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: and
: if...then; implies
: not
: ...from this I conclude
1. Complete the following
a)
b)
c)
,
,
,
,
,
2. How might each of the above be useful in solving Sudoku puzzles?
Answers by tomorrow, please. For full marks, show all working.
8^)
Edit: I was only kidding about having to show all working, honest! My 'evil teacher' persona took over... 8^P
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b) FALSE
c) umm.. yeh... if I had more time and less work to do...
also... how did this happen??? it's the freakiest thing ever... I typed in the italics tags and it showed up backwards...
soluopotoyanaP xelA (spudtater) wrote,
2007-05-29 11:08:00
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b) B
c) ¬D
Not sure what you're getting at with (c). For (a), if an assignment generates a contradiction it's wrong, with (b) if some other assignment comes out either way, it must be the case. Again, (c) doesn't seem to say much, if you must choose exactly 1 option from 3 and you know it's one of two options, then you know it's not the third and you must choose 1 from 2 (B xor C, ie B <-> ¬C, in this case)
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Well, it's useful for Sudoku, anyway.
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1)
a)
If A is false, the statements on both sides of the and will be true as 0->* = 1. If A is true, then one of the statements will be false. 1->B will be false if B is false and 1->¬B is false if B is true. Thus, the statement boils down to:
¬A
b)
As the implies truth table is only false where 1->0, we can tell that this will only be the case when B is false, meaning that B must be true for the statement to be true. Further, when B is false, either the left or right side will be false depending on A's state (as the right side merely has the negation of A on teh left side). Thus, the statement boils down to:
B
c)
I took the liberty of doing a big truth table. The columns W to Z correspond to the 5 statements in the C section, R is the result as a whole, assuming that all 5 statements are effectively connected with logical and statements.
A BCD W V X Y Z R
0 000 1 1 1 1 0 0
0 001 1 1 1 1 0 0
0 010 1 1 1 1 1 1
0 011 1 0 0 1 1 0
0 100 1 1 1 1 0 0
0 101 0 1 0 1 0 0
0 110 0 0 1 1 1 0
0 111 0 0 0 1 1 0
1 000 1 1 1 0 1 0
1 001 1 1 1 0 1 0
1 010 1 1 1 0 1 0
1 011 1 0 0 0 1 0
1 100 1 1 1 1 1 1
1 101 0 1 0 1 1 0
1 110 0 0 1 1 1 0
1 111 0 0 0 1 1 0
As you can see, the statement is only true in two states, ¬A¬BC¬D, and AB¬C¬D. From this I conclude very little.
2)
I haven't the faintest clue.
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c) would be a lot prettier if we had XORs.
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b) B
c) ¬ D∧¬(B∧C)∧¬(¬B∧¬C) (since we don't seem to have xor in this universe).
2) If a figure (in a Sudoko) leads to a contradiction; it is wrong: if a figure arises on all assumptions whatsoever, it is correct: if a figure leads to further figures some one of which must be wrong, the original figure is wrong.
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In sudoku, this third fact is useful because if you've got two squares, say {3|5} and {1|7|9}, you can test out what happens when you try the two numbers. Choose {3} in the former and you end up, say, with {9} in the latter. Choose {5} and you end up with {1}. You conclude that {7} is an impossible value for that square, and erase it.
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1)A→B - premise
2)A→¬B - premise
3)A - assumption
4)B - modus ponens, 1,3
5)¬B - modus ponens, 2,3
6)B∧¬B - 4,5
7)¬A; - proof by contradiction, 3, 7.
Pretty!
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*Crash...*
*THUD*
*Spring goes boing* *
And left in a twittering state...* Me congratulate you.
Me no long able to work... Me not sure can answer quiz questions, tho me will be there tonight... me think...
*screech... *
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1.(a) ¬A (because if A then B & -B)
(b) B (B is true regardless of A)
(c) Either, AB¬C¬D or ¬A¬BC¬D, so I can conclude ¬D and B xor C ?
No full proofs because well, because really. It's been a while and I can't remember all the individual steps one would usually put in a logical proof. I also can't find my logic notes.
2. (a) If by placing a number in a specific position you generate a contradiction, that number is incorrect. I can imagine where that would occur, but at the moment I can't see where both B & -B would flow directly from A.
(b) Sometimes a square can only take one number. Take for example the circumstance where you need to position 1, 4 & 9 to complete a 3x3. It may be the case that you know that 2 squares could take any of the three numbers, but the third square can only be 9. The placing of that 9 is correct irrespective of the placing of the 1 & 4.
(c) Again I can see how an iterative process would lead you to this pattern of options. If you have 'D' then you've gone wrong somewhere. Beyond that...*shrug*
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1c): B xor C is already fairly obvious, so ¬D is the interesting bit.
2a): You're mostly there. Try pencilling in a '5' in a square as a possibility. If you end up having to put in, say, a '7' in a row which already contains a 7, then you've got your contradiction. (Square is 7) & ¬(Square is 7). The initial assumption is wrong, and you can forget '5' as a possible value.
2b): A little more complex than this. Say you have a square which can only take values {1|9}. Let's call '1' A and '9' ¬A. Work out both possibilities simultaneously. You might find that when working from '1', a different square has to be a '6'. Working from '9', the same square is also a '6'. Using fact b), we can ink in '6' as being the only possible value.
2c): I refer you to my answer to