned and do not change the six-exercise count.

GB.01CSolution

For one logistic example x=(1,−1)x=(1,-1), y=0y=0, weights (0,0)(0,0), bias zero, no regularizer, compute loss, gradients, and a simultaneous step at learning rate .2.2. Verify the sign of the new logit against the target.

GB.02CSolution

For a one-hidden-unit network with x=1,W=0,b=0,v=2,c=0,y=0x=1,W=0,b=0,v=2,c=0,y=0, derive all four gradients under binary logit loss. Explain which value the backward pass must cache to use 1−a21-a^2.

GB.03TSolution

Three causal self-attention positions have zero scores and scalar values (2,4,12)(2,4,12). Compute all outputs. Change only the last value to 120; which outputs may change? Explain why a fixed causal mask supports that answer.

GB.04TSolution

A row contains a user’s session features, a later refund decision, and an aggregate calculated using every session that month. The team randomly splits rows and predicts refunds for new users at session start. Identify three potential leakage mechanisms and redesign the split and feature-availability contract.

GB.05TSolution

A consumer commits one result but the worker dies before producer acknowledgment. On retry, a newly deployed model is available. State what GP2 should do, how the model identity is preserved, and what fault test would demonstrate the local claim without overstating power-loss evidence.

GB.06TSolution

A benchmark tests 20 parameter settings and reports only the fastest single run. A model evaluation reports 98% accuracy without class counts. Write a review that separates the two evidence failures, requests a better protocol, and states what remains unknown rather than assuming misconduct or success.

Smallest useful hints#

For .01–02, begin at p−yp-y. For .03, count the allowed values in each row. For .05, the submitted model digest is part of the immutable payload, not a suggestion to use a similar artifact.

Complete solutions: GB#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

GB.01Show solution

The negative target should lower the logit

Initially z=0,p=1/2z=0,p=1/2, loss log⁡2\log2. The weight gradient is (1/2)(1,−1)=(.5,−.5)(1/2)(1,-1)=(.5,-.5) and bias gradient .5.5. Subtract .2.2 times these to obtain weights (−.1,.1)(-.1,.1) and bias −.1-.1. The new logit is −.1−.1−.1=−.3-.1-.1-.1=-.3, giving probability about .42556.42556 and loss log⁡(1+e−.3)≈.55436\log(1+e^{-.3})\approx.55436, smaller than log⁡2\log2. The sign is consistent with moving probability toward target zero on this example.

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GB.02Show solution

Follow the hidden sensitivity separately from the output weight

Forward values are u=a=z=0,p=.5u=a=z=0,p=.5. Thus δz=.5\delta_z=.5, ∇vℓ=0\nabla_v\ell=0, ∇cℓ=.5\nabla_c\ell=.5, δu=.5(2)(1−0)=1\delta_u=.5(2)(1-0)=1, ∇Wℓ=1\nabla_W\ell=1, and ∇bℓ=1\nabla_b\ell=1. Cache the hidden activation a=tanh⁡(u)a=\tanh(u), or deliberately recompute it from the same old parameters, to form 1−a21-a^2. Updating parameters before completing this pass changes the point at which derivatives are evaluated.

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GB.03Show solution

Only allowed values contribute

Outputs are 22, (2+4)/2=3(2+4)/2=3, and (2+4+12)/3=6(2+4+12)/3=6. Replacing the last value by 120 changes only the third to (2+4+120)/3=42(2+4+120)/3=42. The first two rows exclude the future position before normalization and value mixing. This result is for the fixed given scores/mask and changed value; changing an earlier input representation in a full network can affect other allowed keys, queries, or values according to that network’s graph.

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GB.04Show solution

Align the dataset with the decision time and target population

The later refund decision is the label, not an available input; using it or a derivative of it as a feature leaks the outcome. A full-month aggregate may include sessions after the decision time. Row splitting can place the same user in training and test, contrary to the new-user evaluation target. Split by user before fitting preprocessing, define every feature from information available at session start, and keep later outcomes only as labels. Use a time-aware evaluation as well when forecasting a changing future population. Record missing coverage and label delays rather than assuming every row is equally complete.

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GB.05Show solution

Retry the original operation, not a new model choice

The producer remains pending while the consumer already has a receipt and result for the original payload/model digest. Supply that exact artifact, validate the digest, return the stored result on duplicate delivery, and then acknowledge the producer. Do not substitute the new model. A subprocess test should exit after the consumer commit, confirm the pending producer and one consumer result using new connections, retry, and assert one result plus acknowledgment. This demonstrates a local process-crash window; it does not test real power failure or distributed replicated durability.

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GB.06Show solution

Request evidence appropriate to each claim

The benchmark combines parameter selection, single-run noise, and selective reporting. Request the full candidate set, equal-output contract, workload distribution, build/timing boundary, paired repetitions with order control, and a predeclared or transparently exploratory selection rule. The model score lacks class balance and error costs; request counts, split units, held-out protocol, confusion matrix, probability-sensitive metrics, and uncertainty at a justified unit. Neither omission proves fraud or a useless method. The present information is insufficient to endorse a broad speedup or reliable predictive utility.

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How to use an assessment result#

For each question, record setup, reasoning, execution, verification, and communication as missing, guided, independent, or transferred. Preserve wrong attempts and name the earliest error. A misunderstanding of an invariant or transaction boundary requires conceptual repair even when arithmetic is correct. Reserve a changed problem for a later day before declaring readiness for a larger system. A score on this book is not an accredited degree assessment.