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Problem solution · Turing

CCC 2003 J2 - Perfect Picture

CCC 2003 J2 - Perfect Picture: 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
28 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 2003 J2 - Perfect Picture, 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

  • 28 lines of Turing from the credited upstream file ccc03j2.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 2003 J2 - Perfect Picture · TuringTuring
Use this to learn the idea, then write your own version.
% CCC 2003% Problem J2: Picture Perfect%% A rectangle of a given area has% has a minimum perimeter if arranged as a square (or closest to it)% so find the square root of the area (number of pictures)% and then find the largest integer <= that square root % That solves the problem%% keyboard and screen I/O var a, l, w : intloop    put "Enter number of pictures:"    get a    exit when a = 0    l := round (sqrt (a))    loop        exit when a mod l = 0        l := l - 1    end loop    w := a div l    put "Minimum perimeter is ", 2 * l + 2 * w, " with dimensions ", w,         " X ", l    put ""end loop

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