{"pubkey":"8c9e5a47d62921e8cd34b571cd95acc8be198b65460296ac53dad041339a8c9c","posts":[{"contentHash":"6fb219f80ca87196401689821d04f3d95132a3ab298c893e26120fe0a8fb277b","protocol":"bitcoin","authorPubkey":"8c9e5a47d62921e8cd34b571cd95acc8be198b65460296ac53dad041339a8c9c","text":"What's the smart way to solve it? The logic I think is necessary is: Every alloy ID should have a list of alloy IDs that are weaker than it. For every pair of alloys N and M, where N is stronger than M, update list N to include alloy M, update any list K that includes N to also include M, update any list J that includes K to also include M, repeat as necessary. Is there a simpler way to do it?","topicHash":"913c62c2ee31793c7e8517600dbc8f0c337afae133700a128a7d8ffb8a21c93e","topicName":"technology","parentHash":"152da294875013b9b9aa5a1c8867f3dad74707354c8ecd22c76b947df80bcc8d","costSunk":"0","biasSupport":0,"biasOppose":0,"biasVouch":0,"biasNull":0,"blockHeight":null,"confirmations":0,"timestamp":"2026-11-04T05:09:39.223Z","tripFp":"297b4983bb1b","viewCount":0,"powDifficulty":0,"relation":0,"ephemeral":true,"nonce":"e42f736e503c8eca","sig":"f9cde7031cd0bc2b8e1cc81569c6aad574ad1d042ed8065a5f8f3f31a01f84d80835a81603a56244e6ab460abe73adefbeec74c3362b464a3064ae5d42da9158","ephStamp":{"blockHashHex":"00000000000000000001db3692f60243aaa2b07feb7a376b6263ab3598410c92","nonceHex":"ca2f000000000000"},"replyCount":1,"chainReplyCount":0,"vaporReplyCount":1,"clusterSize":2,"receiptHeight":974484,"expiryHeight":978804,"authorMarks":{"since":"2026-11-03T19:04:47.254Z","via":"sight","inscribed":0,"likes":0,"biases":0},"lang":"en"}],"nextCursor":null,"pageReceipt":{"host":"6af58b1822e0cabfe57d76540f745d059b438f489a796ec6a705c6540233845e","t":"2026-11-08T20:49:07.925Z","sig":"55bb8cca35402a54f8bcef61f3a490d1dc286d744b2964a3b78dd5f9b18e08c48df274722100b883cbe2bf49272205504fcf53f06fdd8eeb8acfdae928a383e6"},"marks":{"since":"2026-11-03T19:04:47.254Z","via":"sight","inscribed":0,"likes":0,"biases":0}}