Mixing
How can I know the length of sublists in a code without running the code?
Clash Royale CLAN TAG#URR8PPP
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I have a list called eaData which contains sublists of the form a,b, number, where each sublist gives the number of steps in the Euclidean algorithm for numbers $a$ and $b$.
I have the following code:
eaSteps[a_, b_] :=
NestWhileList[
#[[2]], Mod[#[[1]], #[[2]]] &,
a, b,
#[[2]] != 0 &
]
So for example
eaSteps[1736, 1333] = 1736,1333,1333,403,403,124,124,31,31,0`
I also have
eaData =
SortBy[
Flatten[
Table[
a, b, Length[eaSteps[a, b]] - 1,
a, 1, 100,
b, 1, a(* the b value cannot exceed the a value *)
],
1
],
Part[#,3] &
]
My question is that my eaData list contains 5050 sublists. How could I know there are 5050 subentries before I even run the code?
I know that for the EA, the maximum number of steps is at most 5 times the minimum number of digits of either a or b, how does that help me here?
And if $b leq a leq 1000$ then how can I find the a and b values that require the most steps in the EA?
I'm thinking that I know a has at most 4 digits, so there are at most 20, steps. But now I'm stuck. Any help? thank you!
list-manipulation
edited 1 hour ago
kglr
167k8188390
asked 4 hours ago
user130306
233
New contributor
user130306 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
4
down vote
favorite
I have a list called eaData which contains sublists of the form a,b, number, where each sublist gives the number of steps in the Euclidean algorithm for numbers $a$ and $b$.
I have the following code:
eaSteps[a_, b_] :=
NestWhileList[
#[[2]], Mod[#[[1]], #[[2]]] &,
a, b,
#[[2]] != 0 &
]
So for example
eaSteps[1736, 1333] = 1736,1333,1333,403,403,124,124,31,31,0`
I also have
eaData =
SortBy[
Flatten[
Table[
a, b, Length[eaSteps[a, b]] - 1,
a, 1, 100,
b, 1, a(* the b value cannot exceed the a value *)
],
1
],
Part[#,3] &
]
My question is that my eaData list contains 5050 sublists. How could I know there are 5050 subentries before I even run the code?
I know that for the EA, the maximum number of steps is at most 5 times the minimum number of digits of either a or b, how does that help me here?
And if $b leq a leq 1000$ then how can I find the a and b values that require the most steps in the EA?
I'm thinking that I know a has at most 4 digits, so there are at most 20, steps. But now I'm stuck. Any help? thank you!
list-manipulation
edited 1 hour ago
kglr
167k8188390
asked 4 hours ago
user130306
233
New contributor
user130306 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
shouldn't you changea, 1, 25toa, 1, 100to get 5050 sublists?
– kglr
4 hours ago
yes sorry, I meant a,1,100, i just mean, how do I know without running the code itself that I'd get 5050. what is the logic behind it?
– user130306
4 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I have a list called eaData which contains sublists of the form a,b, number, where each sublist gives the number of steps in the Euclidean algorithm for numbers $a$ and $b$.
I have the following code:
eaSteps[a_, b_] :=
NestWhileList[
#[[2]], Mod[#[[1]], #[[2]]] &,
a, b,
#[[2]] != 0 &
]
So for example
eaSteps[1736, 1333] = 1736,1333,1333,403,403,124,124,31,31,0`
I also have
eaData =
SortBy[
Flatten[
Table[
a, b, Length[eaSteps[a, b]] - 1,
a, 1, 100,
b, 1, a(* the b value cannot exceed the a value *)
],
1
],
Part[#,3] &
]
My question is that my eaData list contains 5050 sublists. How could I know there are 5050 subentries before I even run the code?
I know that for the EA, the maximum number of steps is at most 5 times the minimum number of digits of either a or b, how does that help me here?
And if $b leq a leq 1000$ then how can I find the a and b values that require the most steps in the EA?
I'm thinking that I know a has at most 4 digits, so there are at most 20, steps. But now I'm stuck. Any help? thank you!
list-manipulation
edited 1 hour ago
kglr
167k8188390
asked 4 hours ago
user130306
233
New contributor
user130306 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a list called eaData which contains sublists of the form a,b, number, where each sublist gives the number of steps in the Euclidean algorithm for numbers $a$ and $b$.
I have the following code:
eaSteps[a_, b_] :=
NestWhileList[
#[[2]], Mod[#[[1]], #[[2]]] &,
a, b,
#[[2]] != 0 &
]
So for example
eaSteps[1736, 1333] = 1736,1333,1333,403,403,124,124,31,31,0`
I also have
eaData =
SortBy[
Flatten[
Table[
a, b, Length[eaSteps[a, b]] - 1,
a, 1, 100,
b, 1, a(* the b value cannot exceed the a value *)
],
1
],
Part[#,3] &
]
My question is that my eaData list contains 5050 sublists. How could I know there are 5050 subentries before I even run the code?
I know that for the EA, the maximum number of steps is at most 5 times the minimum number of digits of either a or b, how does that help me here?
And if $b leq a leq 1000$ then how can I find the a and b values that require the most steps in the EA?
I'm thinking that I know a has at most 4 digits, so there are at most 20, steps. But now I'm stuck. Any help? thank you!
list-manipulation
list-manipulation
edited 1 hour ago
kglr
167k8188390
asked 4 hours ago
user130306
233
New contributor
user130306 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
kglr
167k8188390
asked 4 hours ago
user130306
233
New contributor
user130306 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
kglr
167k8188390
edited 1 hour ago
kglr
167k8188390
edited 1 hour ago
kglr
167k8188390
167k8188390
asked 4 hours ago
user130306
233
New contributor
user130306 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
user130306
233
asked 4 hours ago
user130306
233
233
New contributor
user130306 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user130306 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user130306 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
shouldn't you changea, 1, 25toa, 1, 100to get 5050 sublists?
– kglr
4 hours ago
yes sorry, I meant a,1,100, i just mean, how do I know without running the code itself that I'd get 5050. what is the logic behind it?
– user130306
4 hours ago
add a comment |Â
shouldn't you changea, 1, 25toa, 1, 100to get 5050 sublists?
– kglr
4 hours ago
yes sorry, I meant a,1,100, i just mean, how do I know without running the code itself that I'd get 5050. what is the logic behind it?
– user130306
4 hours ago
shouldn't you change
a, 1, 25 to a, 1, 100 to get 5050 sublists?– kglr
4 hours ago
shouldn't you change
a, 1, 25 to a, 1, 100 to get 5050 sublists?– kglr
4 hours ago
yes sorry, I meant a,1,100, i just mean, how do I know without running the code itself that I'd get 5050. what is the logic behind it?
– user130306
4 hours ago
yes sorry, I meant a,1,100, i just mean, how do I know without running the code itself that I'd get 5050. what is the logic behind it?
– user130306
4 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Update: FWIW it is also possible to find a pair of numbers a, b less than n with maximum length for the sequence eaSteps[a,b] without "running the code". It seems that Fibonacci sequence can be used to identify the pairs with maximum eaSteps length (I have no idea why though):
ClearAll[maxEALengthPair]
maxEALengthPair[n_] := Module[i = 0, While[Fibonacci[++i] < n]; Fibonacci[i-1, i-2]];
TeXForm @ Grid[#, ## & @@ #2 & @@@ Transpose["n", "pair",
#, maxEALengthPair /@ # &@50, 100, 500, 1000, 10000, 100000, 1000000, 10^7],
Dividers -> All]
$tinybeginarrayc
hline
textn & 50 & 100 & 500 & 1000 & 10000 & 100000 & 1000000 & 10000000 \
hline
textpair & 34,21 & 89,55 & 377,233 & 987,610 & 6765,4181 & 75025,46368 & 832040,514229 &
9227465,5702887 \
hline
endarray$
Brute-force approach gives the same results for n ∈ 50, 100, 500, 1000:
nl = 50, 100, 500, 1000 ;
#, MaximalBy[Reverse /@ Subsets[Range[#], 2], Length[eaSteps[#]] &][[1]] & /@ #&@nl
Grid[#, ## & @@ #2 & @@@ Transpose["n", "pair", #,
MaximalBy[Reverse /@ Subsets[Range[#], 2], Length[eaSteps[#]] &][[1]] & /@ # &@nl],
Dividers -> All]
$tinybeginarrayc
hline
textn & 50 & 100 & 500 & 1000 \
hline
textpair & 34,21 & 89,55 & 377,233 & 987,610 \
hline
endarray$
Note: For some n, there are multiple pairs with maximum eaSteps length. For example, for n=50 there are 7 pairs that give a length 7 list:
MaximalBy[Reverse /@ Subsets[Range[50], 2], Length[eaSteps[#]] &]
34, 21, 47, 29, 50, 29, 49, 30, 49, 31, 50, 31, 47,
34
Length[eaSteps @ #] - 1 &/@ %
7, 7, 7, 7, 7, 7, 7
The table above gives only the first of multiple pairs.
Original answer:
You can get the length of Flatten[Table[a, b, foo, a, 1, n, b, 1, a], 1] using
n (n + 1)/2
(which is Sum[1, i, 1, n, j, 1, i])
or
Length[Subsets[Range[n], 1, 2]]
For n = 100:
100 101 /2
5050
Length[Subsets[Range[100], 1, 2]]
5050
table[n_Integer] := Flatten[Table[a, b, foo, a, 1, n, b, 1, a], 1];
Length[table[100]]
5050
For the second part of the question, using a brute force approach
MaximalBy[Reverse/@Subsets[Range[1000],2], Length[eaSteps[#]]&]
987, 610
Length[Length[eaSteps[987, 610]]] -1
14
answered 3 hours ago
kglr
167k8188390
thank you! so basically the reason I would know its 5050 before I even run the code is if I useSubsetsright?
– user130306
3 hours ago
@user130306, that's correct.
– kglr
3 hours ago
great, thank you again. and for the second part of the question, do the values a = 987, b = 610, n = 14 work and is that the right answer?
– user130306
3 hours ago
using a brute force approachMaximalBy[Reverse/@Subsets[Range[1000],2], Length[eaSteps[#]]&]does give987, 610andLength[eaSteps[987, 610]]gives 15.
– kglr
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Update: FWIW it is also possible to find a pair of numbers a, b less than n with maximum length for the sequence eaSteps[a,b] without "running the code". It seems that Fibonacci sequence can be used to identify the pairs with maximum eaSteps length (I have no idea why though):
ClearAll[maxEALengthPair]
maxEALengthPair[n_] := Module[i = 0, While[Fibonacci[++i] < n]; Fibonacci[i-1, i-2]];
TeXForm @ Grid[#, ## & @@ #2 & @@@ Transpose["n", "pair",
#, maxEALengthPair /@ # &@50, 100, 500, 1000, 10000, 100000, 1000000, 10^7],
Dividers -> All]
$tinybeginarrayc
hline
textn & 50 & 100 & 500 & 1000 & 10000 & 100000 & 1000000 & 10000000 \
hline
textpair & 34,21 & 89,55 & 377,233 & 987,610 & 6765,4181 & 75025,46368 & 832040,514229 &
9227465,5702887 \
hline
endarray$
Brute-force approach gives the same results for n ∈ 50, 100, 500, 1000:
nl = 50, 100, 500, 1000 ;
#, MaximalBy[Reverse /@ Subsets[Range[#], 2], Length[eaSteps[#]] &][[1]] & /@ #&@nl
Grid[#, ## & @@ #2 & @@@ Transpose["n", "pair", #,
MaximalBy[Reverse /@ Subsets[Range[#], 2], Length[eaSteps[#]] &][[1]] & /@ # &@nl],
Dividers -> All]
$tinybeginarrayc
hline
textn & 50 & 100 & 500 & 1000 \
hline
textpair & 34,21 & 89,55 & 377,233 & 987,610 \
hline
endarray$
Note: For some n, there are multiple pairs with maximum eaSteps length. For example, for n=50 there are 7 pairs that give a length 7 list:
MaximalBy[Reverse /@ Subsets[Range[50], 2], Length[eaSteps[#]] &]
34, 21, 47, 29, 50, 29, 49, 30, 49, 31, 50, 31, 47,
34
Length[eaSteps @ #] - 1 &/@ %
7, 7, 7, 7, 7, 7, 7
The table above gives only the first of multiple pairs.
Original answer:
You can get the length of Flatten[Table[a, b, foo, a, 1, n, b, 1, a], 1] using
n (n + 1)/2
(which is Sum[1, i, 1, n, j, 1, i])
or
Length[Subsets[Range[n], 1, 2]]
For n = 100:
100 101 /2
5050
Length[Subsets[Range[100], 1, 2]]
5050
table[n_Integer] := Flatten[Table[a, b, foo, a, 1, n, b, 1, a], 1];
Length[table[100]]
5050
For the second part of the question, using a brute force approach
MaximalBy[Reverse/@Subsets[Range[1000],2], Length[eaSteps[#]]&]
987, 610
Length[Length[eaSteps[987, 610]]] -1
14
answered 3 hours ago
kglr
167k8188390
thank you! so basically the reason I would know its 5050 before I even run the code is if I useSubsetsright?
– user130306
3 hours ago
@user130306, that's correct.
– kglr
3 hours ago
great, thank you again. and for the second part of the question, do the values a = 987, b = 610, n = 14 work and is that the right answer?
– user130306
3 hours ago
using a brute force approachMaximalBy[Reverse/@Subsets[Range[1000],2], Length[eaSteps[#]]&]does give987, 610andLength[eaSteps[987, 610]]gives 15.
– kglr
2 hours ago
add a comment |Â
up vote
2
down vote
accepted
Update: FWIW it is also possible to find a pair of numbers a, b less than n with maximum length for the sequence eaSteps[a,b] without "running the code". It seems that Fibonacci sequence can be used to identify the pairs with maximum eaSteps length (I have no idea why though):
ClearAll[maxEALengthPair]
maxEALengthPair[n_] := Module[i = 0, While[Fibonacci[++i] < n]; Fibonacci[i-1, i-2]];
TeXForm @ Grid[#, ## & @@ #2 & @@@ Transpose["n", "pair",
#, maxEALengthPair /@ # &@50, 100, 500, 1000, 10000, 100000, 1000000, 10^7],
Dividers -> All]
$tinybeginarrayc
hline
textn & 50 & 100 & 500 & 1000 & 10000 & 100000 & 1000000 & 10000000 \
hline
textpair & 34,21 & 89,55 & 377,233 & 987,610 & 6765,4181 & 75025,46368 & 832040,514229 &
9227465,5702887 \
hline
endarray$
Brute-force approach gives the same results for n ∈ 50, 100, 500, 1000:
nl = 50, 100, 500, 1000 ;
#, MaximalBy[Reverse /@ Subsets[Range[#], 2], Length[eaSteps[#]] &][[1]] & /@ #&@nl
Grid[#, ## & @@ #2 & @@@ Transpose["n", "pair", #,
MaximalBy[Reverse /@ Subsets[Range[#], 2], Length[eaSteps[#]] &][[1]] & /@ # &@nl],
Dividers -> All]
$tinybeginarrayc
hline
textn & 50 & 100 & 500 & 1000 \
hline
textpair & 34,21 & 89,55 & 377,233 & 987,610 \
hline
endarray$
Note: For some n, there are multiple pairs with maximum eaSteps length. For example, for n=50 there are 7 pairs that give a length 7 list:
MaximalBy[Reverse /@ Subsets[Range[50], 2], Length[eaSteps[#]] &]
34, 21, 47, 29, 50, 29, 49, 30, 49, 31, 50, 31, 47,
34
Length[eaSteps @ #] - 1 &/@ %
7, 7, 7, 7, 7, 7, 7
The table above gives only the first of multiple pairs.
Original answer:
You can get the length of Flatten[Table[a, b, foo, a, 1, n, b, 1, a], 1] using
n (n + 1)/2
(which is Sum[1, i, 1, n, j, 1, i])
or
Length[Subsets[Range[n], 1, 2]]
For n = 100:
100 101 /2
5050
Length[Subsets[Range[100], 1, 2]]
5050
table[n_Integer] := Flatten[Table[a, b, foo, a, 1, n, b, 1, a], 1];
Length[table[100]]
5050
For the second part of the question, using a brute force approach
MaximalBy[Reverse/@Subsets[Range[1000],2], Length[eaSteps[#]]&]
987, 610
Length[Length[eaSteps[987, 610]]] -1
14
answered 3 hours ago
kglr
167k8188390
thank you! so basically the reason I would know its 5050 before I even run the code is if I useSubsetsright?
– user130306
3 hours ago
@user130306, that's correct.
– kglr
3 hours ago
great, thank you again. and for the second part of the question, do the values a = 987, b = 610, n = 14 work and is that the right answer?
– user130306
3 hours ago
using a brute force approachMaximalBy[Reverse/@Subsets[Range[1000],2], Length[eaSteps[#]]&]does give987, 610andLength[eaSteps[987, 610]]gives 15.
– kglr
2 hours ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Update: FWIW it is also possible to find a pair of numbers a, b less than n with maximum length for the sequence eaSteps[a,b] without "running the code". It seems that Fibonacci sequence can be used to identify the pairs with maximum eaSteps length (I have no idea why though):
ClearAll[maxEALengthPair]
maxEALengthPair[n_] := Module[i = 0, While[Fibonacci[++i] < n]; Fibonacci[i-1, i-2]];
TeXForm @ Grid[#, ## & @@ #2 & @@@ Transpose["n", "pair",
#, maxEALengthPair /@ # &@50, 100, 500, 1000, 10000, 100000, 1000000, 10^7],
Dividers -> All]
$tinybeginarrayc
hline
textn & 50 & 100 & 500 & 1000 & 10000 & 100000 & 1000000 & 10000000 \
hline
textpair & 34,21 & 89,55 & 377,233 & 987,610 & 6765,4181 & 75025,46368 & 832040,514229 &
9227465,5702887 \
hline
endarray$
Brute-force approach gives the same results for n ∈ 50, 100, 500, 1000:
nl = 50, 100, 500, 1000 ;
#, MaximalBy[Reverse /@ Subsets[Range[#], 2], Length[eaSteps[#]] &][[1]] & /@ #&@nl
Grid[#, ## & @@ #2 & @@@ Transpose["n", "pair", #,
MaximalBy[Reverse /@ Subsets[Range[#], 2], Length[eaSteps[#]] &][[1]] & /@ # &@nl],
Dividers -> All]
$tinybeginarrayc
hline
textn & 50 & 100 & 500 & 1000 \
hline
textpair & 34,21 & 89,55 & 377,233 & 987,610 \
hline
endarray$
Note: For some n, there are multiple pairs with maximum eaSteps length. For example, for n=50 there are 7 pairs that give a length 7 list:
MaximalBy[Reverse /@ Subsets[Range[50], 2], Length[eaSteps[#]] &]
34, 21, 47, 29, 50, 29, 49, 30, 49, 31, 50, 31, 47,
34
Length[eaSteps @ #] - 1 &/@ %
7, 7, 7, 7, 7, 7, 7
The table above gives only the first of multiple pairs.
Original answer:
You can get the length of Flatten[Table[a, b, foo, a, 1, n, b, 1, a], 1] using
n (n + 1)/2
(which is Sum[1, i, 1, n, j, 1, i])
or
Length[Subsets[Range[n], 1, 2]]
For n = 100:
100 101 /2
5050
Length[Subsets[Range[100], 1, 2]]
5050
table[n_Integer] := Flatten[Table[a, b, foo, a, 1, n, b, 1, a], 1];
Length[table[100]]
5050
For the second part of the question, using a brute force approach
MaximalBy[Reverse/@Subsets[Range[1000],2], Length[eaSteps[#]]&]
987, 610
Length[Length[eaSteps[987, 610]]] -1
14
answered 3 hours ago
kglr
167k8188390
Update: FWIW it is also possible to find a pair of numbers a, b less than n with maximum length for the sequence eaSteps[a,b] without "running the code". It seems that Fibonacci sequence can be used to identify the pairs with maximum eaSteps length (I have no idea why though):
ClearAll[maxEALengthPair]
maxEALengthPair[n_] := Module[i = 0, While[Fibonacci[++i] < n]; Fibonacci[i-1, i-2]];
TeXForm @ Grid[#, ## & @@ #2 & @@@ Transpose["n", "pair",
#, maxEALengthPair /@ # &@50, 100, 500, 1000, 10000, 100000, 1000000, 10^7],
Dividers -> All]
$tinybeginarrayc
hline
textn & 50 & 100 & 500 & 1000 & 10000 & 100000 & 1000000 & 10000000 \
hline
textpair & 34,21 & 89,55 & 377,233 & 987,610 & 6765,4181 & 75025,46368 & 832040,514229 &
9227465,5702887 \
hline
endarray$
Brute-force approach gives the same results for n ∈ 50, 100, 500, 1000:
nl = 50, 100, 500, 1000 ;
#, MaximalBy[Reverse /@ Subsets[Range[#], 2], Length[eaSteps[#]] &][[1]] & /@ #&@nl
Grid[#, ## & @@ #2 & @@@ Transpose["n", "pair", #,
MaximalBy[Reverse /@ Subsets[Range[#], 2], Length[eaSteps[#]] &][[1]] & /@ # &@nl],
Dividers -> All]
$tinybeginarrayc
hline
textn & 50 & 100 & 500 & 1000 \
hline
textpair & 34,21 & 89,55 & 377,233 & 987,610 \
hline
endarray$
Note: For some n, there are multiple pairs with maximum eaSteps length. For example, for n=50 there are 7 pairs that give a length 7 list:
MaximalBy[Reverse /@ Subsets[Range[50], 2], Length[eaSteps[#]] &]
34, 21, 47, 29, 50, 29, 49, 30, 49, 31, 50, 31, 47,
34
Length[eaSteps @ #] - 1 &/@ %
7, 7, 7, 7, 7, 7, 7
The table above gives only the first of multiple pairs.
Original answer:
You can get the length of Flatten[Table[a, b, foo, a, 1, n, b, 1, a], 1] using
n (n + 1)/2
(which is Sum[1, i, 1, n, j, 1, i])
or
Length[Subsets[Range[n], 1, 2]]
For n = 100:
100 101 /2
5050
Length[Subsets[Range[100], 1, 2]]
5050
table[n_Integer] := Flatten[Table[a, b, foo, a, 1, n, b, 1, a], 1];
Length[table[100]]
5050
For the second part of the question, using a brute force approach
MaximalBy[Reverse/@Subsets[Range[1000],2], Length[eaSteps[#]]&]
987, 610
Length[Length[eaSteps[987, 610]]] -1
14
answered 3 hours ago
kglr
167k8188390
edited 29 mins ago
answered 3 hours ago
kglr
167k8188390
answered 3 hours ago
kglr
167k8188390
answered 3 hours ago
kglr
167k8188390
167k8188390
thank you! so basically the reason I would know its 5050 before I even run the code is if I useSubsetsright?
– user130306
3 hours ago
@user130306, that's correct.
– kglr
3 hours ago
great, thank you again. and for the second part of the question, do the values a = 987, b = 610, n = 14 work and is that the right answer?
– user130306
3 hours ago
using a brute force approachMaximalBy[Reverse/@Subsets[Range[1000],2], Length[eaSteps[#]]&]does give987, 610andLength[eaSteps[987, 610]]gives 15.
– kglr
2 hours ago
add a comment |Â
thank you! so basically the reason I would know its 5050 before I even run the code is if I useSubsetsright?
– user130306
3 hours ago
@user130306, that's correct.
– kglr
3 hours ago
great, thank you again. and for the second part of the question, do the values a = 987, b = 610, n = 14 work and is that the right answer?
– user130306
3 hours ago
using a brute force approachMaximalBy[Reverse/@Subsets[Range[1000],2], Length[eaSteps[#]]&]does give987, 610andLength[eaSteps[987, 610]]gives 15.
– kglr
2 hours ago
thank you! so basically the reason I would know its 5050 before I even run the code is if I use
Subsets right?– user130306
3 hours ago
thank you! so basically the reason I would know its 5050 before I even run the code is if I use
Subsets right?– user130306
3 hours ago
@user130306, that's correct.
– kglr
3 hours ago
@user130306, that's correct.
– kglr
3 hours ago
great, thank you again. and for the second part of the question, do the values a = 987, b = 610, n = 14 work and is that the right answer?
– user130306
3 hours ago
great, thank you again. and for the second part of the question, do the values a = 987, b = 610, n = 14 work and is that the right answer?
– user130306
3 hours ago
using a brute force approach
MaximalBy[Reverse/@Subsets[Range[1000],2], Length[eaSteps[#]]&] does give 987, 610 and Length[eaSteps[987, 610]] gives 15.– kglr
2 hours ago
using a brute force approach
MaximalBy[Reverse/@Subsets[Range[1000],2], Length[eaSteps[#]]&] does give 987, 610 and Length[eaSteps[987, 610]] gives 15.– kglr
2 hours ago
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shouldn't you change
a, 1, 25toa, 1, 100to get 5050 sublists?– kglr
4 hours ago
yes sorry, I meant a,1,100, i just mean, how do I know without running the code itself that I'd get 5050. what is the logic behind it?
– user130306
4 hours ago