{"version":1,"lectureId":"01M14TZDF7NPG7EJ913X21HSZK","attempt":0,"publication":{"slug":"pigeonhole-principle-boxes","title":"The Pigeonhole Principle: Choosing the Boxes","subject":"mathematics","summary":"The pigeonhole principle takes one line to state and one sentence to prove, and then proves things that look nothing like it. This lecture states it once, over four boxes and five items, and spends the rest of its time on three consequences: that any group of people contains two who know the same number of others in the group, that any ten distinct numbers contain four that rise or four that fall, and Dirichlet's theorem that every irrational number is chased by fractions to within one over the denominator squared. Each proof is a single sentence once the boxes have been chosen, so each time the boxes are built on screen, including the first choice that fails and has to be repaired. For a first year student meeting combinatorial argument for the first time.","metaDescription":"State the pigeonhole principle in one line, then watch it prove three things that look nothing like it. The trick is choosing the boxes.","transcript":"Here is a piece of mathematics you already know. If you have more things than places to put them, some place gets two things. That is it. It sounds like it could not possibly prove anything, and over the next few minutes I would like to change your mind about that. So here it is, properly. Four boxes, five items. Put them away however you like, however cleverly, and one of these boxes ends up holding two. Say the same thing with a letter in it. Put n plus one items into n boxes, and some box holds at least two of them. And the proof is a single sentence. If every box held one item or none, then adding up over the boxes you would have at most n items in total. But you have more than n. So some box was holding two. That sentence is the easy half, and I will never have to say anything harder than it. The hard half never appears in it at all, because the hard half is deciding what the items are and what the boxes are. So let me put three statements on the board. Not one of them mentions a box, or an item, or anything that looks like counting. Two people at any party know the same number of people there. Any ten numbers, all different, contain four that go steadily up or four that go steadily down. And every irrational number has fractions chasing it far more closely than it has any right to expect. Every one of those is the sentence you just heard, applied once. So watch the boxes each time, and not the sentence. The boxes are the argument. Six people at a party. Some of them know each other and some do not, and knowing is mutual: if I know you, you know me. That is the only rule there is, and here is one particular evening. Now go round and count. A knows exactly one person here. B knows two. And C also knows two. D, E and F each know three of the others. So the six counts are one, two, two, three, three, three, and they have already collided twice. You might reasonably think that was luck, and that I drew the lines to make it happen. I did not. Whatever this picture had looked like, two of the counts would have had to agree, and I want to show you why. Take the people themselves as the items. For the boxes, take the possible answers to the question: how many people here do you know? Nobody can know more than the other n minus one, and nobody can know fewer than none at all, so every count lands somewhere in this list. Now count the boxes. Zero, one, two, and so on up to n minus one. That is n boxes for n people, and pigeonhole gives us absolutely nothing. Six items in six boxes can sit one to a box quite happily. So the obvious choice of boxes fails, and I want you to sit with that for a moment, because this is exactly where these proofs are won or lost. Look at the two boxes on the ends. Can they both be used at once? Suppose somebody at this party knows nobody. Then no one in the room can know everybody, because knowing everybody would include knowing that person. Box zero and box n minus one can never both be occupied. So one of the two ends is always empty, whichever way it falls. The boxes were never n. There were only ever n minus one of them. Six people, five boxes. And now the sentence from the beginning does all the remaining work, without being asked twice. Two of them land together. Two people at any party, anywhere, know the same number of people at that party. Notice how little of that was cleverness. One honest look at which boxes could actually be used, and the theorem fell out. Ten numbers, all different, in whatever order somebody wrote them down. Position along the bottom, value up the side. I want to hunt for runs. A run means this: read from left to right, skipping whatever you like, and take terms that keep going up, or terms that keep going down. They do not have to be neighbours. Start at the second term, which is seven. Going up from there I can reach nine, and then ten. A rising run of three. Going down from that same seven I can reach five, and then four. A falling run of three as well. So attach two numbers to every position. Call them u and d: the length of the longest rise that starts there, and the length of the longest fall that starts there. At the second term, both of them are three. Now, does this sequence contain a run of four? It does. Two, five, eight, ten, at positions three, five, seven and nine, rising the whole way. And that was not luck either. Any ten distinct numbers contain four that rise, or four that fall. To see it, suppose they do not. Suppose there is no rising run of four anywhere, and no falling run of four either. Then every u is one, two or three, and so is every d. Two numbers, three choices each. Three times three is nine possible pairs, and there are the nine of them, one box apiece. But there are ten positions, and each position hands you one pair. Ten items, nine boxes. So two positions, i somewhere to the left of j, carry the identical pair. Same u, same d. And now it breaks. The numbers are all different, so either a i is below a j or above it. Suppose it is below. Take the longest rise starting at j, and stick a i on the front of it. That is a rise starting at i, and it is one term longer. So u i beats u j, when they were supposed to be equal. If a i is the larger one instead, the very same trick runs downhill and d i beats d j. Either way the two pairs could not have matched, so the assumption is dead. There has to be a monotone run of four somewhere in that sequence. And ten was not an arbitrary number. Nine positions fit into nine boxes without any fight at all, so ten is the first length that forces the issue. Three by three, plus one. One more, and this time the boxes are not objects at all. They are stretches of a line, and that turns out to change nothing about the argument and everything about what it can prove. The subject is approximating an irrational number by a fraction. Getting close to the square root of two is easy: take enough decimal places. The real question is how close you can get while keeping the denominator small. Dirichlet's answer is startling. For every whole number N there is a fraction p over q, with q no bigger than N, that sits within one over q times N of your number. And since q is at most N, that is within one over q squared. A denominator of five buying an error under one twenty fifth is not what randomly chosen fractions do for you. So, the boxes. Let N be five, and take the first six multiples of root two: nought, one point four one, two point eight three, four point two four, five point six six, and seven point zero seven. Now throw away the whole number part of each one and keep only what comes after the point. Six numbers, every one of them somewhere between nought and one, and here they are. And cut that stretch from nought to one into five equal pieces. Those are the boxes. Six numbers, five intervals, and I do not have to look at the numbers to know what happens next. Two of the six share an interval. Here they are: the very first one, which is nought exactly, and the last one, seven point zero seven with the seven thrown away. Both of them inside the leftmost box. And two numbers sitting in one interval of width a fifth are less than a fifth apart. That is everything the boxes were ever for. From here it is arithmetic. Write it out in general. The items are these N plus one fractional parts, one for each multiple of alpha from nought up to N. The boxes are the N intervals. More numbers than intervals, so two of them land in the same one: call their indices i and j, with i the smaller. The two fractional parts differ by less than one over N. Now unpack what a fractional part is. It is the number itself, minus some whole number. So that small difference is j alpha minus i alpha, with an integer taken off it. Give those two things names. Let q be j minus i, which is between one and N, and let p be the integer that came off. Then q alpha minus p is less than one over N in size. Divide the whole line through by q. Alpha minus p over q is less than one over q N. And q is at most N, so one over q N is at most one over q squared, which is the theorem. Put our own numbers in. The two indices were nought and five, so q is five, and p works out at seven. Seven fifths, which is one point four. The true error is a hundredth and a bit, comfortably inside the one twenty fifth we were promised. And run the whole thing again with a larger N and you get a different fraction, a better one. Which means an irrational number has infinitely many fractions chasing it this closely, forever. And the boxes that proved it were five stretches of a line. Three theorems, then, and three choices of box. In the first, the items were people and the boxes were the possible answers to a question, once we had noticed that two of those answers could never both be used. In the second, the items were positions in a sequence, and the boxes were pairs of numbers we had to invent from nothing. In the third, the items were multiples of alpha, and the boxes were stretches of a line. And the sentence at the end was the same all three times. More items than boxes, so two share a box. It was never the hard part. The boxes were.","watch":{"version":1,"scenes":[{"title":"One Line, Then the Hard Part","start":0,"end":94.0981875,"objects":{"bins":"a Polygon [gray] drawn in pens (vertices=((0.4, 0.5), (2.0, 0.5), (2.0, 1.9), (0.4, 1.9)), filled=False)","bins_2":"a Polygon [gray] drawn in pens (vertices=((2.2, 0.5), (3.8000000000000003, 0.5), (3.8000000000000003, 1.…, filled=False)","bins_3":"a Polygon [gray] drawn in pens (vertices=((4.0, 0.5), (5.6, 0.5), (5.6, 1.9), (4.0, 1.9)), filled=False)","bins_4":"a Polygon [gray] drawn in pens (vertices=((5.8, 0.5), (7.4, 0.5), (7.4, 1.9), (5.8, 1.9)), filled=False)","card":"a Title that says \"Discrete Mathematics — The Pigeonhole Principle: Choosing the Boxes\"","chips":"a Point [blue] drawn in pens (location=(1.2, 1.2))","chips_2":"a Point [blue] drawn in pens (location=(3.0, 1.2))","chips_3":"a Point [yellow] drawn in pens (location=(4.5, 1.2))","chips_4":"a Point [yellow] drawn in pens (location=(5.1, 1.2))","chips_5":"a Point [blue] drawn in pens (location=(6.6, 1.2))","head_plan":"a Heading that says \"Three Things It Proves\"","head_state":"a Heading that says \"The Whole Principle\"","note":"a Panel that says \"Nothing in that line tells you what the items are or what the boxes are. Choosing them is the whole of the work.\"","oneline":"a Math [text] that says \"$lt.eq 1 thin upright(\"per box\") arrow.r.double lt.eq n thin upright(\"items\")$\"","pens":"a Figure (x_range=(0.0, 8.0), y_range=(0.0, 2.6), aspect=(8.0, 2.6))","plan":"a Block [text] that says \"Two people at any party know the same number of people there. Any ten distinct numbers hide four that rise, or four that fall. Every irrational number is chased by fractions closer than it deserves.\"","statement":"a Math [text] that says \"$n + 1 thin upright(\"items,\") thin n thin upright(\"boxes\") arrow.r.double upright(\"two share\")$\""},"beats":[{"start":0,"say":"Here is a piece of mathematics you already know. If you have more things than places to put them, some place gets two things. That is it. It sounds like it could not possibly prove anything, and over the next few minutes I would like to change your mind about that.","live":[],"does":[[0,"card is shown on the screen, written out."],[1.5,"card: enter:write-left-to-right."],[14.6635,"card is hidden from the screen — left the board."]]},{"start":15.8635,"say":"So here it is, properly. Four boxes, five items. Put them away however you like, however cleverly, and one of these boxes ends up holding two.","live":null,"does":[[15.8635,"head_state is shown on the screen, written out."],[15.8635,"pens is shown on the screen, written out."],[17.802,"bins is shown on the screen, written out."],[17.922,"bins_2 is shown on the screen, written out."],[18.041999999999998,"bins_3 is shown on the screen, written out."],[18.162,"bins_4 is shown on the screen, written out."],[19.195,"chips is shown on the screen, written out."],[19.375,"chips_2 is shown on the screen, written out."],[19.555,"chips_3 is shown on the screen, written out."],[19.735,"chips_4 is shown on the screen, written out."],[19.915,"chips_5 is shown on the screen, written out."]]},{"start":26.2855,"say":"Say the same thing with a letter in it. 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So some box was holding two.","live":["statement","pens","head_state","bins","bins_2","bins_3","bins_4","chips","chips_2","chips_3","chips_4","chips_5"],"does":[[35.987,"oneline is shown on the screen, written out."],[43.011,"oneline (the \"lt.eq n thin upright(\"items\")\" part) is emphasized."],[46.68,"oneline (the \"lt.eq n thin upright(\"items\")\" part) is no longer emphasized."]]},{"start":48.370999999999995,"say":"That sentence is the easy half, and I will never have to say anything harder than it. 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Not one of them mentions a box, or an item, or anything that looks like counting.","live":[],"does":[[59.50149999999999,"head_plan is shown on the screen, written out."],[60.336999999999996,"plan is shown on the screen, written out."]]},{"start":67.82199999999999,"say":"Two people at any party know the same number of people there. Any ten numbers, all different, contain four that go steadily up or four that go steadily down. 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Whatever this picture had looked like, two of the counts would have had to agree, and I want to show you why.","live":["party","seen","head_party","folk","folk_2","folk_3","folk_4","folk_5","folk_6","links","links_2","links_3","links_4","links_5","links_6","links_7"],"does":[[134.1071875,"head_party is hidden from the screen — left the board."],[134.1071875,"party is hidden from the screen — left the board."],[134.1071875,"folk is hidden from the screen — party left the board."],[134.1071875,"folk_2 is hidden from the screen — party left the board."],[134.1071875,"folk_3 is hidden from the screen — party left the board."],[134.1071875,"folk_4 is hidden from the screen — party left the board."],[134.1071875,"folk_5 is hidden from the screen — party left the board."],[134.1071875,"folk_6 is hidden from the screen — party left the board."],[134.1071875,"links is hidden from the screen — party left the board."],[134.1071875,"links_2 is hidden from the screen — party left the board."],[134.1071875,"links_3 is hidden from the screen — party left the board."],[134.1071875,"links_4 is hidden from the screen — party left the board."],[134.1071875,"links_5 is hidden from the screen — party left the board."],[134.1071875,"links_6 is hidden from the screen — party left the board."],[134.1071875,"links_7 is hidden from the screen — party left the board."],[134.1071875,"seen is hidden from the screen — left the board."],[134.1071875,"slots is shown on the screen, written out."],[134.1071875,"bins is shown on the screen, written out."],[134.2071875,"bins_2 is shown on the screen, written out."],[134.4071875,"bins_3 is shown on the screen, written out."],[134.7071875,"bins_4 is shown on the screen, written out."],[135.1071875,"bins_5 is shown on the screen, written out."]]},{"start":135.3071875,"say":"Take the people themselves as the items. 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And now the sentence from the beginning does all the remaining work, without being asked twice.","live":["slots","head_boxes","bins","bins_2","bins_3","bins_4","bins_5","tags","tags_2","tags_3","tags_4","tags_5"],"does":[[207.1771875,"guests is shown on the screen, written out."],[207.3371875,"guests_2 is shown on the screen, written out."],[207.4971875,"guests_3 is shown on the screen, written out."],[207.65718750000002,"guests_4 is shown on the screen, written out."],[207.8171875,"guests_5 is shown on the screen, written out."],[207.9771875,"guests_6 is shown on the screen, written out."]]},{"start":213.8961875,"say":"Two of them land together. Two people at any party, anywhere, know the same number of people at that party. Notice how little of that was cleverness. One honest look at which boxes could actually be used, and the theorem fell out.","live":["slots","head_boxes","bins","bins_2","bins_3","bins_4","bins_5","tags","tags_2","tags_3","tags_4","tags_5","guests","guests_2","guests_3","guests_4","guests_5","guests_6"],"does":[[214.9991875,"guests_3 is indicated — a transient flash."],[215.29918750000002,"guests_4 is indicated — a transient flash."],[220.1071875,"verdict is shown on the screen, written out."],[222.9751875,"A box is drawn around verdict."],[229.02652083333334,"head_boxes is hidden from the screen — left the board."],[229.02652083333334,"slots is hidden from the screen — left the board."],[229.02652083333334,"bins is hidden from the screen — slots left the board."],[229.02652083333334,"bins_2 is hidden from the screen — slots left the board."],[229.02652083333334,"bins_3 is hidden from the screen — slots left the board."],[229.02652083333334,"bins_4 is hidden from the screen — slots left the board."],[229.02652083333334,"bins_5 is hidden from the screen — slots left the board."],[229.02652083333334,"tags is hidden from the screen — slots left the board."],[229.02652083333334,"tags_2 is hidden from the screen — slots left the board."],[229.02652083333334,"tags_3 is hidden from the screen — slots left the board."],[229.02652083333334,"tags_4 is hidden from the screen — slots left the board."],[229.02652083333334,"tags_5 is hidden from the screen — slots left the board."],[229.02652083333334,"guests is hidden from the screen — slots left the board."],[229.02652083333334,"guests_2 is hidden from the screen — slots left the board."],[229.02652083333334,"guests_3 is hidden from the screen — slots left the board."],[229.02652083333334,"guests_4 is hidden from the screen — slots left the board."],[229.02652083333334,"guests_5 is hidden from the screen — slots left the board."],[229.02652083333334,"guests_6 is hidden from the screen — slots left the board."],[229.02652083333334,"verdict is hidden from the screen — left the board."],[229.02652083333334,"work is hidden from the screen — left the board."]]}]},{"title":"Ten Numbers, and a Run of Four","start":230.0681875,"end":395.78841666666665,"objects":{"cells":"a Polygon [gray] drawn in grid (vertices=((0.75, 0.75), (1.75, 0.75), (1.75, 1.75), (0.75, 1.75)), filled=False)","cells_2":"a Polygon [gray] drawn in grid (vertices=((1.9, 0.75), (2.9, 0.75), (2.9, 1.75), (1.9, 1.75)), filled=False)","cells_3":"a Polygon [gray] drawn in grid (vertices=((3.05, 0.75), (4.05, 0.75), (4.05, 1.75), (3.05, 1.75)), filled=False)","cells_4":"a Polygon [gray] drawn in grid (vertices=((0.75, 1.9), (1.75, 1.9), (1.75, 2.9), (0.75, 2.9)), filled=False)","cells_5":"a Polygon [gray] drawn in grid (vertices=((1.9, 1.9), (2.9, 1.9), (2.9, 2.9), (1.9, 2.9)), filled=False)","cells_6":"a Polygon [gray] drawn in grid (vertices=((3.05, 1.9), (4.05, 1.9), (4.05, 2.9), (3.05, 2.9)), filled=False)","cells_7":"a Polygon [gray] drawn in grid (vertices=((0.75, 3.05), (1.75, 3.05), 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(location=(9, 10))","example":"a Math [text] that says \"$u_2 = 3, quad d_2 = 3$\"","fall_legs":"a Line [red] drawn in plot (start=(2, 7), end=(5, 5))","fall_legs_2":"a Line [red] drawn in plot (start=(5, 5), end=(8, 4))","four_legs":"a Line [yellow] drawn in plot (start=(3, 2), end=(5, 5))","four_legs_2":"a Line [yellow] drawn in plot (start=(5, 5), end=(7, 8))","four_legs_3":"a Line [yellow] drawn in plot (start=(7, 8), end=(9, 10))","grid":"a Figure (x_range=(-0.1, 4.6), y_range=(-0.1, 4.6), aspect=(1.0, 1.0))","head_grid":"a Heading that says \"Nine Boxes for Ten Positions\"","head_seq":"a Heading that says \"Ten Numbers in Some Order\"","plot":"an Axes (x_range=(0.0, 11.0), y_range=(0.0, 11.0), x_ticks_every=1.0)","proof":"a Derivation [text] that says \"$u_i, d_i &in brace.l 1, 2, 3 brace.r \\ 10 thin upright(\"positions\") &arrow.r 9 thin upright(\"pairs\") \\ (u_i, d_i) &= (u_j, d_j), quad i < j \\ a_i < a_j &arrow.r.double u_i > u_j \\ a_i > a_j &arrow.r.double d_i > d_j$\"","rise_legs":"a Line [green] drawn in plot (start=(2, 7), end=(4, 9))","rise_legs_2":"a Line [green] drawn in plot (start=(4, 9), end=(9, 10))","row_tags":"a Math [text] that says \"$1$\" drawn in grid","row_tags_2":"a Math [text] that says \"$2$\" drawn in grid","row_tags_3":"a Math [text] that says \"$3$\" drawn in grid","u_tag":"a Math [green] that says \"$u$\" drawn in grid","verdict":"a Tex [text] that says \"So a monotone run of four always exists.\""},"beats":[{"start":230.0681875,"say":"Ten numbers, all different, in whatever order somebody wrote them down. Position along the bottom, value up the side.","live":[],"does":[[230.0681875,"head_seq is shown on the screen, written out."],[230.0681875,"plot is shown on the screen, written out."],[235.10718749999998,"dots is shown on the screen, written out."],[235.22718749999999,"dots_2 is shown on the screen, written out."],[235.3471875,"dots_3 is shown on the screen, written out."],[235.4671875,"dots_4 is shown on the screen, written out."],[235.5871875,"dots_5 is shown on the screen, written out."],[235.7071875,"dots_6 is shown on the screen, written out."],[235.82718749999998,"dots_7 is shown on the screen, written out."],[235.94718749999998,"dots_8 is shown on the screen, written out."],[236.0671875,"dots_9 is shown on the screen, written out."],[236.1871875,"dots_10 is shown on the screen, written out."]]},{"start":238.7601875,"say":"I want to hunt for runs. A run means this: read from left to right, skipping whatever you like, and take terms that keep going up, or terms that keep going down. They do not have to be neighbours.","live":["plot","head_seq","dots","dots_2","dots_3","dots_4","dots_5","dots_6","dots_7","dots_8","dots_9","dots_10"],"does":[]},{"start":253.4551875,"say":"Start at the second term, which is seven. Going up from there I can reach nine, and then ten. A rising run of three.","live":null,"does":[[258.4701875,"rise_legs is shown on the screen, drawn."],[258.7701875,"rise_legs_2 is shown on the screen, drawn."]]},{"start":262.8901875,"say":"Going down from that same seven I can reach five, and then four. A falling run of three as well.","live":["plot","head_seq","dots","dots_2","dots_3","dots_4","dots_5","dots_6","dots_7","dots_8","dots_9","dots_10","rise_legs","rise_legs_2"],"does":[[265.5951875,"fall_legs is shown on the screen, drawn."],[265.8951875,"fall_legs_2 is shown on the screen, drawn."]]},{"start":270.5261875,"say":"So attach two numbers to every position. Call them u and d: the length of the longest rise that starts there, and the length of the longest fall that starts there. At the second term, both of them are three.","live":["plot","head_seq","dots","dots_2","dots_3","dots_4","dots_5","dots_6","dots_7","dots_8","dots_9","dots_10","rise_legs","rise_legs_2","fall_legs","fall_legs_2"],"does":[[275.6461875,"plot moves to a new place on the board."],[275.6461875,"def_u is shown on the screen, written out."],[279.4081875,"def_d is shown on the screen, written out."],[282.5541875,"example is shown on the screen, written out."]]},{"start":284.8721875,"say":"Now, does this sequence contain a run of four? It does. 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Three times three is nine possible pairs, and there are the nine of them, one box apiece.","live":["claim","grid","head_grid","cells","cells_2","cells_3","cells_4","cells_5","cells_6","cells_7","cells_8","cells_9","col_tags","col_tags_2","col_tags_3","row_tags","row_tags_2","row_tags_3","u_tag","d_tag"],"does":[[323.2291875,"cells is indicated — a transient flash."],[323.3091875,"cells_2 is indicated — a transient flash."],[323.38918750000005,"cells_3 is indicated — a transient flash."],[323.46918750000003,"cells_4 is indicated — a transient flash."],[323.5491875,"cells_5 is indicated — a transient flash."],[323.6291875,"cells_6 is indicated — a transient flash."],[323.7091875,"cells_7 is indicated — a transient flash."],[323.7891875,"cells_8 is indicated — a transient flash."],[323.8691875,"cells_9 is indicated — a transient flash."]]},{"start":328.0781875,"say":"But there are ten positions, and each position hands you one pair. 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Same u, same d.","live":["claim","grid","head_grid","cells","cells_2","cells_3","cells_4","cells_5","cells_6","cells_7","cells_8","cells_9","col_tags","col_tags_2","col_tags_3","row_tags","row_tags_2","row_tags_3","u_tag","d_tag","chips","chips_2","chips_3","chips_4","chips_5","chips_6","chips_7","chips_8","chips_9","chips_10"],"does":[[339.5451875,"proof is shown on the screen, written out."],[341.14718750000003,"chips_5 is indicated — a transient flash."],[341.54718750000006,"chips_6 is indicated — a transient flash."]]},{"start":343.5646875,"say":"And now it breaks. The numbers are all different, so either a i is below a j or above it. Suppose it is below.","live":null,"does":[]},{"start":352.88918750000005,"say":"Take the longest rise starting at j, and stick a i on the front of it. That is a rise starting at i, and it is one term longer. So u i beats u j, when they were supposed to be equal.","live":null,"does":[[362.83918750000004,"proof is shown on the screen, written out."]]},{"start":366.6901875,"say":"If a i is the larger one instead, the very same trick runs downhill and d i beats d j. Either way the two pairs could not have matched, so the assumption is dead. There has to be a monotone run of four somewhere in that sequence.","live":null,"does":[[370.6371875,"proof is shown on the screen, written out."],[377.1501875,"verdict is shown on the screen, written out."],[380.32018750000003,"A box is drawn around verdict."]]},{"start":382.7196875,"say":"And ten was not an arbitrary number. Nine positions fit into nine boxes without any fight at all, so ten is the first length that forces the issue. Three by three, plus one.","live":["claim","verdict","grid","head_grid","cells","cells_2","cells_3","cells_4","cells_5","cells_6","cells_7","cells_8","cells_9","col_tags","col_tags_2","col_tags_3","row_tags","row_tags_2","row_tags_3","u_tag","d_tag","chips","chips_2","chips_3","chips_4","chips_5","chips_6","chips_7","chips_8","chips_9","chips_10"],"does":[[394.74675,"claim is hidden from the screen — left the board."],[394.74675,"grid is hidden from the screen — left the board."],[394.74675,"cells is hidden from the screen — grid left the board."],[394.74675,"cells_2 is hidden from the screen — grid left the board."],[394.74675,"cells_3 is hidden from the screen — grid left the board."],[394.74675,"cells_4 is hidden from the screen — grid left the board."],[394.74675,"cells_5 is hidden from the screen — grid left the board."],[394.74675,"cells_6 is hidden from the screen — grid left the board."],[394.74675,"cells_7 is hidden from the screen — grid left the board."],[394.74675,"cells_8 is hidden from the screen — grid left the board."],[394.74675,"cells_9 is hidden from the screen — grid left the board."],[394.74675,"col_tags is hidden from the screen — grid left the board."],[394.74675,"col_tags_2 is hidden from the screen — grid left the board."],[394.74675,"col_tags_3 is hidden from the screen — grid left the board."],[394.74675,"row_tags is hidden from the screen — grid left the board."],[394.74675,"row_tags_2 is hidden from the screen — grid left the board."],[394.74675,"row_tags_3 is hidden from the screen — grid left the board."],[394.74675,"u_tag is hidden from the screen — grid left the board."],[394.74675,"d_tag is hidden from the screen — grid left the board."],[394.74675,"chips is hidden from the screen — grid left the board."],[394.74675,"chips_2 is hidden from the screen — grid left the board."],[394.74675,"chips_3 is hidden from the screen — grid left the board."],[394.74675,"chips_4 is hidden from the screen — grid left the board."],[394.74675,"chips_5 is hidden from the screen — grid left the board."],[394.74675,"chips_6 is hidden from the screen — grid left the board."],[394.74675,"chips_7 is hidden from the screen — grid left the board."],[394.74675,"chips_8 is hidden from the screen — grid left the board."],[394.74675,"chips_9 is hidden from the screen — grid left the board."],[394.74675,"chips_10 is hidden from the screen — grid left the board."],[394.74675,"head_grid is hidden from the screen — left the board."],[394.74675,"proof is hidden from the screen — left the board."],[394.74675,"verdict is hidden from the screen — left the board."]]}]},{"title":"Boxes That Are Intervals","start":395.78841666666665,"end":651.5731458333332,"objects":{"answer":"a Math [text] that says \"$q = 5, quad p = 7, quad abs(sqrt(2) - 7 / 5) < 1 / 25$\"","closing":"a Tex [text] that says \"The last sentence was the same all three times.\"","g0":"a Math [text] that says \"$brace.l k alpha brace.r, quad k = 0, 1, dots, N$\"","g1":"a Math [text] that says \"$N + 1 thin upright(\"numbers,\") quad N thin upright(\"intervals\")$\"","g2":"a Math [text] that says \"$abs(brace.l j alpha brace.r - brace.l i alpha brace.r) < 1 / N$\"","g3":"a Math [text] that says \"$q = j - i, quad p = floor(j alpha) - floor(i alpha)$\"","g4":"a Math [text] that says \"$abs(q alpha - p) < 1 / N$\"","g5":"a Math [text] that says \"$abs(alpha - p / q) < 1 / (q N) lt.eq 1 / q^2$\"","head_int":"a Heading that says \"Cutting Up the Unit Interval\"","head_proof":"a Heading that says \"From Two Dots to a Fraction\"","head_recap":"a Heading that says \"Three Choices of Box\"","head_thm":"a Heading that says \"What Dirichlet Promised\"","marks":"a Math [text] that says \"$0$\" drawn in unit","marks_2":"a Math [text] that says \"$0.2$\" drawn in unit","marks_3":"a Math [text] that says \"$0.4$\" drawn in unit","marks_4":"a Math [text] that says \"$0.6$\" drawn in unit","marks_5":"a Math [text] that says \"$0.8$\" drawn in unit","marks_6":"a Math [text] that says \"$1$\" drawn in unit","pens":"a Polygon [gray] drawn in unit (vertices=((0.0, -0.09), (0.2, -0.09), (0.2, 0.09), (0.0, 0.09)), filled=False)","pens_2":"a Polygon [gray] drawn in unit (vertices=((0.2, -0.09), (0.4, -0.09), (0.4, 0.09), (0.2, 0.09)), filled=False)","pens_3":"a Polygon [gray] drawn in unit (vertices=((0.4, -0.09), (0.6, -0.09), (0.6, 0.09), (0.4, 0.09)), filled=False)","pens_4":"a Polygon [gray] drawn in unit (vertices=((0.6, -0.09), (0.8, -0.09), (0.8, 0.09), (0.6, 0.09)), filled=False)","pens_5":"a Polygon [gray] drawn in unit (vertices=((0.8, -0.09), (1.0, -0.09), (1.0, 0.09), (0.8, 0.09)), filled=False)","promise":"a Math [text] that says \"$abs(alpha - p / q) < 1 / q^2$\"","recap":"a Block [text] that says \"Party: items are people, boxes are the possible answers. Sequence: items are positions, boxes are pairs $(u_i, d_i)$. Approximation: items are multiples of $alpha$, boxes are intervals.\"","seeds":"a Point [blue] drawn in unit","seeds_2":"a Point [blue] drawn in unit (location=(0.41421356237309515, 0.0))","seeds_3":"a Point [blue] drawn in unit (location=(0.8284271247461903, 0.0))","seeds_4":"a Point [blue] drawn in unit (location=(0.24264068711928566, 0.0))","seeds_5":"a Point [blue] drawn in unit (location=(0.6568542494923806, 0.0))","seeds_6":"a Point [blue] drawn in unit (location=(0.0710678118654755, 0.0))","setup":"a Tex [text] that says \"Six numbers, five intervals.\"","thm":"a Panel that says \"For every real $alpha$ and every whole number $N gt.eq 1$ there are integers $p$ and $q$ with $1 lt.eq q lt.eq N$ and $abs(alpha - p / q) < 1 / (q N)$.\"","unit":"a Figure (x_range=(-0.08, 1.08), y_range=(-0.34, 0.3), aspect=(1.16, 0.64))"},"beats":[{"start":395.78841666666665,"say":"One more, and this time the boxes are not objects at all. They are stretches of a line, and that turns out to change nothing about the argument and everything about what it can prove.","live":[],"does":[[395.78841666666665,"head_thm is shown on the screen, written out."]]},{"start":406.9649166666666,"say":"The subject is approximating an irrational number by a fraction. Getting close to the square root of two is easy: take enough decimal places. The real question is how close you can get while keeping the denominator small.","live":["head_thm"],"does":[]},{"start":421.5899166666666,"say":"Dirichlet's answer is startling. For every whole number N there is a fraction p over q, with q no bigger than N, that sits within one over q times N of your number.","live":null,"does":[[422.50741666666664,"thm is shown on the screen, written out."]]},{"start":433.72991666666667,"say":"And since q is at most N, that is within one over q squared. A denominator of five buying an error under one twenty fifth is not what randomly chosen fractions do for you.","live":["thm","head_thm"],"does":[[437.9794166666666,"promise is shown on the screen, written out."],[442.19441666666665,"promise (the \"1 / q^2\" part) is emphasized."],[443.71541666666667,"promise (the \"1 / q^2\" part) is no longer emphasized."],[446.16491666666667,"head_thm is hidden from the screen — left the board."],[446.16491666666667,"promise is hidden from the screen — left the board."],[446.16491666666667,"thm is hidden from the screen — left the board."],[446.16491666666667,"unit is shown on the screen, written out."]]},{"start":447.36491666666666,"say":"So, the boxes. Let N be five, and take the first six multiples of root two: nought, one point four one, two point eight three, four point two four, five point six six, and seven point zero seven.","live":["unit"],"does":[[447.36491666666666,"head_int is shown on the screen, written out."]]},{"start":463.81291666666664,"say":"Now throw away the whole number part of each one and keep only what comes after the point. Six numbers, every one of them somewhere between nought and one, and here they are.","live":["unit","head_int"],"does":[[472.64841666666666,"seeds is shown on the screen, written out."],[472.80841666666663,"seeds_2 is shown on the screen, written out."],[472.96841666666666,"seeds_3 is shown on the screen, written out."],[473.1284166666667,"seeds_4 is shown on the screen, written out."],[473.28841666666665,"seeds_5 is shown on the screen, written out."],[473.4484166666666,"seeds_6 is shown on the screen, written out."]]},{"start":474.23541666666665,"say":"And cut that stretch from nought to one into five equal pieces. Those are the boxes. Six numbers, five intervals, and I do not have to look at the numbers to know what happens next.","live":["unit","head_int","seeds","seeds_2","seeds_3","seeds_4","seeds_5","seeds_6"],"does":[[476.7894166666666,"pens is shown on the screen, written out."],[476.9294166666666,"pens_2 is shown on the screen, written out."],[477.06941666666665,"pens_3 is shown on the screen, written out."],[477.20941666666664,"pens_4 is shown on the screen, written out."],[477.3494166666666,"pens_5 is shown on the screen, written out."],[477.3494166666666,"marks is shown on the screen, written out."],[477.44941666666665,"marks_2 is shown on the screen, written out."],[477.64941666666664,"marks_3 is shown on the screen, written out."],[477.94941666666665,"marks_4 is shown on the screen, written out."],[478.3494166666666,"marks_5 is shown on the screen, written out."],[478.8494166666666,"marks_6 is shown on the screen, written out."],[481.7004166666666,"setup is shown on the screen, written out."]]},{"start":486.64191666666665,"say":"Two of the six share an interval. Here they are: the very first one, which is nought exactly, and the last one, seven point zero seven with the seven thrown away. Both of them inside the leftmost box.","live":["unit","setup","head_int","seeds","seeds_2","seeds_3","seeds_4","seeds_5","seeds_6","pens","pens_2","pens_3","pens_4","pens_5","marks","marks_2","marks_3","marks_4","marks_5","marks_6"],"does":[[487.8384166666666,"pens is indicated — a transient flash."],[490.7754166666666,"seeds: give the object a name where it had none (show_label)."],[493.32941666666665,"seeds_6: give the object a name where it had none (show_label)."]]},{"start":500.91891666666663,"say":"And two numbers sitting in one interval of width a fifth are less than a fifth apart. That is everything the boxes were ever for. From here it is arithmetic.","live":null,"does":[[512.0179166666667,"unit moves to a new place on the board."],[512.0179166666667,"head_int is hidden from the screen — left the board."],[512.0179166666667,"setup is hidden from the screen — left the board."]]},{"start":513.2179166666666,"say":"Write it out in general. The items are these N plus one fractional parts, one for each multiple of alpha from nought up to N.","live":["unit","seeds","seeds_2","seeds_3","seeds_4","seeds_5","seeds_6","pens","pens_2","pens_3","pens_4","pens_5","marks","marks_2","marks_3","marks_4","marks_5","marks_6"],"does":[[513.2179166666666,"head_proof is shown on the screen, written out."],[515.3894166666666,"g0 is shown on the screen, written out."]]},{"start":522.5834166666666,"say":"The boxes are the N intervals. More numbers than intervals, so two of them land in the same one: call their indices i and j, with i the smaller. The two fractional parts differ by less than one over N.","live":["unit","seeds","seeds_2","seeds_3","seeds_4","seeds_5","seeds_6","pens","pens_2","pens_3","pens_4","pens_5","marks","marks_2","marks_3","marks_4","marks_5","marks_6","g0","head_proof"],"does":[[523.1174166666666,"g1 is shown on the screen, written out."],[534.5764166666667,"g2 is shown on the screen, written out."]]},{"start":537.3939166666667,"say":"Now unpack what a fractional part is. It is the number itself, minus some whole number. So that small difference is j alpha minus i alpha, with an integer taken off it.","live":["unit","seeds","seeds_2","seeds_3","seeds_4","seeds_5","seeds_6","pens","pens_2","pens_3","pens_4","pens_5","marks","marks_2","marks_3","marks_4","marks_5","marks_6","g0","g1","g2","head_proof"],"does":[]},{"start":550.4634166666666,"say":"Give those two things names. Let q be j minus i, which is between one and N, and let p be the integer that came off. Then q alpha minus p is less than one over N in size.","live":null,"does":[[551.9724166666666,"g3 is shown on the screen, written out."],[560.0764166666667,"g4 is shown on the screen, written out."]]},{"start":564.8904166666666,"say":"Divide the whole line through by q. Alpha minus p over q is less than one over q N. And q is at most N, so one over q N is at most one over q squared, which is the theorem.","live":["unit","seeds","seeds_2","seeds_3","seeds_4","seeds_5","seeds_6","pens","pens_2","pens_3","pens_4","pens_5","marks","marks_2","marks_3","marks_4","marks_5","marks_6","g0","g1","g2","g3","g4","head_proof"],"does":[[565.3664166666666,"g5 is shown on the screen, written out."],[571.1134166666666,"g5 (the \"1 / (q N)\" part) is emphasized."],[577.5684166666666,"g5 (the \"1 / (q N)\" part) is no longer emphasized."],[577.5684166666666,"g5 (the \"1 / q^2\" part) is emphasized."],[578.7414166666666,"g5 (the \"1 / q^2\" part) is no longer emphasized."]]},{"start":580.1194166666667,"say":"Put our own numbers in. The two indices were nought and five, so q is five, and p works out at seven. Seven fifths, which is one point four.","live":["unit","seeds","seeds_2","seeds_3","seeds_4","seeds_5","seeds_6","pens","pens_2","pens_3","pens_4","pens_5","marks","marks_2","marks_3","marks_4","marks_5","marks_6","g0","g1","g2","g3","g4","g5","head_proof"],"does":[[582.3474166666666,"answer is shown on the screen, written out."]]},{"start":590.8194166666666,"say":"The true error is a hundredth and a bit, comfortably inside the one twenty fifth we were promised. And run the whole thing again with a larger N and you get a different fraction, a better one.","live":["unit","seeds","seeds_2","seeds_3","seeds_4","seeds_5","seeds_6","pens","pens_2","pens_3","pens_4","pens_5","marks","marks_2","marks_3","marks_4","marks_5","marks_6","g0","g1","g2","g3","g4","g5","answer","head_proof"],"does":[[595.3584166666666,"A box is drawn around answer."]]},{"start":602.5764166666665,"say":"Which means an irrational number has infinitely many fractions chasing it this closely, forever. And the boxes that proved it were five stretches of a line.","live":null,"does":[[612.1779166666665,"answer is hidden from the screen — left the board."],[612.1779166666665,"g0 is hidden from the screen — left the board."],[612.1779166666665,"g1 is hidden from the screen — left the board."],[612.1779166666665,"g2 is hidden from the screen — left the board."],[612.1779166666665,"g3 is hidden from the screen — left the board."],[612.1779166666665,"g4 is hidden from the screen — left the board."],[612.1779166666665,"g5 is hidden from the screen — left the board."],[612.1779166666665,"head_proof is hidden from the screen — left the board."],[612.1779166666665,"unit is hidden from the screen — left the board."],[612.1779166666665,"seeds is hidden from the screen — unit left the board."],[612.1779166666665,"seeds_2 is hidden from the screen — unit left the board."],[612.1779166666665,"seeds_3 is hidden from the screen — unit left the board."],[612.1779166666665,"seeds_4 is hidden from the screen — unit left the board."],[612.1779166666665,"seeds_5 is hidden from the screen — unit left the board."],[612.1779166666665,"seeds_6 is hidden from the screen — unit left the board."],[612.1779166666665,"pens is hidden from the screen — unit left the board."],[612.1779166666665,"pens_2 is hidden from the screen — unit left the board."],[612.1779166666665,"pens_3 is hidden from the screen — unit left the board."],[612.1779166666665,"pens_4 is hidden from the screen — unit left the board."],[612.1779166666665,"pens_5 is hidden from the screen — unit left the board."],[612.1779166666665,"marks is hidden from the screen — unit left the board."],[612.1779166666665,"marks_2 is hidden from the screen — unit left the board."],[612.1779166666665,"marks_3 is hidden from the screen — unit left the board."],[612.1779166666665,"marks_4 is hidden from the screen — unit left the board."],[612.1779166666665,"marks_5 is hidden from the screen — unit left the board."],[612.1779166666665,"marks_6 is hidden from the screen — unit left the board."]]},{"start":613.3779166666666,"say":"Three theorems, then, and three choices of box. In the first, the items were people and the boxes were the possible answers to a question, once we had noticed that two of those answers could never both be used.","live":[],"does":[[613.3779166666666,"head_recap is shown on the screen, written out."],[615.3164166666667,"recap is shown on the screen, written out."],[618.8224166666666,"recap (the \"boxes are the possible answers\" part) is emphasized."]]},{"start":626.9459166666666,"say":"In the second, the items were positions in a sequence, and the boxes were pairs of numbers we had to invent from nothing. In the third, the items were multiples of alpha, and the boxes were stretches of a line.","live":["recap","head_recap"],"does":[[628.9664166666666,"recap (the \"boxes are pairs\" part) is emphasized."],[628.9664166666666,"recap (the \"boxes are the possible answers\" part) is no longer emphasized."],[636.1184166666667,"recap (the \"boxes are intervals\" part) is emphasized."],[636.1184166666667,"recap (the \"boxes are pairs\" part) is no longer emphasized."],[639.8334166666666,"recap (the \"boxes are intervals\" part) is no longer emphasized."]]},{"start":640.4334166666666,"say":"And the sentence at the end was the same all three times. More items than boxes, so two share a box. It was never the hard part. The boxes were.","live":null,"does":[[641.0604166666667,"closing is shown on the screen, written out."],[650.5314791666666,"closing is hidden from the screen — left the board."],[650.5314791666666,"head_recap is hidden from the screen — left the board."],[650.5314791666666,"recap is hidden from the screen — left the board."]]}]}]},"durationSeconds":652,"chapters":[{"title":"One Line, Then the Hard Part","startSeconds":0,"narration":"Here is a piece of mathematics you already know. If you have more things than places to put them, some place gets two things. That is it. It sounds like it could not possibly prove anything, and over the next few minutes I would like to change your mind about that. So here it is, properly. Four boxes, five items. Put them away however you like, however cleverly, and one of these boxes ends up holding two. Say the same thing with a letter in it. Put n plus one items into n boxes, and some box holds at least two of them. And the proof is a single sentence. If every box held one item or none, then adding up over the boxes you would have at most n items in total. But you have more than n. So some box was holding two. That sentence is the easy half, and I will never have to say anything harder than it. The hard half never appears in it at all, because the hard half is deciding what the items are and what the boxes are. So let me put three statements on the board. Not one of them mentions a box, or an item, or anything that looks like counting. Two people at any party know the same number of people there. Any ten numbers, all different, contain four that go steadily up or four that go steadily down. And every irrational number has fractions chasing it far more closely than it has any right to expect. Every one of those is the sentence you just heard, applied once. So watch the boxes each time, and not the sentence. The boxes are the argument."},{"title":"Six People at a Party","startSeconds":94.0981875,"narration":"Six people at a party. Some of them know each other and some do not, and knowing is mutual: if I know you, you know me. That is the only rule there is, and here is one particular evening. Now go round and count. A knows exactly one person here. B knows two. And C also knows two. D, E and F each know three of the others. So the six counts are one, two, two, three, three, three, and they have already collided twice. You might reasonably think that was luck, and that I drew the lines to make it happen. I did not. Whatever this picture had looked like, two of the counts would have had to agree, and I want to show you why. Take the people themselves as the items. For the boxes, take the possible answers to the question: how many people here do you know? Nobody can know more than the other n minus one, and nobody can know fewer than none at all, so every count lands somewhere in this list. Now count the boxes. Zero, one, two, and so on up to n minus one. That is n boxes for n people, and pigeonhole gives us absolutely nothing. Six items in six boxes can sit one to a box quite happily. So the obvious choice of boxes fails, and I want you to sit with that for a moment, because this is exactly where these proofs are won or lost. Look at the two boxes on the ends. Can they both be used at once? Suppose somebody at this party knows nobody. Then no one in the room can know everybody, because knowing everybody would include knowing that person. Box zero and box n minus one can never both be occupied. So one of the two ends is always empty, whichever way it falls. The boxes were never n. There were only ever n minus one of them. Six people, five boxes. And now the sentence from the beginning does all the remaining work, without being asked twice. Two of them land together. Two people at any party, anywhere, know the same number of people at that party. Notice how little of that was cleverness. One honest look at which boxes could actually be used, and the theorem fell out."},{"title":"Ten Numbers, and a Run of Four","startSeconds":230.0681875,"narration":"Ten numbers, all different, in whatever order somebody wrote them down. Position along the bottom, value up the side. I want to hunt for runs. A run means this: read from left to right, skipping whatever you like, and take terms that keep going up, or terms that keep going down. They do not have to be neighbours. Start at the second term, which is seven. Going up from there I can reach nine, and then ten. A rising run of three. Going down from that same seven I can reach five, and then four. A falling run of three as well. So attach two numbers to every position. Call them u and d: the length of the longest rise that starts there, and the length of the longest fall that starts there. At the second term, both of them are three. Now, does this sequence contain a run of four? It does. Two, five, eight, ten, at positions three, five, seven and nine, rising the whole way. And that was not luck either. Any ten distinct numbers contain four that rise, or four that fall. To see it, suppose they do not. Suppose there is no rising run of four anywhere, and no falling run of four either. Then every u is one, two or three, and so is every d. Two numbers, three choices each. Three times three is nine possible pairs, and there are the nine of them, one box apiece. But there are ten positions, and each position hands you one pair. Ten items, nine boxes. So two positions, i somewhere to the left of j, carry the identical pair. Same u, same d. And now it breaks. The numbers are all different, so either a i is below a j or above it. Suppose it is below. Take the longest rise starting at j, and stick a i on the front of it. That is a rise starting at i, and it is one term longer. So u i beats u j, when they were supposed to be equal. If a i is the larger one instead, the very same trick runs downhill and d i beats d j. Either way the two pairs could not have matched, so the assumption is dead. There has to be a monotone run of four somewhere in that sequence. And ten was not an arbitrary number. Nine positions fit into nine boxes without any fight at all, so ten is the first length that forces the issue. Three by three, plus one."},{"title":"Boxes That Are Intervals","startSeconds":395.78841666666665,"narration":"One more, and this time the boxes are not objects at all. They are stretches of a line, and that turns out to change nothing about the argument and everything about what it can prove. The subject is approximating an irrational number by a fraction. Getting close to the square root of two is easy: take enough decimal places. The real question is how close you can get while keeping the denominator small. Dirichlet's answer is startling. For every whole number N there is a fraction p over q, with q no bigger than N, that sits within one over q times N of your number. And since q is at most N, that is within one over q squared. A denominator of five buying an error under one twenty fifth is not what randomly chosen fractions do for you. So, the boxes. Let N be five, and take the first six multiples of root two: nought, one point four one, two point eight three, four point two four, five point six six, and seven point zero seven. Now throw away the whole number part of each one and keep only what comes after the point. Six numbers, every one of them somewhere between nought and one, and here they are. And cut that stretch from nought to one into five equal pieces. Those are the boxes. Six numbers, five intervals, and I do not have to look at the numbers to know what happens next. Two of the six share an interval. Here they are: the very first one, which is nought exactly, and the last one, seven point zero seven with the seven thrown away. Both of them inside the leftmost box. And two numbers sitting in one interval of width a fifth are less than a fifth apart. That is everything the boxes were ever for. From here it is arithmetic. Write it out in general. The items are these N plus one fractional parts, one for each multiple of alpha from nought up to N. The boxes are the N intervals. More numbers than intervals, so two of them land in the same one: call their indices i and j, with i the smaller. The two fractional parts differ by less than one over N. Now unpack what a fractional part is. It is the number itself, minus some whole number. So that small difference is j alpha minus i alpha, with an integer taken off it. Give those two things names. Let q be j minus i, which is between one and N, and let p be the integer that came off. Then q alpha minus p is less than one over N in size. Divide the whole line through by q. Alpha minus p over q is less than one over q N. And q is at most N, so one over q N is at most one over q squared, which is the theorem. Put our own numbers in. The two indices were nought and five, so q is five, and p works out at seven. Seven fifths, which is one point four. The true error is a hundredth and a bit, comfortably inside the one twenty fifth we were promised. And run the whole thing again with a larger N and you get a different fraction, a better one. Which means an irrational number has infinitely many fractions chasing it this closely, forever. And the boxes that proved it were five stretches of a line. Three theorems, then, and three choices of box. In the first, the items were people and the boxes were the possible answers to a question, once we had noticed that two of those answers could never both be used. In the second, the items were positions in a sequence, and the boxes were pairs of numbers we had to invent from nothing. In the third, the items were multiples of alpha, and the boxes were stretches of a line. And the sentence at the end was the same all three times. More items than boxes, so two share a box. It was never the hard part. The boxes were."}]}}