Can A* be used to generate algorithm sourcecode?


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?


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.

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Node1 是根

节点 1 子节点 2

节点 1 子节点 3

示例 1



Node1 是根

节点 1 子节点 2

节点 1 子节点 3

节点 1 子节点 4


Node4 插入到 node1 子节点中


Node1 是根

针对 node1、node4 运行算子

Len(node1.children) <= Maxsize

示例 2

节点拆分 - 插入 node5



节点 1 子节点 2

节点 1 子节点 3

节点 1 子节点 4



节点 3 子节点 1

节点 3 子节点 2

节点 3 子节点 4

节点 4 子节点 5

这表示一个节点拆分,因为每个节点里面只能有 3 个项目。

我们如何学习一个节点只能有 3 个节点的规则?


最大尺寸为 3 在系统中插入运算符 Len(node.children) 最大尺寸 == 等于 >= 大于或等于 <= 小于或等于 > 大于 < 小于


补丁说明 Node1 不再是根节点 Node3 是根节点 Node2 的子节点是 node1 从 node1 中删除 node4 为什么从 node1 中删除 node4 为什么node4被添加到node3

理论。什么事实是真的。 将 node1 属性与 node4 进行比较 最终.... Len(node1.children) == MaxSize 将node4添加到node3 为什么? 理论。什么事实是真的 节点 4 >= 节点 3 真 >= 是否适用于所有示例?还是更复杂 扭动操作。一个下降,一个上升,下降连接一个下降。 根 = A 节点 1 是 A 新根 = B

扭曲操作是 - 根 = B B.儿童 = A


应该有图案。因此,对象中通常有一个属性可以进行比较以进行 btree 拆分。

因此,在 btree 中最多有一个字段或一个比较来确定对象的目的地。




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


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





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.