Olympiad Maths Prep

Track / Stage 6 / 51 of 400 #1051 of 2000

Problem 1051

National olympiad, first round
Combinatorics Difficulty 6.0 Prove it

| [Algorithm Theory (Miscellaneous). ]] |
| :--- |
| [ Semi-invariants ]\left[\begin{array}{ll}\text { Semi-invariants }\end{array}\right] |
| [ Backtracking ]\left[\begin{array}{l}\text { Backtracking }\end{array}\right] |

Authors: Mednikov L.E., Shapovalov A.V.

Captain Vrungel laid out a shuffled deck of 52 cards in a circle in his cabin, leaving one space free. Seaman Fuchs, from the deck, without leaving the helm and without knowing the initial layout, names a card. If this card is next to the free space, Vrungel moves it there, without telling Fuchs. Otherwise, nothing happens. Then Fuchs names another card, and so on as many times as he wants, until he says "stop." Can Fuchs ensure that after saying "stop"

a) every card is definitely not in the same place it was initially?

b) the ace of spades is definitely not next to the free space?

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

a) Here is one possible way. Fuchs over 52 turns names 52252^{2} cards - in each turn, all cards once in the same order, after which he says "stop". Note that every time the cards move in one direction: if card aa is moved, the next one to be moved will be card bb on the other side of the free space: it will be named earlier than aa. In each turn, at least one card will move, so there will be at least 52 moves, and each card will leave its place. On the other hand, each card has moved no more than 52 times, so it will not have time to make a full circle (53 moves) and return to its place.

b) We will show that for any strategy of Fuchs (sequence of named cards), there is a "losing" initial position that will result in a position with the ace next to the free space.

By naming a card again, we "undo" the previous move. Take any position with the ace next to the free space. By naming the entire sequence of cards from the given strategy in reverse order, we will "undo" this position to the required "losing" initial one.

## Answer

a) Can; b) cannot.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.