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Good point.
Can't blame comcast ???
Okay, possibly, maybe, the Landlord decided, "We will only have one Internet provider, we don't want prospective residents to have a choice of what they like"
At least here where I am (leaving soon) I could choose between the two.
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AT&T is not an option.
I saw that EPB (The electric utility) had a Fiber signal for sale. I got happy, for a moment.
Nope, the agent on the phone specifically told me "we are not allowed to go in there".
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New digs in Tennessee for 3 months.
Need Electric utility account.
Is EPB the only game in town ?
Tried asking Google, ha ha.
modified 17-Sep-15 16:24pm.
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Homework problem.
Q: On the (x,y) coordinate system, you start at (0,0) and want to get to (14,14). At each step n, you move exactly n steps either right or up. So initially, you can only move 1 unit right or up, then 2, then 3, etc. How many paths are there to (14,14)?
A: He got one on his own and he understood the symmetry and got 2. I told him my answer and why.
Anyone wants to try?
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7 8 9
Mongo: Mongo only pawn... in game of life.
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I get 8 paths. You can move up to the point where the sum of 1-n = 28. This happens at n = 7. (1 + 2 + 3 + 4 + 5 + 6 + 7 = 28).
So the question is how many ways can you divide numbers 1-7 into two groups of 14? I count 8 ways.
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Didn't think I had to spell that one out:
761
752
743
7412
653
6521
6431
5432
For each of those they add up to 14. The missing numbers also add to 14.
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But you can only start with one step.
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That's completely irrelevant.
761 => 1U, 2R, 3R, 4R, 5R, 6U, 7U.
752 => 1R, 2U, 3R, 4R, 5U, 6R, 7U.
743 => 1R, 2R, 3U, 4U, 5R, 6R, 7U.
7412 => 1U, 2U, 3R, 4U, 5R, 6U, 7U.
....
Follow my previous post. Simply put the 1 first instead of the order I have. I have them in a more mathematical order that I used to split the numbers into the two groups. Order does not matter.
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My bad. I missed a pair.
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Ha ha, I did too. I was with you all the way. I was all, "Yeah, you tell 'im, Bassam!"
Until you said that and I rechecked my answers.
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To err is human. To umm is herman.
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That's assuming you make a wrong turn at Albuquerque.
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Oh never mind, reread the question..
edit: 8 then. And since you seem to want to have them listed:
R1 R2 U3 R4 U5 U6 R7
R1 R2 U3 U4 R5 R6 U7
R1 U2 R3 R4 U5 R6 U7
R1 U2 U3 U4 U5 R6 R7
U1 R2 R3 R4 R5 U6 U7
U1 R2 U3 U4 R5 U6 R7
U1 U2 R3 R4 U5 U6 R7
U1 U2 R3 U4 R5 R6 U7
the numbers are superfluous of course, but they prevent taking a wrong turn at Albuquerque..
modified 17-Sep-15 13:38pm.
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I guess I missed something. I thought 'n' was a specific number to be repeated at each step
so the answer would be only the factors of 14. Therefore only these 'n' steps would be allowed.
1,2,7, and 14 - 2R, 2U, etc... or 7R,7U ...
So the number of paths would be 4
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Shouldn't it be 16? If there are 8 distinct groups of 3-4 numbers that sum to 14, one path will start with the first move going up and the next will start with the first move going right.
Incidentally, here is my code:
function combine(a, min) {
var fn = function(n, src, got, all) {
if (n == 0) {
if (got.length > 0) {
all[all.length] = got;
}
return;
}
for (var j = 0; j < src.length; j++) {
fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all);
}
return;
}
var all = [];
for (var i = min; i < a.length; i++) {
fn(i, a, [], all);
}
all.push(a);
return all;
}
function arr_sum(arr) {
var n = 0;
for (var i = 0; i < arr.length; i++) {
n += arr[i];
}
return n;
}
k=[1,2,3,4,5,6,7];
combine(k, 1).filter(function(v, i) {
return 14 == arr_sum(v)
})
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No. You always start with 1 step.
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I think there are 16 distinct paths, as below
R1U2R3R4U5R6U7 U1R2U3U4R5U6R7
R1R2U3U4R5R6U7 U1U2R3R4U5U6R7
R1R2U3R4U5U6R7 U1U2R3U4R5R6U7
U1U2R3U4R5R6U7 R1R2U3R4U5U6R7
U1U2R3R4U5U6R7 R1R2U3U4R5R6U7
U1R2U3U4R5U6R7 R1U2R3R4U5R6U7
R1U2U3U4U5R6R7 U1R2R3R4R5U6U7
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Some dupes. First and third to last, for example.
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