2nd Solve--LaGard 3330

[CPT1911]

Glad you are finding a method that works for you!

Yes, you are correct to absolutely suspect that the low point you find during your check of the wheelpack every ten increments may turn out to be a gate. That's why I also do this:

5. Use step 3 low for W3, step 4 low point for W2, and brute force W1 in increments of 2 beginning with low spots from step 1

If you consider that each gate is approximately 3 increments wide, and you assume that your lock will read a single wheel when moving the entire wheel pack, you actually have a 30% change of catching a gate when you do the initial wheel pack map (you are actually checking 99,0,1 & 9,10,11 & 19,20,21, etc for a total of 30 increments checked). If, on your lock, two wheels can be read when the entire wheel pack moves, you can double that to 60% of finding a gate. Or to think of it another way, we use 2.5 increments as the standard when searching for a gate...that's 40 readings on the dial...so searching 10 increments is actually doing 25% of the total work required for a full spin around. Naturally, the world doesn't always follow these convenient probabilities, but the point is that those first ten readings are worth paying attention to even after you've found a gate!

Isn't it weird how it works sometimes?

Right?! I start a manipulation with a picture in my head of what is going to happen, and then all of a sudden the lock does something unexpected and the plan has to change! What's the expression? "No plan survives initial contact with the enemy..."

[LockManipulator]

If you consider that each gate is approximately 3 increments wide, and you assume that your lock will read a single wheel when moving the entire wheel pack, you actually have a 30% change of catching a gate when you do the initial wheel pack map

I don't think I quite understand this. Are you assuming there is a 100% chance a lock will indicate a gate on the first graph and so there is a 30% chance for each wheel? Or a 30% chance of finding a gate at all? Because if there is a 100% chance then there would still not be an equal chance for the wheels to indicate, the lock will be biased towards the 3rd wheel usually.

[CPT1911]

I don't think I quite understand this. Are you assuming there is a 100% chance a lock will indicate a gate on the first graph and so there is a 30% chance for each wheel? Or a 30% chance of finding a gate at all? Because if there is a 100% chance then there would still not be an equal chance for the wheels to indicate, the lock will be biased towards the 3rd wheel usually.

Hi Daggers! Flywheel's experince of finding a gate during his wheel mapping was something that happened to me a few times and that is what prompted my response.

I was suggesting that those 10 readings, actually succeed in searching more of the dial than we might intuitively think, and therefore, any extreme low point that you find in this wheel mapping procedure, should be revisited immediately and eliminated when searching for other gates. My calculation was as follows:

Actual Coverage
99,0,1
9,10,11
19,20,21
29,30,31
39,40,41
49,50,51
59,60,61
69,70,71
79,80,81
89,90,91

As you can see, dialing AWx by discrete 10s actually covers 30 numbers. 30/100 possibilities = 30% of dial is covered by dialing 10 numbers during wheel mapping sequence, assuming, as you pointed out, that ANY wheel could actually read in that initial AWx rotation.

In actual practice, I guess we usually dial with a 2.5 increment variance to be more conservative...that's 40 readings. But still, those 10 mapping readings constitute 25% of the readings we would otherwise take in a normal spin.... 25% ≈ 30%

I'm not the sharpest math guy in the world so please tell me if I am missing something here!

Edit: When I use the term "wheel mapping" in this context, I am not talking about exhaustively searching the wheels by 2s or 2.5s only taking a reading every 10.

[LockManipulator]

Ok I follow your line of thinking here! I'm not too much of a math person myself either but it'd be cool to see someone actually try to apply this to a large number of LaGards and see how well the percentages actually fit In reality

[flywheel]

Do we really have a 30% chance of finding a gate when mapping by 10?

First we make the assumption that only one wheel in the pack is returning information. If true, all graphs of a lock's full wheel pack would be identical no matter what the combination. This is true with the exception of the 2-3 increment space where the fence lies in a gate and rests on a lower wheel.
How about the gate being three increments wide? The variance is +/- 1.25 (or gates being 2.5 increments wide) which seems to hold true when testing the combinations I've found. I might even be inclined to say the variance is closer to +/- 1 increment.

For the sake of argument we'll consider the above assumptions okay. The question now becomes how much of the wheel pack are we really testing? I would argue that we only have 20% chance of finding a gate when mapping by 10. The actual length covered by three numbers is only two increments wide.

0 1 2
| | |

If we do that ten times then only 20 increment lengths have been explored out of the possible 100.

[flywheel]

Not done yet. Instead I tried approaching this from the standpoint of probability. Statistics can lead to some screwy results if calculated incorrectly. So, here we go, for better or worse...

Assuming the fence only tests one wheel and the gate is 3 increments wide the probability of finding a gate is 3/100. The probability of NOT finding a gate is 97/100. If we were to test the wheel pack 10 times the chances of missing the gate 9 times and finding it once yields the following:

22.8%

If requested, I'll attempt to type up the equation used for this type of problem. If I bungled it up let me know. Good day!

[Oldfast]

http://youtu.be/TCKOI24k_UY

[flywheel]

Changed the combination and got the darn thing open again. This is quite a humbling hobby. After the past couple success's I thought I had this lock "solved". A few saw tooth graphs later I had to ask, "what am I looking for again?"

No luck finding a gate when mapping the wheel pack in increments of ten this time. After parking W1 and W2 at the low spot I found a continuous high spot over 20 increments. I wouldn't be able to apply Altashot's 1/4 dial theory with that. I tried the second lowest spot and mapped W3 again. It gave me enough indication for a single high spot that I could apply Altashot's theorem.

In the end it all worked out. So, I guess if parking the first two wheels at the low spot (after mapping the wheel pack by 10) doesn't give any good info park the wheels at the second lowest spot and map W3 again.

After starting with a LaGard will it be noticeably easier to open an S&G?

The next goal is to disassemble the lock and put it back together. Putting it back together properly will be an added bonus. Enough rambling for now...

Good night and good luck!

[Oldfast]

Great stuff! Gotta say, I've been impressed with a lot of your thoughts since you came around.

....After starting with a LaGard will it be noticeably easier to open an S&G?

There's ALWAYS exceptions, but for the most part... yeah, I wouldn't be a
bit surprised to see you plow through a fair number of the standard S&G's.

[CPT1911]

So, I guess if parking the first two wheels at the low spot (after mapping the wheel pack by 10) doesn't give any good info park the wheels at the second lowest spot and map W3 again.

Interesting idea, flywheel! I have never tried using the "second" lowest spot, but anything is possible. If I don't find W3's gate with W1+W2 at the low, I immediately switch to AWR (my preference) or AWL. I've always found a gate on this second run. I park W1+W2 at this gate indication and then take W3 in isolation a second time. I've always found W3's gate on this second run...basically, the initial park of W1+W2 on an increment of 10 wasn't "low enough" to expose W3 in my experience. If you haven't changed the combo yet on your lock, I would be really interested in hearing if this method works for you. Naturally, this is a more conventional approach at a time when it seems you are trying to take advantage of alta's shortcut...I haven't gotten there yet.

After starting with a LaGard will it be noticeably easier to open an S&G?

Man, I'd say it depends! My experience is that it is EASIER to find gate indications on the S&G but HARDER to place them on the correct wheel. I've had lots of S&Gs give me total garbage on high low tests and even iso tests whereas I tend to trust those tests more on the LaGard. I've had very few indicate W3 with W1+W2 parked (can't explain why). Could just be my batch of S&G's but man I've blown a lot of opens by putting gates on the wrong wheels! I don't do the wheel map technique we have been talking about with the LaGard on the S&Gs. I use a totally different approach.

[flywheel]

If you haven't changed the combo yet on your lock, I would be really interested in hearing if this method works for you. Naturally, this is a more conventional approach at a time when it seems you are trying to take advantage of alta's shortcut...I haven't gotten there yet.

This time I did AWL and had to decide between two low spots, 80 and 100. I double checked and decided 100 was lower by <1/8. So I proceeded W1+W2 @L100 and W3 AR. As I descended through the numbers I saw the high point come up where I expected it (@55), but it was much better defined than before. Looking good so far I continued on and soon found a spot significantly higher at 37.5 and 27.5. This threw the 1/4 dial theory out the window, but that was okay. Instead, I had an ideal gate signature at 32.5 that can only be described as "reaching out of the page to slap my face".

Maybe when AWR (or vice versa) fails to yield a gate or obvious "1/4 dial high point" I should go AWL and try again.

Take it easy!

[CPT1911]

Thanks for trying it and sharing your results, flywheel!