Assignment 2 - Question

[kwong ho]

Hi All,

I have some problem on Forall and exists.

Do I understand the "forall" meaning correctly? 

ForAll

My interpretation is : "For all element in given set, they have to pass the given conditions. " => " any 1 element in the given set fail the condition is false, return false"

Forall ( given sets, condition) return true if all element 

Exists:

My interpretation is : "For any element in given set, that pass the given conditions. " => "any 1 element in the given set pass the condition is true, return true"

how do I relate to Forall

Thanks,

Kelvin

[Xianming Mu]

刚才可能说得不太清楚，还是用中文吧，既然是中文小组。

举个例子，“存在一个硬币是正面” 等价于 “不是所有的硬币都是反面”。 我觉得这是个关键点。

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Regards,

Samuel

[kwong ho]

謝~~ 我想我清楚 the relationship.

Forall : all element in s: Set is pass p-condition return true.  

In other words, if find 1 false, return false 

 

Exists : only 1 element in s: Set is pass the p-condition return true 

In other words, if find 1 true, return true  

the relationship of exists is flipped result and flipped condition

[BZ]

就是这个意思。如果你的exist()里不调用forall()会被扣分。

[Leonard Siu]

I believe your understanding is correct for both cases, however I don't think Exists is related to ForAll, at least not with my implementation. For my implementation, they both follow the same structure outline in the assignment:

def forall(s: Set, p: Int => Boolean): Boolean = {
 def iter(a: Int): Boolean = {
   if (???) ???
   else if (???) ???
   else iter(???)
 }
 iter(???)

[kwong ho]

After all the hints, I able to make a one liner of exists using forall. 

I got it from https://class.coursera.org/progfun-2012-001/forum/thread?thread_id=889

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Regards,
Kwong W. Ho

[Xianming Mu]

For the exists implementation, I was also inspired by some comment from the forum. I just count it as the most significante obstacle between me and full score :)

Samuel

[kwong ho]

Forall : Only all element in s: Set pass p-condition return true. || any 1 element fail p-condition, return false In other words, if find 1 false, return false

Exists : Only all element in s: Set fail the p-condition, return false || any 1 element pass the p-condition return true In other words, if find 1 true, return true

the relationship between exists and forall , flipped condition and flipped the result will be the other one