{"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":44708,"title":"Sherlock Holmes: The Spy Problem","description":"A guest at a party is a spy if this person is not known by any other guest, but knows all of them. There is at most one spy at a party, for if there were two spies, they would not know each other. A particular party may have no spy. To find the spy, if one exists, at a party, you are allowed to ask only one type of question—asking a guest whether they know a second guest. It is assumed that everyone answers your question truthfully.\r\n\r\nIf _n_ guests attend the party, what is the maximum number of questions required to solve the problem for the  *worst case*?\r\n\r\n*THEORY:* The solution can be obtained by using the decrease-by-one algorithm to initially reduce the group of possible spy. If _n_ = 1, that one person is a spy by definition. If _n_  \u003e 2, select two guests, say Alice and Bob. Ask Alice whether she knows Bob. If Alice does not know Bob, remove Alice from the group of possible spy. Otherwise, remove Bob from this group. Solve the problem recursively for the remaining group of _n - 1_ guests who can still be spy until only one guest is left in the group. Finally, in the *worst case*, you will then need to verify from all the guest at the party if the person left in the group is actually a spy.","description_html":"\u003cp\u003eA guest at a party is a spy if this person is not known by any other guest, but knows all of them. There is at most one spy at a party, for if there were two spies, they would not know each other. A particular party may have no spy. To find the spy, if one exists, at a party, you are allowed to ask only one type of question—asking a guest whether they know a second guest. It is assumed that everyone answers your question truthfully.\u003c/p\u003e\u003cp\u003eIf \u003ci\u003en\u003c/i\u003e guests attend the party, what is the maximum number of questions required to solve the problem for the  \u003cb\u003eworst case\u003c/b\u003e?\u003c/p\u003e\u003cp\u003e\u003cb\u003eTHEORY:\u003c/b\u003e The solution can be obtained by using the decrease-by-one algorithm to initially reduce the group of possible spy. If \u003ci\u003en\u003c/i\u003e = 1, that one person is a spy by definition. If \u003ci\u003en\u003c/i\u003e  \u0026gt; 2, select two guests, say Alice and Bob. Ask Alice whether she knows Bob. If Alice does not know Bob, remove Alice from the group of possible spy. Otherwise, remove Bob from this group. Solve the problem recursively for the remaining group of \u003ci\u003en - 1\u003c/i\u003e guests who can still be spy until only one guest is left in the group. Finally, in the \u003cb\u003eworst case\u003c/b\u003e, you will then need to verify from all the guest at the party if the person left in the group is actually a spy.\u003c/p\u003e","function_template":"function q = Spy(x)\r\n  y = x;\r\nend","test_suite":"%%\r\nassert(isequal(Spy(0),0))\r\n\r\n%%\r\nassert(isequal(Spy(1),0))\r\n\r\n%%\r\nassert(isequal(Spy(666),1994))\r\n\r\n%%\r\nassert(isequal(Spy(3),5))\r\n\r\n%%\r\nassert(isequal(Spy(1234),3698))\r\n\r\n%%\r\nassert(isequal(Spy(59),173))\r\n\r\n%%\r\nassert(isequal(Spy(17),47))\r\n\r\n%%\r\nassert(isequal(Spy(101),299))\r\n\r\n%%\r\nassert(isequal(Spy(555),1661))\r\n\r\n%%\r\nassert(isequal(Spy(2),2))\r\n\r\n%%\r\nassert(isequal(Spy(200),596))","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":178544,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":19,"test_suite_updated_at":"2018-08-14T13:44:04.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2018-08-03T10:14:10.000Z","updated_at":"2026-01-29T15:46:12.000Z","published_at":"2018-08-03T10:14:51.000Z","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\u003eA guest at a party is a spy if this person is not known by any other guest, but knows all of them. There is at most one spy at a party, for if there were two spies, they would not know each other. A particular party may have no spy. To find the spy, if one exists, at a party, you are allowed to ask only one type of question—asking a guest whether they know a second guest. It is assumed that everyone answers your question truthfully.\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\u003eIf\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003en\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e guests attend the party, what is the maximum number of questions required to solve the problem for the \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003eworst case\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\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:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003eTHEORY:\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e The solution can be obtained by using the decrease-by-one algorithm to initially reduce the group of possible spy. If\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003en\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e = 1, that one person is a spy by definition. If\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003en\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u0026gt; 2, select two guests, say Alice and Bob. Ask Alice whether she knows Bob. If Alice does not know Bob, remove Alice from the group of possible spy. Otherwise, remove Bob from this group. Solve the problem recursively for the remaining group of\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003en - 1\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e guests who can still be spy until only one guest is left in the group. Finally, in the\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003eworst case\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e, you will then need to verify from all the guest at the party if the person left in the group is actually a spy.\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":44708,"title":"Sherlock Holmes: The Spy Problem","description":"A guest at a party is a spy if this person is not known by any other guest, but knows all of them. There is at most one spy at a party, for if there were two spies, they would not know each other. A particular party may have no spy. To find the spy, if one exists, at a party, you are allowed to ask only one type of question—asking a guest whether they know a second guest. It is assumed that everyone answers your question truthfully.\r\n\r\nIf _n_ guests attend the party, what is the maximum number of questions required to solve the problem for the  *worst case*?\r\n\r\n*THEORY:* The solution can be obtained by using the decrease-by-one algorithm to initially reduce the group of possible spy. If _n_ = 1, that one person is a spy by definition. If _n_  \u003e 2, select two guests, say Alice and Bob. Ask Alice whether she knows Bob. If Alice does not know Bob, remove Alice from the group of possible spy. Otherwise, remove Bob from this group. Solve the problem recursively for the remaining group of _n - 1_ guests who can still be spy until only one guest is left in the group. Finally, in the *worst case*, you will then need to verify from all the guest at the party if the person left in the group is actually a spy.","description_html":"\u003cp\u003eA guest at a party is a spy if this person is not known by any other guest, but knows all of them. There is at most one spy at a party, for if there were two spies, they would not know each other. A particular party may have no spy. To find the spy, if one exists, at a party, you are allowed to ask only one type of question—asking a guest whether they know a second guest. It is assumed that everyone answers your question truthfully.\u003c/p\u003e\u003cp\u003eIf \u003ci\u003en\u003c/i\u003e guests attend the party, what is the maximum number of questions required to solve the problem for the  \u003cb\u003eworst case\u003c/b\u003e?\u003c/p\u003e\u003cp\u003e\u003cb\u003eTHEORY:\u003c/b\u003e The solution can be obtained by using the decrease-by-one algorithm to initially reduce the group of possible spy. If \u003ci\u003en\u003c/i\u003e = 1, that one person is a spy by definition. If \u003ci\u003en\u003c/i\u003e  \u0026gt; 2, select two guests, say Alice and Bob. Ask Alice whether she knows Bob. If Alice does not know Bob, remove Alice from the group of possible spy. Otherwise, remove Bob from this group. Solve the problem recursively for the remaining group of \u003ci\u003en - 1\u003c/i\u003e guests who can still be spy until only one guest is left in the group. Finally, in the \u003cb\u003eworst case\u003c/b\u003e, you will then need to verify from all the guest at the party if the person left in the group is actually a spy.\u003c/p\u003e","function_template":"function q = Spy(x)\r\n  y = x;\r\nend","test_suite":"%%\r\nassert(isequal(Spy(0),0))\r\n\r\n%%\r\nassert(isequal(Spy(1),0))\r\n\r\n%%\r\nassert(isequal(Spy(666),1994))\r\n\r\n%%\r\nassert(isequal(Spy(3),5))\r\n\r\n%%\r\nassert(isequal(Spy(1234),3698))\r\n\r\n%%\r\nassert(isequal(Spy(59),173))\r\n\r\n%%\r\nassert(isequal(Spy(17),47))\r\n\r\n%%\r\nassert(isequal(Spy(101),299))\r\n\r\n%%\r\nassert(isequal(Spy(555),1661))\r\n\r\n%%\r\nassert(isequal(Spy(2),2))\r\n\r\n%%\r\nassert(isequal(Spy(200),596))","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":178544,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":19,"test_suite_updated_at":"2018-08-14T13:44:04.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2018-08-03T10:14:10.000Z","updated_at":"2026-01-29T15:46:12.000Z","published_at":"2018-08-03T10:14:51.000Z","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\u003eA guest at a party is a spy if this person is not known by any other guest, but knows all of them. There is at most one spy at a party, for if there were two spies, they would not know each other. A particular party may have no spy. To find the spy, if one exists, at a party, you are allowed to ask only one type of question—asking a guest whether they know a second guest. It is assumed that everyone answers your question truthfully.\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\u003eIf\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003en\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e guests attend the party, what is the maximum number of questions required to solve the problem for the \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003eworst case\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\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:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003eTHEORY:\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e The solution can be obtained by using the decrease-by-one algorithm to initially reduce the group of possible spy. If\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003en\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e = 1, that one person is a spy by definition. If\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003en\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u0026gt; 2, select two guests, say Alice and Bob. Ask Alice whether she knows Bob. If Alice does not know Bob, remove Alice from the group of possible spy. Otherwise, remove Bob from this group. Solve the problem recursively for the remaining group of\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003en - 1\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e guests who can still be spy until only one guest is left in the group. 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