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https://github.com/srush/GPU-Puzzles
Solve puzzles. Learn CUDA.
https://github.com/srush/GPU-Puzzles
cuda machine-learning puzzles
Last synced: 11 days ago
JSON representation
Solve puzzles. Learn CUDA.
- Host: GitHub
- URL: https://github.com/srush/GPU-Puzzles
- Owner: srush
- License: mit
- Created: 2022-07-11T18:26:08.000Z (over 2 years ago)
- Default Branch: main
- Last Pushed: 2024-09-01T16:14:47.000Z (2 months ago)
- Last Synced: 2024-10-29T15:40:08.712Z (12 days ago)
- Topics: cuda, machine-learning, puzzles
- Language: Jupyter Notebook
- Homepage:
- Size: 581 KB
- Stars: 9,784
- Watchers: 196
- Forks: 853
- Open Issues: 18
-
Metadata Files:
- Readme: README.md
- License: LICENSE
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README
# GPU Puzzles
- by [Sasha Rush](http://rush-nlp.com) - [srush_nlp](https://twitter.com/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:
[NUMBA CUDA Guide](https://numba.readthedocs.io/en/stable/cuda/index.html)
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.
[](https://colab.research.google.com/github/srush/GPU-Puzzles/blob/main/GPU_puzzlers.ipynb)
(If you are into this style of puzzle, also check out my [Tensor
Puzzles](https://github.com/srush/Tensor-Puzzles) for PyTorch.)
[Walkthrough Guide](https://www.youtube.com/watch?v=K4T-YwsOxrM)
```python
!pip install -qqq 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.py
```
```python
import numba
import numpy as np
import warnings
from lib import CudaProblem, Coord
```
```python
warnings.filterwarnings(
action="ignore", category=numba.NumbaPerformanceWarning, module="numba"
)
```
## Puzzle 1: Map
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.*
```python
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)
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [0. 0. 0. 0.]
Spec : [10 11 12 13]
## Puzzle 2 - Zip
Implement a kernel that adds together each position of `a` and `b` and stores it in `out`.
You have 1 thread per position.
```python
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)
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 |
| 0 | 0 | 0 | 0 |
```python
```
```python
problem.check()
```
Failed Tests.
Yours: [0. 0. 0. 0.]
Spec : [0 2 4 6]
## Puzzle 3 - Guards
Implement a kernel that adds 10 to each position of `a` and stores it in `out`.
You have more threads than positions.
```python
def map_guard_test(cuda):
def call(out, a, size) -> None:
local_i = cuda.threadIdx.x
# FILL ME IN (roughly 2 lines)
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [0. 0. 0. 0.]
Spec : [10 11 12 13]
## Puzzle 4 - Map 2D
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.
```python
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)
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [[0. 0.]
[0. 0.]]
Spec : [[10 11]
[12 13]]
## Puzzle 5 - Broadcast
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.
```python
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)
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [[0. 0.]
[0. 0.]]
Spec : [[0 1]
[1 2]]
## Puzzle 6 - Blocks
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.*
```python
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)
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [0. 0. 0. 0. 0. 0. 0. 0. 0.]
Spec : [10 11 12 13 14 15 16 17 18]
## Puzzle 7 - Blocks 2D
Implement the same kernel in 2D. You have fewer threads per block
than the size of `a` in both directions.
```python
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)
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]]
Spec : [[11. 11. 11. 11. 11.]
[11. 11. 11. 11. 11.]
[11. 11. 11. 11. 11.]
[11. 11. 11. 11. 11.]
[11. 11. 11. 11. 11.]]
## Puzzle 8 - Shared
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.)
```python
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)
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 | 0 | 0 | 1 |
```python
problem.check()
```
Failed Tests.
Yours: [0. 0. 0. 0. 0. 0. 0. 0.]
Spec : [11. 11. 11. 11. 11. 11. 11. 11.]
## Puzzle 9 - Pooling
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.*
```python
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)
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [0. 0. 0. 0. 0. 0. 0. 0.]
Spec : [ 0. 1. 3. 6. 9. 12. 15. 18.]
## Puzzle 10 - Dot Product
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.*
```python
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)
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [0.]
Spec : 140
## Puzzle 11 - 1D Convolution
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.
```python
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
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)
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [0. 0. 0. 0. 0. 0.]
Spec : [ 5. 8. 11. 14. 5. 0.]
Test 2
```python
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
Spec : [14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74. 80. 41. 14. 0.]
## Puzzle 12 - Prefix Sum
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](https://en.wikipedia.org/wiki/Prefix_sum) algorithm in shared memory.
That is, each step of the algorithm should sum together half the remaining numbers.
Follow this diagram:
```python
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)
return call
# Test 1
SIZE = 8
out = np.zeros(1)
inp = np.arange(SIZE)
problem = CudaProblem(
"Sum (Simple)",
sum_test,
[inp],
out,
[SIZE],
Coord(1, 1),
Coord(TPB, 1),
spec=sum_spec,
)
problem.show()
```
# Sum (Simple)
Score (Max Per Thread):
| Global Reads | Global Writes | Shared Reads | Shared Writes |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [0.]
Spec : [28.]
Test 2
```python
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [0. 0.]
Spec : [28. 77.]
## Puzzle 13 - Axis Sum
Implement a kernel that computes a sum over each column of `a` and stores it in `out`.
```python
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)
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [[0.]
[0.]
[0.]
[0.]]
Spec : [[ 15.]
[ 51.]
[ 87.]
[123.]]
## Puzzle 14 - Matrix Multiply!
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.
```python
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)
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [[0. 0.]
[0. 0.]]
Spec : [[ 1 3]
[ 3 13]]
Test 2
```python
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 |
| 0 | 0 | 0 | 0 |
```python
problem.check()
```
Failed Tests.
Yours: [[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]
Spec : [[ 140 364 588 812 1036 1260 1484 1708]
[ 364 1100 1836 2572 3308 4044 4780 5516]
[ 588 1836 3084 4332 5580 6828 8076 9324]
[ 812 2572 4332 6092 7852 9612 11372 13132]
[ 1036 3308 5580 7852 10124 12396 14668 16940]
[ 1260 4044 6828 9612 12396 15180 17964 20748]
[ 1484 4780 8076 11372 14668 17964 21260 24556]
[ 1708 5516 9324 13132 16940 20748 24556 28364]]