{"group":{"id":1,"name":"Community","lockable":false,"created_at":"2012-01-18T18:02:15.000Z","updated_at":"2025-12-14T01:33:56.000Z","description":"Problems submitted by members of the MATLAB Central community.","is_default":true,"created_by":161519,"badge_id":null,"featured":false,"trending":false,"solution_count_in_trending_period":0,"trending_last_calculated":"2025-12-14T00:00:00.000Z","image_id":null,"published":true,"community_created":false,"status_id":2,"is_default_group_for_player":false,"deleted_by":null,"deleted_at":null,"restored_by":null,"restored_at":null,"description_opc":null,"description_html":null,"published_at":null},"problems":[{"id":42639,"title":"Find the Final State of an Abelian Sandpile","description":"Let us define an \u003chttp://nautil.us/issue/23/dominoes/the-amazing-autotuning-sandpile Abelian sand pile\u003e as a matrix that is only in a stable and final state when all of its elements are less than 4. When any of its elements is greater than 4, distribute four sand grains, one each to the elements above, below, to the left, and to the right of the offending element. Continue doing this until the matrix is stable.\r\n\r\nSo\r\n\r\n 0 0 0 \r\n 0 4 1\r\n 1 0 0\r\n\r\nwould become\r\n\r\n 0 1 0\r\n 1 0 2\r\n 1 1 0\r\n\r\nbecause the 4 in the middle is unstable and gets distributed.\r\n\r\nWhat makes this process Abelian is that the order in which you distribute the \"sand grains\" doesn't matter. This gives you considerable algorithmic freedom.\r\n\r\nSo, given an input matrix A, return the resulting stable Abelian sand pile matrix B.\r\n\r\nNote: you won't have to worry about sand grains \"falling off the edge of the table.\" The given matrix A will always be big enough to handle the resulting stable matrix B.\r\n\r\nExamples\r\n\r\n A = [ 0 0 0 0\r\n       0 4 4 0\r\n       0 0 0 0 ]\r\n\r\n B = [ 0 1 1 0\r\n       1 1 1 1\r\n       0 1 1 0 ]\r\n\r\nand\r\n\r\n A = [ 0 0 0\r\n       0 9 0\r\n       0 0 0 ] \r\n\r\n B = [ 0 2 0\r\n       2 1 2\r\n       0 2 0 ]\r\n\r\n\r\n","description_html":"\u003cp\u003eLet us define an \u003ca href = \"http://nautil.us/issue/23/dominoes/the-amazing-autotuning-sandpile\"\u003eAbelian sand pile\u003c/a\u003e as a matrix that is only in a stable and final state when all of its elements are less than 4. When any of its elements is greater than 4, distribute four sand grains, one each to the elements above, below, to the left, and to the right of the offending element. Continue doing this until the matrix is stable.\u003c/p\u003e\u003cp\u003eSo\u003c/p\u003e\u003cpre\u003e 0 0 0 \r\n 0 4 1\r\n 1 0 0\u003c/pre\u003e\u003cp\u003ewould become\u003c/p\u003e\u003cpre\u003e 0 1 0\r\n 1 0 2\r\n 1 1 0\u003c/pre\u003e\u003cp\u003ebecause the 4 in the middle is unstable and gets distributed.\u003c/p\u003e\u003cp\u003eWhat makes this process Abelian is that the order in which you distribute the \"sand grains\" doesn't matter. This gives you considerable algorithmic freedom.\u003c/p\u003e\u003cp\u003eSo, given an input matrix A, return the resulting stable Abelian sand pile matrix B.\u003c/p\u003e\u003cp\u003eNote: you won't have to worry about sand grains \"falling off the edge of the table.\" The given matrix A will always be big enough to handle the resulting stable matrix B.\u003c/p\u003e\u003cp\u003eExamples\u003c/p\u003e\u003cpre\u003e A = [ 0 0 0 0\r\n       0 4 4 0\r\n       0 0 0 0 ]\u003c/pre\u003e\u003cpre\u003e B = [ 0 1 1 0\r\n       1 1 1 1\r\n       0 1 1 0 ]\u003c/pre\u003e\u003cp\u003eand\u003c/p\u003e\u003cpre\u003e A = [ 0 0 0\r\n       0 9 0\r\n       0 0 0 ] \u003c/pre\u003e\u003cpre\u003e B = [ 0 2 0\r\n       2 1 2\r\n       0 2 0 ]\u003c/pre\u003e","function_template":"function B = sandpile(A)\r\n  B = zeros(size(A));\r\nend","test_suite":"%%\r\nA = 1;\r\nB_correct = 1;\r\nassert(isequal(sandpile(A),B_correct))\r\n\r\n%%\r\nA = [0 0 0;0 4 1;1 0 0];\r\nB_correct  = [0 1 0;1 0 2;1 1 0];\r\nassert(isequal(sandpile(A),B_correct))\r\n\r\n%%\r\nA = [0 0 0 0;0 4 4 0;0 0 0 0];\r\nB_correct = [0 1 1 0;1 1 1 1;0 1 1 0];\r\nassert(isequal(sandpile(A),B_correct))\r\n\r\n%%\r\nA = [0 0 0 0 0;0 0 0 0 0;0 0 17 0 0;0 0 0 7 0;0 0 0 0 0];\r\nB_correct = [0 0 1 0 0;0 2 1 2 0;1 1 1 3 1;0 2 3 1 2;0 0 1 2 0];\r\nassert(isequal(sandpile(A),B_correct))\r\n\r\n%%\r\nA = [0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0;0 0 0 16 2 3 13 0 0 0 0;0 0 0 5 11 10 8 0 0 0 0;0 0 0 9 7 6 12 0 0 0 0;0 0 0 4 14 15 1 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0];\r\nB_correct = [0 0 0 0 0 0 0 0 0 0 0;0 0 1 3 3 3 3 1 0 0 0;0 2 3 2 3 3 2 3 2 0 0;1 1 1 3 2 3 1 1 1 1 0;1 3 1 1 3 3 1 2 3 1 0;1 3 0 3 3 3 3 1 3 1 0;1 0 3 2 2 3 0 3 0 1 0;0 2 2 3 2 2 3 2 2 0 0;0 0 2 0 3 3 0 2 0 0 0;0 0 0 1 1 1 1 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0];\r\nassert(isequal(sandpile(A),B_correct))\r\n\r\n","published":true,"deleted":false,"likes_count":4,"comments_count":1,"created_by":7,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":74,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":24,"created_at":"2015-09-29T20:59:19.000Z","updated_at":"2026-04-01T07:15:49.000Z","published_at":"2015-09-29T23:01:28.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eLet us define an\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:hyperlink w:docLocation=\\\"http://nautil.us/issue/23/dominoes/the-amazing-autotuning-sandpile\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eAbelian sand pile\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e as a matrix that is only in a stable and final state when all of its elements are less than 4. When any of its elements is greater than 4, distribute four sand grains, one each to the elements above, below, to the left, and to the right of the offending element. Continue doing this until the matrix is stable.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eSo\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[ 0 0 0 \\n 0 4 1\\n 1 0 0]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ewould become\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[ 0 1 0\\n 1 0 2\\n 1 1 0]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ebecause the 4 in the middle is unstable and gets distributed.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eWhat makes this process Abelian is that the order in which you distribute the \\\"sand grains\\\" doesn't matter. This gives you considerable algorithmic freedom.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eSo, given an input matrix A, return the resulting stable Abelian sand pile matrix B.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eNote: you won't have to worry about sand grains \\\"falling off the edge of the table.\\\" The given matrix A will always be big enough to handle the resulting stable matrix B.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eExamples\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[ A = [ 0 0 0 0\\n       0 4 4 0\\n       0 0 0 0 ]\\n\\n B = [ 0 1 1 0\\n       1 1 1 1\\n       0 1 1 0 ]]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eand\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[ A = [ 0 0 0\\n       0 9 0\\n       0 0 0 ] \\n\\n B = [ 0 2 0\\n       2 1 2\\n       0 2 0 ]]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"}],"problem_search":{"errors":[],"problems":[{"id":42639,"title":"Find the Final State of an Abelian Sandpile","description":"Let us define an \u003chttp://nautil.us/issue/23/dominoes/the-amazing-autotuning-sandpile Abelian sand pile\u003e as a matrix that is only in a stable and final state when all of its elements are less than 4. When any of its elements is greater than 4, distribute four sand grains, one each to the elements above, below, to the left, and to the right of the offending element. Continue doing this until the matrix is stable.\r\n\r\nSo\r\n\r\n 0 0 0 \r\n 0 4 1\r\n 1 0 0\r\n\r\nwould become\r\n\r\n 0 1 0\r\n 1 0 2\r\n 1 1 0\r\n\r\nbecause the 4 in the middle is unstable and gets distributed.\r\n\r\nWhat makes this process Abelian is that the order in which you distribute the \"sand grains\" doesn't matter. This gives you considerable algorithmic freedom.\r\n\r\nSo, given an input matrix A, return the resulting stable Abelian sand pile matrix B.\r\n\r\nNote: you won't have to worry about sand grains \"falling off the edge of the table.\" The given matrix A will always be big enough to handle the resulting stable matrix B.\r\n\r\nExamples\r\n\r\n A = [ 0 0 0 0\r\n       0 4 4 0\r\n       0 0 0 0 ]\r\n\r\n B = [ 0 1 1 0\r\n       1 1 1 1\r\n       0 1 1 0 ]\r\n\r\nand\r\n\r\n A = [ 0 0 0\r\n       0 9 0\r\n       0 0 0 ] \r\n\r\n B = [ 0 2 0\r\n       2 1 2\r\n       0 2 0 ]\r\n\r\n\r\n","description_html":"\u003cp\u003eLet us define an \u003ca href = \"http://nautil.us/issue/23/dominoes/the-amazing-autotuning-sandpile\"\u003eAbelian sand pile\u003c/a\u003e as a matrix that is only in a stable and final state when all of its elements are less than 4. When any of its elements is greater than 4, distribute four sand grains, one each to the elements above, below, to the left, and to the right of the offending element. Continue doing this until the matrix is stable.\u003c/p\u003e\u003cp\u003eSo\u003c/p\u003e\u003cpre\u003e 0 0 0 \r\n 0 4 1\r\n 1 0 0\u003c/pre\u003e\u003cp\u003ewould become\u003c/p\u003e\u003cpre\u003e 0 1 0\r\n 1 0 2\r\n 1 1 0\u003c/pre\u003e\u003cp\u003ebecause the 4 in the middle is unstable and gets distributed.\u003c/p\u003e\u003cp\u003eWhat makes this process Abelian is that the order in which you distribute the \"sand grains\" doesn't matter. This gives you considerable algorithmic freedom.\u003c/p\u003e\u003cp\u003eSo, given an input matrix A, return the resulting stable Abelian sand pile matrix B.\u003c/p\u003e\u003cp\u003eNote: you won't have to worry about sand grains \"falling off the edge of the table.\" The given matrix A will always be big enough to handle the resulting stable matrix B.\u003c/p\u003e\u003cp\u003eExamples\u003c/p\u003e\u003cpre\u003e A = [ 0 0 0 0\r\n       0 4 4 0\r\n       0 0 0 0 ]\u003c/pre\u003e\u003cpre\u003e B = [ 0 1 1 0\r\n       1 1 1 1\r\n       0 1 1 0 ]\u003c/pre\u003e\u003cp\u003eand\u003c/p\u003e\u003cpre\u003e A = [ 0 0 0\r\n       0 9 0\r\n       0 0 0 ] \u003c/pre\u003e\u003cpre\u003e B = [ 0 2 0\r\n       2 1 2\r\n       0 2 0 ]\u003c/pre\u003e","function_template":"function B = sandpile(A)\r\n  B = zeros(size(A));\r\nend","test_suite":"%%\r\nA = 1;\r\nB_correct = 1;\r\nassert(isequal(sandpile(A),B_correct))\r\n\r\n%%\r\nA = [0 0 0;0 4 1;1 0 0];\r\nB_correct  = [0 1 0;1 0 2;1 1 0];\r\nassert(isequal(sandpile(A),B_correct))\r\n\r\n%%\r\nA = [0 0 0 0;0 4 4 0;0 0 0 0];\r\nB_correct = [0 1 1 0;1 1 1 1;0 1 1 0];\r\nassert(isequal(sandpile(A),B_correct))\r\n\r\n%%\r\nA = [0 0 0 0 0;0 0 0 0 0;0 0 17 0 0;0 0 0 7 0;0 0 0 0 0];\r\nB_correct = [0 0 1 0 0;0 2 1 2 0;1 1 1 3 1;0 2 3 1 2;0 0 1 2 0];\r\nassert(isequal(sandpile(A),B_correct))\r\n\r\n%%\r\nA = [0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0;0 0 0 16 2 3 13 0 0 0 0;0 0 0 5 11 10 8 0 0 0 0;0 0 0 9 7 6 12 0 0 0 0;0 0 0 4 14 15 1 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0];\r\nB_correct = [0 0 0 0 0 0 0 0 0 0 0;0 0 1 3 3 3 3 1 0 0 0;0 2 3 2 3 3 2 3 2 0 0;1 1 1 3 2 3 1 1 1 1 0;1 3 1 1 3 3 1 2 3 1 0;1 3 0 3 3 3 3 1 3 1 0;1 0 3 2 2 3 0 3 0 1 0;0 2 2 3 2 2 3 2 2 0 0;0 0 2 0 3 3 0 2 0 0 0;0 0 0 1 1 1 1 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0];\r\nassert(isequal(sandpile(A),B_correct))\r\n\r\n","published":true,"deleted":false,"likes_count":4,"comments_count":1,"created_by":7,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":74,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":24,"created_at":"2015-09-29T20:59:19.000Z","updated_at":"2026-04-01T07:15:49.000Z","published_at":"2015-09-29T23:01:28.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eLet us define an\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:hyperlink w:docLocation=\\\"http://nautil.us/issue/23/dominoes/the-amazing-autotuning-sandpile\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eAbelian sand pile\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e as a matrix that is only in a stable and final state when all of its elements are less than 4. When any of its elements is greater than 4, distribute four sand grains, one each to the elements above, below, to the left, and to the right of the offending element. Continue doing this until the matrix is stable.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eSo\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[ 0 0 0 \\n 0 4 1\\n 1 0 0]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ewould become\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[ 0 1 0\\n 1 0 2\\n 1 1 0]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ebecause the 4 in the middle is unstable and gets distributed.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eWhat makes this process Abelian is that the order in which you distribute the \\\"sand grains\\\" doesn't matter. This gives you considerable algorithmic freedom.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eSo, given an input matrix A, return the resulting stable Abelian sand pile matrix B.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eNote: you won't have to worry about sand grains \\\"falling off the edge of the table.\\\" The given matrix A will always be big enough to handle the resulting stable matrix B.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eExamples\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle 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