Welcome to our gaming guide for “A Little to the Left: Daily Tidy Guide”! In this article, we will provide a step-by-step logic for solving the Calendar type Daily Tidy. We will also include helpful examples and explanations to make the process easier to understand. Let’s get started!

### Intro

(the one where you have to place stickers on a calendar)

NOTE: If you need more help than the general principles and examples below can solve for you, then your best bet is to post a SCREENSHOT in the Discussions in the game forum. And remember that we all get different variations on the Daily Theme. So your specific Calendar type Daily Tidy will __not__ be identical to someone else’s for that same date.

———————————————————-

**I’ll go through the main principles first, with some examples.And then I’ll show ‘how’ on an actual screenshot of a Calendar Daily Tidy.**

(*I’ll try to add more screenshots of different variations later, but will have to wait for that Daily type to come around again a couple of times. *)

### Main principles

2) It’s a simple recurring pattern of X days between repeats of the same icon/sticker.

3) So the point of the puzzle is to find the ‘X’ that works for that specific puzzle.

NOTE that it’s the __same ‘X’ for ALL your sticker types__. So once you’ve worked out the correct ‘X’ for one of your sticker types for that Calendar, then you’ve done all the hard work. And just need to keep a steady count for the rest.

4) All stickers must be placed on the calendar.

5) How to work out your ‘X’ , step-by-step:

OK – so start by looking at the calendar you’ve got.

There are various stickers on the calendar already, and a bunch of stickers to the left and right of your calendar that need to be placed.

Already on the calendar there is AT LEAST one pair of identical stickers already placed. Count how many days are between them.

Now some slightly differing examples, depending on the type of pattern you happen to get. I’ve put the most details into Example A in the step-by-step. And in the other examples just show some variations in what you may be seeing, and how to figure out that ‘X’.

EXAMPLE A:

- You see a pink rose on the 2nd and the same pink rose on 15th. That’s 13 days in between.
- Now look at your unplaced stickers – how many pink roses are there. Ok – so 1 more to place of the pink roses.
- Perhaps there is also another pair of stickers already on the calendar, and they are 26 days apart. Let’s say, a red dot on 3rd and the same red dot on 29th. And only 1 more to place of that sticker type too.
- Now – the purpose of the puzzle is to place the stickers with the same ‘X’ days between all stickers of the same type.
- There is only one way to do that in this example, which is 13 days between them all.

(BECAUSE – Roses: 13 days between 2nd and 15th , and then 13 days on for the last sticker. Red dot: 26 days between 3rd and 29th. Can’t place the last one 26 days on or back, because the calendar isn’t that long. So the extra one must go midway between the two, with 13 days between each. Therefore : X=13). - OK – then you’ve found your ‘X’ for that calendar – which is 13. So you now know that your stickers have to be placed so that there are 13 days between each sticker of the same design.
- So – for the pink roses: on 2nd, 15th and the last one you put on 28th (2+13=15, 15+13=28). And for the red dot: on 3rd, 29th and the last one you put midway on 16th (3+13=16, 16+13=29).
- For any sticker that is on the calendar only once already, you count 13 days forwards, or 13 backwards, or 13 both ways, depending on how many stickers you have to place, and where on the calendar the first one is (beginning, end, or middle (
*this is easier to explain with a concrete image, so see the picture and comments below.*). The puzzle generator works so that there is only one way that it’s possible to place them (unambiguous), so it should be obvious what is correct.

EXAMPLE B:

Already placed pairs of same sticker types are all 24 days apart, with only __one__ more sticker to be placed of those types. Then your ‘X’ for that calendar is 12. For all the sticker types.

As an example – a bird on 4th and 28th, and then the last one on 16th. (4+12=16, 16+12=24)

EXAMPLE C:

An already placed pair of same sticker type is placed 24 days apart, but there are __two !__ more stickers of the same type to be placed.

Then your ‘X’ is 8. For all the sticker types.

As an example – a sun on 6th and 30th, with two more needing to be placed: put them on 14th and 22th. (6+8=14, 14+8=22, 22+8=30)

And of course, there may be other variations, but they all work along these same rules of internal logic.

**Now, if you found that still a bit confusing just reading all this writing – here’s a picture and explanations to go with it:**

### Picture, with explanatory comments

On here – you can see that there are 4

__pairs__of stickers that are already placed on the calendar: The red rocks, grey rocks, pinky-orange squares and cat’s paws. And they all have

__one__more sticker of each as yet unplaced.

We can, for example, pick the red rock as starting point (

*works the same whichever we pick of these*):

- Red rock on 2nd and 26th, and one extra sticker to place. The two already placed are 24 days apart. Obviously not possible to place the remaining sticker another 24 days further on, or back, as the calendar doesn’t go that far. So the only possible is in the middle of the other two – so that they are all evenly spaced out. I.e. 12 days between them. So, red rocks on 2nd, 14th and 26th.
**And as simply as that – you’ve found your ‘X’ for this calendar : It’s 12.** - It works exactly the same way for the grey rocks on 3rd and 27th, and one extra to place: 24 days between them. Not possible to have the last sticker 24 days further on, or back, so has to go midway = 12 days between each. So, grey rocks on 3rd, 15th and 27th.
- Same for pinky-orange square on 4th and 28th, and one extra to place. Must go in the middle, 12 days between each. So, pinky-orange squares on 4th, 16th and 28th.
- Same for cat’s paw on 6th and 30th, and one extra to place. Must go in the middle, 12 days between each. So, cat’s paws on 6th, 18th and 30th.
Now, on to the stickers that are on the calendar only once from the puzzle starts.

**The thing to remember here is that your ‘X’ is the same for all the stickers**, so you’re still on 12 days between stickers of the same pattern: - Yellow rock on 11th, and one extra sticker to place. You can’t go 12 days back with your extra sticker, because you’re only on the 11th. So you must go forward 12 days, to the 23rd. So, yellow rocks on 11th and 23rd.
- Squiggle on the 20th, and one extra sticker to place. You can’t go forward 12 days, since the calendar doesn’t go that far. So you have to go back 12 days, to the 8th. So, squiggles on 8th and 20th.
- Cat’s face on 21st, and one extra to place. Same as for the squiggle, you can’t go forward 12 days, so must go back. To the 9th. So, cat’s faces on 9th and 21st.
- Green-black-red rock on 22nd, and one extra to place. Same as above, can’t go forward, so must go back 12 days. To the 10th. So, green-red-black stones on the 10th and 22nd.
**And there you have it: All your stickers correctly placed. And Daily Tidy solved đź™‚**

———Note 1: I find it easiest to solve by finding the ‘X’ and then just counting for all the sticker types. Like I’ve done here.

But others may perhaps find it just as easy to first solve the ‘what distance apart are they’ (the ‘X’) for one of the sticker types (like for example the red rocks in this puzzle here). And then place the rest according to how far away from that sticker type the others are. Like: a grey rock comes right after a red rock, and then pinky-orange square right after that again, and so on. That’s based on exactly the same logical premise for the solution – that all the stickers follow the same simple repetetive pattern and same distance between them.Note 2: At present, there is nothing stopping you from placing more than one sticker on the same date on the calendar. I myself have never had a Calendar Daily where the correct solution meant having to place more than one sticker on one day. I had heard that other people had done that sometimes. But I am unsure now whether this is part of intended solutions, or just some sort of ‘short-circuiting’ of the puzzle. Devs are onto the issue now of ‘doubling the pattern’, which was not intended as a possible solution to the puzzle type. So at this point, I don’t know if the fix that is coming for this issue, also will remove the possibility of placing more than one sticker on a day. Or not.

### Work in progress, and please leave a comment if needs correcting anywhere

I’m an old lady with clunky hands on the keyboard, so I apologise for any typos that may have crept in despite my best efforts.

And if I have made any mistakes in here – or you see anything that needs amending or appending, then please leave a message in the Comments section below.

Hope this can be of help to people struggling to figure out the internal logic to this puzzle type. It’s really not very hard, once you’ve got your head round it. So I hope I have been able to explain things in an understandable way!

And that wraps up our share on “A Little to the Left: Daily Tidy Guide”. If you have any additional insights or tips to contribute, don’t hesitate to drop a comment below. For a more in-depth read, you can refer to the original article here by oldladygreybun, who deserves all the credit. Happy gaming!