This answer originally reported a hand-made X = 22 pounds solution but,
while I was finding better solutions and preparing an analysis,
Nopalaa’s program found a
likely-maximal solution
that can balance up to X = 38 pounds.
So this answer now applies my analysis to Nopalaa’s solution.
Here are Nopalaa’s weights
and the only two weighings that absolutely require
the mystery weight to sit on the long side of the scale,
where everything weighs double what it would on the short side.
2 5.5 8.5 9 9.5
Short side . Long side (multiply by 2)
.
Reference weights . Mystery weight Reference weight
.
8.5 + 9.5 . 3.5 + 5.5
.
2 + 5.5 + 9 + 9.5 . 13
.
. (Every other weighing can be made with
. the mystery weight on the short side.)
That mystery weights 3.5 and 13 require placing them
on the long side, while all other weighings do not,
is reflected in a tree diagram of weight combinations,
a portion of which is previewed here.
M is on X X
short side 'v'v'v'v'v'v'v'''v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'''v'v'v'v'v'v'v'v'v'
0 1 2 3.5 5 6 7 8 9 10 11 12 13 14 15 16 17
M is ^^^.^^^^^^^^^^^^^.^^^^^^^^..^^^^.^^^^^^^^^^....^^^.^^^.^..^^......^...^
on R = 2 /\/_\/\\/|\\|/||/_|||\///\__\|||_///|||\///____|||_///_|__\/______|___/
long = 5.5 _|__|____/__/_\\\_|||________///_\/_|||________///_____|__________/
side = 8.5 _|__\_____________/||_______________//|________________/
= 9 _|_________________/|_________________/
= 9.5 _|__________________/
0 || 1 space = 1/4 pound
The story behind the diagram begins with a table of six ways
that each reference weight R can contribute to
balancing mystery weight M.
Amount of mystery weight M balanced by reference weight R
R is on the R is R is on the
short side not used long side
M is on the short side - R 0 + 2R
M is on the long side + R/2 0 - R
The choices shown in this table
are diagrammed with a pair of decision trees that meet at a number line.
The top tree grows downward as it represents
the possibility that M is on the short side;
the bottom tree grows upward as it represents
the possibility that M is on the long side.
MAGNIFIED:
R is on same R is R is on other
Branches when (short) side not used (long) side
M is on the __________|_________________________
short side / | \
/ | \
Number line . . . -R . . . +0 . +R/2 . . . . . +2R . . . .
\ | /
Branches \__________|_____/
when M is on |
the long side R is on same R is R is on other
(long) side not used (short) side
UNMAGNIFIED:
M is on _____________|___________________________
short side / | \
v'''''''''''''v'''''''''''''''''''''''''''v
(e.g, R = 7) -7 0 +3.5 +14
^.............^......^
M is on \_____________|______/
long side | || 1 space = 1/4 pound
Here is the tree diagram for 2 reference
weights — .5 and
1 pound — that
can balance up to X = 2 pounds.
Note that M = 1, 1.5 or 2 pounds
may balanced only with M on the short side
while M = .5 requires M to sit on the long side.
M is on .5 1.5 (Only the leaves of the
short side v'v'v'v'''v'v'v'''v top tree are displayed
0 : 1 2 because that tree is
M is on ^.^^^.^^^^ a double-sized mirror
long side R = .5 \_|/\_\/|/ copy of the lower tree.)
R = 1 \___|_/
0 || 1 space = 1/4 pound
SUPER-MAGNIFIED,
WITH FULL TOP: 0
_____________|_____________________________
R = 1 / -1 | +2 \
/ _____|_____________ \
M is ______/_________/____ | +1 \ ______\______________
on R = .5 / -.5 | +1 / \| \ / -.5 | +1 \
short / | / \ \ / | \
side v ' v ' v ' v ' ' ' v ' v ' v ' ' ' v
0 .5 1 1.5 2 (3)
M is ^ . ^ ^ ^ . ^ ^ ^ ^
on \ -.5 | / \ \ /-.5| /
long R = .5 \______|__/ \ -.5 |\_/____|__/
side \ \_____|_/ /
\ -1 | +.5 /
R = 1 \_____________|_____/ || 1 space = 1/16 pound
|
0
Having these diagrams in mind made it easy
to manipulate the resultant number line alone,
without rendering full trees,
and easy enough to find a solution with X = 29 by hand.
With the above layouts as reference, though,
here instead are Nopalaa’s weights
balancing up to X = 38 pounds.
Much detail is obscured by overlapped branches
but it is clear that many irregular mystery-weight gaps
exist before reaching decision tree leaves.
(The top line displayed has the leaves of the top tree (M on short side),
identical to the bottom tree's leaves, but in reverse order and stretched
twice as wide. The bottom tree's streak of 1/4 weights from -19 to -6.75
corresponds to the top longest streak of 1/2 weights, from 13.5 to 38.)
M is on X X
short side v'''''''v'''''''''''''v'v'''''v'''v'v'v'''v'v'v'v'''''''v'v'v'v'v'v'v'v'v'v'''v'v'v'v'''''v'v'v'v'v'v'v'v'''v'v'v'v'v'v'v'v'v'v'v'v'v'''v'v'v'v'v'v'v'v'''v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'''v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'''v'v'v'v'v'''v'''''v'v'''v'''v'v'v'''v'v'''''v'''v'''v'v'''''v'''''''''''''''v'''''''''''''''''''v'''''''v'''''''''''''''v
-19 -6.75 0 1 2 3.5 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38
M is ^.......^...^.........^.......^..^^.^.^..^^.^^^.^.^^..^.^^^^^.^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.^^^^^^^^^^^^^^^^^^.^^^^^^^^.^^^^^^^^^^^^^.^^^^^^^^..^^^^.^^^^^^^^^^....^^^.^^^.^..^^......^...^
on R = 2 \_______|___/ \_______|__\/_\_\__||_|//_/ /\__\_\\\|\_|/|||\/\/\/\\|\||/||//\//\/\\/|\\|/||/|/|\/\/\/\\|_|\/||||\///\\/\|||_/|/|//\/_\/\\/|\\|/||/_|||\///\__\|||_///|||\///____|||_///_|__\/______|___/
long = 5.5 \_____________________|__________/\_\_\____________\__\_|\|_|______/_/\/\|\_|__|____/__\\\/\|\|_|______/_\_/\|||_|||____///_/\\|__|__|____/__/_\\\_|||________///_\/_|||________///_____|__________/
side = 8.5 \_________________________________|_\_\____________/__\__\____________|_|_\____________/|/_|\\___________/__|__\_____________/||_______________//|________________/
= 9 \___________________________________|_\_______________/__\________________|_________________/|_________________/
= 9.5 \_____________________________________|__________________/
0
(scroll sideways for most of diagram) || 1 space = 1/4 pound
