The word "tourism" recalls, besides a rest at the seaside or in the mountain,
visits to museums.
Everywhere one happens to be one can find a museum about a particular subject,
with the purpose to atone in a display of cultural engagement, every
possible concession to laziness and enjoyment, almost it was about a guilt.
Naturally one cannot generalize, but sometimes really really think a museum is only
an excuse to give oneself airs of intellectuality, while it's not really needed.
And maybe it's almost as if one justifies some museums, like for example the
torture instruments in San Gimignano, just to mention one.
Torture instruments!
Shouldn't one be a kind of maniac to visit such a museum?
And what about some fetishism for the visitor of the museum of Umbrella at
Gignese?
It obviously depends on the location of the holiday. I am not saying for sure
that the Uffizi wouldn't deserve to be visited if one is in Florence, or the
same for the Vatican Museums in Rome, or the Louvre in Paris. But to me, in
general, museum is kind of boring, i have to admit in a guilty tone.
In general when we are in holiday we prefer to spend our energies in the
methodical research of an authentic osteria, a genuine trattoria, an artisan
cheese farm. And so, during our holiday in the Cuneo area some times ago (here
the excursions), we ended up shopping at La Formaggeria, in Torre Pellice.
Visiting a shop like that one must conclude that, from the cultural point of
view, it's not something really different from a museum. Philosophy of Bruna
Magnano, that runs the shop, is to refuse commercial mass products and to
offer instead niche quality, linked to the territory, trying to divulge the
tradition, preserving it from the homologation imposed by the outside world.
I believe that this is the right goal of a museum. The only difference is that
a museum reminds me of a "moldy" smell, while La Formaggeria, instead, it welcomes the
visitor with an attractive scent.
The counter exposes dozens of cheeses from that area, besides other products:
fresh pasta, olive oils, various type of preserves... Among our purchases there
are the cheese shown in these photos:
 Pagliettina, smooth cheese of mixed cow, goat and sheep milk, strong and
acidulous taste
 Aged Goat Tomino, demistrong, intense and resolute savour
 Pecorino aged in grape must, demistrong, resolute notes of flavor coming
from the must that colors the surface of the crust
 Ricotta aged in hay, demistrong, salted, taste and smell of hay
La Formaggeria Di Magnano Bruna
Via Arnaud, 6, 10066 Torre Pellice (TO)
Friday, May 29, 2009
Tuesday, May 26, 2009
Appello in difesa della democrazia, in difesa della Costituzione
(sorry for the post in Italian)
Sono giornate molto pesanti, in cui le parole gravano come macigni, e se l’argomento di queste parole sono la Democrazia, il Diritto, la Giustizia, il rischio è che questi macigni si trasformino in frane, di quelle che travolgono interi paesi cancellandone la storia, cancellandone la civiltà, rinnegandone l’etica.
Mancano due settimane alle elezioni europee, nel nostro Paese questo appuntamento, a causa delle parolemacigno del capo del governo, rischia di assumere caratteristiche che vanno ben al di là del risultato puramente elettorale.
Una cosa soprattutto assume un importante valore politico: la coesione che travalica le sigle, di un fronte di difesa democratico della Costituzione e delle Istituzioni.
Attualmente sono cinque i soggetti politici che partecipando alla competizione europea possono rappresentare questo fronte: i due cartelli elettorali di sinistra, il PD, IDVDi Pietro e UDC.
Dei cinque partiti o movimenti il PD è l’unico che, ad oggi, sostiene la campagna dei referendum di riforma della legge elettorale. Nell’eventualità che il referendum passi ci ritroveremmo con un sistema che prevederà premio di maggioranza al partito di maggioranza relativa (non alla coalizione) e innalzamento della soglia minima di sbarramento. Risultano evidenti due cose: che una minoranza del paese, ma in possesso di una maggioranza relativa, avrebbe uno strapotere e una consistente porzione di elettori non avrebbero rappresentanza parlamentare.
In questi giorni è davanti gli occhi di tutti l’inaudito attacco alle istituzioni da parte del capo del Governo. Credo che proseguire sulla strada del referendum sarebbe come iniettare cellule malate in un corpo che già sano non è.
Il PD deve uscire dall’equivoco e riconoscere che il tema del referendum è di fatto superato da una evidente emergenza democratica e che sarebbe un suicidio della democrazia anche solo ipotizzare leggi che diano maggiori poteri agli organismi di governo.
La democrazia è un sistema di governo con evidenti imperfezioni, ma anche con importanti anticorpi che normalmente impediscono la degenerazione. Il nostro compito è quello di far sì che non calino le difese immunitarie insite nella nostra Costituzione.
Una rinuncia da parte del PD ad appoggiare e sostenere il referendum potrebbe inoltre raccogliere il consenso di molti compagni che non riconoscendosi nell’area dei due cartelli elettorali di sinistra, si troverebbero nell’imbarazzo di un voto all’Italia dei Valori, che pur essendo un partito di sicura opposizione a Berlusconi, non rappresenta la cultura di sinistra, o di una astensione, in quanto non si sentirebbero sufficientemente tutelati proprio in funzione del referendum liberticida.
Blog promotori:
A sinistra
il Russo
La Mente persa
L’eco dell’Appennino
Vengo da lontano ma so dove andare
PS. Chi condivide questa richiesta copi e incolli sul proprio blog il post senza aggiungere o togliere nulla possibilmente segnalando l’adesione a uno dei cinque blog promotori o alla seguente mail indemocrazia@yahoo.it
Sono giornate molto pesanti, in cui le parole gravano come macigni, e se l’argomento di queste parole sono la Democrazia, il Diritto, la Giustizia, il rischio è che questi macigni si trasformino in frane, di quelle che travolgono interi paesi cancellandone la storia, cancellandone la civiltà, rinnegandone l’etica.
Mancano due settimane alle elezioni europee, nel nostro Paese questo appuntamento, a causa delle parolemacigno del capo del governo, rischia di assumere caratteristiche che vanno ben al di là del risultato puramente elettorale.
Una cosa soprattutto assume un importante valore politico: la coesione che travalica le sigle, di un fronte di difesa democratico della Costituzione e delle Istituzioni.
Attualmente sono cinque i soggetti politici che partecipando alla competizione europea possono rappresentare questo fronte: i due cartelli elettorali di sinistra, il PD, IDVDi Pietro e UDC.
Dei cinque partiti o movimenti il PD è l’unico che, ad oggi, sostiene la campagna dei referendum di riforma della legge elettorale. Nell’eventualità che il referendum passi ci ritroveremmo con un sistema che prevederà premio di maggioranza al partito di maggioranza relativa (non alla coalizione) e innalzamento della soglia minima di sbarramento. Risultano evidenti due cose: che una minoranza del paese, ma in possesso di una maggioranza relativa, avrebbe uno strapotere e una consistente porzione di elettori non avrebbero rappresentanza parlamentare.
In questi giorni è davanti gli occhi di tutti l’inaudito attacco alle istituzioni da parte del capo del Governo. Credo che proseguire sulla strada del referendum sarebbe come iniettare cellule malate in un corpo che già sano non è.
Il PD deve uscire dall’equivoco e riconoscere che il tema del referendum è di fatto superato da una evidente emergenza democratica e che sarebbe un suicidio della democrazia anche solo ipotizzare leggi che diano maggiori poteri agli organismi di governo.
La democrazia è un sistema di governo con evidenti imperfezioni, ma anche con importanti anticorpi che normalmente impediscono la degenerazione. Il nostro compito è quello di far sì che non calino le difese immunitarie insite nella nostra Costituzione.
Una rinuncia da parte del PD ad appoggiare e sostenere il referendum potrebbe inoltre raccogliere il consenso di molti compagni che non riconoscendosi nell’area dei due cartelli elettorali di sinistra, si troverebbero nell’imbarazzo di un voto all’Italia dei Valori, che pur essendo un partito di sicura opposizione a Berlusconi, non rappresenta la cultura di sinistra, o di una astensione, in quanto non si sentirebbero sufficientemente tutelati proprio in funzione del referendum liberticida.
Blog promotori:
A sinistra
il Russo
La Mente persa
L’eco dell’Appennino
Vengo da lontano ma so dove andare
PS. Chi condivide questa richiesta copi e incolli sul proprio blog il post senza aggiungere o togliere nulla possibilmente segnalando l’adesione a uno dei cinque blog promotori o alla seguente mail indemocrazia@yahoo.it
Friday, May 22, 2009
Pick's theorem and youthful loves
I am a faithful husband.
But there are some old loves that every now and then come back to stimulate and fill my mind, making me live again those emotions that i felt when i used to date them. And so, here i am to savour, once again with renewed strength, those pleasures i was granted of, almost like in the candour of a teenage flirt.
One of those loves is Mathematics. ;)
It's thanks to Giovanna and her blog Matematicamedie that i found something of this matter still unknown to me. Infact, i have never heard before about Pick's theorem.
As the teacher Giovanna well explains om her very interesting post, that tells about a wonderful work her class made, Pick's theorem computes in an easy and concise way the area of a polygon built under some particular rules.
To clarify better the problem, let me try with some definitions.
Given a regular grid of points (that we can imagine as the crossings of the lines on a squared paper), a "Pick's segment" is a segment that join two points of the grid without intersecting any other on its path. If we need a segment that lays on an extra point, it is considered like the concatenation of two consecutive segments lying on the same line.
A "Pick's polygon" is a closed concatenation of Pick's segments not selfintersecting.
It's right the area of Pick's polygon that is evaluated by Pick's theorem.
Please note that the nonselfintersection condition is demanded for any area computation.
figure 1
It doesn't make much sense infact to evaluate the area of a selfintersecting polygon.
Moreover the condition for which a segment doesn't have to contain any other point of the grid more than its extrema, doesn't reduce the representation power of the definition. If a polygon has a side cut by a point of the grid, well, in that point one can also consider an additional vertex that generate a straight angle. For example figure 1 shows a Pick's polygon with four vertex and four sides, in different colors, even if it is perfectly superimposable on a triangle.
As Giovanna reveals, Pick's theorem asserts that area A of a Pick's polygon is given by the formula
A = PI + PC/2 1
dove PI is the number of the "internal" points (or the points of the regular grid that fall in teh internal part of the polygon) and PC is the number of the "contour" points: the ones that fall on the vertices of the Pick's segments  sides of the polygon, or, in other words, the points that lay on the edge.
Clearly this formula is valid if the obtained value is then multiplied by the area of the square that defines the regular grid. To make it more simple, on this post, i consider a grid in which the area of the square is one.
With reference to figure 1, for example, PI is the number of the points painted in blue (7), PC the ones in yellow (4), so the area of the polygon is
A = 7 + 4/2  1 = 8
It works. Try and see!
In a comment of the discussion that follows the quoted blog, Giovanna suggests me a couple of links to web pages where there is also the "official" demostration of Pick's theorem. Nevertheless i didn't follow those links, nor i have any intention to do it before submitting this post, because what enchants me of mathematics is not so much the result itself, even when in cases like this the formula has a really charming elegance. And not even i need a concrete proof to believe in the truth of the formula: i trust in Giovanna!
No, what i find irresistible is to use mathematics in order to find a demonstration my own.
Naturally this task is simplified by the fact that i already know which is the result i have to reach and i am already convinced that that result is true. So, i have only to find a right path to reach the goal (without a GPS navigator system ;)).
Trying to wear the pants of an eventual reader that doesn't know Pick's theorem, i warn now that the rest of this post shows my own demonstration and so, if one is interested to find another one, this is a good point to stop reading.
So...
First of all i find the way to count the internal points in the polygon, the geometrical informations on the vertices given.
I start defining a Cartesian axis system centered in any point of the grid lower and more on the left of the whole polygon (i believe that these last conditions are irrelevant, but, since i can choose, i simplify my own life!).
figure 2
I concentrate now to find an easy way to express the number of the points that are "under"
a Pick's segment. With the word "under" i mean that their ordinates are greater or equal to
zero and lower than the intersection point between the vertical line and the segment, and,
moreover, their abscissa falls between or equal to the abscissae of the segment extrema.
In the example shown in figure 2 that is the number of the points painted in yellow.
figure 3
In order to do so, i consider the figure obtained by all the points that are "under" the segment
plus all the ones that i obtain rotating the figure of a 180deg angle around the center
of the segment. These points are disposed as a rectangle that i name
R (an example is given in figure 3).
The number of points belonging to this rectangle si clearly given by the number of points at the base multipled by the number of point in the height.
Naming P1 and P2 the extrema of the segment, and (x1, y1) and (x2, y2) their coordinates, respectively, the computation of the number of points in the rectangle is given by this following formula:
R = (x2  x1 +1) * (y2 + y1 + 1)
The comuted value R is an even number.
figure 4
Infact, as it is shown in figure 4, R can be divided in three
rectangles, marked by colors red, green and blue. Clearly the red and blue rectangles are
equal, and so they contain the same number of points. The sum of all those points, so,
is even. To show that R is even, then, it is enough to
show that the nubmer of points in the green rectangle is also even.
Numbers dx = x2  x1 and dy = y2  y1 are prime between each other. Infact if it was not like that, segment between tra P1 and P2 would meet other points of the grid. So, the segment wouldn't match the definition which a point of the grid cannot belong to a Pick's segment, unless one of its extrema. Since dx and dy are prime each other, atleast one of them is an odd number (if both were even, they would have 2 as a common divisor). So, atleast one value between dx+1 and dy+1 is even. It follows that their product is necessarily even, and incidentally that product is right the number of points of the green rectangle. So, R &is even.
Number R can be divided by 2, and the result of that division is given by the number of point "under" the segment plus one. Infact for construction the number of points belonging to R that are "upon" the segment is the same as the number of the ones that are "under", and equal to the half of R subtracted by the ones that are right "on" the segment (the 2 extrema).
figure 5
I am interested now to find the number of points that are under the segment except the
last column (as shown, in figure 5, by the points marked in green). This number, that i
name Q, is clearly given by
Q = (R2)/2  y2
since y2 is the number of points of the last column, and so
Q = (x2  x1 +1) * (y2 + y1 + 1) / 2  1  y2
figure 6
It easy to verify that this formula is valid also if point
P1 lies in a position higher than point
P2 or if they are at the same height
(that is y1
= y2).
In case P1 lies at the right of P2 (that is x1 > x2), the formula is still valid, except for the result that changed its sign and that also the involved extrema is counted. This result is shown in figure 6.
In this case, infact, the base of the rectangle R is the projection of the segment excluding the extrema, signinverted. The formula to count the points
R = (x2  x1 +1) * (y2 + y1 + 1)
represents so the inverse of the number of points in that rectangle. Note that the area is 0 if x1 = x2 + 1. The first part of the formula Q, (x2  x1 +1) * (y2 + y1 + 1) / 2, denotes the number of points that are under the segment, excluding the first and the last column, as shown by the points marked in green in figure 6.
figure 7
The second part of the formula Q,
1 y2, instead, clearly denotes the left column of the
points under the segment signinverted, counting, this time, also point
P2 itself, as shown by the blue points
in figure 6. The algebric sum of the two parts, given by gormula Q
Q = (x2  x1 +1) * (y2 + y1 + 1) / 2  1  y2
then denotes the number of points that are "under" the segment, excluding the rightmost column and counting also the left extrema, the all inverted of sign.
Formula Q is useful to compute, in a concatenation of segments, the number of points that are "under" for the ones that go from left to right, but in the same time "upon" the ones that run right to left. Figure 7 shows the concatenation of two segments left to right: the number of points under the red segment are summed to the number of points under the blue one.
In figures 81, 82, 83 is shown the procedure to algebrically sum the points of two consecutive
segments, one lefttoright and the other righttoleft. In 81 are shown in red the points
for the red segment, positive, in 82 in blue the ones for the blue segment, in 83 their algebric
sum.
In figure 91 to 96 sequence it is shown the iteration for segments from
P1P2
to P6P1
(closing the path).
Figures 91 and 92 show how to sum the contributions of two segments lefttoright, figure 93 shows the subtraction of points from the amount obtained in the previous step. In blue it's shown the "negative" points, that is the ones that are subtracted in excess. Those points undo some of the positive ones at point 94. In 95 and 96 are then subtracted the points of the last two, both righttoleft, segments.
The value obtained by the iteration is the number of the polygon's internal points, except two points that are subtracted in excess (extrema P1 and P4), and one that is summed in excess (P3). Considering the reasons for this happening, the conclusion is that the exceeding summed points are the vertices that form "left concavities" (the ones that form a concavity between a righttoleft segment and the next lefttoright one) and, viceversa, the exceeded subtracted points are the vertices that form "left convexities" (the ones that form a convexity from the previous righttoleft segment and the next lefttoright one).
It's easy to show that, going along the polygon clockwise, the total number of leftconcavity vertices is equal to the number of leftconvexity vertices minus one, infact, for each "change of direction", sooner or later there must be another change of direction in the opposite way. So, to compute the exact number of points internal to the polygon, if N is the number of leftconvexities, we must sum N and subtract N1, or, in other words, subtract 1.
Considering again the formula for Q, and generalizing on the indices of the points:
Qi = (xi+1  xi +1) * (yi+1 + yi + 1) / 2  1  yi+1
This is valid for all the n segments of the polygon, varying i on values 1, 2, ..., n. Index n+1 identifies point P1, so that the last segment is the one that join Pn to P1.
Modifying a little this formula, multiplying and collecting:
Qi = (xi+1  xi) * (yi+1 + yi) / 2 + (xi+1  xi + yi+1 + yi + 1)/2  1  yi+1
obtaining, then
Qi = (xi+1  xi) * (yi+1 + yi) / 2 + (xi+1  xi yi+1 + yi  1)/2
I can cut this formula, naming
Ai = (xi+1  xi) * (yi+1 + yi) / 2
Bi = (xi+1  xi yi+1 + yi  1)/2
and so
Qi = Ai + Bi
The number of polygon's internal points is than given by
PI = (Σi = 1..nQi) +1 = [Σi = 1..n(Ai + Bi)] +1
or:
PI = (Σi = 1..nAi) + (Σi = 1..nBi) +1
Said
A = Σi = 1..nAi
B = Σi = 1..nBi
the conclusion is that
PI = A + B + 1
The portion B can be simplified. Infact
B = Σi = 1..nBi = Σi = 1..n[(xi+1  xi yi+1 + yi  1)/2]
or, in other words
B = [(x2x1y2+y11) + (x3x2y3+y21) + ... + (xnxn1yn+yn11) + (x1xny1+yn1)]/2
Observing the formula written in this way, we note that all the terms xi and yi are canceled each other, and what remains is 1 summed n times, where n is the number of the Pick's polygon sides, which is clearly equal to the number of the points that lie on the edge of the polygon. So
B = PC/2
and so
PI = A  PC/2 + 1
Now i try to compute the area of the Pick's polygon.
figure 10
In a similar way as above, i consider the Pick's polygon segment by segment.
Every segment, if covered leftto right, gives a positive contribution
given by the area of the trapezoid formed by the segment itself and its projection
on the xaxis, as shown in figure 10. The are of the trapezoid is givne by the
sum of the bases (y1 + y2)
multiplied by the height (x2  x1) divided by 2.
Generalizing, then
Ai = (xi+1  xi) * (yi+1 + yi) / 2
This formula is valid also for the segment that are covered righttoleft, except for the fact that the obtained area is negative, since the height (xi+1  xi) is negative. The algebric sum of the contribution of all the segment, any verse covered, generates the area of the polygon, as shown in the sequence of figures 111 to 116.
At any step of the sequence the portion Ai
relative to the considered segment, is algebrically summed.
In figure 113 it is marked in blue the portion that is subtracted in excess, and that is then
recovered by the positive area summed at step 114.
Summation
Σi=1..n Ai
That computes the area of the polygon, is exactly the portion A of the formula obtained above
PI = A  PC/2 + 1
From this previos we obtain then
A = PI + PC/2  1
which is the thesis of Pick's theorem itself.
When i was a student, at this point, one concludes the exercise, enjoying the moment of suffered glory, with the achronym
CVD
("come volevasi dimostrare" = "as one wanted to demonstrate")
But there are some old loves that every now and then come back to stimulate and fill my mind, making me live again those emotions that i felt when i used to date them. And so, here i am to savour, once again with renewed strength, those pleasures i was granted of, almost like in the candour of a teenage flirt.
One of those loves is Mathematics. ;)
It's thanks to Giovanna and her blog Matematicamedie that i found something of this matter still unknown to me. Infact, i have never heard before about Pick's theorem.
As the teacher Giovanna well explains om her very interesting post, that tells about a wonderful work her class made, Pick's theorem computes in an easy and concise way the area of a polygon built under some particular rules.
To clarify better the problem, let me try with some definitions.
Given a regular grid of points (that we can imagine as the crossings of the lines on a squared paper), a "Pick's segment" is a segment that join two points of the grid without intersecting any other on its path. If we need a segment that lays on an extra point, it is considered like the concatenation of two consecutive segments lying on the same line.
A "Pick's polygon" is a closed concatenation of Pick's segments not selfintersecting.
It's right the area of Pick's polygon that is evaluated by Pick's theorem.
Please note that the nonselfintersection condition is demanded for any area computation.
figure 1
Moreover the condition for which a segment doesn't have to contain any other point of the grid more than its extrema, doesn't reduce the representation power of the definition. If a polygon has a side cut by a point of the grid, well, in that point one can also consider an additional vertex that generate a straight angle. For example figure 1 shows a Pick's polygon with four vertex and four sides, in different colors, even if it is perfectly superimposable on a triangle.
As Giovanna reveals, Pick's theorem asserts that area A of a Pick's polygon is given by the formula
A = PI + PC/2 1
dove PI is the number of the "internal" points (or the points of the regular grid that fall in teh internal part of the polygon) and PC is the number of the "contour" points: the ones that fall on the vertices of the Pick's segments  sides of the polygon, or, in other words, the points that lay on the edge.
Clearly this formula is valid if the obtained value is then multiplied by the area of the square that defines the regular grid. To make it more simple, on this post, i consider a grid in which the area of the square is one.
With reference to figure 1, for example, PI is the number of the points painted in blue (7), PC the ones in yellow (4), so the area of the polygon is
A = 7 + 4/2  1 = 8
It works. Try and see!
In a comment of the discussion that follows the quoted blog, Giovanna suggests me a couple of links to web pages where there is also the "official" demostration of Pick's theorem. Nevertheless i didn't follow those links, nor i have any intention to do it before submitting this post, because what enchants me of mathematics is not so much the result itself, even when in cases like this the formula has a really charming elegance. And not even i need a concrete proof to believe in the truth of the formula: i trust in Giovanna!
No, what i find irresistible is to use mathematics in order to find a demonstration my own.
Naturally this task is simplified by the fact that i already know which is the result i have to reach and i am already convinced that that result is true. So, i have only to find a right path to reach the goal (without a GPS navigator system ;)).
Trying to wear the pants of an eventual reader that doesn't know Pick's theorem, i warn now that the rest of this post shows my own demonstration and so, if one is interested to find another one, this is a good point to stop reading.
So...
First of all i find the way to count the internal points in the polygon, the geometrical informations on the vertices given.
I start defining a Cartesian axis system centered in any point of the grid lower and more on the left of the whole polygon (i believe that these last conditions are irrelevant, but, since i can choose, i simplify my own life!).
figure 2
figure 3
The number of points belonging to this rectangle si clearly given by the number of points at the base multipled by the number of point in the height.
Naming P1 and P2 the extrema of the segment, and (x1, y1) and (x2, y2) their coordinates, respectively, the computation of the number of points in the rectangle is given by this following formula:
R = (x2  x1 +1) * (y2 + y1 + 1)
The comuted value R is an even number.
figure 4
Numbers dx = x2  x1 and dy = y2  y1 are prime between each other. Infact if it was not like that, segment between tra P1 and P2 would meet other points of the grid. So, the segment wouldn't match the definition which a point of the grid cannot belong to a Pick's segment, unless one of its extrema. Since dx and dy are prime each other, atleast one of them is an odd number (if both were even, they would have 2 as a common divisor). So, atleast one value between dx+1 and dy+1 is even. It follows that their product is necessarily even, and incidentally that product is right the number of points of the green rectangle. So, R &is even.
Number R can be divided by 2, and the result of that division is given by the number of point "under" the segment plus one. Infact for construction the number of points belonging to R that are "upon" the segment is the same as the number of the ones that are "under", and equal to the half of R subtracted by the ones that are right "on" the segment (the 2 extrema).
figure 5
Q = (R2)/2  y2
since y2 is the number of points of the last column, and so
Q = (x2  x1 +1) * (y2 + y1 + 1) / 2  1  y2
figure 6
In case P1 lies at the right of P2 (that is x1 > x2), the formula is still valid, except for the result that changed its sign and that also the involved extrema is counted. This result is shown in figure 6.
In this case, infact, the base of the rectangle R is the projection of the segment excluding the extrema, signinverted. The formula to count the points
R = (x2  x1 +1) * (y2 + y1 + 1)
represents so the inverse of the number of points in that rectangle. Note that the area is 0 if x1 = x2 + 1. The first part of the formula Q, (x2  x1 +1) * (y2 + y1 + 1) / 2, denotes the number of points that are under the segment, excluding the first and the last column, as shown by the points marked in green in figure 6.
figure 7
Q = (x2  x1 +1) * (y2 + y1 + 1) / 2  1  y2
then denotes the number of points that are "under" the segment, excluding the rightmost column and counting also the left extrema, the all inverted of sign.
Formula Q is useful to compute, in a concatenation of segments, the number of points that are "under" for the ones that go from left to right, but in the same time "upon" the ones that run right to left. Figure 7 shows the concatenation of two segments left to right: the number of points under the red segment are summed to the number of points under the blue one.
figure 81 figure 82 figure 83 
figure 91 figure 92 figure 93 figure 94 figure 95 figure 96 
Figures 91 and 92 show how to sum the contributions of two segments lefttoright, figure 93 shows the subtraction of points from the amount obtained in the previous step. In blue it's shown the "negative" points, that is the ones that are subtracted in excess. Those points undo some of the positive ones at point 94. In 95 and 96 are then subtracted the points of the last two, both righttoleft, segments.
The value obtained by the iteration is the number of the polygon's internal points, except two points that are subtracted in excess (extrema P1 and P4), and one that is summed in excess (P3). Considering the reasons for this happening, the conclusion is that the exceeding summed points are the vertices that form "left concavities" (the ones that form a concavity between a righttoleft segment and the next lefttoright one) and, viceversa, the exceeded subtracted points are the vertices that form "left convexities" (the ones that form a convexity from the previous righttoleft segment and the next lefttoright one).
It's easy to show that, going along the polygon clockwise, the total number of leftconcavity vertices is equal to the number of leftconvexity vertices minus one, infact, for each "change of direction", sooner or later there must be another change of direction in the opposite way. So, to compute the exact number of points internal to the polygon, if N is the number of leftconvexities, we must sum N and subtract N1, or, in other words, subtract 1.
Considering again the formula for Q, and generalizing on the indices of the points:
Qi = (xi+1  xi +1) * (yi+1 + yi + 1) / 2  1  yi+1
This is valid for all the n segments of the polygon, varying i on values 1, 2, ..., n. Index n+1 identifies point P1, so that the last segment is the one that join Pn to P1.
Modifying a little this formula, multiplying and collecting:
Qi = (xi+1  xi) * (yi+1 + yi) / 2 + (xi+1  xi + yi+1 + yi + 1)/2  1  yi+1
obtaining, then
Qi = (xi+1  xi) * (yi+1 + yi) / 2 + (xi+1  xi yi+1 + yi  1)/2
I can cut this formula, naming
Ai = (xi+1  xi) * (yi+1 + yi) / 2
Bi = (xi+1  xi yi+1 + yi  1)/2
and so
Qi = Ai + Bi
The number of polygon's internal points is than given by
PI = (Σi = 1..nQi) +1 = [Σi = 1..n(Ai + Bi)] +1
or:
PI = (Σi = 1..nAi) + (Σi = 1..nBi) +1
Said
A = Σi = 1..nAi
B = Σi = 1..nBi
the conclusion is that
PI = A + B + 1
The portion B can be simplified. Infact
B = Σi = 1..nBi = Σi = 1..n[(xi+1  xi yi+1 + yi  1)/2]
or, in other words
B = [(x2x1y2+y11) + (x3x2y3+y21) + ... + (xnxn1yn+yn11) + (x1xny1+yn1)]/2
Observing the formula written in this way, we note that all the terms xi and yi are canceled each other, and what remains is 1 summed n times, where n is the number of the Pick's polygon sides, which is clearly equal to the number of the points that lie on the edge of the polygon. So
B = PC/2
and so
PI = A  PC/2 + 1
Now i try to compute the area of the Pick's polygon.
figure 10
Generalizing, then
Ai = (xi+1  xi) * (yi+1 + yi) / 2
This formula is valid also for the segment that are covered righttoleft, except for the fact that the obtained area is negative, since the height (xi+1  xi) is negative. The algebric sum of the contribution of all the segment, any verse covered, generates the area of the polygon, as shown in the sequence of figures 111 to 116.
figure 111 figure 112 figure 113 figure 114 figure 115 figure 116 
Summation
Σi=1..n Ai
That computes the area of the polygon, is exactly the portion A of the formula obtained above
PI = A  PC/2 + 1
From this previos we obtain then
A = PI + PC/2  1
which is the thesis of Pick's theorem itself.
When i was a student, at this point, one concludes the exercise, enjoying the moment of suffered glory, with the achronym
CVD
("come volevasi dimostrare" = "as one wanted to demonstrate")
Etichette:
Mathematics,
Pick's theorem
Friday, May 8, 2009
Monviso
Last week we enjoyed a much deserved vacation. We went to an area that we never visited before:
the Monviso (or mount Viso) area.
Unfortunately during the first three days the weather was ugly, very ugly. We had a lot of rain and down to the bottom of the valley they had also some flooding problems. Fortunately the next days were much better, even if we couldn't completely enjoy the excursions we had planned, both because the hiking paths were full of mud, and because the ugly weather caused little landslides that made it dangerous if not impossible to pass by on the paths. Moreover we were a little ingenuously unprepared to all of that snow (it can be for the microclimate of the lake if in our area, at the same altitude, snow is pretty much all melted).
Anyway we were able to do at least some hikes.
The first one is named "Tumpi la pisso", which in the local dialect i think it means something like "little waterfall in the pond". The path starts at the village Rore, in Varaita valley. Parking the car we missed the first deviation, although it was well indicated by a wooden sign, and we followed the white and red marks, on a skinny road, till we understood we were on the wrong one. Backtracking on our own steps we finally entered the right path, on the side of the brook, but here we couldn't go very far because one of the fords on the stream was impassable at all for the heavy rain of the past days.
Attention! This path is populated by "savarnots", troublemaking elfs of the Occitan mythology. Unfortunately we didn't have the chance to meet a real one: Maddie and Mr. Bentley had to be content to bark only towards one of the puppets that can be met along the path, settled by the local cultural association.
Another nice excursion was the one on mount Bracco. Parked the car on the road next to a convent and an inn, we started to walk entering the woods. Often the water flowed on the path itself making it difficult to walk because of the mud. Every now and then there were some buildings made out of stones that had been used for agricultural purposes. Fortunately we were next to one of them when all of the sudden we were surprised by a hailstorm, and we could use it as a shelter. Nothing terrible, but concerned that the situation could become worse we preferred to go back. Needless to say that the dogs were excited for the unusual weather conditions.
The entire tour of mount Bracco doesn't look very difficult, if the weather is reasonable. It is anyway kind of long: it asks several hours hiking.
The last excursion of this holiday was the one at Torrette towards Tenou. We parked the car on the side of a secondary road, right where there is the start point of the path.
The beginning is an enough wide white road. The path then goes into the forest, always uphill, goes next to some agricultural house and a little waterfall of a torrent affluent to Varaita river. It opens then in a meadow at the end of which it is impossible to keep walking because of a snow avalanche.
To tell the truth we had been warned, in the beginning of the hike, by a really gentle old man who we asked for directions. After warning us to use caution, he signaled us a detour from the main path, right next to a votive small chapel in the beginning of the meadow, that would have allowed us to pass the interruption.
Anyway the sight of some roe deer grazing was enough reward for the effort of the climb, and so, after a stop to eat a slice of pizza and a sip of water, we decided to go back to the bottom of the valley, giving back the "landlords" their peace. Unfortunately, infact, we were not able to pass by unseen, despite our attempts not to make too much noise and to convince the dogs that we were already settled for dinner, and so there was no need to exercise their hunting instinct.
For sure we will go back in this area to try some other itineraries. We couldn't infact
be able to try the one for which we were more stimulated, around the sources of the Po
river, between Pian della Regina and Pian del Re.
We lodged at Bed and Breakfast Il Bosco delle Terrecotte, at Barge, a restructured farmhouse in the peace atmosphere of the vineyards.
Unfortunately during the first three days the weather was ugly, very ugly. We had a lot of rain and down to the bottom of the valley they had also some flooding problems. Fortunately the next days were much better, even if we couldn't completely enjoy the excursions we had planned, both because the hiking paths were full of mud, and because the ugly weather caused little landslides that made it dangerous if not impossible to pass by on the paths. Moreover we were a little ingenuously unprepared to all of that snow (it can be for the microclimate of the lake if in our area, at the same altitude, snow is pretty much all melted).
Anyway we were able to do at least some hikes.
The first one is named "Tumpi la pisso", which in the local dialect i think it means something like "little waterfall in the pond". The path starts at the village Rore, in Varaita valley. Parking the car we missed the first deviation, although it was well indicated by a wooden sign, and we followed the white and red marks, on a skinny road, till we understood we were on the wrong one. Backtracking on our own steps we finally entered the right path, on the side of the brook, but here we couldn't go very far because one of the fords on the stream was impassable at all for the heavy rain of the past days.
Attention! This path is populated by "savarnots", troublemaking elfs of the Occitan mythology. Unfortunately we didn't have the chance to meet a real one: Maddie and Mr. Bentley had to be content to bark only towards one of the puppets that can be met along the path, settled by the local cultural association.
in red the first part on the asphalted road, in green the right (correct) path, unfortunately cut short. A: parking lot. 
Round trip on the first part:
Round trip on the second part:

Another nice excursion was the one on mount Bracco. Parked the car on the road next to a convent and an inn, we started to walk entering the woods. Often the water flowed on the path itself making it difficult to walk because of the mud. Every now and then there were some buildings made out of stones that had been used for agricultural purposes. Fortunately we were next to one of them when all of the sudden we were surprised by a hailstorm, and we could use it as a shelter. Nothing terrible, but concerned that the situation could become worse we preferred to go back. Needless to say that the dogs were excited for the unusual weather conditions.
The entire tour of mount Bracco doesn't look very difficult, if the weather is reasonable. It is anyway kind of long: it asks several hours hiking.

Outward:
Backward (on the same path): 
The last excursion of this holiday was the one at Torrette towards Tenou. We parked the car on the side of a secondary road, right where there is the start point of the path.
The beginning is an enough wide white road. The path then goes into the forest, always uphill, goes next to some agricultural house and a little waterfall of a torrent affluent to Varaita river. It opens then in a meadow at the end of which it is impossible to keep walking because of a snow avalanche.
To tell the truth we had been warned, in the beginning of the hike, by a really gentle old man who we asked for directions. After warning us to use caution, he signaled us a detour from the main path, right next to a votive small chapel in the beginning of the meadow, that would have allowed us to pass the interruption.
Anyway the sight of some roe deer grazing was enough reward for the effort of the climb, and so, after a stop to eat a slice of pizza and a sip of water, we decided to go back to the bottom of the valley, giving back the "landlords" their peace. Unfortunately, infact, we were not able to pass by unseen, despite our attempts not to make too much noise and to convince the dogs that we were already settled for dinner, and so there was no need to exercise their hunting instinct.

Outward (including the roe deers meadow tour):

We lodged at Bed and Breakfast Il Bosco delle Terrecotte, at Barge, a restructured farmhouse in the peace atmosphere of the vineyards.
Statistics of all the excursions during year 2009:
Total time: 11 hours and 44 minutes
Distance: 34.73km [21.58mi]
Difference of level: 2286m [7500']
Minimum altitude: 715m [2346']
Maximum altitude: 1550m [5085']
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