https://github.com/zdimension/stablewall
Solution to the GKS 2020 Round C problem "Stable Wall"
https://github.com/zdimension/stablewall
Last synced: 11 months ago
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Solution to the GKS 2020 Round C problem "Stable Wall"
- Host: GitHub
- URL: https://github.com/zdimension/stablewall
- Owner: zdimension
- License: mit
- Created: 2020-06-28T08:59:49.000Z (over 5 years ago)
- Default Branch: master
- Last Pushed: 2020-07-27T02:39:51.000Z (over 5 years ago)
- Last Synced: 2025-02-04T18:52:14.239Z (about 1 year ago)
- Language: Python
- Size: 8.79 KB
- Stars: 0
- Watchers: 3
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
- License: LICENSE
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README
# stablewall
Solution to the GKS 2020 Round C problem "Stable Wall"
## How it works
### Step 1
Let 
be the set of letters and ,(M,O),(O,Z)\})
the set of directed dependencies between letters.
A dependency )
is *redundant* (noted )
) iff ,(x,b)\}\subset{E})
.
Let 
where 
refers to the "ground".
Let \}\cup\{(a,*)\mid{a}\in\Sigma,\nexists(x,y)\in{E},x=a\})
. In this case, ,(M,O),(O,Z),(Z,*)\})
. In this case, there were no *redundant* dependencies. The dependency )
was added to allow easier computation of the node degrees.
Let 
denote the directed graph )
.
The input is valid iff is acyclic.
The number of valid solutions is obtained using the expression !})
.
### Step 2
Let 
denote the transitive closure of 
. We have ,(A,Z),(M,O),(M,Z),(O,Z)\})
. Let \notin{R^{+}})
The valid solutions 
are the permutations of 
ordered under 
.
In other words, we have _{\left\[1,\left|\Sigma\right|\right\]}\in{S}(\Sigma)\mid\forall{i}\in\left\[1,\left|\Sigma\right|-1\right\],\sigma_{i}\preceq\sigma_{i+1}\})
(note that here 
is an ordered tuple and not a set).
### Step 3: Preprocessing
Let )
which returns the set of all nodes reachable from the specified node, i.e. all supporting letters for a specified letter with =\{b\mid(a,b)\in{R}^{+}\})
(this can be done using DFS or recursive graph traversal), here we have
| 
| )
|
| ------------- | ------------- |
| 
| 
|
| 
| 
|
| 
| 
|
| 
| |
### Step 4: Processing
1. While there are letters yet to be added to the answer:
1. Find the first one whose supporting letters are all already present, and add it.
2. If no such letter is found, halt. There is a cycle in the directed graph, which means two letters are codependent. Thus, there is no solution.