Please answer to this new question. This question is totally new. Consider a gam
ID: 3754266 • Letter: P
Question
Please answer to this new question. This question is totally new.
Consider a game where there are two piles of pebbles. On a player’s turn, he may remove between 1 and 3 pebbles from the first pile or between 1 and 4 pebbles from the second pile. HOWEVER, PEBBLES MAY ONLY BE RE- MOVED FROM THE SECOND PILE AFTER ALL PEBBLES ARE GONE FROM THE FIRST PILE. The person who removes the last pebble wins. Label a game of this form as (x, y) where x is the initial number of pebbles in the first pile and y is the initial number of pebbles in the second pile. How many values of (x, y) are there with x [1..20], y [1..20] that are first player wins assuming that both players play optimally.
Explanation / Answer
Answer:
Presently, as some of you have understood, this issue would be simple on the off chance that we could utilize the Sprague-Grundy hypothesis. Yet, that is not the purpose of the issue. The purpose of allotting this issue before we see the SG hypothesis is that I needed you to investigate and play with the conceivable positions, figuring out subtraction recreations, and entirety diversions.
What I am searching for is an investigation of examples that give a P-position. Maybe the most ideal approach to do this is to compose the positions in a lattice, and after that you can attract bolts to where they can go. You realize that the P-positions in the first subtraction diversion are of the frame 5 k, k N, and the P-positions in the second capacity are of the shape 6 k, k N. It is not all that simple to perceive what happens when you set up them together as depicted in the issue.
Let us consider pile one contain any no of pebble 1, 2, 3, 7, 8, 9, 13, 14, 15, 19, 20 with pile two containing any no of pebbles 5, 6, 7, 8, 13, 14, 15, 16 then A .
So, total no of cases in which A wins are
Total no. of cases of pebbles in pile one * total no. of cases given above in pile two
11 x 8 = 88.