DIY frames analysis

I think I found the way the system counts the directional inputs during the struggle.

To get the stagger I used Wolf's 62K where you need 4 inputs (directional ones or buttons) to break it.
So I tried several combinations of directional inputs only to determine the exact "struggling value" associated at each of them.

Keeping it short in a string of directional inputs without any neutral position in it the system does not count any diagonals except for the final one when the string ends and there's a neutral after that.

This model is consistent even with KoD's results.


Here's what I did in details.

From the failure or success in blocking the following up P (after Wolf's drop kick stagger) by a succession of particualar directional inputs sequences I was able to determine when an inputs does count or not.

If you want to replicate the test keep in mind that you have to hold the guard button before the struggle window (to be sure just hold G even before being hit by Wolf's drop kick).

The first string I looked into it was the succesfull shortest one among those with continuous and consecutive inputs (no pause and with the stick going clockwise/counterwise).

Knowing that four cardinal directions suffice to break this stagger I started out with 6321478 (it contains four cardinal directions) and it did work.

Then I tried 632147 and it worked too.

Tried 63214 but this time it didn't work

Now I did the same process but I started the string with a diagonal instead.

Tried 3214789 and it worked.

Tried 321478 and it failed.
This last one was the "equivalent" of 632147. Both have three cardinal directions and three diagonals but they are out of phase for just one cardinal direction or diagonal (it depends on how you look at them).
From this I got that the "struggling value" (from now on SV) depends somehow on the inputs' order.

Then I looked at the strings in the form of a "quarter circle motion" (236), a neutral (5) and a last part consisting of continuous and consecutive inputs.
The objective was to find the minimum amount of additional inputs needed in the last part to break the stagger.

It worked with ("~" means Neutral"):

- 632~47
- 632~63
- 632~69

It failed with

- 632~32
- 632~8

Being the first part (632) equal in bot cases the difference is in the second part.
From the failure of 632~8, being a string of four inputs (what is needed to break the drop kick' stagger) with three cardinal directions, we can deduce that the diagonal 3 between two cardinal inputs in the first part has a SV < 1 (every cardinal directions has a SV = 1).

From the failure of 632~32 we can deduce that the diagonal in the second part (32) has a SV < 1 and that the sum of the SVs of both diagonals is < 1 and if they have the same weight on average the SV is < 1/2

From the succesfull cases instead we can deduce that the sum of the SV of the two diagonals is >= 1 and in particular that the SV of the diagonal in the second part (47,63, 69) is > of the SV of the diagonal in the first part (632).

Even if we swap the first part (632) with the second part (47) the string (47~632) works the same.
In the case of a not working string swapping the two parts (632~32 -> 32~632) doesn't change the result too

Now to find out the exact SV for the three type of diagonals (632, 32, 47) I looked into those string in the form of cardinal direction, a neutral, a cardinal direction, a neutral and a last part as the second part in the succesfull examples (47, 63, 69)

So I tried with 6~4~47, 6~4~63, 6~4~69 and all of them worked.
From this we can deduce that the SV of a diagonal before a neutral is = 1 (as stated before you can swap the string's parts and the result will not change).

Now we look back at the strings 3214789 (that worked) and 321478 (that did not work).
From these we can get that the SV of the first diagonal (3214789, 321478) is < 1 and that the sum of the SVs of the first three diagonals (3214789, 321478) is < 1 and if they have the same weight on average the SV is < 1/3.
As we can see with longer strings we get that the average SV for all the diagonals except the ones before a netrual goes down.
Because there isn't a constructive proof to demonstrate that this value is 0 an educated guess is that these type of diagonals have SV = 0

To see if this theory is consistent I tried several other sequences even with buttons (the SV of a button is = 1).

The successfull notable onea are:

- 23~23
- 63~63
- 21~21
- 623P (aka the shoryuken)

The failed notable onea are:

- 6~4~36
- 32~36
- 32~63

Other successfull ones are:

- 212~21
- 632~47
- 632147
- 214~21
- 236P9
- 32~478
- 232~63
-632~63
- 23~63~63
- 23~7~32
- 74~32~4
- 3214~8~2
- 236P32
- 632126
- 6326P
- 6321P
- 421P
- 63~63~6
- 3~63236
- 232~3~63
-3~3~63
- 6~4~23
-632696
- 632~69
- 232123
- 3212321
- 2321232
- 63269
- 212321
- 232323
- 632123
- 696321
- 63P21
- 32P~632
- 23632P
- 32~63P
- 2~63P
- 3P2~3P
- 21~63~47
- 47~632
- 23~6~4
- 6~4~63
- 63~4~6
- 23~23
- 63~63
- 21~21
- 21~212
- 3~47~2
- 3~63P2


Other failed ones are:

- 32~478
- 32~32
- 3~32P
- 323P
- 63214
- 632~8
- 63236
- 36P9
- 321478
- 96214
- 963214
- 14P7
- 63P2
- 6~4~36
- 32123
- 321232
- 23632
- 3632P
-363P2
- 32~36
- 32~63