Post History
#7: Post edited
Encode and decode floating point integers [FINALIZED]
Imagine you have only one byte (8 bits) to store a value, but need to store values from $0$ to $4032$. Impossible, until you are also told that an error of 1/64 of the exact value does not matter. With that knowledge, you decide to design a floating point integer type, which is defined as follows: * All representable numbers are of the form $m\cdot 2^e$, where $m$ is the mantissa and $e$ is the exponent. There are no NaNs and no infinities. * The upper three bits of the byte are the exponent, the lower five are the mantissa. Since no negative numbers are stored, no sign bit is needed. I'll refer to the three bit number represented by the exponent bits as $E$, and the 5 bit value represented by the mantissa bits as $M$ * If $E=0$, then $e=0$ and $m=M$ (“denormalized integer”). * Otherwise, $e=E-1$ and $m=M+32$ Your task is now to write code that converts a positive integer to that floating point format and back. In particular: * Encoding: - Your code takes an integer $0\le n\le 4032$, and outputs an integer $0\le r\le 255$. You can assume that the input integer is always in the allowed range (that is, no overflow handling is required) - If $n$ can be represented exactly, $r$ is the corresponding representation. - Otherwise, if $n$ is exactly in the middle of two consecutive exactly representable numbers, $r$ is the choice with the even mantissa. - Otherwise, $r$ is the representation of the closest exactly representable number. * Decoding: Your code takes an integer $0\le r\le 255$ and outputs the integer $0\le n\le 4032$ it represents. You can write either two separate routines for encoding and decoding, or a single routine that takes an extra argument telling whether to encode or decode. Here “routine” refers to any form of executable code taking input and giving output (program, function, macro, …). This is <span class="badge is-tag">code-golf</span>, that is the shortest code wins. If you choose separate encoding and decoding routines, the score is the sum of the sizes of the encoding and decoding routines. An ungolfed (and unoptimized) Python implementation is available [here][Python implementation] Test cases: Encoding: input output 0 0 1 1 31 31 32 32 63 63 64 64 65 64 66 65 67 66 68 66 69 66 70 67 123 94 124 94 125 94 126 95 127 96 128 96 129 96 130 96 131 97 256 128 2021 223 4032 255 Decoding: input output 0 0 1 1 31 31 32 32 63 63 64 64 65 66 66 68 93 122 94 124 95 126 96 128 97 132 128 256 160 512 223 2016 224 2048 255 4032 [Python implementation]: https://tio.run/##lVZdb9owFH3Pr7jqtCmhVAohoQ0qSGxjElrTTYOHalWFUmJGpuBEibNuYv3t7NqOIQGDtAiZ2Of6fp1jQ/aHrVLa3W6D0f1sMp2O5u8nsykMwDN2K8Fo@hlXHGi1oGl2BR0jmNzP7798C0Z3k@/jj1o7dPUwHz@MPsw43GruMAwjIksgdJFGxKR9iCmz4GrIv/sG4BMvgcItNLdJiD85YWVOgRpi5Q1EhJF8HVMCbEWA/M5SSigTIMH4tnh7WcUJQbdDnlCLAM@4ynHvmcDlACsU82W4YGkO0lyFKrIkZkDL9TPJeb4psLyki5CRCNYhZXFRhBDSCJK0YPAcs0Js5LM5n6E7Cu/ArJxjOy1hEMz3fgZgYppDIBZaNjhRWYTRzxLdVwEJLNFVlqcZ5pSnJY1i@kP10WntYw9VTfgx68BgoBCeej2Xt0guoh1r36MAE6ybXGLDRO@Sgpy0UpnPkKBd9xp0iWZBXADOYJ3mnMxQmmBqZZg0iR2jfyJiV56xVlU7CqQoE54bxtixkv4i@TJJX9qqfTnJ0BAdhixOqWpYwOs9pbyx1Ee9SFsmUGnSHMPtbfMsWPAXgkrzERGazzWa5wXlnPbmQVJhcr0WMGEMiRnbVl3FSvT8WXMmNAxxq7E40A1L7OnRca0VuMaDIw7EdonU8sZxTW@EyYU80hf96my35aosGlfli1h8rRrCSMFM5WlOwzXpQ8HyNvYlK1n1npZMTaoqJcEYeJfEY8PJkyn2W7smqQ0D5WxXdJYjBeby4utoOkWqYdNw9GpuZCZD99XihUpHfHphaZqqvH0aTe7@y1sbilVaJhE8E9hUOcoghsGbtAgLwjv9KEKZqtVtPrPlaLU1WEeOOqzbkaMWc@Sow3pdOWoxV45azDuD9aSFFruWFlrs5gzmn8aubelZg3UcUZ/vajH3DOadwUR9vqfFRH1@T4vdnMH801jXPoMJ3n1d7Y7Xk1GPMcd2@D7H0fDu2kIvjsfrq8Dq1GsFWscOBVrDjgRaxw4FWsOOBFrHDgVax7wDwdQxKdAbHeZ3RdO0ufiulI0W86Q0tNghEXXsWtCoiycFgzTqsB7nwevo9jlc9I7d6Wkxl2OuLhekXNIvL8Mnw@D/RdSFJX7j1eUlL0lx4bfUomVs/wE
#6: Post edited
Encode and decode floating point integers
- Encode and decode floating point integers [FINALIZED]
#5: Post edited
- Imagine you have only one byte (8 bits) to store a value, but need to store values from $0$ to $4032$. Impossible, until you are also told that an error of 1/64 of the exact value does not matter.
- With that knowledge, you decide to design a floating point integer type, which is defined as follows:
- * All representable numbers are of the form $m\cdot 2^e$, where $m$ is the mantissa and $e$ is the exponent. There are no NaNs and no infinities.
- * The upper three bits of the byte are the exponent, the lower five are the mantissa. Since no negative numbers are stored, no sign bit is needed.
- I'll refer to the three bit number represented by the exponent bits as $E$, and the 5 bit value represented by the mantissa bits as $M$
- * If $E=0$, then $e=0$ and $m=M$ (“denormalized integer”).
- * Otherwise, $e=E-1$ and $m=M+32$
- Your task is now to write code that converts a positive integer to that floating point format and back. In particular:
- * Encoding:
- - Your code takes an integer $0\le n\le 4032$, and outputs
- an integer $0\le r\le 255$. You can assume that the input
- integer is always in the allowed range (that is, no overflow
- handling is required)
- - If $n$ can be represented exactly, $r$ is the
- corresponding representation.
- - Otherwise, if $n$ is exactly in the middle of two
- consecutive exactly representable numbers, $r$ is the
- choice with the even mantissa.
- - Otherwise, $r$ is the representation of the closest
- exactly representable number.
- * Decoding:
- Your code takes an integer $0\le r\le 255$ and outputs
- the integer $0\le n\le 4032$ it represents.
- You can write either two separate routines for encoding and decoding, or a single routine that takes an extra argument telling whether to encode or decode. Here “routine” refers to any form of executable code taking input and giving output (program, function, macro, …).
This is code golf, the shortest code wins. If you choose separate encoding and decoding routines, the score is the sum of the sizes of the encoding and decoding routines.- An ungolfed (and unoptimized) Python implementation is available [here][Python implementation]
- Test cases:
- Encoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 64
- 66 65
- 67 66
- 68 66
- 69 66
- 70 67
- 123 94
- 124 94
- 125 94
- 126 95
- 127 96
- 128 96
- 129 96
- 130 96
- 131 97
- 256 128
- 2021 223
- 4032 255
- Decoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 66
- 66 68
- 93 122
- 94 124
- 95 126
- 96 128
- 97 132
- 128 256
- 160 512
- 223 2016
- 224 2048
- 255 4032
- [Python implementation]: https://tio.run/##lVZdb9owFH3Pr7jqtCmhVAohoQ0qSGxjElrTTYOHalWFUmJGpuBEibNuYv3t7NqOIQGDtAiZ2Of6fp1jQ/aHrVLa3W6D0f1sMp2O5u8nsykMwDN2K8Fo@hlXHGi1oGl2BR0jmNzP7798C0Z3k@/jj1o7dPUwHz@MPsw43GruMAwjIksgdJFGxKR9iCmz4GrIv/sG4BMvgcItNLdJiD85YWVOgRpi5Q1EhJF8HVMCbEWA/M5SSigTIMH4tnh7WcUJQbdDnlCLAM@4ynHvmcDlACsU82W4YGkO0lyFKrIkZkDL9TPJeb4psLyki5CRCNYhZXFRhBDSCJK0YPAcs0Js5LM5n6E7Cu/ArJxjOy1hEMz3fgZgYppDIBZaNjhRWYTRzxLdVwEJLNFVlqcZ5pSnJY1i@kP10WntYw9VTfgx68BgoBCeej2Xt0guoh1r36MAE6ybXGLDRO@Sgpy0UpnPkKBd9xp0iWZBXADOYJ3mnMxQmmBqZZg0iR2jfyJiV56xVlU7CqQoE54bxtixkv4i@TJJX9qqfTnJ0BAdhixOqWpYwOs9pbyx1Ee9SFsmUGnSHMPtbfMsWPAXgkrzERGazzWa5wXlnPbmQVJhcr0WMGEMiRnbVl3FSvT8WXMmNAxxq7E40A1L7OnRca0VuMaDIw7EdonU8sZxTW@EyYU80hf96my35aosGlfli1h8rRrCSMFM5WlOwzXpQ8HyNvYlK1n1npZMTaoqJcEYeJfEY8PJkyn2W7smqQ0D5WxXdJYjBeby4utoOkWqYdNw9GpuZCZD99XihUpHfHphaZqqvH0aTe7@y1sbilVaJhE8E9hUOcoghsGbtAgLwjv9KEKZqtVtPrPlaLU1WEeOOqzbkaMWc@Sow3pdOWoxV45azDuD9aSFFruWFlrs5gzmn8aubelZg3UcUZ/vajH3DOadwUR9vqfFRH1@T4vdnMH801jXPoMJ3n1d7Y7Xk1GPMcd2@D7H0fDu2kIvjsfrq8Dq1GsFWscOBVrDjgRaxw4FWsOOBFrHDgVax7wDwdQxKdAbHeZ3RdO0ufiulI0W86Q0tNghEXXsWtCoiycFgzTqsB7nwevo9jlc9I7d6Wkxl2OuLhekXNIvL8Mnw@D/RdSFJX7j1eUlL0lx4bfUomVs/wE
- Imagine you have only one byte (8 bits) to store a value, but need to store values from $0$ to $4032$. Impossible, until you are also told that an error of 1/64 of the exact value does not matter.
- With that knowledge, you decide to design a floating point integer type, which is defined as follows:
- * All representable numbers are of the form $m\cdot 2^e$, where $m$ is the mantissa and $e$ is the exponent. There are no NaNs and no infinities.
- * The upper three bits of the byte are the exponent, the lower five are the mantissa. Since no negative numbers are stored, no sign bit is needed.
- I'll refer to the three bit number represented by the exponent bits as $E$, and the 5 bit value represented by the mantissa bits as $M$
- * If $E=0$, then $e=0$ and $m=M$ (“denormalized integer”).
- * Otherwise, $e=E-1$ and $m=M+32$
- Your task is now to write code that converts a positive integer to that floating point format and back. In particular:
- * Encoding:
- - Your code takes an integer $0\le n\le 4032$, and outputs
- an integer $0\le r\le 255$. You can assume that the input
- integer is always in the allowed range (that is, no overflow
- handling is required)
- - If $n$ can be represented exactly, $r$ is the
- corresponding representation.
- - Otherwise, if $n$ is exactly in the middle of two
- consecutive exactly representable numbers, $r$ is the
- choice with the even mantissa.
- - Otherwise, $r$ is the representation of the closest
- exactly representable number.
- * Decoding:
- Your code takes an integer $0\le r\le 255$ and outputs
- the integer $0\le n\le 4032$ it represents.
- You can write either two separate routines for encoding and decoding, or a single routine that takes an extra argument telling whether to encode or decode. Here “routine” refers to any form of executable code taking input and giving output (program, function, macro, …).
- This is <span class="badge is-tag">code-golf</span>, that is the shortest code wins. If you choose separate encoding and decoding routines, the score is the sum of the sizes of the encoding and decoding routines.
- An ungolfed (and unoptimized) Python implementation is available [here][Python implementation]
- Test cases:
- Encoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 64
- 66 65
- 67 66
- 68 66
- 69 66
- 70 67
- 123 94
- 124 94
- 125 94
- 126 95
- 127 96
- 128 96
- 129 96
- 130 96
- 131 97
- 256 128
- 2021 223
- 4032 255
- Decoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 66
- 66 68
- 93 122
- 94 124
- 95 126
- 96 128
- 97 132
- 128 256
- 160 512
- 223 2016
- 224 2048
- 255 4032
- [Python implementation]: https://tio.run/##lVZdb9owFH3Pr7jqtCmhVAohoQ0qSGxjElrTTYOHalWFUmJGpuBEibNuYv3t7NqOIQGDtAiZ2Of6fp1jQ/aHrVLa3W6D0f1sMp2O5u8nsykMwDN2K8Fo@hlXHGi1oGl2BR0jmNzP7798C0Z3k@/jj1o7dPUwHz@MPsw43GruMAwjIksgdJFGxKR9iCmz4GrIv/sG4BMvgcItNLdJiD85YWVOgRpi5Q1EhJF8HVMCbEWA/M5SSigTIMH4tnh7WcUJQbdDnlCLAM@4ynHvmcDlACsU82W4YGkO0lyFKrIkZkDL9TPJeb4psLyki5CRCNYhZXFRhBDSCJK0YPAcs0Js5LM5n6E7Cu/ArJxjOy1hEMz3fgZgYppDIBZaNjhRWYTRzxLdVwEJLNFVlqcZ5pSnJY1i@kP10WntYw9VTfgx68BgoBCeej2Xt0guoh1r36MAE6ybXGLDRO@Sgpy0UpnPkKBd9xp0iWZBXADOYJ3mnMxQmmBqZZg0iR2jfyJiV56xVlU7CqQoE54bxtixkv4i@TJJX9qqfTnJ0BAdhixOqWpYwOs9pbyx1Ee9SFsmUGnSHMPtbfMsWPAXgkrzERGazzWa5wXlnPbmQVJhcr0WMGEMiRnbVl3FSvT8WXMmNAxxq7E40A1L7OnRca0VuMaDIw7EdonU8sZxTW@EyYU80hf96my35aosGlfli1h8rRrCSMFM5WlOwzXpQ8HyNvYlK1n1npZMTaoqJcEYeJfEY8PJkyn2W7smqQ0D5WxXdJYjBeby4utoOkWqYdNw9GpuZCZD99XihUpHfHphaZqqvH0aTe7@y1sbilVaJhE8E9hUOcoghsGbtAgLwjv9KEKZqtVtPrPlaLU1WEeOOqzbkaMWc@Sow3pdOWoxV45azDuD9aSFFruWFlrs5gzmn8aubelZg3UcUZ/vajH3DOadwUR9vqfFRH1@T4vdnMH801jXPoMJ3n1d7Y7Xk1GPMcd2@D7H0fDu2kIvjsfrq8Dq1GsFWscOBVrDjgRaxw4FWsOOBFrHDgVax7wDwdQxKdAbHeZ3RdO0ufiulI0W86Q0tNghEXXsWtCoiycFgzTqsB7nwevo9jlc9I7d6Wkxl2OuLhekXNIvL8Mnw@D/RdSFJX7j1eUlL0lx4bfUomVs/wE
#4: Post edited
- Imagine you have only one byte (8 bits) to store a value, but need to store values from $0$ to $4032$. Impossible, until you are also told that an error of 1/64 of the exact value does not matter.
- With that knowledge, you decide to design a floating point integer type, which is defined as follows:
- * All representable numbers are of the form $m\cdot 2^e$, where $m$ is the mantissa and $e$ is the exponent. There are no NaNs and no infinities.
- * The upper three bits of the byte are the exponent, the lower five are the mantissa. Since no negative numbers are stored, no sign bit is needed.
- I'll refer to the three bit number represented by the exponent bits as $E$, and the 5 bit value represented by the mantissa bits as $M$
- * If $E=0$, then $e=0$ and $m=M$ (“denormalized integer”).
- * Otherwise, $e=E-1$ and $m=M+32$
- Your task is now to write code that converts a positive integer to that floating point format and back. In particular:
- * Encoding:
- - Your code takes an integer $0\le n\le 4032$, and outputs
- an integer $0\le r\le 255$. You can assume that the input
- integer is always in the allowed range (that is, no overflow
- handling is required)
- - If $n$ can be represented exactly, $r$ is the
- corresponding representation.
- - Otherwise, if $n$ is exactly in the middle of two
- consecutive exactly representable numbers, $r$ is the
- choice with the even mantissa.
- - Otherwise, $r$ is the representation of the closest
- exactly representable number.
- * Decoding:
- Your code takes an integer $0\le r\le 255$ and outputs
- the integer $0\le n\le 4032$ it represents.
- You can write either two separate routines for encoding and decoding, or a single routine that takes an extra argument telling whether to encode or decode. Here “routine” refers to any form of executable code taking input and giving output (program, function, macro, …).
- This is code golf, the shortest code wins. If you choose separate encoding and decoding routines, the score is the sum of the sizes of the encoding and decoding routines.
An ungolfed (an unoptimized) Python implementation is available [here][Python implementation]- Test cases:
- Encoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 64
- 66 65
- 67 66
- 68 66
- 69 66
- 70 67
- 123 94
- 124 94
- 125 94
- 126 95
- 127 96
- 128 96
- 129 96
- 130 96
- 131 97
- 256 128
- 2021 223
- 4032 255
- Decoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 66
- 66 68
- 93 122
- 94 124
- 95 126
- 96 128
- 97 132
- 128 256
- 160 512
- 223 2016
- 224 2048
- 255 4032
- [Python implementation]: https://tio.run/##lVZdb9owFH3Pr7jqtCmhVAohoQ0qSGxjElrTTYOHalWFUmJGpuBEibNuYv3t7NqOIQGDtAiZ2Of6fp1jQ/aHrVLa3W6D0f1sMp2O5u8nsykMwDN2K8Fo@hlXHGi1oGl2BR0jmNzP7798C0Z3k@/jj1o7dPUwHz@MPsw43GruMAwjIksgdJFGxKR9iCmz4GrIv/sG4BMvgcItNLdJiD85YWVOgRpi5Q1EhJF8HVMCbEWA/M5SSigTIMH4tnh7WcUJQbdDnlCLAM@4ynHvmcDlACsU82W4YGkO0lyFKrIkZkDL9TPJeb4psLyki5CRCNYhZXFRhBDSCJK0YPAcs0Js5LM5n6E7Cu/ArJxjOy1hEMz3fgZgYppDIBZaNjhRWYTRzxLdVwEJLNFVlqcZ5pSnJY1i@kP10WntYw9VTfgx68BgoBCeej2Xt0guoh1r36MAE6ybXGLDRO@Sgpy0UpnPkKBd9xp0iWZBXADOYJ3mnMxQmmBqZZg0iR2jfyJiV56xVlU7CqQoE54bxtixkv4i@TJJX9qqfTnJ0BAdhixOqWpYwOs9pbyx1Ee9SFsmUGnSHMPtbfMsWPAXgkrzERGazzWa5wXlnPbmQVJhcr0WMGEMiRnbVl3FSvT8WXMmNAxxq7E40A1L7OnRca0VuMaDIw7EdonU8sZxTW@EyYU80hf96my35aosGlfli1h8rRrCSMFM5WlOwzXpQ8HyNvYlK1n1npZMTaoqJcEYeJfEY8PJkyn2W7smqQ0D5WxXdJYjBeby4utoOkWqYdNw9GpuZCZD99XihUpHfHphaZqqvH0aTe7@y1sbilVaJhE8E9hUOcoghsGbtAgLwjv9KEKZqtVtPrPlaLU1WEeOOqzbkaMWc@Sow3pdOWoxV45azDuD9aSFFruWFlrs5gzmn8aubelZg3UcUZ/vajH3DOadwUR9vqfFRH1@T4vdnMH801jXPoMJ3n1d7Y7Xk1GPMcd2@D7H0fDu2kIvjsfrq8Dq1GsFWscOBVrDjgRaxw4FWsOOBFrHDgVax7wDwdQxKdAbHeZ3RdO0ufiulI0W86Q0tNghEXXsWtCoiycFgzTqsB7nwevo9jlc9I7d6Wkxl2OuLhekXNIvL8Mnw@D/RdSFJX7j1eUlL0lx4bfUomVs/wE
- Imagine you have only one byte (8 bits) to store a value, but need to store values from $0$ to $4032$. Impossible, until you are also told that an error of 1/64 of the exact value does not matter.
- With that knowledge, you decide to design a floating point integer type, which is defined as follows:
- * All representable numbers are of the form $m\cdot 2^e$, where $m$ is the mantissa and $e$ is the exponent. There are no NaNs and no infinities.
- * The upper three bits of the byte are the exponent, the lower five are the mantissa. Since no negative numbers are stored, no sign bit is needed.
- I'll refer to the three bit number represented by the exponent bits as $E$, and the 5 bit value represented by the mantissa bits as $M$
- * If $E=0$, then $e=0$ and $m=M$ (“denormalized integer”).
- * Otherwise, $e=E-1$ and $m=M+32$
- Your task is now to write code that converts a positive integer to that floating point format and back. In particular:
- * Encoding:
- - Your code takes an integer $0\le n\le 4032$, and outputs
- an integer $0\le r\le 255$. You can assume that the input
- integer is always in the allowed range (that is, no overflow
- handling is required)
- - If $n$ can be represented exactly, $r$ is the
- corresponding representation.
- - Otherwise, if $n$ is exactly in the middle of two
- consecutive exactly representable numbers, $r$ is the
- choice with the even mantissa.
- - Otherwise, $r$ is the representation of the closest
- exactly representable number.
- * Decoding:
- Your code takes an integer $0\le r\le 255$ and outputs
- the integer $0\le n\le 4032$ it represents.
- You can write either two separate routines for encoding and decoding, or a single routine that takes an extra argument telling whether to encode or decode. Here “routine” refers to any form of executable code taking input and giving output (program, function, macro, …).
- This is code golf, the shortest code wins. If you choose separate encoding and decoding routines, the score is the sum of the sizes of the encoding and decoding routines.
- An ungolfed (and unoptimized) Python implementation is available [here][Python implementation]
- Test cases:
- Encoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 64
- 66 65
- 67 66
- 68 66
- 69 66
- 70 67
- 123 94
- 124 94
- 125 94
- 126 95
- 127 96
- 128 96
- 129 96
- 130 96
- 131 97
- 256 128
- 2021 223
- 4032 255
- Decoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 66
- 66 68
- 93 122
- 94 124
- 95 126
- 96 128
- 97 132
- 128 256
- 160 512
- 223 2016
- 224 2048
- 255 4032
- [Python implementation]: https://tio.run/##lVZdb9owFH3Pr7jqtCmhVAohoQ0qSGxjElrTTYOHalWFUmJGpuBEibNuYv3t7NqOIQGDtAiZ2Of6fp1jQ/aHrVLa3W6D0f1sMp2O5u8nsykMwDN2K8Fo@hlXHGi1oGl2BR0jmNzP7798C0Z3k@/jj1o7dPUwHz@MPsw43GruMAwjIksgdJFGxKR9iCmz4GrIv/sG4BMvgcItNLdJiD85YWVOgRpi5Q1EhJF8HVMCbEWA/M5SSigTIMH4tnh7WcUJQbdDnlCLAM@4ynHvmcDlACsU82W4YGkO0lyFKrIkZkDL9TPJeb4psLyki5CRCNYhZXFRhBDSCJK0YPAcs0Js5LM5n6E7Cu/ArJxjOy1hEMz3fgZgYppDIBZaNjhRWYTRzxLdVwEJLNFVlqcZ5pSnJY1i@kP10WntYw9VTfgx68BgoBCeej2Xt0guoh1r36MAE6ybXGLDRO@Sgpy0UpnPkKBd9xp0iWZBXADOYJ3mnMxQmmBqZZg0iR2jfyJiV56xVlU7CqQoE54bxtixkv4i@TJJX9qqfTnJ0BAdhixOqWpYwOs9pbyx1Ee9SFsmUGnSHMPtbfMsWPAXgkrzERGazzWa5wXlnPbmQVJhcr0WMGEMiRnbVl3FSvT8WXMmNAxxq7E40A1L7OnRca0VuMaDIw7EdonU8sZxTW@EyYU80hf96my35aosGlfli1h8rRrCSMFM5WlOwzXpQ8HyNvYlK1n1npZMTaoqJcEYeJfEY8PJkyn2W7smqQ0D5WxXdJYjBeby4utoOkWqYdNw9GpuZCZD99XihUpHfHphaZqqvH0aTe7@y1sbilVaJhE8E9hUOcoghsGbtAgLwjv9KEKZqtVtPrPlaLU1WEeOOqzbkaMWc@Sow3pdOWoxV45azDuD9aSFFruWFlrs5gzmn8aubelZg3UcUZ/vajH3DOadwUR9vqfFRH1@T4vdnMH801jXPoMJ3n1d7Y7Xk1GPMcd2@D7H0fDu2kIvjsfrq8Dq1GsFWscOBVrDjgRaxw4FWsOOBFrHDgVax7wDwdQxKdAbHeZ3RdO0ufiulI0W86Q0tNghEXXsWtCoiycFgzTqsB7nwevo9jlc9I7d6Wkxl2OuLhekXNIvL8Mnw@D/RdSFJX7j1eUlL0lx4bfUomVs/wE
#3: Post edited
- Imagine you have only one byte (8 bits) to store a value, but need to store values from $0$ to $4032$. Impossible, until you are also told that an error of 1/64 of the exact value does not matter.
- With that knowledge, you decide to design a floating point integer type, which is defined as follows:
- * All representable numbers are of the form $m\cdot 2^e$, where $m$ is the mantissa and $e$ is the exponent. There are no NaNs and no infinities.
- * The upper three bits of the byte are the exponent, the lower five are the mantissa. Since no negative numbers are stored, no sign bit is needed.
- I'll refer to the three bit number represented by the exponent bits as $E$, and the 5 bit value represented by the mantissa bits as $M$
- * If $E=0$, then $e=0$ and $m=M$ (“denormalized integer”).
- * Otherwise, $e=E-1$ and $m=M+32$
- Your task is now to write code that converts a positive integer to that floating point format and back. In particular:
- * Encoding:
- - Your code takes an integer $0\le n\le 4032$, and outputs
- an integer $0\le r\le 255$. You can assume that the input
integer is always in the allowed range (that is, no error- handling is required)
- - If $n$ can be represented exactly, $r$ is the
- corresponding representation.
- - Otherwise, if $n$ is exactly in the middle of two
- consecutive exactly representable numbers, $r$ is the
- choice with the even mantissa.
- - Otherwise, $r$ is the representation of the closest
- exactly representable number.
- * Decoding:
- Your code takes an integer $0\le r\le 255$ and outputs
- the integer $0\le n\le 4032$ it represents.
- You can write either two separate routines for encoding and decoding, or a single routine that takes an extra argument telling whether to encode or decode. Here “routine” refers to any form of executable code taking input and giving output (program, function, macro, …).
- This is code golf, the shortest code wins. If you choose separate encoding and decoding routines, the score is the sum of the sizes of the encoding and decoding routines.
- An ungolfed (an unoptimized) Python implementation is available [here][Python implementation]
- Test cases:
- Encoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 64
- 66 65
- 67 66
- 68 66
- 69 66
- 70 67
- 123 94
- 124 94
- 125 94
- 126 95
- 127 96
- 128 96
- 129 96
- 130 96
- 131 97
- 256 128
- 2021 223
- 4032 255
- Decoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 66
- 66 68
- 93 122
- 94 124
- 95 126
- 96 128
- 97 132
- 128 256
- 160 512
- 223 2016
- 224 2048
- 255 4032
- [Python implementation]: https://tio.run/##lVZdb9owFH3Pr7jqtCmhVAohoQ0qSGxjElrTTYOHalWFUmJGpuBEibNuYv3t7NqOIQGDtAiZ2Of6fp1jQ/aHrVLa3W6D0f1sMp2O5u8nsykMwDN2K8Fo@hlXHGi1oGl2BR0jmNzP7798C0Z3k@/jj1o7dPUwHz@MPsw43GruMAwjIksgdJFGxKR9iCmz4GrIv/sG4BMvgcItNLdJiD85YWVOgRpi5Q1EhJF8HVMCbEWA/M5SSigTIMH4tnh7WcUJQbdDnlCLAM@4ynHvmcDlACsU82W4YGkO0lyFKrIkZkDL9TPJeb4psLyki5CRCNYhZXFRhBDSCJK0YPAcs0Js5LM5n6E7Cu/ArJxjOy1hEMz3fgZgYppDIBZaNjhRWYTRzxLdVwEJLNFVlqcZ5pSnJY1i@kP10WntYw9VTfgx68BgoBCeej2Xt0guoh1r36MAE6ybXGLDRO@Sgpy0UpnPkKBd9xp0iWZBXADOYJ3mnMxQmmBqZZg0iR2jfyJiV56xVlU7CqQoE54bxtixkv4i@TJJX9qqfTnJ0BAdhixOqWpYwOs9pbyx1Ee9SFsmUGnSHMPtbfMsWPAXgkrzERGazzWa5wXlnPbmQVJhcr0WMGEMiRnbVl3FSvT8WXMmNAxxq7E40A1L7OnRca0VuMaDIw7EdonU8sZxTW@EyYU80hf96my35aosGlfli1h8rRrCSMFM5WlOwzXpQ8HyNvYlK1n1npZMTaoqJcEYeJfEY8PJkyn2W7smqQ0D5WxXdJYjBeby4utoOkWqYdNw9GpuZCZD99XihUpHfHphaZqqvH0aTe7@y1sbilVaJhE8E9hUOcoghsGbtAgLwjv9KEKZqtVtPrPlaLU1WEeOOqzbkaMWc@Sow3pdOWoxV45azDuD9aSFFruWFlrs5gzmn8aubelZg3UcUZ/vajH3DOadwUR9vqfFRH1@T4vdnMH801jXPoMJ3n1d7Y7Xk1GPMcd2@D7H0fDu2kIvjsfrq8Dq1GsFWscOBVrDjgRaxw4FWsOOBFrHDgVax7wDwdQxKdAbHeZ3RdO0ufiulI0W86Q0tNghEXXsWtCoiycFgzTqsB7nwevo9jlc9I7d6Wkxl2OuLhekXNIvL8Mnw@D/RdSFJX7j1eUlL0lx4bfUomVs/wE
- Imagine you have only one byte (8 bits) to store a value, but need to store values from $0$ to $4032$. Impossible, until you are also told that an error of 1/64 of the exact value does not matter.
- With that knowledge, you decide to design a floating point integer type, which is defined as follows:
- * All representable numbers are of the form $m\cdot 2^e$, where $m$ is the mantissa and $e$ is the exponent. There are no NaNs and no infinities.
- * The upper three bits of the byte are the exponent, the lower five are the mantissa. Since no negative numbers are stored, no sign bit is needed.
- I'll refer to the three bit number represented by the exponent bits as $E$, and the 5 bit value represented by the mantissa bits as $M$
- * If $E=0$, then $e=0$ and $m=M$ (“denormalized integer”).
- * Otherwise, $e=E-1$ and $m=M+32$
- Your task is now to write code that converts a positive integer to that floating point format and back. In particular:
- * Encoding:
- - Your code takes an integer $0\le n\le 4032$, and outputs
- an integer $0\le r\le 255$. You can assume that the input
- integer is always in the allowed range (that is, no overflow
- handling is required)
- - If $n$ can be represented exactly, $r$ is the
- corresponding representation.
- - Otherwise, if $n$ is exactly in the middle of two
- consecutive exactly representable numbers, $r$ is the
- choice with the even mantissa.
- - Otherwise, $r$ is the representation of the closest
- exactly representable number.
- * Decoding:
- Your code takes an integer $0\le r\le 255$ and outputs
- the integer $0\le n\le 4032$ it represents.
- You can write either two separate routines for encoding and decoding, or a single routine that takes an extra argument telling whether to encode or decode. Here “routine” refers to any form of executable code taking input and giving output (program, function, macro, …).
- This is code golf, the shortest code wins. If you choose separate encoding and decoding routines, the score is the sum of the sizes of the encoding and decoding routines.
- An ungolfed (an unoptimized) Python implementation is available [here][Python implementation]
- Test cases:
- Encoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 64
- 66 65
- 67 66
- 68 66
- 69 66
- 70 67
- 123 94
- 124 94
- 125 94
- 126 95
- 127 96
- 128 96
- 129 96
- 130 96
- 131 97
- 256 128
- 2021 223
- 4032 255
- Decoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 66
- 66 68
- 93 122
- 94 124
- 95 126
- 96 128
- 97 132
- 128 256
- 160 512
- 223 2016
- 224 2048
- 255 4032
- [Python implementation]: https://tio.run/##lVZdb9owFH3Pr7jqtCmhVAohoQ0qSGxjElrTTYOHalWFUmJGpuBEibNuYv3t7NqOIQGDtAiZ2Of6fp1jQ/aHrVLa3W6D0f1sMp2O5u8nsykMwDN2K8Fo@hlXHGi1oGl2BR0jmNzP7798C0Z3k@/jj1o7dPUwHz@MPsw43GruMAwjIksgdJFGxKR9iCmz4GrIv/sG4BMvgcItNLdJiD85YWVOgRpi5Q1EhJF8HVMCbEWA/M5SSigTIMH4tnh7WcUJQbdDnlCLAM@4ynHvmcDlACsU82W4YGkO0lyFKrIkZkDL9TPJeb4psLyki5CRCNYhZXFRhBDSCJK0YPAcs0Js5LM5n6E7Cu/ArJxjOy1hEMz3fgZgYppDIBZaNjhRWYTRzxLdVwEJLNFVlqcZ5pSnJY1i@kP10WntYw9VTfgx68BgoBCeej2Xt0guoh1r36MAE6ybXGLDRO@Sgpy0UpnPkKBd9xp0iWZBXADOYJ3mnMxQmmBqZZg0iR2jfyJiV56xVlU7CqQoE54bxtixkv4i@TJJX9qqfTnJ0BAdhixOqWpYwOs9pbyx1Ee9SFsmUGnSHMPtbfMsWPAXgkrzERGazzWa5wXlnPbmQVJhcr0WMGEMiRnbVl3FSvT8WXMmNAxxq7E40A1L7OnRca0VuMaDIw7EdonU8sZxTW@EyYU80hf96my35aosGlfli1h8rRrCSMFM5WlOwzXpQ8HyNvYlK1n1npZMTaoqJcEYeJfEY8PJkyn2W7smqQ0D5WxXdJYjBeby4utoOkWqYdNw9GpuZCZD99XihUpHfHphaZqqvH0aTe7@y1sbilVaJhE8E9hUOcoghsGbtAgLwjv9KEKZqtVtPrPlaLU1WEeOOqzbkaMWc@Sow3pdOWoxV45azDuD9aSFFruWFlrs5gzmn8aubelZg3UcUZ/vajH3DOadwUR9vqfFRH1@T4vdnMH801jXPoMJ3n1d7Y7Xk1GPMcd2@D7H0fDu2kIvjsfrq8Dq1GsFWscOBVrDjgRaxw4FWsOOBFrHDgVax7wDwdQxKdAbHeZ3RdO0ufiulI0W86Q0tNghEXXsWtCoiycFgzTqsB7nwevo9jlc9I7d6Wkxl2OuLhekXNIvL8Mnw@D/RdSFJX7j1eUlL0lx4bfUomVs/wE
#2: Post edited
- Imagine you have only one byte (8 bits) to store a value, but need to store values from $0$ to $4032$. Impossible, until you are also told that an error of 1/64 of the exact value does not matter.
- With that knowledge, you decide to design a floating point integer type, which is defined as follows:
- * All representable numbers are of the form $m\cdot 2^e$, where $m$ is the mantissa and $e$ is the exponent. There are no NaNs and no infinities.
- * The upper three bits of the byte are the exponent, the lower five are the mantissa. Since no negative numbers are stored, no sign bit is needed.
- I'll refer to the three bit number represented by the exponent bits as $E$, and the 5 bit value represented by the mantissa bits as $M$
- * If $E=0$, then $e=0$ and $m=M$ (“denormalized integer”).
- * Otherwise, $e=E-1$ and $m=M+32$
- Your task is now to write code that converts a positive integer to that floating point format and back. In particular:
- * Encoding:
- Your code takes an integer $0\le n\le 1984$, and outputs- an integer $0\le r\le 255$. You can assume that the input
- integer is always in the allowed range (that is, no error
- handling is required)
- - If $n$ can be represented exactly, $r$ is the
- corresponding representation.
- - Otherwise, if $n$ is exactly in the middle of two
- consecutive exactly representable numbers, $r$ is the
- choice with the even mantissa.
- - Otherwise, $r$ is the representation of the closest
- exactly representable number.
- * Decoding:
- Your code takes an integer $0\le r\le 255$ and outputs
- the integer $0\le n\le 4032$ it represents.
- You can write either two separate routines for encoding and decoding, or a single routine that takes an extra argument telling whether to encode or decode. Here “routine” refers to any form of executable code taking input and giving output (program, function, macro, …).
- This is code golf, the shortest code wins. If you choose separate encoding and decoding routines, the score is the sum of the sizes of the encoding and decoding routines.
- An ungolfed (an unoptimized) Python implementation is available [here][Python implementation]
- Test cases:
- Encoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 64
- 66 65
- 67 66
- 68 66
- 69 66
- 70 67
- 123 94
- 124 94
- 125 94
- 126 95
- 127 96
- 128 96
- 129 96
- 130 96
- 131 97
- 256 128
- 2021 223
- 4032 255
- Decoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 66
- 66 68
- 93 122
- 94 124
- 95 126
- 96 128
- 97 132
- 128 256
- 160 512
- 223 2016
- 224 2048
- 255 4032
- [Python implementation]: https://tio.run/##lVZdb9owFH3Pr7jqtCmhVAohoQ0qSGxjElrTTYOHalWFUmJGpuBEibNuYv3t7NqOIQGDtAiZ2Of6fp1jQ/aHrVLa3W6D0f1sMp2O5u8nsykMwDN2K8Fo@hlXHGi1oGl2BR0jmNzP7798C0Z3k@/jj1o7dPUwHz@MPsw43GruMAwjIksgdJFGxKR9iCmz4GrIv/sG4BMvgcItNLdJiD85YWVOgRpi5Q1EhJF8HVMCbEWA/M5SSigTIMH4tnh7WcUJQbdDnlCLAM@4ynHvmcDlACsU82W4YGkO0lyFKrIkZkDL9TPJeb4psLyki5CRCNYhZXFRhBDSCJK0YPAcs0Js5LM5n6E7Cu/ArJxjOy1hEMz3fgZgYppDIBZaNjhRWYTRzxLdVwEJLNFVlqcZ5pSnJY1i@kP10WntYw9VTfgx68BgoBCeej2Xt0guoh1r36MAE6ybXGLDRO@Sgpy0UpnPkKBd9xp0iWZBXADOYJ3mnMxQmmBqZZg0iR2jfyJiV56xVlU7CqQoE54bxtixkv4i@TJJX9qqfTnJ0BAdhixOqWpYwOs9pbyx1Ee9SFsmUGnSHMPtbfMsWPAXgkrzERGazzWa5wXlnPbmQVJhcr0WMGEMiRnbVl3FSvT8WXMmNAxxq7E40A1L7OnRca0VuMaDIw7EdonU8sZxTW@EyYU80hf96my35aosGlfli1h8rRrCSMFM5WlOwzXpQ8HyNvYlK1n1npZMTaoqJcEYeJfEY8PJkyn2W7smqQ0D5WxXdJYjBeby4utoOkWqYdNw9GpuZCZD99XihUpHfHphaZqqvH0aTe7@y1sbilVaJhE8E9hUOcoghsGbtAgLwjv9KEKZqtVtPrPlaLU1WEeOOqzbkaMWc@Sow3pdOWoxV45azDuD9aSFFruWFlrs5gzmn8aubelZg3UcUZ/vajH3DOadwUR9vqfFRH1@T4vdnMH801jXPoMJ3n1d7Y7Xk1GPMcd2@D7H0fDu2kIvjsfrq8Dq1GsFWscOBVrDjgRaxw4FWsOOBFrHDgVax7wDwdQxKdAbHeZ3RdO0ufiulI0W86Q0tNghEXXsWtCoiycFgzTqsB7nwevo9jlc9I7d6Wkxl2OuLhekXNIvL8Mnw@D/RdSFJX7j1eUlL0lx4bfUomVs/wE
- Imagine you have only one byte (8 bits) to store a value, but need to store values from $0$ to $4032$. Impossible, until you are also told that an error of 1/64 of the exact value does not matter.
- With that knowledge, you decide to design a floating point integer type, which is defined as follows:
- * All representable numbers are of the form $m\cdot 2^e$, where $m$ is the mantissa and $e$ is the exponent. There are no NaNs and no infinities.
- * The upper three bits of the byte are the exponent, the lower five are the mantissa. Since no negative numbers are stored, no sign bit is needed.
- I'll refer to the three bit number represented by the exponent bits as $E$, and the 5 bit value represented by the mantissa bits as $M$
- * If $E=0$, then $e=0$ and $m=M$ (“denormalized integer”).
- * Otherwise, $e=E-1$ and $m=M+32$
- Your task is now to write code that converts a positive integer to that floating point format and back. In particular:
- * Encoding:
- - Your code takes an integer $0\le n\le 4032$, and outputs
- an integer $0\le r\le 255$. You can assume that the input
- integer is always in the allowed range (that is, no error
- handling is required)
- - If $n$ can be represented exactly, $r$ is the
- corresponding representation.
- - Otherwise, if $n$ is exactly in the middle of two
- consecutive exactly representable numbers, $r$ is the
- choice with the even mantissa.
- - Otherwise, $r$ is the representation of the closest
- exactly representable number.
- * Decoding:
- Your code takes an integer $0\le r\le 255$ and outputs
- the integer $0\le n\le 4032$ it represents.
- You can write either two separate routines for encoding and decoding, or a single routine that takes an extra argument telling whether to encode or decode. Here “routine” refers to any form of executable code taking input and giving output (program, function, macro, …).
- This is code golf, the shortest code wins. If you choose separate encoding and decoding routines, the score is the sum of the sizes of the encoding and decoding routines.
- An ungolfed (an unoptimized) Python implementation is available [here][Python implementation]
- Test cases:
- Encoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 64
- 66 65
- 67 66
- 68 66
- 69 66
- 70 67
- 123 94
- 124 94
- 125 94
- 126 95
- 127 96
- 128 96
- 129 96
- 130 96
- 131 97
- 256 128
- 2021 223
- 4032 255
- Decoding:
- input output
- 0 0
- 1 1
- 31 31
- 32 32
- 63 63
- 64 64
- 65 66
- 66 68
- 93 122
- 94 124
- 95 126
- 96 128
- 97 132
- 128 256
- 160 512
- 223 2016
- 224 2048
- 255 4032
- [Python implementation]: https://tio.run/##lVZdb9owFH3Pr7jqtCmhVAohoQ0qSGxjElrTTYOHalWFUmJGpuBEibNuYv3t7NqOIQGDtAiZ2Of6fp1jQ/aHrVLa3W6D0f1sMp2O5u8nsykMwDN2K8Fo@hlXHGi1oGl2BR0jmNzP7798C0Z3k@/jj1o7dPUwHz@MPsw43GruMAwjIksgdJFGxKR9iCmz4GrIv/sG4BMvgcItNLdJiD85YWVOgRpi5Q1EhJF8HVMCbEWA/M5SSigTIMH4tnh7WcUJQbdDnlCLAM@4ynHvmcDlACsU82W4YGkO0lyFKrIkZkDL9TPJeb4psLyki5CRCNYhZXFRhBDSCJK0YPAcs0Js5LM5n6E7Cu/ArJxjOy1hEMz3fgZgYppDIBZaNjhRWYTRzxLdVwEJLNFVlqcZ5pSnJY1i@kP10WntYw9VTfgx68BgoBCeej2Xt0guoh1r36MAE6ybXGLDRO@Sgpy0UpnPkKBd9xp0iWZBXADOYJ3mnMxQmmBqZZg0iR2jfyJiV56xVlU7CqQoE54bxtixkv4i@TJJX9qqfTnJ0BAdhixOqWpYwOs9pbyx1Ee9SFsmUGnSHMPtbfMsWPAXgkrzERGazzWa5wXlnPbmQVJhcr0WMGEMiRnbVl3FSvT8WXMmNAxxq7E40A1L7OnRca0VuMaDIw7EdonU8sZxTW@EyYU80hf96my35aosGlfli1h8rRrCSMFM5WlOwzXpQ8HyNvYlK1n1npZMTaoqJcEYeJfEY8PJkyn2W7smqQ0D5WxXdJYjBeby4utoOkWqYdNw9GpuZCZD99XihUpHfHphaZqqvH0aTe7@y1sbilVaJhE8E9hUOcoghsGbtAgLwjv9KEKZqtVtPrPlaLU1WEeOOqzbkaMWc@Sow3pdOWoxV45azDuD9aSFFruWFlrs5gzmn8aubelZg3UcUZ/vajH3DOadwUR9vqfFRH1@T4vdnMH801jXPoMJ3n1d7Y7Xk1GPMcd2@D7H0fDu2kIvjsfrq8Dq1GsFWscOBVrDjgRaxw4FWsOOBFrHDgVax7wDwdQxKdAbHeZ3RdO0ufiulI0W86Q0tNghEXXsWtCoiycFgzTqsB7nwevo9jlc9I7d6Wkxl2OuLhekXNIvL8Mnw@D/RdSFJX7j1eUlL0lx4bfUomVs/wE
#1: Initial revision
Encode and decode floating point integers
Imagine you have only one byte (8 bits) to store a value, but need to store values from $0$ to $4032$. Impossible, until you are also told that an error of 1/64 of the exact value does not matter. With that knowledge, you decide to design a floating point integer type, which is defined as follows: * All representable numbers are of the form $m\cdot 2^e$, where $m$ is the mantissa and $e$ is the exponent. There are no NaNs and no infinities. * The upper three bits of the byte are the exponent, the lower five are the mantissa. Since no negative numbers are stored, no sign bit is needed. I'll refer to the three bit number represented by the exponent bits as $E$, and the 5 bit value represented by the mantissa bits as $M$ * If $E=0$, then $e=0$ and $m=M$ (“denormalized integer”). * Otherwise, $e=E-1$ and $m=M+32$ Your task is now to write code that converts a positive integer to that floating point format and back. In particular: * Encoding: - Your code takes an integer $0\le n\le 1984$, and outputs an integer $0\le r\le 255$. You can assume that the input integer is always in the allowed range (that is, no error handling is required) - If $n$ can be represented exactly, $r$ is the corresponding representation. - Otherwise, if $n$ is exactly in the middle of two consecutive exactly representable numbers, $r$ is the choice with the even mantissa. - Otherwise, $r$ is the representation of the closest exactly representable number. * Decoding: Your code takes an integer $0\le r\le 255$ and outputs the integer $0\le n\le 4032$ it represents. You can write either two separate routines for encoding and decoding, or a single routine that takes an extra argument telling whether to encode or decode. Here “routine” refers to any form of executable code taking input and giving output (program, function, macro, …). This is code golf, the shortest code wins. If you choose separate encoding and decoding routines, the score is the sum of the sizes of the encoding and decoding routines. An ungolfed (an unoptimized) Python implementation is available [here][Python implementation] Test cases: Encoding: input output 0 0 1 1 31 31 32 32 63 63 64 64 65 64 66 65 67 66 68 66 69 66 70 67 123 94 124 94 125 94 126 95 127 96 128 96 129 96 130 96 131 97 256 128 2021 223 4032 255 Decoding: input output 0 0 1 1 31 31 32 32 63 63 64 64 65 66 66 68 93 122 94 124 95 126 96 128 97 132 128 256 160 512 223 2016 224 2048 255 4032 [Python implementation]: https://tio.run/##lVZdb9owFH3Pr7jqtCmhVAohoQ0qSGxjElrTTYOHalWFUmJGpuBEibNuYv3t7NqOIQGDtAiZ2Of6fp1jQ/aHrVLa3W6D0f1sMp2O5u8nsykMwDN2K8Fo@hlXHGi1oGl2BR0jmNzP7798C0Z3k@/jj1o7dPUwHz@MPsw43GruMAwjIksgdJFGxKR9iCmz4GrIv/sG4BMvgcItNLdJiD85YWVOgRpi5Q1EhJF8HVMCbEWA/M5SSigTIMH4tnh7WcUJQbdDnlCLAM@4ynHvmcDlACsU82W4YGkO0lyFKrIkZkDL9TPJeb4psLyki5CRCNYhZXFRhBDSCJK0YPAcs0Js5LM5n6E7Cu/ArJxjOy1hEMz3fgZgYppDIBZaNjhRWYTRzxLdVwEJLNFVlqcZ5pSnJY1i@kP10WntYw9VTfgx68BgoBCeej2Xt0guoh1r36MAE6ybXGLDRO@Sgpy0UpnPkKBd9xp0iWZBXADOYJ3mnMxQmmBqZZg0iR2jfyJiV56xVlU7CqQoE54bxtixkv4i@TJJX9qqfTnJ0BAdhixOqWpYwOs9pbyx1Ee9SFsmUGnSHMPtbfMsWPAXgkrzERGazzWa5wXlnPbmQVJhcr0WMGEMiRnbVl3FSvT8WXMmNAxxq7E40A1L7OnRca0VuMaDIw7EdonU8sZxTW@EyYU80hf96my35aosGlfli1h8rRrCSMFM5WlOwzXpQ8HyNvYlK1n1npZMTaoqJcEYeJfEY8PJkyn2W7smqQ0D5WxXdJYjBeby4utoOkWqYdNw9GpuZCZD99XihUpHfHphaZqqvH0aTe7@y1sbilVaJhE8E9hUOcoghsGbtAgLwjv9KEKZqtVtPrPlaLU1WEeOOqzbkaMWc@Sow3pdOWoxV45azDuD9aSFFruWFlrs5gzmn8aubelZg3UcUZ/vajH3DOadwUR9vqfFRH1@T4vdnMH801jXPoMJ3n1d7Y7Xk1GPMcd2@D7H0fDu2kIvjsfrq8Dq1GsFWscOBVrDjgRaxw4FWsOOBFrHDgVax7wDwdQxKdAbHeZ3RdO0ufiulI0W86Q0tNghEXXsWtCoiycFgzTqsB7nwevo9jlc9I7d6Wkxl2OuLhekXNIvL8Mnw@D/RdSFJX7j1eUlL0lx4bfUomVs/wE