Problem 090
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Comments, questions and clarifications about PE problems.
- Animus
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Re: Problem 090
To me it looks like your validity test is faulty. The second example in the problem description will IMHO return false with your code while testing for 09 (erroneously).
Please check this.
Please check this.
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itteerde
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Re: Problem 090
Animus wrote: Sun Apr 23, 2017 11:11 am To me it looks like your validity test is faulty. The second example in the problem description will IMHO return false with your code while testing for 09 (erroneously).
Please check this.
That is correct, had the correct test for 69 and did not apply it to 09 but only to 36, 49 and 64. Also my understanding of the problem would yield twice the number of solutions because I think the identity of each cube should matter - but it does not. But I suspected that and tried my solution and half of it anyways
So try your result /2 if you get stuck, too...
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hamsterofdeath
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Re: Problem 090
i think the problem description should clarify what exactly "unique arrangement" means
i wasn't sure if die1 + die2 is the same arrangement as die2 + die1 and ended up calculating a (somewhat correct) solution that was twice as large as the official one
i wasn't sure if die1 + die2 is the same arrangement as die2 + die1 and ended up calculating a (somewhat correct) solution that was twice as large as the official one
- kenbrooker
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Re: Problem 090
I think this prior Post addresses your recommendation,
particularly the last sentence, as demonstrated in
the Problem description:
particularly the last sentence, as demonstrated in
the Problem description:
Re: Problem 090
Post by Animus » Tue Jan 24, 2017 4:43 am
You are right, they are considered as duplicates.srinathmkce wrote: ↑Mon Jan 23, 2017 8:24 pm
Consider we are able to form all the squares with these two combinations
(1) { 0, 1, 2, 6, 7, 9 } { 1, 3, 4, 5, 8, 9 }
(2) { 1, 3, 4, 5, 8, 9 } { 0, 1, 2, 6, 7, 9 }
The 2nd one is not a duplicate ? I guess earlier post in the forum says its duplicate.
Some of the comments above are a little misleading. You can't reach all squares with the same sequence of
the two cubes, so the first and the second have to be interchangeable.
"Good Judgment comes from Experience;
Experience comes from Bad Judgment..."

Experience comes from Bad Judgment..."

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jimfan
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Re: Problem 090
As a non-English speaker allow me to elaborate how I understand the problem statement and why I am having (present tense!) trouble with it.
I cannot properly grasp the meaning of the compound noun "distinct arrangement".
It appears only twice throughout. First occurrence:
Now the trouble. The last sentence:
Then it must be referring to the integer set discussed above. Regardless if {0, 5, 6, 7, 8, 9}, {1, 2, 3, 4, 8, 9} is considered as duplicate of {1, 2, 3, 4, 8, 9}, {0, 5, 6, 7, 8, 9} or not, here we are instead talking about two sets - each of which is a set of integer.
I hope you see my point: The two occurrences of "distinct arrangement" are qualitatively different.
To further complicate the matter, I find the noun "arrangement" in another sentence:
Anyone could point out if my concept / understanding / use of English is mistaken?
Jim
I cannot properly grasp the meaning of the compound noun "distinct arrangement".
It appears only twice throughout. First occurrence:
Which suggests a "distinct arrangement" refers to certain attribute of a single cube. The following two lines:In determining a distinct arrangement we are interested in the digits on each cube, not the order.
Equivocally assert that a "distinct arrangement" is a set of integer. It is also the set of integer printed on a cube. Because it is a set, order of its member is immaterial.{1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5}
{1, 2, 3, 4, 5, 6} is distinct from {1, 2, 3, 4, 5, 9}
Now the trouble. The last sentence:
Second occurrence of "distinct arrangement". Wait, of the two cubes? Can I arrange two objects in more than two ways? Obviously no...... And I tried to enter 2 as answer, which is wrong (for sure).How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?
Then it must be referring to the integer set discussed above. Regardless if {0, 5, 6, 7, 8, 9}, {1, 2, 3, 4, 8, 9} is considered as duplicate of {1, 2, 3, 4, 8, 9}, {0, 5, 6, 7, 8, 9} or not, here we are instead talking about two sets - each of which is a set of integer.
I hope you see my point: The two occurrences of "distinct arrangement" are qualitatively different.
To further complicate the matter, I find the noun "arrangement" in another sentence:
And this "arrangement" seems to mean two set of integer in a row.However, for this problem we shall allow the 6 or 9 to be turned upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3, 4, 6, 7} allows for all .........
Anyone could point out if my concept / understanding / use of English is mistaken?
Jim

- RobertStanforth
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Re: Problem 090
For the purposes of this problem, an "arrangement of two cubes" means a set of two cubes, where each cube is a set of six digits. Thus your "set of sets" interpretation is correct.
If two "arrangements of two cubes" are found to differ from one another only be reordering the cubes, or by reordering the digits within a cube, then they are considered the same arrangement. "Distinct arrangements" are arrangements that are not considered the same in that sense.
If two "arrangements of two cubes" are found to differ from one another only be reordering the cubes, or by reordering the digits within a cube, then they are considered the same arrangement. "Distinct arrangements" are arrangements that are not considered the same in that sense.
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jimfan
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Re: Problem 090
Thanks RobertStanforth. Turns out my understanding to "distinct arrangement" was mistaken. I wrongly treated the two set-of-set {{0, 1, 2, 3, 4, 5}, {0, 1, 2, 3, 6, 8}} and {{0, 1, 2, 3, 6, 8}, {0, 1, 2, 3, 4, 5}} as distinct.
Going through the whole thread again, people were reminded for this earlier.
Yet I still feel using the word "set" is much more precise compare to "distinct arrangement", though part of the true fun of PE is to pinpoint and de-puzzle the problem statement.
And I managed to solve 090 just now - it was a bug and too many candidate sets were rejected due to that naughty zero.
Going through the whole thread again, people were reminded for this earlier.
Yet I still feel using the word "set" is much more precise compare to "distinct arrangement", though part of the true fun of PE is to pinpoint and de-puzzle the problem statement.
And I managed to solve 090 just now - it was a bug and too many candidate sets were rejected due to that naughty zero.

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RichardDL
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Re: Problem 090
"However, for this problem we shall allow the 6 or 9 to be turned upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3, 4, 6, 7} allows for all nine square numbers to be displayed; otherwise it would be impossible to obtain 09."
Does it mean impossible with this arrangement or completely impossible? For instance, with {4,0,8,6,7,5},{2,1,4,3,7,9} I can see all the square numbers. I'm using 7s as invalid digits while trying solutions and I still have two spare places.
As I see it, that the 6,9 equivalance is unnecessary, but alright that's how the puzzle is set. Am I missing something?
Does it mean impossible with this arrangement or completely impossible? For instance, with {4,0,8,6,7,5},{2,1,4,3,7,9} I can see all the square numbers. I'm using 7s as invalid digits while trying solutions and I still have two spare places.
As I see it, that the 6,9 equivalance is unnecessary, but alright that's how the puzzle is set. Am I missing something?
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Junglemath
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Re: Problem 090
If you don't allow for the 6 and 9 to be flipped upside down, then you won't be able to make as many arrangements that the problem is asking for.RichardDL wrote: Fri Dec 10, 2021 12:09 pm "However, for this problem we shall allow the 6 or 9 to be turned upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3, 4, 6, 7} allows for all nine square numbers to be displayed; otherwise it would be impossible to obtain 09."
Does it mean impossible with this arrangement or completely impossible? For instance, with {4,0,8,6,7,5},{2,1,4,3,7,9} I can see all the square numbers. I'm using 7s as invalid digits while trying solutions and I still have two spare places.
As I see it, that the 6,9 equivalance is unnecessary, but alright that's how the puzzle is set. Am I missing something?
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gbohus
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Problem 090
It's not clear to me what "distinct arrangement" means.
The problem says {1, 2, 3, 4, 5, 6} is different from {1, 2, 3, 4, 5, 9} but they represent the same extended set.
So, an arrangement where {1, 2, 3, 4, 5, 6} is on cube #1 and an arrangement when {1, 2, 3, 4, 5, 9} is on cube #1, cube #2's being the same, is this one or two arrangements?
Thanks.
The problem says {1, 2, 3, 4, 5, 6} is different from {1, 2, 3, 4, 5, 9} but they represent the same extended set.
So, an arrangement where {1, 2, 3, 4, 5, 6} is on cube #1 and an arrangement when {1, 2, 3, 4, 5, 9} is on cube #1, cube #2's being the same, is this one or two arrangements?
Thanks.
- SAG145
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PierrotLeFou
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Re: Problem 090
I didn't solve this problem yet but my understanding is as follows:
Mathematically speaking, the two arrangements are differencs.
But since we allow that 6 is the same as 9 reversed, for the purpose of this problem, they are the same.
For the first cube, we need 0, 1, 2, 3, 4, 6, 8. But the 6 can be a 9.
Since a square can't end with 2, 3, 7, 8, and 0 in this problem,
we need 1, 4, 9, 6, 5 for the second cube.
and anything for the sixth digit.
I guess we can have [6, 6], [6, 9], [9, 9] as being the same for the second cube?
edit: I solved the problem.
The arrangements are not the same and they have to be counted separately but treated somehow ... as being the same.
Mathematically speaking, the two arrangements are differencs.
But since we allow that 6 is the same as 9 reversed, for the purpose of this problem, they are the same.
For the first cube, we need 0, 1, 2, 3, 4, 6, 8. But the 6 can be a 9.
Since a square can't end with 2, 3, 7, 8, and 0 in this problem,
we need 1, 4, 9, 6, 5 for the second cube.
and anything for the sixth digit.
I guess we can have [6, 6], [6, 9], [9, 9] as being the same for the second cube?
edit: I solved the problem.
The arrangements are not the same and they have to be counted separately but treated somehow ... as being the same.
I'm always right ... until I'm wrong
