The Folded Rouble

After the war, in $1947$, in the Soviet Union new ban­knotes were intro­duced, which were still in cir­cu­la­tion in $1956$.

In that year, Vladimir Igore­vich Arnold posed the prob­lem of the folded rou­ble. Can you build a pla­nar poly­gon by fold­ing a sheet of paper (a rou­ble bill) so that the perime­ter of this poly­gon is larger than the perime­ter of the orig­i­nal sheet?

In 1961 in Rus­sia the coins and ban­knotes changed again, so that also the design of the rou­ble changed, becom­ing a much smaller piece of paper. At that time, how­ever, the prob­lem was not resolved.

Fur­ther­more, although a pos­i­tive answer “It is pos­si­ble” con­tra­dicts the intu­ition, there must be a math­e­mat­i­cal rea­son for a neg­a­tive answer. If we fold a rec­tan­gle along a straight line, then the perime­ter can only be reduced: to the already exist­ing edge we have to add the seg­ment of the line along which we have folded, but we have to remove a bro­ken line that has the same end­points as that seg­ment. If we do some­thing sim­i­lar, that is if we fold again the obtained poly­gon along a straight line, the sit­u­a­tion is the same: the perime­ter increases by the length of a seg­ment, but decreases by the length of a bro­ken line. This type of fold­ing — along a straight line — is called “sim­ple” and can only decrease the perime­ter. This how­ever is only one rea­son, but not yet proof.

So can we or not increase the ini­tial perime­ter of the rec­tan­gle? In the years $1991$ and $1993$, yet the ban­knotes were changed, and that of the rou­ble of $1961$ came out of cir­cu­la­tion. But the Arnold prob­lem still remained unsolved.

Since that time a Russ­ian rou­ble worth, unfor­tu­nately, so lit­tle that there are no more notes, but only coins, of that value.

At the begin­ning of the twenty-first cen­tury, how­ever, the prob­lem was solved. The first math­e­mat­i­cally rig­or­ous solu­tion was pro­vided by a stu­dent of N. P. Dol­bilin, Alexei Tarasov. He invented an algo­rithm for fold­ing a square so that in total you get a pla­nar poly­gon with a greater perime­ter.

Who only wishes to enjoy the movie, can skip the next part, which has been added for those who want to under­stand well how the sheet has to be folded.

Take a square sheet of paper, and let us divide it into square cells, for exam­ple, $4 \times 4$. Colour the cells with two colours like a chess­board and draw from the cen­tre of each square a defined num­ber of rays. Then add in every red square a green star, so that the size of the stars grow in the direc­tion of one diag­o­nal. Now fold the paper in a strip, and then into a rec­tan­gle, and, even­tu­ally, into a tri­an­gle. The result­ing object is done as fol­lows. One half of it con­tains sev­eral blue lay­ers, the other con­tains red-green lay­ers. The way of draw­ing the stars was such that after fold­ing they increase in size as one passes from one layer to another. Let us start to fold each tri­an­gle so that the par­al­lel blue lay­ers go to one side, and the green and red to the other. We get a sur­face that even­tu­ally is folded again into a pla­nar poly­gon.

This poly­gon has one red part (the blue tri­an­gles result to be hid­den inside) and a green comb. We observe that the num­ber of teeth of this comb is exactly the num­ber of green stars, that is, of red squares that were ini­tially in the square.

But did the perime­ter increase, com­pared to that of the ini­tial square? Is the prob­lem solved? If we com­pare the fig­ures, we imme­di­ately see that the perime­ter is decreased. Why then we made such com­pli­cated fold­ing?

In the con­crete exam­ple we have con­sid­ered, we used a gen­eral algo­rithm. And in this algo­rithm there are two para­me­ters: the num­ber of cells in the ini­tial square and the num­ber of rays in each square. Let us look at what hap­pens if we change these para­me­ters.

For the same sub­di­vi­sion in $4 \times 4$ cells we will increase the num­ber of rays within each cell. This leads to a nar­row­ing of the teeth of the comb, to a lower inter­sec­tion of them, and con­se­quently, to a mod­er­ate growth of the perime­ter.

There is still another para­me­ter, the num­ber of cells that fill the ini­tial square. If we increase this para­me­ter, then the comb of the result­ing con­struc­tion will have a big­ger num­ber of teeth.

The simul­ta­ne­ous growth of the two para­me­ters, both the num­ber of cells and the amount of rays, gives an increase in the perime­ter. But how much it may increase? It results, up to infin­ity. But this means that at some moment it will exceed the perime­ter of the ini­tial square!

The prob­lem of the folded rou­ble — to fold a rec­tan­gle so that its perime­ter increases — is solved. But how many times we need to fold? Rather a lot. From the work of A. Tarasov we can get an esti­mate: for a square divided in $16 \times 16$ cells and $16^2 \cdot 30$ rays in each cell, the perime­ter of the obtained poly­gon will be greater than the perime­ter of the ini­tial square.

This can­not be shown in the film, but can it be achieved in real­ity? Surely you well remem­ber that fold­ing a sheet of paper, albeit very thin, is pos­si­ble no more than $7—8$ times. If you do not do this for a long time, try that with a sim­ple exper­i­ment. So, what the same prob­lem posed by V. I. Arnold, and its solu­tion with an algo­rithm “not fea­si­ble”, give us? Cer­tainly an instru­ment of progress in sci­ence, which cer­tainly will be use­ful in future devel­op­ments.