Teacher answer key

Decisions and Flags

Gather four kinds of evidence: prediction of relevant state; behavior across varied cases including boundaries; explanation using trace/state evidence; and a reasoned correction of one mismatch. Mark each as emerging, with support or independent using paper or your school’s existing approved process. A challenge pass is machine-behavior evidence, not proof of authorship or complete mastery. Accept equivalent correct programs; the public key is a model, not the only possible answer.

Public teaching material. Solutions are examples; equivalent correct programs may also pass the actual checks.

10. Conditional Logic: Making Decisions

Compare values, then branch on a stored flag.

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1. Predict and trace

Before running the Try program, predict its output and trace the first three executed instructions. Track the relevant registers, flags or memory as needed. Then step the program to compare.

LOAD 7
CMP 7
JUMP_IF_ZERO same
LOAD 0
JUMP show
same: LOAD 1
show: PRINT
HALT

Queue input: None

Initialize memory: All bytes initially zero

Try prediction key

Output: 1

Instruction	PC before	PC after	R0 before	R0 after	Relevant state	Output
LOAD	0	1	0	7	`{"registers":[7,0,0,0,0,0,0,0],"flags":{"zero":false,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[]}`	None
CMP	1	2	7	7	`{"registers":[7,0,0,0,0,0,0,0],"flags":{"zero":true,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[]}`	None
JUMP_IF_ZERO	2	5	7	7	`{"registers":[7,0,0,0,0,0,0,0],"flags":{"zero":true,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[]}`	None

2. Build and check

Read two integers. Print 1 when they are equal, otherwise print 0. Handle both equal and unequal inputs.

Required instruction types: INPUT, CMP, JUMP_IF_ZERO

Example challenge solution

INPUT
COPY R0 R1
INPUT
CMP R1
JUMP_IF_ZERO same
LOAD 0
JUMP show
same: LOAD 1
show: PRINT
HALT

Actual checker fixtures

Case 1

Input: 5, 5
Initial memory: All bytes initially zero
Expected output: 1

Case 2

Input: 2, 3
Initial memory: All bytes initially zero
Expected output: 0

Case 3

Input: -2, -2
Initial memory: All bytes initially zero
Expected output: 1

3. Explain the machine

Can Z=true when R0 is nonzero?

Reasoning and teaching note

Yes: CMP 7 with R0=7 sets zero from the difference while R0 remains7.

11. Even or Odd

Use remainder and a branch to classify input.

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1. Predict and trace

Before running the Try program, predict its output and trace the first three executed instructions. Track the relevant registers, flags or memory as needed. Then step the program to compare.

LOAD -3
MOD 2
JUMP_IF_ZERO even
LOAD 0
JUMP show
even: LOAD 1
show: PRINT
HALT

Queue input: None

Initialize memory: All bytes initially zero

Try prediction key

Output: 0

Instruction	PC before	PC after	R0 before	R0 after	Relevant state	Output
LOAD	0	1	0	-3	`{"registers":[-3,0,0,0,0,0,0,0],"flags":{"zero":false,"negative":true,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[]}`	None
MOD	1	2	-3	-1	`{"registers":[-1,0,0,0,0,0,0,0],"flags":{"zero":false,"negative":true,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[]}`	None
JUMP_IF_ZERO	2	3	-1	-1	`{"registers":[-1,0,0,0,0,0,0,0],"flags":{"zero":false,"negative":true,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[]}`	None

2. Build and check

Read an integer. Print 1 for an even integer and 0 for an odd integer.

Required instruction types: INPUT, MOD, JUMP_IF_ZERO

Example challenge solution

INPUT
MOD 2
JUMP_IF_ZERO even
LOAD 0
JUMP show
even: LOAD 1
show: PRINT
HALT

Actual checker fixtures

Case 1

Input: -3
Initial memory: All bytes initially zero
Expected output: 0

Case 2

Input: 8
Initial memory: All bytes initially zero
Expected output: 1

Case 3

Input: 0
Initial memory: All bytes initially zero
Expected output: 1

Case 4

Input: 7
Initial memory: All bytes initially zero
Expected output: 0

3. Explain the machine

Is−3 odd even though its MOD 2 result is−1?

Reasoning and teaching note

Yes. Any nonzero signed remainder is odd; zero remainder identifies even values including0.

12. A Memory-backed Decision

Make a decision from stored data and record the result.

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1. Predict and trace

Before running the Try program, predict its output and trace the first three executed instructions. Track the relevant registers, flags or memory as needed. Then step the program to compare.

STORE 23 40
LOAD [40] R1
LOAD 23
CMP R1
JUMP_IF_ZERO match
LOAD 0
JUMP record
match: LOAD 1
record: STORE 41
PRINT
HALT

Queue input: None

Initialize memory: 40: 23

Try prediction key

Output: 1

Instruction	PC before	PC after	R0 after	Relevant state	Output
STORE	0	1	0	`{"registers":[0,0,0,0,0,0,0,0],"flags":{"zero":false,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[40]}`	None
LOAD	1	2	0	`{"registers":[0,23,0,0,0,0,0,0],"flags":{"zero":false,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[40],"memoryWrites":[]}`	None
LOAD	2	3	23	`{"registers":[23,23,0,0,0,0,0,0],"flags":{"zero":false,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[]}`	None

2. Build and check

Memory 40 holds a supplied ticket code. Read one guess, then print and store at address 41 a 1 for a match or a 0 otherwise.

Required instruction types: INPUT, CMP, STORE, JUMP_IF_ZERO

Example challenge solution

LOAD [40] R1
INPUT
CMP R1
JUMP_IF_ZERO match
LOAD 0
JUMP record
match: LOAD 1
record: STORE 41
PRINT
HALT

Actual checker fixtures

Case 1

Input: 23
Initial memory: 40: 23
Expected output: 1
Expected final memory: 40: 23, 41: 1
Required memory reads: 40

Case 2

Input: 22
Initial memory: 40: 23
Expected output: 0
Expected final memory: 40: 23, 41: 0
Required memory reads: 40

Case 3

Input: 0
Initial memory: 40: 0
Expected output: 1
Expected final memory: 40: 0, 41: 1
Required memory reads: 40

3. Explain the machine

Why should the program still work when memory 40 changes?

Reasoning and teaching note

The decision rule reads stored state instead of embedding a fixed ticket code; record/display same1 or0 at41.