Wordle

[Eddie]

1 hour ago, ketteringram said:

A one! I'm still waiting for that.

Baffled how you got Nerdle the other day or yesterday , from only having the Equal sign in the right place, to complete on the next line. 

I usually spend between 30 seconds and two minutes between lines. I'd say three minutes max. 

It took me about half an hour of analysis, and I think that there was only one possible solution at that stage.. I'll have a look for a nerdle archive and see if I can reverse-engineer my starting guess, then I'll see if I can work out from that what my thought processes would have been. I posted at 9:36 am on Monday, so there was no alcohol involved.

[TigerTedd]

1 hour ago, Eddie said:

It took me about half an hour of analysis, and I think that there was only one possible solution at that stage.. I'll have a look for a nerdle archive and see if I can reverse-engineer my starting guess, then I'll see if I can work out from that what my thought processes would have been. I posted at 9:36 am on Monday, so there was no alcohol involved.

That’s what I do. After a couple of incorrect guesses I just keep looking at different options and will normally find there is only 1 that fits. Takes me a while to really think about it. If it was a timed game I’d probably never complete it. 

[Eddie]

On 21/03/2022 at 09:36, Eddie said:

 

nerdlegame 61 2/6

?⬛?⬛??⬛?
????????

 

Here's my logic brain-dump, @ketteringram.

My first guess - 23+67=90. Note - I ALWAYS use that as my first guess because it uses 6 different digits, and the addition of two 2-digit numbers giving a third 2-digit number.

The guess gave me the 2, + and = in the right place and, critically, the 7 and 0 in the wrong place. I have also confirmed the equation as being of the form ab+cd=ef, so no -, no *, no /.

At this stage, I am yet to use 1, 4, 5 and 8 (and I have eliminated 3, 6 and 9).

We can determine immediately that position 4 cannot possibly be '8' because that would give us a 3-digit solution. Equally it can't be the '7' because this would, as a minimum, make position 7 a '9' - and we have eliminated that with the first guess.

My first guess put a '0 in position 8 - in the equation but in the wrong place. Convention, at least as far as nerdle goes, is that a 2-digit number does not start with '0'. This means that the misplaced '0' cannot go in positions 2, 4, 7 or 8 (the priority order is green-purple-black) - so it has to go in position 5, the only one left, and there is only one '0' (two or three zeroes is impossible because of the purple in position 8).

So let's put it ('0') in position 5 - fixed. (Edit: see footnote.)

If that's the case, then our equation now reads 2?+?0=??.

This assumption means that positions 2 and 8 must be the same digit (x+0=x), and we know that one of the other unknowns is a '7'.

Of the numbers which could go in position 4 (1, 2, 4 or 5), we can eliminate '1' straight away because that would make position 7 a '3'. Similarly, it cannot be a '4' (that would make position 7 a '6' - already determined as not possible).

Could it be a '5'?             2?+50=7?. Placing the '5' means we must put the '7' in position 7. Now we've successfully moved the '7' and '0' into new positions. Positions 2 and 8 could then each be '1', '2', '4' or '5' - a really hazardous situation because there's just no way of knowing which without entering a completely new equation (and costing a guess) to eliminate some of the possibilities, so I 'parked' it for a moment.

Could it be a '2'?            2?+20=4?. Placing the '2' means that the other two digits must be '7'.

So I had 5 possibilities - one using a '2' in position 4 or four using a '5' in position 4. I used to play poker, so the choice was obvious at this stage. If I'm wrong, then I have a 1-in-4 or a dummy guess - or both. It's a bit like betting against the 'big blind' when everyone else has folded and you're sitting A-K suited or a pair - raise, and if he calls, go all-in.

27+20=47 - w00t, and a nice two-go solution.

Edit: I have just realised that 20+27=47 also works, and that my elimination logic about placing the '0' in position 5 was in error. It could have been position 2 or 5 - but is 27+20=47 the same or different to 20+27=47? 

Edited March 23, 2022 by Eddie

[Eddie]

From the nerdle site...

"If the answer we're looking for is 10+20=30, then we will accept 20+10=30 too (unless you turn off 'commutative answers' in settings)."

I didn't even know there was a 'commutative answers' setting.

[ketteringram]

8 minutes ago, Eddie said:

Here's my logic brain-dump, @ketteringram.

My first guess - 23+67=90. Note - I ALWAYS use that as my first guess because it uses 6 different digits, and the addition of two 2-digit numbers giving a third 2-digit number.

The guess gave me the 2, + and = in the right place and, critically, the 7 and 0 in the wrong place. I have also confirmed the equation as being of the form ab+cd=ef, so no -, no *, no /.

At this stage, I am yet to use 1, 4, 5 and 8 (and I have eliminated 3, 6 and 9).

We can determine immediately that position 4 cannot possibly be '8' because that would give us a 3-digit solution. Equally it can't be the '7' because this would, as a minimum, make position 7 a '9' - and we have eliminated that with the first guess.

My first guess put a '0 in position 8 - in the equation but in the wrong place. Convention, at least as far as nerdle goes, is that a 2-digit number does not start with '0'. This means that the misplaced '0' cannot go in positions 2, 4, 7 or 8 (the priority order is green-purple-black) - so it has to go in position 5, the only one left, and there is only one '0' (two or three zeroes is impossible because of the purple in position 8).

So let's put it ('0') in position 5 - fixed.

If that's the case, then our equation now reads 2?+?0=??.

This assumption means that positions 2 and 8 must be the same digit (x+0=x), and we know that one of the other unknowns is a '7'.

Of the numbers which could go in position 4 (1, 2, 4 or 5), we can eliminate '1' straight away because that would make position 7 a '3'. Similarly, it cannot be a '4' (that would make position 7 a '6' - already determined as not possible).

Could it be a '5'?             2?+50=7?. Placing the '5' means we must put the '7' in position 7. Now we've successfully moved the '7' and '0' into new positions. Positions 2 and 8 could then each be '1', '2', '4' or '5' - a really hazardous situation because there's just no way of knowing which without entering a completely new equation (and costing a guess) to eliminate some of the possibilities, so I 'parked' it for a moment.

Could it be a '2'?            2?+20=4?. Placing the '2' means that the other two digits must be '7'.

So I had 5 possibilities - one using a '2' in position 4 or four using a '5' in position 4. I used to play poker, so the choice was obvious at this stage. If I'm wrong, then I have a 1-in-4 or a dummy guess - or both. It's a bit like betting against the 'big blind' when everyone else has folded and you're sitting A-K suited or a pair - raise, and if he calls, go all-in.

27+20=47 - w00t, and a nice two-go solution.

That's impressive method ? but not the puzzle I was on about. Unless you just quoted the wrong image. I was talking about puzzle number 62. 

But it's fine, you don't need to explain that one too ?

I do it very differently. I don't really try to get it in as few goes as possible. Though I suppose that's the aim, and probably the way most people attempt it. I just like to solve it before running out of lines. So if my first go doesn't really reveal much, I pretty much instantly fill in line two with a load of stuff I've not used in line one. I'm not the patient type! 

 

 

 

[Eddie]

2 hours ago, ketteringram said:

That's impressive method ? but not the puzzle I was on about. Unless you just quoted the wrong image. I was talking about puzzle number 62. 

But it's fine, you don't need to explain that one too ?

I do it very differently. I don't really try to get it in as few goes as possible. Though I suppose that's the aim, and probably the way most people attempt it. I just like to solve it before running out of lines. So if my first go doesn't really reveal much, I pretty much instantly fill in line two with a load of stuff I've not used in line one. I'm not the patient type! 

 

 

 

I'm retired - I have all the time in the world, and I am trying very hard to retain my mental faculties so I try to keep my mind occupied doing mathematical puzzles and games. The short-term memory is going though, and that, together with the loss of my short-term memory and sense of humour, is a worry.

I'll re-visit #62 and perform the same type of analysis tomorrow (If I remember - did I tell you I have a short-term memory problem?)

[Eddie]

Average...

Wordle 278 4/6

?⬜⬜⬜?
⬜⬜⬜⬜?
??⬜⬜⬜
?????

nerdlegame 64 3/6

??⬛?⬛?⬛⬛
?⬛????⬛⬛
????????

https://nerdlegame.com #nerdle

Daily Quordle #59
5️⃣6️⃣
8️⃣7️⃣
quordle.com
⬜⬜??? ⬜⬜⬜??
⬜⬜⬜⬜⬜ ⬜⬜⬜⬜⬜
⬜⬜⬜⬜⬜ ⬜⬜⬜⬜?
⬜?⬜⬜? ⬜?⬜⬜⬜
????? ⬜??⬜⬜
⬛⬛⬛⬛⬛ ?????

⬜⬜⬜⬜⬜ ⬜⬜⬜⬜⬜
⬜⬜⬜?⬜ ⬜⬜??⬜
⬜⬜??? ⬜⬜?⬜?
⬜⬜?⬜⬜ ⬜⬜?⬜⬜
⬜⬜⬜⬜⬜ ?⬜⬜⬜⬜
⬜⬜⬜⬜? ⬜⬜⬜⬜?
⬜??⬜⬜ ?????
????? ⬛⬛⬛⬛⬛
 

[TigerTedd]

9 hours ago, ketteringram said:

That's impressive method ? but not the puzzle I was on about. Unless you just quoted the wrong image. I was talking about puzzle number 62. 

But it's fine, you don't need to explain that one too ?

I do it very differently. I don't really try to get it in as few goes as possible. Though I suppose that's the aim, and probably the way most people attempt it. I just like to solve it before running out of lines. So if my first go doesn't really reveal much, I pretty much instantly fill in line two with a load of stuff I've not used in line one. I'm not the patient type! 

 

 

 

That’s funny, that’s exactly my thoughts on quordle and wordle. I’m not interested in doing it in as few guesses as possible, just doing it within the guesses. But with nerdle, I like to challenge myself to do it in as few guesses as possible.

[TigerTedd]

9 hours ago, Eddie said:

Here's my logic brain-dump, @ketteringram.

My first guess - 23+67=90. Note - I ALWAYS use that as my first guess because it uses 6 different digits, and the addition of two 2-digit numbers giving a third 2-digit number.

The guess gave me the 2, + and = in the right place and, critically, the 7 and 0 in the wrong place. I have also confirmed the equation as being of the form ab+cd=ef, so no -, no *, no /.

At this stage, I am yet to use 1, 4, 5 and 8 (and I have eliminated 3, 6 and 9).

We can determine immediately that position 4 cannot possibly be '8' because that would give us a 3-digit solution. Equally it can't be the '7' because this would, as a minimum, make position 7 a '9' - and we have eliminated that with the first guess.

My first guess put a '0 in position 8 - in the equation but in the wrong place. Convention, at least as far as nerdle goes, is that a 2-digit number does not start with '0'. This means that the misplaced '0' cannot go in positions 2, 4, 7 or 8 (the priority order is green-purple-black) - so it has to go in position 5, the only one left, and there is only one '0' (two or three zeroes is impossible because of the purple in position 8).

So let's put it ('0') in position 5 - fixed. (Edit: see footnote.)

If that's the case, then our equation now reads 2?+?0=??.

This assumption means that positions 2 and 8 must be the same digit (x+0=x), and we know that one of the other unknowns is a '7'.

Of the numbers which could go in position 4 (1, 2, 4 or 5), we can eliminate '1' straight away because that would make position 7 a '3'. Similarly, it cannot be a '4' (that would make position 7 a '6' - already determined as not possible).

Could it be a '5'?             2?+50=7?. Placing the '5' means we must put the '7' in position 7. Now we've successfully moved the '7' and '0' into new positions. Positions 2 and 8 could then each be '1', '2', '4' or '5' - a really hazardous situation because there's just no way of knowing which without entering a completely new equation (and costing a guess) to eliminate some of the possibilities, so I 'parked' it for a moment.

Could it be a '2'?            2?+20=4?. Placing the '2' means that the other two digits must be '7'.

So I had 5 possibilities - one using a '2' in position 4 or four using a '5' in position 4. I used to play poker, so the choice was obvious at this stage. If I'm wrong, then I have a 1-in-4 or a dummy guess - or both. It's a bit like betting against the 'big blind' when everyone else has folded and you're sitting A-K suited or a pair - raise, and if he calls, go all-in.

27+20=47 - w00t, and a nice two-go solution.

Edit: I have just realised that 20+27=47 also works, and that my elimination logic about placing the '0' in position 5 was in error. It could have been position 2 or 5 - but is 27+20=47 the same or different to 20+27=47? 

When you have eliminated all which is impossible, then whatever remains, however improbable, must be the truth.

[Eddie]

3 hours ago, TigerTedd said:

When you have eliminated all which is impossible, then whatever remains, however improbable, must be the truth.

Elementary, my dear TigerTedd.

[May Contain Nuts]

Just did an Octordle and 4 of the words were 'Aloft Pudgy Groin Flute'

Thought it'd fit in with some of the willy waving here ?

Edited March 24, 2022 by Coconut's Beard

[TigerTedd]

For once I got something worth a share on quordle: 

Daily Quordle #59
4️⃣7️⃣
6️⃣5️⃣
quordle.com
⬜??⬜? ⬜?⬜⬜?
⬜⬜⬜⬜⬜ ⬜⬜⬜⬜?
⬜???⬜ ⬜??⬜⬜
????? ⬜??⬜⬜
⬛⬛⬛⬛⬛ ⬜⬜?⬜⬜
⬛⬛⬛⬛⬛ ?⬜⬜⬜⬜
⬛⬛⬛⬛⬛ ?????

⬜⬜⬜⬜⬜ ?⬜⬜⬜⬜
⬜?⬜?? ⬜⬜⬜??
⬜⬜⬜⬜? ⬜⬜⬜⬜⬜
⬜⬜⬜⬜⬜ ?⬜⬜⬜⬜
⬜??⬜⬜ ?????
????? ⬛⬛⬛⬛⬛

[Eddie]

Not straightforward tonight by any means...

Wordle 279 3/6

⬜?⬜⬜?
⬜?⬜⬜?
?????

nerdlegame 65 4/6

?⬛⬛⬛???⬛
??⬛⬛??⬛⬛
??⬛⬛???⬛
????????

https://nerdlegame.com #nerdle

Daily Quordle #60
4️⃣8️⃣
7️⃣6️⃣
quordle.com
?⬜⬜⬜⬜ ??⬜⬜⬜
⬜?⬜⬜⬜ ⬜⬜???
⬜⬜⬜?? ⬜⬜?⬜⬜
????? ?⬜⬜⬜⬜
⬛⬛⬛⬛⬛ ⬜⬜?⬜⬜
⬛⬛⬛⬛⬛ ⬜⬜?⬜⬜
⬛⬛⬛⬛⬛ ⬜⬜⬜⬜⬜
⬛⬛⬛⬛⬛ ?????

⬜⬜⬜⬜? ⬜⬜⬜⬜⬜
⬜⬜⬜⬜⬜ ⬜⬜⬜?⬜
??⬜?⬜ ???⬜⬜
⬜⬜⬜?⬜ ⬜⬜⬜⬜⬜
??⬜⬜⬜ ????⬜
??⬜⬜⬜ ?????
????? ⬛⬛⬛⬛⬛
 

 

[Mrs Cone]

That 2nd one was hard x

Daily Quordle #60
7️⃣8️⃣
5️⃣4️⃣
quordle.com
⬜⬜??⬜ ⬜?⬜⬜⬜
⬜⬜⬜⬜⬜ ⬜⬜⬜⬜⬜
⬜⬜⬜⬜? ?⬜?⬜⬜
⬜⬜⬜⬜⬜ ⬜⬜?⬜⬜
⬜⬜?⬜⬜ ⬜⬜⬜⬜⬜
?⬜??? ??⬜⬜⬜
????? ?⬜⬜⬜⬜
⬛⬛⬛⬛⬛ ?????

⬜⬜⬜?⬜ ⬜⬜⬜⬜?
⬜⬜?⬜⬜ ⬜?⬜⬜⬜
⬜?⬜?⬜ ⬜???⬜
??⬜⬜⬜ ?????
????? ⬛⬛⬛⬛⬛
 

[uttoxram75]

48 minutes ago, Mrs Cone said:

That 2nd one was hard x

Daily Quordle #60
7️⃣8️⃣
5️⃣4️⃣
quordle.com
⬜⬜??⬜ ⬜?⬜⬜⬜
⬜⬜⬜⬜⬜ ⬜⬜⬜⬜⬜
⬜⬜⬜⬜? ?⬜?⬜⬜
⬜⬜⬜⬜⬜ ⬜⬜?⬜⬜
⬜⬜?⬜⬜ ⬜⬜⬜⬜⬜
?⬜??? ??⬜⬜⬜
????? ?⬜⬜⬜⬜
⬛⬛⬛⬛⬛ ?????

⬜⬜⬜?⬜ ⬜⬜⬜⬜?
⬜⬜?⬜⬜ ⬜?⬜⬜⬜
⬜?⬜?⬜ ⬜???⬜
??⬜⬜⬜ ?????
????? ⬛⬛⬛⬛⬛
 

[uttoxram75]

Just now, uttoxram75 said:

On behalf of the forum I would like to apologise for the childish puerility of my post and would like to assure the Cone family I only did it to stop other posters from posting equally ridiculous 1970s type rubbish.

??

[TigerTedd]

Wordle and quordle both nearly did for me today, tricky ones. 

but the nerdle is a thing of beauty. 

nerdlegame 66 2/6

⬛⬛?⬛⬛???
????????

Wordle 280 5/6

?⬛⬛⬛?
⬛?⬛?⬛
⬛???⬛
⬛⬛⬛⬛⬛
?????

Daily Quordle #61
9️⃣5️⃣
8️⃣3️⃣
quordle.com
⬜⬜⬜⬜? ⬜⬜⬜⬜⬜
⬜⬜⬜⬜⬜ ?⬜⬜?⬜
⬜⬜⬜⬜? ⬜⬜⬜??
⬜⬜?⬜? ???⬜⬜
⬜⬜?⬜⬜ ?????
⬜⬜?⬜? ⬛⬛⬛⬛⬛
⬜⬜⬜?⬜ ⬛⬛⬛⬛⬛
⬜⬜??⬜ ⬛⬛⬛⬛⬛
????? ⬛⬛⬛⬛⬛

⬜?⬜⬜? ⬜??⬜⬜
⬜⬜?⬜⬜ ⬜⬜⬜?⬜
⬜?⬜⬜⬜ ?????
⬜⬜⬜⬜? ⬛⬛⬛⬛⬛
⬜⬜⬜⬜⬜ ⬛⬛⬛⬛⬛
⬜???? ⬛⬛⬛⬛⬛
⬜?⬜?? ⬛⬛⬛⬛⬛
????? ⬛⬛⬛⬛⬛

[Eddie]

Likewise, a struggle. Wordle was one of those occasions when I possibly benefited from a 'dummy' guess 2, even though I had a 'right place' letter. With quordle, I thought a failure was nearly certain.

Wordle 280 4/6

⬜?⬜⬜?
⬜?⬜⬜⬜
??⬜⬜?
?????

Daily Quordle #61
9️⃣7️⃣
6️⃣8️⃣
quordle.com
⬜⬜⬜⬜? ⬜⬜⬜⬜⬜
⬜⬜⬜?⬜ ????⬜
⬜⬜⬜⬜⬜ ⬜⬜⬜⬜⬜
⬜⬜??⬜ ⬜⬜⬜?⬜
⬜⬜?⬜⬜ ⬜????
⬜⬜??⬜ ⬜⬜⬜⬜⬜
⬜⬜?⬜⬜ ?????
⬜⬜⬜⬜? ⬛⬛⬛⬛⬛
????? ⬛⬛⬛⬛⬛

⬜⬜⬜?? ⬜⬜⬜?⬜
⬜⬜⬜⬜⬜ ⬜⬜??⬜
⬜?⬜?⬜ ⬜⬜??⬜
?⬜?⬜⬜ ⬜⬜⬜?⬜
⬜⬜⬜⬜⬜ ⬜??⬜⬜
????? ⬜⬜⬜⬜?
⬛⬛⬛⬛⬛ ⬜??⬜⬜
⬛⬛⬛⬛⬛ ?????
 

nerdlegame 66 3/6

⬛??⬛??⬛⬛
??⬛???⬛?
????????

https://nerdlegame.com #nerdle

[Mostyn6]

Quordle top right did me today. My last two guesses had the correct final four letters. I opted for the practical object as last guess.

[ketteringram]

43 minutes ago, Mostyn6 said:

Quordle top right did me today. My last two guesses had the correct final four letters. I opted for the practical object as last guess.

I did try the top right one first, got it on the fourth line.

It was the top left one that did me. ?