I have queue q of integer numbers stored in an array in the circular fashion
from front to rear that is..
f: 1
r: 8
array Q: 0 1 2 3 4 5 6 7 8 9
2 8 4 4 3 5 4
What is the array representation of queue q after i perform the following?
while q.front() is an even number do q.enqueue(q.dequeue()).
It will loop until it reaches number 3. So the array will be
0 1 2 3 4 5 6 7 8 9
4 4 3 5 4 2 8
Related
I have a complete binary tree of height 'h'.
How do I find 'h' number of unrelated partitions for this ?
NOTE:
Unrelated partition means no child can be present with its immediate parent.
There is a constraint on the number of nodes in each partition.
The difference of the maximum number nodes in a partition and the minimum number of nodes in the partition can either be 0 or 1.
Also, root is excluded from including in the partitions.
Who devised the problem probably had a more elegant solution in mind, but the following works.
Let's say we have h partitions numbered 1 to h, and that the nodes of partition n have value n. The root node has value 0, and does not participate in the partitions. Let's call a partition even if nis even, and odd if n is odd. Let's also number the levels of the complete binary tree, ignoring the root and starting from level 1 with 2 nodes. Level n has 2n nodes, and the complete tree has 2h+1-1 nodes, but only P=2h+1-2 nodes belong to the partitions (because the root is excluded). Each partition consists of p=⌊P/h⌋ or p=⌈P/h⌉ nodes, such that ∑ᵢpᵢ=P.
If the height h of the tree is even, put all even partitions into the even levels of the left subtree and the odd levels of the right subtee, and put all odd partitions into the odd levels of the left subtree and the even levels of the right subtree.
If h is odd, distribute all partitions up to partition h-1 like in the even case, but distribute partition h evenly into the last level of the left and right subtrees.
This is the result for h up to 7 (I wrote a tiny Python library to print binary trees to the terminal in a compact way for this purpose):
0
1 1
0
1 2
2 2 1 1
0
1 2
2 2 1 1
1 1 3 3 2 2 3 3
0
1 2
2 2 1 1
1 1 1 1 2 2 2 2
2 4 4 4 4 4 4 4 1 3 3 3 3 3 3 3
0
1 2
2 2 1 1
1 1 1 1 2 2 2 2
2 2 2 2 2 2 4 4 1 1 1 1 1 1 3 3
3 3 3 3 3 3 3 3 3 3 5 5 5 5 5 5 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5
0
1 2
2 2 1 1
1 1 1 1 2 2 2 2
2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1
1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 4 4 4 4 4 4 4 4 4 4
4 4 4 4 4 4 4 4 4 4 4 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 3 3 3 3 3 3 3 3 3 3 3 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
0
1 2
2 2 1 1
1 1 1 1 2 2 2 2
2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
3 3 3 3 3 3 3 3 3 3 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 4 4 4 4 4 4 4 4 4 4 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7
And this is the code that generates it:
from basicbintree import Node
for h in range(1, 7 + 1):
root = Node(0)
P = 2 ** (h + 1) - 2 # nodes in partitions
p = P // h # partition size (may be p or p + 1)
if h & 1: # odd height
t = (p + 1) // 2 # subtree tail nodes from split partition
n = (h - 1) // 2 # odd or even partitions in subtrees except tail
else: # even height
t = 0 # no subtree tail nodes from split partition
n = h // 2 # odd or even partitions in subtrees
s = P // 2 - t # subtree nodes excluding tail
r = s - n * p # partitions of size p + 1 in subtrees
x = [p + 1] * r + [p] * (n - r) # nodes indexed by subtree partition - 1
odd = [1 + 2 * i for i, c in enumerate(x) for _ in range(c)] + [h] * t
even = [2 + 2 * i for i, c in enumerate(x) for _ in range(c)] + [h] * t
for g in range(1, h + 1):
start = 2 ** (g - 1) - 1
stop = 2 ** g - 1
if g & 1: # odd level
root.set_level(odd[start:stop] + even[start:stop])
else: # even level
root.set_level(even[start:stop] + odd[start:stop])
print('```none')
root.print_tree()
print('```')
All trees produced up to height 27 have been programmatically confirmed to meet the specifications.
Some parts of the algorithm would need a proof, like, e.g., that it's always possible to choose an even size for the split partition in the odd height case, but this and other proofs are left as an exercise to the reader ;-)
I need to partition a set S={1, 2, 3, … , n} consisting of consecutive numbers such that each subset has has at least 2 elements (rule 1) and it consists of consecutive numbers (rule 2).
The rules are:
Each subset has at least two elements.
All elements of all subsets are consecutive.
All elements of S are included in the partition.
Examples:
There is 1 subset for n = 2:
1 2
There is 1 subset for n = 3:
1 2 3
There are 2 subset combinations for n = 4:
1 2 3 4
1 2 - 3 4
There are 3 subset combinations for n = 5:
1 2 3 4 5
1 2 - 3 4 5
1 2 3 - 4 5
There are 5 subset combinations for n = 6:
1 2 3 4 5 6
1 2 - 3 4 5 6
1 2 3 - 4 5 6
1 2 3 4 - 5 6
1 2 - 3 4 - 5 6
There are 8 subset combinations for n = 7:
1 2 3 4 5 6 7
1 2 - 3 4 5 6 7
1 2 3 - 4 5 6 7
1 2 3 4 - 5 6 7
1 2 3 4 5 - 6 7
1 2 - 3 4 - 5 6 7
1 2 - 3 4 5 - 6 7
1 2 3 - 4 5 - 6 7
There are 13 subset combinations for n = 8:
1 2 3 4 5 6 7 8
1 2 - 3 4 5 6 7 8
1 2 3 - 4 5 6 7 8
1 2 3 4 - 5 6 7 8
1 2 3 4 5 - 6 7 8
1 2 3 4 5 6 - 7 8
1 2 - 3 4 - 5 6 7 8
1 2 - 3 4 5 - 6 7 8
1 2 - 3 4 5 6 - 7 8
1 2 3 - 4 5 - 6 7 8
1 2 3 - 4 5 6 - 7 8
1 2 3 4 - 5 6 - 7 8
1 2 - 3 4 - 5 6 - 7 8
There are 21 subset combinations for n = 9:
1 2 3 4 5 6 7 8 9
1 2 - 3 4 5 6 7 8 9
1 2 3 - 4 5 6 7 8 9
1 2 3 4 - 5 6 7 8 9
1 2 3 4 5 - 6 7 8 9
1 2 3 4 5 6 - 7 8 9
1 2 3 4 5 6 7 - 8 9
1 2 - 3 4 - 5 6 7 8 9
1 2 - 3 4 5 - 6 7 8 9
1 2 - 3 4 5 6 - 6 7 9
1 2 - 3 4 5 6 7 - 8 9
1 2 3 - 4 5 - 6 7 8 9
1 2 3 - 4 5 6 - 7 8 9
1 2 3 - 4 5 6 7 - 8 9
1 2 3 4 - 5 6 - 7 8 9
1 2 3 4 - 5 6 7 - 8 9
1 2 3 4 5 - 6 7 - 8 9
1 2 - 3 4 - 5 6 - 7 8 9
1 2 - 3 4 - 5 6 7 - 8 9
1 2 - 3 4 5 - 6 7 - 8 9
1 2 3 - 4 5 - 6 7 - 8 9
There are 34 subset combinations for n = 10:
1 2 3 4 5 6 7 8 9 10
1 2 - 3 4 5 6 7 8 9 10
1 2 3 - 4 5 6 7 8 9 10
1 2 3 4 - 5 6 7 8 9 10
1 2 3 4 5 - 6 7 8 9 10
1 2 3 4 5 6 - 7 8 9 10
1 2 3 4 5 6 7 - 8 9 10
1 2 3 4 5 6 7 8 - 9 10
1 2 - 3 4 - 5 6 7 8 9 10
1 2 - 3 4 5 - 6 7 8 9 10
1 2 - 3 4 5 6 - 6 7 9 10
1 2 - 3 4 5 6 7 - 8 9 10
1 2 - 3 4 5 6 7 8 - 9 10
1 2 3 - 4 5 - 6 7 8 9 10
1 2 3 - 4 5 6 - 7 8 9 10
1 2 3 - 4 5 6 7 - 8 9 10
1 2 3 - 4 5 6 7 8 - 9 10
1 2 3 4 - 5 6 - 7 8 9 10
1 2 3 4 - 5 6 7 - 8 9 10
1 2 3 4 - 5 6 7 8 - 9 10
1 2 3 4 5 - 6 7 - 8 9 10
1 2 3 4 5 - 6 7 8 - 9 10
1 2 3 4 5 6 - 7 8 - 9 10
1 2 - 3 4 - 5 6 - 7 8 9 10
1 2 - 3 4 - 5 6 7 - 8 9 10
1 2 - 3 4 - 5 6 7 8 - 9 10
1 2 - 3 4 5 - 6 7 - 8 9 10
1 2 - 3 4 5 - 6 7 8 - 9 10
1 2 - 3 4 5 6 - 7 8 - 9 10
1 2 3 - 4 5 - 6 7 - 8 9 10
1 2 3 - 4 5 - 6 7 8 - 9 10
1 2 3 - 4 5 6 - 7 8 - 9 10
1 2 3 4 - 5 6 - 7 8 - 9 10
1 2 - 3 4 - 5 6 - 7 8 - 9 10
I didn't write them down here but there are 55 subset combinations for n = 11 and 89 subset combinations for n = 12.
I need to write a Visual Basic code listing all possible subset groups for n. I have been thinking on the solution for days but it seems that the solution of the problem is beyond my capacity. The number of required nested loops increases with n and I could not figure out how to program the nested loops with increasing number. Any help will be greatly appreciated.
After some research, I found out this is the problem of "compositions of n with all parts >1" and the total number of possible compositions are Fibonacci numbers (Fn-1 for n).
We already know the answer for these cases (as you wrote in your examples):
n=2
n=3
n=4
For n=5:
You can partition from 2: 1 2 - 3 4 5. This is like dividing the 5 member set into two sets, first one n=2, and second one n=3. We can now continue dividing each half, but we already know the solutions when n=2 and n=3!
You can partition from 3: 1 2 3 - 4 5. This is like dividing the 5 member set into two sets, first one n=3, and second one n=2. We can now continue dividing each half, but we already know the solution when n=2 and n=3!
For n=6:
You can partition into two sets from 2: 1 2 - 3 4 5 6. This is like dividing the 6 member set into two sets, first one n=2, and second one n=4. We can now continue dividing each half, but we already know the solution when n=2. Solve the second half by assuming n=4!
You can partition into two sets from 3: 1 2 3 - 4 5 6. This is like dividing the 6 member set into two sets, first one n=3, and second one n=3, We can now continue dividing each half, but we already know the solution when n=3 and n=3!
You can partition into two sets from 3: 1 2 3 4 - 5 6. This is like dividing the 6 member set into two sets, first one n=4, and second one n=2, We can now continue dividing each half. Solve the first half by setting n=4. For the second half, we already know the solution when n=2!
This is a simple recursion relationship. The general case:
Partition (S): (where |S|>4)
- For i from 2 to |S|-2, partition the given set into two halves: s1 and s2 from i (s1={1,...,i}, s2={i+1,...,n}), and print the two subsets as a solution.
- Recursively continue for each half by calling Partition(s1) and Partition(s2)
---
Another perhaps harder solution is to assume we are dividing the numbers 1 to n into n sections, where the length of each section can be either 0, 2, or a number greater than 2. In other words let xi be the length of each section:
x1 + x2 + ... xn = n, where the range of xi is: {0} + [2,n]
This is a system of linear non-equalities can can be solved by methods described here.
My answer to you is to try and come up with a recurrence relation of the given pattern. Think recursively. How can I break this problem down into smaller subproblems until reaching the smallest problem. Solve that smallest problem. After solving that smallest problem, think induction. Hypothesize on the what the nth step will be and how you will reach the (n+1)th step. Try to solve that (n+1)th step. Once you have come up with a recurrence relation on the pattern being given, it should not be too difficult to think about how to solve this pattern recursively. Instead of trying to use nested loops, this approach may be more intuitive.
In a programming event they asked this question.
Question:
1 2 3 4 5 6 7 8 9 10 J Q K.
Write and algorithm to sort the cards in reverse order in 7 steps.
K Q J 10 9 8 7 6 5 4 3 2 1.
I couldn't trace out the sorting alogrithm.
Which algoritm is used here?
I think it's something like quicksort, but with block moves:
1 2 3 4 5 6 7 8 9 10 J Q K
8 9 10 J Q K 7 1 2 3 4 5 6
J Q K 8 9 10 7 1 2 3 4 5 6
J Q K 8 9 10 7 4 5 6 1 2 3
K Q J 8 9 10 7 4 5 6 1 2 3
K Q J 10 9 8 7 4 5 6 1 2 3
K Q J 10 9 8 7 6 5 4 1 2 3
K Q J 10 9 8 7 6 5 4 3 2 1
Update: actually it might be even simplier: just swap 1 and K and 2 and Q, 3 and J etc. Just seven steps :)
If a step is a swap then Selection sort will order the list in 6 steps.
K Q J 10 9 8 7 6 5 4 3 2 1
1 Q J 10 9 8 7 6 5 4 3 2 K
1 2 J 10 9 8 7 6 5 4 3 Q K
1 2 3 10 9 8 7 6 5 4 J Q K
1 2 3 4 9 8 7 6 5 10 J Q K
1 2 3 4 5 8 7 6 9 10 J Q K
1 2 3 4 5 6 7 8 9 10 J Q K
Consider the following example (values in vectors are target practice results and I'm trying to automagically sort by shooting score). We generate three vectors. We sort values in columns 1:20 in ascending order and rows in descending order based on out.tot column.
# Generate data
shooter1 <- round(runif(n = 20, min = 1, max = 10))
shooter2 <- round(runif(n = 20, min = 1, max = 10))
shooter3 <- round(runif(n = 20, min = 1, max = 10))
out <- data.frame(t(data.frame(shooter1, shooter2, shooter3)))
colnames(out) <- 1:ncol(out)
out.sort <- t(apply(out, 1, sort, na.last = FALSE))
out.tot <- apply(out , 1, sum)
colnames(out.sort) <- 1:ncol(out.sort)
out2 <- cbind(out.sort, out.tot)
out3 <- apply(out2, 2, sort, decreasing = TRUE, na.last = FALSE)
out2 has row names attached while out3 lost them. The only difference is that I used MARGIN = 2, which is probably the culprit (because it takes in column by column). I can match rows by hand, but is there a way I can keep row names in out3 from disappearing?
> out2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 out.tot
shooter1 1 2 2 3 3 3 4 5 5 5 6 6 6 6 6 7 8 9 9 10 106
shooter2 1 3 3 3 3 4 4 4 5 5 5 5 5 6 7 8 8 9 9 10 107
shooter3 1 1 2 2 2 3 3 4 5 5 5 6 6 6 6 7 8 8 8 9 97
> out3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 out.tot
[1,] 1 3 3 3 3 4 4 5 5 5 6 6 6 6 7 8 8 9 9 10 107
[2,] 1 2 2 3 3 3 4 4 5 5 5 6 6 6 6 7 8 9 9 10 106
[3,] 1 1 2 2 2 3 3 4 5 5 5 5 5 6 6 7 8 8 8 9 97
If I understand your example, going from out2 to out3 you are sorting each column independently - meaning that the values on row 1 may not all come from the data generated from shooter1. It makes sense then that the rownames are dropped in as much as rownames are names of observations and you are no longer keeping data from one observation on one row.
I was asked to find a 11x11-grid containing the digits such that one can read the squares of 1,...,100. Here read means that you fix the starting position and direction (8 possibilities) and if you can find for example the digits 1,0,0,0,0,4 consecutively, you have found the squares of 1, 2, 10, 100 and 20. I made a program (the algorithm is not my own. I modified slightly a program which uses best-first search to find a solution but it is too slow. Does anyone know a better algorithm to solve the problem?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <vector>
#include <algorithm>
using namespace std;
int val[21][21];//number which is present on position
int vnum[21][21];//number of times the position is used - useful if you want to backtrack
//5 unit borders
int mx[4]={-1,0,1,0};//movement arrays
int my[4]={0,-1,0,1};
int check(int x,int y,int v,int m)//check if you can place number - if you can, return number of overlaps
{
int c=1;
while(v)//extract digits one by one
{
if(vnum[x][y] && (v%10)!=val[x][y])
return 0;
if(vnum[x][y])
c++;
v/=10;
x+=mx[m];
y+=my[m];
}
return c;
}
void apply(int x,int y,int v,int m)//place number - no sanity checks
{
while(v)//extract digits one by one
{
val[x][y]=v%10;
vnum[x][y]++;
v/=10;
x+=mx[m];
y+=my[m];
}
}
void deapply(int x,int y,int v,int m)//remove number - no sanity checks
{
while(v)
{
vnum[x][y]--;
v/=10;
x+=mx[m];
y+=my[m];
}
}
int best=100;
void recur(int num)//go down a semi-random path
{
if(num<best)
{
best=num;
if(best)
printf("FAILED AT %d\n",best);
else
printf("SUCCESS\n");
for(int x=5;x<16;x++) // 16 and 16
{
for(int y=5;y<16;y++)
{
if(vnum[x][y]==0)
putchar('.');
else
putchar(val[x][y]+'0');
}
putchar('\n');
}
fflush(stdout);
}
if(num==0)
return;
int s=num*num,t;
vector<int> poss;
for(int x=5;x<16;x++)
for(int y=5;y<16;y++)
for(int m=0;m<4;m++)
if(t=check(x,y,s,m))
poss.push_back((x)|(y<<8)|(m<<16)|(t<<24));//compress four numbers into an int
if(poss.size()==0)
return;
sort(poss.begin(),poss.end());//essentially sorting by t
t=poss.size()-1;
while(t>=0 && (poss[t]>>24)==(poss.back()>>24))
t--;
t++;
//t is now equal to the smallest index which has the maximal overlap
t=poss[rand()%(poss.size()-t)+t];//select random index>=t
apply(t%256,(t>>8)%256,s,(t>>16)%256);//extract random number
recur(num-1);//continue down path
}
int main()
{
srand((unsigned)time(0));//seed
while(true)
{
for(int i=0;i<21;i++)//reset board
{
memset(val[i],-1,21*sizeof(int));
memset(vnum[i],-1,21*sizeof(int));
}
for(int i=5;i<16;i++)
{
memset(val[i]+5,0,11*sizeof(int));
memset(vnum[i]+5,0,11*sizeof(int));
}
recur(100);
}
}
Using a random search so far I only got to 92 squares with one unused spot (8 missing numbers: 5041 9025 289 10000 4356 8464 3364 3249)
1 5 2 1 2 9 7 5 6 9 5
6 1 0 8 9 3 8 4 4 1 2
9 7 2 2 5 0 0 4 8 8 2
1 6 5 9 6 0 4 4 7 7 4
4 4 2 7 6 1 2 9 0 2 2
2 9 6 1 7 8 4 4 0 9 3
6 5 5 3 2 6 0 1 4 0 6
4 7 6 1 8 1 1 8 2 8 1
8 0 1 3 4 8 1 5 3 2 9
0 5 9 6 9 8 8 6 7 4 5
6 6 2 9 1 7 3 9 6 9
The algorithm basically uses as solution encoding a permutation on the input (search space is 100!) and then places each number in the "topmost" legal position. The solution value is measured as the sum of the squares of the lengths of the numbers placed (to give more importance to long numbers) and the number of "holes" remaining (IMO increasing the number of holes should raise the likehood that another number will fit in).
The code has not been optimized at all and is only able to decode a few hundred solutions per second. Current solution has been found after 196k attempts.
UPDATE
Current best solution with this approach is 93 without free holes (7 missing numbers: 676 7225 3481 10000 3364 7744 5776):
9 6 0 4 8 1 0 0 9 3 6
6 4 0 0 2 2 5 6 8 8 9
1 7 2 9 4 1 5 4 7 6 3
5 8 2 3 8 6 4 9 6 5 7
2 4 4 4 1 8 2 8 2 7 2
1 0 8 9 9 1 3 4 4 9 1
2 1 2 9 6 1 0 6 2 4 1
2 3 5 5 3 9 9 4 0 9 6
5 0 0 6 1 0 3 5 2 0 3
2 7 0 4 2 2 5 2 8 0 9
9 8 2 2 6 5 3 4 7 6 1
This is a solution (all 100 numbers placed) however using a 12x12 grid (MUCH easier)
9 4 6 8 7 7 4 4 5 5 1 7
8 3 0 5 5 9 2 9 6 7 6 4
4 4 8 3 6 2 6 0 1 7 8 4
4 8 4 2 9 1 4 0 5 6 1 4
9 1 6 9 4 8 1 5 4 2 0 1
9 4 4 7 2 2 5 2 2 5 0 0
4 6 2 2 5 8 4 2 7 4 0 2
0 3 3 3 6 4 0 0 6 3 0 9
9 8 0 1 2 1 7 9 5 5 9 1
6 8 4 2 3 5 2 6 3 2 0 6
9 9 8 2 5 2 9 9 4 2 2 7
1 1 5 6 6 1 9 3 6 1 5 4
It has been found using a truly "brute force" approach, starting from a random matrix and keeping randomly changing digits when that improved the coverage.
This solution has been found by an highly unrolled C++ program automatically generated by a Python script.
Update 2
Using an incremental approach (i.e. keeping a more complex data structure so that when changing a matrix element the number of targets covered can be updated instead than recomputed) I got a much faster search (about 15k matrices/second investigated with a Python implementation running with PyPy).
In a few minutes this version was able to find a 99 quasi-solution (a number is still missing):
7 0 5 6 5 1 1 5 7 1 6
4 6 3 3 9 8 8 6 7 6 1
3 9 0 8 2 6 1 1 4 7 8
1 1 0 8 9 9 0 0 4 4 6
3 4 9 0 4 9 0 4 6 7 1
6 4 4 6 8 6 3 2 5 2 9
9 7 8 4 1 1 4 0 5 4 2
6 2 4 1 5 2 2 1 2 9 7
9 8 2 5 2 2 7 3 6 5 0
3 1 2 5 0 0 6 3 0 5 4
7 5 6 9 2 1 6 5 3 4 6
UPDATE 3
Ok. After a some time (no idea how much) the same Python program actually found a complete solution (several ones indeed)... here is one
6 4 6 9 4 1 2 9 7 3 6
9 2 7 7 4 4 8 1 2 1 7
1 0 6 2 7 0 4 4 8 3 4
2 1 2 2 5 5 9 2 9 6 5
9 2 5 5 2 0 2 6 3 9 1
1 6 3 6 0 0 9 3 7 0 6
6 0 0 4 9 0 1 6 0 0 4
9 8 4 4 8 0 1 4 5 2 3
2 4 8 2 8 1 6 8 6 7 5
1 7 6 9 2 4 5 4 2 7 6
6 6 3 8 8 5 6 1 5 2 1
The searching program can be found here...
You've got 100 numbers and 121 cells to work with, so you'll need to be very efficient. We should try to build up the grid, so that each time we fill a cell, we attain a new number in our list.
For now, let's only worry about 68 4-digit numbers. I think a good chunk of the shorter numbers will be in our grid without any effort.
Start with a 3x3 or 4x4 set of numbers in the top-left of your grid. It can be arbitrary, or fine-tune for slightly better results. Now let's fill in the rest of the grid one square at a time.
Repeat these steps:
Fill an empty cell with a digit
Check which numbers that knocked off the list
If it didn't knock off any 4-digit numbers, try a different digit or cell
Eventually you may need to fill 2 cells or even 3 cells to achieve a new 4-digit number, but this should be uncommon, except at the end (at which point, hopefully there's a lot of empty space). Continue the process for the (few?) remaining 3-digit numbers.
There's a lot room for optimizations and tweaks, but I think this technique is fast and promising and a good starting point. If you get an answer, share it with us! :)
Update
I tried my approach and only got 87 out of the 100:
10894688943
60213136008
56252211674
61444925224
59409675697
02180334817
73260193640
.5476685202
0052034645.
...4.948156
......4671.
My guess is that both algorithms are too slow. Some optimization algorithm might work like best-first search or simulated annealing but my I don't have much experience on programming those.
Have you tried any primary research on Two-Dimensional Bin Packing (2DBP) algorithms? Google Scholars is a good start. I did this a while ago when building an application to generate mosaics.
All rectangular bin packing algorithms can be divided into 4 groups based on their support for the following constraints:
Must the resulting bin be guillotine cuttable? I.e. do you have to later slice the bin(s) in half until all the pieces are unpacked?
Can the pieces be rotated to fit into the bin? Not an issue with square pieces, so this makes more algorithms available to you.
Out of all the algorithms I looked into, the most efficient solution is an Alternate Directions (AD) algorithm with a Tabu Search optimization layer. There are dissertations which prove this. I may be able to dig-up some links if this helps.
Some ideas off the top of my head, without investing much time into thinking about details.
I would start by counting the number of occurrences of each digit in all squares 1..100. The total number of digits will be obviously larger than 121, but by analyzing individual frequencies you can deduce which digits must be grouped on a single line to form as many different squares as possible. For example, if 0 has the highest frequency, you have to try to put as many squares containing a 0 on the same line.
You could maintain a count of digits for each line, and each time you place a digit, you update the count. This lets you easily compute which square numbers have been covered by that particular line.
So, the program will still be brute-force, but it will exploit the problem structure much better.
PS: Counting digit frequencies is the easiest way to decide whether a certain permutation of digits constitutes a square.