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Problem solution · Turing

CCC 2008 J2 - Do the Shuffle

CCC 2008 J2 - Do the Shuffle: a Turing solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Direct simulation
Source
CCCSolutions
Length
37 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For CCC 2008 J2 - Do the Shuffle, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 37 lines of Turing from the credited upstream file ccc08j2.t.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2008 J2 - Do the Shuffle · TuringTuring
Use this to learn the idea, then write your own version.
% CCC 2008% J2: Do the Shuffle%% playlist starts at  ABCDE%% button 1 moves first to last% button 2 moves last to first% button 3 switches 1 and 2% button 4 output list%% this is a simple string assignment% var playlist : string (5)var button, presses : int playlist := "ABCDE"loop    put "Button number: " ..    get button    put "Number of presses: " ..    get presses    exit when button = 4 and presses = 1    for i : 1 .. presses        if button = 1 then            playlist := playlist (2 .. 5) + playlist (1)        elsif button = 2 then            playlist := playlist (5) + playlist (1 .. 4)        else            playlist := playlist (2) + playlist (1) + playlist (3 .. 5)        end if    end forend loop put playlist (1) + " " + playlist (2) + " " + playlist (3) + " " +    playlist (4) + " " + playlist (5)

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