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<!doctype html><html lang="zh-CN"><meta charset="utf-8"><meta name="viewport" content="width=device-width,initial-scale=1"><title>六边融合:双色与墙 · 设计评审</title><style>body{max-width:920px;margin:24px auto;padding:0 18px;background:#f7f5fb;color:#303047;font:15px/1.7 system-ui,'Microsoft YaHei'}article{background:white;border:1px solid #ddd5ed;border-radius:16px;padding:20px;margin:20px 0;break-inside:avoid}.boards{display:flex;gap:24px;flex-wrap:wrap}figure{margin:0;width:380px;max-width:100%}svg{max-height:450px;width:100%}details{cursor:pointer}a{color:#6552aa}@media print{body{background:white}article{border:none;page-break-after:always}} </style><a href="play.html">进入试玩</a><h1>六边融合:双色与墙 · 8关设计评审</h1><p>六向共边相邻,每格一步。蓝格 → 橙格,每步换颜色;黑黄永久墙另行封住共边。同色或有墙都不能通过。○是蓝格,△是橙格。</p><p>原始盘面和作者解分开展示。标题是设计任务,真人推理路径与难度仍需试玩验证。数字、窗口、方向、类别和开关显示在格内;步差与先后关系见逐关推理依据。</p><article><h2>01 · 入门 · 六向双色交替,再看一堵墙</h2><div class="boards"><figure><svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 276.71361036774664 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fill="#665c78">c5</text></svg><figcaption>原始盘面 · 格id用于对照JSON</figcaption></figure><details><summary>展开作者解与路线</summary><figure><svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 276.71361036774664 450.00000000000006" role="img" aria-label="六向双色交替,再看一堵墙"><polygon points="155.26492021014093,78.57142857142858 121.4486901576057,137.14285714285717 53.816230052535246,137.14285714285717 20,78.57142857142858 53.816230052535204,20.000000000000007 121.4486901576057,20" fill="#64acf3" stroke="#aaa1c7" stroke-width="2"/><polygon points="256.71361036774664,137.14285714285717 222.8973803152114,195.71428571428575 155.26492021014093,195.71428571428575 121.4486901576057,137.14285714285717 155.2649202101409,78.57142857142858 222.8973803152114,78.57142857142858" fill="#ffae55" stroke="#aaa1c7" stroke-width="2"/><polygon points="155.26492021014093,195.71428571428575 121.4486901576057,254.28571428571433 53.816230052535246,254.28571428571433 20,195.71428571428575 53.816230052535204,137.14285714285717 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font-size="10" fill="#665c78">c7 · △</text></svg></figure><p>c6 → c7 → c5 → c4 → c2 → c0 → c1 → c3</p></details></div><h3>推理依据</h3><ul><li>○/△每步交替,按节点数量推断端点类别</li><li>永久墙 c2—c3 断开表面邻接,需绕行</li></ul><p>关键分叉:第 4 步应走 c4,诱惑路线走 c3。排除依据:蓝橙颜色必须交替。</p><details><summary>逐项机制必要性反例</summary><p>解除 alternation 后:c6 → c7 → c5 → c3 → c4 → c2 → c1 → c0</p><p>解除 wall:0 后:c6 → c7 → c5 → c4 → c2 → c3 → c1 → c0</p></details></article><article><h2>03 · 普通 · 颜色数量与墙共同决定端点</h2><div class="boards"><figure><svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 178.0829393811855 450" role="img" aria-label="颜色数量与墙共同决定端点"><polygon points="98.90453678924885,54.166666666666664 79.17840259193665,88.33333333333333 39.72613419731222,88.33333333333333 20,54.166666666666664 39.72613419731219,20.000000000000004 79.17840259193665,20" fill="#64acf3" stroke="#aaa1c7" stroke-width="2"/><polygon points="158.0829393811855,88.33333333333333 138.3568051838733,122.5 98.90453678924885,122.5 79.17840259193665,88.33333333333333 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→ c5 → c6 → c7 → c9 → c10 → c8</p></details></div><h3>推理依据</h3><ul><li>○/△每步交替,按节点数量推断端点类别</li><li>永久墙 c7—c8 断开表面邻接,需绕行</li></ul><p>关键分叉:第 10 步应走 c10,诱惑路线走 c8。排除依据:蓝橙颜色必须交替。</p><details><summary>逐项机制必要性反例</summary><p>解除 alternation 后:c0 → c1 → c2 → c3 → c4 → c5 → c6 → c7 → c9 → c8 → c10</p><p>解除 wall:0 后:c0 → c1 → c2 → c3 → c4 → c5 → c6 → c7 → c8 → c10 → c9</p></details></article><article><h2>04 · 普通 · 两堵墙配合六向出口选择</h2><div class="boards"><figure><svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 222.08739259057427 450" role="img" aria-label="两堵墙配合六向出口选择"><polygon points="92.83495703622971,51.53846153846153 74.62621777717229,83.07692307692307 38.20873925905744,83.07692307692307 20,51.53846153846154 38.20873925905741,20.000000000000004 74.62621777717229,20" fill="#ffae55" stroke="#aaa1c7" stroke-width="2"/><polygon points="147.461174813402,83.07692307692307 129.25243555434457,114.61538461538461 92.83495703622971,114.61538461538461 74.62621777717229,83.07692307692307 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断开表面邻接,需绕行</li><li>永久墙 c6—c8 断开表面邻接,需绕行</li></ul><p>关键分叉:第 9 步应走 c9,诱惑路线走 c8。排除依据:蓝橙颜色必须交替。</p><details><summary>逐项机制必要性反例</summary><p>解除 alternation 后:c5 → c3 → c1 → c0 → c2 → c4 → c6 → c7 → c8 → c10 → c9 → c11</p><p>解除 wall:0 后:c5 → c3 → c1 → c0 → c2 → c4 → c6 → c7 → c9 → c8 → c10 → c11</p><p>解除 wall:1 后:c5 → c3 → c1 → c0 → c2 → c4 → c6 → c8 → c10 → c11 → c9 → c7</p></details></article><article><h2>05 · 进阶 · 区域绕行同时满足颜色和墙</h2><div class="boards"><figure><svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 255.19419124340598 450" role="img" aria-label="区域绕行同时满足颜色和墙"><polygon points="106.07767649736239,57.27272727272727 84.55825737302179,94.54545454545455 41.51941912434061,94.54545454545455 20,57.27272727272727 41.51941912434058,20.000000000000004 84.55825737302179,20" fill="#ffae55" stroke="#aaa1c7" stroke-width="2"/><polygon points="170.63593387038418,94.54545454545455 149.11651474604358,131.8181818181818 106.07767649736239,131.8181818181818 84.55825737302179,94.54545454545455 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c2</p></details></div><h3>推理依据</h3><ul><li>○/△每步交替,按节点数量推断端点类别</li><li>永久墙 c8—c10 断开表面邻接,需绕行</li><li>永久墙 c6—c9 断开表面邻接,需绕行</li></ul><p>关键分叉:第 9 步应走 c10,诱惑路线走 c13。排除依据:蓝橙颜色必须交替。</p><details><summary>逐项机制必要性反例</summary><p>解除 alternation 后:c1 → c0 → c3 → c6 → c4 → c7 → c9 → c12 → c13 → c10 → c11 → c8 → c5 → c2</p><p>解除 wall:0 后:c1 → c0 → c3 → c6 → c4 → c2 → c5 → c7 → c9 → c12 → c10 → c8 → c11 → c13</p><p>解除 wall:1 后:c1 → c0 → c3 → c6 → c9 → c12 → c10 → c7 → c4 → c2 → c5 → c8 → c11 → c13</p></details></article><article><h2>06 · 进阶 · 两处区域出口互相牵制</h2><div class="boards"><figure><svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 237.26134197312211 450" role="img" aria-label="两处区域出口互相牵制"><polygon points="98.90453678924885,54.166666666666664 79.17840259193665,88.33333333333333 39.72613419731222,88.33333333333333 20,54.166666666666664 39.72613419731219,20.000000000000004 79.17840259193665,20" fill="#64acf3" stroke="#aaa1c7" stroke-width="2"/><polygon points="158.0829393811855,88.33333333333333 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c0 → c3 → c6 → c9 → c12 → c15 → c13 → c14 → c11 → c8 → c10 → c7 → c5 → c2</p></details></div><h3>推理依据</h3><ul><li>○/△每步交替,按节点数量推断端点类别</li><li>永久墙 c9—c10 断开表面邻接,需绕行</li><li>永久墙 c2—c4 断开表面邻接,需绕行</li></ul><p>关键分叉:第 7 步应走 c12,诱惑路线走 c7。排除依据:蓝橙颜色必须交替。</p><details><summary>逐项机制必要性反例</summary><p>解除 alternation 后:c4 → c1 → c0 → c3 → c6 → c9 → c7 → c10 → c12 → c15 → c13 → c14 → c11 → c8 → c5 → c2</p><p>解除 wall:0 后:c4 → c1 → c0 → c3 → c6 → c7 → c10 → c9 → c12 → c15 → c13 → c14 → c11 → c8 → c5 → c2</p><p>解除 wall:1 后:c4 → c2 → c5 → c7 → c10 → c8 → c11 → c14 → c13 → c15 → c12 → c9 → c6 → c3 → c0 → c1</p></details></article><article><h2>07 · 难 · 三片区域、颜色与多墙联动</h2><div class="boards"><figure><svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 209.0811502626761 450" role="img" aria-label="三片区域、颜色与多墙联动"><polygon points="87.63246010507044,49.28571428571428 70.72434507880283,78.57142857142856 36.908115026267616,78.57142857142856 20,49.285714285714285 36.908115026267595,20.000000000000004 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c6 → c3 → c1 → c0</p><p>解除 wall:1 后:c15 → c16 → c17 → c14 → c11 → c8 → c5 → c2 → c4 → c7 → c10 → c13 → c12 → c9 → c6 → c3 → c1 → c0</p><p>解除 wall:2 后:c15 → c16 → c17 → c14 → c11 → c8 → c10 → c13 → c12 → c9 → c6 → c3 → c4 → c7 → c5 → c2 → c1 → c0</p></details></article><article><h2>08 · 难 · 起点未知:用颜色与墙反推全路</h2><div class="boards"><figure><svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 209.0811502626761 450" role="img" aria-label="起点未知:用颜色与墙反推全路"><polygon points="87.63246010507044,49.28571428571428 70.72434507880283,78.57142857142856 36.908115026267616,78.57142857142856 20,49.285714285714285 36.908115026267595,20.000000000000004 70.72434507880283,20" fill="#ffae55" stroke="#aaa1c7" stroke-width="2"/><polygon points="138.35680518387326,78.57142857142856 121.44869015760565,107.85714285714285 87.63246010507044,107.85714285714285 70.72434507880283,78.57142857142856 87.63246010507042,49.285714285714285 121.44869015760565,49.28571428571428" fill="#64acf3" stroke="#aaa1c7" 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y="410.71428571428567" text-anchor="middle" font-family="system-ui" font-size="10" fill="#665c78">c18 · △</text></svg></figure><p>c14 → c17 → c16 → c18 → c15 → c12 → c9 → c10 → c13 → c11 → c8 → c5 → c2 → c4 → c1 → c0 → c3 → c6 → c7</p></details></div><h3>推理依据</h3><ul><li>○/△每步交替,按节点数量推断端点类别</li><li>永久墙 c4—c7 断开表面邻接,需绕行</li><li>永久墙 c5—c7 断开表面邻接,需绕行</li><li>永久墙 c12—c13 断开表面邻接,需绕行</li></ul><p>关键分叉:第 14 步应走 c4,诱惑路线走 c1。排除依据:蓝橙颜色必须交替。</p><details><summary>逐项机制必要性反例</summary><p>解除 alternation 后:c14 → c17 → c16 → c18 → c15 → c12 → c9 → c10 → c13 → c11 → c8 → c5 → c2 → c1 → c0 → c3 → c4 → c6 → c7</p><p>解除 wall:0 后:c14 → c17 → c16 → c18 → c15 → c12 → c9 → c10 → c13 → c11 → c8 → c5 → c2 → c4 → c7 → c6 → c3 → c0 → c1</p><p>解除 wall:1 后:c14 → c17 → c16 → c18 → c15 → c12 → c9 → c10 → c13 → c11 → c8 → c5 → c7 → c6 → c3 → c0 → c1 → c4 → c2</p><p>解除 wall:2 后:c15 → c18 → c16 → c17 → c14 → c11 → c13 → c12 → c9 → c10 → c8 → c5 → c2 → c4 → c1 → c0 → c3 → c6 → c7</p></details></article></html>