bushindo gets 20:16 A vs B, 24:12 B vs C, and 24:12 C vs A. am i missing something? perhaps a tweak could bring A:B at the same odds as the others

A conjugate descent algorithm allows me to improve the odd a bit further, now it's 21:36. It doesn't seem possible to get the lowest odd up much further.

     [,1] [,2] [,3] [,4] [,5] [,6]

[1,]    3   16    2   15    4   17

[2,]    6   18    5   10    9    7

[3,]    1   13   12   11   14    8

I got the same odds as DeeGee with slightly different dice can anything better be achieved?

Die 1 die 2 die 3

2 3 1

5 6 4

9 7 8

12 10 11

13 14 15

16 17 18

I can get 20:16 odds, but even better odds might be possible.

This is an extension but my earlier smaller example from 1-9.

Dice A: 9,3,2

Dice B: 8,7,1

Dice C: 6,5,4

We can see from above that the smallest edge is 5:4 (Dice A over dice B). We can extend this to 1-18 by letting the numbers in the smaller example be consective numbers in the larger example (i.e. smaller example number 1 = larger example number 1 and 2, smaller example number 8 = larger example number 16 and 15). So this configuration would have the same 20:16 odds as the 1-9 example

Dice A: 18, 17, 6, 5, 4, 3

Dice B: 16, 15, 14, 13, 2, 1

Dice C: 12, 11, 10, 9, 8, 7

But better odds might be posssible by taking advantage of the finer granularity of the dice. Still working on this, but what I would really like is a conclusive proof of optimality, as opposed to just an example though.

one player does have an advantage...

If only I get paid for working on puzzles...

It is possible to construct 3 dice so that dice A beats dice B, B beats C, and C beats A. Then the one who gets to construct the die always win. Here's a smaller example from 1-9.

Dice A: 9,3,2

Dice B: 8,7,1

Dice C: 6,5,4

We can extend this to the case of 1-18 easily. The optimality case will have to wait. I'm in a meeting right now, so I have to pretend to be working.

Six words were written one below the

other in a 6x6 square. The numbers

0 through 5 were written in scrambled

order down the side and the words

shifted to the right (cyclically)

these amounts. The same sequence of

numbers was written across the top

of the resulting square and the

various columns shifted down

(cyclically) these amounts. The

result was:

     Y R O O N Y

     O L A O P O

     A S T N T C

     E S T U R M

     I E N E S G

     B S P E R R

What were the words?

Excellent puzzle. I resisted the urge to brute force this puzzle. The paper-and-pencil method, surprisingly, is faster and a lot more satisfying.

What's needed now is a little intelligent guessing. For example, what letter is likely to be in front of NG? Work out all of the consequences of this guess then do some more intelligent guessing. So far, you're doin' good!

The poet is right, it is such a joy.

I'll keep the plaintext hidden for other solvers. This is a good puzzle. I find that the guessing process goes a lot faster once I write a function that takes in an incomplete key and construct the plaintext. The whole process then just involves adding guesses to the key and see how the reconstructed plaintext looks

Agreed.. that is the first part. Are there any other possible solutions?

Well, I tried writing a matrix, but logic is so much easier.

Imagine the cell values for the chess board for such a configuration that satisfies the original post. There must be a maximum value for the entire 64 cells. Select the cell with this maximum value, call it M. By definition of maximum, since its neighbors' values can not exceed M's value, all the neighbors must be equal to M in order for M to be the average of its neighbors. Extend this argument for the entire chessboard and we'll see that all cells have to be equal.

Thanks for the clarification. So, my understanding is that

1) The sequence of 26 alphabet letters are scrambled. We call this the key. A plaintext is chosen

2) The 1st letter of the plaintext is encoded as the right-hand neighbor of the same letter in the key.

3) The key is advanced 1 position.

4) Repeat step 2 and 3 with the 2nd letter of the plain text, then the 3rd, and so forth.

I just need one last clarification. If the key is a scramble of the alphabet, then the last letter of the this sequence does not have a right-hand neighbor. Does the wrap-around rule, i.e. right-hand neighbor of last letter is the first letter in sequence, apply here?

Each letter of a message is replaced

by its right-hand neighbor in a

certain scrambled sequence. After

each step all the letters in this

scrambled sequence are advanced one

position in the normal alphabetical

sequence.

I'm not sure about the instructions. My impression is that

1) There is a certain sentence of unknown length (Maybe this key is the same as the plaintext, but scrambled. I'm not sure about this). Spaces are not present. Call this the key.

2) Each letter in the plain-text is replaced with the right-hand neighbor of the same letter in the key.

3) After encryption once, the letters of the keys are advanced one position.

I'm not too clear about step 3. Does it mean that step 1-3 are repeated several times on the same plain-text?

I understood this problem a little differently. I figured you just keep going around and around and around until 1 of 2 things happens. Either you are on 'square 4' and you flip "heads" to move 2 spaces (or whichever of heads/tails was chosen to be 2), OR you are on 'square 5' and you flip "tails" to move only 1 square.

Well...maybe I understood the question the same, but this is what I got:

There is a 1/2 shot that you will get exactly 6 points because the first time you go around, you've got 50/50 that you'll land on 'square 6'. You've got a 1/4 shot that you will get exactly 12, landing on 'square 6' on exactly the 2nd round. 1/8 says you'll land on square 6 exactly the 3rd time around, giving you 18 points. 1/16 that you get 24 points, etc.

So...the formula in my mind is (1/2 * 6) + (1/4 * 12) + (1/8 * 18) + (1/16 * 24) + (1/32 * 30) + (1/64 * 36) + (1/128 * 42) ...... + [1/(2?n) * (6*n)]

Unfortunately, I got a "D-" in Calculus about a million years ago in high school, so I can't be positive; but after manually going out to about 12 powers of '2', it seems that the number is approaching 12.

So, 12 is my guess.

I've still yet to be right at one of these, so if I'm wrong AGAIN, I've always appreciated someone pointing out where I went wrong!

You have the right idea, except that you started with the wrong value for the change of landing on a 6, which isn't 1/2. The post above described one way to get that value. Here's another way to get that value

The chance of landing on a 6 changes depending on what square you are presently on. If you are on square 5, your chance of landing on a 6 is 1/2. If you are on square 4, your chance of landing on a 6 is 3/4. You want to compute the chance of landing on a 6 from square 1. Let P(N ) be the chance of landing on square 6, given that you are on square N, N != 6. The following system of equation becomes apparent.

P(1) = (1/2) P( 2 ) + (1/2) P(3)

P(2) = (1/2) P( 3 ) + (1/2) P(4)

P(3) = (1/2) P( 4 ) + (1/2) P(5)

P(4) = (1/2) P( 5 ) + (1/2)

P(5) = (1/2)

If you start from P(4) and solves upwards, you'll get P(1), which you can plug into your existing equation to give the right answer

I started by figuring the most likely letters to begin a word and worked from there. Once the first word was figured, well the saying became quite clear.

Laugh and the world laughs with you Weep and you weep alone

The brute force method is conceptually clear, but is a devil to program...

The full quote can be exhaustively searched in 4¹² searches. However, if we only want the top row and bottom row in plain text, we only need to search a 4⁴= 256 space, since the cardboard must include 4 and only 4 holes in the top and bottom row. Because of the rotation and reflection, to specify all possible configuration we only need to list 4 holes. Each hole can either occur in one of four quadrants, giving the space 4⁴ states. The reconstruction of the plain text, given a 4-length vector of holes-configuration is a devil to program. Once we construct a set of all 256 possible top-bottom row of plain text, this is the only one that stands out

[1] "LAUGHAND"

[1] "EEPALONE"

Of course, once we have the top row and the bottom row, we can apply the same method and get 256 possible combination of row 2 and row 5. However, the existing plaintext is a great help, and makes finishing a lot easier.

Circular hopscotch is played on six

squares arranged in a circle. A coin

is tossed and a player starting on

square 1 jumps, advancing either 1 or

2 squares according as the coin falls

head or tail. When he lands on square

6, he is 'out'. Each player gets a

score equal to the number of squares

stood on or passed over during his

turn. (Thus the only possible scores

are multiples of 6.) What is the

average number of points a player may

expect to score?

SUPERPRISMATIC CLARIFICATION:

Each advancement ( 1 or 2 squares)

is preceded by a separate coin toss.

That is to say, a single coin toss

does not control all the jumps a

player makes.

Is it weird of me to say that I start to experience withdrawal symptoms if I go too long without seeing a new Walter Penney puzzle?

average number of points is 9.142857. This is a geometric distribution, or distribution of time to first success (a success being defined in this case as getting a 6). This game can be discretized into rounds of trying to get to 6 from 1. If you pass by the 6 without landing on it, then you must be at square 1 and another round starts. The number of points for every round is 6. A system of recursive equations shows that the probability of successfully getting to 6 from 1 is always 0.6562500. The expected number of round before getting a 6 is 1/0.6562500 = 1.523810. Multiplying that by 6 gives the expected score of 9.142857.

I just reviewed this, and I found that the construct proof can be simplified greatly.

For any sequence S of length L after the decimal point (i.e. S = .12345678), we would like to construct an N such that sqrt(N) contains the subsequence S.

Let K = 10^L+1. The real number x that contains the sub-sequence S in its square root is

x = (K + S)^L+1 = K² + 2*K*S + S²

We approximate this with the integer N = K² + 2*K*S, which essentially amount to applying the floor function to x. Note that because we choose K = 10^L+1, N is guaranteed to be an integer. The error term is S²/(K + S)². Note that this error term is very small, even smaller than the smallest digit in S. We can arbitrarily make the error term even smaller by choosing a even larger K.

AWAKE and CYCLE would have worked. And so would BADGE, CABOB, and CABLE.

Claim: For any finite decimal sequence S of length L (i.e. S = .34566554, S = .23123124, ...), there exist a integer number N such that the square root of N includes the subsequence S.

Proof: Imagine the plot of sqrt(x), x being a real number. This plot is continuous, and monotonically increasing from 0 to infinity. If we take g(x) = sqrt(x) modulo 1 (basically taking only the part behind the decimal point), then the range of g(x) goes between 0 and 1. g(x) increases from 0 to 1, then cycles over back to 0 and increases again, this continues ad infinitum. By continuity, it is not difficult to see that we can easily find a real number, x, such that sqrt(x) modulo 1 includes the subsequence S.

Our problem is that only integers, N, are allowed. So the graph of g(N) looks very much like g(x), except that it is only evaluated at discrete integers. The overall properties of g(N) remains the same in that it increases from 0 to 1, and then cycles back to 0 and increases again. Without loss of generality, let's say that our subsequence S = .12345678. In order for the claim to be false, g(N) must not fall within the interval I = [.12345678, .12345679) for any N. As N increases, g(N) must cycle through (0,1) monotonically, but with ever smaller step size increase. Eventually, the step size will become smaller than the interval length I, and will have to fall inside I, producing a N that satisfies the claim.

This is an existence proof, meaning that while we know such an integer N exist for any subsequence S, we can't produce the number N. My guess is that if we quantity the step size as some function of N, then we can work out some sort of method to predict how many steps it would take before g(N) falls inside the desired interval I.

That said, it seems that any word constructed from dms172's list might have been a password, since we just showed that for any arbitrary subsequence we can find a integer N that satisfies the original post. There are 5^10 possible words, but here are the non-gibberish words.

fudge

fugle

fable

facia

fadge

faena

etape

ethic

eying

eagle

dwine

dying

dacha

dagga

cubic

cuing

cycle

cable

cabob

cache

cadge

budge

bugle

build

bwana

babka

badge

asana

asdic

awake

awing

azine

azine

abaka

abele

wacke

Here's your second number, and a third, and a four, and so on...

In the last post we have an existence proof, namely that for any sequence S there exist an integer N such that sqrt(N) includes the subsequence S. This is the construction proof for deriving such an N.

Let S be a finite sequence of decimals after the decimal point (S= .12345678, S = .45324543, etc. ). Let L be the length of S. We wish to find an integer K such that

(K + S )² = 0 mod 1

Of course, for finite S, there is no such integer K so that the equation above is exactly true. But we can get the differences between the two terms to be artitrarily small, which is good enough. Expand the equation above

K² + 2*K*S + S² = 0 modulo 1

2*K*S + S² = 0 modulo 1 since K² is an integer (Equation 1)

Expand this back into a normal equation

K = i * ( 1 - S^² )/( 2*S) where i is some integer (Equation 2)

Let i = 10^L+1 * ( 2*S)/( 1 - S^². That is, we let i equal to the the floor function of the inverse of ( 1 - S^² )/( 2*S) multiplied by 10^L+1. In that case, K = 10^L+1. We can plug K and S back into equation 1, and derive an N = (K + S )² as the integer that satisfies the original post.

Because of the truncation we did to i to make it an integer, these K and i do not make Equation 2 exactly equal. However, the error is very small, and let's examine it.

Let R, 0 < R < 1, be the error remainder term such that (K + R + S)^2 is an integer. We can expand this term as

(K + R + S)² = ( K² + 2*K*S + S² ) <--- this is our estimate N

+ R( 2K + R + 2S ) <---- this is the integer distance from N to the next perfect square.

The error in decimal term is R( 2K + R + 2S )/(K + R + S)², which reduces down to 2R/K = 2R/10^{L + 1}. As it is, the error is small enough in magnitude that it doesn't affect the subsequence S after the decimal point.

So, the summary is

1) Select the sequence S of length L.

2) Construct N = ( 10^{L + 1} + S )². Apply the floor function if needed.

3) This N is your desired number.

First one is correct (and that is the only method I know). I think the second one may not be a solution..

1/p in binary may not terminate.

Let me refine bonanova's approach a bit

Let head=1, tail = 0. Suppose that you flipped a coin three times, and the number x turn out as .101. If you continue flipping until infinity, the number x must belong in the interval [ .10100000000..., .1011111111111...], which has a width of 1/2ⁿ, n being the number of flips. So we revise bonanova's strategy as follow

1) Divide the interval into p parts [0, 1/p), [1/p, 2/p), ... [(p-1)/p ].

2) Flip the coin as many times as it takes. Update the value of x at every turn.

3) Stop flipping once the interval [x, x + 1/2ⁿ ] belongs entirely within one of the p intervals from part 1).

I'll work on the optimal part later. Need to do other work that actually paids for now.

I essentially brute forced it with a program I wrote in python.

I first wrote out all the letters of the alphabet and put the numbers given next to each one.

Since we are dealing with modulo N, we know that a number can never appear here that is larger than N.

Therefore N must be greater than 377.

For numbers in the Fibonacci sequence that are less than N, a modulo N operation will have no effect.

Therefore each of the numbers in the sequence before 377 must appear at least once.

You can definitely assign the letter M to 21 since it is the only possible place to put 21, and 21 was not given yet must appear at least once.

For letters like F, you cannot be sure if a 2 or a 3 belongs there.

There are 10 missing letters, and one duplicate (the 34), so you know the sequence continues 11 elements past 377.

To solve the problem, you should look at these numbers within the Fibonacci sequence, perform the modulo N operation on them,

and see if they can fill these categories without contradiction.

You can come up with groups of numbers that you should look for.

Since there are an A and a B slot, which are surrounded by a 0 and 1, we can define group 1;

Group 1 will keep count the 0's and 1's we get from the modulo N operation. There should be exactly 2.

Letters F and H are more of a problem since they share 3. If you find a 3, you cannot be sure of where to put it until you have

found a number satisfying the remaining letter.

We will make 3 different groups therefore, one to count 2's, one to count 3's, and one to count (4's and 5's).

Group 2 will count 2's. There should be at most 1 of these.

Group 3 will count 3's, The sum of the counts of group 2, group 3, and group 4 should not exceed 2.

Group 4 will contain 4's and 5's. There should be at most 1 of these.

Group 5 will contain a count of numbers found that lie between 55 and 89 inclusive (for the letter Q).

There should be 1 of these.

Group 6 will contain a count of numbers found that are equal or greater than 377.

There should be 5 of these.

Group 7 will account for the term that produces that duplicate 34.

It will count the number of 34's. There should be exactly one of these.

I first generated the first 26 numbers in the Fibonacci sequence.

I then selected a range of numbers to test N for. If I don't find anything at first, I can make this range wider to cover more numbers.

For each of these N, I go through the latter 11 (I think) numbers in my generated list of Fibonacci numbers (the ones greater than 377) and take the number modulo N. I classify the number into one of the groups mentioned above and increase the counter of that group by 1. If it does not fit into any of those groups, we move on to the next N since that N is incompatible with the key.

I then check to see if any of the groups grew too large (Group 1's count exceeding 2 for example, or group 6 exceeding 5).

If any of those conditions fail, we move on to the next N.

If we can get through all 11 numbers from the Fibonacci sequence without breaking any of these rules, it is very likely we have found the right solution. The program stops and informs me of a potential match. I then double check everything and there we go, we have an answer. It just happened to be 521.

Thanks for the detailed description. It's good to see a fellow Python user on the forum. And a better welcome to the forum, by the way.

F B I P I

E A H O H

D Z G N G

C Y F M F

B X E L E

A W D K D

Z V C J C

Y U B I B

X T A H A

W S Z G Z

How do you know it isn't AWAKE? I'm sure there are many words you can form by taking one letter from each column.

How about CYCLE?

AWAKE and CYCLE would have worked. And so would BADGE, CABOB, and CABLE.

Claim: For any finite decimal sequence S of length L (i.e. S = .34566554, S = .23123124, ...), there exist a integer number N such that the square root of N includes the subsequence S.

Proof: Imagine the plot of sqrt(x), x being a real number. This plot is continuous, and monotonically increasing from 0 to infinity. If we take g(x) = sqrt(x) modulo 1 (basically taking only the part behind the decimal point), then the range of g(x) goes between 0 and 1. g(x) increases from 0 to 1, then cycles over back to 0 and increases again, this continues ad infinitum. By continuity, it is not difficult to see that we can easily find a real number, x, such that sqrt(x) modulo 1 includes the subsequence S.

Our problem is that only integers, N, are allowed. So the graph of g(N) looks very much like g(x), except that it is only evaluated at discrete integers. The overall properties of g(N) remains the same in that it increases from 0 to 1, and then cycles back to 0 and increases again. Without loss of generality, let's say that our subsequence S = .12345678. In order for the claim to be false, g(N) must not fall within the interval I = [.12345678, .12345679) for any N. As N increases, g(N) must cycle through (0,1) monotonically, but with ever smaller step size increase. Eventually, the step size will become smaller than the interval length I, and will have to fall inside I, producing a N that satisfies the claim.

This is an existence proof, meaning that while we know such an integer N exist for any subsequence S, we can't produce the number N. My guess is that if we quantity the step size as some function of N, then we can work out some sort of method to predict how many steps it would take before g(N) falls inside the desired interval I.

That said, it seems that any word constructed from dms172's list might have been a password, since we just showed that for any arbitrary subsequence we can find a integer N that satisfies the original post. There are 5^10 possible words, but here are the non-gibberish words.

fudge

fugle

fable

facia

fadge

faena

etape

ethic

eying

eagle

dwine

dying

dacha

dagga

cubic

cuing

cycle

cable

cabob

cache

cadge

budge

bugle

build

bwana

babka

badge

asana

asdic

awake

awing

azine

azine

abaka

abele

wacke

N = 521

A full period of the Fibonacci sequence modulo 521 is:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 89, 466, 34, 500, 13, 513, 5, 518, 2, 520, 1

When placed in ascending order, you get

0, 1, 1, 1, 2, 2, 3, 5, 5, 8, 13, 13, 21, 34, 34, 55, 89, 89, 144, 233, 377, 466, 500, 513, 518, 520

Assigning to each letter,

A 0

B 1

C 1

D 1

E 2

F 2

G 3

H 5

I 5

J 8

K 13

L 13

M 21

N 34

O 34

P 55

Q 89

R 89

S 144

T 233

U 377

V 466

W 500

X 513

Y 518

Z 520

This is the correct key for encoding and decoding the text presented in the question.

Nice work. I'd like to know your approach, if you'd like to share.

The value of the sequence 1, 2, 3, 5,

8, ... , where each member is the sum

of the two preceding, are reduced

mod N; i.e., if any value is greater

than N, N is subtracted. The

resulting sequence has period 26, and

the values, in numerical order, are

assigned to the letters A to Z. Note

that by this scheme the same number

may represent more than one letter.

The message, "Gardening is just a soil

sport," converted into numbers by this

process becomes: 3 0 89 1 2 34 5 34 3

/ 5 144 / 8 377 144 233 / 0 / 144 34

5 13 / 144 55 34 89 233. Find N.

SUPERPRISMATIC CORRECTION: "greater

than" should be "greater than or equal

to" in the explanation of "mod N" in

the first sentence. Walter Penney

was fallible.

I'm not sure that I understand the instruction in this puzzle. Is the following interpretation of the problem correct?

1) A number N is chosen, where N is larger than the largest number in the cipher code.

2) The fibonnaci series F(i) modulo N has a period of 26, meaning that for any index i, F(i) mod N == F(i + 26) mod N

3) The list of 26 values from within 1 period is sorted in ascending order, and each number assigned a letter according to its ordinal index (smallest number is A, second smallest is B, and so on).

4) This transformation is then applied to "Gardening is just a soil sport" to get ( 3 0 89 1 2 34 5 34 3/ 5 144 / 8 377 144 233 / 0 / 144 34 5 13 /144 55 34 89 233 )

5) Find N

PS. Also, does this sequence start at 1 or 0? Many references for the fibonnacci series start at 0.

Using this representation of the function, F = (x + 1) / ((y + 1) / F)), I have exhausted all possible distinct quadruplets of positive integers. Answer: none exists. If this formula is wrong please post a spoiler for me. Thanks.

This may be the reason for your results

F = x +1/ (y+ 1/(x + 1/(y+ ....))) = x + 1/(y + 1/F), which is an entirely different expression than the one in your post, F = (x + 1) / ((y + 1) / F)).

Making the groups 2, 3, 4 does not necessarily produce a win. The girl can take 2 from the group of 4 to make groups of (2,3,2).

Start analysis from the bottom up,

The final winning move will occur with a single group of 1 or 2 petals.

(1) and (2) are winning positions.

Let the the numbers in the parenthesis represent separated groups of the designated sizes.

The positions: (1,1) and (3) are losing positions because they put the other player in the winning position no matter what.

The position (2,1) is a winning position because it can put the other player in a losing position (1,1).

The position (2,2) is a losing position because for any possible action, it will put the other player in a winning position,

since (2) and (1,2) are both winning positions.

Following this line of reasoning: these are all winning positions:

(1), (2), (1,2), (1,3), (1,1,1), (1,1,2), (4), (5), (2,3), (1,2,2), (2,2,2), (2,4), along with more

By adding the petals that could be removed in one turn, and double checking, the following configurations are shown to all be losing positions:

(1,4), (1,1,3), (1,1,1,1), (6), (2,5), (2,2,3), (3,3), (2,5).

I have not built on my list further, but this is enough to prove that putting the girl in (2,3,4) is putting the girl in a winning position (meaning if she makes all the right choices, she will win regardless of your choices).

She would simply need to take two petals from the group of 4 to put you in the losing position (2,3,2) or (2,2,3).

The results of any possible move you could make would put the girl in another winning position:

(2,3) which is a winning position since it can put the enemy in losing position (2,2)

(1,2,3) which is a winning position since it can put the enemy in losing position (1,1,3)

(2,2,2) which is a winning position since it can put the enemy in losing position (2,2)

(2,2,1) which is a winning position since it can put the enemy in losing position (2,2).

The girl will make her appropriate move and put you in position (2,2) or position (1,1,3).

Let's examine position (2,2), possible results after your move are to place the girl in positions:

(1,2) or (2)

If you place her in (1,2)

she will place you in (1,1), you will take 1, and she will win.

If you place her in (2), she will take both petals and win.

Let's examine position (1,1,3):

The possible results you can put her in are:

(1,3)

(1,1,2)

(1,1,1)

If you place her in (1,3), she will put you in (1,1) and she will win.

If you place her in (1,1,2), she will put you in (1,1) and she will win.

If you place her in (1,1,1), she will put you in (1,1) and she will win.

Therefore if she can put you in position (2,3,2) she can win no matter what you do if she makes the right moves.

By putting her in position (2,3,4), you allow her to place you in position (2,3,2) and win.

I have not found the correct answer, but I believe I have shown that making groups of 2, 3, and 4 will not necessarily secure a win.

To solve the puzzle, we would just need to find a losing position that can be created from (2,8) in one move.

There is a mistake in your reasoning.

You correct showed that (2,2,2) is a winning position. The mistake is that you assumed that if the boy gets to the state (2,3,2), he would remove 1 from the edge of the group of 3, hence handing the winning position of (2,2,2) to the girl. However, if the boy were to remove the middle petal from the group of 3, he would hand the losing position of (2,1,1,2) to the girl.

In general, if the boy can reduce the configurations to an even number of groups of size 1, an even number of group of size 2, or an even number of groups of size 3, then he can win. He can also win if he can produce any combination of even groups of size 1 to 3 (i.e (1,1,2,2) ). Induction will show that group size doesn't matter, he can win as long as he has an even number of groups for any size.

If there are three petals all adjacent (in a row), the boy could pick the middle petal, making it impossible for the girl to pick the last two. In this case, the boy would win.

I need my day's allowance of Walter Penney puzzle!

Lemma: if the boy can reduce the groups on the daisy to an even number of 1's, an even number of 2's, an even number of 3's, or any combination of even groups of 1 to 3, he can win. Likewise, if the girl can reduce the groups to combination of even groups of 1 to 3, she can win.

Start from the group of 8, let the boy pick the fourth petal, leaving the daisy with groups of 2, 3, and 4. You can easily see that doesn't matter what the girl pick next, the boy can reduce it to his winning combination within 1 move or 2 moves.

(if you could call it a method <g>)

I play a *lot* of scrabble. (OK, I may have been slightly exaggerating on the speed...it took about 15 mins or so )

For this one, the Qu is pretty obvious...

In addition, you could see that you run out of vowels really quickly. I was operating under the assumption that no letter was repeating. so we're talking (discounting the U paired with the Q) 1 long word, 2 words (1 with 2 vowels and 1 with 1) or 3 words, each with one vowel (i discarded the "scrabble words" that are silly in the context of something like a walter penney puzzle <g>)... In any of the cases you realize that you will have more than one few consonant connectors. You're talking stuff like NT, LF, FT, CH, TH, LT, RK etc.

I operated under the assumption that it was at least 2 words. When dealing with Qu, I tend (just habit, I guess) I look for words that START with QU.

You can see rather quickly that

QUI

QUO

QUE

are your starters. Plus I had the puzzle in mind, and it happened that I was able to get quirk quickly (I ran through Quilt, Quick, Quench,

(eliminated because of two vowels + u) and hit Quirk, which fit the puzzle. (although I was thrown and didn't see it right away because of the lack of "Y" I was looking for an adjective)

that gave me:

E O B C F H L N T

Still operating under the assumption that there's two words left, I was still looking for consonant connectors.... CH, NT, FT, TH. I'm also looking for a word that has something to do with church seat. At first I thought pew, but obviously no P or W (<g>)...Some quick rearranging gave me BENCH (actually I saw BENCH because I had previously seen QUENCH)

So now I have O L F T - Only thing that makes sense is LOFT.

Now, I wouldn't have gotten it without the letters (qualify: it would have taken me a LOT longer...you did the "hard" part

That's a wonderful approach. Thanks for the description, I learned a lot from it.

The word "face" in the last sentence should be "faces"

Any day that I can solve a Walter Penney puzzle without having a crash course in group theory or berlekamp factorization algorithm is a good day.

Top : 1, 9

Bottom: 3,7

Left: 6,8

Right: 2,12

Front: 4, 11

Back: 5, 10

Belch Fork Quint

Belch Font Quirk

Blench Fork Quit

Blench Oft Quirk

Bench Folk Quirt

Bench Fork Quilt

Bench Loft Quirk

Blotch Fen Quirk

Herb Flock Quint

Bilk Quench Fort

Brink Clef Quoth

Cob Elf Nth Quirk

bonanova, I'd like to ask you the same question I asked tpaxatb. Is there a non-computer particular algorithm or approach that you use to construct/deconstruct anagrams?

loft/bench/quirk

Good teamwork. I have never been good at anagrams, so I'd like to know how you did it so quick. Is there a particular algorithm or approach that you use to construct/deconstruct anagrams?