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#251 Re: Help Me ! » Variation of mastermind game » 2014-08-14 02:56:30
Ah, I think I get the hang of the game. 
Make your first move. If you have two or more black pegs, we are done (1 step). Otherwise you have either 0 or 1 black peg.
Case 1: 0 black pegs.
Try again, using a different colour for each position.
… (i) If you get 0 black pegs again, you know the correct combination (it’s the one where each position has the colour you haven’t tried for it) – and you can play it on your third move. In this case we achieve two or more black pegs in 3 steps.
…(ii) If you get 2 black pegs, we are done (2 steps).
…(iii) If you get 1 black peg, leave two colours and change one.
… … (a) If this gives you 0 black pegs, the colour you’ve changed was right before. In this case repeat step 1(i) and we are done in 4 steps.
… … (b) If this gives you 2 black pegs we are done (3 steps).
… … (c) If this gives you 1 black peg, the colour you’ve changed is still wrong – AND was wrong before. In this case you know the right colour for that position so you can put the right colour in that position for the rest of your movves. The black peg is for one of the other two positions. Change one of them. If the one you change is not the black-peg one, we are done (4 steps). Otherwise restore this colour and we are done (5 steps).
Case 2: 1 black peg.
Leave two colours and change one.
… (i) If you get 2 black pegs, we are done (2 steps).
… (ii) If you get 0 black pegs, the colour you’ve changed was right before. Restore it and change one of the other two.
… … (a) If this gives you 2 black pegs we are done (3 steps).
… … (b) If this gives you 1 black peg the colour you’ve changed is still wrong – AND was wrong before. So you know the correct colour for that position and we are done in 4 steps.
… … (c) You can’t get 0 black pegs because at this stage you already definitely the colour of one position.
… (iii) If you still get 1 black peg, the colour you’ve changed is still wrong – AND was wrong before. In this case you know the right colour for that position so you can put the right colour in that position for the rest of your moves. The black peg is for one of the other two positions. Change one of them. If the one you change is not the black-peg one, we are done (3 steps). Otherwise restore this colour and we are done (4 steps).
So the most number of steps you need to take is actually 5 (Case 1(iii)(c)) if you reason correctly.
#252 Introductions » Olinguito » 2014-08-14 02:08:39
- Olinguito
- Replies: 6
It is exactly a year since I was discovered in the forests of South America. 
#253 Re: Help Me ! » Functions » 2014-08-14 01:53:45
Examine their domains (i.e. values for which each function is defined). Take, for example, x = −4. Is f and/or g defined at this value?
#254 Re: Help Me ! » Challenging proof. » 2014-08-14 00:36:11
#255 Re: Help Me ! » Challenging proof. » 2014-08-13 22:32:33
This is known as the sandwich theorem or squeeze theorem.
#256 Re: Help Me ! » Variation of mastermind game » 2014-08-13 20:26:06
Can you please explain "Hence there are 27 − 6 − 1 = 20 permutations giving at most one black peg"?
There are 6+1 permutations that earn at least two black pegs in the game so the number of those that give you at most one black peg is 27−(6+1).
#257 Re: Help Me ! » Inverse trig functions » 2014-08-13 20:20:33
Indeed:
[list=*]
[*]
[/list]
Proof:
[list=*]
[*]
[/list]Thus in the interval .
#259 Re: Help Me ! » Variation of mastermind game » 2014-08-12 19:55:53
Well, just from the way you phrase the question, the answer is 1. You might strike it lucky on your first move.
But I think you mean what is the minimum number of moves needed to be guaranteed at least two black pegs, assuming a worst-case scenario. Since repetition of colours are allowed there are 3³ = 27 permutations altogether. To get exactly two black pegs, the wrong position can be any of the 2 wrong colours, and there are 3 positions for the wrong colour to be, so there are 3 × 2 = 6 such permutations. And there is only 1 permutation for three black pegs (the winning sequence). Hence there are 27 − 6 − 1 = 20 permutations giving at most one black peg.
Therefore the answer to your question (assuming you don’t duplicate any of your moves) is 21. In the worst-case scenario you would try all the 20 permutations giving fewer than two black pegs first before you get warmer.
#260 Re: Help Me ! » Help with logic question please » 2014-08-12 11:58:03
You need to use the fact that there are some stones:
#261 Re: Jokes » Anagrams » 2014-08-12 11:34:33
Butterfly = Flutter by
#262 Re: Help Me ! » ans » 2014-08-12 11:10:37
#263 Re: Help Me ! » Geometric Proof » 2014-08-12 10:50:24
You have to do two things:
[list=1]
[*]Show that CD is twice as long as AB.[/*]
[*]Show that AC is perpendicular to BD[/*]
[/list]
1. is straightforward. Let E be the foot of the perpendicular from A to CP and F the foot of the perpendicular from B to PD. Then AEFB is a rectangle, AE bisects CP and BF bisects PD, and |CD| = |CP| + |PD| = |CE| + |EP| + |PF| + |FD| = 2|EP| + 2|PF| = 2|EF| = 2|AB|.
2. is more complicated. I’m afraid I can’t do it geometrically but I can do it using co-ordinate algebra. The working is too complicated so I’ll omit it here.
Using 1. and 2. we can solve the problem. Let Q be the midpoint of CD (so |CQ| = |QD| = |AB|). Then QB is perpendicular to BD since AC is perpendicular to BD and ACQB is a parallelogram. Thus QB is the perpendicular bisector in triangle QDG, where G is the point on BD extended on the side of B such that |DB| = |BG|. Also (by similar triangles) QG intersects AB at R, the midpoint of AB. Now triangle RBG is similar to triangle QDG and so is an isoscles triangle. Thus |RB| = |RG|. But by similar triangles |RG| = |RQ|. Thus |RB| = |RQ|. QED.
#264 Re: This is Cool » How to find whether a given number is prime or not? » 2014-08-12 09:33:27
We don’t actually need K. We just have to test whether A is divisible by any prime less than √A.