solutions_hw1 - Solutions to homework 1 1.5#1 Mis`re...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Solutions to homework 1 1.5#1 Mis` ere version of the take-away game. There are 21 chips, we can remove 1, 2, or 3. Last player to move loses, hence position 1 is a P-position, from positions 2,3, and 4 we can move to 1, hence these are N-positions. Now, 5 is a P-position again, since we can only move to N-positions from it, and so on... We can guess the pattern by the first couple sample points: if x 1 mod 4 then it is P-position, else an N-position. We can prove this by showing that (i) the terminal position is an N-position (mis` ere); (ii) If we make a move from a P-position ( x 1 mod 4), then x - 1 6≡ 1 mod 4, x - 2 6≡ 1 mod 4, nor x - 3 6≡ 1 mod 4; (iii) Similarly, if y 6≡ 1 mod 4, then one of y - 1, y - 2, y - 3 will be congruent to 1 mod 4. Since, 21 1 mod 4, the second player can always move to P-position during the game, and force the first player to lose. 1.5#4 Subtraction games (a) S = { 1 , 3 , 5 , 7 } . In this case P = { 0 , 2 , 4 , . . . , 2 k, . . . } = { 2 k | k N } , the even numbers. To prove this we only have to note that using a step from S we always jump from even to odd numbers. (b) S = { 1 , 3 , 6 } . After computing a couple of positions, we guess that P = { x | x 0 , 2 , or 4 mod 9 } . We need to check that p + s / P , p P , s S , and also that s S such that n / P : n + s P . (c) S = { 1 , 2 , 4 , 8 , 16 , . . . } . P = { 3 k | k N } . Note that powers of 2 are not divisible by 3 and that subtracting 1 or 2 of any number not divisible by 3 will result in a number divisible by 3.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern