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README.md
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README.md
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# What every scientist should know about computer architecture
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**Important**: these are instructor notes, remove this file before showing the materials to the students. The notes can be added after the lecture, of course.
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## Introduction
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- [Puzzle](puzzle.ipynb) (how swapping two nested for-loops makes out for a >27× slowdown
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- Let students play around with the notebook and try to find the "bug"
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- A more thorough benchmark using the same code is [here](benchmark_python/)
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## A digression in CPU architecture and the memory hierarchy
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- Go to [A Primer in CPU architecture](architecture)
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- The need for a hierarchical access to data for the CPU should be clear now ➔ the "starving" CPU problem
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- Have a look at the historical evolution of [speeds](speed/) of different components in a computer:
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- the CPU clock rate
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- the memory (RAM) bandwidth, latency clock rate
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- the storage media access rates
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- Measure size and timings for the memory hierarchy on my machine with a low level [C benchmark](benchmark_low_level)
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## Back to the Python benchmark (second try)
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- can we explain what is happening?
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- it must have to do with the good (or bad) use of cache properties
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- but how are numpy arrays laid out in memory?
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## Anatomy of a numpy array
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- [memory layout of numpy arrays](numpy)
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## Back to the Python benchmark (third try)
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- can we explain what is happening now? Yes, more or less ;-)
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- quick fix for the [puzzle](puzzle.ipynb): try and add `order='F'` in the "bad" snippet and see that is "fixes" the bug ➔ why?
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Notes on the [Python benchmark](benchmark_python/):
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- while running it attached to the P-core (`cpu0`), the P-core was running under a constant load of 100% (almost completely user-time) and at a fixed frequency of 3.8 GHz, where the theoretical max would be 5.2 GHz
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- while running it attached to the E-core (`cpu10`), the E-core was running under a constant load of 100% (almost completely user-time) and at a fixed requency of 2.5 GHz, where the theoretical max would be 3.9 GHz
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- ... ➔ the CPU does not "starve" because it scales its speed down to match the memory throughput? Or I am misinterpreting this? This problem which at first sight should be perfectly memory-bound, becomes CPU-bound, or actually, exactly balanced? ;-)
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## Excerpts of parallel Python
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- [The dangers and joys of automatic parallelization](parallel) (like in numpy linear algebra routines) and the use of clusters/schedulers (but also on your laptop)
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## Concluding remarks
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- how is all of this relevant for the users of a computing cluster?
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puzzle.ipynb
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puzzle.ipynb
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{
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"cells": [
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"execution": {
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"iopub.execute_input": "2024-03-04T09:40:28.904Z",
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"iopub.status.busy": "2024-03-04T09:40:28.896Z",
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"iopub.status.idle": "2024-03-04T09:40:28.978Z",
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"shell.execute_reply": "2024-03-04T09:40:28.967Z"
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}
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},
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"outputs": [],
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"source": [
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"import numpy as np"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"execution": {
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"iopub.execute_input": "2024-03-04T10:02:39.062Z",
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"iopub.status.busy": "2024-03-04T10:02:39.057Z",
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"iopub.status.idle": "2024-03-04T10:02:39.068Z",
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"shell.execute_reply": "2024-03-04T10:02:39.071Z"
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}
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},
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"outputs": [],
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"source": [
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"# create a collection of time series\n",
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"# in real life, this data comes from an experiment/simulation\n",
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"n_series = 30\n",
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"len_one_series = 2**21 # ➔ 2^21 ≈ 2 millions (8Bytes x 2^21/2^20 [MB] = 16 MB)\n",
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"time_series = []\n",
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"for idx in range(n_series):\n",
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" time_series.append(np.zeros((len_one_series,1), dtype='float64'))"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"execution": {
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"iopub.execute_input": "2024-03-04T10:02:41.027Z",
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"iopub.status.busy": "2024-03-04T10:02:41.020Z",
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"iopub.status.idle": "2024-03-04T10:02:41.036Z",
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"shell.execute_reply": "2024-03-04T10:02:41.040Z"
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}
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},
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"outputs": [],
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"source": [
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"# how much memory does one time series need?\n",
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"ts_size = time_series[0].nbytes/2**20 # -> 2^20 is 1MB\n",
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"print('Size of one time series (MB):', ts_size)\n",
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"print('Size of collection (MB):', n_series*ts_size)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"execution": {
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"iopub.execute_input": "2024-03-04T10:06:08.461Z",
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"iopub.status.busy": "2024-03-04T10:06:08.459Z",
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"iopub.status.idle": "2024-03-04T10:06:08.466Z",
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"shell.execute_reply": "2024-03-04T10:06:08.468Z"
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}
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},
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"outputs": [],
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"source": [
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"# let's load the collection in one big array\n",
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"def load_data_row(x, time_series):\n",
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" \"\"\"Store one time series per raw\"\"\"\n",
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" for row, ts in enumerate(time_series):\n",
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" x[row,:] = ts\n",
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" return x"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"execution": {
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"iopub.execute_input": "2024-03-04T10:06:10.280Z",
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"iopub.status.busy": "2024-03-04T10:06:10.277Z",
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"iopub.status.idle": "2024-03-04T10:06:10.284Z",
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"shell.execute_reply": "2024-03-04T10:06:10.288Z"
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}
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},
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"outputs": [],
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"source": [
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"# let's load the collection in one big array\n",
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"def load_data_column(x, time_series):\n",
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" \"\"\"Store one time series per column\"\"\"\n",
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" for column, ts in enumerate(time_series):\n",
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" x[:,column] = ts\n",
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" return x"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"execution": {
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"iopub.execute_input": "2024-03-04T10:06:14.959Z",
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"iopub.status.busy": "2024-03-04T10:06:14.956Z",
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"iopub.status.idle": "2024-03-04T10:06:17.437Z",
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"shell.execute_reply": "2024-03-04T10:06:17.443Z"
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}
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},
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"outputs": [],
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"source": [
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"x = np.zeros((n_series, len_one_series, 1), dtype='float64')\n",
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"%timeit load_data_row(x, time_series)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"execution": {
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"iopub.execute_input": "2024-03-04T10:06:20.056Z",
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"iopub.status.busy": "2024-03-04T10:06:20.053Z",
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"iopub.status.idle": "2024-03-04T10:06:21.695Z",
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"shell.execute_reply": "2024-03-04T10:06:21.700Z"
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}
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},
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"outputs": [],
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"source": [
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"x = np.zeros((len_one_series, n_series, 1), dtype='float64')\n",
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"%timeit load_data_column(x, time_series)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.12.4"
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},
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"nteract": {
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"version": "0.28.0"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}
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