%%lockres%% collision probability magic marker: 16,777,215May 29th, 2009
@jrowlandjones blogged about a dubious deadlock case. I recommend this article as is correct and presents a somewhat esoteric case of deadlock: the resource hash collision. The lock manager in SQL Server doesn’t know what it locks, it just locks ‘resources’ (basically strings). It is the job of higher level components like the the access methods of the storage engine to present the ‘resource’ to the lock manager and ask for the desired lock. When locking rows in a heap or a b-tree the storage engine will synthesize a ‘resource’ from the record identifier. Since these resources have a limited length, the storage engine has to reduce the effective length of a key to the maximum length is allowed to present to the lock manager, and that means that the record’s key will be reduced to 6 bytes. This is achieved by hashing the key into a 6 byte hash value. Nothing spectacular here.
But if you have a table with a key of length 50 bytes and its reduced to 6 bytes, you may hit a collision. So how likely is this to happen?
On 6 bytes there are 281,474,976,710,656 distinct possible values. Its a pretty big number? Not that big actually. If we meet at a party and I say ‘I bet somebody in the room shares your birthday’ would you bet against me? You probably should What if I change my question to ‘I bet there are two people in this room that share the birthday!’? Now I will probably take your money. This is called a ‘meet-in-the-middle‘ attack in cryptography and basically it says that you get a 50% collision probability at half the hash length. So the SQL %%lockres%% hash will produce two records with same hash, with a 50% probability, out of the table, any table, of only 16,777,215 record. That suddenly doesn’t look like a cosmic constant, does it? And keep in mind that this is the absolutely maximum value, where the key has a perfectly random distribution. In reality keys are not that random after all. Take Jame’s example: a datetime (8 bytes), a country code (1 byte), a group code (2 bytes) and a random id (4 bytes). From these 15 bytes quite a few are actually constant: eg. every date between 2000 and 2010 has the first 4 bytes identical (0×00) and the 5th byte only has two possible values (0×08 or 0×09). If from the other codes (country, group) we use only 50% of the possible values, then in effect we use, generously, just 10 bytes of the 15 bytes of the key. This means the table has a 50% collision probability at only about 11 million records. considering that he was doing a ‘paltry’ 900 million records upload, no wonder he got collisions…