| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To calculate the required memory size in Mebibytes (MiB) for storing a frame in 720p using the Y'CbCr 4:2:0 sampling scheme, we need to follow these steps: 1. Calculate the number of bytes required for storing the luminance (Y') component. 2. Calculate the number of bytes required for storing the chrominance (Cb and Cr) components. 3. Sum the bytes from steps 1 and 2 to get the total bytes required. 4. Convert the total bytes to Mebibytes (MiB). Step 1: Calculate the number of bytes for the luminance (Y') component. Since the Y' component is sampled at a 4:4:4 ratio, we need one byte per pixel for Y'. The total number of pixels is 1280 x 720. Thus, the number of bytes required for the Y' component is: \[ \text{Bytes for Y'} = 1280 \times 720 \times 1 = 921,600 \, \text{bytes} \] Step 2: Calculate the number of bytes for the chrominance (Cb and Cr) components. The Cb and Cr components are sampled at a 4:2:0 ratio. This means that for every 4 pixels, we store the chrominance values for 2 pixels. Hence, the number of Cb and Cr samples required is half the number of pixels: \[ \text{Samples for Cb and Cr} = \frac{1280 \times 720}{2} = 460,800 \] Since each sample requires 1 byte, the total number of bytes for the Cb and Cr components combined is: \[ \text{Bytes for Cb and Cr} = 460,800 \times 2 = 921,600 \, \text{bytes} \] Step 3: Sum the bytes from steps 1 and 2. The total bytes required for storing the frame is: \[ \text{Total bytes} = \text{Bytes for Y'} + \text{Bytes for Cb and Cr} = 921,600 + 921,600 = 1,843,200 \, \text{bytes} \] Step 4: Convert the total bytes to Mebibytes (MiB). To convert bytes to Mebibytes, we use the conversion factor \(1 \, \text{MiB} = 1024^2 \, \text{bytes}\): \[ \text{Total MiB} = \frac{1,843,200}{1024^2} = \frac{1,843,200}{1,048,576} \approx 1.76 \, \text{MiB} \] Therefore, the required memory size for storing a frame in 720p with the Y'CbCr 4:2:0 sampling scheme is approximately 1.76 MiB (rounded to 3 significant figures). |
1.32 |
| deepseekr1 | To calculate the required memory size for a 720p frame using Y'CbCr 4:2:0: 1. **Determine the number of samples per component:** - **Y' (Luma):** Full resolution = \(1280 \times 720 = 921,600\) samples. - **Cb and Cr (Chroma):** Subsampled by ½ in both dimensions. Each has \( \frac{1280}{2} \times \frac{720}{2} = 640 \times 360 = 230,400\) samples. Combined, they total \(230,400 \times 2 = 460,800\) samples. 2. **Total memory in bytes:** \[ 921,600\,(\text{Y'}) + 460,800\,(\text{Cb+Cr}) = 1,382,400\,\text{bytes}. \] 3. **Convert to Mebibytes (1 MiB = \(1024^2\) bytes):** \[ \frac{1,382,400}{1,048,576} \approx 1.31836\,\text{MiB}. \] 4. **Round to 3 significant figures:** \(1.32\,\text{MiB}\). **Answer:** \(\boxed{1.32}\) MiB. |
1.32 |
| gpto1 | To calculate the required memory size in Mebibytes (MiB) for storing a 720p frame using the Y'CbCr 4:2:0 sampling scheme, we'll follow these steps: **1. Understand the Frame Resolution and Components:** A 720p frame has a resolution of **1280 × 720 pixels**. Each pixel consists of three primary colour components: **Y'** (luminance), **Cb**, and **Cr** (chrominance components). **2. Comprehend the Y'CbCr 4:2:0 Sampling Scheme:** In the Y'CbCr 4:2:0 sampling scheme: - The **luminance (Y')** component is sampled at **full resolution**. - The **chrominance (Cb and Cr)** components are **subsampled by a factor of 2** in both the horizontal and vertical dimensions. This means for every **2×2 block of pixels**, there are: - **4 Y' samples** (one for each pixel). - **1 Cb sample** and **1 Cr sample** (shared among the 4 pixels). **3. Calculate the Number of Samples for Each Component:** - **Y' (Luminance) Samples:** \[ \text{Total Y' samples} = 1280 \times 720 = 921,\!600 \text{ samples} \] - **Cb and Cr (Chrominance) Samples:** \[ \text{Total Cb samples} = \text{Total Cr samples} = \left( \frac{1280}{2} \right) \times \left( \frac{720}{2} \right) = 640 \times 360 = 230,\!400 \text{ samples each} \] **4. Calculate the Total Number of Samples:** \[ \begin{align*} \text{Total samples} &= \text{Y' samples} + \text{Cb samples} + \text{Cr samples} \\ &= 921,\!600 + 230,\!400 + 230,\!400 \\ &= 1,\!382,\!400 \text{ samples} \end{align*} \] **5. Determine the Total Memory in Bytes:** Each sample requires **1 byte** of storage. \[ \text{Total memory in bytes} = \text{Total samples} \times 1 \, \text{byte} = 1,\!382,\!400 \, \text{bytes} \] **6. Convert Bytes to Mebibytes (MiB):** 1 Mebibyte (MiB) is equal to \( 1024^2 = 1,\!048,\!576 \) bytes. \[ \begin{align*} \text{Total memory in MiB} &= \frac{\text{Total memory in bytes}}{1,\!048,\!576 \, \text{bytes/MiB}} \\ &= \frac{1,\!382,\!400}{1,\!048,\!576} \\ &\approx 1.3184 \, \text{MiB} \end{align*} \] **7. Round to Three Significant Figures:** \[ \text{Total memory in MiB} \approx 1.32 \, \text{MiB} \] **Answer:** The required memory size is **1.32 MiB**. |
1.32 |
| GPT-4o | To calculate the required memory size in Mebibytes (MiB) for storing a 720p frame using the Y'CbCr 4:2:0 sampling scheme, follow these steps: 1. **Determine the total number of pixels in the frame:** \[ 1280 \times 720 = 921,600 \text{ pixels} \] 2. **Understand the Y'CbCr 4:2:0 sampling scheme:** - Y' (luminance) component: Full resolution (1280 × 720) - Cb (blue-difference chroma) and Cr (red-difference chroma) components: Each is quarter resolution because the chroma is subsampled by a factor of 2 in both dimensions. Therefore, each chroma component has dimensions: \[ \frac{1280}{2} \times \frac{720}{2} = 640 \times 360 \] 3. **Calculate the total number of bytes required for each component:** - Y' component: \[ 1280 \times 720 \times 1 \text{ byte} = 921,600 \text{ bytes} \] - Cb component: \[ 640 \times 360 \times 1 \text{ byte} = 230,400 \text{ bytes} \] - Cr component: \[ 640 \times 360 \times 1 \text{ byte} = 230,400 \text{ bytes} \] 4. **Sum the bytes required for all the components:** \[ 921,600 \text{ bytes (Y')} + 230,400 \text{ bytes (Cb)} + 230,400 \text{ bytes (Cr)} = 1,382,400 \text{ bytes} \] 5. **Convert the total bytes to Mebibytes (MiB):** - 1 MiB = 1,024^2 bytes = 1,048,576 bytes \[ \frac{1,382,400 \text{ bytes}}{1,048,576 \text{ bytes/MiB}} \approx 1.318 \text{ MiB} \] 6. **Round the result to three significant figures:** \[ 1.32 \text{ MiB} \] **Solution: The required memory size for storing a 720p frame using the Y'CbCr 4:2:0 sampling scheme is approximately 1.32 MiB.** |
1.32 |
| llama3.1-70B-Instruct | To calculate the required memory size in Mebibytes (MiB) for storing a frame in 720p using the Y'CbCr 4:2:0 sampling scheme, we need to understand how this sampling scheme affects the storage requirements. In the Y'CbCr 4:2:0 sampling scheme, for every 4x2 block of pixels (8 pixels), there are: - 8 Y' (luminance) samples, each requiring 1 byte. - 2 Cb (blue-difference) samples, each requiring 1 byte. - 2 Cr (red-difference) samples, each requiring 1 byte. This means that for every 8 pixels, instead of needing 24 bytes (3 bytes per pixel for Y'CbCr), we only need 12 bytes (8 for Y' + 2 for Cb + 2 for Cr). Given that a 720p frame has 1280 × 720 pixels, we first calculate the total number of pixels: 1280 pixels/frame × 720 lines/frame = 921,600 pixels/frame Since the sampling scheme is applied in 2x2 blocks, we divide the total number of pixels by 4 (since each block is 2x2, or 4 pixels) to find out how many blocks there are: 921,600 pixels / 4 = 230,400 blocks Each block requires 12 bytes for storage (as calculated above), so the total memory required in bytes is: 230,400 blocks/frame × 12 bytes/block = 2,764,800 bytes/frame To convert this to Mebibytes (MiB), we divide by 1024^2 (since 1 MiB = 1024^2 bytes): 2,764,800 bytes / (1024^2) = 2,764,800 / 1,048,576 ≈ 2.64 MiB Therefore, the required memory size in Mebibytes (MiB) for storing a frame in 720p using the Y'CbCr 4:2:0 sampling scheme is approximately 2.64 MiB, to three significant figures. |
1.32 |
| llama3.1-405B-Instruct | To calculate the required memory size in Mebibytes (MiB) for storing a frame in 720p using the Y'CbCr 4:2:0 sampling scheme, we need to follow these steps: 1. Calculate the total number of pixels in one frame: - There are 1280 × 720 pixels in one 720p frame. 2. Calculate the total number of bytes required to store one frame: - Each pixel contains three primary-colour components (Y', Cb, and Cr). - In the Y'CbCr 4:2:0 sampling scheme, for every 4 luminance (Y') samples, there are 2 chrominance (Cb and Cr) samples. This means that for every 4 pixels, we have 4 Y' samples, 2 Cb samples, and 2 Cr samples, resulting in a total of 4 + 2 + 2 = 8 samples. - Since each sample requires 1 byte of memory for storage, the total number of bytes required to store one frame is: - (1280 x 720) * (4 + 2 + 2) / 4 = 921600 * 2 = 1843200 bytes 3. Convert the total number of bytes to Mebibytes (MiB): - 1 Mebibyte has 1024^2 bytes. - To convert bytes to MiB, divide the total number of bytes by 1024^2: - 1843200 bytes / (1024^2) = 1843200 / 1048576 ≈ 1.76 MiB Therefore, the required memory size in Mebibytes (MiB) for storing a frame in 720p using the Y'CbCr 4:2:0 sampling scheme is approximately 1.76 MiB. |
1.32 |
| DeepSeek-Coder-V2 | To calculate the required memory size in Mebibytes (MiB) for storing a frame in 720p with the Y'CbCr 4:2:0 sampling scheme, we need to follow these steps: 1. **Determine the total number of pixels in the frame:** The resolution of 720p is 1280 × 720 pixels. \[ \text{Total pixels} = 1280 \times 720 = 921,600 \text{ pixels} \] 2. **Understand the Y'CbCr 4:2:0 sampling scheme:** In Y'CbCr 4:2:0, each pixel has one Y' (luma) component, but the Cb (chroma blue) and Cr (chroma red) components are sub-sampled. Specifically, for every 4 Y' components, there is 1 Cb and 1 Cr component. This means that for every 2x2 block of pixels, there are 4 Y' components, 1 Cb component, and 1 Cr component. 3. **Calculate the number of Y', Cb, and Cr components:** - Each pixel has 1 Y' component. - For every 4 pixels, there is 1 Cb and 1 Cr component. Therefore, the number of Y' components is: \[ \text{Number of Y' components} = 921,600 \] The number of Cb and Cr components is: \[ \text{Number of Cb components} = \text{Number of Cr components} = \frac{921,600}{4} = 230,400 \] 4. **Calculate the total number of bytes required:** Each component (Y', Cb, Cr) requires 1 byte of memory. \[ \text{Total bytes for Y'} = 921,600 \text{ bytes} \] \[ \text{Total bytes for Cb} = 230,400 \text{ bytes} \] \[ \text{Total bytes for Cr} = 230,400 \text{ bytes} \] \[ \text{Total bytes} = 921,600 + 230,400 + 230,400 = 1,382,400 \text{ bytes} \] 5. **Convert the total number of bytes to Mebibytes (MiB):** 1 Mebibyte (MiB) is \(1024^2\) bytes. \[ \text{Total Mebibytes} = \frac{1,382,400}{1024^2} = \frac{1,382,400}{1,048,576} \approx 1.319 \text{ MiB} \] Therefore, the required memory size in Mebibytes (MiB) for storing a frame in 720p with the Y'CbCr 4:2:0 sampling scheme is: \[ \boxed{1.32} \] |
1.32 |
| Qwen2-72B-Instruct | To find the required memory size in Mebibytes (MiB) for storing a frame in 720p with the sampling scheme Y'CbCr 4:2:0, we need to follow these steps: 1. Calculate the number of bytes required for storing the luminance (Y') component. 2. Calculate the number of bytes required for storing the chrominance (Cb and Cr) components. 3. Sum the bytes from steps 1 and 2 to get the total bytes required. 4. Convert the total bytes to Mebibytes (MiB). Step 1: Calculate the number of bytes for the luminance (Y') component. Since the Y' component is sampled at a 4:4:4 ratio, we need one byte per pixel for Y'. The total number of pixels is 1280 x 720. Thus, the number of bytes required for the Y' component is: \[ \text{Bytes for Y'} = 1280 \times 720 \times 1 = 921,600 \, \text{bytes} \] Step 2: Calculate the number of bytes for the chrominance (Cb and Cr) components. The Cb and Cr components are sampled at a 4:2:0 ratio. This means that for every 4 pixels, we store the chrominance values for 2 pixels. Hence, the number of Cb and Cr samples required is half the number of pixels: \[ \text{Samples for Cb and Cr} = \frac{1280 \times 720}{2} = 460,800 \] Since each sample requires 1 byte, the total number of bytes for the Cb and Cr components combined is: \[ \text{Bytes for Cb and Cr} = 460,800 \times 2 = 921,600 \, \text{bytes} \] Step 3: Sum the bytes from steps 1 and 2. The total bytes required for storing the frame is: \[ \text{Total bytes} = \text{Bytes for Y'} + \text{Bytes for Cb and Cr} = 921,600 + 921,600 = 1,843,200 \, \text{bytes} \] Step 4: Convert the total bytes to Mebibytes (MiB). To convert bytes to Mebibytes, we use the conversion factor \(1 \, \text{MiB} = 1024^2 \, \text{bytes}\): \[ \text{Total MiB} = \frac{1,843,200}{1024^2} = \frac{1,843,200}{1,048,576} \approx 1.76 \, \text{MiB} \] Therefore, the required memory size for storing a frame in 720p with the Y'CbCr 4:2:0 sampling scheme is approximately 1.76 MiB (rounded to 3 significant figures). |
1.32 |
| baichuan4 | To calculate the required memory size for storing a frame in 720p using the Y'CbCr 4:2:0 sampling scheme, we need to consider the following information: - A 720p frame has 1280 × 720 pixels. - Each pixel contains three primary-colour components (Y', Cb, and Cr). - In the Y'CbCr 4:2:0 sampling scheme, for every four pixels, only one chroma sample (Cb or Cr) is stored. This means that for every two pixels, one chroma sample is shared. - Each primary-colour component requires 1 byte of memory for storage. First, let's calculate the total number of Y' samples: Total Y' samples = 1280 × 720 Next, let's calculate the total number of Cb and Cr samples. Since we have a 4:2:0 sampling scheme, we need to divide the total number of pixels by 2 to get the number of Cb and Cr samples: Total Cb and Cr samples = (1280 × 720) / 2 Now, let's calculate the total memory required for one frame: Total memory for one frame = (Total Y' samples × 1 byte) + (Total Cb and Cr samples × 1 byte) Total memory for one frame = (1280 × 720 × 1) + ((1280 × 720) / 2 × 1) Total memory for one frame = 921600 + 460800 Total memory for one frame = 1382400 bytes Finally, let's convert the total memory for one frame into Mebibytes (MiB): 1 MiB = 1024^2 bytes Total memory in MiB = Total memory for one frame / 1024^2 Total memory in MiB ≈ 1382400 / 1048576 Total memory in MiB ≈ 1.318 Therefore, the required memory size in Mebibytes (MiB) for storing a frame in 720p with the Y'CbCr 4:2:0 sampling scheme is approximately 1.32 MiB (rounded to three significant figures). |
1.32 |
| doubao | In Y'CbCr 4:2:0, for every 4 pixels, there is 1 Cb and 1 Cr component. So for 1280×720 pixels, the number of Y' components is 1280×720. The number of Cb and Cr components is 1280×720/4 = 240×720. The total number of bytes for Y' is 1280×720×1 = 921600 bytes. The total number of bytes for Cb and Cr is 2×240×720×1 = 345600 bytes. The total memory required for one frame is 921600 + 2×345600 = 1612800 bytes. Converting to MiB: 1612800 / (1024^2) ≈ 1.54 MiB (to 3 significant figures). |
1.32 |
| minimax | To calculate the required memory size for storing a frame in 720p with the Y'CbCr 4:2:0 sampling scheme, we need to understand how this color space works and how it affects the memory requirements. In Y'CbCr 4:2:0, for every four Y' (luma) samples, there is one Cb (blue-difference chroma) and one Cr (red-difference chroma) sample. This means that the chroma samples are at half the resolution of the luma samples in both the horizontal and vertical directions. Here's how we calculate the memory: 1. Calculate the total number of Y' samples: Since there are 1280 × 720 pixels and each pixel has one Y' sample, we have: 1280 × 720 = 921,600 Y' samples 2. Calculate the total number of Cb and Cr samples: Since the chroma samples are at half the resolution, we divide the number of pixels by 2 in both dimensions for each chroma component: (1280 / 2) × (720 / 2) = 640 × 360 = 230,400 Cb samples Since the Cb and Cr samples are the same, we have another 230,400 Cr samples. 3. Add up the total number of samples: Y' samples + Cb samples + Cr samples = 921,600 + 230,400 + 230,400 = 1,382,400 samples 4. Since each sample requires 1 byte of memory, the total memory required in bytes is: 1,382,400 bytes 5. Convert bytes to Mebibytes (MiB): 1 MiB = 1024^2 bytes Memory in MiB = 1,382,400 bytes / (1024^2 bytes/MiB) Now, let's do the math: Memory in MiB = 1,382,400 / (1024 × 1024) Memory in MiB ≈ 1,382,400 / 1,048,576 Memory in MiB ≈ 1.318 MiB Rounded to three significant figures, the required memory size for storing a frame in 720p with the Y'CbCr 4:2:0 sampling scheme is approximately 1.32 MiB. |
1.32 |
| kimi | To calculate the required memory size in Mebibytes (MiB) for storing a frame in 720p with the Y'CbCr 4:2:0 sampling scheme, we need to consider the following: 1. The frame resolution is 1280 x 720 pixels. 2. The Y'CbCr 4:2:0 sampling scheme uses 1 byte for each of the Y (luminance) components and shares the Cb and Cr (chrominance) components between groups of pixels. 3. For every 4 pixels, there are 4 Y values, 1 Cb value, and 1 Cr value. Let's break it down step by step: 1. Calculate the total number of pixels in the frame: \(1280 \text{ pixels (width)} \times 720 \text{ pixels (height)} = 921,600 \text{ pixels}\) 2. Determine the number of Y values: Since there is 1 Y value per pixel, we have 921,600 Y values. 3. Determine the number of Cb and Cr values: In the 4:2:0 scheme, every 4 pixels share 1 Cb and 1 Cr value. Therefore, we divide the total number of pixels by 4 to get the number of Cb and Cr values: \(921,600 \text{ pixels} / 4 = 230,400 \text{ Cb and Cr values each}\) 4. Calculate the total number of bytes required for Y, Cb, and Cr: \(921,600 \text{ Y values} \times 1 \text{ byte/Y value} + 230,400 \text{ Cb values} \times 1 \text{ byte/Cb value} + 230,400 \text{ Cr values} \times 1 \text{ byte/Cr value} = 921,600 \text{ bytes} + 230,400 \text{ bytes} + 230,400 \text{ bytes} = 1,382,400 \text{ bytes}\) 5. Convert the total number of bytes to Mebibytes (MiB): \(1 \text{ MiB} = 1024^2 \text{ bytes} = 1,048,576 \text{ bytes}\) \(1,382,400 \text{ bytes} / 1,048,576 \text{ bytes/MiB} \approx 1.318 \text{ MiB}\) 6. Round to three significant figures: \(1.318 \text{ MiB} \approx 1.32 \text{ MiB}\) So, the required memory size for storing a frame in 720p with the Y'CbCr 4:2:0 sampling scheme is approximately 1.32 MiB. |
1.32 |
| wenxinyiyan | To calculate the required memory size in Mebibytes (MiB) for storing a 720p frame with the Y'CbCr 4:2:0 sampling scheme, we need to consider the number of pixels and the amount of data required for each pixel. In the Y'CbCr 4:2:0 sampling scheme, the luma component (Y') is sampled for every pixel, while the chroma components (Cb and Cr) are subsampled. Specifically, for every 4x2 block of pixels, there are 4 luma samples and 2 chroma samples (one Cb and one Cr) for the entire block. Given that there are 1280x720 pixels in one 720p frame, we can calculate the total memory required as follows: 1. Luma component (Y'): - Total luma samples: 1280x720 = 921,600 - Memory required for luma: 921,600 bytes 2. Chroma components (Cb and Cr): - Since Cb and Cr are subsampled in 4:2:0, we have half the resolution for each component. - Total chroma samples for Cb or Cr: (1280/2) x (720/2) = 640x360 = 230,400 - Memory required for chroma: 230,400 bytes x 2 = 460,800 bytes (for both Cb and Cr) Total memory required for one frame: - Luma + Chroma = 921,600 bytes + 460,800 bytes = 1,382,400 bytes To convert this to Mebibytes (MiB), we divide by \(1024^2\) (since 1 MiB = \(1024^2\) bytes): - Total memory in MiB: \(\frac{1,382,400}{1024^2}\) ≈ 1.318 MiB Rounding to 3 significant figures, we get ≈ 1.32 MiB. Therefore, the required memory size in Mebibytes (MiB) for storing a frame in 720p with the Y'CbCr 4:2:0 sampling scheme is approximately 1.32 MiB. |
1.32 |
| zhipu | To calculate the required memory size in Mebibytes (MiB) for storing a 720p frame using the Y'CbCr 4:2:0 sampling scheme, we need to follow these steps: 1. **Determine the total number of pixels in the frame:** - A 720p frame has a resolution of 1280 × 720 pixels. - Total pixels = 1280 × 720 = 921,600 pixels. 2. **Understand the Y'CbCr 4:2:0 sampling scheme:** - In Y'CbCr 4:2:0, for every 4 pixels, there is 1 Y' (luminance) component for each pixel, but only 1 Cb (chrominance blue) and 1 Cr (chrominance red) component shared among the 4 pixels. - This means for every 4 pixels, we have 4 Y' components, 1 Cb component, and 1 Cr component. 3. **Calculate the total number of bytes required for the frame:** - Each Y' component requires 1 byte. - Each Cb and Cr component also requires 1 byte each. - For every 4 pixels: - Y' components: 4 bytes (1 byte per pixel) - Cb component: 1 byte - Cr component: 1 byte - Total bytes for every 4 pixels = 4 (Y') + 1 (Cb) + 1 (Cr) = 6 bytes. 4. **Calculate the total bytes for the entire frame:** - Total pixels = 921,600. - Groups of 4 pixels = 921,600 / 4 = 230,400 groups. - Total bytes = 230,400 groups × 6 bytes per group = 1,382,400 bytes. 5. **Convert bytes to Mebibytes (MiB):** - 1 Mebibyte (MiB) = 1024^2 bytes = 1,048,576 bytes. - Required memory in MiB = 1,382,400 bytes / 1,048,576 bytes/MiB ≈ 1.318 MiB. 6. **Round to 3 significant figures:** - 1.318 MiB rounded to 3 significant figures is 1.32 MiB. ### Solution: The required memory size in Mebibytes (MiB) for storing a 720p frame using the Y'CbCr 4:2:0 sampling scheme is **1.32 MiB** (to 3 significant figures). |
1.32 |
| gpto1 | To calculate the required memory size in Mebibytes (MiB) for storing a 720p frame using the Y'CbCr 4:2:0 sampling scheme, we'll follow these steps: **1. Understand the Frame Resolution and Components:** A 720p frame has a resolution of **1280 × 720 pixels**. Each pixel consists of three primary colour components: **Y'** (luminance), **Cb**, and **Cr** (chrominance components). **2. Comprehend the Y'CbCr 4:2:0 Sampling Scheme:** In the Y'CbCr 4:2:0 sampling scheme: - The **luminance (Y')** component is sampled at **full resolution**. - The **chrominance (Cb and Cr)** components are **subsampled by a factor of 2** in both the horizontal and vertical dimensions. This means for every **2×2 block of pixels**, there are: - **4 Y' samples** (one for each pixel). - **1 Cb sample** and **1 Cr sample** (shared among the 4 pixels). **3. Calculate the Number of Samples for Each Component:** - **Y' (Luminance) Samples:** \[ \text{Total Y' samples} = 1280 \times 720 = 921,\!600 \text{ samples} \] - **Cb and Cr (Chrominance) Samples:** \[ \text{Total Cb samples} = \text{Total Cr samples} = \left( \frac{1280}{2} \right) \times \left( \frac{720}{2} \right) = 640 \times 360 = 230,\!400 \text{ samples each} \] **4. Calculate the Total Number of Samples:** \[ \begin{align*} \text{Total samples} &= \text{Y' samples} + \text{Cb samples} + \text{Cr samples} \\ &= 921,\!600 + 230,\!400 + 230,\!400 \\ &= 1,\!382,\!400 \text{ samples} \end{align*} \] **5. Determine the Total Memory in Bytes:** Each sample requires **1 byte** of storage. \[ \text{Total memory in bytes} = \text{Total samples} \times 1 \, \text{byte} = 1,\!382,\!400 \, \text{bytes} \] **6. Convert Bytes to Mebibytes (MiB):** 1 Mebibyte (MiB) is equal to \( 1024^2 = 1,\!048,\!576 \) bytes. \[ \begin{align*} \text{Total memory in MiB} &= \frac{\text{Total memory in bytes}}{1,\!048,\!576 \, \text{bytes/MiB}} \\ &= \frac{1,\!382,\!400}{1,\!048,\!576} \\ &\approx 1.3184 \, \text{MiB} \end{align*} \] **7. Round to Three Significant Figures:** \[ \text{Total memory in MiB} \approx 1.32 \, \text{MiB} \] **Answer:** The required memory size is **1.32 MiB**. |
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| deepseekv3 | To calculate the required memory size for storing a 720p frame in the Y'CbCr 4:2:0 sampling scheme, follow these steps: --- ### **Step 1: Understand the Y'CbCr 4:2:0 Sampling Scheme** In Y'CbCr 4:2:0: - The **Y' (luma)** component is stored at full resolution (1280 × 720). - The **Cb and Cr (chroma)** components are subsampled at half resolution in both horizontal and vertical directions (640 × 360 each). --- ### **Step 2: Calculate the Memory for Each Component** 1. **Y' Component:** - Resolution: 1280 × 720 - Memory size = \( 1280 \times 720 \times 1 \, \text{byte} = 921,600 \, \text{bytes} \) 2. **Cb and Cr Components:** - Resolution: 640 × 360 each - Memory size for Cb = \( 640 \times 360 \times 1 \, \text{byte} = 230,400 \, \text{bytes} \) - Memory size for Cr = \( 640 \times 360 \times 1 \, \text{byte} = 230,400 \, \text{bytes} \) --- ### **Step 3: Total Memory in Bytes** Add the memory sizes of all components: \[ 921,600 \, \text{bytes} + 230,400 \, \text{bytes} + 230,400 \, \text{bytes} = 1,382,400 \, \text{bytes} \] --- ### **Step 4: Convert to Mebibytes (MiB)** 1 Mebibyte (MiB) = \( 1024^2 \) bytes = \( 1,048,576 \) bytes. \[ \text{Memory size in MiB} = \frac{1,382,400 \, \text{bytes}}{1,048,576 \, \text{bytes/MiB}} \approx 1.32 \, \text{MiB} \] --- ### **Final Answer** The required memory size is **1.32 MiB** (to 3 significant figures). |
1.32 |