{"id":34573,"date":"2023-03-31T02:15:35","date_gmt":"2023-03-30T20:45:35","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-what-would-be-the-source-code-like-to-add-two-n-x-n-x-n-x-n-tensors-closed\/"},"modified":"2023-03-31T02:15:35","modified_gmt":"2023-03-30T20:45:35","slug":"solved-what-would-be-the-source-code-like-to-add-two-n-x-n-x-n-x-n-tensors-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-what-would-be-the-source-code-like-to-add-two-n-x-n-x-n-x-n-tensors-closed\/","title":{"rendered":"[Solved] What would be the source code like to add two `N x N x N x N` tensors? [closed]"},"content":{"rendered":"<p> [ad_1]<br \/>\n<\/p>\n<div id=\"answer-75838217\" class=\"answer js-answer accepted-answer js-accepted-answer\" data-answerid=\"75838217\" data-parentid=\"75837651\" data-score=\"1\" data-position-on-page=\"1\" data-highest-scored=\"1\" data-question-has-accepted-highest-score=\"1\" itemprop=\"acceptedAnswer\" itemscope itemtype=\"https:\/\/schema.org\/Answer\">\n<div class=\"post-layout\">\n<div class=\"votecell post-layout--left\"><\/div>\n<div class=\"answercell post-layout--right\">\n<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<blockquote>\n<p>What would be the source code like to add two N x N x N x N tensors?<br \/>\nIf we have four dimensions in our problem, how can we write the add() function as we don&#8217;t have more than three dimensions?<\/p>\n<\/blockquote>\n<p>As is usual in computer programming, there are many ways to do it.<\/p>\n<p>Here is a demonstration of 4 methods<\/p>\n<ul>\n<li>using a 1D grid (x), striding<\/li>\n<li>using a 2D grid (x,y), with the grid &#8220;striding&#8221; in z, and w<\/li>\n<li>using a 3D grid (x,y,z) with the grid &#8220;striding&#8221; in w<\/li>\n<li>using a 3D grid (x,y,z) but &#8220;encoding&#8221; the w dimension in the x grid variable, therefore not requiring a loop in kernel code<\/li>\n<\/ul>\n<p>example:<\/p>\n<pre><code>$ cat t2231.cu\n#include &lt;iostream&gt;\n\nusing it = int;\nconst it NPOW = 6;         \/\/ 2^6 = 64\nconst it N = (1&lt;&lt;NPOW);    \/\/ the 4th power of this must fit in type it!\nusing dt = int;\n\n\/\/1D\n__global__ void k1(dt *a, dt *b, dt *c, it n){\n\n  for (it i = blockIdx.x*blockDim.x+threadIdx.x; i &lt; n*n*n*n; i += gridDim.x*blockDim.x) \/\/ 1D grid-stride loop\n    c[i] = a[i] + b[i];\n}\n\n\/\/2D, \"striding\" in z, w, must launch threads to cover x,y\n__global__ void k2(dt *a, dt *b, dt *c, it n){\n\n  it x = blockIdx.x*blockDim.x+threadIdx.x;\n  it y = blockIdx.y*blockDim.y+threadIdx.y;\n  if (x &lt; n &amp;&amp; y &lt; n)\n    for (it w = 0; w &lt; n; w++)\n      for (it z = 0; z &lt; n; z++){\n        it idx = (((((w*n)+z)*n)+y)*n)+x;\n        c[idx] = a[idx]+b[idx];}\n}\n\n\/\/3D, \"striding\" in w, must launch threads to cover x,y,z\n__global__ void k3(dt *a, dt *b, dt *c, it n){\n\n  it x = blockIdx.x*blockDim.x+threadIdx.x;\n  it y = blockIdx.y*blockDim.y+threadIdx.y;\n  it z = blockIdx.z*blockDim.z+threadIdx.z;\n  if (x &lt; n &amp;&amp; y &lt; n &amp;&amp; z &lt; n)\n    for (it w = 0; w &lt; n; w++){\n      it idx = (((((w*n)+z)*n)+y)*n)+x;\n      c[idx] = a[idx]+b[idx];}\n}\n\n\/\/4D, extracting w from x, must launch threads to cover w*x,y,z\n__global__ void k4(dt *a, dt *b, dt *c, it n){\n\n  it px = blockIdx.x*blockDim.x+threadIdx.x;\n  it w = px&gt;&gt;NPOW;\n  it x = px-(w&lt;&lt;NPOW);\n  it y = blockIdx.y*blockDim.y+threadIdx.y;\n  it z = blockIdx.z*blockDim.z+threadIdx.z;\n  if (x &lt; n &amp;&amp; y &lt; n &amp;&amp; z &lt; n &amp;&amp; w &lt; n){\n    it idx = (((((w*n)+z)*n)+y)*n)+x;\n    c[idx] = a[idx]+b[idx];}\n}\n\n\nint main(){\n  it sz = N*N*N*N;\n  dt *a = new dt[sz];\n  dt *b = new dt[sz];\n  dt *c = new dt[sz];\n  dt *d_a, *d_b, *d_c;\n  for (it i = 0; i &lt; sz; i++) {a[i] = i+1; b[i] = i+1;}\n  cudaMalloc(&amp;d_a, sizeof(d_a[0])*sz);\n  cudaMalloc(&amp;d_b, sizeof(d_a[0])*sz);\n  cudaMalloc(&amp;d_c, sizeof(d_a[0])*sz);\n  cudaMemcpy(d_a, a, sizeof(d_a[0])*sz, cudaMemcpyHostToDevice);\n  cudaMemcpy(d_b, b, sizeof(d_b[0])*sz, cudaMemcpyHostToDevice);\n  \/\/ choose \"arbitrary\" dimensions for 1D case\n  k1&lt;&lt;&lt;80, 1024&gt;&gt;&gt;(d_a, d_b, d_c, N);\n  cudaMemcpy(c, d_c, sizeof(d_c[0])*sz, cudaMemcpyDeviceToHost);\n  for (it i = 0; i &lt; sz; i++) if (c[i] != a[i]+b[i]) std::cout &lt;&lt; \"Error1\" &lt;&lt; std::endl;\n  cudaMemset(d_c, 0, sizeof(d_c[0])*sz);\n  \/\/ choose x,y to at least cover N,N for 2D case\n  dim3 block2(8,8);\n  dim3 grid2((N+block2.x-1)\/block2.x, (N+block2.y-1)\/block2.y);\n  k2&lt;&lt;&lt;grid2, block2&gt;&gt;&gt;(d_a, d_b, d_c, N);\n  cudaMemcpy(c, d_c, sizeof(d_c[0])*sz, cudaMemcpyDeviceToHost);\n  for (it i = 0; i &lt; sz; i++) if (c[i] != a[i]+b[i]) std::cout &lt;&lt; \"Error2\" &lt;&lt; std::endl;\n  cudaMemset(d_c, 0, sizeof(d_c[0])*sz);\n  \/\/ choose x,y,z to at least cover N,N,N for 3D case\n  dim3 block3(8,8,8);\n  dim3 grid3((N+block3.x-1)\/block3.x, (N+block3.y-1)\/block3.y, (N+block3.z-1)\/block3.z);\n  k3&lt;&lt;&lt;grid3, block3&gt;&gt;&gt;(d_a, d_b, d_c, N);\n  cudaMemcpy(c, d_c, sizeof(d_c[0])*sz, cudaMemcpyDeviceToHost);\n  for (it i = 0; i &lt; sz; i++) if (c[i] != a[i]+b[i]) std::cout &lt;&lt; \"Error3\" &lt;&lt; std::endl;\n  cudaMemset(d_c, 0, sizeof(d_c[0])*sz);\n  \/\/ choose x,y,z to at least cover N*N,N,N for 4D case\n  dim3 grid4((N*N+block3.x-1)\/block3.x, (N+block3.y-1)\/block3.y, (N+block3.z-1)\/block3.z);\n  k4&lt;&lt;&lt;grid4, block3&gt;&gt;&gt;(d_a, d_b, d_c, N);\n  cudaMemcpy(c, d_c, sizeof(d_c[0])*sz, cudaMemcpyDeviceToHost);\n  for (it i = 0; i &lt; sz; i++) if (c[i] != a[i]+b[i]) std::cout &lt;&lt; \"Error4\" &lt;&lt; std::endl;\n}\n\n$ nvcc -o t2231 t2231.cu\n$ compute-sanitizer .\/t2231\n========= COMPUTE-SANITIZER\n========= ERROR SUMMARY: 0 errors\n$\n<\/code><\/pre>\n<\/p><\/div>\n<div class=\"mt24\"><\/div>\n<\/div>\n<p>            <span class=\"d-none\" itemprop=\"commentCount\">2<\/span> <\/p><\/div>\n<\/div>\n<p>[ad_2]<\/p>\n<p>solved What would be the source code like to add two `N x N x N x N` tensors? [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"<p>[ad_1] What would be the source code like to add two N x N x N x N tensors? If we have four dimensions in our problem, how can we write the add() function as we don&#8217;t have more than three dimensions? As is usual in computer programming, there are many ways to do it. &#8230; <a title=\"[Solved] What would be the source code like to add two `N x N x N x N` tensors? [closed]\" class=\"read-more\" href=\"https:\/\/jassweb.com\/solved\/solved-what-would-be-the-source-code-like-to-add-two-n-x-n-x-n-x-n-tensors-closed\/\" aria-label=\"More on [Solved] What would be the source code like to add two `N x N x N x N` tensors? [closed]\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[1564],"class_list":["post-34573","post","type-post","status-publish","format-standard","hentry","category-solved","tag-cuda"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>[Solved] What would be the source code like to add two `N x N x N x N` tensors? 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