Kara Levis Period 3

A Collection of Mind Games Kara Levis Period 3

Purpose • I intend to present a variety of problems designed to make students think outside of the box • These brain teasers range in difficulty and cover a variety of indirect mathematical subjects • The problems are presented with a “Questions” page • Each subsequent page is numbered in accordance to the question it is solving

Table of Contents • Trains ----- Slides 4-9 • Amida Kuji ----- Slides 10-18 • Subsets in Sequence ----- Slides 19-25 • Classic Hats ----- Slides 26-29 • Number Hats ----- Slides 30-35 • Game of Set ----- Slides 36-40 • Other Recommendations ----- Slide 41

Trains- Problem • You can use rods of integer sizes to build "trains" that all share a common length. A "train of length 5" is a row of rods whose combined length is 5. Here are some examples: • Notice that the 1-2-2 train and the 2-1-2 train contain the same rods but are listed separately. If you use identical rods in a different order, this is a separate train

Trains- Questions • How many trains of length 5 are there? • Come up with a formula for the number of trains of length n. (Assume you have rods of every possible integer length available.) Prove that your formula is correct.

Trains-1 • Write out the possibilities, so we can look for a pattern later: 1-1-1-1-1, 2-1-1-1, 1-2-1-1, 1-1-2-1, 1-1-1-2, 3-1-1, 1-3-1, 1-1-3, 4-1, 1-4, 2-2-1, 2-1-2, 1-2-2, 3-2, 2-3, 5 • Clearly, there are 16 possible trains of length 5

Trains-2 • I solved this problem by listing the number of trains possible for lengths 2 to 5: • Clearly, with each new length, the number of solutions equals twice the number of previous solutions • This pattern yields the formula: Number of solutions=2length of train-1

Trains- 2 • Now we need to prove this formula • Let’s say x=the length of the train, and y=the number of solutions • When x increases by one, half of the solutions can be found simply by adding a 1 to the end of the existing trains from the previous length • Eg. For length 4 we have 1-1-2, and for length five we have 1-1-2-1

Trains-2 • Now, we need to prove this formula • Let’s say x=the length of the train, and y=the number of solutions • There will be x types of blocks that can be used, with two solutions always being “11-12-13-…1x”(with x blocks) and one solution being “x” (with one block) • When x>2, there will always be a solution with “(x-1)-1” and “1-(x-1)” • Following this pattern, one can see that there will always be a total of 2x-1 solutions

Amida-kuji- Problem • Amida-kuji is a type of networking theory in which people are assigned numbers. These numbers are placed at the top of lines connected to prizes at the other end. • Horizontal lines are randomly drawn to connect adjacent vertical lines, with the only parameter being that no adjacent horizontal lines lie on the same longitude • The winners of each prize are determined by following the path created by both vertical and horizontal lines until the end of the vertical line, and the associated prize, are reached

Amida-kuji- Questions • Will each person get a different prize no matter what horizontal lines are added? • Is there a way to place horizontal lines to make any assignment of prizes you want?

Amida-kuji-1 • For each horizontal line added, two “players” will switch places: • In this case, players 1 and 2 switch • Based on this concept, there will always be only one player on each vertical line, and each player will receive a different prize 1 2 3

Amida-kuji-2 • With the exception that no two players can receive the same prize, it would seem that all combinations of players and prizes are possible • There is no limit to the number of horizontal lines used to connect players, and so any path desired should be available

Amida-kuji-2 • After sampling several random sequences of the digits one thru 5, I developed a strategy that seems able to solve any combination: approaching the set-up in the order of the end product • Ie., if the desired order is 31452, then the lines to transfer “3” would be drawn first, then “1,” then “4,” and so forth

Amida-kuji-2 • Taking this example of 31452: • First draw the lines to transfer “3” to the prize at the end of vertical line “1” • After this transfer, “1” is already on vertical line “2,” so we will leave it there 1 2 3 4 5 3 1 2

Amida-kuji-2 • Taking this example of 31452: • With “3” and “1” in the proper end points, transfer “4” to vertical line “3,” and “5” to vertical line “4” • The dotted green line shows how this allows “2” to end up on vertical line “5,” completing the puzzle 1 2 3 4 5 3 1 4 5 2

Amida-kuji-2 • Though this solution was fairly simple, infinite other solutions are possible to solve the 31452 order, such as the one shown above, where “5” and “2” switch back and forth between their destinations • Obviously, this also insinuates that solutions can become infinitely complex 1 2 3 4 5 3 1 4 5 2

Amida-kuji-2 • Sample of a more complex amida-kuji puzzle: 73516824 1 2 3 4 5 6 7 8 7 3 5 1 6 8 2 4

Subsets in Sequence- Problem & Questions The set {a, b, c} has exactly eight subsets. • List all the subsets of {a, b, c}. • Now arrange those subsets in a sequence so that each subset in the sequence differs from the one before it in one of two ways: • One new element is inserted: Example 1: {a}, {a, b} Example 2: {a, b}, {a, b, c} • One old element removed Example 1: {a, b, c}, {a, c} Example 2: {b}, Ø • Can you create more than one sequence that fits these criteria? • Create a listing like this for the subsets of {a, b, c, d}. • Describe a general method (or explain why there isn’t one) for creating such a sequence of the subsets of any set {a1, a2, a3,...an}.

Subsets in Sequence-1 • The set consists of {a,b,c} • Thus, the possible subsets are {a,b,c}, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, and { }.

Subsets in Sequence-2 • The subsets are stated again here: {a,b,c}, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, and { }. • This is one possible solution for the sequence of the subsets that fits the requirements: { }, {a}, {a,b}, {b}, {a,c}, {c}, {b,c}, {a,b,c} • In terms of the number of elements, the general solution is 0, 1, 2, 1, 2, 1, 2, 3, or else the opposite

Subsets in Sequence-3 • It appears that there are only two possible solutions for the sequence. Both will either start with the empty set, or the largest possible set, and works toward the opposite goal • This works because there were three subsets with 1 elements and three with 2 elements, allowing them to alternate nicely, however, the empty set and initial set were left unmatched. Starting with a subset other than one of these mismatched examples would lead to a faulty solution

Subsets in Sequence-4 • For the set of {a,b,c,d} the possible subsets are: {a,b,c,d}, {a}, {b}, {c}, {d}, {a,b}, {a,c}, {a,d}, {b,c}, {b,d}, {c,d}, {a,b,c}, {a,b,d}, {a,c,d}, {b,c,d}, { } • In sequence, these subsets are: { }, {a}, {a,b}, {b}, {a,c}, {c}, {a,d}, {d}, {b,c}, {a,b,c}, {b,d}, {a,b,d}, {c,d}, {a,c,d}, {a,b,c,d}, {b,c,d} • In terms of the number of elements, this sequence is 0, 1, 2, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, or the opposite

Subsets in Sequence-4 • However, unlike the sequence for a set with three elements, the sets with four has an alternate solution. If starting with a subset with four elements and working toward the null set this sequence emerges: 4, 3, 2, 3, 2, 3, 2, 3, 2, 1, 2, 1, 2, 1, 0, 1 • This gives another viable solution, with the opposite also being in accordance with the requirements

Subsets in Sequence-5 • For a set of n numbers, there are 2n possible subsets, always giving an even number • Arrange the subsets into two categories: those with an odd number of elements, and those with an even number of elements (null set is even) • Then, order each group by number of elements from least to greatest • Merge the two groups by starting at one extreme (greatest or least number of elements) and working toward the other end one subset at a time • If there is one subset left over after the alternating pattern stops, try stating from the opposite extreme and starting over

Classic Hats- Problem & Questions • Three men are brought before the king. The king places a hat on each man’s head without them seeing the color of their hat. The king then tells the men “Each hat is either black or white.” and asks them to raise their hand if they can see a black hat. At this, all three men raise their hands. The king then says “Raise your hand if you know the color of your hat” and all hands go down. Then, one man raises his hand. • What color is the man’s hat? Explain.

Classic Hats-1 • Each man can see the other two hats, but not his own, and each sees at least one Black hat. A A 1 2 C C B B • Clearly, option 1 is not correct because A cannot see a Black hat. Thus, there are at least two Black hats, meaning only option 2 or 3 are possible solutions A 3 C B

Classic Hats-1 • Having ruled out option 1, assume that option 2 is correct. • B can see a White hat (A) and a Black hat (C). • B knows that option A is not possible, and there must be at least two Black hats. Therefore, B can immediately guess that his hat is Black. A A 1 2 C C B B

Classic Hats-1 • BUT B DOESN’T GUESS RIGHT AWAY! • Thus, the man who speaks realized that none of the others can see a Black hat and a White hat, or they would have guessed their own hat as Black right away (as B did in the hypothetical) • Thus, the man who speaks knows that option 3 is the only possible solution, and realized that his hat must be Black A 3 C B

Number Hats- Problem & Question • Three men are in a room. A fourth man places a hat on each of their heads, without allowing them to see their own hat. The fourth man then tells them that each hat has a positive integer written on it and that the sum of two of the integers is equal to the third. The three men are then asked what the number is that is on their own hat: • Man 1: I cannot determine my number • Man 2: I am unable to determine my number • Man 3: I too cannot determine my number • Man 1: My number is 50 • How did Man 1 determine his number?

Number Hats-1 • In general, this problem cannot be solved except for certain cases of numbers. For this problem let’s assume that the hats used were labeled either as 20, 30, and 50 or 25, 25, and 50 • Consider the second possibility. Man 1 begins the dialogue. When he looks, he sees Man 2 and 3 both wearing hats labeled as 25. He knows that all the hats are labeled with positive numbers, and thus his hat cannot be labeled zero. Therefore, on the first guess, he would know that his hat was labeled as 50, the sum of 25+25

Number Hats-1 • Now consider the first scenario. • Man 1 begins. He sees hats labeled 20 (Man 2) and 30 (Man 3). Thus, he already know that his own hat can only be 10 or 50 (20+10=30 or 20+30=50). However, based on this information, he cannot determine which number it is. • Man 2 and 3 then fail to guess their numbers, allowing Man 1 to narrow his options

Number Hats-1 • Assume Man 1 has the hat labeled 10 (his only other option than 50). He sees Man 2 wearing the label 20 and Man 3 wearing the label 30. • Man 3 sees hats labeled 10 (Man 1) and 20 (Man 2). He knows that his own hat must be labeled either 30 or 10. • However, as Man 3 is thinking, he is able to solve for his number.

Number Hats-1 • Man 3 realizes that ifhis number was ten, and he can see 10 (Man 1) and 20 (Man 2), then Man 2 would have seen 10 (Man 1) and 10 (Man 3). • Given that zero is not an option, Man 2 would have known that his hat was the sum of the other two and guessed his number. But he didn’t. Therefore, Man 3 knows his number is not 10. • Man 3 can then guess that his number is 30 (his only other option than 10)

Number Hats-1 • BUT MAN 3 COULDTN’T GUESS HIS NUMBER! • Man 1 now knows that his number cannot be 10, as this would have allowed either Man 2 or Man 3 to have guessed their own numbers. Therefore, Man 1 knows his number must be 50.

Game of Set • The cards in Set represent all of the possible combinations of three types of each of the following attributes: • Shape (circle, squiggle, diamond) • Shading (Solid, hashed, no shading) • Color (Red, green, blue/purple) • Number (1, 2, 3) • The goal of the game is to form a set

Game of Set • A set consists of three cards. For each for the attributes (color, shape, number, shading) all three cards must be either all the same or all different. There cannot be an “odd card out” with respect to any trait. • The following group of cards forms a set.

Game of Set • In contrast to the above set (from the previous slide), the following is not a set • Notice how the shading is neither the same or different for all three cards. Card 2 is the “odd card out"

Game of Set • To play Set, 12 cards are placed face-up on the table. Two players then try to find sets within these cards, as defined by the rules in the previous slides. The game continues until no more sets can be found, and the player to find the most sets wins.

Game of Set- Questions • The math connection: • If you pull out 3 cards from the whole deck, what is the probability that they form a Set? • If we pull 9 cards from the deck, what is the maximum number of Sets possible among them? • How many cards can be left on the table at the end of the game? • What is the maximum number of cards in a Set-less pile? Can you establish upper or lower bounds for this maximum?

Also Try… • This site has a bunch of fun and interactive math games (I highly recommend trying some): • http://www.mathplayground.com/logicgames.html • This site has the basis the Trains, Subsets, and Amida Kuji problems, as well as other ideas for those interested: • http://www2.edc.org/makingmath/mathproj.asp • This was my source for the hats puzzles, and also contains some other logic problems: • http://people.sc.fsu.edu/~burkardt/fun/puzzles/hats_puzzle.html

VS Lex Luthor, Darkseid, Doomsday, Brainiac, Bizarro, Zod, Mister Mxyzptlk, Toyman, Metallo, Cyborg Superman, Prankster, Parasite, Intergang… Superman

Superman Wins!!!

Why so SERIOUS?

Kara Levis Period 3