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unreality
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I is the second letter since piece, pique and niece all an I and E in common and fines doesn't end in E therefore I must be the second letter.

Not sound logic. The I and ending E could be part of PIQUE, (or even the C in) PIECE, and NIECE and FINES have FI_E_ left.

Framm +5

Edited by Framm 18
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Not sound logic. The I and ending E could be part of PIQUE, (or even the C in) PIECE, and NIECE and FINES have FI_E_ left.

The I and E are the ONLY 2 letters that ALL THREE words have in common and if they all only have 1 letter correct. I or E must be the letters that could be correct. If it were the P that was correct, the number would be 0 at Niece. If it were the C then it again would have been 0 in pique. If it were the 3rd E, again it would have been 0 in Pique. That leaves only I or E to be the possible correct letters. And with Fines, the I is the only letter still in the same place as the other 3 words that have 1 correct letter, leaving I to be correct.

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The I and E are the ONLY 2 letters that ALL THREE words have in common and if they all only have 1 letter correct. I or E must be the letters that could be correct. If it were the P that was correct, the number would be 0 at Niece. If it were the C then it again would have been 0 in pique. If it were the 3rd E, again it would have been 0 in Pique. That leaves only I or E to be the possible correct letters. And with Fines, the I is the only letter still in the same place as the other 3 words that have 1 correct letter, leaving I to be correct.

I see where you are coming from really I do. But let's break this down.

PIECE (1)

NIECE (1)

P and N are eliminated

Middle E has been eliminated by GREAT Leaving _I_CE

PIQUE (1) Leaving _IQUE

FINES (1) N has been eliminated by GANJA

S has been eliminated by DRATS

Leaving FI_E_

While the I is the only common letter, it can not be concluded that it is the letter that is in the right spot.

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My logic is quite sound here.

Piece, Niece and Pique all have ONLY have 2 letters in common and all have 1 letter in the correct place.

It cannot be the P or N Piece and Niece eliminate them.

Moving to the second letter I. All three have this in the same place meaning it's possible.

Third letter. E and Q. If E were the only correct place, Pique would have 0. If Q were correctly placed, Piece and Niece would have 0.

Forth letter. C and U. If it were C Pique would be 0. If it were U, Piece and Niece would be 0.

Final letter. E. All words have this letter in common.

Since Second letter, I, and Final Letter, E, are the only 2 letters all three words share and have a 1 score for a correct placement. Since in the word Fines, the I is the second letter and S is the final letter. I must be correct since Fines still has 1 correct letter in place. If the F were correct, all other words would have 0 since no other word uses F. If it were the N Ganja would have 1 since Fines and Ganja share the letter N. if it were the E, all other words would have 0. If it were the S, Grats would have 1 since they share the last S. That leaves ONLY I to be in the correct slot.

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I must be the second letter.

Looking at NIECE, N can't be right since PIECE only has one letter different and they both have 1.

The first E can't be since GREAT has 0, and the last E is wrong because of SPREE having 0.

That leaves the I or the C.

With the P in Pique being wrong because of the PIECE/NIECE thing, that leaves _IQU_ (since the E is still wrong).

With SINUS, we can eliminate both S's with SPREE and DRATS and the N is gone because of GANJA. That leaves _I_U_.

Now we have:

_I_U_ from SINUS, _I_C_ from NIECE and _IQU_ from PIQUE. All of these only have 1, so it can't be the U and the C since they're in the same position. If it were the Q and the C, then SINUS should have 0. If it were the U, NIECE would have 0. The only way that all three numbers can have 1 is if the I is the second letter.

I'll read through PM's logic, but I'll put this up in case something comes up... :mellow:

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I think that you still have a problem in your logic. It starts here:

Third letter. E and Q. If E were the only correct place, Pique would have 0. If Q were correctly placed, Piece and Niece would have 0.

Forth letter. C and U. If it were C Pique would be 0. If it were U, Piece and Niece would be 0.

If you look at each letter in isolation, then what you say makes sense, but based on what you said here, there is nothing that prevents both Q and C from being correct. That would satisfy PIQUE, NIECE and PIECE.

The logic to actually prove it is I is quite convoluted with the current word set. :wacko:

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I think that you still have a problem in your logic. It starts here:

If you look at each letter in isolation, then what you say makes sense, but based on what you said here, there is nothing that prevents both Q and C from being correct. That would satisfy PIQUE, NIECE and PIECE.

The logic to actually prove it is I is quite convoluted with the current word set. :wacko:

As stated in my initial post though I and E are the only 2 letters that ALL 3 words (with a correct letter) have in common. And I is the only letter that fines also has in common with those.

Piece

Niece

Pique

Fines

The second letter I is the ONLY letter that ALL 4 words have in the same place and have a correct letter.

You eliminate Q and C due to Fines having a correct letter but no Q or C. My last paragraph sums that part up...As it states why each letter in fines can't be the correct one other the I.

Edited by Prince_Marth85
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As stated in my initial post though I and E are the only 2 letters that ALL 3 words (with a correct letter) have in common. And I is the only letter that fines also has in common with those.

Piece

Niece

Pique

Fines

The second letter I is the ONLY letter that ALL 4 words have in the same place and have a correct letter.

You eliminate Q and C due to Fines having a correct letter but no Q or C. My last paragraph sums that part up...As it states why each letter in fines can't be the correct one other the I.

But based on what you said, I still think that you could satisfy all of those words if the F in FINES is correct, the Q in PIQUE and the C in PIECE/NIECE. Obviously, F_QC_ won't make (m)any words, but based on the logical analysis, there is nothing to prevent that from being the case. Logic doesn't understand English. If you had a way to prove that the F couldn't be right, then you'd have it, but as it stands, since they are all in different positions, they can all be considered correct without creating a contradiction.

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But based on what you said, I still think that you could satisfy all of those words if the F in FINES is correct, the Q in PIQUE and the C in PIECE/NIECE. Obviously, F_QC_ won't make (m)any words, but based on the logical analysis, there is nothing to prevent that from being the case. Logic doesn't understand English. If you had a way to prove that the F couldn't be right, then you'd have it, but as it stands, since they are all in different positions, they can all be considered correct without creating a contradiction.

F can't be correct since ALL other words would be 0 since on other words contain F...As stated in the paragraph.

Since Second letter, I, and Final Letter, E, are the only 2 letters all three words share and have a 1 score for a correct placement. Since in the word Fines, the I is the second letter and S is the final letter. I must be correct since Fines still has 1 correct letter in place. If the F were correct, all other words would have 0 since no other word uses F. If it were the N Ganja would have 1 since Fines and Ganja share the letter N. if it were the E, all other words would have 0. If it were the S, Grats would have 1 since they share the last S. That leaves ONLY I to be in the correct slot.
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Prince Marth you've fallen into the same trap that others have - in general the phrase "have in common" is a sign of an illogical proof. Sometimes by accident it works out but you have to show why. In the simplest case:

GRAPE - 1

SPARK - 1

the only thing they have in common is the center A. Does that mean A is the middle letter? Not necessarily. It COULD mean that. But it could also be the G in GRAPE and the K from SPARK. Or a whole bunch of other combinations.

~~~

In general the best way to prove a word is to start with a word that's in "one excess", that is, of its grade (1,2,3, or 4), all of those numbers are accounted for as previously-filled-in-letters EXCEPT FOR ONE. Then you look at remaining letters - if N is the grade it got, you'll have to look at the remaining 6-N letters and eliminate 5-N of them, leaving just the single letter that could logically be the desired letter.

To eliminate a letter you find a word with "no excess" - either a Zero or a number with all the numbers accounted for previously-filled-in-letters. This word has the letter you want to eliminate in the spot you want to eliminate it from.

Sometimes proofs are more convoluted (like dawh's just now) but the above is the simplest way

edit ~ I haven't acutally looked at your proof. Yours could be one of those that works out properly. I will right now though

Edited by unreality
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