当前位置:网站首页>Différences entre les contraintes monotones et anti - monotones
Différences entre les contraintes monotones et anti - monotones
2022-07-07 10:40:00 【Xminyang】
▚ 01 Monotone Constraints Contraintes monotones
1.1 Definitions Définition
【Monotone Constraints | SpringerLink 】
A constraint C is monotone if and only if for all itemsets S and S′:
if S ⊇ S′ and S violates C, then S′ violates C.
1.2 Key Points Points saillants
Monotone constraints possess the following property. If an itemset S violates a monotone constraint C, then any of its subsets also violates C. Equivalently, all supersets of an itemset satisfying a monotone constraint C also satisfy C (i.e., C is upward closed). By exploiting this property, monotone constraints can be used for reducing computation in frequent itemset mining with constraints. As frequent itemset mining with constraints aims to find frequent itemsets that satisfy the constraints, if an itemset S satisfies a monotone constraint C, no further constraint checking needs to be applied to any superset of S because all supersets of S are guaranteed to satisfy C. Examples of monotone constraints include min(S. Price) ≤ $30, which expresses that the minimum price of all items in an itemset S is at most $30. Note that, if the minimum price of all items in S is at most $30, adding more items to S would not increase its minimum price (i.e., supersets of S would also satisfy such a monotone constraint).

▚ 02 Anti-monotone Constraints Contrainte anti - monotone
2.1 Definitions Définition
【Anti-monotone Constraints | SpringerLink】
A constraint C is anti-monotone if and only if for all itemsets S and S′:
if S ⊇ S′and S satisfies C, then S′ satisfies C.
2.2 Key Points Points saillants
Anti-monotone constraints possess the following nice property. If an itemset S satisfies an anti-monotone constraint C, then all of its subsets also satisfy C (i.e., C is downward closed). Equivalently, any superset of an itemset violating an anti-monotone constraint C also violates C. By exploiting this property, anti-monotone constraints can be used for pruning in frequent itemset mining with constraints. As frequent itemset mining with constraints aims to find itemsets that are frequent and satisfy the constraints, if an itemset violates an anti-monotone constraint C, all its supersets (which would also violate C) can be pruned away and their frequencies do not need to be counted. Examples of anti-monotone constraints include min(S. Price) ≥ $20 (which expresses that the minimum price of all items in an itemset S is at least $20) and the usual frequency constraint support(S) ≥ minsup (i.e., frequency(S) ≥ minsup). For the former, if the minimum price of all items in S is less than $20, adding more items to S would not increase its minimum price (i.e., supersets of S would not satisfy such an anti-monotone constraint). For the latter, it is widely used in frequent itemset mining, with or without constraints. It states that (i) all subsets of a frequent itemset are frequent and (ii) any superset of an infrequent itemset is also infrequent. This is also known as the Apriori property.

▚ 03 Explanation Explication
Hypothèses:On vaS violates CEn tant qu'événementA,S′ violates CEn tant qu'événementB;Et S satisfies CPour l'événementnot A,then S′ satisfies CPour l'événementnot B.
En ce moment,SelonMonotone Constraints Définition (A → B),C'est - à - dire(not B → not A);
SelonAnti-monotonicity Constraints Définition (not A → not B),C'est - à - dire(B → A);
Parce que(A → B)Cela ne signifie pas nécessairement(B → A), Donc les deux (Monotone Constraints & Anti-monotonicity Constraints) La déclaration est différente .

▚ 04 Example Exemple
For an example. Consider-
C1 = Sum of elements is greater than 5
C2 = Sum of elements is at most 5
U(universe) = Set of non-negative real numbers
In case of C1,
If S violates C1, then S’ obviously violates C1 as well (S being a superset of S’)
Eg. S = {1, 2}, S’ = {2}
Hence C1 is monotonic.
In case of C2,
If S satisfies C2, then S’ obviously satisfies C2 as well (S being a superset of S’)
Eg. S = {1, 2}, S’ = {2}
Hence C2 is anti-monotonic.

▚ 05 Contraintes de l'exploration des données
Constraint-Based Mining — A General Picture

Article de référence
边栏推荐
- php \n 换行无法输出
- Multisim--软件相关使用技巧
- leetcode-304:二维区域和检索 - 矩阵不可变
- Those confusing concepts (3): function and class
- MONAI版本更新到 0.9 啦,看看有什么新功能
- 【PyTorch 07】 动手学深度学习——chapter_preliminaries/ndarray 习题动手版
- 优雅的 Controller 层代码
- 小程序跳转H5,配置业务域名经验教程
- Deeply analyze the main contents of erc-4907 agreement and think about the significance of this agreement to NFT market liquidity!
- A small problem of bit field and symbol expansion
猜你喜欢

How much review time does it usually take to take the intermediate soft exam?

Mendeley--免费的文献管理工具,给论文自动插入参考文献

P2788 math 1 - addition and subtraction
![[actual combat] transformer architecture of the major medical segmentation challenges on the list --nnformer](/img/de/0cf12132216ffbde896a7b12022184.png)
[actual combat] transformer architecture of the major medical segmentation challenges on the list --nnformer

openinstall与虎扑达成合作,挖掘体育文化产业数据价值

What does intermediate software evaluator test

SQL Server 知识汇集9 : 修改数据

CSAPP Bomb Lab 解析

Using tansformer to segment three-dimensional abdominal multiple organs -- actual battle of unetr

CSAPP bomb lab parsing
随机推荐
P1223 排队接水/1319:【例6.1】排队接水
软考一般什么时候出成绩呢?在线蹬?
IIC Basics
01 use function to approximate cosine function (15 points)
Unable to open kernel device '\.\vmcidev\vmx': operation completed successfully. Reboot after installing vmware workstation? Module "devicepoweron" failed to start. Failed to start the virtual machine
无法打开内核设备“\\.\VMCIDev\VMX”: 操作成功完成。是否在安装 VMware Workstation 后重新引导? 模块“DevicePowerOn”启动失败。 未能启动虚拟机。
IIC基本知识
Deeply analyze the main contents of erc-4907 agreement and think about the significance of this agreement to NFT market liquidity!
1321: [example 6.3] deletion problem (noip1994)
如何顺利通过下半年的高级系统架构设计师?
对word2vec的一些浅层理解
Monai version has been updated to 0.9. See what new functions it has
Network engineer test questions and answers in May of the first half of 2022
1323:【例6.5】活动选择
使用Tansformer分割三维腹部多器官--UNETR实战
P2788 数学1(math1)- 加减算式
The width of table is 4PX larger than that of tbody
[牛客网刷题 Day6] JZ27 二叉树的镜像
[牛客网刷题 Day5] JZ77 按之字形顺序打印二叉树
Socket communication principle and Practice