Such as input condition is:
B> A, c> A,
If such save
If (b> A)
B=2, a=1;
If (c> A)
C=a + 1;
The c=2;
If (b==c)//b==2, c==2
Output size b and c uncertain.
The else
B and c size
O
But there is a problem here
If c> B> A, e> D> A
Set a=1 so
B=2, c=3
D=2, e=3
If judged according to equal
Judgment method:
If (c==e)
Print (" relationship between c and e not sure ")
The else
Print (" relationship between c and e can determine ")
But so b and e can also determine the relationship between, because b!=e,
In fact the relationship between b and e not sure
If the collection to save this relationship, the said upper limit of the set?
Input: b> A, c> A
Set a=100,
B=[101, infinity]
C=[101, infinity]
Input increases d> B is to: d> B> A, c> A
Assignment to d:
D=[[0] b + 1, infinity] is d=[102, infinity]
So whether d is greater than a just judge d [0] is greater than a [0], is a collection of inclusion relation is used to judge the
If the input
e> C is a: d> B> A, e> C> A
E assignment:
E should be [[0] c + 1, infinity] : [101 + 1, infinity] [102, infinity]
A collection of e and d, e and d uncertain
Question:
If added a f> E: d> B> A, f> e> C> A
For the f value assignment should be: f=[e [0] + 1, infinity] or [103, infinity]
F=[103, infinity] and d=[102, infinity] f [0] & gt; D [0], f is larger than d! In fact the relationship between f and d not sure! This contradiction
If such improvements:
Condition: b> A, c> A
Set a=100,
Int x=a specific number
B=[101] x
C=[101] x
So save
If the input:
D> B d> namely; B> A, c> A
If set d=[x + 1, x * 2] being each number is greater than b, d
In fact that is problematic,
Determine a specific number of x? Set a freely? Such as 10000, but it still has bugs
Can't write here.. This bug is easy to think of here don't write for a while.
So how to save such a logic relation?
CodePudding user response:
Didn't understand your purpose, the relationship between b and c should be uncertain, so it should be the or to describe the relationship betweenC> namely;=b | | c<=b
CodePudding user response:
According to your description, to save the logic relationship, my advice is to not die can use tree?CodePudding user response:
The building Lord, you have a logical problem:If (c==e)
Print (" relationship between c and e not sure ")
The else
Print (" relationship between c and e can determine ")
C==e not sure when unexpectedly is relationship? This logic to tell I can't understand, could you go into more detail about?