{"group":{"id":1,"name":"Community","lockable":false,"created_at":"2012-01-18T18:02:15.000Z","updated_at":"2026-04-06T14:01:22.000Z","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":"2026-04-06T00:00:00.000Z","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":49845,"title":"Decay Constant","description":null,"description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(0, 0, 0); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none solid rgb(0, 0, 0); white-space: normal; \"\u003e\u003cdiv style=\"block-size: 42px; display: block; min-width: 0px; padding-block-start: 0px; padding-top: 0px; perspective-origin: 407px 21px; transform-origin: 407px 21px; vertical-align: baseline; \"\u003e\u003cdiv style=\"font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; perspective-origin: 384px 21px; text-align: left; transform-origin: 384px 21px; white-space: pre-wrap; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; margin-right: 10px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; \"\u003e\u003cspan style=\"\"\u003eThe half-life of a decaying radioactive isotope is given. Please determine its decay constant in the unit of per-second. Note: one year is defined as 365 days not 365.25 days.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function y = decay(halfl)\r\n  y = exp(halfl);\r\nend","test_suite":"%%\r\nhalfl='5.27y'\r\ny_corr=4.1678e-09;\r\nassert(abs(decay(halfl)-y_corr)\u003c1e-5)\r\n%%\r\nhalfl='4.5656h'\r\ny_corr=4.2172e-5;\r\nassert(abs(decay(halfl)-y_corr)\u003c1e-5)\r\n%%\r\nhalfl='1.345s'\r\ny_corr=0.515351;\r\nassert(abs(decay(halfl)-y_corr)\u003c1e-5)\r\n%%\r\nhalfl='3.1415d'\r\ny_corr=2.5537e-06;\r\nassert(abs(decay(halfl)-y_corr)\u003c1e-5)\r\n%%\r\nhalfl='12345s'\r\ny_corr=5.6148e-05;\r\nassert(abs(decay(halfl)-y_corr)\u003c1e-5)\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":2,"created_by":180632,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":17,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2021-01-17T21:08:00.000Z","updated_at":"2025-12-07T18:03:03.000Z","published_at":"2021-01-17T21:08:00.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\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\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe half-life of a decaying radioactive isotope is given. Please determine its decay constant in the unit of per-second. Note: one year is defined as 365 days not 365.25 days.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":49870,"title":"A Simple Shielding Problem","description":null,"description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(0, 0, 0); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none solid rgb(0, 0, 0); white-space: normal; \"\u003e\u003cdiv style=\"block-size: 126px; display: block; min-width: 0px; padding-block-start: 0px; padding-top: 0px; perspective-origin: 407px 63px; transform-origin: 407px 63px; vertical-align: baseline; \"\u003e\u003cdiv style=\"font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; perspective-origin: 384px 63px; text-align: left; transform-origin: 384px 63px; white-space: pre-wrap; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; margin-right: 10px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; \"\u003e\u003cspan style=\"\"\u003eA Half Value Layer (HVL) is defined as the thickness of the material at which the intensity of radiation entering it is reduced by one half. The HVL is a function of material and type of radiation source. The HVL of material 1, 2, 3, and 4 are 44.5 mm, 12.7 mm, 4.8 mm, and 3.3 mm, respectively, against the gamma rays coming out of an unknown material called Inurdrium. Your task is to arrange these shielding materials with varying thickness to get an effective reduction of the intensity of the gamma rays by 94.62%. The answer is not unique but it will be examined against the requirement. Using all four materials with non-zero thickness is part of the requirement.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function y = hvl()\r\n  y = [1 2 3 4;5 6 7 8];\r\nend","test_suite":"%%\r\nn_mat=4;\r\nyy=hvl();\r\nassert(all(yy(2,:)~=0))\r\nassert(sum(yy(1,:))==10)\r\nhvl_ref=[44.5 12.7 4.8 3.3];\r\nhh=1;\r\nfor i=1:4\r\n    hh=hh*(0.5)^(yy(2,i)/hvl_ref(yy(1,i)));\r\nend\r\nassert(abs(hh-0.0538)\u003c1e-3)\r\n","published":true,"deleted":false,"likes_count":2,"comments_count":1,"created_by":180632,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":14,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2021-01-18T22:44:07.000Z","updated_at":"2025-12-07T18:20:31.000Z","published_at":"2021-01-18T22:44:07.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\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\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eA Half Value Layer (HVL) is defined as the thickness of the material at which the intensity of radiation entering it is reduced by one half. The HVL is a function of material and type of radiation source. The HVL of material 1, 2, 3, and 4 are 44.5 mm, 12.7 mm, 4.8 mm, and 3.3 mm, respectively, against the gamma rays coming out of an unknown material called Inurdrium. Your task is to arrange these shielding materials with varying thickness to get an effective reduction of the intensity of the gamma rays by 94.62%. The answer is not unique but it will be examined against the requirement. Using all four materials with non-zero thickness is part of the requirement.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"}],"problem_search":{"errors":[],"problems":[{"id":49845,"title":"Decay Constant","description":null,"description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(0, 0, 0); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none solid rgb(0, 0, 0); white-space: normal; \"\u003e\u003cdiv style=\"block-size: 42px; display: block; min-width: 0px; padding-block-start: 0px; padding-top: 0px; perspective-origin: 407px 21px; transform-origin: 407px 21px; vertical-align: baseline; \"\u003e\u003cdiv style=\"font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; perspective-origin: 384px 21px; text-align: left; transform-origin: 384px 21px; white-space: pre-wrap; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; margin-right: 10px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; \"\u003e\u003cspan style=\"\"\u003eThe half-life of a decaying radioactive isotope is given. Please determine its decay constant in the unit of per-second. Note: one year is defined as 365 days not 365.25 days.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function y = decay(halfl)\r\n  y = exp(halfl);\r\nend","test_suite":"%%\r\nhalfl='5.27y'\r\ny_corr=4.1678e-09;\r\nassert(abs(decay(halfl)-y_corr)\u003c1e-5)\r\n%%\r\nhalfl='4.5656h'\r\ny_corr=4.2172e-5;\r\nassert(abs(decay(halfl)-y_corr)\u003c1e-5)\r\n%%\r\nhalfl='1.345s'\r\ny_corr=0.515351;\r\nassert(abs(decay(halfl)-y_corr)\u003c1e-5)\r\n%%\r\nhalfl='3.1415d'\r\ny_corr=2.5537e-06;\r\nassert(abs(decay(halfl)-y_corr)\u003c1e-5)\r\n%%\r\nhalfl='12345s'\r\ny_corr=5.6148e-05;\r\nassert(abs(decay(halfl)-y_corr)\u003c1e-5)\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":2,"created_by":180632,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":17,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2021-01-17T21:08:00.000Z","updated_at":"2025-12-07T18:03:03.000Z","published_at":"2021-01-17T21:08:00.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\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\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe half-life of a decaying radioactive isotope is given. Please determine its decay constant in the unit of per-second. Note: one year is defined as 365 days not 365.25 days.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":49870,"title":"A Simple Shielding Problem","description":null,"description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(0, 0, 0); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none solid rgb(0, 0, 0); white-space: normal; \"\u003e\u003cdiv style=\"block-size: 126px; display: block; min-width: 0px; padding-block-start: 0px; padding-top: 0px; perspective-origin: 407px 63px; transform-origin: 407px 63px; vertical-align: baseline; \"\u003e\u003cdiv style=\"font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; perspective-origin: 384px 63px; text-align: left; transform-origin: 384px 63px; white-space: pre-wrap; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; margin-right: 10px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; \"\u003e\u003cspan style=\"\"\u003eA Half Value Layer (HVL) is defined as the thickness of the material at which the intensity of radiation entering it is reduced by one half. The HVL is a function of material and type of radiation source. The HVL of material 1, 2, 3, and 4 are 44.5 mm, 12.7 mm, 4.8 mm, and 3.3 mm, respectively, against the gamma rays coming out of an unknown material called Inurdrium. Your task is to arrange these shielding materials with varying thickness to get an effective reduction of the intensity of the gamma rays by 94.62%. The answer is not unique but it will be examined against the requirement. Using all four materials with non-zero thickness is part of the requirement.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function y = hvl()\r\n  y = [1 2 3 4;5 6 7 8];\r\nend","test_suite":"%%\r\nn_mat=4;\r\nyy=hvl();\r\nassert(all(yy(2,:)~=0))\r\nassert(sum(yy(1,:))==10)\r\nhvl_ref=[44.5 12.7 4.8 3.3];\r\nhh=1;\r\nfor i=1:4\r\n    hh=hh*(0.5)^(yy(2,i)/hvl_ref(yy(1,i)));\r\nend\r\nassert(abs(hh-0.0538)\u003c1e-3)\r\n","published":true,"deleted":false,"likes_count":2,"comments_count":1,"created_by":180632,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":14,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2021-01-18T22:44:07.000Z","updated_at":"2025-12-07T18:20:31.000Z","published_at":"2021-01-18T22:44:07.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\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\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eA Half Value Layer (HVL) is defined as the thickness of the material at which the intensity of radiation entering it is reduced by one half. The HVL is a function of material and type of radiation source. The HVL of material 1, 2, 3, and 4 are 44.5 mm, 12.7 mm, 4.8 mm, and 3.3 mm, respectively, against the gamma rays coming out of an unknown material called Inurdrium. Your task is to arrange these shielding materials with varying thickness to get an effective reduction of the intensity of the gamma rays by 94.62%. The answer is not unique but it will be examined against the requirement. 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