CSE 202 Homework 1 Matthias Springer, A Yes, there does always exist a perfect matching without a strong instability.

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1 CSE 0 Homework 1 Matthias Spriger, A Problem 1 Notatio a b meas that a is matched to b. a < b c meas that b likes c more tha a. Equality idicates a tie. Strog istability Yes, there does always exist a perfect matchig without a strog istability. Ituitio We ca resolve ties by orderig the persos ivolved arbitrarily. I.e. whe m 1 ad m are tied for w 1, we ca just assume that w 1 likes m 1 better tha m. Sice m 1 ad m were tied before, w 1 will ever complai about our choice. The Gale-Shapley algorithm could the be ru without modificatios ad is guarateed to retur a perfect stable matchig. Proof We show that the Gale-Shapley algorithm always geerates a perfect matchig without a strog istability, eve if there may be ties. We will prove this by cotradictio. We make a small modificatio to the algorithm: if there are multiple choices for a woma to propose to a ma (because there are multiple me that she likes best), she radomly proposes to oe. Also, ote that me do ot switch for a wome that are tied with the curret parter. We assume, that the output of the algorithm cotais a strog istability. I.e. m 1 w 1 ad w w, but m 1 < w1 m ad w < m w 1. Therefore, w 1 must have proposed to m before she proposed to m 1 1. m 1 w 1, so oe of the followig must be true. m already had a better parter w i. Evetually, m switched for w 1. Therefore, w m w i > m... > m w 1. This violates our assumptio that w < m w 1. m accepts w 1, but evetually switches to w. Therefore, there must be wome, icludig w, that m likes better. Therefore, w m w i > m... > m w 1. This violates our assumptio that w < m w 1. 1 m 1 ad m are ot tied (see assumptio).

2 CSE 0 Homework 1 Matthias Spriger, A Note that w 1 ad w caot be tied, sice this would ot impose a strog istability. Our assumptio is wrog. There ca be o strog istability. Weak istability No, there is ot always a perfect matchig without a weak istability. We show this by providig a example i which all perfect matchigs have a weak istability. w 1 = m1 w w 1 = m w m 1 > w1 m m 1 > w m We examie the two possible matchigs. m 1 w 1 ad m w cotais a weak istability: w likes m 1 more tha her curret parter m ad m 1 does ot care (both wome are tied). m 1 w ad m w 1 cotais a weak istability: w 1 likes m 1 more tha her curret parter m ad m 1 does ot care (both wome are tied). Therefore, there does ot always exist a perfect matchig without a weak istability. Problem For a woma, it is ot possible to ed up with a true better ma by lyig about her prefereces. Let w be a woma with the true preferece m 1 > w m. Let us assume that the Gale-Shapley algorithm creates m truth w ad that w preteds that m > lie w m 1. We compare the results of the algorithm with ad without the lie.

3 CSE 0 Homework 1 Matthias Spriger, A Case 1: m 1 < m truth ad m < m truth Origial proposig sequece: (..., m truth,..., m 1,..., m,...) New proposig sequece: (..., m truth,..., m,..., m 1,...) Both m 1 ad m are actually ad after lyig lower o w s preferece list tha her matched parter m truth. Therefore, w is matched to m truth before ad after the lie, because, i both cases, w does ot eve propose to m 1 or m, sice she was matched with a better parter before ad he did ot leave w. Case : m 1 > m truth ad m > m truth Origial proposig sequece: (..., m 1,..., m,..., m truth,...) New proposig sequece: (..., m,..., m 1,..., m truth,...) Both m 1 ad m are actually ad after lyig higher o w s preferece list tha her matched parter m truth. Therefore, w is matched to m truth before ad after the lie, because, i both cases, w proposes to m 1 ad m ad both will either reject her or first accept her ad leave her later. At some poit, w proposes to m. We kow that w m truth i the first ru of the algorithm, although w proposed to m before. Therefore, m either already had a better parter or he accepted w ad the switched to aother woma. This will also happe after lyig, sice w s lie does ot affect the proposig sequece of the other wome. The same argumet holds true for m 1. Case 3: m 1 > m truth ad m < m truth Origial proposig sequece: (..., m 1,..., m truth,..., m,...) New proposig sequece: (..., m,..., m truth,..., m 1,...) Whe w lies, the followig is true. w will ot be matched with a ma she proposed to before m, because these me evetually rejected her i the first ru of the algorithm. Whe w proposes to m, he might accept her ad stay with her. I that case, w eds up with a worse parter tha without lyig. I case m rejects w or leaves her later, w will propose to all me util m truth will evetually accept her ad stay with her. All the I geeral, the orderig of me o w s preferece list below m truth does ot matter.

4 CSE 0 Homework 1 Matthias Spriger, A me i betwee will reject her or leave her, because that is what they did i the first ru of the algorithm. Case 4: m 1 < m truth ad m > m truth Not possible, sice m 1 < m truth < m violates our assumptio that m 1 > m. Case 5: m = m truth ad m 1 > m truth Origial proposig sequece: (..., m 1,..., m truth,...) New proposig sequece: (..., m truth,..., m 1,...) w chages the orderig i such a way that she proposes to her matched parter earlier ad to a better parter later. m truth s predecessors will (evetually) reject w, but m truth will accept w ad stay with her, sice he did ot get a better proposal i the first ru. Case 6: m = m truth ad m 1 < m truth Not possible, sice m 1 < m violates our assumptio that m 1 > m. Case 7: m 1 = m truth ad m < m truth Origial proposig sequece: (..., m truth,..., m,...) New proposig sequece: (..., m,..., m truth,...) All me before m will reject w s proposal or leave her later, sice that is what they did i the first ru. w proposes to m before she proposes to m truth. m might accept her proposal ad stay with her. I that case, w eded up with a worse parter. I case m rejects her (evetually), w will propose to all me betwee m ad m truth. These me will reject her, because they did the same i the first ru. m truth will accept w ad stay with her. Case 8: m 1 = m truth ad m = m truth w did ot lie. I every case, w did either ed up with the same parter agai or get a worse parter. Therefore, w caot improve her matchig by lyig.

5 CSE 0 Homework 1 Matthias Spriger, A Problem 3 Summary The key idea is to calculate all values F j together istead of separately. Whe we take a look at the formula for F j, we see that C ad C j are costats that are multiplied with the two sums. The difficult part is to calculate q i i<j q i (j i) i>j efficietly. We ca geerate two polyoms Q(x) (for (j i) the q i ) ad M(x) (for the deomiators) of less tha degree such that their multiplicatio cotais the values F j. This ca be doe efficietly with the algorithm discussed i the lecture. The we multiply every coefficiet of the resultig polyom with the costats C ad q j. Basic Idea Calculate all F j together istead of separately. Geerate polyoms Q(x) = i=1 q ix i M(x) = 1 i=1 x 1+i + x 1 i i i Calculate F (x) = Q(x) M(x) usig the algorithm discussed i the lecture. Degree of both polygos is less tha. Rutime O( log ). Extract coefficiets F j = f j from F (x) at positios j = 1 i for i = Full Algorithm We ca rewrite F j as follows. ( q i F j = C q j (j i) i>j i<j q i (j i) Now, we geeate the polyoms Q(x) ad M(x) as preseted i the previous sectio. For istace, for = 4, we geerate the followig polyoms. ) Q(x) = q 1 x 3 + q x + q 3 x + q 4

6 CSE 0 Homework 1 Matthias Spriger, A M(x) = 1 9 x6 1 4 x5 x 4 + x x We use the method discussed i the lecture to multiply these two polyoms. For polyoms of degree m this takes O(m log m) time. deg(q(x)) = 1 ad deg(m(x)) = ( 1) <. Therefore, the multiplicatio takes O( log ) time. Note: the algorithm for multiplyig two polyomials was desiged for polyoms whose degree is a power of. Therefore, we might have to exted the polyoms, resultig i a polyom of degree. Therefore, the multiplicatio takes O(4 log 4) = O( log ) time. For the multiplicatio, we assume that Q(x) has the same degree as M(x) by addig the missig powers of x with a coefficiet of zero (e.g. 0x 6 i the example). We also add missig powers of x iside M(x) (e.g. 0x 3 i the example). M(x) Q(x) =... + x 6 ( q q 3 4 q 4 9 ) + x5 (q 1 q 3 q 4 4 )+ x 4 ( q q q 4 ) + x 3 ( q q 4 + q 3) +... The example above shows oly the coefficiets of x k with k = I geeral, we are iterested oly i the coefficiets of x k with k = 1 i ad i = The other coefficiets are calcuated by the algorithm, but we do ot eed them. We ca ow geerate the term F j = C q j c 1 j, where c i is the coefficiet of x i i M(x) Q(x). This requires O() multiplicatios i total. Pseudo Code Q.size() M Array[ 1] result Array[] for i = 1 to do M[ 1 + i] i M[ 1 i] i ed for d log deg M Q Q with d coefficiets (fill with zeros)

7 CSE 0 Homework 1 Matthias Spriger, A M M with d coefficiets (fill with zeros) p mult(m, Q) for i = 1 to do result[i 1] = p[ 1 i] ed for retur result Rutime complexity Geerate polyom Q(x): O() Geerate polyom M(x): O() Covert polyoms Q(x) ad M(x) to polyom with same umber of coefficiets, where the umber of coefficiets is a power of : O(4) each (ext power of might be almost ) Multiply polyoms (usig FFT): O(4 log 4) Extract coefficiets from resultig polyom: O() Therefore, the rutime complexity is O(4 log 4) = O( log ). Problem 4 Basic Idea Give a arbitrary startig vertex, a local miimum is the ed of a decreasig path, i.e. every ext vertex is smaller tha the curret vertex. Fid the smallest vertex v i the middle colum. Check if this smallest vertex is a local miimum. If ot, there must be a local miimum o that side of the colum where v s (left/right) eighbor is smaller tha v (could be both sides). A smallest path startig from v s eighbor will ever cross this colum. Fid the smallest vertex w i the middle row of the side that we chose i the last step.

8 CSE 0 Homework 1 Matthias Spriger, A Check if w is a local miimum. If ot, there must be a local miimum iside the quarter sector cotaiig v s chose eighbor if w > v. Otherwise, there must be a local miimum i the quarter sector where w s (upper/lower) eighbor is smaller tha w. We cotiue recursively o a smaller grid. Full Algorithm 1. Fid cell (, i), 1 i that has a miimal value x,i.. Check (, i) s eighbors. If they are both greater tha x,i, we retur the local miimum. 3. Otherwise, there must be a local miimum o the side that cotais the eighbor vertex P that has a smaller value tha x,i. If both eighbors have a smaller value, the both sides cotai a local miimum ad we may choose P arbitrarily. Proof: Startig with a vertex of value x i, we ca fid a local miimum by choosig a eighbor that has a smaller value x j < x i. We must evetually reach a local miimum, because all values x k are distict. Therefore, there must always be a smaller eighbor or, i case we reached the local miimum, all eighbors are greater. The path we cosider right ow starts at P with a value that is smaller tha x,i. This x P is smaller tha the smallest umber of the colum. We ca be sure that this (decreassig-value) path will ever cross this colum agai. Otherwise, the path would hit a greater value, sice all vertices o the colum have a greater umber tha x P. 4. Fid cell ( j, ), with 1 j 1 if we just chose the left side, or otherwise + 1 j (if we chose the right side), such that x j, is miimal. 5. Check ( j, ) s eighbors. If they are both greater tha x j,, we retur the local miimum. 6. We compare x,i ad x j,. (a) If x,i < x j,, the the path startig at P will ot cross the row, because all vertices o the row have a greater value tha the

9 CSE 0 Homework 1 Matthias Spriger, A x P. Therefore, we kow that there must be a local miimum i the (square) sector that cotais P. We cotiue the algorithm i that sector recursively. The ew sector s size is equal to or smaller tha. (b) If x,i > x j,, the we take oe of ( j, ) s eighbors whose value is smaller tha x j, ad call it Q 3. We kow that the path startig from Q will cross either the half-row or the colum (same reaso as i the proof above). Therefore, we kow that there must be a local miimum i the (square) sector that cotais Q. We cotiue the algorithm i that sector recursively. The ew sector s size is equal to or smaller tha. = 1 is the base case for the recursio. I that case, the grid cosists of oly oe cell that must be a local miimum. Pseudo Code This is the pseudo code for fidmi(g, ), where G is the grid graph ad is its size i oe dimesio. if = 1 the base case output(1, 1) keep track of acutal idices, see commet below ed if c x c v for i = 1 to do x G[c x, i] if x < c v the step 1 c v x c y i ed if ed for l G[c x 1, c y ] r G[c y + 1, c y ] if l > x r > x the step output(c x, c y ) keep track of actual idices 3 There must always exist such a vertex. Otherwise we would already have retured the local miimum.

10 CSE 0 Homework 1 Matthias Spriger, A else directio x 1 if l < x else 1 ed if r y r v for i = + directio x; i > 0 i < + 1; i i + directio x do x G[i, r y ] if x < r v the step 4 r v x r x i ed if ed for u G[r x, r y 1] d G[r x, r y + 1] c ed 1 if l < x else if u > x d > x the step5 output(r x, r y ) keep track of actual idices else if c v < r v the step 6a r ed 1 if c y < else else step 6b r ed 1 if d < x else ed if retur fidmi(g[ : c ed, : r ed], ) I the pseudo code, G[a : b, c : d] meas that we copy the array rectagle horizotally from idex a to b ad vertically from c to d. The resultig array begis at idex 1. I the actual implemetatio of the algorithm, we would ot do this because it is too expesive. We ca just provide the coordiates of the ew array segmet that we are workig o. The oly reaso for choosig this way i the pseudo code is to keep it readable ad uderstadable. I the rutime complexity aalysis, I will ot accout for copyig the array. It is also importat to remember, that the output of the idices of the pseduo code algorithm does ot keep track of the actual idices. I.e. every recursive call of the algorithm gets a array whose first elemet is at positio (1, 1), although the real positio of this first elemet might be somethig else (because it was copied i the example i order to keep the pesudo code free of idex umber cruchig). All out-of-bouds accesses will retur. I.e. G[0, ] = G[ + 1, 1] =.

11 CSE 0 Homework 1 Matthias Spriger, A Rutime complexity We assume that the grid G is a square grid with a height ad legth of. Fidig the miimum of G s middle colum: probes. Checkig if the colum s miimum is a local miimum: probes. Fidig the miimum of the middle colum: 1 = O() probes4. Checkig if the row s miimum is a local miimum: probes. We eed O() probes per recursio/coquer step. Every divide step divides by half 5. Therefore this is the rutime complexity for the algorithm: T () = T ( ) + O(). The base case takes a costat amout of time (T (1) = c) We ca show that T () = O(c ) by usig the recursio tree method. Every level cotais exactly oe problem ad the problem size is cut i half o every ext level. Therefore, the problem complexity ca be described by the decreasig geometric series: T () = log c i=1 c i. 6 Therefore, the rutime complexity of the algorithm is O(). Problem 5 Basic Idea We sort all lies by their slopes. We use divide-ad-coquer ad geerate two equal-sized subproblems by splittig the lies i the middle. Note, that the sortig is preserved. The algorithm is supposed to retur the evelope that cotais the visible lies oly. 4 Remember that we already split the grid vertically at this poit 5 If the is odd, we take. This does ot affect the rutime complexity asymptotically. 6 We ca assume that it is a ifiite decreasig geometric series.

12 CSE 0 Homework 1 Matthias Spriger, A I the coquer step, we take two evelopes ad combie them. Therefore, we fid the sigle itersectio poit of the evelopes ad from there o, we take the lie segmets that are o top. Before, we take the lie segmets from the other evelope. The itersectio poit ca be foud i liear time with a sca lie algorithm that begis at ad stops at the begiig of every ew lie segmet i either oe of the two evelopes. Full Algorithm The algorithm gets a array of lie equatios as iput, ad the lie equatios must already be sorted by their slope value. If that is ot the case, this ca be doe i a preprocessig step i O( log ) if is the umber of lie equatios. The algorithm returs a list of visible lie equatios ad their itersectio poits. Together, these form lie segmets, where each lie equatio is oly used betwee the two itersectio poits with the predecessor ad the successor lie equatio. Note, that the left-most ad the right-most lies have oly oe itersectio ad go to ad respectively. We divide the list of equatios i the middle ad work o both list recursively (divide ad coquer). The coquer step works as follows. We eed to merge the two sequeces of lie segmets (also called evelopes) by fidig their itersectio poit ad creatig oe sigle list of lie segmets. Durig this step, some of the lies might become ivisible ad are therefore discarded. This ca be doe i O(). Let A be the lie segmets from the first recursive divide call ad B be the lie segmets from the secod recursive divide call. At some poit, a segmet from A might itersect with a segmet from B. At that poit, we combie A ad B by savig the itersectio poit ad takig all previous lie segmets from A ad

13 CSE 0 Homework 1 Matthias Spriger, A takig all followig lie segmets from B, if A was o top before ad B is o top later 7. Note, that there ca oly be oe sigle itersectio of A ad B, because A ad B are sorted by their slopes ad all slopes i A are lower tha all slopes i B. If there were a secod itersectio poit, the two evelopes had to itersect a secod time. Therefore, A s or B s slopes would have to decrease agai or A s slopes would have to become greater tha a slope i B. This cotradicts the fact, that A ad B are sorted. We ca merge both evelopes i liear time by usig a sca lie algorithm. We start at x = ad determie the value of the first lie equatio i both evelopes 8. By doig this, we kow which of the two evelopes is o top. The we move to the ext closest itersectio poit (o the x-axis) i the two evelopes 9. Note, that the slope of a evelope ca oly chage at these itersectio poits. We evaluate the other evelope at this itersectio poit 10. If from ow o, the other lie segmets/the other evelope are o top (e.g. before, A was o top, but ow B is o top) 11, the A ad B itersect ad we combie them as described before. We retur the combied evelopes. Rutime Complexity Sortig by slopes: O( log ) Sca lie algorithm for combiig two evelopes (coquer step): O() Full recursio with divide ad coquer: T () = T ( ) + O() ad the base cases T (1) = T () = cost. We kow that T () = O( log ). Therefore, the overall rutime complexity is O( log ). 7 Otherwise, we take all previous lie segmets from B, add the itersectio poit, ad take all followig lie segmets from A. 8 Note, that the first equatio does ot have a itersectio poit to its left. 9 Both evelopes have differet itersectio poits ad we do ot care where the itersectio poit comes from. 10 We kow the segmets of the other evelope ad where they begi ad ed. 11 We have to evaluate both evelopes.

Infinite Sequences and Series

Infinite Sequences and Series Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet