# Tag Info

65

You can see from clue 4 that there are no 7, 3, or 8. You can see from clue 5 that there is a 0 in either the first or second slot, and from clue 3 that it must be in the first slot. That means the code is [0][?][?]. 6 cannot be the correct number from clue 1, since it would have to be in the wrong position for clue 2. We already know 8 is incorrect, so 2 ...

47

The code is: Explanations: One number is correct and is also correctly placed: One number is correct but wrongly placed: Two numbers are correct and wrongly placed: Nothing is correct: One number is correct but wrongly placed: Approach/Thought Process Chosen 682 - One number is correct and well placed 614 - One number is correct but wrong placed ...

40

My solution: Clues 1 & 2 say number 6 is not included as it cannot be in both the wrong and the right place at the same time Clue 3 then says 2 and 0 are included Clue 1 then says 2 must be in the last position Clue 3 then says 0 must be in the first position Clue 2 then has 1 & 4 as possible values, but only 4 is not in the middle position 042, ...

36

31

Wikipedia has the nice section on optimal Mastermind strategies: In 1977, Donald Knuth demonstrated that the codebreaker can solve the pattern in five moves or fewer, using an algorithm that progressively reduced the number of possible patterns. The algorithm works as follows: Create the set S of 1296 possible codes, 1111,1112,.., 6666. ...

18

We first present a complete description and analysis of an approach that takes Furthermore, we will show that with fewer guesses one cannot guarantee success. Description and analysis of the algorithm With a first guess 00000, the six possible answers are as follows: Score 0: Then the hidden number is 11111. Done. Score 1: Then we succeed with a total ...

16

376 and 370 must share one number and not another. based on possibilities of first number and can't be three or the last number would be repeated twice in one of them. seeing as 3 isn't the first number 7 must be the second number because of 376 and 370. can't be 6 or 0 for the third number or else there would be two numbers with 2 correct guesses so ...

16

The final word is: I had missed three of the clues, but Silenius has helped me to fill in the blanks. #1 #2 #3 #4 #5 Final

16

The secret code is: Explanation: At first glance, my gut instinct (soon to be proved wrong) told me this must be impossible! After all: However, then I realised I had completely overlooked the possibility that: Which colour could this be? Well, we know from guess#3 and guess#4 that we have to have: So: It now follows that the other three must be ordered ...

15

I see no unique solution for this: Unless you assume the solution must uniquely defined by the information given. (Or hopefully simpler terms all the numbers appear in the correct position in the responses.) Naming the friends A,B,C,D gives A: 280, B: 376 , C: 304 , D: 370 Position 1 has possibilities 2,3,3,3 Assuming 3 is right leads to no unique ...

14

It is impossible with fewer than six guesses, because it could happen that the correct color is one that your friend never guesses. This strategy gets every combination with six guesses: Always guess a combination that has not been ruled out yet. If possible, choose a combination with a color that has not been guessed before. If still possible, choose a ...

14

Start by guessing the following N digit numbers: One of these guesses must give us a 50% response because If we have a 50% guess, all we have to do is After this, there are only 2 possible values for the hidden number: This was $(N+1)+(N-1)+2\le2002$ guesses in total.

14

I think the correct password is Reasoning

12

[5, 0, 1]: 2 [1, 3, 5]: 1 [4, 8, 3]: none. [1, 6, 7]: 1 [4, 3, 0]:

11

This is a tricky one, because one of the solution digits is never seen! It must be deduced by elimination. So the solution is

10

• PESTLE 2r0w • VIOLIN 0r0w • PUZZLE 1r1w • CAMERA 0r2w • HELIUM 1r1w • OXYGEN 0r1w • BANANA 0r0w How about:

9

I play Mastermind with numbers instead of colours, because I first learned it in the second grade as Bagel Pico Fermi which uses numbers. For the rest of my answer, I will refer to red pegs as "bagels", and white pegs as "picos" (and holes without pegs as "fermis"). The system I tend to use is suboptimal but very easy to follow. It goes as follows: Start ...

9

9

Using the old Mastermind game colors of B = correct in correct place and W = correct in wrong place, we have the following clues: a. 2657 WW b. 0415 W c. 4268 d. 1749 BB Solution: From c, we know that From 1 and a, we know that From 2 and b, we know that From 1 and 3, we know that From 3 and 4, we know that At this point, we know that the ...

7

According to codebreaker-mastermind-superhirn.blogspot.co.uk The exact year when the number guessing game Bulls and Cows was invented is not known. "Bulls and Cows has been played as a paper-and-pencil game for a century or more. I first played a computer version in 1968 on Titan, the Cambridge University Atlas" (John Francis, 2010 [dead link to "Bulls ...

7

From the first clue, From the fourth clue, Hence there must be Now by the third clue, the code must contain By the second clue, the code contains As it turns out,

7

Your friend needs to find the correct plane out of 72 possibilities. Assuming each question he asks has to be a yes/no question (otherwise he could just ask "what colour is the plane?" etc., making it trivial!), each answer he gets gives him one bit of information. 72 in binary is 1001000, which is 7 bits, so at least 7 questions are needed. He can get the ...

7

Call the three clues 1, 2, 3. Looking at 1 and 3, we can deduce that Therefore two of are in the number. But it can't be Looking at 2 we deduce that Hence Therefore looking at 1, Thus the second last digit must be a since A+B+C+D+E=10D+E. Now the third digit must therefore be a and the sum of the first four digits must be Therefore we must have ...

6

A simple strategy which is good and computationally much faster than Knuth's is the following (I have programmed both) Create the list 1111,...,6666 of all candidate secret codes Start with 1122. Repeat the following 2 steps: 1) After you got the answer (number of red and number of white pegs) eliminate from the list of candidates all codes that would ...

6

Here are some of the cryptics: 1 States of mind of a weird sinful city Branches tableware Rumble of a Greek nocturnal bird First, a mountain combined with the sound of laughter 2 Reasoning hidden in catalog icons Polishes Sam & Sue initially 3 Heartless kissers fly by night A flightless bird in France produces a bird 4 Bill is a potential mate ...

6

I don't know why you have concluded that the scoring ambiguity is not resolved by the Wikipedia page. There is text on that page which, to me, clearly addresses and resolves this issue: If there are duplicate colours in the guess, they cannot all be awarded a key peg unless they correspond to the same number of duplicate colours in the hidden code. ...

5

I really enjoyed this Here is how I deduced this: I started off with clue #4 (nothing is correct). Knowing this, I moved onto clue #5 (one number is correct but wrong placed) We now look at clue #3 (two are correct but wrong placed). Now if we look at clues #1 and #2. Clue #1 states that one number is correct and correctly placed, and clue #2 states ...

5

Any first move is equal, as long as it follows the pattern XXYY. The way you number the holes in the first move is only a base reference point for the subsequent moves. So, choose an arbitrary first move, name those first choices as they are named at the start of the stragey and then apply the complete strategy. The article basically says "start with 1122,...

5

Let's say the number is ABC and let's pick 376 as our starting point. Case 1: C = 6 376 => A != 3 and B !=7 370 => A == 3 or B == 7 Contradiction, so C cannot be 6 Case 2: B = 7 304 => A == 3 or C == 4 376 => A != 3 and C != 6 => C = 4 280 => A == 2 Solution: 274 Case 3: A = 3 280 => B == 8 or C == 0 370 => B != 7 and C != 0 =...

Only top voted, non community-wiki answers of a minimum length are eligible