Solutions

Tic-Tac-Toe

If we number the squares like a push button phone, the first move was at 2, the last at 6

 3tsmall

If O played last, then X can force a win by playing at 4. Since O is an expert, he would not have allowed this. Thus it was X that played last. Since each played has moved three times, this means that O went first.

The only non-losing reponse to a corner opening is to play in the middle. Since the middle is not taken, the opening move could not have been in a corner and must therefore have been at 2.

With that established, trial and error reveals only two possible orders in which the moves could have been played with neither playing giving the other a chance to force the win. 2,1,9,8,3,6 or 2,8,3,1,9,6.


Dots & Boxes

Original position

 boxessol2

Since there are 40 lines in the game, and 16 boxes, and since completing a box gives one an extra line to draw, the number of turns in the game should be 40-15-D, where D is the number of times a single line completed 2 boxes or “doublecrosses”. (the last box completed is on the last move of the game, and thus no extra is conferred) Since Leon went both first and last, there must have been an odd number of moves in the game, and hence an even number of doublecorsses.

The last move must have completed a box for Leon, but not one for Noel, and it was therefor one of the four lines in the central +. It was a double cross, so there must have been at least one more. In a game where players always take as many boxes as they can, the only way that a doublecross can be set up is as part of a loop of boxes taken all at once. The last move of the game was part of the loop of four boxes in the center of the board, the other doublecross must have been a part of the loop of boxes around the border.Since they were all taken at once, and must have been given away the turn before that, the position before the last 16 lines must have contained just the two empty loops.

 


Scrabble

 scrabble1

GAY at a5

It is possible to determine the contents of your opponents hand.

The word APOCALYPSE at c11 must have been played all at once since no substring of its characters forms a word except for OCA, which couldn’t have been there ahead of time, as it would not have been attached to the start square. because at most seven tiles can be played at once, the A in AT, the P in POOR, and the E in ELIXIR must have all been there before APOCALYSPE was played.

Using this, and the simple fact that all words must have been connected to the start square at all times, we can narrow down the list of possible last moves to

  • AY (b5) scoring 5
  • ODD (e10) scoring 5
  • IT (e2) scoring 2
  • RUT (j15) scoring 5
  • ASKS (j1) scoring 22
  • A,K, and S of ASKS (j1) scoring 13
  • K and S of ASKS (j1) scoring 13
  • S of ASKS (j1) scoring 8
  • HEY (k8) scoring 9
  • Y of HEY (k8) scoring 9
  • R of SAUCER (L10) scoring 5
  • JIVE (m1) scoring 28
  • NO (m4) scoring 4

Most of these, however, would never have been used by your opponent, who always plays the highest scoring word that she can.

  • The A in AY could be played at d1 for 6 points
  • The O and D in ODD could make ODE at b13 for 8 points
  • The T in IT could be used for TASKS at I1 for 9 points
  • The R and U in RUT could be used for RUM at b14 for 10 points
  • ASKS could have been played down at a8 for 30 points
  • Just the S from ASKS could have been in JIVES at m1 for 15 points
  • Just the Y in HEY could have been in WEEDY and YA at h4 for 25 points
  • The R in SAUCER could have been in RAY at a5 for 6 points
  • The O in NO could have been in YO at k7 for 5 points

Since all of the other plays could have been outscored, your opponents last play was JIVE. So her hand at the start of the last play was JIV????. What could the other four letters have been?

Well, the only letters in the set which aren’t either in your hand,or on the board are 5 N’s, 2 G’s, F, Z, V, and I. She could not have had a F, or she would have played JIFF (FA and IT) down at f14 for 40 points. She could not have had a G or she would have played JIG (and WING) down at n6 for 35 points. And she could not have had more than one N or she would have played DJINN down at o12 for 39 points. Hence her had and the start of last round must have been JIVNZVI.

Thus her current hand is ???NZVI. The ?’s have to be N’s, G’s, or F’s because that’s all that’s left.

Finally, now you play GAY (scoring 7) at a5 knowing that our opponent, who always plays the highest scoring word will play ZING (scoring 42) at a8. When you pick your next letter, it must be an N, F, or G (since that’s all that’s left) and you will be able to play FORMALIZING (scoring 137 and leaving your oppoinent with 12 points in tiles) , NORMALIZING (scoring 128 and leaving your oppoinent with 15 points in tiles) , or GLAMORIZING (scoring 131 and leaving your oppoinent with 14 points in tiles) at a1. Ending the game, and giving you a net gain of at least 123 points. Nice comeback!

 


Checkers

 chksol
Starting at the position above (which is easy to get to) the game continued:

 

1. i1-j2
……i5-h4
2. f6-e7
……g4-f6
3. j2-h4

Arriving at the problem position.


Bridge

Play the Queen.

Let’s assume that your remaining opponent has The King of clubs. That would mean that everybody had a club this hand.
Since everybody followed suit when you cashed the three Aces, this means that no one had a void when you came into the game

When no one has a void, that mean that everyone followed suit on all previous tricks, and thus the cards have come out in groups of four from the same suit. The number of cards of each suit left in the players hands must therefore be either 13, 9, or 5. When we came to the table, there must have been at least 5 spades, 5 hearts, 5 diamonds, and 9 clubs (five isn’t enough, because we have four and we must allow at least one for each of our opponents). This makes a total of 24 cards. But since each player only has 5 cards when we start, this is impossible.

Our original assumption must have been wrong. East doesn’t have the King of Clubs, and playing the queen will win both remaining tricks.



In one hand of bridge, all four deuces took tricks. Game was bid and made. How many overtricks?

None

In order for a deuce to take a trick, all other players must be void in that suit.
Consider the last time during the game that a non-trump deuce takes a trick.
1. No trump is played, or this deuce would not win the trick.
2. No two cards played in this hand can be the same suit. For if they are, that means that at least two players are not void in that suit and the deuce of that suit could not have taken a trick.

We have four diffent suits (from 2) and none of them are trump (from 1) so we must be playing in no trump.

Again, since in order for a deuce to take a trick, all other players must be void in that suit, we can deduce that each player must have won with the deuce of the suit that he played on this trick.

A deuce can only take a trick if it is on lead, and it cannot win opening lead, so the defenders must have won the lead twice in addition to winning with the two deuces. So four tricks went to the defense. Since gane was bid and made, that means that the winners must have taken all remaining trciks, exactly making the contract of 3 No Trump.


Life

 Consider a (5n-2) by (5n-2) square life pattern. To determine if it has a predecessor, we would need to look at the possible configurations of the 5n by 5n square that surrounds pattern with one extra box on each side. If any of the n-squared 5 by 5 boxes that make up our prospective predecessor is completely empty, we can be replaced with a 5 by 5 square that is empty except for a single cell in the center without effected subsequent generations. Consequently, we need only consider (2^25 – 1)^(n^2) of the 2^25(n^2) possible predecessor patterns.

Since there are 2^((5n-2)^2) different possible configurations of our square, that means that if (2^25 – 1)^(n^2) is less than 2^25(n^2) that there won’t be enough predecessors to go around, and some of them will have to be “orphans.”
This happens when n=465163200. So there is a pattern with no predecessor that will fit into a 2325816000 by 2325816000 square, guaranteed.
If each cell were 1/16 of and inch, this square would be nearly the size of the United States.

goe
Patterns with no predecessors are called Garden of Eden patterns. So far 3 have been found (using different ideas and tons of computer time).
The smallest known GOE pattern (pictured at above) is only 14 by 14. To find out more about it and how it was found, go to the discoverers home page.

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