{ "id": 667046, "title": "Reverse engineering the sine formula with high numerical precision", "content": "

In this competition, I learnt a lot about the importance of exploratory data analysis and high precision in numerical computing.

\n

Note: I did not use a multi-layer perceptron to generate my test set y label submission.

\n

Let's walk through how I built my model from the beginning.

\n

At the beginning, I used AutoGluon (an automatic machine learning library) to build a baseline model. I used medium_quality, which trains several tree-based models (LightGBM, XGBoost, and Random Forest) and neural networks and takes a weighted average of the predictions as the final result for better performance compared to one model. Note that the model choice is not important; you can train an MLP or Gradient Boosting Regressor as well.

\n

I submitted to the leaderboard and I got 0.03318. That was quite good, but clearly not the highest possible score I could get as evidenced by the leaderboard ranking at that time. Given that the leaderboard's highest score at that time was around 0.003, a 10% error compared to my submission, I guessed something interesting might lie inside the data.

\n

So I quickly ran some data visualizations, and here is what I got:

\n

\"\"

\n

Hmm, the dataset definitely looks synthetically generated. And there is some sort of symmetry going on. Without a clue, I asked Gemini 3.0 Pro if we could hand-craft a formula by reparameterizing based on x1 and x2. It turns out that by plotting y against \\(r = \\sqrt{x_1^2 + x_2^2}\\), I got:

\n

\"\"

\n

That looked a lot clearer. I was quite certain that the dataset had something to do with sine curves. And it looked like the sine curves were generated with an increasing frequency.

\n

That was an interesting property. After brainstorming, I came up with the following formulation of the equation:

\n

$$y = A \\cdot \\sin(B \\cdot r^2 + C \\cdot r + D) + E$$

\n

Where:

\n\n

Let's throw it into scipy and fit the formula:

\n
import matplotlib.pyplot as plt\nimport numpy as np\nimport pandas as pd\nfrom scipy.optimize import curve_fit\nfrom sklearn.metrics import mean_squared_error\nfrom sklearn.model_selection import KFold\n\ntrain_df = pd.read_csv(\"train.csv\")\ntest_df = pd.read_csv(\"test.csv\")\n\ntrain_df[\"r\"] = np.sqrt(train_df[\"x1\"] ** 2 + train_df[\"x2\"] ** 2)\ntest_df[\"r\"] = np.sqrt(test_df[\"x1\"] ** 2 + test_df[\"x2\"] ** 2)\n\nX_r = train_df[\"r\"].values\ny = train_df[\"y\"].values\n\n\ndef model(r, A, B, C, D, E):\n    return A * np.sin(B * r**2 + C * r + D) + E\n\n\np0 = [0.5, 5.0, 2.0, 0.0, 1.0]\nkf = KFold(n_splits=5, shuffle=True, random_state=42)\n\ncv_mses = []\nfor train_idx, val_idx in kf.split(train_df):\n    X_tr, y_tr = X_r[train_idx], y[train_idx]\n    X_va, y_va = X_r[val_idx], y[val_idx]\n\n    popt, _ = curve_fit(model, X_tr, y_tr, p0=p0)\n\n    y_pred = model(X_va, *popt)\n    cv_mses.append(mean_squared_error(y_va, y_pred))\n\nprint(f\"Model CV MSE: {np.mean(cv_mses):.6f} +/- {np.std(cv_mses):.6f}\")\n\npopt_final, _ = curve_fit(model, X_r, y, p0=p0)\nprint(f\"Final Parameters: {popt_final}\")\n\nr_grid = np.linspace(0, 1.5, 1000)\ny_fit = model(r_grid, *popt_final)\nplt.figure(figsize=(12, 6))\nplt.scatter(X_r, y, s=1, alpha=0.1, color=\"blue\", label=\"Training Data\")\nplt.plot(r_grid, y_fit, \"r-\", lw=3, label=\"Curve Fit\")\nplt.xlabel(\"Radius r\")\nplt.ylabel(\"y\")\nplt.title(f\"Global Curve Fit (CV MSE: {np.mean(cv_mses):.6f})\")\nplt.legend()\nplt.show()\n\ny_test_pred = model(test_df[\"r\"].values, *popt_final)\nsubmission_df = pd.DataFrame({\"id\": np.arange(1, 2001), \"y\": y_test_pred})\nsubmission_df.to_csv(\"submission.csv\", index=False)\n
\n

This gave me

\n

$$ y = 0.5008 \\sin(6.0419 r^2 - 0.0602 r + 0.0175) + 1.0001 $$

\n
Model CV MSE: 0.022197 +/- 0.000680\nFinal Parameters: [ 0.50083101  6.04194113 -0.06023903  0.01753956  1.00005388]\n
\n

I submitted to the leaderboard, and it gave me 0.00300.

\n

While the results are great, the decimal places of the variables look a little bit off. Given that I assumed the dataset was likely synthetically generated, I made a few inductive biases:

\n\n

That way, I simplified the formula into:

\n

$$ y = 0.5 \\sin(6 r^2) + 1 $$

\n

This gave me a leaderboard score of 0.00247.

\n

Now, I notice that I'm still not the highest in the leaderboard ranking. This could be due to several reasons, for example, Kaggle ranks the first person to submit higher if 2 users have the same score. However, Kaggle also evaluates the leaderboard metric with full precision!

\n

I looked into the submission.csv I had generated:

\n
id,y\n1,0.5015993834619267\n2,1.467676273657998\n3,0.6799769398701817\n4,1.3485892841618887\n5,1.4317098100334664\n6,1.4830281438178594\n
\n

Given that the formula was a sine curve, which computers can only approximate instead of precisely calculating, I guessed the 16-decimal-place precision in the generated CSV was not enough.

\n

From what I've learnt in my Computer Organization class, we can use higher-precision data types to calculate the equation more accurately.

\n

So I came up with code that uses the mpmath library to calculate the equation with a precision of 1000 decimal places.

\n
import pandas as pd\nimport numpy as np\nimport mpmath as mp\n\nmp.mp.dps = 1000\n\ntest_df = pd.read_csv(\"test.csv\", dtype={\"x1\": \"string\", \"x2\": \"string\"})\ny_str = []\nfor s1, s2 in zip(test_df[\"x1\"], test_df[\"x2\"]):\n    x1 = mp.mpf(s1)\n    x2 = mp.mpf(s2)\n    r2 = x1 * x1 + x2 * x2\n    y = mp.mpf(\"1.0\") + mp.mpf(\"0.5\") * mp.sin(mp.mpf(\"6.0\") * r2)\n    y_str.append(mp.nstr(y, n=1000))\n\nsubmission_df = pd.DataFrame({\"id\": np.arange(1, len(y_str) + 1), \"y\": y_str})\nsubmission_df.to_csv(\"submission.highprec.csv\", index=False)\n
\n
id,y\n1,0.5020010035360700362206597987786111594123629203358911339986548017850158905600859624002760847100101825697287740091135184314121656099284776859060497464055326926497526579695155903427397807547335046591328437414112053491467098697123170383443093995673538365205906606598644579196408660325326207702152674767992697337469445205249467267461159328532940235671828697134392104897840619129686044876977490052755418447038765050002790158456973136828515036168950085646898585581163114283837881893419888916048993719924760483147981640195213664576822883955370330805606427682016319769403913699298048954607869454199696624623522313392965222617573250637046702242516828514221318261959544818487150798003601533338542222064801402146597805840478120953298110953341944499988669822935832411005456448468153095677237213760359000128768411674740636908858852447189469698118394833456613496772417505835189486470543416150138239264288877096831448733942264472455617271948744732050852303308487799689916152250772489866587741914356802089109051589033\n
\n

This looked good! I then submitted this CSV as my final submission on the leaderboard. My leaderboard ranking went slightly higher. So I declared this a success and stopped working on it.

\n

So, who are you?

\n

I am a random undergraduate student from the University of Hong Kong, currently on winter break. I was looking for fun Kaggle competitions. Thank you to the competition host for providing such an interesting dataset!

", "comments": [], "comment_count": 0, "messages": [ { "id": 3389281, "postDate": "2026-01-10T19:10:22.763Z", "content": "

In this competition, I learnt a lot about the importance of exploratory data analysis and high precision in numerical computing.

\n

Note: I did not use a multi-layer perceptron to generate my test set y label submission.

\n

Let's walk through how I built my model from the beginning.

\n

At the beginning, I used AutoGluon (an automatic machine learning library) to build a baseline model. I used medium_quality, which trains several tree-based models (LightGBM, XGBoost, and Random Forest) and neural networks and takes a weighted average of the predictions as the final result for better performance compared to one model. Note that the model choice is not important; you can train an MLP or Gradient Boosting Regressor as well.

\n

I submitted to the leaderboard and I got 0.03318. That was quite good, but clearly not the highest possible score I could get as evidenced by the leaderboard ranking at that time. Given that the leaderboard's highest score at that time was around 0.003, a 10% error compared to my submission, I guessed something interesting might lie inside the data.

\n

So I quickly ran some data visualizations, and here is what I got:

\n

\"\"

\n

Hmm, the dataset definitely looks synthetically generated. And there is some sort of symmetry going on. Without a clue, I asked Gemini 3.0 Pro if we could hand-craft a formula by reparameterizing based on x1 and x2. It turns out that by plotting y against \\(r = \\sqrt{x_1^2 + x_2^2}\\), I got:

\n

\"\"

\n

That looked a lot clearer. I was quite certain that the dataset had something to do with sine curves. And it looked like the sine curves were generated with an increasing frequency.

\n

That was an interesting property. After brainstorming, I came up with the following formulation of the equation:

\n

$$y = A \\cdot \\sin(B \\cdot r^2 + C \\cdot r + D) + E$$

\n

Where:

\n\n

Let's throw it into scipy and fit the formula:

\n
import matplotlib.pyplot as plt\nimport numpy as np\nimport pandas as pd\nfrom scipy.optimize import curve_fit\nfrom sklearn.metrics import mean_squared_error\nfrom sklearn.model_selection import KFold\n\ntrain_df = pd.read_csv(\"train.csv\")\ntest_df = pd.read_csv(\"test.csv\")\n\ntrain_df[\"r\"] = np.sqrt(train_df[\"x1\"] ** 2 + train_df[\"x2\"] ** 2)\ntest_df[\"r\"] = np.sqrt(test_df[\"x1\"] ** 2 + test_df[\"x2\"] ** 2)\n\nX_r = train_df[\"r\"].values\ny = train_df[\"y\"].values\n\n\ndef model(r, A, B, C, D, E):\n    return A * np.sin(B * r**2 + C * r + D) + E\n\n\np0 = [0.5, 5.0, 2.0, 0.0, 1.0]\nkf = KFold(n_splits=5, shuffle=True, random_state=42)\n\ncv_mses = []\nfor train_idx, val_idx in kf.split(train_df):\n    X_tr, y_tr = X_r[train_idx], y[train_idx]\n    X_va, y_va = X_r[val_idx], y[val_idx]\n\n    popt, _ = curve_fit(model, X_tr, y_tr, p0=p0)\n\n    y_pred = model(X_va, *popt)\n    cv_mses.append(mean_squared_error(y_va, y_pred))\n\nprint(f\"Model CV MSE: {np.mean(cv_mses):.6f} +/- {np.std(cv_mses):.6f}\")\n\npopt_final, _ = curve_fit(model, X_r, y, p0=p0)\nprint(f\"Final Parameters: {popt_final}\")\n\nr_grid = np.linspace(0, 1.5, 1000)\ny_fit = model(r_grid, *popt_final)\nplt.figure(figsize=(12, 6))\nplt.scatter(X_r, y, s=1, alpha=0.1, color=\"blue\", label=\"Training Data\")\nplt.plot(r_grid, y_fit, \"r-\", lw=3, label=\"Curve Fit\")\nplt.xlabel(\"Radius r\")\nplt.ylabel(\"y\")\nplt.title(f\"Global Curve Fit (CV MSE: {np.mean(cv_mses):.6f})\")\nplt.legend()\nplt.show()\n\ny_test_pred = model(test_df[\"r\"].values, *popt_final)\nsubmission_df = pd.DataFrame({\"id\": np.arange(1, 2001), \"y\": y_test_pred})\nsubmission_df.to_csv(\"submission.csv\", index=False)\n
\n

This gave me

\n

$$ y = 0.5008 \\sin(6.0419 r^2 - 0.0602 r + 0.0175) + 1.0001 $$

\n
Model CV MSE: 0.022197 +/- 0.000680\nFinal Parameters: [ 0.50083101  6.04194113 -0.06023903  0.01753956  1.00005388]\n
\n

I submitted to the leaderboard, and it gave me 0.00300.

\n

While the results are great, the decimal places of the variables look a little bit off. Given that I assumed the dataset was likely synthetically generated, I made a few inductive biases:

\n\n

That way, I simplified the formula into:

\n

$$ y = 0.5 \\sin(6 r^2) + 1 $$

\n

This gave me a leaderboard score of 0.00247.

\n

Now, I notice that I'm still not the highest in the leaderboard ranking. This could be due to several reasons, for example, Kaggle ranks the first person to submit higher if 2 users have the same score. However, Kaggle also evaluates the leaderboard metric with full precision!

\n

I looked into the submission.csv I had generated:

\n
id,y\n1,0.5015993834619267\n2,1.467676273657998\n3,0.6799769398701817\n4,1.3485892841618887\n5,1.4317098100334664\n6,1.4830281438178594\n
\n

Given that the formula was a sine curve, which computers can only approximate instead of precisely calculating, I guessed the 16-decimal-place precision in the generated CSV was not enough.

\n

From what I've learnt in my Computer Organization class, we can use higher-precision data types to calculate the equation more accurately.

\n

So I came up with code that uses the mpmath library to calculate the equation with a precision of 1000 decimal places.

\n
import pandas as pd\nimport numpy as np\nimport mpmath as mp\n\nmp.mp.dps = 1000\n\ntest_df = pd.read_csv(\"test.csv\", dtype={\"x1\": \"string\", \"x2\": \"string\"})\ny_str = []\nfor s1, s2 in zip(test_df[\"x1\"], test_df[\"x2\"]):\n    x1 = mp.mpf(s1)\n    x2 = mp.mpf(s2)\n    r2 = x1 * x1 + x2 * x2\n    y = mp.mpf(\"1.0\") + mp.mpf(\"0.5\") * mp.sin(mp.mpf(\"6.0\") * r2)\n    y_str.append(mp.nstr(y, n=1000))\n\nsubmission_df = pd.DataFrame({\"id\": np.arange(1, len(y_str) + 1), \"y\": y_str})\nsubmission_df.to_csv(\"submission.highprec.csv\", index=False)\n
\n
id,y\n1,0.5020010035360700362206597987786111594123629203358911339986548017850158905600859624002760847100101825697287740091135184314121656099284776859060497464055326926497526579695155903427397807547335046591328437414112053491467098697123170383443093995673538365205906606598644579196408660325326207702152674767992697337469445205249467267461159328532940235671828697134392104897840619129686044876977490052755418447038765050002790158456973136828515036168950085646898585581163114283837881893419888916048993719924760483147981640195213664576822883955370330805606427682016319769403913699298048954607869454199696624623522313392965222617573250637046702242516828514221318261959544818487150798003601533338542222064801402146597805840478120953298110953341944499988669822935832411005456448468153095677237213760359000128768411674740636908858852447189469698118394833456613496772417505835189486470543416150138239264288877096831448733942264472455617271948744732050852303308487799689916152250772489866587741914356802089109051589033\n
\n

This looked good! I then submitted this CSV as my final submission on the leaderboard. My leaderboard ranking went slightly higher. So I declared this a success and stopped working on it.

\n

So, who are you?

\n

I am a random undergraduate student from the University of Hong Kong, currently on winter break. I was looking for fun Kaggle competitions. Thank you to the competition host for providing such an interesting dataset!

", "rawMarkdown": "In this competition, I learnt a lot about the importance of exploratory data analysis and high precision in numerical computing.\n\nNote: I did not use a multi-layer perceptron to generate my test set `y` label submission.\n\nLet's walk through how I built my model from the beginning.\n\nAt the beginning, I used [AutoGluon](https://auto.gluon.ai/stable/index.html) (an automatic machine learning library) to build a baseline model. I used `medium_quality`, which trains several tree-based models (LightGBM, XGBoost, and Random Forest) and neural networks and takes a weighted average of the predictions as the final result for better performance compared to one model. Note that the model choice is not important; you can train an MLP or Gradient Boosting Regressor as well.\n\nI submitted to the leaderboard and I got `0.03318`. That was quite good, but clearly not the highest possible score I could get as evidenced by the leaderboard ranking at that time. Given that the leaderboard's highest score at that time was around `0.003`, a 10% error compared to my submission, I guessed something interesting might lie inside the data.\n\nSo I quickly ran some data visualizations, and here is what I got:\n\n![](https://www.googleapis.com/download/storage/v1/b/kaggle-user-content/o/inbox%2F29566806%2Fdf7cd62be6450369a9c593f288fa5807%2Fcontour.png?generation=1768072409620546&alt=media)\n\nHmm, the dataset definitely looks synthetically generated. And there is some sort of symmetry going on. Without a clue, I asked Gemini 3.0 Pro if we could hand-craft a formula by reparameterizing based on x1 and x2. It turns out that by plotting `y` against \\\\(r = \\sqrt{x_1^2 + x_2^2}\\\\), I got:\n\n![](https://www.googleapis.com/download/storage/v1/b/kaggle-user-content/o/inbox%2F29566806%2F547d26a6e02372a7c3371dab0dca9648%2Fradius-transformation.png?generation=1768072951525915&alt=media)\n\nThat looked a lot clearer. I was quite certain that the dataset had something to do with sine curves. And it looked like the sine curves were generated with an *increasing frequency*.\n\nThat was an interesting property. After brainstorming, I came up with the following formulation of the equation:\n\n$$y = A \\cdot \\sin(B \\cdot r^2 + C \\cdot r + D) + E$$\n\nWhere:\n* \\\\(A\\\\): Amplitude\n* \\\\(B, C\\\\): Frequency and frequency-rate\n* \\\\(D\\\\): Phase\n* \\\\(E\\\\): Vertical offset\n\nLet's throw it into `scipy` and fit the formula:\n\n```\nimport matplotlib.pyplot as plt\nimport numpy as np\nimport pandas as pd\nfrom scipy.optimize import curve_fit\nfrom sklearn.metrics import mean_squared_error\nfrom sklearn.model_selection import KFold\n\ntrain_df = pd.read_csv(\"train.csv\")\ntest_df = pd.read_csv(\"test.csv\")\n\ntrain_df[\"r\"] = np.sqrt(train_df[\"x1\"] ** 2 + train_df[\"x2\"] ** 2)\ntest_df[\"r\"] = np.sqrt(test_df[\"x1\"] ** 2 + test_df[\"x2\"] ** 2)\n\nX_r = train_df[\"r\"].values\ny = train_df[\"y\"].values\n\n\ndef model(r, A, B, C, D, E):\n return A * np.sin(B * r**2 + C * r + D) + E\n\n\np0 = [0.5, 5.0, 2.0, 0.0, 1.0]\nkf = KFold(n_splits=5, shuffle=True, random_state=42)\n\ncv_mses = []\nfor train_idx, val_idx in kf.split(train_df):\n X_tr, y_tr = X_r[train_idx], y[train_idx]\n X_va, y_va = X_r[val_idx], y[val_idx]\n\n popt, _ = curve_fit(model, X_tr, y_tr, p0=p0)\n\n y_pred = model(X_va, *popt)\n cv_mses.append(mean_squared_error(y_va, y_pred))\n\nprint(f\"Model CV MSE: {np.mean(cv_mses):.6f} +/- {np.std(cv_mses):.6f}\")\n\npopt_final, _ = curve_fit(model, X_r, y, p0=p0)\nprint(f\"Final Parameters: {popt_final}\")\n\nr_grid = np.linspace(0, 1.5, 1000)\ny_fit = model(r_grid, *popt_final)\nplt.figure(figsize=(12, 6))\nplt.scatter(X_r, y, s=1, alpha=0.1, color=\"blue\", label=\"Training Data\")\nplt.plot(r_grid, y_fit, \"r-\", lw=3, label=\"Curve Fit\")\nplt.xlabel(\"Radius r\")\nplt.ylabel(\"y\")\nplt.title(f\"Global Curve Fit (CV MSE: {np.mean(cv_mses):.6f})\")\nplt.legend()\nplt.show()\n\ny_test_pred = model(test_df[\"r\"].values, *popt_final)\nsubmission_df = pd.DataFrame({\"id\": np.arange(1, 2001), \"y\": y_test_pred})\nsubmission_df.to_csv(\"submission.csv\", index=False)\n```\n\nThis gave me\n\n$$ y = 0.5008 \\sin(6.0419 r^2 - 0.0602 r + 0.0175) + 1.0001 $$\n\n```\nModel CV MSE: 0.022197 +/- 0.000680\nFinal Parameters: [ 0.50083101 6.04194113 -0.06023903 0.01753956 1.00005388]\n```\n\nI submitted to the leaderboard, and it gave me `0.00300`.\n\nWhile the results are great, the decimal places of the variables look a little bit off. Given that I assumed the dataset was likely synthetically generated, I made a few [inductive biases](https://en.wikipedia.org/wiki/Inductive_bias):\n\n* The dataset is hand-crafted by a human\n* The true formula responsible for generating the samples must be simple\n* For the values in the formula, prefer integers over floating point numbers\n\nThat way, I simplified the formula into:\n\n$$ y = 0.5 \\sin(6 r^2) + 1 $$\n\nThis gave me a leaderboard score of `0.00247`.\n\nNow, I notice that I'm still not the highest in the leaderboard ranking. This could be due to several reasons, for example, [Kaggle ranks the first person to submit higher if 2 users have the same score. However, Kaggle also evaluates the leaderboard metric with *full precision!*](https://www.kaggle.com/docs/competitions#leaderboard:~:text=In%20the%20event%20of%20an%20exact%20score%20tie%2C%20the%20tiebreaker%20is%20the%20team%20which%20submitted%20earlier.%20Kaggle%20always%20uses%20full%20precision%20when%20determining%20rankings%2C%20not%20just%20the%20truncated%20precision%20shown%20on%20the%20Leaderboard.)\n\nI looked into the `submission.csv` I had generated:\n\n```\nid,y\n1,0.5015993834619267\n2,1.467676273657998\n3,0.6799769398701817\n4,1.3485892841618887\n5,1.4317098100334664\n6,1.4830281438178594\n```\n\nGiven that the formula was a sine curve, which computers can only approximate instead of precisely calculating, I guessed the 16-decimal-place precision in the generated CSV was not enough.\n\nFrom what I've learnt in my Computer Organization class, we can use higher-precision data types to calculate the equation more accurately.\n\nSo I came up with code that uses the `mpmath` library to calculate the equation with a precision of `1000` decimal places.\n\n```\nimport pandas as pd\nimport numpy as np\nimport mpmath as mp\n\nmp.mp.dps = 1000\n\ntest_df = pd.read_csv(\"test.csv\", dtype={\"x1\": \"string\", \"x2\": \"string\"})\ny_str = []\nfor s1, s2 in zip(test_df[\"x1\"], test_df[\"x2\"]):\n x1 = mp.mpf(s1)\n x2 = mp.mpf(s2)\n r2 = x1 * x1 + x2 * x2\n y = mp.mpf(\"1.0\") + mp.mpf(\"0.5\") * mp.sin(mp.mpf(\"6.0\") * r2)\n y_str.append(mp.nstr(y, n=1000))\n\nsubmission_df = pd.DataFrame({\"id\": np.arange(1, len(y_str) + 1), \"y\": y_str})\nsubmission_df.to_csv(\"submission.highprec.csv\", index=False)\n```\n\n```\nid,y\n1,0.5020010035360700362206597987786111594123629203358911339986548017850158905600859624002760847100101825697287740091135184314121656099284776859060497464055326926497526579695155903427397807547335046591328437414112053491467098697123170383443093995673538365205906606598644579196408660325326207702152674767992697337469445205249467267461159328532940235671828697134392104897840619129686044876977490052755418447038765050002790158456973136828515036168950085646898585581163114283837881893419888916048993719924760483147981640195213664576822883955370330805606427682016319769403913699298048954607869454199696624623522313392965222617573250637046702242516828514221318261959544818487150798003601533338542222064801402146597805840478120953298110953341944499988669822935832411005456448468153095677237213760359000128768411674740636908858852447189469698118394833456613496772417505835189486470543416150138239264288877096831448733942264472455617271948744732050852303308487799689916152250772489866587741914356802089109051589033\n```\n\nThis looked good! I then submitted this CSV as my final submission on the leaderboard. My leaderboard ranking went slightly higher. So I declared this a success and stopped working on it.\n\n# So, who are you?\n\nI am a random undergraduate student from the University of Hong Kong, currently on winter break. I was looking for fun Kaggle competitions. Thank you to the competition host for providing such an interesting dataset!" } ], "complete": true, "source": "live", "fetched_at": "2026-10-20T05:53:47Z", "competition_id": 125114, "competition_slug": "function-approximation-using-multilayer-perceptron", "team_id": 15055531, "solution_url": "/competitions/function-approximation-using-multilayer-perceptron/writeups/reverse-engineering-the-sine-formula-with-high-num" }