03-02-2016, 09:33 PM

(02-29-2016, 04:23 PM)agenta Wrote: Let`s do the math:That is incorrect. Perhaps you used the --keyspace option? That isn't usable for these calculations because the output figure is scaled with a "hidden multiplier", and that multiplier is different for Mask1/2 vs. Mask3 so you can't compare the numbers at all.

According to hashcat:

Mask1: “?a?l?l?l?l?l?l?l” has 308 915 776 combinations.

Mask2: “?a?u?l?l?l?l?l?l” has 308 915 776 combinations.

Mask3: “?a?a?a?a?a?a?a?a” has 735 091 890 625 combinations.

Let assume that I want “?a” to float in mask1 on all position it is equal to run 8 different masks (?a?l?l?l?l?l?l?l, ?l?a?l?l?l?l?l?l, ?l?l?a?l?l?l?l?l, ?l?l?l?a?l?l?l?l ….. ?l?l?l?l?l?l?l?a). The result will be:

308 915 776 * 8= 2 471 326 208 (~0,33 % from Mask3).

If I want “?a” and “?u” to float then we have 8 * 7 = 56 different masks for execution. Total of combination will be:

308 915 776*56= 17 299 283 456 (~ 2,353% from Mask3).

I don’t see your logic magnum.

Here's math:

Mask1: “?a?l?l?l?l?l?l?l” has 95*26^7 = 763 021 966 720 combinations.

Mask2: “?a?u?l?l?l?l?l?l” has the same numbers.

Mask3: “?a?a?a?a?a?a?a?a” has 95^8 = 6 634 204 312 890 625 combinations.

So your exact figures are not correct BUT other than that you are right, sorry! I was thinking of the opposite case (that Phil mentioned) to try to reduce ?a?a?a?a?a?a?a?a given that at least one of them is ?u and one is ?l. That will end up more expensive than just biting the bullet.