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Question Number 12883 by FilupS last updated on 05/May/17
for −128≤x≤127 and −127≤y≤128 where x,y∈Z Point P(x,y) is a point on the cartesian plane. From the origin, angle θ is made counter −clockwise with the positive x−axis. (1) How many unique angles θ exist if x,y∈P? (2) Furthermore, how many unique angles θ exist for the full range of x,y∈Z?
$$\mathrm{for}\:\:\:\:\:−\mathrm{128}\leqslant{x}\leqslant\mathrm{127} \\ $$$$\mathrm{and}\:\:\:−\mathrm{127}\leqslant{y}\leqslant\mathrm{128} \\ $$$$\mathrm{where}\:\:\:{x},{y}\in\mathbb{Z} \\ $$$$\: \\ $$$$\mathrm{Point}\:{P}\left({x},{y}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{cartesian}\:\mathrm{plane}. \\ $$$$\: \\ $$$$\mathrm{From}\:\mathrm{the}\:\mathrm{origin},\:\mathrm{angle}\:\theta\:\mathrm{is}\:\mathrm{made}\:\mathrm{counter} \\ $$$$−\mathrm{clockwise}\:\mathrm{with}\:\mathrm{the}\:\mathrm{positive}\:{x}−\mathrm{axis}. \\ $$$$\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{How}\:\mathrm{many}\:\mathrm{unique}\:\mathrm{angles}\:\theta\:\mathrm{exist} \\ $$$$\mathrm{if}\:{x},{y}\in\mathbb{P}? \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Furthermore},\:\mathrm{how}\:\mathrm{many}\:\mathrm{unique} \\ $$$$\mathrm{angles}\:\theta\:\mathrm{exist}\:\mathrm{for}\:\mathrm{the}\:\mathrm{full}\:\mathrm{range}\:\mathrm{of}\:{x},{y}\in\mathbb{Z}? \\ $$
Answered by mrW1 last updated on 10/May/17
This is an interesting question. We don′t neet to treat the case when the points lie on the x− or y−axis, the angles are 0° (x>0, y=0) 90° (x=0, y>0) 180° (x<0, y=0) 270° (x=0, y<0) Let′s begin with the case when the points lie in the first quadrant. x=i ∈ P, y=j ∈ P 1≤i≤m 1≤j≤n The angle θ for P(x,y) is defined by tan θ = (y/x)=(j/i) The number of unique angles is the number of unique values of (j/i) with 1≤i≤n and 1≤j≤m. There are m×n points, we have also m×n values of (j/i), but some of them are not unique, e.g. points (2,3), (4,6) have the same angle θ, since (3/2)=(6/4). To find the number of unique values of (j/i), we should simplify (j/i) to ((j′)/(i′)) with i′ = ((LCM(i,j))/j) and j′ = ((LCM(i,j))/i), since (j/i)=((LCM(i,j)×j)/(LCM(i,j)×i))=(((LCM(i,j))/i)/((LCM(i,j))/j))=((j′)/(i′)) e.g. for point (2,3): LCM(2,3)=6, i′=(6/3)=2, j′=(6/2)=3, (j/i)=((j′)/(i′))=(3/2) and for point (4,6): LCM(4,6)=12, i′=((12)/6)=2, j′=((12)/4)=3, (j/i)=((j′)/(i′))=(3/2) An other way to reduce the fraction (j/i) to ((j′)/(i′)) is: i′=(i/(GCD(i,j))) and j′=(j/(GCD(i,j))) e.g. for point (2,3): GCD(2,3)=1, i′=(2/1)=2, j′=(3/1)=3, (j/i)=((j′)/(i′))=(3/2) and for point (4,6): GCD(4,6)=2, i′=(4/2)=2, j′=(6/2)=3, (j/i)=((j′)/(i′))=(3/2) For all points (i,j) with 1≤i≤m and 1≤j≤n, we can construct m×n values of (j/i)=((j′)/(i′)). Now we only need to count how many of these values are unique. It is difficult to give a general formula for calculating the number of unique values ((j′)/(i′)), but for concrete numbers m and n, we can easily get the result with help of Excel. For m=10, n=20, for example, the result is 127 unique angles, see fig. in comment. For m=127, and n=128, the result is 9979 unique angles. But the table in this case is too big to be shown here.
$${This}\:{is}\:{an}\:{interesting}\:{question}. \\ $$$$ \\ $$$${We}\:{don}'{t}\:{neet}\:{to}\:{treat}\:{the}\:{case}\:{when} \\ $$$${the}\:{points}\:{lie}\:{on}\:{the}\:{x}−\:{or}\:{y}−{axis},\: \\ $$$${the}\:{angles}\:{are} \\ $$$$\mathrm{0}°\:\left({x}>\mathrm{0},\:{y}=\mathrm{0}\right) \\ $$$$\mathrm{90}°\:\left({x}=\mathrm{0},\:{y}>\mathrm{0}\right) \\ $$$$\mathrm{180}°\:\left({x}<\mathrm{0},\:{y}=\mathrm{0}\right) \\ $$$$\mathrm{270}°\:\left({x}=\mathrm{0},\:{y}<\mathrm{0}\right) \\ $$$$ \\ $$$${Let}'{s}\:{begin}\:{with}\:{the}\:{case}\:{when}\:{the} \\ $$$${points}\:{lie}\:{in}\:{the}\:{first}\:{quadrant}. \\ $$$${x}={i}\:\in\:\mathbb{P},\:{y}={j}\:\in\:\mathbb{P} \\ $$$$\mathrm{1}\leqslant{i}\leqslant{m} \\ $$$$\mathrm{1}\leqslant{j}\leqslant{n} \\ $$$$ \\ $$$${The}\:{angle}\:\theta\:{for}\:{P}\left({x},{y}\right)\:{is}\:{defined}\:{by} \\ $$$$\mathrm{tan}\:\theta\:=\:\frac{{y}}{{x}}=\frac{{j}}{{i}} \\ $$$${The}\:{number}\:{of}\:{unique}\:{angles}\:{is} \\ $$$${the}\:{number}\:{of}\:{unique}\:{values}\:{of} \\ $$$$\frac{{j}}{{i}}\:{with}\:\mathrm{1}\leqslant{i}\leqslant{n}\:{and}\:\mathrm{1}\leqslant{j}\leqslant{m}. \\ $$$$ \\ $$$${There}\:{are}\:{m}×{n}\:{points},\:{we}\:{have}\:{also} \\ $$$${m}×{n}\:{values}\:{of}\:\frac{{j}}{{i}},\:{but}\:{some}\:{of}\:{them} \\ $$$${are}\:{not}\:{unique},\:{e}.{g}.\:{points}\:\left(\mathrm{2},\mathrm{3}\right),\:\left(\mathrm{4},\mathrm{6}\right) \\ $$$${have}\:{the}\:{same}\:{angle}\:\theta,\:{since}\:\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{6}}{\mathrm{4}}. \\ $$$$ \\ $$$${To}\:{find}\:{the}\:{number}\:{of}\:{unique}\:{values} \\ $$$${of}\:\frac{{j}}{{i}},\:{we}\:{should}\:{simplify}\:\frac{{j}}{{i}}\:{to}\:\frac{{j}'}{{i}'}\:{with} \\ $$$${i}'\:=\:\frac{{LCM}\left({i},{j}\right)}{{j}}\:{and}\:{j}'\:=\:\frac{{LCM}\left({i},{j}\right)}{{i}},\:{since} \\ $$$$\frac{{j}}{{i}}=\frac{{LCM}\left({i},{j}\right)×{j}}{{LCM}\left({i},{j}\right)×{i}}=\frac{\frac{{LCM}\left({i},{j}\right)}{{i}}}{\frac{{LCM}\left({i},{j}\right)}{{j}}}=\frac{{j}'}{{i}'} \\ $$$$ \\ $$$${e}.{g}.\:{for}\:{point}\:\left(\mathrm{2},\mathrm{3}\right):\:{LCM}\left(\mathrm{2},\mathrm{3}\right)=\mathrm{6}, \\ $$$${i}'=\frac{\mathrm{6}}{\mathrm{3}}=\mathrm{2},\:{j}'=\frac{\mathrm{6}}{\mathrm{2}}=\mathrm{3},\:\frac{{j}}{{i}}=\frac{{j}'}{{i}'}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${and}\:{for}\:{point}\:\left(\mathrm{4},\mathrm{6}\right):\:{LCM}\left(\mathrm{4},\mathrm{6}\right)=\mathrm{12}, \\ $$$${i}'=\frac{\mathrm{12}}{\mathrm{6}}=\mathrm{2},\:{j}'=\frac{\mathrm{12}}{\mathrm{4}}=\mathrm{3},\:\frac{{j}}{{i}}=\frac{{j}'}{{i}'}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$${An}\:{other}\:{way}\:{to}\:{reduce}\:{the}\:{fraction}\:\frac{{j}}{{i}} \\ $$$${to}\:\frac{{j}'}{{i}'}\:{is}: \\ $$$${i}'=\frac{{i}}{{GCD}\left({i},{j}\right)}\:{and}\:{j}'=\frac{{j}}{{GCD}\left({i},{j}\right)} \\ $$$${e}.{g}.\:{for}\:{point}\:\left(\mathrm{2},\mathrm{3}\right):\:{GCD}\left(\mathrm{2},\mathrm{3}\right)=\mathrm{1}, \\ $$$${i}'=\frac{\mathrm{2}}{\mathrm{1}}=\mathrm{2},\:{j}'=\frac{\mathrm{3}}{\mathrm{1}}=\mathrm{3},\:\frac{{j}}{{i}}=\frac{{j}'}{{i}'}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${and}\:{for}\:{point}\:\left(\mathrm{4},\mathrm{6}\right):\:{GCD}\left(\mathrm{4},\mathrm{6}\right)=\mathrm{2}, \\ $$$${i}'=\frac{\mathrm{4}}{\mathrm{2}}=\mathrm{2},\:{j}'=\frac{\mathrm{6}}{\mathrm{2}}=\mathrm{3},\:\frac{{j}}{{i}}=\frac{{j}'}{{i}'}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$${For}\:{all}\:{points}\:\left({i},{j}\right)\:{with}\:\mathrm{1}\leqslant{i}\leqslant{m}\:{and} \\ $$$$\mathrm{1}\leqslant{j}\leqslant{n},\:{we}\:{can}\:{construct}\:{m}×{n}\:{values} \\ $$$${of}\:\frac{{j}}{{i}}=\frac{{j}'}{{i}'}. \\ $$$${Now}\:{we}\:{only}\:{need}\:{to}\:{count}\:{how}\:{many} \\ $$$${of}\:{these}\:{values}\:{are}\:{unique}. \\ $$$$ \\ $$$${It}\:{is}\:{difficult}\:{to}\:{give}\:{a}\:{general}\:{formula} \\ $$$${for}\:{calculating}\:{the}\:{number}\:{of}\:{unique} \\ $$$${values}\:\frac{{j}'}{{i}'},\:{but}\:{for}\:{concrete}\:{numbers} \\ $$$${m}\:{and}\:{n},\:{we}\:{can}\:{easily}\:{get}\:{the}\:{result} \\ $$$${with}\:{help}\:{of}\:{Excel}. \\ $$$$ \\ $$$${For}\:{m}=\mathrm{10},\:{n}=\mathrm{20},\:{for}\:{example},\:{the} \\ $$$${result}\:{is}\:\mathrm{127}\:{unique}\:{angles},\:{see}\:{fig}. \\ $$$${in}\:{comment}. \\ $$$$ \\ $$$${For}\:{m}=\mathrm{127},\:{and}\:{n}=\mathrm{128},\:{the}\:{result} \\ $$$${is}\:\mathrm{9979}\:{unique}\:{angles}.\:{But}\:{the}\:{table}\:{in} \\ $$$${this}\:{case}\:{is}\:{too}\:{big}\:{to}\:{be}\:{shown}\:{here}. \\ $$
Commented by mrW1 last updated on 07/May/17
Commented by mrW1 last updated on 07/May/17
this picture shows which (j/i) value is unique and which is duplicate (yellow).
$${this}\:{picture}\:{shows}\:{which}\:\frac{{j}}{{i}}\:{value}\:{is}\:{unique}\: \\ $$$${and}\:{which}\:{is}\:{duplicate}\:\left({yellow}\right). \\ $$
Commented by FilupS last updated on 07/May/17
This is amazing! Thank you!
$$\mathrm{This}\:\mathrm{is}\:\mathrm{amazing}!\:\mathrm{Thank}\:\mathrm{you}! \\ $$
Commented by mrW1 last updated on 11/May/17
Here the explanation how it works. We have: (j/i)=((j′)/(i′)) with i′=((LCM(i,j))/j) [or (i/(GCD(i,j)))] j′=((LCM(i,j))/i) [or (j/(GCD(i,j)))] For each point (i,j) it can be one of following two cases: Case 1: i′=i and j′=j ⇒point (i,j) descripts a new angle. e.g. point (2,7): LCM(2,7)=14 i′=((14)/7)=2=i j′=((14)/2)=7=j ⇒point (2,7) has a new angle. Case 2: i′<i and j′<j ⇒point (i,j) descripts no new angle, its angle is the same as for point(i′,j′). e.g. point (8,20): LCM(8,20)=40 i′=((40)/(20))=2<i=8 j′=((40)/8)=5<j=20 ⇒point (8,20) has no new angle, since it has the same angle as point (2,5). Case 1 means also GCD(i,j)=1. Case 2 means also GCD(i,j)>1.
$${Here}\:{the}\:{explanation}\:{how}\:{it}\:{works}. \\ $$$$ \\ $$$${We}\:{have}: \\ $$$$\frac{{j}}{{i}}=\frac{{j}'}{{i}'} \\ $$$${with}\:{i}'=\frac{{LCM}\left({i},{j}\right)}{{j}}\:\left[{or}\:\frac{{i}}{{GCD}\left({i},{j}\right)}\right] \\ $$$${j}'=\frac{{LCM}\left({i},{j}\right)}{{i}}\:\:\left[{or}\:\frac{{j}}{{GCD}\left({i},{j}\right)}\right] \\ $$$$ \\ $$$${For}\:{each}\:{point}\:\left({i},{j}\right)\:{it}\:{can}\:{be}\:{one}\:{of} \\ $$$${following}\:{two}\:{cases}: \\ $$$$ \\ $$$${Case}\:\mathrm{1}: \\ $$$${i}'={i}\:{and}\:{j}'={j} \\ $$$$\Rightarrow{point}\:\left({i},{j}\right)\:{descripts}\:{a}\:{new}\:{angle}. \\ $$$${e}.{g}.\:{point}\:\left(\mathrm{2},\mathrm{7}\right): \\ $$$${LCM}\left(\mathrm{2},\mathrm{7}\right)=\mathrm{14} \\ $$$${i}'=\frac{\mathrm{14}}{\mathrm{7}}=\mathrm{2}={i} \\ $$$${j}'=\frac{\mathrm{14}}{\mathrm{2}}=\mathrm{7}={j} \\ $$$$\Rightarrow{point}\:\left(\mathrm{2},\mathrm{7}\right)\:{has}\:{a}\:{new}\:{angle}. \\ $$$$ \\ $$$${Case}\:\mathrm{2}: \\ $$$${i}'<{i}\:{and}\:{j}'<{j} \\ $$$$\Rightarrow{point}\:\left({i},{j}\right)\:{descripts}\:{no}\:{new}\:{angle}, \\ $$$${its}\:{angle}\:{is}\:{the}\:{same}\:{as}\:{for}\:{point}\left({i}',{j}'\right). \\ $$$${e}.{g}.\:{point}\:\left(\mathrm{8},\mathrm{20}\right): \\ $$$${LCM}\left(\mathrm{8},\mathrm{20}\right)=\mathrm{40} \\ $$$${i}'=\frac{\mathrm{40}}{\mathrm{20}}=\mathrm{2}<{i}=\mathrm{8} \\ $$$${j}'=\frac{\mathrm{40}}{\mathrm{8}}=\mathrm{5}<{j}=\mathrm{20} \\ $$$$\Rightarrow{point}\:\left(\mathrm{8},\mathrm{20}\right)\:{has}\:{no}\:{new}\:{angle},\:{since} \\ $$$${it}\:{has}\:{the}\:{same}\:{angle}\:{as}\:{point}\:\left(\mathrm{2},\mathrm{5}\right). \\ $$$$ \\ $$$${Case}\:\mathrm{1}\:{means}\:{also}\:{GCD}\left({i},{j}\right)=\mathrm{1}. \\ $$$${Case}\:\mathrm{2}\:{means}\:{also}\:{GCD}\left({i},{j}\right)>\mathrm{1}. \\ $$
Commented by mrW1 last updated on 08/May/17
Commented by mrW1 last updated on 08/May/17
Commented by mrW1 last updated on 09/May/17
Now we create a table with m columns (for i=1 to m) and n rows (for j=1 to n). We give the cell (i,j) the value 1 if it is case 1 and the value 0 if it is case 2. In Excel you can do this by entering following formula into the first cell, let′s say it is B5: =1−SIGN(GCD($A5,B$4)−1) and then copy this cell to all cells of the table. In this way we get a table like this:
$${Now}\:{we}\:{create}\:{a}\:{table}\:{with}\:{m}\:{columns} \\ $$$$\left({for}\:{i}=\mathrm{1}\:{to}\:{m}\right)\:{and}\:{n}\:{rows}\:\left({for}\:{j}=\mathrm{1}\:{to}\:{n}\right). \\ $$$$ \\ $$$${We}\:{give}\:{the}\:{cell}\:\left({i},{j}\right)\:{the}\:{value}\:\mathrm{1}\:{if}\:{it}\:{is} \\ $$$${case}\:\mathrm{1}\:{and}\:{the}\:{value}\:\mathrm{0}\:{if}\:{it}\:{is}\:{case}\:\mathrm{2}. \\ $$$$ \\ $$$${In}\:{Excel}\:{you}\:{can}\:{do}\:{this}\:{by}\:{entering} \\ $$$${following}\:{formula}\:{into}\:{the}\:{first}\:{cell}, \\ $$$${let}'{s}\:{say}\:{it}\:{is}\:{B}\mathrm{5}: \\ $$$$=\mathrm{1}−{SIGN}\left({GCD}\left(\${A}\mathrm{5},{B\$}\mathrm{4}\right)−\mathrm{1}\right) \\ $$$${and}\:{then}\:{copy}\:{this}\:{cell}\:{to}\:{all}\:{cells}\:{of}\:{the}\: \\ $$$${table}. \\ $$$$ \\ $$$${In}\:{this}\:{way}\:{we}\:{get}\:{a}\:{table}\:{like}\:{this}: \\ $$
Commented by mrW1 last updated on 08/May/17
Commented by mrW1 last updated on 09/May/17
Cells with value 1 have unique (j/i) and cells with value 0 are duplicates. The sum of all values in this table is the number of unique angles for 1≤i≤m and 1≤j≤n.
$${Cells}\:{with}\:{value}\:\mathrm{1}\:{have}\:{unique}\:\frac{{j}}{{i}}\:{and} \\ $$$${cells}\:{with}\:{value}\:\mathrm{0}\:{are}\:{duplicates}. \\ $$$$ \\ $$$${The}\:{sum}\:{of}\:{all}\:{values}\:{in}\:{this}\:{table}\:{is} \\ $$$${the}\:{number}\:{of}\:{unique}\:{angles}\:{for} \\ $$$$\mathrm{1}\leqslant{i}\leqslant{m}\:{and}\:\mathrm{1}\leqslant{j}\leqslant{n}. \\ $$
Commented by mrW1 last updated on 09/May/17
From this table you can easily get the number of unique angles for any number of points, e.g. 7×22 points.
$${From}\:{this}\:{table}\:{you}\:{can}\:{easily}\:{get}\:{the} \\ $$$${number}\:{of}\:{unique}\:{angles}\:{for}\:{any}\: \\ $$$${number}\:{of}\:{points},\:{e}.{g}.\:\mathrm{7}×\mathrm{22}\:{points}. \\ $$
Commented by mrW1 last updated on 09/May/17
Commented by mrW1 last updated on 09/May/17
The big question now is: is it possible to find a formula to calculate the number of unique angles for m×n points?
$${The}\:{big}\:{question}\:{now}\:{is}: \\ $$$${is}\:{it}\:{possible}\:{to}\:{find}\:{a}\:{formula}\:{to} \\ $$$${calculate}\:{the}\:{number}\:{of}\:{unique}\:{angles} \\ $$$${for}\:{m}×{n}\:{points}? \\ $$
Commented by FilupS last updated on 09/May/17
This is fascinating!
$$\mathrm{This}\:\mathrm{is}\:\mathrm{fascinating}! \\ $$
Commented by FilupS last updated on 09/May/17
Creating a formula for LCM(i,j) wouldn′t be very straight forward. It would be extremely difficult and possibly involve prime number distributions, which is going into number theory and the prime numbers
$$\mathrm{Creating}\:\mathrm{a}\:\mathrm{formula}\:\mathrm{for}\:{LCM}\left({i},{j}\right) \\ $$$$\mathrm{wouldn}'\mathrm{t}\:\mathrm{be}\:\mathrm{very}\:\mathrm{straight}\:\mathrm{forward}. \\ $$$$ \\ $$$$\mathrm{It}\:\mathrm{would}\:\mathrm{be}\:\mathrm{extremely}\:\mathrm{difficult}\:\mathrm{and}\:\mathrm{possibly} \\ $$$$\mathrm{involve}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{distributions}, \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{going}\:\mathrm{into}\:\mathrm{number}\:\mathrm{theory}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{prime}\:\mathrm{numbers} \\ $$
Commented by FilupS last updated on 09/May/17
The best function I can think of would be something like (correct if wrong): i′=((LCM(i,j))/j) j′=((LCM(i,j))/i) δ_(ij) = { ((1 ⇔ i′=i ∧ j′=j)),((0 ⇔ i′<i ∧ j′<j)) :} Φ=Σ_i ^m Σ_j ^n δ_(ij) Φ=total unique angles
$$\mathrm{The}\:\mathrm{best}\:\mathrm{function}\:\mathrm{I}\:\mathrm{can}\:\mathrm{think}\:\mathrm{of}\:\mathrm{would} \\ $$$$\mathrm{be}\:\mathrm{something}\:\mathrm{like}\:\left(\mathrm{correct}\:\mathrm{if}\:\mathrm{wrong}\right): \\ $$$${i}'=\frac{\mathrm{LCM}\left({i},{j}\right)}{{j}}\:\:\:\:\:\:\:{j}'=\frac{\mathrm{LCM}\left({i},{j}\right)}{{i}} \\ $$$$\delta_{{ij}} =\begin{cases}{\mathrm{1}\:\:\:\:\:\Leftrightarrow\:\:\:\:\:{i}'={i}\:\:\wedge\:\:{j}'={j}}\\{\mathrm{0}\:\:\:\:\:\Leftrightarrow\:\:\:\:\:{i}'<{i}\:\:\wedge\:\:{j}'<{j}}\end{cases} \\ $$$$\Phi=\underset{{i}} {\overset{{m}} {\sum}}\underset{{j}} {\overset{{n}} {\sum}}\delta_{{ij}} \\ $$$$\: \\ $$$$\Phi=\mathrm{total}\:\mathrm{unique}\:\mathrm{angles} \\ $$
Commented by mrW1 last updated on 09/May/17
I agree with you. We can write: δ_(ij) =1−sign(i−((LCM(i,j))/j)) ⇒Φ=Σ_(i=1) ^m Σ_(j=1) ^n δ_(ij) ⇒Φ=mn−Σ_(i=1) ^m Σ_(j=1) ^n sign(i−((LCM(i,j))/j))
$${I}\:{agree}\:{with}\:{you}. \\ $$$$ \\ $$$${We}\:{can}\:{write}: \\ $$$$\delta_{{ij}} =\mathrm{1}−{sign}\left({i}−\frac{{LCM}\left({i},{j}\right)}{{j}}\right) \\ $$$$\Rightarrow\Phi=\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\delta_{{ij}} \\ $$$$\Rightarrow\Phi={mn}−\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}{sign}\left({i}−\frac{{LCM}\left({i},{j}\right)}{{j}}\right) \\ $$
Commented by FilupS last updated on 09/May/17
I think thats the best solution!
$$\mathrm{I}\:\mathrm{think}\:\mathrm{thats}\:\mathrm{the}\:\mathrm{best}\:\mathrm{solution}! \\ $$
Commented by mrW1 last updated on 09/May/17
An other simple formula is: δ_(ij) =1−sign(GCD(i,j)−1) Φ=mn−Σ_(i=1) ^m Σ_(j=1) ^n sign(GCD(i,j)−1)
$${An}\:{other}\:{simple}\:{formula}\:{is}: \\ $$$$\delta_{{ij}} =\mathrm{1}−{sign}\left({GCD}\left({i},{j}\right)−\mathrm{1}\right) \\ $$$$\Phi={mn}−\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}{sign}\left({GCD}\left({i},{j}\right)−\mathrm{1}\right) \\ $$
Commented by FilupS last updated on 09/May/17
can you explain how you got this?
$$\mathrm{can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{how}\:\mathrm{you}\:\mathrm{got}\:\mathrm{this}? \\ $$
Commented by mrW1 last updated on 10/May/17
(j/i)=((j′)/(i′))=((j/(GCD(i,j)))/(i/(GCD(i,j)))) when i′=i and j′=j, it means (j/i) can not be simplified any more, i.e. GCD(i,j)=1. E.g. for point (2,7): GCD(2,7)=1, (7/2) is a fraction which is already simplified to lowest terms. when i′<i and j′<j, it means (j/i) can be simplified further, i.e. GCD(i,j)>1. E.g. for point (8,20): GCD(8,20)=4, ((20)/8) can be simplified to ((20/4)/(8/4))=(5/2). if GCD(i,j)=1, ⇒1−sign(GCD(i,j)−1)=1 ⇒ unique if GCD(i,j)>1, ⇒1−sign(GCD(i,j)−1)=0,⇒ duplicate
$$\frac{{j}}{{i}}=\frac{{j}'}{{i}'}=\frac{\frac{{j}}{{GCD}\left({i},{j}\right)}}{\frac{{i}}{{GCD}\left({i},{j}\right)}} \\ $$$${when}\:{i}'={i}\:{and}\:{j}'={j},\:{it}\:{means}\:\frac{{j}}{{i}}\:{can} \\ $$$${not}\:{be}\:{simplified}\:{any}\:{more},\:{i}.{e}. \\ $$$${GCD}\left({i},{j}\right)=\mathrm{1}.\:{E}.{g}.\:{for}\:{point}\:\left(\mathrm{2},\mathrm{7}\right):\: \\ $$$${GCD}\left(\mathrm{2},\mathrm{7}\right)=\mathrm{1},\:\frac{\mathrm{7}}{\mathrm{2}}\:{is}\:{a}\:{fraction}\:{which} \\ $$$${is}\:{already}\:{simplified}\:{to}\:{lowest}\:{terms}. \\ $$$$ \\ $$$${when}\:{i}'<{i}\:{and}\:{j}'<{j},\:{it}\:{means}\:\frac{{j}}{{i}}\:{can} \\ $$$${be}\:{simplified}\:{further},\:{i}.{e}.\:{GCD}\left({i},{j}\right)>\mathrm{1}. \\ $$$${E}.{g}.\:{for}\:{point}\:\left(\mathrm{8},\mathrm{20}\right):\:{GCD}\left(\mathrm{8},\mathrm{20}\right)=\mathrm{4}, \\ $$$$\frac{\mathrm{20}}{\mathrm{8}}\:{can}\:{be}\:{simplified}\:{to}\:\frac{\mathrm{20}/\mathrm{4}}{\mathrm{8}/\mathrm{4}}=\frac{\mathrm{5}}{\mathrm{2}}. \\ $$$$ \\ $$$${if}\:{GCD}\left({i},{j}\right)=\mathrm{1},\: \\ $$$$\:\Rightarrow\mathrm{1}−{sign}\left({GCD}\left({i},{j}\right)−\mathrm{1}\right)=\mathrm{1}\:\Rightarrow\:{unique} \\ $$$${if}\:{GCD}\left({i},{j}\right)>\mathrm{1},\: \\ $$$$\Rightarrow\mathrm{1}−{sign}\left({GCD}\left({i},{j}\right)−\mathrm{1}\right)=\mathrm{0},\Rightarrow\:{duplicate} \\ $$
Commented by mrW1 last updated on 10/May/17
I didn′t expect that the solution could be expressed in such a simple formula, eventhough it is no formula with which one can directly get the result. Thank you for this interesting and challenging question. To solve it has made a lot of fun.
$${I}\:{didn}'{t}\:{expect}\:{that}\:{the}\:{solution}\:{could}\:{be} \\ $$$${expressed}\:{in}\:{such}\:{a}\:{simple}\:{formula}, \\ $$$${eventhough}\:{it}\:{is}\:{no}\:{formula}\:{with} \\ $$$${which}\:{one}\:{can}\:{directly}\:{get}\:{the}\:{result}. \\ $$$$ \\ $$$${Thank}\:{you}\:{for}\:{this}\:{interesting}\:{and} \\ $$$${challenging}\:{question}.\:{To}\:{solve}\:{it} \\ $$$${has}\:{made}\:{a}\:{lot}\:{of}\:{fun}. \\ $$
Commented by FilupS last updated on 10/May/17
Thank you for solving it
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{solving}\:\mathrm{it} \\ $$

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