{"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":42698,"title":"Why the heck are they blinking!?!?","description":"Merry Christmas everyone!  Sadly, the lights you've had on your tree for so many years burned out, and it's time to get a new set.  Being a skilled (and cheap!) mathematician, you realize that you can estimate the total length of the strings of lights you'll need for your tree with two simple parameters:\r\n\r\n* The radius of the base of the tree (in feet)\r\n* The Number of rows of lights you want on your tree, from top to bottom.\r\n\r\nThe rows of lights are equally spaced vertically around the tree.  Given these two variables, calculate how long your string of lights has to be in order to wrap around your tree.  You want to buy the minimum possible length of lights, because NOBODY likes having to untangle any more lights than they have to!\r\n\r\nHelpful hints - The answers calculated below model the lights on your tree as a two-dimensional Spiral of Archimedes, so the number of rows of lights is equal to the number of times your spiral makes a full 360-degree loop.  The final Cartesian coordinate of your spiral should be (width, 0).  If someone far smarter than I am wants to make the full three-dimensional version of this problem, knock yourself out!","description_html":"\u003cp\u003eMerry Christmas everyone!  Sadly, the lights you've had on your tree for so many years burned out, and it's time to get a new set.  Being a skilled (and cheap!) mathematician, you realize that you can estimate the total length of the strings of lights you'll need for your tree with two simple parameters:\u003c/p\u003e\u003cul\u003e\u003cli\u003eThe radius of the base of the tree (in feet)\u003c/li\u003e\u003cli\u003eThe Number of rows of lights you want on your tree, from top to bottom.\u003c/li\u003e\u003c/ul\u003e\u003cp\u003eThe rows of lights are equally spaced vertically around the tree.  Given these two variables, calculate how long your string of lights has to be in order to wrap around your tree.  You want to buy the minimum possible length of lights, because NOBODY likes having to untangle any more lights than they have to!\u003c/p\u003e\u003cp\u003eHelpful hints - The answers calculated below model the lights on your tree as a two-dimensional Spiral of Archimedes, so the number of rows of lights is equal to the number of times your spiral makes a full 360-degree loop.  The final Cartesian coordinate of your spiral should be (width, 0).  If someone far smarter than I am wants to make the full three-dimensional version of this problem, knock yourself out!\u003c/p\u003e","function_template":"function l = Length_of_Lights(w,n)\r\n  y = w*n;\r\nend","test_suite":"%%\r\nw=1;n=1;\r\nLOL=Length_of_Lights(w,n)\r\ntest=abs(LOL-3.3830)\r\nassert(test\u003c=0.001)\r\n%%\r\nw=4;n=5;\r\nLOL=Length_of_Lights(w,n)\r\ntest=abs(LOL-63.1273)\r\nassert(test\u003c=0.001)\r\n%%\r\nw=20;n=15;\r\nLOL=Length_of_Lights(w,n)\r\ntest=abs(LOL-943.086)\r\nassert(test\u003c=0.01)\r\n%%\r\nr=[31.4584 62.9167 94.3751 125.8335 157.2919 188.7502 220.2086 251.6670 283.1253 314.5837];\r\nk=ceil(10*rand)\r\nLOL=Length_of_Lights(k,10)\r\ntest=abs(LOL-r(k))\r\nassert(test\u003c=0.001)\r\n%%\r\nr=[23.6813 45.0198 66.7403 88.5798 110.4727 132.3946 154.3341 176.2850 198.2439 220.2086];\r\nk=ceil(10*rand)\r\nLOL=Length_of_Lights(7,k)\r\ntest=abs(LOL-r(k))\r\nassert(test\u003c=0.001)\r\n%%\r\nw = floor(sqrt(Length_of_Lights(9,4)))\r\nn=floor(sqrt(Length_of_Lights(5,8)))\r\nLOL=Length_of_Lights(w,n)\r\ntest=abs(LOL-345.9679)\r\nassert(test\u003c=0.01)","published":true,"deleted":false,"likes_count":2,"comments_count":9,"created_by":1615,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":34,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":37,"created_at":"2015-12-29T16:44:33.000Z","updated_at":"2026-02-27T09:52:37.000Z","published_at":"2015-12-29T16:44:33.000Z","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\u003eMerry Christmas everyone! Sadly, the lights you've had on your tree for so many years burned out, and it's time to get a new set. Being a skilled (and cheap!) mathematician, you realize that you can estimate the total length of the strings of lights you'll need for your tree with two simple parameters:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe radius of the base of the tree (in feet)\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe Number of rows of lights you want on your tree, from top to bottom.\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\u003eThe rows of lights are equally spaced vertically around the tree. Given these two variables, calculate how long your string of lights has to be in order to wrap around your tree. You want to buy the minimum possible length of lights, because NOBODY likes having to untangle any more lights than they have to!\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\u003eHelpful hints - The answers calculated below model the lights on your tree as a two-dimensional Spiral of Archimedes, so the number of rows of lights is equal to the number of times your spiral makes a full 360-degree loop. The final Cartesian coordinate of your spiral should be (width, 0). If someone far smarter than I am wants to make the full three-dimensional version of this problem, knock yourself out!\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":42698,"title":"Why the heck are they blinking!?!?","description":"Merry Christmas everyone!  Sadly, the lights you've had on your tree for so many years burned out, and it's time to get a new set.  Being a skilled (and cheap!) mathematician, you realize that you can estimate the total length of the strings of lights you'll need for your tree with two simple parameters:\r\n\r\n* The radius of the base of the tree (in feet)\r\n* The Number of rows of lights you want on your tree, from top to bottom.\r\n\r\nThe rows of lights are equally spaced vertically around the tree.  Given these two variables, calculate how long your string of lights has to be in order to wrap around your tree.  You want to buy the minimum possible length of lights, because NOBODY likes having to untangle any more lights than they have to!\r\n\r\nHelpful hints - The answers calculated below model the lights on your tree as a two-dimensional Spiral of Archimedes, so the number of rows of lights is equal to the number of times your spiral makes a full 360-degree loop.  The final Cartesian coordinate of your spiral should be (width, 0).  If someone far smarter than I am wants to make the full three-dimensional version of this problem, knock yourself out!","description_html":"\u003cp\u003eMerry Christmas everyone!  Sadly, the lights you've had on your tree for so many years burned out, and it's time to get a new set.  Being a skilled (and cheap!) mathematician, you realize that you can estimate the total length of the strings of lights you'll need for your tree with two simple parameters:\u003c/p\u003e\u003cul\u003e\u003cli\u003eThe radius of the base of the tree (in feet)\u003c/li\u003e\u003cli\u003eThe Number of rows of lights you want on your tree, from top to bottom.\u003c/li\u003e\u003c/ul\u003e\u003cp\u003eThe rows of lights are equally spaced vertically around the tree.  Given these two variables, calculate how long your string of lights has to be in order to wrap around your tree.  You want to buy the minimum possible length of lights, because NOBODY likes having to untangle any more lights than they have to!\u003c/p\u003e\u003cp\u003eHelpful hints - The answers calculated below model the lights on your tree as a two-dimensional Spiral of Archimedes, so the number of rows of lights is equal to the number of times your spiral makes a full 360-degree loop.  The final Cartesian coordinate of your spiral should be (width, 0).  If someone far smarter than I am wants to make the full three-dimensional version of this problem, knock yourself out!\u003c/p\u003e","function_template":"function l = Length_of_Lights(w,n)\r\n  y = w*n;\r\nend","test_suite":"%%\r\nw=1;n=1;\r\nLOL=Length_of_Lights(w,n)\r\ntest=abs(LOL-3.3830)\r\nassert(test\u003c=0.001)\r\n%%\r\nw=4;n=5;\r\nLOL=Length_of_Lights(w,n)\r\ntest=abs(LOL-63.1273)\r\nassert(test\u003c=0.001)\r\n%%\r\nw=20;n=15;\r\nLOL=Length_of_Lights(w,n)\r\ntest=abs(LOL-943.086)\r\nassert(test\u003c=0.01)\r\n%%\r\nr=[31.4584 62.9167 94.3751 125.8335 157.2919 188.7502 220.2086 251.6670 283.1253 314.5837];\r\nk=ceil(10*rand)\r\nLOL=Length_of_Lights(k,10)\r\ntest=abs(LOL-r(k))\r\nassert(test\u003c=0.001)\r\n%%\r\nr=[23.6813 45.0198 66.7403 88.5798 110.4727 132.3946 154.3341 176.2850 198.2439 220.2086];\r\nk=ceil(10*rand)\r\nLOL=Length_of_Lights(7,k)\r\ntest=abs(LOL-r(k))\r\nassert(test\u003c=0.001)\r\n%%\r\nw = floor(sqrt(Length_of_Lights(9,4)))\r\nn=floor(sqrt(Length_of_Lights(5,8)))\r\nLOL=Length_of_Lights(w,n)\r\ntest=abs(LOL-345.9679)\r\nassert(test\u003c=0.01)","published":true,"deleted":false,"likes_count":2,"comments_count":9,"created_by":1615,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":34,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":37,"created_at":"2015-12-29T16:44:33.000Z","updated_at":"2026-02-27T09:52:37.000Z","published_at":"2015-12-29T16:44:33.000Z","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\u003eMerry Christmas everyone! Sadly, the lights you've had on your tree for so many years burned out, and it's time to get a new set. Being a skilled (and cheap!) mathematician, you realize that you can estimate the total length of the strings of lights you'll need for your tree with two simple parameters:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe radius of the base of the tree (in feet)\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe Number of rows of lights you want on your tree, from top to bottom.\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\u003eThe rows of lights are equally spaced vertically around the tree. Given these two variables, calculate how long your string of lights has to be in order to wrap around your tree. You want to buy the minimum possible length of lights, because NOBODY likes having to untangle any more lights than they have to!\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\u003eHelpful hints - The answers calculated below model the lights on your tree as a two-dimensional Spiral of Archimedes, so the number of rows of lights is equal to the number of times your spiral makes a full 360-degree loop. The final Cartesian coordinate of your spiral should be (width, 0). If someone far smarter than I am wants to make the full three-dimensional version of this problem, knock yourself out!\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\"}]}"}],"term":"tag:\"blinking 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