Is there a way for recurring Todoist tasks to repeat a task on a specific day of the week that is longer than a month per part?

For example, a task that repeats every 4 months on the 1st Saturday.

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# Tag: day

## Todo is recurring task every x months on a specific day of the week

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## Can I configure my day view in Google Calendar to show only certain hours?

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## Unity – How can I tell if a game object is touching another game object on a particular day?

## Achievement – Advent of Code 2019 Day 3: Crossed wires

## dnd 5e – Can my artisan use an infusion to make a hold bag every day to make a profit?

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## Excel counting numbers considering today's day

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Is there a way for recurring Todoist tasks to repeat a task on a specific day of the week that is longer than a month per part?

For example, a task that repeats every 4 months on the 1st Saturday.

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I have a lot of 15-minute meetings and blocks of time that I reserve for the small task schedule. However, when I look at my day in the Google Calendar, they all overlap. This is because there is not enough vertical space for rows with a granularity of 15 minutes. At least half of the storage space is wasted, however, when I never / rarely have anything on the calendar (e.g. from 8 to 9 p.m.). If my calendar only went from 9 a.m. to 8 p.m., a 15 minute appointment would be the same size as a 30 minute appointment and nothing would have to overlap.

Look at the attached picture to see what I mean. Is there a way to show only certain hours in the day view and thus enlarge the vertical space that every minute occupies?

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I've seen similar questions here on Stack Exchange – Game Development, but none of the answers to these questions helped my problem. I work in Unity 2019.2.18f1 with the Visual Studios 2019 community.

What I want to know to know when a game object (the player) is touching another game object with the tag name "Block". If the player collides with a "block", he should be placed: `jumpPossible = true;`

But when it leaves the block (by jumping or falling) `jumpPossible = false;`

What I currently have in mine `PlayerController`

Script (the important stuff anyway):

```
private void OnCollisionEnter(Collision collision) {
if (collision.gameObject.CompareTag("Block")) {
jumpPossible = true;
}
}
private void OnCollisionExit(Collision collision) {
if (collision.gameObject.CompareTag("Block")) {
jumpPossible = false;
}
}
```

The problem I have is when the player is in one block and then touching another block. If it leaves one of the blocks, the following is set: `jumpPossible = false;`

and you can't jump like that.

I want this script to run whenever the player touches *any* Gameobject with the tag "Block" should allow jumping.

Find the intersection of two paths closest to the origin in a Manhattan grid.`input_3.txt`

You will find here

The prompts can be found in the original problem description

```
import re
with open("input_3.txt", "r") as f:
coordinates = (wire.split(",") for wire in f.read().splitlines())
class Point:
def __init__(self, x: int, y: int, cumulative_steps: int):
self.x = x
self.y = y
self.cumulative_steps = cumulative_steps
def __eq__(self, other):
return (self.x, self.y) == (other.x, other.y)
def __hash__(self):
return hash((self.x, self.y))
def __repr__(self):
return f"x:{self.x}, y:{self.y}"
class WirePath:
def __init__(self, **kwargs):
self.origin = Point(0, 0, 0)
self.path = kwargs.get("path", None)
self.points = (
self.origin,
)
self.current_point = None
if self.path:
self.generate_travelled_points(self.path)
def travel(self, direction, velocity):
start = self.origin if not self.current_point else self.current_point
# calculate cumulative steps for each point added
if direction in "RL":
# if R subtract negative (add), if L subtract (minus)
pos = -1 if direction == "R" else 1
# add all points between the two segments
new_points = (
Point(start.x - (pos * p), start.y, start.cumulative_steps + p)
for p in range(1, abs(velocity) + 1)
)
else:
# if U subtract negative (add), if D subtract (minus)
pos = -1 if direction == "U" else 1
new_points = (
Point(start.x, start.y - (pos * p), start.cumulative_steps + p)
for p in range(1, abs(velocity) + 1)
)
self.points.extend(new_points)
self.current_point = self.points(-1)
def generate_travelled_points(self, vectors: list):
for vector in vectors:
# extract the direction & velocity
r = re.compile("((a-zA-Z)+)((0-9)+)")
m = r.match(vector)
direction = m.group(1)
velocity = int(m.group(2))
# add the point to self.points
self.travel(direction, velocity)
def manhattan_distance_between_points(a, b):
return abs(a(0) - b(0)) + abs(a(1) - b(1))
wire1 = WirePath(path=coordinates(0))
wire2 = WirePath(path=coordinates(1))
intersections = set((x for x in wire1.points if x.cumulative_steps != 0)).intersection(
(x for x in wire2.points if x.cumulative_steps != 0)
)
intersections_with_distance_from_origin = (
((x.x, x.y), manhattan_distance_between_points((x.x, x.y), (0, 0)))
for x in intersections
if x.cumulative_steps != 0
)
# part 1
closest_intersection = min(intersections_with_distance_from_origin, key=lambda t: t(1))
print(closest_intersection)
intersection_pairs = ()
for intersection in intersections:
# find points from each wire that matches the hash
wire1_intersects = min(
(i for i in wire1.points if i == intersection), key=lambda t: t.cumulative_steps
)
wire2_intersects = min(
(i for i in wire2.points if i == intersection), key=lambda t: t.cumulative_steps
)
intersection_pairs.append((wire1_intersects, wire2_intersects))
min_steps = min(
intersection_pairs, key=lambda t: t(0).cumulative_steps + t(1).cumulative_steps
)
min_combined_steps_wire1 = min_steps(0).cumulative_steps
min_combined_steps_wire2 = min_steps(1).cumulative_steps
comb = min_combined_steps_wire1 + min_combined_steps_wire2
# part 2
print(
f"Min combined steps: {comb}, wire1: {min_combined_steps_wire1}, wire2: {min_combined_steps_wire2} at {min_steps}"
)
```

My code is pretty slow (~ 7s on my PC) and I feel like it's a combination

a) how to store all points in memory instead of using derived points by storing line fragments instead, and

b) my wrong handling of the set iteration in the last part, trying to find the respective pair of points (one for each wire) in order to get the respective one `cumulative_steps`

.

I was a little confused about how best to handle this because a wire can cross an intersection multiple times (though I believe in this record that ultimately didn't appear?), And as a result, I didn't just have to find the intersections, but also the intersections and the minimum steps required for each wire to get to that intersection.

in the *Eberron: Uprising from the last war*, p. 57, under the Artificer class description, is in the Infuse Item section:

## Infuse an object

Whenever you have finished a long pause, you can touch a non-magical object and infuse it with one of your artificial infusions to turn it into a magical object. (…)

Your infusion remains in one item indefinitely (…)

You can infuse more than one non-magical object at the end of a long pause. The maximum number of objects is shown in the Infused Elements column of the Artificer table. You have to touch each of the objects and

Each of your infusions can only be in one object. (…) If you try to exceed your maximum number of infusions, the oldest infusion ends immediately and then the new infusion applies.

It becomes clear that you cannot use the same infusion for one object, at most one *Bag of holding* at a time and if you created another, the first would cease to be one *Bag of holding*.

If you skipped the city after selling one before making a new one (making the one you sell back an everyday bag), you could theoretically make a profit, but your reputation could soon catch up with you …

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I have an Excel spreadsheet that looks like this:

In the "Total" column, I have the total number of minutes that have to be run.

In the days after the "Todo" column, I have the number of minutes that must be run on that day. Now I want to calculate the number of minutes remaining from TODAY in the "Todo" column.

For example. on today (26.03.2020) I would like to see 65 in the todo column (= 100 – 35). How can I do that?

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