{"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":3065,"title":"Cycling — Critical Power","description":"From Training and Racing with a Power Meter by Allen and Coggan:\r\n\r\n\"A number of equations have been presented in the scientific literature describing human power output as a function of time, some derived from modeling based on the underlying physiology, and some simply derived empirically. One of the simplest and most robust, though, is the original \"critical power\" concept first proposed by H. Monod around 1960. Various formulations of this idea have been presented, but the original equation is a hyperbolic of the form: t = AWC / (P – CP), where t is time to exhaustion [in seconds], P is current power [in Watts], CP is work rate (i.e., power) asymptote, and AWC is degree of curvature of the relationship.\"\r\n\r\nYou will be given values for AWC and CP. Write a function to return the time that the cyclist can maintain for an array of power (P) values. The times should be rounded to the nearest second. If P \u003c= CP, the cyclist can theoretically maintain that power indefinitely (Inf).","description_html":"\u003cp\u003eFrom Training and Racing with a Power Meter by Allen and Coggan:\u003c/p\u003e\u003cp\u003e\"A number of equations have been presented in the scientific literature describing human power output as a function of time, some derived from modeling based on the underlying physiology, and some simply derived empirically. One of the simplest and most robust, though, is the original \"critical power\" concept first proposed by H. Monod around 1960. Various formulations of this idea have been presented, but the original equation is a hyperbolic of the form: t = AWC / (P – CP), where t is time to exhaustion [in seconds], P is current power [in Watts], CP is work rate (i.e., power) asymptote, and AWC is degree of curvature of the relationship.\"\u003c/p\u003e\u003cp\u003eYou will be given values for AWC and CP. Write a function to return the time that the cyclist can maintain for an array of power (P) values. The times should be rounded to the nearest second. If P \u0026lt;= CP, the cyclist can theoretically maintain that power indefinitely (Inf).\u003c/p\u003e","function_template":"function [t] = cycling_crit_power(AWC,CP,P)\r\n\r\nt = zeros(size(P));\r\n\r\nend\r\n","test_suite":"%%\r\nAWC = 5e4;\r\nCP = 200;\r\nP = [150 200 225 250 275 300 350 400 500 1000];\r\nt_corr = [Inf,Inf,2000,1000,667,500,333,250,167,63];\r\nassert(isequal(cycling_crit_power(AWC,CP,P),t_corr))\r\n\r\n%%\r\nAWC = 5.3e4;\r\nCP = 222;\r\nP = [150 200 225 250 275 300 350 400 500 1000];\r\nt_corr = [Inf,Inf,17667,1893,1000,679,414,298,191,68];\r\nassert(isequal(cycling_crit_power(AWC,CP,P),t_corr))\r\n\r\n%%\r\nAWC = 4.6e4;\r\nCP = 250;\r\nP = [150 200 225 250 275 300 350 400 500 1000];\r\nt_corr = [Inf,Inf,Inf,Inf,1840,920,460,307,184,61];\r\nassert(isequal(cycling_crit_power(AWC,CP,P),t_corr))\r\n\r\n%%\r\nAWC = 5e4;\r\nCP = 300;\r\nP = 250:50:1500;\r\nt_corr = [Inf,Inf,1000,500,333,250,200,167,143,125,111,100,91,83,77,71,67,63,59,56,53,50,48,45,43,42];\r\nassert(isequal(cycling_crit_power(AWC,CP,P),t_corr))\r\n\r\n%%\r\nind = randi(4);\r\nswitch ind\r\n\tcase 1\r\n\t\tAWC = 5e4;\r\n\t\tCP = 200;\r\n\t\tP = [150 200 225 250 275 300 350 400 500 1000];\r\n\t\tt_corr = [Inf,Inf,2000,1000,667,500,333,250,167,63];\r\n\tcase 2\r\n\t\tAWC = 5.3e4;\r\n\t\tCP = 222;\r\n\t\tP = [150 200 225 250 275 300 350 400 500 1000];\r\n\t\tt_corr = [Inf,Inf,17667,1893,1000,679,414,298,191,68];\r\n\tcase 3\r\n\t\tAWC = 4.6e4;\r\n\t\tCP = 250;\r\n\t\tP = [150 200 225 250 275 300 350 400 500 1000];\r\n\t\tt_corr = [Inf,Inf,Inf,Inf,1840,920,460,307,184,61];\r\n\tcase 4\r\n\t\tAWC = 5e4;\r\n\t\tCP = 300;\r\n\t\tP = 250:50:1500;\r\n\t\tt_corr = [Inf,Inf,1000,500,333,250,200,167,143,125,111,100,91,83,77,71,67,63,59,56,53,50,48,45,43,42];\r\nend\r\nassert(isequal(cycling_crit_power(AWC,CP,P),t_corr))\r\n","published":true,"deleted":false,"likes_count":2,"comments_count":0,"created_by":26769,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":43,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2015-03-05T04:31:29.000Z","updated_at":"2026-02-16T11:45:09.000Z","published_at":"2015-03-05T04:31:29.000Z","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\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"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\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\u003eFrom Training and Racing with a Power Meter by Allen and Coggan:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\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\u003e\\\"A number of equations have been presented in the scientific literature describing human power output as a function of time, some derived from modeling based on the underlying physiology, and some simply derived empirically. One of the simplest and most robust, though, is the original \\\"critical power\\\" concept first proposed by H. Monod around 1960. Various formulations of this idea have been presented, but the original equation is a hyperbolic of the form: t = AWC / (P – CP), where t is time to exhaustion [in seconds], P is current power [in Watts], CP is work rate (i.e., power) asymptote, and AWC is degree of curvature of the relationship.\\\"\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\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\u003eYou will be given values for AWC and CP. Write a function to return the time that the cyclist can maintain for an array of power (P) values. The times should be rounded to the nearest second. If P \u0026lt;= CP, the cyclist can theoretically maintain that power indefinitely (Inf).\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":3065,"title":"Cycling — Critical Power","description":"From Training and Racing with a Power Meter by Allen and Coggan:\r\n\r\n\"A number of equations have been presented in the scientific literature describing human power output as a function of time, some derived from modeling based on the underlying physiology, and some simply derived empirically. One of the simplest and most robust, though, is the original \"critical power\" concept first proposed by H. Monod around 1960. Various formulations of this idea have been presented, but the original equation is a hyperbolic of the form: t = AWC / (P – CP), where t is time to exhaustion [in seconds], P is current power [in Watts], CP is work rate (i.e., power) asymptote, and AWC is degree of curvature of the relationship.\"\r\n\r\nYou will be given values for AWC and CP. Write a function to return the time that the cyclist can maintain for an array of power (P) values. The times should be rounded to the nearest second. If P \u003c= CP, the cyclist can theoretically maintain that power indefinitely (Inf).","description_html":"\u003cp\u003eFrom Training and Racing with a Power Meter by Allen and Coggan:\u003c/p\u003e\u003cp\u003e\"A number of equations have been presented in the scientific literature describing human power output as a function of time, some derived from modeling based on the underlying physiology, and some simply derived empirically. One of the simplest and most robust, though, is the original \"critical power\" concept first proposed by H. Monod around 1960. Various formulations of this idea have been presented, but the original equation is a hyperbolic of the form: t = AWC / (P – CP), where t is time to exhaustion [in seconds], P is current power [in Watts], CP is work rate (i.e., power) asymptote, and AWC is degree of curvature of the relationship.\"\u003c/p\u003e\u003cp\u003eYou will be given values for AWC and CP. Write a function to return the time that the cyclist can maintain for an array of power (P) values. The times should be rounded to the nearest second. If P \u0026lt;= CP, the cyclist can theoretically maintain that power indefinitely (Inf).\u003c/p\u003e","function_template":"function [t] = cycling_crit_power(AWC,CP,P)\r\n\r\nt = zeros(size(P));\r\n\r\nend\r\n","test_suite":"%%\r\nAWC = 5e4;\r\nCP = 200;\r\nP = [150 200 225 250 275 300 350 400 500 1000];\r\nt_corr = [Inf,Inf,2000,1000,667,500,333,250,167,63];\r\nassert(isequal(cycling_crit_power(AWC,CP,P),t_corr))\r\n\r\n%%\r\nAWC = 5.3e4;\r\nCP = 222;\r\nP = [150 200 225 250 275 300 350 400 500 1000];\r\nt_corr = [Inf,Inf,17667,1893,1000,679,414,298,191,68];\r\nassert(isequal(cycling_crit_power(AWC,CP,P),t_corr))\r\n\r\n%%\r\nAWC = 4.6e4;\r\nCP = 250;\r\nP = [150 200 225 250 275 300 350 400 500 1000];\r\nt_corr = [Inf,Inf,Inf,Inf,1840,920,460,307,184,61];\r\nassert(isequal(cycling_crit_power(AWC,CP,P),t_corr))\r\n\r\n%%\r\nAWC = 5e4;\r\nCP = 300;\r\nP = 250:50:1500;\r\nt_corr = [Inf,Inf,1000,500,333,250,200,167,143,125,111,100,91,83,77,71,67,63,59,56,53,50,48,45,43,42];\r\nassert(isequal(cycling_crit_power(AWC,CP,P),t_corr))\r\n\r\n%%\r\nind = randi(4);\r\nswitch ind\r\n\tcase 1\r\n\t\tAWC = 5e4;\r\n\t\tCP = 200;\r\n\t\tP = [150 200 225 250 275 300 350 400 500 1000];\r\n\t\tt_corr = [Inf,Inf,2000,1000,667,500,333,250,167,63];\r\n\tcase 2\r\n\t\tAWC = 5.3e4;\r\n\t\tCP = 222;\r\n\t\tP = [150 200 225 250 275 300 350 400 500 1000];\r\n\t\tt_corr = [Inf,Inf,17667,1893,1000,679,414,298,191,68];\r\n\tcase 3\r\n\t\tAWC = 4.6e4;\r\n\t\tCP = 250;\r\n\t\tP = [150 200 225 250 275 300 350 400 500 1000];\r\n\t\tt_corr = [Inf,Inf,Inf,Inf,1840,920,460,307,184,61];\r\n\tcase 4\r\n\t\tAWC = 5e4;\r\n\t\tCP = 300;\r\n\t\tP = 250:50:1500;\r\n\t\tt_corr = [Inf,Inf,1000,500,333,250,200,167,143,125,111,100,91,83,77,71,67,63,59,56,53,50,48,45,43,42];\r\nend\r\nassert(isequal(cycling_crit_power(AWC,CP,P),t_corr))\r\n","published":true,"deleted":false,"likes_count":2,"comments_count":0,"created_by":26769,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":43,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2015-03-05T04:31:29.000Z","updated_at":"2026-02-16T11:45:09.000Z","published_at":"2015-03-05T04:31:29.000Z","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\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"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\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\u003eFrom Training and Racing with a Power Meter by Allen and Coggan:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\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\u003e\\\"A number of equations have been presented in the scientific literature describing human power output as a function of time, some derived from modeling based on the underlying physiology, and some simply derived empirically. One of the simplest and most robust, though, is the original \\\"critical power\\\" concept first proposed by H. Monod around 1960. Various formulations of this idea have been presented, but the original equation is a hyperbolic of the form: t = AWC / (P – CP), where t is time to exhaustion [in seconds], P is current power [in Watts], CP is work rate (i.e., power) asymptote, and AWC is degree of curvature of the relationship.\\\"\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\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\u003eYou will be given values for AWC and CP. Write a function to return the time that the cyclist can maintain for an array of power (P) values. The times should be rounded to the nearest second. If P \u0026lt;= CP, the cyclist can theoretically maintain that power indefinitely (Inf).\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:\"threshold\"","current_player_id":null,"fields":[{"name":"page","type":"integer","callback":null,"default":1,"directive":null,"facet":null,"facet_method":"and","operator":null,"param":null,"static":null,"prepend":true},{"name":"per_page","type":"integer","callback":null,"default":50,"directive":null,"facet":null,"facet_method":"and","operator":null,"param":null,"static":null,"prepend":true},{"name":"sort","type":"string","callback":null,"default":null,"directive":null,"facet":null,"facet_method":"and","operator":null,"param":null,"static":null,"prepend":true},{"name":"body","type":"text","callback":null,"default":"*:*","directive":null,"facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":false},{"name":"group","type":"string","callback":null,"default":null,"directive":"group","facet":true,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"difficulty_rating_bin","type":"string","callback":null,"default":null,"directive":"difficulty_rating_bin","facet":true,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"id","type":"integer","callback":null,"default":null,"directive":"id","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"tag","type":"string","callback":null,"default":null,"directive":"tag","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"product","type":"string","callback":null,"default":null,"directive":"product","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"created_at","type":"timeframe","callback":{},"default":null,"directive":"created_at","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"profile_id","type":"integer","callback":null,"default":null,"directive":"author_id","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"created_by","type":"string","callback":null,"default":null,"directive":"author","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"player_id","type":"integer","callback":null,"default":null,"directive":"solver_id","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"player","type":"string","callback":null,"default":null,"directive":"solver","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"solvers_count","type":"integer","callback":null,"default":null,"directive":"solvers_count","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"comments_count","type":"integer","callback":null,"default":null,"directive":"comments_count","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"likes_count","type":"integer","callback":null,"default":null,"directive":"likes_count","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"leader_id","type":"integer","callback":null,"default":null,"directive":"leader_id","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"leading_solution","type":"integer","callback":null,"default":null,"directive":"leading_solution","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true}],"filters":[{"name":"asset_type","type":"string","callback":null,"default":null,"directive":null,"facet":null,"facet_method":"and","operator":null,"param":null,"static":"\"cody:problem\"","prepend":true},{"name":"profile_id","type":"integer","callback":{},"default":null,"directive":null,"facet":null,"facet_method":"and","operator":null,"param":"author_id","static":null,"prepend":true}],"query":{"params":{"per_page":50,"term":"tag:\"threshold\"","current_player":null,"sort":"map(difficulty_value,0,0,999) asc"},"parser":"MathWorks::Search::Solr::QueryParser","directives":{"term":{"directives":{"tag":[["tag:\"threshold\"","","\"","threshold","\""]]}}},"facets":{"#\u003cMathWorks::Search::Field:0x00007f10287e44d0\u003e":null,"#\u003cMathWorks::Search::Field:0x00007f10287e4430\u003e":null},"filters":{"#\u003cMathWorks::Search::Field:0x00007f10287e3b70\u003e":"\"cody:problem\""},"fields":{"#\u003cMathWorks::Search::Field:0x00007f10287e4750\u003e":1,"#\u003cMathWorks::Search::Field:0x00007f10287e46b0\u003e":50,"#\u003cMathWorks::Search::Field:0x00007f10287e4610\u003e":"map(difficulty_value,0,0,999) asc","#\u003cMathWorks::Search::Field:0x00007f10287e4570\u003e":"tag:\"threshold\""},"user_query":{"#\u003cMathWorks::Search::Field:0x00007f10287e4570\u003e":"tag:\"threshold\""},"queried_facets":{}},"query_backend":{"connection":{"configuration":{"index_url":"http://index-op-v2/solr/","query_url":"http://search-op-v2/solr/","direct_access_index_urls":["http://index-op-v2/solr/"],"direct_access_query_urls":["http://search-op-v2/solr/"],"timeout":10,"vhost":"search","exchange":"search.topic","heartbeat":30,"pre_index_mode":false,"host":"rabbitmq-eks","port":5672,"username":"search","password":"J3bGPZzQ7asjJcCk","virtual_host":"search","indexer":"amqp","http_logging":"true","core":"cody"},"query_connection":{"uri":"http://search-op-v2/solr/cody/","proxy":null,"connection":{"parallel_manager":null,"headers":{"User-Agent":"Faraday v1.0.1"},"params":{},"options":{"params_encoder":"Faraday::FlatParamsEncoder","proxy":null,"bind":null,"timeout":null,"open_timeout":null,"read_timeout":null,"write_timeout":null,"boundary":null,"oauth":null,"context":null,"on_data":null},"ssl":{"verify":true,"ca_file":null,"ca_path":null,"verify_mode":null,"cert_store":null,"client_cert":null,"client_key":null,"certificate":null,"private_key":null,"verify_depth":null,"version":null,"min_version":null,"max_version":null},"default_parallel_manager":null,"builder":{"adapter":{"name":"Faraday::Adapter::NetHttp","args":[],"block":null},"handlers":[{"name":"Faraday::Response::RaiseError","args":[],"block":null}],"app":{"app":{"ssl_cert_store":{"verify_callback":null,"error":null,"error_string":null,"chain":null,"time":null},"app":{},"connection_options":{},"config_block":null}}},"url_prefix":"http://search-op-v2/solr/cody/","manual_proxy":false,"proxy":null},"update_format":"RSolr::JSON::Generator","update_path":"update","options":{"url":"http://search-op-v2/solr/cody"}}},"query":{"params":{"per_page":50,"term":"tag:\"threshold\"","current_player":null,"sort":"map(difficulty_value,0,0,999) asc"},"parser":"MathWorks::Search::Solr::QueryParser","directives":{"term":{"directives":{"tag":[["tag:\"threshold\"","","\"","threshold","\""]]}}},"facets":{"#\u003cMathWorks::Search::Field:0x00007f10287e44d0\u003e":null,"#\u003cMathWorks::Search::Field:0x00007f10287e4430\u003e":null},"filters":{"#\u003cMathWorks::Search::Field:0x00007f10287e3b70\u003e":"\"cody:problem\""},"fields":{"#\u003cMathWorks::Search::Field:0x00007f10287e4750\u003e":1,"#\u003cMathWorks::Search::Field:0x00007f10287e46b0\u003e":50,"#\u003cMathWorks::Search::Field:0x00007f10287e4610\u003e":"map(difficulty_value,0,0,999) asc","#\u003cMathWorks::Search::Field:0x00007f10287e4570\u003e":"tag:\"threshold\""},"user_query":{"#\u003cMathWorks::Search::Field:0x00007f10287e4570\u003e":"tag:\"threshold\""},"queried_facets":{}},"options":{"fields":["id","difficulty_rating"]},"join":" "},"results":[{"id":3065,"difficulty_rating":"easy"}]}}