This question is follow up version of What's the fewest weights you need to balance any weight from 0,5 to 40 pounds, on a modified balance scale with one arm twice as long as the other.
(Picture courtesy of Jamal Senjaya)
In the original question, the answer is 6 weights and lots of solutions are possible, it is somehow shown that it is not possible to get 5 weights from 0 to 40.
In this question, you have 5 weights only (any positive value for each weight is possible) to weigh from 0 to X pounds exactly with the 0.5 pound increment as the original question.
What is the maximum value of X? and you need to provide all 5 weights' weight with it.
Answer
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
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