The first challenge should be interpreted as follows:
There is a logistic robot (demo: https://miau.my-x.hu/miau/304/robotkar.MOV).
This robot is functioning - seemingly correctly.
Question: are we capable of deriving all of the rules behind the visible surface?
Testing scenarios are already prepared: https://miau.my-x.hu/miau/320/testing_task1.xlsx
Tasks: interpreting all test scenarios (XLSX: 1;...;12), understanding data/structures, deriving the rules of the black box system, defining substitution-characters/colours for the used pattern: ?(????) <-- detailed solutions for place-ID (A;B;C;D)
Detailed demo-solutions for Scenario#9 (in Experiment#1):
***
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=green
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
***
Basic question: Is it correct? If not, which solution-layer why not?
Parallel question: How should we store all solution layers in case of all Students and assumped that everybody may store unlimited guesses for each experiment, scenario, layer? (It means: each layer can have different number of guesses in case of a particular Student. / The expected database-structure should support the evaluation of the best Student based e.g. on pivot-tables/queries...)
Cell(E27)_old_all=?(????) <--Cell(E27)_new_ color =green
looks like should be:
Cell(E27)_old_all=?(????) <--Cell(E27)_new_ color =yellow
besides this, the solution seem to be correct overall. Deriving from: https://miau.my-x.hu/miau/320/testing_task1.xlsx
looks like should be:
Cell(E27)_old_all=?(????) <--Cell(E27)_new_ color =yellow
besides this, the solution seem to be correct overall. Deriving from: https://miau.my-x.hu/miau/320/testing_task1.xlsx
Question: are we capable of deriving all of the rules behind the visible surface?
Answer: not all of the rules from the visible surface. Only the fact that the robot seem to be picking up the "cube"s from first in line and then scanning(most likely the color) them using using the machine on the left(from the camera point) then stacking them on the left in color order.
Task:
EXPT.9
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_ color =yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_ color = blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_ color =red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_ color =not-used
***
Cell(E27)_old_all= ? (????) <--Cell(E27)_new_number=1
Cell(E28)_old_all= ? (????) <--Cell(E28)_new_number=2
Cell(E29)_old_all= ? (????) <--Cell(E29)_new_number=1
Cell(E30)_old_all= ? (????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?( ???? ) <--Cell(E27)_new_letters=(b)
Cell( E28)_old_all=?( ???? ) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?( ???? ) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?( ???? ) <--Cell(E30)_new_letters=not-used=()
***
EXPT.10
***
Cell(K27)_old_all=?(????) <--Cell(K27)_new_ color =green
Cell(K28)_old_all=?(????) <--Cell(K28)_new_ color = red
Cell(K29)_old_all=?(????) <--Cell(K29)_new_ color =yellow
Cell(K30)_old_all=?(????) <--Cell(K30)_new_ color =not-used
***
Cell(K27)_old_all= ? (????) <--Cell(K27)_new_number=2
Cell(K28)_old_all= ? (????) <--Cell(K28)_new_number=1
Cell(K29)_old_all= ? (????) <--Cell(K29)_new_number=1
Cell(K30)_old_all= ? (????) <--Cell(K30)_new_number=not-used=0
***
Cell(K27)_old_all=?( ???? ) <--Cell(K27)_new_letters=(ab)
Cell(K28)_old_all=?( ???? ) <--Cell(K28)_new_letters=(c)
Cell(K29)_old_all=?( ???? ) <--Cell(K29)_new_letters=(d)
Cell(K30)_old_all=?( ???? ) <--Cell(K30)_new_letters=not-used=()
***
EXPT.11
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_ color =red
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_ color = green
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_ color =yellow
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_ color =not-used
***
Cell(Q27)_old_all= ? (????) <--Cell(Q27)_new_number=1
Cell(Q28)_old_all= ? (????) <--Cell(Q28)_new_number=2
Cell(Q29)_old_all= ? (????) <--Cell(Q29)_new_number=1
Cell(Q30)_old_all= ? (????) <--Cell(Q30)_new_number=not-used=0
***
Cell(Q27)_old_all=?( ???? ) <--Cell(Q27)_new_letters=(b)
Cell(Q28)_old_all=?( ???? ) <--Cell(Q28)_new_letters=(ad)
Cell(Q29)_old_all=?( ???? ) <--Cell(Q29)_new_letters=(c)
Cell(Q30)_old_all=?( ???? ) <--Cell(Q30)_new_letters=not-used=()
***
EXPT.12
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_ color =yellow
Cell(W28)_old_all=?(????) <--Cell(W28)_new_ color = green
Cell(W29)_old_all=?(????) <--Cell(W29)_new_ color =blue
Cell(W30)_old_all=?(????) <--Cell(W30)_new_ color =not-used
***
Cell(W27)_old_all= ? (????) <--Cell(W27)_new_number=1
Cell(W28)_old_all= ? (????) <--Cell(W28)_new_number=1
Cell(W29)_old_all= ? (????) <--Cell(W29)_new_number=2
Cell(W30)_old_all= ? (????) <--Cell(W30)_new_number=not-used=0
***
Cell(W27)_old_all=?( ???? ) <--Cell(W27)_new_letters=(a)
Cell(W28)_old_all=?( ???? ) <--Cell(W28)_new_letters=(b)
Cell(W29)_old_all=?( ???? ) <--Cell(W29)_new_letters=(cd)
Cell(W30)_old_all=?( ???? ) <--Cell(W30)_new_letters=not-used=()
***
Answer: not all of the rules from the visible surface. Only the fact that the robot seem to be picking up the "cube"s from first in line and then scanning(most likely the color) them using using the machine on the left(from the camera point) then stacking them on the left in color order.
Task:
EXPT.9
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_ color =yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_ color = blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_ color =red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_ color =not-used
***
Cell(E27)_old_all= ? (????) <--Cell(E27)_new_number=1
Cell(E28)_old_all= ? (????) <--Cell(E28)_new_number=2
Cell(E29)_old_all= ? (????) <--Cell(E29)_new_number=1
Cell(E30)_old_all= ? (????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?( ???? ) <--Cell(E27)_new_letters=(b)
Cell( E28)_old_all=?( ???? ) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?( ???? ) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?( ???? ) <--Cell(E30)_new_letters=not-used=()
***
EXPT.10
***
Cell(K27)_old_all=?(????) <--Cell(K27)_new_ color =green
Cell(K28)_old_all=?(????) <--Cell(K28)_new_ color = red
Cell(K29)_old_all=?(????) <--Cell(K29)_new_ color =yellow
Cell(K30)_old_all=?(????) <--Cell(K30)_new_ color =not-used
***
Cell(K27)_old_all= ? (????) <--Cell(K27)_new_number=2
Cell(K28)_old_all= ? (????) <--Cell(K28)_new_number=1
Cell(K29)_old_all= ? (????) <--Cell(K29)_new_number=1
Cell(K30)_old_all= ? (????) <--Cell(K30)_new_number=not-used=0
***
Cell(K27)_old_all=?( ???? ) <--Cell(K27)_new_letters=(ab)
Cell(K28)_old_all=?( ???? ) <--Cell(K28)_new_letters=(c)
Cell(K29)_old_all=?( ???? ) <--Cell(K29)_new_letters=(d)
Cell(K30)_old_all=?( ???? ) <--Cell(K30)_new_letters=not-used=()
***
EXPT.11
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_ color =red
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_ color = green
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_ color =yellow
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_ color =not-used
***
Cell(Q27)_old_all= ? (????) <--Cell(Q27)_new_number=1
Cell(Q28)_old_all= ? (????) <--Cell(Q28)_new_number=2
Cell(Q29)_old_all= ? (????) <--Cell(Q29)_new_number=1
Cell(Q30)_old_all= ? (????) <--Cell(Q30)_new_number=not-used=0
***
Cell(Q27)_old_all=?( ???? ) <--Cell(Q27)_new_letters=(b)
Cell(Q28)_old_all=?( ???? ) <--Cell(Q28)_new_letters=(ad)
Cell(Q29)_old_all=?( ???? ) <--Cell(Q29)_new_letters=(c)
Cell(Q30)_old_all=?( ???? ) <--Cell(Q30)_new_letters=not-used=()
***
EXPT.12
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_ color =yellow
Cell(W28)_old_all=?(????) <--Cell(W28)_new_ color = green
Cell(W29)_old_all=?(????) <--Cell(W29)_new_ color =blue
Cell(W30)_old_all=?(????) <--Cell(W30)_new_ color =not-used
***
Cell(W27)_old_all= ? (????) <--Cell(W27)_new_number=1
Cell(W28)_old_all= ? (????) <--Cell(W28)_new_number=1
Cell(W29)_old_all= ? (????) <--Cell(W29)_new_number=2
Cell(W30)_old_all= ? (????) <--Cell(W30)_new_number=not-used=0
***
Cell(W27)_old_all=?( ???? ) <--Cell(W27)_new_letters=(a)
Cell(W28)_old_all=?( ???? ) <--Cell(W28)_new_letters=(b)
Cell(W29)_old_all=?( ???? ) <--Cell(W29)_new_letters=(cd)
Cell(W30)_old_all=?( ???? ) <--Cell(W30)_new_letters=not-used=()
***
Argumentation = there is no green cube in the set... :-)
There is a lot of potential errors in the solutions for #10-11-12 (c.f. https://miau.my-x.hu/miau/320/testing_task1_guesses.xlsx). Please, derive: which scenarios are pros and which scenarios are cons for each particular solution-layer? Please, try to answer layer-by-layer! (e.g. EXPT.10***Cell(K27)_old_all=?(????) <--Cell(K27)_new_ color =green<--pros? & cons?)
Each experiment (EXPT.9, EXPT.10, etc.) contains multiple guesses for different cell references (E27, K27, etc.), with three key attributes:
New Color (e.g., yellow, blue, red)
New Number (e.g., 1, 2, 0)
New Letters (e.g., (a), (b), etc.)
Data entry(EXPT.9):
INSERT INTO Guesses (student_id, experiment_id, layer_id, cell_reference, old_value, new_color, new_number, new_letters, timestamp)
VALUES
(1, 9, 1, 'E27', '?(????)', 'yellow', 1, '(b)', CURRENT_TIMESTAMP),
(1, 9, 1, 'E28', '?(????)', 'blue', 2, '(ad)', CURRENT_TIMESTAMP),
(1, 9, 1, 'E29', '?(????)', 'red', 1, '(c)', CURRENT_TIMESTAMP),
(1, 9, 1, 'E30', '?(????)', 'not-used', 0, '()', CURRENT_TIMESTAMP);
Query to find the "best student"
SELECT student_id, COUNT(*) AS correct_guesses
FROM Evaluations
WHERE correct = 1
GROUP BY student_id
ORDER BY correct_guesses DESC;
retrieving guesses from other experiments:
SELECT * FROM Guesses WHERE experiment_id = 10;(11; etc.)
New Color (e.g., yellow, blue, red)
New Number (e.g., 1, 2, 0)
New Letters (e.g., (a), (b), etc.)
Data entry(EXPT.9):
INSERT INTO Guesses (student_id, experiment_id, layer_id, cell_reference, old_value, new_color, new_number, new_letters, timestamp)
VALUES
(1, 9, 1, 'E27', '?(????)', 'yellow', 1, '(b)', CURRENT_TIMESTAMP),
(1, 9, 1, 'E28', '?(????)', 'blue', 2, '(ad)', CURRENT_TIMESTAMP),
(1, 9, 1, 'E29', '?(????)', 'red', 1, '(c)', CURRENT_TIMESTAMP),
(1, 9, 1, 'E30', '?(????)', 'not-used', 0, '()', CURRENT_TIMESTAMP);
Query to find the "best student"
SELECT student_id, COUNT(*) AS correct_guesses
FROM Evaluations
WHERE correct = 1
GROUP BY student_id
ORDER BY correct_guesses DESC;
retrieving guesses from other experiments:
SELECT * FROM Guesses WHERE experiment_id = 10;(11; etc.)
It would be nice to see an Excel-demo with at least one appropriate pivot-output to demonstrate the power of the suggested (seemingly robust) data-structure...
Colour Layer: b is mapped to green while there is no green
Number Layer: Seems valid if numbers indicate stacking height.
Letters Layer: No, d or a appears in the layer
Which solution-layer is incorrect and why?
Colour Layer: unused color green and yellow
Letters Layer: Inconsistent with the number layer.
Correct solution looks something like this:
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(a)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=(d)
***
Number Layer: Seems valid if numbers indicate stacking height.
Letters Layer: No, d or a appears in the layer
Which solution-layer is incorrect and why?
Colour Layer: unused color green and yellow
Letters Layer: Inconsistent with the number layer.
Correct solution looks something like this:
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(a)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=(d)
***
Answer is: Not correct, why?
Because
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=green, this is the error, because it should be yellow instead of green.
Correct solution looks like this:
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
***
Because
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=green, this is the error, because it should be yellow instead of green.
Correct solution looks like this:
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
***
If we assume that every student can submit unlimited guesses for each experiment, scenario, and solution layer, we need a structured and organized way to store this data. The system should be capable of handling:
1. Multiple students working on the same problem.
2. Multiple attempts per student for each scenario.
3. Different layers of solutions as students refine their understanding.
4. Comparisons between correct and incorrect answers to track learning progress.
Proposed Storage Structure
We can use a database table to store the solutions efficiently. Here is a suggested schema:
Column Name (Description)
Student_ID (Identifies the student who submitted the answer.)
Experiment_ID (The experiment number (e.g., #9, #10, #11, #12).)
Layer_ID (Represents the version of a student’s solution for a given experiment.)
Timestamp (The exact date and time of submission (helps track progress).)
Input_Data (The initial setup of cubes/colors (e.g., a, b, c, d with specific colors).)
Generated_Output_Data (The student's proposed output (e.g., "2(ab), 1(c), 1(d)").)
Correctness_Status (Whether the answer is correct, partially correct, or incorrect.)
Teacher_Feedback (Optional field where teachers can give hints or explanations.)
1. Multiple students working on the same problem.
2. Multiple attempts per student for each scenario.
3. Different layers of solutions as students refine their understanding.
4. Comparisons between correct and incorrect answers to track learning progress.
Proposed Storage Structure
We can use a database table to store the solutions efficiently. Here is a suggested schema:
Column Name (Description)
Student_ID (Identifies the student who submitted the answer.)
Experiment_ID (The experiment number (e.g., #9, #10, #11, #12).)
Layer_ID (Represents the version of a student’s solution for a given experiment.)
Timestamp (The exact date and time of submission (helps track progress).)
Input_Data (The initial setup of cubes/colors (e.g., a, b, c, d with specific colors).)
Generated_Output_Data (The student's proposed output (e.g., "2(ab), 1(c), 1(d)").)
Correctness_Status (Whether the answer is correct, partially correct, or incorrect.)
Teacher_Feedback (Optional field where teachers can give hints or explanations.)
How This Structure Helps?
1. Allows unlimited guesses – Since each attempt is stored with a unique Layer_ID, students can refine their solutions without overwriting previous attempts.
2. Supports collaboration – If students compare their Generated_Output_Data with peers, they can learn from each other’s mistakes.
3. Provides a learning timeline – With timestamps, we can track whether students improve over time.
4. Facilitates automated evaluation – The system can automatically compare Generated_Output_Data with correct patterns and mark correctness (Correct, Incorrect, Partial).
5. Helps teachers analyze common mistakes – If many students make the same error, the teacher can adjust explanations accordingly.
It would also be nice to see an Excel-demo with at least one appropriate pivot-output to demonstrate the power of the suggested (seemingly robust) data-structure... (Remark: In a final thesis, if texts are integrated into the own documents, but they come from conversations with the ChatGPT/Copilot/etc. - it is always necessary to use quotation signs and the source must also be defined. A plagiat-problem is the last, what somebody do need...
If this solution (for scenario #9) will be accepted as a fact (similar to the scenarios #1-2-3-4-5-6-7-8, are we already prepared to solve the problems presented in scenarios #10-11-12?
Please, try to declare potential rules based on the TESTING_TASK1.XLSX!
Rule#1: Based on the scenarios #4-5-6-7-8: green towers of cubes will consequently be built on the place-id "A".
Rule#i: (your turns in new entries below)
(Based on a lot of these partial rules, we have to derive the "hermeneutic trap" created for your in this task!)
(This task is a kind of magic performance, where your mind will be influenced to think/to see in an irrational way...)
(You have the necessary details to be capable of interpreting the entire system correctly, but one disturbing impulse is present - even in multiply copies...)
Rule#1: Based on the scenarios #4-5-6-7-8: green towers of cubes will consequently be built on the place-id "A".
Rule#i: (your turns in new entries below)
(Based on a lot of these partial rules, we have to derive the "hermeneutic trap" created for your in this task!)
(This task is a kind of magic performance, where your mind will be influenced to think/to see in an irrational way...)
(You have the necessary details to be capable of interpreting the entire system correctly, but one disturbing impulse is present - even in multiply copies...)
In the excel file, yellow and green are always in the first place, maybe it's not a coincidence?
maybe they are related in some way, because then how could you decide what the order will be in #10 #11 #12?
yellow and green never met before the #10.
In addition, it was written "accidentally" that there is green instead of yellow.
Anyway, my first answer to #9 is what I wrote, but it's going to be more complicated than we think.
The rules are:
The system is a "Black Box" that produces an output according to specific rules
The columns contain different input values (Inputx) (e.g. a, b, c, d).
The outputs are organized according to certain rules, arranging the colored cubes in towers. (Green and yellow in 1st place, blue in 2nd place and finally red in 3rd place)
Their shape looks like this ?(????), the falling question mark is replaced by how many pieces of the given color are found (if there are more pieces, then by definition there will not be 1 but more than 1, this can also be seen in the robot video, because it stacks the cubes of the same color on top of each other. The question marks in the brackets indicate which letters are assigned to the given color.
And the task is to answer #9 #10 #11 #12 and write something instead of the many question marks.
The system is a "Black Box" that produces an output according to specific rules
The columns contain different input values (Inputx) (e.g. a, b, c, d).
The outputs are organized according to certain rules, arranging the colored cubes in towers. (Green and yellow in 1st place, blue in 2nd place and finally red in 3rd place)
Their shape looks like this ?(????), the falling question mark is replaced by how many pieces of the given color are found (if there are more pieces, then by definition there will not be 1 but more than 1, this can also be seen in the robot video, because it stacks the cubes of the same color on top of each other. The question marks in the brackets indicate which letters are assigned to the given color.
And the task is to answer #9 #10 #11 #12 and write something instead of the many question marks.
Excellent interpretations!
Recommendation for this task: the rules should be formulated as simple as possible! Long text streams do not really have a clear structure. The rules are correct formulated, if they can be transformed into source codes by all of you without any comlications (cf. Knuth)...
The general requirement of the black box system in this case is to sort colored blocks into 4 slots (A, B, C, D). Each slot should hold blocks of one color only.
The Problem is happening in scenarios #4, #9–12; the system makes a mistake: it assigns the color green to slot A in the current logic. However, slot A was already assigned to the color yellow in scenario#1
This mix-up (green and yellow in slot A) breaks the rule of “one color per slot.” The system can’t stack blocks correctly because of this conflict.
Example:
In Experiment #2 (Scenario #9), the system mistakenly places a yellow block in slot A. However, in Scenario #5, a green block is placed in the same slot A. This inconsistency causes multiple colors to be stacked in one column, leading to sorting errors.
>>>>The detailed solution for scenarios #9- 12 and the corrected logic on experiment#1 are in the attached EXCEL file.
The issue is that two different colors are being assigned to a single placeholder, which indicates a flaw in the system's logic. This may be due to outdated internal software or a malfunctioning sensor. Additionally, the color recognition algorithm might not be working properly.
Further possible experiments:
By adding more colors and placeholders to give more randomized inputs to test the performance with the correct logic.
Testing color recognition algorithm by real-time monitoring.
Instead of RGB, we can use HSV for multi-dimensional color analysis. In real-life scenarios, it is crucial to identify items without any mistakes In any condition.
Excellent focus points can be found in the above-presented interpretation. BUT the offered xlsx-version is still in the hermeneutical trap it means, the solution can not be accepted. The above-listed conclusions are more complex, than the reality is. :-)
Detailed argumentation: GREEN towers may not built everywher, only on the slot/place-id "A": the initial scenarios (#1-2-3-4-5-6-7-8) present clear examples: green cubes must be to slot "A". We do not have other "instruction" and the instructions (rules) MUST BE FOLLOWED: a software do make each step based on predefinied rules. We are searching for such an interpretation for the black-box-system, where EACH previous rule is valid for ever. It is forbidden to create/assume hidden rules, which are given, but the impacts of these hidden rules could still not be observed. More observations/scenarios are not needed to interpret the black box system as a very-very-simple white-box-system. We have however to ignore the very impulse leading to the hermeneutical trapping... The interpretation above did already highlight this sensitive point... :-)
Detailed argumentation: GREEN towers may not built everywher, only on the slot/place-id "A": the initial scenarios (#1-2-3-4-5-6-7-8) present clear examples: green cubes must be to slot "A". We do not have other "instruction" and the instructions (rules) MUST BE FOLLOWED: a software do make each step based on predefinied rules. We are searching for such an interpretation for the black-box-system, where EACH previous rule is valid for ever. It is forbidden to create/assume hidden rules, which are given, but the impacts of these hidden rules could still not be observed. More observations/scenarios are not needed to interpret the black box system as a very-very-simple white-box-system. We have however to ignore the very impulse leading to the hermeneutical trapping... The interpretation above did already highlight this sensitive point... :-)
Sub-task helping to focus correctly: Which sentence is the most relevant?
here: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p164988
here: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p164988
Sub-task helping to focus correctly: Which rules (coming from scenarios #1-2-3-4-5-6-7-8 are definitely not followed? here in case of scenarios #10-11-12: https://moodle.kodolanyi.hu/pluginfile.php/444774/mod_forum/attachment/164988/testing_task1_Shagai.xlsx?forcedownload=1
Basic question answer: Not correct.
WHY?
In Experiment#1, Scenario#9, there is not any green colour in "Input9" column, which makes first solution layer wrong (Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=green).
The correct answer is
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
***
WHY?
In Experiment#1, Scenario#9, there is not any green colour in "Input9" column, which makes first solution layer wrong (Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=green).
The correct answer is
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
***
Basic question answer is incorrect.
why? there is no green.
Experiment#1 Scenario#9
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
Scenario#10
Cell(K27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(K28)_old_all=?(????) <--Cell(E28)_new_colour=green
Cell(K29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(K30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(K27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(K28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(K29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(K30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(K27)_old_all=?(????) <--Cell(E27)_new_letters=(d)
Cell(K28)_old_all=?(????) <--Cell(E28)_new_letters=(ab)
Cell(K29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(K30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
Scenario#11
Cell(Q27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(Q28)_old_all=?(????) <--Cell(E28)_new_colour=green
Cell(Q29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(Q30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(Q27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(Q28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(Q29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(Q30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(Q27)_old_all=?(????) <--Cell(E27)_new_letters=(c)
Cell(Q28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(Q29)_old_all=?(????) <--Cell(E29)_new_letters=(b)
Cell(Q30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
Scenario#12
Cell(W27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(W28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(W29)_old_all=?(????) <--Cell(E29)_new_colour=green
Cell(W30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(W27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(W28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(W29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(W30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(W27)_old_all=?(????) <--Cell(E27)_new_letters=(a)
Cell(W28)_old_all=?(????) <--Cell(E28)_new_letters=(cd)
Cell(W29)_old_all=?(????) <--Cell(E29)_new_letters=(b)
Cell(W30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
why? there is no green.
Experiment#1 Scenario#9
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
Scenario#10
Cell(K27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(K28)_old_all=?(????) <--Cell(E28)_new_colour=green
Cell(K29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(K30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(K27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(K28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(K29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(K30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(K27)_old_all=?(????) <--Cell(E27)_new_letters=(d)
Cell(K28)_old_all=?(????) <--Cell(E28)_new_letters=(ab)
Cell(K29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(K30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
Scenario#11
Cell(Q27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(Q28)_old_all=?(????) <--Cell(E28)_new_colour=green
Cell(Q29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(Q30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(Q27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(Q28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(Q29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(Q30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(Q27)_old_all=?(????) <--Cell(E27)_new_letters=(c)
Cell(Q28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(Q29)_old_all=?(????) <--Cell(E29)_new_letters=(b)
Cell(Q30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
Scenario#12
Cell(W27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(W28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(W29)_old_all=?(????) <--Cell(E29)_new_colour=green
Cell(W30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(W27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(W28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(W29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(W30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(W27)_old_all=?(????) <--Cell(E27)_new_letters=(a)
Cell(W28)_old_all=?(????) <--Cell(E28)_new_letters=(cd)
Cell(W29)_old_all=?(????) <--Cell(E29)_new_letters=(b)
Cell(W30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
Not correct!
In Scenario #9 (Experiment #1), the input is:
- a = blue
- b = yellow
- c = red
- d = blue
Since there is no green cube in the input, there is no doubt that green should not appear as the first output in this Scenario. In other words, with no green present, the only possible color for the first output (cell E27) is yellow this time.
The correct solution is:
Color Layer:
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
Number Layer:
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
Letters Layer:
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
Is this scenario (#9) with the above-mentioned correct solution a relevant information unit for the further cases? (scenarios #10-11-12)
Unfortunately, these solutions for the scenarios #10-11-12 are not acceptable: e.g. because of the 5-times declared rule (see scenarios #4-5-6-7-8): green cubes must definitely be placed to spot "A"! Each rule must be followed in any rate...
Is this correct-interpreted scenario (#9) helpful for the further (unsolved) scenarios (see #10-11-12)?
Answer for basic question: WRONG
There is no green, instead we have yellow, blue and red colors. According to the other scenarios, when there is no green, yellow takes place A, in this case Cell(E27). So, correct answer would looks like:
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(a)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=(d)
There is no green, instead we have yellow, blue and red colors. According to the other scenarios, when there is no green, yellow takes place A, in this case Cell(E27). So, correct answer would looks like:
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(a)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=(d)
IF green is exclusive to ID A. Therefore, yellow cannot appear on ID A.
If the first test/scenario
If the first test/scenario
shows yellow on ID A, it must be an error or an exception that needs to be corrected.
By replacing yellow with green on ID A and assigning yellow to ID D, we ensure that all rules are followed.
By replacing yellow with green on ID A and assigning yellow to ID D, we ensure that all rules are followed.
Rules based on observation on the TESTING_TASK1.XLSX and logistic robot's demo video:
Rule#1: The robot will pick one cube at a time.
Rule#2: The robot will only pick the cubes in the given order.
Rule#3: Based on the scenarios, GREEN cubes built on place-id "A", BLUE cubes on place-id "B" and RED cubes build on place-id "C".
Rule#4: Based on the scenario #1 and scenario #9 of the experiment #2 tables YELLOW cubes will build a tower on the place-id "A".
Rule#1: The robot will pick one cube at a time.
Rule#2: The robot will only pick the cubes in the given order.
Rule#3: Based on the scenarios, GREEN cubes built on place-id "A", BLUE cubes on place-id "B" and RED cubes build on place-id "C".
Rule#4: Based on the scenario #1 and scenario #9 of the experiment #2 tables YELLOW cubes will build a tower on the place-id "A".
In the rules 4 to 8 scenario, GREEN cube must be placed to spot "A". BLUE cube must be placed to spot "B", RED cube must be placed to spot "C" according to the rule 2 to 8. Each spot already occupied by the cubes except YELLOW cube. Logically, A for GREEN, B for BLUE, C for RED and there is one spot that empty. However, The rule of scenario 1, YELLOW cubes must be on spot "A", if YELLOW cube meet GREEN cube, Rule 1 cannot be followed due to Rule 4-8 scenario. So, if GREEN cube and YELLOW cubes matched, GREEN one MUST be on spot "A" and YELLOW one cannot be on neither "B" nor "C" spots, because of Rule 2-8 scenario they are already occupied by BLUE and RED cubes. Finally there is single spot that empty is "D" and we can put YELLOW cube to the spot "D".
Experiment#1, Scenario#10,11,12 answers according to the rules.
Experiment#1 Scenario#10
Cell(K27)_old_all=?(????) <--Cell(K27)_new_colour=green
Cell(K28)_old_all=?(????) <--Cell(K28)_new_colour=not-used
Cell(K29)_old_all=?(????) <--Cell(K29)_new_colour=red
Cell(K30)_old_all=?(????) <--Cell(K30)_new_colour=yellow
***
Cell(K27)_old_all=?(????) <--Cell(K27)_new_number=2
Cell(K28)_old_all=?(????) <--Cell(K28)_new_number=not-used=0
Cell(K29)_old_all=?(????) <--Cell(K29)_new_number=1
Cell(K30)_old_all=?(????) <--Cell(K30)_new_number=1
***
Cell(K27)_old_all=?(????) <--Cell(K27)_new_letters=(ab)
Cell(K28)_old_all=?(????) <--Cell(K28)_new_letters=not-used=()
Cell(K29)_old_all=?(????) <--Cell(K29)_new_letters=(c)
Cell(K30)_old_all=?(????) <--Cell(K30)_new_letters=(d)
***
Scenario#11
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_colour=green
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_colour=not-used
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_colour=red
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_colour=yellow
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_number=2
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_number=not-used=0
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_number=1
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_number=1
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_letters=(ad)
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_letters=not-used=()
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_letters=(b)
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_letters=(c)
***
Scenario#12
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_colour=green
Cell(W28)_old_all=?(????) <--Cell(W28)_new_colour=blue
Cell(W29)_old_all=?(????) <--Cell(W29)_new_colour=not-used
Cell(W30)_old_all=?(????) <--Cell(W30)_new_colour=yellow
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_number=1
Cell(W28)_old_all=?(????) <--Cell(W28)_new_number=2
Cell(W29)_old_all=?(????) <--Cell(W29)_new_number=not-used=0
Cell(W30)_old_all=?(????) <--Cell(W30)_new_number=1
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_letters=(b)
Cell(W28)_old_all=?(????) <--Cell(W28)_new_letters=(cd)
Cell(W29)_old_all=?(????) <--Cell(W29)_new_letters=not-used=()
Cell(W30)_old_all=?(????) <--Cell(W30)_new_letters=(a)
***
Experiment#1, Scenario#10,11,12 answers according to the rules.
Experiment#1 Scenario#10
Cell(K27)_old_all=?(????) <--Cell(K27)_new_colour=green
Cell(K28)_old_all=?(????) <--Cell(K28)_new_colour=not-used
Cell(K29)_old_all=?(????) <--Cell(K29)_new_colour=red
Cell(K30)_old_all=?(????) <--Cell(K30)_new_colour=yellow
***
Cell(K27)_old_all=?(????) <--Cell(K27)_new_number=2
Cell(K28)_old_all=?(????) <--Cell(K28)_new_number=not-used=0
Cell(K29)_old_all=?(????) <--Cell(K29)_new_number=1
Cell(K30)_old_all=?(????) <--Cell(K30)_new_number=1
***
Cell(K27)_old_all=?(????) <--Cell(K27)_new_letters=(ab)
Cell(K28)_old_all=?(????) <--Cell(K28)_new_letters=not-used=()
Cell(K29)_old_all=?(????) <--Cell(K29)_new_letters=(c)
Cell(K30)_old_all=?(????) <--Cell(K30)_new_letters=(d)
***
Scenario#11
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_colour=green
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_colour=not-used
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_colour=red
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_colour=yellow
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_number=2
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_number=not-used=0
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_number=1
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_number=1
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_letters=(ad)
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_letters=not-used=()
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_letters=(b)
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_letters=(c)
***
Scenario#12
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_colour=green
Cell(W28)_old_all=?(????) <--Cell(W28)_new_colour=blue
Cell(W29)_old_all=?(????) <--Cell(W29)_new_colour=not-used
Cell(W30)_old_all=?(????) <--Cell(W30)_new_colour=yellow
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_number=1
Cell(W28)_old_all=?(????) <--Cell(W28)_new_number=2
Cell(W29)_old_all=?(????) <--Cell(W29)_new_number=not-used=0
Cell(W30)_old_all=?(????) <--Cell(W30)_new_number=1
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_letters=(b)
Cell(W28)_old_all=?(????) <--Cell(W28)_new_letters=(cd)
Cell(W29)_old_all=?(????) <--Cell(W29)_new_letters=not-used=()
Cell(W30)_old_all=?(????) <--Cell(W30)_new_letters=(a)
***
Green is exclusive to ID A for smaller towers (1-3 cubes).
Yellow is exclusive to ID A for larger towers (4 cubes) when all inputs are the same color.
This means:
If the input cubes are different colors, the system builds smaller towers, and green is used for ID A.
If the input cubes are the same color, the system builds larger towers, and yellow is used for ID A.
Since IDs can only have one colored tower at a time, the system must choose between green and yellow based on the input colors.
If the inputs are different colors, use green for ID A.
If the inputs are the same color, use yellow for ID A.
Or yellow is on ID A if there no green in input :-)
Yellow is exclusive to ID A for larger towers (4 cubes) when all inputs are the same color.
This means:
If the input cubes are different colors, the system builds smaller towers, and green is used for ID A.
If the input cubes are the same color, the system builds larger towers, and yellow is used for ID A.
Since IDs can only have one colored tower at a time, the system must choose between green and yellow based on the input colors.
If the inputs are different colors, use green for ID A.
If the inputs are the same color, use yellow for ID A.
Or yellow is on ID A if there no green in input :-)
Based on the above rules that I have drawn from my observations, I solved the scenarios #10, #11 and #12
So, If yellow and green must be placed to spot "A" for example experiment#1 scenario1-8 and experiment#2 scenario 1-8, So both of them must be placed to spot "A". Can we stack green and yellow on spot "A", according to the rules it does not break any rules, there is not rules about 2 different colors cannot stack on each other.
Scenario#10
Cell(K27)_old_all=?(????) <--Cell(E27)_new_colour=green,yellow
Cell(K28)_old_all=?(????) <--Cell(E28)_new_colour=not-used
Cell(K29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(K30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(K27)_old_all=?(????) <--Cell(E27)_new_number=3
Cell(K28)_old_all=?(????) <--Cell(E28)_new_number=not-used=0
Cell(K29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(K30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(K27)_old_all=?(????) <--Cell(E27)_new_letters=(acd)
Cell(K28)_old_all=?(????) <--Cell(E28)_new_letters=not-used=()
Cell(K29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(K30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
Scenario#11
Cell(Q27)_old_all=?(????) <--Cell(E27)_new_colour=green,yellow
Cell(Q28)_old_all=?(????) <--Cell(E28)_new_colour=not-used
Cell(Q29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(Q30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(Q27)_old_all=?(????) <--Cell(E27)_new_number=3
Cell(Q28)_old_all=?(????) <--Cell(E28)_new_number=not-used=0
Cell(Q29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(Q30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(Q27)_old_all=?(????) <--Cell(E27)_new_letters=(acd)
Cell(Q28)_old_all=?(????) <--Cell(E28)_new_letters=not-used=()
Cell(Q29)_old_all=?(????) <--Cell(E29)_new_letters=(b)
Cell(Q30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
Scenario#12
Cell(W27)_old_all=?(????) <--Cell(E27)_new_colour=yellow,green
Cell(W28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(W29)_old_all=?(????) <--Cell(E29)_new_colour=not-used
Cell(W30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(W27)_old_all=?(????) <--Cell(E27)_new_number=2
Cell(W28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(W29)_old_all=?(????) <--Cell(E29)_new_number=not-used=0
Cell(W30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(W27)_old_all=?(????) <--Cell(E27)_new_letters=(ab)
Cell(W28)_old_all=?(????) <--Cell(E28)_new_letters=(cd)
Cell(W29)_old_all=?(????) <--Cell(E29)_new_letters=not-used=()
Cell(W30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
Scenario#10
Cell(K27)_old_all=?(????) <--Cell(E27)_new_colour=green,yellow
Cell(K28)_old_all=?(????) <--Cell(E28)_new_colour=not-used
Cell(K29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(K30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(K27)_old_all=?(????) <--Cell(E27)_new_number=3
Cell(K28)_old_all=?(????) <--Cell(E28)_new_number=not-used=0
Cell(K29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(K30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(K27)_old_all=?(????) <--Cell(E27)_new_letters=(acd)
Cell(K28)_old_all=?(????) <--Cell(E28)_new_letters=not-used=()
Cell(K29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(K30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
Scenario#11
Cell(Q27)_old_all=?(????) <--Cell(E27)_new_colour=green,yellow
Cell(Q28)_old_all=?(????) <--Cell(E28)_new_colour=not-used
Cell(Q29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(Q30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(Q27)_old_all=?(????) <--Cell(E27)_new_number=3
Cell(Q28)_old_all=?(????) <--Cell(E28)_new_number=not-used=0
Cell(Q29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(Q30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(Q27)_old_all=?(????) <--Cell(E27)_new_letters=(acd)
Cell(Q28)_old_all=?(????) <--Cell(E28)_new_letters=not-used=()
Cell(Q29)_old_all=?(????) <--Cell(E29)_new_letters=(b)
Cell(Q30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
Scenario#12
Cell(W27)_old_all=?(????) <--Cell(E27)_new_colour=yellow,green
Cell(W28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(W29)_old_all=?(????) <--Cell(E29)_new_colour=not-used
Cell(W30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(W27)_old_all=?(????) <--Cell(E27)_new_number=2
Cell(W28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(W29)_old_all=?(????) <--Cell(E29)_new_number=not-used=0
Cell(W30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(W27)_old_all=?(????) <--Cell(E27)_new_letters=(ab)
Cell(W28)_old_all=?(????) <--Cell(E28)_new_letters=(cd)
Cell(W29)_old_all=?(????) <--Cell(E29)_new_letters=not-used=()
Cell(W30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
It is correct except for the first row. We don't have a green cube, so green which is in the first row should be replaced with yellow.
Cell(E27)_old_all = ? (????) <-- Cell(E27)_new_color = yellow
Cell(E28)_old_all = ? (????) <-- Cell(E28)_new_color = blue
Cell(E29)_old_all = ? (????) <-- Cell(E29)_new_color = red
Cell(E30)_old_all = ? (????) <-- Cell(E30)_new_color = not used.
Cell(E27)_old_all = ? (????) <-- Cell(E27)_new_color = yellow
Cell(E28)_old_all = ? (????) <-- Cell(E28)_new_color = blue
Cell(E29)_old_all = ? (????) <-- Cell(E29)_new_color = red
Cell(E30)_old_all = ? (????) <-- Cell(E30)_new_color = not used.
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=(d) <--Why is this particular definition wrong? Please, deliver argumentations based on the entire documentation here!
IF green is exclusive to ID A. Therefore, yellow cannot appear on ID A.<--this declaration is wrong, if each scenario (#1-2-3-4-5-6-7-8) and its results should be accepted for ever! Yellow cubes do appear in ID A (see scenario #1). <--There is no error, there is no exception! The facts (scenarios #1-2-3-4-5-6-7-8) must be accepted as facts. The ancient and unfortunatelly the modern "science" try to interpret the world so, that a lot of facts are excluded in order to have a nive therory... :-) But this process is not to follow in general...
Rule#1-2 = logistic rule and not a logical rule about the process logic behing the logistic...
Rule#3 = it is more rules parallel: #3a for green cubes, #3b for blue cubes, #3c for red cubes, BUT the #3b and #3c are more strong than #3a! Why? Please, use the mirroring technique: the opposite declaration should also be interpreted (e.g. there are positive scenarios for #3b and there is no scenario with opposite conclusions/risk potentials)...
(Experiment #2 is in this moment quasi still not existing:-)
Rule#3 = it is more rules parallel: #3a for green cubes, #3b for blue cubes, #3c for red cubes, BUT the #3b and #3c are more strong than #3a! Why? Please, use the mirroring technique: the opposite declaration should also be interpreted (e.g. there are positive scenarios for #3b and there is no scenario with opposite conclusions/risk potentials)...
(Experiment #2 is in this moment quasi still not existing:-)
https://moodle.kodolanyi.hu/pluginfile.php/444774/mod_forum/attachment/165003/testing_task1%20-%20Amin-Erdene.xlsx?forcedownload=1 <--WOW!
+
https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p164988 <-- see first sentence (Each slot should hold blocks of one color only.)<--Important declarion - but true or wrong compared of the above-highlighted WOW-XLSX?
Who is finally capable of seeing the expected complexity with the expected clarity?!
What is the hermeneutical trap in the definition of this task (see: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p164960)
(((BTW: such kind of logical constructions will later be necessary, when you are writing the final thesis, especially the chapter about the literature behind your thesis-title...)))
+
https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p164988 <-- see first sentence (Each slot should hold blocks of one color only.)<--Important declarion - but true or wrong compared of the above-highlighted WOW-XLSX?
Who is finally capable of seeing the expected complexity with the expected clarity?!
What is the hermeneutical trap in the definition of this task (see: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p164960)
(((BTW: such kind of logical constructions will later be necessary, when you are writing the final thesis, especially the chapter about the literature behind your thesis-title...)))
if YELLOW cube meet GREEN cube, Rule 1 cannot be followed due to Rule 4-8 scenario. <-- wrong association (wrong solutions for #10-11-12). See: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p165010
The rule of the scenario#1 must be followed for ever...
The rule of the scenario#1 must be followed for ever...
Interesting approach: but it can not be accepted! The robot does not have any information units about ALL cubes before starting with the tower-building-process! The count of the cubes having the same colour is an information unit, but never existing in this system - we observers can interpret such a phenonemon (count) - later (after closing a scenario). The robot will never know, how many cubes will still be set into the process and/or when is a process closed at all... :-)
So, If yellow and green must be placed to spot "A" for example experiment#1 scenario1-8 and experiment#2 scenario 1-8, So both of them must be placed to spot "A". Can we stack green and yellow on spot "A", according to the rules it does not break any rules, there is not rules about 2 different colors cannot stack on each other. <--WOW (still see: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p165010)
It is worth to read all previous entries! :-)
Especially: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p165013
Go, please, from this link backward!
Especially: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p165013
Go, please, from this link backward!
The system takes input values (e.g., a, b, c, d) and determines the number and color of the cubes.
The output is organized into towers, following a fixed color order:
1st place: Green and Yellow cubes
2nd place: Blue cubes
3rd place: Red cubes
The output format is ?(????), where:
The "?" is replaced by the number of cubes of that color.
The brackets contain the corresponding input values for that color.
If there are multiple cubes of the same color, they are stacked on top of each other (e.g., if there are 2 green cubes, the first place will show "2Green").
The output is organized into towers, following a fixed color order:
1st place: Green and Yellow cubes
2nd place: Blue cubes
3rd place: Red cubes
The output format is ?(????), where:
The "?" is replaced by the number of cubes of that color.
The brackets contain the corresponding input values for that color.
If there are multiple cubes of the same color, they are stacked on top of each other (e.g., if there are 2 green cubes, the first place will show "2Green").
What is the conclusion based on these declarations concerning the scenarios #10-11-12! The KNUTH-principle says: knowledge/science is, what can be transformed/transferred/translated/transscripted into source codes = what can be used for operative steps in an objective/consistent/consequent way...
This is very good!
Please, always try to use the "very-good-materials" for the next step of the concluding process...
New task: please, try to create a rel. small, but complex prompt for the ChatGPT and/or Copilot, etc. in order to involve it into this project. This task has two layers: what is a good prompt? Parallel: how good is the interpretation potential of ChatGPT/Copilot/etc. - in caseof a good prompt!
According to all scenario rules, there is not any rules about not stacking different color. It seems bit tricky but, in logically, robot can put one cube in the spot at time in order. So, in scenario 10, green block must be on the spot A firstly twice before the yellow one. In scenario 11, green, yellow and green again in order. In the scenario 12, green one is placed after yellow in order.
Here are the correct answers following all the rules in scenario 1 to 8.
Experiment#1, Scenario#10,11,12 answers according to the rules.
Experiment#1 Scenario#10
Cell(K27)_old_all=?(????) <--Cell(K27)_new_colour=green, yellow
Cell(K28)_old_all=?(????) <--Cell(K28)_new_colour=not-used
Cell(K29)_old_all=?(????) <--Cell(K29)_new_colour=red
Cell(K30)_old_all=?(????) <--Cell(K30)_new_colour=not used
***
Cell(K27)_old_all=?(????) <--Cell(K27)_new_number=3
Cell(K28)_old_all=?(????) <--Cell(K28)_new_number=not-used=0
Cell(K29)_old_all=?(????) <--Cell(K29)_new_number=1
Cell(K30)_old_all=?(????) <--Cell(K30)_new_number=not-used=0
***
Cell(K27)_old_all=?(????) <--Cell(K27)_new_letters=(abd)
Cell(K28)_old_all=?(????) <--Cell(K28)_new_letters=not-used=()
Cell(K29)_old_all=?(????) <--Cell(K29)_new_letters=(c)
Cell(K30)_old_all=?(????) <--Cell(K30)_new_letters=not-used=()
***
Scenario#11
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_colour=green, yellow
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_colour=not-used
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_colour=red
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_colour=not-used
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_number=3
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_number=not-used=0
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_number=1
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_number=not-used=0
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_letters=(acd)
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_letters=not-used=()
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_letters=(b)
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_letters=not-used=()
***
Scenario#12
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_colour=yellow, green
Cell(W28)_old_all=?(????) <--Cell(W28)_new_colour=blue
Cell(W29)_old_all=?(????) <--Cell(W29)_new_colour=not-used
Cell(W30)_old_all=?(????) <--Cell(W30)_new_colour=not-used
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_number=2
Cell(W28)_old_all=?(????) <--Cell(W28)_new_number=2
Cell(W29)_old_all=?(????) <--Cell(W29)_new_number=not-used=0
Cell(W30)_old_all=?(????) <--Cell(W30)_new_number=not-used=0
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_letters=(ab)
Cell(W28)_old_all=?(????) <--Cell(W28)_new_letters=(cd)
Cell(W29)_old_all=?(????) <--Cell(W29)_new_letters=not-used=()
Cell(W30)_old_all=?(????) <--Cell(W30)_new_letters=not-used=()
***
Here are the correct answers following all the rules in scenario 1 to 8.
Experiment#1, Scenario#10,11,12 answers according to the rules.
Experiment#1 Scenario#10
Cell(K27)_old_all=?(????) <--Cell(K27)_new_colour=green, yellow
Cell(K28)_old_all=?(????) <--Cell(K28)_new_colour=not-used
Cell(K29)_old_all=?(????) <--Cell(K29)_new_colour=red
Cell(K30)_old_all=?(????) <--Cell(K30)_new_colour=not used
***
Cell(K27)_old_all=?(????) <--Cell(K27)_new_number=3
Cell(K28)_old_all=?(????) <--Cell(K28)_new_number=not-used=0
Cell(K29)_old_all=?(????) <--Cell(K29)_new_number=1
Cell(K30)_old_all=?(????) <--Cell(K30)_new_number=not-used=0
***
Cell(K27)_old_all=?(????) <--Cell(K27)_new_letters=(abd)
Cell(K28)_old_all=?(????) <--Cell(K28)_new_letters=not-used=()
Cell(K29)_old_all=?(????) <--Cell(K29)_new_letters=(c)
Cell(K30)_old_all=?(????) <--Cell(K30)_new_letters=not-used=()
***
Scenario#11
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_colour=green, yellow
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_colour=not-used
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_colour=red
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_colour=not-used
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_number=3
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_number=not-used=0
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_number=1
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_number=not-used=0
***
Cell(Q27)_old_all=?(????) <--Cell(Q27)_new_letters=(acd)
Cell(Q28)_old_all=?(????) <--Cell(Q28)_new_letters=not-used=()
Cell(Q29)_old_all=?(????) <--Cell(Q29)_new_letters=(b)
Cell(Q30)_old_all=?(????) <--Cell(Q30)_new_letters=not-used=()
***
Scenario#12
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_colour=yellow, green
Cell(W28)_old_all=?(????) <--Cell(W28)_new_colour=blue
Cell(W29)_old_all=?(????) <--Cell(W29)_new_colour=not-used
Cell(W30)_old_all=?(????) <--Cell(W30)_new_colour=not-used
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_number=2
Cell(W28)_old_all=?(????) <--Cell(W28)_new_number=2
Cell(W29)_old_all=?(????) <--Cell(W29)_new_number=not-used=0
Cell(W30)_old_all=?(????) <--Cell(W30)_new_number=not-used=0
***
Cell(W27)_old_all=?(????) <--Cell(W27)_new_letters=(ab)
Cell(W28)_old_all=?(????) <--Cell(W28)_new_letters=(cd)
Cell(W29)_old_all=?(????) <--Cell(W29)_new_letters=not-used=()
Cell(W30)_old_all=?(????) <--Cell(W30)_new_letters=not-used=()
***
WOW! What can we identify as hermeneutical trap in the definition of this entire task? Why are a lot of wrong solutions/interpretations? How should have been formulated this task as such in order to minimize the misunderstanding potential of this task?
We assumed each slot could hold only one color, ignoring the possibility of stacking. This is a hermeneutical trap a thinking error where our biases (like assuming "one color, one slot") blind us to simpler logic. The robot logic worked correctly; the problem was our false interpretation.
Hermeneutical traps happen when we force our beliefs onto situations instead of seeking the real explanation.
I can see/experience that biases distort understanding. Always test assumptions against evidence.
to minimize the potential misunderstanding of this task, we can state the rules explicitly.
e.g. Multiple colors can occupy the same slot.
Hermeneutical traps happen when we force our beliefs onto situations instead of seeking the real explanation.
I can see/experience that biases distort understanding. Always test assumptions against evidence.
to minimize the potential misunderstanding of this task, we can state the rules explicitly.
e.g. Multiple colors can occupy the same slot.
The basic question answer is: Incorrect.
Reason:
In Experiment #1, Scenario #9, there is no green color present in the "Input9" column. This means that the first solution layer is incorrect, specifically:
Cell(E27)_old_all=?(????) ← Cell(E27)_new_colour=green.
The correct answer:
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
***
Reason:
In Experiment #1, Scenario #9, there is no green color present in the "Input9" column. This means that the first solution layer is incorrect, specifically:
Cell(E27)_old_all=?(????) ← Cell(E27)_new_colour=green.
The correct answer:
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
***
Summary of the Rules (IN031: Corrupted logistic robot | KJE Moodle + IN031: Corrupted logistic robot | KJE Moodle)
Logistic Rules:
- The robot picks one cube at a time.
- The robot processes cubes in their given order on the conveyor belt.
Color (Logical) Rules:
3. Color Assignment (Parallel):- GREEN cubes → place-id "A" (Rule #3a)
- BLUE cubes → place-id "B" (Rule #3b, stronger)
- RED cubes → place-id "C" (Rule #3c, stronger)
The mirroring technique confirms that deviations from the blue and red assignments would lead to risk scenarios that are not observed.
- Exception for YELLOW Cubes:
- When there is no green input, YELLOW cubes are built in place-id "A" (as observed in Scenario #1 and Scenario #9).
This reveals the hermeneutical trap—we assumed a more complex rule than necessary. The correct rule is that green always goes to slot A if present, and the other colors shift accordingly. Towers are not strictly limited to one color per slot; multiple towers can appear in the same slot when needed. (IN031: Corrupted logistic robot | KJE Moodle)
Exact interpretation! Parallel: the robot does not have information about more than 3 colours (red, blue, green), AND for each undefinied colour-code, slot A is dedicated! Congratulation! Project is closed! :-)
FYI: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p165023
Please, compare your solution with this one: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p165023
The first solution layer (Colours) is incorrect because:
- In Experiment #1, Scenario #9, there is no green colour present in the "Input9" column.
- This means the assignment:
Cell(E27)_new_colour = green is incorrect.
Solution
Cell(E27)_old_all=?(????) <-- Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <-- Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <-- Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <-- Cell(E30)_new_colour=not-used
Numbers
CCell(E27)_old_all=?(????) <-- Cell(E27)_new_number=1 Cell(E28)_old_all=?(????) <-- Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <-- Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <-- Cell(E30)_new_number=not-used=0
Letters
Cell(E27)_old_all=?(????) <-- Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <-- Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?(????) <-- Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <-- Cell(E30)_new_letters=not-used=()
Final Answer:
The first solution layer (colours) was incorrect due to the green value. The correct assignment should be yellow instead. Numbers and letters layers were correct.
Please, follow the closing interpretations here: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p165023
Not correct ,this error happened because it is yellow instead of green (Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=green).
Correct anwser:
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
***
Correct anwser:
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_colour=yellow
Cell(E28)_old_all=?(????) <--Cell(E28)_new_colour=blue
Cell(E29)_old_all=?(????) <--Cell(E29)_new_colour=red
Cell(E30)_old_all=?(????) <--Cell(E30)_new_colour=not-used
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_number=1
Cell(E28)_old_all=?(????) <--Cell(E28)_new_number=2
Cell(E29)_old_all=?(????) <--Cell(E29)_new_number=1
Cell(E30)_old_all=?(????) <--Cell(E30)_new_number=not-used=0
***
Cell(E27)_old_all=?(????) <--Cell(E27)_new_letters=(b)
Cell(E28)_old_all=?(????) <--Cell(E28)_new_letters=(ad)
Cell(E29)_old_all=?(????) <--Cell(E29)_new_letters=(c)
Cell(E30)_old_all=?(????) <--Cell(E30)_new_letters=not-used=()
***
Do/Did we need the information units from the experiment#2 in order to derive the hermeneutical trap as such? Which experiment (#1 or #2) has more (relevant) information units if at all?
Experiment #2 contains more relevant information units compared to Experiment #1. In Experiment #1, the colors green, red, and blue follow a known order, while yellow is the unknown variable. Yellow could appear first, between any of the other colors, or last. However, in Experiment #2, yellow is clearly placed first in input 9, which provides a more definite and relevant piece of information.
I would use a database structure with separate tables for students, experiments, solution layers, student solutions, and evaluations. This setup would allow for unlimited guesses per student per layer. In Excel, I would use pivot tables to analyze total scores, the number of guesses, and the best guesses per student. This structure would support detailed evaluation and help rank students based on performance and correctness.
This is my answer to experiment # 1 EXP 9-12 and experiment # 2 9-12
Please let me know if there's anything I’m missing or if I’ve made any errors in my approach.
This is my answer to experiment # 1 EXP 9-12 and experiment # 2 9-12
Please let me know if there's anything I’m missing or if I’ve made any errors in my approach.
Please, try to interpret the enteri communication - especialy this one: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p165027
https://miau.my-x.hu/miau/320/moodle_cubes_logic/testing_task1_solutions.xlsx <--please, compare your solution with this one. Question: can we define a structure for storing unlimized guesses WITHOUT clarifying the final visualisation for the common accepted solutions for scenarios #10-11-12?
https://miau.my-x.hu/miau/320/moodle_cubes_logic/testing_task1_solutions.xlsx <-- worth interpreting in order to derive the final structure for storing unlimited guesses!
A demo structure for storing unlimited guesses in cases where all student, experiment, scenario, place ID, and layers can be found in the attached EXCEL file.
Please let me know if there are any suggestions for improving its efficiency.
Please let me know if there are any suggestions for improving its efficiency.
The structuring as such is not the direct problem (although we are seraching for the appropriate storing-structure - but the appropriate structure is not a technical phenomenon rather a way e.g. to suspicion interpretation): we are searching for a storing structure where the reports can be interpreted in a useful way. Selected records are not the expected reports in general. The so-called cross-tabs (in Excel: pivots) create from the selected raw records a multidimensional report with appropriate report-structures (row-headers, column-headers, aggregation-rules, etc.). But selected (raw) records as such can be interesting IF THE INTERPRETATION RULES FOR THEM ARE GIVEN! Therefore: the develpment steps are in ranked form: 0. Suspicion/hypothesis --> 1. interpretation rules --> 2. needed e.g. pivot-structure (output) ---> 3. storing strucutre (input). The file "guesses_demo.xlsx" presents steps 3. and 2. (if the selected raw records are accepted as reports) but these reports cover no planned needs (c.f. step 0), therefore there are no interpretation rules given to explain, whether a suspicion/hypothesis is wrong or true?! Demo: Suspicion = The wrong guesses are using more the 3 slots! Rule: If the count of the affected slots in case of a given (or each) Student is more than 3 for all scenarios (#1-12), than the solution must be wrong! Report: row-header(s): Student-ID(s), Column-headers: Slot-IDs, cells = count of guesses concerning the slots, Filter: All scenarios. Expectation: the Slot-ID=D may never exit in the reports in case of Students having the expected solution about the hermeneutical trap! (Parallel demo: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p165068)
This parallel task about the storing structure should have been a kind of hint helping to identify the hermeneutical trap (see: PARALLEL task:-)
This (https://miau.my-x.hu/miau/320/moodle_cubes_logic/summary_report_20250212-0535.xlsx) is the most recent information package about the activities of the affected Students: the well-known question is, WHO IS THE BEST? c.f. https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p165066 (Suspicion = Can we evauate EACH Student with the same evaluation index? / Interpretation rule = IF the COCO Y0-model does deliver for each Student the same norm value of 1000, THEN the suspicion/hypothesis is true / Needed report: OAM (<--identical with this XLSX = https://miau.my-x.hu/miau/320/moodle_cubes_logic/summary_report_20250212-0535.xlsx - where objects = Students, attributes = statistical phenomena with appropriate DIRECTION for a ranked OAM as direct input for COCO-Y0)... Direction demo: e.g. the more is the Number of discussions posted and/or the Number of replies posted and/or the Number of attachments and/or the Number of views and/or the Word count and/or the Character count THE BETTER is the performance + the younger the Earliest post THE BETTER is the performance + the later the Most recent post THE BETTER is the performance...
More: https://miau.my-x.hu/miau/320/moodle_cubes_logic/?C=M;O=D - e.g. Which kind of new attributes could we define with which direction to improve the derivation of the BEST STUDENT based on this structure: https://miau.my-x.hu/miau/320/moodle_cubes_logic/discussion.html
I have prepared an Excel demo file that organizes student guesses in a structured format and includes a pivot table for analysis.
The pivot table summarizes student attempts for each cell reference.
Rows: Student_ID
Columns: Cell_Reference (E27, E28, etc.)
Values: Count of correct attempts per cell.
The "Overall Correct?" column indicates whether the latest valid attempt matches the expected solution.
This allows for easy evaluation of students' progress while tracking multiple guesses. Let me know if any adjustments are needed.
The pivot table summarizes student attempts for each cell reference.
Rows: Student_ID
Columns: Cell_Reference (E27, E28, etc.)
Values: Count of correct attempts per cell.
The "Overall Correct?" column indicates whether the latest valid attempt matches the expected solution.
This allows for easy evaluation of students' progress while tracking multiple guesses. Let me know if any adjustments are needed.
https://miau.my-x.hu/miau/320/moodle_cubes_logic/guessing_demo.xlsx - Please, see the yellow sheet and the yellow cell! The interpretation as such seems to be smart! :-) But the pivot-table presents 2 counts in the yellow cell and one of them is a green-value in the background. If all two raw data are green-values, then the count=2-status is also given - and the monitoring effect does become irrational?! Excel-pivots can be defined with filters, where the expected rules might probably be enforced?!
I have updated my Excel demo. Here’s a summary of the changes I made:
Filtering Incorrect Attempts:
- I added a new column called Is_Correct in the raw_data sheet. This column flags an attempt as “Yes” only if the New_Color exactly matches the expected value for the corresponding cell (for example, for E27 the expected color is “yellow”, for E28 “blue”, for E29 “red”, and for E30 “not-used”).
- This ensures that if a student submits an incorrect guess (e.g., “green” for E27), it is marked “No” and will not be counted in the pivot table.
Refined Pivot Table:
- In the pivot table, I now aggregate only those rows where Is_Correct is “Yes”.
- Additionally, I added an Overall Correct? column that indicates “Yes” only if the student has at least one correct attempt for each of the required cells (E27, E28, E29, and E30).
- This change prevents the irrational monitoring effect where an incorrect (green) value might inflate the count.
Unlimited Guesses:
- Each guess is stored as a separate record with a unique Attempt_Number, so the system can handle unlimited guesses without overwriting previous attempts. This allows for a complete analysis of the students’ progression toward the correct solution.
As you said,
"We are waiting for an Excel-demo with real or realistic data AND first of all with at least a meaningful pivot-output, where we have rules for relevant interpretations concerning this output."
I just tried to figure out how we could store all this data, allowing for unlimited potential guesses per student, in the DEMO.xlsx file using a Pivot Table to create a simple output. I believe all necessary explanations are included in the attached file.
https://miau.my-x.hu/miau/320/moodle_cubes_logic/DEMO.xlsx <--the filtered approaches could already be useful, but the prepared report seems to be still not matured enough? (see: https://moodle.kodolanyi.hu/mod/forum/discuss.php?d=63498#p165066)