- by Sasha Rush - srush_nlp
GPU architectures are critical to machine learning, and seem to be becoming even more important every day. However, you can be an expert in machine learning without ever touching GPU code. It is hard to gain intuition working through abstractions.
This notebook is an attempt to teach beginner GPU programming in a completely interactive fashion. Instead of providing text with concepts, it throws you right into coding and building GPU kernels. The exercises use NUMBA which directly maps Python code to CUDA kernels. It looks like Python but is basically identical to writing low-level CUDA code. In a few hours, I think you can go from basics to understanding the real algorithms that power 99% of deep learning today. If you do want to read the manual, it is here:
I recommend doing these in Colab, as it is easy to get started. Be
sure to make your own copy, turn on GPU mode in the settings (Runtime / Change runtime type, then set Hardware accelerator to GPU), and
then get to coding.
(If you are into this style of puzzle, also check out my Tensor Puzzles for PyTorch.)
!pip install -qqq git+https://github.com/chalk-diagrams/planar git+https://github.com/danoneata/chalk@srush-patch-1
!wget -q https://github.com/srush/GPU-Puzzles/raw/main/robot.png https://github.com/srush/GPU-Puzzles/raw/main/lib.pyimport numba
import numpy as np
import warnings
from lib import CudaProblem, Coordwarnings.filterwarnings(
action="ignore", category=numba.NumbaPerformanceWarning, module="numba"
)Implement a "kernel" (GPU function) that adds 10 to each position of vector a
and stores it in vector out. You have 1 thread per position.
Warning This code looks like Python but it is really CUDA! You cannot use standard python tools like list comprehensions or ask for Numpy properties like shape or size (if you need the size, it is given as an argument). The puzzles only require doing simple operations, basically +, *, simple array indexing, for loops, and if statements. You are allowed to use local variables. If you get an error it is probably because you did something fancy :).
Tip: Think of the function call as being run 1 time for each thread.
The only difference is that cuda.threadIdx.x changes each time.
def map_spec(a):
return a + 10
def map_test(cuda):
def call(out, a) -> None:
local_i = cuda.threadIdx.x
# FILL ME IN (roughly 1 lines)
out[local_i] = a[local_i] + 10
return call
SIZE = 4
out = np.zeros((SIZE,))
a = np.arange(SIZE)
problem = CudaProblem(
"Map", map_test, [a], out, threadsperblock=Coord(SIZE, 1), spec=map_spec
)
problem.show()# Map
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 1 | 1 | 0 | 0 |
problem.check()Passed Tests!
Implement a kernel that adds together each position of a and b and stores it in out.
You have 1 thread per position.
def zip_spec(a, b):
return a + b
def zip_test(cuda):
def call(out, a, b) -> None:
local_i = cuda.threadIdx.x
# FILL ME IN (roughly 1 lines)
out[local_i] = a[local_i] + b[local_i]
return call
SIZE = 4
out = np.zeros((SIZE,))
a = np.arange(SIZE)
b = np.arange(SIZE)
problem = CudaProblem(
"Zip", zip_test, [a, b], out, threadsperblock=Coord(SIZE, 1), spec=zip_spec
)
problem.show()# Zip
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 2 | 1 | 0 | 0 |
problem.check()Passed Tests!
Implement a kernel that adds 10 to each position of a and stores it in out.
You have more threads than positions.
def map_guard_test(cuda):
def call(out, a, size) -> None:
local_i = cuda.threadIdx.x
# FILL ME IN (roughly 2 lines)
if local_i < size:
out[local_i] = a[local_i] + 10
return call
SIZE = 4
out = np.zeros((SIZE,))
a = np.arange(SIZE)
problem = CudaProblem(
"Guard",
map_guard_test,
[a],
out,
[SIZE],
threadsperblock=Coord(8, 1),
spec=map_spec,
)
problem.show()# Guard
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 1 | 1 | 0 | 0 |
problem.check()Passed Tests!
Implement a kernel that adds 10 to each position of a and stores it in out.
Input a is 2D and square. You have more threads than positions.
def map_2D_test(cuda):
def call(out, a, size) -> None:
local_i = cuda.threadIdx.x
local_j = cuda.threadIdx.y
# FILL ME IN (roughly 2 lines)
if local_i < size and local_j < size:
out[local_i, local_j] = a[local_i, local_j] + 10
return call
SIZE = 2
out = np.zeros((SIZE, SIZE))
a = np.arange(SIZE * SIZE).reshape((SIZE, SIZE))
problem = CudaProblem(
"Map 2D", map_2D_test, [a], out, [SIZE], threadsperblock=Coord(3, 3), spec=map_spec
)
problem.show()# Map 2D
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 1 | 1 | 0 | 0 |
problem.check()Passed Tests!
Implement a kernel that adds a and b and stores it in out.
Inputs a and b are vectors. You have more threads than positions.
def broadcast_test(cuda):
def call(out, a, b, size) -> None:
local_i = cuda.threadIdx.x
local_j = cuda.threadIdx.y
# FILL ME IN (roughly 2 lines)
if local_i < size and local_j < size:
out[local_i, local_j] = a[local_i, 0] + b[0, local_j]
return call
SIZE = 2
out = np.zeros((SIZE, SIZE))
a = np.arange(SIZE).reshape(SIZE, 1)
b = np.arange(SIZE).reshape(1, SIZE)
problem = CudaProblem(
"Broadcast",
broadcast_test,
[a, b],
out,
[SIZE],
threadsperblock=Coord(3, 3),
spec=zip_spec,
)
problem.show()# Broadcast
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 2 | 1 | 0 | 0 |
problem.check()Passed Tests!
Implement a kernel that adds 10 to each position of a and stores it in out.
You have fewer threads per block than the size of a.
Tip: A block is a group of threads. The number of threads per block is limited, but we can
have many different blocks. Variable cuda.blockIdx tells us what block we are in.
def map_block_test(cuda):
def call(out, a, size) -> None:
i = cuda.blockIdx.x * cuda.blockDim.x + cuda.threadIdx.x
# FILL ME IN (roughly 2 lines)
if i < size:
out[i] = a[i] + 10
return call
SIZE = 9
out = np.zeros((SIZE,))
a = np.arange(SIZE)
problem = CudaProblem(
"Blocks",
map_block_test,
[a],
out,
[SIZE],
threadsperblock=Coord(4, 1),
blockspergrid=Coord(3, 1),
spec=map_spec,
)
problem.show()# Blocks
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 1 | 1 | 0 | 0 |
problem.check()Passed Tests!
Implement the same kernel in 2D. You have fewer threads per block
than the size of a in both directions.
def map_block2D_test(cuda):
def call(out, a, size) -> None:
i = cuda.blockIdx.x * cuda.blockDim.x + cuda.threadIdx.x
# FILL ME IN (roughly 4 lines)
j = cuda.blockIdx.y * cuda.blockDim.y + cuda.threadIdx.y
if i < size and j < size:
out[i, j] = a[i, j] + 10
return call
SIZE = 5
out = np.zeros((SIZE, SIZE))
a = np.ones((SIZE, SIZE))
problem = CudaProblem(
"Blocks 2D",
map_block2D_test,
[a],
out,
[SIZE],
threadsperblock=Coord(3, 3),
blockspergrid=Coord(2, 2),
spec=map_spec,
)
problem.show()# Blocks 2D
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 1 | 1 | 0 | 0 |
problem.check()Passed Tests!
Implement a kernel that adds 10 to each position of a and stores it in out.
You have fewer threads per block than the size of a.
Warning: Each block can only have a constant amount of shared
memory that threads in that block can read and write to. This needs
to be a literal python constant not a variable. After writing to
shared memory you need to call cuda.syncthreads to ensure that
threads do not cross.
(This example does not really need shared memory or syncthreads, but it is a demo.)
TPB = 4
def shared_test(cuda):
def call(out, a, size) -> None:
shared = cuda.shared.array(TPB, numba.float32)
i = cuda.blockIdx.x * cuda.blockDim.x + cuda.threadIdx.x
local_i = cuda.threadIdx.x
if i < size:
shared[local_i] = a[i]
cuda.syncthreads()
# FILL ME IN (roughly 2 lines)
out[i] = shared[local_i] + 10
return call
SIZE = 8
out = np.zeros(SIZE)
a = np.ones(SIZE)
problem = CudaProblem(
"Shared",
shared_test,
[a],
out,
[SIZE],
threadsperblock=Coord(TPB, 1),
blockspergrid=Coord(2, 1),
spec=map_spec,
)
problem.show()# Shared
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 1 | 1 | 1 | 1 |
problem.check()Passed Tests!
Implement a kernel that sums together the last 3 position of a and stores it in out.
You have 1 thread per position. You only need 1 global read and 1 global write per thread.
Tip: Remember to be careful about syncing.
def pool_spec(a):
out = np.zeros(*a.shape)
for i in range(a.shape[0]):
out[i] = a[max(i - 2, 0) : i + 1].sum()
return out
TPB = 8
def pool_test(cuda):
def call(out, a, size) -> None:
shared = cuda.shared.array(TPB, numba.float32)
i = cuda.blockIdx.x * cuda.blockDim.x + cuda.threadIdx.x
local_i = cuda.threadIdx.x
# FILL ME IN (roughly 8 lines)
if i < size:
shared[i] = a[i]
cuda.syncthreads()
if i < 1:
out[i] = shared[i]
elif i < 2:
out[i] = shared[i] + shared[i - 1]
else:
out[i] = shared[i - 2] + shared[i - 1] + shared[i]
return call
SIZE = 8
out = np.zeros(SIZE)
a = np.arange(SIZE)
problem = CudaProblem(
"Pooling",
pool_test,
[a],
out,
[SIZE],
threadsperblock=Coord(TPB, 1),
blockspergrid=Coord(1, 1),
spec=pool_spec,
)
problem.show()# Pooling
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 1 | 1 | 3 | 1 |
problem.check()Passed Tests!
Implement a kernel that computes the dot-product of a and b and stores it in out.
You have 1 thread per position. You only need 2 global reads and 1 global write per thread.
Note: For this problem you don't need to worry about number of shared reads. We will handle that challenge later.
def dot_spec(a, b):
return a @ b
TPB = 8
def dot_test(cuda):
def call(out, a, b, size) -> None:
shared = cuda.shared.array(TPB, numba.float32)
i = cuda.blockIdx.x * cuda.blockDim.x + cuda.threadIdx.x
local_i = cuda.threadIdx.x
# FILL ME IN (roughly 9 lines)
if i < size:
shared[i] = a[i] * b[i]
cuda.syncthreads()
threshold = size // 2
while threshold > 0:
j = i + threshold
if i < threshold:
shared[i] = shared[i] + shared[j]
cuda.syncthreads()
threshold = threshold // 2
if i == 0:
out[0] = shared[0]
return call
SIZE = 8
out = np.zeros(1)
a = np.arange(SIZE)
b = np.arange(SIZE)
problem = CudaProblem(
"Dot",
dot_test,
[a, b],
out,
[SIZE],
threadsperblock=Coord(SIZE, 1),
blockspergrid=Coord(1, 1),
spec=dot_spec,
)
problem.show()# Dot
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 2 | 1 | 7 | 4 |
problem.check()Passed Tests!
Implement a kernel that computes a 1D convolution between a and b and stores it in out.
You need to handle the general case. You only need 2 global reads and 1 global write per thread.
def conv_spec(a, b):
out = np.zeros(*a.shape)
len = b.shape[0]
for i in range(a.shape[0]):
out[i] = sum([a[i + j] * b[j] for j in range(len) if i + j < a.shape[0]])
return out
MAX_CONV = 4
TPB = 8
TPB_MAX_CONV = TPB + MAX_CONV
SHARE_MEM_FULL_SIZE = 16
def conv_test(cuda):
def call(out, a, b, a_size, b_size) -> None:
i = cuda.blockIdx.x * cuda.blockDim.x + cuda.threadIdx.x
local_i = cuda.threadIdx.x
# FILL ME IN (roughly 17 lines)
shared_a = cuda.shared.array(SHARE_MEM_FULL_SIZE, numba.float32)
shared_b = cuda.shared.array(SHARE_MEM_FULL_SIZE, numba.float32)
for idx in range(i, a_size, TPB):
shared_a[idx] = a[idx]
for idx in range(i, b_size, TPB):
shared_b[idx] = b[idx]
cuda.syncthreads()
for idx in range(i, a_size, TPB):
ans = 0
for j in range(b_size):
if idx + j < a_size:
ans += shared_a[idx + j] * shared_b[j]
out[idx] = ans
return call
# Test 1
SIZE = 6
CONV = 3
out = np.zeros(SIZE)
a = np.arange(SIZE)
b = np.arange(CONV)
problem = CudaProblem(
"1D Conv (Simple)",
conv_test,
[a, b],
out,
[SIZE, CONV],
Coord(1, 1),
Coord(TPB, 1),
spec=conv_spec,
)
problem.show()# 1D Conv (Simple)
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 2 | 1 | 6 | 2 |
problem.check()Passed Tests!
Test 2
out = np.zeros(15)
a = np.arange(15)
b = np.arange(4)
problem = CudaProblem(
"1D Conv (Full)",
conv_test,
[a, b],
out,
[15, 4],
Coord(2, 1),
Coord(TPB, 1),
spec=conv_spec,
)
problem.show()# 1D Conv (Full)
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 3 | 2 | 16 | 3 |
problem.check()Passed Tests!
Implement a kernel that computes a sum over a and stores it in out.
If the size of a is greater than the block size, only store the sum of
each block.
We will do this using the parallel prefix sum algorithm in shared memory. That is, each step of the algorithm should sum together half the remaining numbers. Follow this diagram:
TPB = 8
def sum_spec(a):
out = np.zeros((a.shape[0] + TPB - 1) // TPB)
for j, i in enumerate(range(0, a.shape[-1], TPB)):
out[j] = a[i : i + TPB].sum()
return out
def sum_test(cuda):
def call(out, a, size: int) -> None:
cache = cuda.shared.array(TPB, numba.float32)
i = cuda.blockIdx.x * cuda.blockDim.x + cuda.threadIdx.x
local_i = cuda.threadIdx.x
# FILL ME IN (roughly 12 lines)
if i < size:
cache[local_i] = a[i]
cuda.syncthreads()
# stride = TPB // 2
# while stride > 0:
# if local_i < stride and local_i + stride < size:
# cache[local_i] = cache[local_i] + cache[local_i + stride]
# cuda.syncthreads()
# stride = stride // 2
stride = 1
while stride < TPB:
if local_i % (stride * 2) == 0 and local_i + stride < size:
cache[local_i] = cache[local_i] + cache[local_i + stride]
cuda.syncthreads()
stride *= 2
if local_i == 0:
out[cuda.blockIdx.x] = cache[local_i]
return call
# Test 1
SIZE = 6
out_num = (SIZE // TPB) + 1
out = np.zeros(out_num)
inp = np.arange(SIZE)
problem = CudaProblem(
"Sum (Simple)",
sum_test,
[inp],
out,
[SIZE],
Coord(out_num, 1),
Coord(TPB, 1),
spec=sum_spec,
)
problem.show()# Sum (Simple)
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 1 | 1 | 7 | 4 |
problem.check()Passed Tests!
Test 2
SIZE = 15
out = np.zeros(2)
inp = np.arange(SIZE)
problem = CudaProblem(
"Sum (Full)",
sum_test,
[inp],
out,
[SIZE],
Coord(2, 1),
Coord(TPB, 1),
spec=sum_spec,
)
problem.show()# Sum (Full)
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 1 | 1 | 7 | 4 |
problem.check()Passed Tests!
Implement a kernel that computes a sum over each column of a and stores it in out.
TPB = 8
def sum_spec(a):
out = np.zeros((a.shape[0], (a.shape[1] + TPB - 1) // TPB))
for j, i in enumerate(range(0, a.shape[-1], TPB)):
out[..., j] = a[..., i : i + TPB].sum(-1)
return out
def axis_sum_test(cuda):
def call(out, a, size: int) -> None:
cache = cuda.shared.array(TPB, numba.float32)
i = cuda.blockIdx.x * cuda.blockDim.x + cuda.threadIdx.x
local_i = cuda.threadIdx.x
batch = cuda.blockIdx.y
# FILL ME IN (roughly 12 lines)
if local_i < size:
cache[local_i] = a[batch, local_i]
cuda.syncthreads()
# stride = TPB // 2
# while stride > 0:
# if local_i < stride and local_i + stride < size:
# cache[local_i] = cache[local_i] + cache[local_i + stride]
# cuda.syncthreads()
# stride = stride // 2
stride = 1
while stride < TPB:
if local_i % (stride * 2) == 0 and local_i + stride < size:
cache[local_i] = cache[local_i] + cache[local_i + stride]
cuda.syncthreads()
stride *= 2
if local_i == 0:
out[batch, 0] = cache[0]
return call
BATCH = 4
SIZE = 6
out = np.zeros((BATCH, 1))
inp = np.arange(BATCH * SIZE).reshape((BATCH, SIZE))
problem = CudaProblem(
"Axis Sum",
axis_sum_test,
[inp],
out,
[SIZE],
Coord(1, BATCH),
Coord(TPB, 1),
spec=sum_spec,
)
problem.show()# Axis Sum
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 1 | 1 | 7 | 4 |
problem.check()Passed Tests!
Implement a kernel that multiplies square matrices a and b and
stores the result in out.
Tip: The most efficient algorithm here will copy a block into shared memory before computing each of the individual row-column dot products. This is easy to do if the matrix fits in shared memory. Do that case first. Then update your code to compute a partial dot-product and iteratively move the part you copied into shared memory. You should be able to do the hard case in 6 global reads.
def matmul_spec(a, b):
return a @ b
TPB = 3
def mm_oneblock_test(cuda):
def call(out, a, b, size: int) -> None:
a_shared = cuda.shared.array((TPB, TPB), numba.float32)
b_shared = cuda.shared.array((TPB, TPB), numba.float32)
i = cuda.blockIdx.x * cuda.blockDim.x + cuda.threadIdx.x
j = cuda.blockIdx.y * cuda.blockDim.y + cuda.threadIdx.y
local_i = cuda.threadIdx.x
local_j = cuda.threadIdx.y
# FILL ME IN (roughly 14 lines)
sum = 0
for tile in range(0, size, TPB):
if i < size and tile + local_j < size:
a_shared[local_i, local_j] = a[i, tile + local_j]
if j < size and tile + local_i < size:
b_shared[local_i, local_j] = b[tile + local_i, j]
cuda.syncthreads()
for k in range(TPB):
if tile + k < size:
sum += a_shared[local_i, k] * b_shared[k, local_j]
if i < size and j < size:
out[i, j] = sum
return call
# Test 1
SIZE = 2
out = np.zeros((SIZE, SIZE))
inp1 = np.arange(SIZE * SIZE).reshape((SIZE, SIZE))
inp2 = np.arange(SIZE * SIZE).reshape((SIZE, SIZE)).T
problem = CudaProblem(
"Matmul (Simple)",
mm_oneblock_test,
[inp1, inp2],
out,
[SIZE],
Coord(1, 1),
Coord(TPB, TPB),
spec=matmul_spec,
)
problem.show(sparse=True)# Matmul (Simple)
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 2 | 1 | 4 | 2 |
problem.check()Passed Tests!
Test 2
SIZE = 8
out = np.zeros((SIZE, SIZE))
inp1 = np.arange(SIZE * SIZE).reshape((SIZE, SIZE))
inp2 = np.arange(SIZE * SIZE).reshape((SIZE, SIZE)).T
problem = CudaProblem(
"Matmul (Full)",
mm_oneblock_test,
[inp1, inp2],
out,
[SIZE],
Coord(3, 3),
Coord(TPB, TPB),
spec=matmul_spec,
)
problem.show(sparse=True)# Matmul (Full)
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 6 | 1 | 16 | 6 |
problem.check()Passed Tests!