# "*": "https://raw.githubusercontent.com/wefindx/schema/master/method/oo-item.yaml" # "base:title": "0oo - Can A* be used to generate algorithm sourcecode?" "og:title": "Can A* be used to generate algorithm sourcecode?" "og:description": "If you project the desired code variables onto a multidimensional space and with a time dimension, where each variable has a assignment algebra you can imagine the input code causing movements to the output algebra for each variable How do you measure that a solution gets nearer to the solution when you haven't planned what you could do It should be a multidimensional walk toward a solution that solves the programming problem. I want to generate the btree algorithm and other algorithms with this approach I should provide example inputs to output mappings and join the Data between the input …" "og:image": "https://avatars0.githubusercontent.com/u/28134655" "og:url": "/method/139001/" "base:css": "/static/css/bootstrap.min.9c25540d6272.css" "base:extra-css": "/static/css/base.57997aeac1df.css" "base:favicon": "/static/favicon.acaa334f0136.ico" "base:body_class": "" "layout:logo": "/static/0oo.8d2a8bbef612.svg" "layout:index": "/" "layout:menu": "/menu/" "layout:categories": "/intents/" "layout:ideas": "/methods/" "layout:projects": "/projects/" "layout:users": "/users/" "layout:about": "/about/" "layout:help": "/help/" "layout:bug_report": "https://github.com/wefindx/0oo" "layout:login": "/accounts/login/" "layout:light-off": "/darken/?darken=true" "layout:set-multilingual": "/mulang/?mulang=true" "layout:lang": "言語" "layout:set-language-post-action": "/i18n/setlang/" "layout:csrf-token": "gbl7caLzUWLIApgmlts3P8thcqWQXBUPbkXDhihe4SYqmUsKUBjEZAgjzYHh70J0" "layout:input-next": "/method/139001/" "layout:languages": [{"code": "ja", "is-active": "true", "name": "日本語"}, {"code": "lt", "is-active": "false", "name": "Lietuviškai"}, {"code": "zh-hans", "is-active": "false", "name": "简体中文"}, {"code": "en", "is-active": "false", "name": "English"}, {"code": "ru", "is-active": "false", "name": "Русский"}, {"code": "oo", "is-active": "false", "name": "O;o,"}] # "item:parent:intents": [{"url": "/intent/19001/", "title": "Programming too difficult"}] "item:title": "Can A* be used to generate algorithm sourcecode?" "item:summary": "I thought of an idea to use a* algorithm to "walk" toward a solution that solves a programming problem. Each neighbour can be a weighted list of programming tasks such as if statement, for each, assignment, recursive call. How do you map example data structures transformations to code that fulfils the changes to the data structure? You need a cost function that decreases when the output gets nearer to the solution. What does the cost function for code solving a problem look similar to?" "item:voting": 0 "item:voting:add": "/admin/hlog/voting/add/?method=139001" "item:voting:csrf_token": "gbl7caLzUWLIApgmlts3P8thcqWQXBUPbkXDhihe4SYqmUsKUBjEZAgjzYHh70J0" "item:voting:submit-value-option": {"selected": "[-]", "value": "-"} "item:base-administration": false "item:body": | If you project the desired code variables onto a multidimensional space and with a time dimension, where each variable has a assignment algebra you can imagine the input code causing movements to the output algebra for each variable How do you measure that a solution gets nearer to the solution when you haven't planned what you could do It should be a multidimensional walk toward a solution that solves the programming problem. I want to generate the btree algorithm and other algorithms with this approach I should provide example inputs to output mappings and join the Data between the input and output and try recalculate the processing steps to go from one to the other but I am having problems deciding what the cost function looks like. "item:source-date": "" "item:permalink": "/method/139001/?l=ja" "item:owner": "chronological" "item:created": "2022-05-18T08:06:56.151167" "item:ownerlink": "/user/198/chronological" # "item:link:add": "/admin/hlog/link/add/?parent=139001" "item:link:items": "item:project:items": # "item:comment:items": - "id": "a-269001" "mtrans": | 考えました。出力で生成されたコードをスーパーセットとして想像すると、コードのすべての行が出力コードのサブセットの一部になります。 したがって、出力セットのサブセットを生成する何らかの方法が必要です。 各セットは、回答を生成する順序付けられた命令のセットです。ブルートフォース検索は避けたいので、コード行を生成しようとするたびに出力に近づく必要があります。 コードが何をすべきかについてのヒューリスティックを提供する必要があると思います。 "text": | I had a thought. If you imagine the output generated code as a superset then every line of code is part of a subset of the output code. So you need some way of generating subsets of the output set. Each set is an ordered set of instructions that generate the answer. I want to avoid brute force search, every attempt of generating a line of code should get nearer to the output. I think we have to provide heuristics of approximately what the code should do. "owner": "chronological" "ownerlink": "/user/198/chronological" "permalink": "/method/139001/?l=ja#a-269001" "created": "2022-05-18T12:56:47.315685" - "id": "a-270001" "mtrans": | 出力構造をスキャンして、データの各関係のファクトを生成できます。 ソースデータから宛先データへのパストラバーサルに基づいて、データが移動した場所を比較できます。パッチを生成します。 これには、このデータへのこのオブジェクトへのポインタがあります。 入力データ Node1はルートです Node1の子node2 Node1の子node3 例1 ノードnode4を挿入します 必要な出力データ Node1はルートです Node1の子node2 Node1の子node3 Node1の子node4 パッチ node1の子に挿入されたNode4 なぜnode1？理論。 Node1はルートです node1、node4に対して演算子を実行します Len（node1.children）&lt;=最大サイズ 例2 ノード分割-ノード5を挿入 入力データ ルートノードはnode1です Node1の子node2 Node1の子node3 Node1の子node4 必要なターゲットデータ ルートノードはnode3です Node3の子node1 Node3の子node2 Node3の子node4 Node4の子node5 これは、各ノード内に3つのアイテムしか含めることができないため、ノードの分割を表します。 1つのノードが持つことができるノードは3つだけであるというルールをどのように学習しますか？ ファクトのシステムに定数を挿入します MaxSizeは3です システムに演算子を挿入します Len（node.children） MaxSize ==等しい &gt;=以上 &lt;=以下 &gt;より大きい &lt;未満 各演算子のすべての順列を実行して、パッチコマンドステップを実行することを決定します。 パッチの説明 Node1はルートノードではなくなりました Node3はルートノードです Node2の子はnode1です node1からnode4を削除します node4がnode1から削除されたのはなぜですか node4がnode3に追加されたのはなぜですか 理論。どんな事実が真実でしたか。 node1のプロパティをnode4と比較します 最終的.... Len（node1.children）== MaxSize node4をnode3に追加します なんで？ 理論。本当の事実 Node4&gt; = Node3 true &gt; =はすべての例に当てはまりますか？それとももっと複雑ですか ツイスト操作。 1つは下降し、1つは上昇し、1つは下降し、もう1つは下降します。 ルート=A Node1はAです 新しいルート=B ツイスト操作は- ルート=B B.children = A そして、データがどこに移動するかを確認します。適切な場所に同じデータを使用して同じ構造を決定論的に作成するステップを生成する必要があります。 パターンがあるはずです。したがって、通常、btree分割を行うために比較されるプロパティがオブジェクトにあります。 したがって、btree内のオブジェクトの宛先を決定するフィールドまたは比較は最大で1つです。 入力オブジェクトごとに条件ステートメントをランダムに生成できると思いました。 パッチをウォークし、条件オブジェクトを挿入します。 グラフ変換の問題です "text": | We can scan the output structure and generate facts of each relationship of data. We can compare where the data moved through based on a path traversal of source data to destination data. Generate a patch. This has a pointer to this object to this piece of data. Input data Node1 is root Node1 children node2 Node1 children node3 Example 1 Insert a node node4 Desired Output data Node1 is root Node1 children node2 Node1 children node3 Node1 children node4 Patch Node4 inserted into node1 children Why node1? Theorise. Node1 is root Run operators against node1, node4 Len(node1.children) <= Maxsize Example 2 Node split - insert node5 Input data Root node is node1 Node1 children node2 Node1 children node3 Node1 children node4 Desired target data Root node is node3 Node3 children node1 Node3 children node2 Node3 children node4 Node4 children node5 This represents a node split as each node can only have 3 items inside it. How do we learn the rule that a node can only have 3 nodes? Insert a constant into the system of facts MaxSize is 3 Insert operators in system Len(node.children) MaxSize == Equal to >= Greater than or equal to <= Less than or equal to > Greater than < Less than Run every permutation of each operator to decide to do patch command steps. Patch instructions Node1 is no longer root node Node3 is root node Node2 children is node1 Remove node4 from node1 Why was node4 removed from node1 Why was node4 added to node3 Theorise. What fact was true. Compare node1 properties to node4 Eventually.... Len(node1.children) == MaxSize Add node4 to node3 Why? Theorise. What fact is true Node4 >= Node3 true Does >= hold for all examples? Or is it more complicated Twist operation. One goes down, one goes up and going down joins one going down. Root = A Node1 is A New root = B Twist operation is - Root = B B.children = A And see where the data moves. We need to generate the steps that deterministically creates the same structure with the same data in the right places. There shall be patterns. So there is usually a property in the object that is compared against to do a btree split. So there is at most one field or a comparison that determines the destination of the object in the btree. I thought we can randomly generate condition statements for each input object. Walk the patch and insert condition objects. It's a graph transformation problem "owner": "chronological" "ownerlink": "/user/198/chronological" "permalink": "/method/139001/?l=ja#a-270001" "created": "2022-05-18T15:04:07.215363" "base:js": "/static/js/base.c7357c06cc89.js"