Question

Problem 4: Misere Empty and Divide. There are two boxes. Initially, one box contains m chips and the other contains n chips. Such a position is denoted by (m, n), where m > 0 and n > 0. The two players alternate moving. A move consists of emptying one of the boxes, and dividing the contents of the other between the two boxes with at least one chip in each box. There is a unique terminal position, namely (1,1). Consider Misere rules: last player to move loses. Show that the only P-positions are of the form , or (3k - 1,3 1) (3k 1,1), (1,3k 1) where k,l > 0 are arbitrary natural numbers.

Answer 1

* Solution Given that data "Misere Emply and Devide" consider Misere reeles: last player to Move losses, show that the only p- position are of the form (BK-1, 1) (1,3K-1), or 3K-1,(36-1) where K, iso are arbitary naluoy number So, * Find to position of we will by an winning for player IR those for a- players Here, P→ position for that winning for player of previous N → position that winning for next player. & these PRA position of ax As the only position of Appear 1 of in game * The PRN ax, AS Main statement * N- position from every at least on max to 'p' position * Every 'p' positions of foom More as to n position * The game of all positions are an as where m Rn ax-odd et m=5 n=7 (517) + As positions * The a- player as tatagy of winning * the 1 st player as first box of f chips b/w a- boxes (4,3) - stratage

* The a-player as empty box as an odd chips. Then, the Divide for other way of chips Then * the odd number of chips AS box (1,3) * The 1- player as only of remove by 1- chip from 1-box as considered by (211) & the a-plagers as remove for a single chip as other of divid by a- boxes wher m Corn by all position of 2x even axw positions * N- player of a box show by staratagy.