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authorAndreas Grois <andi@grois.info>2024-07-22 23:53:26 +0200
committerAndreas Grois <andi@grois.info>2024-07-22 23:53:26 +0200
commita4734cad1ef5f69cf9505c923eb2990c5403616c (patch)
treefeb1bf9545bfafdcc7b4abc8034141d411764aac /TODO
parent1ff5b4e46f52fa05624523b17a5188991ab82cf3 (diff)
Add GetElem instances.
Diffstat (limited to 'TODO')
-rw-r--r--TODO31
1 files changed, 19 insertions, 12 deletions
diff --git a/TODO b/TODO
index 3f2db67..382afc6 100644
--- a/TODO
+++ b/TODO
@@ -1,5 +1,7 @@
This is a rough outline of upcoming tasks:
+[ ] Prove that an index exists such that after CompleteTree.heapPush the pushed element can be obtained by
+ CompleteTree.get
[ ] Prove that CompleteTree.heapUpdateAt returns the element at the given index
[ ] Prove that CompleteTree.heapRemoveLastWithIndex and CompleteTree.heapRemoveLast yield the same tree
[ ] Prove that CompleteTree.heapRemoveLastWithIndex and CompleteTree.heapRemoveLast yield the same element
@@ -9,19 +11,24 @@ This is a rough outline of upcoming tasks:
- A potential approach is to show that any index in the resulting tree can be converted into an index
in the original tree (by adding one if it's >= the returned index of heapRemoveLastWithIndex), and getting
elements from both trees.
- - **EASIER**: Or just show that except for the n=1 case the value is unchanged.
- The type signature proves the rest.
+ - Maybe easier: Or show that except for the n=1 case the value is unchanged.
+ The type signature proves the rest, so try, like induction maybe?
[ ] Prove that if CompleteTree.indexOf returns some, CompleteTree.get with that result fulfills the predicate.
-[ ] Prove that CompleteTree.heapUpdateRoot indeed removes the value at the root.
- [x] A part of this is to show that the new root is not the old root, but rather the passed in value or
- the root of one of the sub-trees.
- [ ] The second part is to show that the recursion (if there is one) does not pass on the old root.
- - Basically that heapUpdateRoot either does not recurse, or recurses with the value it started out with.
- - This should be relatively straightforward. (left unchanged or left.heapUpdateRoot value and right
- unchanged or right.heapUpdateRoot value)
- [x] The last part would be to show that only one element gets removed
- - Enforced by type signature.
-[ ] Prove that CompleteTree.heapUpdateAt indeed removes the value at the given index.
+[ ] Prove that CompleteTree.heapUpdateRoot indeed exchanges the value at the root.
+ - Straightforward, but hard to prove:
+ [ ] Show that for each index in the original heap (except 0) the element is now either the new root, or
+ in the left or in the right subtree.
+ - Maybe a new predicate? CompleteTree.contains x => root = x OR left.contains x OR right.contains x
+ - Try unfolding both functions like in heapRemoveLastWithIndexReturnsItemAtIndex.
+ - Alternative approach, but less convincing
+ [x] A part of this is to show that the new root is not the old root, but rather the passed in value or
+ the root of one of the sub-trees.
+ [ ] The second part is to show that the recursion (if there is one) does not pass on the old root.
+ - Basically that heapUpdateRoot either does not recurse, or recurses with the value it started out with.
+ - This should be relatively straightforward. (left unchanged or left.heapUpdateRoot value and right
+ unchanged or right.heapUpdateRoot value)
+[ ] Prove that CompleteTree.heapUpdateAt indeed updates the value at the given index.
+ - Use contains again (if that works), or
- Basically making sure that if index != 0 the recursion does either pass on the passed-in value, or the current
node's value.
- This sounds so trivial, that I am not sure if it's even worth the effort.