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Problem solution · Turing

CCC 2010 J2 - Up and Down

CCC 2010 J2 - Up and Down: 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
59 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 2010 J2 - Up and Down, 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

  • 59 lines of Turing from the credited upstream file ccc10j2.t.
  • The implementation keeps its working state in language-native values and containers.
  • 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 2010 J2 - Up and Down · TuringTuring
Use this to learn the idea, then write your own version.
% J2: 2010: Up and Down%% loops and decisions and keeping track of things. %% Process each person separately.% Track both the distance and steps for each.% Quit JUST before the last set % and then add (or subtract) the remaining steps. % var a, b, c, d, s : intvar nikkySteps, nikkyDistance : intvar byronSteps, byronDistance : intvar next, sgn : int get a, b, c, d, s nikkySteps := 0nikkyDistance := 0next := asgn := 1loop    exit when nikkySteps + next >= s    nikkySteps := nikkySteps + next    nikkyDistance := nikkyDistance + sgn * next    if sgn = 1 then	next := b    else	next := a    end if    sgn := sgn * -1end loopnikkyDistance := nikkyDistance + sgn * (s - nikkySteps) byronSteps := 0byronDistance := 0next := csgn := 1loop    exit when byronSteps + next >= s    byronSteps := byronSteps + next    byronDistance := byronDistance + sgn * next    if sgn = 1 then	next := d    else	next := c    end if    sgn := sgn * -1end loopbyronDistance := byronDistance + sgn * (s - byronSteps) if nikkyDistance > byronDistance then    put "Nikky"elsif nikkyDistance < byronDistance then    put "Byron"else    put "Tied"end if

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