I think I'm going to get sub 900 since i just realized i can in-parallel compute whether stage 5 of the hash is odd just by looking at bits 16 and 0 of stage 4 with less delay.....
I think I'm going to get sub 900 since i just realized i can in-parallel compute whether stage 5 of the hash is odd just by looking at bits 16 and 0 of stage 4 with less delay.....
Let me put down my thought process: You have to start to think of designing a 6-slot x8-len vector pipeline doing 48 hashes in parallel first which needs at least 10 steps —- if you convert three stages to multiply adds and do parallel XORs for the other three) —- the problem with 10 cycle hashing is you need to cram 96 scalar xors along side your vector pipeline, so that will use all 12 ALUs for 8 of those cycles. Leaving you only 24 more scalar ops per hash cycle which isn’t enough for the 48 tree value xors..
so you must use at least 11 steps per hash, with 96 xors (including the tree value xor) done in the scalar alus using 8 steps, and giving 3*12 Alu ops per hash cycle. You need 12 more ops per hash to do odd/even, so you must be 12 stages, and just do all of the hash ops in valu, 4 cycles of 12 alus doing modulo, 8 cycles x 12 alus free
With 12 steps and 48 parallel you’re absolute minimum could be 4096/48 x 12 = 1,024 cycles, since stage 10 can be optimized (you don’t need the odd/even modulo cycle, and can use some of those extra scalar cycles to pre-xor the constant can save you ~10 cycles. 1024 gonna be real hard, but I can imagine shenanigans to get it down to 1014, sub-1000 possible by throwing more xor to the scalar alus.
I performed a similar analysis to you and found it very difficult to imagine sub-1000. Your comment I think convinced me that it may be possible, though. Interesting.
I'm below the threshold for recruiting but not below Claude at the moment. Not sure where I am going wrong.
For the first several rounds (when every tree value is in use) Combine the stage 5 XOR with the subsequent round’s tree XORs. You can determine even/odd in hash stage 5 starting with a ^ (a>>16) without Xoring the constant, then you can only need one XOR, this saves you a ton of XORs
Create separate instruction bundles for the first round, rounds 1-5 (combining hash stages 5 XOR with next round tree XORs) and 6-9 (not every tree node is used anymore), round 10 round 11-14 and round 15 and combine them.
you can use add_imm in parallel to load consts. stage 0 you have to do load the tree first and the vals, by later stages when everything is in scratch, you could use 12 scalar XORs and 6 vector XORs on scratch. once you vload vals, you can start to do XORs but can only advance so much at a time, so I’m starting to work on getting hash stages moving to different rounds faster to hide the initial vloads and get to the heavy load section sooner and spread the load pain.
BROADCAST LOAD SCHEDULE
======================================================================
Round | Unique | Load Strategy
------|--------|------------------------------------------
0 | 1 | 1 broadcast → all 256 items
1 | 2 | 2 broadcasts → groups
2 | 4 | 4 broadcasts → groups
3 | 8 | 8 broadcasts → groups
4 | 16 | 16 broadcasts → groups
5 | 32 | 32 broadcasts → groups
6 | 63 | 63 loads (sparse, use indirection)
7 | 108 | 108 loads (sparse, use indirection)
8 | 159 | 159 loads (sparse, use indirection)
9 | 191 | 191 loads (sparse, use indirection)
10 | 224 | 224 loads (sparse, use indirection)
11 | 1 | 1 broadcast → all 256 items
12 | 2 | 2 broadcasts → groups
13 | 4 | 4 broadcasts → groups
14 | 8 | 8 broadcasts → groups
15 | 16 | 16 broadcasts → groups
Total loads with grouping: 839Total loads naive: 4096
Load reduction: 4.9x