Exercise 1: Three Siblings, Sorted Purely by z-index — Possible Solution
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THE TEST
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    s1 = block({'background-color': 'red',    'z-index': '2'})
    s2 = block({'background-color': 'green',  'z-index': '0'})
    s3 = block({'background-color': 'blue',   'z-index': '1'})
    container = block({}, children=[s1, s2, s3])   # source order: red, green, blue

    layout_block(container, root_cb)
    [cmd.color for cmd in build_display_list_zindex(container)]

RESULT
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    ['green', 'blue', 'red']

Sorted purely by z-index ascending (0, 1, 2) -- green (z=0) paints
first, blue (z=1) second, red (z=2) last (on top) -- completely
independent of the original source order (red, green, blue).

WHY THE RESULT IS SORTED BY Z-INDEX, NOT SOURCE POSITION
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build_display_list_zindex's own relevant line is:

    ordered_children = sorted(layout_box.children, key=get_z_index)

sorted() with a key function reorders the input list according to
whatever get_z_index() returns for each element -- it does NOT
preserve the original list's own order except among elements whose
key values are equal (Python's stable-sort guarantee). Here, all three
z-index values (2, 0, 1) are genuinely distinct, so there are no ties
to preserve -- the output order is determined ENTIRELY by the numeric
comparison of z-index values: 0 < 1 < 2, giving [s2, s3, s1] regardless
of the fact that s1 appeared first in the original children list.

WHY THIS WORKS AS AN ANSWER
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This confirms z-index is a genuine override of source order, not
merely a tie-breaker that only matters when source order is somehow
ambiguous. The chapter's own two-sibling example already demonstrated
this with two boxes; testing three, with z-index values deliberately
NOT matching either ascending or descending source order (2, 0, 1
rather than, say, 0, 1, 2 which could coincidentally match a "reverse
source order" pattern), confirms the sort genuinely keys off z-index
alone -- there's no hidden dependency on the boxes' own original
positions sneaking into the result.