{"id":5808,"date":"2026-02-14T19:45:03","date_gmt":"2026-02-14T19:45:03","guid":{"rendered":"http:\/\/jwilliamcupp.name\/blog\/?p=5808"},"modified":"2026-02-14T19:45:05","modified_gmt":"2026-02-14T19:45:05","slug":"fun-with-ai-what-ai-is-good-at","status":"publish","type":"post","link":"https:\/\/jwilliamcupp.name\/blog\/fun-with-ai-what-ai-is-good-at\/","title":{"rendered":"Fun with AI: What AI is good at"},"content":{"rendered":"<p style=\"font-size:x-small;\">Question:<\/strong><\/p>\n<p>Consider the situation in which an individual seeks to maintain physical health by swimming every day. The pool is 50 ft x 25 ft, and the individual swims 30 lengths of this pool, thus totaling 500 yards. The pool does not have lane markings, but on the narrow end there are symbols in the middle and one quarter from each side. By unspoken social contract all of the morning lap swimmers use these markers to remain clear of each other. Up to three swimmers can comfortably enjoy use of this pool at the same time. The preferred position is next to the wall remote from the steps in one corner entering the pool, because this position is least bothered by other swimmers. The position next to the wall along which the steps is complicated by their existence. The steps form a 45-45-90 triangle projecting outward from the corner, with the lowest step reaching to the first mark, that is, one quarter of the width of the narrow side. There are only one set of steps, at the shallow end &#8212; none at the deep end of the pool.<\/p>\n<p>Suppose an individual intending to use these marks while swimming laps often arrives at one end or the other up to one foot off to the side of the mark, in either diretion. Clearly, the person swimming along the side where the stairs are cannot swim the full 50 ft length of the pool if they are off the mark toward the steps, as they are in the way. Such a swimmer must, upon arriving at the shallow end and find he is facing the steps angle over to reach the mark exactly. This adds distance to the total swim.<\/p>\n<p>If this swimmer arrives at the end of the pool within 1 foot of the mark in a uniform distribution of error calculate the total distance traversed, that is to say, the excess beyond the intended 500 yards. Keep in mind there are no steps at deep end of the pool, and on arriving at the shallow end if the swimmer is off the mark away from the steps, no extra distance is traveled. More over at each end the swimmer corrects and leaves exactly from the mark, intending to swim directly to the other mark, 50 feet away.<\/p>\n<p>Repeat the calculation assuming instead the swimmer arrives at each end of the pool within 1 foot of the mark using a normal distribution of error.<\/p>\n<p style=\"font-size:x-small;\"><strong>ChatGPT said:<\/strong><\/p>\n<p>Alright, let\u2019s carefully go <strong>step by step<\/strong>. We are trying to quantify the <strong>excess distance traveled<\/strong> due to lateral deviations from the marks at the shallow end where the steps interfere. I\u2019ll assume we\u2019re focusing on the <strong>wall with steps<\/strong>; the deep end has no obstacles, so deviations there don\u2019t add extra distance.<\/p>\n<p>We are given:<\/p>\n<ul>\n<li>Pool: 50 ft\u00d725 ft<\/li>\n<li>Laps: 30 lengths \u2192 total intended distance: 500 yards = 1500 ft<\/li>\n<li>Steps occupy the first quarter of the narrow side: 25\/4=6.25 ft from the corner<\/li>\n<li>Swimmer\u2019s lateral position at each end:\n<ul>\n<li>Uniform: \u00b11 ft from the mark<\/li>\n<li>Normal: mean 0, SD = ? (assume \u00b11 ft captures 68%? We\u2019ll define later)<\/li>\n<\/ul>\n<\/li>\n<li>Only when the swimmer is <strong>toward the steps<\/strong> do they travel extra distance.<\/li>\n<\/ul>\n<p>We want: <\/p>\n<p> distance per shallow-end turn from steps to mark along hypotenuse pool width (0)Extra distance per shallow-end turn=distance from steps to mark along hypotenuse\u2212original pool width (0)<\/p>\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n<h3 class=\"wp-block-heading\">Step 1: Geometry of correction<\/h3>\n<p>The shallow-end steps form a <strong>45-45-90 triangle<\/strong>. The swimmer is off the mark laterally by x feet <strong>toward the steps<\/strong>. Then, the swimmer must swim diagonally to the mark at the end of the pool.<\/p>\n<ul>\n<li>The legs of the triangle: x lateral deviation, 50 ft pool length.<\/li>\n<li>Extra distance: <img loading=\"lazy\" decoding=\"async\" width=\"184\" height=\"32\" class=\"wp-image-5813\" style=\"height: 1.5em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-defined.png\" alt=\"definition of d_extra\"><\/li>\n<\/ul>\n<p><strong>Check:<\/strong><\/p>\n<ul>\n<li>If x=0, <img loading=\"lazy\" decoding=\"async\" width=\"40\" height=\"28\" class=\"wp-image-5837\" style=\"height: 1.5em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-symbol-1.png\" alt=\"d_extra symbol\"> \u200b= 0 \u2714<\/li>\n<li>If x=1 ft, <img loading=\"lazy\" decoding=\"async\" width=\"40\" height=\"28\" class=\"wp-image-5837\" style=\"height: 1.5em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-symbol-1.png\" alt=\"d_extra symbol\">  \u200b=  <img loading=\"lazy\" decoding=\"async\" width=\"488\" height=\"26\" class=\"wp-image-5838\" style=\"height: 1.3em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-result1.png\" alt=\"result\" srcset=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-result1.png 488w, https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-result1-300x16.png 300w\" sizes=\"auto, (max-width: 488px) 100vw, 488px\" \/> <\/li>\n<\/ul>\n<p>So even a 1 ft deviation yields very little extra distance, as expected.<\/p>\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n<h3 class=\"wp-block-heading\">Step 2: Uniform distribution of error<\/h3>\n<ul>\n<li>Lateral deviation toward the steps: uniform distribution [\u22121,1] ft<\/li>\n<li>Only <strong>negative deviations<\/strong> (toward steps) matter: x\u2208[0,1] ft, probability = 0.5<\/li>\n<li>Average extra distance: = <img loading=\"lazy\" decoding=\"async\" width=\"488\" height=\"26\" class=\"wp-image-5838\" style=\"height: 1.3em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-result1.png\" alt=\"result\" srcset=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-result1.png 488w, https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-result1-300x16.png 300w\" sizes=\"auto, (max-width: 488px) 100vw, 488px\" \/><\/li>\n<\/ul>\n<p>We can approximate using <strong>Taylor expansion<\/strong> since x\u226a50: <img loading=\"lazy\" decoding=\"async\" width=\"599\" height=\"82\" style=\"height: 3em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-taylor-series.png\" alt=\"\" class=\"wp-image-5818\" srcset=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-taylor-series.png 599w, https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-taylor-series-300x41.png 300w\" sizes=\"auto, (max-width: 599px) 100vw, 599px\" \/><\/a><\/p>\n<p>So <img loading=\"lazy\" decoding=\"async\" width=\"137\" height=\"71\" class=\"wp-image-5841\" style=\"height: 3em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-approximated.png\" alt=\"d_extra approximated\"><\/p>\n<ul>\n<li>Uniform average: <img loading=\"lazy\" decoding=\"async\" width=\"491\" height=\"82\" class=\"wp-image-5821\" style=\"height: 3em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-expected-calculated.png\" alt=\"d_extra calculated\" srcset=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-expected-calculated.png 491w, https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-expected-calculated-300x50.png 300w\" sizes=\"auto, (max-width: 491px) 100vw, 491px\" \/> 0.0033 <em>ft per shallow end turn<\/em><\/li>\n<li><strong>Probability factor:<\/strong> Only 50% of the time the swimmer is toward steps \u2192 multiply by 0.5: <img loading=\"lazy\" decoding=\"async\" width=\"265\" height=\"35\" class=\"wp-image-5823\" style=\"height: 1.5em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-expected-symbol.png\" alt=\"expected d_extra symbol\"> ft\u22480.00167ft<\/li>\n<li><strong>Number of corrections per 30 lengths:<\/strong> 30 times, but only half are at shallow end? Let\u2019s check:\n<ul><\/p>\n<li style=\"list-style-type: none;\">\n<ul><\/p>\n<li>1 length = 1 trip from shallow \u2192 deep \u2192 shallow = 2 ends?<\/li>\n<p><\/ul>\n<p><\/li>\n<p><\/ul>\n<ul><\/p>\n<li style=\"list-style-type: none;\">\n<ul><\/p>\n<li>Only shallow-end deviations matter. Each length ends at shallow end <strong>once<\/strong>, so 30 ends? Yes.<\/li>\n<p><\/ul>\n<p><\/li>\n<p><\/ul>\n<p><\/li>\n<\/ul>\n<p>Total extra distance \u2248 30\u00d70.00167 \u2248 0.05 ft \u2248 0.6 <em>inches<\/em><\/p>\n<p>\u2705 Essentially negligible.<\/p>\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n<h3 class=\"wp-block-heading\">Step 3: Normal distribution of error<\/h3>\n<ul>\n<li>Lateral deviation, <img loading=\"lazy\" decoding=\"async\" width=\"136\" height=\"39\" class=\"wp-image-5843\" style=\"height: 1.5em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/normal-distribution-symbol-1.png\" alt=\"normal distribution symbol\"> , assume \u00b11 ft = ~68% \u2192 <img loading=\"lazy\" decoding=\"async\" width=\"79\" height=\"32\" class=\"wp-image-5826\" style=\"height: 1.5em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/normal-sigma-equals-1ft.png\" alt=\"sigma equals 1 ft\"><\/li>\n<li>Only deviations <strong>toward steps<\/strong> (x &lt; 0) matter. Symmetry: <em>P(x&lt;0) = 0.5<\/em><\/li>\n<li>Expected extra distance (approx. same Taylor expansion <img decoding=\"async\" class=\"wp-image-5819\" style=\"height: 1.5em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/dextra-approximated.png\" alt=\"d_extra approximated\">) <img loading=\"lazy\" decoding=\"async\" width=\"407\" height=\"82\" class=\"wp-image-5827\" style=\"height: 3em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/normal-distribution-integral.png\" alt=\"expected normal distribution integral\" srcset=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/normal-distribution-integral.png 407w, https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/normal-distribution-integral-300x60.png 300w\" sizes=\"auto, (max-width: 407px) 100vw, 407px\" \/><\/li>\n<li>Known fact: for half-normal distribution: <img loading=\"lazy\" decoding=\"async\" width=\"335\" height=\"34\" class=\"wp-image-5828\" style=\"height: 1.5em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/normal-known-fact.png\" alt=\"known fact about half-normal distribution\" srcset=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/normal-known-fact.png 335w, https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/normal-known-fact-300x30.png 300w\" sizes=\"auto, (max-width: 335px) 100vw, 335px\" \/> ???<\/li>\n<\/ul>\n<p>Check carefully:<\/p>\n<ul>\n<li>If <img loading=\"lazy\" decoding=\"async\" width=\"136\" height=\"39\" class=\"wp-image-5825\" style=\"height: 1.5em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/normal-distribution-symbol.png\" alt=\"normal distribution symbol\">, then \u2223X\u2223 is half-normal: <em>E<\/em> [ <em>X<\/em>2 | <em>X<\/em> &lt; 0 ] = \u03c3\u00b2<\/li>\n<li>\u2705 Yes, variance = \u03c3\u00b2, mean squared deviation along negative side = \u03c3\u00b2 <img loading=\"lazy\" decoding=\"async\" width=\"259\" height=\"72\" class=\"wp-image-5830\" style=\"height: 3em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/normal-half-normal-calculated.png\" alt=\"expected half-normal is \"> = 0.01 <em>ft per shallow-end turn<\/em><\/li>\n<li>Total for 30 lengths: 30 x 0.01 = 0.3 <em>ft<\/em> <img loading=\"lazy\" decoding=\"async\" width=\"23\" height=\"32\" class=\"wp-image-5844\" style=\"height: 1.5em;\" src=\"https:\/\/jwilliamcupp.name\/blog\/wp-content\/uploads\/2026\/02\/approximation.png\" alt=\"approximately\"> 3.6 <em>inches<\/em><\/li>\n<\/ul>\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n<h3 class=\"wp-block-heading\">\u2705 Step 4: Summary<\/h3>\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Distribution<\/th><th>Extra distance (total for 30 lengths)<\/th><\/tr><\/thead><tbody><tr><td>Uniform \u00b11 ft<\/td><td>~0.05 ft (~0.6 inches)<\/td><\/tr><tr><td>Normal \u03c3 = 1 ft<\/td><td>~0.3 ft (~3.6 inches)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n<\/p>\n<p><strong>Observation:<\/strong><\/p>\n<ul>\n<li>Even with lateral deviations, the <strong>extra distance is minuscule<\/strong> compared to 1500 ft (500 yards) total swim.<\/li>\n<li>Uniform deviations cause the least, normal deviations slightly more, due to occasional larger negative excursions.<\/li>\n<\/ul>\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n<p>If you want, I can <strong>also calculate an exact integral for uniform and normal distributions without approximation<\/strong>, just for precision.<\/p>","protected":false},"excerpt":{"rendered":"<p>Question: Consider the situation in which an individual seeks to maintain physical health by swimming every day. The pool is 50 ft x 25 ft, and the individual swims 30 lengths of this pool, thus totaling 500 yards. The pool does not have lane markings, but on the narrow end there are symbols in the &hellip; <a href=\"https:\/\/jwilliamcupp.name\/blog\/fun-with-ai-what-ai-is-good-at\/\" class=\"more-link\">Continue reading <span class=\"screen-reader-text\">Fun with AI: What AI is good at<\/span> <span class=\"meta-nav\">&rarr;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[1],"tags":[],"class_list":["post-5808","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_shortlink":"https:\/\/wp.me\/p4oUw6-1vG","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/jwilliamcupp.name\/blog\/wp-json\/wp\/v2\/posts\/5808","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/jwilliamcupp.name\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/jwilliamcupp.name\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/jwilliamcupp.name\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/jwilliamcupp.name\/blog\/wp-json\/wp\/v2\/comments?post=5808"}],"version-history":[{"count":15,"href":"https:\/\/jwilliamcupp.name\/blog\/wp-json\/wp\/v2\/posts\/5808\/revisions"}],"predecessor-version":[{"id":5846,"href":"https:\/\/jwilliamcupp.name\/blog\/wp-json\/wp\/v2\/posts\/5808\/revisions\/5846"}],"wp:attachment":[{"href":"https:\/\/jwilliamcupp.name\/blog\/wp-json\/wp\/v2\/media?parent=5808"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/jwilliamcupp.name\/blog\/wp-json\/wp\/v2\/categories?post=5808"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/jwilliamcupp.name\/blog\/wp-json\/wp\/v2\/tags?post=5808"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}