# 5 weights on modified balance scale

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.

• By adjustable weight you mean one can modify the weights in an increment of 0.5? Sep 10, 2017 at 11:06
• @Netham You can change your weight for each weights. Any weight is possible since they are positive values.
– Oray
Sep 10, 2017 at 11:14
• In your recent update (second para), that problem has a limitation of fixed weights, but if you make them adjustable/dynamic then it will be possible. Am I missing something here? Sep 10, 2017 at 11:20
• @Netham You are going to choose 5 weights, then you are not going to change it :) I will fix that part, I see what you meant. just deleted the adjustable part, I hope it is clear now?
– Oray
Sep 10, 2017 at 11:22

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

• Solutions with X > 22 seem to be working out.
– humn
Sep 11, 2017 at 6:50
• i have found more than 22 too.
– Oray
Sep 11, 2017 at 9:34

I found a solution:

2, 5.5, 8.5, 9, 9.5.

It can be used to weight from 0 to 38. I can't prove it is the max, but i am pretty sure (used programming to find it).

Bonus: 38.5 cannot be measured with it, but 39, 39.5, 40 can, so 38.5 is the only one which can not be measured from 0-40.

• Great find, Nopalaa, I'm changing my answer to an analysis of these weights
– humn
Sep 11, 2017 at 15:02
• yes this is what i have found as well.
– Oray
Sep 11, 2017 at 15:16