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By Miguel de Guzman

This publication, constructed from the author's sequence of lectures introduced in Japan in 1995, identifies and describes present effects and concerns in convinced parts of computational fluid dynamics, mathematical physics and linear algebra the math of a sandwich; nim; the bridges of Konigsberg; "solitaire" confinement; the mathematician as a naturalist; 4 shades are sufficient; bounce frog; abridged chess; the key of the oval room

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Sample text

When we reach the first vertex S\ where untraveled arcs start, let's follow them. As before, we can't stop unless it's at S\. Now, when all the arcs that start at S\ have been followed, we continue from Si along the initial path, C, until we reach the first vertex, S2, where there are untraveled arcs either Uhe LJjridges of LKbnigsoerg 37 from path C or in the extension we just made. This way we end up traveling along all the arcs. Thus, if all the arcs have an even valence, it's possible to follow the path of the proposed figure and, what's more, we have the solution for tracing our path.

Are you able to trace the following figures without lifting pencil from paper and without repeating any line? Could you do the same thing starting and ending at the same point? Fig. 2 Fig. 3 Une l/jridges Fig. 4 of LKbnigsSerg 33 Fig. 5 Try and try again. You're almost sure to be familiar with Figure 3 and have probably done it many times, but you'll have a hard time finishing where you started. Figure 4 is so easy to trace that unless you try to do it wrong, you can start at any point and return to the same point without repeating any arcs, covering the entire path almost effortlessly.

5 Try and try again. You're almost sure to be familiar with Figure 3 and have probably done it many times, but you'll have a hard time finishing where you started. Figure 4 is so easy to trace that unless you try to do it wrong, you can start at any point and return to the same point without repeating any arcs, covering the entire path almost effortlessly. Figure 2 looks easier, with fewer segments, but it, just like Figure 6 below, which just has three segments, is clearly Fig. 6 impossible ("trivial," as some people would say almost insultingly).

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