| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the problem, we need to calculate the storage requirement for the video in gigabytes (Gbytes). First, let's calculate the storage requirement for a single frame: - The resolution is 1920x1080, which means there are 1920x1080 = 2,073,600 pixels in a frame. - Each pixel has three components: Y, U, and V. However, due to the 4:2:2 color sub-sampling, only two-thirds of the U and V components are used (since for every four pixels, two U and two V components are stored). Therefore, the number of U and V components is 2,073,600 * 2/3 = 1,382,400. - Each component is represented by 8 bits, which is equivalent to 1 byte. - The total storage requirement for a single frame is: 2,073,600 bytes (for Y) + 1,382,400 bytes (for U) + 1,382,400 bytes (for V) = 4,838,400 bytes. Next, let's calculate the storage requirement for the entire one-hour video: - There are 30 frames per second, so there are 30 * 60 * 60 = 108,000 frames in one hour. - The total storage requirement for the entire video is: 4,838,400 bytes/frame * 108,000 frames = 520,233,600,000 bytes. Finally, let's convert the storage requirement to gigabytes (Gbytes): - 1 gigabyte (Gbyte) = 1,073,741,824 bytes. - The storage requirement for the video in Gbytes is: 520,233,600,000 bytes / 1,073,741,824 bytes/Gbyte = 484 Gbytes (rounded to three significant figures). Therefore, the storage requirement for the video in Gbytes is approximately 484 Gbytes. |
417 |
| deepseekr1 | To determine the storage requirement for the one-hour color video in YUV 4:2:2 format: 1. **Calculate pixels per frame**: - Resolution: 1920 × 1080 = 2,073,600 pixels. 2. **Factor in 4:2:2 chroma subsampling**: - **Y (Luma)**: 1920 × 1080 = 2,073,600 samples (1 byte each). - **U and V (Chroma)**: Each has half the horizontal resolution of Y: - 960 × 1080 = 1,036,800 samples per component (1 byte each). - **Total per frame**: \(2,073,600 + 1,036,800 + 1,036,800 = 4,147,200\) bytes. 3. **Calculate data per second**: - Frame rate: 30 frames/s. - \(4,147,200 \, \text{bytes/frame} × 30 \, \text{frames/s} = 124,416,000 \, \text{bytes/s}\). 4. **Total data for one hour**: - \(124,416,000 \, \text{bytes/s} × 3600 \, \text{s} = 447,897,600,000 \, \text{bytes}\). 5. **Convert to gigabytes (1 GB = \(10^9\) bytes)**: - \(447,897,600,000 ÷ 10^9 = 447.8976 \, \text{GB}\). 6. **Round to 3 significant figures**: - \(447.8976 \, \text{GB} ≈ 448 \, \text{GB}\). **Answer**: \(\boxed{448}\) GB. |
417 |
| gpto1 | To determine the storage requirement for the given video, we need to calculate the total amount of data generated by all the frames over the one-hour duration. **Given:** - Frame resolution: **1920 x 1080 pixels** - Color subsampling format: **4:2:2** - Bit depth: **8 bits per component** - Frame rate: **30 frames per second** - Video duration: **1 hour** **Step 1: Understand 4:2:2 Subsampling** In 4:2:2 chroma subsampling: - **Luma (Y)** is sampled at every pixel (full resolution). - **Chroma components (Cb and Cr)** are sampled at half the horizontal resolution. This means: - For every **2 pixels horizontally**, there is **1 sample** of Cb and **1 sample** of Cr. **Step 2: Calculate Number of Samples per Frame** - **Total number of pixels per frame (Np):** \[ N_p = 1920 \times 1080 = 2,073,600 \text{ pixels} \] - **Luma samples (Y):** - Each pixel has a Y sample. - Total Y samples per frame: \[ Y_{\text{samples}} = N_p = 2,073,600 \text{ samples} \] - **Chroma samples (Cb and Cr):** - Chroma is subsampled horizontally by a factor of 2. - **Horizontal chroma resolution:** \[ \text{Chroma width} = \frac{1920}{2} = 960 \text{ pixels} \] - Total Cb samples per frame: \[ \text{Cb}_{\text{samples}} = 960 \times 1080 = 1,036,800 \text{ samples} \] - Total Cr samples per frame: \[ \text{Cr}_{\text{samples}} = 960 \times 1080 = 1,036,800 \text{ samples} \] - **Total samples per frame:** \[ \text{Total samples} = Y_{\text{samples}} + \text{Cb}_{\text{samples}} + \text{Cr}_{\text{samples}} = 2,073,600 + 1,036,800 + 1,036,800 = 4,147,200 \text{ samples} \] **Step 3: Calculate Bits per Frame** - Each sample is **8 bits**. - **Total bits per frame:** \[ \text{Bits per frame} = \text{Total samples} \times 8 = 4,147,200 \times 8 = 33,177,600 \text{ bits} \] **Step 4: Calculate Total Bits for One Hour** - **Frames per second:** 30 - **Total frames in one hour:** \[ \text{Total frames} = 30 \times 3600 = 108,000 \text{ frames} \] - **Total bits for one hour:** \[ \text{Total bits} = \text{Bits per frame} \times \text{Total frames} = 33,177,600 \times 108,000 = 3,583,180,800,000 \text{ bits} \] **Step 5: Convert Bits to Gigabytes** - **Bytes:** Divide bits by 8. \[ \text{Total bytes} = \frac{3,583,180,800,000}{8} = 447,897,600,000 \text{ bytes} \] - **Gigabytes:** Divide bytes by \(1,000,000,000\) (since 1 GB = \(10^9\) bytes). \[ \text{Total gigabytes} = \frac{447,897,600,000}{1,000,000,000} = 447.898 \text{ GB} \] **Step 6: Round to Three Significant Figures** - Rounded storage requirement: \[ \text{Storage requirement} \approx 448 \text{ GB} \] **Answer:** **Approximately 448 Gbytes of storage are required for the video.** |
417 |
| GPT-4o | To determine the storage requirement for the given video, we need to follow these steps: 1. **Calculate the number of frames in one hour of video.** 2. **Determine the size of each frame in bytes.** 3. **Multiply the size of one frame by the total number of frames to get the total storage requirement in bytes.** 4. **Convert the total storage requirement from bytes to gigabytes.** Let's go through these steps in detail. ### Step 1: Calculate the Number of Frames The video has a frame rate of 30 frames per second. There are 3600 seconds in one hour. \[ \text{Number of frames} = 30 \, \text{frames/second} \times 3600 \, \text{seconds} = 108,000 \, \text{frames} \] ### Step 2: Determine the Size of Each Frame The video is in YUV 4:2:2 format with 8 bits per component. In 4:2:2 format: - The Y component has a sample for each pixel. - The U and V components have one sample for every two pixels horizontally. Given the resolution is 1920x1080: - The number of Y samples per frame is \(1920 \times 1080\). - The number of U samples per frame is \(1920/2 \times 1080 = 960 \times 1080\). - The number of V samples per frame is \(1920/2 \times 1080 = 960 \times 1080\). Each sample is 8 bits (1 byte). \[ \text{Size of Y component} = 1920 \times 1080 \times 1 \, \text{byte} = 2,073,600 \, \text{bytes} \] \[ \text{Size of U component} = 960 \times 1080 \times 1 \, \text{byte} = 1,036,800 \, \text{bytes} \] \[ \text{Size of V component} = 960 \times 1080 \times 1 \, \text{byte} = 1,036,800 \, \text{bytes} \] Total size of one frame: \[ \text{Total size per frame} = 2,073,600 \, \text{bytes} (Y) + 1,036,800 \, \text{bytes} (U) + 1,036,800 \, \text{bytes} (V) = 4,147,200 \, \text{bytes} \] ### Step 3: Calculate the Total Storage Requirement Multiply the size of one frame by the total number of frames: \[ \text{Total storage in bytes} = 108,000 \, \text{frames} \times 4,147,200 \, \text{bytes/frame} = 447,897,600,000 \, \text{bytes} \] ### Step 4: Convert Bytes to Gigabytes 1 Gigabyte (GB) = \(2^{30}\) bytes, which is approximately \(1,073,741,824\) bytes. \[ \text{Total storage in GB} = \frac{447,897,600,000 \, \text{bytes}}{1,073,741,824 \, \text{bytes/GB}} \approx 417 \, \text{GB} \] Thus, the storage requirement for the one-hour video is approximately **417 GB** (to 3 significant figures). |
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| llama3.1-70B-Instruct | To determine the storage requirement for the video, we need to calculate the total number of bits required to store one hour of video. First, let's calculate the number of bits required to store one frame: * Y component: 1920 x 1080 x 8 bits = 16,777,280 bits * U component: 1920 x 540 x 8 bits = 8,388,160 bits (since U and V components are sub-sampled by a factor of 2 in the horizontal direction) * V component: 1920 x 540 x 8 bits = 8,388,160 bits Total number of bits per frame = 16,777,280 + 8,388,160 + 8,388,160 = 33,553,600 bits Since the frame rate is 30 frames/s, the total number of bits required to store one second of video is: 33,553,600 bits/frame x 30 frames/s = 1,006,608,000 bits/s To store one hour of video, we need to multiply the number of bits per second by the number of seconds in an hour (3600 s): 1,006,608,000 bits/s x 3600 s = 3,623,808,000,000 bits To convert this to Gbytes, we divide by 8 (since there are 8 bits in a byte) and then divide by 1,000,000,000 (since there are 1,000,000,000 bytes in a Gbyte): 3,623,808,000,000 bits / 8 = 452,976,000,000 bytes 452,976,000,000 bytes / 1,000,000,000 = 452.98 Gbytes Rounded to three significant figures, the storage requirement for the video is approximately 453 Gbytes. |
417 |
| llama3.1-405B-Instruct | To determine the storage requirement for the video, we need to calculate the total number of bits required to store the video. First, let's calculate the number of bits required to store one frame: * The frame resolution is 1920x1080, so the total number of pixels is 1920 x 1080 = 2,073,600 pixels. * Each pixel has 3 components (Y, U, and V), and each component is represented by 8 bits. Therefore, the total number of bits required to store one pixel is 3 x 8 = 24 bits. * Since the color sub-sampling format is 4:2:2, the U and V components are sub-sampled by a factor of 2 in the horizontal direction. This means that for every 4 pixels, there are only 2 U and V components. Therefore, the total number of bits required to store one frame is: 2,073,600 pixels x 24 bits/pixel = 49,766,400 bits However, due to the 4:2:2 sub-sampling, the actual number of bits required to store one frame is: 49,766,400 bits x (4/4 + 2/4 + 2/4) = 49,766,400 bits x 0.75 = 37,324,800 bits Next, we need to calculate the total number of bits required to store the entire video: * The frame rate is 30 frames/s, and the video is 1 hour long. Therefore, the total number of frames is: 30 frames/s x 3600 s (1 hour) = 108,000 frames * The total number of bits required to store the entire video is: 37,324,800 bits/frame x 108,000 frames = 4,031,481,600,000 bits Finally, we can convert the total number of bits to Gbytes: 4,031,481,600,000 bits / (8 bits/byte) = 503,935,200,000 bytes 503,935,200,000 bytes / (1,073,741,824 bytes/Gbyte) ≈ 469 Gbytes Therefore, the storage requirement for the video is approximately 469 Gbytes. |
417 |
| DeepSeek-Coder-V2 | To determine the storage requirement for the one-hour color video in YUV format, we need to follow these steps: 1. **Calculate the number of frames in the video:** - The video duration is 1 hour, which is 3600 seconds. - The frame rate is 30 frames per second. - Therefore, the total number of frames is: \[ 3600 \text{ seconds} \times 30 \text{ frames/second} = 108000 \text{ frames} \] 2. **Determine the size of each frame:** - The frame resolution is 1920x1080. - The color sub-sampling format is 4:2:2. - Each component (Y, U, V) is 8 bits. For the 4:2:2 format: - The Y component has 1920x1080 pixels. - The U and V components each have half the number of pixels in the horizontal direction, so they have 1920/2 = 960 pixels horizontally and 1080 pixels vertically. Therefore: - The number of Y samples is 1920x1080. - The number of U samples is 960x1080. - The number of V samples is 960x1080. The total number of samples per frame is: \[ 1920 \times 1080 + 960 \times 1080 + 960 \times 1080 = 1920 \times 1080 + 2 \times 960 \times 1080 \] Simplifying: \[ 1920 \times 1080 + 2 \times 960 \times 1080 = 1920 \times 1080 + 1920 \times 1080 = 2 \times 1920 \times 1080 \] The total number of samples per frame is: \[ 2 \times 1920 \times 1080 \] Each sample is 8 bits, so the total number of bits per frame is: \[ 2 \times 1920 \times 1080 \times 8 \text{ bits} \] Converting bits to bytes (since 1 byte = 8 bits): \[ 2 \times 1920 \times 1080 \text{ bytes} \] 3. **Calculate the total storage requirement:** - The total number of bytes for all frames is: \[ 108000 \text{ frames} \times 2 \times 1920 \times 1080 \text{ bytes/frame} \] Simplifying: \[ 108000 \times 2 \times 1920 \times 1080 \text{ bytes} \] Converting bytes to gigabytes (since 1 GB = 10^9 bytes): \[ \frac{108000 \times 2 \times 1920 \times 1080}{10^9} \text{ GB} \] Calculating the numerical value: \[ \frac{108000 \times 2 \times 1920 \times 1080}{10^9} \approx 443.392 \text{ GB} \] Rounding to 3 significant figures: \[ 443 \text{ GB} \] Therefore, the storage requirement for the video is: \[ \boxed{443} \] |
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| Qwen2-72B-Instruct | To solve the problem, we need to calculate the storage requirement for the video in gigabytes (Gbytes). First, let's calculate the storage requirement for a single frame: - The resolution is 1920x1080, which means there are 1920x1080 = 2,073,600 pixels in a frame. - Each pixel has three components: Y, U, and V. However, due to the 4:2:2 color sub-sampling, only two-thirds of the U and V components are used (since for every four pixels, two U and two V components are stored). Therefore, the number of U and V components is 2,073,600 * 2/3 = 1,382,400. - Each component is represented by 8 bits, which is equivalent to 1 byte. - The total storage requirement for a single frame is: 2,073,600 bytes (for Y) + 1,382,400 bytes (for U) + 1,382,400 bytes (for V) = 4,838,400 bytes. Next, let's calculate the storage requirement for the entire one-hour video: - There are 30 frames per second, so there are 30 * 60 * 60 = 108,000 frames in one hour. - The total storage requirement for the entire video is: 4,838,400 bytes/frame * 108,000 frames = 520,233,600,000 bytes. Finally, let's convert the storage requirement to gigabytes (Gbytes): - 1 gigabyte (Gbyte) = 1,073,741,824 bytes. - The storage requirement for the video in Gbytes is: 520,233,600,000 bytes / 1,073,741,824 bytes/Gbyte = 484 Gbytes (rounded to three significant figures). Therefore, the storage requirement for the video in Gbytes is approximately 484 Gbytes. |
417 |
| baichuan4 | To calculate the storage requirement for the video, we need to determine the size of each frame and then multiply it by the total number of frames in one hour. First, let's calculate the size of one frame: The frame resolution is 1920x1080, which means there are 1920 pixels horizontally and 1080 pixels vertically. In a 4:2:2 color sub-sampling format, for every 4 Y (luminance) samples, there are 2 U (blue-difference chroma) and 2 V (red-difference chroma) samples. This means that for every 4 pixels, there are 4 Y bytes, 2 U bytes, and 2 V bytes. So, for one frame, the total number of bytes is: (1920 / 4) * (1080 * 4) = 480 * 1080 * 4 = 2073600 bytes for Y (1920 / 4) * (1080 * 2) = 480 * 1080 * 2 = 1036800 bytes for U (1920 / 4) * (1080 * 2) = 480 * 1080 * 2 = 1036800 bytes for V Adding these together, we get: 2073600 + 1036800 + 1036800 = 4147200 bytes for one frame. Now, let's calculate the total number of frames in one hour: 30 frames/s * 60 s/min * 60 min/hour = 108000 frames/hour. Finally, let's calculate the total storage requirement for the video: 4147200 bytes/frame * 108000 frames/hour = 44785600000 bytes/hour. Converting this to Gbytes (1 Gbyte = 10^9 bytes), we get: 44785600000 bytes/hour / 10^9 bytes/Gbyte = 44.7856 Gbytes. Rounding to 3 significant figures, the storage requirement for the video is 44.8 Gbytes. |
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| doubao | For each pixel, there are 1920 x 1080 pixels. For the Y component, it's 1920 x 1080 pixels. For the U and V components (due to 4:2:2 sub-sampling), it's 1920 x 1080 / 2 pixels each. The total number of pixels for Y is 1920 x 1080. The total number of pixels for U and V is 1920 x 1080 / 2 each. The total number of bits for Y is 1920 x 1080 x 8. The total number of bits for U and V is 1920 x 1080 / 2 x 8 each. The total number of bits for one frame is 1920 x 1080 x 8 + 1920 x 1080 / 2 x 8 + 1920 x 1080 / 2 x 8 = 1920 x 1080 x 8 x (1 + 1/2 + 1/2) = 1920 x 1080 x 16 bits. Since the frame rate is 30 frames/s, the number of bits per second is 1920 x 1080 x 16 x 30 bits/s. In one hour (3600 seconds), the total number of bits is 1920 x 1080 x 16 x 30 x 3600 bits. To convert to gigabytes, divide by 8 x 1024 x 1024 x 1024. The storage requirement is approximately 13.5 Gbytes. |
417 |
| minimax | To calculate the storage requirement for the video, we need to determine the number of bytes required to store one second of video and then multiply that by the total number of seconds in an hour. First, let's calculate the number of bytes per frame: - The frame resolution is 1920x1080 pixels. - With 4:2:2 color sub-sampling, the chrominance components (U and V) are sampled at half the rate of the luminance component (Y) in the horizontal direction. This means that for every two pixels, there are four Y samples and two U and two V samples. - Each component (Y, U, V) is 8 bits (1 byte) per sample. For the luminance (Y) component: - There are 1920 pixels horizontally and 1080 vertically, so there are 1920 * 1080 Y samples per frame. - Each Y sample is 1 byte, so the Y component takes 1920 * 1080 bytes per frame. For the chrominance (U and V) components: - There are half as many U and V samples horizontally, so 1920 / 2 = 960 samples horizontally for each chrominance component. - The number of U and V samples vertically remains the same, so 960 * 1080 for each U and V component. - Each U and V sample is 1 byte, so each chrominance component takes 960 * 1080 bytes per frame. Now, let's calculate the total bytes per frame: Bytes per frame = (Bytes for Y) + (Bytes for U) + (Bytes for V) Bytes per frame = (1920 * 1080) + (960 * 1080) + (960 * 1080) Bytes per frame = (1920 * 1080) + 2 * (960 * 1080) Bytes per frame = (1920 * 1080) + 2 * (1036800) Bytes per frame = 2073600 + 2073600 Bytes per frame = 4147200 bytes Since there are 30 frames per second, we need to calculate the bytes per second: Bytes per second = Bytes per frame * 30 Bytes per second = 4147200 * 30 Bytes per second = *** bytes Now, convert bytes to Gbytes (1 Gbyte = 2^30 bytes): Gbytes per second = Bytes per second / (2^30) Gbytes per second = *** / (2^30) Gbytes per second ≈ *** / *** Gbytes per second ≈ 0.116 Gbytes Finally, since there are 3600 seconds in an hour, we calculate the total storage requirement for one hour: Total storage requirement = Gbytes per second * 3600 Total storage requirement ≈ 0.116 Gbytes * 3600 Total storage requirement ≈ 417.6 Gbytes So, the storage requirement for the video is approximately 417.6 Gbytes (rounded to three significant figures). |
417 |
| kimi | To determine the storage requirement for the video, we need to calculate the size of one frame and then multiply it by the total number of frames in one hour. 1. **Calculate the size of one frame:** - The frame resolution is 1920x1080 pixels. - The color sub-sampling format is 4:2:2, which means for every four Y (luminance) samples, there are two U (chrominance) and two V (chrominance) samples. - Each component (Y, U, V) uses 8 bits. Since it's 4:2:2, for every two horizontal pixels, there is one U and one V sample. So, for a 1920x1080 frame: - Number of Y samples = 1920 pixels * 1080 rows = 2,073,600 - Number of U samples = 1920/2 pixels * 1080 rows = 1,036,800 - Number of V samples = 1920/2 pixels * 1080 rows = 1,036,800 Total number of samples per frame = Y + U + V = 2,073,600 + 1,036,800 + 1,036,800 = 4,147,200 samples Since each sample is 8 bits, the total number of bits per frame is: 4,147,200 samples * 8 bits/sample = 33,177,600 bits Convert bits to bytes (1 byte = 8 bits): 33,177,600 bits / 8 = 4,147,200 bytes Convert bytes to gigabytes (1 GB = 1,073,741,824 bytes): 4,147,200 bytes / 1,073,741,824 bytes/GB ≈ 0.003862 GB 2. **Calculate the total storage requirement for one hour:** - There are 30 frames per second. - There are 3600 seconds in one hour. Total number of frames in one hour = 30 frames/s * 3600 s = 108,000 frames Total storage requirement = 0.003862 GB/frame * 108,000 frames ≈ 416.236 GB 3. **Round to three significant figures:** 416.236 GB rounded to three significant figures is 416 GB. So, the storage requirement for the video is approximately 416 GB. |
417 |
| wenxinyiyan | To calculate the storage requirement for the given video, we need to consider the frame resolution, color sub-sampling format, bits per component, and the frame rate. The video is in YUV format with a 4:2:2 color sub-sampling, which means that the U and V components are sampled at half the rate of the Y component. The frame resolution is 1920x1080, and there are 8 bits for each component. The video is one hour long, and the frame rate is 30 frames per second. First, let's calculate the storage requirement for one frame: * The Y component takes 1920 * 1080 * 8 bits. * The U and V components, due to 4:2:2 sub-sampling, take half the resolution of Y each, so they take (1920/2) * (1080) * 8 bits each. Adding these components together gives us the total bits for one frame: 1920 * 1080 * 8 + (1920/2) * (1080) * 8 * 2 = 1920 * 1080 * 8 * (1 + 0.5 * 2) = 1920 * 1080 * 8 * 2 Now, we need to calculate the total number of frames in one hour: * There are 30 frames per second, so in one hour (3600 seconds), there are 30 * 3600 frames. Finally, we multiply the storage requirement for one frame by the total number of frames and convert the result to Gbytes (1 Gbyte = 8 Gbits): (1920 * 1080 * 8 * 2 bits/frame) * (30 * 3600 frames) / (8 * 1024 * 1024 * 1024 bits/Gbyte) = (1920 * 1080 * 2 * 30 * 3600) / (1024 * 1024 * 1024) = 44.237 Gbytes Rounded to three significant figures, the storage requirement is 44.2 Gbytes. |
417 |
| zhipu | To determine the storage requirement for the video, we need to calculate the size of each frame and then multiply that by the number of frames in the video. Here's the step-by-step solution: 1. **Calculate the size of one frame:** - The resolution is 1920x1080 pixels. - The color format is YUV 4:2:2, which means for every 4 pixels, there are 4 Y (luminance) components and 2 each of U and V (chrominance) components. - Each component is 8 bits (1 byte). **Luminance (Y) component:** - 1920x1080 pixels = 2,073,600 pixels. - Each pixel has one Y component. - Size of Y component = 2,073,600 pixels * 1 byte/pixel = 2,073,600 bytes. **Chrominance (U and V) components:** - For 4:2:2 sub-sampling, each pair of pixels shares one U and one V component. - Therefore, the number of U and V components is half the number of pixels. - Size of U component = 1920x1080 / 2 pixels * 1 byte/pixel = 1,036,800 bytes. - Size of V component = 1,036,800 bytes (same as U). **Total size of one frame:** - Total size = Size of Y + Size of U + Size of V - Total size = 2,073,600 bytes + 1,036,800 bytes + 1,036,800 bytes - Total size = 4,147,200 bytes. 2. **Calculate the total number of frames in one hour:** - Frame rate = 30 frames/s. - Duration = 1 hour = 3600 seconds. - Total frames = 30 frames/s * 3600 s = 108,000 frames. 3. **Calculate the total storage requirement:** - Total storage = Total size of one frame * Total number of frames - Total storage = 4,147,200 bytes/frame * 108,000 frames - Total storage = 449,064,960,000 bytes. 4. **Convert bytes to gigabytes (GB):** - 1 GB = 1,073,741,824 bytes (since 1 GB = 1024^3 bytes). - Total storage in GB = 449,064,960,000 bytes / 1,073,741,824 bytes/GB - Total storage in GB ≈ 418.37 GB. Rounded to three significant figures, the storage requirement for the video is approximately **418 GB**. |
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| gpto1 | To determine the storage requirement for the given video, we need to calculate the total amount of data generated by all the frames over the one-hour duration. **Given:** - Frame resolution: **1920 x 1080 pixels** - Color subsampling format: **4:2:2** - Bit depth: **8 bits per component** - Frame rate: **30 frames per second** - Video duration: **1 hour** **Step 1: Understand 4:2:2 Subsampling** In 4:2:2 chroma subsampling: - **Luma (Y)** is sampled at every pixel (full resolution). - **Chroma components (Cb and Cr)** are sampled at half the horizontal resolution. This means: - For every **2 pixels horizontally**, there is **1 sample** of Cb and **1 sample** of Cr. **Step 2: Calculate Number of Samples per Frame** - **Total number of pixels per frame (Np):** \[ N_p = 1920 \times 1080 = 2,073,600 \text{ pixels} \] - **Luma samples (Y):** - Each pixel has a Y sample. - Total Y samples per frame: \[ Y_{\text{samples}} = N_p = 2,073,600 \text{ samples} \] - **Chroma samples (Cb and Cr):** - Chroma is subsampled horizontally by a factor of 2. - **Horizontal chroma resolution:** \[ \text{Chroma width} = \frac{1920}{2} = 960 \text{ pixels} \] - Total Cb samples per frame: \[ \text{Cb}_{\text{samples}} = 960 \times 1080 = 1,036,800 \text{ samples} \] - Total Cr samples per frame: \[ \text{Cr}_{\text{samples}} = 960 \times 1080 = 1,036,800 \text{ samples} \] - **Total samples per frame:** \[ \text{Total samples} = Y_{\text{samples}} + \text{Cb}_{\text{samples}} + \text{Cr}_{\text{samples}} = 2,073,600 + 1,036,800 + 1,036,800 = 4,147,200 \text{ samples} \] **Step 3: Calculate Bits per Frame** - Each sample is **8 bits**. - **Total bits per frame:** \[ \text{Bits per frame} = \text{Total samples} \times 8 = 4,147,200 \times 8 = 33,177,600 \text{ bits} \] **Step 4: Calculate Total Bits for One Hour** - **Frames per second:** 30 - **Total frames in one hour:** \[ \text{Total frames} = 30 \times 3600 = 108,000 \text{ frames} \] - **Total bits for one hour:** \[ \text{Total bits} = \text{Bits per frame} \times \text{Total frames} = 33,177,600 \times 108,000 = 3,583,180,800,000 \text{ bits} \] **Step 5: Convert Bits to Gigabytes** - **Bytes:** Divide bits by 8. \[ \text{Total bytes} = \frac{3,583,180,800,000}{8} = 447,897,600,000 \text{ bytes} \] - **Gigabytes:** Divide bytes by \(1,000,000,000\) (since 1 GB = \(10^9\) bytes). \[ \text{Total gigabytes} = \frac{447,897,600,000}{1,000,000,000} = 447.898 \text{ GB} \] **Step 6: Round to Three Significant Figures** - Rounded storage requirement: \[ \text{Storage requirement} \approx 448 \text{ GB} \] **Answer:** **Approximately 448 Gbytes of storage are required for the video.** |
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| deepseekv3 | To determine the storage requirement for the one-hour color video in YUV 4:2:2 format, follow these steps: --- ### Step 1: Calculate the size of one frame The YUV 4:2:2 format has the following characteristics: - **Luma (Y)**: Full resolution (1920x1080), 8 bits per pixel. - **Chrominance (U, V)**: Half horizontal resolution (960x1080), 8 bits per pixel for each component. #### Size of Y component: \[ \text{Y size} = 1920 \times 1080 \times 8 = 16,588,800 \, \text{bits} \] #### Size of U component: \[ \text{U size} = 960 \times 1080 \times 8 = 8,294,400 \, \text{bits} \] #### Size of V component: \[ \text{V size} = 960 \times 1080 \times 8 = 8,294,400 \, \text{bits} \] #### Total size of one frame: \[ \text{Frame size} = \text{Y size} + \text{U size} + \text{V size} = 16,588,800 + 8,294,400 + 8,294,400 = 33,177,600 \, \text{bits} \] --- ### Step 2: Calculate the size of one second of video The video has a frame rate of 30 frames per second. \[ \text{Size per second} = 33,177,600 \, \text{bits/frame} \times 30 \, \text{frames/s} = 995,328,000 \, \text{bits/s} \] --- ### Step 3: Calculate the size of one hour of video One hour has 3600 seconds. \[ \text{Size per hour} = 995,328,000 \, \text{bits/s} \times 3600 \, \text{s} = 3,583,180,800,000 \, \text{bits} \] Convert bits to bytes (1 byte = 8 bits): \[ \text{Size per hour in bytes} = \frac{3,583,180,800,000}{8} = 447,897,600,000 \, \text{bytes} \] Convert bytes to gigabytes (1 GB = \(10^9\) bytes): \[ \text{Size per hour in GB} = \frac{447,897,600,000}{10^9} = 447.898 \, \text{GB} \] --- ### Final Answer: The storage requirement for the one-hour video is **448 GB** (rounded to 3 significant figures). |
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