# Quiz with answers — KV cache, recurrent memory, MLA, and low rank ### Question 1 Solve the worked-case variant: Double context length from 4,096 to 8,192 tokens while keeping 32 layers, 8 KV heads, d_head=128, and BF16. Recompute the simplified raw cache. - A. With 32 layers, 8 KV heads, d_head=128, BF16, and 4,096 tokens, simplified raw cache is 4,096×32×2×8×128×2 bytes = 536,870,912 bytes = 512 MiB (0.5 GiB). Quartering latent width reduces the per-token term, not linear growth. - B. Cache grows from 536,870,912 bytes (512 MiB) to 1,073,741,824 bytes (1 GiB). Growth is linear in token count; MLA or recurrent state can reduce different dimensions but also changes fidelity or retrieval behavior. - C. Simplified formulas omit alignment, quantization, buffers, and sharing details. They compare trends, not promise real footprint. - D. The state S — 32 layers × 128 × 128 × BF16 = 1 MiB — summarizes the whole past: 4,096 or 524,288 tokens, still 1 MiB. Growth disappears; this is where the previous sessions’ memories were heading. **Answer: B.** Cache grows from 536,870,912 bytes (512 MiB) to 1,073,741,824 bytes (1 GiB). Growth is linear in token count; MLA or recurrent state can reduce different dimensions but also changes fidelity or retrieval behavior. The correct answer executes the requested change and gives a checkable result; the other texts do not close this calculation or trace. --- ### Question 2 Which causal order correctly connects the first three stages of “KV cache, recurrent memory, MLA, and low rank”? - A. MLA → Fixed recurrent state → Standard KV cache - B. Fixed recurrent state → Standard KV cache → MLA - C. Standard KV cache → Fixed recurrent state → MLA - D. Standard KV cache → MLA → Fixed recurrent state **Answer: C.** Standard KV cache → Fixed recurrent state → MLA The chain follows the taught progression; reversing stages consumes a representation or state before it is produced. --- ### Question 3 If “MLA” is removed, which diagnostic method is defensible? - A. Keep the same input, predict the first output that depends on “MLA,” then compare the before/after trace. - B. Also change the data to amplify the difference. - C. Observe only the final output and invent the cause. - D. Conclude that the whole system fails before measuring. **Answer: A.** Keep the same input, predict the first output that depends on “MLA,” then compare the before/after trace. One intervention and a prior prediction make the delta attributable to the removed mechanism. --- ### Question 4 Which verdict respects this session’s validity boundary? - A. The mechanism guarantees accuracy, speed, and stability for every workload. - B. One successful example proves the whole architecture is superior. - C. The mechanism name alone is enough for a production choice. - D. Simplified formulas omit alignment, quantization, buffers, and sharing details. They compare trends, not promise real footprint. **Answer: D.** Simplified formulas omit alignment, quantization, buffers, and sharing details. They compare trends, not promise real footprint. The correct answer bounds the conclusion; the others turn a local relation into a global guarantee. --- ### Question 5 Which evidence best matches the stated status of “KV cache, recurrent memory, MLA, and low rank”? - A. The route loads without an error. - B. Mixed: established mechanisms + source-reported Kimi K3-style choices. - C. Every learner opened the file. - D. The same result is assumed on every hardware target. **Answer: B.** Mixed: established mechanisms + source-reported Kimi K3-style choices. Product or mechanism evidence must remain attributed and measured; availability and completion do not prove value. --- ### Question 6 When should a simpler baseline be preferred to “MLA is not fixed memory”? - A. Never: the newest mechanism wins by default. - B. As soon as one memory metric falls, regardless of quality. - C. When a controlled test shows equivalent quality with lower memory, latency, or complexity. - D. As soon as the diagram contains fewer components. **Answer: C.** When a controlled test shows equivalent quality with lower memory, latency, or complexity. The choice depends on a measured trade-off on the real workload, not novelty or one isolated metric. --- ### Question 7 A learner gets the right result but cannot explain “Fixed recurrent state.” Which remediation is most useful? - A. Rebuild the first missing transformation, label its inputs and outputs, then test a neighboring case. - B. Accept the answer because the final number is correct. - C. Provide the final result a second time. - D. Change several variables and ask for an intuition. **Answer: A.** Rebuild the first missing transformation, label its inputs and outputs, then test a neighboring case. The remediation targets the first causal break and then requires transfer instead of rewarding a guessed result. --- ### Question 8 Which submission actually demonstrates the outcome “Distinguish architectural low rank from LoRA.”? - A. A list of terms without causal relations. - B. A screenshot without values or interpretation. - C. A confident claim without a baseline or threshold. - D. A trace with starting data, transformations, observed result, boundary, and next experiment. **Answer: D.** A trace with starting data, transformations, observed result, boundary, and next experiment. The correct submission makes the reasoning reproducible and the verdict revisable by future measurement.