3
It is Restricted Integer Partitions, but with maximum number.
Question
Three positive integers are given. First number is number to divide, second number is length of partition, and third number is maximum number. First number is always largest, and bigger than other two.
For example, 5, 2, 3
. Then, make partition of 5
which have 2
parts, and maximum number you can use is 3
. Note that you don't have to use 3
: maximum number can be 2
or 1
.
In this case, there is only one partition : 3, 2
.
Partition is unordered : which means 3 + 2
and 2 + 3
is same.
But in case like 7, 3, 3
, there are two partition : 3, 3, 1
and 3, 2, 2
.
To make it sure, You can use third number as largest number, but you don't have to use it. So 5, 4, 3
is true.
Question is : Is there are more than one partition, given length and maximum number?
Output is True
or 1
or whatever you want when there is only one partition, and False
or 0
or whatever you want where there are more than one partition, or no partition.
Winning condition
This is code golf, so code with shortest byte wins.
Examples
Input -> Output
7, 6, 2 -> True (2+1+1+1+1+1) : 1 partitions
5, 4, 4 -> True (2+1+1+1) : 1 partitions
5, 4, 3 -> True (2+1+1+1) : 1 partitions
5, 4, 2 -> True (2+1+1+1) : 1 partitions
5, 3, 2 -> True (2+2+1) : 1 partitions
7, 2, 3 -> False no partitions
7, 2, 2 -> False no partitions
7, 2, 1 -> False no partitions
9, 5, 3 -> False (3+3+1+1+1), (3+2+2+1+1), (2+2+2+2+1) : 3 partitions
6, 3, 3 -> False (3+2+1), (2+2+2) : 2 partitions
Are the input number guaranteed to be positive integers? – xnor – 2019-05-26T06:34:33.663
Ah, I forgot that. Yes. – LegenDUST – 2019-05-26T06:35:23.293
It would be good for the challenge to have test cases that include subtle cases where a solution might fail. Also, I'd suggest saying that you're talking about unordered partitions, meaning the 2+3 and 3+2 don't count as two different partitions. – xnor – 2019-05-26T06:38:33.413
@xnor I added some examples, but I couldn't find subtle cases. – LegenDUST – 2019-05-26T06:49:20.377
How do you get
7, 6, 2 -> True
? With dividing 7 into 6 parts, I only see 2+1+1+1+1+1. Also, you say that the first number is always the largest, but it's not with2, 7, 3
. – xnor – 2019-05-26T06:51:50.600@xnor because there are only one partition. In
2+1+1+1+1+1
, 2 is maximum number. And I mistyped2, 7, 3
. I'll fix it. – LegenDUST – 2019-05-26T06:55:53.823shouldn't the second test case be
5, 4, 2
? – attinat – 2019-05-26T07:08:21.453@attinat You don't have to use third number as largest number : it is largest number that you 'can' use, not you 'must' use. Sorry if it is not clear. – LegenDUST – 2019-05-26T07:11:44.630
When you say that the first number is the largest, is it always strictly bigger than the other two, or can it be equal? – xnor – 2019-05-26T07:19:01.750
@xnor It can't be equal. Sorry for unclear question, and I'll fix it. – LegenDUST – 2019-05-26T07:21:13.263
Currently a rule that fits all your test cases is "the first number is exactly one bigger than the second number", which makes it easy for a solver to mistakenly think an incomplete solution of theirs is right because it passes the test cases. – xnor – 2019-05-26T08:06:40.977
Doesn't
9, 5, 3
also have(2+2+2+2+1)
? – Jonathan Allan – 2019-05-26T15:23:14.280@JonathanAllan My fault. I'll fix it. – LegenDUST – 2019-05-27T08:13:12.780