If you're a learner, try the check questions in each part first, then come here to check. Each question is worth 20 points (100 per part); you'll find grading points and sample answers for the Colab assignments. Numbers in the sample answers may differ slightly depending on your environment.
Back to the Educator TIP on the Special 1 page
Answer · Sum = 1×0.8 + 0×(−0.5) = 0.8. Since 0.8 ≥ 0.5, the output is 1.
20 points · Grading: working 10, result 10
Answer · A perceptron divides space with only one straight line (plane). A band needs two boundaries, one on each side, so a single line can't separate it. A hidden layer creates several boundaries and combines them.
20 points · Grading: "a single line" 10, "needs two boundaries" 10
Answer · Overfitting: the model memorized the training data. Remedies: any two of collecting more data, using fewer hidden neurons, or stopping training early.
20 points · Grading: diagnosing overfitting 10, each remedy 5
Answer · Fixed: projection neurons → Kenyon cells (random sparse connections). Learned: Kenyon cells → output neurons (changed by dopamine).
20 points · Grading: 10 each
Answer · With just a wide, sparse, fixed representation, simple image problems can be solved well enough by training only the output layer. However, since the representation itself can't improve, it has limits on harder problems.
20 points · Grading: "a fixed representation is enough" 10, mentioning the limits 10
Assignment · In Code 1, change TOP (the fraction of cells that switch on) from 0.08 to 0.5. Report how training accuracy and the number of active Kenyon cells change, and explain in two sentences why sparsity is useful.
Sample answer · Raising TOP to 0.5 switches on half the Kenyon cells, so the codes for different inputs overlap a lot. Because the toy data is easy, training accuracy can still come out high, but the number of active cells rises from about 80 to about 500, making memories more likely to blend together. A sparse code uses only a few non-overlapping cells per input, so it can tell similar smells apart, and editing one memory disturbs the others less.
Grading points · Confirms the change in active count (5) · Reports accuracy (5) · Two benefits of sparsity (less overlap, less interference) (10)
Assignment · In Code 2, set the MLP hidden size to 8, 64 and 512, and compare validation accuracy with two seeds. Interpret whether overfitting appears, using the learning curves (cell 7).
Sample answer · A larger hidden size makes training loss drop faster, but validation accuracy doesn't rise, or it wobbles depending on the seed. With 160 example images (96 for training), a 512-unit hidden layer easily memorizes the training data. Judge overfitting by pointing to the moment in the learning curves where the gap between training loss and validation accuracy starts to widen.
Grading points · Results table for three sizes × two seeds (8) · Evidence from learning curves (6) · Interpretation tied to data size (6)
Back to the Educator TIP on the Special 2 page
Answer · A memory signal retrieved from the mushroom body must be passed to the convergence neurons (PD2a1/b1) in the lateral horn for the memory to become behavior.
20 points · Grading: "mushroom body → lateral horn" 10, "the convergence neurons are required" 10
Answer · Temperature itself can change behavior, so we must tell whether the memory score dropped because the neurons were blocked or because of the temperature.
20 points · Grading: purpose of the control 20
Answer · 0.2 + 0.6 × (−1.0) = −0.4 (toward avoid). If blocked, 0.2 (only instinct remains, toward approach).
20 points · Grading: 10 each
Answer · Because that neuron is a shared layer that receives both pathways (the memory signal from the mushroom body and the instinct signal from food-odor projection neurons) and outputs them together.
20 points · Grading: the concept of "sharing" 20
Answer · The instinct values are all 0, so every class gets the same score and the model always picks the first class. 0.09 is the fraction of the test data that belongs to the first class.
20 points · Grading: "the scores become equal" 10, "fraction of the first class" 10
Assignment · In Code 1, change innate to [0.0, 0.5] (instinct leaning toward avoid). Report in a table how the before-training, after-training and blocked results change, and compare the cases where instinct and memory point the same way versus opposite ways.
Sample answer · With innate = [0.0, 0.5], the instinct score points to "avoid" even before training. Punishment training lowers the "approach" score on the memory side, so the fly still avoids after training, and blocking the memory input still gives avoid from instinct alone. With the original setting (innate = [0.5, 0.0]), blocking returned the fly to approach. When instinct and memory point the same way, behavior stays the same even after blocking, so memory's contribution stays hidden; only when they point opposite ways does the blocking experiment reveal memory's role.
Grading points · Results table for the three conditions (8) · Comparison with the original setting (6) · Interpretation that "it shows only when they point opposite ways" (6)
Assignment · In Code 2's MemoryInnate, put a value into innate (e.g., 2.0 for the X class), train, then block the memory input. Compare the confusion matrices before and after blocking, and describe how instinct pulls on the decision.
Sample answer · Putting 2.0 for the X class into innate makes the memory pathway learn to push back against this pull during training. When blocked, only instinct remains and most images get classified as X, so predictions pile up in the X column of the confusion matrix. Point out that instinct acts like a bias that isn't learned, and memory is a correction on top of it.
Grading points · Confusion matrices before and after blocking (8) · Explains the pile-up in the X column (6) · Bias + correction interpretation (6)
Back to the Educator TIP on the Special 3 page
Answer · 30 × 0.275 × 0.65 ≈ 5.36 mV.
20 points · Grading: formula 10, value 10
Answer · 7 mV to reach threshold. 7 / 0.1575 ≈ 44.4 mV. With gain 1.0 that takes about 162 synapses at once; with gain 0.65, about 249.
20 points · Grading: computing 7 mV 10, value of g 10
Answer · The stage just before the motor output. The bitter brake (Scapula) blocks the feeding-command pathway.
20 points · Grading: "just before motor output" 20
Answer · Within 200 ms, spikes per tick multiply fivefold and Kenyon cells wake up at 39 Hz, and the resulting inhibition shuts off even the grooming command neurons (runaway).
20 points · Grading: runaway activity 10, grooming command shuts off 10
Answer · Learning only about 4% of the connections still gets a fair number right, but because it can't learn the representation itself, it scores lower than the MLP. Also mention that with so little data (32 test examples), the difference can wobble by ±0.1.
20 points · Grading: comparing the ratio 10, limits and uncertainty 10
Assignment · In Code 1, change the gain from 0.65 to 0.55 and 0.75, report the number of neurons that fired and the mean firing rate in a table, and interpret the results in two sentences, linking them to the runaway in the "Tuning the gain" section.
Sample answer · Lowering the gain (0.55) reduces the current each input adds, so the number of firing neurons and the mean firing rate drop compared with 0.65; raising it (0.75) increases them. Even in Code 1's small network, firing that spreads through feedback grows quickly as the gain rises. In the mod's whole male nervous system, this growth shows up as runaway activity (five times the spikes per tick, Kenyon cells at 39 Hz), which is why 0.65 was chosen.
Grading points · Results table for the three gains (8) · Explains the direction of increase (6) · Link to runaway (6)
Assignment · In Code 2's ConnectomeLIF, change every inhibitory neuron sign (−0.5) to +0.18 (removing inhibition) and retrain. Compare the spike raster (cell 8) and accuracy with two seeds, and describe what inhibition does in the network.
Sample answer · Without inhibition, most hidden neurons fire more often and more uniformly, so the spike raster becomes dense. Firing patterns differ less from input to input, leaving the readout layer less information to tell them apart, so accuracy drops or wobbles depending on the seed. Inhibition limits activity, making patterns sparse and distinguishable (the same role as the Kenyon cell top-8% rule in Special 1).
Grading points · Spike raster comparison (6) · Accuracy for two seeds (6) · Interpretation of inhibition's role (8)