Back to the future Doc. Brown's 7 String of Gears puzzle

[Tito_H]

The other day, Doc replied to my solving method to the Accelerating Reels puzzle, saying, "Well done!"

He then told me that he had come up with another puzzle after seeing some kids in the past world playing with a string threaded through a line of pulleys, spinning them around.

He also sent me a diagram outlining the puzzle, which looks like this:



Puzzle Overview:
1. A total of 7 main gears are fixed to the wood board. The gear number, number of teeth, and numbers written clockwise on the top (circumference) of each gear are as follows:

Gear Teeth Numbering
A1 100 0 to 9 (10 numbers)
B1 90 0 to 8 (9 numbers)
C1 80 0 to 7 (8 numbers)
D1 70 0 to 6 (7 numbers)
C2 80 0 to 7 (8 numbers)
B2 90 0 to 8 (9 numbers)
A2 100 0 to 9 (10 numbers)

2. A total of 6 idler gears (16 teeth) are placed between the main gears, and are fixed by meshing with A1, C1, A2, and C2.

3. B1 and B2 are sliding gears that slide left and right, and D1 slides up and down, meshing with the idler gears.

4. The way to play is as follows:
1) Rotate the main gears randomly and stop them where one of the numbers is at the top. This is a scramble, and the play starts here.

2) Slide one of the sliding gears (B1, B2, D1) and rotate it while meshing with the adjacent idler gear. Rotating 2 main gears at the same time (it is illegal to rotate a main gear alone or three main gears at the same time).

3) Return the slid sliding gear to its original position.

4) Repeat steps 2) and 3) above until the numbers at the top of all main gears become same (from 0 and 6), and the problem is solved.

5) The winner is the person who completes solve without rotating the gears as much as possible. The number of moves is calculated by how many times the number on the top row changes.

Let's try solving this using an example.
The numbers on the top row of each main gear are as follows. (From now on, I will omit the term "top row.")

A1B1C1D1C2B2A2
1 2 3 4 2 2 2 → scrambled

In this case, I rotated the gears as follows:

A1B1C1D1C2B2A2
1 2 3 4 2 2 2

↓ Connect B1 to A1 and turn B1 slightly to change 2 to 3. At this time, A1 is 2.

2 3 3 4 2 2 2

↓ Connect B1 to C1 and turn B1 slightly to change 3 to 2. At this time, C1 becomes 2.

2 2 2 4 2 2 2

↓ Connect D1 to C1, rotate D1 clockwise two times, then rotate it a little more to set it to 2. At this time, C1 will remain at 2.

2 2 2 2 2 2 2 → solved

Total: 18 moves

Alternatively, I could solve it by doing it like this:

A1B1C1D1C2B2A2
1 2 3 4 2 2 2

↓ Connect D1 to C1, rotate D1 clockwise twice, then rotate it a little more to make it 2. At this time, C1 will remain at 3.

1 2 3 2 2 2 2

↓ Connect B1 to C1 and turn B1 slightly to change 2 to 1. At this time, C1 becomes 2.

1 1 2 2 2 2 2

↓ Connect B1 to A1 and turn B1 slightly to change 1 to 2. At this time, A1 becomes 2.

2 2 2 2 2 2 2 → solved

Total: 18 moves

In this example, the number of the three bottom row main gears is 2, so it's relatively easy to know that you just need to rotate the other gears to make them all equal.

But wait a minute!

After thinking about it again, I realized I could solve it in much fewer moves!

So, readers, can you guess how I solved it?

 

[Tito_H]

Here is the solve I found:

A1B1C1D1C2B2A2
1 2 3 4 2 2 2 → scrambled

↓ Connect B1 to A1 and turn B1 slightly to change 2 to 1. At this time, A1 becomes 0.

0 1 3 4 2 2 2

↓ Connect B1 to C1 and turn B1 slightly to change 1 to 0. At this time, C1 becomes 2.

0 0 2 4 2 2 2

↓ Connect D1 to C1 and turn D1 slightly to change 4 to 2. At this time, C1 becomes 0.

0 0 0 2 2 2 2

↓ Connect D1 to C2 and turn D1 slightly to change 2 to 0. At this time, C2 becomes 0.

0 0 0 0 0 2 2

↓ Connect B2 to A2 and turn B2 slightly to change 2 to 0. At this time, A2 becomes 0.

0 0 0 0 0 0 0 → solved

Total: 8 moves

Initially, I thought it would be best to set all the main gears to 2, since those on the bottom row are all 2, but I realized that in this case the problem could be solved with far fewer moves if I set them to 0. Interesting!

As usual, I will see if I can get FMC solution for this puzzle.

 

[Tito_H]

I would like to write about my thoughts on how to solve this puzzle.

First, I will look for a solution in a hypothetical version where all the main gears have the same number of teeth and the numbers 0 to 9 are written on their top surfaces (circumferences).

<Premise>
1 After scrambling the main gears, the top numbers of them A1, B1, C1, D1, C2, B2, A2 are a1, b1, c1, d1, c2, b2, a2 respectively.

2 The rotation amounts of main gears A1, B1, C1, D1, C2, B2, A2 are A1, B1, C1, D1, C2, B2, A2 respectively.

3 When the solve is complete, the top number of all main gears is "s".

<Calculation>
A1 is the amount of rotation required to turn a1 into s, so the following formula holds:

A1 = s - a1 •••(1)

When the A1 gear is rotated, it must be meshed with the B1 gear one gear ahead, so B1 also rotates the same amount. Therefore, the amount of rotation of B1 must take into account the amount of rotation of the A1 gear, and the following formula holds true:

B1 = s - b1 - A1 •••(2)

Based on the same idea, the following equations also hold for C1, D1, C2, B2, and A2.

C1 = s - c1 - B1 •••(3)
D1 = s - d1 - C1 •••(4)
C2 = s - c2 - D1 •••(5)
B2 = s - b2 - C2 •••(6)
A2 = s - a2 - B2 •••(7)

Substitute each equation into the others to finally get "s".

Substitute (1) into (2).
B1 = s - b1 - (s - a1)
= a1 - b1 •••(8)

Substitute (8) into (3).
C1 = s - c1 - (a1 - b1) •••(9)

Substitute (9) into (4).
D1 = s - d1 - s + c1 + (a1 - b1)
= (a1 - b1) + (c1 - d1) •••(10)

Substitute (10) into (5).
C2 = s - c2 - (a1 - b1) - (c1 - d1) •••(11)

Substitute (11) into (6).
B2 = s - b2 - s + c2 + (a1 - b1) + (c1 - d1)
= (a1 - b1) + (c1 - d1) + (c2 -b2) •••(12)

Substitute (12) into (7).
A2 = s - a2 - (a1 - b1) - (c1 - d1) - (c2 -b2) •••(13)

Since it is against the rules to rotate only the A2 gear, A2 = 0 is a necessary condition in equation (13). Substituting A2 = 0 into equation (13) converts it into an equation to obtain s.

A2 = s - a2 - (a1 - b1) - (c1 - d1) - (c2 -b2)
↓ Substitute 0 into A2.
0 = s - a2 - (a1 - b1) - (c1 - d1) - (c2 -b2)
↓ Convert to an expression to get s
s = (a1 - b1) + (c1 - d1) + (c2 -b2) + a2 •••(14)

Now we have the formula for "s". If we look closely, we can see that s is the sum of the difference between the numbers of main gears A1 and B1, C1 and D1, and C2 and B2, plus the number of A2.

Let's consider the case where the result of calculation (14) above is s > 9. Since the numbers of the main gear are 0 to 9, if the amount of gear rotation increases or decreases by 10 (one rotation of the gear), it will return to the original number. Therefore, s can be rewritten as follows:

s = ((a1 - b1) + (c1 - d1) + (c2 -b2) + a2) mod 10 •••(15)

From the above, if you turn the main gears A1, B1, C1, D1, C2, B2, and A2 in order starting from A1 so that the numbers become s, the solve will be completed. (If you turn B2 to make A2 s, it should automatically become s.)

Let's try it out. First, the example we gave earlier:

A1B1C1D1C2B2A2
1 2 3 4 2 2 2 → scrambled

Substitute the scrambled numbers into formula (15) to find s.

s = ((a1 - b1) + (c1 - d1) + (c2 -b2) + a2) mod 10
= ((1 - 2) + (3 - 4) + (2 - 2) + 2) mod 10
= (-1 - 1 + 0 + 2) mod 10
= 0 mod 10
= 0

Solve with s = 0.

A1B1C1D1C2B2A2
1 2 3 4 2 2 2
↓ -1
0 1 3 4 2 2 2
↓ -1
0 0 2 4 2 2 2
↓ -2
0 0 0 2 2 2 2
↓ -2
0 0 0 0 0 2 2
↓ -2
0 0 0 0 0 0 0 → solved

Total: 8 moves

Let's try solving some other examples.

A1B1C1D1C2B2A2
5 2 4 2 7 1 8 → scrambled

Calculate s.

s = ((a1 - b1) + (c1 - d1) + (c2 -b2) + a2) mod 10
= ((5 - 2) + (4 - 2) + (7 - 1) + 8) mod 10
= (3 + 2 + 6 + 8) mod 10
= 19 mod 10
= 9

Solve with s = 9.

A1B1C1D1C2B2A2
5 2 4 2 7 1 8
↓ +4
9 6 4 2 7 1 8
↓ +3
9 9 7 2 7 1 8
↓ +2
9 9 9 4 7 1 8
↓ +5
9 9 9 9 2 1 8
↓ -3
9 9 9 9 9 8 8
↓ +1
9 9 9 9 9 9 9 → solved

Total: 18 moves

Now we have a solution for the hypothetical version. Now let's consider the solution for the real version. Unlike the fictional version, the real version "skips" some numbers before 0, except for the A1 and A2 gears, as shown below.

Gear: Skipping numbers
A1: No skip
B1: number 9 (8⇄0)
C1: number 8, 9 (7⇄0)
D1: number 7, 8, 9 (6⇄0)
C2: number 8, 9 (7⇄0)
B2: number 9 (8⇄0)
A2: No skip

For this reason, there are cases where the A2 gear does not automatically become "s" when calculated using the hypothetical version of the formula. Let's take a look at the example below.

A1B1C1D1C2B2A2
6 0 5 1 1 1 1 → scrambled

Calculate s.

s = ((a1 - b1) + (c1 - d1) + (c2 -b2) + a2) mod 10
= ((6 - 0) + (5 - 1) + (1 - 1) + 1) mod 10
= (6 + 4 + 0 + 1) mod 10
= 11 mod 10
= 1

Solve with s = 1.

A1B1C1D1C2B2A2
6 0 5 1 1 1 1
↓ -5
1 4 5 1 1 1 1
↓ -3
1 1 2 1 1 1 1
↓ -1
1 1 1 0 1 1 1
↓ +1
1 1 1 1 2 1 1
↓ -1
1 1 1 1 1 0 1
↓ +1
1 1 1 1 1 1 2 → The A2 gear does not automatically become 1. To solve this, we need to mesh the B2 gear with the A2 gear, rotate the B2 gear once, and reduce the A2 gear by 1.
↓ +9 (B2 & A2)
1 1 1 1 1 1 1 → solved

Total: 21 moves

Although the solution itself can be completed using the current approach, it is not the best solution from the perspective of FMC. So, how do you change the calculation formula to get the correct "s"?

Actually, I've been thinking about it for the past few days, but I still haven't found a solution!

I may have to ask Doc for help.

====added on 21-Feb-2026====
Regarding the example above, I found another solution, and I don't know why I didn't think of it at the time.
Here is the solution:

A1B1C1D1C2B2A2
6 0 5 1 1 1 1 → scrambled

Solve with s = 1.

A1B1C1D1C2B2A2
6 0 5 1 1 1 1
↓ +5
1 5 5 1 1 1 1
↓ -4
1 1 1 1 1 1 1 → Solved!

Total 9 moves

In this example, the main gears B1, C1, etc. can be rotated without passing through the zones that skip numbers (8⇄0, 7⇄0, etc.), so it is not a good example. An example that passes through those zones should have been provided.

 

[Tito_H]

While I was struggling for days to find a solution to this puzzle that took FMC into account, Superman's grandson came to visit me.

He asked me, "Grandpa Tito, what are you having trouble with?" I explained the details of the puzzle and why I was having trouble with it, and he said, "I see. It would be easy if the numbers on each gear were the same, but they're different, so it's hard to know which numbers to match when you start scrambling." I replied, "That's right," and sighed.

Then he took out his paper and pencil and began to draw something…

After a while, he said, "Grandpa, if you think about this puzzle in a slightly different way, you might find a better solution. For example, take a look at this," and showed me the picture he had been drawing.

His idea was to create a slightly unusual imaginary elevator, as shown below.



<Imaginary Elevator Details>
1. Seven elevators are drawn side by side, numbered from left to right: A1, B1, C1, D1, C2, B2, A2.

2. The ground floor is floor 0, and the top floors are floors 9 for A1 and A2, 8 for B1 and B2, 7 for C1 and C2, and 6 for D1. Floors 9 for B1 and B2, floors 8 and above for C1 and C2, and floors 7 and above for D1 are painted black.

3. Each elevator goes to the ground floor when it goes above its top floor, and goes to its top floor when it goes below ground floor. This is what's strange about imaginary elevators; in reality, this is not possible.

4. On each elevator lane on paper, 'erasers' representing gondolas are placed at different floors.

5. The rule is that the gondolas in two adjacent lanes must be moved the same amount at the same time.

6. The puzzle is solved when all the gondolas are lined up in a horizontal row.

I see, so instead of gears we adjust the elevator gondolas. It's true that it's easier to grasp the relative positions of the gondolas, which can be moved in a straight line, than to try to match the numbers with round gears. That's a good idea.

Well, I'll play around with him for a while, moving the gondolas of this imaginary elevator together, and look for an FMC-like solution...

 

[Tito_H]

Following Superman's grandson's suggestion, I replaced this puzzle with a group of elevators, raising and lowering the gondolas, and really found a solution.

I noticed that if you line up two gondolas, you can combine them to go to any floor (except for Not-usable floors), and that the gondola with more Not-usable floors takes less time to go around all the floors.

My solution, which took these into consideration, is below.

<Premise>
1. The main gears are called A1, B1, C1, D1, C2, B2, and A2 gears.

2. The numbers on the top row of the main gears are a1, b1, c1, d1, c2, b2, and a2, respectively.

3. The amount of rotation of the main gears is A1, B1, C1, D1, C2, B2, and A2, respectively.

4. The direction of rotation of the main gears is CCW if the amount of rotation is positive, and CW if it is negative.
(Note that in this puzzle, dears rotate instead of hour hands, so the rotation is opposite to that of a clock.)

5. Since the number of D1 gear is 0 to 6, the numbers of all gears must be within that range in the solve. From now on, this range of gears will be called "the valid range".

<Step>
1. Engage gear B1 with gear C1 and align b1 with a1. However, if a1 = 9, set b1 to 8.

The rotation direction of gear B1 will be the one with the smaller amount of rotation.

2. Engage gear B2 with gear C2 and align b2 with a2. However, if a2 = 9, set b2 to 8.

The rotation direction of gear B2 will be the one with the smaller amount of rotation.

3. Engage gear D1 with gear C1 or C2 and perform substeps 1) and 2) below to align c1, d1, and c2.
The rotation direction of gear D1 will be the one with the smaller amount of rotation.

Sub-step 1) Engage gear D1 with gear C2 and align d1 with c1.

Sub-step 2) Engage gear D1 with gear C1 and align d1•c1 with c2.

However, in the following cases, c1 and c2 are outside "the valid range" and d1 cannot be aligned with them, so exception handling is performed.

Exception 1) If (c1 = 7) & (c2 = 7):
Engage D1 gear with C1 gear, and shift D1 gear by one position in either the CCW or CW direction to set c1 = 0 or 6. Then, perform Sub-steps 1) and 2).
However, if d1 = 0 or 6, rotate it in a direction that will prevent c1 = d1 after the gear rotation.

Exception 2) If (c1 = 7) & (c2 ≠ 7) & (d1 = c2):
Engage D1 gear with C1 gear, and rotate D1 gear 360 degrees in either CCW or CW, whichever is convenient, to set c1 = 0 or 6. Then, perform Sub-step 1).

Exception 3) If (c1 = 7) & (c2 ≠ 7) & (d1 ≠ c2):
Switch the order of Sub-steps 1) and 2), and perform them.

Exception 4) If (c1 = d1) & (c2 = 7):
Engage D1 gear with C2 gear, and rotate D1 gear 360 degrees CCW or CW, whichever is convenient, to set c2 = 0 or 6. Then, perform Sub-step 2).

Exception 5) If exception 4) occurs after performing Sub-step 1):
In Sub-step 1), turn the D1 gear in the opposite direction and align d1 with c1. Then, perform Sub-step 2).

4. Engage B1 gear with A1 gear, and align a1 and b1 with c1.

5. Engage B2 gear with A2 gear, and align a2 and b2 with c2.

===Added on 5-Mar-2026===
6. After solving Steps 1 to 5 above, perform the following optimization to see if the total number of moves can be reduced.

<Optimization>
1. In Steps 1, 2, and 3, try reversing the rotation direction of gears B1, B2, and D1.

2. In Steps 1 and 2, instead of setting b1 = a1 and b2 = a2, try setting b1 = a1 - 1 and b2 = a2 - 1. (In Steps 4 and 5, if you rotate gears B1 and B2 from 8 to 0, the numbers will line up from 0 onwards, with b1 = a1 and b2 = a2.)

Please refer to the post for more details.
=======

That's all!

Personally, I'm not happy with the large number of exception handling steps in Step 3, but for now it's unavoidable.

Let's take a look at some examples and try solving them. First, let's look at the number combinations we solved previously in the hypothetical version.

A1B1C1D1C2B2A2
5 2 4 2 7 1 8 → Scrambled
↓ Step1. B1/C1 +3
5 5 7 2 7 1 8
↓ Step2. B2/C2 -2
5 5 7 2 5 8 8
↓ Step 3. Now that we have the Exception 3), switch the order of Sub-steps 1) and 2).
↓ Sub-step 2). D1/C1 +3
5 5 2 5 5 8 8
↓ Sub-step 1). D1/C2 -3
5 5 2 2 2 8 8
↓ Step4. A1/B1 -3
2 2 2 2 2 8 8
↓ Step5. A2/B2 -6
2 2 2 2 2 2 2 → Solved

Total 20 moves

Surely, it was solved.

Let's look at another example.

A1B1C1D1C2B2A2
3 8 3 2 0 4 6 → Scrambled
↓ Step1. B1/C1 +4
3 3 7 2 0 4 6
↓ Step2. B2/C2 +2
3 3 7 2 2 6 6
↓ Step 3. Now that you have the Exception 2), mesh D1 gear with C1 gear, rotate D1 gear 360 degrees in either the CCW or CW direction that is convenient, set c1 to 0 or 6, and then perform Sub-step 2).
↓ Step3 Exception 2). D1/C1 +6
3 3 6 2 2 6 6
↓ Sub-step 2). D1/C2 +4
3 3 6 6 6 6 6
↓ Step4. A1/B1 +3
6 6 6 6 6 6 6 → Solved (Step5 skip)

Total 19 moves

This has also been solved.

By the way, let's try solving another example that was solved the other day in 9 moves using this solution method.

A1B1C1D1C2B2A2
6 0 5 1 1 1 1 → Scrambled
↓ Step1. B1/C1 -3
6 6 2 1 1 1 1
↓ Step2. skip
↓ Step3. D1/C2 +1
6 6 2 2 2 1 1
↓ Step4. A1/B1 -4
2 2 2 2 2 1 1
↓ Step5. A2/B2 +1
2 2 2 2 2 2 2 → Solved

Total 9 moves

The form of the solve has changed, but it was still solved in 9 moves.

Readers, please make up any 7 numbers (except for A1/A2 gears, be careful of "the valid range") and try to solve it accordingly.

I'll ask Doc what he thinks about this solution from the FMC point of view.

 

[Tito_H]

I received a reply from Doc right away.

The letter read, "Dear Tito, I know my puzzle gave you a lot of trouble, but I'm glad you found a solution. However, your solution is a little problematic from the perspective of FMC. I'd like you to study the Exception handling procedures in Step 3 a little more."

I thought, "I thought so." I hadn't felt quite satisfied with my own solution, so I had half expected Doc's criticism.

The letter continued, "For example, in the example below you posted, your solution requires 20 moves, but if you do it right, you can solve it in 18 moves. Good luck. From Doc."

A1B1C1D1C2B2A2
5 2 4 2 7 1 8 → Scrambled
↓ Step1. B1/C1 +3
5 5 7 2 7 1 8
↓ Step2. B2/C2 -2
5 5 7 2 5 8 8
↓ Step 3: Now that we have the Exception 3), switch the order of Sub-steps 1) and 2).
↓ Sub-step 2) D1/C1 +3
5 5 2 5 5 8 8
↓ Sub-step 1) D1/C2 -3
5 5 2 2 2 8 8
↓ Step4 A1/B1 -3
2 2 2 2 2 8 8
↓ Step5 A2/B2 -6
2 2 2 2 2 2 2 → Solved

Total 20 moves → Doc said this could be 18 moves.

I was impressed and thought, "Doc is amazing," and decided to review Step 3 of my solution…

 

[Tito_H]

Following Doc's instructions, I recently solved the string 5 2 4 2 7 1 8 in 20 moves, but when I changed the steps to the following, I was able to solve it in 18 moves. The change was to change B2/C2 -3 instead of B2/C2 -2 in Step 2. Although a2 is 8, b2 is 7, so the two numbers do not line up, but in Step 5, the B2 gear rotates from 8 to 0, skipping the 9, so a2 and b2 line up from 0 onwards.

(Improved procedure)
A1B1C1D1C2B2A2
5 2 4 2 7 1 8 → Scrambled
↓ Step1. B1/C1 +3
5 5 7 2 7 1 8
↓ Step2. B2/C2 -3 ← Changed!
5 5 7 2 4 7 8
↓ Step3: Now that we have the Exception 3), switch the order of Sub-steps 1) and 2).
↓ Sub-step 2) D1/C1 +2
5 5 1 4 4 7 8
↓ Sub-step 1) D1/C2 -3
5 5 1 1 1 7 8
↓ Step4 A1/B1 -4
1 1 1 1 1 7 8
↓ Step5 A2/B2 +3
1 1 1 1 1 1 1 → Solved

Total 18 moves

It's difficult to decide whether to choose B2/C2 -2 or B2/C2 -3 when doing Step 2, so I think it's best to first solve it using my method and count the total number of moves, then solve it using a different procedure and compare the total number of moves. Step 1 and Step 2 have the same concept, so the above also applies to Step 1.

I also reviewed the procedure for solving another sequence, 3 8 3 2 0 4 6, which I solved the other day in 19 moves. In the procedure from the other day, as shown below, the sequence takes the form of Exception 2) in Step 3, so it takes 6 moves to rotate the D1 gear once.

(Steps from the other day)
A1B1C1D1C2B2A2
3 8 3 2 0 4 6 → Scrambled
↓ Step1. B1/C1 +4
3 3 7 2 0 4 6
↓ Step2. B2/C2 +2
3 3 7 2 2 6 6
↓ Step3 Exception 2) D1/C1 +6 ←Too much!
3 3 6 2 2 6 6
↓ Step3 Sub-step 2) D1/C2 +4
3 3 6 6 6 6 6
↓ Step4 A1/B1 +3
6 6 6 6 6 6 6 → Solved (Step5 skip)

Total 19 moves

So, in Step 1, instead of rotating the B1/C1 gear by +4, I decided to rotate it in the opposite direction, rotating it by -5. Then, as shown below, I was able to solve it in just 14 moves! Although Step 1 requires one more move, Step 3 does not become an Exception, and as a result, the total number of moves is fewer.

(Improved procedure)
A1B1C1D1C2B2A2
3 8 3 2 0 4 6 → Scrambled
↓ Step1. B1/C1 -5 ←Changed!
3 3 6 2 0 4 6
↓ Step2. B2/C2 +2
3 3 6 2 2 6 6
↓ Step 3. D1/C2 +4
3 3 6 6 6 6 6
↓ Step4 A1/B1 +3
6 6 6 6 6 6 6 → Solved (Step5 skip)

Total 14 moves

Based on the above, I will add a new step, Number 6, to the solution procedure.

———————————-
Step 6. After solving Steps 1 to 5 above, perform the following optimization to see if the total number of moves can be reduced.

<Optimization >
1. In Steps 1, 2, and 3, try reversing the rotation direction of gears B1, B2, and D1.

2. In Steps 1 and 2, instead of setting b1 = a1 and b2 = a2, try setting b1 = a1 - 1 and b2 = a2 - 1. (In Steps 4 and 5, if you rotate gears B1 and B2 from 8 to 0, the numbers will line up from 0 onwards, with b1 = a1 and b2 = a2.)
———————————-

It would be ideal if you could look ahead <Optimization> when executing Steps 1, 2, and 3, but I think that's a high hurdle. And, since implementing <Optimization> takes a lot of time, it might be a good idea to only implement it if the resolution falls under one of the Exceptions in step 3.

I'll report this to the doc.

 

[Tito_H]

I have made a web program with which you can enjoy this puzzle
☟

https://scratch.mit.edu/projects/1311781586