Solving Logic Puzzles Using R

If efficiency is not your primary concern, then here is a pretty straightforward way to brute force the results: just generate all possible combinations and then filter out the ones that do not satisfy the conditions.

library(dplyr)

dt <- purrr::cross_df(list(
  name = list(c("Bill", "Ernie", "Oscar", "Sammy", "Tony")),
  nickname = combinat::permn(c("Slats", "Stretch", "Tiny", "Tower", "Tree")), 
  height = combinat::permn(c(6.6, 6.5, 6.3, 6.1, 6))
))

dt %>%  
  group_by(id = (seq_len(n()) - 1L) %/% 5L) %>% 
  filter(
    height[name == "Oscar"] > height[nickname == "Tree"], 
    height[nickname == "Tree"] > height[name == "Tony"], 
    height[name == "Bill"] > height[name == "Sammy"], 
    height[name == "Bill"] < height[nickname == "Slats"], 
    nickname[name == "Tony"] != "Tiny",
    height[nickname == "Stretch"] > height[name == "Oscar"], 
    height[nickname == "Stretch"] < 6.6
  )

dt looks like this

# A tibble: 72,000 x 3
   name  nickname height
   <chr> <chr>     <dbl>
 1 Bill  Slats       6.6
 2 Ernie Stretch     6.5
 3 Oscar Tiny        6.3
 4 Sammy Tower       6.1
 5 Tony  Tree        6  
 6 Bill  Slats       6.6
 7 Ernie Stretch     6.5
 8 Oscar Tiny        6.3
 9 Sammy Tree        6.1
10 Tony  Tower       6  
# ... with 71,990 more rows

Output is

# A tibble: 5 x 4
# Groups:   id [1]
  name  nickname height    id
  <chr> <chr>     <dbl> <int>
1 Bill  Stretch     6.5 14398
2 Ernie Slats       6.6 14398
3 Oscar Tiny        6.3 14398
4 Sammy Tree        6.1 14398
5 Tony  Tower       6   14398

Here is a tidyverse solution which does the heavy lifting and leaves the very last bit to us.

Lets start with the setup.

library(tidyverse)

players = c("Bill", "Ernie", "Oscar", "Sammy", "Tony")
nick_nms = c("Slats", "Stretch", "Tiny", "Tower", "Tree")
heights = c(6.6, 6.5, 6.3, 6.1, 6.0)

In the first step we combine conditions 1 and 4:

# Result 1: Based on condition 1 & 4
res1 <- cross_df(list(Oscar = heights,
              Tree = heights,
              Tony = heights,
              Stretch = heights)
         ) %>% 
  filter(Oscar > Tree,
         Tree > Tony,
         Stretch > Oscar,
         Stretch < 6.6)
res1 
#> # A tibble: 1 x 4
#>   Oscar  Tree  Tony Stretch
#>   <dbl> <dbl> <dbl>   <dbl>
#> 1   6.3   6.1     6     6.5

In the second step we look at the result of condition 2 and also take result 1 into account:

# Result 2: Based on conditon 2
res2 <- cross_df(list(Slats = heights,
                      Bill = heights,
                      Sammy = heights)) %>% 
  filter(Slats > Bill,
         Bill > Sammy,
         # after result 1 add:
         !Bill %in% c(6.3, 6),
         !Sammy %in% c(6.3, 6)) 
res2
#> # A tibble: 1 x 3
#>   Slats  Bill Sammy
#>   <dbl> <dbl> <dbl>
#> 1   6.6   6.5   6.1

Now all we need to do is to reshape our results 1 and 2:

# reshape data:
res1_long <- res1 %>%
  pivot_longer(cols = everything(),
               values_to = "height")

res2_long <- res2 %>%
  pivot_longer(cols = everything(),
               values_to = "height")

dat2 <- res1_long %>% 
  bind_rows(res2_long) %>% 
  mutate(type = ifelse(name %in% players, "player", "nick")) %>% 
  pivot_wider(names_from = type,
              values_from = name)
dat2
#> # A tibble: 5 x 3
#>   height player nick   
#>    <dbl> <chr>  <chr>  
#> 1    6.3 Oscar  <NA>   
#> 2    6.1 Sammy  Tree   
#> 3    6   Tony   <NA>   
#> 4    6.5 Bill   Stretch
#> 5    6.6 <NA>   Slats

At this point we can infer the final result by keeping condition 3 in mind: Tony is not Tiny. Then the rest falls into place. Of course it would be nice to have an approach which calculates the final result. But anything I've come up with was so much more complicated compared to just taking condition 3 into account and fill the tibble manually:

dat2 %>% 
  # Fill in the last 
  mutate(player = ifelse(is.na(player),
                         "Ernie",
                         player),
         
         nick = case_when(is.na(nick) & player == "Oscar" ~ "Tiny",
                          is.na(nick) ~ "Tower",
                          TRUE ~ nick)
         )
#> # A tibble: 5 x 3
#>   height player nick   
#>    <dbl> <chr>  <chr>  
#> 1    6.3 Oscar  Tiny  
#> 2    6.1 Sammy  Tree   
#> 3    6   Tony   Tower  
#> 4    6.5 Bill   Stretch
#> 5    6.6 Ernie  Slats

^{Created on 2022-01-01 by the reprex package (v0.3.0)}

1. Write a function that translates nickname to name for a given allocation, e.g. realname <- function(nickname, i) ... where i is the allocation you're testing.

2. Write a function that translates name to height for a given allocation in that list, e.g. height <- function(name, j) ... where j is the entry in that list.

3. Write the 4 conditions in terms of these functions. E.g. condition 1 is

   height("Oscar", j) > height(realname("Tree", i), j) &&
   height(realname("Tree", i), j) > height("Tony", j)
   

4. Do for loops over all possible allocations i and j and run all 4 tests. If you ever find all tests coming out TRUE, you've got the solution.

Edited to add: A slightly simpler approach: just randomly permute the nicknames and the heights, and run the tests on each random selection. You'll probably run more tests than the loops (because you'll test some configurations more than once), but it shouldn't take long to cover all 14400 possibilities.