{"id":25815,"date":"2022-12-13T14:08:28","date_gmt":"2022-12-13T08:38:28","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-prolog-define-predicates-closed\/"},"modified":"2022-12-13T14:08:28","modified_gmt":"2022-12-13T08:38:28","slug":"solved-prolog-define-predicates-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-prolog-define-predicates-closed\/","title":{"rendered":"[Solved] Prolog Define Predicates [closed]"},"content":{"rendered":"<p> [ad_1]<br \/>\n<\/p>\n<div id=\"answer-49816653\" class=\"answer js-answer downvoted-answer accepted-answer js-accepted-answer\" data-answerid=\"49816653\" data-parentid=\"49814655\" data-score=\"-5\" data-position-on-page=\"1\" data-highest-scored=\"1\" data-question-has-accepted-highest-score=\"1\" itemprop=\"acceptedAnswer\" itemscope itemtype=\"https:\/\/schema.org\/Answer\">\n<div class=\"post-layout\">\n<div class=\"votecell post-layout--left\"><\/div>\n<div class=\"answercell post-layout--right\">\n<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Try this one \ud83d\ude42 The main idea is to accumulate number of hours in the predicate min by subtracting 60 from total time and when the total time is lower than 60 we can unify counted hours from accumulator with H variable and rest of time with M variable.<\/p>\n<pre><code>minutes(X,H,M):-                     %main predicate\n    min(X,0,H,M).\n\nmin(X,H,ResultH,ResultM):-           %ending condition of recursion\n    X&lt;60,\n    ResultH is H,                    %unification with counted hours\n    ResultM is X.\n\nmin(X,H,ResultM,ResultH):-           %recursive predicate\n    X &gt;= 60,\n    X2 is X-60, H2 is H+1,           %subtracting from total time\n    min(X2,H2,ResultM,ResultH).\n<\/code><\/pre>\n<\/p><\/div>\n<div class=\"mt24\"><\/div>\n<\/div>\n<p>            <span class=\"d-none\" itemprop=\"commentCount\">2<\/span> <\/p><\/div>\n<\/div>\n<p>[ad_2]<\/p>\n<p>solved Prolog Define Predicates [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"<p>[ad_1] Try this one \ud83d\ude42 The main idea is to accumulate number of hours in the predicate min by subtracting 60 from total time and when the total time is lower than 60 we can unify counted hours from accumulator with H variable and rest of time with M variable. minutes(X,H,M):- %main predicate min(X,0,H,M). min(X,H,ResultH,ResultM):- &#8230; <a title=\"[Solved] Prolog Define Predicates [closed]\" class=\"read-more\" href=\"https:\/\/jassweb.com\/solved\/solved-prolog-define-predicates-closed\/\" aria-label=\"More on [Solved] Prolog Define Predicates [closed]\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[2171],"class_list":["post-25815","post","type-post","status-publish","format-standard","hentry","category-solved","tag-prolog"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>[Solved] Prolog Define Predicates [closed] - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-prolog-define-predicates-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Prolog Define Predicates [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Try this one \ud83d\ude42 The main idea is to accumulate number of hours in the predicate min by subtracting 60 from total time and when the total time is lower than 60 we can unify counted hours from accumulator with H variable and rest of time with M variable. minutes(X,H,M):- %main predicate min(X,0,H,M). min(X,H,ResultH,ResultM):- ... 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