Nested Loops: A Repeat Inside a Repeat

Answer key

1. 1. It starts again and runs all of its repeats
   Each outer pass restarts the inner loop from the beginning. Running only once or only on the first pass forgets the restart, and carrying on from where it stopped is how a single long loop works, not a nested one.
2. 2. 12 times
   The outer loop makes 3 passes and each runs the inner loop 4 times: 3 × 4 = 12. Choosing 7 adds the repeat numbers, and choosing 3 or 4 ignores one of the loops.
3. 3. A B B B A B B B
   Each outer pass says A once and then B three times, and there are two passes. A A B B B B B B puts both A steps first, A B B B ignores the outer loop, and A B A B A B treats the inner loop as 1 repeat.
4. 4. 1 1 2 1 1 2
   In each outer pass the inner loop says 1 1 and then say 2 runs once. Two passes give 1 1 2 1 1 2. 1 1 2 is only one pass. 1 2 1 2 treats the inner loop as one repeat. 1 1 1 1 2 2 puts say 2 after both loops.
5. 5. 20 squares
   The F step runs 5 × 4 = 20 times, and each F moves one square. 9 comes from adding 5 + 4. Choosing 4 or 5 uses only one of the loops.
6. 6. 3 rows of 5 stars
   The inner loop draws 5 stars and next line ends the row, once per outer pass, so there are 3 rows of 5. 5 rows of 3 swaps the loops. One row of 8 adds the numbers, and one row of 15 leaves out the next line step, so the stars never break into rows.
7. 7. Repeat 2 [Repeat 4 [say *]; next line]
   The inner loop makes the 4 stars and next line in the outer loop ends each row. The first wrong option makes 4 rows of 2 stars. The second has no next line, so it makes one row of 8 stars. The third has next line inside the inner loop, so every star lands in its own row.
8. 8. 24
   The step total = total + 2 runs 4 × 3 = 12 times, so the total is 12 × 2 = 24. 14 adds the repeats first, (4 + 3) × 2. 12 forgets to add 2 each time, and 9 adds 4, 3 and 2.
9. 9. On G, facing left
   Pass 1: F F to the top right corner, then R to face down. Pass 2: F F down to G, then R to face left. One square above G is where the robot is after only one F of the second pass. Back on S would need four passes around a ring. On G, facing down forgets the last R, which still runs at the end of pass 2.
10. 10. Repeat 4 [Repeat 3 [F] R]
    Each side of the ring is 3 moves long and a ring has 4 sides, so the program is Repeat 4 [Repeat 3 [F] R]. Four forward moves in a row would leave the grid. Repeat 3 [Repeat 3 [F] R] stops after three sides, on the bottom left square, one side short of S.
11. 11. * / ** / ***
    On pass r the inner loop repeats r times. Pass 1 draws 1 star, pass 2 draws 2 and pass 3 draws 3. The option with 3 stars in every row uses a fixed inner count. The option that shrinks reverses the order, because r counts up.
12. 12. 10
    The rows have 1, 2, 3 and 4 stars: 1 + 2 + 3 + 4 = 10. 16 would be right only if every row had 4 stars. 14 is a slip when adding the row counts, and 4 is just the last row.
13. 13. 1 2 3 2 4 6
    While a is 1, b goes 1, 2, 3 and says 1 2 3. Then a becomes 2 and b starts again: 2 4 6. Putting 1 2 2 4 3 6 first holds b still and changes a, which swaps the loops. 1 2 3 4 5 6 keeps counting instead of multiplying, and 1 2 3 1 2 3 ignores a.
14. 14. Move total = 0 to before the outer Repeat
    The step total = 0 inside the outer loop wipes the count at the start of every pass, so only the last pass is left: 2. Setting total to 0 once, before both loops, keeps all 3 × 2 = 6 counts. The other fixes keep the reset: Repeat 3 inside would still say 3, total + 3 would say 3, and say total inside the inner loop would say 1 2 1 2 1 2.
15. 15. 12 times
    Three nested loops multiply all three numbers: 3 × 2 × 2 = 12. 7 adds them. 10 is 3 × 2 + 2 + 2, which multiplies only the first two loops. 6 uses only two of the loops.